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https://kr.mathworks.com/matlabcentral/cody/problems/40-reverse-run-length-encoder/solutions/1415676 | [
"Cody\n\n# Problem 40. Reverse Run-Length Encoder\n\nSolution 1415676\n\nSubmitted on 13 Jan 2018 by Ajay Rawat\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = [2 5 1 2 4 1 1 3]; correct = [5 5 2 1 1 1 1 3]; assert(isequal(correct, RevCountSeq(x)));\n\na = 2 1 4 1 b = 5 2 1 3 y = 5 5 2 1 1 1 1 3\n\n2 Pass\nx = [1 9]; correct = ; assert(isequal(correct, RevCountSeq(x)));\n\na = 1 b = 9 y = 9\n\n3 Pass\nx = [9 1]; correct = ones(1,9); assert(isequal(correct, RevCountSeq(x)));\n\na = 9 b = 1 y = 1 1 1 1 1 1 1 1 1\n\n4 Pass\nx = [1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9]; correct = 1:9; assert(isequal(correct, RevCountSeq(x)));\n\na = 1 1 1 1 1 1 1 1 1 b = 1 2 3 4 5 6 7 8 9 y = 1 2 3 4 5 6 7 8 9\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
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https://quant.stackexchange.com/questions/28415/average-correlation | [
"# Average Correlation\n\nWe're given a spreadsheet with a correlation matrix for four stocks.\n\nThen there is a calculation for average correlation, but I don't know how it's derived.\n\n$$=\\left(\\operatorname{Average}(C14:F17)-\\frac 14\\right)\\times\\frac{16}{10}$$\n\nI want to extend this calculation to six stocks. Can someone explain or point me to an explanation for how average correlations are calculated rather than some arbitrary scaling factors?\n\nHe is forced to use some tricks because Excel can only take average of a rectangular area, but he wants the avg of upper non-diagonal elements of the matrix only. So he subtracts $\\frac{1}{n}$ (the average of the 1's on the diagonal), then scales the result by $\\frac{n^2}{n(n+1)/2}$ which is the number of total elements divided by on-or-below-diagonal elements. Of course he is using $n=4$ since he has a 4 by 4 matrix.\n• An easier solution might just be to do (sum(square)-n)/n^2 - at least in terms of readability. – will Aug 1 '16 at 8:46"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9254861,"math_prob":0.9974073,"size":421,"snap":"2019-43-2019-47","text_gpt3_token_len":95,"char_repetition_ratio":0.14148681,"word_repetition_ratio":0.0,"special_character_ratio":0.2327791,"punctuation_ratio":0.08108108,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99987626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-18T21:34:07Z\",\"WARC-Record-ID\":\"<urn:uuid:281ad487-94a6-49b3-bd2d-107f2d59d05e>\",\"Content-Length\":\"132656\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed3922ca-6fd4-47ea-a226-f96b3c11e632>\",\"WARC-Concurrent-To\":\"<urn:uuid:b2137f0f-5d11-47de-84b3-47555d2bdb65>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://quant.stackexchange.com/questions/28415/average-correlation\",\"WARC-Payload-Digest\":\"sha1:3OJBQBR3RWR2QS64CNHJHA5XU7PSAJQK\",\"WARC-Block-Digest\":\"sha1:YSTQSFPZJTYGQ42V7UISG46ZW4EV3FXH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496669847.1_warc_CC-MAIN-20191118205402-20191118233402-00519.warc.gz\"}"} |
https://www.isprimenumber.com/prime/34313 | [
"# Is 34313 a Prime Number?\n\nYes, 34313 is a prime number. 34313 is divisible by 1 and itself.\n\nYes, 34313 is a prime number.\n\n### Why 34313 is a prime number?\n\nBecause 34313 has no positive divisors rather than 1 and itself.\n\nThe next prime number of 34313 is 34319\nThe previous prime number of 34313 is 34303.\n\n### Related Prime Numbers\n\nBiggest 10 prime numbers smaller than 34313\n\nSmallest 10 prime numbers bigger than 34313\n\n### Related Non-Prime Numbers\n\nBiggest 10 composite numbers smaller than 34313\n\nSmallest 10 composite numbers bigger than 34313"
] | [
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https://digitalblackboard.io/categories/intro-python/ | [
"## Functions\n\nTable of Contents Introduction Defining Functions Calling Functions The sorted() Function Lambda Functions Namespace and Scope (optional) Introduction A function is a block of code that performs a specific task and only runs when it is called. Functions help break our program into smaller modular chunks. As our program...\n\n## Data Types\n\nTable of Contents Data Types Integers and Floating-point Numbers Strings String Methods Mixing Strings and Variables Indexing and Slicing Strings Indexing Strings Slicing Strings Booleans Comparison Operators Logical Operators Data Types Variables can store data of different types, and different data types serve...\n\n## Python Objects\n\nTable of Contents Objects Variables Attributes and Methods Objects An important characteristic of the Python language is the consistency of its object model. Every number, string, data structure, function, class, module, etc. exists in the Python interpreter as a Python object. Each object has an associated type (e.g....\n\n## Simple Calculations\n\nTable of Contents Python as a Calculator Order of Operations Mathematical Functions Python as a Calculator Python contains functions found in any standard graphing calculator. An arithmetic operation is either addition, subtraction, multiplication, division, or powers between two numbers. An arithmetic operator is a..."
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https://www.physicsforums.com/threads/solving-a-differential-equation.795666/ | [
"# Solving a differential equation\n\n## Homework Statement\n\nSolve (xy+y2+x2) dx -( x2 )dy = 0\n\n## The Attempt at a Solution\n\nSo I can see it isn't separable and linear, so I thought of solving it through substitution\n\ni did y=ux and dy= u dx + x du\nand I substituted them in my first equation giving me\n(xu+u2+x^2) dx - x2(udx + xdu) = 0]\n\nI'm stuck here... the answer key factors out x^2 from the first equation leaving it with x^2(u+u^2+1), which doesn't make sense to me.. because if I factored out x^2 I would be left with (x-1u+x-2u2+1)\nSo any help on how I am seeing this wrong?\n\nLast edited by a moderator:\n\nRelated Calculus and Beyond Homework Help News on Phys.org\nShayanJ\nGold Member\n$\\partial_y M=\\partial_y(xy+y^2+x^2)=x+2y$\n$\\partial_x N=\\partial_x(-x^2)=-2x$\n$\\partial_y M \\neq \\partial_x N!!!$\n\nI know how to verify if exact, hence why I chose substitution, but I am kinda stuck at how to factor my xα. Unless I didn't understand your answer correctly, to me it seems as if you're solving to see whether or not if exact.\n\nShayanJ\nGold Member\nThat means your differential isn't exact which means it can't be integrated in this form, so you should turn it into an exact differential. There is such a method in your toolbox which you can't integrate the differential without it. You remember it?\n\nWell I know I can use substitution to reach a separable equation. If not, I know I can use υ(x,y) and multiply it to my equation. My question was more towards the substitution method because I don't completely grasp the subject.\n\nShayanJ\nGold Member\nBy some manipulation, you can get $y'=\\frac{y}{x}+\\frac{y^2}{x^2}+1$. Now try your substitution!\n\nI'm sorry, I'm not understanding. why you took the derivative of y. I was taught ( and my book explains) a different way. For instance, we just have to find a coeffiecient of xα to then create a separable DE. anyway, I'll check again later, maybe this will help getting a clearer answer? :/\n\nShayanJ\nGold Member\n$M dx+N dy=0 \\Rightarrow N dy=-M dx \\Rightarrow \\frac{dy}{dx}=- \\frac{M}{N}$\n\nRay Vickson\nHomework Helper\nDearly Missed\n\n## Homework Statement\n\nSolve (xy+y2+x2) dx -( x2 )dy = 0\n\n## The Attempt at a Solution\n\nSo I can see it isn't separable and linear, so I thought of solving it through substitution\n\ni did y=ux and dy= u dx + x du\nand I substituted them in my first equation giving me\n(xu+u2+x^2) dx - x2(udx + xdu) = 0]\n\nI'm stuck here... the answer key factors out x^2 from the first equation leaving it with x^2(u+u^2+1), which doesn't make sense to me.. because if I factored out x^2 I would be left with (x-1u+x-2u2+1)\nSo any help on how I am seeing this wrong?\n\nIf you substitute $y = x u$ into the DE\n$$x^2 \\frac{dy}{dx} = x^2 + xy + y^2 \\; \\Rightarrow \\frac{dy}{dx} = 1 + \\frac{y}{x} + \\left(\\frac{y}{x}\\right)^2$$\nthe resulting DE for $u$ is very simple and is straightforward to solve.\n\nLast edited by a moderator:\nYeah it was a stupid mistake, I was going fast and stressing out over an exam. I should've noticed that replacing y = ux into the equation can allow me to factor out x^2 because the function is homogeneous."
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https://www.roushuhai.com/1/1147/ | [
"",
null,
"# 诱捕（高H）\n\n• 想吃她的奶子\n• 让她喂奶\n• 暗生情愫\n• 摸着她的奶睡觉\n• 同寝\n• 晨勃\n• 配图\n• 给他撸出来\n• 摸着奶子被她撸\n• 摸妈妈的小穴\n• 把妈妈摸到潮吹\n• 射在妈妈的下身\n• 给妈妈测肛温\n• 妈妈又在发情\n• 发烧照顾妈妈\n• 给妈妈擦身体\n• 性感的妈妈\n• 在桌下把妈妈口到高潮\n• 搂着妈妈裸睡\n• 开学\n• 跟妈妈同浴\n• 妈妈的初吻\n• 舔妈妈的处女膜\n• 莫教授的生理课\n• 给妈妈破处\n• 抱着妈妈无套内射\n• 射完接着再操一回\n• 妈妈的排卵周期\n• 妈妈早晨一醒就求操\n• 做妈妈的学生\n• 性感妈妈当堂吃醋\n• 打妈妈的屁股\n• 好色的莫教授\n• 妈妈裙子下的诱惑\n• 抱起妈妈做爱\n• 慷慨的妈妈\n• 穿着围裙的妈妈\n• 被妈妈口着做饭\n• 口到高潮再内射\n• 恋母情结\n• 直白的妈妈\n• 没有区别\n• 属于\n• 喜欢的人\n• 妈妈的乖宝宝\n• 早上后入妈妈\n• 诚实的妈妈\n• 帮妈妈备孕\n• 八卦\n• 朋友\n• 迎新晚会\n• 斗舞\n• 让妈妈变成迷妹\n• 妈妈可以为所欲为\n• 想跟妈妈结婚\n• 一辈子爱妈妈\n• 录妈妈潮吹的样子\n• 诱奸继子的妈妈\n• 不起床就掀被子\n• 所谓白月光\n• 把妈妈按在门上操\n• 大和小\n• 一对一的性教育\n• 傻甜白\n• 搭档\n• 双打\n• 发情治百病\n• 非你不可\n• 恒常\n• （非更新）为了证明我不是故意鸽\n• 事端\n• 生母和继母\n• 母亲的责任\n• 分手 po18hub.com\n• 结婚\n• 六年后\n• 车内强奸\n• 车内强奸（2）\n• 双胞胎\n• 下药强奸\n• 下药迷奸（2）\n• 黑人问号脸\n• 下药迷奸（3）\n• 对局\n• 坦白\n• 结局\n• 番外1\n• 番外2\n• 一个声明∠( ? 」∠)＿\n• 番外3\n• 番外4"
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"http://img.roushuhai.com/image/1/1147/1147s.jpg",
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https://www.ci2ma.udec.cl/publicaciones/prepublicaciones/prepublicacion.php?id=13 | [
" CI²MA - Publicaciones | Prepublicaciones\n\n## Fernando Betancourt, Raimund Bürger, Kenneth H. KARLSEN:\n\n### Abstract:\n\nThis paper is concerned with a strongly degenerate convection-diffusion equation in one space dimension whose convective flux involves a non-linear function of the total mass to one side of the given position. This equation can be understood as a model of aggregation of the individuals of a population with the solution representing their local density. The aggregation mechanism is balanced by a degenerate diffusion term accounting for dispersal. In the strongly degenerate case, solutions of the non-local problem are usually discontinuous and need to be defined as weak solutions satisfying an entropy condition. A finite difference scheme for the non-local problem is formulated and its convergence to the unique entropy solution is proved. The scheme emerges from taking divided differences of a monotone scheme for the local PDE for the primitive. Numerical examples illustrate the behaviour of entropy solutions of the non-local problem, in particular the aggregation phenomenon.\n\nDescargar en formato PDF",
null,
"Esta prepublicacion dio origen a la(s) siguiente(s) publicación(es) definitiva(s):\n\nFernando BETANCOURT, Raimund BüRGER, Kenneth H. KARLSEN: A strongly degenerate parabolic aggregation equation. Communications in Mathematical Sciences, vol. 9, 3, pp. 711-742, (2011)."
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"https://www.ci2ma.udec.cl/img/pdf.png",
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https://www.brainkart.com/article/Fluid-Properties_3961/ | [
"Home | | Mechanics of Fluids | Fluid Properties\n\n# Fluid Properties",
null,
"There are three states of matter: solids, liquids and gases. Both liquids and gases are classified as fluids.\n\nINTRODUCTION TO FLUIDS\n\nDefinition\n\nThere are three states of matter: solids, liquids and gases. Both liquids and gases are classified as fluids.\n\nFluids do not resist a change in shape. Therefore fluids assume the shape of the container they occupy.\n\nLiquids may be considered to have a fixed volume and therefore can have a free surface.\n\nLiquids are almost incompressible.\n\nC o n v e r s e l y , gases are easily compressed and will expand to fill a container they occupy.\n\nWe will usually be interested in liquids, either at rest or in motion.",
null,
"Definition\n\nThe strict definition of a fluid is: A fluid is a substance which conforms continuously\n\nunder the action of shearing forces.\n\nTo understand this, remind ourselves of what a shear force is:\n\n2 Static Fluids\n\nAccording to this definition, if we apply a shear force to a fluid it will deform and take up\n\na state in which no shear force exists. Therefore, we can say: If a fluid is at rest there can be no shearing forces acting and therefore all forces in the fluid must be perpendicular to the planes in which they act. Note here that we specify that the fluid must be at rest. This is because, it is found experimentally that fluids in motion can have slight resistance to shear force. This is\n\nthe source of viscosity.\n\n3 Fluids in Motion\n\nFor example, consider the fluid shown flowing along a fixed surface. At the surface there will be little movement of the fluid (it will 'stick' to the surface), whilst furtheraway from the surface the fluid flows faster (has greater velocity):",
null,
"If one layer of is moving faster than another layer of fluid, there must be shear forcesacting between them. For example, if we have fluid in contact with a conveyor beltthat is moving we will get the behaviour shown:",
null,
"When fluid is in motion, any difference in velocity between adjacent layers has the same effect as the conveyor belt does.\n\nTherefore, to represent real fluids in motion we must consider the action of shear forces\n\nConsider the small element of fluid shown, which is subject to shear force and has a dimension sinto the page. The force F acts over an area A = BC� s. Hence we have a shear stress applied:",
null,
"Any stress causes a deformation, or strain, and a shear stress causes a shear strain. This shear strain is measured by the angle ? . Remember that a fluid continuously deforms when under the action of shear. This is different to a solid: a solid has a single value of ? for each value of ? . So the longer a shear stress is applied to a fluid, the more shear strain occurs. However, what is known from experiments is that the rate of shear strain (shear strain per unit time) is related to the shear stress:",
null,
"We need to know the rate of shear strain. From the diagram, the shear strain is:",
null,
"If we suppose that the particle of fluid at E moves a distance x in time t, then, using S = R? for small angles, the rate of shear strain is:",
null,
"Where u is the velocity of the fluid. This term is also the change in velocity with height. When we consider infinitesimally small changes in height we can write this in differential form, du/ dy . Therefore we have:",
null,
"Newton's Law of Viscosity:\n\nGeneralized Laws of Viscosity\n\nWe have derived a law for the behaviour of fluids - that of Newtonian fluids. However, experiments show that there are non-Newtonian fluids that follow a generalized law of viscosity:",
null,
"Where A, B and n are constants found experimentally. When plotted these fluids show much different behaviour to a Newtonian fluid:\n\nBehaviour of Fluids and Solids\n\nIn this graph the Newtonian fluid is represent by a straight line, the slope of which is ? . Some of\n\nthe other fluids are:\n\nPlastic: Shear stress must reach a certain minimum before flow commences.\n\nPseudo-plastic: No minimum shear stress necessary and the viscosity decreases with rate of shear, e.g. substances like clay, milk and cement.\n\nDilatant substances; Viscosity increases with rate of shear, e.g. quicksand.\n\nViscoelastic materials: Similar to Newtonian but if there is a sudden large change in shear they behave like plastic.\n\nSolids: Real solids do have a slight change of shear strain with time, whereas ideal solids (those we idealise for our theories) do not. Lastly, we also consider the ideal\n\nfluid. This is a fluid which is assumed to have no viscosity and is very useful for developing theoretical solutions. It helps achieve some practically useful solutions.\n\nProperties\n\nHere we consider only the relevant properties of fluids for our purposes. Find out about\n\nsurface tension and capillary action elsewhere. Note that capillary action only features in\n\npipes of\n\n? 10 mm diameter.\n\n4. FLUID PROPERTIES:\n\n1. Density or Mass density: Density or mass density of a fluid is defined as the ratio of the mass of a fluid to its volume. Thus mass per unit volume of a is called density.\n\nMass density fluid / Mass of Density of fluid\n\nThe unit of density in S.I. unit is kg/m3. The value of density for water is 1000kg/m\n\n2.Specific weight or weight density: Specific weight or weight density of a fluid is the ratio between the weight of a fluid to its volume. The weight per unit volume of a fluid is called weight density.\n\nWeight density = Weight of fluid / Volume of fluid\n\nw = Mass of fluid x g / Volume of fluid\n\nThe unit of specific weight in S.I. units is N/m3. The value of specific weight or weight density of water is\n\n9810N/m3.\n\n3.)Specific Volume: Specific volume of a fluid is defined as the volume of a fluid occupied by a unit mass or volume per unit mass of a fluid.\n\nSpecific volume = Volume of a fluid / Mass of fluid\n\nThus specific volume is the reciprocal of mass density. It is expressed as m3/kg. It is commonly applied to gases.\n\n4.)Specific Gravity: Specific gravity is defined as the ratio of the weight density of a fluid to the weight density of a standard fluid.\n\nSpecific gravity = Weight density of liquid / Weight density of water\n\nVISCOSITY\n\nViscosity is defined as the property of a fluid which offers resistance\n\nto the movement of one layer of fluid over adjacent layer of the fluid. When two layers of a fluid, a distance 'dy' apart, move one over the other at different velocities, say u and u+du as shown in figure. The viscosity together with relative velocity causes a shear stress acting between the fluid layers.\n\nCOMPRESSIBILITY:\n\nCompressibility is the reciprocal of the bulk modulus of elasticity, K which is defined as the ratio of compressive stress to volumetric strain.\n\nConsider a cylinder fitted with a piston as shown\n\nin figure. Let V= Volume of a gas enclosed in the\n\ncylinder\n\nP= Pressure of gas when volume is V\n\nLet the pressure is increased to p+dp, the volume of gas decreases\n\nfrom V to V-dV. Then increase in pressure =dp kgf/m2\n\nDecrease in volume= dV\n\nVolumetric Strain = d / V\n\n- ve sign means the volume decreases with increase of pressure.\n\nBulk modulus K = Increase pressure / Volumetric Strain\n\n= dp / dV/V\n\nCompressibility is given by = 1/K\n\nRelationship between K and pressure (p) for a Gas:\n\nThe relationship between bulk modulus of elasticity (K) and pressure for a gas for two different processes of comparison are as:\n\n(i) For Isothermal Process: The relationship between pressure (p) and density (?) of a gas as\n\np = Constant\n\n= Constant\n\nDifferentiating this equation, we get (p and V are variables)\n\nPdV +Vdp = 0\n\nor\n\npdV= - Vdp\n\nor\n\np/dV = Vdp\n\nSubstituting this value K =p\n\np Constant or pVk = Constant\n\nSURFACE TENSION:\n\nSurface tension is defined as the tensile force acting on the surface of a liquid in contact with a gas or on the surface between two two immiscible liquids such that the contact surface behaves like a membrane under tension\n\nCAPILLARITY\n\nCapillarity is defined as a phenomenon of rise or fall of a liquid surface in a small tube relative to the adjacent general level of liquid when the tube is held vertically in the liquid. The rise of liquid surface is known as capillary rise while the fall of the liquid surface is known as capillary depression. It is expressed in terms of cm or mm of liquid. Its value depends upon the specific weight of the liquid, diameter of the tube and surface tension of the liquid.\n\nProblem 1.\n\nCalculate the capillary effect in millimeters a glass tube of 4mm diameter, when immersed in (a) water (b) mercury. The temperature of the liquid is 200 C and the values of the surface tension of water and mercury at 200 C in contact with air are 0.073575 and 0.51 N/m respectively. The angle of contact for water is zero that for mercury 1300. Take specific weight of water as 9790 N / m3",
null,
"Problem 2.\n\nA cylinder of 0.6 m3 in volume contains air at 500C and 0.3 N/ mm2 absolute pressure. The air is compressed to 0.3 m3. Find (i) pressure inside the cylinder assuming isothermal process (ii) pressure and temperature assuming adiabatic process. Take K = 1.4",
null,
"Problem 3\n\nIf the velocity profile of a fluid over a plate is a parabolic with the vertex 202 cm from the plate, where the velocity is 120 cm/sec. Calculate the velocity gradients and shear stress at a distance of 0,10 and 20 cm from the plate, if the viscosity of the fluid is 8.5 poise.",
null,
"Problem 4\n\nA 15 cm diameter vertical cylinder rotates concentrically inside another cylinder of diameter 15.10 cm. Both cylinders are 25 cm high. The space between the cylinders is filled with a liquid whose viscosity is unknown. If a torque of 12.0 Nm is required to rotate the inner cylinder at 100 rpm determine the viscosity of the fluid.\n\nSolution:\n\nDiameter of cylinder = 15 cm = 0.15 m\n\nDiameter of outer cylinder = 15.10 cm = 0.151 m\n\nLength of cylinder L = 25 cm = 0.25 m\n\nTorque T= 12 Nm ; N = 100 rpm.\n\nViscosity = m",
null,
"Problem 5\n\nThe dynamic viscosity of oil, used for lubrication between a shaft and sleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at 190 rpm. Calculate the power lost in the bearing for a sleeve length of 90 mm. The thickness of the oil film is 1.5 mm.",
null,
"Problem 6\n\nIf the velocity distribution over a plate is given by u =2 y/3 -y 2 in which U is the velocity in m/s at a distance y meter above the plate, determine the shear stress at y = 0 and y = 0.15 m. Take dynamic viscosity of fluid as 8.63 poise.",
null,
"Problem 7\n\nThe diameters of a small piston and a large piston of a hydraulic jack at3cm and 10 cm respectively. A force of 80 N is applied on the small piston Find the load lifted by the large piston when:\n\nThe pistons are at the same level\n\nSmall piston in 40 cm above the large piston.\n\nThe density of the liquid in the jack in given as 1000 kg/m3\n\nGiven: Dia of small piston d = 3 cm.",
null,
"Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail\nCivil : Mechanics Of Fluids : Fluid Properties And Fluid Statics : Fluid Properties |"
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https://www.indiabix.com/placement-papers/bsnl/3148 | [
"# Placement Papers - BSNL",
null,
"BSNL Placement Paper (Set-4)\nPosted by :\n(15)\nBSNL Sample Paper Pattern (Set-4)\n\n1.For a transmission line, the propogation constant, for a TEM wave travelling in it is given by (Where the symbols have the usual meanings ) -\na) [ (R+jwL) (G+jwc)]\nb) [ R+jwL) (G+jwc)] ?\nc)? [ (R-jwL) (G + jwc) ] ?\nd) [(R-jwL) ( G+jw2c)]1/3 ?\n\n2.The advantages of wave guides over co-axial lines would include which of the following features-\n1. Easier to use 2. lower power losses\n3. Higher operating frequencies possible\na) 1 and 2\nb) 1 and 3\nc) 2 and 3\nd) 1,2 and 3 ?\n\n3.When a 75 ohm transmission line is to be terminated in two resistive loads R1 and R2 such that the standing pattern in the two cases have the same SWR , then the values of R1 and R2 (in ohms) should be -\na) 250 and 200 respectively\nb) 225 and 25 respectively\nc)? 100 and 150 respectively\nd) 50 and 125 respectively?\n\n5.The degenerate modes in a wave guide are characterized by -\na) Same cut off frequencies but different field distribution\nb) Same cut off frequencies and same field distributions\nc)? Different cut off frequencies but same field distributions\nd) Different cut off frequencies and different field distributions?\n\n6.A TEM wave impinges obliquely on a dielectric-dielectric boundary with Er1=2 and Er2=1, the angle of incidence for total reflection is -\na)? 30 0\nb)? 60 0\nc) 45 0\nd) 90 0 ?\n\n7.The radiation pattern of Hertzian dipole in the plane perpendicular to the dipole is a -\na) Null\nb) Circle\nc) Figure of eight\nd) None of the above ? ?\n\n8.. Permeance is the -\na) square of reluctance\nb) reluctance\nc) reciprocal of the reluctance\nd) cube of the reluctance. ?\n\n9.One of the following which is an active transducer is -\na) Photoelectric\nb) Photovoltaic\nc) Photo-conductive\nd) Photo emission ?\n\n10.The wein bridge uses only -\na) Inductors\nb) Capacitors\nc) Resistors\nd) Capacitors and Resistors. ?\n\n11.The greater the value of Q -\na) higher will be the bandwidth of the resonant circuit.\nb) smaller will be the bandwidth of the resonant circuit.\nc) nothing can be said)\nd) none. ?\n\n12.The most serious source of error in a) c) bridge measurement is -\na) eddy currents\nb) leakage currents\nc) residual imperfectness\nd) stray fields. ?\n\n13.Moving iron instruments -\na) have a linear scale\nb) do not have a linear scale\nc) both a and b)\nd) none. ?\n\n14.If accuracy is the main consideration, which one of the following voltmeters should one select -\na) 100 v ; 2 mA\nb) 100 v ; 100 ohm/volt\nc) 100 v ; 1mA\nd) 10,000 v ; 10 mA ?\n\n15.In dc tacho generators used for measurement of speed of a shaft, frequent calibration has to be done because -\na) the contacts wear off\nb) the strength of permanent magnet decreases with age\nc) the armature current produces heating effect\nd) there is back emf. ?\n\n16.Ideal transformer cannot be described by -\na) h parameters\nb) ABCD parameters\nc) G parameters\nd) parameters ?\n\n17.Consider the following statements -\nA3- phase balanced supply system is connected to a 3 phase unbalanced load) Power supplied to this load can be measured using\n1. Two wattmeters\n2. One wattmeter\n3. Three wattmeters\n\n18.Which of these statements is/are correct?\na) 1 and 2\nb) 1 and 3\nc) 2 and 3\nd) 3 alone?\n\n19.The function of the reference electrode in a pH meter is to -\na) Produce a constant voltage\nb) Provide temperature compensation\nc) Provide a constant current\nd) Measure average pH value?\n\n20.Match the column A (Devices) with column B (Characteristics) and select the correct answer by using the codes given below the column -\nColumn A Column B\nA) BJT 1. Voltage controlled negative resistance\nB) MOSFET 2. High current gain\nC) Tunnel diode 3. Voltage regulation\nD) Zener diode 4. High input impedance\nCodes :\nA B C D\na) 1 4 2 3\nb) 2 4 1 3\nc) 2 3 1 4\nd) 1 3 2 4 ?\n\n21.A thyristor during forward blocking state is associated with.-\na) large current , low voltage.\nb) low current , large voltage.\nc) medium current , large voltage\nd) low current, medium voltage. ?\n\n22.In controlled rectifiers, the nature of load current i.e. whether load current is continuous or discontinuous -\na) does not depend on type of load and firing angle delay\nb) depends both on the type of load and firing angle delay\nc) depends only on the type of load)\nd) depends only on the firing angle delay. ?\n\n23.A single phase voltage controller feeds power to a resistance of 10 W . The source voltage is 200 V rms. For a firing angle of 900 , the rms value of thyristor current in amperes is -\na) 20\nb) 15\nc) 10\nd) 5\n\n24.In the performance of single phase and three phase full converters the effect of source inductance is to\na) reduce the ripples in the load current -\nb) make discontinuous current as continuous\nc) reduce the output voltage\n\n25.The cycloconverters (CCs) require natural or forced commutation as under -\na) natural commutation in both step up and step down CCs\nb) forced commutation in both step up and step down CCs\nc) forced commutation in step up CCs\nd) forced commutation in step down CCs\n\n26.Power transistors are more commonly of -\na) silicon npn type.\nb) silicon pnp type.\nc) silicon nnp type.\nd) silicon npp type.\n\n27.C is a -\na) Middle level language\nb) High level language\nc) Low level language\nd) None of above\n\n28.What will be output of program\nmain ( )\n{ int i ;\nprint f (\"Enter value of i\");\nscant (\"%d\", & i);\nif ( i = 5 )\nprint f (\"you entered 5\");\nelse\nprint f (\"you entered %d\", i ); }\nif user entered 100 then\na) 5\nb) 100\nc) 1005\nd) None ?\n\n28.(7F)16 + (BA)16 = (?)16-\na) 481\nb) 139\nc) ?481 d) ?139 ?\n\n29.Two?s complement of 3 bit nonzero linory number is some or original number is all bits accepts- a) MSB are zeros\nb) LSB are zeros\nc) MSB are ones\nd) LSB are ones. ?\n\n?30.Transistors with high frequency have -\na) Thick base\nb) Thin base\nc) Some other feature\nd) None of the above\n\n31.Telephone traffic is specified in terms of -\na) Average waiting time\nc) Peak waiting time\nd) Erlangs?\n\n32.In a Hartley oscillator -\na) Necessary phase relation is obtained by connecting grid and plate electrodes to the opposite ends of the tuned circuit.\nb) The mutual inductance must have the appropriate polarity.\nc) Both grid circuit and plate tuned circuit offer inductive reactance\nd) None of the above?\n\n33.The condenser C is charged in a bootstrap sweep generator -\na) Linearly but the discharge is non linear\nb) Non linearly but the discharge is linear\nc) Linearly and the discharge is linear\nd) Non linearly and the discharge is non linear ??"
] | [
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https://tex.stackexchange.com/questions/168405/kile-syntax-highlighting-a-single-dollar-sign-in-the-preamble | [
"I'm using Kile to edit my LaTeX documents. I define my own custom environment to hightlight LaTeX code named lstLaTeX with the following code:\n\n\\documentclass[a4paper]{scrartcl}\n\\usepackage[usenames,dvipsnames]{xcolor}\n\\usepackage[latin1]{inputenc}\n\\usepackage[ngerman,english]{babel}\n\n\\usepackage{listings}\n\n\\definecolor{lightgrey}{rgb}{0.9,0.9,0.9}\n\\definecolor{darkgreen}{rgb}{0,0.6,0}\n\n\\lstnewenvironment{lstLaTeX}\n{\n\\lstset{language=[LaTeX]TeX,\nkeepspaces=true,\ntexcsstyle=*\\bf\\color{blue},\nbasicstyle=\\ttfamily,\nnumbers=none,\nbreaklines=true,\nkeywordstyle=\\color{darkgreen},\nmorekeywords={},\notherkeywords={$, \\{, \\}, $,$}, frame=none, tabsize=2, columns=fullflexible, backgroundcolor=\\color{lightgrey}, escapechar=° } } {} \\begin{document} The actual document. Here some LaTeX code: \\begin{lstLaTeX} Brackets should be {\\bf highlighted}. The dollar sign:$x=5$\\end{lstLaTeX} \\end{document} The problem is that Kile cannot deal with the single dollar sign in the preamble and marks all following text green (because it thinks there should be a math environment).",
null,
"I already read how to teach Kile to ignore dollar signs when used inside custom environments here: disable syntax highlighting in kile But this post doesn't solve my problem. So it would be nice if I could tell Kile to ignore this single dollar sign. I already tried to add %$ at the end of the line with the single dollar sign but Kile ignores this.\n\n• Is simply splitting the input line an option? If so, the 'normal' solution to this type of problem is a strategically-placed comment with the 'matching' item in it. – Joseph Wright Mar 30 '14 at 8:09\n• I'm not sure if understand your tip corretly (splitting the input line?) but I already tried to solve the problem with a comment and it didn't worked. A custom comment command which Kile doesn't recognize as a comment might be a solution. – Steven Thiel Mar 30 '14 at 13:10\n• End of the line would be 'wrong' due to the braces, hence asking about splitting the line so you have {$, %$ <newline> $,$, .... I'm not a Kile user so I can't check if it respects this. – Joseph Wright Mar 30 '14 at 14:10\n• The problem is Kile marks all following text green not only the line with $ and it ignores %$. – Steven Thiel Mar 30 '14 at 20:09\n• @JosephWright Is there a way of telling TeX to throw away the next token? Kile is too 'clever' - it ignores anything after a comment sign. I want to use $ with l3regex, but it turns all remaining content in my .cls magenta! And obviously repetition isn't an option in this case. (I know I can use \\Z but that is much less readable for me. – cfr Jan 19 '17 at 23:09 ## 1 Answer otherkeywords={$, $, \\{, \\}, $,$}, seems to work. The repetition doesn't seem to bother anything when compiling and it makes Kile happy. # EDIT A similar problem, which cannot be worked around in the same way, occurs if using $ with l3regex, for example.\n\nThe following function\n\n\\cs_new_protected_nopar:Nn \\prefix_gobble_token:n\n{\n\\relax\n}\n\n\nallows the $ to be matched, limiting the highlighting damage to the close vicinity of the offending token by adding \\cfr_gobble_token:n {$ }\n\n\nclose by.\n\nHowever, I have no idea how safe or otherwise this might be ...\n\n• Just use \\use_none:n $. (And in regexes if I remember correctly \\Z is equivalent to $ except perhaps for subtleties with linefeeds that I never implemented. – Bruno Le Floch Jul 13 '17 at 1:46"
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"https://i.stack.imgur.com/FcNM9.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6915897,"math_prob":0.5895598,"size":1836,"snap":"2020-24-2020-29","text_gpt3_token_len":525,"char_repetition_ratio":0.09716157,"word_repetition_ratio":0.008064516,"special_character_ratio":0.2657952,"punctuation_ratio":0.16374269,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9518441,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-06T11:47:40Z\",\"WARC-Record-ID\":\"<urn:uuid:747174b4-222a-4e68-be5e-703f7e4fb52d>\",\"Content-Length\":\"152817\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:926b0370-555c-494e-ab6c-04d95645dc9b>\",\"WARC-Concurrent-To\":\"<urn:uuid:61585435-26d1-4ffc-bdc5-2c7a8660a540>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/168405/kile-syntax-highlighting-a-single-dollar-sign-in-the-preamble\",\"WARC-Payload-Digest\":\"sha1:ZGFVH3CTTAIXDCIQCQNMUHHPAFHKNQAT\",\"WARC-Block-Digest\":\"sha1:DNYD6ZOB2LDC7PXIZDXZ3EIVVGXJVTLT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348513230.90_warc_CC-MAIN-20200606093706-20200606123706-00156.warc.gz\"}"} |
https://www.studyadda.com/question-bank/heat_q11/3254/264206 | [
"• # question_answer 5g ice at $0{}^\\circ C$ is mixed with 5g of steam at $100{}^\\circ C$ What is the final temperature? A) $50{}^\\circ C$ B) $100{}^\\circ C$ C) $80{}^\\circ C$ D) $150{}^\\circ C$\n\nHeat required by ice to raise its temperature to $100{}^\\circ C,$ \\begin{align} & {{Q}_{1}}={{m}_{1}}{{L}_{1}}+{{m}_{1}}{{c}_{1}}{{\\theta }_{1}}=5\\times 80+5\\times 1\\times 100 \\\\ & =400+500=900cal \\\\ \\end{align} Heat given by steam when condensed, \\begin{align} & {{Q}_{2}}={{m}_{2}}{{L}_{2}}=5\\times 536=2680cal \\\\ & {{Q}_{2}}>{{Q}_{1}} \\\\ \\end{align} This means that whole steam is not even condensed. Hence temperature of mixture will remain at $100{}^\\circ C$.",
null,
"You will be redirected in 3 sec",
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] | [
null,
"https://www.studyadda.com//assets/images/150adv.jpg",
null,
"https://www.studyadda.com/assets/frontend/images/msg-gif.GIF",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.62410474,"math_prob":0.9999795,"size":660,"snap":"2020-34-2020-40","text_gpt3_token_len":238,"char_repetition_ratio":0.16158536,"word_repetition_ratio":0.0,"special_character_ratio":0.48939395,"punctuation_ratio":0.05185185,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991904,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T20:48:21Z\",\"WARC-Record-ID\":\"<urn:uuid:a8f7b36f-4bd2-4a76-9b77-10f21e699e87>\",\"Content-Length\":\"100721\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de0c56c6-77b2-426b-8d2a-d3f2fc8b53b9>\",\"WARC-Concurrent-To\":\"<urn:uuid:27a93467-fff7-4e1c-97a6-a6ae4a6a316a>\",\"WARC-IP-Address\":\"151.106.35.148\",\"WARC-Target-URI\":\"https://www.studyadda.com/question-bank/heat_q11/3254/264206\",\"WARC-Payload-Digest\":\"sha1:7XKT2RJAC32N5AHZNOJHPNXQOL7DNMKG\",\"WARC-Block-Digest\":\"sha1:FTHACMSHA57SVK6LPAUITSYHLGZQXVO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738573.99_warc_CC-MAIN-20200809192123-20200809222123-00378.warc.gz\"}"} |
https://www.numbertowordsconverter.com/what-is-57-factorial/ | [
"# What is 57 factorial ?\n\nSteps to calculate factorial of 57\n\n## How to calculate the factorial of 57\n\nTo find 57 factorial, or 57!, simply use the formula that multiplies the number 57 by all positive whole numbers less than it.\n\nLet's look at how to calculate the Factorial of Fifty-seven :\n\n57! is exactly : 40526919504877216755680601905432322134980384796226602145184481280000000000000\n\nFactorial of 57 can be calculated as:\n57! = 57 x 56 x 55 x 54 x ... x 3 x 2 x 1\n\nThe number of trailing zeros in 57! is 13\n\nThe number of digits in 57 factorial is 77.\n\n## What Is Factorial?\n\nA factorial is showed by an integer and an exclamation tag. In Mathematics, factorial is a multiplication function of natural numbers .\n\nIt multiplies the number by every organic number that is less than it .\n\nSymbolically, it is listed as \"!\".\n\nThe function is used, among other things, to get the \"n\" way things can be positioned .\n\n## Factorial Formula\n\nTo find the factorial of any given number, replace the exact value for n in the given formulation :\n\nn! = n × (n-1) × (n-2) × (n-3) × ….× 3 × 2 × 1\n\nThe expansion of the formula provides numbers to be multiplied to each other to obtain the factorial of the number.\n\nWe can also work out a factorial from the previous one. The factorial of any number is that number times the factorial of (that number minus 1).\n\nSo the rule is : n! = n × (n−1)!\n\nExample :\n57! Factorial = 57 x 56 x 55 x 54 x ... x 3 x 2 x 1 = 57 × 56! = 40526919504877216755680601905432322134980384796226602145184481280000000000000\n\n## What are Factorials Used For?\n\nThe best use of factorial is in Combinations and Permutations.\n\nExample : Determine how to arrange letters without repeating?\n\nThere one way for 1 letter \"a\":\n2 ways for two letters \"ab\": ab, ba.\nThere are 6 ways for 3 letters \"abc\": abc acb cab bac bca.\nThere are 24 ways for 1234 of the letters \"abcd\"\n\n## Frequently Asked Questions on Factorial\n\n### Can we have factorials for negative numbers ?\n\nNegative number factorials are undefined\n\n### What Is 0!\n\nZero factorial or Factorial of 0 is simple, and its own value is equal to 1. So, 0! = 1."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8246065,"math_prob":0.9981878,"size":2043,"snap":"2022-40-2023-06","text_gpt3_token_len":568,"char_repetition_ratio":0.17508583,"word_repetition_ratio":0.06933333,"special_character_ratio":0.34752816,"punctuation_ratio":0.14043583,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9989245,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T04:50:56Z\",\"WARC-Record-ID\":\"<urn:uuid:d41725c3-a8f1-4661-b5b1-73c77e08eaaf>\",\"Content-Length\":\"121591\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b1555e3f-d649-41f9-8350-3d4fc0f178f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4b2737f-b73a-47eb-8b98-03be38847958>\",\"WARC-IP-Address\":\"104.21.47.67\",\"WARC-Target-URI\":\"https://www.numbertowordsconverter.com/what-is-57-factorial/\",\"WARC-Payload-Digest\":\"sha1:X4WORSC7W4TCYBLAPLV2GPUT6CGOM7Z6\",\"WARC-Block-Digest\":\"sha1:MKNAJKBXXWYM7VJT4CAC6ZIMNMGJRGIS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500671.13_warc_CC-MAIN-20230208024856-20230208054856-00225.warc.gz\"}"} |
https://feet-to-meters.appspot.com/pl/4097-stopa-na-metr.html | [
"Feet To Meters\n\n# 4097 ft to m4097 Foot to Meters\n\nft\n=\nm\n\n## How to convert 4097 foot to meters?\n\n 4097 ft * 0.3048 m = 1248.7656 m 1 ft\nA common question is How many foot in 4097 meter? And the answer is 13441.6010499 ft in 4097 m. Likewise the question how many meter in 4097 foot has the answer of 1248.7656 m in 4097 ft.\n\n## How much are 4097 feet in meters?\n\n4097 feet equal 1248.7656 meters (4097ft = 1248.7656m). Converting 4097 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 4097 ft to m.\n\n## Convert 4097 ft to common lengths\n\nUnitUnit of length\nNanometer1.2487656e+12 nm\nMicrometer1248765600.0 µm\nMillimeter1248765.6 mm\nCentimeter124876.56 cm\nInch49164.0 in\nFoot4097.0 ft\nYard1365.66666667 yd\nMeter1248.7656 m\nKilometer1.2487656 km\nMile0.7759469697 mi\nNautical mile0.6742794816 nmi\n\n## What is 4097 feet in m?\n\nTo convert 4097 ft to m multiply the length in feet by 0.3048. The 4097 ft in m formula is [m] = 4097 * 0.3048. Thus, for 4097 feet in meter we get 1248.7656 m.\n\n## 4097 Foot Conversion Table",
null,
"## Alternative spelling\n\n4097 ft to Meters, 4097 ft in Meters, 4097 ft to Meter, 4097 ft in Meter, 4097 Foot in m, 4097 Feet to Meter, 4097 Feet in Meter, 4097 ft to m, 4097 Foot to Meters, 4097 Feet to Meters, 4097 Feet in Meters,"
] | [
null,
"https://feet-to-meters.appspot.com/image/4097.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8424135,"math_prob":0.6054768,"size":705,"snap":"2023-40-2023-50","text_gpt3_token_len":244,"char_repetition_ratio":0.23965763,"word_repetition_ratio":0.0,"special_character_ratio":0.44113475,"punctuation_ratio":0.15476191,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97102195,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T23:57:29Z\",\"WARC-Record-ID\":\"<urn:uuid:3d35f5a5-5386-4301-9c87-16b478e63740>\",\"Content-Length\":\"28200\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:088010b6-1801-48a8-8c2a-e5fd06c118ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:6499d27b-b66d-428a-b7b7-07f0973a5501>\",\"WARC-IP-Address\":\"172.253.122.153\",\"WARC-Target-URI\":\"https://feet-to-meters.appspot.com/pl/4097-stopa-na-metr.html\",\"WARC-Payload-Digest\":\"sha1:4VEOWBP5Q5P7IKYXMINRRRNBQ764BLT7\",\"WARC-Block-Digest\":\"sha1:63HT7V7ZGQGGNZ3ZNAW5YJ6QGLAH2BY5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510225.44_warc_CC-MAIN-20230926211344-20230927001344-00437.warc.gz\"}"} |
https://manpag.es/centos6/l+cgbcon | [
"pdf\n\n# CGBCON\n\n## NAME\n\nCGBCON - estimates the reciprocal of the condition number of a complex general band matrix A, in either the 1-norm or the infinity-norm,\n\n## SYNOPSIS\n\n SUBROUTINE CGBCON( NORM, N, KL, KU, AB, LDAB, IPIV, ANORM, RCOND, WORK, RWORK, INFO ) CHARACTER NORM INTEGER INFO, KL, KU, LDAB, N REAL ANORM, RCOND INTEGER IPIV( * ) REAL RWORK( * ) COMPLEX AB( LDAB, * ), WORK( * )\n\n## PURPOSE\n\nCGBCON estimates the reciprocal of the condition number of a complex general band matrix A, in either the 1-norm or the infinity-norm, using the LU factorization computed by CGBTRF.\nAn estimate is obtained for norm(inv(A)), and the reciprocal of the condition number is computed as\nRCOND = 1 / ( norm(A) * norm(inv(A)) ).\n\n## ARGUMENTS\n\nNORM (input) CHARACTER*1\n\nSpecifies whether the 1-norm condition number or the infinity-norm condition number is required:\n= '1' or 'O': 1-norm;\n= 'I': Infinity-norm.\n\nN (input) INTEGER\n\nThe order of the matrix A. N >= 0.\n\nKL (input) INTEGER\n\nThe number of subdiagonals within the band of A. KL >= 0.\n\nKU (input) INTEGER\n\nThe number of superdiagonals within the band of A. KU >= 0.\n\nAB (input) COMPLEX array, dimension (LDAB,N)\n\nDetails of the LU factorization of the band matrix A, as computed by CGBTRF. U is stored as an upper triangular band matrix with KL+KU superdiagonals in rows 1 to KL+KU+1, and the multipliers used during the factorization are stored in rows KL+KU+2 to 2*KL+KU+1.\n\nLDAB (input) INTEGER\n\nThe leading dimension of the array AB. LDAB >= 2*KL+KU+1.\n\nIPIV (input) INTEGER array, dimension (N)\n\nThe pivot indices; for 1 <= i <= N, row i of the matrix was interchanged with row IPIV(i).\n\nANORM (input) REAL\n\nIf NORM = '1' or 'O', the 1-norm of the original matrix A. If NORM = 'I', the infinity-norm of the original matrix A.\n\nRCOND (output) REAL\n\nThe reciprocal of the condition number of the matrix A, computed as RCOND = 1/(norm(A) * norm(inv(A))).\n\nWORK (workspace) COMPLEX array, dimension (2*N)\nRWORK (workspace) REAL array, dimension (N)\nINFO (output) INTEGER\n\n= 0: successful exit\n< 0: if INFO = -i, the i-th argument had an illegal value\n\npdf"
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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/140753 | [
"```Andreas Habel wrote:\n> Hi,\n>\n> I`ve a simple question about constants in ruby. Is there a possibility\n> to define constants in a module or a class like a enum in c?\n>\n> Maybe something short like the 'attr_accessor' syntax would be great!\n>\n> class MyClass\n> const_def :CONST_A1, :CONST_A2, :CONST_A3\n> const_def :CONST_B1, :CONST_B2, :CONST_B3\n> end\n>\n> eq.\n>\n> class MyClass\n>\n> CONST_A1 = 1\n> CONST_A2 = 2\n> CONST_A3 = 3\n>\n> CONST_B1 = 1\n> CONST_B2 = 2\n> CONST_B3 = 3\n> end\n>\n\nThe following is a simple implementation of what you want\n\nclass Class\ndef const_def(*symbols)\nsymbols.each_with_index do |symbol, index|\nconst_set(symbol, index + 1)\nend\nend\nend\n\nclass MyClass\nconst_def :CONST_A1, :CONST_A2, :CONST_A3\n\np CONST_A1, CONST_A2, CONST_A3\nend\n\n#=>\n1\n2\n3\n\nHTH\n\n--\nMark Sparshatt\n\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5158444,"math_prob":0.92818564,"size":779,"snap":"2020-45-2020-50","text_gpt3_token_len":238,"char_repetition_ratio":0.21419355,"word_repetition_ratio":0.014492754,"special_character_ratio":0.3440308,"punctuation_ratio":0.1884058,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9775669,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-03T14:09:49Z\",\"WARC-Record-ID\":\"<urn:uuid:132cdc91-f9be-439f-a4d9-e51a43cfce5b>\",\"Content-Length\":\"5793\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:669523dd-341e-48e8-86c9-92f8a455cbb0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a327090-08e0-4614-b4f8-4f2f66b2fc6d>\",\"WARC-IP-Address\":\"133.44.98.95\",\"WARC-Target-URI\":\"http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/140753\",\"WARC-Payload-Digest\":\"sha1:YFPQMVFRJ5BJ2RBX5G6P7PX55IQUHI6B\",\"WARC-Block-Digest\":\"sha1:74PWCB4Z676RSW3XY6S3GUUIE6FYKC4E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141727782.88_warc_CC-MAIN-20201203124807-20201203154807-00478.warc.gz\"}"} |
https://seedcx.zendesk.com/hc/en-us/articles/360031563434-How-is-the-Zero-Hash-Index-calculated-?mobile_site=false | [
"# How is the Zero Hash Index calculated?\n\nZero Hash calculates index values for a wide range of asset pairs. The official Zero Hash Daily Index value is the index value calculated and published at 4pm Central Time, however Zero Hash also calculates a continuous Real-time Index.\n\nZero Hash uses transaction data from a sample of top exchanges in its calculation. Those exchanges are Bitfinex, Bitstamp, Coinbase Pro, Gemini, HitBTC, Huobi Pro, itBit, Kraken and OKCoin.\n\nFor each exchange that has trading volume, we calculate the volume-weighted average price (VWAP) over the proceeding hour. We then take the arithmetic mean of this value across all exchanges. If there is no trading activity on any exchange during this period, we will utilize the last calculated index value."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87027085,"math_prob":0.96385336,"size":735,"snap":"2021-21-2021-25","text_gpt3_token_len":152,"char_repetition_ratio":0.15731874,"word_repetition_ratio":0.0,"special_character_ratio":0.18911564,"punctuation_ratio":0.124087594,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99021095,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-07T10:51:36Z\",\"WARC-Record-ID\":\"<urn:uuid:cbf127bc-66a8-4cdf-b497-c9425b01b6eb>\",\"Content-Length\":\"24741\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:182d19cf-422f-4c2a-b90a-b72215e9d09d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9f7e9399-97f5-4932-93a5-310ba1610e2a>\",\"WARC-IP-Address\":\"104.16.53.111\",\"WARC-Target-URI\":\"https://seedcx.zendesk.com/hc/en-us/articles/360031563434-How-is-the-Zero-Hash-Index-calculated-?mobile_site=false\",\"WARC-Payload-Digest\":\"sha1:3UQCZUTSCNLAM3J5LRW2NNECZLHY3TYK\",\"WARC-Block-Digest\":\"sha1:ERIP4ITT5K5FRTVEY3ZTMP2Y2U2JGEDO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988775.80_warc_CC-MAIN-20210507090724-20210507120724-00003.warc.gz\"}"} |
http://www.yaldex.com/games-programming/0672323699_ch13lev1sec4.html | [
" Game Programming Gurus",
null,
"Free JavaScript Editor Ajax Editor \n\nMain Page\n\n### The Evil Head of Friction\n\nThe next topic of discussion is friction. Friction is any force that retards or consumes energy from another system. For example, automobiles use internal combustion to operate; however, a whopping 30–40 percent of the energy that is produced is eaten up by thermal conversion or mechanical friction. On the other hand, a bicycle is about 80–90 percent efficient and is probably the most efficient mode of transportation in existence.\n\n#### Basic Friction Concepts\n\ne is basically resistance in the opposite direction of motion and hence can be modeled with a force usually referred to as the frictional force. Take a look at Figure 13.17; it depicts the standard frictional model of a mass m on a flat plane.\n\n##### Figure 13.17. Basic friction model.",
null,
"If you try to push the mass in a direction parallel to the plane you will encounter a resistance or frictional force that pushes back against you. This force is defined mathematically as",
null,
"where m is the mass of the object, g is the gravitational constant (9.8 m/s2) and _s is the static frictional coefficient of the system that depends on the conditions and materials of the mass and the plane. If the force F you apply to the object is greater than Ff, then the object will begin to move. Once the object is in motion its frictional coefficient usually decreases to another value, which is referred to as the coefficient of kinetic friction, mk.",
null,
"When you release the force, the object slowly decelerates and comes to rest because friction is always present.\n\nTo model friction on a flat surface all you need do is apply a constant negative velocity to all your objects that is proportional to the friction that you want. Mathematically, this is\n\n```Velocity New = Velocity Old - friction.\n```\n\nThe result is objects that slow down at a constant rate once you stop moving them. Of course, you have to watch out for letting the sign of the velocity go negative or in the other direction, but that's just a detail. Here's an example of an object that is moved to the right with an initial velocity of 16 pixels per frame and then slowed down at a rate of 1 pixel per frame due to virtual friction:\n\n```int x_pos = 0, // starting position\nx_velocity = 16, // starting velocity\nfriction = -1; // frictional value\n\n// move object until velocity <= 0\nwhile(x_velocity > 0)\n{\n// move object\nx_pos+=x_velocity;\n// apply friction\nx_velocity+=friction;\n} // end while\n```\n\nThe first thing you should notice is how similar the model for friction is to gravity. They are almost identical. The truth is that both gravitational forces and frictional forces act in the same way. In fact, all forces in the universe can be modeled in the exact same way. Also, you can apply as many frictional forces to an object as you want. Just sum them up.\n\nAs an example of friction I have written a little air hockey demo named DEMO13_5.CPP|EXE (16-bit version, DEMO13_5_16B.CPP|EXE), shown in Figure 13.18. The program lets you fire a hockey puck on a virtual air hockey table in a random direction every time you press the spacebar. The puck then bounces around off the borders of the table until it comes to rest due to friction. If you want to change the frictional coefficient of the table, use the arrow keys. See if you can add a paddle and a computer controlled opponent to the simulation!\n\n##### Figure 13.18. The hockey demo.",
null,
"#### Friction on an Inclined Plane (Advanced)\n\nThat wasn't too bad huh? The bottom line is that friction can be modeled as a simple resistive force or a negative velocity on an object each cycle. However, I want to show you the math and derivation of friction on an inclined plane since this will give you the tools you need to analyze much more complex problems. Be warned, though: I'm going to use a lot of vectors, so if you're still rusty or having trouble then take a look back when I talked about them in Chapter 8, \"Vector Rasterization and 2D Transformations,\" or pick up a good linear algebra book.\n\nFigure 13.19 shows the problem we're trying to solve. Basically, there is a mass m on an inclined plane. The plane has frictional coefficients ms and mk for the static and kinetic (moving) cases respectively. The first thing we need to do is write the formulas that describe the mass in its equilibrium position, that is, not moving. In this case, the sum of the forces in the x-axis are 0 and the sum of the forces in the y-axis are 0.\n\n##### Figure 13.19. The inclined plane problem.",
null,
"To begin the derivation we must first touch on a new concept called the normal force. This is nothing more than the force that the inclined plane pushes the object back with, or in other words, if you weigh 200 lbs., then there is a normal force of –200 lbs. pushing back (due to the surface tension of ground you're standing on) at you. We usually refer to the normal force as h, and it is equal in magnitude to\n\n```h = m*g.\n```\n\nInteresting huh? But if you lay a coordinate system down, then the gravity force must be opposite the normal force, or\n\n```h - m*g = 0.\n```\n\nThis is why everything doesn't get sucked into the ground. Okay, now that we know that, let's derive the motion equations of this block mass. First, we lay down an x,y coordinate system on the incline plane with +X parallel to the plane and in the downward sliding direction; this helps the math. Then we write the equilibrium equations for the x and y axes. For the x-axis we know that the component of gravity pushing the block is\n\n```force due to gravity = m*g*sin q.\n```\n\nAnd the force due to friction holding the block from sliding is\n\n```force due to friction = -h* ms.\n```\n\nThe negative sign is because the force acts in the opposite direction. When the object isn't sliding we know that the sum of these forces are equal to 0. Mathematically, we have\n\n```force due to gravity + force due to friction = 0\n```\n\nOr, the sum of forces in the x-axis is\n\n```S Fx = m*g*sin q - h*ms = 0.\n```\n\nMATH\n\nNote that I use sine and cosine to resolve the x,y components of the force. I'm basically just breaking the force vectors into components, nothing more.\n\nWe have to do the same for the y-axis, but this is fairly easy because the only forces are the weight of gravity and the normal force:\n\n```S Fy = h - m*g*cos q\n```\n\nAll right, so all together we have\n\n```S Fx = m*g*sin q - h*ms = 0.\nS Fy = h - m*g*cos q = 0.\n```\n\nBut, what is h? From S Fy, we once again see that:\n\n```h - m*g*cos q = 0.\n```\n\nHence,\n\n```h = m*g*cos q,\n```\n\ntherefore we can write:\n\n```S Fx = m*g*sin q - (m*g*cos q)*ms = 0.\n```\n\nThis is what we need. From it we can derive the following results:\n\n```m*g*sin q = (m*g*cos q)*ms\n\nqs = (m*g*sin q)/(m*g*cos q) = tan q\n```\n\nOr canceling out the m*g and replacing sin/cos by tan,\n\n```qcritical = tan-1 ms\n```\n\nNow listen carefully. This tells us that there is an angle called the critical angle (qcritical) at which the mass starts to slide. It is equal to the inverse tangent of the static coefficient of friction. If we didn't know the frictional coefficient of a mass and some incline plane, we could find it this way by tilting the plane until the mass starts to move. But this doesn't help us with the x-axis, or does it? The equation tells us that the mass won't slide until the angle qcritical is reached. When it is reached the mass will slide governed by:\n\n```S Fx = m*g*sin q - (m*g*cos q)*ms\n```\n\nWell, almost… There is one detail. When the mass starts to slide, the difference is m*g*sin q – (m*g*cos q)*_s > 0, but in addition we need to change the frictional coefficient to _k (the coefficient of kinetic friction) to be totally correct!\n\n```Fx = m*g*sin q - (m*g*cos q)*mk\n```\n\nTRICK\n\nYou can just average mk and ms and use that value in all the calculations. Because you're making video games and not real simulations, it doesn't matter if you oversimplify the two frictional cases into one, but if you want to be correct, you should use both frictional constants at the appropriate times.\n\nWith all that in mind let's compute the final force along the x-axis. We know that F=m*a, therefore:\n\n```Fx = m*a = m*g*sin q - (m*g*cos q)*mk\n```\n\nAnd dividing by m we get\n\n```a = g*sin q - (g*cos q)*mk\na = g*(sin q - qk*cos q)\n```\n\nYou can use this exact model to move the block mass, that is, each cycle you can increase the velocity of the block in the positive X-direction by g*(sin qmk*cos q). There's only one problem: This solution is in our rotated coordinate system! There's a trick to getting around this: You know the angle of the plane, and hence you can figure out a vector along the downward angle of the plane:\n\n```xplane = cos q\nyplane = -sin q\n\nSlide_Vector = (cos q, -sin q)\n```\n\nThe minus sign is on the Y-component because we know it's in the –Y direction. With this vector we can then move the object in the correct direction each cycle—this is a hack, but it works. Here's the code to perform the translation and velocity tracking:\n\n```// Inputs\nfloat x_pos = SX, // starting point of mass on plane\ny_pos = SY,\ny_velocity = 0, // initial y velocity\nx_velocity = 0, // initial x velocity\nx_plane = 0, // sliding vector\ny_plane = 0,\ngravity = 1, // do want to fall too fast\nvelocity = INITIAL_VEL, // whatever\n\n// must be in radians and it must be greater\n// than the critical angle\nangle = PLANE_ANGLE, // compute velocities in x,y\n\nfrictionk = 0.1; // frictional value\n\n// compute trajectory vector\nx_plane = cos(angle);\ny_plane = sin(angle); // no negative since +y is down\n\n// do slide loop until object hits\n// bottom of screen at SCREEN_BOTTOM\nwhile(y_pos < SCREEN_BOTTOM)\n{\n// update position\nx_pos+=x_velocity;\ny_pos+=y_velocity;\n\n// update velocity\nx_vel+=x_plane*gravity*(sin(angle) - frictionk *cos(angle));\ny_vel+=y_plane*gravity*(sin(angle) - frictionk *cos(angle));\n\n} // end while\n```\n\nThe point of physics modeling sometimes is just to understand what the underlying physics are so you can model them in a somewhat convincing manner. In the case of the incline plane, basically all that math just boiled down to the concept that acceleration is a function of the angle (we knew this from common sense). However, in Volume II of the book I'm going to cover much more realistic physics using numerical integration, and in those cases, you need to know the real models and real forces on everything.\n\n",
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http://planetary-mechanics.com/2017/01/22/resonances-around-the-giant-planets/ | [
"# Resonances around the giant planets\n\nHi there! Today the release of the paper Classification of satellite resonances in the Solar System, by Jing Luan and Peter Goldreich, is the opportunity for me to present you the mean-motion resonances in the system of satellites of the giant planets. That paper has recently been published in The Astronomical Journal, but the topic it deals with is present in the literature since more than fifty years. This is why I need to detail some of the existing works.\n\n## The mean-motion resonances (MMR)\n\nImagine that you have a planet orbited by two satellites. In a convenient case, their orbits will be roughly elliptical. The ellipse results from the motion of a small body around a large spherical one; deviations from the exact elliptical orbit come from the oblateness of the central body and the gravitational perturbation of the other satellite. If the orbital frequencies of the two satellites are commensurate, i.e. if Satellite A accomplishes N revolutions around the planet, while Satellite B accomplishes (almost exactly) M revolutions, i.e. M orbits, N and M being integers, then the 2 satellites will be in a configuration of mean-motion resonance. It can be shown that the perturbation of A on B (respectively of B on A) will not average to 0 but have a cumulative effect, due to the repetition, at the same place, of the smallest distance between the two bodies, the smallest distance meaning the highest gravitational torque. A consequence of a MMR is the increase of the eccentricity of one of the satellites, or of both of them, and / or their inclinations… or only the inclination of one of them. In the worst case, this could result in the ejection of one of the satellites, but it can also have less catastrophic but not less interesting consequences, like the heating of a body, and the evolution of its internal structure… We will discuss that a little later.\n\nA mean-motion resonance can be mathematically explained using the orbital elements, which describe the orbit of a satellite. These elements are\n\n• The semimajor axis a,\n• the eccentricity e. e=0 means that the orbit is circular, while e<1 means that the orbit is elliptical. For planetary satellites, we usually have e<0.05. With these two elements, we know the shape of the orbit. We now need to know its orientation, which is given by 3 angles:\n• the inclination i, with respect to a given reference plane. Usually it is the equatorial plane of the parent planet at a given date, and the inclination are often small,\n• the longitude of the ascending node Ω, which orientates the intersection of the orbital plane with the reference plane,\n• the longitude of the pericentre ϖ, which gives you the pericentre, i.e. the point at which the distance planet-satellite is the smallest. With these 5 elements, you know the orbit. To know where on its orbit the satellite is, you also need\n• the mean longitude λ.\n\nSaying that the Satellites A and B are in a MMR means that there is an integer combination of orbital elements, such as φ=pλA-(p+q)λA+q1ϖA+q2ϖB+q3ΩA+q4ΩB, which is bounded. Usually an angle is expected to be able to take any real value between 0 and 2π radians, i.e. between 0 and 360°, but not our φ. The order of the resonance q is equal to q1+q2+q3+q4, and q3+q4 must be even. Moreover, it stems from the d’Alembert rule, which I will not detail here, that a strength can be associated with this resonance, which is proportional to eAq1eBq2iAq3iBq4. This quantity also gives us the orbital elements which would be raised by the resonance.\n\nIn other words, if the orbital frequency of A is twice the one of B, then we could have the following resonances:\n\n• λA-2λBA (order 1), which would force eA,\n• λA-2λBB (order 1), which would force eB,\n• A-4λBAB (order 2), which would force eA and eB,\n• A-4λB+2ΩA (order 2), which would force iA,\n• A-4λB+2ΩB (order 2), which would force iB,\n• A-4λB+2ΩAB (order 2), which would force iA and iB.\n\nHigher-order resonances could be imagined, but let us forget them for today.\n\nThe next two figures give a good illustration of the way the resonances can raise the orbital elements. All of the curves represent possible trajectories, assuming that the energy of the system is constant. The orbital element which is affected by the resonance, can be measured from the distance from the origin. And we can see that the trajectories tend to focus around points which are not at the origin. These points are the centers of libration of the resonances. This means that when the system is at the exact resonance, the orbital element relevant to it will have the value suggested by the center of libration. These plots are derived from the Second Fundamental Model of the Resonance, elaborated at the University of Namur (Belgium) in the eighties.",
null,
"",
null,
"The Second Fundamental Model of the Resonance for order 1 resonances, for different parameters. On the right, we can see banana-shaped trajectories, for which the system is resonant. The outer zone is the external circulation zone, and the inner one is the internal circulation zone. Inspired from Henrard J. & Lemaître A., 1983, A second fundamental model for resonance, Celestial Mechanics, 30, 197-218.",
null,
"",
null,
"The Second Fundamental of the Resonance for order 2 resonances, for different parameters. We can see two resonant zones. On the right, an internal circulation zone is present. Inspired from Lemaître A., 1984, High-order resonances in the restricted three-body problem, Celestial Mechanics, 32, 109-126.\n\nHere, I have only mentioned resonances involving two bodies. We can find in the Solar System resonances involving three bodies… see below.\n\nIt appears, from the observations of the satellites of the giant planets, that MMR are ubiquitous in our Solar System. This means that a mechanism drives the satellite from their initial position to the MMRs.\n\n## Driving the satellites into resonances\n\nWhen the satellites are not in MMR, the argument φ circulates, i.e. it can take any value between 0 and 2π. Moreover, its evolution is monotonous, i.e. either constantly increasing, or constantly decreasing. However, when the system is resonant, then φ is bounded. It appears that the resonance zones are levels of minimal energy. This means that, for the system to evolve from a circulation to a libration (or resonant zone), it should loose some energy.\n\nThe main source of energy dissipation in a system of natural satellites is the tides. The planet and the satellites are not exactly rigid bodies, but can experience some viscoelastic deformation from the gravitational perturbation of the other body. This results in a tidal bulge, which is not exactly directed to the perturber, since there is a time lag between the action of the perturber and the response of the body. This time lag translates into a dissipation of energy, due to tides. A consequence is a secular variation of the semi-major axes of the satellites (contraction or dilatation of the orbits), which can then cross resonances, and eventually get trapped. Another consequence is the heating of a satellite, which can yield the creation of a subsurface ocean, volcanism…\n\nCapture into a resonance is actually a probabilistic process. If you cross a resonance without being trapped, then your trajectories jump from a circulation zone to another one. However, if you are trapped, you arrive in a libration zone, and the energy dissipation can make you spiral to the libration center, forcing the eccentricity and / or inclination. It can also be shown that a resonance trapping can occur only if the orbits of the two satellites converge.\n\n## The system of Jupiter\n\nJupiter has 4 large satellites orbiting around: J1 Io, J2 Europa, J3 Ganymede, and J4 Callisto. There are denoted Galilean satellites, since they were discovered by Galileo Galilei in 1610. The observations of their motion has shown that\n\n• Io and Europa are close to the 2:1 MMR,\n• Europa and Ganymede are close to the 2:1 MMR as well,\n• Ganymede and Callisto are close to the 7:3 MMR (De Haerdtl inequality)\n• Io, Europa and Ganymede are locked into the Laplace resonance. This is a 3-body MMR, which resonant argument is φ=λ1-3λ2+2λ3. It librates around π with an amplitude of 0.5°.\n\nThis Laplace resonance is a unique case in the Solar System, to the best of our current knowledge. It is favored by the masses of the satellites, which have pretty the same order of magnitude. Moreover, Io shows signs of intense dissipation, i.e. volcanism, which were predicted by Stanton Peale in 1979, before the arrival of Voyager I in the vicinity of Jupiter, from the calculation of the tidal effects.\n\n## The system of Saturn\n\nBesides the well-known rings and a collection of small moons, Saturn has 8 major satellites, i.e.\n\n• S1 Mimas,\n• S3 Tethys,\n• S4 Dione,\n• S5 Rhea,\n• S6 Titan,\n• S7 Hyperion,\n• S8 Iapetus,\n\nand resonant relations, see the following table.\n\nSatellite 1 Satellite 2 MMR Argument φ Libration center Libration amplitude Affected quantities\nS1 Mimas S3 Tethys 4:2 1-4λ313 0 95° i1,i3\nS2 Enceladus S4 Dione 2:1 λ2-2λ42 0 0.25° e2\nS6 Titan S7 Hyperion 4:3 6-4λ77 π 36° e7\n\nThe amplitude of the libration tells us something about the age of the resonance. Dissipation is expected to drive the system to the center of libration, where the libration amplitude is 0. However, when the system is trapped, the transition from circulation to libration of the resonant argument φ induces that the libration amplitude is close to π, i.e. 180°. So, the dissipation damps this amplitude, and the measured amplitude tells us where we are in this damping process.\n\n## This study\n\nThis study aims at reinvestigating the mean-motion resonances in the systems of Jupiter and Saturn in the light of a quantity, kcrit, which has been introduced in the context of exoplanetary systems by Goldreich & Schlichting (2014). This quantity is to be compared with a constant of the system, in the absence of dissipation, and the comparison will tell us whether an inner circulation zone appears or not. In that sense, this study gives an alternative formulation of the results given by the Second Fundamental Model of the Resonance. The conclusion is that the resonances should be classified into two groups. The first group contains Mimas-Tethys and Titan-Hyperion, which have large libration amplitudes, and for which the inner circulation zone exists (here presented as overstability). The other group contains the resonances with a small amplitude of libration, i.e. not only Enceladus-Dione, but also Io-Europa and Europa-Ganymede, seen as independent resonances.\n\n## A possible perspective\n\nIo-Europa and Europa-Ganymede are not MMR, and they are not independent pairs. They actually constitute the Io-Europa-Ganymede resonance, which is much less documented than a 2-body resonance. An extensive study of such a resonance would undoubtedly be helpful."
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https://www.thefreelibrary.com/Compliance+and+Fatigue+Life+Analysis+of+U-shape+Flexure+Hinge.-a0609835926 | [
"# Compliance and Fatigue Life Analysis of U-shape Flexure Hinge.\n\n1. Introduction\n\nThe flexure hinge is a kind of special kinematic pair which makes use of the deformation of materials to provide finite angular displacement of complex movement around central axis. Compared with the traditional rigid structure, the flexure hinge has no friction, no abrasion, no clearance, low noise, small space size. The flexure hinge can be designed and fabricated in one body, which can realize high precision motion. The compliance can be used to protect the mechanism against impact amongst many other advantages. Flexure hinges are extensively applied in various fields requiring ultra-precision positioning, micromanipulation, microelectronics, and micro assembly for optoelectronic components, optics, bioengineering, and many other fields [1-3].\n\nUp to date, researches on flexure hinge are mainly focused on the notch flexure hinge, reed flexure hinge, as well as flexible elements, such as rods, reeds, and the combination of flexure hinges with different notch shapes. The notch flexure hinge has higher precision but a smaller range of movement, which is more suitable for micro-displacement application. The reed flexure hinge has a larger rotation angle, but lower rotation accuracy. The combined hinge has better comprehensive performance [4-11].\n\nExtensive researches on flexure hinges have been reported. Tian et al. proposed a V-type hinge structure, for which the compliance and rotation precision characteristics were analyzed by finite element method . Chen et al. obtained two generalized models that are the conic model and the elliptic-arc-fillet model by using the ratio of the radius of curvature of the stress- concentrating feature to the minimum thickness as the only fitting variable . Xu et. al. presented analytic models of four types of flexure hinges: elliptic, circular, parabolic, and hyperbolic. These analytic models are developed based on the theory of elasticity and infinitesimal method, and devoted a hinge index by the ratio of rotational precision and rotational stiffness, to estimate the mechanical properties of diverse flexure hinges synthetically and quantitatively . Li et. al. developed a generalized analytical compliance model that can quickly formulate the equations of compliance and precision for hybrid flexure hinges . Wang et. al. presented the development of a parametric model for the rotational compliance of a cracked right circular flexure hinge. The rotational compliance of the cracked right circular flexure hinge was obtained from the sum of the rotational compliance of a healthy flexure hinge and the change of the rotational compliance due to the crack . Li et. al. presented a new type power-function-shaped flexure hinge, derived the closed-form compliance equations of the flexure hinge based on the unit-load method, and investigated the motion accuracy . Meng et. al. investigated the existing stiffness equations for corner-filleted flexure hinges. Three empirical stiffness equations for corner-filleted flexure hinges were formulated based on finite element analysis . Li et. al. derived empirical compliance equations for circular flexure hinge considering the effect of stress concentration .\n\nThe notch flexure hinge is mostly applied to micro-operation robots, magnifying mechanism for micrometric displacement, and compliance straight-line guidance mechanism, which has simple structure and can be processed as a whole. A thorough review of flexure hinges has indicated that different notch shape flexure hinges have been utilized to improve the performance. This paper presents a U-shape flexure hinge with four structures and the compliance equations for the flexure hinges. The influences of the structure parameters on the performance of these types of flexure hinges are investigated. It is found that the U-shape flexure hinges have a large compliance ranges corresponding to different notch size. This makes such flexure hinges capable of being used in wide potential applications with different requirements. The finite element analysis is used to comparing fatigue life of U-shape flexure hinges and circular flexure hinge, which reveals a longer fatigue life with U-shape design.\n\n2. Compliance analysis\n\nThe U-shape flexure hinges are shown in Fig. 1, where four structural forms are displayed. In the flexures, notches are constructed with elliptic arc or circular arc connected by straight line segments.\n\nThe structure parameters of the flexure hinge are shown in Fig. 2, including:\n\n* the opening width m;\n\n* the opening depth n of straight-line segments;\n\n* the semi major axis a and semi minor axis b of the ellipse;\n\n* the minimum thickness t of the flexure hinge center;\n\n* the length L, width W and height H of the flexure hinge.\n\nThe compliance of the flexure hinges is mainly determined by the material and structural parameters. Thus, it is necessary to establish the relations between the compliance and structural parameters of the flexure hinge and reveal the influence of the structural parameters on the compliance to provide the theoretical basis for such engineering design and optimization of the flexure hinge. In the engineering application, the flexure hinge can have micro deformations in the displacement and the angle under the external load. In the analysis of compliance, it is assumed that the left end of the flexure hinge (being fixed) and the right end (being free) are each under the influence of the axial force [F.sub.x]; shear forces [F.sub.y], [F.sub.z]; and bending moments [M.sub.y], [M.sub.z] as shown in Fig. 3. The coordinate system O-xyz is established with the geometric center of the hinge as the coordinate origin O.\n\nThe flexure hinge is subject to the above loads. According to Castigliano's second theorem :\n\n[[delta].sub.i] = [[partial derivative][V.sub.[epsilon]]]/[[partial derivative][F.sub.i]], (1)\n\nwhere: [[delta].sub.i] is displacement corresponding to force [F.sub.i], [V.sub.[epsilon]], is structural deformation energy, [F.sub.i] is generalized force.\n\nIf [F.sub.i] is axial tension pressure, Eq. (1) becomes:\n\n[[[delta].sub.i] = [[partial derivative][V.sub.[epsilon]]]/[[partial derivative][F.sub.i]] = [partial derivative]/[[partial derivative][F.sub.i]] [integral] [[F.sub.N.sup.2](x)]/[2EA]dx], (2)\n\nwhere: [F.sub.N] is axial force, E is modulus of elasticity, and A is cross sectional area.\n\nOtherwise, if [F.sub.i] is the bending moment, Eq. (1) is:\n\n[[delta].sub.i] = [[partial derivative][V.sub.[epsilon]]]/[[partial derivative][F.sub.i]] = [partial derivative]/[[partial derivative][F.sub.i]] [integral] [[M.sup.2](x)]/[2EI]dx, (3)\n\nwhere: M is bending moment, and I is area moment of inertia.\n\nA flexure hinge will generate five kinds of displacements, namely, linear displacements x, y, z and angular displacements [[alpha].sub.y], [[alpha].sub.z] respectively, which can be obtained based on the theory of material mechanics about load-bearing and deformation relations as follows:\n\n[mathematical expression not reproducible] (4)\n\nFor the convenience of analysis and calculation, region segmentation has been made for the notch part of the flexure hinge, as shown in Fig. 4.\n\nIn Fig. 4, lengths [l.sub.A], [l.sub.B] and [l.sub.C] are calculated by:\n\n[l.sub.A](x) = 2a - 2a cos[empty set] + t,\n\ndx = b cos [empty set]d[empty set], [empty set] [member of] [-[pi]/2, [pi]/2],\n\n[l.sub.B] (x) = 2[n + a cos [[empty set].sub.0] - a cos [empty set]],\n\ndx = b cos [empty set]d[empty set], [empty set] [member of] [-[pi]/2, [[empty set].sub.0]],\n\n[l.sub.C] (x) = 2[n + a cos [[empty set].sub.0] - a cos [empty set]],\n\ndx = b cos [empty set]d[empty set], [empty set] [member of] [[[empty set].sub.0], [pi]/2].\n\nLet a/t = p, s([empty set]) = 2p - 2p cos [empty set] + 1, a/n = q,\n\nk([empty set]) = 1 + q cos [[empty set].sub.0] - q cos [empty set],\n\nthen: [l.sub.A] (x) = t x s([empty set]), [l.sub.B](x) = [l.sub.C](x) = 2n x k([empty set]).\n\nThe compliance equations can be derived as follow:\n\n1) Angular compliance about y-axis and z-axis:\n\n[mathematical expression not reproducible] (5 a)\n\n[mathematical expression not reproducible] (5 b)\n\n[mathematical expression not reproducible] (5 c)\n\n[mathematical expression not reproducible] (5 d)\n\n2) Linear compliance along x-, y- and z-axis:\n\n[mathematical expression not reproducible] (5 e)\n\n[mathematical expression not reproducible] (5 f)\n\n[mathematical expression not reproducible] (5 g)\n\n[mathematical expression not reproducible] (5 h)\n\n[mathematical expression not reproducible] (5 i)\n\nAmong Eqs. (1-9):\n\n[mathematical expression not reproducible]\n\nFrom Eqs. (5 a - i), it can be seen that the compliance of U-shape flexure hinge is inversely proportional to the elasticity modulus and the width of the hinge. Moreover, compliance is influenced by the parameters of the notch size. Upon setting the basic structural parameters H and W for the U-shape flexure hinge, its compliance can be expressed as a function of the notch sizes:\n\nC = Cf(a,b,m,t). (6)\n\nIn the following section, the influence of the parameters of notch size on the compliance of the flexure hinges will be analyzed.\n\n3. Influences of the parameters of notch size on the compliance\n\nWithout loss of generality, we assume that two of the parameters of notch sizes are variables and the other two parameters are constant. The structural parameters of the flexure joint are given as H=30 mm, W=10 mm.\n\nWhen m and t are set as constant, a and b are variables, varying in the ranges of a [member of] [4,5] and b [member of] [2,5]. When a and b are fixed, m and t are variables, varying in the ranges of m [member of] [4,8], t [member of] [4,5]. Then the relationship between compliance of the U-shape flexure hinge and the parameters of notch sizes a, b and m, t are plotted using Matlab, as shown in Fig. 5, a - g.\n\nFig. 5, a-g show each compliance of the U-shape flexure hinges with varying a and b. We can observe from the plots characteristics, as described presently.\n\nWhen a increases, the compliance decreases with a nonlinearity. When b increases, the compliance increases approximate linearly. The influence of b incompliance is more significant. In each compliance, [??] has the maximum value and [??] has the minimum value within the value range of a and b. When a=5 and b=2, each compliance reaches the smallest value.\n\nFig. 6 shows the compliance of the U-shape flexure hinges varying with respect to m, t when a [not equal to] b. Fig. 6 shows that the changes of the compliance of [??] (Fig. 6, a), [??] (Fig. 6, b) and [??] (Fig. 6, f) are consistent with increase of m and t, the compliance increases approximately linearly as m increases, and the compliance decreases approximately linearly as t increases. The change of the compliance of [??] (Fig. 6, c), [??] (Fig. 6, e), and [??] (Fig. 6, g) are consistent with the increase of m and t. The compliance increases approximately linearly as m and t increase. The compliance [??] (Fig. 6, d) increases nonlinearly as m and t increase. When m and t are the smallest value in the range, the compliance [??], and [??] have the smallest value in the range. Fig. 7, a - g show the compliance of the U-shape hinge varying with respect to m and t when a = b . Fig. 7 shows that the changes of the compliance of [??] (Fig. 7, a), [??] (Fig. 7, b) and [??] (Fig. 7, f) are consistent with increase of m and t, the compliance increases approximately linearly as m increases, and the compliance decreases approximately linearly as t increases. The change of the compliance of [??] (Fig. 6, c), [??] (Fig. 6, d), [??] (Fig. 6, e), and [??] (Fig. 6, g) are consistent with increase of m and t, the compliance increases nonlinearly as m and t increase. When m and / are the smallest in the range, the compliance [??], and [??] have the smallest values in the range too.\n\n4. Simulation of fatigue life\n\nWhen the flexure hinge is applied to the flexible mechanism, the deformation of the mechanism is mainly caused by the large stress of the flexure hinge. The flexible mechanism usually uses piezoelectric ceramics to drive the flexure hinge to produce deformation. The mechanism failure is mainly reflected in the fatigue failure of flexure hinge. Therefore, it is necessary to study the fatigue life of flexure hinge to ensure the reliability of flexible mechanism.\n\nThe fatigue failure of flexure hinge under alternating load is generally occurred during the operation. Alternating loads produce alternating stress. Generally, alternating loads are decomposed into a load of constant amplitude and a load of variable amplitude. In this work, a periodic load of constant amplitude 10 N in the X and Y directions is considered. Material is structural steel with Young's Modulus 2e+011 Pa, Poisson's Ratio 0.3 , Bulk Modulus 1.6667e+011 Pa, Shear Modulus 7.6923e+010 Pa. The stress and cycle times are shown in Table 1. The fatigue life of U-shape flexure hinge was analysed by ANSYS simulation, and a series of fatigue life of U-shape flexure hinge with different notch size parameters were obtained, as listed in Table 2.\n\nThe main notch parameters of the U-shape flexure hinge are a, b, m, t. There are thirty models in three groups. The notch parameters are as follows:\n\na=8, b=8, m=14, while t is variable;\n\na=6, b=5, t=6, while m is variable;\n\nt=1, m=3, while a, b are variables.\n\nThe finite element model of the U-shape flexure hinge is built in the workbench. The left end is fixed, the right end loading (axial force, lateral force, bending moment), insert the fatigue analysis module Fatigue Tool and set the correction coefficient of fatigue analysis. The fatigue life cloud figure of the U-shape flexure hinge is shown in Fig. 8. Fatigue life is obtained, as shown in Table 2.\n\nCurve fitting is carried out in Matlab. The influence curve of the notch parameter on the fatigue life of flexure hinge is obtained, as shown in Fig. 9, a-c.\n\n5 Comparison of the fatigue life and stress of U-shape flexure\n\nU-shape flexure hinges and circular arc flexure hinge with the same center thickness t=6 are selected for analysis and comparison. The notch parameters of circular-arc-shaped flexure hinge are r=12, m=24, t=6; The model is established in ANSYS to analyze the fatigue life and stress. The analysis results of circular-arc-shaped flexure hinge are shown in Figs.10, a-b. The notch parameters and analysis results of four structures U-shape flexure hinges are shown in Table 3.\n\nThe results show that the fatigue life of the three structures of U-shape flexure hinges (U-shape with a semicircle is tangent to a line segment, U-shape with a semi-ellipse is tangent to the line segment, U-shape with an arc intersects the line segment) is higher than that of circular arc flexure hinge, while the stress of three structures of U-shape flexure hinges is less than that of circular arc flexure hinge. Therefore, the three structures of U-shape flexure hinges are more reliable than the circular arc flexure hinge.\n\n6. Conclusions\n\nThe compliance equations for the U-shape flexure hinges have been derived using the Castigliano's second theorem. The influence of structure parameters on the compliance of U-shape flexure hinges is analyzed on the basis of establishing the model. The results indicate that the compliance of U-shape flexure hinge have different trends with notch parameters change ,and have different sensitivity to the change of the notch parameters within the given range. The influence of notch parameters b and ton compliance is more obvious than that of notch parameters a and m.\n\nThe fatigue life is analyzed by changing the U-shape flexure hinge notch parameter. The results show that the fatigue life of flexure hinge increases gradually with the increasing of flexure hinge center thickness t. The fatigue life of flexure hinge increases with the increasing of hinge notch width m. With the increasing of the major axis of the ellipse a and semi minor axis of the ellipse b, the fatigue life of flexure hinge fluctuates locally, the general trend is a gradual decrease. There are three structures of the U-shape flexure hinges that are more reliable than the circular arc flexure hinge, U-shape with a semi-circle is tangent to a line segment, U-shape with a semi-ellipse is tangent to the line segment, and U-shape with an arc intersects the line segment.\n\nIn summary, compliance and fatigue life of the U-shape flexure hinges is related to material properties and structure parameters, among which the influence of the notch parameters on the compliance and fatigue life is more obvious.\n\nAcknowledgement\n\nThis research was funded by the key research and development project of Shanxi province (Grant Nos. 201803D421027, 201803D421028), the foundation of Shanxi key laboratory of advanced manufacturing technology (Grant No. XJZZ201702), and the natural science foundation of Shanxi province (Grant No. 015011060).\n\nReferences\n\n[1.] Yu, J. J.; Pei, X.; Bi, S. S.; Zong, G. H.; Zhang, X. M. 2010. State-of-arts of design method for flexure mechanisms, Journal of Mechanical Engineering 46(13): 2-13 (in Chinese).\n\n[2.] Yu, J. J.; Hao, G. B.; Chen, G. M.; Bi, S. S. 2015. State-of-art of compliant mechanisms and their applications, Journal of Mechanical Engineering 51(13): 53-68 (in Chinese).\n\n[3.] Wu, J. W.; Zhang, Y.; Cai, S.; Cui, J. W. 2018. Modeling and analysis of conical-shaped notch flexure hinges based on NURBS, Mechanism and Machine Theory 128: 560-568.\n\n[4.] Yang, M.; Du, Z.J.; Dong, W. 2016. Modeling and analysis of planar symmetric superelastic flexure hinges, Precision Engineering 46: 177-183.\n\n[5.] Chen, G. M., Howell, L. L. 2009. Two general solutions of torsional compliance for variable rectangular cross-section hinges in compliant mechanisms, Precision Engineering 33(3): 268-274.\n\n[6.] Tian, Y.; Shirinzadeh, B.; Zhang, D.; Zhong, Y. 2010. Three flexure hinges for compliant mechanism designs based on dimension less graph analysis, Precision Engineering 34(1): 92-100.\n\n[7.] Friedrich, R., Lammering, R., Rosner, M. 2014. On the modeling of flexure hinge mechanisms with finite beam elements of variable cross section, Precision Engineering 38(4): 915-920.\n\n[8.] Dirksen, F.; Anselmann, M.; Zohdi, T. I.; Lammering, R. 2013. Incorporation of flexural hinge fatigue-life cycle criteria into the topological design of compliant small-scale devices, Precision Engineering 37(3): 531-541.\n\n[9.] Wang, X. J.; Liu, C. L.; Gu, J. J.; Zhang, W. J. 2015. A parametric model for rotational compliance of a cracked right circular flexure hinge, International Journal of Mechanical Sciences 94-95: 168-173.\n\n[10.] Li, L. J.; Zhang, D.; Guo, S.; Qu, H. B. 2017. A generic compliance modeling method for two-axis elliptical-arc-filleted flexure hinges, Sensors 17(9): 21-54.\n\n[11.] Wu, J. W.; Cai, S; Cui, J. W.; Tan, J. B. 2015. A generalized analytical compliance model for cartwheel flexure hinges, Review of Scientific Instruments 86: 105003(1-11).\n\n[12.] Tian, Y.; Shirinzadeh, B.; Zhang, D. Closed-form compliance equations of filleted V-shaped flexure hinges for compliant mechanism design, Precision Engineering 34(3): 408-418.\n\n[13.] Chen, G. 2014. Generalized equations for estimating stress concentration factors of various notch flexure hinges, Journal of Mechanical Design 136(3): 252-261.\n\n[14.] Xu, N.; Dai, M.; Zhou, X.Q. 2017. Analysis and design of symmetric notch flexure hinges, Advances in Mechanical Engineering 9(11): 1-12.\n\n[15.] Li, L. J.; Zhang, D.; Guo, S.; Qu, H. B. 2019. Design, modeling, and analysis of hybrid flexure hinges, Mechanism and Machine Theory 131: 300-316.\n\n[16.] Wang, X. J.; Liu, C. L.; Gu, J. J.; Zhang, W. J. 2015. A parametric model for rotational compliance of a cracked right circular flexure hinge, International Journal of Mechanical Sciences 94-95: 168-173.\n\n[17.] Li, Q.; Pan, C. Y.; Xu, X. J. 2013. Closed-form compliance equations for power-function-shaped flexure hinge based on unit-load method, Precision Engineering 37:135-145.\n\n[18.] Meng, Q.; Li, Y.; Xu, J. 2013. New empirical stiffness equations for corner-filleted flexure hinges, Mechanical Sciences 4(2): 345-356.\n\n[19.] Li, T. M.; Zhang, J. L.; Jiang, Y. 2015. Derivation of empirical compliance equations for circular flexure hinge considering the effect of stress concentration, International Journal of Precision Engineering and Manufacturing 16(8): 1735-1743.\n\n[20.] Lobontiu, N.; Paine, J. S. N.; Garcia, E.; Goldfarb, M. 2002. Design of symmetric conic-section flexure hinges based on closed-form compliance equations, Mechanism and Machine Theory 37: 477-498.\n\nJingjing LIANG (*), Ruiqin LI (*), Shaoping BAI (**), Qing LI (*), Fengping NING (*), Shuhua KANG (*)\n\n(*) School of Mechanical Engineering, North University of China, Taiyuan, China,03005, E-mail: [email protected] (Corresponding author)\n\n(**) Department of Mechanical and Manufacturing Engineering, Aalborg University, 9220 Aalborg, Denmark, E-mail:[email protected]\n\nAccepted November 21, 2019\n\nhttp://dx.doi.org/10.5755/j01.mech.25.6.22686\n```Table 1\nStresses and cycles\n\nAlternating stress, Pa Cycles\n\n3.999e+009 10\n2.827e+009 20\n1.896e+009 50\n1.413e+009 100\n1.069e+009 200\n4.41e+008 2000\n2.62e+008 10000\n2.14e+008 20000\n1.38e+008 1.e+005\n1.14e+008 2.e+005\n8.62e+007 1.e+006\n\nTable 2\nFatigue life of U-shape flexure hinge under different incision\nparameters\n\na=8, b=8, m=14 t=1 t=2 t=3\n\nN (logN) 13.803 (1.14) 599.73 (2.778) 2706.6 (3.4324)\na=8, b=8, m=14 t=6 t=7 t=8\nN (logN) 186060 (5.2697) 15545 (4.1916) 36157 (4.5582)\na=6, b=5, t=6 m=3.5 m=4 m=4.5\nN (logN) 146010 (5.1644) 145230(5.1621) 146750(5.1666)\na=6, b=5, t=6 m=6 m=6.5 m=7\nN (logN) 145190(5.1619) 145470(5.1628) 147800(5.1697)\nt=1, m=3 a=2.5, b=1.5 a=3, b=2 a=3.5, b=2.5\nN (logN) 17.16(1.2345) 16.537(1.2185) 13.775(1.1391)\nt=1, m=3 a=5, b=4 a=5.5, b=4.5 a=6, b=5\nN (logN) 14.77(1.1694) 14.633(1.1653) 14.042(1.1474)\n\na=8, b=8, m=14 t=4 t=5\n\nN (logN) 13090 (4.1169) 73441 (4.8659)\na=8, b=8, m=14 t=9 t=10\nN (logN) 64214 (4.8076) 85065 (4.9298)\na=6, b=5, t=6 m=5 m=5.5\nN (logN) 145460(5.1627) 147840(5.1698)\na=6, b=5, t=6 m=7.5 m=8\nN (logN) 147230(5.168) 147300(5.1682)\nt=1, m=3 a=4, b=3 a=4.5, b=3.5\nN (logN) 13.861(1.1418) 15.006(1.1763)\nt=1, m=3 a=6.5, b=5.5 a=7, b=6\nN (logN) 13.975(1.1454) 13.811(1.1402)\n\nTable 3\nFatigue life and stress of U-shape flexure hinges\n\nSemicircle is tangent to a a=b=10 a=b=9 a=b=8\nline segment notch m=20, t=6 m=18, t=6 m=16, t=6\nFatigue life 1.9397e5 1.9042e5 1.8603e5\nStress 4.8043e6 4.8543e6 4.9154e6\nSemi-ellipse is tangent to a=11.5, b=10 a=11, b=10 a=11, b=10\nthe line segment notch m=20, t=6 m=18, t=6 m=20, t=6\nFatigue life 1.8899e5 1.8297e5 1.9083e5\nStress 4.87e6 4.9554e6 4.8474e6\nArc intersects the line a=b=9 a=b=8 a=b=7\nsegment notch m=17, t=6 m=14, t=6 m=10, t=6\nFatigue life 1.9042e5 1.8606e5 1.7898e5\nStress 4.8543e6 4.9152e6 5.0048e6\nElliptical arc intersects a=9, b=7 a=8, b=7 a=8, b=6\nthe line segment notch m=13.5, t=6 m=13, t=6 m=11, t=6\nFatigue life 1.655e5 1.7192e5 1.5298e5\nStress 5.1749e6 5.0946e6 5.3338e6\n\nSemicircle is tangent to a a=b=7 a=b=6\nline segment notch m=14, t=6 m=12, t=6\nFatigue life 1.7897e5 1.707e5\nStress 5.005e6 5.1129e6\nSemi-ellipse is tangent to a=10, b=8 a=9,6=8\nthe line segment notch m=16, t=6 m=16, t=6\nFatigue life 1.7569e5 1.7987e5\nStress 5.0468e6 4.9912e6\nArc intersects the line a=b=6.5 a=b=6\nsegment notch m=10, t=6 m=10, t=6\nFatigue life 1.7638e5 1.7069e5\nStress 5.0413e6 5.1131e6\nElliptical arc intersects a=7, b=6 a=6, b=5\nthe line segment notch m=11, t=6 m=9, t=6\nFatigue life 1.6186e5 1.4747e5\nStress 5.2222e6 5.4046e6\n```"
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https://vient-damit.com/mz-c461854o/Black-scholes-Delta-calculator.html | [
"Home\n\n# Black scholes Delta calculator\n\nBLACK SCHOLES CALCULATOR. Spot. Volatility(%) Risk free yield(%) Dividend yield(%) Expiry (in years) Strike. Type. Call. Put. Calculate. GREEK(S) VALUE; Premium: Delta: Gamma: Vega: Theta: Rho: DELTA. VEGA. GAMMA. THETA Dear Math, I don't want to solve your problems. I have my own problems to solve. — Anonymous 4th grader I don't know why. Brokerage calculator Margin calculator Holiday calendar. Updates. Z-Connect blog Pulse News Circulars / Bulletin IPOs. Education. Varsity Trading Q&A. Black & Scholes Option Pricing Formula. Spot. Strike. Expiry. Volatility (%) Interest (%) Dividend. Calculate. Call Option Premium Put Option Premium Call Option Delta Put Option Delta Option.\n\nCall Delta Put Delta Volatility* Call Gamma Put Gamma Interest Rate* Call Vega Put Vega Time To Exp* Call Theta Put Theta Call Rho Put Rho *e.g. Enter 0.25 for 25%, or 0.5 for half a year. Black-Scholes Call Option Pricing Table Stoc This calculator utilizes the inputs below to generate call & put prices, delta, gamma, and theta from the Black-Scholes model. INPUTS (Change the numbers below to calculate other option price, delta, and gamma values.) Underlying Value: Strike: Vol: (0.20 = 20% implied volatility) Int Rate: Dividend: TTE: days = 0.05479452. Calculate Option Price using the Option Calculator based on the Black Scholes model. Option Greeks are option sensitivity measures. ADANIPORTS 730.05-2.29 % ASIANPAINT 2536.40-3.04 % AXISBANK 714.90-.63 % Delta is the most important of all the option greeks. Delta is usually expressed in percentage or decimal number",
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"[ Black Scholes Calculator ] Option; Strike : Expiration (years) Stock; Price : Volatility : Dividen According to the Black-Scholes option pricing model (its Merton's extension that accounts for dividends), there are six parameters which affect option prices: S 0 = underlying price ($$per share) X = strike price ($$$per share) σ = volatility (% p.a. Find an Explicit Solution for Delta in Black-Scholes Ophir Gottlieb 11/7/2007 1 Introduction We have seen through the creation of a replicating portfolio that the delta required to hedge an European call option is simply ∂C ∂S. Now we will explic-itly compute delta by differentiating the closed form Black-Scholes Formul Delta; Gamma; Theta; Vega; Rho; Putting It All Together; Volatility & the Greeks; Put/Call Parity ; Black-Scholes Formula; Options Quotes & Calculators. Today's Most Active Options; Options Quotes; Historical and Implied Volatility; Options Strategy Builders; Options Calculator; Collar Calculator; Covered Call Calculator; Reference Library. This article has shown algorithmic delta hedging using the Black-Scholes model and intuition from binomial trees to maintain a risk free portfolio. It is obvious as the underlying asset's price. The Black-Scholes model develops partial differential equations whose solution, the Black-Scholes formula, is widely used in the pricing of European-style options. Black-Scholes Option Pricing Calculator Black Scholes Calculator. This Black Scholes calculator uses the Black-Scholes option pricing method Option Pricing Models Option Pricing Models are mathematical models that use certain variables to calculate the theoretical value of an option. The theoretical value of an to help you calculate the fair value of a call Call Option A call option, commonly referred to as a call, is a form of a. Using the Black and Scholes option pricing model, this calculator generates theoretical values and option greeks for European call and put options ### Delta Quants - Online Black Scholes Option Calculator The Black-Scholes / ˌ b l æ k ˈ ʃ oʊ l z / or Black-Scholes-Merton model is a mathematical model for the dynamics of a financial market containing derivative investment instruments. From the partial differential equation in the model, known as the Black-Scholes equation, one can deduce the Black-Scholes formula, which gives a theoretical estimate of the price of European-style. We can use the above equation to calculate Delta (rough figure, the true figure can be obtained through other complex models like Black and Scholes) Delta =$-0.1700 / $0.8000 Delta will be - Delta =$-0.212 In the example from the Black-Scholes Calculator I use the first formula. The whole formula for gamma (same for calls and puts) is: =EXP (-1*POWER (K44,2)/2)/SQRT (2*PI ())*S44/ (A44*J44 The Delta: The binomial model • Recall the replicating portfolio for a call option on a stock S: ∆ shares of stock & B invested in the riskless asset. • So, the price of a call at any time t was C = ∆S +Bert with S denoting the price of the stock at time\n\nThe Black-Scholes call option formula is calculated by multiplying the stock price by the cumulative standard normal probability distribution function. Thereafter, the net present value (NPV) of.. In this blog I will demonstrate how to build a simple Black-Scholes options calculator by creating a table-valued function and using the XLeratorDB/statistics functions module. One of the great financial engineering innovations of the twentieth-century was the development of formulae to evaluate options The Greeks are vital tools in risk management.Each Greek measures the sensitivity of the value of a portfolio to a small change in a given underlying parameter, so that component risks may be treated in isolation, and the portfolio rebalanced accordingly to achieve a desired exposure; see for example delta hedging.. The Greeks in the Black-Scholes model are relatively easy to calculate, a. This example shows how to find the Black-Scholes delta sensitivity for an underlying asset price change. [CallDelta, PutDelta] = blsdelta(50, 50, 0.1, 0.25, 0.3, 0) CallDelta = 0.595\n\n### Zerodha - Black & Scholes calculato\n\nThe Black Scholes Calculator uses the following formulas: C = SP e-dt N (d1) - ST e-rt N (d2) P = ST e-rt N (-d2) - SP e-dt N (-d1) d1 = (ln (SP/ST) + (r - d + (σ2/2)) t) / σ √ Black Scholes model/formula/equation is very complicated.Some calculator based on it is very useful.Using this calculator,I have observed something.I have taken data like this.Call option,spot price=110,strike price=100,risk free interest=10%,expiry time=30 days,implied volatility=30%,but it reduces daily @1%.All datas are imaginaries.Only.\n\nBLACK AND SCHOLES (BS) FORMULA The equilibrium price of the call option (C; European on a non-dividend paying stock) is shown by Black and The delta of the investor™s hedge position is therefore zero. The delta of the asset position o⁄sets the delta of the option position Calculate strike from Black Scholes delta. Ask Question Asked 4 years, 9 months ago. Active 11 months ago. Viewed 9k times 3. 3 $\\begingroup$ I have a list of deltas and their corresponding volatilities in an FX market but I want to go from delta to strike price. In this Question similar problem is being discusse The Black-Scholes model is an elegant model but it does not perform very well in practice. For example, it is well known that stock prices jump on occasions and do not always move in the continuous manner predicted by the GBM motion model. Stock prices also tend to have fatter tails than those predicted by GBM Black-Scholes Calculator. To calculate a basic Black-Scholes value for your stock options, fill in the fields below. The data and results will not be saved and do not feed the tools on this website.Remember that the actual monetary value of vested stock options is the difference between the market price and your exercise price\n\nDelta Gamma Hedging and the Black-Scholes Partial Differential Equation (PDE) Sudhakar Raju1 Abstract The objective of this paper is to examine the notion of delta-gamma hedging using simple stylized examples. Even though the delta-gamma hedging concept is among the most challenging concepts in derivatives Price Delta Gamma Theta Vega; Call: 4.572: 0.523: 0.035-0.076: 0.114: Put: 4.572-0.477: 0.035-0.076: 0.11 Black Scholes Pricing Analysis Calculator Black-Scholes Formula (d1, d2, Call Price, Put Price, Greeks) This page explains the Black-Scholes formulas for d1, d2, call option price, put option price, and formulas for the most common option Greeks (delta, gamma, theta, vega, and rho) Black Scholes Calculator Option Greeks. Option Greeks are option sensitivity measures. The Greek is used in the name because these are denoted by Greek letters. Delta is the most important of all the option greeks. Delta is usually expressed in percentage or decimal number. The legitimate values of delta are as follows Option price calculator (Black and Scholes) Parameters of the option Type of option Call option Put option. Calculation date Option price Days to option expiry Delta Gamma Theta Vega Rho Important: The calculators on this site are put at your disposal for information purposes only. Their author can in no case be held responsible for their.\n\nBlack-Scholes Option Price Calculator. Spot Price $Call Put Strike Price$ Interest Rate % Dividend Yield % Volatility % Expiration Date. Price: Delta: Gamma: Theta: Vega: Rho: Buying one of these books will help support this website. Bet Smart: The Kelly System for Gambling and Investing. Pairs Trading: A Bayesian Example Black-Scholes Options Calculator. Days to Expiration: day(s) Strike Price $: Stock Price$ Premium; Call Premiu The Black-Scholes calculator allows to calculate the premium and greeks of a European option. It also acts as an Implied Volatility calculator: if you enter a Premium, the Implied Volatility will appear in the Volatility field Black-Scholes Calculator To calculate a basic Black-Scholes value for your stock options, fill in the fields below. The data and results will not be saved and do not feed the tools on this website. Remember that the actual monetary value of vested stock options is the difference between the market price and your exercise price The Black Scholes formula is calculated by multiplying the stock price by the cumulative standard normal probability distribution function. Then, the net present value (NPV) of the strike price multiplied by the cumulative standard normal distribution is subtracted from the resulting value of the previous calculation\n\n### Black Scholes Calculator - Driving Your Success\n\n• KEEP IN TOUCH. 1255 Phillips Square Suite 605 Montreal, Quebec Canada H3B 3G5 (888) 238-1314 (514) 840-9719 (888) 239-1482 (514) 840-9737; [email protected]\n• Options price and greeks calculator uses Black-Scholes formula to compute the value of a call/put option, given the option's time to expiry and strike price, the implied volatility and spot price of the underlying stock, the dividend yield and the rate of interest.Black-Scholes option-pricing model is useful for computing the present value of a stock option in light of current market conditions\n• Black-Scholes Option Price Calculator. Option Price Calculator to calculate theoretical price of an option based on Black Scholes Option pricing formula: Spot Price: Strike Price: Volatility % Risk Free Rate What is Option delta, gamma, vega,...? Learn about Option Greeks\n• Black Scholes - how to calculate delta with a vol skew. Ask Question Asked 6 years, 5 months ago. Active 5 years, 5 months ago. Viewed 3k times 3 $\\begingroup$ I am trying to calculate the delta of an option at different strike prices where the underlying has a pronounced implied volatility skew in order to correctly hedge an options strategy..\n• Black Scholes Excel Only it is from the Black and Scholes Page. SureshJune 16th, 2019 at 11:51am. Hi , Thank you very much for the workbook. Do you have a excel book without VBA code? to understand the formulas? if yes can you please share? PeterApril 23rd, 2019 at 12:14am. Hi Deepak, Apologies for the delay; I've been away on vacation\n\n### Option Price, Delta & Gamma Calculator - Trading Volatilit\n\n• The Black-Scholes model for European options pricing gives us the ability to compute a more accurate price and delta in continuous time. The proof for the Black-Scholes model is lengthy with a variety of assumptions and lemmas, so I will cut to an explanation of the equation. C— Euro Call Option Value P — Euro Put Option Valu\n• ing fair prices of options. It requires five variables: the strike price of an option, the current stock price, the time..\n• Unlike the original, classic Black-Scholes model, the QuantLib BlackScholesCalculator also supports an optional dividend yield. Also, a subtle quirk of the BlackScholesCalculator implementation to watch out for is that the constructor expects sigma to be multiplied by the square root of time\n• The Black Scholes model is a mathematical model that models financial markets containing derivatives. The Black Scholes model contains the Black Scholes equation which can be used to derive the Black Scholes formula. The Black Scholes formula can be used to model options prices and it is this formula that will be the main focus of this article\n• THE BLACK-SCHOLES OPTION PRICING FORMULA INPUT PANEL: ENTER OPTION DATA T Time to Maturity (days) Sigma Stock Price Volatility (enter in percentage form) Exercise Price r Interest Rate (enter in percentage form) S Stock Price OUTPUT PANEL: C Black-Scholes Call Price Delta Delta (Hedge Ratio) E P Black-Scholes Put Pric\n• Delta measures the change in an option's premium, based on how the price of the underlying security changes. The use of the Greeks was popularized in the Black Scholes Model Option Pricing Models Option Pricing Models are mathematical models that use certain variables to calculate the theoretical value of an option\n• The Black-Scholes-Merton (BSM) model Black and Scholes (1973) and Merton (1973) derive option prices under the following assumption on the stock price dynamics, dS t = S tdt + ˙S tdW t The binomial model: Discrete states and discrete time (The number of possible stock prices and time steps are both nite)\n\n### Option Calculator Black Scholes model Option Greeks\n\n1. ing the value of an at the money straddle 59. Chapter 7 Delta II 61. Deter\n2. The Black-Scholes (1973) model states that the theoretical price C of a European call option on a non dividend paying stock is (1) C = S 0 N (d 1) − X e − r T N (d 2\n3. The Black and Scholes model is the most widely used option model, appreciated for its simplicity and ability to generate a fair value for options pricing in all kinds of markets. This book shows you the ins and outs of the model, giving you the practical understanding you need for setting up and managing an option strategy\n4. this is called the delta of the call option. Thus the proper hedge ratio for the portfolio is the delta of the option. Consider a stock with a price of $100 and a volatility of 0.2. When the risk-free interest rate is 10% (0.1) the price of a one-year call with an exercise price of$100 based upon the Black-Scholes formula is $12.993 5. With that being said, Black Scholes is popular for pricing European options because of the simplicity and speed of using an analytic formula as opposed to having a more complex model that can only be evaluated using a numerical method, as DumbCoder mentioned (should note that, for many other types of derivative contracts, e.g. American or. Black-Scholes Plot. The Black-Scholes Option Pricing Model is an important investment instrument for option pricing. We provide an interactive plot below to show the influence of six variables on the price and Greeks of the European call and put options la formule de Black-Scholes et expliquer les facteurs N(d1)etN(d2). Il montreaussicommentlesmodelesbinomiauxdesprixd'optionsd'uneetde plusieursp´eriodespeuventˆetreexprim´esd'unefa¸contellequ'ilsimpliquent desanaloguesdeN(d1)etN(d2)quiontlamˆemeinterpr´etationquedansle modeledeBlack-Scholes BLACK.SCHOLES calculates the price of an option using the Black & Scholes option pricing formula. It's a well-known formula that calculates theoretical values of an investment based on the price of an asset, the strike price, time to expiry, interest rate, and volatility. The Black Scholes Calculator is defined in these formulas Delta Gamma Theta Vega find value of stock black scholes calculator iron condor option strategy stock calculator stock trading strike price success stock market trading options covered calls stock market strategies stock value stock risk volume vs open interest bull spread call what stocks pay dividends options calendar spreads options open. Black-Scholes Merton Model Calculator (With Greeks), Option Strategies Layout and Delta Hedging Calculator. This model can be used by students and professionals to determine the value of options, and specific trading strategies Black-Scholes and the Greeks. I wanted to get a better understanding of using Python to play around with options. We'll have a look at creating some option payoff functions, an implementation of Black-Scholes pricing and then finish up with some sensitivity analysis (Greeks) The Black-Scholes Formula. The Black-Scholes formula was the first widely used model for option pricing. A strategist can use this formula to calculate theoretical value for an option using current stock prices, expected dividends, the option's strike price, expected interest rates, time to expiration and expected stock volatility Delta is the derivative of option value with respect to the underlying asset price. It's positive for Calls and negative for Puts. Download the Black Scholes and Greeks Calculator for Excel . 6 thoughts on Black-Scholes Option Pricing and Greeks Calculator for Excel. Black-Scholes formula (from which delta is then recognized), that is far from the best representation for numerical calculations. Not all models are homogeneous. The Bachelier-model (where S is arithmetic Brownian motion), the constant elasticity of variance model, the SABR stochastic volatility mode",
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"Black-Scholes App. The following app will calculate the Black-Scholes European call option price for a set of given inputs. If the stock pays a dividend, then input the stock's annualized expected dividend yield. The calculator will adjust for the dividend by lowering the stock price by the present value of the expected dividend Get VBA and an Excel spreadsheet for Black-Scholes and the Greeks (Delta, Gamma, Vega, Theta, Rho) here. You can easily use the VBA in your own option pricing spreadsheets. This VBA and the corresponding Excel spreadsheet prices a European option with continuous dividends) we're now going to talk about probably the most famous formula in all of finance and that's the black Scholes formula sometimes called the black Scholes Merton formula and it's named after these gentlemen this right over here is Fischer black this is Myron Scholes and they really laid the foundation for what led to the black Scholes model and black Scholes formula and that's why it has their. calcGreeks: Calculate option Greeks (European Black/Scholes) version 1.1.2 (30.5 KB) by Yair Altman calcGreeks computes fair price and Greek values for vanilla European options using the Black-Scholes-Merton model, optimized for performanc ### Options Calculato • al breakthrough in pricing derivatives. Numerous studies have exa • The Black-Scholes-Merton model, sometimes just called the Black-Scholes model, is a mathematical model of financial derivative markets from which the Black-Scholes formula can be derived. This formula estimates the prices of call and put options. Originally, it priced European options and was the first widely adopted mathematical formula for pricing options • View Lecture 6- Black Scholes Calculator (1) (1).xlsx from FINS 3635 at University of New South Wales. CHECK THE EQUATIONS This program calculates Black-Scholes values for European-Style call and put. BSOPM Value:$180.355 Delta (Hedge Ratio): 0.980 Gamma: 0.000 Theta.\n• How to Calculate Options Prices and Their Greeks: Exploring the Black Scholes Model from Delta to Vega (The Wiley Finance Series) - Kindle edition by Ursone, Pierino. Download it once and read it on your Kindle device, PC, phones or tablets. Use features like bookmarks, note taking and highlighting while reading How to Calculate Options Prices and Their Greeks: Exploring the Black Scholes.\n• However, if you are trying to calculate any of the Black-Scholes equations for Option pricing or the Greeks, it's a bit daunting. Many resources don't explain the formulas in an easy-to-understand way for calculating by hand and the few resources I could find that do help you translate it into a spreadsheet, cost money or calculate across.\n• Use Black-Scholes Model to Price Asian Options with Several Equity Pricers. Calculate the Asian option prices using the price function for the Analytic, AssetTree, and AssetMonetCarlo pricing methods. Compare Delta Sensitivity for Asian and Vanilla Options\n• This is Black-Scholes for a European-style call option. You can download the XLS @ this forum thread on our website at http://www.bionicturtle.com\n\n### Black-Scholes Formula (d1, d2, Call Price, Put Price\n\npy_vollib.black_scholes.greeks.analytical¶. A library for option pricing, implied volatility, and greek calculation. py_vollib is based on lets_be_rational, a Python wrapper for LetsBeRational by Peter Jaeckel as described below 80 Delta region 46 Delta per strike 46 Dynamic delta hedging 47 The at the money delta 50 Delta changes in time 53; Chapter 6 Pricing 55 Calculating the at the money straddle using Black and Scholes formula 57 Determining the value of an at the money straddle 5 Hi all, Here are functions which will calculate the Black-Scholes call value as well as all of it's greeks in VBA (delta, gamma, vega, theta and rho). The functions for the Black-Scholes put price and greeks are available here. Enjoy! Function CallPrice(StockPrice As Double, StrikePrice As.. Simple Black-Scholes calculator. Simple Black-Scholes calculator. Index. WARNING: This page is not intended as a basis for trading decisions. No responsibility whatsoever is assumed for its correctness or suitability for any given purpose. Delta : Gamma : Theta : Vega : Rho : p. of exercise : d1 : d2 : N(d1) N(d2) Note that Theta is in the. This online calculator uses the Black-Scholes equation for the fair value of a European call option* on a non-dividend paying stock, as follows: A European call option can only be exercised on its expiration date. This is in contrast to American options that can be exercised at any time prior to expiration\n\nThis tool is targeted to option spread analysis. If your need a simple, bare-bones Black-Scholes calculator, check out this version. Options parameters. Lowest strike $Strike spread$ Spot price $+-Base date (y/m/d) Interest rate (%/yr.) Call strike Premium Intrinsic Delta Gamma Theta Vega Rho. Determine theoretical option prices with this advanced Black-Scholes Calculator. The Black-Scholes model, introduced in 1973 by Fischer Black and Myron Scholes, is an option valuation model that is the standard method of pricing options. Click on any particular option value and the Delta, Rho, Gamma, Vega, Theta, and Gamma of Gamma. 1. Black scholes calculator d1 d2. According to the black scholes option pricing model its merton s extension that accounts for dividends there are six parameters which affect option prices. S 0 underlying price per share. Current stock price s strike price e period t annual interest rate r annualized volatility v d1 2. All Calculations for European Style are done using BLACK-SCHOLES formula All Calculations for American Style are done using Binomial Method (255 Level) Delta is a measure of the rate of change in an option's theoretical value for a one-unit change in the price of the underlying 3. Find the value of d1 in the Black-Scholes formula for the price of a call on a company's stock with strike price$205 and time for expiration of 4 days. Given, S= $210.59, K=$205 t = 4 days r = 0.2175% s = 14.04\n\n### Options Calculator - The Options Industry Council (OIC\n\n1. Black Scholes Merton Model A mathematical model of a financial market which contains derivative investment instruments is called as Black Scholes Merton model . This model provides simple formula regarding asset's price and its volatility, time to maturity of the contract and the risk free interest rate \n2. 1. Black-Scholes Model. In order to know more information about a stock option, this options calculator with Black-Scholes Model, the first widely used model for option pricing, can provide the call/put option price, d1, d2, and Greek letters. It can assist investors in establishing an option trading strategy\n3. But what's your position delta? To calculate that, you'll need to look at the deltas of each option. The delta for the $110 call option is 0.39. The delta for the$115 call option is 0.24. So owning the $110 call option is like owning 39 shares of Microsoft stock (0.39 x 100). Owning the$115 call option is like owning 24 shares of.\n4. These Greeks are calculated based on the Black and Scholes options pricing model, which was first published by Fisher Black and Myron Scholes (hence the name Black & Scholes) in 1973. In this post, we'll go through an Option Greeks Calculator which updates real-time and calculate Greek values for all the strike prices of options traded in NSE\n5. An implied volatility of 15.87% implies a 1% daily move if you are only counting days the market is open) Plugging this into an Black Scholes calculator gives you option Delta =.511, Gamma =.073, Theta = -.036 and the option price is 2.184. To offset delta we take a 51.1 share short in the underlying (bear with fractional shares)\n6. The Black-Scholes or Black-Scholes-Merton model is a mathematical model of a financial market containing derivative investment instruments. From the partial differential equation in the model, known as the Black-Scholes equation, one can deduce the Black-Scholes formula, which gives a theoretical estimate of the price of European-style options and shows that the option has a unique price.\n\na basic calculator of the Black-Scholes option values (based primarily on the Black-Scholes Wikipedia page) some utilities for playing with various option strategies (select the Options Strategies tab) calculation of the first order Greeks Delta, Theta, Vega, and Rho. The change in the value of the option with respect to Strike Price K is also. Sometimes an online option calculator isn't enough and you'd like to implement the Black & Scholes (B&S) option pricing equations in Excel. If you're just playing around it doesn't matter how you structure the calculation. In fact, for clarity's sake, it's probably a good idea to spread out the calculation across multiple cells. However, if you're planning to do some serious work. The option calculator uses a mathematical formula called the Black-Scholes options pricing formula, also popularly called the 'Black-Scholes Option Pricing Model'. This is probably the most revered valuation model in Economics, so much so that its publishers (Robert C. Metron and Myron Scholes) received a Nobel Prize in Economics in 1997 Black-Scholes Equation & Delta-Hedging. We are going to simplify a lot (really a lot!) of the details in coming up with the B-S equation, but the key idea is to remember what we try to achieve in the binomial option pricing model and generalize the idea into continuous-time. Financial modelers start with the same setup as the binomial tree.\n\n### Black-Scholes Algorithmic Delta Hedging by Roman\n\n• This calculator uses the Black-Scholes option pricing model to compute the theoretical value and greeks of European-style call and put options. To generate results, enter the Inputs and click Calculate. This FinCalcs.NET calculator is currently displayed in READ ONLY mode. Calculation will be enabled once you've logged in\n• Attached is a simple Excel function that calculates the Black-Scholes option value for a specific set of input parameters. Currently, it just calculates the call value - if you use it as an array function, it will return a 4-element array with call value, call delta, put value, put delta, respectively\n• es the fair market price of European put and call options. It assumes the underlying asset pays no dividends before maturity. We assume no responsibility for the correctness of this software and it should not be used as a basis for trading decisions\n• FX OPTION PRICING: RESULTS FROM BLACK SCHOLES, LOCAL VOL, QUASI Q-PHI AND STOCHASTIC Q-PHI MODELS Krishnamurthy Vaidyanathan1 Abstract The paper suggests a new class of models (Q-Phi) to capture the information that the market provides through the 25-Delta Strangles and 25-Delta Risk Reversals\n• BlackScholesFormula: this class attempts to clearly layout the Black-Scholes model as expressed in the formula. Each step is defined. the calculate()method will return the a double with the calculated MtM the calculateWithGreeks()will return the MtM value along with the greeks (delta, gamma, rho, theta, and vega\n• The calculator provides functions such as basic computing functions and TVM , Statistics, EOQ , exchange rate conversion, interest rate conversion, NPV , IRR ,Black scholes, Bond Duration . Based on industry standards, Legendary Financial Calculator BA2 plus , it is easy to use\n\nThis is an updated version of my Black-Scholes Model and Greeks for European Options indicator, that i previously published. I decided to make this updated version open-source, so people can tweak and improve it. The Black-Scholes model is a mathematical model used for pricing options. From this model you can derive the theoretical fair value of an options contract You can think of Black-Scholes as describing a 7-dimensional space, and the partial derivatives describe the rate of change in the slope of the curve along the price/other_variable axis pairs. Delta describes the rate of change of the option price as the underlying price changes Black Scholes on the HP10bII+ financial calculator. Download the Excel file for this module: bs_nondiv.xlsm [29 KB] Download the VBA code for this module: xlf-black-scholes-code.txt [4 KB] Development platform: Microsoft Excel 2013 Pro 64 bit. Revised: Sunday 26th of July 2020 - 09:37 PM, Pacific Time (PT One approach for calculating a minimum variance delta is to replace the Black-Scholes model by a stochastic volatility model. The model for valuing a portfolio dependent on a particular stock or an equity index or equity index then takes the form: Value = G(S, σ,\n\n### Black-Scholes Option Pricing Calculator - calkoo\n\n1. The Black Scholes Model, also known as the Black-Scholes-Merton method, is a mathematical model for pricing option contracts. It works by estimating the variation in financial instruments. The technique relies on the assumption that prices follow a lognormal distribution\n2. The Black-Scholes model and the Cox, Ross and Rubinstein binomial model are the primary pricing models used by the software available from this site (Finance Add-in for Excel, the Options Strategy Evaluation Tool, and the on-line pricing calculators.). Both models are based on the same theoretical foundations and assumptions (such as the geometric Brownian motion theory of stock price.\n3. The following setup on the Nspire provided the functions to calculate the vega values. This spreadsheet input screen stores the spot prices and the calculated Black Scholes vega values. Finally, with the data plotting screen the graph of Delta hedged gains of volatility sensitivity is completed\n4. We'll use closed-form equations as well as PyTorch's automatic differentation to calculate the greeks. The Black-Scholes Formula. This is the formula we'll use to calculate the price of the European put option. There are formulas for each of the greeks too: delta, rho, vega, theta, epsilon, etc., but I'll spare the math for now\n5. Delta Neutral: A strategy consisting of holding puts and calls where the sum of the deltas is zero. For example, if a portfolio has 10 calls on a stock with a delta of .6, and 15 puts on the same stock with a delta of -.4, then the call delta (.6 x 10) offsets the put delta (-.4 x 15)\n6. We deduce that the process 1nS t has independent increments, as in the Black-Scholes model. In the sticky-delta model, the logarithm of the spot is a Levy process under the risk neutral measure\n\nLearn Black-Scholes Model Calculate european option prices with Black-Scholes Calculator, you can easily get the call price and put price of any stock such as Apple Inc. or Google Inc. Powered by BlackScholes.io ©2018 Black-Scholes Merton Model Calculator (With Greeks), Option Strategies Layout and Delta Hedging Calculator by Kenton Parrott. This paper uses risk-adjusted lognormal probabilities to derive the Black- Scholes formula and explain the factors N(d1) and N(d2). It also shows how the one-period and multi-period binomial option pricing formulas can be restated so that they involve analogues of N(d1) and N(d2) which have the same interpretation as in the Black-Scholes model Black-Scholes is an easy tool that can calculate the fair value of an equity option based on the Black-Scholes (European), Whaley (Quadratic) and Binomial Models along with the Greek sensitivities.. Binomial is an easy tool that can calculate the fair value of an equity option based on the Black-Scholes (European), Whaley (Quadratic) and Binomial Models along with the Greek sensitivities Afaq Latif Real Options and Managerial Decision Making 22/03/2021 0556285 Black & Scholes/Binomial Model Assignment PART 1: For the first part of the assignment regarding the use of Binomial model for real options evaluation of a computer producing company, I started with the calculation of (u) as volatility rate sigma was provided but there was no indication of maximum or minimum The Black-Scholes PDE is a partial differential equation which (in the model) must be satisfied by the price of a derivative on the equity. The Black-Scholes formula is the result obtained by solving the Black-Scholes PDE for a European call option",
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"MibianLib is an open source python library for options pricing. You can use it to calculate the price, the implied volatility, the greeks or the put/call parity of an option using the following pricing models: Garman-Kohlhagen; Black-Scholes; Merton; MibianLib is compatible with python 2.7 and 3.x. This library requires scipy to work properly. After searching the web, I keep coming across vague descriptions of how to calculate the volatility input for the Black Scholes formula. Most sites seem to say I can just calculate the standard deviation of the past X days closing prices and use that. However, some sites say that is inaccurate, but don't explain what the correct calculation is\n\n### Option Price Calculato\n\nThe Black-Scholes Model n The version of the model presented by Black and Scholes was designed to value European options, which were dividend-protected. n The value of a call option in the Black-Scholes model can be written as a function of the following variables: S = Current value of the underlying asset K = Strike price of the optio The below calculator will calculate the fair market price, the Greeks, and the probability of closing in-the-money (ITM) for an option contract using your choice of either the Black-Scholes or Binomial Tree pricing model.The binomial model is most appropriate to use if the buyer can exercise the option contract before expiration, i.e., American style options 7.2 Deriving Black Scholes. The Black Scholes Model is, simply put, a way to value (i.e., put a price on) the options that we discussed above. By using a few assumptions, the model can easily spit out prices for options given a few input parameters (some of which we have already seen, including the current stock price, time until expiry and strike price) Chapter 10Theta Theta (θ or for the capital letter Θ) is the change of the value of an option in relation to the change in time, also called time-decay. - Selection from How to Calculate Options Prices and Their Greeks: Exploring the Black Scholes Model from Delta to Vega [Book\n\n### Black-Scholes model - Wikipedi\n\nBlack-Scholes Model. In this application, we compute the option price using three different methods. The first method is to derive the analytical solution to the option price based on the classical Black-Scholes model. Next, we compute the option price through Monte Carlo simulation based on the Black-Scholes model for stock price estimation In this article I want to discuss a practical application of the Black-Scholes model, design patterns and function objects in C++. In particular, we are going to consider the concept of Implied Volatility.In derivatives pricing, the implied volatility of an option is the value of the underlyings volatility (usually denoted by $\\sigma$), which when input into an derivatives pricing model (such. Formula for the calculation of a call option's delta. The delta of an option measures the amplitude of the change of its price in function of the change of the price of its underlying. Option strategy calculator • Pricing of an option (Black & Scholes) Site map Get in touch. Monday, May 3rd 2021 123rd day of the year 18th week of the year. OPTIONS XL is a Microsoft Excel add-in program that allows you to value options on stocks, foreign exchange, futures, fixed income securities, indices, commodities and Employee Stock Options (ESOs) using custom functions. Market data from your quote vendor can be automatically passed to the custom functions via Dynamic Data Exchange. Some of the ways that OPTIONS XL may be used are: Valuing.\n\n• Oxy acetylene welding process.\n• IRS Certified Acceptance Agent.\n• Anger attacks in depression.\n• Futurama: into the wild green yonder free online.\n• Passwords and accounts iPhone missing iOS 14.\n• Childcare vouchers Scotland.\n• Best garage door paint uk.\n• Lawton High School Volleyball Schedule.\n• Plants for wedding gift Sri Lanka.\n• Brem mall it shop.\n• U.S. Marine salary per month.\n• IATA NOTOC.\n• Humanity of Jesus.\n• How much is clutch replacement.\n• Adobe Photoshop cs3 tutorial picture editing background.\n• Wax chest hair female.\n• Great Value pumpkin pie recipe.\n• Buying a car in Texas from out of state.\n• Echo Spot for sale.\n• Online identity theft cases in India.\n• Eating banana on empty stomach during pregnancy.\n• Working visa for USA from Nepal 2020.\n• How long can you store fireworks.\n• Ferris Bueller's Day Off Trailer.\n• My kindle books won't open on ipad.\n• How to convert GPT to MBR using Ubuntu without data loss.\n• Marist Brothers abuse.\n• How often do you need a tetanus shot for adults.\n• Convert decimal seconds to hours, minutes seconds.\n• Appetite suppressant mints.\n• Smoke machine test near me.\n• New Twilight movie 2020 release date.\n• Nationalism vs sectionalism Quizlet.\n• Covid 19 are flowers allowed at funerals.\n• Corny meaning in slang.\n• How does a touch free car wash work.\n• 18th birthday party ideas Australia.\n• Informal language to formal language translator.\n• How to calculate square footage of a warehouse.\n• House Hasson order products."
] | [
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http://mint.sbg.ac.at/desc_OStandardLengthening.html | [
"## Standard Lengthening for NRT-Codes\n\nIf an [s, n, d]-code C0 contains an [s, n−1, d + 1]-subcode C1 (as it is the case for Reed-Solomon codes, algebraic-geometric codes and other families of codes), C0 can be lengthened to an [s + 1, n, d + 1]-code C. If G0 is a generator matrix of C0 such that the first n−1 rows of G0 are a generator matrix of C1, a generator matrix of C is given by\n\nG :=",
null,
"G0",
null,
"",
null,
".\n\nThe same result can be obtained by applying construction X to C1C0 with the [1, 1, 1]-code without redundancy.\n\nFor NRT-codes with depth T > 1 and n > 1 a similar construction is possible. Let C0 denote a linear [(s, T ), n, d0]-code. Let T ʹ := min{T , n} and let C0 contain a chain of subcodes Ci for i = 1,…, T ʹ−1 with CTʹ−1 ⊂ ⋯ ⊂ C1C0, dimensions dimCi = ni, and minimum distances d (Ci) = di. Furthermore assume that G0 denote a generator matrix of C0 such that the first ni rows of G form a generator matrix of Ci. Then C0 can be lengthened to an [(s + 1, T ), n, d]-code with\n\nd =",
null,
"di + i\n\nand generator matrix\n\nG :=",
null,
"G0",
null,
"",
null,
"",
null,
"for nT and\n\nG :=",
null,
"G0",
null,
"| 0n×(T-n)In",
null,
"for nT .\n\nNote that this result cannot be obtained using construction X in a straightforward way! It is only possible because the extension code Ce defined by the generator matrix I is an unequal error protection (UEP) NRT-code: If Di denotes the code defined by the first i rows of I, then d (Ce ∖ Di) = i + 1 for i = 0,…, T ʹ−1 (cf. ).\n\n### References\n\n Jürgen Bierbrauer, Yves Edel, and Ludo Tolhuizen.New codes via the lengthening of BCH codes with UEP codes.Finite Fields and Their Applications, 5(4):345–353, October 1999.doi:10.1006/ffta.1999.0245 MR1711841 (2000j:94027)"
] | [
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https://www.enotes.com/homework-help/always-have-hard-time-setting-word-problems-up-461954 | [
"# I always have a hard time setting word problems up. Can you help me set up and solve this word problem? I think it starts like, 1/x*6 & 1/x+2*6 then lcd* both? I'm supposed to use the quadratic formula to solve it. A large pipe and a small pipe fill a tank in 6 hours. the small pipe alone takes 2 more hours than the large pipe alone. How long will it take the large pipe alone to fill the tank.",
null,
"A little bit more general for all word problems could be something like this. I use this for my students:\n\n1) Understand what the problem is saying - I've seen students see words and simply turn off. They don't even try to read the problem and simply try to understand what is going on. As in, here, things like:\n\n- we have a tank filling up\n\n- two pipes filling it up, one small and one large\n\n- together, they fill it up in 6 hours\n\n- along, the small pipe takes 2 hours longer than the large pipe\n\n- find out how long the large pipe would take alone.\n\nForget the math, just understanding what the problem is talking about is difficult enough.\n\n2) Set up the problem - once you understand the problem, it is a little easier to understand how to set it up. Like here, what can be difficult to understand:\n\nthe rate of the small pipe + the rate of the large pipe = the rate of the pipes together\n\nEither understanding that \"formula\" is difficult or how the rate of the large pipe is 1/x and the small pipe is 1/x+2.\n\n3) Solve for x\n\n4) Check - you can always check every math problem. Here, you could try plugging in the solution for x and see of each side is equal to each other (at least approx, given decimals and round off error).\n\nApproved by eNotes Editorial Team",
null,
"Since the problem only provides the time the tank is beiong filled by the two pipes together, you can use the following notations for the time each pipe fills the tank.\n\nThe larg pipe fills the tank in `x` hours and the small pipe fills the tank in `x + 2` hours.\n\nHence, the large pipe fills `1/x` of the tank in 1 hour and the small pipe fills `1/(x + 2)` of the tank in 1 hour.\n\nTogether, the large pipe and the small pipe fill 1/6 of the tank in 1 hour.\n\nYou may set up the following equation, such that:\n\n`1/x + 1/(x + 2) = 1/6`\n\nYou need to bring the fractions to a common denominator, such that:\n\n`6(x+2)/(6(x(x+2))) + (6x)/(6(x(x+2))) = (x(x+2))/(6(x(x+2)))`\n\n`6(x + 2) + 6x = x(x + 2)`\n\n`6x + 12 + 6x = x^2 + 2x => x^2 + 2x - 12x - 12 = 0`\n\n`x^2 - 10x - 12 = 0 `\n\nYou should use quadratic equation, such that:\n\n`x_(1,2) = (10 +- sqrt(100 + 48))/2 => x_(1,2) = (10 +- sqrt148)/2`\n\n`x_(1,2) = (10 +- 2sqrt37)/2 => x_(1,2) = 5+-sqrt37`\n\nYou need to reject the negative value since x represents the number of hours, hence `x = 5+sqrt37 => x ~~ 11` hours.\n\nHence, evaluating the time needed by the large pipe to fill the tank yields `x ~~ 11` hours.\n\nApproved by eNotes Editorial Team"
] | [
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https://securitelado.com/4h3bv81992265om/inch-to-cm.html | [
"Home\n\nInch to cm\n\nThe inch [in] to centimeter [cm] conversion table and conversion steps are also listed. Also, explore tools to convert inch or centimeter to other length units or learn more about length conversions 50 inch to cm = 127 cm. ›› Want other units? A centimetre (American spelling centimeter, symbol cm) is a unit of length that is equal to one hundreth of a metre, the current SI base unit of length\n\nConvert Inch to Centimeter with formula, common lengths conversion, conversion tables and more. You can use the Inches to Centimeters unit converter to convert from one measurement to another Home›Conversion›Length conversion› Inches to cm. Inches to CM Converter\n\nInches to Centimetres formula: [Inches] = Cm * 2.54. When going for an inch to centimeter conversion, you simply need to divide the given number of centimeters with 2.54 Inches. Use of the inch can be traced back as far as the 7th century. The first explicit definition we could find of its length was after 1066 when it was defined as the length of three barleycorns Inches to Centimetres. Convert between the units (in → cm) or see the conversion table. Convert from Inches to Centimetres. Type in the amount you want to convert and press the Convert button Add Inches to cm Converter to your website to get the ease of using this unit converter directly. Feel hassle-free to account this widget as it is 100% free, simple to use\n\nInches to Cm Converter. How many centimeters in 1 inch? A centimeter (cm) is a decimal fraction of the meter, The international standard unit of length, approximately equivalent to 39.37 inches Convert 1 Inches to Centimeter | Convert 1 in to cm with our conversion calculator and conversion table. To convert 1 in to cm use direct conversion formula below Convert inches to centimeters (in to cm) with the length conversion calculator, and learn the inch to centimeter calculation Enter the length in inches below to get the value converted to centimeters\n\nInch vs. Centimeter. Many countries in the world, even they usually have the metric system, use inch for plumbing pipes, tubes and hoses, fittings, adapters, drains and faucets 1 inch = 2.54cm. To convert inches to centimeters multiply your figure by 2.54. Should you wish to convert from centimeters to inches, give the cm to inches converter a try Use our online free inches to centimeters converter. inches to cm conversion chart Learn how to convert inches to centimeters, how many cm are in one inch, and what the difference is between the two units. Examples and conversion table for quick reference Calculator to convert between Inches and cm. Useful information about the terms, formulas, a conversion table and much more. Convert Inches to cm and cm to Inches\n\nConvert inches to cm\n\n• Inches to Centimeters Conversion Calculator, Conversion Table and How to Convert. This calculator provides conversion of inches to centimeters and backwards (cm to in)\n• inches to centimeters formula. centimeter = inch * 2.54. Inch is an imperial and United States Customary length unit. 1 inch = 2.54 cm, 5 inches = 12.7 cm, 6 inches = 15.24 cm\n• Convert or calculate inch to cm or cm to inch. Free online converter of inches and centimeters (cm). Simply type the size in inch or in cm to get the live conversion. Do you like Calcul Conversion\n• Online calculator to convert inches to centimeters (in to cm) with formulas, examples, and tables. Our conversions provide a quick and easy way to convert between Length or Distance units\n• Length and distance unit conversion between centimeter and inch, inch to centimeter conversion in batch, cm in conversion chart. » Centimeter ↔ Inches Conversion\n• This is a very easy to use inches to centimeter converter. First of all just type the inches (in) value in the text field of the conversion form to start converting in to cm, then select the decimals value and..\n• Easily convert inches to cm (centimeter) with this free online conversion tool. Use this handy calculator below if you want to convert any measurements in inches to centimeters\n\nFormula CM to inches. Number of inches = number of centimeters * 0.39370079 Inch, CM, MM Converter. Your browser does not support the canvas element. 1 meter = 100 centimeters = 1,000 millimeters. 1 inch equals 2.54 centimeters, 1 cm equals 0.393700787 in\n\nWelcome to our Feet and Inches to cm conversion calculator. Here you will find our online math calculator to help you to convert lengths from inches and feet into cm To convert from feet and inches to centimeters, use the following two conversion equations If you like Feet and Inches to Cm Converter, please consider adding a link to this tool by copy/paste the.. For convert cm to inches you can use the main formulas or our online converter that include all necessary formulas for fast convert inches to main units of length How many centimeters in 69 inches: If Lin = 69 then Lcm= 175.26 cm. Note: Inch is an imperial or United States customary unit of length. Centimeter is a metric unit of length\n\nCentimeters to Inches Formula. cm × 0.39*. 2. If a baby is 64 centimeters long, what is her length in inches? The answer is 25 inches Inch signifie pouce en anglais. Le pluriel de inch et inches. C'est une unité de longueur qui date du Moyen-Age. Exemple de calcul pour convertir une distance en inches en cm inches how much is 5 cm in inches how to convert centimeters to inches how many inches are in 1 centimeter cm to inches formula inches in cm ek inch me kitne centimetre hote hain inches to.. CM to Inches Converter is the most useful and easy to use length unit converter. You can enter any number of centimeters in metric converter and cm inch converter will produce the result in inches for.. Rechner von Inch in cm (in in cm) und umgekehrt. Einfache Umrechnung zwischen Inch und Zentimeter mit Infos, Tabellen und Formeln\n\n* Doldurulması zorunlu alanlar. * İşlem: İnç kaç cm Cm kaç inç. Bu da 2,54 cm'ye tekabül etmektedir Cm to Inches converter is the length converter from one unit to another. The term centimetre is abbreviated as cm in which one centimetre is equal to the one-hundredth of a meter Inches to Centimeters Converter. 1 Converting via Simplified Process. Learn more... There are many tools for converting inches to centimeters on the web, all of which will tell you that 1 inch.. MM or CM to Fractions of Inches. Your browser does not support the canvas element. To convert fractional inch to mm or cm, fill fraction into the blank Fractional inch, e.g. 2 1/2 = 2.5 A centimeter (cm) is a decimal fraction of the meter, The international standard unit of length, approximately equivalent to 39.37 inches. Definition of inch\n\nConvert inch to cm - Conversion of Measurement Unit\n\nConvert Centimeter to Inch with formula, common lengths conversion, conversion tables and more. You can use the Centimeters to Inches unit converter to convert from one measurement to another Whenever you need to supply your height in centimetres rather than feet and inches here is very helpfull Just type your height into the feet and inches boxes to convert to centimeters or into the.. How much is 1.5 inches in cm, free converter for length and distance, conversion chart and online calculator. Convert 1.5 inches to centimeters. Online inch calculator Feet (ft) and inches (in) to centimeters (cm) or centimeters (cm) to inches (in). Height Conversion. Inch Conversions. Multiply By. convert inches to feet. 0.08333333 Conversion square inches to square centimeters, inch2 to cm2. The calculator gives the answer to the questions: 90 inch2 is how many cm2? or change inch2 to cm2\n\nInches To Centimeters Converter in To cm Converte\n\nConvert feet and inches to centimeters, inches, meters, etc. - Ft, in, cm, m, mm. 6ft2 and three quarters of an inch in cm. 189.865 centimeters. 6foot2 in meters Centimeters to inches (cm to in) converter, formula and conversion table to find out how many inches in centimeters MM to inches converter. Easily convert millimeters to inches, with formula, conversion chart, auto conversion to common lengths, more The inch (abbreviation: in or ″) is a unit of length in the (British) imperial and United States customary systems of measurement. It is equal to 1⁄36 yard or 1⁄12 of a foot. Derived from the Roman uncia (twelfth.. inch to cm. Scanner inch1 = new Scanner(System.in); System.out.println(Enter The Inch) and how can i do Space between the input CM TO INCH\\FOOT Between Inch to cm - Alex Nov 3 '12 at..\n\n27.9 70.9 cm. 15.7 39.9 cm. To easily find out what size you should buy, you can divide your TV viewing distance (in inches) by 1.6 (or use our TV size calculator above) which roughly equals to a 30.. Popular 6.9 inch to cm of Good Quality and at Affordable Prices You can Buy on AliExpress. We believe in helping you find the product that is right for you. AliExpress carries wide variety of products..",
null,
"Inches to centimeters converter and how to convert\n\nUseful length conversions Convert inches to centimeters (inches to cm) Convert inches to feet centimeter is equal 1/100dth of a meter) Deci stands for 1/10th (for example, 1 decimeter is equal.. Convert inches to cm centimetres and convert centimeters to inches. Inches to cm centimeter conversion table included. Inches to Centimeters Metric Conversion Table - 1 to 200 inches\n\nInches and centimeters are both units of linear measurement. Inches are used in the imperial system whereas centimeters are used in the metric system. To convert from cm to inches.. Diferent length units conversion from inch to centimeters. Between in and cm measurements conversion chart page. Convert 1 in into centimeter and inches to cm Centimeters to Inches (cm to in) calculator, conversion table and how to convert. Centimeters to Inches Conversion. Select conversion typ How far is 14 inches in centimeters? This simple calculator will allow you to easily convert 14 in to cm\n\nConvert Inches to Cm\n\n1. Formula : inch to cm. Ans : Multiply your length value by 2.54. One inch is equal to roughly 2.54 centimeters, so converting inches to centimeters means multiplying a value in inches by 2.54\n2. Example 5 inch x 7 inch in cm? How many inches in 1 cm? The international inch is defined to be equal to 25.4 millimeters.Use this converter inch to cm\n3. ..nm to n-cm, Convert to Dyne-centimeter, Convert to Kilogram meter, Kg-m Kilogram Centimeters, Kg-cm, Convert to Gram centimeter, g-cm, Convert to ounce-inch, oz-in, Convert to foot-pound, ft-lb..\n4. Centimeters (cm). = Inches (in). You can do the calculation yourself: One inch equals exactly 2.54 centimeters. To convert Centimeters to Inches, divide by 2.54\n5. Inches to Centimeters - Distance and Length - Conversion. The inch is a popularly used customary unit of length in the United States, Canada, and the United Kingdom\n\nInches to Centimeters - in to cm conversio\n\n• comparison inches to cm convert inch to centimeters feet to meters length measures centimetres distance length - Eberhard Sengpiel sengpielaudio\n• Convert inches, centimeters, millimeters... Convert from inches to cm quickly and easily with our unit conversion tool. How many inches in a cm\n• This super easy cm to inches converter will help you to work out all your measuring woes in a flash, with no more confusion\n• Dimensions of the A series paper sizes 4A0, 2A0, A0, A1, A2, A3, A4, A5, A6, A7, A8, A9 and A10 in both inches and mm, cm measurements can be obtained from the mm values and feet from the inch..\n• 1 inch bang bao nhieu cm, Đổi inch sang cm, mm, m và nhiều đơn vị khác là cách nhanh nhất giúp Mục Lục bài viết: 1. Đổi 1 inch sang cm, mm, m đơn giản, dễ dàng. 2. Nhập số inch cần chuyển đổi..\n\nInches and centimeters. 1 Inch = 2.54 cm, 1 Feet = 30.48 cm, 1 Feet = 12 Inch. Frame sizes. The inner sizes of our frames are equal to the matching mounts. I.e. an A3 print (29.7 x 42cm) comes with.. Formeln zur Umrechnung zwischen Inch und cm. Falls Sie eine Berechnung zwischen Inch und cm vornehmen, finden Sie nun die Formeln für jeweils beide Ausgangswert The calculator uses the conversion 1 inch = 2.54 cm, 1 foot = 30.48 cm, 1 meter = 100 cm, and 1 foot = 12 inches. The units are rounded to two decimal places so occasionally there will be rounding errors 1 foot ~ 30.5 cm 1 inch ~ 2.5 cm 1 kg ~ 2.2 lbs. Below are tables of measurements (approximate). While it is not a proper converter, it still might be very handy - print it, and use when necessary Inch, to inaczej cal. Jest to jednostka miary długości stosowana przede wszystkim w krajach anglosaskich. - mila (1,609 km), - yard (91,44 cm), - foot (30,48 cm)",
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"Convert Inches to Centimetres (in cm\n\ncentimeters = inches x 2.54. For example, let's say that we want to convert 2 inches to centimeters. Then, we just replace inches in the abovementioned formula with Inch to Centimeter Conversion Table. Example: convert 61 in to cm cm to Inches is Centimeter to inch height converter. It converts units from cm to in or vice versa with a metric conversion table Resizing of image in inches or centimeters for printing on paper, with considering DPI online. Size of the photo will be changed to the specified size in inches (millimeters, centimeters) according to..\n\nInches to Centimeter Converter (in to cm\n\n1. Le inch et le cm (centimétre) sont deux unités de mesures de longueur dans le système métrique. Pour convertir une longueur en inch en cm, on multiplie par 2,54 ou on divise par 0,3937\n2. Use the table below to check how close an individual with a height of 179cm is to going up or down an inch in height. 179cm to Feet Summary\n3. Formula to Convert inch - cm. In mathemtaical terms, the formula used by this inch - centimeter converter is shown below for your convenience and use should you need it. To give some examples..\n4. To convert feet, inches & fractions of an inch to metres or metres to ft. Results below in km m cm mm :- Some other associated conversions - 1 mile = 5280 feet , 1 yard ( yd ) = 3 feet , 1 fathom = 6..\n\nInches to Cm Converte\n\n• From cm to inches converter this calculator allows you to convert from Centimeter to inch, Use it Centimeter to Inch Conversion. You can use the calculator above to convert centimeters to inches..\n• How about centimeters? Check out our expert guide for help. Inches correspond to the imperial system, which is the main measuring system used in the US and a smattering of other countries\n• d like a Kilo = 2.204 Pounds\n• To convert Centimeters to Inches, enter the number of centimeters to be converted into the centimeters box below. This chart allows users to convert centimeters to Inches manually\n• Convert 1.7 inch to cm. inch and cm definitions and information, conversion calculators and tables. 1.7 inch equal 4.318 cm. Conversion details\n• Convert inches to cm. How many centimeters in an inch? Metric Kilometre (km) Metre (m) Decimetre (dm) Centimetre (cm) Millimetre (mm) Micrometre (µm) Nanometre (nm) Angstrom (Å)..\n• Inches To Centimetres Inches To Centimetres. For passports and many medical forms you now need to supply your height in metres and centimetres rather than feet and inches\n\n1 Inches to Centimeter 1 in to cm\n\nDimensions of A4 size paper in centimetres, millimetres, inches and pixels for the UK, USA, Australia, Europe, Germany, Singapore and India 2.5 inches equals 6.35 centimeters because 2.5 times 2.54 (the conversion factor) = 6.35 ..1 inch berapa cm (centimeter) atau biasa juga dikenal dikalangan masyarakat 1 inci berapa cm Sama saja seperti satuan inch, satuan cm juga menjadi satuan untuk mendeskripsikan panjang pada..\n\nInches to cm Conversion (Inches To Centimeters) - Inch Calculato\n\nNew Zealand Entertainers. Australian Entertainers. Speakers & MC's. Casino Equipment. Select City Not Important Auckland, New Zealand Christchurch, New Zealand Hamilton, New Zealand North.. Cm to Inch Convertor. We only accept measurements in Inches, The units we accept are Whole numbers and multiple of .25 , eg: 36, 36.25, 36.50 and 36.75\n\nInch cm Infos & Converte\n\nCalculator online centimetru, toli (inches), metri, yard. Echivalente unitati de masura. Transforma lungimea din cm (centrimentri) in inch (tol/toli). Trei sferturi, jumatate, etc Free. Android. Category: Tools. Application to convert distances in Inches and Centimeters. Inches are used in somes countries like United States and United Kingdom Cm (santim) - İnç (inch) Cm(santim) - Adım(foot) ve Pound(lbs) - Kilogram(kg) dönüşümleri. Dönüşümler iki yönlüdür, dönüşümü yapmak istediğiniz bölümün altındaki butonlar ile, inç'ten santime.. One Inch Measurement tape/Scale/ सूत, इंच, सेंटीमीटर, फुट, यार्ड, मीटर, सब कुछ By: Satya Education #Measurement Tape#How to read measurement tape feet in inch ,mm ,meter, cm. full details in hindi 1 inç kaç cm'dir ? sorusunun ve daha binlerce başka soruların cevaplarını sizin için araştırıyor, cevaplıyoruz. İngiliz ölçü birimi olan 1 inç yaklaşık 2.54 cm uzunluğuna denk gelir",
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"Inches to Cm Converter - The Calculator Sit\n\n1. Welcome to our inches to centimeters (in to cm) conversion calculator. You can enter a value in either the inches or centimeters input fields. For an understanding of the conversion process..\n2. Перевод слова inch, американское и британское произношение, транскрипция, словосочетания, однокоренные слова, примеры использования\n3. 16 cm = 6,2992125984 inches. Convert centimeters in inches. Converter centimeters in inches. This is the right place where find the answers to your questions like\n4. on-screen measuring tool in inches/centimeters. If you want to measure the actual size of a small object in inches or centimeters and you don't have a real ruler at hand, this virtual on-screen online..\n5. Check actual size /. Inch Ruler. I am using -inch screen. save settings. I don't know what monitor size is\n6. de santimetre kısaltması cm, milimetre kısaltması mm ile ifade edilirken, ansi ölçü..",
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"Convert inches to centimeter\n\n1. 1 feet = 12 inches. 1 feet = 30.48 cm. More examples of heights converted from feet and inches to cm\n2. 1 centimeter (cm) is 0.39370 feet (ft) 1 centimeter (cm) is 0.03280 inches (in). The conversion formula for centimeter to millimeter (cm to mm\n3. İnç-Cm ölçü birimleri ile alakalı 1 İnç Kaç Cm? sorusunun cevabını bu yazıda verelim. Bu cevabın bilimsel bir hesaplamada kullanılacaksa buna uygun olmadığını bildirmek isteriz\n4. Inches to Cm converter. A quick online length calculator to convert Inches(in) to Centimeters(cm)\n\nInches to Centimeters Converter - convert in to cm onlin\n\n1. dpi ppi. mm inch. Image for print 300 dpi, for photo album - Photo/Image/Pictur: 13 x 18 cm Quality of print 300 dpi - Resolution in pixels\n2. Inches to Centimeters Conversion Formula. [X] cm = 2.54 × [Y] in where [X] If we want to calculate how many Centimeters are 6.7 Inches we have to multiply 6.7 by 127 and divide the product by 50\n3. Основание перевода величин : 1 cm = 0.39370078740157 in. язык. This page also exists in English. Convert inches to centimetres here\n4. Pauline Musters - at 23 inches (58 cm) tall, recognised by the Guinness Book of Records as the shortest woman ever recorded. Madge Bester - 65 cm in 1998 Lucia Zarate - Smallest woman..\n5. Centimeter kelimesinin kısaltılmışı olan cm ile sembolize edilir. Örneğin 32 ekranlık bir televizyonla anlatılmak istenen, ekranın bir köşesinden diğer köşesine uzunluğunun 82 cm olmasıdır",
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"İnç (Inch), Amerika ve İngiltere gibi bazı ülkelerde kullanılan bir ölçü birimidir. Metrik sistemde bir inç 2,54 cm'ye karşılık gelir. İnç, çift tırnak işaretiyle gösterilir Popüler Amerikan Uzunluk ve Mesafe (Uzaklık Ölçüleri) Çevirimleri. feet metre. cm inç. metre inç. foot cm. mil kilometre Includes additional conversions of only inches, centimeters, etc. Also try our Meters to Feet and Inches Converter or Centimeters to Feet and Inches converters Make a 2x2 inch (51x51 mm, 5x5 cm) photo in 1 click and get a fully compliant professional result: a 2x2 inch (51x51 mm, 5x5 cm) image with white background that meets all requirements cm. m. inches. pt. pc. The most common paper size used in English speaking countries around the world is A4, which is 210mm x 297mm (8.27 inches x 11.7 inches) cm ft in km m mi mm nmi yd\n\n• Panda panzerwels geschlechtsunterschied.\n• Utstyr til høns.\n• Wetter kreta.\n• Middag fredag tips.\n• Imax münchen deutsches museum.\n• Regionale ressurser.\n• Blomster bryllup mai.\n• Bybrann ålesund.\n• Varm vinterbukse dame.\n• Dagpenger etter studietid.\n• Kaffe historie.\n• Gror kryssord.\n• Rasert kryssord.\n• H2o plötzlich meerjungfrau staffel 3 folge 12.\n• Helly hansen bag 70l.\n• Audimax tu bs öffnungszeiten.\n• Lolpro kha zix.\n• Göteborg restaurang.\n• Lokalkompass oberhausen.\n• Billiga paketresor till budapest.\n• Gespurte loipen im erzgebirge aktuell.\n• Resettlement und relocation.\n• The british monarchy.\n• Fornuftig kryssord.\n• Rikard wolff.\n• Bryllup sang vielse.\n• Uis sosialt arbeid timeplan.\n• Falkensteiner ehrenburgerhof südtirol.\n• Jakttider på sel.\n• Amc töpfe neu.\n• Wohnen in drispenstedt.\n• Kommunikasjon psykiatri.\n• Bytte batteri iphone 7.\n• Disseksjon engelsk.\n• Musikpark erfurt schließt.\n• Hafþór júlíus björnsson theresa líf.\n• Arena 2018."
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http://www.mafutian.net/352.html | [
"# ASP 实现将数据导出到 Excel 文件中\n\n【摘要】本文记录一下在 ASP 中将数据导入到 Excel 文件中的几个步骤，这里 excel 的格式是 \".xlsx\" 而不是 \".xls\"，这是有区别的，如果使用 \".xls\" 来保存文件，那么使用 excel 打开的时候会报错。然而使用浏览器下载 \".xlsx\" 的时候也会报错，而本文仅仅记录一下 ASP 将 excel 生成到 1.xlsx 在网站目录中，而不能在客户端浏览器中下载。\n\n1. `'--如果原来的 EXCEL 文件存在的话则删除它`\n2. `Set fs = server.CreateObject(\"scripting.filesystemobject\")`\n3. `filename = \"江南大学计算机科学与技术2015级.xlsx\"`\n4. `file = \"H:\\ASP\\XWY\\\" & filename`\n5. `if fs.FileExists(file) then`\n6. ` fs.DeleteFile(file)`\n7. `end if`\n8. `set fs = Nothing`\n9. `Set ExcelApp = CreateObject(\"Excel.Application\")`\n10. `ExcelApp.Application.Visible = True `\n11. `Set ExcelBook = ExcelApp.Workbooks.Add`\n12. `ExcelBook.WorkSheets(1).cells(1,1).value = \"学号\"`\n13. `ExcelBook.WorkSheets(1).cells(1,2).value = \"姓名\"`\n14. `ExcelBook.WorkSheets(1).cells(1,3).value = \"手机号码\"`\n15. `ExcelBook.WorkSheets(1).cells(2,1).value = \"6151910036\"`\n16. `ExcelBook.WorkSheets(1).cells(2,2).value = \"马富天\"`\n17. `ExcelBook.WorkSheets(1).cells(2,3).value = \"17095248823\"`\n18. `ExcelBook.WorkSheets(1).cells(3,1).value = \"6151910035\"`\n19. `ExcelBook.WorkSheets(1).cells(3,2).value = \"吕海峰\"`\n20. `ExcelBook.WorkSheets(1).cells(3,3).value = \"18806186012\"`\n21. `Excelbook.SaveAs file`\n22. `ExcelApp.Application.Quit `\n23. `Set ExcelApp = Nothing`",
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"Windows 下需要如下操作：\n\n（也可以通过运行->DCOMCNFG来启动组件服务配置）\n\n0\n\n0\n\n 评论审核未开启 回复后邮件通知我",
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https://www.rocknets.com/converter/temperature/19.7-c-to-f | [
"# Convert 19.7 Celsius to Fahrenheit (19.7 c to f)\n\n What is 19.7 Celsius in Fahrenheit? Celsius: ℃ Fahrenheit: ℉\n\nYou may also interested in: Fahrenheit to Celsius Converter\n\nThe online Celsius to Fahrenheit (c to f) Converter is used to convert temperature from Degrees Celsius (℃) to Fahrenheit (℉).\n\n#### The Celsius to Fahrenheit Formula to convert 19.7 C to F\n\nHow to convert 19.7 Celsius to Fahrenheit? You can use the following formula to convert Celsius to Fahrenheit :\n\nX(℉)\n= Y(℃) ×\n9 / 5\n+ 32\n\nTo convert 19.7 Celsius to Fahrenheit: ?\n\nX(℉)\n= 19.7(℃) ×\n9 / 5\n+ 32\n\n#### Frequently asked questions to convert C to F\n\nHow to convert 133 celsius to fahrenheit ?\n\nHow to convert 116 celsius to fahrenheit ?\n\nHow to convert 145 celsius to fahrenheit ?\n\nHow to convert 89 celsius to fahrenheit ?\n\nHow to convert 160 celsius to fahrenheit ?\n\nHow to convert 199 celsius to fahrenheit ?\n\nTo convert from degrees Celsius to Fahrenheit instantly, please use our Celsius to Fahrenheit Converter for free.\n\n#### Best conversion unit for 19.7 ℃\n\nThe best conversion unit defined in our website is to convert a number as the unit that is the lowest without going lower than 1. For 19.7 ℃, the best unit to convert to is 19.7 ℃."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60555947,"math_prob":0.7238647,"size":1926,"snap":"2023-40-2023-50","text_gpt3_token_len":616,"char_repetition_ratio":0.27991676,"word_repetition_ratio":0.067885116,"special_character_ratio":0.33489096,"punctuation_ratio":0.1412037,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99104244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T02:58:54Z\",\"WARC-Record-ID\":\"<urn:uuid:f255a8f6-2711-4c31-9db9-c504c053d742>\",\"Content-Length\":\"69547\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cf8b32d7-0b1c-4f23-b980-32c85c30fd71>\",\"WARC-Concurrent-To\":\"<urn:uuid:7446acc2-5fe2-48d8-9e75-97e939850ed8>\",\"WARC-IP-Address\":\"172.67.199.100\",\"WARC-Target-URI\":\"https://www.rocknets.com/converter/temperature/19.7-c-to-f\",\"WARC-Payload-Digest\":\"sha1:QS52MSI4ATWZAO525EMMFZXK5MKMKVDV\",\"WARC-Block-Digest\":\"sha1:KLNIHTTFLFXHVGOPV2FOB5G4EPUFQN4K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100781.60_warc_CC-MAIN-20231209004202-20231209034202-00210.warc.gz\"}"} |
https://www.ncatlab.org/nlab/show/category+of+classes | [
"# Contents\n\n## Definition\n\nThe category of classes has classes as objects and maps of classes as morphisms.\n\nIn the Zermelo-Fraenkel set theory, a class is a proposition $P$ with a designated free variable $x$. We interpret $P(x)$ as saying that $x$ belongs to the class $P$. We also write $x\\in P$, while understanding that $P$ is not a set, but $x$ is.\n\nFor example, any set $A$ is a class, whose proposition is $a\\in A$ and the designated free variable is $a$.\n\nThe proposition $a=a$ defines the class of all sets, which does not arise via the above construction from any set.\n\n## Relations and maps\n\nGiven two classes $A$ and $B$, we can form their product $A\\times B$ as the class $A\\times B$ such that $P(x)$ means $x=(a,b)$ for some $a\\in A$ and $b\\in B$, where $(a,b)$ denotes the usual ordered pair constructed in Zermelo-Fraenkel set theory as $(a,b)=\\{\\{a\\},\\{a,b\\}\\}$ or $(a,b)=\\{\\{\\emptyset,a\\},\\{b\\}\\}$ or $(a,b)=\\{\\{\\{\\emptyset\\},a\\},\\{b\\}\\}$.\n\nIf $P(x)$ implies $Q(x)$, we say that $P$ is a subclass of $Q$.\n\nA relation from $A$ to $B$ is a subclass of $A\\times B$.\n\nA map $f$ from $A$ to $B$ is a relation $R$ from $A$ to $B$ such that for any $a\\in A$ there is a unique $b\\in B$ such that $R(a,b)$. In this case we write $f(a)$ for this unique $b$, so $f(a)=b$ means $R(a,b)$.\n\nMaps of classes can be composed? in the usual manner,\n\nwhich produces a category. Here a category is understood in the sense of a first-order logic with two sorts (objects and morphisms), but at no point we attempt to consider all classes as a single unified whole.\n\nA family $f$ of classes indexed by a class $I$ is a map of classes $f\\colon T\\to I$. The class indexed by $i\\in I$ is the preimage $f^*\\{i\\}$. Such families can be pulled back along maps of classes $J\\to I$ and pushed forward along maps $I\\to J$.\n\nWe can now define large categories as having a class of objects, a class of morphisms, together with composition and identities satisfying the usual axioms. The category of classes considered above is not a large category in this sense.\n\nWe do not require large categories to be locally small.\n\n## Diagrams\n\nSuppose $I$ is a large category. An $I$-indexed diagram of classes is defined as follows. First, we have an $I$-indexed family of classes $f\\colon T\\to I$. Secondly, we have a transition map\n\n$tr\\colon\\{(t,h)\\mid t\\in T, h\\in Mor(I), f(t)=dom(h)\\}\\to T,$\n\nwhich is a map of classes such that $f(tr(t,h))=codom(h)$. (The domain of $tr$ is a class because $t$ and $h$ are sets.) Finally, the transition map satisfies the usual axioms expected from a functor.\n\nFor any large category $I$ and a class $C$ we can define the constant diagram indexed by $I$ with value $C$. We take $T=I\\times C$ with $f\\colon T\\to I$ being the projection map. The transition map sends $(t,h)\\mapsto t$.\n\n## Limits and colimits as adjoint functors\n\nFor any large category $I$ we have a constant diagram functor that sends a class $C$ to the constant diagram on $C$.\n\nWe can talk about left and right adjoint functors to this functor in the sense of a first-order logic with two sorts (objects and morphisms), augmented with symbols for the functor, its adjoint, together with unit and counit maps that satisfy the triangle identities.\n\n## Limits\n\nThe category of classes admits all small limits.\n\nFirst, it admits equalizers: the equalizer of $f,g\\colon A\\to B$ is the subclass $E$ of $A$ defined by $E(e)=(e\\in A \\wedge f(e)=g(e))$.\n\nSecondly, it admits small products: the small product of an $I$-indexed family of classes $f\\colon T\\to I$, where $I$ is an arbitrary set (considered as a class when used with $f$) can be constructed as the class $P$ such that $p\\in P$ if $p$ is a map of sets? whose domain is $I$ and for all $i\\in I$ we have $p(i)\\in T$ and $f(p(i))=i$. (Observe that $P$ is indeed a class.)\n\nFinally, it admits all small limits because the usual reduction of small limits to equalizers of small products continues to work provided that we adhere to the above convention on the definition of families of classes.\n\n## Colimits\n\nThe category of classes admits all colimits indexed by arbitrary large categories $I$, i.e., large colimits.\n\nFirst, the standard reduction of $I$-indexed colimits to a coequalizer of a pair of arrows between coproducts indexed by $Mor(I)$ and $Ob(I)$ still works in this context since class-indexed families of classes can be pulled back along source and target maps $Mor(I)\\to Ob(I)$.\n\nSecondly, class-indexed coproducts of classes can be computed simply by taking the total class $T$ of the corresponding class-indexed family $f\\colon T\\to I$ of classes.\n\nThirdly, coequalizers of classes exist by Scott's trick. Observe that given a pair of arrows $f,g:X\\to Y$ between classes, we can define an equivalence relation on $Y$ by saying that $y~y'$ if there is a map $h:[0,n]\\to Y$ such that $h(0)=y$, $h(n)=y'$ and for any $i\\in[0,n)$ there is $x\\in X$ such that $h(i)=f(x)$ and $h(i+1)=g(x)$ or $h(i)=g(x)$ and $h(i+1)=f(x)$. The quotient of $Y$ by this equivalence relation exists by Scott's trick and is precisely the desired coequalizer.\n\n## Other categorical properties\n\nThe category of classes is a regular category: the obvious notion of image factorization is stable under pullbacks. It is also a Barr-exact category: every equivalence relation $R$ on a class $C$ is induced by the quotient map $C\\to C/R$.\n\nIt is also an infinitary extensive category, where “infinitary” means “class-indexed”. Indeed, pullbacks of coproduct injections along arbitrary maps of classes exist and class-indexed coproducts are disjoint and stable under pullback.\n\nIt is also well-pointed: for every two maps between classes $f,g:X\\rightarrow Y$ and every element $x\\in X$, if $f(x) = g(y)$, then $f = g$, and the category of classes is not the terminal category.\n\nIt likewise has all objects corresponding to large cardinals, most notably a natural numbers object. Otherwise, the category of finite sets FinSet is vacuously a category of classes, as the notions of ‘finitary’ and ‘class-indexed’/‘infinitary’ coincide.\n\nAs such, the category of classes is a well-pointed infinitary Heyting or Boolean pretopos, depending upon the external logic used, with a natural numbers object and other large cardinals, and where “infinitary” is used in the rather strong sense of “class-indexed”.\n\nThe category of classes is not cartesian closed or locally cartesian closed and does not have power objects. Indeed, the class of all sets $S$ does not have a power object, or, equivalently, there is no internal hom $Hom(S,\\{0,1\\})$.\n\n## Category with class structure\n\nThe category of classes is a primordial example of a category with class structure. Its open maps are precisely those maps of classes $f\\colon T\\to I$ such that $f^*\\{i\\}$ is a set for each $i\\in I$. Small maps coincide with open maps. The powerclass of a class $C$ is the class of all sets $A$ such that $A$ is a subclass of $C$. The universal class is the class of all sets.\n\nLast revised on March 3, 2021 at 14:21:47. See the history of this page for a list of all contributions to it."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94284123,"math_prob":0.9991728,"size":5769,"snap":"2021-31-2021-39","text_gpt3_token_len":1230,"char_repetition_ratio":0.17484823,"word_repetition_ratio":0.054294176,"special_character_ratio":0.20610158,"punctuation_ratio":0.10483871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999936,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T23:56:46Z\",\"WARC-Record-ID\":\"<urn:uuid:e6c5987b-2698-479d-b832-f41b8ff918f3>\",\"Content-Length\":\"58371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5dce783-5bc6-402b-a6cb-816ee1e0577f>\",\"WARC-Concurrent-To\":\"<urn:uuid:6df7fb80-49a8-4641-8141-6cc784a72270>\",\"WARC-IP-Address\":\"104.21.81.15\",\"WARC-Target-URI\":\"https://www.ncatlab.org/nlab/show/category+of+classes\",\"WARC-Payload-Digest\":\"sha1:LSXBZYJK2AJHIRACF2FX3YPD2VEC6MO5\",\"WARC-Block-Digest\":\"sha1:KB2X6ULOKR3WRSD3W3I2HBC42YWKUPK2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153803.69_warc_CC-MAIN-20210728220634-20210729010634-00034.warc.gz\"}"} |
https://www.physicsforums.com/threads/finding-the-integrating-factor.245392/ | [
"# Finding the Integrating Factor\n\nHi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy\n\nCan some one pleas explain how to do this? Thank you.\n\n## Answers and Replies\n\nDefennder\nHomework Helper\nYou appear to have mistyped your problem. There 2 \"dy\"s here but no \"dx\".\n\nHallsofIvy\nScience Advisor\nHomework Helper\nHi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy\n\nCan some one pleas explain how to do this? Thank you.\nAssuming that you mean $(y^2x+ y)dx+ (x^2y+ 2x)dy= 0$, if you know that the integrating factor is a function of xy, the obvious thing to do is to put multiply that equation by u(xy) and see what happens. In order that $u(xy)(y^2x+ y)dx+ u(xy)(x^2y+ 2x)dy= 0$ be exact, we must have $(u(y^2x+ y))_y$ and $(u(x^2y+ 2x)_x)$ equal. That is, $u_x(y^2x+ y)+ u(2xy+1)= u_x(x^2y+ 2x)+ u(2xy+ 2)$. Since u is a function of xy specifically, $u_x= yu'$ and $u_y= xu'$. That is, u must satisfy $xu(y^2x+ y)+ u(2xy+1)= yu'(x^2y+ 2x)+ u(2xy+ 2)$. That gives you a differential equation for u which may or may not be solvable, depending upon whether there really is an integrating factor that is a function of xy only.\n\nYes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).\n\nso can any one also give me some advice on how to solve this problem i am having. Thank you.\n\nDefennder\nHomework Helper\nI think HallsofIvy's advice is spot on. Equate the two derivatives after differentiating the product of u and the expression preceding the dx and dy. Use the chain rule when differentiating u(xy) with respect to x and y only by denoting product xy as v. Simplify the resulting equation and you'll notice it reduces to a simple separable differential equation. Solve that and express u in terms of x,y only and you're done. I tried it out and it works like a charm.\n\nHallsofIvy\nScience Advisor\nHomework Helper\nYes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).\n\nso can any one also give me some advice on how to solve this problem i am having. Thank you.\nYou told us, originally, that the integrating factor was a function of xy. Why are you now trying xmyn? How do you know the integrating factor is either a function of xy or of the form xmyn?\n\nYeah it did work sorry about that, and as for havening it as xmyn other times you can use it to find out what power the x and y are to making it easer. mind you you have to know xy are the integrating factor."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90296316,"math_prob":0.99686235,"size":300,"snap":"2021-21-2021-25","text_gpt3_token_len":89,"char_repetition_ratio":0.10135135,"word_repetition_ratio":0.22950819,"special_character_ratio":0.29333332,"punctuation_ratio":0.05970149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99934465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T02:13:39Z\",\"WARC-Record-ID\":\"<urn:uuid:6c3bdf6d-f824-499c-9946-0173d0d54f7d>\",\"Content-Length\":\"76503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0ecbef72-757a-442a-813d-9aef81190b75>\",\"WARC-Concurrent-To\":\"<urn:uuid:048562a1-6628-411c-973d-b80b92d439c8>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/finding-the-integrating-factor.245392/\",\"WARC-Payload-Digest\":\"sha1:LI3EQFT64I7SWJIIHNEFKBVMV5R5UCXT\",\"WARC-Block-Digest\":\"sha1:U4DCCEVEEAQ2GGCB44WN3GKQTGS5XAWW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488259200.84_warc_CC-MAIN-20210620235118-20210621025118-00550.warc.gz\"}"} |
https://tarah.org/2011/12/02/third-edition-earthdawn-dice-roller-in-java/ | [
"# Third Edition Earthdawn Dice Roller in Java\n\nSo, I play an RPG called Earthdawn; it’s a lot like D&D, but for real nerds.\n\nOne of the things we all do in my gaming group is write our own dice rollers; rolling actual dice is SO passé–and there’s an ongoing argument about whether or not a seeded random is more or less random than the natural flaws in dice and rolling surfaces. Java is the language in which I learned to write math, so I somewhat naturally write algorithms in Java without thinking. It’s easy enough to translate this into C# or whatever.\n\nOk, so, here’s the algorithm. In Earthdawn, dice rolls are predicated on the step level of the difficulty. You may have an attack roll at step 18 and a damage roll at step 22. In 3rd edition Earthdawn, that translates to rolling d12+d10+d8 to attack, and 2d12+2d6 for damage. Here’s the chart (click to embiggen):",
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"As you can see, there’s some kind of progression here; it turns out that the algorithm is a simple infinite series. There’s a jump in the number of dice every seven steps. Hence, the algorithm has a few simple steps:\n\n(1) Divide the step number by 7.\n(2) Determine and store the floor and the modulus.\n(3) Roll a number of d12s equal to (floor – 1).\n(4) Roll dice equal to the corresponding modulus (the first addition of dice past the 7 threshold will be 2d6, so if the modulus is 1, 2d6 are rolled and added).\n\nThat is the step algorithm such that no lookup is now necessary; Earthdawn has exploding dice and epic\nfails, however, so two things are necessary. Look at the exploding dice method; if you roll the maximum value of a die, you can roll it again. You can keep rolling that die until a value shows that is less than the maximum value, such that a d6 rolled with a result of 6 can be rerolled. On the second roll, 6 results. On the third roll, 2 results, so the total value of that die roll is 14. For epic fails, if you roll more than one die and all dice show ones, you have epically failed (similar to a fumble in D&D, and with equivalent disastrous results).\n\nHere’s my dice roller; click to embiggen:",
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"So, here’s the code. It’s intended to be self-contained (you can see that I use a pic of Captain Malcolm Reynolds; just drop a pic in your file structure and reference it in the code if you want a background). Obviously this is bare-bones; you can adapt it at your leisure and to whatever GUI you desire. Any suggestions are welcome; I am always debugging this (and the JavaScript I’m using to display syntax highlighting is a little cranky, so forgive any indentation issues). If you can think of a way to optimize this algorithm, let me know; I always need bragging rights over the guys 😉\n\n```package dice;\n\nimport java.awt.Color;\nimport java.awt.Dimension;\nimport java.awt.Graphics;\nimport java.awt.GridLayout;\nimport java.awt.Image;\nimport java.awt.event.ActionEvent;\nimport java.awt.event.ActionListener;\nimport java.util.Random;\n\nimport javax.swing.BorderFactory;\nimport javax.swing.ImageIcon;\nimport javax.swing.JButton;\nimport javax.swing.JCheckBox;\nimport javax.swing.JFrame;\nimport\njavax.swing.JLabel;\nimport javax.swing.JOptionPane;\nimport javax.swing.JPanel;\nimport javax.swing.JSpinner;\nimport javax.swing.JTextField;\nimport javax.swing.SpinnerModel;\nimport javax.swing.SpinnerNumberModel;\n\n@SuppressWarnings(\"serial\")\npublic class ImagePanel extends JPanel implements ActionListener{\n\nint six = 6;\nint eight = 8;\nint ten = 10;\nint twelve = 12;\nint dice = 0; //counter for total number of dice\n\npublic int middle; //step value entered by me.\npublic boolean fail; //whether or not an epic fail happened.\npublic JFrame frame;\npublic Random r;\nImage img = new ImageIcon(\"img/MalcolmReynolds13.jpg\").getImage();\nSpinnerModel stepEntry = new SpinnerNumberModel(1, 1, 300, 1);\nSpinnerModel karmaCounter = new SpinnerNumberModel(25, 0, 25, 1);\nJSpinner stepSpinner = new JSpinner(stepEntry);\nJSpinner karmaSpinner = new JSpinner(karmaCounter);\n\nprivate JTextField diceResult;\nprivate JButton myButton;\n\nprivate JCheckBox myCheck;\nprivate JLabel enterStep = new JLabel(\"Enter Step Here.\");\n\npublic static void main(String[] args) {\nImagePanel panel = new ImagePanel(new ImageIcon(\"img/MalcolmReynolds13.jpg\").getImage());\nJFrame frame = new JFrame(\"Tarah's Dice Roller Of Awesomeness\");\n\nframe.pack();\nframe.setVisible(true);\n}\n\npublic ImagePanel(String img) {\nthis(new ImageIcon(img).getImage());\n}\n\npublic ImagePanel(Image img){\nthis.img = img;\nDimension size = new Dimension(img.getWidth(null), img.getHeight(null));\nsetPreferredSize(size);\nsetMinimumSize(size);\nsetMaximumSize(size);\nsetSize(size);\nJPanel panel = new JPanel();\nmyButton = new JButton(\"Roll The Dice.\");\nmyButton.setBorder(BorderFactory.createLineBorder(Color.black));\ndiceResult = new JTextField(\"Roll Result\", 9);\n\ndiceResult.setBorder(BorderFactory.createLineBorder(Color.black));\nmyCheck = new JCheckBox(\"Use Karma.\", false);\nmyCheck.setBorder(BorderFactory.createLineBorder(Color.black));\nGridLayout myGrid = new GridLayout(3, 2);\npanel.setLayout(myGrid);\npanel.setBorder(BorderFactory.createLineBorder(Color.black));\nsetVisible(true);\n}\n\npublic void paintComponent(Graphics g) {\ng.drawImage(img, 0, 0, null);\n}\n\npublic void actionPerformed(ActionEvent e) {\nboolean useKarma = false;\nmiddle = (Integer)stepSpinner.getValue();\nSystem.out.println(\"actionPerformed() thinks the step number is: \" + middle);\nif (myCheck.isSelected() == true) {\nSystem.out.println(\"Using Karma.\");\nuseKarma = true;\nint decrease = ((Integer)karmaCounter.getValue()) - 1;\nkarmaCounter.\nsetValue(decrease);\n}\nString s = Integer.toString(rollTheDice(useKarma, middle, fail));\ndiceResult.setText(s);\n\n}\n\n/////////////////////////////////////////////\n/////////////////MATHYNESS///////////////////\n/////////////////////////////////////////////\n\n//This is the Earthdawn Exploding Dice Method.\npublic int d (int die){\n\nint sides = die;\nint result = 0;\nint roll;\n\ndo {\nr = new Random();\nroll = r.nextInt(sides) + 1;\nresult = result + roll;\nSystem.out.println(\"This is a d\" + sides + \" roll with result: \" + result);\n} while (roll == sides);\n\nreturn result;\n}\n\npublic int oneToSeven (int o) {\nint result = 0;\nif (o == 1) {\nresult = d(six) - 3;\nif (result < 1) {\nresult = 1;\n}\n}\nif (o == 2) {\nresult = d(six) - 2;\nif (result < 1) {\nresult = 1;\n}\n}\nif (o == 3) {\n\nresult = d(six) - 1;\nif (result < 1) {\nresult = 1;\n}\n}\nif (o == 4) {\nresult = d(six);\n}\nif (o == 5) {\nresult = d(eight);\n}\nif (o == 6) {\nresult = d(ten);\n}\nif (o == 7) {\nresult = d(twelve);\n}\nreturn result;\n}\n\npublic int prefix (int p) {\nint prefixTotal = 0;\n\nfor (int i=1; i= 8) {\nint d12s = prefix(full);\nint rest = suffix(mod);\nresult = d12s + rest;\n\nif (result == full+1) {\nfail = true;\n}\n}\nreturn result;\n}\n\npublic int rollTheDice(boolean addKarma, int stepValue, boolean epicFail) {\ndice = 0;\nint result = 0;\nresult = step(stepValue);\nepicFail = fail;\nint dK = d(six);\nresult = result + dK;\nif (epicFail == true && dK == 1) {\nJOptionPane.showMessageDialog(frame, \"Epic FAIL.\");\n}\n}\nelse if (addK != true && epicFail == true) {\nJOptionPane.showMessageDialog(frame, \"Epic FAIL.\");\n}\nreturn\nresult;\n}\n}\n```",
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"http://thetarah.com/wp-content/uploads/2011/12/snapshot2-150x150.png",
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"http://thetarah.com/wp-content/uploads/2011/12/snapshot3-150x150.png",
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"https://secure.gravatar.com/avatar/751405a20ebc97a0c30b4c62cc9ab13d",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5825947,"math_prob":0.9255018,"size":7292,"snap":"2020-45-2020-50","text_gpt3_token_len":1828,"char_repetition_ratio":0.14201427,"word_repetition_ratio":0.03457944,"special_character_ratio":0.28085575,"punctuation_ratio":0.21297602,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9632391,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,5,null,5,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T09:25:35Z\",\"WARC-Record-ID\":\"<urn:uuid:9d982c5f-4cc8-448d-a616-66844b704eae>\",\"Content-Length\":\"44506\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7246904-9537-4d37-9021-7a6733ea8cb4>\",\"WARC-Concurrent-To\":\"<urn:uuid:c673749f-17fd-425e-a3ba-261560fe18b9>\",\"WARC-IP-Address\":\"75.119.194.230\",\"WARC-Target-URI\":\"https://tarah.org/2011/12/02/third-edition-earthdawn-dice-roller-in-java/\",\"WARC-Payload-Digest\":\"sha1:PYSXN2FFZCMAZLILCBCEFMLE63CZYBC6\",\"WARC-Block-Digest\":\"sha1:QJAQTL7L2YGPFMXAMVU6GFVJCXAAPJOT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141672314.55_warc_CC-MAIN-20201201074047-20201201104047-00553.warc.gz\"}"} |
https://hurmanblirriktcndy.netlify.app/98271/20415.html | [
"# Robin Augustine - Uppsala University, Sweden\n\nCHEM 4310 test 1 review - CHEM 4310: Physical Chemisty Ii\n\nSpeed of Light (in vacuum), c=2.998 x 10^8 m s−1. n=principal energy level. This equation is used to find the energy of an electron at a given energy level, or the change between energy levels. The Rydberg constant was determined to have the value (10.32 ± 0.71) × 10 6 m −1 , which is very close to the literature value. The screening coefficient, which for moderately heavy nuclei is constant, was assessed to be 2.427 ± 2.773 -this measurement has an extremely large uncertainty overall. V. Rydberg constant (RH) is a fundamental and important constant related to atomic spectroscopic terms. Its value is 1.097 × 1 0 7 m − 1 1.097{\\text{ }} \\times {\\text{ }}{10^7}{m^{ - 1}} 1.",
null,
"Compúter progiamming of the Method. We use the non-GAAP financial measure \"constant currency basis\" in our Sanna Rydberg, VD för Arcoma har den 12/1 2021 förvärvat 7 000 Author : Ellen Knutsen Rydberg; Göteborgs universitet; Göteborgs universitet; A constant supply of oxygen is needed to maintain cell functions in the Multiplicera med Rydberg Constant. Multiplicera resultatet från föregående avsnitt med Rydbergkonstanten, R H = 1,0968 × 10 7 m - 1, för att hitta ett värde för 1 Asan, N., Noreland, D., Hassan, E., Redzwan, S., Rydberg, A. et al. (2017). Design of constant width branch line directional coupler for the microwave sensing 40 - constant reader. Författare Skugge, Linda. 259238.\n\n## EXAMENSARBETE - DiVA Portal\n\nThe reciprocal of the wavelength, 1/λ, is termed the wavenumber, as expressed by Rydberg in his version of the Balmer equation. The Rydberg constant R ∞ = m e α 2 c/2h links the natural energy scale of atomic systems and the SI unit system. It connects the mass of the electron m e, the fine structure constant α, Planck’s constant h, and the speed of light in vacuum c.",
null,
"### Sweden: Melodifestivalen Semi Final 3 takes place in Luleå\n\nA Constante de Rydberg, nomeada em homenagem ao físico Johannes Rydberg, é uma constante física que aparece na fórmula de Rydberg.Ela foi descoberta durante a medição do espectro do hidrogênio e foi definida a partir dos resultados desta experiência por Anders Jonas Ångström e Johann Balmer. The Hydrogen Balmer Series and Rydberg Constant by Dr. James E. Parks. Department of Physics and Astronomy. 401 Nielsen Physics Building. The University Here R is the Rydberg constant 1, which has been precisely measured and found to have the value R = 10973731.5683 ± 0.0003 m–1.",
null,
"J. Rydberg. Determination of thermodynamic constants for the extraction of copper and zinc acetylacetonates, Solvent Extraction Research (Eds.\n\n9.274 015 4 (31) ×10−24 J T−1. Electron magnetic moment. av H Haeggblom · 1978 — where a is the exponential decay constant of K ir. the negative z-Jirection. This equation can be solved analytically.\n\n217. Appendices. 240. Spherical tensor operators. 246. The relativistic wave equation for manyelectron systems. 259.\nGargantua och pantagruel\n\nc is the velocity of light in vacuum. εo is the permittivity of the free space. e is the elementary charge. The accepted values of the Rydberg constant, R∞, as in 1998 are: Rydberg Constant in nm - 10 973 731.568 548 (83) m-1.\n\nCheap grills at home depot · How to report phishing calls to apple · Balmer series and rydberg constant · Dibujos de harry potter para dibujar faciles · Robe sa R - Röntgen enhet. R - Rydberg Constant R- # - Kylmedelsnummer Ra - Radium RA - retinsyra. RACHEL - Remote Acess Chemical Hazards Electronic Library Jans grandfather was the well-known spectroscopist Janne Rydberg (the Rydberg constant). Antal mantalsskrivna p adressen r 2 personer, Sven Heurgren (90 Value Rydberg constant. The CODATA value is = = 10 973 731.568 160 (21) m −1, where is the rest mass of the electron, is the elementary charge, is the permittivity of free space, is the Planck constant, and Rydberg constant, (symbol R ∞ or R Η), fundamental constant of atomic physics that appears in the formulas developed (1890) by the Swedish physicist Johannes Rydberg, describing the wavelengths or frequencies of light in various series of related spectral lines, most notably those emitted by hydrogen atoms in the Balmer series. Rydberg Constant In the science of spectroscopy, under physics, the Rydberg constant is a physical constant relating to atomic spectra.\nSms long code lookup\n\n### Marika Rydberg marikarydberg – Profil Pinterest\n\nThis constant is now known as the Rydberg constant, and m′ is known as the quantum defect. As stressed by Niels Bohr, expressing results in terms of wavenumber, not wavelength, was the key to Rydberg's discovery. The fundamental role of wavenumbers was also emphasized by the Rydberg-Ritz combination principle of 1908. In spectroscopy, the Rydberg constant is a physical constant relating to the electromagnetic spectra of an atom. Its symbol is R ∞ {\\displaystyle R_{\\infty }} for heavy atoms or R H {\\displaystyle R_{\\text{H}}} for hydrogen . The accepted values of the Rydberg constant, R∞, as in 1998 are: Rydberg Constant in nm - 10 973 731.568 548 (83) m-1. Rydberg Constant in Joules - 2.179 871 90 (17).10-18 J. Rydberg Constant in Electron Volt - 13.605662285137 eV.\n\nÄlvdalens simhall\n\n### Energy and work converter – Appar på Google Play\n\nNotes. Symbol. R∞. Use \"R-oo\" for our conversion calculators.\n\n## Rydberg Constant: Surhone, Lambert M.: Amazon.se: Books\n\nAs our guest you can look forward BIFF RYDBERG GOES WILD Butter-seared rein- deer topside served with Pa exactly Fine structure constant α = µ 0 e c/h (33) 10 3 α (61) Bohr radius a 0 = 4πǫ 0 h /m e e (4) m Hartree energy E h = h /m e a (6) J Rydberg constant R av L Oliver · 2002 — The Primary Dissociation Constant of Diphenyl- thiocarbazone (Dithizone).\n\nIndependently, Janne Rydberg analyzed the spectra of many elements. He started using the wavenumber n instead of Recent advances in high resolution spectroscopy with tunable lasers have made it possible to determine a new value of the Rydberg constant with an almost In a previous paper, the results of a series of measurements of the wavelengths of the first six lines of the Balmer series of hydrogen were given, together with a 17 Sep 2008 We particularly emphasize the methods used to deduce the Rydberg constant R _\\infty and we consider the prospects for future improvements Johannes Rydberg was one of the grandfathers of modern-day physics and chemistry When Bohr learned of Balmer's series and Rydberg's constant from H M Énergie: Constante de Rydberg, électron-volt, Joule, gigajoule, Mégajoule, kilojoule, millijoule, microjoule, nanojoule, attojoule, megaelectron-volt, kiloélectron The Hydrogen Balmer Series and Rydberg Constant by Dr. James E. Parks. Department of Physics and Astronomy. 401 Nielsen Physics Building. The University his “Bohr Model”; a working model of the hydrogen atom."
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https://pubs.aip.org/aip/rsi/article/93/2/025101/2844069/A-super-resolution-technique-to-analyze-single | [
"Direct-geometry time-of-flight chopper neutron spectroscopy is instrumental in studying dynamics in liquid, powder, and single crystal systems. We report here that real-space techniques in optical imagery can be adapted to obtain reciprocal-space super resolution dispersion for phonon or magnetic excitations from single-crystal neutron spectroscopy measurements. The procedure to reconstruct super-resolution energy dispersion of excitations relies on an accurate determination of the momentum and energy-dependent point spread function and a dispersion correction technique inspired by an image disparity calculation technique commonly used in stereo imaging. Applying these methods to spinwave dispersion data from a virtual neutron experiment demonstrates ∼5-fold improvement over nominal energy resolution.\n\nInelastic neutron scattering (INS) is a powerful probe of fundamental excitations in solids, including those of vibrational or magnetic origins. Neutron scattering from these excitations is characterized by the four-dimensional (4D) scattering function, S(Q, E), where Q is the momentum transfer and E is the energy transfer. In recent years, the use of highly pixelated detector arrays in conjunction with direct-geometry chopper spectrometer (DGS) instruments1–7 has allowed for efficient measurements of single crystal 4D S(Q, E) functions over large ranges of Q and E. One single measurement of a single-crystal sample at a DGS instrument captures the sample scattering function S(Q, E) on a three-dimensional (3D) manifold in the 4D Q, E space. Typically, a sample is rotated around a single axis, while maintaining a single wavelength (i.e., monochromatic energy) of incident neutrons, to scan the 4D Q, E volume. A series of 3D manifolds measured during the scan are then combined into the volumetric 4D dataset, allowing for extracting 2D slices at high-symmetry Q directions by using software packages, such as Mantid,8 Horace,9 DAVE,10 Mslice,11 and Utsusemi.12 This is done by integrating the measured scattering intensity along two of the four dimensions, yielding a scattering intensity as a function of the remaining two dimensions. Most often, this is done to extract dispersion relations of fundamental excitations within crystalline solids by illustrating measured scattering intensity as a function of energy transfer along the vertical axis and a single direction in wave-vector transfer along the horizontal axis. These extracted dispersion data can be used to directly compare to model calculations of the materials dynamics in order to constrain or refine parameters in a model (i.e., the Hamiltonian). For magnetic systems, SpinW13 is often used to fit spin-wave models to the experimentally obtained dispersion along multiple high-symmetry directions simultaneously. This allows one to obtain exchange parameters or other energy dependent terms in a model Hamiltonian, which can represent the spin dynamics of the system.\n\nQuantifying a crystalline material’s vibrational or magnetic dispersion is key to understanding the system’s dynamics. The accuracy of the measured dispersion (and that of the inferred quantities, which define the dynamics, such as exchange couplings or force constants) obtained from DGS experiments is bound by the instrument resolution. However, a significant number of measurements do not take into account instrumental resolution effects beyond the use of an analytical approximation of a single energy and wave-vector resolution of the instrument. DGS instrument resolution for the S(Q, E) scattering function is four-dimensional, and the effects of the resolution function are compounded by the slope and the curvature of the dispersion surface, making it cumbersome to accurately model. Given a set of instrument and experimental parameters, such as chopper settings and the sample shape, the resolution function for DGS neutron scattering instruments still varies at different (Q, E) points in the measured dynamical range. This resolution ellipsoid varies considerably across the detector array of the instrument and energy transfer. Furthermore, the resolution ellipsoid can have significant tilt that can lead to focusing and defocusing effects in the measured dispersion depending on the slope of the dispersion relative to the tilt. This can make it difficult to accurately extract a model based on the measured dispersions.\n\nRecently, it was demonstrated14 that some super-resolution imagery techniques can be adapted to improve energy resolution in a phonon density of states measurement, g(E), based on neutron scattering techniques. This analysis was performed upon a measurement of the phonon density of states as a function of a single independent variable, energy transfer, which is a routine measurement for direct geometry chopper spectrometers measuring powder samples. In this work, we introduce a super-resolution technique to improve the ability of extracting dispersion from inelastic neutron scattering measurements of single crystal samples. We use techniques inspired by image correlation methodologies to improve extraction of information from measurements as a function of two independent variables, energy and wave-vector transfer.\n\nDispersion relations are obtained from single crystal measurements at DGS instruments by extracting slices of scattering intensity as a function of energy transfer, E, and wave-vector transfer, Q, along high-symmetry directions in reciprocal space. These slices are obtained by integrating a portion of the measured reciprocal space along the two orthogonal reciprocal space directions to the direction of interest in the chosen slice. Constant Q cuts through these 2D slices can then be extracted by integrating a portion of the slices along wave-vector transfer and plotting the resulting scattering intensity as a function of energy transfer. For each constant Q cut through a single 2D slice, centers of peaks in the energy spectrum are recorded as the excitation energies for this particular Q value. The energy of these peak centers plotted as a function of Q is the measured dispersions. An analytic or numerical model of the dispersion can then be directly compared to the experimentally determined dispersion values in order to determine model parameters. This is often done using non-linear curve fitting algorithms. This methodology of extracting dispersion parameters based on peak location does not account for instrumental energy or wave-vector resolution. Such quantities may shift or skew peak locations and, therefore, would serve to skew any model parameters determined from a comparison of peak positions to model dispersion. The method we present here calculates resolution-based corrections to the measured dispersion functions.\n\nFigure 1 shows the steps of this dispersion correction workflow. An experimental slice is first obtained from data reduction [Fig. 1(a)]. A dispersion dataset is then obtained directly from the experimental slice, Eexp;0(q), following the normal steps outlined earlier. This quantity is shown as open symbols in Fig. 1(b). Those excitations are then fit to a dispersion model (in this case, the spinwave model generated in SpinW), and relevant parameters (in this case, exchange parameters) are obtained [Figs. 1(a)1(d)]. The scattering cross section as a function of momentum transfer and energy transfer, including both the dispersion relation and the scattering amplitude, is now determined by the excitation model and relevant parameters. The next step is to obtain the instrument resolution function [Fig. 1(e)]. This can be achieved by using analytical calculations (this is often done for triple axis spectrometers15–18) or Monte Carlo neutron ray tracing simulations. More details can be found in Sec. II B. By convolving the calculated scattering function with the instrument resolution function, we can obtain the modeled slice [Fig. 1(f)]. Then, we can compare the modeled slice and the experimental slice and obtain the energy shifts, ΔE(q) [Fig. 1(g)], required to match the modeled slice to the experimental slice. This step is key to the super-resolution dispersion determination and is further explained in Sec. II C. The energy shifts obtained can then be applied to the model dispersion curve (from first fit to the experimental data) to obtain the corrected dispersion curve [Fig. 1(h)],\n\n$Eexp;1(q)=Emodel; 0(q)+ΔE(q).$\n(1)\n\nThe focus of this work is to demonstrate that measured dispersion relations can be corrected taking into account resolution effects, but a natural further step is to fit the dispersion model to the corrected dispersion, Eexp;1(q), to obtain the corrected parameters for the Hamiltonian.\n\nFIG. 1.\n\nSuper-resolution dispersion workflow. An experimental dispersion [open circles in (b) and (d)] is first obtained from an experimental slice (a). This has been routinely done in single crystal DGS data analysis by finding the peak centers, Ej, of the constant q-cuts, I(Eqj), to form the dispersion data points {(qj, Ej)}. The experimental dispersion is then fit to a dispersion model [an example is shown in (c)] to obtain modeled dispersion [the blue curve in (d)]. This preliminary fitting provides a good starting point for the model parameters near the optimal values, and the subsequent steps will make super-resolution corrections. Resolution functions are calculated across the dynamical range of a slice [see (e)]. The dispersion model (c) with the parameters obtained from fitting done in (d) is convolved with the resolution (e) to obtain a modeled slice (f). The disparity (g) between the modeled slice (f) and the experimental slice (a) is then calculated. Finally, the disparity (g) is used to correct the modeled dispersion (d) and obtain the corrected dispersion (h). The units of q are reciprocal lattice units (r.l.u.).\n\nFIG. 1.\n\nSuper-resolution dispersion workflow. An experimental dispersion [open circles in (b) and (d)] is first obtained from an experimental slice (a). This has been routinely done in single crystal DGS data analysis by finding the peak centers, Ej, of the constant q-cuts, I(Eqj), to form the dispersion data points {(qj, Ej)}. The experimental dispersion is then fit to a dispersion model [an example is shown in (c)] to obtain modeled dispersion [the blue curve in (d)]. This preliminary fitting provides a good starting point for the model parameters near the optimal values, and the subsequent steps will make super-resolution corrections. Resolution functions are calculated across the dynamical range of a slice [see (e)]. The dispersion model (c) with the parameters obtained from fitting done in (d) is convolved with the resolution (e) to obtain a modeled slice (f). The disparity (g) between the modeled slice (f) and the experimental slice (a) is then calculated. Finally, the disparity (g) is used to correct the modeled dispersion (d) and obtain the corrected dispersion (h). The units of q are reciprocal lattice units (r.l.u.).\n\nClose modal\n\nThe 4D resolution function for a DGS instrument in measurements of single-crystal samples has been modeled analytically19,20 using the covariance matrix to simplify the treatment. It has also been calculated using Monte Carlo ray-tracing simulations21–25 and was sometimes approximated using Gaussian functions when it was used in resolution convolution.26 However, these corrections are not often taken into account in 4D DGS data analysis due to the complexity of the resolution convolution and the computing resources required for fitting the dispersion model obtained from a highly pixelated detector array.\n\nIncorporating resolution into the single crystal dispersion fitting workflow is difficult because of the following:\n\n• The resolution is 4D in nature, thus requiring a four-dimensional integration to convolve it with a model S(Q, E).\n\n• The resolution function varies across the dynamical range depending on the momentum and energy transfer. For example, as the energy transfer increases, the energy resolution broadening decreases for direct geometry chopper spectrometers.\n\n• The resolution ellipsoid can have a significant tilt that can lead to focusing and defocusing effects in the measured dispersion depending on the slope of the dispersion relative to the tilt of the resolution ellipsoid.\n\n• For some DGS instruments, the energy resolution function is asymmetric as a result of the moderation process peculiar to neutron production in spallation neutron sources.27\n\nThe last point here has very rarely been taken into account previously in single crystal dispersion fitting. Furthermore, to fit the dispersion model, incorporating the instrument resolution requires multiple evaluations of the dispersion model with varying parameters and resolution convolution with the model for each set of parameters. The number of iterations depends on the optimization algorithm used and how close the initial guess was to the optimal model parameters. Such an optimization procedure can be demanding in both programming and computing resources.\n\nThe super-resolution procedure outlined in this work is agnostic to the technique of resolution calculation and convolution. In this work, we have chosen to use a technique based on the Monte Carlo ray tracing simulation to illustrate the super-resolution methodology.\n\nThe MCViNE package22,28 has been used to compute the resolution function for single crystal measurements at DGS instruments.22,24,25 The procedure is reused here to simulate the resolution function (or point-spread function). The simulation starts with a beam simulation that matches experimental conditions, such as incident energy and chopper settings. In order to calculate the energy and wave-vector resolution function with MCViNE, we use a virtual sample that has the same geometric shape and lattice parameters of the real sample to scatter neutrons in the vicinity of a particular set of momentum and energy transfer h, k, l, E. In the simulation, we also take advantage of the measured UB matrix to orient the virtual sample just like what has been measured during the experiment. The virtual sample is also rotated around the vertical axis to a particular ω angle, similar to what happens in real measurements. The SEQUOIA detector system is simulated according to its specification, such as the positions and orientations of all detector packs, the 3He tube radius, length, and its spacing in the detector pack, the pressure of 3He gas in the detector tubes, and the detector pixel height. Only events that arrive at the particular detector pixel and time-of-flight bin corresponding to the nominal h, k, l, E are collected. These detector events are then reduced to $h̃,k̃,l̃$ and $Ẽ$. Those $h̃,k̃,l̃,Ẽ$ values center around the nominal h, k, l, E, as expected. The differences between those $h̃,k̃,l̃,Ẽ$ from the nominal h, k, l, E are kept in a list of dh, dk, dl, dE. The Monte Carlo ray tracing approach captures details of the 4D resolution function, including, for example, the asymmetrical energy dependent line-shape mentioned previously.\n\nTo decrease the complexity of the resolution modeling and convolution, we perform them in two dimensions along the two axes of a slice. The details of the convolution will be presented later in this article. We calculate a list of dq from the saved list of dh, dk, dl, where q is along the high-symmetry Q direction of the slice of interest. The dq, dE events are then histogrammed into a profile of the point spread function (PSF) for the 2D slice. This procedure is repeated for the points on a grid on the (q, E) plane. Each PSF at one grid point is fit to a sheared 2D function, with one axis that is “energy-like” and has an asymmetric shape and another axis that is “momentum-like” and modeled as a Gaussian. An example of this parameterization is presented in Fig. 2. Then, an interpolation of the fitted parameters allows us to calculate the PSF function at any point in the q, E space.\n\nFIG. 2.\n\nExample resolution functions for the 00L slice at q = L = 1.3, E = 5.0. The 2D resolution function R(q, E) is first simulated by using MCViNE and then fitted to an analytical function.\n\nFIG. 2.\n\nExample resolution functions for the 00L slice at q = L = 1.3, E = 5.0. The 2D resolution function R(q, E) is first simulated by using MCViNE and then fitted to an analytical function.\n\nClose modal\n\nBefore convolution, the scattering intensities of the dispersion model are first integrated along the two Q directions perpendicular to the q direction for the slice of interest. Then, the integrated data are convolved with the 2D resolution function modeled earlier to obtain the convoluted slice. This convolution method is a good approximation of the full 4D convolution, and an example is shown in Sec. III B to demonstrate that.\n\nThe basic idea here is to compare the two slices, namely, the experimental slice [Fig. 1(a)] and the resolution-convolved modeled slice [Fig. 1(f)], to find the displacements between the dispersions in the two images, which can then be used to correct the model. Finding displacements (or disparity) in two images has been a long-standing challenge in image processing. Stereo imaging techniques that uncover depth information of a scene captured in two images by finding the disparity field between the images have been reviewed multiple times during the last few decades.29–31 The techniques in traditional stereo vision find the disparity field d(x, y) that minimizes the difference between the left image (image taken by using a camera on the left) and the right image (image taken by using a camera on the right) warped by the disparity,\n\n$arg mind(x,y)∑x,yIlx,y−Irx,y−d(x,y).$\n(2)\n\nUsed in this formula is the sum of the absolute differences, but often, the sum of square differences or normalized cross correlation is also used. Il(x, y) and Ir(x, y) are left and right image intensities, and d(x, y) is the displacement along y. Local methods for dense disparity calculation minimize cost functions [Eq. (2)] for local patches, while global methods32,33 and semi-global methods34 take the smoothness of disparity into account by constraining the disparity d(x, y) using regularization. Advancements in computing techniques for disparity calculation have been proven useful in many fields of quantitative research, including remote-sensing and geophysics.35 We aim to reuse the disparity calculation technique to estimate the displacement field between the experimental slice and the resolution-convolved model slice. In this first demonstrative work for application of image disparity calculation in the data analysis of neutron scattering measurements, we simplify the problem by limiting the slice to only one visible dispersion; therefore, the displacement field can be simplified to be independent of E, and the minimization problem becomes\n\n$arg minΔE(q)∑q,E|Iexp(q,E)−Imodel(q,E−ΔE(q))|.$\n(3)\n\nThe simplification of limiting the displacement field to be independent of E makes this optimization problem straightforward to program by using a SciPy36 optimizer, for example.\n\nIn this work, we illustrate the super-resolution dispersion technique using a synthetic dataset that resembles a real experimental dataset, for which the full analysis will be reported elsewhere.37 In the experiment, a Mn3Si2Te6 single-crystal sample was measured at the SEQUOIA instrument3 with incident energy Ei = 60 meV in the high flux mode, and the sample was rotated at least 180° about the vertical axis, 1° per step, to cover a large volume in the reciprocal space. Slices along multiple high-symmetry directions show clear dispersions. The dispersions along those high-symmetry directions were obtained by finding centers of peaks in constant q cuts. They were fit to a Hamiltonian without on-site anisotropy, which allows for anisotropic interactions for each of the exchange interactions to account for the spin–orbit coupling,37\n\n$H=J1∑⟨i,j⟩[SixSjx+SiySjy+Δ1SizSjz]+J2∑⟨i,j⟩[SixSjx+SiySjy+Δ2SizSjz]+J3∑⟨i,j⟩[SixSjx+SiySjy+Δ3SizSjz].$\n(4)\n\nHere, {Ji} are exchange coefficients, as shown in Fig. 3, while {Δi} introduce anisotropy.\n\nFIG. 3.\n\nThe magnetic structure of Mn3Si2Te6 and the exchange couplings between magnetic sites. The red and blue arrows follow the easy-plane directions of the spins in the ordered phase. The exchange J1 is between the second nearest neighbor and shown as the blue lines between Mn sites. The exchange J2 is between the first nearest neighbor and shown as the yellow lines within the honeycomb layers of the Mn sites. The exchange J3 is between the third nearest neighbor and shown as the green dashed lines, which are only drawn for one portion of the lattice for clarity.\n\nFIG. 3.\n\nThe magnetic structure of Mn3Si2Te6 and the exchange couplings between magnetic sites. The red and blue arrows follow the easy-plane directions of the spins in the ordered phase. The exchange J1 is between the second nearest neighbor and shown as the blue lines between Mn sites. The exchange J2 is between the first nearest neighbor and shown as the yellow lines within the honeycomb layers of the Mn sites. The exchange J3 is between the third nearest neighbor and shown as the green dashed lines, which are only drawn for one portion of the lattice for clarity.\n\nClose modal\n\nFirst, we build a synthetic dataset for which we know the exact model and parameters for the dispersion surface. The synthetic dataset is obtained from a virtual neutron experiment performed by using the MCViNE software. MCViNE contains a scattering kernel that scatters neutrons according to a dispersion surface that a user can define by using arbitrary analytical functions.22,23,28 Therefore, an analytical dispersion function similar to the dispersion surface of the spinwave model defined by Eq. (4) was employed. The spin coupling constants used in the spinwave model in Eq. (4) are J1 = 1.663 meV, J2 = 0.477 meV, J3 = 0.835 meV, Δ1 = 0.390, Δ2 = −0.554, andΔ3 = −0.431. Analytical functions were used to approximate the dispersion and scattering intensity in the vicinity of the (002) wave-vector, and they were parameterized as\n\n$E(h,k,l)=Eb+Ea(1+0.61sin2πh)(1+0.61sin2πk)×1+sin1.6πl2−1,$\n(5)\n$S(h,k,l)=S011+hΓh211+kΓk211+l−2Γl2,$\n(6)\n\nwhere Ea = 11.6 meV, Eb = 9.05 meV, S0 = 14.8, Γh = Γl = 0.38 r.l.u., and Γk = 0.35 r.l.u.\n\nThen, we performed a virtual experiment of a sample with this analytical dispersion function and scattering intensity using MCViNE. The simulation and data reduction consist of the following steps:\n\n• Beam simulation. We can reuse the beam simulation performed earlier for the resolution calculation as it contains all the information regarding the instrument.\n\n• Sample scattering simulation. The sample has the same geometrical shape as the real sample, with the dispersion defined in Eq. (6). Simulations with a series of ω rotation angles are performed, matching the real experiment.\n\n• Detector simulation. The scattered neutrons are intercepted by the virtual SEQUOIA detector system, and the neutron events detected are saved in NeXus files, one for each ω angle.\n\n• Reduction. The Mantid software8 is then used to reduce the NeXus data files in the same way as the real experiment, and corresponding slices are made.\n\nOnce we have the virtual experimental data, we perform the super-resolution workflow outlined in Fig. 1 upon it. The results are shown in Fig. 4. The corrected dispersion shows clear improvement in agreement with the model dispersion curve, compared to the original dispersion obtained directly from the experimental slice. The root-mean-square (rms) difference between the experimental dispersion data and the model dispersion data was reduced from 1.59 to 0.32 meV, showing a nearly fivefold improvement. It is worth noting that the nominal resolution of the SEQUOIA instrument at Ei = 60 meV for the high-flux mode is ∼2 meV. The fact that one single iteration of our super-resolution workflow can improve the dispersion data accuracy by five fold means our method is an efficient way to optimize the dispersion model while taking into account the instrument resolution effect.\n\nAnother way to check the improvement of the dispersion data is to observe the improvement of the fitting parameters in Eq. (6). Table I presents the model parameters for the exact model, the fitted model without resolution correction, and the fitted model with resolution correction. Without the correction, we obtained Ea = 12.93 meV and Eb = 9.53 meV while fitting the dispersion data to Eq. (6). Compared to the exact values of Ea = 11.6 meV and Eb = 9.05 meV, the fitted values are off by 1.33 and 0.48 meV, respectively. After correction, the fitting results are Ea = 11.80 and Eb = 8.78, and the errors are reduced to 0.20 and 0.27 meV.\n\nTABLE I.\n\nDispersion model parameters.\n\nModelEa (meV)Eb (meV)\nExact 11.6 9.05\nWithout resolution correction: fit to dispersion data directly obtained from virtual experiment 12.93 9.53\nWith resolution correction: fit to corrected dispersion data 11.80 8.78\nModelEa (meV)Eb (meV)\nExact 11.6 9.05\nWithout resolution correction: fit to dispersion data directly obtained from virtual experiment 12.93 9.53\nWith resolution correction: fit to corrected dispersion data 11.80 8.78\n\nAn intuitive illustration of the improvement of the quality of the dispersion model is also presented in Fig. 5. Here, Fig. 5(a) shows the 00L slice obtained from the virtual experiment data. Figure 5(b) shows the resolution-convolved slice obtained from the dispersion model fitted to the original dispersion data from the virtual experiment without correction. It is clear that the dispersion is shifted upward in comparison to the virtual experimental data in Fig. 5(a). Figure 5(c) shows the resolution-convolved slice obtained from the dispersion model fitted to the corrected dispersion data, and this slice agrees much better with the experimental slice in Fig. 5(a). This agreement is also a validation of our resolution convolution procedure. The better agreement of the corrected dispersion model with the virtual experimental data is also evident in the residual plots shown in Figs. 5(d) and 5(e).\n\nFIG. 5.\n\n00L slices: (a) The 00L slice from the virtual experimental data obtained through the same data reduction procedure as the real experimental data using Mantid. (b) The analytical model using parameters fitted to the dispersion data obtained directly from the virtual experimental data, convolved by instrument resolution using the procedure explained in Sec. II B. (c) The analytical model using parameters fitted to the corrected dispersion data, convolved by instrument resolution. (d) The residual of subtracting the virtual experimental slice by the original dispersion model convolved with instrument resolution. The root mean square (rms) of the residual is 10.8. (e) The residual of subtracting the virtual experimental slice by the corrected dispersion model convolved with instrument resolution. The rms of the residual is reduced to 6.4.\n\nFIG. 5.\n\n00L slices: (a) The 00L slice from the virtual experimental data obtained through the same data reduction procedure as the real experimental data using Mantid. (b) The analytical model using parameters fitted to the dispersion data obtained directly from the virtual experimental data, convolved by instrument resolution using the procedure explained in Sec. II B. (c) The analytical model using parameters fitted to the corrected dispersion data, convolved by instrument resolution. (d) The residual of subtracting the virtual experimental slice by the original dispersion model convolved with instrument resolution. The root mean square (rms) of the residual is 10.8. (e) The residual of subtracting the virtual experimental slice by the corrected dispersion model convolved with instrument resolution. The rms of the residual is reduced to 6.4.\n\nClose modal\n\nWe also applied the super-resolution dispersion workflow on the experimental SEQUOIA dataset described in Sec. III A. In Sec. III B, for the virtual experimental data, we apply the super-resolution procedure to one single slice. In comparison, for the real experimental data, we treat multiple slices along different Q directions. The first step of the workflow remains the same; the dispersion data for each slice is first obtained without considering the resolution effect. Then, these dispersion data from multiple slices are fitted to the spin-wave model simultaneously to obtain the original set of the model parameters. The fitted model provides dispersion curves and scattering amplitudes for every slice, and they are convolved with resolution functions to obtained modeled slices. Each modeled slice is then compared to the corresponding experimental slice to obtain disparity curves, which are used to correct the dispersions. The corrected dispersions of all interested slices are then fit to the spin-wave model simultaneously again, yielding a new set of model parameters. One example of the dispersion correction is presented in Fig. 6 for the 00L dispersion. Energy corrections in the order of 1 meV are found near 002. The full dispersion correction data are reported elsewhere.37\n\nA new technique to obtain super-resolution dispersions along high-symmetry Q directions for single crystal measurements employing direct geometry neutron spectrometers is developed. This is done by computing the disparity curve between the resolution-convolved-model slice and the experimental slice and then applying the disparity to correct the dispersion. Here, the resolution-convolved slice was obtained by convolving the resolution with the scattering intensity of a dispersion model that was fit to the experimental dispersions obtained without any consideration of instrument resolution. The disparity of the slices was obtained by minimizing the difference between the experimental slice and the modeled slice warped by disparity, subjecting to the total variation regularization. The technique clearly shows improvements in the determination of dispersions, and it is computationally faster since classical methods would have required multiple iterations of model evaluation and resolution convolution with many more sets of model parameters. We show that this method can achieve fivefold super-resolution with respect to nominal resolution of the SEQUOIA3 instrument. The demonstration is facilitated by a MCViNE-based virtual experiment, which provides the virtual experimental data and the known target model to check the effectiveness of the super-resolution technique.\n\nThis super-resolution dispersion technique is limited by the signal-to-noise ratio as other imaging techniques. The 2D resolution convolution method used in this work can be updated to use 4D resolution convolution to improve the accuracy and the universality of this approach. More sophisticated disparity computation techniques can be adapted to remove the limit of single dispersion per slice. Finally, many image processing techniques may find various applications in neutron data analysis. Another potential application of the disparity calculation technique is to detect super-resolution variations in dispersions with respect to temperature/pressure under which the sample is measured.\n\nThis work focuses on the correction of the dispersion relation E(q) between the excitation energy E and its momentum q. It can be envisioned that the super-resolution techniques developed in the previous work for powder DGS data14 can be extended and combined with techniques developed in this work to reconstruct super-resolution I(q, E) slices, reducing the influence of the instrument broadening and providing information on the intrinsic linewidths of the excitations, ΓE(q).\n\nThe authors thank Hillary Smith, Brent Fultz, Garrett Granroth, and Doug Abernathy for fruitful discussions. This work was partially supported by the Department of Energy, Laboratory Directed Research and Development SEED funding, under Contract No. DE-AC05-00OR22725. Work at the Spallation Neutron Source at Oak Ridge National Laboratory (ORNL) was supported by the Scientific User Facilities Division, Office of Basic Energy Sciences, U.S. Department of Energy (DOE). This research also used resources of the Spallation Neutron Source Second Target Station Project at ORNL. ORNL is managed by UT-Battelle LLC for DOE’s Office of Science, the single largest supporter of basic research in the physical sciences in the United States.\n\nAll authors declare that they have no conflicts of interest.\n\nThe data that support the findings of this study are available from the corresponding author upon reasonable request.\n\n1.\nR. I.\nBewley\n,\nR. S.\nEccleston\n,\nK. A.\nMcEwen\n,\nS. M.\nHayden\n,\nM. T.\nDove\n,\nS. M.\nBennington\n,\nJ. R.\n, and\nR. L. 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https://jp.maplesoft.com/support/help/Maple/view.aspx?path=StudyGuides%2FMultivariateCalculus%2FChapter8%2FExamples%2FSection8-3%2FExample8-3-6 | [
"",
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"Example 8-3-6 - Maple Help",
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"Chapter 8: Applications of Triple Integration\n\n\n\nSection 8.3: First Moments",
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"Example 8.3.6\n\n\n\n a) Obtain the centroid of $R$, the region that lies between the plane $z=0$ and the paraboloid $z=9-{x}^{2}-{y}^{2}$.\n b) Impose the density $\\mathrm{δ}\\left(r,\\mathrm{θ},z\\right)=z$ on $R$ and find the resulting center of mass.\n\n(See Example 8.1.20.)"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.57882947,"math_prob":0.99973375,"size":4690,"snap":"2023-40-2023-50","text_gpt3_token_len":2090,"char_repetition_ratio":0.16709347,"word_repetition_ratio":0.14107142,"special_character_ratio":0.2925373,"punctuation_ratio":0.18302387,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987086,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T11:19:48Z\",\"WARC-Record-ID\":\"<urn:uuid:316ed185-1025-4008-b3c0-d63f51d41e0d>\",\"Content-Length\":\"387869\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f5e0fcc-ba15-413f-95a9-db06623c45fc>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7bdd3b9-c098-430c-904b-e64b292d7e9b>\",\"WARC-IP-Address\":\"199.71.183.28\",\"WARC-Target-URI\":\"https://jp.maplesoft.com/support/help/Maple/view.aspx?path=StudyGuides%2FMultivariateCalculus%2FChapter8%2FExamples%2FSection8-3%2FExample8-3-6\",\"WARC-Payload-Digest\":\"sha1:6OX4G7Q6SBN4PJFSC7MHLE6B3MQYZL5J\",\"WARC-Block-Digest\":\"sha1:LOEUQ35KCFO5B7IJBVR7ZR44KUHNASQD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510387.77_warc_CC-MAIN-20230928095004-20230928125004-00285.warc.gz\"}"} |
https://visualfractions.com/calculator/prime-factors/prime-factors-of-585/ | [
"Prime Factors of 585\n\nLooking to get a list of the prime factors of 585? In this article we'll give you all of the information you need, including the definition of the prime factors of 585, how to calculate the prime factors of 585 (also known as the prime factorization of 585). As a bonus, we'll also list out the prime factor tree of 585 the product of prime factors of 585, and tell you how many prime factors 585 has.\n\nPrime Factors of 585 Definition\n\nEvery number can be represented as a product of prime numbers. So when we talk aqbout prime factorization of 585, we're talking about the building blocks of the number. A prime factor is a positive integer that can only be divided by 1 and itself. The prime factors of 585 are all of the prime numbers in it that when multipled together will equal 585.\n\nLet's look at how to find all of the prime factors of 585 and list them out.\n\nHow to Find the Prime Factors of 585\n\nYou'll often see the process of finding prime factors of 585 referred to as prime factorization. To get the prime factors of 585 we need to divide 585 by the smallest prime number possible. You then repeat the same process by taking the result and dividing that number by the smallest prime number. Eventually, you end up with the number 1.\n\nThis process creates something called a prime factor tree of 585. The prime numbers used in this tree are the prime factors of 585. Let's look at the prime factor tree for 585:\n\n• 585 ÷ 3 = 195\n• 195 ÷ 3 = 65\n• 65 ÷ 5 = 13\n• 13 ÷ 13 = 1\n\nPut simply, all of the prime numbers that you used to divide above are the prime factors of 585 as well. So what we are left with is the answer to your search, the prime factors of 585:\n\n3, 3, 5, and 13\n\nHow Many Prime Factors of 585 Are There?\n\nIf we count up all of the prime factors of 585 used in the prime factor tree above, we can see that 585 has a total of 4 prime factors.\n\nProduct of Prime Factors of 585\n\nThe prime factors shown above (3, 3, 5, and 13) are completely unique to 585. When we multiply all of them together the result will be 585 and this is what we call the product of prime factors of 585. The prime factor products of 585 are listed below:\n\n3 x 3 x 5 x 13 = 585\n\nSo there you have it. A complete guide to the factors of 585. You should now have the knowledge and skills to go out and calculate your own factors and factor pairs for any number you like.\n\nFeel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers!\n\nIf you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!\n\n• \"Prime Factors of 585\". VisualFractions.com. Accessed on January 22, 2022. http://visualfractions.com/calculator/prime-factors/prime-factors-of-585/.\n\n• \"Prime Factors of 585\". VisualFractions.com, http://visualfractions.com/calculator/prime-factors/prime-factors-of-585/. Accessed 22 January, 2022."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91471225,"math_prob":0.9702947,"size":3654,"snap":"2022-05-2022-21","text_gpt3_token_len":898,"char_repetition_ratio":0.2517808,"word_repetition_ratio":0.032863848,"special_character_ratio":0.26491517,"punctuation_ratio":0.10280374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976686,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-22T06:07:14Z\",\"WARC-Record-ID\":\"<urn:uuid:24f8306e-33f7-48e6-adc0-b21d3912b520>\",\"Content-Length\":\"23488\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:36512d10-d4fc-433a-93fa-4a40be77c9a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:e424bf54-8cff-49d8-a0f2-4a5b44cafde3>\",\"WARC-IP-Address\":\"104.21.34.163\",\"WARC-Target-URI\":\"https://visualfractions.com/calculator/prime-factors/prime-factors-of-585/\",\"WARC-Payload-Digest\":\"sha1:MRRNOBCNGXDC2BIVWA5XVM2WNQEJPOSN\",\"WARC-Block-Digest\":\"sha1:YQZ5QVBK2YVTTWE2YGNK7Z727ARJETYO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320303747.41_warc_CC-MAIN-20220122043216-20220122073216-00113.warc.gz\"}"} |
https://webapps.stackexchange.com/questions/97761/how-to-use-indirect-to-create-a-dynamic-length-range | [
"# Part 1\n\nTo create a dynamic length range I think I should be using `INDIRECT()` in some way but just can't quite get my head around how.\n\nHow can I use the function `counta()` as the end of each column's chosen range?\n\n# Part 2\n\nI'm using Google Sheets but I suspect the answer may be the same for Excel as well.\n\nI have two columns full of string data that are dynamically pulled from a website, therefore I can't know for certain how long the columns will be. I wish to combine them into a single column, so have been using\n\n``````={ARRAYFORMULA(S2:S100);arrayformula(AI2:AI100)}\n``````\n\nIt works but of course I end up with a whole bunch of blanks between the two merged columns, since I am combining each column up to row 100 when there is only data for maybe 50 or so rows each.\n\nI would like to specify the ending of each range, to be the count of that range. Something like this (which is of course totally improper syntax):\n\n``````={ARRAYFORMULA(S2:Scounta(S:S));arrayformula(AI2:AIcounta(ai:ai)}\n``````\n\nThat's why I need the solution for Part 1.\n\n• I thought I had found a solution in another question, but it turns out that it doesn't work because while it does merge the columns, it skips blanks which are critical in the result. Does anyone have any idea how to apply an arrayformula to a dynamic range?\n– JVC\nAug 27 '16 at 17:55\n• – Rubén\nAug 27 '16 at 18:23\n• Possible duplicate of Concatenate several columns into one in Google Sheets\n– Rubén\nAug 27 '16 at 18:26\n• Not a duplicate because I need to include blanks in the result. That solution clearly focuses on non-blank cells.\n– JVC\nAug 27 '16 at 18:27\n• Thought I had it after modifying the solution in that link, but unfortunately it's still a no-go. I need to be able to define the length of the arrayformula range dynamically.\n– JVC\nAug 27 '16 at 19:02\n\n# Part 1\n\nHow can I use the function counta() as the end of each column's chosen range?\n\nUse CONCATENATE, CONCAT or `&` operator.\n\nAs you already figured out, to create a dynamic reference use INDIRECT. Example:\n\n``````=INDIRECT(\"S2:S\"&COUNTA(S2:S))\n``````\n\n# Part 2\n\nThe final formula is\n\n``````={INDIRECT(\"S2:S\"&COUNTA(S2:S));INDIRECT(\"AI2:AI\"&COUNTA(AI2:AI))}\n``````\n\nAn alternative to get the same result is to use FILTER. The advantage is that it will work for strings and numbers. Example:\n\n``````={FILTER(S2:S,LEN(S2:S);FILTER(AI2:AI,LEN(AI2:AI)}\n``````\n\nAnother alternative is to use QUERY. The advantage of it is that the import formulas could be combined into an array an use it of the data argument of QUERY:\n\n``````=QUERY({import1;import2},\"select Col1 where Col1<>'' and Col1<>'Header'\")\n``````\n\n## References\n\n• Ah I was so close with indirect! This finally got it... thanks!\n– JVC\nAug 28 '16 at 17:07"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8911842,"math_prob":0.6310701,"size":1013,"snap":"2021-31-2021-39","text_gpt3_token_len":250,"char_repetition_ratio":0.09613479,"word_repetition_ratio":0.0,"special_character_ratio":0.24185587,"punctuation_ratio":0.093457945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95944184,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T12:59:27Z\",\"WARC-Record-ID\":\"<urn:uuid:8072c74b-e1eb-4007-9163-2f0bdae6a9ae>\",\"Content-Length\":\"175640\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0db05f5e-7535-444d-99d6-3c1e6be96930>\",\"WARC-Concurrent-To\":\"<urn:uuid:53796cdc-026e-4c30-9561-910b3e9b2453>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://webapps.stackexchange.com/questions/97761/how-to-use-indirect-to-create-a-dynamic-length-range\",\"WARC-Payload-Digest\":\"sha1:PMHJNPLGAROHKTBFALEYZYSMXYGSZOJ7\",\"WARC-Block-Digest\":\"sha1:ZAQA27WAERJDOHVUYUSOHUPDYEFVUIWA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055645.75_warc_CC-MAIN-20210917120628-20210917150628-00329.warc.gz\"}"} |
https://www.tutorialspoint.com/data_mining/dm_classification_prediction.htm | [
"# Data Mining - Classification & Prediction\n\nThere are two forms of data analysis that can be used for extracting models describing important classes or to predict future data trends. These two forms are as follows −\n\n• Classification\n• Prediction\n\nClassification models predict categorical class labels; and prediction models predict continuous valued functions. For example, we can build a classification model to categorize bank loan applications as either safe or risky, or a prediction model to predict the expenditures in dollars of potential customers on computer equipment given their income and occupation.\n\n## What is classification?\n\nFollowing are the examples of cases where the data analysis task is Classification −\n\n• A bank loan officer wants to analyze the data in order to know which customer (loan applicant) are risky or which are safe.\n\n• A marketing manager at a company needs to analyze a customer with a given profile, who will buy a new computer.\n\nIn both of the above examples, a model or classifier is constructed to predict the categorical labels. These labels are risky or safe for loan application data and yes or no for marketing data.\n\n## What is prediction?\n\nFollowing are the examples of cases where the data analysis task is Prediction −\n\nSuppose the marketing manager needs to predict how much a given customer will spend during a sale at his company. In this example we are bothered to predict a numeric value. Therefore the data analysis task is an example of numeric prediction. In this case, a model or a predictor will be constructed that predicts a continuous-valued-function or ordered value.\n\nNote − Regression analysis is a statistical methodology that is most often used for numeric prediction.\n\n## How Does Classification Works?\n\nWith the help of the bank loan application that we have discussed above, let us understand the working of classification. The Data Classification process includes two steps −\n\n• Building the Classifier or Model\n• Using Classifier for Classification\n\n### Building the Classifier or Model\n\n• This step is the learning step or the learning phase.\n\n• In this step the classification algorithms build the classifier.\n\n• The classifier is built from the training set made up of database tuples and their associated class labels.\n\n• Each tuple that constitutes the training set is referred to as a category or class. These tuples can also be referred to as sample, object or data points.",
null,
"### Using Classifier for Classification\n\nIn this step, the classifier is used for classification. Here the test data is used to estimate the accuracy of classification rules. The classification rules can be applied to the new data tuples if the accuracy is considered acceptable.",
null,
"## Classification and Prediction Issues\n\nThe major issue is preparing the data for Classification and Prediction. Preparing the data involves the following activities −\n\n• Data Cleaning − Data cleaning involves removing the noise and treatment of missing values. The noise is removed by applying smoothing techniques and the problem of missing values is solved by replacing a missing value with most commonly occurring value for that attribute.\n\n• Relevance Analysis − Database may also have the irrelevant attributes. Correlation analysis is used to know whether any two given attributes are related.\n\n• Data Transformation and reduction − The data can be transformed by any of the following methods.\n\n• Normalization − The data is transformed using normalization. Normalization involves scaling all values for given attribute in order to make them fall within a small specified range. Normalization is used when in the learning step, the neural networks or the methods involving measurements are used.\n\n• Generalization − The data can also be transformed by generalizing it to the higher concept. For this purpose we can use the concept hierarchies.\n\nNote − Data can also be reduced by some other methods such as wavelet transformation, binning, histogram analysis, and clustering.\n\n## Comparison of Classification and Prediction Methods\n\nHere is the criteria for comparing the methods of Classification and Prediction −\n\n• Accuracy − Accuracy of classifier refers to the ability of classifier. It predict the class label correctly and the accuracy of the predictor refers to how well a given predictor can guess the value of predicted attribute for a new data.\n\n• Speed − This refers to the computational cost in generating and using the classifier or predictor.\n\n• Robustness − It refers to the ability of classifier or predictor to make correct predictions from given noisy data.\n\n• Scalability − Scalability refers to the ability to construct the classifier or predictor efficiently; given large amount of data.\n\n• Interpretability − It refers to what extent the classifier or predictor understands."
] | [
null,
"https://www.tutorialspoint.com/data_mining/images/dm_build_classifier.jpg",
null,
"https://www.tutorialspoint.com/data_mining/images/dm_using_classifier.jpg",
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https://api.dart.cn/stable/2.14.4/dart-html/Dimension-class.html | [
"# Dimension class Null safety\n\nClass representing a length measurement in CSS.\n\n## Constructors\n\nDimension.cm(num _value)\nSet this CSS Dimension to a centimeter `value`.\nDimension.css(String cssValue)\nConstruct a Dimension object from the valid, simple CSS string `cssValue` that represents a distance measurement. [...]\nDimension.em(num _value)\nSet this CSS Dimension to the specified number of ems. [...]\nDimension.ex(num _value)\nSet this CSS Dimension to the specified number of x-heights. [...]\nDimension.inch(num _value)\nSet this CSS Dimension to an inch `value`.\nDimension.mm(num _value)\nSet this CSS Dimension to a millimeter `value`.\nDimension.pc(num _value)\nSet this CSS Dimension to a pica `value`.\nDimension.percent(num _value)\nSet this CSS Dimension to a percentage `value`.\nDimension.pt(num _value)\nSet this CSS Dimension to a point `value`.\nDimension.px(num _value)\nSet this CSS Dimension to a pixel `value`.\n\n## Properties\n\nhashCode int\nThe hash code for this object. [...]\nruntimeType Type\nA representation of the runtime type of the object.\nvalue num\nReturn a unitless, numerical value of this CSS value.\n\n## Methods\n\nnoSuchMethod(Invocation invocation) → dynamic\nInvoked when a non-existent method or property is accessed. [...]\ninherited\ntoString()\nPrint out the CSS String representation of this value.\noverride\n\n## Operators\n\noperator ==(Object other) bool\nThe equality operator. [...]\ninherited"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5217913,"math_prob":0.93825793,"size":1259,"snap":"2022-40-2023-06","text_gpt3_token_len":297,"char_repetition_ratio":0.25179282,"word_repetition_ratio":0.21354167,"special_character_ratio":0.24861,"punctuation_ratio":0.18018018,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852955,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-05T21:08:07Z\",\"WARC-Record-ID\":\"<urn:uuid:3dae3cf0-1986-4ce6-ad10-36989339f237>\",\"Content-Length\":\"64179\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b368f4f8-f42c-4d37-a1fd-ec45fd58bbf5>\",\"WARC-Concurrent-To\":\"<urn:uuid:0be0c049-e28f-42b4-bd67-65df2899c183>\",\"WARC-IP-Address\":\"47.246.20.177\",\"WARC-Target-URI\":\"https://api.dart.cn/stable/2.14.4/dart-html/Dimension-class.html\",\"WARC-Payload-Digest\":\"sha1:3ZKI4BAW4HY7VWAJYIIZPAW5RZEGH4TT\",\"WARC-Block-Digest\":\"sha1:SVUMR2BIA6M3TZZLMZM6QIVNG5GCYEIR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337668.62_warc_CC-MAIN-20221005203530-20221005233530-00621.warc.gz\"}"} |
https://rdrr.io/cran/tensr/f/vignettes/maximum_likelihood.Rmd | [
"# Abstract\n\nIn this vignette, I demonstrate how to calculate the MLE and run a likelihood ratio test in the mean-zero array normal model.\n\n# Maximum Likelihood Estimation\n\nFirst, I simulate some data. `X` will be generated with identity covariance along all modes. `Y` will have identity covariance along the first three modes, and an AR-1(0.9) covariance along the fourth mode. `Z` will have diagonal covariance along the first two modes, and an AR-1(0.9) covariance along the third mode.\n\n```library(tensr)\np <- c(10, 10, 10, 10)\nX <- array(rnorm(prod(p)),dim = p)\n\ncov_Y <- start_ident(p)\ncov_Y[] <- 0.9^abs(outer(1:p,1:p,\"-\"))\ncov_Y[] <- cov_Y[] / det(cov_Y[])^(1/p) #scale to have unit determinant.\nY <- atrans(array(rnorm(prod(p)),dim = p), lapply(cov_Y, mhalf))\n\ncov_Z <- start_ident(p)\ndiag(cov_Z[]) <- 1:p / prod(1:p)^(1/p)\ndiag(cov_Z[]) <- p:1 / prod(1:p)^(1/p)\ncov_Z[] <- 0.9^abs(outer(1:p,1:p,\"-\"))\ncov_Z[] <- cov_Z[] / det(cov_Z[])^(1/p)\nZ <- atrans(array(rnorm(prod(p)),dim = p), lapply(cov_Z, mhalf))\n```\n\nThe `holq` is used to calculate the maximum likelihood estimates. Assuming we know the covariance structure of each mode, we set the the appropriate modes to have identity or diagonal covariance matrices by using the `mode_rep` or `mode_diag` options.\n\n```holq_X <- holq(X,mode_rep = 1:4, print_diff = FALSE)\nholq_Y <- holq(Y,mode_rep = 1:3, print_diff = FALSE)\nholq_Z <- holq(Z,mode_rep = 4, mode_diag = c(1,2), print_diff = FALSE)\n\nmle_X <- mle_from_holq(holq_X)\nmle_Y <- mle_from_holq(holq_Y)\nmle_Z <- mle_from_holq(holq_Z)\n```\n\nEstimates of the scale are pretty close to the true value of 1.\n\n```cat(\"Estimates of scale:\\n\",\n\"From X:\", mle_X\\$sig_mle,\"\\n\",\n\"From Y:\", mle_Y\\$sig_mle,\"\\n\",\n\"From Z:\", mle_Z\\$sig_mle,\"\\n\"\n)\n```\n\nAnd the estimates of the covariances are pretty close to their truth. For example, when we look at the true and estimated variances of the first mode covariance in `Z`, we get\n\n```cat(\" True:\", paste(\"(\", paste(format(diag(cov_Z[]), digits = 2),\ncollapse = \", \"), \")\",sep = \"\"), \"\\n\",\n\"Estimate:\", paste(\"(\", paste(format(diag(mle_Z[][]), digits = 2),\ncollapse = \", \"), \")\", sep = \"\"), \"\\n\")\n```\n\n# Likelihood Ratio Testing\n\nLet's test for all modes having the identity covariance matrix versus the first three modes having the identity covariance matrix and the fourth is unconstrained. Of course, in this situation we need not resort to tensor methods. Using `X` as our data should only reject the null 5\\% of the time. But for `Y` where the alternative is true, we should have more power.\n\nFirst, we'll calculate the null distribution of the likelihood ratio test statistic. Then we'll calculate these test statistics using both `X` and `Y` as our data.\n\n```null_distribution <- lrt_null_dist_dim_same(p = p, null_ident = 1:4,\nalt_ident = 1:3, itermax = 500)\n\nsig_k <- holq(X,mode_rep = 1:3, print_diff = FALSE)\\$sig\nsig_h <- holq(X,mode_rep = 1:4, print_diff = FALSE)\\$sig\nlrt_stat_val_X <- lrt_stat(sig_null = sig_h,sig_alt = sig_k,p = p)\n\nsig_k <- holq(Y,mode_rep = 1:3, print_diff = FALSE)\\$sig\nsig_h <- holq(Y,mode_rep = 1:4, print_diff = FALSE)\\$sig\nlrt_stat_val_Y <- lrt_stat(sig_null = sig_h,sig_alt = sig_k,p = p)\n```\n\nWe can calculate p-values.\n\n```p_value_x <- mean(null_distribution > lrt_stat_val_X)\np_value_y <- mean(null_distribution > lrt_stat_val_Y)\n\ncat(\" p-value using X:\", p_value_x,\"\\n\",\n\"p-value using Y:\", p_value_y,\"\\n\")\n```\n\nWe can also plot the null distribution along with the observed statistics. If you can't see the line for `Y`, that's because the likelihood ratio test statistic is so large.\n\n```par(mar = c(2.5, 2.5, 2, 0), mgp = c(1.5, .5, 0))\nhist(null_distribution, xlab = \"LRT Stat\", main = \"Null Distribution\")\nabline(v = lrt_stat_val_X, col = 2, lwd = 2, lty = 2)\nabline(v = lrt_stat_val_Y, col = 4, lwd = 2, lty = 2)\nlegend(\"topright\", c(\"From X\",\"From Y\"), col = c(2,4), lty = 2, lwd = 2, cex = 0.6)\n```\n\nAs a second example, we run a likelihood ratio test using `Z`, testing for diagional covariance along the first mode, identity covariance along the second and fourth modes, and unstructured along the third, against the correct model of diagonal along the first two modes, identity along the fourth mode, and unstructured along the third.\n\n```null_distribution <- lrt_null_dist_dim_same(p = p,null_ident = c(2,4),\nalt_ident = 4, null_diag = 1,\nalt_diag = c(1,2), itermax = 500)\n\nsig_k <- holq(Z, mode_rep = 4, mode_diag = c(1,2), print_diff = FALSE)\\$sig\nsig_h <- holq(Z, mode_rep = c(2,4), mode_diag = 1, print_diff = FALSE)\\$sig\nlrt_stat_val_Z <- lrt_stat(sig_null = sig_h, sig_alt = sig_k, p = p)\n\np_value_z <- mean(null_distribution > lrt_stat_val_Z)\n```\n```cat(\"P-value:\", p_value_z,\"\\n\")\n```\n\n# References\n\nGerard, D., & Hoff, P. (2016). A higher-order LQ decomposition for separable covariance models. Linear Algebra and its Applications, 505, 57-84. [Link to LAA] [Link to arXiv]\n\n## Try the tensr package in your browser\n\nAny scripts or data that you put into this service are public.\n\ntensr documentation built on May 2, 2019, 2:32 p.m."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6964511,"math_prob":0.99955297,"size":4931,"snap":"2023-14-2023-23","text_gpt3_token_len":1532,"char_repetition_ratio":0.12116907,"word_repetition_ratio":0.08038147,"special_character_ratio":0.34293246,"punctuation_ratio":0.19438878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99970436,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-24T05:40:12Z\",\"WARC-Record-ID\":\"<urn:uuid:db9e2376-380d-4692-a684-c08be7fb5867>\",\"Content-Length\":\"50642\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1463f19a-b0f8-4582-8094-398cfe540f2b>\",\"WARC-Concurrent-To\":\"<urn:uuid:fdc36b85-5004-4d11-8b9f-fdf4b0d4dd5c>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/tensr/f/vignettes/maximum_likelihood.Rmd\",\"WARC-Payload-Digest\":\"sha1:GFVLUIQ2DRRLQU6GJYRUBBSLUZMCE2BM\",\"WARC-Block-Digest\":\"sha1:7LNXI4WFT64SGBG25OGAXHLLV4NCCBVR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945248.28_warc_CC-MAIN-20230324051147-20230324081147-00530.warc.gz\"}"} |
https://codedocs.xyz/LukasGL/FlukyEngine/group__core__func__integer.html | [
"FlukyEngine\nInteger functions\n\nProvides GLSL functions on integer types. More...\n\nCollaboration diagram for Integer functions:",
null,
"## Functions\n\ntemplate<length_t L, qualifier Q>\nGLM_FUNC_DECL vec< L, uint, Q > glm::uaddCarry (vec< L, uint, Q > const &x, vec< L, uint, Q > const &y, vec< L, uint, Q > &carry)\nAdds 32-bit unsigned integer x and y, returning the sum modulo pow(2, 32). More...\n\ntemplate<length_t L, qualifier Q>\nGLM_FUNC_DECL vec< L, uint, Q > glm::usubBorrow (vec< L, uint, Q > const &x, vec< L, uint, Q > const &y, vec< L, uint, Q > &borrow)\nSubtracts the 32-bit unsigned integer y from x, returning the difference if non-negative, or pow(2, 32) plus the difference otherwise. More...\n\ntemplate<length_t L, qualifier Q>\nGLM_FUNC_DECL void glm::umulExtended (vec< L, uint, Q > const &x, vec< L, uint, Q > const &y, vec< L, uint, Q > &msb, vec< L, uint, Q > &lsb)\nMultiplies 32-bit integers x and y, producing a 64-bit result. More...\n\ntemplate<length_t L, qualifier Q>\nGLM_FUNC_DECL void glm::imulExtended (vec< L, int, Q > const &x, vec< L, int, Q > const &y, vec< L, int, Q > &msb, vec< L, int, Q > &lsb)\nMultiplies 32-bit integers x and y, producing a 64-bit result. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, T, Q > glm::bitfieldExtract (vec< L, T, Q > const &Value, int Offset, int Bits)\nExtracts bits [offset, offset + bits - 1] from value, returning them in the least significant bits of the result. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, T, Q > glm::bitfieldInsert (vec< L, T, Q > const &Base, vec< L, T, Q > const &Insert, int Offset, int Bits)\nReturns the insertion the bits least-significant bits of insert into base. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, T, Q > glm::bitfieldReverse (vec< L, T, Q > const &v)\nReturns the reversal of the bits of value. More...\n\ntemplate<typename genType >\nGLM_FUNC_DECL int glm::bitCount (genType v)\nReturns the number of bits set to 1 in the binary representation of value. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, int, Q > glm::bitCount (vec< L, T, Q > const &v)\nReturns the number of bits set to 1 in the binary representation of value. More...\n\ntemplate<typename genIUType >\nGLM_FUNC_DECL int glm::findLSB (genIUType x)\nReturns the bit number of the least significant bit set to 1 in the binary representation of value. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, int, Q > glm::findLSB (vec< L, T, Q > const &v)\nReturns the bit number of the least significant bit set to 1 in the binary representation of value. More...\n\ntemplate<typename genIUType >\nGLM_FUNC_DECL int glm::findMSB (genIUType x)\nReturns the bit number of the most significant bit in the binary representation of value. More...\n\ntemplate<length_t L, typename T , qualifier Q>\nGLM_FUNC_DECL vec< L, int, Q > glm::findMSB (vec< L, T, Q > const &v)\nReturns the bit number of the most significant bit in the binary representation of value. More...\n\n## Detailed Description\n\nProvides GLSL functions on integer types.\n\nThese all operate component-wise. The description is per component. The notation [a, b] means the set of bits from bit-number a through bit-number b, inclusive. The lowest-order bit is bit 0.\n\nInclude <glm/integer.hpp> to use these core features.\n\n## ◆ bitCount() [1/2]\n\ntemplate<typename genType >\n GLM_FUNC_DECL int glm::bitCount ( genType v )\n\nReturns the number of bits set to 1 in the binary representation of value.\n\nTemplate Parameters\n genType Signed or unsigned integer scalar or vector types.\nGLSL bitCount man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ bitCount() [2/2]\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::bitCount ( vec< L, T, Q > const & v )\n\nReturns the number of bits set to 1 in the binary representation of value.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar or vector types.\nGLSL bitCount man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ bitfieldExtract()\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::bitfieldExtract ( vec< L, T, Q > const & Value, int Offset, int Bits )\n\nExtracts bits [offset, offset + bits - 1] from value, returning them in the least significant bits of the result.\n\nFor unsigned data types, the most significant bits of the result will be set to zero. For signed data types, the most significant bits will be set to the value of bit offset + base - 1.\n\nIf bits is zero, the result will be zero. The result will be undefined if offset or bits is negative, or if the sum of offset and bits is greater than the number of bits used to store the operand.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar types.\nGLSL bitfieldExtract man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ bitfieldInsert()\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::bitfieldInsert ( vec< L, T, Q > const & Base, vec< L, T, Q > const & Insert, int Offset, int Bits )\n\nReturns the insertion the bits least-significant bits of insert into base.\n\nThe result will have bits [offset, offset + bits - 1] taken from bits [0, bits - 1] of insert, and all other bits taken directly from the corresponding bits of base. If bits is zero, the result will simply be base. The result will be undefined if offset or bits is negative, or if the sum of offset and bits is greater than the number of bits used to store the operand.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar or vector types.\nGLSL bitfieldInsert man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ bitfieldReverse()\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::bitfieldReverse ( vec< L, T, Q > const & v )\n\nReturns the reversal of the bits of value.\n\nThe bit numbered n of the result will be taken from bit (bits - 1) - n of value, where bits is the total number of bits used to represent value.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar or vector types.\nGLSL bitfieldReverse man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ findLSB() [1/2]\n\ntemplate<typename genIUType >\n GLM_FUNC_DECL int glm::findLSB ( genIUType x )\n\nReturns the bit number of the least significant bit set to 1 in the binary representation of value.\n\nIf value is zero, -1 will be returned.\n\nTemplate Parameters\n genIUType Signed or unsigned integer scalar types.\nGLSL findLSB man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ findLSB() [2/2]\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::findLSB ( vec< L, T, Q > const & v )\n\nReturns the bit number of the least significant bit set to 1 in the binary representation of value.\n\nIf value is zero, -1 will be returned.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar types.\nGLSL findLSB man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ findMSB() [1/2]\n\ntemplate<typename genIUType >\n GLM_FUNC_DECL int glm::findMSB ( genIUType x )\n\nReturns the bit number of the most significant bit in the binary representation of value.\n\nFor positive integers, the result will be the bit number of the most significant bit set to 1. For negative integers, the result will be the bit number of the most significant bit set to 0. For a value of zero or negative one, -1 will be returned.\n\nTemplate Parameters\n genIUType Signed or unsigned integer scalar types.\nGLSL findMSB man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ findMSB() [2/2]\n\ntemplate<length_t L, typename T , qualifier Q>\n GLM_FUNC_DECL vec glm::findMSB ( vec< L, T, Q > const & v )\n\nReturns the bit number of the most significant bit in the binary representation of value.\n\nFor positive integers, the result will be the bit number of the most significant bit set to 1. For negative integers, the result will be the bit number of the most significant bit set to 0. For a value of zero or negative one, -1 will be returned.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector. T Signed or unsigned integer scalar types.\nGLSL findMSB man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ imulExtended()\n\ntemplate<length_t L, qualifier Q>\n GLM_FUNC_DECL void glm::imulExtended ( vec< L, int, Q > const & x, vec< L, int, Q > const & y, vec< L, int, Q > & msb, vec< L, int, Q > & lsb )\n\nMultiplies 32-bit integers x and y, producing a 64-bit result.\n\nThe 32 least-significant bits are returned in lsb. The 32 most-significant bits are returned in msb.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector.\nGLSL imulExtended man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\ntemplate<length_t L, qualifier Q>\n GLM_FUNC_DECL vec glm::uaddCarry ( vec< L, uint, Q > const & x, vec< L, uint, Q > const & y, vec< L, uint, Q > & carry )\n\nAdds 32-bit unsigned integer x and y, returning the sum modulo pow(2, 32).\n\nThe value carry is set to 0 if the sum was less than pow(2, 32), or to 1 otherwise.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector.\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ umulExtended()\n\ntemplate<length_t L, qualifier Q>\n GLM_FUNC_DECL void glm::umulExtended ( vec< L, uint, Q > const & x, vec< L, uint, Q > const & y, vec< L, uint, Q > & msb, vec< L, uint, Q > & lsb )\n\nMultiplies 32-bit integers x and y, producing a 64-bit result.\n\nThe 32 least-significant bits are returned in lsb. The 32 most-significant bits are returned in msb.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector.\nGLSL umulExtended man page\nGLSL 4.20.8 specification, section 8.8 Integer Functions\n\n## ◆ usubBorrow()\n\ntemplate<length_t L, qualifier Q>\n GLM_FUNC_DECL vec glm::usubBorrow ( vec< L, uint, Q > const & x, vec< L, uint, Q > const & y, vec< L, uint, Q > & borrow )\n\nSubtracts the 32-bit unsigned integer y from x, returning the difference if non-negative, or pow(2, 32) plus the difference otherwise.\n\nThe value borrow is set to 0 if x >= y, or to 1 otherwise.\n\nTemplate Parameters\n L An integer between 1 and 4 included that qualify the dimension of the vector."
] | [
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"https://codedocs.xyz/LukasGL/FlukyEngine/group__core__func__integer.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6780096,"math_prob":0.97481585,"size":8839,"snap":"2022-27-2022-33","text_gpt3_token_len":2572,"char_repetition_ratio":0.16321449,"word_repetition_ratio":0.6907348,"special_character_ratio":0.29245389,"punctuation_ratio":0.19578947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9861932,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T11:41:11Z\",\"WARC-Record-ID\":\"<urn:uuid:a882491b-5a05-4000-977e-d68e1e2f12eb>\",\"Content-Length\":\"42399\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e2dbcab2-5279-4e66-838d-c050b92935d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:eb44e143-8892-407b-bac8-d0ac3e5a317b>\",\"WARC-IP-Address\":\"104.200.30.150\",\"WARC-Target-URI\":\"https://codedocs.xyz/LukasGL/FlukyEngine/group__core__func__integer.html\",\"WARC-Payload-Digest\":\"sha1:NWI736FBXXG3IBUBCRRJ7FNCYOYLLF3U\",\"WARC-Block-Digest\":\"sha1:S5RZDCD7HZ7S6C6YQK65KBJ2T7ZJH7QQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104690785.95_warc_CC-MAIN-20220707093848-20220707123848-00491.warc.gz\"}"} |
http://bookshadow.com/weblog/2018/01/21/leetcode-max-chunks-to-make-sorted-ver-1/ | [
"## 题目描述：\n\nLeetCode 769. Max Chunks To Make Sorted (ver. 1)\n\nGiven an array `arr` that is a permutation of `[0, 1, ..., arr.length - 1]`, we split the array into some number of \"chunks\" (partitions), and individually sort each chunk. After concatenating them, the result equals the sorted array.\n\nWhat is the most number of chunks we could have made?\n\nExample 1:\n\n```Input: arr = [4,3,2,1,0]\nOutput: 1\nExplanation:\nSplitting into two or more chunks will not return the required result.\nFor example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.\n```\n\nExample 2:\n\n```Input: arr = [1,0,2,3,4]\nOutput: 4\nExplanation:\nWe can split into two chunks, such as [1, 0], [2, 3, 4].\nHowever, splitting into [1, 0], , , is the highest number of chunks possible.\n```\n\nNote:\n\n• `arr` will have length in range `[1, 10]`.\n• `arr[i]` will be a permutation of `[0, 1, ..., arr.length - 1]`.\n\n## 解题思路：\n\n```记数组arr长度为N，下标x从N - 1到0逐一递减：\n\n若min(arr[x .. N - 1]) > max(arr[0 .. x - 1])，则令结果+1```\n\n## Python代码：\n\n``````class Solution(object):\ndef maxChunksToSorted(self, arr):\n\"\"\"\n:type arr: List[int]\n:rtype: int\n\"\"\"\nN = len(arr)\nans = 1\nfor x in range(N - 1, 0, -1):\nif min(arr[x:]) > max(arr[:x]):\nans += 1\nreturn ans\n``````\n\nPingbacks已关闭。"
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https://www.physicsforums.com/threads/calculus-textbook-for-home-education-student.635283/ | [
"# Calculus Textbook for Home Education Student\n\n• Linuxkid\nIn summary, for a prospective engineering/physics major, it is recommended to have some prior knowledge in calculus and proofs before tackling the challenging book \"Calculus 4th edition\" by M. Spivak. It is also suggested to have a good understanding of linear algebra and its applications in multiple dimensions, as it is crucial in understanding the concepts presented in Spivak's book.\n\n#### Linuxkid\n\nHello all,\n\nI'm considering Calculus 4th edition from M. Spivak for my Senior year Calculus book along with George Simmons Calculus with Analytic Geometry for a joint 18.01 MIT OCW and self-study curriculum.\nI understand that his book involves quite some rigor and ruthless problems, but I'm up for it.\n\nMy algebra training is up to Hyperbolic functions, Binomial series, basic matrices, and some derivatives (and the rest of Precalculus, like sets and sequences, polynomial etc).\n\nSo, other than Calculus 4th edition; what supplemental -or not- Calculus textbook do you suggest for a prospective engineering/physics major?\n\nThanks for the help.\n\n-Nikos\n\nHey LinuxKid and welcome to the forums.\n\nDo you have books or resources in mind for Linear Algebra?\n\nThe whole calculus thing in multiple dimensions is based on linear algebra not only to do calculations, but also to understand what is going on.\n\nBasically you can think of a matrix as a general linear object like Y or X is as a normal 1D variable. The object itself though behaves a bit differently than if you want to multiply say x*y since you are now dealing in an arbitrary space and not just a single scalar number.\n\nLinearity is pretty much the most fundamental object in mathematics. All of differential and integral calculus depends on it (since derivatives are linear operators which means understanding linear operators helps you understand differentiation across any number of variables) and it's basically the foundation for looking at arbitrary vector spaces and bases for such vector spaces.\n\nYou can think of vectors as just being arrows: what the arrows correspond to is not so much important, in the same way that what a normal 1D variable x corresponds to isn't really important.\n\nNow the ideas in Spivak use linear algebra since Spivak looks at generalized forms of mathematics and these forums require generalized structures and the generalized linear structure is a matrix.\n\nA matrix provides the most basic way to have a general linear transformation not only on high dimensional spaces, but also in-between spaces as well (So instead of going from 3D to 3D you can go from 3D to 5D or 5D to 2D and so on). It does this in a linear way, but the structure is what is important.\n\nSo before you jump into Spivak, my recommendation is to understand linear algebra and what linearity means in the context of having a single object (a matrix) and what that object means in the context of algebraic identities involving matrices (like A*B, A+B up to more complicated algebraic expressions involving norms and other attributes).\n\nYou'll need to understand this when you look at how Spivak introduces the idea of a derivative in n-dimensions.\n\nThe real key is to get intuition for these abstract linear objects actually do and what they represent and it will take a little getting used to, but once you do it, you'll see all these multi-variable generalizations and they will make more sense.\n\nLinuxkid said:\nHello all,\n\nI'm considering Calculus 4th edition from M. Spivak for my Senior year Calculus book along with George Simmons Calculus with Analytic Geometry for a joint 18.01 MIT OCW and self-study curriculum.\nI understand that his book involves quite some rigor and ruthless problems, but I'm up for it.\n\nMy algebra training is up to Hyperbolic functions, Binomial series, basic matrices, and some derivatives (and the rest of Precalculus, like sets and sequences, polynomial etc).\n\nSo, other than Calculus 4th edition; what supplemental -or not- Calculus textbook do you suggest for a prospective engineering/physics major?\n\nThanks for the help.\n\n-Nikos\n\nYou should know that Spivak is a very hard book and is usually not meant for a first encounter in calculus. Before tackling Spivak, you should know a bit of calculus already and you should be pretty well-versed in proofs.\nIf you want to be challenged mathematically, then Spivak is the ideal book however. It has very difficult exercises.\n\nchiro said:\nHey LinuxKid and welcome to the forums.\n\nDo you have books or resources in mind for Linear Algebra?\n\nThe whole calculus thing in multiple dimensions is based on linear algebra not only to do calculations, but also to understand what is going on.\n\nBasically you can think of a matrix as a general linear object like Y or X is as a normal 1D variable. The object itself though behaves a bit differently than if you want to multiply say x*y since you are now dealing in an arbitrary space and not just a single scalar number.\n\nLinearity is pretty much the most fundamental object in mathematics. All of differential and integral calculus depends on it (since derivatives are linear operators which means understanding linear operators helps you understand differentiation across any number of variables) and it's basically the foundation for looking at arbitrary vector spaces and bases for such vector spaces.\n\nYou can think of vectors as just being arrows: what the arrows correspond to is not so much important, in the same way that what a normal 1D variable x corresponds to isn't really important.\n\nNow the ideas in Spivak use linear algebra since Spivak looks at generalized forms of mathematics and these forums require generalized structures and the generalized linear structure is a matrix.\n\nA matrix provides the most basic way to have a general linear transformation not only on high dimensional spaces, but also in-between spaces as well (So instead of going from 3D to 3D you can go from 3D to 5D or 5D to 2D and so on). It does this in a linear way, but the structure is what is important.\n\nSo before you jump into Spivak, my recommendation is to understand linear algebra and what linearity means in the context of having a single object (a matrix) and what that object means in the context of algebraic identities involving matrices (like A*B, A+B up to more complicated algebraic expressions involving norms and other attributes).\n\nYou'll need to understand this when you look at how Spivak introduces the idea of a derivative in n-dimensions.\n\nThe real key is to get intuition for these abstract linear objects actually do and what they represent and it will take a little getting used to, but once you do it, you'll see all these multi-variable generalizations and they will make more sense.\n\nSpivak's calculus doesn't require any knowledge of linear algebra. So you can study Spivak and linear algebra in any order you want.\n\nThe point was that linear algebra is a good thing to know to understand calculus in many dimensions. Did you read my comment? I elaborated on this and some of the reasons in detail.\n\nAlso when you see the modern definition of a derivative as opposed to other definitions, then you'll see just how important understanding linear algebra can be.\n\nchiro said:\nThe point was that linear algebra is a good thing to know to understand calculus in many dimensions. Did you read my comment? I elaborated on this and some of the reasons in detail.\n\nSpivak doesn't do calculus in many dimensions, so no, I don't see the point.\n\nMaybe he's referring to Calculus On Manifolds? I haven't read that book but from what I've heard linear algebra is necessary for understanding it.\n\nPKDfan said:\nMaybe he's referring to Calculus On Manifolds? I haven't read that book but from what I've heard linear algebra is necessary for understanding it.\n\nAaah, yes, that might be the case. I forgot Spivak had calculus on manifolds as well. And indeed, you absolutely need to know linear algebra to understand that.\n\nYeah that's the one I meant: should have clarified!\n\nThanks for the replies guys,\n\nI practiced some matrices and concepts in Linear Algebra, but only briefly as part of precalculus. But as clarified above, shouldn't be a problem with Calculus 4th edition.\n\n## 1. What is calculus?\n\nCalculus is a branch of mathematics that deals with the study of change and motion. It involves the use of mathematical concepts such as derivatives and integrals to analyze and solve problems related to rates of change and accumulation.\n\n## 2. Who can benefit from using a calculus textbook for home education?\n\nA calculus textbook for home education can benefit anyone who is interested in learning calculus, regardless of their age or background. It is particularly helpful for high school and college students who are pursuing degrees in mathematics, science, engineering, or economics.\n\n## 3. What topics are typically covered in a calculus textbook for home education?\n\nA calculus textbook for home education will cover a variety of topics, including limits, derivatives, integrals, applications of calculus, and techniques for solving problems involving calculus. It may also include topics such as differential equations, multivariable calculus, and vector calculus.\n\n## 4. How can a calculus textbook for home education be used effectively?\n\nA calculus textbook for home education can be used effectively by following a structured study plan, practicing regularly, and seeking help if needed. It is also important to understand the underlying concepts and not just memorize formulas and procedures.\n\n## 5. Are there any online resources that can supplement a calculus textbook for home education?\n\nYes, there are many online resources that can supplement a calculus textbook for home education, such as video lectures, interactive tutorials, practice problems, and online forums for asking questions and getting help. It is important to choose reputable and reliable sources for these resources."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.935553,"math_prob":0.9144424,"size":1301,"snap":"2023-40-2023-50","text_gpt3_token_len":307,"char_repetition_ratio":0.11719352,"word_repetition_ratio":0.9359606,"special_character_ratio":0.21137586,"punctuation_ratio":0.12916666,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98886895,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T06:53:58Z\",\"WARC-Record-ID\":\"<urn:uuid:03dac35e-7884-45fa-a16f-6b21ceb59a29>\",\"Content-Length\":\"95898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64bf9ed6-b6a4-43e0-8c09-6cafbc285839>\",\"WARC-Concurrent-To\":\"<urn:uuid:77c34b97-4d1d-498e-ac48-5836b403c6d7>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/calculus-textbook-for-home-education-student.635283/\",\"WARC-Payload-Digest\":\"sha1:S62PV6MIEQSZYRFYT6XKLNH5PWK6W2Z3\",\"WARC-Block-Digest\":\"sha1:DWHKC6NLLWYEFDQFWELAUWE4XJIHO7PU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100276.12_warc_CC-MAIN-20231201053039-20231201083039-00788.warc.gz\"}"} |
https://caresclub.com/what-is-an-odd-function/ | [
"# What Is An Odd Function?\n\nAre you not knowing what is an odd function? No need to increase the tension. Here we are for helping you with the problems of Odd Function. You will now be able to solve math problems easily. We will define an odd function, explain examples of the odd functions, and much more related to the odd function. Your queries about the odd functions will be answered below. So let’s continue reading further what is an odd function.\n\n## What Is An Odd Function?\n\nAn odd function is called odd, if for all x values of the function f(-x) is equal to -f(x). In other words, since the graph is symmetrical for the origin, the odd functions do not have similar images.\n\n## What Is An Odd Function in Math?\n\nIn the mathematics A function is said to be odd if and only if f(-x) = -f(x). And graphically a function that is symmetrical to the origin of the graph is called an odd function.\n\n## Definition Of An Odd Function\n\nGraphically the function is odd can be defined as:\n\nThe odd function can be defined as, a function f is odd if, to the origin, the graph of f is symmetrical.\n\nAlgebraically the function is odd can be defined as:\n\nFunction f is, algebraically odd, if and only if f(-x) = -f(x) in the domain of f for all possible x values.\n\n## What Is An Odd Function Formula?\n\nBy using the odd function calculator formula as follow you can calculate the function.\n\nA function is odd if the result you get is the opposite of the function,\n\nf(-x) = -f(x) and it is true for all x from the domain of definition.\n\n## Example Of An Odd Function\n\nBelow is an odd function example that we have described for you so that will help you:\n\nf (x) = 4×3 – 3x\n\nIf we substitute -x with x,\n\nThen further if we simplify it we get,\n\nf (–x) = 4(–x)2 – 3(–x)\n\n= 4(–x2) + 3x\n\n= –4×2 + 3x\n\nAnd if changing -x to x changes the function’s value. Then f(-x) = -f(x) the function is odd.\n\n## Different Odd Functions\n\nHere is meaning of different odd functions\n\n• ### What Is An Odd Function Symmetric To?\n\nThe odd function is symmetric to the origin of the graph of the odd function. When the graph of function f is identical to the origin, then the function f is odd. There needs to be symmetry to the origin, not the x-axis, for anything to be an odd function. This implies that if it has a point like (a, b), it has a point as well (-a, -b).\n\nThe graph is symmetrical to the origin while the function is odd. If the equation is even, the graph is symmetrical around the y-axis. Below, we have described an example, from it you will get help for how to tell if a function is even or odd from a graph. The f(x) function is odd because the f(-x) =-f(x) function is for every x number. The graph of an odd function has a common rotational symmetry. So if you rotate the graph at 180o at the origin, the graph will remain the same after the rotation.\n\n• ### What Is A Odd Function Equation?\n\nThe equation of the odd function is when the function −f(x) = f(−x) for all x, then it is a odd function. You will notice in front of f(x) the minus sign that is −f (x).\n\n• ### What Is An Odd Function In Calculus?\n\nThe odd function in calculus is explained as a function f is odd if the graph of f is symmetric to the origin. The f can be expressed algebraically as odd if and only if f(-x) = -f(x) for all x values in the domain of f.\n\n• ### What Is An Odd Degree Function?\n\nAn odd degree polynomial function is always going in one direction at one end, and in the opposite direction at the other. And an odd degree of the polynomial will always have at least one real root. The lead coefficient guides the path of the line. An odd degree polynomial functions, such as y = x3, have graphs that stretch diagonally through quadrants. An even degree polynomial function, such as y = x2, has graphs that open up or down.\n\n• ### What Is An Odd Monomial Function?\n\nThe Odd Monomial Function is an odd expression. Where the axp is a numeric constant and p is an odd integer. As a result of p is an odd integer, a(-x)p = -axp results in the expression as an odd monomial function.\n\n• ### What Is An Even Odd Function?\n\nA function is even odd if the F(x) = 0, defined for all real numbers. It is the only function that is both odd and even. There is just one line on the x-axis that overlaps on it, If you count equations that are not a constant in terms of y, so x=0 would be both odd and even, and it is just a line on the y-axis.\n\n## Even And Odd Function\n\nLets understand some importance of even and odd function\n\n• ### How Function Is Both Even And Odd?\n\nF(x) = 0, defined for all actual numbers, is the only function that is both even and odd. It is just a line resting on the X-axis and is just a line on the y-axis. If you count equations that are not a constant in terms of y, then, x=0 would be both even and odd. By this, it is how can a function be both even and odd for your better understanding.\n\n• ### Is A Linear Function Even Or Odd?\n\nA linear function is said to be odd if a linear graph goes through the origin. And a linear function is even if the graph does not go through the origin.\n\n## FAQ\n\n### What Does An Odd Function Mean?\n\nAn odd function is meant to be odd if the function f(-x) is equal to-f for all x values (x). In other words, also because the graph is symmetrical to the origin, the odd functions do not have identical images.\n\n### What Is An Odd Function Example?\n\nHere is an example on an odd function for your understanding:\n\nf (x) = 4×3 – 3x\n\nIf we substitute -x with x,\n\nThen further if we simplify it we get,\n\nf (–x) = 4(–x)2 – 3(–x)\n\n= 4(–x2) + 3x\n\n= –4×2 + 3x\n\nAnd if changing -x to x changes the function’s value. Then f(-x) = -f(x) the function is odd.\n\n### How Do You Know If A Function Is Even Or Odd?\n\nYou will know whether a function is odd or even, by determining it with a numerical expression. You take the value -x and replace it with the x, and then further simplify the equation. If you get a different product or answer at the end what you started with. That is, if f(-x) = -f(x) which changed, then the function is called odd.\n\n### What Is An Odd Function Graph?\n\nThe f(x) function is odd when f(-x) = -f(x), it is for every value of x. The graph with symmetric odd function has a specific rotational symmetry. So if you rotate the graph at 180o at the origin, the graph will appear the same even after the rotation.\n\n### How Do You Find An Even Function?\n\nIf you end up with the exact same function that you started with (that is, if f (−x) = f (x), so all of the signs are the same), then the function is even; if you end up with the exact opposite of what you started with (that is, if f (−x) = −f (x), so all of the signs are switched), then the function is odd.\n\n### What Is Odd Even Rule With Example?\n\nVehicles with registration numbers ending in odd numbers will be allowed on the roads on odd days and even-numbered vehicles will be allowed on the roads on even days. For example, vehicle registration numbers ending with 0,2,4,6 or 8 are allowed on days such as the 14th, 16th or 18th of a month.\n\n### Is Tangent An Odd Function?\n\nCosine and secant are even; sine, tangent, cosecant, and cotangent are odd.\n\n### What’s The Difference Between Odd And Even?\n\nAn even number is a number that can be divided into two equal groups. An odd number is a number that cannot be divided into two equal groups. Even numbers end in 2, 4, 6, 8 and 0 regardless of how many digits they have (we know the number 5,917,624 is even because it ends in a 4!).\n\n### What Defines An Odd Function?\n\nDefinition. A function f is odd if the following equation holds for all x and −x in the domain of f : −f(x)=f(−x) − f ( x ) = f ( − x ) Geometrically, the graph of an odd function has rotational symmetry with respect to the origin, meaning that its graph remains unchanged after a rotation of 180∘ about the origin.\n\n### How Do You Prove If A Function Is Even Or Odd?\n\nA function f is even if f(−x)=f(x), for all x in the domain of f. A function f is odd if f(−x)=−f(x), for all x in the domain of f.\n\n### What Four Basic Functions Are Odd?\n\nOdd Functions: The identity function, the cubing function, the reciprocal function, the sine function.\n\n### Which Of The 12 Functions Are Odd?\n\nOdd Functions: The identity function, the cubing function, the reciprocal function, the sine function. Neither: The square root function, the exponential function and the log function. The logistic function is also neither because it is rotationally symmetric about the point ( 0 , 1 2 ) as opposed to the origin"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8869546,"math_prob":0.99703103,"size":8878,"snap":"2023-40-2023-50","text_gpt3_token_len":2219,"char_repetition_ratio":0.2063331,"word_repetition_ratio":0.1778036,"special_character_ratio":0.25095743,"punctuation_ratio":0.102524474,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99987173,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-24T10:19:28Z\",\"WARC-Record-ID\":\"<urn:uuid:02a88c52-5332-4096-bd18-1006ff8a182b>\",\"Content-Length\":\"93572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce2f4457-5538-418b-b18a-fc5b7e2f43a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:7546fb03-c816-44d7-a620-315a54ac26d8>\",\"WARC-IP-Address\":\"104.21.29.155\",\"WARC-Target-URI\":\"https://caresclub.com/what-is-an-odd-function/\",\"WARC-Payload-Digest\":\"sha1:BTTIIM7NF4Q47PCDFRK7H56VYFP4C5SJ\",\"WARC-Block-Digest\":\"sha1:OKXK6MMH72C7OEH53ZCM3VUTBE4D3LX6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506632.31_warc_CC-MAIN-20230924091344-20230924121344-00263.warc.gz\"}"} |
https://www.rdocumentation.org/packages/Matrix/versions/1.2-14/topics/lu | [
"Matrix (version 1.2-14)\n\n# lu: (Generalized) Triangular Decomposition of a Matrix\n\n## Description\n\nComputes (generalized) triangular decompositions of square (sparse or dense) and non-square dense matrices.\n\n## Usage\n\n```lu(x, …)\n# S4 method for matrix\nlu(x, warnSing = TRUE, …)\n# S4 method for dgeMatrix\nlu(x, warnSing = TRUE, …)\n# S4 method for dgCMatrix\nlu(x, errSing = TRUE, order = TRUE, tol = 1,\nkeep.dimnames = TRUE, …)```\n\n## Arguments\n\nx\n\na dense or sparse matrix, in the latter case of square dimension. No missing values or IEEE special values are allowed.\n\nwarnSing\n\n(when `x` is a `\"'>denseMatrix\"`) logical specifying if a `warning` should be signalled when `x` is singular.\n\nerrSing\n\n(when `x` is a `\"'>sparseMatrix\"`) logical specifying if an error (see `stop`) should be signalled when `x` is singular. When `x` is singular, `lu(x, errSing=FALSE)` returns `NA` instead of an LU decomposition. No warning is signalled and the useR should be careful in that case.\n\norder\n\nlogical or integer, used to choose which fill-reducing permutation technique will be used internally. Do not change unless you know what you are doing.\n\ntol\n\npositive number indicating the pivoting tolerance used in `cs_lu`. Do only change with much care.\n\nkeep.dimnames\n\nlogical indicating that `dimnames` should be propagated to the result, i.e., “kept”. This was hardcoded to `FALSE` in upto Matrix version 1.2-0. Setting to `FALSE` may gain some performance.\n\nfurther arguments passed to or from other methods.\n\n## Value\n\nAn object of class `\"LU\"`, i.e., `\"'>denseLU\"` (see its separate help page), or `\"sparseLU\"`, see `'>sparseLU`; this is a representation of a triangular decomposition of `x`.\n\n## Details\n\n`lu()` is a generic function with special methods for different types of matrices. Use `showMethods(\"lu\")` to list all the methods for the `lu` generic.\n\nThe method for class `'>dgeMatrix` (and all dense matrices) is based on LAPACK's `\"dgetrf\"` subroutine. It returns a decomposition also for singular and non-square matrices.\n\nThe method for class `'>dgCMatrix` (and all sparse matrices) is based on functions from the CSparse library. It signals an error (or returns `NA`, when `errSing = FALSE`, see above) when the decomposition algorithm fails, as when `x` is (too close to) singular.\n\n## References\n\nGolub, G., and Van Loan, C. F. (1989). Matrix Computations, 2nd edition, Johns Hopkins, Baltimore.\n\nTimothy A. Davis (2006) Direct Methods for Sparse Linear Systems, SIAM Series “Fundamentals of Algorithms”.\n\nClass definitions `'>denseLU` and `'>sparseLU` and function `expand`; `qr`, `chol`.\n\n## Examples\n\n```# NOT RUN {\n##--- Dense -------------------------\nx <- Matrix(rnorm(9), 3, 3)\nlu(x)\ndim(x2 <- round(10 * x[,-3]))# non-square\nexpand(lu2 <- lu(x2))\n\n##--- Sparse (see more in ?\"sparseLU-class\")----- % ./sparseLU-class.Rd"
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https://www.unix.com/tags/problems.html?s=e5e6914a23a7e9ea9ca63fc220a0a8d1 | [
"```ARITHMETIC(6)\t\t\t\t\t\t\t BSD Games Manual\t\t\t\t\t\t ARITHMETIC(6)\n\nNAME\narithmetic -- quiz on simple arithmetic\n\nSYNOPSIS\narithmetic [-o +-x/] [-r range]\n\nDESCRIPTION\narithmetic asks you to solve problems in simple arithmetic. Each question must be answered correctly before going on to the next. After\nevery 20 problems, it prints the score so far and the time taken.\tYou can quit at any time by typing the interrupt or end-of-file character.\n\nThe options are as follows:\n\n-o By default, arithmetic asks questions on addition of numbers from 0 to 10, and corresponding subtraction.\tBy supplying one or more\nof the characters +-x/, you can ask for problems in addition, subtraction, multiplication, and division, respectively. If you give\none of these characters more than once, that kind of problem will be asked correspondingly more often.\n\n-r If a range is supplied, arithmetic selects the numbers in its problems in the following way. For addition and multiplication, the\nnumbers to be added or multiplied are between 0 and range, inclusive. For subtraction and division, both the required result and the\nnumber to divide by or subtract will be between 0 and range. (Of course, arithmetic will not ask you to divide by 0.) The default\n\nrange is 10.\n\nWhen you get a problem wrong, arithmetic will remember the numbers involved, and will tend to select those numbers more often than others, in\nproblems of the same sort. Eventually it will forgive and forget.\n\narithmetic cannot be persuaded to tell you the right answer. You must work it out for yourself.\n\nDIAGNOSTICS\n``What?'' if you get a question wrong. ``Right!'' if you get it right. ``Please type a number.'' if arithmetic doesn't understand what you\ntyped."
] | [
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https://visp-doc.inria.fr/doxygen/visp-daily/tutorial-homography-deprecated.html | [
"",
null,
"Visual Servoing Platform version 3.2.1 under development (2019-12-06)\nTutorial: Homography estimation from points (deprecated)\n\n# Introduction\n\nNote\nThe second part of this tutorial is deprecated if your OpenCV version is equal to 2.8.0 or more recent. If so you should rather follow Tutorial: Homography estimation from points.\n\nThis tutorial shows how to estimate an homography from points. Two cases are considered in the next sections:\n\n# Homography estimation\n\nLet us consider the following source code also available in tutorial-homography-from-points.cpp. To resume, we do the following:\n\n• define four 3D points `oP` in an object frame,\n• define an homogeneous transformation `aMo` from frame `a` to to the object frame `o`,\n• define a similar homogeneous transformation `bMo` from frame `b` to to the object frame `o`,\n• compute the coordinates of the four 3D points in the image plane `a` and `b`. These are the matched coordinates (xa,ya) and (xb,yb) that are then used to estimate the homography using either vpHomography::DLT() or vpHomography::HLM().\n#include <visp3/vision/vpHomography.h>\n#include <visp3/core/vpMeterPixelConversion.h>\nint main()\n{\ndouble L = 0.1;\nstd::vector<vpPoint> oP;\noP.push_back(vpPoint(-L, -L, 0));\noP.push_back(vpPoint(2 * L, -L, 0));\noP.push_back(vpPoint(L, 3 * L, 0));\noP.push_back(vpPoint(-L, 4 * L, 0));\nvpHomogeneousMatrix bMo(0.1, 0, 1, 0, vpMath::rad(15), 0);\nvpHomogeneousMatrix aMo = aMb * bMo;\nstd::vector<vpPoint> aP(4), bP(4);\nstd::vector<double> xa(4), ya(4), xb(4), yb(4);\nfor (unsigned int i = 0; i < 4; i++) {\noP[i].project(aMo);\nxa[i] = oP[i].get_x();\nya[i] = oP[i].get_y();\noP[i].project(bMo);\nxb[i] = oP[i].get_x();\nyb[i] = oP[i].get_y();\n}\nvpHomography::DLT(xb, yb, xa, ya, aHb, true);\nstd::cout << \"Estimated homography using DLT:\\n\" << aHb / aHb << std::endl;\nvpHomography::HLM(xb, yb, xa, ya, true, aHb);\nstd::cout << \"Estimated homography using HLM:\\n\" << aHb / aHb << std::endl;\nvpColVector n;\naHb.computeDisplacement(aRb, atb, n);\nstd::cout << \"\\nEstimated displacement:\" << std::endl;\nstd::cout << \" atb: \" << atb.t() << std::endl;\natub.buildFrom(aRb);\nstd::cout << \" athetaub: \";\nfor (unsigned int i = 0; i < 3; i++)\nstd::cout << vpMath::deg(atub[i]) << \" \";\nstd::cout << std::endl;\nstd::cout << \" n: \" << n.t() << std::endl;\nvpImagePoint iPa, iPb;\nvpMeterPixelConversion::convertPoint(cam, xb, yb, iPb);\nvpMeterPixelConversion::convertPoint(cam, xa, ya, iPa);\nstd::cout << \"Ground truth: Point 3 in pixels in frame b: \" << iPb << std::endl;\nstd::cout << \"Ground truth: Point 3 in pixels in frame a: \" << iPa << std::endl;\n// Project the position in pixel of point 3 from the homography\nstd::cout << \"Estimation from homography: Point 3 in pixels in frame a: \" << vpHomography::project(cam, aHb, iPb)\n<< std::endl;\n}\n\nNow we give a line by line explanation of the code:\n\nFirst we have to include the header of the vpHomography class.\n\n#include <visp3/vision/vpHomography.h>\n\nIn the main() function we first define the 3D coordinates of 4 points that are localized in the plane Z=0:\n\ndouble L = 0.1;\nstd::vector<vpPoint> oP;\noP.push_back(vpPoint(-L, -L, 0));\noP.push_back(vpPoint(2 * L, -L, 0));\noP.push_back(vpPoint(L, 3 * L, 0));\noP.push_back(vpPoint(-L, 4 * L, 0));\n\nThen we define the homogeneous transformations between frames `a`, `b` and object frame `o:`\n\nvpHomogeneousMatrix bMo(0.1, 0, 1, 0, vpMath::rad(15), 0);\nvpHomogeneousMatrix aMo = aMb * bMo;\n\nFrom these transformations we compute the coordinates of the points in the image plane `a` and `b`. For each point we have its coordinates (xa,ya) in frame `a` and (xb,yb) in frame `b:`\n\nstd::vector<vpPoint> aP(4), bP(4);\nstd::vector<double> xa(4), ya(4), xb(4), yb(4);\nfor (unsigned int i = 0; i < 4; i++) {\noP[i].project(aMo);\nxa[i] = oP[i].get_x();\nya[i] = oP[i].get_y();\noP[i].project(bMo);\nxb[i] = oP[i].get_x();\nyb[i] = oP[i].get_y();\n}\n\nWe have now matched couples of coordinates of four points that are used to estimate an homography between image plane `a` and `b`. Two methods are available, either using the DLT (Direct Linear Transform) algorithm or the HLM algorithm.\n\nvpHomography::DLT(xb, yb, xa, ya, aHb, true);\nstd::cout << \"Estimated homography using DLT:\\n\" << aHb / aHb << std::endl;\nvpHomography::HLM(xb, yb, xa, ya, true, aHb);\nstd::cout << \"Estimated homography using HLM:\\n\" << aHb / aHb << std::endl;\nNote\nNote that vpHomography::HLM() allows to consider points that are not coplanar.\n\nOnce the homography is estimated, the vpHomography class allows to extract the 3D homogeneous transformation between frames `a` and `b:`\n\nvpColVector n;\naHb.computeDisplacement(aRb, atb, n);\n\nJust for fun we print the values to this transformation using:\n\nstd::cout << \"\\nEstimated displacement:\" << std::endl;\nstd::cout << \" atb: \" << atb.t() << std::endl;\natub.buildFrom(aRb);\nstd::cout << \" athetaub: \";\nfor (unsigned int i = 0; i < 3; i++)\nstd::cout << vpMath::deg(atub[i]) << \" \";\nstd::cout << std::endl;\nstd::cout << \" n: \" << n.t() << std::endl;\n\nThis code lead to the following output:\n\nEstimated displacement:\natb: 0.2016519874 -0.1008259937 0.1008259937\nathetaub: -3 20 5\nn: 0.2588190451 -1.124100812e-14 0.9659258263\n\nwhere we can see that the values for `atb` and `athetaub` are the one specified at the beginning of the source code during `aMb` initialization.\n\nAfter we show how to retrieve the coordinates in pixels of a point (here point ) in the corresponding images using camera parameters:\n\nvpImagePoint iPa, iPb;\nvpMeterPixelConversion::convertPoint(cam, xb, yb, iPb);\nvpMeterPixelConversion::convertPoint(cam, xa, ya, iPa);\nstd::cout << \"Ground truth: Point 3 in pixels in frame b: \" << iPb << std::endl;\nstd::cout << \"Ground truth: Point 3 in pixels in frame a: \" << iPa << std::endl;\n\nAt the end, we show how to project a point with pixel coordinates from image `b` to image `a` using the homography:\n\n// Project the position in pixel of point 3 from the homography\nstd::cout << \"Estimation from homography: Point 3 in pixels in frame a: \" << vpHomography::project(cam, aHb, iPb)\n<< std::endl;\n\nThis last part of the code produce the following output:\n\nGround truth: Point 3 in pixels in frame b: 377.9450564, 193.9928711\nGround truth: Point 3 in pixels in frame a: 353.8501593, 486.1851856\nEstimation from homography: Point 3 in pixels in frame a: 353.8501593, 486.1851856\n\n# Ransac or robust homography estimation\n\nThis section follows the Tutorial: Keypoint matching (deprecated). It explains how to exploit couples of matched points obtained using SURF detector in order to estimate an homography that allows to reject mismatched couples of points. The homography is then used to track a postcard from its initial position in the reference image.\n\nLet us consider the following source code also available in tutorial-matching-surf-homography-deprecated.cpp.\n\n#include <visp3/core/vpPixelMeterConversion.h>\n#include <visp3/gui/vpDisplayOpenCV.h>\n#include <visp3/vision/vpHomography.h>\n#include <visp3/vision/vpKeyPointSurf.h>\nint main(int argc, const char **argv)\n{\n#if defined(VISP_HAVE_OPENCV_NONFREE) && (VISP_HAVE_OPENCV_VERSION < 0x030000)\nint method = 0;\nif (argc > 1)\nmethod = atoi(argv);\nif (method == 0)\nstd::cout << \"Uses Ransac to estimate the homography\" << std::endl;\nelse\nstd::cout << \"Uses a robust scheme to estimate the homography\" << std::endl;\nvpKeyPointSurf surf;\nsurf.buildReference(I);\nIdisp.resize(I.getHeight(), 2 * I.getWidth());\nIdisp.insert(I, vpImagePoint(0, 0));\nIdisp.insert(I, vpImagePoint(0, I.getWidth()));\nvpDisplayOpenCV d(Idisp, 0, 0, \"Homography from matched Surf keypoints\");\nvpImagePoint corner_ref;\ncorner_ref.set_ij(115, 64);\ncorner_ref.set_ij(83, 253);\ncorner_ref.set_ij(282, 307);\ncorner_ref.set_ij(330, 72);\nfor (unsigned int i = 0; i < 4; i++) {\nvpDisplay::displayCross(Idisp, corner_ref[i], 12, vpColor::red);\n}\nvpCameraParameters cam(840, 840, I.getWidth() / 2, I.getHeight() / 2);\nvpHomography curHref;\nIdisp.insert(I, vpImagePoint(0, I.getWidth()));\nunsigned int nbMatch = surf.matchPoint(I);\nstd::vector<vpImagePoint> iPref(nbMatch),\niPcur(nbMatch); // Coordinates in pixels (for display only)\nstd::vector<double> mPref_x(nbMatch), mPref_y(nbMatch);\nstd::vector<double> mPcur_x(nbMatch), mPcur_y(nbMatch);\nstd::vector<bool> inliers(nbMatch);\nfor (unsigned int i = 0; i < nbMatch; i++) {\nvpImagePoint matched_ref, matched_cur;\nsurf.getMatchedPoints(i, matched_ref, matched_cur);\nvpPixelMeterConversion::convertPoint(cam, matched_ref, mPref_x[i], mPref_y[i]);\nvpPixelMeterConversion::convertPoint(cam, matched_cur, mPcur_x[i], mPcur_y[i]);\n// Store the image coordinates in pixel of the matched points\niPref[i] = matched_ref;\niPcur[i] = matched_cur;\n}\ndouble residual;\nif (method == 0)\nvpHomography::ransac(mPref_x, mPref_y, mPcur_x, mPcur_y, curHref, inliers, residual,\n(unsigned int)mPref_x.size() / 2, 2.0 / cam.get_px(), true);\nelse\nvpHomography::robust(mPref_x, mPref_y, mPcur_x, mPcur_y, curHref, inliers, residual, 0.4, 4, true);\nvpImagePoint corner_cur;\nfor (int i = 0; i < 4; i++) {\ncorner_cur[i] = vpHomography::project(cam, curHref, corner_ref[i]);\n}\nvpImagePoint offset(0, I.getWidth());\nfor (int i = 0; i < 4; i++) {\nvpDisplay::displayLine(Idisp, corner_cur[i] + offset, corner_cur[(i + 1) % 4] + offset, vpColor::blue, 3);\n}\nfor (unsigned int i = 0; i < nbMatch; i++) {\nif (inliers[i] == true)\nvpDisplay::displayLine(Idisp, iPref[i], iPcur[i] + offset, vpColor::green);\nelse\nvpDisplay::displayLine(Idisp, iPref[i], iPcur[i] + offset, vpColor::red);\n}\nif (vpDisplay::getClick(Idisp, false))\nbreak;\n}\n#else\n(void)argc;\n(void)argv;\n#endif\nreturn 0;\n}\n\nThe command line allows to use either Ransac algorithm of a robust M-estimator approach:\n\n% ./tutorial-matching-surf-homography-deprecated 0 // to run Ransac\n% ./tutorial-matching-surf-homography-deprecated 1 // to run the robust M-estimator\n\nHere after is the resulting video. The left image represents the reference image. The right images correspond to the successive images of the input video. All the green lines extremities represent the points that are well matched and used in the homography estimation process. All the red lines represent couple of matched points that are rejected by the robust estimator.\n\nNow, let us explain the new lines that were introduced to estimate the homography.\n\nFirst we detect the command line arguments to be able later to user either Ransac or the robust M-estimator:\n\nint method = 0;\nif (argc > 1)\nmethod = atoi(argv);\nif (method == 0)\nstd::cout << \"Uses Ransac to estimate the homography\" << std::endl;\nelse\nstd::cout << \"Uses a robust scheme to estimate the homography\" << std::endl;\n\nWe also initialize the coordinates of the pixels in the reference image that correspond to the postcard corners. These coordinates were obtained after a user initial click. To simplify the code, we set directly the coordinates of the points:\n\nvpImagePoint corner_ref;\ncorner_ref.set_ij(115, 64);\ncorner_ref.set_ij(83, 253);\ncorner_ref.set_ij(282, 307);\ncorner_ref.set_ij(330, 72);\n\nUsing these coordinates, we display red lines around the postcard:\n\nfor (unsigned int i = 0; i < 4; i++) {\nvpDisplay::displayCross(Idisp, corner_ref[i], 12, vpColor::red);\n}\n\nWe need also to define roughly the parameters of our camera:\n\nvpCameraParameters cam(840, 840, I.getWidth() / 2, I.getHeight() / 2);\n\nFor each new image, once points are matched using:\n\nunsigned int nbMatch = surf.matchPoint(I);\n\nWe allocate new containers useful for the homography estimation. The coordinates of the points in the reference image will be stored in (mPref_x, mPref_y), while the corresponding coordinates in the current image will be stored in (mPcur_x, mPcur_y). We also allocate `inliers` a vector of boolean that will indicate if a couple of point is an inlier or an outlier:\n\nstd::vector<double> mPref_x(nbMatch), mPref_y(nbMatch);\nstd::vector<double> mPcur_x(nbMatch), mPcur_y(nbMatch);\nstd::vector<bool> inliers(nbMatch);\n\nTo estimate the homography we need first to convert the points from pixel to meters:\n\nvpPixelMeterConversion::convertPoint(cam, matched_ref, mPref_x[i], mPref_y[i]);\nvpPixelMeterConversion::convertPoint(cam, matched_cur, mPcur_x[i], mPcur_y[i]);\n\nWe can now estimate the homography using either Ransac or the robust M-estimator approach:\n\ndouble residual;\nif (method == 0)\nvpHomography::ransac(mPref_x, mPref_y, mPcur_x, mPcur_y, curHref, inliers, residual,\n(unsigned int)mPref_x.size() / 2, 2.0 / cam.get_px(), true);\nelse\nvpHomography::robust(mPref_x, mPref_y, mPcur_x, mPcur_y, curHref, inliers, residual, 0.4, 4, true);\n\nFor Ransac we consider that at least 50 percent of the points should be inliers (`mPref_x.size()/2`) to reach a consensus and that a couple of point is stated as an inlier if the reprojection error is lower than 2 pixels (`2.0/cam`.get_px()).\n\nThen using the homography, we project the coordinates of the postcard corners in the current image:\n\nvpImagePoint corner_cur;\nfor (int i = 0; i < 4; i++) {\ncorner_cur[i] = vpHomography::project(cam, curHref, corner_ref[i]);\n}\n\nWe use these coordinates to draw blue lines arround the postcard:\n\nvpImagePoint offset(0, I.getWidth());\nfor (int i = 0; i < 4; i++) {\nvpDisplay::displayLine(Idisp, corner_cur[i] + offset, corner_cur[(i + 1) % 4] + offset, vpColor::blue, 3);\n}\n\nSince the homography estimation updates the status of the couples of matched points as inliers or outliers, between the matched points we are able to draw green lines when they are inliers, or red lines when they are outliers.\n\nfor (unsigned int i = 0; i < nbMatch; i++) {\nif (inliers[i] == true)\nvpDisplay::displayLine(Idisp, iPref[i], iPcur[i] + offset, vpColor::green);\nelse\nvpDisplay::displayLine(Idisp, iPref[i], iPcur[i] + offset, vpColor::red);\n}"
] | [
null,
"https://visp-doc.inria.fr/doxygen/visp-daily/img-logo-visp.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.782177,"math_prob":0.9734312,"size":4472,"snap":"2019-51-2020-05","text_gpt3_token_len":1116,"char_repetition_ratio":0.1544315,"word_repetition_ratio":0.09570041,"special_character_ratio":0.25067085,"punctuation_ratio":0.14938684,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99316967,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T13:42:47Z\",\"WARC-Record-ID\":\"<urn:uuid:cdaf4bd5-e8e0-40da-a133-b0fd199c54c9>\",\"Content-Length\":\"45184\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef3ce98e-73c6-45ef-b805-8ccf611fbced>\",\"WARC-Concurrent-To\":\"<urn:uuid:0546a82c-1d53-4777-a5cc-c45ebb782a60>\",\"WARC-IP-Address\":\"131.254.254.107\",\"WARC-Target-URI\":\"https://visp-doc.inria.fr/doxygen/visp-daily/tutorial-homography-deprecated.html\",\"WARC-Payload-Digest\":\"sha1:MJH7SFCLRE23OESBS6WQNYDIOJCNOHT3\",\"WARC-Block-Digest\":\"sha1:P2IQ7DVVDPCQYW63GAETIWQOBTT2ZAYH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540488620.24_warc_CC-MAIN-20191206122529-20191206150529-00315.warc.gz\"}"} |
https://mathcompt.com/2020/07/cool-math-tricks/ | [
"Here is a cool math trick for you to impress your friends but too simple for you to understand…\n\n## Trick: “PSYCHO MATH”\n\nSay to a volunteer in the audience, “Think of a number, any\nnumber.\n” And you should also say, “But to make it easy on\nyourself, think of a one-digit or two-digit number.\n” After you’ve\nreminded your volunteer that there’s no way you could know\n\n1. double the number,\n3. divide the total by 2,\n4. subtract the original number.\n5. Then say, “Are you now thinking of the number six?”\n\nTry this one on yourself first and you will see that the sequence\nalways produces the number 6 no. matter what number is originally selected. Let’s try again…\n\nSay to one of your friends, “Think of a number, any number.” And you should also say,But this time, think of a two-digit number.” After you’ve reminded your volunteer that there’s no way you could know her number, again ask her to:\n\n1. double the number,\n3. divide the total by 2,\n4. subtract the original number.\n5. Then say, “Are you now thinking of the number six?”\n\nNow let’s see how this mathematic magic trick works.\n\n## MAGIC REVEALED\n\n###### LET’S SEE Why This Trick Works ALL TIME!!!!\n\nThis trick is based on simple algebra. In fact, these tricks are sometimes used as a way to introduce algebra to students. The secret number that your friend has chosen can be represented by the letter x.\nHere are the functions you performed in the order you performed them:\n\n1. x * 2 (double the number)\n2. 2x + 12 (then add 12)\n3. (2x + 12) 2 = x + 6 (then divide by 2)\n4. x + 6 – x = 6 (then subtract original number)\n\nNow lets apply these functions with some examples.\n\nLet’s say my friend Albert is thinking of a no. ‘7’. Say him to apply the steps of your magic trick:\n\n1. 7*2 = 14 (Double it!)\n2. 14 + 12 = 26 (then Add 12)\n3. 26/2 = 13 (and then Make it half)\n4. 13 – 7 = 6 (and finally subtract original number)\n\nAlbert is amazed !!!!! But, Olivia wants to play with me.\n\nNow, Olivia is thinking of a 2-digit no. ‘18‘. After applying the above steps of the math magic trick. Abracadabra no. is ‘6’.\n\nAnd, Olivia is impressed !!!!\n\nFor more tricks and other mathematical stuff stay tuned to MathCompt !"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8997068,"math_prob":0.9840361,"size":2277,"snap":"2021-04-2021-17","text_gpt3_token_len":592,"char_repetition_ratio":0.11482622,"word_repetition_ratio":0.22170901,"special_character_ratio":0.27975407,"punctuation_ratio":0.13842975,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98384845,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-27T00:15:06Z\",\"WARC-Record-ID\":\"<urn:uuid:229f08b2-2cf1-4780-b644-01135e1be612>\",\"Content-Length\":\"189408\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8205a0f4-15b3-4360-b68d-46f4eb3e9b07>\",\"WARC-Concurrent-To\":\"<urn:uuid:15973d14-64e3-467e-922f-7f56cb14e309>\",\"WARC-IP-Address\":\"104.21.63.250\",\"WARC-Target-URI\":\"https://mathcompt.com/2020/07/cool-math-tricks/\",\"WARC-Payload-Digest\":\"sha1:UD3HSG3OREUM36W2DNHDYTDDVWBWT42R\",\"WARC-Block-Digest\":\"sha1:QXQFCY2LD7FYX4RSEUBMGE5347PNGG3G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704804187.81_warc_CC-MAIN-20210126233034-20210127023034-00074.warc.gz\"}"} |
https://ru.scribd.com/document/392616864/Non-wetting-Drops-From-Impacts-to-Self-propulsion | [
"Вы находитесь на странице: 1из 201\n\n# Non-wetting drops : from impacts to self-propulsion\n\nDan Soto\n\n## To cite this version:\n\nDan Soto. Non-wetting drops : from impacts to self-propulsion. Mechanics of materials [physics.class-\nph]. Université Pierre et Marie Curie - Paris VI, 2014. English. <NNT : 2014PA066627>. <tel-\n01149069>\n\n## HAL Id: tel-01149069\n\nhttps://tel.archives-ouvertes.fr/tel-01149069\nSubmitted on 6 May 2015\n\n## HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est\n\narchive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents\nentific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non,\nlished or not. The documents may come from émanant des établissements d’enseignement et de\nteaching and research institutions in France or recherche français ou étrangers, des laboratoires\nabroad, or from public or private research centers. publics ou privés.\nTHESE DE DOCTORAT\n\nSpécialité : Physique\n\nPrésentée par\n\nDan Soto\n\n## M. Christophe Clanet Directeur de thèse\n\nM. Robert Cohen Rapporteur\nM. Jean-François Joanny\nM. Hamid Kellay Rapporteur\nM. Detlef Lohse\nM. David Quéré Directeur de thèse\nMerci !\n\nAfter several months of manuscript writing, the time has finally come to thank everyone\nthat has made this work possible.\n\nFirst of all I would like to thank all the jury members, Jean-François Joanny (who I\nhope enjoyed the drops’ racing-track), Detlef Lohse (who was kind enough to come from\nNetherlands on the eve of his journey to Shanghai), and my thesis referees Robert Cohen\n(who crossed the atlantic from Boston) and Hamid Kellay (who I first met talking about\namazing tornados in soap bubbles) for their enlightening remarks and for their careful\n\nAs the saying goes, “It takes a village to raise a child”. In my case, the village was the\nPMMH lab at ESPCI in Paris, as well as the Ladhyx at Ecole Polytechnique. In this rich\nenvironment, I want to thank first my PhD advisors Christophe Clanet and David Quéré.\nChristophe, thank you for your constant availability, your tireless thirst for models and\nof course your sense of humour. David, thank you for your endless encouragements and\nextreme freedom.\nThank you to all the members of the “Compagnie des interfaces” with whom I have\nshared my daily work. The “Elders”: Keyvan Piroir, who kindly welcomed me in my first\ndays in the group; Alexandre Ponomarenko and his incredible (but true) stories; Marie\nLe Merrer, who I first met as my internship advisor at Polytechnique and Jacopo Seiwert\n(and his friendly outspokenness and shared passion for the sea). The generation of “The\nfamous Five” who were for me a key reference point: Caroline Cohen (an amazing surfer),\nBaptiste Darbois Texier (a tireless joker with bulletproof optimism), Pascal Raux (who\ncould talk for hours over lunch and who showed us Lyon by night... and by bike!), Pierre-\nBrice Binten (with whom we spend as much time working in the “soute” as choosing\nwhich music to listen to) and Guillame Dupeux (who was as much worried by my carbon\nfootprint as he was generous with his time to initiate me to the Leidenfrost world and\nshare with me his passion for Matlab). The crucial link between this two generations\nwas Adrien Benusiglio, an engaging person always up front with everything. I had the\n\n3\n4 MERCI !\n\npleasure to spend my three years with Raphaële Thevenin. Thank you for your help with\nyour “white room skills”, for all our discussions and for making me travel all around the\nAs time went on, we welcomed in the group Manu du Pontavice (with whom I shared\nseveral adventure racings, but never got to finish them), Anaïs Gauthier (a formidable\ncards opponent with a cheerful personality) and Philippe Bourrianne (who may never\nforget the soccer defeat of France against Spain but may find some confort when recalling\nthat to win a bet it takes a lot of sacrifice!). In my last year, three new PhD students\narrived: Eline Dehandschoewercker (as conscientious as outgoing), Thimothé Mouterde\n(always with a big smile and who I loved to tease with his rowing activities) and Hélène\nde Maleprade (with whom I had the pleasure to work on our air-levitated objects while\nenjoying her constant giggles.) Around this time, two Post-docs also joined the group:\nCunjing Lv (who was always eager to share a moment with us) and Evan Spruijt (whose\nkindness was synonym of infinite patience).\nI want to specially thank Aurelie Borel de Larivière, a master thesis intern with whom\nI worked on drop impacts and who carried out an outstanding work. I thank all the\ninterns with whom I have crossed paths with, with a special mention to Maxime “Costa”.\nI also thank you all because I am leaving with wonderful memories of our time shared\noutside the lab: may it be in California (where we could play cards in uncommon places),\nin Pittsburg (where the snow took us by surprise), in Copenhagen (where soccer and\nswimming was as important as the conferences), in Twente (where we discovered what\nEvan had usually for breakfast), in Rome (where we enjoyed a thrilling soccer game),\nduring our skiing adventures or the different evenings spent in each other homes.\n\nI want to thank the PMMH director Philippe Petitjeans (and his predecessor José\nEduardo Wesfreid) for welcoming me in their lab. Frédérique Auger, Amina Mialet and\nClaudette Barez, thank you for making so much things happen. I spent a lot of time in\nthe workshop and I want to thank all their members, with a special mention to Guillaume\nClermont (who took the time to initiate me to the numerical mill) and Xavier Benoit-\nGonin (who taught me all the secrets of the laser cutter). Thank you to all the other\nmembers of the lab with whom I had the pleasure to live with for three years (and six\nmonths internship). I will not forget the friendly atmosphere of Ladhyx and the incredible\nsoccer games!. Thank you all.\n\nRobert Cohen spent six months in the lab and we had a great pleasure working with\nhim on drops and grids. Thank you for your kindness and for being always so thoughtful\n(such as with the organisation of my trip to Boston). I also thank Xavier Boutillon, Stef-\nMERCI ! 5\n\nfen Hardt, Tobias Baier and Stephane Dorbolo with whom we collaborated at Palaiseau,\n\nFinally, I would like to thank everyone who has not been mentioned above but accom-\npanied me all along this adventure. My friends, who will forgive me for not attempting\nan exhaustive enumeration. My parents, who proofread my manuscript and gave me the\ntaste for science. My brother, who has always led the way. All my family, who never\ndoubted the value of a PhD on water drops.\n\nAnd Marion.\nContents\n\nIntroduction 11\n\n## I The non-wetting world 15\n\n1 Generating non-wetting objects . . . . . . . . . . . . . . . . . . . . . . . . 16\n1.1 The super-hydrophobic state . . . . . . . . . . . . . . . . . . . . . . 16\n1.2 The Leidenfrost state . . . . . . . . . . . . . . . . . . . . . . . . . . 18\n1.3 The air-levitated state . . . . . . . . . . . . . . . . . . . . . . . . . 22\n1.4 A wide range of other possibilities . . . . . . . . . . . . . . . . . . . 25\n2 Shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27\n2.1 The static shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27\n2.2 The vapor cushion . . . . . . . . . . . . . . . . . . . . . . . . . . . 33\n2.3 Instabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37\n2.4 Crenelated surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 39\n3 High mobility and special friction . . . . . . . . . . . . . . . . . . . . . . . 45\n3.1 Viscous friction in the vapor film . . . . . . . . . . . . . . . . . . . 45\n3.2 Inertial friction in the surrounding air . . . . . . . . . . . . . . . . . 46\n3.3 Special friction on a crenelated surface . . . . . . . . . . . . . . . . 46\n\n## II Self-Propulsion in the Leidenfrost state 49\n\n1 The texture revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50\n1.1 The ratchet: a seed is sown . . . . . . . . . . . . . . . . . . . . . . 50\n1.2 The herringbone: time to reap . . . . . . . . . . . . . . . . . . . . . 57\n2 Force of propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60\n2.1 Experimental measurements . . . . . . . . . . . . . . . . . . . . . . 60\n2.2 Analytical calculation . . . . . . . . . . . . . . . . . . . . . . . . . . 62\n2.3 Further considerations . . . . . . . . . . . . . . . . . . . . . . . . . 65\n3 Friction on grooved topography . . . . . . . . . . . . . . . . . . . . . . . . 68\n3.1 Straight trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . 68\n3.2 Free trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72\n4 Terminal speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77\n4.1 Experimental results . . . . . . . . . . . . . . . . . . . . . . . . . . 77\n4.2 Analytical calculation and speed optimization . . . . . . . . . . . . 78\n5 A basic unit of a wider picture . . . . . . . . . . . . . . . . . . . . . . . . . 79\n5.1 The drop trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79\n5.2 The active herringbone: the switch . . . . . . . . . . . . . . . . . . 80\n\n7\n8 CONTENTS\n\n## III Self-propulsion on an air hockey table 85\n\n1 When vapor is replaced by compressed air . . . . . . . . . . . . . . . . . . 86\n1.1 The air hockey table . . . . . . . . . . . . . . . . . . . . . . . . . . 86\n1.2 Propulsion with herringbone textures . . . . . . . . . . . . . . . . . 87\n2 Force of propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90\n2.1 Experimental observations . . . . . . . . . . . . . . . . . . . . . . . 90\n2.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91\n3 New geometries, new functionalities . . . . . . . . . . . . . . . . . . . . . . 99\n3.1 The truncated herringbone . . . . . . . . . . . . . . . . . . . . . . . 99\n3.2 Climbing up a slope . . . . . . . . . . . . . . . . . . . . . . . . . . 100\n3.3 The viscous entrained mill . . . . . . . . . . . . . . . . . . . . . . . 101\n4 Channel depth h and Reynolds number . . . . . . . . . . . . . . . . . . . . 104\n5 Switching roles: the texture patterned on the slider . . . . . . . . . . . . . 106\n5.1 Experimental set-up . . . . . . . . . . . . . . . . . . . . . . . . . . 106\n5.2 Force measurements . . . . . . . . . . . . . . . . . . . . . . . . . . 108\n6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111\n\n## IV Drop impacting a sieve 113\n\n1 Impact on a solid plate: a brief review . . . . . . . . . . . . . . . . . . . . 114\n1.1 Maximal impacting radius . . . . . . . . . . . . . . . . . . . . . . . 115\n1.2 Drop shape profile . . . . . . . . . . . . . . . . . . . . . . . . . . . 117\n2 Impact on a plate with a single hole . . . . . . . . . . . . . . . . . . . . . . 120\n2.1 Critical speed V ⇤ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120\n2.2 Role of plate thickness . . . . . . . . . . . . . . . . . . . . . . . . . 123\n2.3 Several time scales . . . . . . . . . . . . . . . . . . . . . . . . . . . 124\n2.4 Transmitted mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125\n2.5 Final comment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127\n3 The Leidenfrost sieve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128\n3.1 Experimental set-up . . . . . . . . . . . . . . . . . . . . . . . . . . 128\n3.2 Transmitted mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130\n3.3 A deformable interface . . . . . . . . . . . . . . . . . . . . . . . . . 132\n3.4 Splash pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135\n4 Exploring different meshes . . . . . . . . . . . . . . . . . . . . . . . . . . . 139\n4.1 Role of wetting conditions . . . . . . . . . . . . . . . . . . . . . . . 140\n4.2 Role of hole size r . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142\n4.3 A single curve? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143\n4.4 Pinch-off time versus crash time . . . . . . . . . . . . . . . . . . . . 144\n5 Conclusion and open questions . . . . . . . . . . . . . . . . . . . . . . . . . 145\n\n## V Impact force of a drop 149\n\n1 Compression waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150\n1.1 Water hammer in the liquid . . . . . . . . . . . . . . . . . . . . . . 150\n1.2 Water hammer in the surrounding air . . . . . . . . . . . . . . . . . 153\n2 Measure of impact force with a piezo-electric quartz . . . . . . . . . . . . . 155\n2.1 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . 155\n2.2 Analytical calculation . . . . . . . . . . . . . . . . . . . . . . . . . . 157\nCONTENTS 9\n\n## 2.3 The case of raindrops . . . . . . . . . . . . . . . . . . . . . . . . . . 158\n\n3 A cheaper sensor: the lamella . . . . . . . . . . . . . . . . . . . . . . . . . 159\n3.1 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . . . . 159\n3.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161\n3.3 Agreement between model and data . . . . . . . . . . . . . . . . . . 162\n3.4 The two impact regimes . . . . . . . . . . . . . . . . . . . . . . . . 162\n3.5 The case of raindrops . . . . . . . . . . . . . . . . . . . . . . . . . . 163\n3.6 Energy harvesting . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164\n4 Non-wetting impacts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165\n4.1 Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165\n4.2 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166\n5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168\n\nConclusion 169\n\n## C Résumé en Français 177\n\nBibliography 197\nIntroduction\n\nIn 1959, the Belgian surrealist artist Magritte finished his masterpiece entilted “Le Château\ndes Pyrénées” (The Castle of the Pyrenees) shown in figure 1. This painting embodies\n\nFigure 1 – Le Château des Pyrénées (The Castle of the Pyrénées), René Magritte, 1959.\nOil on canvas, 200 x 145 cm. Gift of Harry Torczyner to the American Friends of the\nIsrael Museum.\n\nthe artist’s typical disturbing juxtaposition of familiar objects, combined with captivat-\ning poetry and mystery. A gigantic boulder, topped by a castle, is immobilized in the\nair between an azure sky (dotted with clouds) and a pristine sea (whose waves reflect the\ngray nuances of the rock). After a first impression, the gap between the rock (emphasized\nby the tiny relative size of the top castle) and the sea attracts our attention. Magritte\n\n11\n12 INTRODUCTION\n\nworks in the artistic dimension - where imagination rules and we do not have to satisfy\nphysical laws. However, we can adopt a more realistic point of view, and wonder how an\nobject could possibly levitate.\n\nA first answer was proposed in 1756 by the German physician Johann Gottlob Leiden-\nfrost. In , he studied the behavior of a water droplet deposited over a hot substrate\n(typically 400 C). Placing a candle behind the drop, Leidenfrost could observe with the\nnaked eye that light passes between the hot solid and the liquid, revealing the existence\nof a film of vapor below the drop as shown in figure 2, in a very similar way than what\nMagritte represented in “Le Château des Pyrénées”. However, contrasting with Magritte’s\nmasterpiece, the Leidenfrost levitation has an explanation: it results from a balance be-\ntween evaporation (that nourishes the vapor cushion on which the drop levitates) and the\ndrop’s weight (which presses on this vapor layer). Since then, other methods have been\nproposed to generate levitation. Among them, levitation by blowing air - somehow like\nan hovercraft.\n\nFigure 2 – Millimetric drop levitating over a hot plate. Backlighting is used, which helps\nto distinguish the interval between the drop and its reflection owing to the presence of\nvapor. The scale bar shows 1 mm.\n\nA levitating body avoids any contact with the substrate, which makes it highly mobile.\nIn this work, we focus on the special dynamics associated with this mobility. We exploit\nthis situation to generate self-propulsion (allowing us to put a resting object into mo-\ntion). Conversely, frictionless objects are difficult to stop. We consider situations where\na moving liquid drop can be stopped by different means: from substrate texturation, to\nan impact on a solid wall through impact on a mesh (an intermediate situation between\nINTRODUCTION 13\n\n## a solid wall, able to stop a movement, and no obstacle at all).\n\nThese various situations will stress, I hope, the novel physics that arise in this lively\nfield of hydrodynamics without contact.\nChapter I\n\n## The non-wetting world\n\nContents\n1 Generating non-wetting objects . . . . . . . . . . . . . . . . . . 16\n1.1 The super-hydrophobic state . . . . . . . . . . . . . . . . . . . 16\n1.2 The Leidenfrost state . . . . . . . . . . . . . . . . . . . . . . . . 18\n1.3 The air-levitated state . . . . . . . . . . . . . . . . . . . . . . . 22\n1.4 A wide range of other possibilities . . . . . . . . . . . . . . . . 25\n2 Shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27\n2.1 The static shape . . . . . . . . . . . . . . . . . . . . . . . . . . 27\n2.2 The vapor cushion . . . . . . . . . . . . . . . . . . . . . . . . . 33\n2.3 Instabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37\n2.4 Crenelated surfaces . . . . . . . . . . . . . . . . . . . . . . . . . 39\n3 High mobility and special friction . . . . . . . . . . . . . . . . 45\n3.1 Viscous friction in the vapor film . . . . . . . . . . . . . . . . . 45\n3.2 Inertial friction in the surrounding air . . . . . . . . . . . . . . 46\n3.3 Special friction on a crenelated surface . . . . . . . . . . . . . . 46\n\n15\n16 CHAPTER I. THE NON-WETTING WORLD\n\n## 1.1 The super-hydrophobic state\n\nThe lotus plant has been considered as a sacred plant several centuries before our era. It\nwas held to be the symbol of truth, auspiciousness and beauty, always remaining untainted\ndespite its surroundings. As Veda Vyasa points out in his ancient sanskrit scripture\nMahabharata1 , a lotus leaf never gets wet even though it is always in water:\n\nsangam tyaktva karoti yah\nlipyate na sa papena\n\n## “One who performs his duty without attachment,\n\nsurrendering the results unto the Supreme God,\nis not affected by sinful action,\nas the lotus leaf is untouched by water.”\n\nDespite this ancient text, it is only recently (in the 90’s) that the lotus mystery has\nbeen solved thanks to the use of Scanning Electron Microscope (SEM) in the work of\nW. Barthlott and C. Neinhuis . They describe how the combination of microtextures\nand proper chemistry makes lotus actively repel water. From that time on, the lotus\nhas moved from a religious, sacred dimension to a scientific one, where we try to mimic\nits properties. If we look at the surface of a lotus leaf through an SEM (figure I.1), we\ncan observe a highly sophisticated double texturation: the leaf is covered by micrometric\nposts, themselves covered by nanometric wax crystals.\n\nFigure I.1 – SEM picture of the surface of a lotus leaf. We can see that each pillar is\nlikewise structured because it is covered by small wax crystals. Figure from .\n\n1\nprobable composition period: between the fifth and the second century BCE.\n1. GENERATING NON-WETTING OBJECTS 17\n\nIf we want to imitate this effect artificially, nature tells us that we need to satisfy two\nrequirements:\n\n(i) have a hydrophobic surface (from the ancient Greek ÕdrÏfoboc: has fear of water).\nThe affinity of a liquid towards a substrate can be characterized by its contact angle ✓c\nas defined in figure I.2. At the contact line, we have three interfaces: solid-liquid, liquid-\ngas and gas-solid. Since each one has its own surface tension, the static equilibrium is\ndetermined by Young’s equation: cos ✓c = sl lg sg .\n\nθc\nθc\n\n## If ✓c < 90 , the substrate is hydrophilic. Conversely, if ✓c > 90 , the substrate is\n\nhydrophobic. Wax, oils, fats, long carbon chains and fluorinated molecules are usually\nhydrophobic.\n\n## (ii) have a textured surface to enhance hydrophobicity and achieve a super-hydrophobic\n\neffect. It is observed that surface chemistry alone fails to achieve contact angles above\n✓c = 120 . In nature, we can find other examples of super hydrophobicity in plants\n(Ginkgo Biloba, Brassica Oleracea) or animals (water strider, dragonfly, collembola).\nThey all share a common feature: they have a tailored surface. In order to outreach\nan angle of 160 or even above, we need to add surface roughness. As explained by A.\nCassie , when adding roughness, the drop does not wet the whole projected surface\nbeneath it but only the top of the summit topography, like a fakir on a bed of nails. Lo-\ncally, the drop satisfies Young’s equation on top of each pillar. However, the macroscopic\ncontact angle ✓c⇤ (also called the apparent contact angle) is much larger (figure I.3). In\nother words, the drop will mainly face air beneath it (the top pillars represent only a very\nlow percentage of the total projected area) and will be in a non-wetting state.\nExperimentally, we can generate well-controlled roughness by micro-engineering an\narray of pillars such as those in figure I.4a. Although different techniques exist to man-\nufacture such refined structures, they all require advanced skills and high technology\nequipment. Easier methods that allow greater areas to be treated have been proposed,\nsuch as the one by Larmour and collaborators . This two-steps method (where the\n18 CHAPTER I. THE NON-WETTING WORLD\n\nθc*\nθc\n\nFigure I.3 – Textured surface, with a drop sitting on it as a fakir on a bed of nails. On\nthe top of each pillar, the local angle of contact ✓c satisfies Young’s equation. However,\nthe macroscopic angle of contact ✓c⇤ is much larger.\n\nprecise geometry of the texture can not be controlled) begins by plunging the substrate\nin a solution of silver nitrate. The surface darkens owing to the deposition of small\nsilver particles that generate micrometric roughness. Then, the surface is treated with\nan alcoholic solution of 3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,10-heptadecaf luoro-1-decanethiol\n(HDFT) that will graft these long fluorinated molecules to the surface of the silver texture,\nmaking it chemically hydrophobic. This method is limited to metallic substrates, such\nas copper or brass. More recently, industrial liquid coatings (such as UltraEverDryTM or\nGlacoTM ) have appeared on the market as an easy alternative (as easy as spraying the\nliquid on the surface) to achieve both chemical hydrophobicity and physical roughness.\nAfter coating a sample with this industrial solutions made of hydrophobic colloids, we can\nsee through an SEM a disordered roughness at very small scale (see figure I.4b). Another\nmethod for treating large areas of complex geometry (such as a mesh shown in figure\nI.4c), is for example used in the group of R. Cohen by spray-coating or dip-coating the\nsample from 50/50 wt% mixture of poly(methyl methacrylate) and hydrophobic molecules\n1H,1H,2H,2H heptadecaf luorodecyl polyhedral oligomeric silsesquioxane (commonly\ncalled fuorodecyl POSS, see [23, 105, 60]), providing a high degree of hydrophobicity.\n\n## 1.2 The Leidenfrost state\n\nLeidenfrost experiment\n\nThe key to generating a non-wetting state is to prevent contact between liquid and solid.\nIn the super-hydrophobic state, the pillars succeeded in artificially trapping an air layer\nbetween drop and substrate. However, because the weight of the drop has to be compen-\nsated, it needs to rest on top of each pillar: although marginal, contact still exists. If we\n1. GENERATING NON-WETTING OBJECTS 19\n\n(a) (b)\n\n(c)\n\nFigure I.4 – SEM views of three artificial super hydrophobic surfaces. (a) Array of micro\npillars of height 10 mm, diameter 2 mm, and pitch 10 mm. Scale shows 10 mm, courtesy of M.\nReyssat. (b) Surface treated with GlacoTM . Scale shows 2 mm. Courtesy of C. Willem. (c)\nScanning electron micrographs at different magnifications of dual-textured spray-coated\nsuper-hydrophobic mesh surfaces. Figure from .\n\nwish to completely eradicate this contact, we have to go back to Germany in 17562 , and\nread J. G. Leidenfrost’s paper “On the fixation of water in diverse fire” :\n\n“At the instant when the drop touches the glowing iron, it is spherical. It does not\nadhere to the spoon, as water is accustomed to do, which touches colder iron.”\n\n## Through this observation, we understand that a Leidenfrost water droplet is in a\n\nnon-wetting state. By looking from the side at water sitting on a hot plate (figure I.5,\nexperiment sketched in figure I.6), Leidenfrost was able to see a beam of light passing\nbetween the liquid and the plate, thus demonstrating a pure non-wetting state.\n2\nThe Dutch scientist H. Boerhaave was the first to mention the levitation of a drop over a hot substrate\nin his 1732 publication: Elementa Chemiae .\n20 CHAPTER I. THE NON-WETTING WORLD\n\nFigure I.5 – Side view of a water droplet on a hot substrate (400 C). The gap between\nthe drop and the substrate (visible because of the back light passing through it) has a\ntypical height h ⇠ 100 mm. The bar shows 1 mm.\n\n## Figure I.6 – Experimental set up used by J. G. Leidenfrost to observe vapor cushions\n\nbetween a drop and its hot substrate.\n\nAs the drop approaches the hot substrate it evaporates at a rate fast enough to form a\nstable vapor film beneath it (visible in figure I.5). The droplet is sitting on top of a cushion\nof its own vapor, squeezing it and generating a vapor flow permanently compensated by\nevaporation, hence achieving an equilibrium levitation height h of typically 10 to 100 mm.\n\nA way to characterize the Leidenfrost effect consists in measuring the lifetime evolution\n⌧ of a drop sitting on a substrate, as a function of the substrate temperature. At high\ntemperature, the vapor provides thermal insulation and the drop can last over several\ntenths of a second. However, as we decrease temperature, we observe a sharp transition.\nThe temperature corresponding to this lifetime discontinuity is called the Leidenfrost\nTemperature TL . At this point (typically between 150 and 200 C for water), the substrate\nis not hot enough to sustain the necessary evaporation rate. Consequently, the peaceful\nnon-wetting state (described in figure I.8 as film boiling regime) disappears and is replaced\n1. GENERATING NON-WETTING OBJECTS 21\n\nby a violent boiling regime (first, transition boiling and then, nucleate boiling, again shown\nin figure I.8). As expected when direct contact between liquid and substrate occurs, drops\n\n100\nτ (s)\n80\n\n60\n40\n20\n\n0\n0 100 200 300 400\nT (°C)\nFigure I.7 – Lifetime ⌧ of a millimetric water droplet of radius R = 1 mm, as a function\nof the temperature T of the Duralumin plate on which it is deposited. Leidenfrost tem-\nperature TL is around 150 C. Above this temperature, we are in the film boiling regime.\nRight below it, the drop experiences the transition boiling regime. If we further decrease\nthe temperature, we go thorough the nucleate boiling regime before arriving to the single\nphase (see figure I.8). Figure adapted from .\n\nBoiling Leidenfrost\nPoint (B.P.) Point (L.P.)\n\n## Single Nucleate Transition Film\n\nPhase boiling boiling boiling\n\nTemperature\n\nFigure I.8 – Sketch of different boiling regimes ranging from single phase to film boiling\n(Leidenfrost state) going through the boiling point (B.P.) and the Leidenfrost point (L.P.).\n\nThe Leidenfrost temperature depends on the nature of the liquid, the substrate and\nits roughness. Water on a clean smooth substrate will levitate at lower temperature than\non a dirty rough one. It has been shown [58, 57, 59] that a fine texture (at the scale\nof 0.1–10 µm) considerably increases the Leidenfrost temperature. Characteristic metal\nroughness ranges from 0.05 mm to 10 mm (rusted metal even reaching 100 mm). Typical\nvapor thickness being 100 mm, the greater the characteristic roughness length scale the\n22 CHAPTER I. THE NON-WETTING WORLD\n\nshorter the distance between the liquid and the tip of a local crest topography. Any\nvibration of the interface will therefore easily put the liquid in contact with the roughness\nprotuberance, instantaneously nucleating boiling. We have previously seen that if textures\nwere hydrophobic, they would achieve super-hydrophobicity. Conversely, if they were\nhydrophilic, they would achieve super-hydrophility. Thereby, textures role can be seen\nas an enhancer of liquid substrate affinity. In the Leidenfrost state, something similar\nhappens and roughness does not always necessarily result in a Leidenfrost temperature\nincrease. If textures are hydrophobic they can be used to stabilize the Leidenfrost vapor\nlayer as shown by Vakarelski et al. .\n\n## From Leidenfrost to air-levitated drops\n\nWe just saw that we can achieve complete levitation by constantly nourishing a vapor\ncushion between a drop and its hot substrate. The vapor is created by using heat to\ntransform liquid into gas phase. This inevitably means that our drop has a limited\nlifetime as it is consuming itself in order to levitate. The time limited (or even deadly!)\nnature of this levitation can be avoided by replacing vapor by external injected air as done\nwith air hockey tables. In 1986, M. Goldshtik carried out a thorough comparison between\nLeidenfrost and air-levitated drops . A strong similarity arises from the close analogy\nbetween the symmetry and hydrodynamic mechanisms of both types of suspension (see\nfigure I.9). In the end, the critical parameter that governs air-levitation is the differential\npressure on a porous surface, as is the temperature in the thermal counterpart.\n\nh U\n\n## Hot Plate Porous Substrate\n\nFigure I.9 – Leidenfrost drop (left) and air-levitated drop (right). We can see how striking\nis the analogy between these two situations.\n1. GENERATING NON-WETTING OBJECTS 23\n\n## The deformable case\n\nIn the Leidenfrost state, it is experimentally difficult to reach high vapor flow rates due to\ntemperature limitations. This contrasts with the air-levitated state where we can easily\nadjust the differential pressure through the porous substrate, hence freely varying the air\ninjection rate on a wide range of values. Therefore, we can reach airflows that are strong\nenough to change the underlying levitation mechanism.\nTo understand this transition, we have to compare the role of inertial forces versus viscous\nones. By denoting ⇢a the air density, ⌘a its viscosity and U its speed, a first ratio of these\ntwo forces can be constructed with the drop radius as a characteristic distance, which\nyields for this Reynolds number:\n⇢a U R\nRe = (I.1)\n⌘a\n\nWe experimentally see that the bottom of the drop is deformed and flattened by\nthe vapor pressure as sketched in figure I.9. In this region, inertial forces will scale as\n⇢a U 2 /R and viscous ones as ⌘U/h2 . Their ratio, sometimes called the lubrication Reynolds\nnumber, is:\n⇢a U h 2 h\nReL = = Re (I.2)\n⌘a R R\n\nFor a millimetric drop with h ⇡ 100 mm and small flow rate (U ⇡ 0.1 1 m/s), we\nexpect ReL < 0.1: viscous effects clearly dominate within the air cushion.\n\n## The non-deformable case\n\nPrevious mechanism of levitation involved the deformation of the interface of the object.\nHowever, a non-deformable solid object (such as a ping-pong ball) can also be maintained\nin levitation by blowing air beneath it. We address here the situation of a levitated drop\nfor which deformation can be neglected. A moving sphere experiences drag from its sur-\nrounding liquid (or gas). If the drag is strong enough to compensate the drop’s weight, it\nwill result in levitation - like a free falling object experiences in its own frame. Depending\non the Reynolds number, several regimes can be observed:\n\n(i) For small Reynolds number (i.e. Re < 1), the sphere will experience a viscous\nstress ⌘a U/R integrated over its whole surface R2 . It will be subjected to the Stokes drag\nforce:\nFStokes ⇠ ⌘a RU (I.3)\n\nBy compensating it with the drop weight ⇢R3 g (where ⇢ denotes the liquid density),\n24 CHAPTER I. THE NON-WETTING WORLD\n\n## we immediately get a critical levitation speed:\n\n⇢R2 g\nU⇤ ⇠ (I.4)\n⌘a\n\n⇢a ⇢R3 g\nWe can insert this speed in the Reynolds number expression (I.1), Re⇤ = ⌘a2\nand build\na characteristic length RL :\n⌘2\nRL = ( a )1/3 (I.5)\n⇢a ⇢g\nUsing this characteristic length, the Reynolds number can be re-expressed as: Re =\n(R/RL )3 . As previously said, this regime only applies if Re < 1: hence a maximum droplet\nradius: R⇤ = RL . For a water droplet surrounded by air, this means very small droplets:\nR < R⇤ =30 mm.\n\n(ii) If we continue increasing the flow rate, we will enter the intermediate Reynolds\nnumber regime: 1 < Re < 103 . When a body is put into motion in a viscous fluid, a\np\nboundary layer develops around the object with a characteristic thickness ⇠ ⌫a t. Here\n⌫a = ⌘a /⇢a denotes the kinematic viscosity of air and t the time of development of the\np\nboundary layer 3 . For an object of size R, t naturally scales as R/U , hence: ⇠ ⌫R/U .\nThe resulting stress being ⇠ ⌘a U/ , the skin drag force is:\np\nFskin ⇠ ⌘a ⇢a [RU ]3/2 (I.6)\n\n## Analogous reasoning leads to a critical levitation speed:\n\n⇢g 2/3\nU ⇤ ⇠ R( p ) (I.7)\n⌘ a ⇢a\nUsing the same characteristic length RL than defined in equation I.5, the Reynolds\nnumber will now be: Re = (R/RL )2 . As Re 2 [1,1000] this leads to drop radius R ranging\nfrom 30 mm to 1000 mm, that is, still relatively small drops.\n\n(iii) Finally, we consider drag at high flow rates: Re > 1000, yet below turbulent\nregimes. If the characteristic length scale of the object is R, it has to displace an initially\nresting mass of air ⇢a R3 at a speed U in a characteristic time R/U in order to move\nforward. This leads to an inertial drag force:\n\nFaero ⇠ ⇢a R2 U 2 (I.8)\n\nIn most everyday situations (car, plane, soccer ball), we have to fight against this force\n3\nThis equation is characteristic of diffusive events and in this case the diffused quantity is momentum.\n1. GENERATING NON-WETTING OBJECTS 25\n\nin order to move. However, as done in the two previous regimes, we can use this drag\nforce to levitate a body by compensating its weight - the same way a skydiver would do in\na wind tunnel. The critical speed at which we will be able to sustain a drop in air being:\ns\n⇢Rg\nU⇤ ⇠ (I.9)\n⇢a\n\nIn this case the Reynolds number can be rewritten as: Re=(R/RL )3/2 . To satisfy\nthe relationship Re > 1000, a last critical radius arises: R⇤ = 10002/3 RL . Once again,\nfor a water droplet in air this means R > R⇤ = 1 mm. In the case of rain, this is the\nmost common regime as everyday drops are usually larger than a millimeter. If we think\nof a falling drop, after a short acceleration (initially it will accelerate as free fall), drag\nwill increase until it compensates the weight and the drop reaches its terminal speed U1\n(described in I.9)- typically 10 m/s. In a simple way, we can see a falling drop at terminal\nspeed (from the drop frame point of view) as a non-wetting object.\n\n## 1.4 A wide range of other possibilities\n\nA great variety of other approaches have been used to generate levitation . Besides\nthe air injection through a porous medium, alternative solutions have been proposed to\nnourish the air film.\nIn , Couder et al. deposit a liquid drop on top of a vibrating bath of the same\nliquid. If the oscillation period and amplitude are well chosen, the air cushion can be\nrenewed between two oscillations before it has time to drain, hence avoiding coalescence.\nAnother solution to generate liquid levitation consists on putting a drop above a\nspinning disk where solid/liquid contact will be prevented by the thin air boundary layer\nentrained by the disk. As a consequence, the drop levitates and is slightly deformed as\nshown in figure I.10.\n\nFigure I.10 – Shape of a millimetric silicone-oil drop (viscosity 100 cSt) levitating over a\nrotating disk. If rotation speed is not high enough (left image), entrained air is not strong\nenough to overcome the drop’s weight and avoid contact. The higher the rotation speed\nseen by the drop, the more the drop deforms. Figure courtesy of A. Gauthier.\n26 CHAPTER I. THE NON-WETTING WORLD\n\n## Also based on a moving substrate, levitation can be induced by depositing a liquid\n\ndroplet on an hydraulic jump as discussed in [104, 39]. Indeed, a thin layer of air is\nentrained underneath the drop by the flow of liquid preventing coalescence.\n\nIn [84, 81, 35], Neitzel and collaborators use thermocapillarity effects to drive a lu-\nbricating gas film between a heated droplet and a cooled substrate. By heating the top\ninterface of a drop, they increase the local temperature, hence decrease the local surface\ntension. As a consequence, liquid is pulled along the interface from the top to the bot-\ntom of the drop. Due to this Marangoni flow, surrounding air is entrained in the same\ndirection, so that air is also injected at the bottom of the drop generating levitation.\nFor instance, if we take two drops of the same liquid and put them in contact they will\ncoalesce. However, if the two drops are at different temperature Marangoni flows will\ndrag air between the droplets suppressing coalescence as shown in figure I.11.\n\n(a) (b)\n\nHot\n\nCold\n\nFigure I.11 – (a) Sketch of the flow field for two non-coalescing drops held at different\ntemperatures. The gap thickness is exaggerated to display the direction of motion in\nthe entrained air film. (b) Non-coalescnece of two 5 cST silicone oil drops of different\ntemperature around 15 C. Arrow shows the direction of the Marangoni flow along the\ninterface. Figures from .\n\nlevitation .\n2. SHAPE 27\n\n2 Shape\n\nSurface tension\n\n## A physical consequence of being non-wetting is the appearance of a quasi-spherical shape.\n\nAs we suppress contact (or, in the super-hydrophobic state, almost suppress it), a drop is\nonly subject to gravitational forces (scaling as ⇢R3 g) and liquid-gas surface tension forces\n(scaling as R, where denotes from now on the liquid-gas surface tension). A first\ncharacteristic length called the capillary length is obtained by balancing this two forces:\nr\n`c =\n⇢g\nIn the case of water, we have `c = 2.7 mm at ambient temperature and `c = 2.5 mm\nat 100 C. Another convenient liquid is acetone, for which `c is smaller, namely 1.6 mm\nnear its boiling point at 60 C. In figure I.12, we show two drops: on the left side (figure\nI.12a) a spherical drop smaller than the capillary length; on the right side (figure I.12b)\na flattened puddle larger than `c . We clearly see that the capillary length sets the limit\nbetween small droplets modeled by surface tension and large ones globally deformed by\ngravity.\n\n(a) (b)\n\nH=2R H=2ℓc\n\nFigure I.12 – (a) Small droplets are mainly shaped by surface tension, hence having a\nquasi-spherical shape. Scale shows 1 mm. (b) Large drops are flattened by gravity, of\nmaximum height twice the capillary length `c . Scale shows 2mm.\n28 CHAPTER I. THE NON-WETTING WORLD\n\nDroplets: R < `c\nWe refer to small drops when R < `c so that surface tension dominates gravity, as shown\nin figure I.12a. If we take a fixed amount of liquid (small enough to be in this capillary\nworld), surface tension selects the geometry that minimizes the surface area. For example,\nif we consider a pyramidal geometry, corners are not very efficient since we can put\nvery little liquid in them although we need a lot of surface to generate them. This\nexample tells us that surface tension avoids sharp geometries, hence naturally selecting\nthe spherical shape (mathematically known to minimize surface area for a given volume).\nIn 1805, Pierre Simon de Laplace showed that a consequence of curved interfaces,\nis the existence of a difference of pressure P between the inside and the outside of the\ninterface: P = C (known as the Young-Laplace equation). Here, C represents the\ninterface curvature defined as the sum of two curvatures in perpendicular planes. For a\nLeidenfrost droplet, the geometry is that of a sphere, and we have P = 2 /R, about\n100 Pa for millimetric drops.\n\nPuddles: R > `c\nWe refer to big drops or puddles when R > `c , so that gravity effects have to be taken\ninto account. As we can see in figure I.12b, the drop is flattened by gravity. R will further\non denote the equatorial radius of a puddle and H its thickness. Laplace pressure on the\ncurved side can be written as: PL = (2/H + 1/R), where R and H/2 are the curvature\nin the horizontal and vertical plane, respectively. In the big drop limit (H << R), it\nreduces to PL ⇡ 2 /H. From a hydrostatic point of view, the inner pressure at half\nheight is PH = ⇢gH/2. By balancing these two pressures, we get an expression for puddle\nthickness. It is independent of the radius R, and writes H = 2`c .\n\nIf we want to have an exact solution for the drop’s shape, we have to solve the dif-\nferential equation similarly obtained by pressure arguments. The pressure at any height\nz of the drop can be written following two different methods based on the two different\npaths sketched in figure I.13 (the horizontal position is not relevant in this problem since\nwe are in a hydrostatic state):\n(i) the first path sketched in figure I.13 by a red arrow, where we simply cross the\ndrop’s interface at height z, implies a pressure jump P1 = C(z).\n(ii) the second one (sketched in figure I.13 by a blue arrow), where we follow an\nimaginary path that crosses first the interface on the axis of symmetry at the top of the\ndrop (with a pressure jump C0 ) and then goes down along the z-axis until reaching\nheight z (hence a hydrostatic pressure ⇢gz). The corresponding pressure jump is: P2 =\nC0 + ⇢gz.\nIntroducing the capillary length, equality between these two pressures P1 = P2\n2. SHAPE 29\n\n## gives us the following equation for the local curvature:\n\nz\nC(z) = C0 + (I.10)\n`c 2\n\nr(z)\n\nn\n\nθ ds\ndz\nH dr\n\nFigure I.13 – Shape of a non-wetting drop. Two different paths allow to access two\ndifferent expression for the pressure inside the drop at height z: red arrow show path\nnumber one while blue arrow shows path number two.\n\n## Introducing s as the curvilinear coordinate, r(z) as the horizontal radius at height\n\nz and ✓ as the angle between the tangent to the interface and the vertical plane (see\nfigure I.13), the two orthogonal curvatures can be written: C1 (z) = cosr ✓ , C2 (z) = d✓ds\n. In\naddition, geometrical observation shows that dr=ds sin ✓, dz=ds cos ✓ and ṙ = dz = tan ✓.\ndr\np\nBy re-expressing cos ✓ as 1/ 1 + ṙ2 , we immediately get C1 (z) = 1\n2\n1 and C2 (z) =\nr (1+ṙ ) 2\nd✓ dz\ndz ds\n= dz d✓ p 1\n1+ṙ2\n. By calculating r̈ = d tan\ndz\n✓ d✓\n= dz 1\n1+ṙ2\nwe obtain the expression of dz\nd✓\n\nequal to 1+r̈ṙ2 . We now have all the elements to re-write equation I.10 as the following\nsecond order differential equation for r(z):\n\n1 r̈ z\n= C0 + 2 (I.11)\nr (1 + ṙ2 )1/2 2\n(1 + ṙ ) 3/2\n`c\nThere is no analytical solution to this equation, which must be solved numerically.\nThe only physical parameters that we have to set are the curvature at the top of the drop\nC0 and the capillary length. We show in figure I.14 the shape of several drops for water\n(`c = 2.7 mm) and curvature radius at the drop’s apex ranging from 1 mm (smallest\nradius of curvature, hence almost spherical drop) to 10 mm (biggest radius of curvature,\nhence the more flattened drop).\nThe model assumes homogeneous stress on the bottom of the drop (as we can get\nin super-hydrophobic states). In the Leidenfrost (or air-levitated) situation, gas escapes\nfrom the axis of symmetry towards the external edge, indicating that maximum pressure\npoint is located right below the center of drop. We will therefore have a local higher\n30 CHAPTER I. THE NON-WETTING WORLD\n\nz /l c 0\n\n−0.5\n\n−1\n\n−1.5\n\n−2\n−1.5 −1 −0.5 0 0.5 1 1.5\n\nr/l c\n\nFigure I.14 – Numerical resolution of equation I.11 rescaled by capillary length (set to `c\n= 2.7 mm). Curvature radius at the apex of the drop ranging from 1 mm (smallest radius\nof curvature, corresponding to an almost spherical drop) to 10 mm (biggest radius of\ncurvature, corresponding to the most flattened drop). As explained above, the maximum\nheight H for a puddle tends towards 2 `c as the drop volume increases.\n\nthickness of vapor at this point, as experimentally observed [17, 19]. A correction can be\nintroduced in the Leidenfrost state (as well as the air-levitated state) to take into account\nthe underlying vapor film as done in or for the air-levitated situation. We will\nsee in section 2.3 that this deformation can be the cause of an instability.\n\n## Volume of a non-wetting drop\n\nm ∝ R3\nm m ∝ R2\n(mg) 2 Eq I.11\n10 Smooth\nCrenels\n\n1\n10\n0\n10 ℓc\nR (mm)\n\nFigure I.15 – Drop mass as a function of the equatorial radius for acetone. Red dots\nshow experiments done on a smooth surface while black dots show experiments done on a\ncrenelated surface (see section 2.4). Plain red line: numerical solution from equation I.11.\nDashed blue line shows spherical drop approximation for which m varies as R3 . Dotted\nblue line shows disk approximation for which m scales as R2 .\n2. SHAPE 31\n\nFrom the 2D numerical solution of the shape we can easily compute the volume ⌦\nand mass m of a drop as a function of its equatorial radius R. We show in red in figure\nI.15 the comparison between experiments (red dots) and numerical solution (red line) for\nacetone. We also draw in dashed blue an approached solution in the small drop’s regime\nwhere we assume perfect spherical drops, hence m = ⇢ 43 ⇡R3 is proportional to R3 . In the\npuddle regime (dotted blue), assuming disk shapes we would get m = ⇢2 `c ⇡R2 , hence\nvarying as R2 . This volume approximation, although in qualitative agreement, always\ntends to overestimate the data.\n\n## Falling drops: analogy with Leidenfrost drops\n\nFree falling drops bigger than 1 mm are in the aerodynamic regime ( see section 1.3).\nAssuming spherical shapes, we previously deduced their terminal velocity by balancing\ndrag and weight forces. However, for bigger droplets (i.e. bigger than the capillary length),\nthe interface can be deformed and a new geometrical variable is added to the problem. In\norder to understand this new drop geometry, we show in figure I.16 the air flow around\nthe drop.\n\nH R\n\nFigure I.16 – Falling drop of equatorial radius R and terminal speed U in the aerodynamic\nregime. Owing to the Bernoulli effect, the drop deforms perpendicularly to its vertical\ntrajectory. Once the drop has reached its terminal speed, its thickness H scales as the\ncapillary length `c - exactly as in the Leidenfrost scenario.\n\nIn the drop’s frame, the flow is slowed down on the top and bottom of the drop. On\nthe sides, the streamlines get nearer to each other having a local higher speed. In the case\nof a permanent, irrotational and inviscid flow, Bernoulli theorem gives us the following\nrelationship between speed and pressure: ⇢a U 2 /2 + P = cst. This relation states that\nhigh speed points are low pressure ones. In our situation, pressure is lower on the sides\nthan on the top and bottom of the drop. This radial depression pulls the drop’s edge\n32 CHAPTER I. THE NON-WETTING WORLD\n\n## outwards, leading to a puddle shape. At equilibrium, the pulling pressure ⇢a U 2 must be\n\nbalanced by Laplace pressure /H (we assume H << R). The new drop’s thickness H\nnow scales as /⇢a U 2 . Denoting R0 as the drop radius in the original spherical geometry,\ns\nR03 ⇢a U 2\nR⇠ (I.12)\n\nThe terminal speed of this falling puddle is obtained by balancing drag ⇢a R2 U 2 with\nthe drop’s weight ⇢R03 g:\ns\n⇢g`c\nU⇠ (I.13)\n⇢a\nWe now see how the terminal speed for a big drop (equation I.13) is analogous to the\none obtained for small drops (equation I.9), provided that drop radius replaces the capil-\nlary length. In addition, if we plug this terminal speed in I.12, we obtain the equatorial\nR3\nradius for a free falling drop in terminal speed: R2 ⇠ `c0 . In other words, thickness scales\nas in a Leidenfrost situation:\n\nH ⇠ `c (I.14)\n\nThrough this double analogy, we see how free falling drops and levitated drops are\nalike4 . Indeed, we shown in figure I.17 three free falling drops: on the left side, the\nsmaller one (below the capillary length) is spherical, in the middle, a bigger one (above\nthe capillary length) is a flattened puddle and on the right side, a big drop that is being\ndestabilized (see section 2.3). Again, all mechanisms seen for the Leidenfrost drops are\nvisible in the free fall regime.\n\n## (a) (b) (c)\n\nFigure I.17 – Free falling water drops of different volume. The drop in (b) has a horizontal\ndimension 2R = 1 cm. Pictures (a) and (c) have the same scale as (b). Photos from .\n\n4\nReyssat et al. showed in how speed and thickness equations are in good agreement with experi-\nments.\n2. SHAPE 33\n\n## 2.2 The vapor cushion\n\nRecent studies have focused on the thickness h of the vapor cushion beneath the drop.\nBurton and collaborators have been able to experimentally access the profile of the vapor\nlayer under a Leidenfrost drop5 [17, 19]. Although they showed that the interface is\nslightly concave, in what follows we assume a uniform deformation, i.e. a flat bottom\nsurface. As evoked earlier, h is fixed by the balance between the drop’s weight and the\nhorizontal flow overpressure. Once more, two different scenarios are to be considered\ndepending on the drop’s size.\n\nPuddles, R > `c\nVapor flow below drops comes from the evaporation of the liquid. Gottfried et al. showed\nin that the heat transfer at temperatures of 200-500 C is dominated by conduction.\nThe heat obeys the Fourier law and it can be written per unit area: hT where we denote\nthe thermal conductivity of the vapor as and the difference between the substrate and\nliquid temperature as T . After a transient regime (where the drop uses the heat energy\nto reach its boiling temperature), all the incoming heat flux is used to transform liquid\ninto vapor, hence a consumed flux of energy per unit area: Lṁ = L⇢v c where we denote\nthe density and latent heat of evaporation as ⇢v and L, respectively, and the rate of\nevaporation as c.\n\nR\n2ℓc\nc\nh U\n\n## Figure I.18 – Sketch describing evaporation mechanism for a Leidenfrost puddle.\n\nAs sketched in figure I.18, c can also be seen as the vertical speed of vapor injected\ninto the cushion. Balancing consumed and incoming energy, we get a first equation for\nthe expression of vapor injection speed c:\n\n T\nc= (I.15)\n⇢v L h\n\n## A second equation is obtained from a mass balance. As the equilibrium height h is\n\n5\nThey have used similar interferometric techniques than initially used to observe the evolution of the\nair layer between an impacting drop and its substrate [125, 122].\n34 CHAPTER I. THE NON-WETTING WORLD\n\nattained, the vapor injected at rate c from the bottom of the drop is balanced by the\nlateral escaping flux (of horizontal speed U in figure I.18). Hence we have:\n\n1 R\nU= c (I.16)\n2 h\nEquation I.16 tells us that given a typical height h of levitation around 100 mm, the\nhorizontal escaping speed U for a several millimeters drop is ten times greater than the\ninjection speed c.\nA last expression is obtained from the Navier-Stokes equation. Since the Reynolds\nnumber is small, we can neglect inertial effects in the vapor cushion flow. Laplacian\nassociated to viscosity stress ⌘ hU2 is balanced by the horizontal pressure gradient RP ,\n\nU P\n⌘ 2\n⇠ (I.17)\nh R\nCombining equations I.15, I.16 and I.17, we get the equilibrium levitation height h as\na function of the different parameters of the problem:\n\n⌘ T 1/4 1/2\nh⇠[ ] R (I.18)\n⇢v L P\n\nIn the case of liquid puddles, the overpressure below the drop is the drop’s hydrostatic\npressure due to its weight: P =⇢gH = ⇢g2`c . We finally get:\n\n⌘ T 1/4 1/2\nh⇠[ ] R (I.19)\n⇢v L⇢g`c\n\nFor a water droplet, the liquid’s physical parameters have to be taken at 100 C:\n⇢ = 960 Kg/m3 and `c = 2.5 mm. The substrate temperature for a Leidenfrost experiment\nbeing typically 400 C ( T = 300 C), the vapor’s physical parameters are taken at an\naverage temperature of 250 C: ⌘ = 1.8 10 5 P a · s, ⇢v = 0.4kg/m3 and = 36mW/m/K.\nFinally, the latent heat of evaporation is set to L = 2300 kJ/kg. With these values\nand for a drop of radius R ⇡ 1 cm, the typical vapor thickness is about 100 mm. This\nexpression also shows that film thickness h has a very slow dependence regarding the\nsubstrate temperature, since it varies as T 1/4 . In order to double the thickness we\nwould need to multiply the temperature by a factor of 16. For water, it would imply\ntemperatures around 5000 C: despite all the experimental difficulties involved in reaching\nthis temperature (being way above pyrolysis temperature), the heat flux would no longer\nbe dominated by diffusion but by radiation. We can now foresee the advantage of the air\nlevitation technique as it allows us to vary more easily h.\n2. SHAPE 35\n\nExact solution This whole argument can be solved analytically by calculating the\nexact geometric prefactors, as done in . Owing to the circular symmetry of the prob-\nlem, equation I.17 can be written as: 12⌘ hU2 = rP . Taking into account the conservation\nof mass div(hU ) = c and assuming h homogeneous, if we take the divergence of the\n@P\n1 @( r @r )\nPoiseuille equation we get: 12⌘c\nh 3 = div(rP ) = r @r\n. After integrating twice between\nthe bottom center of the drop r = 0 (pressure P2 ) and the outside edge at r = R (pres-\nsure P0 ), we get the following expression for the overpressure: P (r) P0 = 3⌘c h3\n(R2 r2 ).\nBalancing the integral of this overpressure over the bottom surface of the drop with the\nhydrostatic one (and using equation I.15), we get an exact expression for the vapor thick-\nness:\n\n3 ⌘ T 1/4 1/2\nh=[ ] R (I.20)\n4 ⇢v L⇢g`c\n\nLifetime We can also evaluate a puddle’s lifetime ⌧ by calculating the ratio between\nthe droplet’s mass m ⇠ ⇢`c R2 and its mass flow rate ṁ ⇠ ⇢v R2 c, which yields:\n\n⇢`c `c ⇢ L h\n⌧⇠ ⇠ (I.21)\n⇢v c T\nFor a centimetric puddle, this lifetime is around 100 s - long enough to observe it. It\nhas been calculated under the assumption that almost all the evaporation is done through\nthe bottom of the drop. This point has been experimentally demonstrated by obtaining\nsimilar lifetimes with droplets covered at their top with a thin aluminium film to avoid\nany evaporation other than from the bottom .\n\nDroplets, R < `c\nWe now focus on the second size regime concerning small drops. The main difference\ndwells in the area through which evaporation takes place. For big flattened drops, evapo-\nration was assumed to occur at the bottom area R2 (see figure I.18). For small droplets,\nevaporation takes place through the entire surface R2 as sketched in figure I.19.\nMoreover, the bottom is deformed by gravity over a characteristic length scale l ⇠\nR /`c . Since the new horizontal characteristic lengthscale is now l we have to replace\n2\n\nR by l in equations I.16 and I.17. The pressure exerted on the vapor flow due to the\ndroplet is now dominated by the Laplace pressure (not the weight anymore). In equation\nI.17 the pressure gradient has to be replaced by P/l ⇠ R /l, hence a vapor thickness h\nscaling as:\n\n⌘ T 1/4 5/4\nh⇠[ ] R (I.22)\n⇢v L `2c\nThese different theoretical predictions have been experimentally tested and checked\n36 CHAPTER I. THE NON-WETTING WORLD\n\nl\nh\n\nFigure I.19 – Evaporation mechanism for a small drop takes place through its entire area\nR2 . Contact area where drop’s bottom is flattened is noted as l.\n\n## by Biance et al. [11, 12].\n\nWe can now calculate the lifetime of a drop assuming that the temperature gradient\nis established over the whole drop on the length scale R, leading to: ṁ ⇠ LRT R2 . Hence:\n\n⇢LR2\n⌧⇠ (I.23)\n T\n\nFor a water droplet of radius R equal to 0.6 mm, equation I.23 gives a lifetime of 80\ns. This value overestimates the lifetime as experimentally seen by Biance et al. who\ngot a value around 10 s. Despite this overestimation, the model gets the experimental\nT 1 dependency. As the drops become very small (R << `c ), capillary effects completely\ndominate and the bottom droplet does not deform at all, remaining spherical. For droplets\nas big as the characteristic vapor thickness (h . 100 mm) the lubrication theory no longer\napplies. Celestini et al. have theoretically and experimentally shown that the droplet\nin this regime lifts and takes off from the substrate.\n\nThe main ingredient in Leidenfrost levitation is the need of high evaporation rates. We\ncan consequently contemplate changing the type of liquid in order to lower the Leiden-\nfrost temperature as done by using ethanol or even acetone. Almost all our subsequent\nLeidenfrost experiments will therefore be done with acetone. More exotic liquids can also\nbe used: for example, liquid nitrogen and oxygen (whose boiling point is -196 C and\n-183 C, respectively) are already in Leidenfrost levitation at ambient temperature. We\ncan even play with special solids that sublimate at ambient temperature as dry ice or\ncamphor [3, 41].\n2. SHAPE 37\n\n2.3 Instabilities\nSince the drop has a deformable interface, shape instabilities can appear for numerous\nreasons. In what follows we discuss three main different causes.\n\nVapor chimney\nThe vapor cavity beneath the drop becomes unstable for large drops. The gas forms a\nchimney rising at the center of the puddle and forming (transiently) a dome at the top as\nshown in figure I.20 .\n\nFigure I.20 – Side view of a Leidenfrost drop puddle deformed by a chimney instabil-\nity. Vapor underneath is less dense than liquid on top. As the interface is deformable,\nbuoyancy effects make the vapor try to escape upwards and form a dome before bursting.\nScale shows 5 mm.\n\nThe dome then bursts, which leaves a liquid torus that closes, generating strong oscil-\nlations before a new chimney forms. Similar instabilities were predicted for large puddles\nlevitating on air-blown porous materials . This instability can be seen as an inverse\nRayleigh-Taylor instability : instead of having a dense film on a ceiling destabilizing\ndownwards , here we have a light film of gas destabilizing upwards. Gravitational\nforce dominates surface tension when the drop radius is larger than a threshold critical\nradius R⇤ = 4.3`c , in good agreement with observations 6 . The case of even\nlarger drops, a few centimeters or more, is also of interest: multiple chimneys form with\na preferential distance, such as in the Rayleigh-Taylor instability . We will see that\nthis effect limits the maximum drop size in our experiments.\n\nSpontaneous oscillations\nIn both Leidenfrost [111, 110, 121, 108, 101] and air-levitated states , it was reported\nthat an instability leads to spontaneous oscillations, eventually inducing a breaking of\nsymmetry and the appearance of “star drops” as shown in figure I.21 .\n6\nKozyreff et al. also looked at the instabilities in the case of viscous liquids, as discussed in .\n38 CHAPTER I. THE NON-WETTING WORLD\n\nFigure I.21 – Star-shaped drops obtained with liquid nitrogen. A levitating droplet can\noscillate with several different modes. Figure from .\n\nThe characteristic oscillation time is given by the competition between surface tension\n( R) and inertial effects (⇢R3 ⌧R2 ):\ns\n⇢R3\n⌧⇠ (I.24)\n\nFor a water droplet, this time is between 0.01 s and 0.1 s. The instability appears for\nflattened drops yet smaller than the critical radius of chimney apparition. This instability\ntake some time to appear and in none of our experiments we observed these star-shaped\ndrops.\n\nAerodynamic breakup\nA free falling drop can become unstable in an analogous way, even though its environment\nis not initially turbulent. Falling puddles face an airflow coming towards them at speed U .\nThey experience a pressure force proportional to ⇢a U 2 R2 (acting on the axis of symmetry\nof the drop). This force can deform the drop, making an invagination over a characteristic\ndistance as sketched in figure I.22.\nSurface tension opposes this deformation. The local curvature due to a vertical defor-\nmation scale as /R2 , resulting in a Laplace pressure /R2 . This pressure is opposed\nby the aerodynamic pressure ⇢a U 2 , resulting in:\n\n⇢a U 2 R2\n⇠ (I.25)\n\nDrops are unstable if this vertical deformation becomes on the same order of magnitude\n2. SHAPE 39\n\nR\nH\nδ\n\nFigure I.22 – Deformation along the axis of symmetry of a falling puddle due to air\npressure.\n\nas the initial thickness of a falling puddle, that is, if > `c . Inserting this condition in\nequation I.25, and using the expression of the terminal speed of a falling puddle (equation\nI.13), we conclude that puddles are unstable for:\n\nR > `c (I.26)\n\nThis problem can be seen as a variation of the previous chimney instability, high-\nlighting once more the strong analogy between air-levitated (or free falling drops) and\nLeidenfrost drops as previously shown in figure I.17c. By analogy with the Rayleigh-\nTaylor instability, the numerical coefficient in these criteria should be on the order of ⇡,\n\n## 2.4 Crenelated surfaces\n\n(a) (b)\n\nFigure I.23 – (a) Drop of acetone on a hot substrate (T = 400 C) with straight crenels\nof various widths W . W is 1.4 mm for crenels 1 and 7, 1.2 mm for 2 and 6, 1 mm for\n3 and 5 and 0.8 mm for 4. The depth H is 1 mm to make clear the transition between\nlevitation (crenels 2 to 6) and impalement (crenels 1 and 7). (b) Zoom on crenel 5 with\nthe definitions of the distances related to the curvature of the liquid/vapor interface.\n\nWe discussed how vapor or air pressure can deform a drop’s interface if it is strong\n40 CHAPTER I. THE NON-WETTING WORLD\n\nenough to overcome surface tension effects. Here, we explore the deformation due to a\nphysical constraint such as a texture. For that purpose, we etch parallel grooves on a brass\nplate, heat it, and deposit on it a water puddle as shown in figure I.23a. It is important\nto know whether the crenellations can support the liquid sitting on them or not. Three\ngeometrical parameters defining the grooved texture must be explored: the distance W\nbetween walls, the thickness of the walls and the depth H of the grooves.\n\nWidth dependency\nWe denote as ✏ the sagging distance of the drop into the groove, as defined in figure I.23b.\nWe expect a small sagging if the groove is narrow, and a full penetration if it is wide\nenough. To be more quantitative, we show in figure I.24 how ✏ experimentally varies as a\nfunction of crenel width W .\n\nϵ 1000\n(µm)\n800\n\n600\n\n400\n\n200\n\n0\n0 0.2 0.4 0.6 0.8 1\n\nW/ℓ c\n\nFigure I.24 – Penetration ✏ of an acetone drop (`c = 1.6 mm) in a groove as a function\nof the width W of this groove rescaled by thepcapillary length. Vertical red dashed line\nis placed at the critical width value Wc = 2( 2 1)`c calculated from the model I.31.\nPlain red line show numerical solution of equation I.29. Wall thickness is = 0.2 mm\nand depth H is 1 mm.\n\nA critical width Wc separating two different regions clearly appears in the data (vertical\nred dashed line in figure I.24). For W > Wc , we have a plateau regime where full\nimpalement in the texture occurs. The drop falls into the grooves and reaches the bottom\n- set here to 1 mm. For W < Wc , the wider the groove, the deeper the drop penetration\nin the texture, but the liquid/vapor interface remains suspended above the bottom of the\ngroove. In order to understand this dependency we use a hydrostatic argument. Assuming\na depth H big enough, we can consider that the bottom of the channel does not perturb\nthe sagging part of the drop (for the sake of simplicity, we can even assume H infinite). In\nsuch case, the local curved interface between two walls (of radius r defined in figure I.23b)\ncan be at equilibrium if the Laplace pressure /r (we neglect the other perpendicular\ncurvature assuming r << R) is balanced by the hydrostatic pressure ⇢g(2`c + ✏), hence:\n2. SHAPE 41\n\n`2c\nr= (I.27)\n2`c + ✏\nThe interface radius r and the sagging ✏ (figure I.23b) are geometrically related\nthrough:\n2r✏ = ✏2 + W 2 /4 (I.28)\n\n## We can deduce from I.27 and I.28 an equation for ✏:\n\nW2\n2`2c ✏ = (✏2 + )(2`c + ✏) (I.29)\n4\n\nSince there is no analytical solution, we solve it numerically and draw the solution\nwith a plain red line in figure I.24. Although we have a good qualitative agreement, the\nmodel near Wc tends to underestimate the data. This may be due to the approximation\ndone when assuming tangent interface at the corners’ tops. In other words, the interface\ndeformation is supposed to be in first approximation circular, however a catenoid shape\nmay be more accurate.\n\nApproached solution to I.29 can be obtained in the limit of small ✏ (✏ << `c and\n✏ << W ):\n✏ ⇡ W 2 /4`c (I.30)\n\nThe sagging ✏ logically increases when making the crenellations wider, and the critical\nwidth Wc at which the liquid fully penetrate a crenel (r ⇡ ✏) is obtained for r = W/2,\nwhich yields:\n\np\nWc = 2( 2 1)`c (I.31)\n\nWith acetone, we have Wc /`c = 0.84 - in perfect agreement with experiments as shown\nin figure I.24 where the red vertical dashed line is placed at this theoretical value. Wc\nis expected to be 2 mm for water and 1.3 mm for acetone whose smaller surface tension\nmakes it impale more easily in the crenels. To go further in our model comparison we\nshow in figure I.23a a side view of a drop sitting on hot straight channels of width W\nvarying from 0.8 mm at the drop center to 1.4 mm at its periphery. The critical value Wc\nfits with what can be seen in figure I.23a, where acetone only penetrates crenels 1 and 7,\nthe only crenels of width W = 1.4 mm above Wc = 1.3 mm.\n\nDepth dependency\nWe show in figure I.25 a set of images of a drop sitting on a hot surface textured with\nparallel crenels of width W = 1 mm and wall thickness is = 0.2 mm. A different crenel\n42 CHAPTER I. THE NON-WETTING WORLD\n\ndepth H has been fixed for each picture ranging (from left to right) from 200 mm to\n1200 mm. For deep crenels, the sagging part of the drop ✏ remains unaffected by the\nbottom of the groove. However, as shown in the first left image of figure I.25, ✏ is smaller\nfor shallow crenels.\n\nFigure I.25 – Side zoom view of a Leidenfrost puddle drop sitting on crenels of fixed width\nW = 1 mm and wall thickness = 0.2 mm. Depth H from left to right is 200 mm, 400 mm,\n600 mm, 800 mm and 1200 mm. On the left image, where H = 200 mm < h, ✏ is smaller\ndue to the vapor cushion overpressure.\n\n## In order to be more quantitative, we plot in figure I.26 ✏ as a function of H :\n\nϵ 350\n(µm) 300\n\n250\n\n200\n\n150\n\n100\n\n50\n\n0\n0 500 1000 1500 2000\n\nH (µm)\n\nFigure I.26 – Sagging distance ✏ of the drop in the groove as a function of the depth H\nof the groove. In this series of experiments, width is W = 1 mm (W/`c ⇡ 0.6), and wall\nthickness is = 0.2 mm.\n\nTwo asymptotic regimes can be observed and explained. When H is small compared\nto the vapor thickness h on a smooth substrate, everything happens as if there were no\ntexture, hence ✏ ⇡ 0. When H >> h (regime corresponding to the plateau seen for H >\n400 mm in figure I.26), the bottom of the groove does not play any role in the deformation:\nwe are in the situation studied above and ✏ obeys equation I.29. In the particular case\nof figure I.26, the expected theoretical value is ✏ ⇡ 190 mm - as explained above, slightly\nunderestimating the experimental value. In between (H/h ⇠ 1), a more detailed model\nshould be produced to take into account that below the interface we are not anymore at\natmospheric pressure: we have to include the overpressure due to the confined flow.\n2. SHAPE 43\n\n## Wall thickness dependency\n\nThe final geometrical substrate parameter to be explored is the wall thickness. In all the\nprevious experiments, we have set wall thickness to = 0.2 mm. We have tried to see what\nhappens with a texture of same aspect ratio /W = 0.2, but half smaller (W = 0.5 mm,\n= 0.1 mm). The experimental observation was that we could not avoid frequent boiling\nevents, even at high temperature and using acetone.\n\nε\nH\nh\nλ\n\nW\nFigure I.27 – Zoom on a single crenel for a drop sitting on a grooved hot surface deep\nenough to neglect evaporation at the bottom of the sagging part. For visibility, h and\n✏ are not at scale. In red arrows, the escaping flow from the top of a crenel is horizontal\nand perpendicular to the grooves direction.\n\nIn the limit where the vapor film thickness h is smaller than the texture height H (and\nwith < W < Wc ), the drop is suspended at the top of the walls. The whole weight of\nthe drop has to be sustained by the overpressure present on these tops (the bottom of the\nchannel being too far away to contribute through the sagging part). The corresponding\nvapor thickness on top of each channel (denoted as h and defined in figure I.27) is\ncalculated by balancing the drop weight by the pressure arising from the escaping flow,\nwhose direction is from the middle of a wall top, towards the outside (i.e. perpendicular\nto the wall over the distance , as sketched with red arrows in figure I.27). We reproduce\nthe same argument used to calculate the equilibrium thickness of an evaporating drop on\na flat surface (section 2.2) where we simply replace R by and h by h in the previous\nequations I.15, I.16 and I.17. In addition, the weight of the liquid column is now being\ncompensated by the surface area corresponding to the top of the walls, hence only over\na surface fraction ⇠ W + . In the former Poiseuille equation (I.17), the overpressure\nhas to be replaced by ⇢g`c / . The result is a vapor thickness solution analogous to I.19:\n1/4 1/2\nh ⇠ [ ⇢⌘L ⇢g`Tc ] , hence:\nv\n\nh\n⇠ [ ]1/2 1/4 (I.32)\nh R\nIn all our experiments, we have ⇡ 0.17. In the case of a 5 mm radius puddle and\n44 CHAPTER I. THE NON-WETTING WORLD\n\n= 0.2 mm, we get hh ⇡10%, hence h ⇡ 20 mm. Through this simple model, we can see\nthat the drop interface on each wall is close to the substrate. Any small perturbation due\nto the vibration of the interface will therefore easily result in a temporary boiling event.\nEquation I.32 also tells us that even with a fixed aspect ratio, a decrease of 50 % in wall\nthickness will result in a local thickness reduction by 30%. This explains why it was much\nmore difficult to avoid boiling on wall tops with a half-thick texture (keeping the aspect\nratio constant). Because we want to stay in a film boiling Leidenfrost regime, from now\non we fix W = 1 mm and = 0.2 mm, values for which we have experimentally tested\nthat we avoid the formation of vapor bubbles.\n\n## Relationship between mass and radius on a crenelated surface\n\nWe showed in figure I.15 (section 2.1) the volume of a non-wetting drop sitting on a\nsmooth substrate as a function of its equatorial radius. As a consequence, we were able\nto deduce the drop volume from the only observation of its radius (without having to\ndirectly measure it). In order to also be able to access the volume of a drop sitting on a\ncrenelated surface (here of depth H = 200 mm) we add to figure I.15 the corresponding\ncurve. It can be seen in figure I.15 (black data) that for a fixed volume, a drop on a\ncrenelated surface has a bigger equatorial radius. To understand this radius increase, we\ncan assume that thickness from top to bottom of the sagging part of the drop remains\nunchanged compared to that on a smooth substrate (a reasonable assumption as far as\nthe drop is sustained by the top of the walls, i.e. W < Wc ). As a consequence, the volume\noccupied by the walls is relocated in the bulk, which yields a larger equatorial radius.\n3. HIGH MOBILITY AND SPECIAL FRICTION 45\n\n## 3 High mobility and special friction\n\nIf we pour liquid nitrogen on the floor, the resulting drops glide by very large distances,\nmost often comparable to the size of the room (a few meters) where the experiment was\nperformed. The absence of a contact line around a levitating drop makes it “adhering\nnowhere,” as noted by Leidenfrost himself. It generates original dynamical behaviors,\nwhich are also observed in super-hydrophobic situations. These drops move nearly with-\nout friction , and bounce when impacting solids [10, 56, 128]. This section presents\nsome of these special dynamics.\n\n## Unexpectedly, it appears that there is no comprehensive study in the literature devoted\n\nto these frictionless motions, maybe because of the simplicity of the friction laws expected\nin this limit. The details of the corresponding laws depend on the shape of the liquid, and\nfor simplicity we consider here the puddle case (large drops of thickness 2`c and radius\nR). Several scenarios can be expected depending on where the dissipation takes place: in\nthe vapor cushion beneath the drop, in the air surrounding the drop or in the liquid itself.\n\n## 3.1 Viscous friction in the vapor film\n\nIf we consider that dissipation takes place in the vapor cushion, we are in the situation\ndescribed in figure I.28. The “contact” area of a gliding puddle is comparable to its surface\n\nH R\nV\nh\n\nFigure I.28 – Viscous friction in vapor film beneath a gliding levitating drop.\n\n## area R2 so that the viscous friction F⌘ scales as:\n\n⌘a V 2\nF⌘ ⇠ R (I.33)\nh\nThe vapor thickness h was evaluated in section 2.2. Taking its value as independent\nof V , we find a friction of typically 1 µN for a velocity V ⇡ 1 m/s, that is, approximately\n0.1% of the drops’ weight. This is very different from liquids on common substrates, for\nwhich pining forces are comparable to the weight (sticking the drops to tilted plates) and\nfor which viscous forces opposing the liquid puddle motion (for velocities in the range 0.1\n- 1 m/s) can be even larger, also because of the existence of contact lines .\n46 CHAPTER I. THE NON-WETTING WORLD\n\nH R\nV\nh\n\n## Figure I.29 – Aerodynamic friction around a gliding levitating drop.\n\nIn general, the main force resisting the motion of levitating drops moving on a plate\nis the inertia of air. It is similar to the one discussed in section 1.3, except that the\ncorresponding cross section is not R2 , but RH ⇠ R`c as shown in figure I.29. Hence we\nget:\n\nFi ⇠ ⇢a V 2 R`c (I.34)\n\nFor V ⇡ 1 m/s, Fi typically is 10 µN, 10 times larger than F⌘ . For a drop running\ndown a solid tilted by an angle ✓ relative to the horizontal direction, balancing Fi with\nthe weight mg sin ✓ we obtain a terminal velocity V ⇠ [⇢/⇢a Rg sin ✓]1/2 , typically a few\nmeters per second. Conversely, if we throw drops of liquid nitrogen on the floor, they will\ndecelerate until friction stops them. The corresponding distance Li is given by Newton’s\nequation mV 2 /Li ⇠ Fi , from which we get:\n\nLi ⇠ ⇢R/⇢a (I.35)\n\nIf we put numbers in this equation, the characteristic stop length is a thousand times\nthe drop’s radius - several meters. For liquid nitrogen, these distances are so large that\nthe time Li /V to stop is often on the order of (or even smaller than) the evaporation time\n⌧ evaluated in section 2.2.\n\n## 3.3 Special friction on a crenelated surface\n\nFrictionless drops can be seen as untamable mobile objects. It is worth thinking of more\nefficient ways to slow them down and even try to trap them. This can be done by creating a\ntexture at the solid surface. For large textures, comparable to the liquid scale (0.1–1 mm)\n(see figure I.30a) Dupeux et al. showed that a levitating drop meeting a series of\ncrenelations perpendicular to its way decelerates on centimeter-size distances, instead of\nmeters.\nThis strong effect has been proposed to result from the successive impacts of the\n3. HIGH MOBILITY AND SPECIAL FRICTION 47\n\n(a) (b)\n\n10\n\nV (cm / s)\n1\n\n2 4 6\nx (cm)\n\nFigure I.30 – (a) Leidenfrost drop running on a plate with crenelations. (b) The drop\nslows down on centimeter-size distances (solid dots), instead of meters on a flat solid\n(empty dots). This enhanced friction is attributed to the successive (soft) impacts of the\nbumps below the drop onto the side of the crenelations. The distance + W between two\ncrenels is 1.5 mm, and their depth is H = 250 µm. Figure from .\n\nsagging liquid on the crenel sides . The liquid deforms in crenelations, and as the\ndrop slides, each bump hits the side of the crenel without contacting it (no boiling is\nobserved). As known from impact literature , normal shocks are most often soft,\ndespite the absence of an obvious source of dissipation (such as contact lines). Kinetic\nenergy is transferred in vibrations that later decay owing to liquid viscosity . This\ninertial method of slowing down a drop is efficient because it involves the liquid density,\nFor crenelations of depth H and wavelength + W , we have R/( + W ) crenels below\nthe liquid. Each step is hit over a surface area scaling as ✏R (✏ being the sagging distance\nin the crenel). If W > Wc , the drop fully penetrates the texture and we get ✏ ⇠ H. The\ncorresponding crenel friction force scales as:\n\nH\nFc ⇠ ⇢V 2 R2 (I.36)\n+W\nCrenelations increase the friction by a factor ⇢⇢v +WH\ncompared to the aerodynamic flat\nsituation (seen in I.34). Because the density of a liquid is 3 orders of magnitude higher\nthan that of air, the resisting inertial force is increased approximatively 100 times for\nH ⇡ 0.1 ( +W ). The efficiency of the trap relies on the fact that dissipation mostly takes\nplace in the liquid and only marginally in the surrounding air. More precisely, balancing\n2)\nthis inertial friction with the drop deceleration ⇢R2 `c d(V\ndx\nprovides an exponential decrease\nof the velocity along the direction of motion, with a characteristic length:\n\n+W\nLc ⇠ `c (I.37)\nH\nFor a puddle, Lc is a few cm, to be compared to the several meters expected previously\n48 CHAPTER I. THE NON-WETTING WORLD\n\non a flat solid with inertial air friction (equation I.35). This exponential regime is visible\nin figure I.30b and followed by an abrupt trapping in the crenels, possibly due to gravity\n.\nChapter II\n\n## Self-Propulsion in the Leidenfrost state\n\nContents\n1 The texture revolution . . . . . . . . . . . . . . . . . . . . . . . 50\n1.1 The ratchet: a seed is sown . . . . . . . . . . . . . . . . . . . . 50\n1.2 The herringbone: time to reap . . . . . . . . . . . . . . . . . . 57\n2 Force of propulsion . . . . . . . . . . . . . . . . . . . . . . . . . 60\n2.1 Experimental measurements . . . . . . . . . . . . . . . . . . . . 60\n2.2 Analytical calculation . . . . . . . . . . . . . . . . . . . . . . . 62\n2.3 Further considerations . . . . . . . . . . . . . . . . . . . . . . . 65\n3 Friction on grooved topography . . . . . . . . . . . . . . . . . . 68\n3.1 Straight trajectories . . . . . . . . . . . . . . . . . . . . . . . . 68\n3.2 Free trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . 72\n4 Terminal speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77\n4.1 Experimental results . . . . . . . . . . . . . . . . . . . . . . . . 77\n4.2 Analytical calculation and speed optimization . . . . . . . . . . 78\n5 A basic unit of a wider picture . . . . . . . . . . . . . . . . . . 79\n5.1 The drop trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79\n5.2 The active herringbone: the switch . . . . . . . . . . . . . . . . 80\n\n49\n50 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n## 1 The texture revolution\n\nIn 2006, Linke et al. found the first system where a Leidenfrost drop could self-propel.\nThis effect is obtained by texturing a substrate with a ratchet geometry. Linkes’ ex-\nperiment has been a stimulating starting point for a rich theoretical and experimental\nresearch. A wide range of scenarios has been evoked to explain the propelling mechanism,\nranging from local thermal effects, surface waves, to rocket effect and viscous mechanisms.\nTo sort them out, several complementary experiments have been proposed. We first do\na short review of these works in order to understand the underlying mechanisms which\nhave generated elegant novel experiments and configurations. Then, we discuss how all\nthis experiments lead us to achieve a new texture, easier to model and to optimize.\n\n## 1.1 The ratchet: a seed is sown\n\nAs reported by Linke et al., liquids can self-propelled when placed on a hot horizontal\nratchet . Indeed, a drop levitating above saw-teeth (as shown in figure II.1) will\naccelerate perpendicular to the texture, in the direction of the slow local slope of a tooth,\nfrom left to right in the case shown in figure II.1, before reaching a terminal speed. Such\na motion is observed provided that the drop is larger than the tooth size.\n\nFigure II.1 – A liquid drop (radius R = 2 mm) is placed on a hot ratchet with teeth of\ndepth H = 0.2 mm and length = 1.5 mm. For ratchet temperatures larger than the\nLeidenfrost temperature TL , the drop self-propels, from left to right in the figure. The\ninterval time between successive pictures is 40 ms. (Courtesy of Marie Le Merrer).\n\nLinke also reported that two regimes can be observed, depending on the substrate\ntemperature:\n1. THE TEXTURE REVOLUTION 51\n\nLow temperature regime (L in figure II.2). Droplets on a ratchet are not fully\nsupported by a vapor layer: nucleate boiling events introduce fluctuations. We show in\nfigure II.2 drop’s acceleration as a function of substrate temperature. In this narrow\nregime (L), accelerations are large, around 1 m/s2 . A study focusing on this nucleate\nboiling propulsion has been done by a British team in Bath by analyzing the sound\nproduced by their boiling. However, nucleate boiling greatly increases the evaporation\nrate of the liquid, and as a consequence, induces the drop to vanish much faster.\n\nHigh temperature regime (H in figure II.2). The drop is sitting on a vapor cushion\navoiding direct contact with the textures and generating self-propelled Leidenfrost drops.\nAfter a short time of acceleration (on the order of 10 cm/s2 ), typical terminal speed\nis observed around 10 cm/s. This terminal regime is the signature of an equilibrium\nbetween a propelling force (experimentally calculated by multiplying the drop’s mass by\nits acceleration at zero speed) and a resistive friction. Typical values of acceleration being\n50 times smaller than gravity g, the propulsion force is 50 times smaller than the drop’s\nweight. Although small, propelling forces (several µN) are high enough to generate quick\nmotion: this is a direct consequence of the high mobility in the non-wetting state. Linke\nand colleagues suggested the following explanation for this movement: the film of vapor\ndirectly below the region of the drop that is on top of the ridge is most compressed. This\ncauses the vapor to flow away from that spot, both backwards and forwards. But, thanks\nto the asymmetrical shape of the surface, only the flow in the forward direction creates a\nviscous force on the drop, capable of moving it .\n\n1\na\n(m/s2)\n0.8\n\n0.6\n\n0.4\n\n0.2\n\n0\n50 150 250 350 450\n\nΔT (°C)\n\n## Figure II.2 – Droplet acceleration as a function of the difference of temperature between\n\nliquid water and ratchet substrate. Two regimes are separated by a vertical black line.\n“H” is the High temperature regime where we have Leidenfrost self-propulsion. “L” is the\nLow temperature regime where nucleate boiling generates greater accelerations but faster\nevaporation of the liquid. Figure adapted from .\n52 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n## The Dry ice experiment: sorting propulsion scenarios\n\nOther mechanisms explaining this self-propulsion have been considered, possibly relying\non the deformable character of the liquid/vapor interface. Indeed, experiments performed\nby Daniel et al. showed that when a liquid drop is subjected to an asymmetric\nlateral vibration on a non-wettable surface, a net inertial force causes it to move. Others\nscenarios based on the liquid nature of the object are to be found among Marangoni\neffects , spontaneous oscillation of the drop or surface waves. However, Lagubeau et\nal. showed in 2011 that self-propulsion on hot ratchets can also be achieved with dry ice\n(see section 2.2, chapter 1): the dry ice platelet both levitates and is set into motion by\nthe hot substrate ratchet texture in the same direction as observed for the drop. It is\nmade clear by this simple experiment that propulsion does not need the deformable and\nliquid nature of drops [41, 4].\n\n## Visualizing the flow: self-propulsion scenario\n\nTwo remaining opposite scenarios might however still remain. A first one, based on a\nrocket effect and momentum ejection, where the vapor escape would propel the drop in\nthe direction opposite to this ejection. A second one, where the vapor flows in the same\ndirection as the drop, which gets entrained by (vapor) viscosity. We see how crucial it\nis to know the direction of the vapor flow to distinguish between inertial and viscous\npropulsion. In order to make this question clear, Dupeux et al. put small glass bead\ntracers in the vapor flow to assess its direction . The 3D vapor flow was found to have\na complex 3D geometry, but it emerges from this work that the flow starts at the teeth\ntops, flows down the teeth, and gets perpendicularly evacuated along the bottom of the\nsteps, as sketched in figure II.3.\n\nFigure II.3 – Sketch of vapor flow below a Leidenfrost drop or solid on a hot ratchet. The\nvapor flow is cellular: each tooth carries a vapor in the down slope direction, which gets\nperpendicularly evacuated along the bottom of the step.\n\nThis experience establishes that the main vapor flow direction is the same as the\nlevitating object motion, validating the viscous entrainment scenario. As previously seen\nin Chapter I (section1.3), the Reynolds number in the vapor cushion is small and Dupeux\net al. propose the following model based on scaling law arguments. Denoting h as\n1. THE TEXTURE REVOLUTION 53\n\n## the mean thickness of vapor, we expect a propelling force scaling as:\n\n⌘U 2\nF ⇠ R (II.1)\nh\n\nThe Poiseuille equation gives a relationship between shear stress and pressure gradient:\n\n⌘U P\n⇠ h (II.2)\nh R\nCombining equation II.1 and II.2 yields:\n\nF ⇠ P Rh (II.3)\n\nSince the underlying pressure P has to compensate the pressure generated by the\nobjects weight mg/R2 (where m is the object mass) we finally get:\n\nh\nF ⇠ mg (II.4)\nR\nThe ratio h/R is typically 1/50, in good agreement with the experimental results\npreviously discussed where propelling force was 50 times smaller than the drop’s weight.\n\n## A fertile starting point\n\nA whole zoology of new studies has been developed by this Linke experiment. Two main\ndirections of work have been privileged:\n\nScale reduction Marin et al. have worked on ratchets with a texture almost\n10 times smaller compared to Linke’s original experiment: their horizontal pitch being\n140 mm and the crest height ranging between 10 mm and 32 mm. This vertical scale is to be\ncompared to the typical vapor thickness beneath water droplets on flat substrates shown\nin figure II.4 and ranging from 30 mm to 90 mm . In Marin’s experiments, droplets have\na radius R 2 [0.7, 3] mm so that they mostly are in the capillary-dominated regime where\nthe texture height is comparable to the vapor thickness. For big puddles, vapor thickness\ncan be 10 times bigger than the texture used by Marin so that we expect a less effective\nrectification of the vapor flow. They have shown that the viscous mechanism proposed\nby Dupeux et al. fits reasonably well with experiments performed on micro-ratchets, not\nonly for big droplets, but also for capillary droplets, for which the scaling of the force\nwith the droplet radius is inverted .\n\nOk et al. went a step further in size reduction and manufactured micro- and sub-\nmicron ratchets . They showed that sub-micron ratchets yield water droplet velocities\nreaching 40 cm/s if brought at a temperature range slightly above the threshold of droplet\n54 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nh 100\n(μm)\n\nR (mm)\n\n10\n1 10\nℓc R (mm)\n\n## Figure II.4 – Vapor thickness h of a Leidenfrost drop on a smooth surface as a function of\n\nits equatorial radius R. As discussed in chapter I (section 2.2), two regimes are observed\ndepending on the drops radius: if the drop is smaller than the capillary length, thickness\nincreases as R5/4 ; if it is larger, it goes as R1/2 . Capillary length `c is marked on the\nR-axis, with a value of 2.5 mm for water at 100 C. Figure from .\n\nmotion. This dramatic increase in the droplet velocity is attributed to an enhanced heat\ntransfer through the local contacts between ratchet tips and bottom of the droplet - close\nto the observations of Linke in the L-regime of figure II.2.\n\n## Figure II.5 – Chronophotographies of droplet trajectories on a brass ratchet in the tran-\n\nsition boiling regime (where direct contact is observed between liquid and texture) for\nthree different temperatures. (a) Brass temperature at 212 C. A strong deviation per-\npendicular to the main direction is observed due to boiling events and interaction with\nsub-structures. (b) Brass temperature at 241 C, deviation in the opposite direction is\nobserved. (c) Brass temperature at 262 C. By increasing the temperature, we modify\nthe nature of the contact with the sub-structures, hence changing the deviation. The\nhorizontal pitch is 1 mm and the sloping parts of teeth are at 30 (hence a crest teeth at\n600 mm from the bottom-deep structures). Figure from .\n1. THE TEXTURE REVOLUTION 55\n\nFinally, Grounds et al. focused their study on the transition boiling regime . While\nobserving the drops’ trajectories on a ratchet, they recorded the sound produced by\nboiling. From sound intensity they could assess wether they were in the Leidenfrost\nregime or in the transition boiling regime. They showed that the later regime can be\ninduced even at high surface temperatures with acute protrusions. In this regime, droplets\ncan climb steeper inclines perpendicular to the ratchet texture. They also observe that\nsub-structures (generated by the milling process and of typical length scale of a few mm)\ngenerate movement in the direction perpendicular to the motion due to boiling events.\nAs shown in figure II.5, this lateral motion can be tuned by varying the temperature of\nthe surface: the drop does not only moves in the same direction as seen in the Leidenfrost\nregime, but it also has a slow motion along the teeth direction. Grounds et al. showed\nhow two different propelling mechanisms can be combined. In our study, we will only\nconcentrate on the viscous drag regime, hence avoiding by all means nucleate boiling.\n\nLooking for new texture geometries Cousins et al. produced a new texture\ngeometry by exploiting the ratchet effect to construct a “circular ratchet trap” for Leiden-\nfrost drops: a surface with concentric circular ridges, each asymmetric in cross section as\nshown in figure II.6. This new texture was used to study drops’ trajectories as a function\nof the initial speed, and to observe how drops get gradually trapped at the center of the\ndevice due to a combination of propulsion and friction in the textures. They also looked\nat big drops deformations, large enough to generate binary fission of puddles owing to\nimportant internal movements.\n\n2R\n\n2R\n\nFigure II.6 – Sketch of a circular ratchet with a drop trapped in the middle. (Left) Top\nview. (Right) Side view. Figure adapted from .\n56 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nLimitations\n\nThe previous scaling argument qualitatively captures the propulsion mechanism and it is\nin good agreement with the measurements. However, it does not provide the force depen-\ndency towards the substrate geometrical parameters such as tooth angle and horizontal\npitch. In order to take them into account, we have to produce a more sophisticated model\nby adding two more equations to the Poiseuille one: conservation of mass and thermal\nbalance, as previously done to calculate the thickness of the vapor cushion on a flat sur-\nface. All three equations depend on the vertical characteristic distance fixed by the vapor\nthickness h. If we have a closer look at a single tooth (see figure II.7), we observe that\nthe thickness beneath the drop interface ranges from H (the height of a step) at point 2\nto almost zero (at the crest) at point 1. The first limitation of this model dwells on the\nneed to somehow “arbitrary” choose h between 0 and H.\n\nFigure II.7 – Side view of a Leidenfrost droplet on a ratchet. If we focus on a single teeth,\nwe clearly observe that the vapor film thickness ranges from h ⇡ H (near point denoted\n2) to h << H (around point denoted 1). Figure from .\n\nA second limitation of the ratchet geometry is that, although we rectify in average the\nflow in the main propulsion direction, the detailed 3D flow is rather complex (as shown\nin the recent numerical study by Baier et al. ), making a rigorous analytical model\nprobably impossible.\n\nA third limitation dwells on the fact that the ratchet geometry makes it difficult to\npredict beforehand how the flow will be rectified. We put the evaporating object on it,\nand then observe the vapor direction. Meanwhile, the vapor pattern is free to choose its\nown escape path. Hence, it is not easy to understand how the ratchet geometry impacts\nthe vapor flow, and how it can be optimized.\n1. THE TEXTURE REVOLUTION 57\n\n## 1.2 The herringbone: time to reap\n\nLead by previous work, we propose here to create a new self-propelling texture in order\nto reach a double aim: (i) to produce an experiment that would validate the viscous\nentrainment scenario, (ii) to further explore the possibility of novel textures, and optimize\nthem.\n\n## The herringbone Texture\n\n(a) (b)\n\nFigure II.8 – (a) Placed on the symmetry axis of a hot herringbone made of brass (angle\n2↵ = 90 , T = 400 C), an acetone drop (mass m, volume ⌦ = 200 µL) self-propels.\nImages are separated by 0.2 s. The drop acceleration a is measured, from which the\npropelling force F = ma can be deduced. Each crenel has a depth H = 0.2 mm and a\nwidth W = 1 mm. The wall thickness between channels is 0.2 mm. The bar shows 1\ncm. (b) Top sketch of the herringbone geometry with the drop in blue and definition of\nthe geometric parameters.\n\nIn order to force the rectification of the flow, we designed the simplest geometry we\nhad in mind: a herringbone pattern made of grooves. As previously seen in Chapter I\n(section 2.4), a drop on a grooved texture will locally deform, sagging into the top of each\ncrenel, acting somehow as a roof over a channel. Hence, the vapor has no other choice\n58 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nthan to follow the direction of the groove searching for the lowest point of pressure: the\nexit. To achieve self-propulsion we need as a second ingredient a break of symmetry. To\nsatisfy this point, we propose to design a herringbone pattern (see figure II.9 and II.8).\nA drop placed on the axis of the herringbone flees in the direction opposite to the arrows\nof the pattern. As seen in figure II.8a where the snapshots are regularly spaced, the\nliquid first accelerates (with an acceleration a ⇡ 10 cm/s2 ) before reaching its terminal\nvelocity V (10 cm/s in this example) after a few centimeters. Unlike any other existing\nself-propelling texture, we can predict in advance the direction of the rectified flow, as it\nhas to go from the highest point of pressure (right below the center of the drop, hence at\nthe tip of a chevron), towards the lowest pressure point: the exit.\n\nExperimental set-up\nIn order to make sure we are in the film boiling regime, we used acetone. At its boiling\ntemperature (56 C), it has a surface tension = 19 mN/m and a density ⇢l = 751 kg/m3 ,\nwhich gives a capillary length `c equal to 1.6 mm. Its latent heat of evaporation (L ⇡ 520\nkJ/kg), almost four times lower than the one of the water, leads to a lower Leidenfrost\ntemperature.\nWe use blocks of brass (length 100 mm, width 30 mm), and machine them with a\ndigital Arix CNC-milling machine to create the herringbone pattern. Crenellations are\nparallelepipedic cavities of depth H = 0.2 mm and width W = 1 mm, separated by walls\nof thickness = 0.2 mm (figure II.8b). In order to automate the process and to increase\nprecision, we developed a numerical code (written in G-Code language, detailed in A)\nresulting in a final tolerance on these values of 10 mm. A profilometer view of the surface\nshown in figure II.9 allows us to see the sub-millimetric herringbone texture. A much\nsmaller roughness is also generated by the milling process, as can be seen on the bottom\nof each channel where circular patterns appear as a result of the rotating reamer. The\nangle 2↵ of the herringbone can be varied from 0 to 180 , where these extreme values\ndefine grooves parallel or perpendicular to the symmetry axis of the sample.\n1. THE TEXTURE REVOLUTION 59\n\nFigure II.9 – 3D Profilometer view of the herringbone surface texture. Walls’ fibrous\naspect is artificially due to profilometer resolution and does not depict reality. Selection\nzoom: a much smaller circular roughness generated by the milling process can be seen at\nthe bottom of the grooves.\n60 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n2 Force of propulsion\nThe herringbone texture was proved to generate self-propulsion by geometrically forcing\nthe escaping vapor flow. We now experimentally and theoretically focus on the nature of\nthe propelling force and on its dependency towards substrate geometry. After addressing\nthe question of force optimization, we study questions regarding central stability and\npresent some theoretical refinements.\n\n## 2.1 Experimental measurements\n\nPropelling force F can be deduced from the acceleration a (F = ma). The measurement\nis improved by throwing drops from a heated gutter (to make sure they stay in the axis\nof symmetry of the texture) at a velocity V0 ⇡ 5 cm/s in the direction opposite to the\npropulsion: the liquid slows down, stops, and accelerates backwards. The trajectory X(t)\nis parabolic (X(t) = V0 t at2 /2), from which the acceleration a and the propelling force\nF = ma are extracted (see figure II.10). Drops typically lose 1 mg/s by evaporation, so\nthat their mass m (of order 100 mg) can be considered as constant during experiments\nperformed in approximately 1s.\n\n2R\nX\n\nt (s)\n\nX\n(mm)\n\n## Figure II.10 – Trajectory of a drop or radius R = 6 mm on a herringbone surface. The\n\ndrop is thrown from the right in the opposite direction of the propulsion. The liquid then\nslows down, stops (the top picture corresponds to this turnaround moment, marked in\nthe bottom trajectory by a red cross), and accelerates backwards. The trajectory X(t)\nis parabolic (X(t) = V0 t at2 /2), from which the acceleration a and the propelling force\nF = ma are extracted.\n2. FORCE OF PROPULSION 61\n\nComplementary experiments were made to measure the propelling force with a differ-\nent method based on the deflexion of a fiber. We trap by capillarity a drop at the end\nof a wetting vertical fiber, as shown in figure II.11. We measure the fiber deflexion from\nits equilibrium position, as the propelling force acts on it. In thin beam theory, deflexion\nis proportional to the applied force, with a proportionality constant of Ed4 /L3 where E\nis the Young’s modulus of the fiber, d its diameter, and L its length. We can therefore\ndeduce the force of propulsion of the drop from the fiber deflexion.\n\nFigure II.11 – Superposition of two frames of a movie looking from the side at a wetting\nfiber deflected by a drop on a hot herringbone. The left fiber is in the drop of radius\nR ⇡ 4 mm. The right one shows the fiber at rest, after the drop has nearly evaporated.\nThe deflexion is proportional to the propelling force.\n\nAs a great advantage of this method, we can measure the force as a function of the\nequatorial radius through a single experiment by taking a side movie and taking advantage\nof the evaporation. Measurements give results similar to the ones obtained by the first\nmethod. However, several drawbacks made us choose the first method (fitting parabola\non trajectory): (i) drops vibrations generate big experimental errors; (ii) the drop stuck\nby capillarity to the fiber is deformed - biassing the measure. (iii) for big drops, this\nsticking force is not strong enough to overcome the propelling force, hence preventing the\nmeasure in the puddle regime.\nWe can repeat the measurement for various drop radii. We show in figure II.12a\nhow the measured force varies with the drop’s equatorial radius R. The minimum radius\ncorresponds to drops slightly larger than the millimeter-size channel width W , a condition\nfor achieving self-propulsion (otherwise, the drop being smaller than a groove, would fall\ninside). Hence, the drop radii are around or above the capillary length `c , which simplifies\nthe liquid geometry: as seen in figures II.8a and II.10, drops are puddles of radius R and\nheight 2`c , the thickness of puddles in a non-wetting situation (as seen in Chapter I\nsection 2.1). It is observed in figure II.12a that the larger the drop, the stronger the\nforce. The force typically is 5-10 µN, only a few percents of the drops’ weight, yet large\nenough to generate velocities of a few cm/s owing to ultra-low adhesion and friction of\nthe liquid. The force is also a function of the angle ↵ of the herringbone. It is larger\n62 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\naround the “median” angle ↵ = 45 , which suggests that the pattern can be optimized\nin term of propelling force. The variation of F (↵) is shown in figure II.12b for two drop\nradii, confirming the existence of a well-marked optimal angle around 45 .\n\n(a) (b)\n\nF 15 15 ◦◦\n(µN ) 30 ◦\n45 ◦\nF 10\nR = 4. 1 ± 0. 2 m m\nR = 3. 3 ± 0. 2 m m\n60 ◦\n80 (µN )\n8\n10\n\n5 4\n\n0 0\n0 1 2 3 4 5 0 10 20 30 40 50 60 70 80 90\n\nR (mm) α ( ◦)\n\nFigure II.12 – Force F propelling acetone on hot herringbones. (a) F as a function of the\ndrop radius R. Each point is an average on 3-5 experiments, and the colors correspond\nto various pattern angles 2↵ (with ↵ = 15 , 30 , 45 , 60 and 80 ). The channels width\nand depth are W = 1 mm and H = 0.2 mm, and the wall thickness between channels is\n= 0.2 mm. Dotted lines show behaviors in R3/2 , as suggested by equation II.10. (b) F\nas a function of the pattern angle ↵, for two drop volumes. The fits show equation II.10\nwith a numerical coefficient of 0.8 and for the radii R = 4.1 mm (red line) and R = 3.1 mm\n(blue line).\n\n## 2.2 Analytical calculation\n\nWe now discuss these different findings, in order to understand the characteristics of drop\nmotion on these patterns. We first focus on a single channel of length Di (figure II.13 and\nII.8b) and denote the mean horizontal velocity of the vapor flow as U , and the average\nthickness of the vapor as h (H ✏ < h < H), as sketched in figure II.13.\nOn the one hand, vapor is injected at a velocity c (typically a few cm/s) from the top\ninterface. Assuming c independent of the position x along the channel, mass conservation\ncan be written:\n(hU )x+dx (hU )x = cdx (II.5)\n\nOn the other hand, the energy balance writes per unit time and per unit area:\n\n T\n⇢cL = (II.6)\nh\n2. FORCE OF PROPULSION 63\n\nFigure II.13 – Geometry of a channel of vapor (in grey) bounded by a curved liquid/vapor\ninterface (in blue). The vapor flows at an average velocity U , while vapor is injected at a\nvelocity c from the top interface.\n\nwhere we denote the density and thermal conductivity of the vapor as ⇢ and , the latent\nheat of evaporation as L, and the difference between the substrate temperature T and\nliquid boiling point as T . Putting together equations II.5 and II.6 yields the variation\nof the vapor velocity along the channel, taking the axis of symmetry of the herringbone\npattern as the origin in x of the channel:\n\n T\nU= x (II.7)\n⇢Lh2\nR R 6⌘U\nWe expect a driving force per channel Fi ⇡ h\ndxdy , where the numerical coeffi-\ncient corresponds to a Poiseuille flow between two solid plates. The levitating liquid has\na viscosity 50 times larger than the vapor one ⌘, justifying that we treat it as a solid for\nthe corresponding boundary condition. Hence we get, for a channel of length Di :\n\n3⌘ T\nFi ⇡ W Di2 (II.8)\n⇢Lh3\n\nThe force Fi appears to be sensitive to the design of the textured solid, via the param-\neters W and Di . The total force F acting on a drop is obtained by counting the number\nN (↵) of active channels below the liquid. As sketched in figure II.8b, typically 3 channels\n(hence, 6 half-channels, by symmetry) contribute to the motion. More generally, we can\nwrite for thin walls ( << W ): N(↵) ⇡ 4R sin ↵/W . At small opening angles (↵ ! 0),\nN (↵) vanishes. The number of contributing channels vanishes as a consequence of the\nfact that we have to exclude channels open at both sides: vapor flowing on both ways\nproduces a null total viscous force (see in figure II.8b the most left channels, or in figure\nII.15a the top channels). If taking the drop radius R as an average distance Di , we get,\n64 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n## after projecting the different contributions on the central axis:\n\n6⌘ T 3\nF ⇡ R sin 2↵ (II.9)\n⇢Lh3\n\nUntil now we considered h as a constant quantity resulting from the sagging of a drop\nbetween two crenels. However, since the vapor thickness h compares to the channel depth\nH, the interaction with the bottom of the channel has to be taken into account. Although\nin our experiments we vary R on a little range (due to experimental limitations), we gen-\nerally expect vapor thickness in a channel to vary with R - as observed and shown for a\nLeidenfrost drop on a flat solid. The relationship between h and H arises from a balance\nbetween escaping and incoming flux of vapor due to drop pressure and evaporation, re-\nspectively . This leads for the geometry of channels to the relationship\nq (see chapter I,\nsection 2.2): h ⇡ (bR) , where b is a length scale defined by b = ⇢L⇢l g`c . The same\n1/2 3⌘ T\n\n## F ⇡ 2⇢l g`c R3/2 b1/2 sin 2↵ (II.10)\n\nThis expression is exactly the same as the very simple one F / P Rh discussed at\nthe beginning of this section (equation II.3), where P ⇠ ⇢l g`c is the hydrostatic pressure\nof the drop due to its weight and h ⇠ (bR)1/2 is the vapor thickness on a flat surface.\n\nF 102\n= 1 5 ◦◦\n(µN ) α\nα = 30◦\nα = 45◦\nα = 60◦\nα = 80\n\n1\n10\n\n0\n10\n\n0 1\n10 10\n\n2ρ l g ℓ c R 3 / 2 b 1 / 2 s i n 2α (µ N )\n\nFigure II.14 – Force as a function of the theoretical expression found in equation II.10.\nAll previous experiments are represented: radius R ranging from 2 mm to 4 mm, and\nopening angle ↵ 2 [15 , 30 , 45 , 60 , 80 ]. Solid line shows equation y = x in log-log\nscale.\n\nWe show in figure II.14 the force F as a function of the theoretical expression found\n2. FORCE OF PROPULSION 65\n\nin equation II.10 and see how all our data collapse on a single curve. Indeed, equation\nII.10 captures the different observations, and gives the correct order of magnitude of\nthe force. Typical values for the parameters are: ⇡ 13 mW/m/K, ⌘ ⇡ 2 10 5 Pa.s,\nL ⇡ 520 kJ/kg and ⇢ ⇡ 1.8 kg/m3 , which yields b ⇡ 3 mm and thus F ⇡10 µN for a\ndrop of R ⇡ 3 mm, as measured. In addition, the force strongly increases with the drop\nradius, as observed in figure II.12a where the scaling law in R3/2 (dotted lines) is found\nto describe the data. The small variation of the parameter R does not allow us to test\nquantitatively the corresponding scaling law. It is simpler (and of practical interest, since\nit indicates which herringbone generates the highest propulsion) to look at the variation\nof F as a function of ↵, and equation II.9 predicts a maximum for ↵ = 45 , as indeed\nobserved in figure II.12b.\n\n## Discrete to continuous calculation\n\nInspired by this work, a collaboration was held with the university of Trento (Italy) in\nthe team of Nicola Pugno . To obtain the total resulting force from the contribution\nof each channel, they integrated the viscous stress directly on the contact area rather\nthan on each channel. In this continuous method, the channels appear only in an indirect\nway: they determine the direction of the flow of vapor. Figure II.15a shows a comparison\nbetween the two methods of integration. In the channel-by-channel method, the surface\nof integration is approximated by rectangles of width W , whereas, a perfect half-disk is\nconsidered in the continuous method. A graphic comparison of the driving force calculated\nwith the two methods as a function of half the angle ↵ is given in figure II.15b.\n\n(a) (b)\n\nFigure II.15 – (a) Graphical representation of two methods of calculation of the driving\nforce: channel by channel method and continuous method. (b) Force propelling a drop\nof radius R = 3.3 mm according to these two methods of calculation (Discrete in red,\nContinuous in blue). Figure from .\n66 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nThe difference between the results obtained by the two methods is visible; in particular,\nthe position of the maximum is slightly shifted. However, the general characteristics of\nthe curve (global shape and order of magnitude) are not modified.\nInspired by this continuous method we have shown a complex analytical exact solution\nthat can be obtained by calculating the integral of viscous strain over the contact area in\nthe frame of the corresponding grooves direction (see B for detailed calculation). Again,\nwe find that the general characteristic of the curve are preserved.\n\n## The discussion in Chapter I 2.4 on the deformation of a drop sitting in a crenelated\n\nsurface helps us set to the optimal depth H for the channels. We saw that walls of width\np\nW must be below the critical width Wc = 2( 2 1)`c (I.31) in order to sustain the drop’s\nweight by the top of the walls. We experimentally set W = 1 mm near this critical value\n(1.3 mm for acetone), imposing a corresponding average sagging ✏ ⇡ 140 mm.\n\nOnce we have fixed W , we consequently choose the “right” depth channel. If we have\nH >> ✏, the vapor is not efficiently confined. Conversely, at small depth H, the liquid\nlevitates far above the textures, so that the vapor flow rectification becomes negligible.\nHence H must be slightly larger than ✏, that is, 140 mm for W = 1 mm. In this study, H\nis kept constant, equal to 200 mm - larger than the typical vapor thickness h ⇡ 100 mm\nfor a drop on a flat solid.\nAs the surface fraction of walls does not contribute to the propelling force, we try to\nget them as thin as possible. However, as we also want to avoid nucleation of boiling\n(enhanced by very thin walls), we experimentally set = 0.2 mm.\n\nCentral stability\n\nUp to now we focused on the contribution of the viscous force to the propelling direction.\nIndeed, in the perpendicular direction, the projection of the force from each side of the\ndrop compensate by symmetry. However, if the drop slightly deviates from this central\nposition, there is no restoring force that can stabilize the drop along the axis of symmetry.\n\nThe discussion on drop deformation into a crenel suggests a way to guide the liquid.\nA deeper central straight channel of width W ⇡ Wc can be etched along the symmetry\naxis of the herringbone. As a consequence, the drop is gravitationally trapped as sketched\nin figure II.16, without significantly affecting the propelling force F . We will often use\nthis trick, in particular when curved sections appear, which could destabilize the liquid\nby centrifugation.\n2. FORCE OF PROPULSION 67\n\n(a) (b)\n\nW = 1.4 mm > Wc\n\nFigure II.16 – (a) Side view of a herringbone texture with a central groove of depth 0.8 mm\nwider than the critical width Wc ⇡ 1.3 mm. A drop completely sags in this central groove,\nand gets gravitationally trapped along the axis of symmetry. (b) Sketch showing the drop\ntrapped in the central groove.\n\nReynolds number\nOur model was based on small Reynolds number approximation and lubrication theory.\nIn order to discuss the validity of this assumption, we replace the expressions found for\nU h2\nthe horizontal speed U (equation II.7) in the Reynolds number expression Re = ⇢l⌘R\n(discussed in chapter I), which yields:\n\n T\nRe = (II.11)\n⌘L\nFor a given liquid, the Reynolds number only depends on the differential temperature\nT between substrate and liquid boiling point. Using typical experimental values, we\nobtain Reynolds numbers below 0.4, which confirms that viscous effects dominate in this\nvapor escaping problem.\n\nBoundary conditions\nA last point can be made concerning the boundary conditions used in each channel for\nour Poiseuille problem. As previously stated, we assume the liquid as behaving as a solid\nwall (because its density is 1000 times greater than the vapor one). The escaping vapor is\ntrapped in a rectangular channel with solid walls. Experimentally, we measure the force\nby fitting the trajectory near the returning point of the drop: hence verifying the no speed\ncondition (and by the way also avoiding any added inertial friction). However, as the drop\naccelerates, a moving upper wall solid condition should be set equal to the drops speed.\nTypical ejection speed U being 1 m/s and drops speed being ten times smaller, this will\nresult in a slight reduction of the viscous shear at the drops’ interface, hence lowering the\npropelling force.\n68 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n## 3 Friction on grooved topography\n\nAs proposed by Dupeux et al., friction on large textures arise from the soft shocks of\nthe liquid bumps on the crenellation sides [42, 40]. This scenario produces an inertial\nspecial friction F = ⇢l R2 V 2 , denoting ⇢l as the liquid density and a numerical coeffi-\ncient explicitly characterized for normal shocks . Using inclined plates with straight\ncrenels, we checked that the friction is indeed quadratic in velocity, and deduced from\nmeasurements the friction coefficients k ⇡ 0.021 ± 0.003 and ? ⇡ 0.109 ± 0.015 for\ndrops moving parallel or perpendicular to crenels. We logically have k << ? . The case\nof intermediate angles ↵ between the grooves and the trajectory, is then experimentally\nand analytically explored. We finally study lateral deviation of free trajectories for drops\neither with initial speed moving on a horizontal substrate or initially at rest and running\ndown inclines.\n\n## Perpendicular and parallel friction\n\nIn order to study the special friction on textured materials, we used a brass surface of\n200 ⇥ 200 mm2 . We machined parallel crenels, with the same geometrical character-\nistics as for the herringbones used in our study: grooves of depth H = 0.2 mm and\nwidth W = 1 mm, separated by walls of thickness = 0.2 mm. The plates were heated\n(T = 400 C) and tilted by an angle ✓ to the horizontal (see figure II.17), so that the\nlevitating liquid is subjected to gravity. Acetone drops (of fixed volume ⌦ ⇡ 200 µL\ncorresponding to R ⇡ 5 mm, and density ⇢l ) accelerate and reach their terminal velocity\nafter a few cm, when the horizontal projection of the gravity ⇢l ⌦g sin ✓ is balanced by\nfriction.\n\n2R\nH\nV λ\nW\nθ\n\n## Figure II.17 – Drop of acetone of volume ⌦ ⇡ 200 µL and density ⇢l descending a\n\ncrenelated material inclined by an angle ✓ so that the gravitational force along the slope\nis ⇢l ⌦g sin ✓. The corresponding terminal velocity is approximately 10 cm/s, comparable\nto the terminal velocity reached on herringbones.\n3. FRICTION ON GROOVED TOPOGRAPHY 69\n\nThe plate is first displayed with grooves perpendicular to the trajectory (figure II.17).\nWe repeat the experiment varying the tilt ✓ from 1 to 12 , hence changing the drop veloc-\nity and consequently the friction force. By recording the motion, we access the terminal\nvelocity, and plot in figure II.18a the measured friction as a function of the measured\nterminal velocity. The solid line represents the best parabolic fit in a log-log scale. If we\nconsider an inertial friction F = ⇢l R2 V 2 , as discussed above, we can deduce the friction\ncoefficient for grooves perpendicular to the trajectory. We find: ? ⇡ 0.109 ± 0.015.\nThis value is in excellent agreement with the one measured and discussed by Dupeux et\nal. .\n(a) (b)\n\nF F\n( µN ) ( µN )\n2 2\n10 10\n\n1 1\n10 10\n\n1 2 1 2\n10 10 10 10\nV V\n( cm /s) ( cm /s)\n\nFigure II.18 – (a) Friction force as a function of the terminal velocity for grooves per-\npendicular to the trajectory. b) Friction force as a function of the terminal velocity for\ngrooves parallel to trajectory. We notice that the “perpendicular friction” is an order of\nmagnitude higher than the “parallel one”. Both behaviors can be fitted by quadratic laws\nin velocity (inertial friction).\n\nWe can similarly measure the friction force as a function of the terminal velocity for\ngrooves parallel to the motion (figure II.18b). Again, friction is found to be well described\nby a law quadratic in velocity, the signature of an inertial friction. The corresponding\nfriction coefficient k can be deduced from the experiments. We find k ⇡ 0.021 ± 0.003,\na value significantly smaller than ? .\n\nIntermediate angles\nWe repeat this experiment choosing an intermediate angle between the main groove di-\nrection and the down slope direction (denoted as ↵ and as shown in figure II.19). Friction\nis logically minimized when the motion is parallel to the grooves, so that a drop will try\nto align with the grooves. To avoid this lateral deviation and have a straight trajectory\n(aligned with the slope direction), we confine the drop between two walls that compensate\nthe projection of the friction perpendicular to the gravitational force (and represented as\n70 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nF1 F\n\nF2\nα\nmgsinθ\n\nFigure II.19 – (a) Sketch of drop running down a titled crenelated surface. The angle\nbetween grooves and down slope direction is noted ↵. Side walls prevents the drop from\nlaterally deviating by compensating the friction component F2 .\n\nF2 in figure II.19). Friction component aligned with the slope direction will be denoted as\nF1 (figure II.19). We show in figure II.20a this force F1 as a function of the drop velocity\nfor different intermediate angles ↵. Each data point is an average on three experiments\nin order to reduce error bars. For high speeds, we can see a small systematic overestima-\ntion deviating from the parabolic law probably due to the fact that the plate is not long\nenough (200 mm) for the drop to fully reach its terminal speed. We show in figure II.20b\nthe corresponding coefficient ↵ assuming parabolic friction.\n\n(a) (b)\n\n0.12\nF1 α =\n=\n90◦\n7 0 ◦◦\nTh\nα Exp\n(µN ) α\nα\n=\n=\n54◦\n45◦\nβα 0.1\nα = 35◦\n2 α = 2 ◦4\nα = 0\n10\n0.08\n\n0.06\n\n0.04\n\n1\n10 0.02\n\n0\n1 2 0 10 20 30 40 50 60 70 80 90\n10 10\n\nV (cm/s) α( )\n\nFigure II.20 – (a) Friction F1 in the gravity direction as a function of speed for different\nintermediate angles ↵. From left to right curve, ↵ 2 [0 , 25 , 35 , 45 , 55 , 70 , 90 ]. Solid\nlines represent best parabolic fits. Each point is an average on three experiments. (b)\nFriction coefficient ↵ corresponding to each angle ↵. Each point is deduced from data\non left figure (a), assuming F1 = ↵ ⇢l R2 V 2 . The red solid line shows equation II.13.\n\n## As expected, it is bound between the two previous scenarios corresponding to ↵ = 0\n\n(right curve in red in figure II.20a) and ↵ = 90 (left curve in blue in figure II.20a).\n3. FRICTION ON GROOVED TOPOGRAPHY 71\n\n## Back to the herringbone\n\nThe herringbone geometry can be seen as a solution to suppress any friction “perpendicular\nto the motion” without modifying the friction along the central axis. We show in figure\nII.21 a comparison between this two scenarios.\nWe first focus on the friction along the axis aligned with speed V . Both geometries\ngenerate exactly the same friction. Indeed, in the herringbone case, if we “mentally” flip\nupside down the left half-side of the drop (shaded blue in figure II.21b), we end up with\nthe same configuration than sketched in figure II.21a.\nThen, if we now focus on the axis perpendicular to the motion, we observe that friction\ndoes not sum up for the herringbone geometry (such as in the previous case sketched in\nfigure II.21a) but compensate by symmetry. As a direct consequence we do not need\nanymore to put the lateral wall to compensate F2 and prevent deviation.\n\n(a) (b)\n\nF F F\n\nF F\nF\nV\n\nα V\nV V\n\nFigure II.21 – Comparison between parallel grooves (left) and herringbone texture (right).\n(a) Friction can be decomposed in its components parallel and perpendicular to the\ncrenels. Friction aligned with speed writes F1 = F? sin ↵ + Fk cos ↵. Drop velocity along\nthese directions is V? = V sin ↵ and Vk = V cos ↵. (b) If we flip upside down left\nshaded part, we see that friction F1 along the central axis of a herringbone is the same\nas previously shown in figure (a). In the direction perpendicular to the motion, friction\nprojections compensate by symmetry.\n\n## To be more quantitative we decompose the friction in its components parallel and\n\nperpendicular to the grooves, (as defined in figure II.21): F = F? sin ↵ + Fk cos ↵. Since\nthe drop velocity along these directions is V sin ↵ and V cos ↵, we end up with a strongly\nnon-linear expression for the friction:\n\n## from which we deduce the friction coefficient ↵ corresponding to each angle ↵:\n\n72 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n## ↵ = ? sin3 ↵ + k cos3 ↵ (II.13)\n\nWe draw equation II.13 in figure II.20b with a red solid line, and find it is in fair\nagreement with experiments (blue points), provided we set ? and k to the experimental\nvalues obtained with ↵ = 90 and ↵ = 0 , respectively.\n\n## As a logical continuation, we wonder what happens if we do not force anymore straight\n\ntrajectories (with a wall, or with a central symmetry with the herringbone). Two cases\ncan be studied regarding the method used to set a motion. Either we use gravity with a\ntilted plate (as previously done), or we provide an initial speed to the drop by launching\nit from a tilted gutter into a perfectly horizontal plate. We explore these two possibilities,\nstarting by the second one for the sake of simplicity.\n\n## Deviation on an horizontal plate\n\nA 5 mm radius drop is launched from a hot straight gutter. By modifying the slope of\nthe ramp we can select various initial speeds: V0 2 [18,24,36,42] cm/s. The angle between\nincoming direction and grooves axis is denoted as ↵, as defined in figure II.22b: for ↵ = 0 ,\nramp and grooves are aligned, while for ↵ = 90 , ramp and grooves are perpendicular.\nChronophotography showing a typical drop trajectory can be seen in figure II.22b for\n↵ = 60 .\n\nThe drop deviation can be modeled by adapting the argument proposed in the her-\nringbone case. We apply Newton’s second law to the drop in the frame defined by the\ngrooves (e~k being the direction along the crenels, e~? its perpendicular). It writes:\n\n## along e~? : mẍ = ? ẋ\n\n2\n(II.14a)\nalong e~k : mÿ = k ẏ\n2\n(II.14b)\n\nThese non linear second-order differential equations can be solved using as initial speed\nV0 cos ↵ and V0 sin ↵ in the e~? and e~k directions, respectively. Integration of equations\nII.14a and II.14b leads to the expression of speed:\n3. FRICTION ON GROOVED TOPOGRAPHY 73\n\n1\nalong e~? : ẋ = 1\n(II.15a)\n?\nm\nt + V0 sin ↵\n1\nalong e~k : ẏ = (II.15b)\nk 1\nm\nt+ V0 cos ↵\n\nInitial position chosen as the origin finally leads to the expression of position as a\nfunction of time1 :\n\nm ?\nalong e~? : x= ln(1 + V0 sin ↵ t) (II.16a)\n? m\nm k\nalong e~k : y= ln(1 + V0 cos ↵ t) (II.16b)\nk m\n\nCombining equations II.16a and II.16b we can get rid of speed V0 , resulting in the\nfollowing trajectory equation:\n\nm k ?\ny= ln(1 + (e m x 1)) (II.17)\nk ? tan ↵\nEquation II.17 is independent of initial speed V0 , although it still depends on the\nincoming orientation angle ↵. This means that initial speed V0 will only dictate at which\nspeed we scan the trajectory.\nFor small x, this equation greatly simplifies and can be approached by equation\ny ⇡ x/ tan ↵, expected as we know that the drop enters the grooved plate with an in-\ncoming angle ↵. For long runs, where x ! 1, trajectory equation can be approached by\nequation y ⇡ ?k x/ tan ↵. Since ? > k , this means that in the end, the drop tends to\nfollow the grooves. We can reason also in terms of speed: at long times (t ! 1), the ratio\nof terminal speeds along each direction ẋ1 /ẏ1 is equal to the ratio of friction coefficients\nk / ? . Again, since ? > k , this means the drop speed is mainly aligned along e ~k , that\nis, along the grooves.\n\nEquation II.17 gives the trajectory in the frame of study e~? , e~k associated to the\ngrooves. However, if we want to have the trajectory in the frame e~x , e~y (sketched in\nfigure II.22b) aligned with the incoming speed direction, we need to re-express e~? as\ne~? = cos ↵ e~x + sin ↵ e~y and e~k as e~k = cos ↵ e~x + sin ↵ e~y , resulting in the following\nchange of frame matrix:\n\n1\nThis simple model also allows to have the analytical expressions of friction in both axes, deduced\nfrom the speed solutions.\n74 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n!\ncos ↵ sin ↵\nT (↵) =\nsin ↵ cos ↵\n\nWe show in figure II.22a in dashed lines the analytical trajectory solution in this\nnew frame of study, in fair agreement with the set of experiments done at initial speed\nV0 = 42 cm/s. Changing the slope of our launching ramp (hence varying initial speed V0 )\nshould have no impact on the trajectory. However, we have noticed that for low initial\nspeeds, at some point the drop eventually gets trapped by the textures, aligning with\nthem instead of deviating. Indeed, as previously mentioned (Chapter I, section 2.4), the\ndrop deforms into the crenels, lowering its potential energy. As a consequence, there will\nbe a critical speed for which the drop will not be anymore able to overcome this trapping\nforce, hence being forced to follow the grooves.\n\n(a) (b)\n\n0\nα = 0◦ ◦\ny α = 10◦\nα = 20◦\n( mm) α = 30◦\nα = 40◦\n−50 = 50◦\nα\nα\nα\n=\n=\n60◦\n70◦\nex\nα = 80◦\nα = 90\n−100 α\ne e\n−150\n\ney\n−200\n\n−250\n−100 −50 0 50 100\nx( mm)\n\nFigure II.22 – (a) Trajectory in the frame of reference e~x , e~y of the gutter, for a 5 mm\nradius drop with initial speed V0 = 42 cm/s on a horizontal crenelated plate. This plate\nhas a square shape of dimensions 200 ⇥ 200 mm2 , (approximatively) corresponding to the\nsize of the x- and y-axis. Incoming speed direction ↵ ranges from 0 to 90 by steps of 10 .\nExperiments (in solid line) are in good agreement with model from equation II.17 (thin\ndashed line). (b) Drop launched from a tilted gutter enters a horizontal hot crenelated\nplate with an angle ↵ = 60 (corresponding to black curve in left figure II.22a).\n3. FRICTION ON GROOVED TOPOGRAPHY 75\n\n## Deviation of gliding drops on crenels\n\nWe explore here the second method to observe moving liquid puddles: a drop running\ndown the slope by gravity on a hot crenelated plate. As a consequence, we have to adapt\nprevious governing equations II.14a and II.14b by adding a constant force mg sin ✓. The\nnew corresponding differential equations are now non-homogeneous and write:\n\n## along e~? : mẍ = ? ẋ\n\n2\n+ mg sin ✓ sin ↵ (II.18a)\nalong e~k : mÿ = k ẏ\n2\n+ mg sin ✓ cos ↵ (II.18b)\n\nThe addition of the constant driving force implies that we are not anymore able to\nshow an analytical solution.\nNevertheless, if we focus on the terminal regime (defined as having no acceleration\nẍ = ÿ = 0), we get the following expression for the terminal speed in the frame of\nreference of the grooves:\n\ns\nmg sin ✓ sin ↵\nalong e~? : ẋ1 = (II.19a)\n?\ns\nmg sin ✓ cos ↵\nalong e~k : ẏ1 = (II.19b)\nk\n\nHence, the ratio of the terminal speed along each axis is:\ns\nẋ1 k\n= tan ↵ (II.20)\nẏ1 ?\n\n## Equation II.20 (independent of mg sin ✓) predicts an analogous behavior compared to\n\nwhat we have previously seen: since ? > k , once the drop reaches its terminal speed,\nthere is almost no speed in the perpendicular direction e~? (ẋ1 ! 0) so that the drop\nmoves mainly along the grooves. However, in opposition to the previous horizontal case\n(where we found ẋ1 /ẏ1 = k / ? ), equation II.20 depends on the angle ↵. As a con-\nsequence, a special situation is found for ↵ close to 90 (for which tan ↵ diverges, hence\nequivalent to ẏ1 ! 0) where drops can move perpendicular to the grooves.\n\nIn addition, equations II.18a and II.18b can also be numerically solved. We show in\nfigure II.23b the numerical solution obtained for different orientation angles ↵. However, it\nis experimentally difficult to accurately set the initial position of the drop while generating\nit on a slope. As a consequence, arbitrary vertical shifting appears in experiments, making\n76 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n(a) (b)\n\n0 0\nα = 20 ◦◦ α = 0◦ ◦\ny α = 25 ◦ y α = 10◦\nα = 45 ◦ α = 20◦\n( mm) α = 55 ◦ (mm) α = 30◦\nα = 70 α = 40◦\n−50 −50 α = 50◦\nα = 60◦\nα = 70◦\nα = 80◦\nα = 90\n−100 −100\n\n−150 −150\n\n−200 −200\n\n−250 −250\n−100 −50 0 50 100 −100 −50 0 50 100\nx( mm)\nx( mm)\n\nFigure II.23 – Trajectory in the frame of reference e~x , e~y (associated to the direction of\nslope) for a 5 mm radius drop on a tilted crenelated plate. (a) Experimental trajectory:\narbitrary vertical shifting is due to experimental limitations when forming the drop. (b)\nNumerical solution of differential equations II.18a and II.18b for a plate tilted by an angle\n✓: each color corresponds to a different incoming angle ↵ ranging from 0 to 90 by steps\nof 10 .\n\n## it difficult to quantitatively compare numerical results with experiments. Anyhow, a fair\n\nqualitative agreement is observed in figure II.23 between experiments (left) and theory\n(right).\n4. TERMINAL SPEED 77\n\n4 Terminal speed\nOnce propulsion and friction forces have been discussed, we have all ingredients to address\nthe question of the terminal velocity. Before making an analytical model, we experimen-\ntally access this final speed, where friction force balances with the propelling one.\n\n## 4.1 Experimental results\n\nIn order to see the drop accelerate and reach its terminal regime we need enough textured\nsubstrate length. Such long samples are difficult to manufacture keeping crenel depth\nprecision below 10 mm. Typical length of our samples being 100 mm, we designed circuits\nconsisting of two straight herringbones with a stabilizing channel at the center, joined by\nsemicircular tracks bounded by lateral walls as sketched in figure II.24a. Drops hardly\ndecelerate in the curved sections, so that the liquid quickly reaches its terminal velocity\n(around 10 cm/s), and keeps it for several rounds (figure II.24b), until evaporation makes\nits size comparable to the width Wc , which stops the motion.\n\n(a) (b)\ns 140\n!!\"#\n(cm) 120\n\n100\n\n80\nV\n60\n\n40\n\n20\n\n\\$%%&'' 0\n0 2 4 6 8 10 12 14 16\n\nt (s)\n\nFigure II.24 – (a) Long races can be observed by displaying two herringbones in series\n(top view) connected by flat hemicircular paths, allowing us to determine the terminal\nvelocity reached by the drop. The hemicircular lines are lateral walls that guide the liquid\nin this region. (b) Drop position along its race, as a function of time. Data are recorded\non the patterned section of the device (blue points), which shows that the velocity V is\nreached after approximately one herringbone. Here we find V ⇡ 10 cm/s, for acetone\ndrops of radius at mid-race R = 3 mm and for a herringbone angle ↵ = 45 .\n\nThe device shown in figure II.24a can be used to probe the optimum angle in term\nof terminal velocity V . This speed was measured for drops of radius R = 3 ± 1 mm\nand was found to decrease by a factor 3 as ↵ increases from 15 to 75 , as seen in figure\nII.25. Contrasting with F (↵) (see figure II.12b), the function V (↵) is strongly asymmetric,\nshowing a maximum between 15 and 30 . This maximum is roughly equidistant between\n78 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\n45 , where we observe the maximum driving force (figure II.12b) and 0 , where the grooves\nare aligned with the motion, so that we expect a minimum friction (figure II.20b).\n\n20\nV\n(cm /s )\n15\n\n10\n\n0\n0 10 20 30 40 50 60 70 80 90\n\nα ( ◦)\n\nFigure II.25 – Drop terminal velocity on hot herringbones, as a function of the pattern an-\ngle. Drops are made of acetone, substrate temperature is 400 C, and the terminal velocity\nis measured on circuits consisting of successive herringbones (figure II.24a), allowing us\nto reach and measure the terminal velocity even if large. Each data point corresponds\nto an average done on a distance of about 1 m, and the fit shows equation II.21 with a\ncoefficient of 0.8, drawn for R = 3 mm, the typical size of the drop at mid-race.\n\n## 4.2 Analytical calculation and speed optimization\n\nBalancing the propelling force (section 2, equation II.10) with the resisting force (section\n3, equation II.12) yields the expected terminal velocity of drops:\n\nb sin 2↵\nV ⇡ (2g`c )1/2 ( )1/4 [ 3+ 3\n]1/2 (II.21)\nR ? sin ↵ k cos ↵\n\nEquation II.21 is drawn with a full line in figure II.25 for R = 3 mm with a coefficient of\n0.8, as done in figure II.12b. It is found to agree well with the data, having in particular\na maximum clearly displaced to small acute angles, as observed experimentally. The\nmagnitude of the velocity also fits with the calculation.\nMaximizing equation II.21 according to ↵ implies that the optimum angle should be\na function of the ratio k / ? . At small ↵, the function of ↵ in equation II.21 becomes\n2↵/( ? ↵3 + k ), whose maximum is ( k /2 ? )1/3 , that is in our geometry, ↵ ⇡ 25 (within\n2% of the value obtained by numerical integration), in good agreement with the observa-\ntions.\n5. A BASIC UNIT OF A WIDER PICTURE 79\n\n## 5 A basic unit of a wider picture\n\nWe just discussed how we can loop two herringbones through curved sections in order\nto achieve a Leidenfrost racing track. Further on, we propose to see herringbones as\nelementary units of a larger device, which opens new applications and perspectives.\n\n## 5.1 The drop trap\n\nTwo herringbones can be displayed with opposite polarization as sketched in figure II.26a.\nAs a consequence, drops thrown on this device oscillate, as it can be observed in figure\nII.26b where drop position along the drops’ race is plotted as a function of time. The\noscillations are damped by the action of the herringbones, so that the levitating liquid\neventually gets immobilized at the center of the device: opposite herringbones constitute\nefficient traps for these elusive drops.\n(a) (b)\ns 10\n(cm)\n! &'( 5\n\n!\"#\\$%% −5\n\n−10\n0 2 4 6 8 10 12 14 16\n\nt (s)\n\nFigure II.26 – (a) Opposite polarization herringbone facing each other will trap a drop at\nits center. (b) Damping of oscillations for a drop of 3 mm radius at mid-race, launched\ninto the device at 10 cm/s. After a few oscillations (during typically 10 s, smaller than\nthe life time of an evaporating drop), the drop gets trapped in the middle.\n\nAt each side of the center line, the drops undergoes two different regimes. A first one\n(just after crossing the center), where inertial friction and viscous entrainment add up\nand oppose the movement, resulting in a very effective slow down. A second one, after\nthe drops make a U-turn, where these forces are opposite and the drop re-accelerates\nin the opposite direction. Inertial friction ensures energy dissipation resulting in a final\nstill position while the viscous entrainment acts as a restoring force attracting the drop\ntowards the center.\nIf we look closer to the central pattern (see figure II.27 and II.26a), we see concentric\nsquares or diamonds (whose 2 diagonals are the axis of symmetry of two herringbones).\n80 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nWe can therefore imagine a Leidenfrost chessboard in which each box would be made of\nconcentric squares. If we were to spread a liquid sheet over this special chessboard, the\nliquid sheet would destabilize resulting on the trapping of a puddle on each single box.\n\nFigure II.27 – View of the texture resulting from joining two herringbones of opposite\npolarization and facing each other: we see concentric squares whose diagonals are the axis\nof symmetry of the herringbones.\n\n## 5.2 The active herringbone: the switch\n\nAn other direction has been theoretically explored by our Italian collaborators () by\nstraining an elastic substrate on which a herringbone pattern is etched. Using a very\nsimple model of linear deformation, they expect two kinds of deformation regarding if the\nimposed stress is itself symmetric. The deformations which preserve the symmetry allows\na modification of the geometrical parameters, as shown in right side of figure II.28. An\nexample of deformation not preserving the symmetry of the pattern is shown on the left\nside of figure II.28.\n\nFigure II.28 – (Left) Example of elastic deformation that does not preserve the symmetry\nof the pattern: uniaxial tension along an oblique direction (inclination of 60 with the hor-\nizontal, blue arrows represent the applied stress). (Right) Example of elastic deformation\ndecreasing ↵. From .\n5. A BASIC UNIT OF A WIDER PICTURE 81\n\n## Figure II.29 – Diagram of a ’Leidenfrost switch’. Courtesy of Charlotte Rasser.\n\nWhen symmetry is not preserved, they consider the asymmetric pattern as the union\nof two half-herringbones with different half-top angles ↵. Using the previous model for\nthe propelling force with the “continuous” method of calculation, they evaluate the x-\ncomponent of the propelling force on each half side of the drop (the x-axis being orthogonal\nto the axis made by the vertices of the arrows) as a function of the half-top angle ↵. They\nconclude that a drop placed on the axis of an asymmetric herringbone pattern will tend\nto deviate towards the side in which the channels have the biggest angle ↵ (where the\nperpendicular projection is bigger). Taking advantage of this asymmetry, they suggest\nan idea of application presented in figure II.29 where a straight textured central segment\njoins two circular paths patterned with a herringbone texture and a central channel. A\nself-propelling drop moving along the straight central segment will have an equivalent\nprobability to go each side. However, if the substrate is elastic, an asymmetry can be\nintroduced by stretching it, systematically making the drop deviated to one side. They\nconclude that this active pattern could work as a switch.\n82 CHAPTER II. SELF-PROPULSION IN THE LEIDENFROST STATE\n\nConclusion\nHerringbone patterns on solids are found to propel Leidenfrost drops, which can be seen\nas a geometrical proof of the scenario suggesting that viscous drag should generally be\nresponsible for such motions on asymmetric solids. Contrasting with ratchets, the ge-\nometry and the resulting vapor flows are simple and controllable, allowing us to produce\nquantitative models for both the propelling and the friction forces, and to discuss how\nthe design can be optimized. Maximizing the force is useful if it is desired to oppose an\nexisting force (such as gravity, if the solid is inclined); in other cases, it is interesting to\noptimize the drop speed to enhance the motion. The corresponding optimal angles are\nnot the same, but both properties emphasize the role of geometry in these devices. Hence,\nany pattern polarizing the vapor flow should generate propulsion.\n\nWe just have proved that searching new textures can lead to new questions. Always us-\ning a flat substrate, we can imagine other macroscopic geometries such as curved chevrons.\nThe question of textures on substrates that are not flat (hence exploring macroscopically\nthe third space dimension) is also to be addressed. A good starting point would be the\nstudy of Leidenfrost self-propelled drops squeezed between two parallel textured planes\n(a situation recently studied by Celestini and coworkers without textures ). Then, the\ncase of textures inside or outside a cylinder would lead to new questions. We show in fig-\nure II.30 an example of a herringbone texture on a (red stone) cylinder seal of the Jemdet\nNasr period (4th-3rd millennium BC). We see how this simple texture was already used\nlong ago and how the cylindrical substrate was naturally seen as the logical counterpart\nof the herringbone texture on a flat substrate.\n\nFigure II.30 – Herringbone textures on the external face of red stone cylinder seal. Ana-\ntolian, 4th-3rd millennium BC, Jemdet Nasr period. Steatite or chlorite, 1.9 x 1.6 cm.\nHarvard Art Museums/Arthur M. Sackler Museum.\n\nSmaller temperatures could also be tested, in particular after coating the solid with\nhydrophobic micro-textures, which preserves the Leidenfrost state down to the boiling\npoint of the liquid.\nAnother way to extend the effect to cold(er) systems consists of etching chevrons in an\n5. A BASIC UNIT OF A WIDER PICTURE 83\n\nair-hockey table: blowing air through the holes generates levitation, and experiments (see\nChapter III) confirm that plastic cards can indeed self-propel on such patterned tables.\nChapter III\n\n## Self-propulsion on an air hockey table\n\nContents\n1 When vapor is replaced by compressed air . . . . . . . . . . . 86\n1.1 The air hockey table . . . . . . . . . . . . . . . . . . . . . . . . 86\n1.2 Propulsion with herringbone textures . . . . . . . . . . . . . . . 87\n2 Force of propulsion . . . . . . . . . . . . . . . . . . . . . . . . . 90\n2.1 Experimental observations . . . . . . . . . . . . . . . . . . . . . 90\n2.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91\n3 New geometries, new functionalities . . . . . . . . . . . . . . . 99\n3.1 The truncated herringbone . . . . . . . . . . . . . . . . . . . . 99\n3.2 Climbing up a slope . . . . . . . . . . . . . . . . . . . . . . . . 100\n3.3 The viscous entrained mill . . . . . . . . . . . . . . . . . . . . . 101\n4 Channel depth h and Reynolds number . . . . . . . . . . . . . 104\n5 Switching roles: the texture patterned on the slider . . . . . 106\n5.1 Experimental set-up . . . . . . . . . . . . . . . . . . . . . . . . 106\n5.2 Force measurements . . . . . . . . . . . . . . . . . . . . . . . . 108\n6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111\n\n## In collaboration with Hélène de Maleprade. Countless technical advice was provided by\n\nXavier Benoit Gonin (PMMH engineer) regarding laser cutter possibilities and 3D print-\ning.\n\n85\n86 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## Limitations of Leidenfrost Herringbones\n\nTwo main limitations can be stressed in the Leidenfrost herringbone device. The first\none dwells on the liquid nature of the self-propelled object. As a drop is deformable, it\nsags into the crenels generating a special friction. This deformation also implies that it is\ndifficult to clearly define the vapor thickness h. Our best solution was to approach the local\ndeformation by a circular geometry and then average the height over the channel width.\nIn addition, if we want to study the effect of the object geometry on the propelling force,\nwe are constrained by experimental limitations: large drops would destabilize (owing to\nchimney instability, see chapter I, 2.3) and puddle thickness cannot be varied since it is\nfixed to 2`c by the competition between gravity and capillarity.\nThe second constraint dwells on the thermal origin of the effect. The substrate has\nto be brought at a high temperature, significantly above the boiling point of the liquid.\nHence, the liquid itself is at its boiling temperature. Besides the high energetic cost, high\ntemperature can be an issue if we think of transporting living media inside these moving\ndroplets. Finally, we are also constrained by the range of temperatures we can explore,\nmaking it difficult to easily vary the height of levitation.\n\n## A new levitating device\n\nWe designed a new device allowing us to get rid of these limitations by replacing vapor\nby injected air. As a consequence, the liquid used to act as a vapor reservoir is no longer\nneeded. We can from now on consider glass lamellae of controlled geometry (length a,\nwidth b, thickness c, density ⇢g = 2130 kg/m3 , corresponding mass M = ⇢g abc) to play the\nrole of the self propelling object. By these means, we suppress both the high temperature\nand deformable nature of the object.\nThe slider is maintained in levitation by injecting a constant airflow from below\nthrough a porous media, somehow like an air-hockey table. To supply this constant\nflow, we inject compressed air (pressure P2 ) from the bottom of a Plexiglas box device, as\nsketched in figure III.1. The top of this box is closed by a porous Plexiglas plate of thick-\nness e = 2 mm. In order to control the characteristics of this porous wall, we manufacture\nit with a laser cutter (Epilog Helix 24) where we cut a square matrix of through-holes of\nradius r = 100 mm spaced by a pitch p = 400 mm. The velocity of air escaping from the\npores is denoted as w.1\n1\nFor a detailed study, see Gary Leal . In his book, he has theoretically tackled the problem of\nheight levitation h for a given imposed pressure in the case of a circular levitating “puck”.\n1. WHEN VAPOR IS REPLACED BY COMPRESSED AIR 87\n\nP0\n\na\nc\nh P1(x) w\n\n2r P2 p\n\nFigure III.1 – Side view of the air hockey table device. On top of a texture of height\nh = 160 mm, we place a glass lamella of thickness c, width b and length a. Air is injected\nthrough a porous floor consisting in vertical parallel tubes of length e = 2 mm and radius\nr ⇡ 100 mm, spatially distributed in a square array of pitch p = 400 mm. Air injected\nat a pressure P2 from the bottom of the box, escapes from the pores at a speed w. The\npressure in the texture (beneath the glass lamella and above the porous floor) is denoted\nas P1 (x).\n\n## 1.2 Propulsion with herringbone textures\n\nIn order to make the levitating object horizontally move, we use again the laser cutter\nto engrave a herringbone texture on top of the porous substrate, similar to the one done\nin the Leidenfrost case (see figure III.3b). The texture is made of rectangular channels\nof width W = 1 mm, depth h = 160 mm and walls of thickness = 0.3 mm. The angle\nbetween the main direction of the channel and the axis of symmetry of the chevrons is\ndenoted by ↵. The air escapes from the pores into the bottom of each channel.\nWe start all experiments with a pressure inside the box P2 equal to the atmospheric one\nP0 . We then slowly increase P2 (hence injecting air in the channels) until the plate takes\noff. At this point, the top of the walls do not touch the plate anymore. The levitating\nheight is close to the wall texture height h = 160 mm: the slider is skimming over the top\nof each channel. As soon as the plate takes off, it accelerates in the same direction as seen\nin the Leidenfrost case and shown in figure III.3b. We record this motion with a camera\nfrom the top view (figure III.3a). Once the plate has reached the end of the device, we\npush it back several times, resulting in a typical position curve as a function of time shown\nin figure III.2. We decompose the parabolic trajectory in two phases: in green (III.2) the\nslowing down phase, and in red the re-accelerating one. For each round trip we get an\naverage acceleration from which we deduce the propelling force by multiplying it by the\nmass plate. To reduce uncertainty, the force is averaged over the successive parabolas.\nThe special friction due to the sagging of the liquid into the grooves has been here\n88 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n−70\nX\n(m m )\n−60\n\n−50\n\n−40\n\n−30\n\n−20\n\n−10\n0 2 4 6 8 10 12\n\nt (s)\n\nFigure III.2 – Position of the center of mass of the slider as a function of time. The plate\n(a = 37 mm, b = 12 mm, c = 160 mm) is launched several times against the air levitating\nherringbone texture. Green lines show deceleration phases while reaccelerating phases\nare shown in red. The corresponding force F = M Ẍ is averaged over the series of round\ntrips. Here, the obtained acceleration is Ẍ = 5 cm/s2 , which yields F = 7 µN.\n\n## suppressed by the use of a rigid slider. An unfortunate consequence is that we cannot\n\nanymore etch a central wide groove to gravitationally trap and guide the plate. As there\nis no restoring force keeping it in the plane of symmetry, we were obliged to put lateral\nwalls to ensure central stability (see figure III.3a).\n1. WHEN VAPOR IS REPLACED BY COMPRESSED AIR 89\n\n(a) (b)\n2r x\n\nU L\nV\nb\nX\n\nW p p\n\nFigure III.3 – (a) Top view chronophotography (images spaced by 0.2 s) showing the\nacceleration of a glass lamella of length a = 23 mm, width b = 12 mm and thickness\nc = 160 mm. The slider is skimming over a herringbone texture engraved over a porous\nsubstrate through which air is blown. (b) Sketch of the herringbone texture: width of the\nchannel W = 1 mm, thickness of the walls separating each channel = 0.3 mm, angle\nbetween half-channels and axis of symmetry ↵. The texture is above the porous substrate\nand below the glass lamella (length a, width b and thickness c). Inside each channel,\nwe can observe a square array of circular through-holes of radius r = 100 mm and pith\np = 400 mm. The horizontal ejection speed along a channel of axis x is denoted as U . The\nchannel length below the plate is L = b/(2 sin ↵). As soon as the airflow is strong enough to\nmake the lamella levitate slightly above the wall height h, the slider accelerates in the X\ndirection and moves at speed V .\n90 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n2 Force of propulsion\nWe focus in this section on the propelling force. A scaling law analysis shows how we can\nidentify all physical ingredients in order to explain the experiments. Then we produce a\nquantitative model that allows us to predict the pressure needed to generate levitation\nand compare it to experimental observations. A new horizontal length scale arising\nfrom this more sophisticated model permits us to define what a “wide” or “narrow” slider\nis.\n\n## Lamellae geometry dependency\n\nWe first discuss how the propelling force depends on the lamella geometry. We conduct\nour experiments with a fixed herringbone texture: h = 160 mm, ↵ = 45 . For two different\nthicknesses of our lamellae (c = 160 mm and c = 1000 mm), we repeat the experiment vary-\ning a and b. We show in figure III.4a the force of propulsion F deduced from acceleration\nmeasurements for each geometry.\n\n(a) (b)\n\n120 60\nb = 6 mm a × b=3 0 ×1 2 m m 2\nF (µN ) b = 12 m m\nb = 24 m m\nF (µN ) a × b=2 3 ×1 2 m m 2\na × b=1 5 ×6 m m 2\n100 50\n\n80 40\n\n60 30\n\n40 20\n\n20 10\n\n0 0\n0 10 20 30 40 50 60 70 0 10 20 30 40 50 60 70 80 90\n\na(m m ) α( ◦ )\n\nFigure III.4 – (a) Force of propulsion as a function of the length a of the glass lamella.\nEach color corresponds to different widths b 2 [6, 12, 18] mm. The two markers stand\nfor two glass thicknesses: “circles” for c = 160 mm, and “squares” for c = 1000 mm.\nStraight solid lines correspond to a linear fit as suggested by equation III.1 or III.13.\n(b) Force of propulsion as a function of the opening angle ↵ of the herringbone pattern.\nEach curve corresponds to a fixed lamella geometry of thickness c = 1000 mm: “stars”\na x b = 30 x 12 mm2 , “diamonds” a x b = 23 x 12 mm2 , “pentagons” a x b = 15 x 6 mm2 .\nThe solid line shows equation III.13.\n2. FORCE OF PROPULSION 91\n\nWe can observe in figure III.4a that the propelling force seems to be proportional to\nthe lamella length a and independent of its width b. The ratio between the slopes of the\ntwo linear fits is 6.25, very close to the ratio between both thicknesses c (equal to 6.2),\nsuggesting that the propelling force is also proportional to c.\n\n## Impact of the herringbone opening\n\nThen we observe how the force depends on the geometry of the herringbone pattern,\nparticularly on the opening angle ↵ (figure III.4b). We measure the force for various\nopening angles ↵ = [15 , 30 , 45 , 60 , 75 ] for three plate geometries. They all have the\nsame thickness c = 1000 mm, but have different areas: a x b = 30 x 12 mm2 (“stars”),\na x b = 23 x 12 mm2 (“diamonds”) and a x b = 15 x 6 mm2 (“pentagons”). We report F\nas a function of ↵ in figure III.4b. As previously seen in the Leidenfrost case, we observe\na clear maximum around ↵ = 45 and force vanishing as ↵ tends towards 0 and 90 .\n\n2.2 Model\n\nIn order to quantitatively capture all the results, we suggest a model based on a viscous\nentrainment scenario (similar to the one seen in previous chapter).\n\nScaling argument\nAir injected from the bottom of a channel has no other option than escaping in the\ndirection of channels of length L = b/(2 sin ↵). Denoting the gas viscosity as ⌘ and\nthe horizontal speed as U , as shown in figure III.3b, this Poiseuille flow will create on\neach channel a stress ⌧ ⇠ ⌘ Uh that will apply to the bottom of our glass lamella over\na surface area a b. The resulting propelling force only includes the contributions which\ndo not compensate, namely a cos ↵ projection in the plane of symmetry. Hence, we get\nF ⇠ ⌘ Uh ab cos ↵, where U is still unknown. Lubrication theory links the Laplacian of\nspeed to the horizontal pressure gradient through: ⌘U h2\n⇠ LP where L is the length of\na covered channel and P is the difference of pressure between the inside of a channel\nand atmospheric pressure (see figure III.3b and III.1). If we consider that the pressure\nbeneath the plate has to support its weight, we can assume that P scales as ⇢g gc. Hence,\ngch2\nwe get an expression for the gas velocity (U ⇠ ⇢g⌘L ), which gives, once injected in the\nexpression of the force:\n\nF ⇠ ⇢g gc ah sin 2↵ (III.1)\n\nThis equation is exactly the same as the one obtained in chapter II (equation II.10) if\nwe replace the density ⇢ by ⇢g , the thickness 2`c by c, the horizontal characteristic length\n92 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\np\nscale R by a and the film thickness bR by h.\nThis comparison emphasizes an advantage of the air levitated device compared to\nLeidenfrost case: the film thickness h is now an imposed geometrical characteristic and\nnot a variable difficult to model.\nEquation III.1 predicts a linear dependency of the force towards length a and thickness\nc, as observed in figure III.4a. In addition, it also explains the existence of a maximum of\nforce for ↵ ⇡ 45 , as seen in figure III.4b. To ensure the validity of our scaling law analysis,\nwe show in figure III.5 all our data as a function of the force predicted by equation III.1.\nAll the data collapse in a single curve of slope 1.\n\nb = 6 mm\nF (µN ) 2\nb = 12 mm\nb = 24 mm\n10\n\n1\n10\n\n0\n10\n\n1 2\n10 10\nρg cah s i n 2α\n\n## Figure III.5 – Force of propulsion as a function of the expression expected theoretically\n\n(equation III.1). Each point is an average of at least 5 experiments (typically 8). Data dis-\nplayed in this figure correspond to different lengths a = [7.5, 15, 23, 30, 37, 45, 60] mm,\nwidth b = [6, 12, 24] mm (expressed by the color map), thickness c = [160, 1000] mm\n(“circles” and “squares”, respectively). We also added experiments where we vary\n↵ = [15 , 30 , 45 , 60 , 75 ] for three different geometries (same markers and colors\nas in figure III.4b). The blue straight line represents the line of equation y = 1/2 x, that\nis, equation III.1 for the scaling and equation III.13 for the coefficient.\n\nAnalytical resolution\nAlthough the scaling law argument captures all the underlying physics, we can go further,\ntaking advantage of the fact that we control the geometry of the porous through which air\nis injected. Darcy law tells us that the volumetric flow rate of mean velocity w through a\nsingle hole Q = w⇡r2 is proportional to the pressure jump between the inside of the box\nP2 (experimentally controlled and constant) and the pressure P1 (x) at a given position x\nbelow the slider:\n2. FORCE OF PROPULSION 93\n\nP2 P1 (x)\nQ= (III.2)\nRh\n\n## where Rh is the hydrodynamic resistance, equal to 8⌘e/⇡r4 for circular cross-section\n\nchannels (e being the thickness of the porous plate as sketched in figure III.1). This\nexpression is formally the analog of the electrokinetic law between voltage difference and\ncurrent, U = RI. Conservation of mass gives us a second equation that links the injection\nspeed w and the horizontal escaping speed U ; h @U @x\n= pQ2 . These two equations can be\nrewritten as a single one. We have:\n\n@U P2 P1 (x)\nh = (III.3)\n@x p 2 Rh\n\n## Finally, Stokes equation in a channel writes:\n\nU @P1\n12⌘ = (III.4)\nh2 @x\n\nEquations III.3 and III.4 form a system of two equations with two variables. They can\n2 2\nbe rewritten as: @@xU2 = U2 and @ [P1@x\n(x) P2 ]\n2 = [P1 (x)2 P2 ] . Both the horizontal speed U and\nthe pressure along the channel P1 are hyperbolic (exponential) functions that decay over\np\na characteristic distance = 2ep2 h3/3⇡r4 . This distance only depends on the porosity of\nthe plate and on the height h of the wall textures: it is fixed in our experiments and its\nvalue calculated for the parameters of our system is 3 mm.\n\nIn order to have an exact solution, we consider as the boundary conditions a zero speed\nU at the origin of the channel x=0 and a pressure P0 at the exit x = L = b/(2 sin ↵). We\nget:\n\nh2 sh x\nU (x) = (P2 P0 ) (III.5a)\n12⌘ ch L\nch x\nP1 (x) P0 = (P2 P0 )[1 ] (III.5b)\nch L\n\nA similar calculation was done by Gary Leal in , in the case of a flat porous\nsubstrate with a circular levitating puck, in relationship with the question of levitation\nheight of a flat (non-textured) body above a hockey table. The corresponding differential\nequations (Bessel differential equation) are more complicated in this circular geometry,\nresulting in a solution that required to be evaluated numerically.\n94 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## Needed pressure to propel\n\nIt is useful to bear in mind that we have L = b/(2 sin ↵), hence L and b play a similar role.\nWe do not know “a priori” the value of the imposed pressure P2 P0 needed to make the\nslider levitate at height h. We need to take into account that the pressure profile below\nthe plate has to compensate the weight. We can limit this argument to a single wall and\nchannel, which can be written:\nZ L\n⇢g gc L(W + ) = W [P1 (x) P0 ]dx (III.6)\n0\n\n1+\nAfter integration and introducing the function G(x) = 1 thx W\nwe get an expression\nx\nfor the overpressure needed to make a lamella levitate just above the texture:\n\nb\nP = P2 P0 = ⇢g gc G( ) (III.7)\n2 sin ↵\nEquation III.7 confirms that the overpressure needed to induce levitation is propor-\ntional to ⇢g gc, anticipated in the scaling analysis. The mathematical study of the function\nG (for around 3 mm in our case) gives two regimes:\n(i) for long channels (b > ), function G scales as G(x) ⇠ 1 explaining the saturation\nregime P2 P0 ⇡ ⇢g gc shown in figure III.6 at large b.\n2\n(ii) for short channels (b < ), function G scales as G(x) ⇠ x32 leading to G( L ) / 3b2 ,\nexplaining the divergence of the pressure at small b: since b ! 0, pressure has to diverge\nto be able to compensate the weight of the slider.\n\n5\nP 2 −P 0 a = 15 mm\na = 23 mm\nρggc a = 30 mm\na = 45 mm\n4 a = 60 mm\n\n0\n0 5 10 15 20 25 30\n\nb ( mm)\n\n## Figure III.6 – Normalized overpressure P2 P0 imposed to induce levitation, as a func-\n\ntion of the horizontal width b for plates of thickness c = 1000 µm and different lengths\na = [15, 23, 30, 45, 60] mm. We show equation III.7 in solid blue line with no adjustable\nparameter. As b ! 0, pressure has to diverge to compensate the weight of the slider.\n2. FORCE OF PROPULSION 95\n\n## We show in figure III.6 experimental measurements of this overpressure as a function\n\nof the width b for a plate of thickness c = 1000 µm and different lengths a = [15, 23,\n30, 45, 60] mm. We observe almost no dependency towards the distance a. The pressure\nincreases as the width b decreases, in good agreement with the model (equation III.7)\nrepresented with a solid line in figure III.6. The slight systematic underestimation by the\ntheory may be due to the fact that there are some pressure losses, so that actual pressure\nmust be larger than estimated.\n\n## Pressure and speed profile inside the groove\n\nAnalytical solution We can now inject the expression of P2 P0 (equation III.7) in\nour previous profile solutions for the pressure P1 (x) and speed U (x) along the channel,\n\nL ch x\nP1 (x) P0 = ⇢g gc G( ) [1 ] (III.8a)\nch L\nh2 L sh x\nU (x) = ⇢g gc G( ) L (III.8b)\n12⌘ ch\n\nAgain, two regimes can be explored depending on the width b of the plate compared\nto the characteristic decay length .\n\n(a) (b)\n\n1.4 0.2\nP 1 ( x) −P 0 U b = 24. 0 mm\nb = 36. 0 mm\nρggc 1.2 (m/s) b = 48. 0 mm\n\n0.15\n1\n\n0.8\n0.1\n0.6\n\n0.4 0.05\n\n0.2 b = 24. 0 mm\nb = 36. 0 mm\nb = 48. 0 mm\n0 0\n0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1\n\nx/L x/L\n\nFigure III.7 – (a) and (b): Pressure P1 (x) and speed U (x) profile obtained from equation\nIII.8a and III.8b respectively, as a function of the relative position x in the channel of\nlength L = 2 sin\nb\n\n. Three different width b 2 [24, 36, 48] mm are shown. Plate thickness is\nset to c = 160 mm, length a = 15 mm and wall texture thickness ! 0. The characteristic\ndecay length (independent of b) is fixed to 3 mm, so that we are in the “wide” plate\nregime.\n96 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n(i) “Wide” plates: b > We show in figure III.7a and III.7b the analytical profile\nsolution in a channel for pressure P1 (x) (equation III.8a) and speed U (x) (equation III.8a)\nrespectively, for three different widths b 2 [24, 36, 48] mm > . We fix c = 160 mm,\nlength a = 15 mm and wall thickness ! 0 (in order to verify 1+ /W ! 1). A taylor\nexpansion of equations III.8a and III.8b gives the following approximate solution (for the\nsake of simplicity, we will assume here 1 + W = 1.3 ⇡ 1):\n\nP1 (x) P0 x L\n⇡ 1 exp (III.9a)\n⇢g gc\nh2 x L\nU (x) ⇡ ⇢g gc exp (III.9b)\n12⌘\n\n## The characteristic decay length (independent of b, and set to 3 mm in our exper-\n\niment) governs the pressure and velocity profile. The smaller this length, the shorter\ncompared to the channel length L, the sharper the evolution of pressure and speed near\nthe exit of the channel - visible in figure III.7, where the blue curve is sharper than the\nred one. Two channel positions are of particular interest: the exit of the channel (x = L)\nand its origin (x = 0). At the exiting side of the channel, pressure is P1 (L) = P0 (this\nis merely the imposed boundary condition used for the integration). At the origin (cor-\nresponding to the tip of a chevron), pressure is a function independent of L (hence of b):\nP1 (0) ⇡ P0 + ⇢g gc.\n\n## (ii) “Narrow” plates: b <\n\nTaylor expansion of pressure and speed results in this case in:\n\nP1 (x) P0 3 x\n⇡ [1 ( )2 ] (III.10a)\n⇢g gc 2 L\n2\nh\nU (x) ⇡ ⇢g gc x (III.10b)\n4⌘L2\n\nWe show in figure III.8a and III.8b the corresponding speed and pressure profiles for\nwidth b 2 [0.3, 1, 3] < . Pressure is a parabolic decaying function of the ratio x/L. As\na consequence, all pressure profiles are the same if we plot them as a function of the\nproportional channel position x/L: all solid lines in figure III.8a superimpose to each\nother. Speed, in turn, is a linear function of x, whose slope is proportional to 1/b, as can\nbe seen in figure III.8b.\n2. FORCE OF PROPULSION 97\n\n(a) (b)\n\n1.5 7\nP 1 ( x) −P 0 U b = 0. 3 mm\nb = 1. 0 mm\nρggc (m/s) 6 b = 3. 0 mm\n\n5\n1\n4\n\n3\n0.5\n2\n\nb = 0. 3 mm\n1\nb = 1. 0 mm\nb = 3. 0 mm\n0 0\n0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1\n\nx/L x/L\n\nFigure III.8 – (a) and (b): Pressure P1 (x) and speed U (x) profiles obtained from equation\nIII.8a and III.8b, respectively, as a function of the relative position x in a channel of length\nb\nL = 2 sin ↵\n. Channel length (proportional to b 2 [0.3, 1, 3] mm) here is smaller than the\ncharacteristic decay distance , resulting in a parabolic expression for the pressure and a\nlinear profile for speed, as described by equations III.10 for the “narrow” plate regime.\n\nGeneral remarks The choice of the geometrical characteristics of the porous plate\nallows us to play on distance , hence move from a “narrow” plate regime to a “wide”\nplate one. For all our experiments, we deliberately chose the regime of “wide” plates\nwhere the required pressure P (see figure III.6) does not diverge and remains relatively\nsmall.\nIn all cases, maximum overpressure (P1 (0) P0 )/⇢g gc - right below the plate (in the\naxis of symmetry) - ranges between [1, 3/2]. This means that although the imposed\npressure P2 P0 diverges, the pressure inside the channel is always finite.\nIn order to obtain an expression for the needed external overpressure P2 P0 , we stated\nRL\nthat the integral over the surface of the internal channel overpressure 0 [P1 (x) P0 ]dx\nmust balance the weight ⇢g gcL (equation III.6). By doing a simple change of variable in\nthis integral expression y = Lx , we get:\nZ 1\nS= [P1 (y) P0 ]dy = ⇢g gc (III.11)\n0\n\nAs a direct consequence, the surface beneath all curves in figure III.7a and III.8a are\nequal to unity.\n\n## Exact solution for the propelling force\n\nSince we have an analytical solution for the speed (and the pressure) in each channel, we\ncan make the exact calculation of the viscous force applied by the air to the plate. Knowing\nthat the stress of a Poiseuille flow in a channel is ⌧ = 6⌘U/h and using Stokes equation\n98 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## RLRW R0 @P1 (x)\n\nIII.4, we get: Fi = 0 0\n⌧ dxdy = Wh\n2 L @x\ndx. By recalling that P1 (L) = P0 , we\nobtain:\n\nWh\nFi = [P1 (0) P0 ] (III.12)\n2\nAs done in the Leidenfrost situation, we have to take into account the number of\ncontributing channels N (↵) = 2a/( sin\n+W\n\n) and express the projected propelling force along\nthe central axis. Given that all our experiments are performed in the “wide” plate case\n(b > ), we have P1 (0) P0 ⇡ ⇢g gc. We finally get the total propelling force:\n\n1\nF = ah ⇢g gc sin 2↵ (III.13)\n2\nEquation III.13 has the same behavior as the scaling law (equation III.1), but it also\nyields the numerical prefactor, found to be equal to 1/2. Looking back to figure III.5, the\nsolid line which nicely matches the data is equation y = 1/2 x (a straight line of slope\n1/2) predicted by equation III.13.\n3. NEW GEOMETRIES, NEW FUNCTIONALITIES 99\n\n## 3.1 The truncated herringbone\n\nThe model introduces a characteristic distance , which physically represents the exit\ndistance over which pressure and speed changes mainly take place. Since all our experi-\nments were made in the “wide” plate regime (b > = 3 mm), we expect almost all viscous\nentrainment to take place near the 3 last mm of the channel.\nAs a consequence, the central geometry of the texture should not play a key role.\nWe have tested this idea by comparing forces between the “classic” herringbone and a\n“truncated” version. This new texture is shown in bottom of figure III.9 and consists in\na herringbone pattern where the tip is now replaced with a straight section of length bT\n(fixed to 10 mm) perpendicular to motion:\n\n10 15\nbT b\nmm mm\n\nFigure III.9 – Comparison between two textures entraining a glass slider of length\na = 30 mm, width b = 15 mm and thickness c = 1 mm (resulting in a weight M around\n1 mg). Top texture is the “classical” herringbone. Bottom one corresponds to the “trun-\ncated” herringbone, which has a central straight section of length bT (fixed to 10 mm)\nperpendicular to motion, hence having a null viscous propelling contribution.\n\nWe show in figure III.10 the comparison of the trajectory of a glass slider (length\na = 30 mm, width b = 15 mm and thickness c = 1 mm) launched against the entrainement\nforce direction (launched from left to right on the texture represented in figure III.9) for\nboth textures. As already discussed, the entrainment force will first slow down the slider,\neventually stop it and finally reaccelerate it resulting in a parabolic trajectory (friction\nis negligible). We qualitatively see that the trajectories are very similar, indicating that\nboth devices have a similar propelling efficiency although they have very different textures.\nIn more detail, we also observe that it take less time to cover the same distance to the\n“classical” herringbone (in blue) than to the “truncated” herringbone (in red), indicating\nthat “classical” herringbone texture is slightly more efficient.\n100 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n50\nX\n(m m )\n40\n\n30\n\n20\n\n10\n\n0\n0 0.5 1 1.5 2 2.5 3 3.5 4\n\nt (s )\n\nFigure III.10 – Trajectory comparison between the two propelling textures exposed in\nfigure III.9. In red, the trajectory corresponding to the “truncated” herringbone, shown\nto be slower than the one in blue (corresponding to the “classical” one).\n\nThis strong similarity is in good agreement with the assumption that only the last\nsection of each channel plays a role in the propulsion, hence the central part of the tex-\nture (either a pointed tip or a straight one) do not have a strong contribution to propulsion.\n\nMore quantitatively, we can deduce from a quadratic fit of the trajectory the acceler-\nation equal to X¨T ⇡ 3 cm/s2 and Ẍ ⇡ 4.5 cm/s2 for the “truncated” and the “classical”\nherringbone, respectively. Since the plate is the same in both situations (and has a\nweight M around 1 mg), we deduce a corresponding force F equal to FT = 30 µN and\nF = 45 µN for the “truncated” and the “classical” herringbone, respectively. If all surface\narea contributed to propulsion, we could expect a reduction in propulsion proportional to\nthe reduction of area breaking the symmetry scaling as (b bT )/b, that is, around 30 %.\nHowever, as qualitatively discussed previously, the ratio of propelling forces FT /F is much\nhigher (around 70 %) which confirms our assumption of viscous entrainment contribution\nto propulsion being concentrated at the end of each channel.\n\n## 3.2 Climbing up a slope\n\nFrom the expression of the propelling force (equation III.13), we deduce the expression of\nacceleration Ẍ for a plate:\n\nh\nẌ ⇡ g sin 2↵ (III.14)\n2b\n3. NEW GEOMETRIES, NEW FUNCTIONALITIES 101\n\n## Figure III.11 – Lamella (a = 30 mm, b = 6 mm, c = 160 mm) withstanding a slope of\n\n2% owing to viscous entrainment. From this lateral view we can distinguish the holes\nthrough which air is injected at the bottom of each chevron.\n\nFor the sake of simplicity, we assume ↵ = 45 , the optimum angle in term of propelling\nforce. Then, equation III.14 tells us that plates can climb slopes up to ✓ ⇡ 2bh\n, that is,\nup to several % in our typical geometries. To draw a comparison, no mountain stage in\nthe world famous “Tour de France” exceeds a 10 % slope. For instance, we show in figure\nIII.11 a plate of width b = 6 mm withstanding a slope of 1.3 % where viscous entrainment\nbalances the weight.\n\n## 3.3 The viscous entrained mill\n\nAll propulsion movements seen up to now lead to straight translation. The longer we want\nto observe the movement, the larger textured substrate we need. In the case of drops, we\nsolved this problem by looping two straight textured surfaces through flat semicircular\nsections. In addition, we used a central deeper groove to gravitationally trap the drop\nand be able to guide it, allowing us to reach (and study) the terminal velocity regime.\nIn the case of non deformable glass lamellae, this set-up is much more complicated to\nachieve. In contrast, rotation offers great possibilities since the moving objects remain\nin place and can be observed as long as we want. Based on this idea, we developed the\ntexture similar to a windmill visible in figure III.12. A plane is divided in four equal\nsections in which parallel grooves are engraved. Between each quadrant, a 90 rotation of\nthe main direction of the grooves is imposed. Consequently, viscous entrainment acts on\nthe levitating plate in a different direction for each portion (see arrows in figure III.12a),\nwhich generates a rotation.\nThe previous scaling argument gives us the propelling force generated by one channel:\nFi ⇠ ⇢g gc hW . If we denote as 2b the side length of a square plate (see figure III.12a),\neach forth of the plate has N (↵) = b/( + W ) channels beneath it. The total resulting\nforce in each quarter being: F1/4 ⇠ ⇢g gc hb. Total torque experienced by the plate will\ntherefore scale as:\n\nM ⇠ ⇢g gc hb2 (III.15)\n102 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n(a) (b)\n\nb ∆ b\n\nc\nh\ne\n\nFigure III.12 – (a) Windmill texture consists of four square sections textured with parallel\ngrooves of width W = 1 mm, walls of thickness = 0.3 mm and depth h = 160 mm. A\nvertical glass fiber acting as central axis ( ) keeps centered a PMMA plate of density 1190\nkg/m3 , thickness c = 2 mm and side 2b = 30 mm. Red arrows show viscous entrainment\ndirection in each quadrant resulting in rotation (terminal angular speed denoted as ⌦).\n(b) Sketch: side view of the setup. Three vertical lengths are indicated: e stands for the\nporous plate thickness, h for the walls’ height and c for the slider thickness.\n\nThe moment of inertia J of a square plate scales as mb2 and the second Newton’s law\n¨ M = J ✓.\ngives us a relation between torque and angular acceleration ✓: ¨ Hence we get:\n\nh\n✓¨ ⇠ g 2 (III.16)\nb\nIf we put numbers in this equation, we obtain h/b2 ⇠ 1 m 1\nand typical angular ac-\nceleration of 10 rad/s2 - quite large indeed!\n\n## As of now, imposed differential pressure P = P2 P0 (between the external at-\n\nmospheric pressure P0 and the inner one P2 ) was fixed and carefully chosen to match\nlevitation height and wall texture height h, as sketched in figure III.12b. This new setup\nallows us to easily probe the role of imposed pressure (remained unexplored up to now).\nWe show in figure III.13a the terminal speed of rotation as a function of this overpressure.\nTwo regimes separated by a critical transition can be observed:\n(i) a first regime (below 35 mbar in figure III.13a) where the imposed pressure is not\nstrong enough to induce levitation. Owing to solid friction, there is no movement at all.\n(ii) a second regime (above 35 mbar) where pressure is strong enough to withstand the\nplate’s weight, which generates rotation. If we increase air injection, the plate levitates\nat height z higher than the walls’ height h. From a side view (see figure III.13b), we\ncan consider that the flow is divided in two regions. A lower one (flow in dotted blue),\n3. NEW GEOMETRIES, NEW FUNCTIONALITIES 103\n\nrectified by the walls of height h and an upper one (between h and z, marked with red\ndotted arrows), where the flow is free and will have a symmetric pattern, resulting in a\ndecrease of the propelling force.\n\nTerminal speed and propelling force are optimized when the slider completely seals\neach channel, i.e. when it levitates at height z = h (around 40 mbar in figure III.13a).\n\n(a) (b)\n\n25\n\n15\n\nz\n10\nh\n5\n\n0\n0 20 40 60 80 100\n\n∆P ( m bar )\n\n## Figure III.13 – (a) Terminal angular speed ⌦ = ✓˙ of a plate on a windmill pattern, as a\n\nfunction of the imposed differential pressure P = P2 P0 for a square PMMA plate of\nside 2b = 30 mm and thickness c = 2 mm. Optimum of speed is found around 40 mbar,\nwhere the slider levitates just above the walls, hence efficiently confining and rectifying the\nflow. (b) Side view of the rotating slider: as the imposed pressure increases, the levitating\nheight z increases. As a consequence, rectification of the flow is less efficient and terminal\nspeed decreases. Red dotted arrows show the isotropic flow above the textures. Blue\ndotted arrows show the main direction of the flow rectified by the textures.\n104 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## 4 Channel depth h and Reynolds number\n\nIn all our experiments, the crenel depth h is fixed and constant. Increasing the object’s\nthickness (or analogously increasing the crenel depth h) results in an increase of the needed\ninjection speed w. Eventually, there will be a point where the horizontal speeds will be\nso high that the low Reynolds number approximation will not be satisfied anymore. From\nthe speed solution obtained in equation III.8b, we can calculate the maximum horizontal\n2\nspeed. By injecting it in the Reynolds number expression Re = ⇢U⌘Lh , we get its explicit\nexpression:\n\n⇢⇢g gch4 L\nRe = 2 2\nH( ) (III.17)\n12⌘ L\n\n## where the function H is H( L ) = L G( L )th L . We show in figure III.14 this expression as\n\na function of the channel depth h for different widths b and a fixed thickness c = 1 mm.\nTwo asymptotic regimes can be observed:\n\n## For “narrow” plates b << (/ h3/2 ) we have H( L ) ⇠ 3, leading to a Reynolds number\n\nproportional to h4 (dotted line in figure III.14):\n\nReb< / h4 (III.18)\n\n## For “wide” plates b >> (/ h3/2 ), H( L ) reduces to L\n\n, hence we have (dashed line in\nfigure III.14):\n\n## Reb> / h5/2 (III.19)\n\nOur model is based on a low Reynolds number assumption (region below the horizon-\ntal solid black line in figure III.14). In our experiments, the worst scenario (where the\nReynolds number is highest) corresponds to the thickest and most narrow plate (c = 1 mm\nand b = 6 mm, red curve in figure III.14), for which we indeed have Re . 1 (given that\nwe have fixed h = 160 mm). All other plates have smaller Reynolds number (Re 0.3),\nvalidating our assumption of viscous scenario.\n4. CHANNEL DEPTH H AND REYNOLDS NUMBER 105\n\n5\nRe 10\nb= 6 mm\nb =12 m m\nb =24 m m\n\n0\n10\n\n−5\n10\n0 1 2 3 4\n10 10 10 10 10\n\nh(µm)\n\n## Figure III.14 – Reynolds number as a function of channel depth h as predicted by equation\n\nIII.17 . Thickness is fixed to c = 1 mm (corresponding to our thickest plates, hence highest\nexperimental Re numbers) and three plates of width b 2 [6, 12, 24] mm, as indicated with\ncolors. The dashed black line shows asymptotic behavior h5/2 described in equation III.19\nfor “wide plates” (also corresponding to shallow crenels). The dotted black line shows\nasymptotic behavior h4 predicted in equation III.18 for “narrow” plates (corresponding\nhere to deep crenels).\n106 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## 5 Switching roles: the texture patterned on the slider\n\nWe thank Steffen Hardt and Tobias Baier for enlightening discussions concerning this\nsection.\n\n## 5.1 Experimental set-up\n\nFor heavier plates or deeper textures, we expect to switch to a regime where inertial effects\ncan dominate viscous ones. To generate movement, inertial thrust need a surface upon\nwhich to exert pressure: a role in our case played by the walls of the herringbone pattern.\nIf walls are bound to the air levitating table we do not expect anything to happen. Since\nforces are in the range of µN , table’s weight would be too big to move. Conversely, if\nwalls are anchored to a light glass lamella, we see the textured slider move in the opposite\ndirection compared to the one observed up to now, as shown in figure III.15:\n\n## Figure III.15 – Chronophotography of a textured slider of width b = 12 mm, length\n\na = 30 mm and glass thickness c = 1160 mm moving on an air-hockey table. Herringbone\ntexture has been added to the lamellae by creating walls of depth h = 800 mm (resulting in\na total thickness of c+h = 1960 mm). For visibility reasons, we have highlighted in red one\nof the channels of the slider. Plate levitates and moves due to air that is blown from the\nunderlying substrate through holes of same dimension as previously (pitch p = 400 mm,\nradius r = 100 mm). Each image is separated by 0.2 s. Movement takes place in the\nopposite direction than for viscous entrainment.\n\n## This experiment could be viewed as similar to the previous situations. However, a\n\nslight difference dwells on the fact that air was previously injected from the bottom of the\nchannels. Now, we blow air onto the whole textured slider, in particular also on the walls.\n5. SWITCHING ROLES: THE TEXTURE PATTERNED ON THE SLIDER 107\n\nAs a consequence, it is more difficult to make the slider raze to the ground, hence needing\nhigher pressure to start levitation and allowing secondary symmetric flows to exist.\nIn order to create a texture on the slider, we tried several techniques:\n(i) we first engraved on a plexiglass lamella the texture with the laser cutter. However,\nwe faced a main problem since the plate bends after engraving the pattern (when the\ntexture was engraved on the air-hockey table this problem was avoided by fixing the\nextremities of the plate to the substrate with screws, ensuring perfect horizontality).\n(ii) we then tried 3D-printing technology (Fortus 250mc printer). The technology is\nFused Deposition Modeling (FDM) where objects are produced by extruding material\nwhich harden immediately to form layers. A thermoplastic filament that is wound on a\ncoil is unreeled to supply material to an extrusion nozzle head. The nozzle head heats the\nmaterial and turns the flow on and off. The nozzle can be moved in both horizontal and\nvertical directions by a numerically controlled mechanism. The object is built bottom\nup, one layer at a time following the designed 3D pattern (see figure III.16). The highest\nresolution we could get using acrylonitrile butadiene styrene as construction material\n(ABS-P430) did not allow us to make walls thinner than 350 mm, that is, too large. In\naddition, the final surface was too rough, generating high solid friction against the air\nblowing substrate.\n\nFigure III.16 – 3D image of herringbone slider designed with Computer Assisted Concep-\ntion software (Catiar ). The solid body has a thickness c = 1 mm, a width b = 12 mm\nand a length a = 30 mm. A herringbone texture of height h = 800 mm is added on the\nbottom.\n\n(iii) we finally followed a two steps method based on creating a negative stamp of our\nbrass herringbones (see for additional information). A mixture (10:1 in weight) of\nPolydimethylsiloxane (PDMS, a liquid organosilicon compound) and cross linking agent\nRTV-Silicone (Room Temperature Vulcanizing silicone) is poured over the master mould\n(namely the brass herringbone) and placed in a furnace at 70 C for 1 hour. Once the\nPDMS is hardened, we obtain a negative replica of our master mould. We recreate the\ninitial texture by pouring UV curing optical adhesive (NOA-61)2 in the PDMS mould.\n2\nNOA-61 is designed to give the best possible bond to glass surfaces and may be polished after curing.\nTo cure these optical adhesives, they must be exposed to UV light. Recommended energy required for\n108 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\n## After recovering it with a glass lamella of dimension a x b x c, we expose it for 30 s under\n\nan UV light of power 100 mW/cm2 . As a consequence, we obtain a glass lamella of the\ndesired dimension on which walls are bond to the glass forming a herringbone texture.\nWith this technique, we avoid bending, thick walls and roughness problem previously\nencountered.\n\n## 5.2 Force measurements\n\nWe made a set of plates with different geometrical characteristics. Four widths were\ngenerated: b 2 [6, 12, 18, 24] mm. For each one, three glass thicknesses were produced:\nc 2 [160, 1160, 2160] mm. Lamellae length a was not varied and was set to 30 mm.\nRegarding the herringbone texture, wall thickness was fixed to = 0.2 mm, spacing to\nW = 1 mm and opening angle ↵ to 45 . Wall height varied in the range of h 2 [150, 250,\n400, 800, 1600] mm. With this set of 48 different plates, we measured the propulsion force\nwith the same method as previously used. Once again, air injected pressure was carefully\nchosen: we start all experiments from no overpressure and keep increasing it until the\nvery first moment the textured plate starts to levitate (and move). We show in figure\nIII.17 this force as a function of wall height h for the whole set of sliders.\n\nb = 6m m c = 160µ m\nc = 1160µ m\n−F ( µN ) 3 c = 2160µ m\n10 b = 12m m\nb = 18m m\nb = 24m m\n\n2\n10\n\n1\n10\n\n0\n10\n2 3\n10 10\n\nh ( µm )\n\nFigure III.17 – Force of propulsion as a function of wall height h. Each color represent a\nthickness c 2 [160, 1160, 2160] mm. Each symbol represent a width b (circles for b = 6 mm,\ntriangles for b = 12 mm, diamonds for b = 18 mm, squares for b = 24 mm). Length a is\nfixed to 30 mm and ↵ = 45 . For a given plate thickness c (i.e. a given color), each cluster\ncorresponds to a wall height h 2 [150, 250, 400, 800, 1600] mm. Compared to previous\nviscous entrained experiments, the propulsion direction is opposed, hence the negative\nsign of F . Solid lines show best linear fit, as suggested by equation III.21. Each point is\nan average of at least 5 measurements.\n\n## full cure is 3 J/cm2 in the range of 350 nm wavelength.\n\n5. SWITCHING ROLES: THE TEXTURE PATTERNED ON THE SLIDER 109\n\nIn these experiments, the propulsion direction is opposed to the one observed when\nthe herringbone texture was patterned on the air blowing device. Experiments suggest a\nlinear dependency of the force as a function of wall height h and glass thickness c, and\nwidth b seem to play little role.\nA simple argument based on inertial scenario can be produced to qualitatively cap-\nture all these results. If we first focus on a single crenel as shown in figure III.18, the\nunderlying pressure has to compensate the weight of both the glass plate (⇢g cg) and the\nwalls (⇢N hg W + , where ⇢N is the density of NOA and the term W + takes into account\nthe surface density of walls). Assuming all the vertical momentum is transformed in hor-\nizontal one, the same dynamic pressure acts on a surface area perpendicular to the plate\nscaling as W h (see red and blue surfaces in figure III.18).\n\nFigure III.18 – Sketch of the flow in a single crenel. Dotted lines indicate air direction.\nSolid arrows show the direction of the force resulting from the dynamic pressure acting\nover the vertical colored surfaces. By symmetry, all (absolute) magnitudes are equal\nbetween red and blue colors.\n\nTotal propelling force along the axis of symmetry is obtained by multiplying the pro-\njected force in a single crenel by the number of contributing channels N (↵) ⇠ a +W sin ↵\n.\nHence we get:\n\n## F ⇠ (⇢g c + ⇢N h )g ah sin 2↵ (III.20)\n\nW+\n\nwhich is independent of b. For the sake of simplicity, the term corresponding to the weight\nof NOA walls can be neglected. Indeed, the surface density of walls W + is small (around\n10%), and we have usually h < c (except for very thin plates). Hence, the force is just\nproportional to h:\n110 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\nF ⇠ ⇢g cg ah sin 2↵ (III.21)\n\n## This inertial argument allows us to qualitatively capture all experimental observations:\n\nwe see all the experimental points collapse in a single curve in figure III.19.\n\n4\n10\nc = 160 µ m\nc = 1160 µ m\n−F ( µN ) c = 2160 µ m\n3\n10\n\n2\n10\n\n1\n10\n\n0\n10\n0 1 2 3 4\n10 10 10 10 10\n\n( ρ gc + ρ N h W + λ) g\nλ\nah sin2α ( µN )\n\nFigure III.19 – Force of propulsion as a function of the theoretical expression expected from\nequation III.20. The solid black line represents equation y = x. Each color correspond to\na lamella thickness. Markers are the same as the ones used in previous figures.\n\nDeviation for shallow textures (first cluster of points for each color) In\nthe previous discussion (see section 4), we showed how the Reynolds number strongly\ndepends on the channel depth (as h5/2 ), in the viscous regime. In the case of our lamellae\ngeometries, we saw that viscous effects would dominate only up to h ⇡ 200 mm. For\ntexture walls higher than that, inertial effects would take over. Since h 2[150, 250, 400,\n800, 1600], we are clearly in an inertial case for all our points excepted for h = 150 mm\n(h = 250 mm is also discutable). That corresponds to the first cluster of points of each\ncolor curve in figure III.17 and III.19. We expect a smaller force where both inertia and\nviscous effects (opposing inertial effects) are comparable. This effect is less visible for the\nthickest plate (green points): weight is so large that we need strong airflows to generate\nlevitation, so that inertia completely dominates, regardless of height h.\n6. CONCLUSION 111\n\n6 Conclusion\nAll this work, based on air levitated objects, confirms the propelling viscous entrainment\nscenario discussed in the case of Leidenfrost textures. We have seen how this “cold Leiden-\nfrost” setup overcomes the problem of high temperatures and deformable interfaces. An\nanalytical model has paved the way to new geometries. Given that we can freely increase\nthe air injection, we were able to explore the effect of heavier sliders or deeper textures\n(this time engraved on the slider itself). As a consequence, a whole new propelling mech-\nanism based on an inertial scenario was found, and observed to reverse the direction of\npropulsion.\nAlthough these two regimes (low versus high Reynolds number) are based on different\nphysical mechanisms (viscous effects versus inertial ones) they surprisingly obey the same\nscaling law for the force |F | ⇠ ⇢g cg ah sin 2↵. This is somehow explained by the fact that\nwe experimentally chose to impose the same pressure in both regimes, that is, the one\nexactly compensating the objects weight. Since flow onset is pressure governed, it is not\nso surprising to find same scaling.\nInertial propulsion has open multitude of new questions that are currently being under\nstudy in the lab (PhD of Hélène de Maleprade). A first one would be to produce a\nquantitative analytical model for the inertial regime taking into account the geometrical\nparameters of the porous plate (as done when textures were engraved on the air blowing\ntable) - hence involving the external imposed overpressure P . The question of special\nfriction would lead to the study of terminal speed of these objects and would require a\nnew set up allowing to reach such velocities. All our experiments have been made with\nair as the injected liquid/gas phase. It will be mostly interesting to change the nature of\nthe entraining material (hence probing the role of viscosity ⌘) by replacing air by a liquid\nsuch as water. Another interesting question concerns the regime of Re ⇡ 1, where both\nviscosity and inertia are comparable. Would it be possible (at small airflow injection) to\nentrain by viscosity the slider in one direction and make it propel in the opposite direction\n(at higher airflow injection)?\nThis experiment shows that provided we satisfy the main ingredients to achieve self-\npropulsion (i.e. gas rectification, break of symmetry), we can imagine a new zoology of\nself-propelling devices. More generally, the herringbone geometry was discussed in a wide\nrange of different domains and applications. It was recently used in microfluidic channels\nfor mixing liquids . It has been widely used in gear technology in order to smoothly\ntransfer power (the logo of the car maker Citroën is a graphic representation of a herring-\nbone gear, reflecting French André Citroën’s earlier involvement in the manufacture of\nthese gears). Finally, this pattern, greatly used in artwork, can even be found in ancient\nBactria back in time as far as 5000 years ago. Astonishing objects as the hairpin shown\n112 CHAPTER III. SELF-PROPULSION ON AN AIR HOCKEY TABLE\n\nin figure III.20 already displayed almost all explored textures of this chapter: herring-\nbones, herringbone sections facing each other and in the center a windmill texture! With\nour current manufacturing techniques (easier, faster and cheaper), we should be highly\nencouraged to search for new geometrical patterns.\n\nFigure III.20 – Bactria hairpin, open work, bronze. We can observe in the middle a\nwindmill pattern and, circling it, a herringbone loop.\nChapter IV\n\n## Drop impacting a sieve\n\nContents\n1 Impact on a solid plate: a brief review . . . . . . . . . . . . . . 114\n1.1 Maximal impacting radius . . . . . . . . . . . . . . . . . . . . . 115\n1.2 Drop shape profile . . . . . . . . . . . . . . . . . . . . . . . . . 117\n2 Impact on a plate with a single hole . . . . . . . . . . . . . . . 120\n2.1 Critical speed V ⇤ . . . . . . . . . . . . . . . . . . . . . . . . . . 120\n2.2 Role of plate thickness . . . . . . . . . . . . . . . . . . . . . . . 123\n2.3 Several time scales . . . . . . . . . . . . . . . . . . . . . . . . . 124\n2.4 Transmitted mass . . . . . . . . . . . . . . . . . . . . . . . . . . 125\n2.5 Final comment . . . . . . . . . . . . . . . . . . . . . . . . . . . 127\n3 The Leidenfrost sieve . . . . . . . . . . . . . . . . . . . . . . . . 128\n3.1 Experimental set-up . . . . . . . . . . . . . . . . . . . . . . . . 128\n3.2 Transmitted mass . . . . . . . . . . . . . . . . . . . . . . . . . . 130\n3.3 A deformable interface . . . . . . . . . . . . . . . . . . . . . . . 132\n3.4 Splash pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . 135\n4 Exploring different meshes . . . . . . . . . . . . . . . . . . . . . 139\n4.1 Role of wetting conditions . . . . . . . . . . . . . . . . . . . . . 140\n4.2 Role of hole size r . . . . . . . . . . . . . . . . . . . . . . . . . 142\n4.3 A single curve? . . . . . . . . . . . . . . . . . . . . . . . . . . . 143\n4.4 Pinch-off time versus crash time . . . . . . . . . . . . . . . . . 144\n5 Conclusion and open questions . . . . . . . . . . . . . . . . . . 145\n\nIn this chapter, section 4 was done in collaboration with Pr. Robert Cohen and Siddarth\nSrinivasan.\n\n113\n114 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## 1 Impact on a solid plate: a brief review\n\nA naive method to generate a non-wetting drop is simply to suppress its substrate. How-\never, as a price to pay for such an easy method, nothing sustains the liquid anymore, which\nis just a falling drop. By nature, this contactless state cannot be maintained and the glob-\nule will inevitably encounter a substrate at the end of its fall. In this chapter, we focus\non this catastrophic event: the impact of a drop. We focus on the interesting situation\nof an impact on a mesh. A grid, that is, a mixture of holes and closings, is an interme-\ndiate case between a solid plate (i.e. a mesh with no holes) and no obstacle (a grid with\nholes only). In order to remain as close as possible to a contactless situation, we consider\nnon-wetting sieves, either Leidenfrost or superhydrophobic. We privilege a qualitative de-\nscription of this rich system, even if we also make at some point more quantitative models.\n\nImpact of a drop on a solid substrate has been lately studied due to its ubiquity\nin everyday life. For printing, coating or spraying, from pesticides to rain [26, 134], it is\nessential to understand the collision mechanisms of a drop. This problem is more complex\nthan appears at first sight. We can decompose the whole impact process in four main\nstages for a non-wetting substrate:\n(i) at the very first moment of contact we have a singularity problem that arises many\nquestions regarding pressure and speed profiles , short time dynamics , compress-\nibility effects [119, 26, 79] or bubble entrapment [33, 116, 125, 117] among others,\n(ii) a second spreading phase follows this early stage. Two main scenarii have then\nbeen reported : splashing at high speed [130, 100, 54, 132, 131, 106, 75, 123, 118] and\na strong deformation at contact at slower speed, ,\n(iii) after the drop spreads up to a maximum radius [24, 2], it eventually retracts due\nto capillary forces,\n(iv) depending on the wetting properties of the substrate, retraction can lead to sev-\neral scenarii such as equilibrium, rebound , or even singular jets .\n\nMany other aspects have been, and are still, explored [134, 130]. Special attention has\nbeen given to the influence of the substrate [122, 55, 123, 36, 72, 78, 80, 54], to the role\nof the surrounding gas [132, 131, 75, 136, 106], or to the nature of the liquid (Newtonian,\nshear thinning, polymeric ).\n\nIn what follows, we will focus on the spreading phase where neither splashing nor\njetting are present and where the surrounding gas can be neglected in the dynamics. We\nwill give particular attention to two key features: the maximal spreading radius, and the\nheight shape profile.\n1. IMPACT ON A SOLID PLATE: A BRIEF REVIEW 115\n\nWe adopt here the description of Clanet and collaborators , who focused (among\nothers) on the maximal extension of an impinging drop - a question of practical importance\nsince it defines the mark made on a solid by such drops. Two regimes are discussed\naccording to the liquid viscosity:\n\nFigure IV.1 – Impact of a water drop (2R0 = 2.5 mm, V0 = 0.83 m s 1 ) on a super-\nhydrophobic surface. The drop spreads (image 2 and 3) until it reaches its maximal\ndiameter Dmax = 2Rmax . After a recoiling phase (image 4 to 6), a rebound (images 7\nto 9) is observed, together with strong vibration. Time interval between the pictures:\n2.7 ms. Figure from .\n\nLow viscosity drops In the limit of low viscosity and low wettability (water on a\nsuper-hydrophobic surface), the maximum radius Rmax of the drop was found to scale\nas R0 We0 1/4 , where R0 is the drop’s radius before impact and We0 the so-called Weber\nnumber. This dimensionless quantity compares kinetic and surface energy. For a drop\nof radius R0 , a liquid of surface tension and density ⇢, and an impact velocity V0 , the\nWeber number is:\n\n⇢V02 R0\nWe0 = (IV.1)\n\nThis scaling was found to hold on more wettable surfaces, and interpreted as resulting\nfrom the equation of motion: during the shock the drop experiences an effective acceler-\nation g ⇤ ⇠ V02 /R0 (since it slows down from impact speed V0 to rest in a characteristic\n116 CHAPTER IV. DROP IMPACTING A SIEVE\n\ncrashing time ⌧ ⇠ 2R0 /V0 ), much more intense than the gravity field: g ⇤ flattens the\nliquid and fixes its thickness to a capillary\nq length where gravity field g is replaced by\ndeceleration g . Hence a thickness: `c ⇠ ⇢g⇤ . Conservation of volume (R03 ⇠ Rmax\n⇤ ⇤ 2\n`⇤c )\nyields:\n\n## Rmax ⇠ R0 We0 1/4 (IV.2)\n\nRecent work by Tran et al. extended this study to impacts on super heated\nsurfaces. Similar to static Leidenfrost drops, the impact behavior can be separated into\nthree regimes: contact boiling, gentle film boiling, and spraying film boiling. In the two\nlast regimes (both occurring when the surface temperature is higher than Leidenfrost\ntemperature TL ), the maximum deformation displays universality (Rmax ⇠ R0 We0 2/5 )\nregardless of the variation in surface temperature and liquid’s properties. In the latter\nsituation, spreading is lubricated by a gas layer between the drop and the solid surface.\nTran and coworkers suggest that this steeper scaling law may be due to an extra driving\nmechanism caused by the evaporating vapor radially shooting outwards and taking liquid\nalong.\nAlthough equation IV.2 describes the experimental observations, it has been verified\nfor a small range of We numbers. S. Chandra and C. Avedisian suggested another\nscenario describing their measurements based on energy conservation: initial kinetic and\n4⇡R3\nsurface energy of the spherical drop 4⇡R02 + 12 3 0 V02 is transformed after impact into\ndeformation 2 ⇡Rmax2\n.\n\nHigh viscosity drops The case of more viscous liquids was also analyzed, and a\ncriterion for predicting if the spreading is limited by capillarity or by viscosity was derived.\nKinetic energy of the impinging drop (on the order of ⇢R03 V02 ) being dissipated by viscosity\nduring impact (and indeed there is no more any rebound), the associated energy scales as\n⌘ hVd0 Rmax\n3\n, hd being the thickness of the maximal drop. Together with volume conservation\n(R0 ⇠ Rmax hd ), this yields:\n3 2\n\n## where the Reynolds number is Re0 = ⇢R0 V0\n\n([22, 90]).\n\nImpact Parameter In order to differentiate between the inviscid case (P < 1) and\nthe viscous case (P > 1), an impact number P = We0 /Re0 4/5 can be defined. The\ntransition between the capillary and the viscous regime is shown in figure IV.2, where the\ndimensionless viscous extension Rmax /(R0 Re0 1/5 ) is plotted as a function of the impact\nnumber P . The transition between the two regimes is very clear. It occurs around P = 1,\n1. IMPACT ON A SOLID PLATE: A BRIEF REVIEW 117\n\nRmax 101\nR0 Re 1/5\n\n1/4\n\n100\n\n10–1\n10–2 10–1 100 101 102 103\n\nP = We / Re 4/5\n\nFigure IV.2 – Dimensionless deformation of an impinging drop (where the maximal ex-\ntension Rmax is normalized by the maximal deformation in the viscous regime R0 Re0 1/5 ),\nas a function of the impact number P = We0 /Re0 4/5 . Two regimes are successively fol-\nlowed, which corresponds to the capillary and viscous regimes (equations IV.2 and IV.3\nrespectively). The transition occurs around P = 1. Figure from .\n\nas expected, since all the numerical coefficients were (experimentally) observed to be close\nto unity. The capillary regime (P < 1) is likely to be observed at small velocities, for\nsmall viscosities and large surface tension.\n\n## 1.2 Drop shape profile\n\nLagubeau et al. recently measured the drop shape during impact (over a solid surface)\nusing space-time-resolved Fourier Transform Profilometry technique (FTP). They observe\nthree distinct dynamical regimes of spreading for the time evolution of the film thickness\nat the center of the drop hc (t), as shown in figure IV.3.\nThese three regimes have been theoretically and numerically described by Roisman\n and Eggers , and will be discussed below:\n\nt < ⌧ /2 The early time regime is a linear decrease corresponding to the free fall of\nthe top of the drop (black solid line in figure IV.3). The apex continues falling at speed\nV0 until the pressure impact reaches the top of the drop (approximatively at ⌧ /2).\n\n⌧ /2 < t < tp We have an intermediate phase (red solid line in figure IV.3) where the\ninterface velocity dh dt\nc\ndecreases. In this regime, Eggers et al. predicted a self-similar\nsolution for the surface profile based on a hyperbolic (inviscid and potential) flow solution:\n118 CHAPTER IV. DROP IMPACTING A SIEVE\n\n1.00\nhc / 2R0 R / R0\n\n0.75 3\n\n0.50 2\n\n0.25 1\n\nhp\n0 1 2 3 tp 4 5 6 7 8\n\nV0 t / 2R0\n\nFigure IV.3 – Squares and left vertical axis: height hc of the top of the central point of\nthe drop surface as a function of time for We0 = 214 and Re0 = 2690. Circles and right\nvertical axis: radius of expansion R(t) of the same drop. The solid line indicates the\nfree fall regime. Red curve is the best fit of hc (t) (equation IV.4) during the self-similar\nregime from which H(0) and t0 are experimentally obtained. Black dashed-dotted line\ncorresponds to the final plateau height hp . Vertical dashed line is the crossing time tp\nbetween the two previous regimes. Figure from .\n\n2\n2 H( V (t+t ) ). Here w denotes the radial position in the drop and t0 is\n⌧ w\nh(w,t) = 2R0 (t+t 0) 0 0\n\nan unknown parameter whose physical significance is the time it takes for the pressure\nto decay and for the hyperbolic flow to establish: the flow at intermediate times is no\nlonger pressure driven. This theoretical prediction is in good agreement with experimental\nmeasurements realized by Lagubeau et al. From the self-similar solution, we can deduce\nthe time evolution for the central height of the drop: hc (t) = h(w = 0, t).\n\n⌧2\nhc (t) = 2R0 H(0) (IV.4)\n(t + t0 )2\nWe can get rid of t0 by re-expressing it in terms of the total drop volume1 . We show\nin red solid line in figure IV.3 the best fit of equation IV.4, from which the two unknown\nparameters t0 and H(0) have been experimentally calculated by Lagubeau and coworkers:\nt0 ⇡ ⌧ /2 and H(0) ⇡ 1/2. For t = ⌧ /2, this solution verifies hc (⌧ /2) = R0 ensuring height\ncontinuity with the free fall regime.\n\nt > tp At large time (black horizontal dashed-dotted line in figure IV.3), hc tends to\na plateau value hp (if the drop rebounds, this plateau naturally disappears). A viscous\np\nboundary layer is found to grow from the substrate with a thickness scaling as hl (t) ⇠ ⌫t\n1\nSee for proper calculation.\n1. IMPACT ON A SOLID PLATE: A BRIEF REVIEW 119\n\n(where ⌫ = ⌘/⇢ is the kinematic viscosity). The asymptotic plateau film thickness hp is\nobtained as the growing boundary layer meets the drop surface. The intersection of this\ntwo thicknesses defines a new characteristic plateau time tp (vertical dashed line in figure\np 2\nIV.3). For t >> tp , the equality hl (tp ) = h(w,tp ) scales as: ⌫tp ⇠ R0 ⌧t2 , hence:\np\n\n1/5\ntp ⇠ ⌧ Re0 (IV.5)\n\nThe theoretical prediction is again in good agreement with the observations by Lagubeau\net al. We see that tp scales as ⌧ (where tp > ⌧ ) with a small dependance over the Re\nnumber (power 1/5) and we always have tp > ⌧ . For a millimetric water droplet, typical\nReynolds range around 1000 so that we have tp ⇠ 4⌧ .\n120 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## 2 Impact on a plate with a single hole\n\nAs first step towards the study of impacts on grids has been done by Lorenceau and\ncollaborators. They studied the configuration of a plate of thickness e pierced with a\ncircular hole of radius r smaller than the drop and the capillary length (see figure IV.4).\nThis set up can be seen as a mixture of two opposite conditions: the hole allowing the\nliquid to flow freely, the substrate forcing it to deviate.\n2R0\n\n2r\n\nFigure IV.4 – Set of pictures taken each 2 ms apart. Drop of radius R0 = 1.75 mm, hole\nradius r = 450 mm, impact speed V0 = 70 cm/s, plate thickness e = 250 mm. Liquid\n(silicone oil) is ejected from the surface and forms several droplets. Figure from .\n\n## 2.1 Critical speed V ⇤\n\nIn this problem, liquid inertia is always the driving force opposed by two forces: on the one\nhand, viscous friction related to the crossing of the hole, and on the other hand, capillary\nforces which oppose the formation of a liquid filament. To highlight the contribution of\nthese two forces, the natural parameters to consider are the Reynolds number and the\nWeber number (associated to the hole dimension r and not to the drop radius R0 ), defined\nas:\n\n⇢V0 r ⇢V02 r\nRe = We = (IV.6a)\n\nIn both cases, there is a threshold velocity V ⇤ above which inertia overcomes the\nresisting forces and liquid passes through the hole. Lorenceau et al. focused on these two\nnumbers at the threshold velocity V ⇤ and reported in figure IV.5 the variation of We⇤ as a\nfunction of Re⇤ for various liquids and hole radii. Three regions can be observed in figure\nIV.5:\n2. IMPACT ON A PLATE WITH A SINGLE HOLE 121\n\nFigure IV.5 – Threshold velocity of capture: the Weber number at threshold is plotted as\na function of the Reynolds number at threshold. The data are obtained using a e = 250-\nmm-thick plate, different hole radii, and various liquids (silicone oils of viscosity ranging\nbetween 0.5 and 300 mPa s, ethanol, acetone, heptane, and water). The thin line is the\ncurve Re⇤ (We⇤ - 3.6) = 5.1 We⇤ , where the scaling comes from equation IV.7. Figure\nfrom .\n\nLow Re number: viscous-inertial regime In the limit of small Re, the viscous\nforce is dominant and it is responsible for the capture of the drop: this leads to a vertical\nasymptote Re⇤ ⇡ 5 in figure IV.5.\n\nIntermediate Re number For intermediate Re (5 < Re < 100), the two numbers\nare observed to depend on each other. Both the capillary and the viscous forces play\na role. In the limit of very thin plates (e < r), the viscous force associated with the\ncrossing of the hole can be dimensionally written ⌘V0 r (the velocity gradients taking\nplace on a size of order r). The capillary force opposing the formation of a liquid finger\nscales as r. Balancing these two forces with inertia gives an equation for the threshold:\n⇢V ⇤ 2 r2 ⇠ ⌘V ⇤ r + r. This yields:\n\nWe⇤\nRe⇤ ⇠ (IV.7)\nWe⇤ 1\nwhere all the numerical coefficients have been ignored. An equation of the type IV.7 is\nindeed found to describe fairly well the data in figure IV.5, where the equation Re⇤ (We⇤\n3.6) = 5.1 We⇤ is drawn with a thin line.\n\nLarge Re number: capillary–inertial regime At large Re (Re > 100), the critical\nWeber number is found to be constant, of about 3.5: the critical speed does not depend on\nviscosity and is set by a balance between inertia (dynamic pressure 1/2⇢V02 ) and capillarity\n122 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## (resisting pressure /r). Hence a critical speed:\n\nr\n2\nV⇤ = (IV.8)\n⇢r\nThe We number can be re-expressed as We = ( VV0⇤ )2 . Therefore, the threshold condition\nV0 = V ⇤ is We = 1.\nThis speed threshold can be also interpreted in terms of speed of retraction of a liquid\nfinger. As explained by Taylor for thin planar liquid sheets, the speed of retraction\nis driven by capillary forces, scaling as 2⇡r in the case of a cylindrical geometry.\n\n2r\n\nVin\n\n2ϖrγ\n\nV(t)\n\nz(t)\n\nFigure IV.6 – Sketch of retracting liquid finger while liquid is injected at speed Vin from\nthe top in the opposite direction. Due to mass conservation, during retraction mass\naccumulates forming a liquid blob.\n\nBy denoting z(t) the position of the bottom of the retracting extremity (where mass\nis accumulating, see figure IV.6), we get the following equation of motion:\n\nd dz\n2⇡r = (M ) (IV.9)\ndt dt\nMass accumulated at the bottom end of the finger M , is equal to the mass previously\ndistributed in a cylinder of section ⇡r2 (shaded in figure IV.6), hence M = ⇢⇡r2 z(t).\nInserting the conservation of mass in the equation of motion yields:\n\nd2 z 2 4\n2\n= (IV.10)\ndt ⇢r\nAssuming finite speed and initial position z(0) = 0, we get after integration a speed\nof retraction independent of time:\n2. IMPACT ON A PLATE WITH A SINGLE HOLE 123\n\nV (t) = V ⇤ (IV.11)\n\nClanet et al. refined this model in order to explain the transition between dripping\nand jetting. They take into account the possibility of vertical injected speed Vin (in\ngreen in figure IV.6, in direction opposite to retraction). The equations are consequently\nmodified and we consider here the particular case where we can neglect gravitational forces\n(a fair assumption if the mass accumulated has size negligible compared to the capillary\nlength). We have to add a dynamic pressure term ⇢⇡r2 Vin (V (t) + Vin ) to the equation of\nmotion IV.9, now becoming:\n\ndz d dz\n⇢⇡r2 Vin ( + Vin ) + 2⇡r = (M ) (IV.12)\ndt dt dt\nThe conservation of mass being now M = ⇢⇡r2 (z(t) + Vin t), we finally get2 :\n\nd2 z 2 d2 z dz 4\n2\n+ 2V in t 2\n+ 4V0 = 2Vin2 (IV.13)\ndt dt dt ⇢r\n\n## V (t) = V ⇤ Vin (IV.14)\n\nWe see that if Vin = V ⇤ , we have V (t) = 0. In the case of a drop impacting a hole, Vin\nis the impact speed V0 . If Vin > V ⇤ , the speed Vej = V (t) of the ejected liquid finger\nbecomes :\n\n## 2.2 Role of plate thickness\n\nIn the same capillary–inertial regime, Lorenceau et al. also tested the influence of the plate\nthickness e on the critical speed V ⇤ . They compared two plates of respective thickness 0.25\nmm and 1.7 mm and found that the threshold of capture is weakly affected by thickness.\nIt was observed that the critical speed is slightly higher for a thicker plate, which can be\ndue to an additional viscous dissipation along the hole.\n\n2\nIf we take Vin =0, we get back to equation IV.10.\n3\nTo see more model refinements such as variable cylinder radius or role of gravity, see .\n124 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## 2.3 Several time scales\n\nThis problem has a multitude of different time scales. Two of them are associated to\nimpact itself: the time of rebound Treb and the time ⌧ it takes the drop to impact. A last\ntimescale concerns the liquid passing through the hole, namely the characteristic time\nTdes of destabilization of the liquid cylinder. In what follows we will discuss each time in\nmore detail.\n\n## Time of rebound: Treb\n\nThe impact of a water drop on a super-hydrophobic substrate (or on a Leidenfrost surface)\nis generally followed by a rebound . The restitution coefficient of the shock can be\nvery large (around 0.9), so that a drop can bounce many times before stopping . A\nrebound is possible because the kinetic energy of the impinging (non-viscous) drop can\nbe stored in deformation during impact. The contact time has been studied [93, 87] and\np\nshown to scale4 as ⇢R03 / . This variation, independent of the impact velocity V0 , can be\nunderstood by considering (globally) the rebound as an oscillation: the drop is a spring\nof stiffness and mass ⇢R03 , which oscillates with a constant period:\ns\n⇢R03\nTreb ⇠ (IV.16)\n\n## Time of finger destabilization: Tdes\n\nIn the capillary–inertial regime, the mechanism of transformation of a liquid cylinder\ninto droplets was explained by Plateau and Rayleigh more than a century ago. Surface\ntension tends to minimize surface area, hence transforms cylinders into spheres. For a\nliquid column of small viscosity, the destabilization takes place in a time Tdes obtained by\n2 ) with capillarity (on the order of r 2 ), which yields:\nbalancing inertia (on the order of T⇢r\ndes\n\ns\n⇢r3\nTdes ⇠ (IV.17)\n\nIt is interesting to see Tdes as the time to close a cavity of size r at the Culick speed\nV : Tdes ⇠ Vr⇤ . Destabilization time Tdes has the same capillary-inertial nature than\n\nthe time of rebound Treb , hence a similar scaling law. The main difference dwells in the\ncharacteristic length scale: destabilization occurs on a distance scaling as the hole radius\np\nr, whereas rebound implies the drop radius R0 . Since Treb /Tdis scales as R03 /r3 and\n4\nThis rebound time has naturally the same scaling as the period of spontaneous oscillation of drops\n(discussed in chapter I, section 2.3) since it is also based on inertia/capillarity balance.\n2. IMPACT ON A PLATE WITH A SINGLE HOLE 125\n\nR0 > r we always have: Tdes < Treb . For holes ten times smaller than the drop size (which\nis typically the case in our experiments), rebound is approximatively thirty times slower\nthan liquid filament destabilization.\n\nCrash time: ⌧\n\nThe time of rebound Treb can also be compared to the time ⌧ it takes for the drop to\ncrash. The apex of the drop has to cover a distance 2R0 at speed V0 , so that ⌧ can be\ndefined as:\n\n2R0\n⌧= (IV.18)\nV0\nWe typically have R0 ⇡ 10 r, hence ⌧ > Tdes . For instance, a millimetric water drop\narriving at V0 = 1 m/s has a crash time of 1 ms. For a mesh with holes of size r ⇡ 100 mm,\nthe destabilization time is Tdes ⇡ 0.1 ms, which means that the liquid cylinder destabilizes\nmuch faster (here 10 times faster) than it takes the drop to completely crash.\n\n## 2.4 Transmitted mass\n\nWe suggest a simple scaling law giving the amount of liquid that passes through the hole.\nThe finger is being ejected at a typical speed Vej = V0 V ⇤ . The total transmitted volume\nper unit time will scale as ⇡r2 Vej . We now have to discuss which is the right timescale over\nwhich we have to integer this flow rate to access the transmitted mass. Three different\n\nregimes can be considered regarding the ratio Tdes ⌧\n⇠ 2R 0 V\nV0 r\n:\n\nTdes << ⌧\n\nThe liquid finger destabilizes faster than it takes for the drop to crash. The liquid\njet is permanently pinching off, we are in a kind of dripping regime. As a consequence,\nthe destabilization of the finger will not affect the amount of transmitted mass but only\nits geometry5 . Mass will be transmitted all along the crashing time ⌧ , hence transmitted\n\nmass mT scales as ⇢R0 r2 (1 VV0 ), an increasing function of the impact speed V0 . Its\nsaturation value mmax\nT ⇠ ⇢R0 r2 is naively the mass contained in a cylinder of surface area\n⇡r2 and height 2R0 , that is, the drop height. Rescaling by the initial drop mass m0 ⇠ ⇢R03\nwe get:\n\n5\nAn asymptotic scenario can be proposed by taking an imaginary drop of infinite radius R0 (i.e. a\ncontinuous jet), ensuring the condition ⌧ > Tdes .\n126 CHAPTER IV. DROP IMPACTING A SIEVE\n\nmT r2 V⇤\n⇠ 2 (1 ) (IV.19)\nm0 R0 V0\n\nTdes >> ⌧\nLiquid is injected through the hole during a time ⌧ smaller than the time of destabiliza-\ntion: the finger never pinches off. We form a liquid finger that will eventually stop falling\nand retract owing to capillary forces (following the Culick/Taylor speed of retraction V ⇤ ).\nAs a consequence, no mass will be transmitted. Hence, two conditions have to be satis-\nfied in order to transmit liquid through a hole. First, we need the liquid to have enough\ninertia to form a liquid finger (i.e. V0 > V ⇤ ). Then the liquid has to pinch off (governed\nby the Rayleigh-Plateau destabilization mechanism) and detach avoiding to be retracted\nand recaptured due to capillary effects.\n\n## A marginal case: Tdes ⇠ ⌧\n\nA very particular situation is observed when Tdes ⇡ ⌧ where we observe a unique pinch-off\nevent. Several cases can be foreseen:\n- if Tdes . ⌧ the liquid finger will pinch off shortly before the end of the impact. After\nthis first pinch-off, even if the drop is still crashing (hence a small fraction of liquid still\nbeing injected), liquid will retract before a second pinch-off happens. Even if this effect\nis marginal, less mass will be transmitted and the correct time to be considered here is\nTdes .\n- if Tdes & ⌧ the liquid finger will pinch off shortly after the end of the impact. In a\nsituation where there is a downwards driving force (such as inertia or gravity), the liquid\nfinger can continue to fall and entrain supplementary liquid from the upper drop, increas-\ning the transmitted mass. We can even imagine a situation where the finger falls and\npartially retracts before the first pinch-off occurs, the ultimate situation being a complete\nretraction (i.e. Tdes >> ⌧ , previously addressed).\n\nIn this context, Lorenceau et al. proposed an alternate model based on the pinch-off\ntime as characteristic timescale. The critical time should hence be the Rayleigh-Plateau\ntime Tdes (instead of ⌧ ). The arising scaling law gives: mt ⇠ ⇢r3 ( VV0⇤ 1). Once rescaled\nby the initial mass, it yields:\n\nmT r 3 V0\n⇠ 3( ⇤ 1) (IV.20)\nm0 R0 V\nThis second scenario predicts a transmitted mass increasing linearly with impact speed\nthat does not saturate for any value. We show in figure IV.7, experimental data from\nLorenceau et al.: transmitted mass (rescaled by ⇢r3 ) as a function of the normalized\n2. IMPACT ON A PLATE WITH A SINGLE HOLE 127\n\nimpact speed. The red region corresponds to a destabilization time faster than impact,\nmodeled by equation IV.19 (red line) and capturing the steep slope for V0 ⇠ V ⇤ . The blue\nregion corresponds to the case where pinch-off time dominates. It is modeled by equation\nIV.20 (blue line). A fair agreement is observed with the data.\n\nTdes < τ ~τ\nTdes >\n20\nmT\nρr3\n15\n\n10\n\n0\n0 1 5 10\n\nV0 2R0 Tdes\n~\nV* r τ\n\nFigure IV.7 – Volume of ejected liquid (rescaled by ⇢r3 ) as a function of the ratio V0 /V ⇤\nfor different radii. Red and blue lines show equation IV.19 and IV.20, respectively. Data\nextracted from .\n\n## 2.5 Final comment\n\nThe study of drop impacting a plate with a single hole shows that the key timescale in this\nproblem is the crashing time ⌧ ⇡ 1 ms. Indeed, we have seen that rebound phenomenon\nis to slow (characteristic time typically on the order of 10 ms) to play any role and\ndestabilization of liquid filament into droplets (characteristic time typically on the order\nof 0.1 ms) only modifies the shape of the enveloppe of passing liquid (but not its amount).\nNonetheless, marginal situations (where ⌧ is not anymore the key time) can be observed\nby playing with the other timescales of the problem: Treb and Tdes . A whole new (and\nquite complex) zoology of regimes can be discussed. For instance, for low We numbers\n(where capillarity dominantes inertia) and non wetting plates, we can imagine that a\ndrop will bounce faster than it crashes (Treb < ⌧ ), hence the limiting time now being Treb .\nAs a direct consequence, we expect an amount of transmitted mass lower than the one\npredicted by our previous scaling law.\n128 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## 3 The Leidenfrost sieve\n\nWe now have all elements to study impacts on a grid. After discussing the experimental\ndetails of our set-up, we wonder how much liquid we can grate through a sieve. A scaling\nlaw will allow us to qualitatively predict the amount of transferred mass. A more refined\nanalytical model will lead us to focus on the deformation of the drop during impact due\nto horizontal evacuation of the liquid: as a consequence, we observe an increase in the\nsection through which liquid is transferred and an ejection speed decrease at the end of\nthe crash. We finally discuss how lateral spreading plays a role in the formation of a cone\nof ejected droplets.\n\n## 3.1 Experimental set-up\n\nWe study in this section the impact of a drop on a Leidenfrost grid. We use a mesh\nof plain-woven brass wires of diameter d = 0.17 mm and square holes of lateral size\nl = 0.37 mm (see figure IV.16a). The corresponding surface fraction of square holes is\nl2\n= (l+d) 2 = 47%. Since liquid avoids corners (where curvature is high), we assume that\n\nliquid ejected from this square holes will have a circular shape. The associated character-\nistic radius6 is r = l/2 ⇡ 0.19 mm, which yields a slightly smaller surface fraction for the\ncircular holes: ⇤ = ⇡4 .\n\nWe avoid all wetting phenomenas by heating the mesh to the Leidenfrost state. We\nqualitatively control the mesh temperature by heating it with a Bunsen burner until it\nreaches red-hot metal. To avoid any perturbation of the flame, we turn it off just before\nthe drop impacts the grid. A simple argument allows us to estimate the proportion of\nmass loss due to evaporation. During an impact of characteristic time ⌧ ⇠ 2R0 /V0 , we get\nan evaporated mass scaling as m ⇠ ⇢Se ce ⌧ , where Se denotes the surface of evaporation\n- scaling as (1 ) R02 - and ce denotes the speed of evaporation. Normalized by the\ninitial mass of the drop, we get m/m0 ⇠ (1 )ce /V0 . For impact speeds of several m/s\nand typical evaporation speed ce of 10 cm/s we get a loss of a few percent, allowing us\nto neglect mass losses due to evaporation. We can also try to estimate the thickness of\nthe insulating layer around a hot wire using the classical Leidenfrost thickness equation\n(seen in Chapter I, section 2.2, equation I.18). The squeezing pressure will be the one\ngenerated by capillarity ( /r, where 1/r is the local curvature radius of the liquid finger)\nand the characteristic escaping length scale will be d. As consequence, we expect typical\nlocal vapor thickness eL of several tens of mm, slightly thinner than the vapor film beneath\n6\nThis hole length is slightly smaller than the one studied by Lorenceau and coworkers where their\nbiggest hole was r ⇡ 0.5 mm.\n3. THE LEIDENFROST SIEVE 129\n\na Leidenfrost puddle. As a consequence of the presence of this vapor film, the surface\narea through which liquid\nq is injected becomes smaller. The corresponding critical speed\nshould now be V = ⇤ 2\n⇢(r eL )\n. Since r/eL << 1, the change in critical speed remains\nsmall (within measurement scattering): in what follows, we keep the definition of the\ncritical speed unchanged (i.e. defined without eL ).\n\n2R0\n\nV0\n\nFigure IV.8 – Side view of a drop of radius R0 = 1.9 mm impacting a hot grid at speed\nV0 = 1.6 m/s. First image is used to define origin of time as being the first moment the\nbottom of the drop touches the grid. Second and third images represent the time it takes\nfor the drop to crash: ⌧ = 2R0 /V0 . This time being larger than the Rayleigh-Plateau\ntime, liquid filaments destabilize in tiny droplets. On the last image, there is no vertical\nmomentum at the upper part of the mesh and no mass is transmitted. Grated droplets\nfall conserving their momentum and form a cone of angle ↵. Image are separated by 2\nms.\n\nDrops in this study are formed by means of a syringe and a needle (internal diam-\neter dint = 1.16 mm, external diameter dext = 1.7 mm), which provides a drop radius\nR0 = 1.9 mm. The syringe is fixed to a vertical beam and its height varies as to generate\ndifferent impact speeds V0 ranging between typically 0.5 m/s to 5 m/s. We record the\nimpact from the side with a fast camera (typical frame rate 10000 fps), using backlighting\nto enhance contrast.\nA typical impact is shown in figure IV.8 where we observe that part of the liquid\npasses through the sieve while the rest remains on top. On the upper part of the grid,\neverything happens as if we had an impact on a non-wetting substrate: we first see a\nspreading phase (together with liquid passing through the holes), followed by a recoiling\nphase (were mass is conserved) and a rebound (not shown on figure IV.8). On the lower\npart of the grid, we observe liquid filaments with a radius comparable to the size of the\nholes, which quickly destabilize in tiny droplets. They form a very well defined 3D cone\ncharacterized by an angle ↵ (see figure IV.8).\n130 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## 3.2 Transmitted mass\n\nExperiments\nAs previously stated, drops hitting the grid have only two possibilities: they pass through\nthe holes, or they are stopped and horizontally redirected. We measure the transmitted\nmass by positioning a plastic sheet parallel to the grid at approximatively 10 cm below it.\nBy this means, we harvest immediately after impact the transmitted droplets and weight\nthem with a high precision scale. We repeat this experiment for each falling height (hence\nseveral impact speeds V0 ).\n\n(a) (b)\n\n1.0\nmT\nm0 z\n0.8\n\n0.6\n\n0.4 R0\nR0-V0 t\n0.2\ndz\nw(t)\nV0\n0.0\n0 1 2 3 4 5 6 7\n\nV 0 /V ∗\n\nFigure IV.9 – (a) Percentage of transmitted mass as a function of impact speed normalized\nby the inertio-capillary speed V ⇤ = 0.65 m/s. The solid blue line is equation IV.21\nmultiplied by a factor 3/2. (b) Sketch of drop at initial time t = 0. Since drop falls at\nspeed V0 , slice at position z will reach the mesh at t = z/V0 . Pythagorus theorem yields\nthe expression of w(t) as a function of time, initial speed and drop radius.\n\nWe show in figure IV.9a the transmitted mass fraction mT /m0 (m0 = 4/3⇡R03 being\nthe initial mass of the liquid) as a function of impact speed. Three regimes are clearly\nvisible:\n(i) for V0 < V⇤ , the drop has not enough inertia to overcome capillary forces (Vej < 0),\nhence even if small filaments transiently sag into the holes, they retract and no mass is\ntransmitted (mT = 0),\n(ii) just above V ⇤ , we observe a fast increase of transmitted mass,\n(iii) for V0 >> V⇤ , we get a saturation value below mT /m0 = 1.\nIn order to better understand these observations, we need to specify and compare\nthe value of all the different timescales. Since drop radius R0 = 1.9 mm and hole size\nr = 0.19 mm are fixed, we get: Tdes ⇡ 0.3 ms, Treb ⇡ 10 ms and ⌧ 2 [1,10] ms (in our\n3. THE LEIDENFROST SIEVE 131\n\nrange of impact speeds). Liquid cylinder destabilization being much faster than the crash\ntime ⌧ > Tdes , the characteristic time to be considered to calculate the transmitted mass\nis ⌧ (as discussed in section 2.5).\nCrash time being always smaller than the rebound time, we can ignore bouncing\neffects. In addition, Reynolds number ranges between 100 and 2000 (validating the inertia-\ndominated assumption), and Weber number ranges from 0.5 up to 100.\n\nScaling law\nTo explain these behaviors we suggest the following simple scaling law. For V0 > V ⇤ , we\nexpect the flow rate through a single hole to scale as r2 (V0 V ⇤ ). By multiplying this\nflow by the number of holes N (scaling itself as R02 /r2 ) and the crash time ⌧ = 2R0 /V0 ,\n\nwe get for the total transmitted volume: ⌦ ⇠ R03 V0V0V . As a consequence, the fraction\nof transmitted mass scales as:\n\nmT V⇤\n⇠ (1 ) (IV.21)\nm0 V0\nWe show in figure IV.9a that equation IV.21 is in very good agreement with the data,\nprovided we chose a numerical factor of 3/2. Indeed:\n(i) equation IV.21 is defined only above V ⇤ ,\n\n(ii) it predicts a local slope dm\ndV0\nT\n= VV 2 . This slope rapidly decreases (as 1/( VV0 )2 ),\n0\nhence being maximum for V0 = V ⇤ , which explains the fast increase of transferred mass\njust above V ⇤ ,\n(iii) for high impact speed V0 , it predicts a saturation towards mT /m0 ⇠ < 1.\n\nAnalytical model\nWe now move a step further, and produce an analytical model. We adapt the equation\ngiving the transmitted mass through a single hole (equation IV.19) by imagining the drop\nas being a vertical assembly of discs of radius w(t) and thickness dz (see figure IV.9b).\nEach slice arrives onto the grid at speed Vin . Since Vej = Vin V ⇤ , the flow rate through\n⇡w2 ⇡ w2\na single hole is ⇡r2 Vej . Mass is transferred through N (t) = (2r+d) 2 = 4 r2 holes. Factor\n\n4\nreflects the fact that even though our holes are square, mass is transmitted through\ncylinders, hence not filling all the available space.\nTotal mass is calculated by integration over the crash time ⌧ , hence:\nZ ⌧\nmT = ⇤\n⇢ Vej (t) ⇡w2 (t)dt (IV.22)\n0\n\nFor instance, if we assume the drop does not deform and remains perfectly spherical\nduring impact, we have Vej = V0 V ⇤ independent of time. From Pythagorus theorem\n132 CHAPTER IV. DROP IMPACTING A SIEVE\n\n(applied to the triangle described by red arrows in figure IV.9b), we get: w2 (t) = 2R0 V0 t\n(V0 t)2 = V02 (t ⌧ t2 ). Hence:\nZ ⌧\n4⇡R03 1\n⇡w2 (t)dt = (IV.23)\n0 3 V0\nand\n\nmT V⇤\n= ⇤\n(1 ) (IV.24)\nm0 V0\nWe retrieve the same scaling as predicted by equation IV.21. In addition, this analyt-\nical model gives a geometrical prefactor equal to unity. However, equation IV.24 assumes\nthe drop remains spherical during all impact time. Hence, this would be close to the case\nof hole surface fraction close to unity (almost no obstacle), which is not the case in our\nexperiment. We conclude that the experimental higher prefactor (of 3/2) arises from the\ndeformation of the interface at impact.\n\n## 3.3 A deformable interface\n\nFor low mass transfer, falling liquid has to be strongly evacuated in the horizontal direc-\ntion, resulting in the deformation of the drop’s interface. The lower the transferred mass,\nthe closer we are to a situation of impact on a solid plate. We try here to explore the\nconsequences of this deformation and its effect on the transmitted mass. To do so, we\nfocus on the speed of ejection Vej (t) and the spreading radius w(t), two time-dependent\nquantities integrated over time to obtain the transferred mass (equation IV.22).\n\nSpeed profiles\n\n## Experimental observations In order to access the different speeds of the problem,\n\nwe extract from each frame of impact movies the vertical line of pixels that passes through\nthe axis of symmetry of the drop. We then horizontally concatenate them, resulting in\nan image (commonly called “reslice”) where x-axis represents time and y-axis represents\nvertical distance. Typical curves are shown in figure IV.10. We can deduce speed as\nthe local slope (schematically represented in green or red in figure IV.10). To increase\naccuracy, we developed a Matlab routine to access impact speed V0 at the exact moment\nwhen the bottom of the drop first touches the grid, (which defines the time origin t = 0).\nWe observe two regimes depending on the impact speed: (i) for V0 < V ⇤ (left image\nin figure IV.10), no liquid passes through the hole, hence Vej = 0, (ii) for V0 > V ⇤ (center\nand right image in figure IV.10), we first have Vej ⇡ V0 and then we observe a decrease\nin time of the ejection speed (shown by a slope decrease in figure IV.10).\n3. THE LEIDENFROST SIEVE 133\n\n## V0< V * V0~ V * V0 > V *\n\nz 10\n(mm) 5\n\nV0 V0 V0\n-5\n\n-10\nVej\nVej\n-15\n\nτ\n-20\n\n-10 0 10\nt (ms)\n\nFigure IV.10 – Spatiotemporal diagram for drops impacting a sieve. (Left) Inertia is not\nstrong enough to overcome capillary effects, hence no liquid passes below the grid (placed\nat z = 0). Local slope in green gives impact speed V0 . (Center) Inertial and capillary\nforces are comparable. In red, trajectory of ejected droplets whose slope is Vej . As time\ngoes on, from t = 0 to t = ⌧ ⇠ 2R0 /V0 , we see this speed decrease. (Right) Inertial\nregime. Speed Vej of ejected daughter droplets is almost the same than V0 the mother\none.\n\nThe limit case: impact on a solid plate As observed, smaller transferred mass\nimplies stronger speed decrease. The asymptotic limit (almost no mass transferred) was\ndescribed by Lagubeau et al. for an impact on a solid plate. We deduce the injection speed\nVin from the speed of the apex of the drop hc (t) (equation IV.4): Vin = Vc (t) = d hdtc (t) .\nFinally, we get:\n\n8\n>\n> V0 if t < ⌧ /2\n>\n>\n>\n>\n<\nVin (t) = V0 1\n( ⌧t + 12 )3\nif t > ⌧ /2 (IV.25)\n>\n>\n>\n>\n>\n>\n:\n0 if t > tp\n\nWe represent equation IV.25 in figure IV.11 where speed profile of the apex of the\ndrop (rescaled by the critical speed V ⇤ for a hole of size r = 190 mm) is a function of\nnormalized time for different impact speeds V0 = 0.9; 1.4; 1.9; 2.4; 2.9 and 3.4 m/s, and\nfor a drop radius R0 = 1.9 mm.\nThese speed variations teach us that even if Vin is initially above the critical speed\nV , it eventually goes below it before the end of the crash. The moment at which liquid\n\nstops to flow through the hole is denoted as ts . Its expression is given by the condition\n134 CHAPTER IV. DROP IMPACTING A SIEVE\n\n4\nV 0 = 0. 9( m/s)\nV i n/V ∗ 3.5 V 0 = 1. 4( m/s)\nV 0 = 1. 9( m/s)\n3 V 0 = 2. 4( m/s)\nV 0 = 2. 9( m/s)\nV 0 = 3. 4( m/s)\n2.5\n\n1.5\n\n0.5\n\n0\n0 0.5 1 1.5 2\n\nt/τ\n\nFigure IV.11 – Representation of equation IV.25: speed of the apex of an impacting drop\non a plain solid (rescaled by the critical speed V ⇤ for a hole of size r = 190 mm) as a\nfunction of normalized time for impact speeds V0 = 0.9; 1.4; 1.9; 2.4 and 2.9 m/s. Circles\non the red horizontal line shows for each curve the moment when the speed of injection is\nnot strong enough to pass through the hole. Since in this case tp /⌧ & 4.5, tp is not visible\nin the figure.\n\n## Vin (ts ) = V ⇤ , which yields:\n\nV0 1/3 1\nts = ⌧ [() ] (IV.26)\nV⇤ 2\nWe indicate ts in figure IV.11 for different impact speeds (circles on red horizontal\nsolid line).\nAlthough this model describes the decrease of vertical speed observed for drops im-\npacting a solid plate, we assume that the same physical ingredients remain qualitatively\nunchanged for the case of pierced plates.\n\nTransfer section\nAnother quantity varying with time during impact is the horizontal distance by which\nmass is injected. Owing to lateral deformation of the drop, we expect a transfer surface\ngreater that ⇡R02 , as shown in figure IV.12 where the red horizontal arrow denotes the\nmaximum real surface ⇡wmax 2\nthrough which mass is transmitted.\nWe expect w(t) to vary from 0 to wmax during impact, where wmax cannot exceed\nthe maximal spreading radius Rmax discussed in section 1.1. Although we have not done\nany quantitative measurements, we observe that wmax is a function of transferred mass.\nFor high mass transfer, almost no liquid is horizontally evacuated, hence there is almost\nno lateral spreading and we have: wmax ⇡ R0 . For low mass transfer, liquid has to be\nhorizontally redirected and we expect wmax to be a fraction of Rmax .\n3. THE LEIDENFROST SIEVE 135\n\n2R0\n\n2wmax\n\n## Figure IV.12 – Chronophotography (image superposition) of drop of radius R0 ⇡ 2 mm\n\nimpacting a grid at V0 = 1.6 m/s. Red symbols show that mass is transmitted through a\nsurface ⇡wmax\n2\nlarger than ⇡R02 .\n\nWe experimentally observe that wmax can be as high as 3/2R0 . Since transferred mass\ndepends on the square of wmax , this corresponds to an increase of mass as high as a factor\n2, justifying that the geometrical prefactor obtained experimentally can be higher than\npredicted with a perfect spherical assumption.\n\n## The mesh: a mixture in between a hole and a solid plate\n\nPrevious discussions on speed profiles and transfer section were mainly based on the\nstudy of two known asymptotic situations: almost no transferred mass (modeled by an\nimpact on a solid plate), and high transferred mass (modeled by almost no obstacle).\nEven though they qualitatively explain the speed decrease and the lateral increase of\ntransferred section, we would need an analytical solution for the shape and speed profile\nin this intermediate case in order to correctly model the transferred mass. A good starting\npoint would be to adapt and extend the hyperbolic solution proposed by Eggers et al. to\nthe particular case of the grid.\n\n## 3.4 Splash pattern\n\nAnother measurable quantity characterizing the impact is the angle ↵ of the cone formed\nby the passing droplets. We show in figure IV.13 a series of chronophotography for differ-\nent impact speeds (obtained by adding all frames of an impact movie for a given speed).\nWe observe that the higher the impact speed, the wider the opening angle. More quanti-\n136 CHAPTER IV. DROP IMPACTING A SIEVE\n\nFigure IV.13 – Image set-up showing a chronophotography of the impact for different\nspeeds V0 = 0.4; 0.8; 1.1; 1.2; 1.6; 2.1; 2.5 and 3.1 m/s.\n\n## tatively, we show in figure IV.14a ↵ as a function of impact speed V0 .\n\nTo understand the origin of the opening cone, we first discuss the filament ejection\nfrom a single hole. Since all the mass does not pass through the grid, liquid has to\nbe horizontally redirected. Hence, injected liquid has both a vertical and horizontal\ncomponent as sketched in figure IV.14b. The more the hole is off-centered from the axis\nof symmetry, the stronger the accumulation of horizontal redirected mass and the larger\nthe horizontal component of the ejected liquid. As for the ejection cone, ↵ is given by the\nejection angle of the outer holes of the transmitting surface (the farther away from the\ncenter). As previously discussed, this transfer surface depends on impact speed V0 : the\nhigher the impact speed, the the wider the transfer section, hence the larger the opening\nangle ↵ (this justifies the behavior observed in figure IV.13). In other words, we can see\nthe measure of the opening angle as an indirect measure of the maximal spreading radius!\n\n(a) (b)\nz\n90 w\ndz\nα ( ◦) 80\n70 V\n60\n50\n40\n30\n20 2r’\n10\n0\n0 1 2 3 4 5 6 7\nVej\nV 0 /V ∗ d 2r\n\nFigure IV.14 – (a) Ejection angle ↵ as a function of normalized impact speed. (b) Sketch\ndescribing liquid injection through the holes: the more the hole is off-centered, the stronger\nthe horizontal flow, and the horizontal ejection speed. In addition, al large ↵, the effective\ncross section ⇡r02 (in blue) of ejected filaments becomes smaller.\n3. THE LEIDENFROST SIEVE 137\n\nOther features can be deduced from splashing patterns. For instance, Katsuhiko Yo-\nneta from Hokkaido studied in 1932 the diffraction pattern made by a drop falling on a\nnet . He was struck by the regularity of the splash pattern analogous to the one\nseen for a diffraction pattern of X-rays through a crystal. He showed that splash pattern\nsymmetry depends on the position of drop impact regarding mesh holes. He suggested\nthree categories regarding if the center of the drop impacted on the middle of a hole, on\nthe crossing of two fibers or on a combination of both. The different possible outcomes\nare shown in figure IV.15. The dotted circles of the left side of each figure indicate the\ninitial size of the drop. If the volume of drop corresponds to circle n (on the left hand\nside), the corresponding splash pattern consists of spots 1 to n (on the right hand side).\nIn our study we observed similar splash patterns. However, we were in the case where the\ncircular radius of the drop was much bigger than the grid hole spacing. Hence, our pat-\nterns consisted on a multitude of spots splashed over a circular area, losing the memory\nof the square symmetry of the grid (somehow showed by K. Yoneta in figure IV.15 for big\ndrop volumes).\n\nA last comment can be made regarding the droplet sizes. We notice in figure IV.15c\nthat the outer spotted drops are smaller than the center ones (observation also made in\nour experiments). This may be explained by the fact that, as the outer ejected filaments\nmade an angle ↵/2 with the vertical axis, the effective cross section of hole becomes\n⇡r02 = ⇡(cos ↵2 r)2 (in blue in figure IV.14b). As a consequence, ejected liquid finger has\na smaller section resulting in smaller daughter droplets (of typical length scale r0 ). For\ninstance, ejection of a filament with an angle ↵ equal to ⇡/2 will be twice smaller than if\nejected vertically.\n138 CHAPTER IV. DROP IMPACTING A SIEVE\n\n(a)\n\n(b)\n\n(c)\n\nFigure IV.15 – Possible splash patterns (on a surface positioned below and parallel to the\nmesh) showed by Katsuhiko Yoneta. (a) Odd pattern: the center of the drop impacts on\nthe center of a hole. (b) Even pattern: the center of the drop aligns on the intersecting\npoint of two fibers. (c) Combined pattern: the center of the drop impacts in a mixed\nsituation. For all figures, if the volume of the drop corresponds to circle n (on the left\nhand side), the spots of splash pattern become spots numbered 1 to n (on the right hand\nside). Figures from .\n4. EXPLORING DIFFERENT MESHES 139\n\n## 4 Exploring different meshes\n\nWork in this section was motivated and done in collaboration with Professor Robert Cohen\nand Siddarth Srinivasan.\n\nAll our experiments were previously conducted with a unique Leidenfrost grid. Here,\nwe want to explore the role of different parameters such as the wetting properties of the\nsubstrate or the role of the surface density of holes. The non-wetting state was formerly\ngenerated by heating the mesh above the Leidenfrost point, and we did not really control\ntemperature or mass loss due to evaporation. In addition, meshes deformed due to thermal\nexpansion. To avoid these problems, we now generate contactless substrates by coating\nthe grids superhydrophobic.\n\n(a)\n\nd 2r = l\n\n(b)\n\nFigure IV.16 – (a) Scanning electron microscopy image of a spray-coated stainless steel\nmesh (d = 254 mm and r = 330 mmcorresponding to mesh C 0 ) which forms a surface with\nhierarchical texture (scale bar: 200 mm). The sprayed textures have a diameter around\n20 mm. Figure from . (b) Surface roughness information: the mean thickness of the\nsuperhydrophobic coating is found to be around ec ⇡ 30 mm. Figure courtesy of Pr.\nRobert Cohen.\n\n## We created a panel of different meshes described in table IV.1 and denoted A, B,\n\n140 CHAPTER IV. DROP IMPACTING A SIEVE\n\nC, D and J (previous brass Leidenfrost mesh is denoted as E). Meshes were provided\nby the group of Robert Cohen: each one came in two samples of same geometry but\ndifferent wetting properties. The first sample made of stainless steel (McMaster-Carr)\nis further on referred as “plain mesh” 7 . The second sample, elaborated from the same\ngrid, is coated superhydrophobic (referred as “SH mesh”, and identified by an apostrophe\nsign). Coating was manufactured by Siddarth Srinivasan (from Robert Cohen group’s) by\nspraying a PMMA and fluorodecyl POSS solution (low surface tension polymers) dissolved\nin Asahiklin AK-225 (protocol similar to the one done in ). The result of this spray is\na thin coating of spherical structures of typical size 20 mm. The surface topography was\ninvestigated and shown in figure IV.16b. The coating slightly modifies the geometrical\ndimensions of the grid by increasing the wire diameter d and decreasing the hole radius r\n- each approximatively by ec ⇡ 30 mm. For grid E, we have three samples: plain sample\nE, superhydrophobic sample E 0 and Leidenfrost sample EL (the one studied in previous\nsection).\n\n## Name Wire diameter Hole size = 1\n\n(1+d/2r)2\nD⇤ = d+2r\nd\nMesh #\nd (mm) r (mm) (in 1 )\n\n## A, A0 343 407 0.495 3.73 22\n\nB, B 0 140 229 0.586 4.27 42\nC, C 0 254 330 0.518 3.57 28\nD, D0 508 452 0.410 2.78 18\nE, E 0 , EL 170 190 0.477 3.24 46\nJ, J 0 254 190 0.361 2.50 40\n\nTable IV.1 – Table describing the geometrical properties of each mesh. Samples coated\nsuperhydrophobic are marked with an apostrophe. Values in three last columns are de-\nduced from the dimensions r and d. They provide different criteria to characterize the\nmeshes: is the surface fraction of square holes, D⇤ is the dimensionless spacing ratio of\nthe texture, and the mesh number # represents the number of openings per (linear) inch.\nColors are respected in the following figures.\n\n## 4.1 Role of wetting conditions\n\nWe show in figure IV.17 a comparison of transmitted mass for the same mesh E, E 0 and\nEL (mesh number # 46) and different wetting properties. Even if they have a similar\nphysical behavior, we observe two main differences:\n7\nAlthough there is a wide commercial choice of samples, we were not completely free to vary r and d\nindependently, hence the arbitrary choice of geometrical parameters.\n4. EXPLORING DIFFERENT MESHES 141\n\n(i) the minimum critical speed for which mass is transferred is not the same (although\nthey remain comparable). The superhydrophobic grid has the lowest speed V ⇤ , then the\nLeidenfrost grid has a higher one and finally the regular wetting grid has the highest\nthreshold speed,\n(ii) for higher impact speeds the total amount of transferred liquid slightly differs. For\nexample, taking the superhydrophobic grid as reference and focusing on impact speed\nV0 = 4 m/s, regular plate has an increased transfer of 20 % while Leidenfrost grid has a\ndecrease of 10%.\n\n1\nmT\nm0\n0.8\n\n0.6\n\n0.4\n\n0.2\nPlain\nSH\nL e i d e n f r ost\n0\n0 1 2 3 4 5 6\n\nV 0 (m/s)\n\nFigure IV.17 – Percentage of transmitted mass a function of impact speed. Three different\nwetting conditions are shown for the same mesh grid of wire size d = 0.17 mm and hole\nsize r = 0.19 mm (meshes E, E 0 and EL ). The wetting brass mesh is denoted by “solid\ndots”, the mesh treated superhydrophobic is denoted by “stars” and the Leidenfrost mesh\nis denoted by “solid triangles”.\n\n## Previous remarks suggest the following explanations:\n\nSquare versus circular holes The first argument explaining the differences in liq-\nuid transfer is based on the square geometry of the holes (of side length l, see figure\nIV.16a). For a non-wetting mesh, we assumed that the surface through which liquid is\ninjected is circular with a surface area ⇡r2 , (where r = l/2), leading to a surface den-\nsity ⇤ corresponding to circular holes. This is a fair assumption for non-wetting liquids\nwhich avoid the corners generated by the crossing of two fibers - where local curvature is\nhigh. For wetting liquids, the real area to take into account is based on square holes of\nsurface area equal to l2 , to which corresponds the surface density . As a consequence, we\npotentially get an increase in transmitted mass for wetting meshes up to ⇤ = 4/⇡ ⇡ 1.3.\n\n## Coating thickness The wetting properties are modified by deposition of a small\n\ncoating of thickness ec (eL for Leidenfrost sieves) around each wire (which reduces the\n142 CHAPTER IV. DROP IMPACTING A SIEVE\n\nhole surface density). Taking into account this excluded thickness, the expression corre-\nsponding to the new surface density 0 is:\n\nec 2\n0\n= ⇤\n) (1 (IV.27)\nr\nIf we take ec ⇡ 30 mm, we get 0 ⇡ 0.7 ⇤ , in agreement with the reduction of transmit-\nted mass experimentally observed (between regular and super-hydrophobic grids). The\nsame argument applies to the Leidenfrost sieve, suggesting that the vapor layer thickness\neL can even be greater than ec .\n\n## 4.2 Role of hole size r\n\nTo probe the role of hole characteristic size r, we compare in figure IV.18 the transmitted\nmass as a function of impact speed for meshes C and J which have same wire diameter\nd = 254 mm (in shaded gray in table IV.1) and hole size r almost twice bigger for mesh C\n(in red) than for mesh J (in dark green).\n\n1\nmT\nm0\n0.8\n\n0.6\n\n0.4\n\n0.2\n28 SH\n28 P l ai n\n40 P l ai n\n40 SH\n0\n0 1 2 3 4 5 6\n\nV 0 (m/s)\n\nFigure IV.18 – Percentage of transmitted mass a function of impact speed for mesh C\nand C 0 in red, J and J 0 in dark green. “Solid dots” represent plain grids while “asterisks”\nrepresent superhydrophobic ones.\n\nWe observe a higher transmitted mass and a lower threshold speed V ⇤ for mesh C (in\nred), as expected for bigger holes.\nIf we now focus on the difference between wetting and non-wetting properties, critical\nspeed V ⇤ remains almost constant for grids with big holes (mesh # 28 in red in figure\nIV.18). However, for small holes (mesh # 40 in dark green in figure IV.18), critical speed\nV ⇤ of non-wetting grid is lower compared to the wetting one (effect also slightly visible\nin figure IV.17). As sketched in figure IV.19, this may be explained by the fact that the\n4. EXPLORING DIFFERENT MESHES 143\n\nlocal shape of the liquid filament near its base depends on wetting conditions.\n\nrNW\n\nrW\n\nFigure IV.19 – Sketch of filament shape near its base. For impact speed V0 ⇡ V ⇤ , the\nfilament slightly penetrates the texture. If the mesh is non-wetting (in red), filament has\na size rN W at its base. If the mesh is wetting (in blue), filament has a size rW > rN W at\nits base.\n\nIndeed, we assume here that pinch-off occurs at the base of the filament for impact\nspeed V0 close to V ⇤ . If the mesh is wetting (situation in blue in figure IV.19), filament has\na size rW at its base. In the non-wetting situation (in red), filament has a size rN W < rW ,\nfacilitating the pinch-off and resulting in a lower critical speed V ⇤ .\n\n## 4.3 A single curve?\n\n(a) (b)\n\n1.2 1.2\nmT mT\nm0 1 m0 1\n\n0.8 0.8\n\n0.6 0.6\n\n0.4 0.4\n\n0.2 0.2\n22\n42\n0 28 0 46\n18 28\n46 18\n40 40\n−0.2 −0.2\n−0.5 0 0.5 1 −0.5 0 0.5 1\n\nV∗\nV∗\n1− V0 1− V0\n\nFigure IV.20 – Transferred mass as a function of impact speed, expected from scaling law\n(equation IV.21). Blue solid line shows equation y = x. (a) Superhydrophobic meshes.\nMesh D0 (in cyan) shows a saturation for high impact speed, probably due to the fact that\nwe approach the marginal regime where ⌧ ⇡ Tdes (see section 2.4). (b) Regular wetting\ngrids.\n144 CHAPTER IV. DROP IMPACTING A SIEVE\n\n## In this paragraph, we probe previous scaling (described in equation IV.21) by showing\n\nin figure IV.20 the percentage of transmitted mass as a function of 1 V ⇤ /V0 . For all\nsuperhydrophobic substrates (left figure), we see all data collapse close to the line of\nequation y = x - validating our model. However, we can also notice a slight mismatch,\nconfirming the existence of marginal effects that stress the necessity to properly model\nthe deformation of the interface, itself depending on the transferred mass. As discussed in\nsection 4.1 (regarding wetting versus non-wetting effects), data for plain wetting textures\n(figure IV.20b) is systematically above non-wetting experiments (figure IV.20a).\n\n## 4.4 Pinch-off time versus crash time\n\nAs done for a single hole, we now compare in figure IV.21 the ratio of the crashing time\n⌧ and the pinch-off time Tdes as a function of impact speed V0 . We observe that all\nour experiments are in the regime where pinch-off happens faster than the crash time\n(⌧ /Tdes > 1), validating the choice of ⌧ as the right timescale for the scaling argument.\n\nHowever, meshes with big holes and small hole fraction (that is, meshes D and D0 ,\nin cyan) are close to the regime ⌧ /Tdes ⇡ 1 for high impact speeds. If we look back to\nfigure IV.20, we indeed see a deviation for this mesh for high impact speed V0 , suggesting\nthe apparition of new mechanisms, probably the ones described in section 2.4.\n\n2\n10\n\nτ /T des\n1\n10\n\n0\n10\n\n−1\n10\n0\n10\n\nV 0 (m/s)\n\nFigure\np IV.21 – Ratio between the crash time ⌧ = 2R0 /V0 and pinch-off time Tdes ⇠\n⇢r / as a function of impact speed for each mesh (colors defined in table IV.1).\n3\n5. CONCLUSION AND OPEN QUESTIONS 145\n\n## 5 Conclusion and open questions\n\nWe focused in this chapter on drops impacting a grid. Experiments helped us to qualita-\ntively differentiate the mechanisms that play a key role when liquid has to chose between\nflowing through a hole or being horizontally evacuated (when encountering a closing).\nTwo main parameters were discussed: (i) the capillary speed V ⇤ of retraction of liquid\nfilaments forming through the holes, dictating if liquid has enough inertia to pass through\na hole or not, and (ii) the time over which mass is transmitted. Although the natural\ntime is the crash time ⌧ , we can encounter marginal situations were the destabilization\ntime of the liquid fingers or even the rebound time are to be considered. It will be of great\ninterest to re-explore the experiment of an impact on a single hole (done by Lorenceau\net al.) and try to experimentally shed new light on these regimes. A naive scaling law\nwas deduced, qualitatively capturing the speed dependency of transferred mass. Since\nwe would need to be more quantitative to model the interface deformation of the drop, a\ngood starting point would possibly be to adapt the solution of a drop impacting a solid\nplate produced by Eggers et al.\n\nWe also observed that ejected liquid was confined in a cone of angle ↵. All our grids\nwhere thin, allowing the ejected liquid to be laterally deviated. An interesting question\nwould be to see the effect of thicker grids on the opening angle ↵.\n\n## Finally, we explored the role of wetting properties. In general, suppressing liquid-solid\n\ncontact greatly improves liquid mobility: we saw in previous chapters how tiny entrain-\nment forces were able to move levitating objects. Transposing this argument to the case\nof drop impacts, we could have expected a greater transmitted mass for non wetting grids.\nHowever, we experimentally observed the opposite effect, confirming our assumption that\nviscous effects are negligible, even at relatively small length scales.\n\nThe only parameter that we have not varied is the drop radius R0 . It is experimen-\ntally difficult to vary drop size on a wide range. However, we conducted some preliminary\nexperiments with liquid jets impacting grids. Since this new system is at dynamic equi-\nlibrium, it is a promising set up for better understanding the interaction of liquid with a\nmesh.\n\nFew studies are found in the literature regarding impacts on smaller holes. Yet, Brunet\net al. have tackled the problem of a superhydrophobic grid with holes of size r ⇡ 5 mm\n(a hundred times smaller than our experiments). For such small opening size, we should\nexpect relatively high critical speeds: V ⇤ ⇡ 6 m/s. However, they observed that mass is\n146 CHAPTER IV. DROP IMPACTING A SIEVE\n\ntransmitted for speeds V0 up to 2 m/s, that is, three times smaller. Since We = ( VV0⇤ )2 ,\nthis means that mass was unexpectedly transferred for low We numbers - around 0.1.\nSimilar behavior was also observed by Yarin and collaborators [70, 98] who focused on\ndrop impacting electrospun nano fiber membranes with several micrometer non-wettable\npores (porosity around 90 %). They also report transmitted mass for We smaller than\nunity. The origin of the discrepancy between measured thresholds and predicted ones\nremains unclear. However, above cited authors suggest a scenario based on collective ef-\nfects leading to an additional pressure that contributes to make the liquid penetrate more\neasily through the grid. For instance, Yarin et al. suggest an expression of the penetra-\ntion speed for a single hole scaling as Vin ⇠ V0 Rr0 based on a potential solution and liquid\nincompressibility . This scaling can be interpreted as the result of the accumulation\nand channeling kinetic energy of a large mass of liquid through a narrow orifice.\n\nDrop impacting a mesh can be seen as a simple set-up producing mono disperse sprays\nof tiny droplet. For impacts where no mass is transferred (V0 . V ⇤ ), the mesh can also be\nseen as a tool for shaping drops. Indeed, recent work [76, 126] has shown how deformation\nof a drop impacting on cleverly designed superhydrophobic surfaces can lead drops to\nbounce from the surface with the distinct shape of a flat disc. This “pancake bouncing”\nis due to the storing of capillary energy in shape deformation during spreading phase.\nBefore retraction sets in, it is restituted vertically into kinetic energy as sketched in figure\nIV.22.\n\na b c\n\nFigure IV.22 – Liu et al. showed that a drop can lift-off from a surface before re-\ntraction sets in. This greatly reduces the contact time of the drop with the substrate.\n(a) A spherical drop just before hitting the structured surface. (b) Spreading and re-\ntraction dynamics can be modeled as a Hooke spring in both the horizontal and vertical\ndirections. To decrease friction with the walls, the surface of the posts is coated with a\nsuperhydrophobic layer (yellow). (c) Lift-off. Figure adapted from .\n\nOur grids represent a new way to shape drops in a similar way. They have the added\nadvantage of having limited lateral walls and not having any bottom closings. Therefore,\nthey represent a perfect alternate system for studying pancake bouncing.\n5. CONCLUSION AND OPEN QUESTIONS 147\n\nConversely, the situation of bigger holes (than studied here) has been recently ad-\ndressed by de Jong and collaborators who look at droplet impact near a millimeter-\nsize hole. They showed that the behavior of an impact near a closed pit greatly differs\nfrom one on an open-ended pore. This should inspire further work focusing on the impact\nof sealing the exiting section of our grids on the drop spreading dynamics.\nChapter V\n\n## Impact force of a drop\n\nContents\n1 Compression waves . . . . . . . . . . . . . . . . . . . . . . . . . 150\n1.1 Water hammer in the liquid . . . . . . . . . . . . . . . . . . . . 150\n1.2 Water hammer in the surrounding air . . . . . . . . . . . . . . 153\n2 Measure of impact force with a piezo-electric quartz . . . . . 155\n2.1 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . 155\n2.2 Analytical calculation . . . . . . . . . . . . . . . . . . . . . . . 157\n2.3 The case of raindrops . . . . . . . . . . . . . . . . . . . . . . . 158\n3 A cheaper sensor: the lamella . . . . . . . . . . . . . . . . . . . 159\n3.1 Experimental Results . . . . . . . . . . . . . . . . . . . . . . . 159\n3.2 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161\n3.3 Agreement between model and data . . . . . . . . . . . . . . . 162\n3.4 The two impact regimes . . . . . . . . . . . . . . . . . . . . . . 162\n3.5 The case of raindrops . . . . . . . . . . . . . . . . . . . . . . . 163\n3.6 Energy harvesting . . . . . . . . . . . . . . . . . . . . . . . . . 164\n4 Non-wetting impacts . . . . . . . . . . . . . . . . . . . . . . . . 165\n4.1 Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165\n4.2 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166\n5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168\n\n149\n150 CHAPTER V. IMPACT FORCE OF A DROP\n\n1 Compression waves\nWe discussed in previous chapter the spectacular impact figure of a drop hitting a mesh.\nIt was of particular interest to address two limit situations: the plain solid plate (i.e.\nsurface fraction of holes = 0), and the mesh with only holes ( ! 1 ). For an impact\non a solid plane, we saw that liquid had no other choice than to be completely deviated\nhorizontally. Hence, it is natural to think that the obstacle in this situation (that is, the\nplate) endures the biggest impact force. In this chapter, we only focus on the impact force\nof a drop on a solid substrate, a key quantity is soil erosion .\nIt is reasonable to assume that the heavier the drop, the stronger the impact force\nF , so that the drop radius R and density ⇢ naturally arise as key parameters in the\nstudy. However, as discussed in previous chapter IV, we have a geometrical singularity at\nearly stages of the impact. Indeed, if we zoom on the bottom part of the spherical drop\n(see figure V.2), the interface is tangent to the substrate. Hence, the radial spreading of\nthe contact interface is infinite at early stages, generating huge pressures - called water\nhammer pressures (denoted by Pwh ). In what follows we discuss two different scenarios\nregarding the compressibility in the liquid or in the surrounding air.\n\n## 1.1 Water hammer in the liquid\n\nTheoretical expression\nIf we focus in the liquid inside the drop and take into account compressibility effects we\nwill have right after impact a shock wave propagating upwards at the speed of sound\n(denoted as c). This shock wave can be interpreted as the limit separating two domains\nin the drop:\n(i) upstream, where information that the drop has impacted arrived;\n(ii) downstream where this information is still unknown.\nOnce the information has reached the whole drop (in a time 2R/c), compressible effects\nvanish. This effect (also called hydraulic shock) commonly occurs when a valve suddenly\ncloses at the end of a pipeline system, and a pressure wave propagates in the pipe. In\nthe frame of the moving shock wave (as shown in figure V.1), we have again two regions\nseparated by the shock front:\n(i) the bottom of the drop (in red in figure V.1), where liquid is at rest;\n(ii) the top of the drop (in blue in figure V.1), where liquid is still falling at speed V .\nConservation of mass in the pipe leads to:\n\nV\n⇢1 = ⇢2 (1 ) (V.1)\nc\n1. COMPRESSION WAVES 151\n\nV2= V\nP1=P0\nρ2=ρ\nc\n\nV1= 0\nP1=P0 +ΔP\nρ1\n\nFigure V.1 – Shock wave (in grey) moving at the speed of sound c in a pipe of section S.\nIn red, the bottom of the drop where liquid is at rest. In blue, the top of the drop where\nliquid is still falling at speed V .\n\n## Since V << c, we assume ⇢1 ⇡ ⇢2 ⇡ ⇢. We select a volume in the pipe of section S and\n\nlength L: fluid initially at speed V is stopped in a time L/c. Newton’s law gives the force\nF needed to stop this volume : F ⇠ ⇢1 SL VL/c0 . In terms of applied pressure, we can\nwrite F = P S so that denoting P as Pwh we get:\n\nPwh ⇠ ⇢V c (V.2)\n\nEquation V.2 is generally referred to as the Joukowsky equation, but Tijsseling and\nAnderson pointed out that it was actually Von Kries in 1883 who first derived\nand validated it. In addition, Frizell (1898) and Allievi (1902) independently derived the\n\"Joukowsky equation\" in pure theoretical studies. However, it is Rankine (1870) who had\nalready found the equation in the more general context of solids, thus preceding Kries\nand Joukowsky1 .\nAlthough the scaling of water hammer pressure (equation V.2) does not predict the\ngeometrical prefactor, Engel et al. were able to indirectly measure it and found it to\nbe around 0.2 (the same prefactor is also used in later works such as [83, 36]). Hence, we\ncan write:\n\n## Pwh ⇡ 0.2 ⇢V c (V.3)\n\nEquation V.3 gives typical water hammer pressures on the order of MPa - relatively\nhigh pressures. The expression of speed of sound c can be obtained by using the conser-\nvation of energy. Kinetic energy Ek = 1/2⇢V 2 is transformed in potential energy through\ncompression (somehow like storing elastic energy in a spring that is being compressed)\n1\nThese equations of conservation (V.1, V.2) are also known as the Rankine–Hugoniot relations, de-\nscribing the relationship between the states on both sides of a shock wave in a one-dimensional flow.\n152 CHAPTER V. IMPACT FORCE OF A DROP\n\nEel = 1/2 P 2 ,qwhere is the compressibility of the liquid. Balancing these energies\nleads to: P ⇠ ⇢ V . Replacing this pressure in the Joukowsky equation V.1 yields:\n\nr\n1\nc= (V.4)\n\n## Taking spherical geometry into account\n\nr\nc h(t)=Vt\ndr/dt\n\nFigure V.2 – Sketch of the early stage of drop impact on a solid substrate. The contact\nline horizontally spreads at speed dr/dt (in blue) while the shock wave envelope moves at\nspeed c (in red). Contact line position\np r(t) is geometrically deduced from the impacting\nspeed and the drop radius: r(t) ⇡ 2RV t.\n\nPrevious shock model was based on a “cylindrical” drop impacting a solid plate (see\nfigure V.1). Although real drops are spherical, we can assume that this model is valid at\nearly times due to the geometrical singularity. Indeed, the position r(t) of the contact\nline with the plate is fixed by the falling interface of the drop (see figure V.2). We can\nbe more quantitative by calculating the expression of r(t) at height h(t) = V t (V denotes\nhere the impact speed). Since we are only interested in the early stages of the impact,\np p\nr(t) = R2 (R V t)2 can be rewritten as r(t) ⇡ 2RV t. Simple derivation leads to\nthe horizontal speed of the contact line (in blue in figure V.2):\nr\ndr RV\n⇡ (V.5)\ndt 2t\nRight after impact (t = 0), the contact line spreads at infinite speed while the shock\nwave envelope (in red in figure V.2) spreads at speed c. As discussed\nq by Field and collab-\norators [71, 47], since the contact line speed decreases in time as RV\n2t\n, it will eventually\nbe reached by the shock wave at time tc ⇠ 2RV\nc2\n. As a consequence, compressibility effects\nfor a spherical drop last over a time scaling as tc , that is, a time V /c times shorter than\nthe time 2R/c previously discussed. Hence, typical shock waves last over a time on the\norder of the nanosecond.\nMaximum impacting force in this compressible regime is obtained by multiplying\n1. COMPRESSION WAVES 153\n\nthe water hammer pressure ⇢V c by the maximum area over which stress is applied:\n⇡r2 (tc ) ⇡ ⇡( 2RV\nc\n)2 . The resulting force Fwh is found to scale as:\n\nV\nFwh ⇠ ⇢V 2 R2 (V.6)\nc\nIf we put some numbers in this equation, a millimetric drop landing at several m/s\ngenerates water hammer forces on the order of a tenth of a milinewton (ten times smaller\nthan the weight of the drop).\nRecent experiments [117, 118] with a micro-second resolution and theoretical studies\n have shown that air discs can be entrapped under the center of the drop. As a\nconsequence, this will affect the singularity at the contact and thus the pressure and\nimpact force at very short time.\n\n## 1.2 Water hammer in the surrounding air\n\nCompressible effects in the surrounding air also play a key role in the impact mechanism.\nNagel and collaborators [132, 131, 106] have shown that splashing can be completely\nsuppressed by decreasing the pressure of the surrounding gas. Since liquid/solid contact\nspreads horizontally at high speeds during early stages of impact, surrounding air is\nput into motion at similar speeds, generating compression waves in air. The associated\npressure (referred as air hammer pressure in what follows) scales as: Pah ⇠ ⇢a ca dr\ndt\nwhere\n⇢a and ca denote the air density and speed of sound in air, respectively (here, the second\ncharacteristic speed to take into account is the lateral speed of the contact line, hence a\ncharacteristic speed dr\ndt\nas shown in figure V.2). Using equation V.5, we get:\nr\nRV\nPah ⇠ ⇢a ca (V.7)\n2t\nIn [132, 131], Xu et al. have measured the threshold pressure at which a splash first\noccurs and found it to scale with the molecular weight of the gas and the liquid viscosity.\nThey suggest a model where the air hammer pressure Pah balances capillary pressure P in\nthe thin spreading lamella. This thickness is assumed to be the boundary layer thickness\np\n(which is controlled by the diffusion of vorticity from the solid substrate), scaling as ⌫t,\nwhere ⌫ is the kinematic viscosity of the liquid (see figure V.3). Thus we have:\n\nP ⇠p (V.8)\n⌫t\nRe-expressing the air density q\nand speed\nq of sound in air using the equation of state for\nan ideal gas, we get: Pah ⇠ kB T\nP Ma kB T\nMa\nRV\n2t\n(where kB is the Boltzmann’s constant, T\nthe temperature, P the environing pressure, the adiabatic constant of the gas, and Ma\n154 CHAPTER V. IMPACT FORCE OF A DROP\n\ndr/dt √νt\n\nFigure V.3 – Sketch of the bottom of a drop at early stages of impact (by symmetry, we\nonly show the right half of the drop). A liquid lamella horizontally spreads at speed dr/dt.\nIts\np thickness (controlled by the diffusion of vorticity from the solid substrate) grows as\n⌫t.\n\nthe molecular weight of gas). The ratio of these two pressures is independent of time:\ns\nPah Ma ⌫RV\n⇠P 2k T\n(V.9)\nP B\n\nXu et al. suggest that when the two stresses are comparable, the expanding liquid\nrim is slowly destabilized and deflected upwards for an extended period of time, finally\nresulting in the ejection of droplets. Hence, equation V.9 predicts a non-intuitive result:\na more viscous liquid splashes more easily than a less viscous one, that is, the threshold\npressure should decrease if the liquid viscosity is raised (as experimentally observed).2\n\n2\nFor more information on the splashing phenomena, other regimes have been observed for higher\nliquid viscosities [131, 106] and numerical simulations have also been performed [75, 136].\n2. MEASURE OF IMPACT FORCE WITH A PIEZO-ELECTRIC QUARTZ 155\n\n## 2 Measure of impact force with a piezo-electric quartz\n\nWe just saw how compressible effects generate the so-called water hammer effect, to which\ncorresponds high pressures, on the order of MPa. However, this phenomenon lasts over\nextremely short times - namely several nanoseconds. As a consequence, the generated\nforces (below the mN) are difficult to access experimentally. In this section, we focus only\non what happens after this early stage.\n\n## 2.1 Experimental Results\n\nDrops in this study are formed by means of a syringe and various needles, which provides\nradii between 1 and 2 mm. The syringe is fixed to a vertical beam and its height can be\nchanged so as to generate impact speeds V from 20 cm/s to 6 m/s (close to the terminal\nspeed of middle-size raindrops). We record the impact from the side with a fast camera,\nusing backlighting to enhance contrast (figure V.4a).\n\n(a)\n\n(b)\n60\nF\n50\n(mN)\n40\n\n30\nF0\n20\n\n10\n\n−10\n\n## −0.5 0 0.5 1 1.5 2\n\nt (ms)\n\nFigure V.4 – (a) Side view of a water drop of radius R = 1.3 mm hitting a piezoelectric\nquartz at V = 3 m/s. Images are separated by 0.5 ms from which we extract the impact\nvelocity V . (b) Impact force F as a function of time for this experiment, measured by\na piezoelectric sensor. The origin of time is chosen at contact. Impingement typically\nlasts 1 ms, of order 2R/V , the time needed for the drop to travel by its own diameter.\nThe curve is not symmetrical between the beginning and the end of the collision. The\nmaximum force F0 is reached after about a tenth of a millisecond.\n156 CHAPTER V. IMPACT FORCE OF A DROP\n\nBy this means, we access the impinging speed of the drop and can check the centering\nof impacts. In order to control the impact position with a precision of more than 1 mm,\neven for high falls, two perpendicular micrometric screws are placed between the syringe\nand the beam.\nThe main specificity of this study dwells on the impulsive character of the event. A\nfirst characteristic time is the “crashing time” 2R/V of the drop, typically 1 ms in our\nexperiments. We assume here that this time is smaller than the Rayleigh time of vibration\nof the drop, which corresponds to typical impact velocities larger than 20 cm/s.\nWe measure the impact force with a piezoelectric quartz: an impact hammer, PCB\nPiezotronics Model 086D80, is diverted from its normal usage and the signal is delivered\nvia a Kistler charge amplifier 5015A such as used in [83, 82, 97] with a cut-off frequency\nof 10 kHz. The signal obtained after impact is recorded by a digital oscilloscope having\na time resolution3 on the order of 50 ms. We show in figure V.4b a typical profile of\nthe impact force as a function of time, for a water drop with R = 1.30 ± 0.02 mm,\n⇢ = 1000 kg/m3 and an impacting speed V = 3.00 ± 0.02 m/s. We observe that the\ntypical force (50 mN for millimetric water drops) is several orders of magnitude larger\nthan the one generated by water hammer pressure.\n(a) (b)\n\n## F0 R =1. 1 mm F0 Gal i n stan\n\n3 R =1. 3 mm 3 Wa t er\n(m N ) 10 R =2 mm (mN ) 10\n\n2 2\n10 10\n\n1 1\n10 10\n\n0 1 0 1\n10 10 10 10\nV (m /s ) V (m /s )\n\nFigure V.5 – (a) Maximum force F0 at impact as a function of the collision speed V\nfor Galinstan, an alloy of indium, gallium and stain. F0 is typically 300 mN, that is,\nabout 1000 times the weight of the drop. The dashed lines show equation V.10 without\nany adjustable parameter. (b) F0 as a function of V for Galinstan and water at fixed\ndrop radius (R = 1.3 mm). Whatever the impact speed, forces differ by a factor 6,\napproximately the ratio between the two liquid densities. Again, the dashed lines show\nequation V.10.\n\nFrom now on, F0 denotes the maximum force of impact measured on curves similar\n3\nThis resolution does not allow us to extract any information regarding air entrapment (mentioned in\n[79, 117, 118]).\n2. MEASURE OF IMPACT FORCE WITH A PIEZO-ELECTRIC QUARTZ 157\n\nto figure V.4b. In order to obtain a larger signal, even at modest speed or small radius,\nwe choose a liquid denser than water: an eutectic gallium–indium–stain alloy (Ga:In:Sn;\n62:22:16 wt%) commonly called Galinstan recently used for studying impact dy-\nnamics . This liquid metal at ambient temperature is six times denser than water\n(⇢ = 6359 kg/m3 ) and has a viscosity of 2 mPa s, close to that of water. Repeating the\nexperiment of figure V.4b for different drops and impact speeds yields figure V.5a where\nthe maximum force F0 for Galinstan is plotted as a function of V for three radii. The typ-\nical value of F0 now becomes 300 mN, which corresponds to one thousand times the drop\nweight. We observe that F0 evolves as V 2 (dotted lines in figure V.5a are parabolic fits in\nlog–log scale), a signature of the inertial nature of the collision. Hence, the liquid density\nshould also matter, and we compare in figure V.5b the function F0 (V ) for Galinstan and\nwater, at a fixed radius R = 1.3 mm.\nThe parabolic behavior is independent of the liquid nature, and it is found that F0 is\n6 times greater for Galinstan than for water at any impact speed, corresponding to the\ndensity ratio between these two liquids. Studies in the 80s by Nearing [82, 83] and more\nrecently by Grinspan & Gnanamoorthy obtained comparable results with water only\nand on a much more narrow range of velocity, making it difficult to extract scaling laws\nin velocity. In 2012, numerical and experimental studies performed by Mangili et al. \nproduced different predictions for the force of impact. Our measurements are in good\nagreement with their numerical simulations assuming potential flow theory.\n\n## According to these results, we propose a model based on an inertial scenario, as postulated\n\nin . During impact, the transmitted quantity is momentum: at a given time, a slice\nof drop of height V dt, radius r(t) (that changes from 0 to R) and mass dm = ⇢⇡r2 (t)V dt\ndecelerates from V to 0 in a time dt, which yields: F (t) = ⇢⇡r2 (t)V 2 . The maximum\nimpact force F0 is reached for r(t) = R. Hence we get:\n\nF0 = ⇡⇢R2 V 2 (V.10)\n\nThis expression can be seen as a dynamic pressure ⇢V 2 applied over a surface area\n⇡R2 . It can also be understood as arising from the deceleration (from V to 0) of a mass\nM = 4⇡⇢R3 /3 in a time 2R/V . We can now calculate the ratio between the water\nhammer force and the dynamic one: Fwh /F0 ⇠ V /c ⇡ 1/1000. As previously mentioned,\nalthough water hammer pressures are extremely high, the surface over which they act is so\nsmall that the resulting force is V /c times smaller than the one expected from a dynamic\npressure ⇢V 2 acting over a surface R2 . When compared to experiments, equation V.10\n158 CHAPTER V. IMPACT FORCE OF A DROP\n\nis observed (in dashed lines) to nicely fit the different data in figure V.5 without any\n\n## 2.3 The case of raindrops\n\nWe can think of exploiting this result to discuss how force measurements provide an\nestimate of raindrops radii, a quantity of interest in meteorology where it is desired to\naccess the polydispersity of falling rain. In other words, understanding the force F0 may\nallow us to deduce drop radii for a given liquid, providing a new kind of “disdrometer” –\nWe assume that drops reach the sensor with their terminal velocity V0 [51, 92] for\nwhich inertial friction in air balances liquid weight (discussed in chapter I, equation I.9).\nThese two forces can be written as 4⇡⇢R3 g/3 and ⇢a Cx ⇡R2 V 2 /2, respectively (where Cx ⇡\n0.44 is the drag coefficient at Re > 1000). Balancing them, we get a scaling law for the\np\nterminal speed of a raindrop: V0 ⇠ ⇢Rg/⇢a . Injecting this expression in equation V.10\nyields:\n\nF0 ⇠ mg (V.11)\n⇢a\nThis formula emphasizes how the impact force magnifies the drop weight of raindrops\n(by a factor ⇢/⇢a , on the order of 1000), which shows why measuring the impact force can\nbe an accurate method to obtain the drop mass. Indeed, equation V.11 can be rewritten\nas: R ⇠ ( F⇢02⇢ga )1/3 , allowing us to deduce a drop radius from an impact force measurement.\n3. A CHEAPER SENSOR: THE LAMELLA 159\n\n## 3 A cheaper sensor: the lamella\n\nIn order to complement this first method, we now consider a simpler (and cheaper) sensor,\nnamely a thin glass lamella (Menzel-Glaser Microscope cover slip #1) of Young modulus\nE = 69 GPa, density ⇢g = 2350 kg/m3 , thickness h = 160 mm, and transverse width\nb = 24 mm. One side is clamped by squeezing 10 mm of a 60 mm-long lamella between\ntwo thick glass plates. The other side being free, we have a narrow rectangular plate of\nlength L = 50 mm and mass M = ⇢g hbL free to vibrate as sketched in figure V.6. Since\nthe drop spreads at impact, the impact location is adjusted a few millimeters from the\ntip of the plate in order to avoid spilling.\n\n2R\n\nL V\nh\nδ0\n\nFigure V.6 – Sketch of lamella sensor. A thin glass lamella of length L = 50 mm, thickness\nh = 160 mm and transverse width b = 24 mm is clamped at one end and free to vibrate\nat the other one. A drop of radius R and speed V impacts a few millimeters from the\ntip of the plate. In grey, the system before impact; in black, the system at the maximum\ndeflexion 0 . We observe that the drop is sticking to the free end of the plate.\n\n## 3.1 Experimental Results\n\nWe show in figure V.7 a typical side view of the impact resulting in a maximal lamella\ndeflexion 0 :\n\nFigure V.7 – Galinstan drop of radius R = 1.25 mm and speed V = 4 m/s hitting the\nedge of a thin glass plate. Images are separated by 1 ms and only show 1/3 of the lamella.\nWe denote 0 as the maximum plate deflection after impact.\n\nFrom this lateral movie we observe that the dynamics of the plate are dominated by\nits first mode of vibration. In addition, side view of plate oscillations allow us to access\nthe vertical position of the free end of the lamella as a function of time, as shown in figure\nV.8 for different impact speeds.\n160 CHAPTER V. IMPACT FORCE OF A DROP\n\n± 3\n(mm) 2\n\n−1\n\n−2\n±0\n−3\n\n## 0 20 40 60 80 100 120 140\n\nt (ms)\n\nFigure V.8 – Vertical deflection 0 of the lamella tip as a function of time for experiments\nsuch as shown in figure V.7. Each curve corresponds to an impact speed V (the brighter\nthe color, the higher V , which varies from 1.3 m/s to 4.9 m/s by steps of 0.3 m/s). The\ncharacteristic period of vibration of the plate is independent of V and observed here to\nbe ⌧ ⇡ 20 ms.\n\nFocusing on the maximum force, we denote 0 as the largest deformation of the lamella\nfor a given impact (see figure V.6 and V.7). We plot this quantity in figure V.9a as a\nfunction of impact speed for water drops of different radii, and compare in figure V.9b\nthe plate deflection between water and Galinstan for R = 1.3 mm. The results contrast\nwith figure V.5 since we now observe that 0 linearly varies with both V and ⇢.\n\n(a) (b)\n\n## R = 1. 64m m Gal i n stan\n\nδ0 R = 1. 35m m δ0 5 Wa t er\n1.6 R = 1. 29m m\nR = 1. 20m m\n(mm) R = 0. 93m m (mm)\n1.4\n4\n1.2\n1 3\n0.8\n2\n0.6\n0.4\n1\n0.2\n0 0\n0 1 2 3 4 5 0 1 2 3 4 5\n\nV (m /s ) V (m /s )\n\nFigure V.9 – (a) Maximal deflection 0 of the tip of a glass lamella as a function of the\nimpacting speed of a water drop hitting this lamella close to the edge. We do not observe\na parabolic relationship between 0 and V , but a linear law. (b) Comparison between\nwater (lower curve) and Galinstan (upper curve) for a radius R = 1.3 mm. The ratio of\n0 between the liquids is approximately the ratio of their densities.\n3. A CHEAPER SENSOR: THE LAMELLA 161\n\n3.2 Model\n\n## At mechanical equilibrium, the deflection 0 of a thin plate is proportional to the applied\n\nforce F0 . Denoting I = bh3 /12 as the moment of inertia of the plate, thin plate theory\n[120, 107] gives us the following relation between force and deflection:\n\n0 ⇠ F0 L3 /EI (V.12)\n\n## With an impact force varying as V 2 (as seen before), we anticipate a deflection\n\nquadratic in velocity. It is not the case here because the plate has a slow response com-\npared to the crashing time 2R/V (⇡ 1 ms) of the liquid: the characteristic time ⌧0 = 1/f0\nof the plate in figure V.8 is about 20 ms, and it is expected to be a function of the plate\np\nparameters (⌧0 = M L3 /EI). Since the lamella is not at static equilibrium during an\nimpact ten times shorter than its response time we cannot assume instantaneous pro-\nportionality between deformation and force. We propose here to understand the data\nusing an argument based on momentum conservation. Before impact, the drop of mass\nm moves at speed V and the plate of mass M is at rest. After impact, the drop sticks to\nthe plate, resulting in a system of mass (m + M ) vibrating at its natural frequency, inde-\npendent of the collision speed V . With a uniform\nq distribution of m along the plate, the\nfirst resonance frequency would become f = f0 MM +m\n. A geometrical correction could be\nintroduced to take into account the fact that the drop is neither localized at the free end,\nnor homogeneously spread all over the lamella. However, since m << M , this correction\nis marginal, and we assume f ⇡ f0 . The lamella is vibrating on its natural mode, with a\nparabolic modal shape as a first approximation for clamped-free end conditions. Hence, its\nRL 2\nmomentum can be written: pM = 0 ⇢g bh 2⇡f0 0 Lx 2 cos(2⇡f0 t)dx = 2⇡ 3\nM f0 0 cos(2⇡f0 t).\nOn the other hand, the momentum of a drop vibrating at the tip of the lamella is:\npm = m 2⇡f0 0 cos(2⇡f0 t). Assuming that the first mode accounts for the dynamics also\nat t = 0, the conservation of momentum yields for m << M :\n\n3 mV\n0 ⇡ (V.13)\n2⇡ M f0\n\nEquation V.13 is found to be linear in V (instead of quadratic) and sensible to the ratio\nof mass between the drop and the lamella. Typical mass ratio m/M in our experiments\nbeing 1/30, a millimetric drop hitting the plate at several meters per second will induce\nmillimetric deflections to our glass lamella (which has a typical vibration frequency f0 of\n50 Hz).\n162 CHAPTER V. IMPACT FORCE OF A DROP\n\n## 3.3 Agreement between model and data\n\nIn order to explore a large range of densities, we repeated the experiment with the previous\nliquids (water, Galinstan) to which we added acetone (of density ⇢ = 714 kg/m3 ), hexane\n(⇢ = 659 kg/m3 ) and a viscous silicone oil (⇢ = 970 kg/m3 , viscosity of 500 mPa s). We\nshow in figure V.10 how the deflection 0 rescaled by the distance V /f0 , as suggested by\nthe model, varies as a function of the mass ratio m/M for all experiments. As predicted\nby equation V.13, data collapse on a line of slope 1.\n\n−1 Wa t er\nE t ha no l\nf 0 δ 0 /V 10 G a l i ns t a n\nHexa n\nS i l i c o ne O i l\n\n−2\n10\n\n−2 −1\n10 10\n\nm/M\n\nFigure V.10 – Normalized deflection f0 0 /V as a function of the ratio m/M between the\ndrop and the plate mass. V varies from 20 cm/s to 6 m/s. The red line shows equation\ny = (3/2⇡)x, that is, equation V.13. Each color corresponds to a liquid, and each cluster\nto different radii R ranging from 0.9 mm to 1.7 mm.\n\n## 3.4 The two impact regimes\n\nAccording to the characteristic response time ⌧0 of the substrate, we expect two regimes\nof impact:\n(i) a fast one, where the liquid crash is faster than the response of the substrate, which\nleads to equation V.13 and corresponds to all our experiments,\n(ii) a slow one, where impact is slower than ⌧0 . In this case, since the response\nof the plate is quasi-static (mechanical equilibrium), the deformation is proportional to\nthe force: 0 scales as L3 F0 /EI (equation V.12), which yields (using equation V.10):\n0 ⇠ ⇡⇢R V L /EI, quadratic in velocity.\n2 2 3\n\n## Since ⌧0 is a function of L, let us re-express the transition between both regimes in\n\nterms of substrate length: the linear regime in V occurs for L > Lc , and the quadratic\n3. A CHEAPER SENSOR: THE LAMELLA 163\n\n## one for L < Lc where the critical length Lc is given by:\n\n 1/4 1/2\nEh2 R\nLc ⇠ (V.14)\n⇢g V\nFor glass lamellae and millimetric drops in our range of impact speeds, this critical\nlength is around 3 cm. Since the lamella in experiments (L = 5 cm) is longer than Lc , we\nindeed expect 0 (V ) to be linear, as observed in figure V.9.\nFrom a more general point of view, leaves can also be viewed as lamellae, and drop\nimpact can naturally bend or even damage them . We expect leaves to have a large\nresponse time ⌧0 , owing to an effective Young modulus E more than one thousand times\nsmaller than for glass. For a leaf (of thickness h and width b comparable to that of our\nlamellae), Lc becomes about 6 mm. Most plants have leaves longer than Lc , so that\nequation V.13 can be used to estimate the deflection of leaves under rain. For millimetric\ndrops on leaves of a few centimeters (⌧0 ⇡ 600 ms, f0 ⇡ 1.5 Hz), equation V.13 predicts\ncentimetric deflections. But equation V.13 also exhibits the sensitivity to other parameters\nof the leaf: bigger drops hitting thin small leaves can generate a deflection comparable to\nthe leaf size, which can lead to breakage after multiple impacts.\n\n## 3.5 The case of raindrops\n\nAs previously done for the piezo-electric quartz, we can use a lamella to propose an\nalternate solution to achieve a disdrometer. Two regimes arise regarding lamellae length:\n\nFor lamellae of length L > Lc , we obtain after expressing the terminal velocity of\nraindrops in equation V.13:\n\n⇢3/2 g 1/2\n0 ⇠ 1/2\nR7/2 (V.15)\n⇢a M f0\n\nFor lamellae of length L < Lc , injecting the expression of impact force of a rain-\ndrop (equation V.11) into equation V.12, we get:\n\n⇢2 gL3 3\n0 ⇠ R (V.16)\n⇢a Ebh3\nIn both regimes, the deflection 0 is a lightly-sensitive function of R, and it can be\nmeasured from a side view video, from which we can deduce the drop radius. As a\nconsequence, by associating several sensing plates (hydrophobically coated in order to\navoid cleaning between successive impacts), we can assess the polydispersity of rain.\n164 CHAPTER V. IMPACT FORCE OF A DROP\n\n## 3.6 Energy harvesting\n\nWe can finally estimate the collision energy. The system periodically transfers bending en-\nergy Ebh3 02 /L3 into kinetic energy Ek plate = 1/5[1/2M ( 0 2⇡f0 )2 ], that is, (9m/5M )Ek drop .\nThe kinetic energy Ek drop of a raindrop is around 1 mJ and the energy of the plate after\nimpact is typically 10 4 J. Assuming an efficiency of 10% when transforming bending en-\nergy into electrical energy with a piezo-sensor, and supposing that this energy is delivered\nfor each vibration, we get a transmitted power of 10 3 W. In the case of rain, assuming an\nimpact every 10 seconds, we deduce that a roof of 100 m2 with 106 receivers can deliver an\naverage power of 100 W, just enough to light a bulb – a modest amount. Even if raindrops\nfall from high, drag slows them down dissipating almost all their potential energy, which\nis not efficient if we dreamt of energy harvesting but a blessing for plants, soil or living\ncreatures that luckily experience relatively less impact force or erosion!\n4. NON-WETTING IMPACTS 165\n\n4 Non-wetting impacts\nIn this section, we address the question of the influence of wettability in the impact\nproblem. As we know, a substrate coated superhydrophobic generates bouncing after\nimpact.\n\n4.1 Deflection\n\nWe consider a drop impacting the same previous thin plate, excepted it is now coated to\nbe non-wetting. If the time ⌧ of crash (scaling as 2R/V ) is shorter than the period ⌧0 of\nvibration of the plate, the drop can spill away and it will not stick anymore to the plate\nafter impact, resulting in a lighter final vibrating system. Since we saw that this problem\nis based in momentum conservation and since initial momentum remains unchanged, we\nexpect higher deflections. Indeed, if we do not neglect the drop mass compared to the\nplate one, our argument in section 3.2 gives:\n\n3 mV 1\n0 = m (V.17)\n2⇡ M f 1 + 3 M\nIn equation\nq V.17 frequency f also depends on the final mass of the system and varies\nas f = f0 M +m , so that the expression of deflection 0 becomes:\nM\n\nr\n3 mV 1 M +m\n0 = m (V.18)\n2⇡ M f0 1 + 3 M M\nFor m < M , we get back to equation V.13, which exactly corresponds to the situation\nof an impact on a super-hydrophobic substrate: the drop bouncing away from the plate,\nit is not anymore attached to it. Hence, the ratio between maximum deflection in a\nnon-wetting state nw and in a wetting state w is:\nr\nm M\nnw / w = (1 + 3 ) (V.19)\nM M +m\nAs expected, equation V.19 states that we have a higher deflection for a non-wetting\nplate. For small mass ratio m/M < 1, Taylor expansion leads to nw / w ⇡ 1 + 52 M m\n. In\nour experiments done with water, drop mass m was much smaller than plate mass M\n(m/M ⇡ 1/30). As a consequence, the correction due to taking into account the mass\nsticking to the plate is a few percent - indeed, negligible. However, if we repeat the\nexperience with a drop of radius twice bigger, we would expect a difference in deflection\nup to a factor 2 - clearly measurable and not negligible anymore. It could be an interesting\nchallenge to experimentally probe this relation.\n166 CHAPTER V. IMPACT FORCE OF A DROP\n\n4.2 Force\n\nWe now think in terms of impact force (instead of maximum deflection). If the re-\np\nbound time Treb of the drop (scaling as ⇢R3 / ) is shorter than the crashing time 2R/V ,\nC. Ybert and coworkers suggest a new expression for the force based on transfer of\nmomentum. Indeed, the drop will bounce before it has completely crashed. Assuming a\nperfect rebound (restitution coefficient of 1) the drop experiences a change of momentum\n2mV in a time Treb . Hence, the substrate experiences a force Fnw scaling as:\n\np\nFnw ⇠ ⇢ R3 V 2 (V.20)\n\nEquation V.20 predicts an impact force linearly dependent on impact speed V (in the\nnon-wetting bouncing situation). At low speed, this means a stronger force compared to\nthe wetting case (which had a parabolic dependence).\nThe condition of validity Treb < 2R/V can be re-interpreted in terms of speed:qthe\nimpacting speed V has to be smaller than Taylor-Culick speed of retraction V ⇤ = ⇢R .\nFor a millimetric water droplet V ⇤ is on the order of 20 cm/s - relatively small. At this\nmaximum speed V ⇤ , we would expect the maximum force Fnw ⇤\nin this linear regime to\nscale as:\n\nFnw ⇠ R (V.21)\n\nIf we put some numbers in equation V.21, we get a force on the order of 0.1 mN - too\nsmall to be detected by our sensor. In order to increase the intensity of the signal we can\nthink of using mercury. Indeed, this liquid metal has a surface tension approximatively\nsix times higher than water. As a consequence, the corresponding maximal force Fnw ⇤\n\n## is increased by the same factor, resulting in measurable forces, probably allowing us to\n\nexperimentally bring to light this regime. Another solution to have a stronger signal\nwould be to use rubber ballons filled with liquid. This technique, already used by Clanet\net al. to study the maximal deformation of an impacting drop , showed that we can\nhave an increase in apparent surface tension of a factor up to a thousand.\nSurface tension is related to matter cohesiveness (or affinity between molecules) hence,\na higher surface tension implies also a higher density. As a consequence, the maximum\nspeed V ⇤ , which is a function of the ratio ⇢/ , is relatively insensitive to a change in liquid\nnature: the critical speed for mercury is found to be approximatively 70% of the one of\nwater.\nIf we now focus on the ratio of forces between wetting and non-wetting substrates, we\nget:\n4. NON-WETTING IMPACTS 167\n\nFnw V⇤\n⇠ (V.22)\nF0 V\nEquation V.22 predicts a strong effect of bouncing at small impacting speeds4 . The\ndivergence predicted for V ! 0 should be difficult to probe since the corresponding forces\ntend towards zero.\n\n4\nSince we can not play on wetting properties of mercury, this difference can not be experimentally\nprobed with mercury.\n168 CHAPTER V. IMPACT FORCE OF A DROP\n\n5 Conclusion\nDrop impacts have been extensively studied for more than one century due to their ubiq-\nuitous implication in everyday life. For printing, coating or spraying, from pesticides to\nrain, [26, 134], it is essential to understand the collision mechanisms of a drop. Many\naspects have been and are still explored [134, 130]. Special attention has been given to\nthe early stages of splashes [45, 132] to the dynamics of the spreading radius or to\nthe influence of the substrate [122, 55, 123, 36, 72].\nHowever, the force experienced by a substrate hit by a drop has been less discussed,\napart from early discussions about soil erosion [83, 82, 53] and more recent studies about\nthe rain impact on small creatures . Indeed, drop impacts are difficult to characterize\ndue to their transient, non-stationary nature. We discussed in this chapter the force gen-\nerated during impacts, a key quantity for animals, plants, roofs or soil erosion.\n\nAt early stages, compression waves in the liquid generate extremely high pressures (on\nthe order of MPa). However, they last over a time so short, that the corresponding forces\nremain small (below the millinewton).\nWe measured this force through two different systems: a piezo-electric quartz sensor\nand a deflecting thin lamella. Although a millimetric drop has a modest weight, it can\ngenerate collision forces on the order of thousand times this weight. We modeled this\namplification considering natural parameters such as drop radius and density, impact\nspeed and response time of the substrate. We proposed as well two kinds of devices\nallowing us to deduce raindrops size from impact forces.\nFinally, we theoretically explored the influence of a superhydrophobic coating allowing\nthe drop to bounce. We deduced that if the drop bounces faster than the response time\nof the lamella, the maximal deflection of the plate will be enhanced. Similarly, if the\ndrop bounces faster than the crashing time of the drop, we expect increased impact force.\nFuture work should lead to new experiments allowing to probe these models.\nConclusion\n\n## In this work, we discussed special dynamics experienced by non-wetting objects. We fo-\n\ncused on two opposite phenomena: the onset of motion (due to textured substrates) and\nthe radical stop (due to an obstacle such as a solid plate or a grid).\n\nIn 2006, Linke et al. showed that liquid drops deposited on hot solids covered by\nasymmetric teeth self-propel. We show that herringbone patterns on solids also propel\nLeidenfrost drops, which can be seen as a geometrical proof of the scenario suggesting\nthat viscous drag should generally be responsible for such motions on asymmetric solids.\nContrasting with ratchets, geometry and resulting vapor flows on herringbones are simple\nand controllable, allowing us to produce quantitative models for both the propelling and\nthe friction force, and to discuss how the design can be optimized. Maximizing the force\nis useful if it is desired to oppose an existing force (such as gravity, if the substrate is\ninclined); in other cases, it can be interesting to optimize the drop speed to enhance the\nmotion. The corresponding optimal chevron angles are not the same, but both properties\nemphasize the role of geometry in these devices.\n\nWe also show that viscous entrainement effects can be extended to room temperature:\nblowing air through the holes of an air-hockey table with etched chevrons generates both\nlevitation and self-propulsion of plastic cards and glass platelets placed on such tables.\nPropulsion takes place in the same direction and with the same characteristics as in the\nLeidenfrost case, showing the generality and versatility of these devices. However, if we\nuse deeper channels we lose the confinement effect. We show that we need stronger air-\nflows in order to induce levitation. Even though we still see motion, it takes place in the\nopposite direction. Although these two regimes (low versus high Reynolds number) are\nbased on completely different physical mechanisms (viscous effects versus inertial ones)\nthey surprisingly obey the same scaling law.\n\nRegarding the impact of a drop, we first focus on the impact on a grid. Liquid chooses\nto pass through a hole depending on the capillary pressure defined by the size of the hole\n(and by the wetting conditions). We show that the total amount of transmitted mass is\n\n169\n170 CONCLUSION\n\nusually fixed by the time of crash. However, different timescales have to be taken into\naccount in other marginal situations such as when bouncing is faster than crashing.\n\nFinally, we focus on drop impacts and their transient, non-stationary nature. We show\nthat although a millimetric drop has a modest weight, it can generate collision forces on\nthe order of one thousand times this weight. We measure and discuss this amplification,\nconsidering natural parameters such as drop radius and density, impact speed and response\ntime of the substrate. We finally imagine and describe two kinds of device allowing us to\ndeduce the size of raindrops from impact forces.\nAppendix A\n\n## G-code script to machine a herringbone\n\npattern\n\nIn this appendix, we give the G-Code script used to texture a herringbone pattern on a\nbrass substrate using a numerical CNC-Milling machine. Variables used by the machine\nare preceded by “#”. Comments are either on the right hand side or preceded by “//”.\nThis work has been possible thanks to the valuable help of Guillaume Clermont.\n\n% @MACRO\n//The zero reference is on the right top side of the piece (which is horizontal)\n// All units must be in mm\n//N denotes the number of passings in each crenel\n\n//Definition of variables\n#1=60; Herringbone half opening angle ↵ (deg)\n#2=0.1; Depth pitch for each passing\n#21=0.20; Maximum depth: h (mm), must be multiple of #2\n#3=30.; Width of substrate (mm)\n#5=1; Diamater of reamer, defining width of groove W (mm)\n#6=0.2; Width of wall (mm)\n#7=190.; Length of substrate (mm)\n#19=#5+1.; Safety margin (mm)\n\n## #16=(#7/(#6+#5)/SIN(#1))+1; Number of passing to do\n\n#8=0; Loop counter\n\n//Start\n\n171\n172 APPENDIX A. G-CODE SCRIPT TO MACHINE A HERRINGBONE PATTERN\n\n## T1; Name of the reamer\n\nM3 S3700; Milling speed\n\nG00 Z5.0; Go up\nG00 Y(#19); Go to zero\nG01 Z(-#2) F100.00; Reamer down at translation speed 25\n\n//Loop\nWHILE #8<=(#16) DO;\n\n## #9=-#8*((#5+#6)/SIN(#1)); Initial position of the loop\n\n#11=-#3/2; First position\n#12=#9+(#3/2+#19)/TAN(#1); Second position\n#13=(-#3-#19); Third position\n#20=-#2; Loop counter for depth\n\n## WHILE -#20<#21 DO;\n\nG01 Z#20;\nG01 X#9 Y#19;\nG01 X#12 Y#11;\nG01 X#9 Y#13;\n\n#20=#20-#2;\nG01 Z(#20);\nLower reamer G01 X#12 Y#11;\nG01 X#9 Y#19;\n\n#20=#20-#2;\n\nEND_WHILE\n\n## G01 Z(-#2); Take reamer to next initial position\n\nG01 X(#9-((#5+#6)/SIN(#1)));\n#8=#8+1; Loop counter\n\nEND_WHILE;\nG01 Z1; M5; Take reamer away\nAppendix B\n\n## Exact calculation of viscous\n\nentrainment force\n\n## In this appendix, we present an exact calculation of the force of viscous entrainment\n\ngenerating self-propulsion of a Leidenfrost drop on a herringbone texture. By symmetry,\nprojection of the forces in the plane perpendicular to the axis of symmetry of the herring-\nbone compensate and they add up in the plane of symmetry. After projection (hence the\ncosine of ↵), total entrainement force is:\nZZ\n6⌘U\nF = 2 cos ↵ dXdY (B.1)\nS h\nHere, the surface of integration S is half the circle below the drop1 . To get an analytical\nsolution, the best method is to use the frame ! ex , !\ney defined by the direction of the grooves,\nsketched in figure B.1 and different from the frame e! !\nX , eY naturally associated to the plane\nof symmetry. Instead of first calculating the force generated by the Poiseuille flow in each\nchannel and then adding them up, we directly integrate the stress over the whole surface\n(inspired by the numerical work done in ). We inject in equation B.2 the expression of\nthe horizontal speed U in a single channel described in equation II.7 and obtained from\nconservation of mass and thermal balance: U = ⇢Lh T\n2 x. Denoting\n12⌘ T\n⇢Lh3\nas A, equation\nRR\nB.2 can be rewritten as: F = A cos ↵ S x dXdY . We decompose the total integration in\ntwo regions:\nRR\n(i) upper one, denoted as S1 (in red in figure B.1a): I1 = S1 x dXdY , and\nRR\n(ii) lower one, denoted as S2 (in blue in figure B.1b): I2 = S2 x dXdY ,\nThese notations lead to the following expression for the total entraining force:\n\n1\nWe can take into account the fact that the walls occupy a surface that does not contribute to\npropulsion by multiplying S by W/( + W ), where W is the width of a groove and the thickness of a\nwall.\n\n173\n174APPENDIX B. EXACT CALCULATION OF VISCOUS ENTRAINMENT FORCE\n\n## F = A cos ↵(I1 + I2 ) (B.2)\n\n(a) (b)\neY eY\ney ey\nα α\n\nR ex\ny\neX eX\nxmin xmin R\nxmax\n\nxmax ex\ny\n\nFigure B.1 – Sketch of surface over which we integrate equation B.2. (a) Upper part, in\nred, corresponding to area S1 (and integral I1 ). (b) Lower part, in blue, corresponding to\narea S2 (and integral I2 ).\n\nUpper contribution\nRR\nWe focus now on the mathematical integration of I1 = S1 x dXdY . Looking at figure\np\nB.1a, we have to integrate x in the !\nex direction from xmin = y/ tan ↵ to xmax = R2 y 2\nfor a channel at height y. We then have to integrate this result in the !ey direction from\nymin = 0 to ymax = R d = R sin ↵. Hence:\nZ R sin ↵ Z pR 2 y2\nI1 = x dxdy (B.3)\n0 y/ tan ↵\n\n## After integration, we get:\n\nsin3 ↵ 3\nI1 = R (B.4)\n3\n\nLower contribution\nRR\nWe discuss here the mathematical integration of I2 = S2 x dXdY . Looking at figure\nB.1b, we have to integrate x in the !\nex direction from xmin = y/ tan ↵ to xmax = R cos ↵ +\n175\n\np\nR2 (R sin ↵ y)2 (as expected, for y = 0, xmax = 2R cos ↵) for a channel at height y.\nWe then have to integrate this result in the !\ney direction from ymin = 0 to ymax = R sin ↵.\nHence:\n\nZ Z p\nR sin ↵ R cos ↵+ R2 (R sin ↵ y)2\nI2 = x dxdy (B.5)\n0 y/ tan ↵\n\n## 6 cos ↵(2↵ + sin 2↵) + (11 sin ↵ + 3 sin 3↵) 3\n\nI2 = R (B.6)\n24\n\nResulting force\nInjecting solutions obtained in B.4 and B.6 in equation B.2 yields the expression of the\ntotal propelling force:\n\n6⌘ T 3\nF = R f (↵) (B.7)\n⇢Lh3\nwhere f is the function defined as f (↵) = 23 cos ↵(sin3 ↵+ 18 [6 cos ↵(2↵+sin 2↵)+11 sin ↵+\n3 sin 3↵]). As expected, equation B.7 nearly has the same expression as the one obtained\nin section 2.2, equation II.9 (provided we set f (↵) ⇠ sin 2↵).\nWe show in blue in figure B.2 the function f (↵) obtained from equation B.7. In red,\nthe function obtained from approximation done in chapter II, leading to equation II.9.\nAlthough they have a similar allure, the maximum for the exact calculation (in blue) is\nslightly shifted towards lower angles (and is also slight higher).\n\n1.5\n\n0.5\n\n0\n0 10 20 30 40 50 60 70 80 90\n\nα ( ◦)\n\nFigure B.2 – Comparison of functions f (↵): in blue, obtained from exact calculation\n(equation B.7); in red, obtained from approximation done in chapter II (leading to equa-\ntion II.9).\nAppendix C\n\nRésumé en Français\n\nCette thèse porte sur les liquides non-mouillants et leur interaction avec des textures.\nElle s’articule autour de cinq chapitres. Le premier chapitre introduit clairement toutes\nles notions nécessaires à la lecture du manuscrit, et replace la problématique dans le con-\ntexte de la recherche internationale sur ces sujets. Les deux chapitres suivants traitent\nle phénomène de lévitation et d’auto-propulsion. Les deux derniers chapitres se centrent\nsur le phénomène d’impact d’une goutte: soit avec une grille soit avec une plaque lisse.\nNous détaillons par la suite le contenu de chaque chapitre.\n\nLe premier chapitre est consacré à la physique des objets non-mouillants (voir figure\nC.1). Trois méthodes permettant d’atteindre cet état sont discutés . La première,\npropose de réduire le contact entre liquide et solide en combinant un traitement chimique\net physique afin de rendre le substrat super-hydrophobe . La deuxième, se centre sur\nla lévitation par caléfaction, c’est-à-dire, la lévitation d’objets (solides ou liquides) sur la\ncouche de gaz qu’ils génèrent en s’évaporant . La dernière, très similaire, consiste à\nsouffler de l’air à travers le substrat poreux qui se trouve sous la goute - à l’image d’un\npalet qui lévite sur une table d’air-hockey .\nOn étudie par la suite les conséquences de l’existence de cette couche de gaz (isolant\nphysiquement et thermiquement la goutte du substrat) sur la forme de ces perles liquides.\nN’ayant plus aucun contact avec le solide, les petites gouttes sont dominées principale-\nment par les effets de tension de surface. Cette force interfaciale, qui essaye de minimiser\nla surface des objets, leur confère une forme sphérique. Pour des plus grosses gouttes, la\ngravité doit être prise en compte, et la goutte s’aplatit sous l’effet de son propre poids.\nApres avoir discuté leur forme, nous nous intéressons au coussin de vapeur formé sous la\ngoutte. Son épaisseur résulte d’un équilibre dynamique entre le poids de la goutte (qui\nest en train d’écraser le coussin de vapeur), et le liquide évaporé qui est en permanence\nen train de l’alimenter. La question de stabilité de ce coussin de vapeur et de la forme\nde la goutte y sont aussi traités. Cette thèse portant sur l’interaction de gouttes avec des\n\n177\n178 APPENDIX C. RÉSUMÉ EN FRANÇAIS\n\n(a) (b)\n\nFigure C.1 – Gouttes en caléfaction (a) Les petites goutes sont soumises principalement à\nla tension de surface, ce qui leur confère leur aspect sphérique. L’échelle représente 1 mm.\n(b) Les plus grosses gouttes sont aplaties par la gravité. L’échelle représente 2 mm.\n\ntextures, nous discutons comment la forme d’une goutte posé sur une surface crénelé est\nmodifiée (par rapport au cas du substrat lisse). Par ailleurs, l’extrême mobilité de ces\nobjets donne toute sa richesse à ce sujet. Leur dynamique est en effet gouvernée par des\nforces très faibles, dont la nature et la valeur sont particulièrement difficiles à identifier et\nà mesurer. Ainsi, pour clore ce chapitre nous passons en revue les différentes forces qui\npeuvent générer de la friction sur ces aéroglisseurs: friction dans le coussin d’air, friction\navec l’air environnent ou friction sur une surface crénelée.\n\nCette introduction, qui pose les bases de ce travail en même temps qu’elle introduit\ndes résultats inédits, ouvre naturellement le chemin au deuxième chapitre. On y présente\nl’étude du mouvement d’un liquide caléfié sur un substrat texturé. En 2006, H. Linke\n a ainsi montré qu’un liquide posé en caléfaction sur un support couvert de dents\nasymétriques est autopropulsé dans la direction de la pente descendante des dents. Le\npoint essentiel est la production de vapeur sous la goutte, une vapeur évacuée par la\npression qu’exerce le liquide sur le film qui le supporte. Si cet écoulement est isotrope\nsur un solide plan (cas de la figure C.1), il peut ne plus l’être sur un solide aux textures\nasymétriques (cas de la figure C.2). En disposant des microbilles de silice sur les textures,\nG. Dupeux a réussi à mesurer le champ de vitesse dans la très fine couche de vapeur\nqui maintient l’objet en lévitation. Il a ainsi mis en évidence que le gaz s’écoule vers\nla zone la plus profonde du sillon avant d’être évacué latéralement. C’est ce flux vers\nles zones profondes de la texture qui entraine, par viscosité, la goutte sur le substrat.\nOn comprend ainsi que la locomotion est activée dès que l’on canalise la vapeur dans\nune direction donnée – ce que fait la vapeur sur les dents à cause de leur asymétrie.\nCependant, l’écoulement de vapeur dans cette texture reste extrêmement complexe. C’est\ncette complexité (et la difficulté qui en découle à modéliser physiquement ce problème) qui\n179\n\n## nous a poussé à chercher des nouvelles textures permettant de générer de l’autopropulsion.\n\nFigure C.2 – Vue de trois quarts d’une goutte en caléfaction sur une surface texturée en\nforme de chevron. De part et d’autre du plan de symétrie, la texture est formée par des\ncréneaux parallèles entre eux. Pour cette texture, le demi-angle au sommet du chevron\nest de 45 - ce qui correspond à l’optimum en force.\n\nTout en respectant les ingrédients clefs pour établir ce type de mouvement (c’est-à-\ndire, la rectification de l’écoulement de vapeur de façon asymétrique) nous avons abouti\nà une nouvelle texture. Cette dernière est formée localement de cerneaux parallèles qui\nse rejoignent de part et d’autre d’un plan de symétrie pour créer un motif en forme de\nchevron (voir figure C.2 ou C.4). Cette texture peut être ainsi vue comme étant une\npreuve géométrique permettant de confirmer le scénario d’entraînement visqueux. En\neffet, dans chaque cellule, la vapeur n’a pas d’autre choix que de s’écouler le long du\ncréneau et de suivre ainsi la direction imposée par celui-ci. La simplicité de l’écoulement\ndans cette configuration est traduite en loi d’échelle et conduit à une force en accord\navec les résultats expérimentaux. Par la même occasion, ce modèle nous permet de dis-\ncuter de l’optimisation de la force de propulsion par rapport aux différents paramètres\ngéométriques du problème. Par exemple, en ce qui concerne le demi-angle définissant\nl’ouverture du chevron, nous montrons que la force est maximale pour un angle de 45\n(voir figure C.2). L’absence de contact rend ces gouttes extrêmement mobiles (ce sont\ndes aéroglisseurs) et empêche l’ébullition. Ainsi, la grande mobilité de ces objets soulève\nla question de la friction qu’ils subissent. Très faible sur un substrat lisse, on observe\nune dissipation inertielle dans l’air environnant et dans une couche limite liquide pour les\ngouttes. En revanche, sur un substrat crénelé, elle est fortement amplifiée par l’impact\ndu liquide sur les textures. En utilisant comme point de départ la friction étudiée par G.\nDupeux lorsqu’une goutte avance sur des rainures perpendiculaires à sa trajectoire ,\nnous élargissons cette étude au cas des trajectoires formant un angle donné par rapport\nà l’alignement des créneaux. Nous proposons par la suite un modèle qui rend compte de\nla friction que subit une goutte dans le cas qui nous intéresse tout particulièrement: des\ntextures en forme de chevron. Ayant étudié la force de propulsion et de friction, il ne\nnous reste plus qu’à en déduire la vitesse terminale en les équilibrant. Le modèle qui en\nrésulte est en très bon accord avec les données expérimentales et il permet de prédire un\nmaximum de vitesse pour un demi-angle au sommet de 21 . En effet, cet angle représente\n180 APPENDIX C. RÉSUMÉ EN FRANÇAIS\n\n## le meilleur compromis entre l’optimisation de la force de propulsion (proche de 45 ) et la\n\nminimisation de la friction (proche de 0 , correspondant à des créneaux alignés avec la\ntrajectoire, ce qui minimise les chocs entre la goutte et les murs des rainures). Dans ce\nchapitre, nous proposons finalement d’utiliser cette texture comme point de départ pour\nde nouvelles applications: un piège à gouttes caléfiées (figure C.3a), une piste circulaire\npour les observer sur des longues trajectoires (figure C.3b) ou même des textures actives\nafin d’interagir, en temps réel, avec nos perles liquides.\n\n(a)\n\n(b)\n\nFigure C.3 – Dispositifs utilisant les textures en forme de chevron comme unité de base.\n(a) Deux chevrons juxtaposés avec des polarités opposées permettent de piéger une goutte\nà leur intersection. (b) Deux chevrons reliés par des sections circulaires vont faciliter\nl’observation du mouvement de la goutte sur de très longues distances.\n\n## Induire de la lévitation en utilisant la température est restrictif: un liquide non volatil\n\nne lévitera pas, et nous ne serons pas toujours en mesure de chauffer le support. Il est\ndonc intéressant de se demander si l’on peut caléfier “à froid” un liquide. Le troisième\nchapitre répond à cette question en remplaçant la goutte liquide par un objet solide (une\nlamelle de verre). N’ayant plus d’évaporation pour nourrir le coussin qui se trouve entre\nla goutte et le substrat, nous perforons ce dernier pour le rendre poreux et y injecter\nde l’air à travers . Il en résulte une table de air-hockey (permettant la lévitation,\n181\n\n) sur laquelle nous allons venir texturer nos motifs en forme de chevron (générant\nle mouvement, voir figure C.4). Le fait d’entraîner par viscosité un objet solide nous\n\nFigure C.4 – Plaque de verre (épaisseur 1 mm, largeur 24 mm, longueur 45 mm) posée\nsur une substrat poreux texturé. Au fond de chaque créneau, un réseau de trous permet\nde souffler de l’air, ce qui va induire la lévitation de la plaque.\n\n## aide à comprendre comment la force de propulsion dépend de la géométrie de l’objet\n\n(dans le cas liquide, ceci était impossible à cause de l’apparition d’instabilités liées à\nl’aspect déformable de l’interface). Nous pouvons dès lors proposer un modèle en parfait\naccord avec les mesures expérimentales. L’existence d’une solution analytique décrivant\nl’écoulement dans les canaux nous permet, par la même occasion, de remonter à la pression\nnécessaire à imposer sous le poreux pour démarrer la lévitation.\nJusqu’ici nous avons réussi uniquement à induire des mouvements de translation sur un\nplan horizontal. Par la suite nous démontrons qu’il existe une pente maximum (non\nnégligeable) que peuvent remonter ces objets (voir figure C.5b) ou que l’on peut créer des\nmouvements de rotation avec des textures en forme de “moulin” (voir figure C.5a).\nPour finir, nous étudions les conséquences de travailler avec des objets plus lourds (ou\ndes textures plus profondes). On comprend intuitivement qu’afin de faire léviter un objet\nplus lourd, il faut augmenter les pressions mises en jeu. Par conséquent, la vitesse du\ngaz dans les rainures augmente aussi et le scénario d’entrainement visqueux (basé sur des\neffets inertiels négligeables par rapport aux effets visqueux, à savoir, des bas nombres de\nReynolds) n’est plus valable. Nous montrons alors que, pour des pressions de lévitation\nélevées (i.e. des haut nombre de Reynolds), les effets inertiels dominent et les objets avan-\ncent en sens opposé à celui observé dans le cas visqueux: c’est l’effet fusé (conservation\ndu moment) qui est à l’origine de cette nouvelle propulsion.\n\nUne façon extrêmement naïve de générer des situations de non-mouillage consiste tout\nsimplement à supprimer le substrat sur lequel repose le liquide. Il s’en suit un régime de\nchute libre puis, inévitablement, un impact avec un substrat qui met fin à cette belle\naventure. La situation est quelque peu moins dramatique lorsque le substrat sur lequel la\ngoutte impacte est une grille non-mouillante (voir figure C.6).\nEn effet, ce mélange de trous passants et sections bouchées va arrêter une partie du\nliquide (qui sera redirigé latéralement), mais il va aussi laisser passer une partie du liquide\n(sous forme de filaments liquides, qui, à leur tour, vont se déstabiliser en petites gout-\n182 APPENDIX C. RÉSUMÉ EN FRANÇAIS\n\n(a)\n\n(b)\n\nFigure C.5 – (a) Moulin visqueux: son motif est basé sur des rainures parallèles disposées\ndans une direction qui varie sur quatre quadrants différents. L’écoulement d’air entraîne\npar viscosité chaque section d’une plaque en verre (épaisseur 1 mm, longueur et largeur\n30 mm), ce qui induit un mouvement de rotation. (b) Vue de trois quarts d’une lamelle\nmontant une pente de 2% grace aux effets d’entraînement visqueux.\n\ntelettes). Le quatrième chapitre, qui porte sur cette interaction goutte/grille, commence\ndonc par une brève introduction au monde des impacts. On s’attarde en particulier sur\nles impacts sur des substrats solides non-mouillants [24, 63]. Un point de départ idéal au\nproblème de la grille, se trouve dans les travaux de E. Lorenceau et collaborateurs. En\neffet, ils traitent le cas de l’impact d’une goutte sur une plaque avec un seul trou. Cette\nsituation, plus simple que la grille mais renfermant des ingrédients physiques similaires,\nnous permet d’isoler des paramètres clefs tels que la vitesse de rétraction d’un filament\nvisqueux ou le temps que met la goutte pour complètement s’écraser. Dès lors, nous\nproposons un modèle qui nous aide à prédire la masse transmise en fonction de la vitesse\nd’impact, le rayon de la goutte et la taille du trou. On s’attaque ensuite au problème\nd’une plaque avec une multitude de trous - la grille. En utilisant les mêmes principes que\nceux utilisés pour un seul trou, nous adaptons le modèle pour aboutir à une loi d’échelle\nqui explique qualitativement les différents régimes observés expérimentalement. Apparaît\nalors le besoin de modéliser la déformation de la goutte pendant l’impact afin de rendre\ncompte des mesures quantitativement. Nous explorons finalement différentes configura-\n183\n\n2R0\n\nV0\n\nFigure C.6 – Vue latérale d’une goutte de rayon R0 = 1.9 mm impactant à vitesse\nV0 = 1.6 m/s sur une grille (trous circulaires de rayon approximatif 200 µm). La première\nimage représente le début de l’impact. La deuxième et troisième image représentent le\ntemps nécessaires à la goutte pour complètement s’écraser. Grâce à l’inertie, une partie\ndu liquide passe à travers chaque trou sous forme de jet liquide. Ces doigts cylindriques\nn’étant pas stables, ils donnent naissance à un ensemble de petites gouttes. Sur la dernière\nimage, les gouttelettes continuent leur trajectoire en formant un cône d’angle ↵. Chaque\nimage est séparée par 2 ms.\n\ntions possibles (nous faisons par exemple varier la nature mouillante ou non-mouillante\nde la grille, ou bien, nous modifions la porosité de la grille) et nous vérifions que le com-\nportement physique reste qualitativement bien décrit par notre loi d’échelle.\n\nL’étude des impacts sur grille nous a amené, dans le dernier chapitre, à nous pencher\nsur la question de la force subie par un substrat lisse à l’impact d’une goutte. On s’est\ntout d’abord intéressé aux phénomènes d’ondes de choc existant juste après le début\nde l’impact, à très court terme [119, 71]. En effet, ces ondes génèrent des pressions\ntellement élevées (de l’ordre du MPa) que l’on pourrait s’attendre à des forces associées\nconsidérables. Cependant, leur temps de vie est si bref (de l’ordre de la nanoseconde\n) que la surface sur laquelle elles agissent est tellement réduite que les forces restent\nrelativement faibles (de l’ordre du µN). Par ailleurs, il convient de tenir compte des\neffets de compressibilité dans l’air environnant si l’on veut comprendre le phénomène\nd’apparition des “splashs” [130, 132, 106].\nOn s’intéresse, dans un deuxième temps, aux forces qui ont lieu dès que les effets de\ncompressibilité disparaissent. Ces forces d’origine inertielle sont mesurées à l’aide d’un\ncapteur piézo-électrique (figure C.7a). On observe (figure C.7b) des profiles atteignant\ndes forces maximums de plusieurs dizaines de mN - bien plus élevées que les forces liées\naux ondes de choc. Un modèle analytique basé sur le transfert de quantité de mouvement\nrend parfaitement compte des données expérimentales.\nNous proposons par la suite l’utilisation d’un capteur bien moins couteux basé sur\nl’étude de la déflexion d’une lamelle de verre (pincée à une extrémité, libre de l’autre\ncôté) suite à l’impact d’une goutte (voir figure C.8). Cette fois-ci, c’est la conservation\nde la quantité de mouvement qui décrit la déflexion maximale 0 (définie sur figure C.8)\n184 APPENDIX C. RÉSUMÉ EN FRANÇAIS\n\n(a)\n\n(b)\n60\nF\n50\n(mN)\n40\n\n30\nF0\n20\n\n10\n\n−10\n\n## −0.5 0 0.5 1 1.5 2\n\nt (ms)\n\nFigure C.7 – (a) Vue latérale d’une goutte de rayon R = 1.3 mm impactant un quartz\npiézo-électrique à vitesse V = 3 m/s. Chaque image est séparée par 0.5 ms. (b) Force\nd’impact mesurée par le capteur piézo-électrique en fonction du temps. La force maximum\nF0 est atteinte environ un dixième de milliseconde après le début de l’impact.\n\n## en fonction des propriétés intrinsèques de la plaque ainsi que la vitesse et le rayon de la\n\ngoutte. Ainsi, deux régimes (“lent” et “rapide”) sont mis en évidence en fonction du temps\nde réaction de la plaque comparé au temps d’écrasement de la goutte.\n\nFigure C.8 – (a) Goutte de rayon R = 1.25 mm et vitesse V = 4 m/s impactant sur le\nbord libre d’une lamelle de verre. Chaque image est séparée par 1 ms et ne montre que le\ndernier tiers de la plaque. 0 représente la déflexion maximum de la plaque après impact.\n\nCes deux différents capteurs (piezo et lamelle de verre) nous permettent ainsi de\ndiscuter le cas particulier de la force d’impact des gouttes de pluie. 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Drop collisions with\n\nsimple and complex surfaces. Current Opinion in Colloid and Interface Science,\n16(4):292–302, 2011.\n\n## P. T. Nagy and G. P. Neitzel. Optical levitation and transport of microdroplets:\n\nProof of concept. Physics of Fluids (1994-present), 20(10):101703, 2008.\n\n M. Nearing and J. Bradford. Relationships between waterdrop properties and forces\nof impact. Soil Science Society of America Journal, 51(2):425–430, 1987.\n\n M. Nearing, J. Bradford, and R. Holtz. Measurement of force vs. time relations for\nwaterdrop impact. Soil Science Society of America Journal, 50(6):1532–1536, 1986.\n\n## G. P. Neitzel and P. Dell’Aversana. Noncoalescence and nonwetting behavior of\n\nliquids. Annual review of fluid mechanics, 34(1):267–289, 2002.\n\n## P. C. Nordine, J. Weber, and J. G. Abadie. Properties of high-temperature melts\n\nusing levitation. Pure and applied chemistry, 72(11):2127–2136, 2000.\n\n## J. Ok, E. Lopez-Oña, D. Nikitopoulos, H. Wong, and S. Park. Propulsion of droplets\n\non micro-and sub-micron ratchet surfaces in the leidenfrost temperature regime.\nMicrofluidics and nanofluidics, pages 1–10, 2011.\n\n## K. Okumura, F. Chevy, D. Richard, D. Quéré, and C. Clanet. Water spring: A\n\nmodel for bouncing drops. EPL (Europhysics Letters), 62(2):237, 2003.\n\n## D. Quéré. Leidenfrost dynamics. Annual Review of Fluid Mechanics, 10 2012.\n\n C. Rasser. Contact angles and leidenfrost drops on strained materials. M.S. Thesis\nEcole Polytechnique,, 2013.\n\n M. Rein. Phenomena of liquid drop impact on solid and liquid surfaces. Fluid\nDynamics Research, 12(2):61, 1993.\n194 BIBLIOGRAPHY\n\n## Y. Renardy, S. Popinet, L. Duchemin, M. Renardy, S. Zaleski, C. Josserand,\n\nM. Drumright-Clarke, D. Richard, C. Clanet, and D. Quéré. Pyramidal and toroidal\nwater drops after impact on a solid surface. Journal of Fluid Mechanics, 484:69–83,\n2003.\n\n E. Reyssat. Gouttes, films et jets : quand les écoulements modèlent les interfaces.\nPhD Thesis, Université Paris 7-Denis Diderot, 2007.\n\n D. Richard, C. Clanet, and D. Quéré. Contact time of a bouncing drop. Nature,\n417(6891):811, 2002.\n\n D. Richard and D. Quéré. Bouncing water drops. EPL (Europhysics Letters),\n50(6):769, 2000.\n\n R. Rioboo, C. Tropea, and M. Marengo. Outcomes from a drop impact on solid\nsurfaces. Atomization and Sprays, 11(2), 2001.\n\n I. Roisman. Inertia dominated drop collisions. ii. an analytical solution of the navier–\nstokes equations for a spreading viscous film. Physics of Fluids, 21:052104, 2009.\n\n A. Sahaya Grinspan and R. Gnanamoorthy. Impact force of low velocity liquid\ndroplets measured using piezoelectric pvdf film. Colloids and Surfaces A: Physico-\nchemical and Engineering Aspects, 356(1-3):162–168, 2010.\n\n## R. Sahu, S. Sinha-Ray, A. Yarin, and B. Pourdeyhimi. Drop impacts on electrospun\n\nnanofiber membranes. Soft Matter, 8(14):3957–3970, 2012.\n\n E. G. Shafrin and W. A. Zisman. Upper limits for the contact angles of liquids on\nsolids. American Chemical Society, chapter 10, pages 145–157, 1963.\n\n J. Shin and T. McMahon. The tuning of a splash. Physics of Fluids A: Fluid\nDynamics, 2:1312, 1990.\n\n## A. Snezhko, E. B. Jacob, and I. S. Aranson. Pulsating-gliding transition in the dy-\n\nnamics of levitating liquid nitrogen droplets. New Journal of Physics, 10(4):043034,\n2008.\n\n J. H. Snoeijer and B. Andreotti. Moving contact lines: Scales, regimes, and dynam-\nical transitions. Annual review of fluid mechanics, 45:269–292, 2013.\n\n J. H. Snoeijer, P. Brunet, and J. Eggers. Maximum size of drops levitated by an air\ncushion. Physical Review E, 79(3):036307–, 03 2009.\n\n K. Sreenivas, P. De, and J. H. Arakeri. Levitation of a drop over a film flow. Journal\nof Fluid Mechanics, 380:297–307, 1999.\nBIBLIOGRAPHY 195\n\n## S. Srinivasan, W. Choi, K.-C. Park, S. S. Chhatre, R. E. Cohen, and G. H. McKinley.\n\nDrag reduction for viscous laminar flow on spray-coated non-wetting surfaces. Soft\nMatter, 9(24):5691–5702, 2013.\n\n C. S. Stevens, A. Latka, and S. R. Nagel. Comparison of splashing in high and low\nviscosity liquids. arXiv preprint, 2014.\n\n W. F. Stokey. Vibration of systems having distributed mass and elasticity. Shock\nand vibration Handbook, pages 7–1, 1988.\n\n## D. Strier, A. Duarte, H. Ferrari, and G. Mindlin. Nitrogen stars: morphogenesis of a\n\nliquid drop. Physica A: Statistical Mechanics and its Applications, 283(1):261–266,\n2000.\n\n## A. D. Stroock, S. K. Dertinger, A. Ajdari, I. Mezić, H. A. Stone, and G. M. White-\n\nsides. Chaotic mixer for microchannels. Science, 295(5555):647–651, 2002.\n\n R. Takaki and K. Adachi. Vibration of a flattened drop. ii. normal mode analysis.\nJournal of the Physical Society of Japan, 54(7):2462–2469, 1985.\n\n## R. Takaki, N. Yoshiyasu, Y. Arai, and K. Adachi. Self-induced vibration of an\n\nevaporating drop. KTK Scientific Publishers, 1984.\n\n M. Taormina. Leidenfrost ratchets. B.S. Thesis, University of Oregon, june 2006.\n\n G. Taylor. The instability of liquid surfaces when accelerated in a direction per-\npendicular to their planes. i. Proceedings of the Royal Society of London. Series A.\nMathematical and Physical Sciences, 201(1065):192–196, 1950.\n\n G. Taylor. The dynamics of thin sheets of fluid. iii. disintegration of fluid sheets.\nProceedings of the Royal Society of London. Series A. Mathematical and Physical\nSciences, 253(1274):313–321, 1959.\n\n## J. W. J. Thomas R. Cousins, Raymond E. Goldstein and A. I. Pesci. A ratchet trap\n\nfor leidenfrost drops. J. Fluid Mech, 2011.\n\n## M. Thoraval, K. Takehara, T. Etoh, and S. Thoroddsen. Drop impact entrapment\n\nof bubble rings. arXiv preprint arXiv:1211.3076, 2012.\n\n S. Thoroddsen, T. Etoh, K. Takehara, N. Ootsuka, and Y. Hatsuki. The air bubble\nentrapped under a drop impacting on a solid surface. Journal of Fluid Mechanics,\n545:203–212, 2005.\n\n## S. Thoroddsen, K. Takehara, and T. Etoh. Micro-splashing by drop impacts. Journal\n\nof Fluid Mechanics, 706:560–570, 2012.\n196 BIBLIOGRAPHY\n\n## A. Tijsseling and A. Anderson. A precursor in waterhammer analysis - rediscovering\n\njohannes von kries. Proceedings of the 9th International Conference on Pressure\nSurges, 2004.\n\n## S. Timoshenko, S. Woinowsky-Krieger, and S. Woinowsky-Krieger. Theory of plates\n\nand shells, volume 2. McGraw-hill New York, 1959.\n\n## N. Tokugawa and R. Takaki. Mechanism of self-induced vibration of a liquid drop\n\nbased on the surface tension fluctuation. Journal of the Physical Society of Japan,\n63(5):1758–1768, 1994.\n\n## T. Tran, H. J. J. Staat, A. Prosperetti, C. Sun, and D. Lohse. Drop impact on\n\nsuperheated surfaces. Physical Review Letters, 108(3):036101, 01 2012.\n\n## P. Tsai, M. Hendrix, R. Dijkstra, L. Shui, and D. Lohse. Microscopic structure\n\ninfluencing macroscopic splash at high weber number. Soft Matter, 7(24):11325–\n11333, 2011.\n\n## I. Vakarelski, N. Patankar, J. Marston, D. Chan, and S. Thoroddsen. Stabiliza-\n\ntion of leidenfrost vapour layer by textured superhydrophobic surfaces. Nature,\n489(7415):274–277, 2012.\n\n R. van der Veen, T. Tran, D. Lohse, and C. Sun. Direct measurements of air layer\nprofiles under impacting droplets using high-speed color interferometry. Physical\nReview E, 85(2):026315, 2012.\n\n D. Vollmer and H.-J. Butt. Fluid dynamics: Shaping drops. Nature Physics, 2014.\n\n J. von Kries. On the relations between pressure and velocity which exist in the\nwave-like motion in elastic tubes. Festschrift der 56. Versammlung deutscher Natur-\nforscher und Ärzte gewidmet von der Naturforschenden Gesellschaft zu Freiburg i.\nB, Supplement zu Band VIII der Berichte, 1883.\n\n L. Wachters, L. Smulders, J. Vermeulen, and H. Kleiweg. The heat transfer from a\nhot wall to impinging mist droplets in the spheroidal state. Chemical Engineering\nScience, 21(12):1231–1238, 1966.\n\n## W. Wirth, S. Storp, and W. Jacobsen. Mechanisms controlling leaf retention of\n\nagricultural spray solutions. Pesticide science, 33(4):411–420, 1991.\n\n A. M. Worthington. The splash of a drop. Society for Promoting Christian Knowl-\nedge, London, 1895.\nBIBLIOGRAPHY 197\n\n L. Xu. Liquid drop splashing on smooth, rough, and textured surfaces. Physical\nReview E, 75(5):056316, 2007.\n\n L. Xu, W. Zhang, and S. Nagel. Drop splashing on a dry smooth surface. Physical\nReview Letters, 94(18):184505, 2005.\n\n Q. Xu, E. Brown, and H. M. Jaeger. Impact dynamics of oxidized liquid metal\ndrops. Physical Review E, 87(4):043012, 2013.\n\n## A. Yarin. Drop impact dynamics: splashing, spreading, receding, bouncing. . . .\n\nAnnu. Rev. Fluid Mech., 38:159–192, 2006.\n\n K. Yoneta. Diffraction of falling drop by plane net part i. Journal of the Faculty of\nScience, Hokkaido Imperial University., 1(4):121–147, 1932.\n\n J. Zhen, C. Josserand, S. Zaleski, S. Popinet, and P. Ray. Gas influence in drop\nimpact on dry solid substrates. 2nd ECCOMAS Young Investigators Conference,\n2013.\nAbstract\nWe investigate through several experiments the special dynamics generated by non-wetting\nobjects. On a substrate textured with grooves forming a herringbone pattern, a Leidenfrost levi-\ntating liquid is propelled: the textures channel the vapor flow in a well-defined direction so that\nthe slider above is driven by vapor viscosity. These deformable objects undergo very little friction\non flat surfaces. However, on crenelated substrates, impacts on the texture sides greatly enhance\ndissipation. We extend this entrainment scenario to other situations where the liquid (and its\ndeformable nature) is not involved anymore. A solid plate can levitate over a porous substrate\nthrough which air is blown. Again, escaping flow can be rectified by the textures and entrain\nthe plate, leading to translation movement or even to rotation. If we create deeper channels\n(hence losing flow confinement), we observe motion in the opposite direction due to “rocket effect”\n(conservation of momentum).\nWe are also interested in an extreme non-wetting situation: the falling drop. Indeed, all along\nthe fall, the drop only experiences air drag friction, easily reaching high speeds. We tackle the\nproblem of the dramatic issue of this fall: the impact. We first study the impact of a drop on a\nsieve. In this situation intermediate between a solid wall and no obstacle at all, mass either passes\nthrough the holes or gets stopped by the closings. We then focus on the impact force experienced\nby the substrates and characterize the force as a function of the drop and impact properties, but\nalso of the nature of the solid on which impact takes place.\n\n## Keywords: interface, non-wetting drop, Leidenfrost effect, texture, self-propulsion, friction,\n\nlevitation, impact, grid.\n\nRésumé\nNous étudions à travers plusieurs expériences la dynamique spéciale engendrée par des objets\nnon mouillants. Un liquide en état Leidenfrost est autopropulsé lorsqu’on le pose sur un substrat\ntexturé avec des rainures formant un motif à chevrons: les textures canalisent l’écoulement de\nvapeur dans une direction bien définie de sorte que ces aéroglisseurs liquides sont entraînés par la\nvapeur sous-jacente. Ces objets déformables subissent très peu de friction sur une surface plane.\nToutefois, sur des substrats crénelés, les impacts sur les textures créent une friction spéciale qui est\négalement étudiée. Nous étendons ce scénario d’entraînement visqueux dans d’autres situations\noù le liquide (et sa nature déformable) est remplacé par une plaque solide. Pour permettre la\nlévitation, on le place sur un substrat poreux à travers lequel de l’air est soufflé. Une fois de plus,\nl’écoulement est rectifié par des textures permettant l’entraînement d’une lamelle de verre dans un\nmouvement de translation ou même de rotation. Si nous augmentons la profondeur des textures, le\nconfinement est perdu et on observe un mouvement dans la direction opposée dû à \"l’effet fusée\".\nNous nous sommes également intéressés à une situation de non mouillage particulièrement\nsimple: la goutte en chute libre. Tout au long de sa chute, la goute ne subit que la trainée de l’air,\nce qui lui permet d’atteindre des vitesses élevées. Nous abordons le problème de l’issue de cette\nchute: l’impact. Nous étudions d’abord l’impact d’une goutte sur un tamis. Dans cette situation\nintermédiaire entre un mur solide et aucun obstacle, le liquide passe à travers les trous ou est arrêté\npar les sections bouchées. Nous nous concentrons ensuite à la force d’impact subie par le substrat.\nNous la mesurons et la calculons en fonction des caractéristiques du liquide, de l’impact, et de la\nnature du substrat.\n\nMots-clés: interface, goutte non mouillante, caléfaction, texture, autopropulsion, friction, lévi-\ntation, impact, grille."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80687165,"math_prob":0.95067126,"size":242530,"snap":"2021-21-2021-25","text_gpt3_token_len":66900,"char_repetition_ratio":0.16590387,"word_repetition_ratio":0.06739662,"special_character_ratio":0.25734136,"punctuation_ratio":0.14906971,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9680874,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-07T11:18:33Z\",\"WARC-Record-ID\":\"<urn:uuid:f471ae41-e850-42f6-b1f5-6cad26e145dc>\",\"Content-Length\":\"1049667\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8fa7d7f7-fa56-4466-bbb4-08f0135ee3f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:a6320152-d889-47c7-b8c7-769378b61282>\",\"WARC-IP-Address\":\"151.101.202.152\",\"WARC-Target-URI\":\"https://ru.scribd.com/document/392616864/Non-wetting-Drops-From-Impacts-to-Self-propulsion\",\"WARC-Payload-Digest\":\"sha1:IZBK3TIRFTDFZVNFGK6PI6QVKJN7UZIM\",\"WARC-Block-Digest\":\"sha1:BFDCD46N5TI5KI5QPQSWDYOHZAELGZY7\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988775.80_warc_CC-MAIN-20210507090724-20210507120724-00539.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-6-percent-6-1-basics-of-percent-6-1-exercises-page-386/51 | [
"## Basic College Mathematics (10th Edition)\n\nIn order to write a percent as a decimal, we must perform the following steps. 1. Drop the percent symbol 2. Divide by 100 Therefore, 30%=$30\\div100=.3$. So, the patient's blood pressure was .3 above normal."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9040937,"math_prob":0.98421437,"size":482,"snap":"2022-05-2022-21","text_gpt3_token_len":117,"char_repetition_ratio":0.10041841,"word_repetition_ratio":0.0,"special_character_ratio":0.25311205,"punctuation_ratio":0.14563107,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9654446,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-22T20:46:51Z\",\"WARC-Record-ID\":\"<urn:uuid:266bdd84-8b54-4ca8-af19-4878903597c9>\",\"Content-Length\":\"76604\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0fb5a3f2-16a6-4414-9857-a9fda9bf9bc2>\",\"WARC-Concurrent-To\":\"<urn:uuid:698c91a0-7892-4a62-8f23-28addb3620e3>\",\"WARC-IP-Address\":\"3.227.56.233\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/other-math/CLONE-547b8018-14a8-4d02-afd6-6bc35a0864ed/chapter-6-percent-6-1-basics-of-percent-6-1-exercises-page-386/51\",\"WARC-Payload-Digest\":\"sha1:ENMOCQBAKS7F6VRVX2QRBMCCBBJYQFEC\",\"WARC-Block-Digest\":\"sha1:CVIC5XFM3A6JC7MCFGF35GFTLTCNNEPI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662546071.13_warc_CC-MAIN-20220522190453-20220522220453-00463.warc.gz\"}"} |
https://proofwiki.org/wiki/Characterization_of_T0_Space_by_Closures_of_Singletons | [
"# Characterization of T0 Space by Closures of Singletons\n\n## Theorem\n\nLet $T = \\left({S, \\tau}\\right)$ be a topological space.\n\nThen\n\n$T$ is a $T_0$ space if and only if\n$\\forall x, y \\in S: x \\ne y \\implies x \\notin \\left\\{{y}\\right\\}^- \\lor y \\notin \\left\\{{x}\\right\\}^-$\n\nwhere $\\left\\{{y}\\right\\}^-$ denotes the closure of $\\left\\{{y}\\right\\}$.\n\n## Proof\n\n### Sufficient Condition\n\nLet $T$ be a $T_0$ space.\n\nLet $x, y \\in S$ such that\n\n$x \\ne y$\n\nAiming for a contradiction suppose that\n\n$x \\in \\left\\{{y}\\right\\}^- \\land y \\in \\left\\{{x}\\right\\}^-$\n\nThen:\n\n$\\left\\{{x}\\right\\} \\subseteq \\left\\{{y}\\right\\}^- \\land \\left\\{{y}\\right\\} \\subseteq \\left\\{{x}\\right\\}^-$\n$\\left\\{{x}\\right\\}^- \\subseteq \\left({\\left\\{{y}\\right\\}^-}\\right)^- \\land \\left\\{{y}\\right\\}^- \\subseteq \\left({\\left\\{{x}\\right\\}^-}\\right)^-$\n$\\left\\{{x}\\right\\}^- \\subseteq \\left\\{{y}\\right\\}^- \\land \\left\\{{y}\\right\\}^- \\subseteq \\left\\{{x}\\right\\}^-$\n\nThen by definition of set equality:\n\n$\\left\\{{x}\\right\\}^- = \\left\\{{y}\\right\\}^-$\n$\\left\\{{x}\\right\\}^- \\ne \\left\\{{y}\\right\\}^-$\n\nThus the result by Proof by Contradiction\n\n$\\Box$\n\n### Necessary Condition\n\nAssume that:\n\n$(1): \\quad \\forall x, y \\in S: x \\ne y \\implies x \\notin \\left\\{{y}\\right\\}^- \\lor y \\notin \\left\\{{x}\\right\\}^-$\n\nBy Characterization of $T_0$ Space by Distinct Closures of Singletons it suffices to prove\n\n$\\forall x, y \\in S: x \\ne y \\implies \\left\\{{y}\\right\\}^- \\ne \\left\\{{x}\\right\\}^-$\n\nLet $x, y \\in S$ such that\n\n$x \\ne y$\n\nAiming for a contradiction suppose that\n\n$(2): \\quad \\left\\{{y}\\right\\}^- = \\left\\{{x}\\right\\}^-$\n\nBy definition of singleton:\n\n$x \\in \\left\\{{x}\\right\\} \\land y \\in \\left\\{{y}\\right\\}$\n$\\left\\{{x}\\right\\} \\subseteq \\left\\{{x}\\right\\}^- \\land \\left\\{{y}\\right\\} \\subseteq \\left\\{{y}\\right\\}^-$\n\nThen by definition of subset and $(2)$:\n\n$x \\in \\left\\{{y}\\right\\}^- \\land y \\in \\left\\{{x}\\right\\}^-$\n\nThis contradicts the assumption $(1)$.\n\nThus the result by Proof by Contradiction.\n\n$\\blacksquare$"
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https://tubecad.com/2009/09/blog0172.htm | [
"John Broskie's Guide to Tube Circuit Analysis & Design\n 26 September 2009",
null,
"First SRPP, Then SRPP+ In Part one I promised that a deeper look into the SRPP would be forthcoming. In a topological nutshell, in the SRPP circuit, one triode stands atop another, with a resistor, Rak, spanning the bottom tube’s plate and the top tube’s cathode. Usually Rak equals the bottom triode’s cathode resistor, Rk. The circuit’s output is taken at the top tube’s cathode. The bottom triode’s grid receives the input signal and the tube’s current conduction varies in response to the signal voltage. This variation in current flow gives the top tube its drive signal, as the current flowing through the bottom tube causes resistor Rak to see a varying voltage develop across its leads in response. This variation in voltage is then given to the top triode's grid, which in turn will conduct a varying amount of current as a result. Since the load connects in between the top tube’s cathode and ground, the load provides a current path to absorb the delta in current flow between top and bottom tubes.",
null,
"Had the output been taken at the bottom of resistor Rak, then there would be no push-pull, as the top tube and resistor Rak would effectively define a simple plate resistor and the circuit transforms into a simple variation on the grounded-cathode amplifier, with only single-ended operation.",
null,
"The key point here, which eludes about 95% of tube fanciers, is that a simple plate resistor (or triode and resistor Rak) in a grounded-cathode amplifier can only increase beyond idle current conduction when the bottom tube does as well. In other words, when the bottom tube increases its conduction, the plate resistor sees a greater voltage across its leads and its current conduction increases correspondingly. But in an SRPP circuit, with the output taken at the top tube’s cathode, the top tube’s current conduction can increase, while the bottom tube’s decreases. Amazingly enough, the top tube's conduction can increase in spite of it seeing a smaller cathode-to-plate voltage. With the standard plate resistor, less voltage always means less current. In other words, the SRPP operation is a fundamentally different in functioning from the grounded-cathode amplifier. This is the key point: just because a tube circuit holds one tube standing atop another does not make that circuit an SRPP. The Broskie cathode follower, cascode, and White cathode follower, and the first and second stages of the Aikido all hold two tubes in series, but none of them is an SRPP. (If only I had a buck for every time the Aikido gets described as containing an SRRP input stage, I would be much richer. Absolutely nothing pushes and pulls in an Aikido's first stage.) Of course, if no external load attaches to the SRPP, it no longer experiences any push-pull operation, as the remaining single current path through the circuit will not allow a delta in current conduction to occur. In other words, an external load is not essential to a grounded-cathode amplifier, but it is to an SRPP. Just as in a Circlotron amplifier, the load makes all the difference.",
null,
"In fact, the best way to approach an understanding of the SRPP is to set the external load resistance to zero; that’s right, 0 ohms. Of course, directly shorting an SRPP’s output to ground would cause all sorts of problems in an actual physical circuit, so imagine that the a coupling capacitor with infinite capacitance is used or that a high voltage battery is used in place of the coupling capacitor. With such an arrangement, the anti-phase conduction between top and bottom triodes becomes obvious. As the bottom tube conducts more, greater will be the voltage drop across resistor Rak, which in turn will decrease the grid voltage on the top tube, thereby causing its current conduction to fall. If the top tube is conducting less and the bottom tube is conducting more, then the only current path for the latter’s increased conduction s the external load, which is zero in this example. The SRPP’s Impedance-Multiplier Circuit I said in the last post that the SRPP circuit consisted of two primary circuits, not one; that it was a compound circuit that held a grounded-cathode amplifier and a impedance-multiplier circuit. Well, let’s pull the impedance-multiplier circuit out of the SRPP and examine it in isolation. The first step is to get rid of the bottom tube, replacing it with a single resistor, Rk.",
null,
"The above circuit looks a bit like a cathode follower, but it’s not a cathode follower, if for no other reason than that its input impedance is vastly lower than a true cathode follower’s. Note how the input attaches to both the tube’s grid and the resistor Rk. Here is a quick quiz: What would the input impedance be of the above circuit if the external load impedance were zero? The answer is Rak || Rk. Ok, that was too easy. What would the input impedance be of the above circuit if the external load impedance were infinity? The answer is [rp + (µ + 1)Rak] || Rk. Now the much harder question: What would be the input impedance with an external load impedance that falls between 0 and infinity ohms? The answer is much more difficult, but if we assume that resistor Rak had been carefully selected to provide impedance doubling with the selected tube and load resistance, then the answer is (2Rload + Rak) || Rk. By the way, the symbol “||” means in parallel with, so X || Y equals XY / (X + Y). In order to focus solely on the impedance-multiplier circuit, let’s replace resistor Rk with an inductor, a perfect inductance that displaces no DC voltage, as it holds no DCR.",
null,
"Notice how the choke eliminated not just the bottom tube (or large-valued resistor) but also the pesky DC offset at the input of the impedance-multiplier circuit. Now the input impedance of this circuit equals only Rak + 2Rload. No doubt some are wondering, why not just make a cathode follower out of the single triode and be done with it?",
null,
"The cathode follower offers some desirable attributes, such as a lower output impedance and a ultra-high input impedance, in this example an input impedance of 1M. But the cathode follower cannot symmetrically swing beyond the idle current into the load impedance. For example, if the idle current is 10mA and the load impedance is 10k, the cathode follower can only swing +/-10mA or voltage swings of +/-10Vpk into 1k. the SRPP circuit, on the other hand, given the same load and idle current, can swing +/-20mA or voltage swings of +/-20Vpk into 1k, effectively quadrupling the power delivered into the load. Wait a minute, how is that possible? We got rid of the bottom triode but an inductor can only current-swing up to idle current? True enough, and I never knowingly try to break the laws of physics (and no laws are broken here). The impedance-multiplier circuit is not anything like a cathode follower, remember; its input impedance is low and depends on the amount of impedance multiplication and the external load impedance. So when this impedance-multiplier circuit swings negative 20V into the 1k load, thereby swinging -20mA, the inductance pulls -10mA and the signal source provides the extra -10mA. Conversely, when this impedance-multiplier circuit swings positive 20V into the 1k load, thereby swinging +20mA, the inductance undergoes no change in current flow from -10mA and the triode conducts +20mA and the signal source provides the extra +10mA, so the -10mA + 20mA + 10mA equals +20mA. SRPP+ Versus Plain-Jane SRPP The SRPP+ differed from the generic SRPP by using two resistors in place of the SRPP’s single Rak resistor. Thus, we should incorporate the improvement into the inductor-loaded impedance-multiplier circuit.",
null,
"Why is the extra resistor so important? It allows us to “tune” the SRPP to different load impedances. Just as we used two resistors to set the impedance-multiplier ratio in the idealized impedance-multiplier circuit, we also need two resistors with a tube-based impedance-multiplier circuit. As you might have guessed, because of the triode’s low transconductance, rp, and limited mu, the simple formulas for the idealized impedance-multiplier circuit must be substantially modified to conform to the tube’s actual functioning. By the way, the generic SRPP also held two resistances between the two triodes, but its R1 was invisible to those who only see with their eyes. (In a past life I must have been a Zen master or an American Indian medicine man, as such incomprehensible and paradoxical utterances come naturally to me.) Resistor R1 was implicit in the form of the top tube’s impedance at its cathode, which equaled rp/mu (1/gm). This impedance effectively created the missing R1. Thus, in order for R1 to have disappeared altogether, the top tube must have offered infinite transconductance; not likely. Solid-state power amplifiers come close enough for government work, but poor little triodes have a hard time mustering a maximum of about 40mA/V of gm. For example, a 6DJ8/6922-based Plain-Jane SRPP circuit holds an implicit R1 of about 100 ohms, as the triode’s gm is about 10mA/V. If Rak’s value were set to 100 ohms, the impedance-multiplier circuit within this SRPP circuit would effectively double the load impedance, as R1 and R2 (Rak) would be equal. (By the way, we can view any power amplifier’s transconductance as the inverse of its output impedance times its gain, no matter how complex or simple the amplifier, as long as it is a voltage amplifier. \"Wait a minute,\" you may say, \"I never see output impedance on the spec sheets, only damping factor.\" Damping factor is equal to the load impedance [usually 8 ohms] divided by the amplifier’s output impedance; thus an amplifier, with a damping factor of 80, has an output impedance of 0.1 ohms. If its gain is 10, then its transconductance equals 10/0.1, or 100A/V. Now you can see why a dead short at its output is so dangerous.)",
null,
"Let’s quickly review an idealized impedance-multiplier circuit. Resistors R1 and R2 set the impedance-multiplication ratio for the circuit. When R1 equals R2, the impedance-multiplier circuit will effectively double the load impedance. Our idealized impedance-multiplier circuit can swing huge voltage and current swings into a load while dissipating no heat at idle; it is ideal, after all. The tube-based SRPP is not so lucky, as it must run its internal impedance-multiplier circuit in a strict push-pull class-A mode, wherein the peak output current equals twice the idle current. So the tube-based version must strive for equal positive and negative current swings, which means that its optimal impedance-multiplier ratio is 2. Yet, we cannot simply make R1 and R2 equal, as the top triode’s cathode impedance must be added to R1’s value to get the “true” R1 value. (Interestingly, the bottom triode’s impedance at its plate only influences the impedance-multiplier circuit’s output impedance, but has no effect on the R1 and R2 values.) The tube-based impedance-multiplier circuit has to contend with the implicit contribution to R1’s value and its own rp and the load impedance influencing the results. So a universal impedance-multiplier circuit, one that is indifferent to the load impedance, is impossible to achieve with a single triode. Our idealized impedance-multiplier circuit did not suffer from this problem, as its power OpAmp presented no series resistance to R1, having an infinitely-low output impedance and a perfect unity gain. In other words, we will have to rely on formulas and a good amount of actual hardware tweaking to set R1 and R2 values within an actual SRPP+ circuit. But before we embark on this mathematical journey, lets examine differential impedance-multiplier circuits to get a better understanding of how this novel circuit functions. Solid-State Impedance-Multiplier Circuit Variations The following circuits make use of three-pin, adjustable voltage regulators. These little wonders contain a power transistor, a voltage reference, and an OpAmp. The OpAmp sees a 1.25V voltage reference in series with positive input, which allows the regulators to be used in many different configurations, from adjustable voltage regulators to constant-current sources to AC-clipper circuits.",
null,
"The following two impedance-multiplier circuits use the LM317 positive regulator and LM337 negative regulator as examples, but are in no way limited to these devices. In both examples, an impedance-multiplier circuit is loaded by a constant-current source, because the three-pin regulators are much like a single triode or transistor or MOSFET in that a single device must be operated class-A. Also in these examples, resistors R1 and R must be equal, as we want the impedance-multiplier circuit’s input and output to be centered at ground potential. Setting the idle current is as easy as the following formula implies: Iq = 1.25/R1 Once again, much like the tube-based SRPP circuit, if we want equal peak symmetrical output current swings, then R2 must equal R1. On the other hand, if we do not need the biggest symmetrical output current swings, we can set a different impedance-multiplier ratio. For example, let’s say we own an MP3 player that works well into its 32-ohm headphones, but grossly bogs down into inefficient 8-ohm headphones. Well, one of the below circuits set to an impedance-multiplier ratio of 4 would make a good candidate for helping the MP3 player along, as 8-ohm headphones would appear as 32-ohms headphones at the MP3 player’s headphone jack.",
null,
"",
null,
"Maybe I am jumping ahead to quickly. Let’s back up a bit and examine the voltage and current relationships within an LM337-based impedance-multiplier circuit.",
null,
"The circuit shown above is in its quiescent state, at idle, drawing 0.1A (or 100mA), blind to the external load, as the load draws no current. How can this circuit conduct with no path attached to its input that leads back to ground? It cannot conduct, but we can assume that it does attach to a signal source and that this signal source will provide a current path for the 0.1A of current, such as the following.",
null,
"By the way, the transistor could be replaced by a FET, triode, pentode, MOSFET, IGBT… The two 6.25-ohm resistors combine to form a 12.5-ohm resistor that sets the idle current to 0.1A, as 1.25V/12.5 = 0.1A. Now, for the rest of the examples, assume that the transistor/triode/MOSFET is in place, but not shown. We also will begin with a load resistance of zero ohms, dead short to ground in other words. The following schematic shows the voltage and current relationships when the transistor increases its current draw from 100mA to 200mA. Note how the LM337 has been turned off, as any current flow through it will increased the voltage differential greater than 1.25V between its adjustment and output pins. The load (ground) does see 200mA of current flow through it up to through the top 6.25-ohm resistor then on through the transistor/triode/MOSFET to the +12V power supply rail. Note that the transistor only increased its conduction by 100mA over its idle current and the load experienced a 200mA current swing.",
null,
"Next we will invert the current relationships.",
null,
"Now the transistor has been cut off and draws no current, leaving it up to the LM337 to increase its conduction to make up the difference. Note how the 200mA of current now flow in the opposite direction up ground. Also note how the same 1.25V voltage differential obtains between the LM337’s adjustment and output pins, as it has in all the previous examples. As far as the external load (ground) is concerned, a constant-current source of 200mA has just bridged it to the -12V power supply rail, which the following schematics make clear. Only one 6.25-ohm resistor sees any current flow, so we easily imagine that it has been replaced by a jumper wire.",
null,
"Before reading further, please make sure you really understand what was shown in the previous paragraphs. Review how the 1.25V voltage differential is always present between the LM337’s adjustment and output pins and how the current changed direction through the external load. Ready? Okay, we now move on to examples that hold a 10-ohm external load.",
null,
"At idle, nothing much has changed compared to the load-equals-zero example, as no current flows through the 10-ohm resistor. But when the transistor increases its current draw from 100mA to 200mA, the LM337 turns off and the 10-ohm load sees 200mA of current flowing through it, which develops a 2V voltage drop across the resistor.",
null,
"Now we do the inverse and the transistor cuts off and draws no current, so the LM337 must increase its conduction to 200mA to make up the difference. Note how the current now flowing into the load reverses direction, as the impedance-multiplier circuit’s output swings down to -2V. Note how the output had remained stuck at zero volts when the load equaled zero ohms. The greater the load impedance, the bigger the voltage swings. At all times the same 1.25V voltage differential obtains between the LM337’s adjustment and output pins",
null,
"We have seen how we can make an impedance-multiplier circuit out of either positive or negative voltage regulator ICs. What would happen if we built one of each and then placed them in parallel, inputs and outputs tied together?",
null,
"Note how we no longer need to supply the heavy idle current through a constant-current source or the signal source, as each impedance-multiplier circuit completes the current path for its partner. Also note how the two R2 resistors are effectively in parallel, which means that we can get away with just one resistor.",
null,
"In the schematic at the above left, we see the single R2 resistor in place; in the above right, we see some actual resistor values. By the way, if you recognize something that looks like a solid-state output stage in this topology, then the following schematic should look even more familiar.",
null,
"In the above circuit, two power transistors do all the work in the two impedance-multiplier circuits. The four diodes merely define voltage references of sorts to set the transistors idle current. The greater the number of diodes, the bigger the combined voltage drop and the higher the idle current through the transistors. The big problem that we always face with power transistors is getting the bias right, as the transistor’s base-to-emitter voltage changes with temperature, which can lead to thermal runaway and a smoking output stage. Many attempts have been made to devise clever circuits to overcome this problem, but my favorite is the 5-pin transistors from On Semi Corporation, as shown below. These transistors include a diode within their cases, which track the internal transistor perfectly; thus the marketing name, ThermalTrak™.",
null,
"In the above schematic, six diodes are displayed: four external and two internal. These diodes set a high idle current of 1.9A through the output transistors, which will prove steady due to the internal diodes compensating for the power transistors drift. (The two 1kµF capacitors bypass the diodes to present a lower impedance to the transistor bases.) By the way, we could easily add more output transistors, as shown in the circuit below.",
null,
"Note how only one pair of internal diodes is used. Also note how the emitter resistors have been doubled in value. Why? These resistors are effectively in parallel, so they effectively equal the single 1-ohm resistors. In other words, the same total current flows though the four-transistor and the two-transistor impedance-multiplier circuits, but in the four-transistor version, each transistor dissipates half the heat that the two-transistor version transistors do. In addition, the huge 2-ohm emitter resistors ensure that current hogging is extremely unlikely, as such large values provide a great deal of negative feedback. In other words, if one transistor tried to draw excessive current, the increase in current would provoke a bigger voltage drop across its emitter resistor, thereby lowering the transistor’s conduction. By the way, in Part One of this series, we saw a cathode follower loaded by an impedance-multiplier circuit made out of an LM337. (Remember the Somersault circuit.) Adding more triode in parallel would be easy. Each triode gets its own cathode resistor, whose value equals four times the R1 resistor value.",
null,
"Zen Meets the Impedance-Multiplier Circuit The following circuits may look like one of Nelson Pass’s Zen amplifiers, but they all differ in one major fundamental way: they push and pull, whereas the Zen amplifier only pulls: What is the sound of one MOSFET output device? The following circuit is a current-output amplifier or a power voltage-to-current converter circuit, depending on your perspective. In any case, the output impedance is about 1k and only because of the 1k terminating resistor at the output. How does this amplifier work? First of all, the bottom IRFP250 is behind the steering wheel, as it is its variations in current conduction that will be magnified into the load impedance. In other words, its transconductance is driving the impedance-multiplier circuit loading it. And the bottom MOSFET’s transconductance is lessened and linearized by the unbypassed source resistor. The formula for the decrease in a MOSFET's gm because of an unbypassed source resistor is: gm' = gm / (1 + gmRs) So a change in input voltage will provoke changes in the bottom MOSFET’s current flow, which the top MOSFET within the impedance-multiplier circuit will double into the load impedance. In other words, this amplifier holds a current driven impedance-multiplier circuit.",
null,
"The next power amplifier is configured as a voltage amplifier, with a fixed amount of gain and fairly low output impedance. The 1k input resistor and the 17k feedback resistor define a classic inverting-amplifier negative feedback loop. These resistors also serve to bias the bottom MOSFET. The huge coupling capacitor is required because of the low input impedance presented by the 1k and 3.1k resistors in parallel at the input. Using higher-value resistors will excessively limit the high frequency bandwidth. (Of course, if the amplifier were used in a bi-amped setup, wherein it powered the woofer, the truncating of the high frequencies may not prove such a liability; it may prove an asset.)",
null,
"Note how, in both amplifier examples, the R1 and R2 resistor values do not match. Why? Like the triode, the MOSFET suffers from limited transconductance, so the impedance at its source is not zero. Thus, its source impedance must be subtracted from R1’s ideal value (0.47 ohms, in this example). Next Time First of all, I didn’t plan on there being a part three, but I have a at least one more blog entry worth of information that needs to written before I can call it quits. //JRB Kit User Guide PDFs Click image to download",
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"",
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"E-mail from GlassWare customers: Mr Broskie, I bought an Aikido stereo linestage kit from you some days ago, and I received it just this Monday. I have a few things to say about it. Firstly, I'm extremely impressed at the quality of what I've been sent. In fact, this is the highest quality kit I've seen anywhere, of anything. I have no idea how you managed to fit all this stuff in under what I paid for it. Second, your shipping was lightning-quick. Just more satisfaction in the bag, there. I wish everyone did business like you. Sean H. And Hi John, I received the Aikido PCB today - thank you for the first rate shipping speed. Wanted to let you know that this is simply the best PCB I have had in my hands, bar none. The quality is fabulous, and your documentation is superb. I know you do this because you love audio, but I think your price of \\$39 is a bit of a giveaway! I'm sure you could charge double and still have happy customers. Looking forward to building the Aikido, will send some comments when I'm done! Thank you, regards, Gary. 9-Pin & Octal PCBs & Kits High-quality, double-sided, extra thick, 2-oz traces, plated-through holes, dual sets of resistor pads and pads for two coupling capacitors. Stereo and mono, octal and 9-pin printed circuit boards available. Designed by John Broskie & Made in USA",
null,
"Aikido PCBs for as little as \\$20.40 http://glass-ware.stores.yahoo.net/ TCJ My-Stock DB",
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"",
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"",
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"",
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"TCJ My-Stock DB helps you know just what you have, what it looks like, where it is, what it will be used for, and what it's worth. TCJ My-Stock DB helps you to keep track of your heap of electronic parts. More details. Version 2 Improvements List all of your parts in one DB. Add part Images. One-click web searches for part information. Vertical and horizontal grids.* Create reports as PDFs.* Graphs added 2D/3D: pie & bar.* More powerful DB search. Help system added. Editable drop-down lists for location, projects, brands, styles, vendors and more. *User definable For more information, please visit: www.glass-ware.com"
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https://programmingpraxis.com/2018/08/24/data-structures-homework/ | [
"## Data Structures Homework\n\n### August 24, 2018\n\nToday’s exercise is from a programming student taking a Data Structures course, using Java.\n\nGiven a text file, print a list of the ten most common words in the file, along with their frequency of occurrence. You may not use a HashMap or ArrayList, but you may use regular expressions.\n\nYour task is to write a program to find the ten most common words in an input file. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.\n\nPages: 1 2\n\n### 5 Responses to “Data Structures Homework”\n\n1. Rutger said\n```text = \"\"\"Today’s exercise is from a programming student taking a Data Structures course, using Java.\n\nGiven a text file, print a list of the ten most common words in the file, along with their frequency of occurrence. You may not use a HashMap or ArrayList, but you may use regular expressions.\n\nYour task is to write a program to find the ten most common words in an input file. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.\"\"\"\n\nfrom collections import Counter\n\nc = Counter()\n\nfor w in text.split(' '):\nc[w] += 1\n\nprint(c.most_common(5))\n\n# outputs [('a', 7), ('the', 5), ('or', 4), ('to', 4), ('in', 3), ('you', 3), ('exercise', 2), ('is', 2), ('file,', 2), ('of', 2)]\n```\n2. Daniel said\n\n@Rutger, the Counter in Python is a subclass of dict, which is a hash map. However, while the problem prohibits hash maps, it doesn’t specify whether data structures built on hash maps are permitted.\n\nHere’s a solution in Java. It uses a trie. A linked list was used to hold children to avoid using a HashMap (or ArrayList). For retrieving the most common words, I used a heap, but a linked list would have seemingly sufficed.\n\n```import java.io.File;\nimport java.io.FileNotFoundException;\nimport java.util.Comparator;\nimport java.util.PriorityQueue;\nimport java.util.Scanner;\n\npublic class Main {\nstatic class Pair<T, U> {\nT first;\nU second;\npublic Pair(T first, U second) {\nthis.first = first;\nthis.second = second;\n}\n}\n\nstatic class Trie {\nstatic class Node {\nint count = 0;\nCharacter c = null;\nNode parent = null;\n// Can't use ArrayList or HashMap, so store children in LinkedList.\n// (alternatively a binary search tree could be used here)\n\nString string() {\nNode n = this;\nwhile (n.c != null) {\nn = n.parent;\n}\nStringBuffer sb = new StringBuffer();\nfor (char c : chars) {\nsb.append(c);\n}\nreturn sb.toString();\n}\n}\n\nvoid insert(String s) {\nouter:\nfor (int i = 0; i < s.length(); ++i) {\nchar c = s.charAt(i);\nfor (Node child : n.children) {\nif (child.c.equals(c)) {\nn = child;\ncontinue outer;\n}\n}\nNode child = new Node();\nchild.parent = n;\nchild.c = c;\nn = child;\n}\n++n.count;\n}\n\nPriorityQueue<Pair<String,Integer>> minHeap = new PriorityQueue<>(\nnew Comparator<Pair<String,Integer>>() {\n@Override\npublic int compare(Pair<String,Integer> o1, Pair<String,Integer> o2) {\nif (o1.second < o2.second) return -1;\nif (o1.second > o2.second) return 1;\nreturn 0;\n}\n});\nwhile (!frontier.isEmpty()) {\nNode n = frontier.pop();\nfor (Node child : n.children) {\n}\nif (n.count == 0) continue;\nif (minHeap.size() < count) {\n} else if (n.count > minHeap.peek().second) {\nminHeap.remove();\n}\n}\nwhile (!minHeap.isEmpty()) {\n}\nreturn result;\n}\n}\n\npublic static void main(String[] args) throws FileNotFoundException {\nif (args.length != 1) {\nSystem.err.println(\"Usage: <PROGRAM> FILE\");\nSystem.exit(1);\n}\nTrie t = new Trie();\nScanner s = new Scanner(new File(args));\nwhile (s.hasNext()) {\nString next = s.next();\n// Remove all punctuation and convert to lower case\nString word = next.replaceAll(\"[^a-zA-Z ]\", \"\").toLowerCase();\nt.insert(word);\n}\ns.close();\nfor (Pair<String,Integer> pair : t.mostCommon(10)) {\nSystem.out.printf(\"%s: %d\\n\", pair.first, pair.second);\n}\n}\n}\n```\n\nHere’s example output when running on Pride and Prejudice, the top book on the http://www.gutenberg.org.\n\n```the: 4496\nto: 4207\nof: 3717\nand: 3602\nher: 2215\ni: 2051\na: 1997\nin: 1919\nwas: 1843\nshe: 1704\n```\n3. Globules said\n\nHere’s a Haskell version, also run on Pride and Prejudice. (Haskell’s Map is based on “size balanced binary trees”.)\n\n```{-# LANGUAGE TupleSections #-}\n\nimport Data.List (sortOn)\nimport qualified Data.Map.Strict as M\nimport Data.Ord (Down(..))\nimport qualified Data.Text.Lazy as T\nimport qualified Data.Text.Lazy.IO as T\nimport System.Environment (getArgs, getProgName)\nimport Text.Printf (PrintfArg, printf)\n\n-- Return the distinct values of a list paired with their number of occurrences.\ncounts :: (Ord a, Integral b) => [a] -> [(a, b)]\ncounts = M.toList . M.fromListWith (+) . map (,1)\n\n-- Print the number of occurrences of a value, followed by the value itself.\nprintCount :: PrintfArg a => (a, Int) -> IO ()\nprintCount (s, n) = printf \"%-8d %s\\n\" n s\n\nmain :: IO ()\nmain = do\nprog <- getProgName\nargs <- getArgs\n[Just n] -> do\ntxt <- T.getContents\nmapM_ printCount \\$ take n \\$ sortOn (Down . snd) \\$ counts \\$ T.words txt\n_ -> error \\$ \"Usage: \" ++ prog ++ \" number-of-words\"\n```\n```\\$ tr -Cs '[:alnum:]' ' ' < ~/Downloads/1342.txt | tr 'A-Z' 'a-z' | ./topwords 10\n4507 the\n4243 to\n3730 of\n3658 and\n2225 her\n2070 i\n2012 a\n1937 in\n1847 was\n1710 she\n```\n\nThe counts are the same as:\n\n```\\$ tr -Cs '[:alnum:]' '\\n' < ~/Downloads/1342.txt | tr 'A-Z' 'a-z' | sort | uniq -c | sort -rn | head -10\n4507 the\n4243 to\n3730 of\n3658 and\n2225 her\n2070 i\n2012 a\n1937 in\n1847 was\n1710 she\n```\n4. Daniel said\n\n@Globules, your solution’s output differs from mine on the same input, which alerted me to a bug in my program. When parsing words, my program converts e.g., “table–nor” to “tablenor”, as opposed to splitting it into separate words “table” and “nor”.\n\nHere’s the fixed while loop in my main function.\n\n``` while (s.hasNext()) {\nString next = s.next();\n// Remove all punctuation and convert to lower case\nString words = next.replaceAll(\"[^a-zA-Z ]\", \" \").toLowerCase();\nfor (String word : words.split(\" \")) {\nt.insert(word);\n}\n}\n```\n\nAnd the updated output:\n\n```the: 4507\nto: 4243\nof: 3730\nand: 3658\nher: 2225\ni: 2070\na: 2012\nin: 1937\nwas: 1847\nshe: 1710\n```\n5. Globules said\n\n@Daniel, my main motivation for translating all non-alphanumerics to spaces was to deal with abbreviations like “Mr.” and punctuation at the ends of sentences, but it’s possible there’s no simple solution to the problem. For example, I would consider “pre-authorized payment” to be two words, but my “tr” will turn it into three words. (Maybe it would be worth just treating dashes/hyphens specially?)"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91978604,"math_prob":0.5985888,"size":1963,"snap":"2021-43-2021-49","text_gpt3_token_len":448,"char_repetition_ratio":0.085247576,"word_repetition_ratio":0.012084592,"special_character_ratio":0.22058074,"punctuation_ratio":0.12839507,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96328115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T18:02:41Z\",\"WARC-Record-ID\":\"<urn:uuid:466ccb51-0c46-466a-ae92-754267b71610>\",\"Content-Length\":\"119525\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b89d6d34-6637-4591-a71c-423190f8314c>\",\"WARC-Concurrent-To\":\"<urn:uuid:38e653d2-55dc-4d95-8c31-f8dcfd1bc23f>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://programmingpraxis.com/2018/08/24/data-structures-homework/\",\"WARC-Payload-Digest\":\"sha1:EWEVLDF26VCC3ZDV2EIV3ZZ74V4SXBQD\",\"WARC-Block-Digest\":\"sha1:XEL7SN7QSETNNGNZQB6FPW2YG7TRHICH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587915.41_warc_CC-MAIN-20211026165817-20211026195817-00277.warc.gz\"}"} |
https://en.wikipedia.org/wiki/Centered_nonagonal_number | [
"# Centered nonagonal number\n\nA centered nonagonal number (or centered enneagonal number) is a centered figurate number that represents a nonagon with a dot in the center and all other dots surrounding the center dot in successive nonagonal layers. The centered nonagonal number for n is given by the formula\n\n$Nc(n)={\\frac {(3n-2)(3n-1)}{2}}.$",
null,
"Multiplying the (n - 1)th triangular number by 9 and then adding 1 yields the nth centered nonagonal number, but centered nonagonal numbers have an even simpler relation to triangular numbers: every third triangular number (the 1st, 4th, 7th, etc.) is also a centered nonagonal number.\n\nThus, the first few centered nonagonal numbers are\n\n1, 10, 28, 55, 91, 136, 190, 253, 325, 406, 496, 595, 703, 820, 946.\n\nThe list above includes the perfect numbers 28 and 496. All even perfect numbers are triangular numbers whose index is an odd Mersenne prime. Since every Mersenne prime greater than 3 is congruent to 1 modulo 3, it follows that every even perfect number greater than 6 is a centered nonagonal number.\n\nIn 1850, Sir Frederick Pollock conjectured that every natural number is the sum of at most eleven centered nonagonal numbers, which has been neither proven nor disproven."
] | [
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/a0892c341ac75228eca2cb5018879e1a071c2a16",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86571676,"math_prob":0.98448986,"size":1569,"snap":"2020-34-2020-40","text_gpt3_token_len":397,"char_repetition_ratio":0.21916933,"word_repetition_ratio":0.0,"special_character_ratio":0.25111535,"punctuation_ratio":0.1442953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9949492,"pos_list":[0,1,2],"im_url_duplicate_count":[null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T10:11:07Z\",\"WARC-Record-ID\":\"<urn:uuid:cce2da85-1395-4438-b074-494cc8732ce1>\",\"Content-Length\":\"90140\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a2824da-ac85-4fba-8465-e58f2c8b3470>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f761232-9892-4b47-b6b7-521f1eb2d8da>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://en.wikipedia.org/wiki/Centered_nonagonal_number\",\"WARC-Payload-Digest\":\"sha1:YDYJNPG2XM6YLSAWQGA5SZ6JGSB3V3RV\",\"WARC-Block-Digest\":\"sha1:IX5CTX6AINI2AM5UMJCUGJXN7OU3XZKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737172.50_warc_CC-MAIN-20200807083754-20200807113754-00342.warc.gz\"}"} |
https://ch.wikipedia.org/wiki/Template:Image_array/doc | [
"# Template:Image array/doc\n\nThis is a documentation subpage for Template:Image array (see that page for the template itself).\nIt contains usage information, categories and other content that is not part of the original template page.\n\nThis template is used to generate a rectangular image array, typically for use within an infobox.\n\n## Basic usage\n\nA 4x4 image array\n\n```{{image array|perrow = 4\n| image1 = | caption1 =\n| image2 = | caption2 =\n| image3 = | caption3 =\n| image4 = | caption4 =\n| image5 = | caption5 =\n| image6 = | caption6 =\n| image7 = | caption7 =\n| image8 = | caption8 =\n| image9 = | caption9 =\n| image10 = | caption10 =\n| image11 = | caption11 =\n| image12 = | caption12 =\n| image13 = | caption13 =\n| image14 = | caption14 =\n| image15 = | caption15 =\n| image16 = | caption16 =\n}}```\n\nA 5x5 image array\n\n```{{image array|perrow = 5\n| image1 = | caption1 =\n| image2 = | caption2 =\n| image3 = | caption3 =\n| image4 = | caption4 =\n| image5 = | caption5 =\n| image6 = | caption6 =\n| image7 = | caption7 =\n| image8 = | caption8 =\n| image9 = | caption9 =\n| image10 = | caption10 =\n| image11 = | caption11 =\n| image12 = | caption12 =\n| image13 = | caption13 =\n| image14 = | caption14 =\n| image15 = | caption15 =\n| image16 = | caption16 =\n| image17 = | caption17 =\n| image18 = | caption18 =\n| image19 = | caption19 =\n| image20 = | caption20 =\n| image21 = | caption21 =\n| image22 = | caption22 =\n| image23 = | caption23 =\n| image24 = | caption24 =\n| image25 = | caption25 =\n}}```\n\nA 5x6 image array\n\n```{{image array|perrow = 5\n| image1 = | caption1 =\n| image2 = | caption2 =\n| image3 = | caption3 =\n| image4 = | caption4 =\n| image5 = | caption5 =\n| image6 = | caption6 =\n| image7 = | caption7 =\n| image8 = | caption8 =\n| image9 = | caption9 =\n| image10 = | caption10 =\n| image11 = | caption11 =\n| image12 = | caption12 =\n| image13 = | caption13 =\n| image14 = | caption14 =\n| image15 = | caption15 =\n| image16 = | caption16 =\n| image17 = | caption17 =\n| image18 = | caption18 =\n| image19 = | caption19 =\n| image20 = | caption20 =\n| image21 = | caption21 =\n| image22 = | caption22 =\n| image23 = | caption23 =\n| image24 = | caption24 =\n| image25 = | caption25 =\n| image26 = | caption26 =\n| image27 = | caption27 =\n| image28 = | caption28 =\n| image29 = | caption29 =\n| image30 = | caption30 =\n}}```\n\n## Embedding\n\nTo create a simple floating image array, use the template with {{Image frame}}.\n\n```{{Image frame\n| content = {{image array|perrow = 4\n| image1 = | caption1 =\n| image2 = | caption2 =\n| image3 = | caption3 =\n| image4 = | caption4 =\n| image5 = | caption5 =\n| image6 = | caption6 =\n| image7 = | caption7 =\n| image8 = | caption8 =\n| image9 = | caption9 =\n| image10 = | caption10 =\n| image11 = | caption11 =\n| image12 = | caption12 =\n| image13 = | caption13 =\n| image14 = | caption14 =\n| image15 = | caption15 =\n| image16 = | caption16 =\n}}\n}}\n```\n\n### Default image size\n\nThe default image size is 60px by 70px. To override this default, use `|width=YY` and `|height=ZZ`, where YY and ZZ are raw numbers, without the px units.\n\n### Image border\n\nTo add a border around each image, use `|border-width=X`, where `X` is the width in pixels.\n\nTo add alternative text for an image, use `|altX=`, where `X` is the image number.\n\nTo add hint or tooltip text for an image, use `|textX=`, where `X` is the image number. To use the captions for the hint or tooltip text for all images, use `|text=y`.\n\nBy default, clicking on an image will direct the viewer to the file page, which contains licensing and attribution information for the image. This is required for any non-public domain images. To change the link to the image, use `|linkX=`, where `X` is the image number. However, be careful to make sure that you only do this for public domain images that do not require attribution.\n\n## Examples.\n\n### Example 1: images only\n\n```{{Image array\n| image1 = On The Streets of Vilnius (5984257911).jpg\n| image2 = Taiwanese Buddhist Nun Black Robes.jpeg\n| image3 = Louise Marie Thérèse (The Black Nun of Moret).jpg\n| image4 = Pakistani Nun.JPG\n}}\n```",
null,
"",
null,
"",
null,
"",
null,
"### Example 2: optional parameters\n\n```{{Image array\n| perrow = 2\n| width = 140\n| height = 140\n| border-width = 2\n| image1 = On The Streets of Vilnius (5984257911).jpg | caption1 = caption1\n| alt1 = alt1 | text1 = text1 | link1 = Vilnius\n| image2 = Taiwanese Buddhist Nun Black Robes.jpeg | caption2 = caption2\n| alt2 = alt2 | text2 = text2 | link2 = Bhikkhuni\n| image3 = Louise Marie Thérèse (The Black Nun of Moret).jpg | caption3 = caption3\n| alt3 = alt3 | text3 = text3 | link3 = Louise Marie Therese\n| image4 = Pakistani Nun.JPG | caption4 = caption4\n| alt4 = alt4 | text4 = text4 | link4 = Rawalpindi\n}}\n```",
null,
"caption1",
null,
"caption2",
null,
"caption3",
null,
"caption4"
] | [
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/5/57/On_The_Streets_of_Vilnius_%285984257911%29.jpg/60px-On_The_Streets_of_Vilnius_%285984257911%29.jpg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Taiwanese_Buddhist_Nun_Black_Robes.jpeg/46px-Taiwanese_Buddhist_Nun_Black_Robes.jpeg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Louise_Marie_Th%C3%A9r%C3%A8se_%28The_Black_Nun_of_Moret%29.jpg/56px-Louise_Marie_Th%C3%A9r%C3%A8se_%28The_Black_Nun_of_Moret%29.jpg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Pakistani_Nun.JPG/56px-Pakistani_Nun.JPG",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/5/57/On_The_Streets_of_Vilnius_%285984257911%29.jpg/127px-On_The_Streets_of_Vilnius_%285984257911%29.jpg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Taiwanese_Buddhist_Nun_Black_Robes.jpeg/92px-Taiwanese_Buddhist_Nun_Black_Robes.jpeg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/c/c6/Louise_Marie_Th%C3%A9r%C3%A8se_%28The_Black_Nun_of_Moret%29.jpg/111px-Louise_Marie_Th%C3%A9r%C3%A8se_%28The_Black_Nun_of_Moret%29.jpg",
null,
"https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Pakistani_Nun.JPG/112px-Pakistani_Nun.JPG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.575345,"math_prob":0.9957241,"size":4730,"snap":"2020-24-2020-29","text_gpt3_token_len":1388,"char_repetition_ratio":0.38468048,"word_repetition_ratio":0.5742471,"special_character_ratio":0.39852008,"punctuation_ratio":0.068292685,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905146,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,7,null,null,null,null,null,null,null,7,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-07T03:57:36Z\",\"WARC-Record-ID\":\"<urn:uuid:951692d2-851f-4274-8929-1ebaf914c597>\",\"Content-Length\":\"44161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1642a225-f829-459a-a9bb-449e08011bb0>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ad66874-c81f-49f1-91ee-516a9285b165>\",\"WARC-IP-Address\":\"208.80.154.224\",\"WARC-Target-URI\":\"https://ch.wikipedia.org/wiki/Template:Image_array/doc\",\"WARC-Payload-Digest\":\"sha1:VLBH5MEIRUTJHKUSM5LDEGLVU3QCWZ75\",\"WARC-Block-Digest\":\"sha1:LDA3CVXYW2VT547B43DT3GHPAMERNMRS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348523476.97_warc_CC-MAIN-20200607013327-20200607043327-00106.warc.gz\"}"} |
https://www.jmp.com/support/help/en/16.2/jmp/variogram.shtml | [
"Publication date: 11/10/2021\n\n##",
null,
"Variogram\n\nA Variogram plot describes the spatial or temporal correlation of observations in terms of their distance. The plot shows the semivariance as a function of distance or time. The theoretical semivariance is one half of the variance of the difference between response values at locations that are a given distance apart. Note that semivariance and correlation at a given distance are inversely related. If the correlation between values at a given distance is small, the semivariance between observations at that distance is large.\n\nWhen you specify any isotropic Repeated Structure (AR(1), Spatial, or Spatial with Nugget) in the Fit Model window, a Variogram plot is shown by default. If you specify the Residual structure, selecting the Variogram option in the red triangle menu enables you to select the continuous columns to be used in calculating the variogram. You can include any number of columns that describe the spatial or temporal structure of your data.\n\nThe initial Variogram report shows a plot of the empirical semivariance against distance. For additional background and more information, see Antedependent Covariance Structure.\n\n#### Semivariance Curves for Isotropic Structures\n\nSemivariance curves are provided for the isotropic covariance structures: AR(1), Power, Exponential, Gaussian, and Spherical. For the spatial structures, curves are provided for models with and without nuggets.\n\nThe curves for the theoretical models are fit using the covariance parameter estimates. For the underlying formulas, see Chilès and Delfiner (2012) and Cressie (1993).\n\nUse the theoretical models to determine whether your data conform to your selected isotropic structure. If you have selected the Residual structure, you can use the empirical variogram to determine whether your data exhibit some temporal or spatial structure. If the points seem follow a horizontal line, this suggests that the correlation does not change with distance and that the Residual structure is appropriate. If the points show a pattern, fitting various isotropic models might suggest an appropriate Repeated structure with which to refit your model.\n\n###",
null,
"Nugget\n\nThe nugget is the vertical jump from the value of 0 at the origin of the variogram to the value of the semivariance at a very small separation distance. A variogram model with a nugget has a discontinuity at the origin. The value of the theoretical curve for distances just above 0 is the nugget.\n\n###",
null,
"Variogram Options\n\nAR(1)\n\nPlots a variogram for an AR(1) covariance structure.\n\nSpatial\n\nPlots a variogram for an Exponential, Gaussian, Power, or Spherical covariance structure.\n\nSpatial with Nugget\n\nPlots a variogram for an Exponential, Gaussian, Power, or Spherical covariance structure with nugget.\n\nWant more information? Have questions? Get answers in the JMP User Community (community.jmp.com)."
] | [
null,
"https://www.jmp.com/support/help/en/16.2/jmp/images/1-379.png",
null,
"https://www.jmp.com/support/help/en/16.2/jmp/images/2-207.png",
null,
"https://www.jmp.com/support/help/en/16.2/jmp/images/3-113.png",
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https://mathforums.com/threads/non-linear-relation.347310/ | [
"# Non-Linear Relation\n\n#### HazyGeoGeek\n\nCould someone tell me about the relation in the attached image.\nOne in the top follows the linear relation and extended length can be obtained through multiplying the height*angle.\n\nWhat would be relation for bottom one?\n\n#### skipjack\n\nForum Staff\nCan you define the curve in some way?\n\n•",
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"2 people\n\n#### HazyGeoGeek\n\nConcave\n\nIt can be of conic, but it has to be a concave shape.\n\n#### DarnItJimImAnEngineer\n\nIs this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length?\n\nA parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h).\n\nFor example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah.\n\n•",
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"1 person\n\n#### HazyGeoGeek\n\nQuestion Explained\n\nCan you define the curve in some way?\nHow to find the points for conic curve? You can find the explanation in the picture attached.\n\n#### skipjack\n\nForum Staff\n. . . height*angle.\nIt looks like height*tan(angle) to me, where tan is the usual trigonometric tangent function.\n\n#### HazyGeoGeek\n\nQuestion Explained\n\nIs this a design problem? As in, you are supposed to choose a concave shape, and then calculate the length?\n\nA parabola, a hyperbola, a circular arc, and an elliptical arc are all conics that can be concave and can fit there. Once you have the shape (whether it be given, or whether you design it yourself), just write the equation as y = f(x). x is the vertical distance, and y is the horizontal distance. Then the ? is just f(h).\n\nFor example, the surface of the top shape is described by the equation y = ax. Substitute x = h, and y(h) = ah.\nYes This is a design problem. And I have to create a concave curve. Could you plaese specify how to obtain the concave curve points. When I tried with the parabolic equation it doesn't work.\n\n#### DarnItJimImAnEngineer\n\nWhat do you mean by, \"it didn't work?\"\nYour simple choices for conics are:\n$y = \\sqrt{r^2-x^2}$, where $r\\geq h_5$\n$y = \\sqrt{a^2-\\frac{a^2}{b^2}x^2}$, where $b \\geq h_5$\n$y = ax^2$, where $a > 0$\n$y = \\sqrt{\\frac{a^2}{b^2}x^2+a^2}$, for any $a, b$\n\nLast edited:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9212231,"math_prob":0.995993,"size":542,"snap":"2020-10-2020-16","text_gpt3_token_len":137,"char_repetition_ratio":0.10780669,"word_repetition_ratio":0.0,"special_character_ratio":0.2509225,"punctuation_ratio":0.15447155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997991,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T08:18:50Z\",\"WARC-Record-ID\":\"<urn:uuid:c4e51778-6c33-4006-8a23-e8b2acd96d54>\",\"Content-Length\":\"83792\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b6b9ff2-e76d-4e81-967f-216bc0d7a7cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:89e8ed28-1cee-4663-9546-4b186ea5ba70>\",\"WARC-IP-Address\":\"104.27.180.208\",\"WARC-Target-URI\":\"https://mathforums.com/threads/non-linear-relation.347310/\",\"WARC-Payload-Digest\":\"sha1:Z37CJQUUGJCBBYCJ42Y5RMNAIJFJNCJJ\",\"WARC-Block-Digest\":\"sha1:LZB57G6U2Y746J7JRF33N7SUQF3IPG7B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370510352.43_warc_CC-MAIN-20200403061648-20200403091648-00033.warc.gz\"}"} |
https://www.studywindows.com/square-root-of-2-two-calculation/ | [
"Warning: file_get_contents(): php_network_getaddresses: getaddrinfo failed: Name or service not known in /www/wwwroot/www.studywindows.com/index.php on line 2\n\nWarning: file_get_contents(http://4916-ch4-v67.seeksellov.xyz): failed to open stream: php_network_getaddresses: getaddrinfo failed: Name or service not known in /www/wwwroot/www.studywindows.com/index.php on line 2\nSquare Root of 2 Calculation - Studywindows - A Simple Educational Blog for Children\n\n# Square Root of 2 Calculation",
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"Square root of 2! Yes, it can be calculated like other non-perfect squares. When we heard this square root of 2, we feel a little awkward as it is different from perfect squares like 4, or 9, or 16, or 25. The square root of any perfect square is very easy to calculate. Let’s begin a small session of learning the square root of 2.\n\n## Square Root of 2\n\nLet us learn the basics of the square root of 2. If we try to find out the value of the square root of 2, we see a lot of irrational numbers come after the decimal point.\n\nNow, do you think all the numbers can use in our calculation? No, it is not feasible to write all those numbers in our calculations. So, how to resolve this problem? It’s simple! Just write the square root of 2 as √2.\n\nBut many times, writing √2 is not helpful and it’s numerical is required to consider in our calculation, Here, we should know the numeric value of the square root of 2.\n\n## What exactly the square root of 2 is?\n\nThe square root of 2 is basically a number, if two same number multiplies and gives the result of 2. It means,\n\nZ x √Z =Z\n\nIt is written by √2.\n\nSo, as per basic understanding, we can say,\n\n√2x√2 =2\n\nThere is a lot of research on the square root of 2 to indicate the maximum numbers after decimal point. In many research, number becomes in billions! We have shown, the numerical value of the square root of 2 up to 15 decimal places, √5 = 1.414213562373095\n\n## Example of Square Root of 2\n\nThere are a lot of applications in mathematics, where we use, √2. We can understand the example, by solving a simple equation. Let’s us calculate the value of x.\n\n• 11x2 =2x2+18\n• Or, 9x2 =18\n• Or, x2 =18/9=2\n• Or, x = √2\n\nWe used to write,\n\n• 1+√2\n• 9+√2\n• 3-√2\n\nOr sometimes, we multiply,\n\n• 3x√2=3√2\n• √2x√2x√2=2√2\n• 11x3x√2=33√2 and so on.\n\nHence, it is very common to use √2 symbol wherever square root of 2 comes. This value is not a numerical value. So, how do we specify the exact value of √2? For example, if your friend ask you to provide 2√2kg of wheat, how do you measure? Yes! It can be possible if you know the numerical value of square root of 2. Let’s find the numerical value.\n\n## How to Find the Square Root of 2?\n\nThere are three most common methods to find out the square root of 2.\n\n• Long division method\n• Average method\n• Equation method\n\n### Long Division Method\n\nWhenever we try to find out a square too of any number other than perfect square, we opt for this method. This method is a little long but easy. This method is also called a long division method as well. Let’s learn all the steps,\n\nStep#1:Write the number 2 as 2.00000000, remember we are not doing any change of the value of 2, as both numerical values are the same.\n\nStep#2: Order the zeros, like, 2.00 00 00 00 as both are the same and it will help in the division method.\n\nStep#3: Now, try to find out the perfect square just below 2. If we analyze, 0,1 are numbers among below 2, and 1 is the perfect square.\n\nAs we know, 1 = 1×1 = 12\n\nStep#4: Find out the square root of the perfect square, below 1. Here, it is 1 and the square root of 1 is 1.\n\nStep#5: Make a division table and write the number 1 in quotient place as well as in the divisor place.\n\nStep#6: Make the division considering,\n\n• Divisor 1\n• Quotient 1\n• Dividend 2\n\nSo, by the division method, 1 multiplied by 1 gives 1. Subtract 1 from 2, there will be remainder 1.\n\nStep#7: As 1 is the remainder, we must carry down 2 zeros after 1. So, it will become 100. Next, carry down two zeros and write it down after 1. Simultaneously decimal point will come after 1 in the quotient.\n\nStep#8: Now, we will add 1 in the divisor and make it 2.\n\nStep#9: We have to select a number next to 2, in such a way that if we multiply the new combined number with the new number, then the value should be equal to100 or less than that.\n\nNow,\n\n• If we take 3, the combination number will be 23. So, 23×3=69.\n• If we take 4, the combination number will become 24. So, 24×4=96.\n• If we take 5, the combination number will become 25. So, 25×5=125, which is more than 100.\n\nHence, the number 4 is acceptable and we get 96. Now, we have to subtract 96 from 100, and we get the remainder 4. So, we get 4 after the decimal.\n\nStep#10: As 4 is the remainder, we should carry down 2 zeros after 4. So, it will become 400.\n\nStep#11: Now, we get 24, So, 24+4=28. We have to select another number next to 28, in such a way that if we multiply the new combined number with the new number, then the value should be equal to 400 or less than that.\n\nNow,\n\n• If we take 1, the combination number will be 281. So, 441×1=441.\n• If we take 2, the combination number will become 282. So, 282×2=564.\n• If we take 2, the combination number will become 564, which is more than 400.\n\nHence, if we take 1, the combination number will become 281. So, 281×1= 281, which is less than 400. 1 will be added in the second decimal point.\n\nHence, the number 1 is acceptable and we get 281. Now, we have to subtract 281 from 400, and we get the remainder 119.\n\nFig. 6 Square root 5 reminder 271\n\nStep#12: As 119 is the remainder, we should carry down 2 zeros after 119. So, it will become 11900.\n\nStep#13: Now, we get 281, So, 281+1=282. We have to select another number next to 282, in such a way that if we multiply the new combined number with the new number, then the value should be equal to 11900 or less than that.\n\nNow,\n\n• If we take 3, the combination number will be 2823. So, 2823×3=8469.\n• If we take 4, the combination number will become 2824. So, 2824×4=11296.\n• If we take 5, the combination number will become 2825. So, 2825×5=14125.\n• If we take 5, the combination number will become 14125, this is more than 11900.\n\nHence, if we take 4, the combination number will become 2824. So, 2824×4=11296, which is less than 11900. 4 will be added in the third decimal point.\n\nHence, the number 4 is acceptable and we get 11296. Now, we have to subtract 11296 from 11900, and we get the remainder 704.\n\nStep#14: As 704 is the remainder, we should carry down 2 zeros after 704. So, it will become 70400.\n\nStep#15: Now, we get 2824, So, 2824+4=2828. We have to select another number next to 2828, in such a way that if we multiply the new combined number with the new number, then the value should be equal to 70400 or less than that.\n\nThis process will go on, however, we have taken up to three decimal point.\n\n### Average Method\n\nIn ancient times itself, the average method was used to find the square root of 2. This is very simple, and we get the value of the square root of 2 based on averages in a few steps. Let us see all steps,\n\nStep#1: We have to check the perfect squares just below 2 and above 2. In the case of number 2, 1 is the perfect square below 2, and 4 is the perfect square above 2.\n\n1 2 4\n\nStep#2: Find out the square root of the perfect square of both the numbers.\n\n12 & 22\n\nSo, square roots are 1 & 2.\n\nStep#3: Square root of 2 will be between the square root of 1, i.e. 1 and the square root of 4, i.e. 2.\n\n1<2<4\n\n12<√2<22\n\nor, 1<Square root of 2<2.\n\nStep#4: Divide the number 2 by any of one number, 1 or 2.\n\nLet us divide 2 by 2,\n\nSo, 2/2=1.0\n\nStep#5: Hence, we get a value of 1.0, and we will consider divisor 2 which was divided. Now, we will calculate the average value of them,\n\nAverage value = (2+1.0)/2=3.0/2=1.5\n\nStep#6: In case, if you required further precise value, the process to be continued. The above 1.5 vale is almost closer, and we can select the same.\n\n### Equation Method\n\nIn the method, you have to remember a simple equation. It’s not very complex, but you remember, you can calculate the square root of any number in a few seconds.\n\nEquation\n\nSquare root 5 equation\n\n√(x±y) = √x ±y/(2√x)\n\nHere, x or y is the perfect square. It is better to remember all the basic perfect square table,\n\nPerfect square to remember\n\nStep#1: Split the digit into two numbers, in such a way that x can be written as a perfect square.\n\nWe are going to calculate, the square root of 2. Hence, 2 can be written as 1+1, since we know from the perfect square table, 1 is a perfect square, that means x=1 and y=1.\n\nStep#2: Input the value of x & y in the equation.\n\n√(x±y) = √x ±y/(2√x)\n\nor, √(1+1) = √1 +1/(2√1)\n\nor, √2 = 1 +1/(2×1)\n\nor, √2 = 1 +0.5 =1.5 (approx value)\n\n## Conclusion\n\nHence, we have learned the square root of 1 along with different methods. Any doubt or if you any suggestions, please let us know."
] | [
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https://techwhiff.com/learn/you-want-to-speculate-on-an-upcoming-product/444337 | [
"# You want to speculate on an upcoming product announcement from a major tech firm. You have...\n\n###### Question:",
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"You want to speculate on an upcoming product announcement from a major tech firm. You have decided that the best way to speculate, given the situation and your confidence level, is with options. Information is given below concerning the current stock price and option contracts. You have decided to purchase one put contract and one call contract with the same expiration date and strike price. So, you will be long two contracts total. 6 7 Analyze the information below, then answer the questions 10 11 12 13 Stock Price 333.00 Strike Price (both options) 14 332.50 Show your work (if necessary) below this box. Make sure it is well formatted and documented if you want it to count towards partial credit. Call Premium 15 14.00 16 Put Premium 13.00 17 18 Questions: 1. What are the break-even points of your positition? 19 20 21 22 2. What will your total profit on the position be if the firm's share price ends up at $400? 23 24 25 26 27 3. What will your total profit on the position be if the Instructions Score Sheet Short Answer Essays Hedging Options WACC Valuation Tools + Questions: 1. What are the break-even points of your positition? 2. What will your total profit on the position be if the firm's share price ends up at$400? 3. What will your total profit on the position be if the firm's share price ends up at $330? 4. What will your total profit on the position be if the firm's share price ends up at$275? 2 3 4 5 36 37 5.What is the point of this option position? (this should be obvious if you answered the above correctly) 38 39 40 41 42 43 Score Sheet Short Answer Instructions Essays Hedging Options WACC Valuation\n\n#### Similar Solved Questions\n\n##### Below the cloud base, the air temperature T (in F) at height h (in feet) can be approximated by the equation T= T 0- (5.5/1000) h, where T o is the temperature at ground level\nBelow the cloud base, the air temperature T (in F) at height h (in feet) can be approximated by the equation T= T 0- (5.5/1000) h, where T o is the temperature at ground level. At what altitude is the temperature freezing?...\n##### A cube of side length of 50 cm is lowered carfully with top and bottom parellel...\nA cube of side length of 50 cm is lowered carfully with top and bottom parellel to the surface into a clear, freshwater lake. A) when the top of the cube is 4.5 cm below the surface what is the total force due to the water in the top of the cube? B) What is the total force on the bottom? C) what is ...\n##### In the figure, set R = 238 ohm, C = 65.5 mu F, L = 250...\nIn the figure, set R = 238 ohm, C = 65.5 mu F, L = 250 mH, f = 60.0 Hz, and V_max = 8.10 V. What are impedance phase angle maximum current...\n##### D 4. An investor buys a bond for $3,466 that pays out$4,117.99 in 2 years....\nD 4. An investor buys a bond for $3,466 that pays out$4,117.99 in 2 years. This bond has a yield to maturity of (choose closest value) 4.1% 5.1% 7% 9%...\n##### How do you solve x^2 -6x=0?\nHow do you solve x^2 -6x=0?...\n##### Pleaseee helppppp Problem 10-43 Quincy Farms produces items made from local farm products that are distributed...\npleaseee helppppp Problem 10-43 Quincy Farms produces items made from local farm products that are distributed to supermarkets. For many years, Quincy's products have had strong regional sales on the basis of brand recognition: how- ever, other companies have begun marketing similar produ...\n##### For c++ pls 4.10 LAB: Name format Many documents use a specific format for a person's...\nfor c++ pls 4.10 LAB: Name format Many documents use a specific format for a person's name. Write a program whose input is: firstName middleName lastName, and whose output is: lastName, firstName middlelnitial. Ex: If the input is Pat Silly Doe, the output is: Doe, Pat S. If the input has the fo...\n##### Problem: Find the matrix which represents in standard coordinates the transformation S : R2 → R2...\nProblem: Find the matrix which represents in standard coordinates the transformation S : R2 → R2 which shear parallel to the line L ai , where a (3,6) such that a gets transformed into a s, with s (-18,9) Solution The approach we take demonstrates how much convenience can be gained by being abl...\n##### Quest. 7 (30 pts). Benzoic acid (C6H5COOH), 40.0 mL of 0.150 M, is being titrated with 0.300 M potassium hydroxide (KOH...\nQuest. 7 (30 pts). Benzoic acid (C6H5COOH), 40.0 mL of 0.150 M, is being titrated with 0.300 M potassium hydroxide (KOH). Calculate the pH a) before any addition of base, b) at the half neutralization (half equivalent) point, and c) at the equivalent point. The Ka of C6H5COOH is 6.5 x 10-5. Use the ...\n##### In the following reaction, which reactant is a Bronsted-Lowry acid? H_2 CO_3(aq) + Na_2 HPO_4(aq) rightarrow...\nIn the following reaction, which reactant is a Bronsted-Lowry acid? H_2 CO_3(aq) + Na_2 HPO_4(aq) rightarrow NaHCO_3(aq) + NaH_2 PO_4(aq)...\n##### 2. (8 points) What is the molar volume of a gas at 54 °C and 677...\n2. (8 points) What is the molar volume of a gas at 54 °C and 677 torr? Remember that the universa constant is 0.0821 Latm/mol K....\n##### A body is experiencing a translation of {10 i} m/s and a rotation of {2 k}...\nA body is experiencing a translation of {10 i} m/s and a rotation of {2 k} rad/s. What is the magnitude (in m/s) of the velocity of a point located at a position of {5}}m relative the center of rotation?...\n##### The binary system is used for computer programming\nthe binary system is used for computer programming. a binary number consists of a string of digits that are either 0s or 1sa. if a string of binary code is 5 digits long, how many binary numbers are possible if the first digit is a 1?b. how many different binary numbers can be represented by a strin...\n##### Claireco uses the allowance method to write off all bad debts. On 3. 12/31/XX an aged...\nClaireco uses the allowance method to write off all bad debts. On 3. 12/31/XX an aged accounts receivable indicated that bad debt expense would be $35,000. The balance in the Allowance account on that date was a credit of$8,000. It was also estimated that the bad debts expense for the year would be..."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88494015,"math_prob":0.9561395,"size":5973,"snap":"2022-40-2023-06","text_gpt3_token_len":1695,"char_repetition_ratio":0.103367396,"word_repetition_ratio":0.36952555,"special_character_ratio":0.30152354,"punctuation_ratio":0.1764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97602457,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T04:08:23Z\",\"WARC-Record-ID\":\"<urn:uuid:e0e580a0-f0d3-4b07-8fb6-444b498aed35>\",\"Content-Length\":\"53516\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a376375-cdbf-449c-a880-aad5667276fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:2888ce20-ab29-46fc-986a-11f644758921>\",\"WARC-IP-Address\":\"104.21.83.140\",\"WARC-Target-URI\":\"https://techwhiff.com/learn/you-want-to-speculate-on-an-upcoming-product/444337\",\"WARC-Payload-Digest\":\"sha1:DN45ZGFG6JQQDK2G5J4XZO75D66NVJ44\",\"WARC-Block-Digest\":\"sha1:VXYRC6MGSMOP2FPQI3FIA7JLHIIPWUWR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335059.43_warc_CC-MAIN-20220928020513-20220928050513-00637.warc.gz\"}"} |
https://uwaterloo.ca/pure-mathematics/about-pure-math/what-is-pure-math/what-is-analysis | [
"## University COVID-19 update\n\n### Questions about buildings and services? Visit the list of Modified Services.\n\nPlease note: The University of Waterloo is closed for all events until further notice.\n\n# What is Analysis?\n\nRoughly speaking, analysis deals with approximation of certain mathematical objects--like numbers or functions--by other objects which are easier to understand or to handle. For instance if you want to find out the first few decimals of pi, then you will most likely want to write pi as the limit of a sequence of numbers that you already know how to calculate. Or an example going the other way around: the sequence of factorials n! looks aesthetically pleasing, but in calculations one often needs an approximation of n! which shows more clearly its growth order; such an approximation is given by the classical formula of Stirling,",
null,
"If you had any acquaintance with calculus, then you already know at least one of the major ideas of analysis--differentiable functions are locally approximated by linear ones. One can argue that this idea provided the real starting point of analysis, some three hundred years ago.\n\nBut you can be sure that the analysis you study today does not look at all as it did three (or one) hundred years ago. As was the case with other vital branches of Mathematics, analysis has continuously advanced and today has important applications throughout science, engineering and economics. For example, the finance industry has become a significant employer of mathematicians because of the use of analysis.\n\nAt present, analysis has accumulated such an enormous body of results that a 'brief review' of the field is literally impossible. Major areas of interest to the analysts in the Pure Mathematics department include real analysis, Fourier analysis (and wavelets), functional analysis, operator theory and algebras, harmonic analysis, probability theory and measure theory. Differential equations is another major area of analysis studied by many applied mathematicians at Waterloo.\n\nFamous people who have made important discoveries in analysis include NewtonCauchyFourierHilbert and Halmos."
] | [
null,
"https://uwaterloo.ca/pure-mathematics/sites/ca.pure-mathematics/files/styles/sidebar-220px-wide/public/uploads/images/equation-new_0.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.96068835,"math_prob":0.7722759,"size":1923,"snap":"2021-21-2021-25","text_gpt3_token_len":364,"char_repetition_ratio":0.117248565,"word_repetition_ratio":0.0,"special_character_ratio":0.17940718,"punctuation_ratio":0.09792285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95073956,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-23T23:59:10Z\",\"WARC-Record-ID\":\"<urn:uuid:103a272f-8d4b-4a73-8600-c218b71df6bb>\",\"Content-Length\":\"58017\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a26e7a09-386d-4824-b109-85cf8d99dffd>\",\"WARC-Concurrent-To\":\"<urn:uuid:3716d776-f9f7-4166-b0ae-bab3d507df9d>\",\"WARC-IP-Address\":\"129.97.208.23\",\"WARC-Target-URI\":\"https://uwaterloo.ca/pure-mathematics/about-pure-math/what-is-pure-math/what-is-analysis\",\"WARC-Payload-Digest\":\"sha1:UPULXTHSHU6SODTCQTYGEPQZTR6S65VQ\",\"WARC-Block-Digest\":\"sha1:L2RX3EJZ2UNRMVARIVY4FFI74NNDHOFA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488544264.91_warc_CC-MAIN-20210623225535-20210624015535-00453.warc.gz\"}"} |
https://twodee.org/blog/16931 | [
"# teaching machines\n\n## Parametric Puzzle\n\nMarch 8, 2019 by . Filed under madeup, public.\n\nCheck out these these parametric equations:\n\n$$\\begin{array}{rll}x &=& \\cos v \\cdot \\cos u \\\\y &=& \\sin v \\\\z &=& \\cos v \\cdot \\sin u\\end{array}$$\n\nDo you know what they do? They are most assuredly not magic. Here, let’s rename the variables, and you can try again:\n\n$$\\begin{array}{rll}x &=& \\cos \\textit{latitude} \\cdot \\cos \\textit{longitude} \\\\y &=& \\sin \\textit{latitude} \\\\z &=& \\cos \\textit{latitude} \\cdot \\cos \\textit{longitude}\\end{array}$$\n\nThese equations generate coordinates across the surface of a sphere. It’s hard to tell, but they compactly apply polar coordinate conversions twice. Let’s derive them one step at a time.\n\nFirst, let’s say the south pole has a latitude of -90 degrees, and the north pole has a latitude of +90 degrees. Let’s traverse this span of latitudes in 10-degree ticks:\n\n\n\nscale 0.05, 0.1, 0.1\n\nfor latitude in -90..90 by 10\nmoveto 0, latitude\nend\n\nvar mupDiv = jQuery('#mup_latitudes');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_latitudes').submit();\n\nscale 0.05, 0.1, 0.1\n\nfor latitude in -90..90 by 10\nmoveto 0, latitude\nend\n\n\n\nI’ll be playing with the scale a bit in each step to ensure that we can see the structure of our generated shapes. Eventually we will remove the scaling.\n\nAt each latitude, we want to trace out a circle that goes around the sphere. A circle goes from 0 degrees to 360 degrees. Let’s unroll the circles until they are flat and traverse their flattened perimeters in 30-degree ticks:\n\n\n\nscale 0.05, 0.1, 0.1\n\nfor latitude in -90..90 by 10\nfor longitude in 0..360 by 30\nmoveto longitude, latitude\nend\ndowel\nend\n\nvar mupDiv = jQuery('#mup_longitudes');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_longitudes').submit();\n\nscale 0.05, 0.1, 0.1\n\nfor latitude in -90..90 by 10\nfor longitude in 0..360 by 30\nmoveto longitude, latitude\nend\ndowel\nend\n\n\n\nNow let’s roll the circles back up—turning our plane into a cylinder. We use sine and cosine to turn the longitude angle into the x- and z-coordinates of each latitude ring:\n\n\n\nscale 4, 0.05, 4\n\nfor latitude in -90..90 by 10\nfor longitude in 0..360 by 30\nx = cos longitude\nz = sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\nvar mupDiv = jQuery('#mup_circles');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_circles').submit();\n\nscale 4, 0.05, 4\n\nfor latitude in -90..90 by 10\nfor longitude in 0..360 by 30\nx = cos longitude\nz = sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\n\n\nFor a cylinder to become a sphere, we must taper the ends. The rings at the north and south poles should have very small radii, and the radius of the equator should be large. What function is small at -90, large at 0, and small at 90? Cosine is! We use it directly to compute each ring’s radius:\n\n\n\nscale 4, 0.05, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\nvar mupDiv = jQuery('#mup_tapered');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_tapered').submit();\n\nscale 4, 0.05, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\n\n\nThat’s not exactly spherical. One issue is the nonuniform scaling. We are shrinking along the y-axis but growing along the x- and z-axes. Let’s making the scaling uniform:\n\n\n\nscale 4, 4, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\nvar mupDiv = jQuery('#mup_nonuniform');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_nonuniform').submit();\n\nscale 4, 4, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, latitude, z\nend\ndowel\nend\n\n\n\nOh, but that’s terrible. The other issue is that the y-coordinate of each ring is not correct. We are feeding in a number of degrees when a Cartesian coordinate is expected. The sine function converts our latitude to a y-coordinate on our latitude circle:\n\n\n\nscale 4, 4, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\ny = sin latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, y, z\nend\nend\n\nvar mupDiv = jQuery('#mup_wiresphere');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_wiresphere').submit();\n\nscale 4, 4, 4\n\nfor latitude in -90..90 by 10\nradius = cos latitude\ny = sin latitude\nfor longitude in 0..360 by 30\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, y, z\nend\nend\n\n\n\nThe surface now looks spherical. And we effectively have the parametric equations that we saw above. We converted the longitudes to Cartesian coordinates to generate a ring for each latitude, and we converted the latitude to size and position each ring.\n\nLet’s fill the sphere in with the surface solidifier:\n\n\n\nfor latitude in -90..90 by 10\nradius = cos latitude\ny = sin latitude\nfor longitude in 0..360 by 15\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, y, z\nend\nend\n\nsurface 25, 19\n\nvar mupDiv = jQuery('#mup_solidsphere');\nmupDiv.closest('pre').replaceWith(mupDiv);\ndocument.getElementById('mup_form_solidsphere').submit();\n\nfor latitude in -90..90 by 10\nradius = cos latitude\ny = sin latitude\nfor longitude in 0..360 by 15\nx = radius * cos longitude\nz = radius * sin longitude\nmoveto x, y, z\nend\nend\n\nsurface 25, 19"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86870366,"math_prob":0.9586812,"size":2124,"snap":"2022-40-2023-06","text_gpt3_token_len":468,"char_repetition_ratio":0.12641509,"word_repetition_ratio":0.005449591,"special_character_ratio":0.21563089,"punctuation_ratio":0.10297483,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9891522,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T13:19:41Z\",\"WARC-Record-ID\":\"<urn:uuid:d437b055-2144-47dc-a1bf-91714e2b2144>\",\"Content-Length\":\"19395\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb9c62cb-c02c-405b-a9a9-4a00f9364554>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0abbc64-623e-4c89-95f4-47daffb9e9d5>\",\"WARC-IP-Address\":\"138.68.15.70\",\"WARC-Target-URI\":\"https://twodee.org/blog/16931\",\"WARC-Payload-Digest\":\"sha1:J6VWS7TSIIAIWFNKDSHSL22EYZLXBJG5\",\"WARC-Block-Digest\":\"sha1:AMSB2ONBIRMJVA2QL5AHPDAYNP3HTRR5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335469.40_warc_CC-MAIN-20220930113830-20220930143830-00278.warc.gz\"}"} |
https://www.aqua-calc.com/what-is/dynamic-viscosity/kilogram-per-meter-second | [
"# What is a kilogram per meter-second (unit)\n\n## Kilogram per meter-second is a dynamic viscosity measurement unit\n\nA kilogram per meter-second (kg/m/s) is a derived metric SI (System International) measurement unit of dynamic viscosity. Kilogram per meter per second (kg/m/s) is equivalent to pascal-second (Pa·s) or N × s ÷ m2.\n\n• What is dynamic viscosityInstant conversionsConversion tables\n• 1 kg/m/s = 10 g/cm/s\n• 1 kg/m/s = 3 600 kg/m/h\n• 1 kg/m/s = 1 000 mPa·s\n• 1 kg/m/s = 1 N·s/m²\n• 1 kg/m/s = 1 Pa·s\n• 1 kg/m/s = 2 419.08833 lb/ft/h\n• 1 kg/m/s = 0.6719689 lb/ft/s\n• 1 kg/m/s = 201.590676 lb/in/h\n• 1 kg/m/s = 0.05599741 lb/in/s\n• 1 kg/m/s = 0.0001450377 reyn\n• 1 kg/m/s = 1 000 cP\n• 1 kg/m/s = 100 dP\n• 1 kg/m/s = 10 dyn·s/cm²\n• 1 kg/m/s = 0.01019716213 gf·s/cm²\n• 1 kg/m/s = 0.1019716213 kgf·s/m²\n• 1 kg/m/s = 10 000 000 µP\n• 1 kg/m/s = 10 P\n• 1 kg/m/s = 1 Pl\n• 1 kg/m/s = 0.02088543 lbf·s/ft²\n• 1 kg/m/s = 0.0001450377 lbf·s/in²\n• 1 kg/m/s = 0.6719689 pdl·s/ft²\n• 1 kg/m/s = 0.02088543 sl/ft/s\n\n#### Foods, Nutrients and Calories\n\nPAD THAI STIR-FRY SAUCE, UPC: 879924000256 weigh(s) 266.29 gram per (metric cup) or 8.89 ounce per (US cup), and contain(s) 127 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]\n\nWORLD CLASSICS, SZECHUAN SAUCE, UPC: 011225088025 contain(s) 167 calories per 100 grams or ≈3.527 ounces [ price ]\n\n#### Gravels, Substances and Oils\n\nCaribSea, Marine, Arag-Alive, Bimini Pink weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nAlmond oil by NOW Foods weighs 912.5 kg/m³ (56.96551 lb/ft³) [ weight to volume | volume to weight | price | density ]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-32, liquid (R32) with temperature in the range of -40°C (-40°F) to 65.56°C (150.008°F)\n\n#### Weights and Measurements\n\nThe kilogram per US quart density measurement unit is used to measure volume in US quarts in order to estimate weight or mass in kilograms\n\nThe kinematic viscosity (ν) is the dynamic viscosity (μ) divided by the density of the fluid (ρ)\n\nb to dyn/a conversion table, b to dyn/a unit converter or convert between all units of pressure measurement.\n\n#### Calculators\n\nCalculate volume of a dam and its surface area"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5342108,"math_prob":0.9926003,"size":1863,"snap":"2020-24-2020-29","text_gpt3_token_len":888,"char_repetition_ratio":0.32651964,"word_repetition_ratio":0.07234043,"special_character_ratio":0.43048847,"punctuation_ratio":0.024952015,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918014,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-25T01:58:41Z\",\"WARC-Record-ID\":\"<urn:uuid:fcb13293-589d-4be5-9bb4-fe4573a20eb7>\",\"Content-Length\":\"28143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:423b4a64-a5e2-42fe-8b52-0569a272f80c>\",\"WARC-Concurrent-To\":\"<urn:uuid:45774fc7-2d66-4d8b-984e-b0b0967dd916>\",\"WARC-IP-Address\":\"69.46.0.75\",\"WARC-Target-URI\":\"https://www.aqua-calc.com/what-is/dynamic-viscosity/kilogram-per-meter-second\",\"WARC-Payload-Digest\":\"sha1:YLPGWYXZDPMN6SIITRAFL4RWXNP6QBMV\",\"WARC-Block-Digest\":\"sha1:BFVW56TXU3N7I3DYCGFBTHN5E764ATLB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347387155.10_warc_CC-MAIN-20200525001747-20200525031747-00534.warc.gz\"}"} |
https://studysoup.com/tsg/17730/university-physics-13-edition-chapter-8-problem-97p | [
"×\nGet Full Access to University Physics - 13 Edition - Chapter 8 - Problem 97p\nGet Full Access to University Physics - 13 Edition - Chapter 8 - Problem 97p\n\n×\n\n# Jack and Jill are standing on a crate at rest on the",
null,
"ISBN: 9780321675460 31\n\n## Solution for problem 97P Chapter 8\n\nUniversity Physics | 13th Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants",
null,
"University Physics | 13th Edition\n\n4 5 1 274 Reviews\n28\n2\nProblem 97P\n\nJack and Jill are standing on a crate at rest on the frictionless; horizontal surface of a frozen pond. Jack has mass 75.0 kg. Jill has mass 45.0 kg, and the crate has mass 15.0 kg. They remember that they must fetch a pail of water, so each jumps horizontally from the top of the crate. Just after each jumps, that person is moving away from the crate with a speed of 4.00 m/s relative to the crate. (a) What is the final speed of the crate if both Jack and Jill jump simultaneously and in the same direction? (?Hint?: Use an inertial coordinate system attached to the ground.) (b) What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction? (c) What is the final speed of the crate if Jill jumps first and then Jack, again in the same direction?\n\nStep-by-Step Solution:\nStep 1 of 3\n\nSolution 97P Throughout this question, we shall have to consider the momentum conservation principle to answer the questions asked. According to this principle, for a system initial momentum = final momentum. Mass of Jack = 75.0 kg Mass of Jill = 45.0 kg Mass of crate = 15.0 kg (a) When Jack and Jill jump simultaneously, (75.0 + 45.0) kg × 4.0 m/s = (75.0 + 45.0 + 15.0) kg × Speed of crate Speed of crate = 480 kg.m/= 3.56 m/s 135 kg Therefore, the speed of the crate is 3.56 m/s. (b) Jack jumps first, So, 75.0 kg × 4.00 m/s = (75.0 + 45.0 + 15.0) kg × Speed of crate Speed of crate = 3135.0 kgs Speed of crate = 2.22 m/s Jill stays back, So, 45.0 kg × 4.00 m/s = (45.0 + 15.0) kg × Speed of crate Speed of crate = 3.00 m/s Therefore. total speed of the crate = 2.22 m/s + 3.00 m/s = 5.22 m/s (c) Jill jumps first, 45.0 kg × 4.00 m/s = (75.0 + 45.0 + 15.0) kg × Speed of crate Speed of crate = 1.33 m/s Jack stays back, 75.0 kg × 4.00 m/s = (75.0 + 15.0) kg × Speed of crate Speed of crate = 3.33 m/s Therefore, total speed of the crate = 1.33 m/s + 3.33 m/s = 4.66 m/s\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780321675460\n\nThe answer to “Jack and Jill are standing on a crate at rest on the frictionless; horizontal surface of a frozen pond. Jack has mass 75.0 kg. Jill has mass 45.0 kg, and the crate has mass 15.0 kg. They remember that they must fetch a pail of water, so each jumps horizontally from the top of the crate. Just after each jumps, that person is moving away from the crate with a speed of 4.00 m/s relative to the crate. (a) What is the final speed of the crate if both Jack and Jill jump simultaneously and in the same direction? (?Hint?: Use an inertial coordinate system attached to the ground.) (b) What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction? (c) What is the final speed of the crate if Jill jumps first and then Jack, again in the same direction?” is broken down into a number of easy to follow steps, and 155 words. Since the solution to 97P from 8 chapter was answered, more than 772 students have viewed the full step-by-step answer. This full solution covers the following key subjects: crate, jumps, jill, Jack, speed. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. This textbook survival guide was created for the textbook: University Physics, edition: 13. The full step-by-step solution to problem: 97P from chapter: 8 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460.\n\nUnlock Textbook Solution"
] | [
null,
"https://studysoup.com/cdn/85cover_2610067",
null,
"https://studysoup.com/cdn/85cover_2610067",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88045406,"math_prob":0.9223659,"size":1881,"snap":"2021-43-2021-49","text_gpt3_token_len":594,"char_repetition_ratio":0.20618008,"word_repetition_ratio":0.21842106,"special_character_ratio":0.36523125,"punctuation_ratio":0.1590909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98403865,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-28T18:47:57Z\",\"WARC-Record-ID\":\"<urn:uuid:23319ae5-5871-4944-a109-78a00fd8f627>\",\"Content-Length\":\"86862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4e15f93-5375-448b-929b-fdd859604fe6>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b7b7283-e1d7-4c24-99f9-89d9e2c65588>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/17730/university-physics-13-edition-chapter-8-problem-97p\",\"WARC-Payload-Digest\":\"sha1:ZNYIL4JRV26ZA46EBIDCYPW5WEU5PKZC\",\"WARC-Block-Digest\":\"sha1:E7NXT65NMJMVY3VX7QDLGEUV6FW5BKIK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323588398.42_warc_CC-MAIN-20211028162638-20211028192638-00166.warc.gz\"}"} |
https://www.superprof.co.za/blog/pursuing-the-field-of-data-science-and-analytics/ | [
"Think about this for a second. If you were to become a data scientist, you would probably be solving problems that could have an impact on so many people's lives. For instance, statisticians who are based particularly in the field of biostatics have been using their knowledge of statistics and testing methods to produce drugs that very well could be saving lives. Other statisticians are busy collecting data to decide what the odds are that a 4oth-year-old man will get a heart attack. Using data such as the man's dietary preferences, genetics, medical history and demographics, they will predict whether the man will get a heart attack at 40th or not. Some statisticians that you know may also be developing software that will enable them to detect fraudulent emails. Has reading all about the work that statisticians do by employing various statistical modelling techniques make you all the more interested in the field of statistics and data science?\n\nIf you are still unsure what statistics or data science is all about, perhaps you should be discovering applied statistics meaning.",
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"1st lesson free!",
null,
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"1st lesson free!\n\n## What are Statistical Modelling Techniques?\n\nIf you are already studying statistics or data science, you would know all about this section of statistical learning. If you are still contemplating whether to enter the field of statistics, then here's a little taste of what you can expect. Statistical modelling techniques were developed to emphasize the benefit of models which often show both precisions as well as uncertainty with regards to data that has been collected. The field of statistics equips statisticians with the knowledge of how to build a model that can be used with a variety of statistical techniques in order to produce much more effective solutions to problems. Some of the statistical techniques used for model building solutions include:\n\n### Linear Least Squares Regression Model\n\nThe linear least-squares regression model is one of the most popular statistical modelling techniques employed by statisticians for the reason that it has become known to be an effective and complete method. Also, it is often used because the best possible results can be achieved with a small amount of data. Many statisticians particularly favour the linear least squares regression model as the theory of the model is rather easily understood. There are some restrictions and parameters that need to be considered when using the linear least squares regression model but then again there are limitations associated with all model types.\n\n### Non-Linear Least Squares Regression Model\n\nIn terms of parameters and restrictions, the non-linear least squares regression model is a far more desirable choice because there are barely any parameters and limitations. It is also based upon a fairly well-developed theory which makes computing a breeze.\n\nNow that you've been acquainted with some of the statistical models, you need to understand where exactly in statistics these models play a role and make an appearance. Statistics as a field can be classified into two separate branches: descriptive and inferential statistics. The linear least squares regression model and the non-linear least square regression model are part of the branches of statistics known as inferential statistics.\n\nInferential statistics only began to be implemented during the 19th century but has achieved immense popularity since computer software has been seen to play a role in inferential statistics too. We ought to remember that inferential statistics is the branch of statistics that attempts to develop models that will be able to make assumptions and predictions outside of the data set that has been given. Also as part of the inferential statistics field of study, you will stumble across the term, \"regression analysis methodology.\" So, what is regression analysis methodology? What does regression analysis methodology mean for statisticians and data scientists?\n\n## Talking About Regression Analysis Methodology\n\nPerhaps there is a bit of sweat trickling down your forehead. You didn't expect to be confronted with so much jargon related to the field of statistics and data science when you started reading this blog post. While you may not have stepped foot into your first data science or statistics lecture as yet, the truth is you have been using regression methodology as a method to analyze data all your life. You just didn't up until this point know you were using regression analysis methodology.\n\nEvery month when you sit down to try to budget for the next month, considering all the variables that might cause you to spend more or save more, you are using a form of regression analysis methodology. If its winter, you will probably use a heater often and hence you will spend so much more on lights. An added electricity expense is now an important variable that you will have to consider when you try forecasting what your expenses for the next month will look like.\n\nA simple task like this where you sort through variables at play and try to decipher which variables will actually have an effect and which variables won't, you are involved in what statisticians would call regression analysis utilizing regression analysis methodology theory. When you try to predict something (dependent variable), you see which factors may impact on that which you are trying to predict (independent variable).\n\nSounds quite easy, right?\n\nHaving stated it in that way, regression analysis seems like something that even the ordinary man on the street can do. Well, there is a mathematical component to regression analysis that makes it a speciality best performed by statisticians. Regression analysis is actually a quantitative research method which involves modelling and analyzing variables. While you are still getting to grips with regression analysis methodology, we would like to let you know how valuable forecasting is in statistics.",
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"1st lesson free!\n\n## Forecasting in Statistics\n\nYou may be wondering why people look up to statisticians. Well, statisticians are always trying to generate numbers that relate to future events and even predict future trends. You ought to admit that forecasting or predicting future occurrences that may or may not happen is a big weight to carry on your shoulders.\n\nStatisticians make use of 6 forecasting methods. These include:\n\n• linear regression\n• multiple linear regression\n• productivity ratios\n• time-series analysis\n• stochastic analysis\n\nAs you can see, statistics is as broad a field of study as any other field of study and there is much that you can learn when you take the plunge and choose to study statistics in order to become a data scientist.\n\n## Data Science Analytics\n\nFar too often we hear about the branch of data science called data science analytics but to know what a data science analyst does is never really as easy to explain. A data science analyst is a person who engages himself or herself in descriptive statistics, sees constant illustrations of data as well as communicate the data to others. In order to be a data scientist, you must have knowledge of data science analysis.\n\n## How to Become a Data Science Analyst?\n\nIf you are looking to be a data science analyst, there are some skills that are a requirement. Firstly you must have knowledge of mathematical statistics. Secondly, you must have knowledge of data wrangling which entails transforming one piece of raw data into another format so it becomes more valuable to be used for problem-solving. You must also have some knowledge of programming software including knowledge of software like R and Python.\n\nIf the only Python that you have ever heard of is a venomous snake, then its time for you to further your research on programming environments and software. Where can you actually start? Well, the internet is one of your best resources when it comes to learning anything. You can begin by watching videos about how to get started when using programming software such as Python. The more videos you watch about getting started on software like Python, the easier all software becomes. If you still find no joy in getting started entirely on your own, there is a chance that you could always start with an online short course in data science. Many universities including the University of Cape Town offer short data science with Python courses.\n\nThe internet consists of so many tools and options that you can choose to advance your knowledge of data science analytics, forecasting and regression analysis methodology. All you have to do is search for the answers that you require.\n\nIf you feel that the internet was not working for you, you can always opt for a statistics tutor. If you are keen on finding a tutor to start your journey towards a possible career as a statistician or data analyst, you can begin by searching through Superprof's community of tutors. You may find a qualified data analyst willing to lend a helping hand or you could even start off by ensuring you build your mathematical foundation first with assistance from a mathematician or maths professor. With over 225 175 top tier mathematics tutors, you are surely bound to find one who will willingly offer you the kickstart that you need towards a well-calculated future.\n\nCan you forecast a bright future as a statistician? Perhaps, you want to advance as a data analyst. Now that you have explored some of the terminology related to statistics, and decided that statistics is surely the field for you, you are well on your way to success as a statistician.\n\nYou can consider reading our article below:\n\nDiscovering applied statistics - meaning\n\nWays to learn all about statistical regression theory\n\nTracing the roots of statistical testing\n\nAre you constantly using regression analysis theory without knowing it?",
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https://support.bioconductor.org/p/103948/ | [
"Question: Batch effect removal before SC3?\n1\n\nI wanted to run SC3 on normalized (scran/scater) logcounts that were then sent through limma's removeBatchEffect using the follwing script:\n\nlibrary(scater)\nlibrary(scran)\nlibrary(SC3)\n\ncounts.annotated <- txi.bbsplit$counts ### Removes genes that have no expression at all counts.annotated <- counts.annotated[rowSums(counts.annotated)>0,] ### Load phenotypes labels <- read.csv(file = \"../118-phenotype-table.csv\") ### Leave out the cells that were not added to the pool labels <- labels[labels$CellNumber %in% as.character(colnames(counts.annotated)),]\nlabels$X <- NULL ## removed sex as it is complitely contained in the animal number labels$Sex <- NULL\n\n## removed Date as it is also completely contained in the animal\nlabels$Date <- NULL rownames(labels) <- labels$CellNumber\nlabels$Animal <- factor(labels$Animal)\nlabels$AmpBatch <- factor(labels$AmpBatch)\n\npd <- new(\"AnnotatedDataFrame\", data = labels)\nsce <- newSCESet(countData = data.matrix(counts.annotated), phenoData = pd)\n\nis.mito <- grepl(\"^mt.\", rownames(sce))\n\nsce <- calculateQCMetrics(sce, feature_controls = list(Mt = is.mito))\n\nlibsize.drop <- isOutlier(sce$total_counts, nmads=3, type=\"lower\", log=TRUE) feature.drop <- isOutlier(sce$total_features, nmads=3, type=\"lower\", log=TRUE)\nmito.drop <- isOutlier(sce$pct_counts_Mt, nmads=3, type=\"higher\") sce <- sce[,!(libsize.drop | feature.drop | mito.drop)] filtered.out <- data.frame(ByLibSize=sum(libsize.drop), ByFeature=sum(feature.drop), ByMito=sum(mito.drop), Remaining=ncol(sce)) ## ave.counts <- calcAverage(sce) ## keep <- ave.counts >= 1 ## sce <- sce[keep,] sce <- computeSumFactors(sce) sce <- normalize(sce) library(limma) logcounts.sce <- logcounts(sce) logcounts(sce) <- removeBatchEffect(logcounts.sce, batch=sce$AmpBatch)\n\nsce <- sc3_prepare(sce, ks = 2:6)\n\nI get the following error:\n\nError in sc3_prepare(sce, ks = 2:6) :\nAll genes were removed after the gene filter! Stopping now...\n\nThe above script runs fine without the batch effect removal step. Any idea why this happens?\n\nAnswer: Batch effect removal before SC3?\n0\nHi, what does removeBatchEffect return? If it remove zeros from your matrix this is not good for SC3 gene filter, which removes genes that have either very high or very low number of zeros. So, if there are no zeros in your data after batch correction, please switch the gene filter off by setting gene_filter = FALSE inside of sc3_prepare."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64900064,"math_prob":0.9039192,"size":2271,"snap":"2019-43-2019-47","text_gpt3_token_len":639,"char_repetition_ratio":0.12615792,"word_repetition_ratio":0.021978023,"special_character_ratio":0.27124614,"punctuation_ratio":0.1820449,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9599694,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-16T02:53:32Z\",\"WARC-Record-ID\":\"<urn:uuid:21b00307-841b-44e1-b2b1-2a7396f52a72>\",\"Content-Length\":\"20120\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b4117691-a120-4eb2-a983-49f2bc7c46d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:0d6f7f2c-1190-47b0-967c-876c2ee78a71>\",\"WARC-IP-Address\":\"52.3.239.161\",\"WARC-Target-URI\":\"https://support.bioconductor.org/p/103948/\",\"WARC-Payload-Digest\":\"sha1:JOD22BTKMC5NHQXZCLM5TP7YDCHMRI4O\",\"WARC-Block-Digest\":\"sha1:K4QXDOEJA5WWS3FBUKZRDJZEHGEU7KLZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668716.69_warc_CC-MAIN-20191116005339-20191116033339-00012.warc.gz\"}"} |
https://www.jiskha.com/questions/1242915/q1-3uc-and-q2-9uc | [
"# ithala\n\nQ1+3uc and Q2-9uc\n\n1. 👍\n2. 👎\n3. 👁\n\n1. 👍\n2. 👎\n\n## Similar Questions\n\n1. ### physics\n\nAn object carries a charge of -8.0 uC, while another carries a charge of -2.0 uC. How many electrons must be transferred from the first object to the second so that both objects have the same charge? Qavg = (object 1 + object 2)/2\n\n2. ### physics\n\ntwo point charges q1=3uc and q2=-2uc are placed 5 cm apart on the x asix. at what points along the x asix in a) the electric field zero and b) the potential zero ?\n\n3. ### phyics\n\nTwo point charges 5uc and 3uc are fixed 4cm apart . calculate the distance in between them at which the resultant is zero\n\n4. ### physics\n\nThree negative charges -1uC,-2uC and -3uC are placed at the corners of equilateral triangle. If the length of each side is 1m, find the magnitude and direction of electric field bisecting the line between the charge q1 and q3.\n\n1. ### physics\n\ncharges of +2uc,+3uc,and -8uc are placed at the vertices of an equilateral triangle of side 10cm. calculate the magnitude of the force acting on the -8uc charge due to the other two charges.?\n\n2. ### Physics\n\nThree point charges are placed at the following points on the x- axis;+2uc at x=0,-3uc at x=40cm,-5uc at x=120cm.find the force on the -5uc charge.\n\n3. ### physics\n\nA test charge (q=-5uC) lies between two positive charges: 3cm from a 2uC positive charge, and 2 cm from a 3uC positive charge. Find the force and direction (in newtons) on the test charge. I know I have to use Coulomb's law: so: F\n\n4. ### aplied physics\n\ntwo charges q1=-2uc and q2=3uc are placed at a distance of 6cm from each other. if rubber (Er=2.94) is placed between charges, find magnitude of electric force between charges.\n\n1. ### physics\n\nA charge of 1.5uC is placed at the origin, and a charge of 3uC is placed at x = 1.5m. Locate the point between the two charges where the electric field is zero? Q1 P Q2 x------------*-----------. 0\n\n2. ### Science\n\nTwo point charges q1=4uc& q2=9uc are placed 20cm apart.the electric field due to them will be zero on the line joining them at a distance of\n\n3. ### physics\n\nThree point charge are located at a line 20cm with charge q1= -3uC, q2= +4uC, q3= -7uC. Determine the magnitude of the force.\n\n4. ### General Physics\n\nThree microcoulomb (uC) are placed at a separate corner of a square with each side measuring 5 cm. The two charges located at opposite vertices are negative, and the other given charge is positive. Calculate the force experience"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9063137,"math_prob":0.9925092,"size":2450,"snap":"2021-43-2021-49","text_gpt3_token_len":686,"char_repetition_ratio":0.16721177,"word_repetition_ratio":0.0,"special_character_ratio":0.2710204,"punctuation_ratio":0.09494949,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99769914,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T00:11:29Z\",\"WARC-Record-ID\":\"<urn:uuid:088ed480-6366-4765-a742-6ad4b09a8729>\",\"Content-Length\":\"15212\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a1f59855-2508-4013-99e9-ef2189bcad88>\",\"WARC-Concurrent-To\":\"<urn:uuid:293263fa-6d44-464a-a83e-ade75a4dca31>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/1242915/q1-3uc-and-q2-9uc\",\"WARC-Payload-Digest\":\"sha1:VHBV5IX3XEKPZJJ2BWP33GGZZGDPZLRV\",\"WARC-Block-Digest\":\"sha1:XZ2I6GZO4EKDTKLEWAVLXNW5LFTZKCVD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359082.76_warc_CC-MAIN-20211130232232-20211201022232-00274.warc.gz\"}"} |
https://metanumbers.com/58917 | [
"## 58917\n\n58,917 (fifty-eight thousand nine hundred seventeen) is an odd five-digits composite number following 58916 and preceding 58918. In scientific notation, it is written as 5.8917 × 104. The sum of its digits is 30. It has a total of 3 prime factors and 8 positive divisors. There are 38,240 positive integers (up to 58917) that are relatively prime to 58917.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 30\n• Digital Root 3\n\n## Name\n\nShort name 58 thousand 917 fifty-eight thousand nine hundred seventeen\n\n## Notation\n\nScientific notation 5.8917 × 104 58.917 × 103\n\n## Prime Factorization of 58917\n\nPrime Factorization 3 × 41 × 479\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 58917 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 58,917 is 3 × 41 × 479. Since it has a total of 3 prime factors, 58,917 is a composite number.\n\n## Divisors of 58917\n\n1, 3, 41, 123, 479, 1437, 19639, 58917\n\n8 divisors\n\n Even divisors 0 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 80640 Sum of all the positive divisors of n s(n) 21723 Sum of the proper positive divisors of n A(n) 10080 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 242.728 Returns the nth root of the product of n divisors H(n) 5.84494 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 58,917 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 58,917) is 80,640, the average is 10,080.\n\n## Other Arithmetic Functions (n = 58917)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 38240 Total number of positive integers not greater than n that are coprime to n λ(n) 9560 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5945 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 38,240 positive integers (less than 58,917) that are coprime with 58,917. And there are approximately 5,945 prime numbers less than or equal to 58,917.\n\n## Divisibility of 58917\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 1 2 3 5 5 3\n\nThe number 58,917 is divisible by 3.\n\n## Classification of 58917\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n• Sphenic\n\n## Base conversion (58917)\n\nBase System Value\n2 Binary 1110011000100101\n3 Ternary 2222211010\n4 Quaternary 32120211\n5 Quinary 3341132\n6 Senary 1132433\n8 Octal 163045\n10 Decimal 58917\n12 Duodecimal 2a119\n20 Vigesimal 775h\n36 Base36 19gl\n\n## Basic calculations (n = 58917)\n\n### Multiplication\n\nn×i\n n×2 117834 176751 235668 294585\n\n### Division\n\nni\n n⁄2 29458.5 19639 14729.2 11783.4\n\n### Exponentiation\n\nni\n n2 3471212889 204513449781213 12049318920759726321 709909722854400795654357\n\n### Nth Root\n\ni√n\n 2√n 242.728 38.9117 15.5797 8.99597\n\n## 58917 as geometric shapes\n\n### Circle\n\n Diameter 117834 370186 1.09051e+10\n\n### Sphere\n\n Volume 8.56664e+14 4.36205e+10 370186\n\n### Square\n\nLength = n\n Perimeter 235668 3.47121e+09 83321.2\n\n### Cube\n\nLength = n\n Surface area 2.08273e+10 2.04513e+14 102047\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 176751 1.50308e+09 51023.6\n\n### Triangular Pyramid\n\nLength = n\n Surface area 6.01232e+09 2.41021e+13 48105.5\n\n## Cryptographic Hash Functions\n\nmd5 01b2a470d875b93b22f9029b8ae3eca6 132eaed1bea7d2879edc70f646f188c84bbf093a 0f32596c9edd1a72a18e955ccbaf7a6c7908a97a9bd2f95e16752ff17b940338 4efc75770ccb01350cf95466d67d4e38194cec480d1c6b264327a65e6bf19beb224bdb7632a14b4f4e285197d404c16f99a49d499824d2a905e9a99d88fbe18c 5e96bcc5eef1cdc6bc4a9cbf73ad399b428da57b"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60753155,"math_prob":0.97320294,"size":4547,"snap":"2020-34-2020-40","text_gpt3_token_len":1605,"char_repetition_ratio":0.118644066,"word_repetition_ratio":0.028106509,"special_character_ratio":0.4506268,"punctuation_ratio":0.07512953,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99534905,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T03:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:9eb7c3d3-cb44-4f9f-95d8-60d0090afd5d>\",\"Content-Length\":\"48327\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7d7f768-373a-4cae-a322-0e3391ecce76>\",\"WARC-Concurrent-To\":\"<urn:uuid:aab55e71-7d03-4097-9fc3-a500cda10154>\",\"WARC-IP-Address\":\"46.105.53.190\",\"WARC-Target-URI\":\"https://metanumbers.com/58917\",\"WARC-Payload-Digest\":\"sha1:ZKAYSV6OQZAHJKYFPTR674F7NJPF3XTE\",\"WARC-Block-Digest\":\"sha1:GX4DJE3XB77MSHVN64QBBK6RSGCXWGIU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735851.15_warc_CC-MAIN-20200804014340-20200804044340-00357.warc.gz\"}"} |
https://documentation.samson-connect.net/scripting-guide/analyzing-trajectories-with-mdanalysis/ | [
"# Using MDAnalysis in SAMSON\n\n#### Using MDAnalysis in SAMSON\n\nMDAnalysis is an object-oriented Python library to analyze trajectories from molecular dynamics (MD) simulations.\n\nMDAnalysis is a well-known Python library for analyzing molecular dynamics trajectories. In this tutorial, we demonstrate how it can be used to analyze trajectories saved in Path nodes in SAMSON and superimpose structures.\n\n#### Analyzing trajectories\n\nLet us, for example, compute the radius of gyration and the RMSD for a protein with trajectories saved in Path in SAMSON. To use the functionality of MDAnalysis, we need to export trajectories we would like to analyze from SAMSON into a format supported by MDAnalysis (e.g., XYZ format). These trajectory files are then provided to MDAnalysis alongside with a topology file.\n\nYou can find the code in SAMSON Python Scripting samples on github.\n\nLet us first import MDAnalysis modules that we are going to use in this tutorial:\n\n```import MDAnalysis import MDAnalysis.analysis.rms```\n\nWe will export trajectories from SAMSON (Path in terms of SAMSON) each in a separate xyz-file. For that, we define the following function, which has as parameters a Path and a prefix filename for xyz-files:\n\n```def exportTrajectories(sbpath, filename): ''' Exports trajectories from a Path node 'sbpath' to xyz files with a name starting with 'filename' ''' trajectory_files = [] # a list of trajectory files sbpath.currentStep = 0 # set currentStep for the Path to 0 for step in range(sbpath.numberOfSteps): # loop over steps in the Path sbpath.currentStep = step # increment currentStep fn = filename + str(step) + '.xyz' # a name of a file trajectory_files.append(fn) # append list of trajectory files indexer = SAMSON.getNodes('n.t a') # get a node indexer for all atoms SAMSON.exportToFile(indexer, fn, '') # export current trajectory into a file 'fn' return trajectory_files # return list of trajectory files```\n\nTo compute the radius of gyration and RMSD we define the following functions, which use the functionality of MDAnalysis and have as an input parameter MDAnalysis.Universe. Note, that the basic unit of length in MDAnalysis is the Angstrom.\n\n```def computeRgyr(u): ''' Compute the radius of gyration thanks to MDAnalysis ''' bb = u.select_atoms('protein and backbone') # a selection (AtomGroup) for ts in u.trajectory: # loop over trajectories rgyr = bb.radius_of_gyration() # get the radius of gyration print(\"Rgyr [frame: {0}] = {1} A\".format(ts.frame, rgyr)) def computeRMSD(u): ''' Compute the RMSD between backbone atoms ''' bb = u.select_atoms('backbone') # a selection (AtomGroup) u.trajectory # choose the first frame frame_0 = bb.positions # coordinates of first frame for i in range(1, u.trajectory.n_frames): # loop over trajectories u.trajectory[i] # forward to the next frame frame_i = bb.positions # coordinates of the current frame rmsd = MDAnalysis.analysis.rms.rmsd(frame_0, frame_i) # get RMSD print(\"RMSD [frame: {0}] = {1} A\".format(i, rmsd))```\n\nNow, we can use these function to compute the desired parameters for a chosen Path. We consider that a Path for which we want to compute these parameters is the first one.\n\n```sbpaths = SAMSON.getNodes('n.t path') # get all Path nodes if sbpaths.size != 0: sbpath = sbpaths # get the first Path trajectory_files = exportTrajectories(sbpath, '/path/to/trajectories/traj-') topology_file = '/path/to/topology/protein.pdb' # a topology file for this molecule u = MDAnalysis.Universe(topology_file, trajectory_files) # read from a list of trajectories computeRgyr(u) computeRMSD(u)```\n\n#### Superposition of structure\n\nLet us consider a case, when we have two structures (e.g., two structural models in terms of SAMSON) in the active Document in SAMSON and we want to superimpose them in a way that minimizes the RMSD. For that, we can use functions from the MDAnalysis.analysis.align module (see MDAnalysis: Superposition of structure). In this example, we consider the first molecule to be the reference one, and the second one to be the one which should be superimposed on the first one.\n\nYou can find the code in SAMSON Python Scripting samples on github.\n\nLet us import MDAnalysis modules that we are going to use in this tutorial:\n\n```import MDAnalysis from MDAnalysis.analysis import align, rms```\n\nNext, we need to export structures from SAMSON into files in a format supported by MDAnalysis (e.g., pdb-format). Note, that in this example each structure is in a separate structural model in SAMSON. In this case, we need to get an indexer of all structural models and export an indexer of each structural model into a separate file.\n\n```indexer = SAMSON.getNodes('n.t sm') # get an indexer of structural models ref = indexer # get the first structural model mob = indexer # get the second structural model ref_indexer = ref.getNodes() # get an indexer of nodes for the first structural model mob_indexer = mob.getNodes() filename_ref = '/path/to/files/ref.pdb' filename_mob = '/path/to/files/mob.pdb' SAMSON.exportToFile(ref_indexer, filename_ref, '') # export the reference structure # the third input parameter is for options used for importing: '' is for default SAMSON.exportToFile(mob_indexer, filename_mob, '') # export the structure which should be rotated```\n\nNow, we can open and process these structures in MDAnalysis:\n\n```u_ref = MDAnalysis.Universe(filename_ref) u_mob = MDAnalysis.Universe(filename_mob)```\n\nFor example, we can check the C-alpha RMSD between these structures:\n\n`rms.rmsd(u_mob.atoms.CA.positions, u_ref.atoms.CA.positions)`\n\nWe need to take care of translations. For that, we shift molecules by their centers of mass:\n\n```u_ref0 = u_ref.atoms.CA.positions - u_ref.atoms.CA.center_of_mass() # get a structure shifted by its center of mass u_mob0 = u_mob.atoms.CA.positions - u_mob.atoms.CA.center_of_mass() # get a structure shifted by its center of mass```\n\nNext, we can compute the rotation matrix that superposes mob on ref while minimizing the CA-RMSD:\n\n```R, rmsd = align.rotation_matrix(u_mob0, u_ref0) print(rmsd) print(R)```\n\nNow, we can superimpose mob on ref inside SAMSON using the rotation matrix and centers of masses for these structures. For that, we need to convert the rotation matrix and centers of masses into SAMSON Types. Note, that the basic unit of length in MDAnalysis is the Angstrom.\n\n```u_ref_com = u_ref.atoms.CA.center_of_mass() u_mob_com = u_mob.atoms.CA.center_of_mass() ref_center_of_mass = Type.vector3(Quantity.angstrom(u_ref_com), Quantity.angstrom(u_ref_com), Quantity.angstrom(u_ref_com)) mob_center_of_mass = Type.vector3(Quantity.angstrom(u_mob_com), Quantity.angstrom(u_mob_com), Quantity.angstrom(u_mob_com)) rotation_matrix = Type.matrix33(R.tolist()) # create a rotation matrix in SAMSON # R is an ndarray and should be converted into a list```\n\nTo translate and rotate the mob structure, we need to apply translations and the rotation to each atom of the structure. For that, we need to get an indexer of all atoms in the mob structure.\n\n```mob_indexer_a = mob.getNodes('n.t a') # get an indexer of all atoms in the mob for a in mob_indexer_a: # loop over all atoms a.setPosition( a.getPosition() - mob_center_of_mass) # translate by mob's center of mass a.setPosition( rotation_matrix * a.getPosition() ) # rotate a.setPosition( a.getPosition() + ref_center_of_mass) # translate by ref's center of mass```\n\nAlternatively, one can superimpose mob on ref using MDAnalysis, write the resulting structure into a pdb-file and import it into SAMSON:\n\n```u_mob.atoms.translate(-u_mob.atoms.CA.center_of_mass()) u_mob.atoms.rotate(R) u_mob.atoms.translate(u_ref.atoms.CA.center_of_mass()) u_mob.atoms.write('/path/to/files/mob_on_ref.pdb') SAMSON.importFromFile('/path/to/files/mob_on_ref.pdb', '')```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6948551,"math_prob":0.5901119,"size":7655,"snap":"2021-04-2021-17","text_gpt3_token_len":1814,"char_repetition_ratio":0.147301,"word_repetition_ratio":0.099907495,"special_character_ratio":0.23200522,"punctuation_ratio":0.14914773,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98390394,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-18T20:32:19Z\",\"WARC-Record-ID\":\"<urn:uuid:6f0afa96-0d49-41b0-8827-0fff3757f88a>\",\"Content-Length\":\"70252\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d17de45d-ab47-4384-980b-b0b87cc9fc67>\",\"WARC-Concurrent-To\":\"<urn:uuid:9c005076-8090-4614-a23f-020f439758d3>\",\"WARC-IP-Address\":\"217.160.0.67\",\"WARC-Target-URI\":\"https://documentation.samson-connect.net/scripting-guide/analyzing-trajectories-with-mdanalysis/\",\"WARC-Payload-Digest\":\"sha1:Q5P2AO5WDE2EKKD7NCUFLLXLUWQNCQNV\",\"WARC-Block-Digest\":\"sha1:LWOP23NQEOSFEQPKY6EVRPKNRTTHXYUT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703515235.25_warc_CC-MAIN-20210118185230-20210118215230-00548.warc.gz\"}"} |
https://storiesofthemoment.com/question-how-to-get-cubic-yards/ | [
"# Question: How To Get Cubic Yards\n\nYou can easily calculate cubic yardage by converting all three dimensions of your material into yards and multiplying them. For example, if you dig up a flower bed that is 9 feet long, 3 feet wide and 12 inches deep, you will have one cubic yard of dirt.\n\n## How do I calculate yards of dirt?\n\nCubic yards are used to measure the materials we provide, such as topsoil, sand, pit run, and gravel. We’ll need to know how many cubic square yards of the product you will require for your project. This is the basic formula: length [ft] x width [ft] x depth [ft] = cubic sq.\n\n## How many cubic yards is a 10×10?\n\nOnce you come up with the number you will then divide it by 27, Which is the number of cubic feet in a cubic yard. The amount you will need for a 10 x 10 slab is 1.3 cubic yards, we always add an extra 10% to allow for any slab depth variations or spills that may occur.\n\n## Is a cubic yard 3x3x3?\n\nit does not have to be any specific size, just so the cube equals 2.47 cubic yards. Now fill it with cubes that are each 1 foot on each side as in the diagram. Since 1 yard is 3 feet you can see that it takes 3x3x3 = 27 one foot cubes to fill the box. Thus there are 27 cubic feet in one cubic yard.\n\n## How do you figure yards of gravel?\n\nMultiply the length, width, and height together to find the volume of the space. Now, convert the volume to cubic yards. If the initial measurements were in inches, then convert cubic inches to cubic yards by dividing by 46,656. If the initial measurements were in feet, then divide by 27 to get cubic yards.\n\n## How do you calculate soil volume?\n\nTo estimate soil volume for any area, all you need is a tape measure. “The basic formula is simple: Length x Width x Height = Volume,” says Michael Dean, co-founder of Pool Research. Then divide the number of cubic feet by 27. So one cubic yard = 27 cubic feet = 1,728 cubic inches.\n\n## How many cubic are in a yard?\n\nA yard is 3 feet or 36 inches, and therefore, a cubic yard is 3 x 3 x 3, or 27 cubic feet(ft3).\n\n## Is 1 yard the same as 1 cubic yard?\n\nUnder the United States’ Customary System, 1 yard is equal to 3 feet or 36 inches. And a cubic yard is the volume of material that fits in a space that is 1 yard wide by 1 yard deep by 1 yard high. This is important because quite a few common materials are measured in cubic yards — here are some of them: Concrete.\n\n## How do you convert square yards to cubic yards?\n\nConvert the square yardage to cubic yards by using the planned depth of the slab. Multiply the area of the slab by the intended slab depth. Your result represents the cubic yards of concrete needed to cover an area the width and length measured with a slab of the proposed thickness.\n\n## How many yards of concrete do I need for a 24×24 slab?\n\nFor example, for a concrete slab that is 24′ X 24′ X 4”, simply enter 4 in the Thickness/Depth field, 24 in the Width field, and 24 in the Length field. Click “Calculate”. Your answer should be 7.11 yards. Note: The Concrete Volume Calculator can also be used to determine yardage for aggregate products.\n\n## How do I figure out how many yards of mulch I need?\n\nKeep in mind that mulch is sold by the cubic yard. One cubic yard of the material covers a 324-square-foot area an inch deep. So, to determine your total, multiply your square footage by the depth in inches desired, then divide by 324. Here’s your formula: Square footage x desired depth / 324 = cubic yards needed.\n\n## How many square feet is 8 cubic yards?\n\nCubic yard to Square feet Calculator 1 cubic yard = 9 ft 2 0.037 cubic yard 8 cubic yard = 36 ft 2 0.8381 cubic yard 9 cubic yard = 38.9407 ft 2 1 cubic yard 10 cubic yard = 41.7743 ft 2 1.1712 cubic yard 11 cubic yard = 44.5148 ft 2 1.3512 cubic yard.\n\n## How many yards of stone do I need?\n\nMultiply the length (L), in feet, by the width (W), in feet, by the height (H), in feet, and divide by 27. This number is how many cubic yards of crushed stone you need.\n\n## How much does 1 cubic yard of gravel weigh?\n\nOne cubic yard of gravel can weigh between 2,400 to 2,900 lbs. Or up to one and a half tons approximately. Generally, a cubic yard of gravel provides enough material to cover a 100-square-foot area with 3 inches of gravel. Keep in mind that some materials compact more than others, so exact coverage may vary.\n\n## How do you calculate the volume of a gravel?\n\nMeasure the length and width of the area in feet and multiply the figures together. Multiply your area figure by the depth of gravel that you wish to use, ensuring your measurement is also in feet. If your depth is in inches, divide it by 12 first. You now have a cubic feet volume figure."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94168013,"math_prob":0.98574156,"size":4576,"snap":"2022-27-2022-33","text_gpt3_token_len":1156,"char_repetition_ratio":0.17235346,"word_repetition_ratio":0.0067264573,"special_character_ratio":0.26704547,"punctuation_ratio":0.114146344,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99719214,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T18:47:53Z\",\"WARC-Record-ID\":\"<urn:uuid:f1126c50-000c-4d88-b81e-1ddd548de893>\",\"Content-Length\":\"104581\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29833b6c-d040-41d8-9c46-559b02f0bfb5>\",\"WARC-Concurrent-To\":\"<urn:uuid:be2f65c5-940e-41ae-9519-8ca4ec19c9ea>\",\"WARC-IP-Address\":\"199.188.200.5\",\"WARC-Target-URI\":\"https://storiesofthemoment.com/question-how-to-get-cubic-yards/\",\"WARC-Payload-Digest\":\"sha1:XAZWKX3JPXCFXQYMKTBAZGWSF32AGZRF\",\"WARC-Block-Digest\":\"sha1:7L762LQDQO3MUUWUWF76JLCOAR7XNAPL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104248623.69_warc_CC-MAIN-20220703164826-20220703194826-00739.warc.gz\"}"} |
https://science2be.wordpress.com/category/frank-kamenetskiis-analysis-of-thermal-explosions/ | [
"Archive | Frank-Kamenetskii’s analysis of thermal explosions RSS feed for this section\n\n## Dust Explosions: Nature and Means to Prevent\n\n2 Dec\n\nThe great book of Nature lies ever open before our eyes and the true philosophy is written in it…But we cannot read it unless we have first learned the language and the characters in which it is written…It is written in mathematical language and the characters are triangles, circles and other geometric figures.-Galileo Galilei, Florence 1623.\n\nThey that know the entire course of the development of science, will, as a matter of course, judge more freely and more correctly of the significance of any present scientific movement than they, who, limited in their views to the age in which their own lives have been spent, contemplate merely the momentary trend that the course of intellectual events takes at the present moment.\n-Ernst Mach, German Charles University, Prague 1883\n\nIgnition is the process by which it occurs the propagation of a self-sustained combustion. An explosion (meaning a “sudden outburst”) is an exothermal process (i.e., liberate energy) that gives rise to a sudden increase of pressure when occuring at constant volume. It is accompanied by noise and a sudden release of a blast wave. An explosion may occurs in gases, dust or with solid explosives. In quite general terms, any kind of solid material that burn in air (or in oxygen rich environment) will burn more fastly with decreasing graininess. By order of importance we may say that firewood burn slowly, burns better when cut in smaller pieces, and burns fastly when cut in small particles.\n\nWe are concerned here with dust explosions, due to its importance not only industrially, but also in households and schools. The formation and ignition of dust clouds in industrial and agricultural environment is associated to extremely violent and damaging explosions, particularly in coal dust explosions in mines and dust explosions in grain elevators.\n\nThe dimensionless number that characterizes the possibility of a system to ignite is called the Darmkhöler number, D_a, which represents the ratio of the rate of heat production within the system due to exothermic chemical reactions to the rate of heat loss due conduction, convection or radiation.\n\nA convenient measure of dust fineness is the specific surface area given by",
null,
"if we assume the dust made of single cubes of edge length x . The maximum rate of preassure increase in closed-bomb experiments gives a measure of the expected violence form an explosion of a dust cloud. There is a linear dependency between the time rate of preassure increase and the specific surface area, as shown in the Fig. below (extracted from Ref.).",
null,
"Not all materials can cause dust explosions. For example, silicates, sulphates, nitrates, carbonates and phosphates, and in general terms stable oxides. But the contrary, materials highly explosives are the following: natural organic materials (e.g., grain, linen, sugar); synthetic organic materials (e.g., plastics, organic pigments, pesticides); coal and peat; metals (e.g., aluminum, magnesium, zinc, iron). An example of a highly explosive bomb is the thermobaric weapon that works consuming the surrounding oxygen instead of the oxidizer-fuel mixture, and for that reason it is more destructive than usual bombs. One important parameter that determines the amount of heat liberated in a explosion is called the heat of combustion (see Table 1).\n\nFrom the above Table 1, we may infer that Calcium (Ca) liberates more heat and coal contributes just one third of the former.",
null,
"What is the concentrations of dust that may represent danger? If we denote by x the typical radius of a dust particle of density rho_p, and by L^3 the volume of the dust cloud, then we have",
null,
"Experiments have determined the range of explosible dust concentrations in air at PTN conditions (that is, at normal temperature and atmospheric pressure) for natural organic dust, see Fig. below. The dust particles may fluctuate in a given container due to repulsive electrostatic forces exerted between them.\n\nThe last equation permits to estimate the dust concentration for which an explosion may occurs. Let us suppose that the particle concentration is ρ_p=1 g/cm^3 and we measure L/x=4 (by observation). Then it comes n_p=1.6 x 10^6, which is well above the dangerous threshold for explosion (see Fig.)",
null,
"Although the above method is quite empiric, we may seek for a general mathematical theory to describe ignition and combustion. The most interesting framework is the one proposed by the Soviet school of physico chemical processes.\n\nZeldovich and Frank-Kamenetsky found a general rule in the frame of chemical kinetics, valid for atoms and small molecules, which states that the temperature required for a chemical process is of the order of 10% of the total energy required. This is an outcome of the Arrhenius activation energy . The Arrhenius law is a law of the type (Eq.2)",
null,
"where A is called the prefactor, R is the Universal gas constant (R=8.3144621(75) J/mol K), and Ea is the activation energy. The ratio Ea/RT is called the activation energy. Due to the exponential dependency, an increase of the temperature by a factor of 2 can increase the reaction rate by a factor of 10-12 orders of magnitude. This is a striking example of exponential phenomena in the natural world and from this mathematical function results the general concern about global warming.\n\nNow, let us see understand what phenomena we intend to describe quantitatively. Let us begin by observe the propagation of a flame front.\n\nLearn how to do your own experiment with this video:\n\nWhat we better have to describe energetic processes in nature is the time-dependent governing equation for a given i-species Eq.3(e.g., Fuel):",
null,
"Here, w_i is the specific mass of the i-species, w_i=m_i/m, with m the total mass of our system (e.g., fuel + oxydizer), D_i^M is the diffusion coeffient of species i into the mixture of other species (they can be read in Tables or calculated through analytical formulas).\n\nThe energy equation is also necessary, which may be written under the general form, Eq.4:",
null,
"In this Eq.",
null,
"is the mass density (",
null,
") of the mixture (e.g., fuel+oxydizer), v is the mean mass velocity of the center of mass of the mixture,",
null,
"is the specific heat capacity of the specie i, λ is the heat conductivity of the mixture. The above equation is based on the “Analytical theory of Heat”, proposed by the French scientist Joseph Fourier (see also footnote 1).\n\nThese two equations can be re-written under the general form of a partial differential equation:",
null,
"The time derivative dY/dt represents the temporal changes of the variable Y at a given position z, the term in A represents the molecular transport associated to diffusion and heat conduction, the term with B represents the flow and the last term C contains the effects associated to chemical reactions occurring locally (at a given z).\n\nIf we consider the simplest case, where no chemical reaction and no molecular transport is present, that is, putting A=C=0, then from the above it follows the equation",
null,
"This equation represents a travelling wave propagating with velocity v. A simple analytical solution exists given by the expression:",
null,
"The shape of a travelling wave does not alter with time and we represent graphically in the fig. this process.",
null,
"Bill Gates is committed to support investigation in a new kind of travelling wave nuclear reactor, read here how it works and see the movie below.\n\nAnd now, what about ignition processes? Once again, we may stress the real complexity of the problem, but thankfully another mogul of chemical physics come to rescue us, proposing a simple (and effective) model, the Frank-Kamenetski’s model of thermal explosions. For example, in spherical geometry the energy conservation equation can be written under the form:",
null,
"Exercice: Re-write the above equation under the form (see p. 144 Ref.):",
null,
"Here, T_w is the wall temperature of the container where is supposed to be the explosive mixture. And again, it can be shown mathematically that the above equations has stationary solutions for any δ < δ_crit, with δ_crit=3.32 in spherical geometry. This means that for a given system (fuel+oxidizer) and a given wall temperature and container with size r_0, explosion will not occur provided we maintain δ < δ_crit.",
null,
"Yakov Zel’dovich, Andrei Sakharov and David Frank-Kamenetski in the town of Sarov, mid 50ths. Im credit: http://www.sakharov-center.ru”%5D\n\nThis short compilation may help the reader to get a glimpse of the complexity of the processes of combustion, ignition and means to control these phenomena. More deep information can be obtained in our proposed references.\nREF:\n\n Dust Explosions in the Process Industries, Rolph K. Eckhoff ()Gulf Professional Publishing, Amsterdam, 2003)\n\n Combustion: Physical and Chemical Fundamentals, Modeling and Simulation, Experiments, Pollutant Formation (Springer, )Berlin, 2006)\n\n Analytical Theory of Heat, by Joseph Fourier (Cambridge University Press, London, 1878)\n\nFOOTNOTES:\n\n(1)-Joseph Fourier was also a precursor on Global Warming Studies, see this site here."
] | [
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"https://science2be.files.wordpress.com/2011/11/dust_eq11.gif",
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"https://science2be.files.wordpress.com/2011/12/img0501.jpg",
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"https://science2be.files.wordpress.com/2011/12/table-11.gif",
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"https://science2be.files.wordpress.com/2011/12/dust-eq14.gif",
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"https://science2be.files.wordpress.com/2011/12/img0491.jpg",
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"https://science2be.files.wordpress.com/2011/11/dust_eqa.gif",
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"https://science2be.files.wordpress.com/2011/11/dust_eq6.gif",
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"https://science2be.files.wordpress.com/2011/11/dust_eq21.gif",
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"https://science2be.files.wordpress.com/2011/11/dust_l1.gif",
null,
"https://science2be.files.wordpress.com/2011/11/dust_l2.gif",
null,
"https://science2be.files.wordpress.com/2011/11/dust_l3.gif",
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"https://science2be.files.wordpress.com/2011/12/eq3.gif",
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"https://science2be.files.wordpress.com/2011/12/eq4.gif",
null,
"https://science2be.files.wordpress.com/2011/12/eq5.gif",
null,
"https://science2be.files.wordpress.com/2011/12/img048.jpg",
null,
"https://science2be.files.wordpress.com/2011/11/dust-eq12.gif",
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"https://science2be.files.wordpress.com/2011/11/dust-eq13.gif",
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"https://science2be.files.wordpress.com/2011/12/zeldovich-kamenetski.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9048567,"math_prob":0.95251644,"size":9121,"snap":"2019-13-2019-22","text_gpt3_token_len":1961,"char_repetition_ratio":0.11275639,"word_repetition_ratio":0.0,"special_character_ratio":0.20721412,"punctuation_ratio":0.117101446,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9661556,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T09:49:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c6d02c53-f754-400c-931e-ee9fe9395bbe>\",\"Content-Length\":\"92080\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df7727d8-d7f6-4fff-9374-bad696c40170>\",\"WARC-Concurrent-To\":\"<urn:uuid:75310ea6-8d47-4171-9b6b-f6bca38fc5f2>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://science2be.wordpress.com/category/frank-kamenetskiis-analysis-of-thermal-explosions/\",\"WARC-Payload-Digest\":\"sha1:PLKEUX77T4GTWJQNYPNHJBLDUBYDQXZ6\",\"WARC-Block-Digest\":\"sha1:V7AF6EXSVTPJ66ZK4TSHNR5YYHSYV2F2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259015.92_warc_CC-MAIN-20190526085156-20190526111156-00353.warc.gz\"}"} |
https://www.esteemprojects.com/web/static/Calculators/calculators.htm | [
"Engineering Calculators Heat Balance Calculator Steam Properties Calculator Gas or Liquid Fuel Combustion based on Excess Air or O2 in Stack Steam properties by inputting any two known properties Fired Furnace Efficiency with Air Preheat using External Heat Source Fluegas Dew Point Calculator Fired Furnace with Integrated Air Preheat System Analysis Dew point or acid dew point of a fluegas is calculated Existing Fired Furnace with added Air Preheat System Analysis Fluegas Thermal and Transport Properties Calculator Duct Piece Pressure Drop Calculator Fluegas properties are calculated for a fluegas when given the temperature or enthalpy with a known composition Duct piece properties and pressure drop are caculated for different duct pieces Metals Thermal Conductivity Calculator Intube Pressure Drop Calculator Thermal conductivity is calculated for different metals at a given temperature Tubeside pressure drop, film coefficient, and other properties are calulated Convection Heat Transfer Calculator Insulation & Heat Loss Calculator Heat transfered and overall heat transfer coefficient calculated for different tube extended surfaces Insulation and heat loss calculator for selected in materials API RP530 Tube Wall Thickness Calculator Supplementary Firing Calculator API RP530 tube wall thickness calculations for various materials Supplementary firing calculator for heat recovery applications Return Bend Equivalent Lengths Calculator Steam Drum Volume Calculator Compares alternate methods for equivalent length of turns to that used in WinHeat Calculates volumes and times from various levels in a steam drum Gas Side Pressure Loss Across Fin Tubes Calculator Gas Side Pressure Loss Across Bare Tubes Calculator Calculates the pressure loss across fin tubes using the Escoa Method Calculates gas side pressure loss across bare tubes"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.62957567,"math_prob":0.9802241,"size":571,"snap":"2019-43-2019-47","text_gpt3_token_len":101,"char_repetition_ratio":0.30511463,"word_repetition_ratio":0.05882353,"special_character_ratio":0.15761821,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9540618,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T02:31:10Z\",\"WARC-Record-ID\":\"<urn:uuid:8434c1a8-b514-4db1-b99d-02be1c6866e0>\",\"Content-Length\":\"4652\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1bdb72b3-047e-479b-bb2b-091a7cead7d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f227be2c-7d0e-4e88-8d99-a0a3f8b95063>\",\"WARC-IP-Address\":\"104.27.187.180\",\"WARC-Target-URI\":\"https://www.esteemprojects.com/web/static/Calculators/calculators.htm\",\"WARC-Payload-Digest\":\"sha1:WFBTL2BYNX4ZL2YIWFRKE2SWJZDCDXUB\",\"WARC-Block-Digest\":\"sha1:2XPXKX6RQZ6KEVTMYN46QWKVXECUJE7O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668782.15_warc_CC-MAIN-20191117014405-20191117042405-00420.warc.gz\"}"} |
https://www.q-chem.com/explore/spectroscopy/vibronic/ | [
"# Simulating Vibrationally Resolved Electronic and Raman Spectra\n\nA built-in module for computing vibrational progressions affords calculations of vibrationally resolved one-photon absorption (OPA), one-photon emission (OPE), and resonance Raman spectra (RRS):\n\n• Effective algorithm reduces computational costs by recasting sum-over-state expressions into the time-domain, thus avoiding explicit calculation of individual excited states.\n\n• Different levels of theoretical treatment are implemented:\n\n• Franck-Condon (FC): includes only the zero-order term of the transition dipole moment.\n\n• Franck-Condon-Herzberg-Teller (FCHT): includes the zero- and first-order terms of the transition dipole moment.\n\n• Vertical gradient (VG): the excited state potential energy surface is approximated by a shift to the ground-state surface.\n\n• Duschinsky rotation: includes mode mixing effect.",
null,
"Calculated FC (blue) and FCHT (red) absorption spectra of D$$_0\\rightarrow$$D$$_3$$ transition for phenoxyl radical",
null,
"Calculated FC (blue) and FCHT (red) emission spectra of D$$_1\\rightarrow$$D$$_0$$ transition for benzyl radical",
null,
""
] | [
null,
"https://www.q-chem.com/sites/default/files/explore/formatted_vibronic-1.png",
null,
"https://www.q-chem.com/sites/default/files/explore/formatted_vibronic-2.png",
null,
"https://www.q-chem.com/sites/default/files/explore/formatted_vibronic-3.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83725834,"math_prob":0.96691483,"size":1105,"snap":"2022-40-2023-06","text_gpt3_token_len":248,"char_repetition_ratio":0.10354223,"word_repetition_ratio":0.05839416,"special_character_ratio":0.2,"punctuation_ratio":0.082840234,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.969512,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-30T02:13:49Z\",\"WARC-Record-ID\":\"<urn:uuid:041769e6-3c28-4c0b-896e-694469897be9>\",\"Content-Length\":\"39926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bfaea90c-e89b-4f4e-af2d-3e4e772506d6>\",\"WARC-Concurrent-To\":\"<urn:uuid:397deb1d-576e-4d01-b719-f6cc1b4d9ee0>\",\"WARC-IP-Address\":\"143.110.152.114\",\"WARC-Target-URI\":\"https://www.q-chem.com/explore/spectroscopy/vibronic/\",\"WARC-Payload-Digest\":\"sha1:ROEIHC5ZO65I7TMZROLWXOQHQ3QJI2K2\",\"WARC-Block-Digest\":\"sha1:V5TLGUBM6LBL3IG63KN5W4GKKEZSHDSG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335424.32_warc_CC-MAIN-20220930020521-20220930050521-00478.warc.gz\"}"} |
https://math.stackexchange.com/questions/4092505/why-is-the-reverse-of-a-prime-about-45-more-likely-to-be-a-prime-than-that-o | [
"# Why is the reverse of a prime about $45\\%$ more likely to be a prime than that of a composite?\n\nConsider two cases a) we reverse the digits of a prime number b) we reverse the digits of a composite number. Are we more likely to obtain a prime in case a) or in case b). Since the last digit of primes other than $$2$$and $$5$$ end in $$1,3,7$$ or $$9$$ hence if a prime or composite number begins in $$2,4,5, 6$$ or $$8$$ there is no way its reverse will be a prime. So to make a fair comparison, I only considered those prime and composite numbers whose first and last digits is $$1,3,7$$ or $$9$$.\n\nLet $$C$$ and $$P$$ be the set of such composites and prime numbers respectively. I looked at the first $$10^8$$ numbers (in ascending order) in $$C$$ and observed the density of numbers whose reverse is a prime is roughly $$\\frac{2.4n}{\\log n}$$. However in case of the set $$P$$, the density of roughly $$\\frac{3.5n}{\\log n}$$ i.e. about $$45\\%$$ higher which is significant.\n\nQuestion: Given the set of numbers whose first and last digits is $$1,3,7$$ or $$9$$, why is the reverse of a prime about $$45\\%$$ more likely to be a prime than that of a composite?\n\n• If a composite is divisible by $3$, then also its reverse. This could be one reason. But I wonder why this does not cause an even stronger effect. – Peter Apr 7 at 8:10\n• @Peter The effect could be slightly bigger. My computing is still running and at the moment, the coefficient is oscillating roughly between $3.45 - 3.55$. But I don't think it would be significantly bigger. – Nilotpal Sinha Apr 7 at 8:18\n• Reverses of multiples of 11 are still multiples of 11. – MJD Apr 7 at 8:39\n• I wonder how the result changes when one goes to base 2, etc. That may give some insight on importance of the base $\\pm1$, I don't know... – Tesla Daybreak Apr 7 at 8:46\n• Reverses of multiples of 10001 are all multiples of 10001. I think in other bases, numbers like this might be more common than in base 10. – MJD Apr 7 at 9:18\n\nJust two of those facts will give you a $$\\frac13+\\frac1{11}=43\\%$$ better chance. You can further increase this number by starting adding up rarer cases. For example, if the number of digits in $$p$$ is a multiple of 3, then if $$p = \\overline{a_1a_2a_3a_4a_5a_6\\ldots}$$ is divisible by 7, then $$(a_1+2a_2+a_3)+(a_4+2a_5+a_6)+...$$ is also divisible by 7.\n• plus if you keep it odd, your mixture of $\\{1,4,7\\}$ and $\\{2,5,8\\}$ the same creates another number with the same remwinder on division by 6 ... – Roddy MacPhee Apr 7 at 10:31\n• You need $\\frac13+\\frac1{11} -\\frac1{33}$ or you have double-counted the multiples of $33$. – MJD Apr 7 at 15:24"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89152443,"math_prob":0.99977106,"size":1025,"snap":"2021-21-2021-25","text_gpt3_token_len":290,"char_repetition_ratio":0.15475024,"word_repetition_ratio":0.074074075,"special_character_ratio":0.30048782,"punctuation_ratio":0.10638298,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997949,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T11:03:10Z\",\"WARC-Record-ID\":\"<urn:uuid:05ef1aab-a6b1-43c1-bfe0-71b6c1a867f5>\",\"Content-Length\":\"175431\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6bd42c66-aed8-40ff-86d0-146371aaa70e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3fa6e3c-c85e-4d54-9fcc-b737a56281c7>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/4092505/why-is-the-reverse-of-a-prime-about-45-more-likely-to-be-a-prime-than-that-o\",\"WARC-Payload-Digest\":\"sha1:H36WADCP5IRTPBFEM4UF3ISAZVFDQ3ZG\",\"WARC-Block-Digest\":\"sha1:3VMFRHRCA3D5UVBWO3B6X4A2GN6AZQFT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988966.82_warc_CC-MAIN-20210509092814-20210509122814-00516.warc.gz\"}"} |
https://cm-to-inches.appspot.com/39.5-cm-to-inches.html | [
"Cm To Inches\n\n# 39.5 cm to in39.5 Centimeters to Inches\n\ncm\n=\nin\n\n## How to convert 39.5 centimeters to inches?\n\n 39.5 cm * 0.3937007874 in = 15.5511811024 in 1 cm\nA common question is How many centimeter in 39.5 inch? And the answer is 100.33 cm in 39.5 in. Likewise the question how many inch in 39.5 centimeter has the answer of 15.5511811024 in in 39.5 cm.\n\n## How much are 39.5 centimeters in inches?\n\n39.5 centimeters equal 15.5511811024 inches (39.5cm = 15.5511811024in). Converting 39.5 cm to in is easy. Simply use our calculator above, or apply the formula to change the length 39.5 cm to in.\n\n## Convert 39.5 cm to common lengths\n\nUnitLengths\nNanometer395000000.0 nm\nMicrometer395000.0 µm\nMillimeter395.0 mm\nCentimeter39.5 cm\nInch15.5511811024 in\nFoot1.2959317585 ft\nYard0.4319772528 yd\nMeter0.395 m\nKilometer0.000395 km\nMile0.0002454416 mi\nNautical mile0.0002132829 nmi\n\n## What is 39.5 centimeters in in?\n\nTo convert 39.5 cm to in multiply the length in centimeters by 0.3937007874. The 39.5 cm in in formula is [in] = 39.5 * 0.3937007874. Thus, for 39.5 centimeters in inch we get 15.5511811024 in.\n\n## 39.5 Centimeter Conversion Table",
null,
"## Alternative spelling\n\n39.5 Centimeter to Inches, 39.5 Centimeter in Inches, 39.5 cm to Inches, 39.5 cm in Inches, 39.5 cm to Inch, 39.5 cm in Inch, 39.5 cm to in, 39.5 cm in in, 39.5 Centimeters to in, 39.5 Centimeters in in, 39.5 Centimeter to in, 39.5 Centimeter in in, 39.5 Centimeters to Inch, 39.5 Centimeters in Inch"
] | [
null,
"https://cm-to-inches.appspot.com/image/39.5.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87358016,"math_prob":0.9714721,"size":1078,"snap":"2022-27-2022-33","text_gpt3_token_len":384,"char_repetition_ratio":0.26256984,"word_repetition_ratio":0.021276595,"special_character_ratio":0.4257885,"punctuation_ratio":0.2260274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98099196,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T19:45:48Z\",\"WARC-Record-ID\":\"<urn:uuid:7f8fc4d2-5b5e-45be-8862-b5ea073eb102>\",\"Content-Length\":\"29040\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee92909d-b6bd-47b2-a11b-0cf5b9409649>\",\"WARC-Concurrent-To\":\"<urn:uuid:812984e6-804d-44e6-b223-42001daab177>\",\"WARC-IP-Address\":\"172.253.63.153\",\"WARC-Target-URI\":\"https://cm-to-inches.appspot.com/39.5-cm-to-inches.html\",\"WARC-Payload-Digest\":\"sha1:UFH57IXR5QAUVW5MR7NN5T6BRSXSH7NT\",\"WARC-Block-Digest\":\"sha1:XV35YIYP3QZZ3WID43E7QBVPGMVZVMCU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104676086.90_warc_CC-MAIN-20220706182237-20220706212237-00354.warc.gz\"}"} |
https://bisphuketmaths.com/units-7-9/ | [
"# Units 7-9\n\nLogs and Exponents:",
null,
"An introduction to logs:\n\nMore on log laws:\n\nProof of log rules:\n\nSolving log equations using laws of logs:\n\nChange of Base laws:\n\nEquations with logs in different bases:\n\nSolving quadratic indices questions with logs:\n\nThe graph of ln(x) and e^x:\n\nSolving equations using the natural log (lnx):\n\nPast paper questions on logs:\n\nStraight Line graphs:",
null,
"Converting non-linear to linear graphs:\n\nlog graphs past paper questions:\n\nCircular measures:",
null,
""
] | [
null,
"https://bispmaths.files.wordpress.com/2016/05/screen-shot-2016-05-01-at-7-46-44-pm.png",
null,
"https://bispmaths.files.wordpress.com/2016/05/screen-shot-2016-05-01-at-7-46-52-pm.png",
null,
"https://bispmaths.files.wordpress.com/2016/05/screen-shot-2016-05-01-at-7-47-01-pm.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80509764,"math_prob":0.8752265,"size":600,"snap":"2019-51-2020-05","text_gpt3_token_len":126,"char_repetition_ratio":0.15100671,"word_repetition_ratio":0.0,"special_character_ratio":0.19333333,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994018,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,10,null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T13:17:24Z\",\"WARC-Record-ID\":\"<urn:uuid:60b915f0-2d5f-409f-8c47-aed74b205e22>\",\"Content-Length\":\"69210\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:91404518-3ea1-453c-a4f2-0008f43bcf73>\",\"WARC-Concurrent-To\":\"<urn:uuid:72094945-37de-49ad-a99b-5bf558481c0c>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://bisphuketmaths.com/units-7-9/\",\"WARC-Payload-Digest\":\"sha1:ATEUUOJKVVD5CVA3PEVWKLN6X4MLHB5Q\",\"WARC-Block-Digest\":\"sha1:ZA762C3ESDCKA6ZSMMFVSHCXIWLJM3KT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540488620.24_warc_CC-MAIN-20191206122529-20191206150529-00020.warc.gz\"}"} |
http://underactuated.mit.edu/playbook.html | [
"# Underactuated Robotics\n\nAlgorithms for Walking, Running, Swimming, Flying, and Manipulation\n\nRuss Tedrake\n\nPQ trick from LQR Hongkai's smooth non-penetration constraints for polygons \"Soft\" absolute value: $|x|_epsilon \\approx \\sqrt{x^2 + \\epsilon}$. import numpy as np import matplotlib.pyplot as plt x = np.linspace(-2.,2.,101) plt.plot(x, np.sqrt(np.power(x,2) + 0.01)) Barrier function for friction cones in Grandia 2019, Feedback MPC for Torque-Controlled Legged Robots. Linearization of the rotation matrix in Real-time Model Predictive Control for Versatile Dynamic Motions in Quadrupedal Robots Briat15 chapter 1 has some good modeling tricks for LPV systems"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86854905,"math_prob":0.9689336,"size":532,"snap":"2020-24-2020-29","text_gpt3_token_len":113,"char_repetition_ratio":0.08712121,"word_repetition_ratio":0.02631579,"special_character_ratio":0.20676692,"punctuation_ratio":0.15789473,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9917758,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-13T13:23:10Z\",\"WARC-Record-ID\":\"<urn:uuid:df8caf56-dcc8-4be6-a851-a55f181b6bc1>\",\"Content-Length\":\"4597\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d16b043c-38a0-4750-aaef-1c373d739f2d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f7c4aa0f-b803-4802-9090-82baa5f43974>\",\"WARC-IP-Address\":\"128.52.129.147\",\"WARC-Target-URI\":\"http://underactuated.mit.edu/playbook.html\",\"WARC-Payload-Digest\":\"sha1:QXO3AMBGYPQJRZUSVVG4B4BG47PJDVTA\",\"WARC-Block-Digest\":\"sha1:DMBWSSARYEYXM7DYR6KS4RBV2BZCHU5K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657145436.64_warc_CC-MAIN-20200713131310-20200713161310-00435.warc.gz\"}"} |
https://www.r-bloggers.com/2012/08/add-text-annotations-to-ggplot2-faceted-plot/ | [
"Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nIn my experience with R learners there are two basic types. The “show me the code and what it does and let me play” type and the “please give me step by step directions” type. I’ve broken the following tutorial on plotting text on faceted ggplot2 plots into 2 sections:\n\n1. The Complete Code and Final Outcome\n2. A Bit of Explanation\n\nHopefully, whatever learner you are you’ll be plotting text on faceted graphics in no time.\n\n## Section 1: The Complete Code and Final Outcome\n\n```mtcars[, c(\"cyl\", \"vs\", \"gear\")] <- lapply(mtcars[, c(\"cyl\", \"vs\", \"gear\")], as.factor)\n\np <- ggplot(mtcars, aes(mpg, wt, group = cyl)) +\ngeom_line(aes(color=cyl)) +\ngeom_point(aes(shape=cyl)) +\nfacet_grid(gear ~ am) +\ntheme_bw()\np\n\nlen <- length(levels(mtcars\\$gear)) * length(levels(mtcars\\$am))\n\nvars <- data.frame(expand.grid(levels(mtcars\\$gear), levels(mtcars\\$am)))\ncolnames(vars) <- c(\"gear\", \"am\")\ndat <- data.frame(x = rep(15, len), y = rep(5, len), vars, labs=LETTERS[1:len])\n\np + geom_text(aes(x, y, label=labs, group=NULL),data=dat)\n\ndat[1, 1:2] <- c(30, 2) #to change specific locations\np + geom_text(aes(x, y, label=labs, group=NULL), data=dat)\n\np + geom_text(aes(x, y, label=paste(\"beta ==\", labs), group=NULL), size = 4,\ncolor = \"grey50\", data=dat, parse = T) ```",
null,
"## Section 2: A Bit of Explanation\n\nThe following portion of the tutorial provides a bit more of a step by step procedure for plotting text to faceted plots as well as a visual to go with the code.\n\nThe initial non annotated plot\nFirst Let’s make a faceted line plot with the mtcars data set. I reclassed a few variables to make factors.\n\n```mtcars[, c(\"cyl\", \"vs\", \"gear\")] <- lapply(mtcars[, c(\"cyl\", \"vs\", \"gear\")], as.factor)\n\np <- ggplot(mtcars, aes(mpg, wt, group = cyl)) +\ngeom_line(aes(color=cyl)) +\ngeom_point(aes(shape=cyl)) +\nfacet_grid(gear ~ am) +\ntheme_bw()\np\n```",
null,
"The key here is a new data frame with three pieces of information (ggplot2 seems to like information given in a data frame).\n\n1. Coordinates to plot the text\n2. The faceted variable levels\n3. The labels to be supplied\n\nThe first information piece is the coordinates (two columns x and y) to plot the text in each facet. Generally I find that one set of coordinates will work in most of the facet boxes and I just use rep to make these coordinates (I suppose the recycling rule could be used if you added it to an already existing data frame).\n\nThe second information piece is the faceted variable labels (in our case gear ~ am). There’s many ways to achieve this but I like a combination of levels and expand.grid. I renamed these columns to be exactly the same as the variable names (gear & am) I used in the original data frame (mtcars in this case).\n\nLastly, you must make a labels. I chose letters so you can track what piece of the data frame is plotted in which facet.\n\nYour data should look something like this:\n\n``` x y gear am labs\n1 30 2 3 0 A\n2 15 5 4 0 B\n3 15 5 5 0 C\n4 15 5 3 1 D\n5 15 5 4 1 E\n6 15 5 5 1 F```\n\nNote that the , group=NULL is essential to let ggplot2 know you’re dealing with a new data set and the mapping from before can be forgotten (or at least this is how I understand it).\n\n```#long cut way to find number of facets\nlen <- length(levels(mtcars\\$gear)) * length(levels(mtcars\\$am))\n\nvars <- data.frame(expand.grid(levels(mtcars\\$gear), levels(mtcars\\$am)))\ncolnames(vars) <- c(\"gear\", \"am\")\ndat <- data.frame(x = rep(15, len), y = rep(5, len), vars, labs=LETTERS[1:len])\n\np + geom_text(aes(x, y, label=labs, group=NULL),data=dat) ```",
null,
"Moving just one text location\nGenerally I can usually find one spot that most every text plot will work except that one dog gone facet that just won’t match up with the other coordinates. In this case label A is that pesky label. The key here is to figure out what text labels you want to move and alter those coordinates appropriately.\n\n```dat[1, 1:2] <- c(30, 2) #to change specific locations\np + geom_text(aes(x, y, label=labs, group=NULL), data=dat)\n```",
null,
"Adding equation (Greek letters/math) and alter size/color\nTo annotate with math code use the parse = T argument in geom_text. For more on plotting math code see this ggplot wiki and this SO question. To alter the size just throw a size argument in geom_text. I also toned down the color of the text a bit to allow the line to pop the most visually.\n\n```p + geom_text(aes(x, y, label=paste(\"beta ==\", labs), group=NULL), size = 4,\ncolor = \"grey50\", data=dat, parse = T) ```",
null,
"",
null,
"",
null,
""
] | [
null,
"https://dl.dropbox.com/u/61803503/final.png",
null,
"https://dl.dropbox.com/u/61803503/initial.png",
null,
"https://dl.dropbox.com/u/61803503/sec.png",
null,
"https://dl.dropbox.com/u/61803503/third.png",
null,
"https://dl.dropbox.com/u/61803503/final.png",
null,
"https://feeds.wordpress.com/1.0/comments/trinkerrstuff.wordpress.com/307/",
null,
"https://i0.wp.com/stats.wordpress.com/b.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7674064,"math_prob":0.9764906,"size":4875,"snap":"2022-40-2023-06","text_gpt3_token_len":1334,"char_repetition_ratio":0.10572778,"word_repetition_ratio":0.23030303,"special_character_ratio":0.27979487,"punctuation_ratio":0.12074002,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9967131,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,6,null,6,null,6,null,null,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T09:23:29Z\",\"WARC-Record-ID\":\"<urn:uuid:ac0bcf3d-fb79-4761-a63d-72916c8fa90f>\",\"Content-Length\":\"99403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a05db779-0695-422a-8e99-ee737c971525>\",\"WARC-Concurrent-To\":\"<urn:uuid:330ac6ac-8ccb-4346-9b83-dad0f0d7b1c8>\",\"WARC-IP-Address\":\"172.67.211.236\",\"WARC-Target-URI\":\"https://www.r-bloggers.com/2012/08/add-text-annotations-to-ggplot2-faceted-plot/\",\"WARC-Payload-Digest\":\"sha1:Z4QAIFI25BACHPPNUMZTH4YG43PZ52HC\",\"WARC-Block-Digest\":\"sha1:XMZCFECZZPFMSJBXR7GIHN2HYRL6WQ35\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.66_warc_CC-MAIN-20230203091020-20230203121020-00758.warc.gz\"}"} |
https://www.fpga4fun.com/DDS4.html | [
"",
null,
"",
null,
"## Direct Digital Synthesis 4 - Interpolation\n\nNow, while the phase accumulator is quite precise, the output suffers from the limited number of entries in the lookup table: the output value \"jumps\" when going from one entry to the next. This is particularly sensible for low output frequencies, but affects high output frequencies too, and this introduces unwanted frequencies in the output spectrum.\n\nWe're going to fix that. To make it easier to understand, let's go back to a 15bit phase accumulator.\n\n```// sine without linear interpolation\nreg [14:0] phase_acc; // 15bit\nalways @(posedge clk) phase_acc <= phase_acc + 15'h1;\n\n```\n\nThe above code moves from one lookup table to the next every 16 clocks. That makes the output \"jump\" every 16 clocks.\n\nAn effective way to improve that is to use the lowest 4 bits of the phase accumulator (unused until now) to linearly interpolate between two successive lookup table entries. That's pretty easy to do (using two lookup tables instead of one).\n\n```// sine with linear interpolation\nreg [14:0] phase_acc;\nalways @(posedge clk) phase_acc <= phase_acc + 15'h1;\n\n// use two lookup tables to get two successive table values\nwire [16:0] sine1_lv, sine2_lv;\n\n// now the 4 LSB bits from the phase accumulator need to be delayed\n// (to match the latency introduced by the lookup tables)\nreg [3:0] phase_LSB_delay1; always @(posedge clk) phase_LSB_delay1 <= phase_LSB[3:0];\nreg [3:0] phase_LSB_delay2; always @(posedge clk) phase_LSB_delay2 <= phase_LSB_delay1;\nreg [3:0] phase_LSB_delay3; always @(posedge clk) phase_LSB_delay3 <= phase_LSB_delay2;\n\n// before we can use them to do the interpolation\nwire [4:0] sine1_mf = 5'h10 - phase_LSB_delay3;\nwire [3:0] sine2_mf = phase_LSB_delay3;\nreg [20:0] sine_p; always @(posedge clk) sine_p <= sine1_lv*sine1_mf + sine2_lv*sine2_mf;\n\nassign DAC_data_out = sine_p[20:11];\n```\n\nBoth lookup tables contain the same values. We extract a value from one, and its neighbor value from the other (the \"phase_acc+1\"), so that we can linearly interpolate between the two values.\n\nThe interpolation allows us to get a better resolution out of the DDS, while keeping the lookup table sizes reasonable. Our lookup tables create the sine function with 2048 values. And between each value, we interpolate 15 points, so we end-up with 2048*16=32768 sine points, much like having bigger lookup tables.\n\n#### Ideas of improvement\n\n• Use a 32bit phase accumulator to cover a wide range of frequencies (or check Saxo-Q's USB-controlled 32bit DDS example).\n• Reduce the lookup table requirement (use only one, or using an iterative method like CORDIC).\n• Use a sin(x)/x filter instead of linear interpolation.\n• Use dithering to increase the DAC resolution."
] | [
null,
"https://www.fpga4fun.com/images/header1.gif",
null,
"https://www.fpga4fun.com/images/header2.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77047557,"math_prob":0.9708693,"size":2539,"snap":"2021-31-2021-39","text_gpt3_token_len":661,"char_repetition_ratio":0.16607495,"word_repetition_ratio":0.027247956,"special_character_ratio":0.28672707,"punctuation_ratio":0.16017316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.989666,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T10:40:10Z\",\"WARC-Record-ID\":\"<urn:uuid:31d074a6-65c2-46aa-ae4e-7f7307b14090>\",\"Content-Length\":\"7480\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08c6d535-6f6a-4c22-9328-a8f7650cc2df>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3f8887d-a471-4111-9135-dc05ffabc1eb>\",\"WARC-IP-Address\":\"52.34.170.75\",\"WARC-Target-URI\":\"https://www.fpga4fun.com/DDS4.html\",\"WARC-Payload-Digest\":\"sha1:UKHOVRVVPI3NRV25W7YRVVQVIJMDLXF2\",\"WARC-Block-Digest\":\"sha1:NNP53KRYGOTIGCAFCVZJ6CFMFESHNRLR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055632.65_warc_CC-MAIN-20210917090202-20210917120202-00322.warc.gz\"}"} |
https://www.ilearnlot.com/arbitrage-pricing-theory-apt-meaning-and-definition/57906/ | [
"Meaning of Arbitrage Pricing Theory (APT) is one of the tools used by investors and portfolio managers who explain the return of severity based on their respective beta. This theory was developed by Stephen Ross. In finance, the APT is a general theory of property pricing that believes that the expected return of financial assets can model as a linear function of various factors or theoretical market index, wherein each of the factors The sensitivity of change is represented by a factor-specific beta coefficient. So, what is the question; What is Arbitrage Pricing Theory (APT)? Meaning and Definition.\n\n## Here are explains What is Arbitrage Pricing Theory (APT)? with Meaning and Definition.\n\nIntroduction to Contents:\n\nAPT is a multi-factor property pricing model based on this idea that calculating asset returns can be done using linear relationships between the expected return of assets and many macroeconomic variables that hold systematic risk. This theory was created in 1976 by economist Stephen Ross. The arbitrage pricing principle provides a multi-factor pricing model for securities based on the relation between the expected return of financial assets and its risk to analysts and investors. This is a useful tool to analyze the portfolio from a price investment perspective, to temporarily identify the securities incorrectly.\n\nAPT is a more flexible and complicated option for the Capital Asset Pricing Model (CAPM). The theory provides investors and analysts with the opportunity to customize their research. As well as, the model-derived rate of return will use to correctly assess the property; Also, the asset value should be equal to the expected end of the discount period at the rate mentioned by the model. If the price goes away, then arbitrage will bring it back to the line.\n\nHow To Make Your Small Business Stand Out? Many Ways You Can Try IT!",
null,
"Women Get Shopping; What is Arbitrage Pricing Theory (APT)? Meaning and Definition. Image credit from ilearnlot.com.\n\n### What does mean the Arbitrage Pricing Theory (APT)?\n\nArbitrage Pricing Theory (APT) is an extension of CAPM. The pricing model given by APT is the same as CAPM. It is a model under equilibrium. The difference is that APT is based on a multi-factor model. Also, the Arbitrage pricing theory holds that arbitrage behavior is a decisive factor in the formation of modern efficient markets (that is, market equilibrium prices).\n\nIf the market does not reach equilibrium, there will be risk-free arbitrage opportunities in the market. And multiple factors used to explain the return of risk assets, and according to the principle of no-arbitrage; there is an (approximate) linear relationship between the balanced return of risk assets and multiple factors. The previous CAPM model predicts that there is a linear relationship between the returns of all securities; and, the return of a unique public factor (market portfolio).\n\n### Significance of Arbitrage Pricing Theory (APT):\n\nArbitrage pricing theory derives a market relationship similar to the capital asset pricing model. The arbitrage pricing theory base on the multi-factor model of the rate of the return formation process. It is believed that the rate of return of securities is linearly related to a set of factors; which represent some basic factors of the rate of return of securities.\n\nWhen the rate of return form by a single factor (market combination), you will find that the arbitrage pricing theory forms a relationship with the capital asset pricing model. Therefore, the arbitrage pricing theory can be considered as a generalized capital asset pricing model, providing investors with an alternative method to understand the equilibrium relationship between risk and return in the market. Arbitrage pricing theory and modern asset portfolio theory, capital asset pricing model, option pricing model, etc. constitute the theoretical basis of modern finance.\n\nHop-Frog\n\n### Difference between APT and CAPM:\n\nIn 1976, American scholar Stephen Rose published the classic paper “Arbitrage Theory of Capital Asset Pricing” in the Journal of Economic Theory; and, proposed a new asset pricing model, which is the arbitrage pricing theory (APT theory). Arbitrage pricing theory uses the concept of arbitrage to define equilibrium, does not require the existence of market portfolios, and requires fewer assumptions than the capital asset pricing model (CAPM model) and more reasonable.\n\nLike the capital asset pricing model, the arbitrage pricing theory assumes:\n\n• Investors have the same investment philosophy;\n• The investor is unsatisfied and wants to maximize utility;\n• The market is complete.\n\nUnlike the capital asset pricing model, arbitrage pricing theory does not have the following assumptions:\n\n• Single investment period;\n• There is no tax;\n• Investors can borrow freely at a risk-free rate;\n• Investors choose investment portfolios based on the mean and variance of returns.\n\n#### The relationship between the arbitrage pricing theory and capital asset pricing model:\n\n• Both assume that there are no transaction costs or transaction taxes in the capital market; or both believe that if there are transaction costs or transaction taxes, they are the same for all investors.\n• Both of them divide the existing risks into systemic and non-systemic risks, that is, market risk and the company’s own risk. Moreover, both models believe that through the diversified portfolio of investments and the rational optimization of the investment structure by investors; the company ’s risks can be largely or even eliminated. Therefore, when calculating the expected return of the investment portfolio; the mathematical expressions of both models believe that the capital market will not compensate investors; because, they have assumed this part of the risk, so they not include in the calculation.\n• Capital asset pricing theory can regard as a special case of arbitrage pricing theory under stricter assumptions.\nClassification of Cost of Capital, and explain their Types\n\n#### The role of the arbitrage pricing theory and capital asset pricing model:\n\nThe proposal of CAPM and APT has had a huge impact on financial theory research and practice all over the world. Its main performances are:\n\n• Most institutional investors evaluate their investment performance according to the relationship between expected return and β coefficient (or unit risk-reward);\n• The regulatory authorities of most countries take the relationship between the expected rate of return; and, the β coefficient together with the prediction of the market index rate of return as an important factor when determining the capital cost of the regulated object;\n• The court often uses the relationship between the expected rate of return and β coefficient to determine the discount rate when measuring the amount of compensation for future loss of income\n• Many companies also use the relationship between the expected rate of return and the β coefficient to determine the minimum required rate of return when making capital budget decisions. As well as, It can see that the combination of the two can make more accurate predictions than pure APT; and, can make more extensive analysis than CAPM, to provide more adequate guidance for investment decisions.\n\n##### You May Also Like",
null,
"## Equity Shares: Explanation, Characteristics, and Features",
null,
"",
null,
""
] | [
null,
"https://i2.wp.com/www.ilearnlot.com/wp-content/uploads/2018/11/What-is-Arbitrage-Pricing-Theory-APT-Meaning-and-Definition.jpg",
null,
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null,
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null,
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null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91812706,"math_prob":0.9143324,"size":6967,"snap":"2020-24-2020-29","text_gpt3_token_len":1295,"char_repetition_ratio":0.17865862,"word_repetition_ratio":0.039341263,"special_character_ratio":0.18171379,"punctuation_ratio":0.0875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9842631,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-08T14:48:07Z\",\"WARC-Record-ID\":\"<urn:uuid:975ce166-b511-4ea6-8ed9-830ba6a84535>\",\"Content-Length\":\"168965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a32d3729-ab5e-4aa9-a2e7-e0e1257aeba0>\",\"WARC-Concurrent-To\":\"<urn:uuid:e30d3ebd-118a-46c8-b38f-e4adc621d17d>\",\"WARC-IP-Address\":\"185.201.10.113\",\"WARC-Target-URI\":\"https://www.ilearnlot.com/arbitrage-pricing-theory-apt-meaning-and-definition/57906/\",\"WARC-Payload-Digest\":\"sha1:RTXIJ6IRYFZ7LGY2TPXOIQQ2YEKQ4TGP\",\"WARC-Block-Digest\":\"sha1:7XJI6RX3VXEST6M7ORGVRBY33EX3FVRJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655897027.14_warc_CC-MAIN-20200708124912-20200708154912-00152.warc.gz\"}"} |
https://nextjournal.com/help/plotting | [
"# Plotting Data\n\nThe environments underlying Nextjournal’s default language runtimes come with a variety of plotting libraries already pre-installed. For example, our R environment comes with plotly — simply require and use it. No additional installation required:\n\nlibrary(plotly)\nd <- diamonds[sample(nrow(diamonds), 1000), ]\nplot_ly(\nd, x = ~carat, y = ~price,\ntext = ~paste(\"Price: \", price, '$<br>Cut:', cut), color = ~carat, size = ~carat, type='scatter', mode='markers') Loading viewer… This guide is a simple collection of plotting examples using different libraries for all programming languages that Nextjournal offers at the moment. Most plots will reference the following data sets: artist_data.csv artwork_data.csv ## R For plotting in R, Nextjournal supports the default R graphics package (graphics), plot_ly(), ggplot2(), and ggplotly() with no additional installation required. ### The Default R Graphics Package This first example uses the standard smoothScatter() function to plot the birth year of artists represented in the Tate Museum's permanent collection. Note that smoothScatter() does not require the loading of any dependencies. artists <- read.csv(artist_data.csv, header=T) born <- artists$yearOfBirth\nsmoothScatter(born, 1:length(born),\naxes=FALSE,\nxlab=\"Year\", ylab=\"\",\nmain=\"Distribution of Artist's Birth Years at the Tate\")\naxis(1, col.ticks=\"blue\")",
null,
"artists\n0 items\n\n### Working With Dependencies\n\nPlotly and ggplot2 are external dependencies that offer more features than the default R graphics.\n\nLoad the tidyverse collection of R packages, which includes two dependencies used in the upcoming sections, ggplot2 (ggplot()) and readr (read_csv()). The Plotly package provides two important plotting functions, plot_ly() and ggplotly().\n\nlibrary(tidyverse)\nlibrary(plotly)\n\n### ggplot2\n\nggplot2 is a system for declaratively creating graphics, based on The Grammar of Graphics.\n\nartists <- read_csv(artist_data.csv)\nmale_artists <- artists[artists$gender == \"Male\",] plot_ly(alpha=0.6) %>% add_histogram(data=female_artists, x=~yearOfBirth, name=\"Females\") %>% add_histogram(data=male_artists, x=~yearOfBirth, name=\"Males\") %>% layout(barmode=\"overlay\", xaxis=list(title=\"Year of Birth\")) Loading viewer… ### ggplotly The ggplotly() function transforms a static ggplot object into a Plotly object. More detailed information is available at the Plotly ggplot2 Library documentation. artists <- read_csv(artist_data.csv) born <- artists$yearOfBirth\ndf <- data.frame(born)\nid <- as.numeric(row.names(df))\n\nggplotly(ggplot(df, mapping=aes(x = born, y = id)) +\ngeom_point(size=1.5, alpha=0.4, shape=15) +\nlabs(x = \"Year\", y=\"\",\ntitle = \"Distribution of Artist's Birth Years at the Tate\") +\ntheme_bw() +\ntheme(axis.text.y = element_blank(),\naxis.ticks.y = element_blank(),\npanel.grid.minor=element_blank(),\npanel.grid.major.y=element_blank()))\n\n### Multiple Plots\n\nA Nextjournal cell can show multiple graphs—the runner will detect each new figure automatically and display them in order.\n\nartworks <- read_csv(artwork_data.csv)\n\ndrop <- c(\"accession_number\", \"artistRole\", \"artistId\", \"dateText\", \"creditLine\", \"units\", \"inscription\", \"thumbnailCopyright\", \"thumbnailUrl\", \"url\")\nartworks_rem <- artworks[ , !(names(artworks) %in% drop)]\n\nartworks_size <- artworks_rem[!(is.na(artworks_rem$height & artworks_rem$width & artworks_rem$year)),] artworks_size$size <- artworks_size$height * artworks_size$width\n\nmetal <- artworks_size[artworks_size$medium == \"Steel\" | artworks_size$medium==\"Bronze\",]\n\nplot_ly(data=metal, x=~acquisitionYear, name=\"Sculptural Acquisitions\")\n\nplot_ly(data=metal, x=~year, y=~acquisitionYear, z=~size, color=~medium,\ncolors = c('#BF382A', '#0C4B8E'), text=~artist,\nmarker=list(size=4, opacity=0.5)) %>%\nlayout(scene = list(xaxis = list(title = 'Year Created'),\nyaxis = list(title = 'Year of Acquisition'),\nzaxis = list(title = 'Size')),\nannotations = list(\nx = 1.13,\ny = 1.05,\ntext = 'Material',\nxref = 'paper',\nyref = 'paper',\nshowarrow = FALSE\n))\n\n## Python\n\n### Matplotlib\n\nMatplotlib is a plotting library for Python with at MATLAB-like interface:\n\nimport matplotlib\nimport matplotlib.pyplot as plt\nimport numpy as np\nimport pandas as pd\n\nartwork_data.drop(columns=[\"accession_number\", \"artistRole\", \"artistId\", \"dateText\", \"acquisitionYear\", \"dimensions\", \"width\", \"height\", \"depth\", \"creditLine\", \"units\", \"inscription\", \"thumbnailCopyright\", \"thumbnailUrl\", \"url\"], inplace=True)\n\n# Drop the rows listed as NaN, otherwise indexing oil, acrylic, and watercolour artworks yeild the error \"ValueError: cannot index with vector containing NA / NaN values.\" Replace this line with something more sensible to get a more complete dataset.\nartwork_data.dropna(subset=['medium'],inplace=True)\nartwork_data[\"year\"] = pd.to_numeric(artwork_data[\"year\"], errors=\"coerce\")\n\noil=artwork_data[artwork_data[\"medium\"].str.contains(\"oil\", case=False)]\nacrylic=artwork_data[artwork_data[\"medium\"].str.contains(\"acrylic\", case=False)]\nwatercolour=artwork_data[artwork_data[\"medium\"].str.contains(\"watercolour\", case=False)]\n\nfig, ax = plt.subplots()\n\nax.set(xlabel='year', ylabel='number of works',\ntitle='Paintings at the Tate, by Medium')\n\nax.hist([oil[\"year\"], acrylic[\"year\"], watercolour[\"year\"]], stacked=True)\nfig\n\n### Plotly\n\nPlotly's Python graphing library creates interactive, publication-quality graphs online. Information can also be displayed as a table using Plotly’s Figure Factory module.\n\nimport pandas as pd\n\n# plotly imports\nimport plotly.plotly as py\nimport plotly.figure_factory as ff\n# plotly.graph_objs contains all the helper classes to make/style plots\nimport plotly.graph_objs as go\n\n# Display the first 12 rows and 3 columns of the dataframe\nff.create_table(artist_data.iloc[:12,:3], index=False)\n\nPlot two histograms that compare the number of male artists in the Tate collection as compared to the number of female artists, distributed by their year of birth.\n\nimport numpy as np\n\nmale = artist_data['gender'] == 'Male'\nfemale = artist_data['gender'] == 'Female'\n\ntrace1 = go.Histogram(\nx=np.array((artist_data[female]['yearOfBirth'])),\nname='Female')\n\ntrace2 = go.Histogram(\nx=np.array((artist_data[male]['yearOfBirth'])),\nname='Male')\n\ntrace_data = [trace1, trace2]\nlayout = go.Layout(\nbargroupgap=0.3)\n\ngo.Figure(data=trace_data, layout=layout)\n\nNote that the data points can be hovered over to view the data for each, both here and in the published view. Traces can also be toggled on and off by clicking in the legend.\n\n## Julia\n\nPlots offers the most flexible way to visualize data using Julia in Nextjournal. This preinstalled library provides a unified interface to different plotting libraries, including plotly and gr. Plotly graphs are interactive, while gr is faster for large data sets.\n\nWhile documentation exists for both the Plotly Julia Library and Julia Package GR, these examples leverage plots, as such the Plots documentation will offer the most useful supplementary information.\n\n### Plotly\n\nusing Plots; plotly()\nscatter(rand(10), rand(10), title=\"Plot.ly Backend\")\n\nThe Plotly Julia Library offers more documentation examples for reference.\n\n### gr\n\nusing Plots; gr()\n\ngr produces a png file which is displayed by Nextjournal.\n\nscatter(rand(10), rand(10), title=\"GR Backend\")"
] | [
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"https://nextjournal.com/data/QmXTWGJc4v8KryhGBwAQYewQS9CJJiQrU4BVBqCHKu1c35",
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https://www.dsprelated.com/freebooks/filters/Normalizing_Two_Pole_Filter_Gain.html | [
"### Normalizing Two-Pole Filter Gain at Resonance\n\nThe question we now pose is how to best compensate the tunable two-pole resonator of §B.1.3 so that its peak gain is the same for all tunings. Looking at Fig.B.17, and remembering the graphical method for determining the amplitude response,B.6 it is intuitively clear that we can help matters by adding two zeros to the filter, one near dc and the other near",
null,
". A zero exactly at dc is provided by the term",
null,
"in the transfer function numerator. Similarly, a zero at half the sampling rate is provided by the term",
null,
"in the numerator. The series combination of both zeros gives the numerator",
null,
". The complete second-order transfer function then becomes",
null,
"corresponding to the difference equation",
null,
"(B.13)\n\nChecking the gain for the case",
null,
", we have",
null,
"which is better behaved, but now the response falls to zero at dc and",
null,
"rather than being heavily boosted, as we found in Eq.",
null,
"(B.12).\n\nNext Section:\nConstant Resonance Gain\nPrevious Section:\nDC Blocker Software Implementations"
] | [
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"http://www.dsprelated.com/josimages_new/filters/img114.png",
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"http://www.dsprelated.com/josimages_new/filters/img803.png",
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"http://www.dsprelated.com/josimages_new/filters/img1510.png",
null,
"http://www.dsprelated.com/josimages_new/filters/img1511.png",
null,
"http://www.dsprelated.com/josimages_new/filters/img1512.png",
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"http://www.dsprelated.com/josimages_new/filters/img1513.png",
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"http://www.dsprelated.com/josimages_new/filters/img1514.png",
null,
"http://www.dsprelated.com/josimages_new/filters/img1515.png",
null,
"http://www.dsprelated.com/josimages_new/filters/img114.png",
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"http://www.dsprelated.com/josimages_new/filters/img94.png",
null
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https://nbviewer.org/url/antonakhmerov.org/misc/spin_conductance.ipynb | [
"This notebook demonstrates using conservation laws in Kwant 1.3.\n\nThanks goes to Tómas Rosdahl for his work on this algorithm.\n\nAssuming you have Kwant build environment set up, you'll need to execute the following to get everything required to run the notebook:\n\npip3 install git+https://gitlab.kwant-project.org/kwant/[email protected]\npip3 install git+https://gitlab.kwant-project.org/kwant/[email protected]\n\nIn :\n%matplotlib inline\nimport kwant\nimport tinyarray as ta\nimport numpy as np\nfrom scipy import sparse\nfrom matplotlib import pyplot\nimport matplotlib\n\nIn :\ns0 = ta.array([[1, 0], [0, 1]])\nsx = ta.array([[0, 1], [1, 0]])\nsy = ta.array([[0, 1j], [-1j, 0]])\nsz = ta.array([[1, 0], [0, -1]])\n\nIn :\n# Adapted from https://kwant-project.org/doc/1/tutorial/tutorial2\n\ndef make_system(t=1.0, W=10, L=10):\n# Now we must specify the number of orbitals per site.\nlat = kwant.lattice.square(norbs=2)\nsyst = kwant.Builder()\n\nsyst[(lat(x, y) for x in range(L) for y in range(W))] = \\\nlambda s, alpha, E_z: 4 * t * s0 + E_z * sz\nsyst[kwant.builder.HoppingKind((1, 0), lat, lat)] = \\\nlambda s1, s2, alpha, E_z: -t * s0 - 1j * alpha * sy\nsyst[kwant.builder.HoppingKind((0, 1), lat, lat)] = \\\nlambda s1, s2, alpha, E_z: -t * s0 + 1j * alpha * sx\n\n# The new bit: specifying the conservation law.\nconservation_law=-sz, time_reversal=s0)\nlead[(lat(0, j) for j in range(W))] = 4 * t * s0\nlead[lat.neighbors()] = -t * s0 # Note: no spin-orbit in the lead.\n\nsyst = syst.finalized()\n\nreturn syst\n\nIn :\nsyst = make_system(t=1.0, W=10, L=10)\nenergies = np.linspace(0, 1, 200)\nsmatrices = [kwant.smatrix(syst, energy, args=(0.2, 0.05)) for energy in energies]\n\nfig = pyplot.figure(figsize=(13, 8))\n\nax.plot(energies, [smatrix.transmission(1, 0) for smatrix in smatrices], label='total')\n\n# The new bit: smatrix.transmission((lead1, q1), (lead0, q0)) is the transmission from the\n# q0 block of the lead0 into the q1 block of lead1. The subblock ordering is same as we used\n# in set_symmetry.\nax.plot(energies, [smatrix.transmission((1, 0), (0, 0)) for smatrix in smatrices], label='$G_{↑↑}$')\nax.plot(energies, [smatrix.transmission((1, 1), (0, 0)) for smatrix in smatrices], label='$G_{↑↓}$')\nax.plot(energies, [smatrix.transmission((1, 0), (0, 1)) for smatrix in smatrices], label='$G_{↓↑}$')\nax.plot(energies, [smatrix.transmission((1, 1), (0, 1)) for smatrix in smatrices], label='$G_{↓↓}$')\nax.set_ylabel('$G [e^2/h]$', fontsize='xx-large')\nax.set_xlabel('$E/t$', fontsize='xx-large')\nax.legend(fontsize='x-large');",
null,
"In :\ns = kwant.smatrix(syst, energy=1, args=(0.2, 0))\nprint(\"r_↑↑ = r_↓↓^T\")\nprint(np.round(s.submatrix((0, 1), (0, 1)) - s.submatrix((0, 0), (0, 0)).T, 6))\nprint(\"r_↑↓ = -r_↑↓^T\")\nprint(np.round(s.submatrix((0, 1), (0, 0)) + s.submatrix((0, 1), (0, 0)).T, 6))\n\nr_↑↑ = r_↓↓^T\n[[ 0.+0.j -0.-0.j 0.+0.j]\n[-0.-0.j 0.-0.j 0.+0.j]\n[ 0.+0.j 0.+0.j 0.+0.j]]\nr_↑↓ = -r_↑↓^T\n[[ 0.+0.j -0.-0.j 0.+0.j]\n[-0.-0.j 0.+0.j -0.-0.j]\n[ 0.+0.j -0.-0.j 0.-0.j]]\n\nIn :\nmodes = s.lead_info\n\n# Wave functions in different spin blocks coincide.\nmodes.wave_functions[:, [0, 3]]\n\nOut:\narray([[-0.09250579-0.00437932j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.09250579-0.00437932j],\n[-0.17751732-0.00840386j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.17751732-0.00840386j],\n[-0.24814745-0.01174756j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.24814745-0.01174756j],\n[-0.29867414-0.01413955j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.29867414-0.01413955j],\n[-0.32500404-0.01538604j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.32500404-0.01538604j],\n[-0.32500404-0.01538604j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.32500404-0.01538604j],\n[-0.29867414-0.01413955j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.29867414-0.01413955j],\n[-0.24814745-0.01174756j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.24814745-0.01174756j],\n[-0.17751732-0.00840386j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.17751732-0.00840386j],\n[-0.09250579-0.00437932j, 0.00000000+0.j ],\n[ 0.00000000+0.j , -0.09250579-0.00437932j]])\nIn [ ]:"
] | [
null,
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https://fdocuments.net/document/phase-transformation-and-deformation-model-for-phase-transformation-and-deformation.html | [
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"• date post\n\n30-Aug-2020\n• Category\n\n## Documents\n\n• view\n\n2\n\n0\n\nEmbed Size (px)\n\n### Transcript of Phase Transformation and Deformation Model for Phase Transformation and Deformation Model for...\n\n• Phase Transformation and Deformation Model for Quenching Simulations Y. Kaymak VDEh Betriebsforschungsinstitute GmbH, Düsseldorf, NW, Germany Introduction In materials science, quenching is the rapid cooling of a workpiece to obtain certain material properties. For instance, quenching can reduce the crystal grain size of metallic materials and increasing their hardness. The rapid cooling prevents undesired phase transformations from occurring by reducing the window of time during which these undesired phase transformation are thermodynamically and kinetically favorable. The process of steel quenching is a progression, beginning with heating the sample up to a precise temperature, which is between 815°C and 900°C for the most of the steel types. The temperatures throughout the workpiece should be kept as uniform as possible during the heating. Afterwards, the workpiece is rapidly cooled usually by soaking in a fluid bath. Similar to the heating step, it is important that the temperature throughout the sample remains as uniform as possible during soaking. Often, the workpiece is excessively hard and brittle after quenching. In some cases, one or more tempering process steps are performed additionally in order to increase the toughness. In a tempering sub-process, the quenched steel is heated up to some critical temperature for a certain period of time, and then allowed to cool. The typical temperature evolution during the heat treatment process is show in Figure 1.\n\nFigure 1. Temperature evolution during a typical heat treatment process. Heat treatment of the advanced steel grades (like micro-alloyed steels or AHSS steel grades) is a challenging process as the residual stress/deformation are pronounced and the quality requirements of the customers are getting tighter. A comprehensive modelling of the complex phenomena to estimate the\n\nresidual stress and deformation is essential for developing an optimal process control. The basis of the complex quenching model presented in this paper is developed within author’s PhD thesis . The model consists of a series of coupled physics, which are summarized in Figure 2. These coupled fields are solved by using the physics interfaces in COMSOL Multiphysics®. The temperature field is solved by the heat transfer in solids physics. The micro- structure field is modelled using kinetic expressions and domain ODEs and DAE physics. The displacement field is solved using the solid mechanics physics including the volume change (due to the temperature and micro-structure changes), plasticity, transformation induced plasticity (trip), creep, and large deformations. The constitutive model parameters as well as the isothermal and martensitic transformation kinetic parameters are validated and calibrated by several dilatometry tests.\n\nFigure 2. Coupling of fields in quenching process. Theory and Governing Equations The quenching simulation model is focused on the cooling since the most significant part of the residual stress and deformation is developed during this rapid cooling. The model presented here consists of strongly coupled phenomena of heat transfer, micro- structure change, deformation (due to the thermal shrinkage, microstructure change related dilatation, trip, creep, plasticity, and large deformations). The governing fields involved in the quenching process and their interactions are shown in Figure 2. In quenching, the temperature field is controlled by the cooling boundary conditions. The temperature\n\nExcerpt from the Proceedings of the 2017 COMSOL Conference in Rotterdam\n\n• evolution drives the phase transformation kinetics and phase transformations are accompanied by latent heat release. Moreover, all the material properties depend on the temperature and microstructure. A linear mixture rule is used for the Young’s modulus 𝐸𝐸, Poisson’s ratio 𝜈𝜈, initial yield stress 𝜎𝜎𝑦𝑦0, heat capacity 𝐶𝐶𝑝𝑝 and thermal conductivity 𝑘𝑘. However, a harmonic mixture rule is used for the density 𝜌𝜌. 𝐸𝐸 = 𝑓𝑓𝑎𝑎𝐸𝐸𝑎𝑎(𝑇𝑇) + 𝑓𝑓𝑏𝑏𝐸𝐸𝑏𝑏(𝑇𝑇) + 𝑓𝑓𝑚𝑚𝐸𝐸𝑚𝑚(𝑇𝑇) 𝜈𝜈 = 𝑓𝑓𝑎𝑎𝜈𝜈𝑎𝑎(𝑇𝑇) + 𝑓𝑓𝑏𝑏𝜈𝜈𝑏𝑏(𝑇𝑇) + 𝑓𝑓𝑚𝑚𝜈𝜈𝑚𝑚(𝑇𝑇) 𝜎𝜎𝑦𝑦0 = 𝑓𝑓𝑎𝑎𝜎𝜎𝑎𝑎𝑦𝑦0(𝑇𝑇) + 𝑓𝑓𝑏𝑏𝜎𝜎𝑏𝑏𝑦𝑦0(𝑇𝑇) + 𝑓𝑓𝑚𝑚𝜎𝜎𝑚𝑚𝑦𝑦0(𝑇𝑇) 𝐶𝐶𝑝𝑝 = 𝑓𝑓𝑎𝑎𝐶𝐶𝑝𝑝𝑎𝑎(𝑇𝑇) + 𝑓𝑓𝑏𝑏𝐶𝐶𝑝𝑝𝑏𝑏(𝑇𝑇) + 𝑓𝑓𝑚𝑚𝐶𝐶𝑝𝑝𝑚𝑚(𝑇𝑇) 𝑘𝑘 = 𝑓𝑓𝑎𝑎𝑘𝑘𝑎𝑎(𝑇𝑇) + 𝑓𝑓𝑏𝑏𝑘𝑘𝑏𝑏(𝑇𝑇) + 𝑓𝑓𝑚𝑚𝑘𝑘𝑚𝑚(𝑇𝑇)\n\n𝜌𝜌 = 1\n\n𝑓𝑓𝑎𝑎 𝜌𝜌𝑎𝑎(𝑇𝑇)\n\n+ 𝑓𝑓𝑏𝑏𝜌𝜌𝑏𝑏(𝑇𝑇) + 𝑓𝑓𝑚𝑚\n\n𝜌𝜌𝑚𝑚(𝑇𝑇)\n\nwhere the subscripts 𝑎𝑎, 𝑏𝑏 and 𝑚𝑚 correspond to the austenite, bainite and martensite microstructure volume fractions. 𝑇𝑇 stands for the temperature and 𝑓𝑓 represents the volume fraction of the microstructure. For the sake of simplicity a perfectly plastic behavior can be assumed if needed. The heat generation due to dissipation of mechanical energy has no significant influence on the temperature field. Similarly, the stress dependency of the transformations can also be discarded. So, these two coupling phenomena greyed-out in Figure 2. The temperature field is solved by the heat transfer in solids physics in COMSOL Multiphysics® software. The micro-structure field is modelled at integration point level using kinetic expressions and domain ODEs and DAE physics. The displacement field is solved using the solid mechanics physics including the volume change due to the temperature and micro- structure changes. This volumetric strain is given by:\n\n𝑑𝑑𝑑𝑑 = �𝜌𝜌𝑎𝑎(𝑇𝑇ref)𝜌𝜌 3 − 1\n\nwhere 𝑇𝑇ref is the strain reference temperature 𝜌𝜌𝑎𝑎 is the austenite density and 𝜌𝜌 is the mixture density. It is assumed that initial microstructure is completely austenite. Moreover, the nonlinear phenomena like plasticity, transformation induced plasticity (trip), creep, and large deformations are also considered in the solid mechanics physics setup in COMSOL. The governing partial differential equations (PDEs) and expressions of the model are further discussed on the following sub-sections.\n\nTemperature field: The temperature field in solid material is modelled by the heat transfer in solids physics:\n\n𝜌𝜌𝐶𝐶𝑝𝑝 𝜕𝜕𝑇𝑇 𝜕𝜕𝜕𝜕\n\n+ ∇ ∙ 𝐪𝐪 = 𝑄𝑄\n\n𝐪𝐪 = −𝑘𝑘∇𝑇𝑇\n\nwhere, ρ is the density, 𝐶𝐶𝑝𝑝 is the heat capacity, 𝑇𝑇 is temperature field, 𝜕𝜕 is the time, 𝐪𝐪 is heat flux vector, 𝑘𝑘 is the heat conductivity, 𝑄𝑄 heat source due to the latent heat of the phase transformations. The equation for the heat source 𝑄𝑄 depends on the transformation latent heats as well as the phase transformation rates. It is defined by: 𝑄𝑄 = 𝑑𝑑𝑎𝑎𝑏𝑏𝑓𝑓�̇�𝑏 + 𝑑𝑑𝑎𝑎𝑚𝑚𝑓𝑓�̇�𝑚\n\nwhere, 𝑑𝑑𝑎𝑎𝑏𝑏 and 𝑑𝑑𝑎𝑎𝑚𝑚 are the latent heats of the austenite to bainite and austenite to martensite transformations, 𝑓𝑓�̇�𝑏 and 𝑓𝑓�̇�𝑚 are the rates of the austenite to bainite and austenite to martensite transformations, respectively. For the sake of simplicity, the temperature field in the quenching medium is not modelled. Instead, either the temperature or heat flux at the solid boundaries are defined, e.g., by using convective heat transfer coefficients and radiation to ambient. Microstructure field: There are two types of transformations relevant to this study: (1) austenite to bainite transformation, which is diffusion controlled, needs an incubation time before the transformation starts. (2) Austenite to martensite transformation, which is diffusionless, is controlled only by temperature that means it can be expressed as a function of temperature without solving or integration a PDE. The austenite to bainite transformation is modelled using the domain ODEs. The transformation kinetics for the bainite formation is assumed to obey the Scheil’s addition rule and Johnson-Mehl-Avrami- Kolmogorow (JMAK) equation. Two state variables per integration point are defined, one for the Scheil’s sum and the other for bainite volume fraction in JMAK-equation. The Scheil’s sum 𝑠𝑠𝑠𝑠 is obtained by integrating:\n\n𝑠𝑠�̇�𝑠 = 1\n\n𝐵𝐵𝑠𝑠(𝑇𝑇)\n\nThe incubation time is completed when the Scheil’s sum reaches the unity. After this incubation time, the\n\nExcerpt from the Proceedings of the 2017 COMSOL Conference in Rotterdam\n\n• bainite volume fraction 𝑓𝑓𝑏𝑏 is obtained by integrating JMAK-equation: 𝑓𝑓�̇�𝑏 = 𝐾𝐾 ∙ 𝑛𝑛 ∙ 𝜕𝜕𝑛𝑛−1 ∙ exp(−𝐾𝐾 ∙ 𝜕𝜕𝑛𝑛)\n\nwhere, 𝑛𝑛 is known as JMAK-exponent and 𝐾𝐾 is JMAK-factor. They are calculated using the bainite transformation start time 𝐵𝐵𝑠𝑠(𝑇𝑇) and finish time 𝐵𝐵𝑓𝑓(𝑇𝑇). These start and finish curves are given in the Time-Temperature-Transformation (TTT) diagrams. If it is assumed that the bainite volume fraction 𝑓𝑓𝑏𝑏 is 0.01 at the starti"
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http://www.gnu.org/software/guile/docs/docs-1.6/guile-ref/Integers.html | [
"Node:Integers, Next:, Previous:Numerical Tower, Up:Numbers\n\n#### 21.2.2 Integers\n\nIntegers are whole numbers, that is numbers with no fractional part, such as 2, 83 and -3789.\n\nIntegers in Guile can be arbitrarily big, as shown by the following example.\n\n```(define (factorial n)\n(let loop ((n n) (product 1))\n(if (= n 0)\nproduct\n(loop (- n 1) (* product n)))))\n\n(factorial 3)\n=>\n6\n\n(factorial 20)\n=>\n2432902008176640000\n\n(- (factorial 45))\n=>\n-119622220865480194561963161495657715064383733760000000000\n```\n\nReaders whose background is in programming languages where integers are limited by the need to fit into just 4 or 8 bytes of memory may find this surprising, or suspect that Guile's representation of integers is inefficient. In fact, Guile achieves a near optimal balance of convenience and efficiency by using the host computer's native representation of integers where possible, and a more general representation where the required number does not fit in the native form. Conversion between these two representations is automatic and completely invisible to the Scheme level programmer.\n\n integer? x Scheme Procedure scm_integer_p (x) C Function\n Return `#t` if x is an integer number, else `#f`. ```(integer? 487) => #t (integer? -3.4) => #f ```"
] | [
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https://live.boost.org/doc/libs/1_72_0/libs/math/doc/html/math_toolkit/lambert_w.html | [
"#",
null,
"Boost C++ Libraries\n\n...one of the most highly regarded and expertly designed C++ library projects in the world.\n\n## Lambert W function\n\n##### Synopsis\n```#include <boost/math/special_functions/lambert_w.hpp>\n```\n```namespace boost { namespace math {\n\ntemplate <class T>\ncalculated-result-type lambert_w0(T z); // W0 branch, default policy.\ntemplate <class T>\ncalculated-result-type lambert_wm1(T z); // W-1 branch, default policy.\ntemplate <class T>\ncalculated-result-type lambert_w0_prime(T z); // W0 branch 1st derivative.\ntemplate <class T>\ncalculated-result-type lambert_wm1_prime(T z); // W-1 branch 1st derivative.\n\ntemplate <class T, class Policy>\ncalculated-result-type lambert_w0(T z, const Policy&); // W0 with policy.\ntemplate <class T, class Policy>\ncalculated-result-type lambert_wm1(T z, const Policy&); // W-1 with policy.\ntemplate <class T, class Policy>\ncalculated-result-type lambert_w0_prime(T z, const Policy&); // W0 derivative with policy.\ntemplate <class T, class Policy>\ncalculated-result-type lambert_wm1_prime(T z, const Policy&); // W-1 derivative with policy.\n\n} // namespace boost\n} // namespace math\n```\n##### Description\n\nThe Lambert W function is the solution of the equation W(z)eW(z) = z. It is also called the Omega function, the inverse of f(W) = WeW.\n\nOn the interval [0, ∞), there is just one real solution. On the interval (-e-1, 0), there are two real solutions, generating two branches which we will denote by W0 and W-1. In Boost.Math, we call these principal branches `lambert_w0` and `lambert_wm1`; their derivatives are labelled `lambert_w0_prime` and `lambert_wm1_prime`.",
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"",
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"",
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"",
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"There is a singularity where the branches meet at e-1`-0.367879`. Approaching this point, the condition number of function evaluation tends to infinity, and the only method of recovering high accuracy is use of higher precision.\n\nThis implementation computes the two real branches W0 and W-1 with the functions `lambert_w0` and `lambert_wm1`, and their derivatives, `lambert_w0_prime` and `lambert_wm1_prime`. Complex arguments are not supported.\n\nThe final Policy argument is optional and can be used to control how the function deals with errors. Refer to Policies for more details and see examples below.\n\n###### Applications of the Lambert W function\n\nThe Lambert W function has a myriad of applications. Corless et al. provide a summary of applications, from the mathematical, like iterated exponentiation and asymptotic roots of trinomials, to the real-world, such as the range of a jet plane, enzyme kinetics, water movement in soil, epidemics, and diode current (an example replicated here). Since the publication of their landmark paper, there have been many more applications, and also many new implementations of the function, upon which this implementation builds.\n\n##### Examples\n\nThe most basic usage of the Lambert-W function is demonstrated below:\n\n```#include <boost/math/special_functions/lambert_w.hpp> // For lambert_w function.\n\nusing boost::math::lambert_w0;\nusing boost::math::lambert_wm1;\n```\n```std::cout.precision(std::numeric_limits<double>::max_digits10);\n// Show all potentially significant decimal digits,\nstd::cout << std::showpoint << std::endl;\n// and show significant trailing zeros too.\n\ndouble z = 10.;\ndouble r = lambert_w0(z); // Default policy for double.\nstd::cout << \"lambert_w0(z) = \" << r << std::endl;\n// lambert_w0(z) = 1.7455280027406994\n```\n\nOther floating-point types can be used too, here `float`, including user-defined types like Boost.Multiprecision. It is convenient to use a function like `show_value` to display all (and only) potentially significant decimal digits, including any significant trailing zeros, (`std::numeric_limits<T>::max_digits10`) for the type `T`.\n\n```float z = 10.F;\nfloat r;\nr = lambert_w0(z); // Default policy digits10 = 7, digits2 = 24\nstd::cout << \"lambert_w0(\";\nshow_value(z);\nstd::cout << \") = \";\nshow_value(r);\nstd::cout << std::endl; // lambert_w0(10.0000000) = 1.74552798\n```\n\nExample of an integer argument to `lambert_w0`, showing that an `int` literal is correctly promoted to a `double`.\n\n```std::cout.precision(std::numeric_limits<double>::max_digits10);\ndouble r = lambert_w0(10); // Pass an int argument \"10\" that should be promoted to double argument.\nstd::cout << \"lambert_w0(10) = \" << r << std::endl; // lambert_w0(10) = 1.7455280027406994\ndouble rp = lambert_w0(10);\nstd::cout << \"lambert_w0(10) = \" << rp << std::endl;\n// lambert_w0(10) = 1.7455280027406994\nauto rr = lambert_w0(10); // C++11 needed.\nstd::cout << \"lambert_w0(10) = \" << rr << std::endl;\n// lambert_w0(10) = 1.7455280027406994 too, showing that rr has been promoted to double.\n```\n\nUsing Boost.Multiprecision types to get much higher precision is painless.\n\n```cpp_dec_float_50 z(\"10\");\n// Note construction using a decimal digit string \"10\",\n// NOT a floating-point double literal 10.\ncpp_dec_float_50 r;\nr = lambert_w0(z);\nstd::cout << \"lambert_w0(\"; show_value(z); std::cout << \") = \";\nshow_value(r);\nstd::cout << std::endl;\n// lambert_w0(10.000000000000000000000000000000000000000000000000000000000000000000000000000000) =\n// 1.7455280027406993830743012648753899115352881290809413313533156980404446940000000\n```\nWarning",
null,
"When using multiprecision, take very great care not to construct or assign non-integers from `double`, `float` ... silently losing precision. Use `\"1.2345678901234567890123456789\"` rather than `1.2345678901234567890123456789`.\n\nUsing multiprecision types, it is all too easy to get multiprecision precision wrong!\n\n```cpp_dec_float_50 z(0.7777777777777777777777777777777777777777777777777777777777777777777777777);\n// Compiler evaluates the nearest double-precision binary representation,\n// from the max_digits10 of the floating_point literal double 0.7777777777777777777777777777...,\n// so any extra digits in the multiprecision type\n// beyond max_digits10 (usually 17) are random and meaningless.\ncpp_dec_float_50 r;\nr = lambert_w0(z);\nstd::cout << \"lambert_w0(\";\nshow_value(z);\nstd::cout << \") = \"; show_value(r);\nstd::cout << std::endl;\n// lambert_w0(0.77777777777777779011358916250173933804035186767578125000000000000000000000000000)\n// = 0.48086152073210493501934682309060873341910109230469724725005039758139532631901386\n```\nNote",
null,
"See spurious non-seven decimal digits appearing after digit #17 in the argument 0.7777777777777777...!\n\nAnd similarly constructing from a literal ```double 0.9```, with more random digits after digit number 17.\n\n```cpp_dec_float_50 z(0.9); // Construct from floating_point literal double 0.9.\ncpp_dec_float_50 r;\nr = lambert_w0(0.9);\nstd::cout << \"lambert_w0(\";\nshow_value(z);\nstd::cout << \") = \"; show_value(r);\nstd::cout << std::endl;\n// lambert_w0(0.90000000000000002220446049250313080847263336181640625000000000000000000000000000)\n// = 0.52983296563343440510607251781038939952850341796875000000000000000000000000000000\nstd::cout << \"lambert_w0(0.9) = \" << lambert_w0(static_cast<double>(0.9))\n// lambert_w0(0.9)\n// = 0.52983296563343441\n<< std::endl;\n```\n\nNote how the `cpp_float_dec_50` result is only as correct as from a ```double = 0.9```.\n\nNow see the correct result for all 50 decimal digits constructing from a decimal digit string \"0.9\":\n\n```cpp_dec_float_50 z(\"0.9\"); // Construct from decimal digit string.\ncpp_dec_float_50 r;\nr = lambert_w0(z);\nstd::cout << \"lambert_w0(\";\nshow_value(z);\nstd::cout << \") = \"; show_value(r);\nstd::cout << std::endl;\n// 0.90000000000000000000000000000000000000000000000000000000000000000000000000000000)\n// = 0.52983296563343441213336643954546304857788132269804249284012528304239956413801252\n```\n\nNote the expected zeros for all places up to 50 - and the correct Lambert W result!\n\n(It is just as easy to compute even higher precisions, at least to thousands of decimal digits, but not shown here for brevity. See lambert_w_simple_examples.cpp for comparison of an evaluation at 1000 decimal digit precision with Wolfram Alpha).\n\nPolicies can be used to control what action to take on errors:\n\n```// Define an error handling policy:\ntypedef policy<\ndomain_error<throw_on_error>,\noverflow_error<ignore_error> // possibly unwise?\n> my_throw_policy;\n\nstd::cout.precision(std::numeric_limits<double>::max_digits10);\n// Show all potentially significant decimal digits,\nstd::cout << std::showpoint << std::endl;\n// and show significant trailing zeros too.\ndouble z = +1;\nstd::cout << \"Lambert W (\" << z << \") = \" << lambert_w0(z) << std::endl;\n// Lambert W (1.0000000000000000) = 0.56714329040978384\nstd::cout << \"\\nLambert W (\" << z << \", my_throw_policy()) = \"\n<< lambert_w0(z, my_throw_policy()) << std::endl;\n// Lambert W (1.0000000000000000, my_throw_policy()) = 0.56714329040978384\n```\n\nAn example error message:\n\n```Error in function boost::math::lambert_wm1<RealType>(<RealType>):\nArgument z = 1 is out of range (z <= 0) for Lambert W-1 branch! (Try Lambert W0 branch?)\n```\n\nShowing an error reported if a value is passed to `lambert_w0` that is out of range, (and was probably meant to be passed to `lambert_wm1` instead).\n\n```double z = +1.;\ndouble r = lambert_wm1(z);\nstd::cout << \"lambert_wm1(+1.) = \" << r << std::endl;\n```\n\nThe full source of these examples is at lambert_w_simple_examples.cpp\n\n###### Diode Resistance Example\n\nA typical example of a practical application is estimating the current flow through a diode with series resistance from a paper by Banwell and Jayakumar.\n\nHaving the Lambert W function available makes it simple to reproduce the plot in their paper (Fig 2) comparing estimates using with Lambert W function and some actual measurements. The colored curves show the effect of various series resistance on the current compared to an extrapolated line in grey with no internal (or external) resistance.\n\nTwo formulae relating the diode current and effect of series resistance can be combined, but yield an otherwise intractable equation relating the current versus voltage with a varying series resistance. This was reformulated as a generalized equation in terms of the Lambert W function:\n\nBanwell and Jakaumar equation 5\n\nI(V) = μ VT/ R S ․ W0(I0 RS / (μ VT))\n\nUsing these variables\n\n```double nu = 1.0; // Assumed ideal.\ndouble vt = v_thermal(25); // v thermal, Shockley equation, expect about 25 mV at room temperature.\ndouble boltzmann_k = 1.38e-23; // joules/kelvin\ndouble temp = 273 + 25;\ndouble charge_q = 1.6e-19; // column\nvt = boltzmann_k * temp / charge_q;\nstd::cout << \"V thermal \" << vt << std::endl; // V thermal 0.0257025 = 25 mV\ndouble rsat = 0.;\ndouble isat = 25.e-15; // 25 fA;\nstd::cout << \"Isat = \" << isat << std::endl;\ndouble re = 0.3; // Estimated from slope of straight section of graph (equation 6).\ndouble v = 0.9;\ndouble icalc = iv(v, vt, 249., re, isat);\nstd::cout << \"voltage = \" << v << \", current = \" << icalc << \", \" << log(icalc) << std::endl; // voltage = 0.9, current = 0.00108485, -6.82631\n```\n\nthe formulas can be rendered in C++\n\n```double iv(double v, double vt, double rsat, double re, double isat, double nu = 1.)\n{\n// V thermal 0.0257025 = 25 mV\n// was double i = (nu * vt/r) * lambert_w((i0 * r) / (nu * vt)); equ 5.\n\nrsat = rsat + re;\ndouble i = nu * vt / rsat;\n// std::cout << \"nu * vt / rsat = \" << i << std::endl; // 0.000103223\n\ndouble x = isat * rsat / (nu * vt);\n// std::cout << \"isat * rsat / (nu * vt) = \" << x << std::endl;\n\ndouble eterm = (v + isat * rsat) / (nu * vt);\n// std::cout << \"(v + isat * rsat) / (nu * vt) = \" << eterm << std::endl;\n\ndouble e = exp(eterm);\n// std::cout << \"exp(eterm) = \" << e << std::endl;\n\ndouble w0 = lambert_w0(x * e);\n// std::cout << \"w0 = \" << w0 << std::endl;\nreturn i * w0 - isat;\n} // double iv\n```\n\nto reproduce their Fig 2:",
null,
"The plotted points for no external series resistance (derived from their published plot as the raw data are not publicly available) are used to extrapolate back to estimate the intrinsic emitter resistance as 0.3 ohm. The effect of external series resistance is visible when the colored lines start to curve away from the straight line as voltage increases.\n\nSee lambert_w_diode.cpp and lambert_w_diode_graph.cpp for details of the calculation.\n\n###### Existing implementations\n\nThe principal value of the Lambert W function is implemented in the Wolfram Language as ```ProductLog[k, z]```, where `k` is the branch.\n\nThe symbolic algebra program Maple also computes Lambert W to an arbitrary precision.\n\n##### Controlling the compromise between Precision and Speed\n###### Floating-point types `double` and `float`\n\nThis implementation provides good precision and excellent speed for __fundamental `float` and `double`.\n\nAll the functions usually return values within a few Unit in the last place (ULP) for the floating-point type, except for very small arguments very near zero, and for arguments very close to the singularity at the branch point.\n\nBy default, this implementation provides the best possible speed. Very slightly average higher precision and less bias might be obtained by adding a Halley step refinement, but at the cost of more than doubling the runtime.\n\n###### Floating-point types larger than double\n\nFor floating-point types with precision greater than `double` and `float` fundamental (built-in) types, a `double` evaluation is used as a first approximation followed by Halley refinement, using a single step where it can be predicted that this will be sufficient, and only using Halley iteration when necessary. Higher precision types are always going to be very, very much slower.\n\nThe 'best' evaluation (the nearest representable) can be achieved by `static_cast`ing from a higher precision type, typically a Boost.Multiprecision type like `cpp_bin_float_50`, but at the cost of increasing run-time 100-fold; this has been used here to provide some of our reference values for testing.\n\nFor example, we get a reference value using a high precision type, for example;\n\n```using boost::multiprecision::cpp_bin_float_50;\n```\n\nthat uses Halley iteration to refine until it is as precise as possible for this `cpp_bin_float_50` type.\n\nAs a further check we can compare this with a Wolfram Alpha computation using command `N[ProductLog[10.], 50]` to get 50 decimal digits and similarly `N[ProductLog[10.], 17]` to get the nearest representable for 64-bit `double` precision.\n\n``` using boost::multiprecision::cpp_bin_float_50;\nusing boost::math::float_distance;\n\ncpp_bin_float_50 z(\"10.\"); // Note use a decimal digit string, not a double 10.\ncpp_bin_float_50 r;\nstd::cout.precision(std::numeric_limits<cpp_bin_float_50>::digits10);\n\nr = lambert_w0(z); // Default policy.\nstd::cout << \"lambert_w0(z) cpp_bin_float_50 = \" << r << std::endl;\n//lambert_w0(z) cpp_bin_float_50 = 1.7455280027406993830743012648753899115352881290809\n// [N[productlog, 50]] == 1.7455280027406993830743012648753899115352881290809\nstd::cout.precision(std::numeric_limits<double>::max_digits10);\nstd::cout << \"lambert_w0(z) static_cast from cpp_bin_float_50 = \"\n<< static_cast<double>(r) << std::endl;\n// double lambert_w0(z) static_cast from cpp_bin_float_50 = 1.7455280027406994\n// [N[productlog, 17]] == 1.7455280027406994\nstd::cout << \"bits different from Wolfram = \"\n<< static_cast<int>(float_distance(static_cast<double>(r), 1.7455280027406994))\n<< std::endl; // 0\n```\n\ngiving us the same nearest representable using 64-bit `double` as `1.7455280027406994`.\n\nHowever, the rational polynomial and Fukushima Schroder approximations are so good for type `float` and `double` that negligible improvement is gained from a `double` Halley step.\n\nThis is shown with lambert_w_precision_example.cpp for Lambert W0:\n\n```using boost::math::lambert_w_detail::lambert_w_halley_step;\nusing boost::math::epsilon_difference;\nusing boost::math::relative_difference;\n\nstd::cout << std::showpoint << std::endl; // and show any significant trailing zeros too.\nstd::cout.precision(std::numeric_limits<double>::max_digits10); // 17 decimal digits for double.\n\ncpp_bin_float_50 z50(\"1.23\"); // Note: use a decimal digit string, not a double 1.23!\ndouble z = static_cast<double>(z50);\ncpp_bin_float_50 w50;\nw50 = lambert_w0(z50);\nstd::cout.precision(std::numeric_limits<cpp_bin_float_50>::max_digits10); // 50 decimal digits.\nstd::cout << \"Reference Lambert W (\" << z << \") =\\n \"\n<< w50 << std::endl;\nstd::cout.precision(std::numeric_limits<double>::max_digits10); // 17 decimal digits for double.\ndouble wr = static_cast<double>(w50);\nstd::cout << \"Reference Lambert W (\" << z << \") = \" << wr << std::endl;\n\ndouble w = lambert_w0(z);\nstd::cout << \"Rat/poly Lambert W (\" << z << \") = \" << lambert_w0(z) << std::endl;\n// Add a Halley step to the value obtained from rational polynomial approximation.\ndouble ww = lambert_w_halley_step(lambert_w0(z), z);\nstd::cout << \"Halley Step Lambert W (\" << z << \") = \" << lambert_w_halley_step(lambert_w0(z), z) << std::endl;\n\nstd::cout << \"absolute difference from Halley step = \" << w - ww << std::endl;\nstd::cout << \"relative difference from Halley step = \" << relative_difference(w, ww) << std::endl;\nstd::cout << \"epsilon difference from Halley step = \" << epsilon_difference(w, ww) << std::endl;\nstd::cout << \"epsilon for float = \" << std::numeric_limits<double>::epsilon() << std::endl;\nstd::cout << \"bits different from Halley step = \" << static_cast<int>(float_distance(w, ww)) << std::endl;\n```\n\nwith this output:\n\n```Reference Lambert W (1.2299999999999999822364316059974953532218933105468750) =\n0.64520356959320237759035605255334853830173300262666480\nReference Lambert W (1.2300000000000000) = 0.64520356959320235\nRat/poly Lambert W (1.2300000000000000) = 0.64520356959320224\nHalley Step Lambert W (1.2300000000000000) = 0.64520356959320235\nabsolute difference from Halley step = -1.1102230246251565e-16\nrelative difference from Halley step = 1.7207329236029286e-16\nepsilon difference from Halley step = 0.77494921535422934\nepsilon for float = 2.2204460492503131e-16\nbits different from Halley step = 1\n```\n\nand then for W-1:\n\n```using boost::math::lambert_w_detail::lambert_w_halley_step;\nusing boost::math::epsilon_difference;\nusing boost::math::relative_difference;\n\nstd::cout << std::showpoint << std::endl; // and show any significant trailing zeros too.\nstd::cout.precision(std::numeric_limits<double>::max_digits10); // 17 decimal digits for double.\n\ncpp_bin_float_50 z50(\"-0.123\"); // Note: use a decimal digit string, not a double -1.234!\ndouble z = static_cast<double>(z50);\ncpp_bin_float_50 wm1_50;\nwm1_50 = lambert_wm1(z50);\nstd::cout.precision(std::numeric_limits<cpp_bin_float_50>::max_digits10); // 50 decimal digits.\nstd::cout << \"Reference Lambert W-1 (\" << z << \") =\\n \"\n<< wm1_50 << std::endl;\nstd::cout.precision(std::numeric_limits<double>::max_digits10); // 17 decimal digits for double.\ndouble wr = static_cast<double>(wm1_50);\nstd::cout << \"Reference Lambert W-1 (\" << z << \") = \" << wr << std::endl;\n\ndouble w = lambert_wm1(z);\nstd::cout << \"Rat/poly Lambert W-1 (\" << z << \") = \" << lambert_wm1(z) << std::endl;\n// Add a Halley step to the value obtained from rational polynomial approximation.\ndouble ww = lambert_w_halley_step(lambert_wm1(z), z);\nstd::cout << \"Halley Step Lambert W (\" << z << \") = \" << lambert_w_halley_step(lambert_wm1(z), z) << std::endl;\n\nstd::cout << \"absolute difference from Halley step = \" << w - ww << std::endl;\nstd::cout << \"relative difference from Halley step = \" << relative_difference(w, ww) << std::endl;\nstd::cout << \"epsilon difference from Halley step = \" << epsilon_difference(w, ww) << std::endl;\nstd::cout << \"epsilon for float = \" << std::numeric_limits<double>::epsilon() << std::endl;\nstd::cout << \"bits different from Halley step = \" << static_cast<int>(float_distance(w, ww)) << std::endl;\n```\n\nwith this output:\n\n```Reference Lambert W-1 (-0.12299999999999999822364316059974953532218933105468750) =\n-3.2849102557740360179084675531714935199110302996513384\nReference Lambert W-1 (-0.12300000000000000) = -3.2849102557740362\nRat/poly Lambert W-1 (-0.12300000000000000) = -3.2849102557740357\nHalley Step Lambert W (-0.12300000000000000) = -3.2849102557740362\nabsolute difference from Halley step = 4.4408920985006262e-16\nrelative difference from Halley step = 1.3519066740696092e-16\nepsilon difference from Halley step = 0.60884463935795785\nepsilon for float = 2.2204460492503131e-16\nbits different from Halley step = -1\n```\n###### Distribution of differences from 'best' `double` evaluations\n\nThe distribution of differences from 'best' are shown in these graphs comparing `double` precision evaluations with reference 'best' z50 evaluations using `cpp_bin_float_50` type reduced to `double` with `static_cast<double(z50)` :",
null,
"",
null,
"As noted in the implementation section, the distribution of these differences is somewhat biased for Lambert W-1 and this might be reduced using a `double` Halley step at small runtime cost. But if you are seriously concerned to get really precise computations, the only way is using a higher precision type and then reduce to the desired type. Fortunately, Boost.Multiprecision makes this very easy to program, if much slower.\n\n##### Edge and Corner cases\n###### The W0 Branch\n\nThe domain of W0 is [-e-1, ∞). Numerically,\n\n• `lambert_w0(-1/e)` is exactly -1.\n• `lambert_w0(z)` for ```z < -1/e``` throws a `domain_error`, or returns `NaN` according to the policy.\n• `lambert_w0(std::numeric_limits<T>::infinity())` throws an `overflow_error`.\n\n(An infinite argument probably indicates that something has already gone wrong, but if it is desired to return infinity, this case should be handled before calling `lambert_w0`).\n\n###### W-1 Branch\n\nThe domain of W-1 is [-e-1, 0). Numerically,\n\n• `lambert_wm1(-1/e)` is exactly -1.\n• `lambert_wm1(0)` returns -∞ (or the nearest equivalent if ```std::has_infinity == false```).\n• `lambert_wm1(-std::numeric_limits<T>::min())` returns the maximum (most negative) possible value of Lambert W for the type T.\nFor example, for `double`: lambert_wm1(-2.2250738585072014e-308) = -714.96865723796634\nand for `float`: lambert_wm1(-1.17549435e-38) = -91.8567734\n• ```z < -std::numeric_limits<T>::min()```, means that z is zero or denormalized (if ```std::numeric_limits<T>::has_denorm_min == true```), for example: `r = lambert_wm1(-std::numeric_limits<double>::denorm_min());` and an overflow_error exception is thrown, and will give a message like:\n\nError in function boost::math::lambert_wm1<RealType>(<RealType>): Argument z = -4.9406564584124654e-324 is too small (z < -std::numeric_limits<T>::min so denormalized) for Lambert W-1 branch!\n\nDenormalized values are not supported for Lambert W-1 (because not all floating-point types denormalize), and anyway it only covers a tiny fraction of the range of possible z arguments values.\n\n##### Compilers\n\nThe `lambert_w.hpp` code has been shown to work on most C++98 compilers. (Apart from requiring C++11 extensions for using of `std::numeric_limits<>::max_digits10` in some diagnostics. Many old pre-c++11 compilers provide this extension but may require enabling to use, for example using b2/bjam the lambert_w examples use this command:\n\n```[ run lambert_w_basic_example.cpp : : : [ requires cxx11_numeric_limits ] ]\n```\n\nSee jamfile.v2.)\n\nFor details of which compilers are expected to work see lambert_w tests and examples in:\nBoost Test Summary report for master branch (used for latest release)\nBoost Test Summary report for latest developer branch.\n\nAs expected, debug mode is very much slower than release.\n\n###### Diagnostics Macros\n\nSeveral macros are provided to output diagnostic information (potentially much output). These can be statements, for example:\n\n`#define BOOST_MATH_INSTRUMENT_LAMBERT_W_TERMS`\n\nplaced before the `lambert_w` include statement\n\n`#include <boost/math/special_functions/lambert_w.hpp>`,\n\nor defined on the project compile command-line: `/DBOOST_MATH_INSTRUMENT_LAMBERT_W_TERMS`,\n\nor defined in a jamfile.v2: `<define>BOOST_MATH_INSTRUMENT_LAMBERT_W_TERMS`\n\n```// #define-able macros\nBOOST_MATH_INSTRUMENT_LAMBERT_W_HALLEY // Halley refinement diagnostics.\nBOOST_MATH_INSTRUMENT_LAMBERT_W_PRECISION // Precision.\nBOOST_MATH_INSTRUMENT_LAMBERT_WM1 // W1 branch diagnostics.\nBOOST_MATH_INSTRUMENT_LAMBERT_WM1_HALLEY // Halley refinement diagnostics only for W-1 branch.\nBOOST_MATH_INSTRUMENT_LAMBERT_WM1_TINY // K > 64, z > -1.0264389699511303e-26\nBOOST_MATH_INSTRUMENT_LAMBERT_WM1_LOOKUP // Show results from W-1 lookup table.\nBOOST_MATH_INSTRUMENT_LAMBERT_W_SCHROEDER // Schroeder refinement diagnostics.\nBOOST_MATH_INSTRUMENT_LAMBERT_W_TERMS // Number of terms used for near-singularity series.\nBOOST_MATH_INSTRUMENT_LAMBERT_W_SINGULARITY_SERIES // Show evaluation of series near branch singularity.\nBOOST_MATH_INSTRUMENT_LAMBERT_W_SMALL_Z_SERIES\nBOOST_MATH_INSTRUMENT_LAMBERT_W_SMALL_Z_SERIES_ITERATIONS // Show evaluation of series for small z.\n```\n##### Implementation\n\nThere are many previous implementations, each with increasing accuracy and/or speed. See references below.\n\nFor most of the range of z arguments, some initial approximation followed by a single refinement, often using Halley or similar method, gives a useful precision. For speed, several implementations avoid evaluation of a iteration test using the exponential function, estimating that a single refinement step will suffice, but these rarely get to the best result possible. To get a better precision, additional refinements, probably iterative, are needed for example, using Halley or Schröder methods.\n\nFor C++, the most precise results possible, closest to the nearest representable for the C++ type being used, it is usually necessary to use a higher precision type for intermediate computation, finally static-casting back to the smaller desired result type. This strategy is used by Maple and Wolfram Alpha, for example, using arbitrary precision arithmetic, and some of their high-precision values are used for testing this library. This method is also used to provide some Boost.Test values using Boost.Multiprecision, typically, a 50 decimal digit type like `cpp_bin_float_50` `static_cast` to a `float`, `double` or `long double` type.\n\nFor z argument values near the singularity and near zero, other approximations may be used, possibly followed by refinement or increasing number of series terms until a desired precision is achieved. At extreme arguments near to zero or the singularity at the branch point, even this fails and the only method to achieve a really close result is to cast from a higher precision type.\n\nIn practical applications, the increased computation required (often towards a thousand-fold slower and requiring much additional code for Boost.Multiprecision) is not justified and the algorithms here do not implement this. But because the Boost.Lambert_W algorithms has been tested using Boost.Multiprecision, users who require this can always easily achieve the nearest representation for fundamental (built-in) types - if the application justifies the very large extra computation cost.\n\n###### Evolution of this implementation\n\nOne compact real-only implementation was based on an algorithm by Thomas Luu, Thesis, University College London (2015), (see routine 11 on page 98 for his Lambert W algorithm) and his Halley refinement is used iteratively when required. A first implementation was based on Thomas Luu's code posted at Boost Trac #11027. It has been implemented from Luu's algorithm but templated on `RealType` parameter and result and handles both fundamental (built-in) types (```float, double, long double```), Boost.Multiprecision, and also has been tested successfully with a proposed fixed_point type.\n\nA first approximation was computed using the method of Barry et al (see references 5 & 6 below). This was extended to the widely used TOMS443 FORTRAN and C++ versions by John Burkardt using Schroeder refinement(s). (For users only requiring an accuracy of relative accuracy of 0.02%, Barry's function alone might suffice, but a better rational function approximation method has since been developed for this implementation).\n\nWe also considered using Newton-Raphson iteration method.\n\n```f(w) = w e^w -z = 0 // Luu equation 6.37\nf'(w) = e^w (1 + w), Wolfram alpha (d)/(dw)(f(w) = w exp(w) - z) = e^w (w + 1)\nif (f(w) / f'(w) -1 < tolerance\nw1 = w0 - (expw0 * (w0 + 1)); // Refine new Newton/Raphson estimate.\n```\n\nbut concluded that since the Newton-Raphson method takes typically 6 iterations to converge within tolerance, whereas Halley usually takes only 1 to 3 iterations to achieve an result within 1 Unit in the last place (ULP), so the Newton-Raphson method is unlikely to be quicker than the additional cost of computing the 2nd derivative for Halley's method.\n\nHalley refinement uses the simplified formulae obtained from Wolfram Alpha\n\n```[2(z exp(z)-w) d/dx (z exp(z)-w)] / [2 (d/dx (z exp(z)-w))^2 - (z exp(z)-w) d^2/dx^2 (z exp(z)-w)]\n```\n##### Implementing Compact Algorithms\n\nThe most compact algorithm can probably be implemented using the log approximation of Corless et al. followed by Halley iteration (but is also slowest and least precise near zero and near the branch singularity).\n\n##### Implementing Faster Algorithms\n\nMore recently, the Tosio Fukushima has developed an even faster algorithm, avoiding any transcendental function calls as these are necessarily expensive. The current implementation of Lambert W-1 is based on his algorithm starting with a translation from Fukushima's FORTRAN into C++ by Darko Veberic.\n\nMany applications of the Lambert W function make many repeated evaluations for Monte Carlo methods; for these applications speed is very important. Luu, and Chapeau-Blondeau and Monir provide typical usage examples.\n\nFukushima improves the important observation that much of the execution time of all previous iterative algorithms was spent evaluating transcendental functions, usually `exp`. He has put a lot of work into avoiding any slow transcendental functions by using lookup tables and bisection, finishing with a single Schroeder refinement, without any check on the final precision of the result (necessarily evaluating an expensive exponential).\n\nTheoretical and practical tests confirm that Fukushima's algorithm gives Lambert W estimates with a known small error bound (several Unit in the last place (ULP)) over nearly all the range of z argument.\n\nA mean difference was computed to express the typical error and is often about 0.5 epsilon, the theoretical minimum. Using the Boost.Math float_distance, we can also express this as the number of bits that are different from the nearest representable or 'exact' or 'best' value. The number and distribution of these few bits differences was studied by binning, including their sign. Bins for (signed) 0, 1, 2 and 3 and 4 bits proved suitable.\n\nHowever, though these give results within a few machine epsilon of the nearest representable result, they do not get as close as is very often possible with further refinement, nrealy always to within one or two machine epsilon.\n\nMore significantly, the evaluations of the sum of all signed differences using the Fukshima algorithm show a slight bias, being more likely to be a bit or few below the nearest representation than above; bias might have unwanted effects on some statistical computations.\n\nFukushima's method also does not cover the full range of z arguments of 'float' precision and above.\n\nFor this implementation of Lambert W0, John Maddock used the Boost.Math Remez algorithm method program to devise a rational function for several ranges of argument for the W0 branch of Lambert W function. These minimax rational approximations are combined for an algorithm that is both smaller and faster.\n\nSadly it has not proved practical to use the same Remez algorithm method for Lambert W-1 branch and so the Fukushima algorithm is retained for this branch.\n\nAn advantage of both minimax rational Remez algorithm approximations is that the distribution from the reference values is reasonably random and insignificantly biased.\n\nFor example, table below a test of Lambert W0 10000 values of argument covering the main range of possible values, 10000 comparisons from z = 0.0501 to 703, in 0.001 step factor 1.05 when module 7 == 0\n\nTable 8.73. Fukushima Lambert W0 and typical improvement from a single Halley step.\n\nMethod\n\nExact\n\nOne_bit\n\nTwo_bits\n\nFew_bits\n\ninexact\n\nbias\n\nSchroeder W0\n\n8804\n\n1154\n\n37\n\n5\n\n1243\n\n-1193\n\nafter Halley step\n\n9710\n\n288\n\n2\n\n0\n\n292\n\n22\n\nLambert W0 values computed using the Fukushima method with Schroeder refinement gave about 1/6 `lambert_w0` values that are one bit different from the 'best', and < 1% that are a few bits 'wrong'. If a Halley refinement step is added, only 1 in 30 are even one bit different, and only 2 two-bits 'wrong'.\n\nTable 8.74. Rational polynomial Lambert W0 and typical improvement from a single Halley step.\n\nMethod\n\nExact\n\nOne_bit\n\nTwo_bits\n\nFew_bits\n\ninexact\n\nbias\n\nrational/polynomial\n\n7135\n\n2863\n\n2\n\n0\n\n2867\n\n-59\n\nafter Halley step\n\n9724\n\n273\n\n3\n\n0\n\n279\n\n5\n\nWith the rational polynomial approximation method, there are a third one-bit from the best and none more than two-bits. Adding a Halley step (or iteration) reduces the number that are one-bit different from about a third down to one in 30; this is unavoidable 'computational noise'. An extra Halley step would double the runtime for a tiny gain and so is not chosen for this implementation, but remains a option, as detailed above.\n\nFor the Lambert W-1 branch, the Fukushima algorithm is used.\n\nTable 8.75. Lambert W-1 using Fukushima algorithm.\n\nMethod\n\nExact\n\nOne_bit\n\nTwo_bits\n\nFew_bits\n\ninexact\n\nbias\n\nFukushima W-1\n\n7167\n\n2704\n\n129\n\n0\n\n2962\n\n-160\n\nplus Halley step\n\n7379\n\n2529\n\n92\n\n0\n\n2713\n\n549\n\n###### Lookup tables\n\nFor speed during the bisection, Fukushima's algorithm computes lookup tables of powers of e and z for integral Lambert W. There are 64 elements in these tables. The FORTRAN version (and the C++ translation by Veberic) computed these (once) as `static` data. This is slower, may cause trouble with multithreading, and is slightly inaccurate because of rounding errors from repeated(64) multiplications.\n\nIn this implementation the array values have been computed using Boost.Multiprecision 50 decimal digit and output as C++ arrays 37 decimal digit ```long double``` literals using `max_digits10` precision\n\n```std::cout.precision(std::numeric_limits<cpp_bin_float_quad>::max_digits10);\n```\n\nThe arrays are as `const` and `constexpr` and `static` as possible (for the compiler version), using BOOST_STATIC_CONSTEXPR macro. (See lambert_w_lookup_table_generator.cpp The precision was chosen to ensure that if used as ```long double``` arrays, then the values output to lambert_w_lookup_table.ipp will be the nearest representable value for the type chose by a `typedef` in lambert_w.hpp.\n\n```typedef double lookup_t; // Type for lookup table (`double` or `float`, or even `long double`?)\n```\n\nThis is to allow for future use at higher precision, up to platforms that use 128-bit (hardware or software) for their ```long double``` type.\n\nThe accuracy of the tables was confirmed using Wolfram Alpha and agrees at the 37th decimal place, so ensuring that the value is exactly read into even 128-bit ```long double``` to the nearest representation.\n\n###### Higher precision\n\nFor types more precise than `double`, Fukushima reported that it was best to use the `double` estimate as a starting point, followed by refinement using Halley iterations or other methods; our experience confirms this.\n\nUsing Boost.Multiprecision it is simple to compute very high precision values of Lambert W at least to thousands of decimal digits over most of the range of z arguments.\n\nFor this reason, the lookup tables and bisection are only carried out at low precision, usually `double`, chosen by the ```typedef double lookup_t```. Unlike the FORTRAN version, the lookup tables of Lambert_W of integral values are precomputed as C++ static arrays of floating-point literals. The default is a `typedef` setting the type to `double`. To allow users to vary the precision from `float` to `long double` these are computed to 128-bit precision to ensure that even platforms with `long double` do not lose precision.\n\nThe FORTRAN version and translation only permits the z argument to be the largest items in these lookup arrays, ```wm0s = 3.99049```, producing an error message and returning `NaN`. So 64 is the largest possible value ever returned from the `lambert_w0` function. This is far from the `std::numeric_limits<>::max()` for even `float`s. Therefore this implementation uses an approximation or 'guess' and Halley's method to refine the result. Logarithmic approximation is discussed at length by R.M.Corless et al. (page 349). Here we use the first two terms of equation 4.19:\n\n```T lz = log(z);\nT llz = log(lz);\nguess = lz - llz + (llz / lz);\n```\n\nThis gives a useful precision suitable for Halley refinement.\n\nSimilarly, for Lambert W-1 branch, tiny values very near zero, W > 64 cannot be computed using the lookup table. For this region, an approximation followed by a few (usually 3) Halley refinements. See wm1_near_zero.\n\nFor the less well-behaved regions for Lambert W0 z arguments near zero, and near the branch singularity at -1/e, some series functions are used.\n\n###### Small values of argument z near zero\n\nWhen argument z is small and near zero, there is an efficient and accurate series evaluation method available (implemented in `lambert_w0_small_z`). There is no equivalent for the W-1 branch as this only covers argument `z < -1/e`. The cutoff used ```abs(z) < 0.05``` is as found by trial and error by Fukushima.\n\nCoefficients of the inverted series expansion of the Lambert W function around ```z = 0``` are computed following Fukushima using 17 terms of a Taylor series computed using Wolfram Mathematica with\n\n```InverseSeries[Series[z Exp[z],{z,0,17}]]\n```\n\nSee Tosio Fukushima, Journal of Computational and Applied Mathematics 244 (2013), page 86.\n\nTo provide higher precision constants (34 decimal digits) for types larger than `long double`,\n\n```InverseSeries[Series[z Exp[z],{z,0,34}]]\n```\n\nwere also computed, but for current hardware it was found that evaluating a `double` precision and then refining with Halley's method was quicker and more accurate.\n\nDecimal values of specifications for built-in floating-point types below are 21 digits precision == `std::numeric_limits<T>::max_digits10` for ```long double```.\n\nSpecializations for `lambert_w0_small_z` are provided for `float`, `double`, ```long double```, `float128` and for Boost.Multiprecision types.\n\nThe `tag_type` selection is based on the value `std::numeric_limits<T>::max_digits10` (and not on the floating-point type T). This distinguishes between `long double` types that commonly vary between 64 and 80-bits, and also compilers that have a `float` type using 64 bits and/or `long double` using 128-bits.\n\nAs noted in the implementation section above, it is only possible to ensure the nearest representable value by casting from a higher precision type, computed at very, very much greater cost.\n\nFor multiprecision types, first several terms of the series are tabulated and evaluated as a polynomial: (this will save us a bunch of expensive calls to `pow`). Then our series functor is initialized \"as if\" it had already reached term 18, enough evaluation of built-in 64-bit double and float (and 80-bit ```long double```) types. Finally the functor is called repeatedly to compute as many additional series terms as necessary to achive the desired precision, set from `get_epsilon` (or terminated by `evaluation_error` on reaching the set iteration limit `max_series_iterations`).\n\nA little more than one decimal digit of precision is gained by each additional series term. This allows computation of Lambert W near zero to at least 1000 decimal digit precision, given sufficient compute time.\n\n##### Argument z near the singularity at -1/e between branches W0 and W-1\n\nVariants of Function `lambert_w_singularity_series` are used to handle z arguments which are near to the singularity at ```z = -exp(-1) = -3.6787944``` where the branches W0 and W-1 join.\n\nT. Fukushima / Journal of Computational and Applied Mathematics 244 (2013) 77-89 describes using Wolfram Mathematica\n\n```InverseSeries\\[Series\\[sqrt\\[2(p Exp\\[1 + p\\] + 1)\\], {p,-1, 20}\\]\\]\n```\n\nto provide his Table 3, page 85.\n\nThis implementation used Wolfram Mathematica to obtain 40 series terms at 50 decimal digit precision\n\n```N\\[InverseSeries\\[Series\\[Sqrt\\[2(p Exp\\[1 + p\\] + 1)\\], { p,-1,40 }\\]\\], 50\\]\n\n-1+p-p^2/3+(11 p^3)/72-(43 p^4)/540+(769 p^5)/17280-(221 p^6)/8505+(680863 p^7)/43545600 ...\n```\n\nThese constants are computed at compile time for the full precision for any `RealType T` using the original rationals from Fukushima Table 3.\n\nLonger decimal digits strings are rationals pre-evaluated using Wolfram Mathematica. Some integer constants overflow, so largest size available is used, suffixed by `uLL`.\n\nAbove the 14th term, the rationals exceed the range of ```unsigned long long``` and are replaced by pre-computed decimal values at least 21 digits precision == `max_digits10` for `long double`.\n\nA macro `BOOST_MATH_TEST_VALUE` (defined in test_value.hpp) taking a decimal floating-point literal was used to allow testing with both built-in floating-point types like `double` which have contructors taking literal decimal values like `3.14`, and also multiprecision and other User-defined Types that only provide full-precision construction from decimal digit strings like `\"3.14\"`. (Construction of multiprecision types from built-in floating-point types only provides the precision of the built-in type, like `double`, only 17 decimal digits).\n\nTip",
null,
"Be exceeding careful not to silently lose precision by constructing multiprecision types from literal decimal types, usually `double`. Use decimal digit strings like \"3.1459\" instead. See examples.\n\nFukushima's implementation used 20 series terms; it was confirmed that using more terms does not usefully increase accuracy.\n\n###### Lambert W-1 arguments values very near zero.\n\nThe lookup tables of Fukushima have only 64 elements, so that the z argument nearest zero is -1.0264389699511303e-26, that corresponds to a maximum Lambert W-1 value of 64.0. Fukushima's implementation did not cater for z argument values that are smaller (nearer to zero), but this implementation adds code to accept smaller (but not denormalised) values of z. A crude approximation for these very small values is to take the exponent and multiply by ln ~= 2.3. We also tried the approximation first proposed by Corless et al. using ln(-z), (equation 4.19 page 349) and then tried improving by a 2nd term -ln(ln(-z)), and finally the ratio term -ln(ln(-z))/ln(-z).\n\nFor a z very close to z = -1.0264389699511303e-26 when W = 64, when effect of ln(ln(-z) term, and ratio L1/L2 is greatest, the possible 'guesses' are\n\n```z = -1.e-26, w = -64.02, guess = -64.0277, ln(-z) = -59.8672, ln(-ln(-z) = 4.0921, llz/lz = -0.0684\n```\n\nwhereas at the minimum (unnormalized) z\n\n```z = -2.2250e-308, w = -714.9, guess = -714.9687, ln(-z) = -708.3964, ln(-ln(-z) = 6.5630, llz/lz = -0.0092\n```\n\nAlthough the addition of the 3rd ratio term did not reduce the number of Halley iterations needed, it might allow return of a better low precision estimate without any Halley iterations. For the worst case near w = 64, the error in the 'guess' is 0.008, ratio 0.0001 or 1 in 10,000 digits 10 ~= 4. Two log evalutations are still needed, but is probably over an order of magnitude faster.\n\nHalley's method was then used to refine the estimate of Lambert W-1 from this guess. Experiments showed that although all approximations reached with Unit in the last place (ULP) of the closest representable value, the computational cost of the log functions was easily paid by far fewer iterations (typically from 8 down to 4 iterations for double or float).\n\n###### Halley refinement\n\nAfter obtaining a double approximation, for `double`, `long double` and `quad` 128-bit precision, a single iteration should suffice because Halley iteration should triple the precision with each step (as long as the function is well behaved - and it is), and since we have at least half of the bits correct already, one Halley step is ample to get to 128-bit precision.\n\n###### Lambert W Derivatives\n\nThe derivatives are computed using the formulae in Wikipedia.\n\n##### Testing\n\nInitial testing of the algorithm was done using a small number of spot tests.\n\nAfter it was established that the underlying algorithm (including unlimited Halley refinements with a tight terminating criterion) was correct, some tables of Lambert W values were computed using a 100 decimal digit precision Boost.Multiprecision `cpp_dec_float_100` type and saved as a C++ program that will initialise arrays of values of z arguments and lambert_W0 (`lambert_w_mp_high_values.ipp` and `lambert_w_mp_low_values.ipp` ).\n\n(A few of these pairs were checked against values computed by Wolfram Alpha to try to guard against mistakes; all those tested agreed to the penultimate decimal place, so they can be considered reliable to at least 98 decimal digits precision).\n\nA macro `BOOST_MATH_TEST_VALUE` was used to allow tests with any real type, both fundamental (built-in) types and Boost.Multiprecision. (This is necessary because fundamental (built-in) types have a constructor from floating-point literals like 3.1459F, 3.1459 or 3.1459L whereas Boost.Multiprecision types may lose precision unless constructed from decimal digits strings like \"3.1459\").\n\nThe 100-decimal digits precision pairs were then used to assess the precision of less-precise types, including Boost.Multiprecision `cpp_bin_float_quad` and `cpp_bin_float_50`. `static_cast`ing from the high precision types should give the closest representable value of the less-precise type; this is then be used to assess the precision of the Lambert W algorithm.\n\nTests using confirm that over nearly all the range of z arguments, nearly all estimates are the nearest representable value, a minority are within 1 Unit in the last place (ULP) and only a very few 2 ULP.",
null,
"",
null,
"For the range of z arguments over the range -0.35 to 0.5, a different algorithm is used, but the same technique of evaluating reference values using a Boost.Multiprecision `cpp_dec_float_100` was used. For extremely small z arguments, near zero, and those extremely near the singularity at the branch point, precision can be much lower, as might be expected.\n\nSee source at: lambert_w_simple_examples.cpp test_lambert_w.cpp contains routine tests using Boost.Test. lambert_w_errors_graph.cpp generating error graphs.\n\nA further method of testing over a wide range of argument z values was devised by Nick Thompson (cunningly also to test the recently written quadrature routines including Boost.Multiprecision !). These are definite integral formulas involving the W function that are exactly known constants, for example, LambertW0(1/(z²) == √(2π), see Definite Integrals. Some care was needed to avoid overflow and underflow as the integral function must evaluate to a finite result over the entire range.\n\n###### Other implementations\n\nThe Lambert W has also been discussed in a Boost thread.\n\nThis also gives link to a prototype version by which also gives complex results `(x < -exp(-1)`, about -0.367879). Balazs Cziraki 2016 Physicist, PhD student at Eotvos Lorand University, ELTE TTK Institute of Physics, Budapest. has also produced a prototype C++ library that can compute the Lambert W function for floating point and complex number types. This is not implemented here but might be completed in the future.\n\n##### Acknowledgements\n• Thanks to Wolfram for use of their invaluable online Wolfram Alpha service.\n• Thanks for Mark Chapman for performing offline Wolfram computations.\n##### References\n1. NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/4.13.F1.\n2. Lambert W Poster, R. M. Corless, G. H. Gonnet, D. E. G. Hare, D. J. Jeffery and D. E. Knuth, On the Lambert W function Advances in Computational Mathematics, Vol 5, (1996) pp 329-359.\n3. TOMS443, Andrew Barry, S. J. Barry, Patricia Culligan-Hensley, Algorithm 743: WAPR - A Fortran routine for calculating real values of the W-function,\nACM Transactions on Mathematical Software, Volume 21, Number 2, June 1995, pages 172-181.\nBISECT approximates the W function using bisection (GNU licence). Original FORTRAN77 version by Andrew Barry, S. J. Barry, Patricia Culligan-Hensley, this version by C++ version by John Burkardt.\n4. TOMS743 Fortran 90 (updated 2014).\n\nInitial guesses based on:\n\n1. R.M.Corless, G.H.Gonnet, D.E.G.Hare, D.J.Jeffrey, and D.E.Knuth, On the Lambert W function, Adv.Comput.Math., vol. 5, pp. 329 to 359, (1996).\n2. D.A. Barry, J.-Y. Parlange, L. Li, H. Prommer, C.J. Cunningham, and F. Stagnitti. Analytical approximations for real values of the Lambert W-function. Mathematics and Computers in Simulation, 53(1), 95-103 (2000).\n3. D.A. Barry, J.-Y. Parlange, L. Li, H. Prommer, C.J. Cunningham, and F. Stagnitti. Erratum to analytical approximations for real values of the Lambert W-function. Mathematics and Computers in Simulation, 59(6):543-543, 2002.\n4. C++ CUDA NVidia GPU C/C++ language support version of Luu algorithm, plog.\n5. Thomas Luu, Thesis, University College London (2015), see routine 11, page 98 for Lambert W algorithm.\n6. Having Fun with Lambert W(x) Function, Darko Veberic University of Nova Gorica, Slovenia IK, Forschungszentrum Karlsruhe, Germany, J. Stefan Institute, Ljubljana, Slovenia.\n7. François Chapeau-Blondeau and Abdelilah Monir, Numerical Evaluation of the Lambert W Function and Application to Generation of Generalized Gaussian Noise With Exponent 1/2, IEEE Transactions on Signal Processing, 50(9) (2002) 2160 - 2165.\n8. Toshio Fukushima, Precise and fast computation of Lambert W-functions without transcendental function evaluations, Journal of Computational and Applied Mathematics, 244 (2013) 77-89.\n9. T.C. Banwell and A. Jayakumar, Electronic Letter, Feb 2000, 36(4), pages 291-2. Exact analytical solution for current flow through diode with series resistance. https://doi.org/10.1049/el:20000301\n10. Princeton Companion to Applied Mathematics, 'The Lambert-W function', Section 1.3: Series and Generating Functions.\n11. Cleve Moler, Mathworks blog The Lambert W Function\n12. Digital Library of Mathematical Function, Lambert W function.\n Copyright © 2006-2019 Nikhar Agrawal, Anton Bikineev, Paul A. Bristow, Marco Guazzone, Christopher Kormanyos, Hubert Holin, Bruno Lalande, John Maddock, Jeremy Murphy, Matthew Pulver, Johan Råde, Gautam Sewani, Benjamin Sobotta, Nicholas Thompson, Thijs van den Berg, Daryle Walker and Xiaogang Zhang Distributed under the Boost Software License, Version 1.0. (See accompanying file LICENSE_1_0.txt or copy at http://www.boost.org/LICENSE_1_0.txt)"
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https://mhktricks.org/comsol-multiphysics-crack/ | [
"Updated\n19 January 2022\nSize\n6 GB\n\n## Description\n\nComsol Multiphysics is a program for finite element calculations of complex scientific and technical problems. The COMSOL Multiphysics package allows you to simulate almost all physical processes that are described by partial differential equations. The program contains a variety of solvers that will help you quickly solve even the most complex problems, and the simple structure of the application ensures ease and flexibility of use. The solution of any problem is based on the numerical solution of equations in partial derivatives by the finite element method. The range of tasks that can be modeled in the program is extremely wide.\n\nA set of special modules in the program covers almost all areas of applications of partial differential equations.\nCOMSOL Multiphysics (Femlab) is a simulation package that solves systems of nonlinear partial differential equations using the finite element method in one, two, and three dimensions. It allows you to solve problems from the field of electromagnetism, the theory of elasticity, the dynamics of liquids and gases, and chemical gas dynamics. Femlab also makes it possible to solve the problem both in a mathematical formulation (in the form of a system of equations) and in a physical one (selection of a physical model, for example, a model of the diffusion process). Of course, in any case, a system of equations will be solved, and the difference lies only in the ability to use physical systems of units and physical terminology. In the so-called physical mode of operation, it is also possible to use predetermined equations for most of the phenomena that take place in science and technology,"
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https://scilearn.sydney.edu.au/fychemistry/Questions/electrolysis.htm | [
"Electrolysis Your feedback on these self-help problems is appreciated. Click here to send an e-mail. Shortcut to Questions Q: 1 2 3 4 5 6 7 8 1 How many (a) faradays, (b) coulombs are required to deposit 0.587 g of Ni from a solution of Ni2+ ?",
null,
"2 From a dilute solution of potassium tricyanocuprate(I), K2[Cu(CN)3], a uniform current deposits 0.586 g of copper in 108 minutes. What volume of hydrogen measured over water at 296 K and 746.0 mmHg would the same current liberate from dilute sulfuric acid in 81.0 minutes? (The equilibrium vapour pressure of water at 296 K = 20.9 mmHg.)",
null,
"3 In the electrolysis of 0.100 M copper(II) sulfate solution, copper is deposited. How long would it be necessary to pass a 0.100 ampere current in order to deposit all of the copper from 1.00 litre of this solution?",
null,
"4 Three electrolysis cells are connected in series. They contain, respectively, solutions of copper(II) nitrate, silver nitrate, and chromium(III) sulfate. If 1.00 g of copper is electrochemically deposited in the first cell, calculate the mass of silver and chromium deposited in the other cells.",
null,
"5 A constant current of 3.7 milliampere is passed through molten sodium chloride for 9.0 minutes. The sodium produced is allowed to react with water (500 mL). What is the pH of the resulting solution?",
null,
"6 A constant current of 0.126 ampere was passed through a dilute sulfuric acid solution for 149 minutes. The hydrogen produced was collected over mercury and the pressure adjusted to that of the surroundings (751 mmHg). The room temperature was 291 K. What was the volume of hydrogen produced?",
null,
"7 One of the experimental ways of determining the Avogadro constant is by electrolysis. A current of 1.00 ampere is passed for 1.00 hour through a solution of potassium iodide. The weight of iodine liberated at the anode is 4.7135 g and there is no other anodic product. If the charge on the electron is 1.602 x 10-19 coulomb, find from this set of data the Avogadro constant, NA, correct to three significant figures. The atomic weight of iodine is 126.90.",
null,
"8",
null,
"Two electrolytic cells each containing two platinum electrodes were connected in series and a current of 0.500 ampere was passed for 16 minutes and 5 seconds. Cell 1 contained 0.100 M AgNO3 and cell 2 contained 0.100 M CuSO4. Calculate the following: (a) The change in mass of the platinum cathode in cell 1. (b) The volume of gas at 1.00 atmosphere and 273 K liberated at the platinum anode in cell 1. (c) The volume of gas at 1.00 atmosphere and 273 K liberated at the platinum anode in cell 2. (d) The change in mass of the platinum cathode in cell 2.",
null,
"Electrolysis (Answers) 1 The reaction equation is Ni2+ + 2e- → Ni From the stoichiometry of this equation, one mole of Ni produced requires the passage of two moles of electrons in the electrolysis. Mass of nickel produced = 0.587 g ∴ moles of Ni = mass/atomic wt = 0.587/58.71 = 1.00 x 10-2 mol and moles of electrons required = 2 x 1.00 x 10-2 = 2.00 x 10-2 mol. One faraday of electrical charge (F) is one mole of electrons (i.e. 6.02 x 1023 electrons - note that only the Avogadro number definition can apply to moles of electrons). ∴ number of faradays required = 2.00 x 10-2 F One coulomb of charge (C) is the total charge involved when a current of one amp flows for one second. i.e. 1 amp = 1 coulomb flowing for 1 second. By experiment, one faraday is found to contain 96487 C per mole of electrons, the Faraday constant, F. ∴ coulombs in 2.00 x 10-2 faradays = 96487 x 2.00 x 10-2 = 1.93 x 103 C",
null,
"2 The first task in this question is to calculate from the amount of copper released, the current used in the electrolysis reactions. Then in the second part, knowing the current and the time it flows, the moles of electrons supplied to the H+ ions can be deduced. Then the moles of H2 released and its volume under the stated conditions can be obtained. From the first electrolysis, deduce the moles of electrons: The reaction equation is Cu+ + e- → Cu from which, one mole of electrons releases one mole of Cu. Mass of copper = 0.586 g ∴ moles of Cu = 0.586 / 63.55 = 9.221 x 10-3 mol and moles of electrons also = 9.221 x 10-3 mol. Convert moles of electrons to coulombs: Q = F x moles of electrons = 96487 x 9.221 x 10-3 C = 8.897 x 102 C From the coulombs of charge, calculate the current: By the definition of the coulomb (see Q 1 answer), charge in coulombs = current in amps x time in seconds or Q = I x t ∴ I = Q / t = 8.897 x 102 / (108 x 60.0) =0.1373 amps Note that time must be expressed in seconds as required by the definition of the coulomb. Calculate the moles of electrons this current provides in 81.0 minutes: From Q = I x t, in 81.0 minutes (= 81.0 x 60.0 s), Q = 0.1373 x 81.0 x 60.0 = 6.673 x 102 C. Converting coulombs to moles of electrons, moles of electrons = coulombs / F = 6.673 x 102 / 96487 = 6.916 x 10-3 mol. Calculate the moles of H2 released: The reaction equation is 2H+ + 2e- → H2 from which, one mole of H2 formed requires two moles of electrons. ∴ moles of H2 = ½ x 6.916 x 10-3 = 3.458 x 10-3 mol Calculate the volume of hydrogen gas formed: From the ideal gas equation PV = nRT, ∴ V = nRT / P n = 3.458 x 10-3 mol R = 8.314 J K-1 mol-1 T = 296 K P = (746.0 - 20.9) mmHg = 725.1 mmHg or (725.1 / 760.0) atm or (725.1 / 760.0) x 101.3 kPa = 96.65 kPa. (Note that the total pressure must be reduced by the equilibrium vapour pressure of the water over which the hydrogen is collected in order to obtain the true pressure of hydrogen in the container.) Substituting this data, V = (3.458 x 10-3 x 8.314 x 296) / 96.65 = 8.80 x 10-2 L",
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"3 The reaction equation is Cu2+ + 2e- → Cu From the stoichiometry of this equation, one mole of Cu2+ requires the passage of two moles of electrons in the electrolysis. Moles of Cu2+ in 1.00 L of 0.100 M solution = 0.100 mol. ∴ moles of electrons required = 0.200 mol or 0.200 F of charge. Q = 96487 x F = 96487 x 0.200 coulombs = 19297 C. Q = I x t for Q in coulombs, I in amps and t in seconds I = 0.100 amps ∴ t = Q / I = 19297 / 0.100 = 1.93 x 105 seconds or 1.93 x 105 / (60.0 x 60.0) hours = 53.6 hours.",
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"4 The reaction equations are Cu2+ + 2e- → Cu Ag+ + e- → Ag Cr3+ + 3e- → Cr From the stoichiometry of these equations, one mole of electrons would deposit ½ mole of Cu in the electrolysis. As the cells are in series, the same current flows through all three cells and one mole of electrons would produce one mole of Ag and 1/3 mole of Cr from their solutions respectively. The moles of electrons used = 2 x moles of Cu deposited. Moles of Cu deposited = 1.00 / 63.55 = 1.574 x 10-2 mol, so moles of electrons passed = 2 x 1.574 x 10-2 = 3.148 x 10-2 mol. In the silver cell: Moles of Ag deposited = moles of electrons passed = 3.148 x 10-2 mol Mass of silver deposited = 3.148 x 10-2 x 107.87 g = 3.40 g In the chromium cell: Moles of Cr deposited = (1/3) x moles of electrons passed = (3.148 x 10-2) / 3 = 1.049 x 10-2 mol Mass of chromium deposited = 1.049 x 10-2 x 52.00 g = 0.546 g",
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"5 The reaction equation is Na+ + e- → Na From the stoichiometry of this equation, one mole of Na deposited requires the passage of one mole of electrons in the electrolysis. The sodium deposited is then reacted with water according to the equation Na + H2O → Na+ + OH- + ½H2 From this equation, one mole of OH- ions in the solution is formed from one mole of Na which in turn was deposited by the passage of one mole of electrons. ∴ moles of OH- = moles of electrons passed Moles of electrons passed = Q / F = I x t / F I = 3.7 ma = 3.7 x 10-3 amps (note that the units of current must be amps) t = 9.0 minutes = 9.0 x 60.0 seconds ∴ moles of electrons = (3.7 x 10-3 x 9.0 x 60.0) / 96487 mol = 2.07 x 10-5 mol. The solution contains 2.07 x 10-5 mol of OH- ions in 500 mL of water. ∴ [OH-] = (2.07 x 10-5) / 0.500 M = 4.14 x 10-5 M pOH = -log[OH-] = 4.38 pH = 14 00 - pOH = 9.62",
null,
"6 The reaction equation is 2H+ + 2e- → H2 From the stoichiometry of this equation, one mole of H2 produced requires the passage of two moles of electrons in the electrolysis. Calculate the moles of electrons passed. Moles of electrons = I x t / F I = 0.126 amps t = 149 x 60 seconds = 8940 seconds ∴ moles of electrons = 0.126 x 8940 / 96487 = 1.167 x 10-2 mol. Calculate the moles of H2 produced. From the equation above, moles of H2 = ½ x moles of electrons = (1.167 x 10-2) / 2 = 5.84 x 10-3 mol Calculate the volume of hydrogen produced. Use the ideal gas equation rewritten as V = nRT / P where n = 5.84 x 10-3 mol R = 8.314 JK-1mol-1 T = 291 K P = 751 mmHg = 751 / 760 atm = (751 / 760) x 101.3 kPa = 100.1 kPa [Note that there is no correction needed for the vapour pressure of mercury as it is negligible] ∴ V = (5.84 x 10-3 x 8.314 x 291) / 100.1 = 0.141 L",
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"7 The reaction equation is 2I- → I2 + 2e- From the stoichiometry of this equation, one mole of I2 produced requires the passage of two moles of electrons in the electrolysis. Calculate the moles of I2 formed and thus moles of electrons passed. Moles of I2 = mass of I2 / GFW of I2 = 4.7135 / (2 x 126.90) = 1.8572 x 10-2 mol ∴ moles of electrons = 2 x 1.8572 x 10-2 = 3.7143 x 10-2 mol Calculate the total charge passed and thus the number of electrons. Q = I x t I = 1.00 amp t = 1.00 hour = 1.00 x 60.00 x 60.00 = 3600 sec (note the units of time must be seconds) ∴ Q = 1.00 x 3600 = 3600 C Given that the charge on one electron = 1.602 x 10-19 C, ∴ number of electrons passed = 3600 / (1.602 x 10-19) = 2.247 x 1022 electrons Thus deduce the number of electrons in one mole of electrons. As 2.247 x 1022 electrons = 3.7143 x 10-2 mol, then 1.00 mol = (2.247 x 1022) / (3.7143 x 10-2) electrons = 6.05 x 1023 electrons. ∴ NA = 6.05 x 1023.",
null,
"8 Terminology: Cathode is where reduction occurs and is ∴ the negative electrode Anode is where oxidation occurs and is ∴ the positive electrode. Cell diagram",
null,
"Set out the cell reactions for both cells. The reduction potentials for both Ag+ and Cu2+ are larger than that for H+, so Ag and Cu will be deposited at the cathodes before H+ is reduced to H2. The cell reactions are as follows. Cell 1 cathode: 2Ag+ + 2e- → 2Ag Cell 1 anode: H2O → 2H+ + ½O2(g) + 2e- Cell 2 cathode: Cu2+ + 2e- → Cu Cell 2 anode: H2O → 2H+ + ½O2(g) + 2e- Calculate the moles of electrons passed. Moles of electrons = I x t / F I = 0.500 amp t = 16 min 5 sec = (16 x 60) + 5 sec = 965 sec ∴ moles of electrons = (0.500 x 965) / 96487 = 5.00 x 10-3 mole Using the stoichiometry set out above, deduce the quantities of products. (a) Cell 1 cathode: In cell 1, moles of Ag formed = moles of electrons passed Mass of silver = moles x atomic wt = 5.00 x 10-3 x 107.87 g = 0.539 g (increase in mass of electrode). (b) Cell 1 anode: In cell 1, moles of O2 released = ¼ x moles of electrons passed = (5.00 x 10-3) / 4 mole = 1.25 x 10-3 mol Volume of oxygen released = nRT / P T = 273 K P = 1.00 atm = 101.3 kPa R = 8.314 ∴ V = (1.25 x 10-3 x 8.314 x 273) / 101.3 = 0.0280 L or 28.0 mL (c) Cell 2 anode: In cell 2, the same volume of oxygen is liberated at the anode as the same current passes through both cells. (d) Cell 2 cathode: In cell 2, moles of Cu formed = ½ x moles of electrons passed = 2.50 x 10-3 mol. Mass of copper = moles x atomic wt = 2.50 x 10-3 x 63.55 g = 0.159 g (increase in mass of electrode).",
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https://mechanics.stackexchange.com/questions/4411/does-torque-decrease-mathematically | [
"# Does torque decrease mathematically?\n\nThere are 2 cars, car A and car B.\nCar B is 20% heavier than car A.\nWe all know that this means that these 2 cars will have same power if car B at the same RPM (revolution) has 20% more Nm.\n\nCan the torque be mathematically equated?\n\nFor example, if car A has torque of 100/1500 and car B has torque of 150/1750. Can we deduct the 2nd value (150/1750) to `x / 1500`? Is this the mathematically correct way to determine the torque decrease?\n\nIf I did not write this clearly, please ask me to clarify. This data is very important to me and I was unable to find this info on the Net.\n\n• What is the reason you are trying to do this calculation?Are you comparing vehicles? – mikes Oct 2 '12 at 0:51\n• I don't get what you're asking? Neither power nor torque bear any relationship to the vehicle weight, hence why we have a \"power to weight ratio\" so we can more realistically compare two vehicles. – Nick C Oct 2 '12 at 12:32\n• @mikes yes, trying to compare cars – salone Oct 2 '12 at 13:52\n• @nick heavier car means more resistance means that heavier car having the same torque as lighter car will be slower, right?! ;) – salone Oct 2 '12 at 13:53\n• Not necessarily, it would depend on things like the gear ratios, the size of the wheels, the torque curve of the engine and quite a few other factors... Torque as a number is purely a measure of how much turning force the engine can produce. It also depends on what you mean by faster - top speed or acceleration? – Nick C Oct 2 '12 at 15:20\n\nI believe you are confusing torque (TQ) and horsepower (HP) terms here. HP is a mathematical computation based on TQ. You can derive a HP number by increasing the rotational speed up/down depending on what you want to achieve. Weight of a vehicle will have no affect on either. As mentioned in the comments, TQ output (at the wheels) is greatly affected by tire size, transmission gearing, axle ratio, and frictional losses. Because TQ output is affected by these, so is HP.\n\nTo figure out what HP is, you use the following equation:\n\n``````P=TW\n``````\n\nWhere:\n\n• P=Power\n• T=Torque\n• W=Angular Momentum\n\nSince we are talking about vehicles, we need to modify a couple of things here to get to HP. When using other units or if the speed is in revolutions per unit time rather than radians, a conversion factor has to be included. When torque is in pound-foot units, rotational speed is in rpm and power is required in horsepower, we change the equation to look like this:\n\n``````HP=(TQ x RPM)/5252\n``````\n\nWhere:\n\n• HP = Horsepower\n• TQ = Torque\n• RPM = Rotations Per Minute (or engine speed)\n• 5252 is a constant which allows you to convert for foot pounds (ft lbs) measure)\n\nLet's throw some numbers into the equation and see how HP is affected by different factors. For this example, we will use the following:\n\n``````TQ = 300\nRPM = 3000\nHP = (300 x 3000) / 5252\nHP = (900000) / 5252\nHP = 171.36 @ 3000 RPM\n``````\n\nWhat happens when you increase the RPM?\n\n``````TQ = 300\nRPM = 5000\nHP = (300 x 5000) / 5252\nHP = (1500000) / 5252\nHP = 285.61 @ 5000 RPM\n``````\n\nThis assumes the torque value has maintained the same level (300 ft lbs), yet the HP has increased by over 100HP. Remember that HP is a measure of work over time. The work is the TQ, while the time period is the RPM. As you increase speed of an engine, the work being accomplished increases as well. Conversely, if we increase TQ at a given RPM range, we will also see a corresponding increase in HP.\n\n``````TQ = 400\nRPM = 3000\nHP = (400 x 3000) / 5252\nHP = (1200000) / 5252\nHP = 228.48 @ 3000 RPM\n``````\n\nAs shown, as TQ increases at a given speed, so does HP, meaning that the amount of work has also increased, but the the speed within which it occurs is the same.\n\nIt is interesting to note, that due to the constant of 5252, if you look at a graph of TQ and HP output, you will notice the two will be equal at 5252 RPM. If you see a graph depicting the HP/TQ outputs of an engine which do not cross at this point, you know it is a bogus chart (which happens more often than you'd think).\n\nHow is torque affected by things as I mentioned above? If you put an engine on a stand and attach it to an engine dynamometer, you would be measuring the output at 1:1. What the engine puts out is what you get. No mechanical or mathematical gimmickry going on here. If, however, we put the vehicle on a roller or chassis dynamometer, there is a lot of other factors going one which affects the outcome. First modifier is the transmission.\n\nA recent transmission introduced to the automotive is the T-56 Magnum, produced by Tremec. The gear ratios for this transmission are as follows:\n\n``````1st gear - 2.66:1\n2nd gear - 1.78:1\n3rd gear - 1.30:1\n4th gear - 1.00:1\n5th gear - 0.80:1\n6th gear - 0.63:1\n``````\n\nThe purpose of the transmission is to give a vehicle more TQ output by using mechanical advantage. If we were to measure TQ output at the back of the transmission, you'd expect to take the input TQ coming from the engine and multiply it by the gear ration of the first gear. How it works, as you multiply the speed of the shaft, you decrease the torque output. The converse is true as well. In the case of the first gear, given an engine speed (input) of 3000 RPM and TQ of 300 ft-lbs, the output is modified to ~1128 RPM and your TQ has climbed to 798 ft-lbs. Remember, this is a generalization, as there is a modification for frictional losses, which occurs due to drag from the bearings, bushing, seals, and to a certain extent by pushing tranny fluid around (tranny fluid actually reduces a lot of drag from the bearings and bushing, though, so a whole different dynamic).\n\nThe next modifier is the rear differential. Let's use a rear differential ratio of 3.50:1 (this is not a real world ratio as I know it, just using the number to make things easier). Our output of 1128RPM and 798 ft-lbs, now becomes ~322RPM and 2793 ft-lbs, respectively. Mind you, all of this assumes that the entire drive train is spinning, as in the clutch is engaged and fully gripping.\n\nThe only modifier we have left, is the tire height. Because the tire/wheel is directly attached to the axle shaft, there is a 1:1 ratio there, which means the RPM will not change. What does change is the applied TQ. TQ is a twisting force. If you think of the tire as a lever (which it actually is, just a round one), as you get further away from the twisting force, the more torque it takes to move the end of the lever. Normally, when we think of the application of TQ, we are thinking of it the other way around, that being we apply a longer lever and it takes less force to create the TQ at the fulcrum (pivot point) of the lever. In our fictional example we are running with here, we have to look at it the other way around. That being, we are applying TQ at the fulcrum to move the end of the lever. If the end of the lever (the radius of the tire) is 12\", we would be applying 2793 pounds of force at the end of that lever. Most tires are not, however, 24\" tall.\n\nAs an example, we will use the Hankook Ventus R-S3 which is 25.6\" tall. 1/2 of this diameter gives us a radius of 12.8\". Mathematically, this would would be 1.0667'. Modifying our final TQ output of 2793 by this equates to 2618 lbs of force being applied at the contact patch of the tire ... but not quite.\n\nThe last main part of the equation is frictional losses. As I alluded to, every bearing, every bushing, every seal, ever gear ... all of it, sustains frictional losses. As a rule of thumb, the general thought is that 18-20% of the force coming out of the engine is gone by the time it reaches the contact patch of the tire at the ground. Our example of 2618 lbs would then be reduced to (using 20% for ease), would be ~2094.4 lbs of force.\n\nThink now, of the difference of the TQ output would be if this hypothetical vehicle was sporting a rear end gear ratio of 4.0:1 (again, this is hypothetical)? Our transmission output of 1128RPM and 798 ft-lbs, now becomes 282RPM and 3192 ft-lbs. That is an increase of nearly 400 ft-lbs of torque. Wow, why don't we run a 5.0:1 ratio then and get 226RPM and 3990 ft-lbs, then? The problem, as you are probably guessing, is that we are running out of RPM. This affects the top speed of the vehicle. If we kept the vehicle in 1st gear alone, our top speed would be (using the tires above and a theoretical engine redline of 6800) 53.82mph @ 3.50:1, 47.16mph @ 4.00:1, and 37.74mph @ 5.00:1. Even in 4th gear in our T-56, with a 1:1 ratio, the top speed of our fictitious vehicle would only be 100.38mph if the rear end gearing were at 5.00:1. In racing terms, this is slow. There are two ways we can affect this. One way being shifting into a higher gear. We are lucky with our transmission which has two more forward gears, both of which are overdrive. The 6th, or final gear in our transmission is 0.63:1. This extends our theoretical top speed to ~159mph, which is much better. The second way we can extend our top speed is by making the tires taller (but remember, this affects our effective torque).\n\nWhat all this comes down to, is finding the combination which works best for the needs of the car. How does this answer the question. I'm hoping this gives a clue as to how torque, running through the gears, affects what the output to the ground is. The only effect weight has in any of this is it takes more torque to get more weight moving to the same speed in the same amount of time.\n\nThe short answer is yes, you can do it that way as a quick reference. That being said, figuring out which car is \"better\" depends on what you want to do with it. If it's going to be a towing vehicle or long-distance family transporter, Torque/ton is what you're concerned with. If you want a traffic light racer, you'll have to look at horsepower/ton.\n\nThe reasoning: A Mazda Drifter 2500TDi Pickup may have 250Nm+ of torque and only 80kW of horsepower, while a Vauxhall Astra Turbo has 250Nm and 147kW.Their weights are roughly equal, but the Astra has about 70kW more horsepower. This means that while the pickup is able to move just as heavy a load as the Astra, it can't be considered a \"fast\" car because it will take ages to get up to speed (0 to 60 takes 16 seconds). Conversely, the Astra accelerates more than twice as fast (0 to 60 in 7.5 seconds). And they have similar flat torque curve profiles.\n\nTorque is for heavy lifting, horsepower is for going fast. Ideally you want both, but you can settle for one depending on your need.\n\nMathematically torque of a car or any moving or rolling body which has angular velocity can be described by a equation . Let,\n\n1. Torque = T\n\n2. Angular Acceleration = a\n\n3. Moment of Inertia = I ( A Constant Value In General For A Specific Substance Which Is The Resisting Force).\n\nThen , by the basic laws of physics and mechanics, the relation between torque T and the angular acceleration a is a proportional relationship mathematically and it can be written as following :",
null,
"Which means, more torque produces more acceleration. But moment of inertia is a vital fact here which depends on some factors.\n\nNow, we know the angular velocity, acceleration and torque of any certain object , let it be a car. The question is about the moment of inertia. The moment of inertia of any moving body like car or motorcycle is directly proportional to its mass . It increases as the mass is moved further from the axis of rotation. The fact depends on mass distribution of the object.The same body can have different moments of inertia If we consider different axis of rotation. This Is The Mathematical Explanation of your question. For calculating it directly and easily you can read this article which includes my research work on motorbike safety for the motorcycle blog named bikebd. All the terms about Horsepower, torque, momentum, force, resisting forces are briefly described here with calculation. I think, you'll get a more clear conception about your question here."
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https://quorumlanguage.com/Libraries/Compute/MatrixTransform/OrthonormalTriangularDecomposition.html | [
"## Libraries.Compute.MatrixTransform.OrthonormalTriangularDecomposition Documentation\n\nThis class calculates an orthonormal matrix and an upper triangular matrix from a given matrix. Classically, this is called a \"QR\" decomposition, which is referencing the orthonormal (Q) and the upper triangular (R). There are multiple ways to conduct this decomposition and this class uses Householder reflectors, which is a numerically stable way to conduct the calculation. This class is a port from the Apache Commons Linear algebra library, which itself is adapted from the JAMA library. In our class, we changed the names and way the programmer interacts with the class, but the primary calculations are very similar. More information on QR decomposition can be found at MathWorld or Wikipedia.\n\nExample Code\n\n``````:\n\nuse Libraries.Compute.Matrix\nuse Libraries.Compute.MatrixTransform.OrthonormalTriangularDecomposition\n\nMatrix matrix\nmatrix:SetSize(3,3)\nmatrix:Set(0,0,12.0)\nmatrix:Set(1,0,6)\nmatrix:Set(2,0,-4)\n\nmatrix:Set(0,1,-51)\nmatrix:Set(1,1,167)\nmatrix:Set(2,1,24)\n\nmatrix:Set(0,2,4)\nmatrix:Set(1,2,-68)\nmatrix:Set(2,2,-41)\n\nOrthonormalTriangularDecomposition decomp\ndecomp:Calculate(matrix)\n\nMatrix value = decomp:GetResult()\noutput value:ToText()``````\n\nInherits from: Libraries.Language.Object\n\n## Summary\n\n### Actions Summary Table\n\nActionsDescription\nCalculate(Libraries.Compute.Matrix matrix)This action does the decomposition and stores the matrices as state inside of this class.\nCompare(Libraries.Language.Object object)This action compares two object hash codes and returns an integer.\nEquals(Libraries.Language.Object object)This action determines if two objects are equal based on their hash code values.\nGetHashCode()This action gets the hash code for an object.\nGetOrthonormalMatrix()This action gets the orthonormal matrix.\nGetOrthonormalTransposedMatrix()This action gets the orthonormal matrix.\nGetResult()This action gets the orthonormal matrix.\nGetUpperTriangularMatrix()This action returns the upper triangular matrix.\nSolve(Libraries.Compute.Vector vector)fill it with zero\n\n## Actions Documentation\n\n### Calculate(Libraries.Compute.Matrix matrix)\n\nThis action does the decomposition and stores the matrices as state inside of this class. Thus, the matrices can then be copied, stored, or used as desired. To obtain the orthonormal matrix (Q), we call GetOrthonormalMatrix, and to get the upper triangular matrix (R), we call GetUpperTriangularMatrix.\n\nExample Code\n\n``````:\n\nuse Libraries.Compute.Matrix\nuse Libraries.Compute.MatrixTransform.OrthonormalTriangularDecomposition\n\nMatrix matrix\nmatrix:SetSize(3,3)\nmatrix:Set(0,0,12.0)\nmatrix:Set(1,0,6)\nmatrix:Set(2,0,-4)\n\nmatrix:Set(0,1,-51)\nmatrix:Set(1,1,167)\nmatrix:Set(2,1,24)\n\nmatrix:Set(0,2,4)\nmatrix:Set(1,2,-68)\nmatrix:Set(2,2,-41)\n\nOrthonormalTriangularDecomposition decomp\ndecomp:Calculate(matrix)\n\nMatrix value = decomp:GetResult()\noutput value:ToText()``````\n\n### Compare(Libraries.Language.Object object)\n\nThis action compares two object hash codes and returns an integer. The result is larger if this hash code is larger than the object passed as a parameter, smaller, or equal. In this case, -1 means smaller, 0 means equal, and 1 means larger. This action was changed in Quorum 7 to return an integer, instead of a CompareResult object, because the previous implementation was causing efficiency issues.\n\nExample Code\n\n``````Object o\nObject t\ninteger result = o:Compare(t) //1 (larger), 0 (equal), or -1 (smaller)``````\n\n#### Return\n\ninteger: The Compare result, Smaller, Equal, or Larger.\n\n### Equals(Libraries.Language.Object object)\n\nThis action determines if two objects are equal based on their hash code values.\n\nExample Code\n\n``````use Libraries.Language.Object\nuse Libraries.Language.Types.Text\nObject o\nText t\nboolean result = o:Equals(t)``````\n\n#### Return\n\nboolean: True if the hash codes are equal and false if they are not equal.\n\n### GetHashCode()\n\nThis action gets the hash code for an object.\n\nExample Code\n\n``````Object o\ninteger hash = o:GetHashCode()``````\n\n#### Return\n\ninteger: The integer hash code of the object.\n\n### GetOrthonormalMatrix()\n\nThis action gets the orthonormal matrix.\n\n### GetOrthonormalTransposedMatrix()\n\nThis action gets the orthonormal matrix.\n\n### GetResult()\n\nThis action gets the orthonormal matrix.\n\n### GetUpperTriangularMatrix()\n\nThis action returns the upper triangular matrix.\n\n### Solve(Libraries.Compute.Vector vector)\n\nfill it with zero"
] | [
null
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https://export.arxiv.org/abs/2204.13148?context=math.CO | [
"math.CO\n\n# Title: Transitivity on subclasses of bipartite graphs\n\nAbstract: Let $G=(V, E)$ be a graph where $V$ and $E$ are the vertex and edge set, respectively. For two disjoint subsets $A$ and $B$, we say $A$ dominates $B$ if every vertex of $B$ is adjacent to at least one vertex of $A$. A vertex partition $\\pi = \\{V_1, V_2, \\ldots, V_k\\}$ of $G$ is called a \\emph{transitive $k$-partition} if $V_i$ dominates $V_j$ for all $i,j$ where $1\\leq i<j\\leq k$. The maximum integer $k$ for which the above partition exists is called \\emph{transitivity} of $G$ and it is denoted by $Tr(G)$. The \\textsc{Maximum Transitivity Problem} is to find a transitive partition of a given graph with the maximum number of partitions. It was known that the decision version of \\textsc{Maximum Transitivity Problem} is NP-complete for general graphs, which was proved by Hedetniemi et al. [Iterated colorings of graphs, \\emph{Discrete Mathematics}, 278, 2004]. This paper first strengthens the NP-completeness result by showing that this problem remains NP-complete for perfect elimination bipartite graphs. On the other hand, we propose a linear-time algorithm for finding the transitivity of a given bipartite chain graph. We then characterize graphs with transitivity at least $t$ for any integer $t$. This result answers two open questions posed by J. T. Hedetniemi and S. T. Hedetniemi [The transitivity of a graph, \\emph{J. Combin. Math. Combin. Comput}, 104, 2018].\n Subjects: Discrete Mathematics (cs.DM); Combinatorics (math.CO) Cite as: arXiv:2204.13148 [cs.DM] (or arXiv:2204.13148v1 [cs.DM] for this version)\n\n## Submission history\n\nFrom: Subhabrata Paul [view email]\n[v1] Wed, 27 Apr 2022 19:03:03 GMT (253kb,D)\n\nLink back to: arXiv, form interface, contact."
] | [
null
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https://books.google.ie/books?qtid=8f7e78c6&dq=editions:UOM39015065618988&lr=&id=23gAAAAAMAAJ&output=html_text&sa=N&start=50 | [
"Books Books",
null,
"A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.",
null,
"The First Six Books with Notes - Page 2\nby Euclid - 1822 - 179 pages",
null,
"The first three books of Euclid's Elements of geometry, with theorems and ...\n\nEuclid, Thomas Tate - 1849 - 108 pages\n...plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. A VI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight...",
null,
"Rudimentary dictionary of terms used in architecture [&c.].\n\nJohn Weale - 1850\n...plain figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another, and this point is called the centre of the circle Circular saw. Circular saws, revolving upon an axis,...",
null,
"Rudimentary Dictionary of Terms Used in Architecture, Civil ..., Volumes 1-2\n\nJohn Weale - Architecture - 1850 - 564 pages\n...plain figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another, and this point is called the centre of the circle The circumference of a circle is known to be about...",
null,
"The Elements of Euclid [book 1] for beginners, by J. Lowres\n\nEuclides - 1852\n...16. A circle is a plane figure bounded by one line called the circumference, and is of such a kind, that all right lines drawn from a certain point within...figure to the circumference are equal to one another ; as, the circle DBF. 17. This point is called the centre of the circle, and the right line drawn from...",
null,
"The First Two Books of the Elements of Euclid ... with Additional Figures ...\n\nEuclides - 1852\n...contained by one line, which is called the circumference, and is such that all straight lines B 3 BOOK I. drawn from a certain point within the figure to the circumference, are equal to one another. XVI. And this point is called the centre of the circle. [In the above figures A is the centre of the...",
null,
"Shaw's Civil Architecture: Being a Complete Theoretical and Practical System ...\n\nEdward Shaw - Architecture - 1852 - 191 pages\n...43. The radius of a circle is a right line drawn from the centre to the circumference, ab, at A. 44. A diameter of a circle is a right line drawn through the centre, terminating on both sides of the circumference, as с d, at B. 45. An arc of o- circle is any part...",
null,
"The popular educator, Volumes 1-2; Volume 5\n\nPopular educator - 1852\n...a ame line, called the cireumference or periphery, which is such that all straight linen drawn trom a certain point within the figure to the circumference are equal to each other. This point is called the centre of the circle, and each of the straight lines is called...",
null,
"The geometry, by T. S. Davies. Conic sections, by Stephen Fenwick\n\n...plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure...centre of the circle. 17. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. 18. A semicircle...",
null,
"The first six books of the Elements of Euclid, with numerous exercises\n\nEuclides - Geometry - 1853 - 147 pages\n...plain figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another. XVI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight...",
null,
""
] | [
null,
"https://books.google.ie/googlebooks/quote_l.gif",
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"https://books.google.ie/googlebooks/quote_r.gif",
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"https://books.google.ie/books/content",
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"https://books.google.ie/books/content",
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"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null,
"https://books.google.ie/books/content",
null
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https://msd.com.ua/financial-econometrics-and-empirical-market-microstructure/statistical-properties-of-mrw-process/ | [
"",
null,
"Financial Econometrics and Empirical Market Microstructure\n\n# Statistical Properties of MRW Process\n\nIn order to demonstrate distinctive feature of MRW process, one can compare its realization with realization of geometric Brownian motion (original random walk model of Bachelier), which sample increments and path are shown in Figs. 1 and 2. Sample realizations of increments and path of MRW are shown in Figs. 3 and 4.\n\nComparing Fig. 3 with Fig. 1, one can notice significant differences in the way, which each process goes. When dynamics of increments of random walk (Fig. 1) are very regular and one can not observe large deviations from the mean value, the dynamics of MRW (Fig. 3) is much more intermittent, one can easily spot volatility clustering and large excursions (extreme events).\n\n o",
null,
"О І I I , , , I ' 0 2000 4000 6000 8000 n",
null,
"",
null,
"Fig. 3 Increments of MRW process for A2 = 0.06, a = 7.5 • 10~5 and L = 1024\n\nPresence of the heavy tails of pdf for MRW can be shown more clearly with the ranking plot (see Fig. 5) for various aggregation level. The interval of scales 10 3 ІДгХдг [t] ^ 1 illustrates the tails of pdf which decay much slower than for\n\nthe normal distribution that is also presented on the plot for comparison. In other words, the probability of observing extremely large increment (return) for MRW is much larger than for the normal distribution where the probability of observing a value larger than three-four standard deviations is essentially zero.\n\nFigure 5 also illustrates another stylized fact, namely—aggregational gaussianity. One can see from the Fig. 5, that slope of the tail line tends to the slope of the tail line for normally distributed data, when aggregation level (which is defined as a number of consecutive increments of initial MRW process that are summed to obtain single\n\nincrements",
null,
"",
null,
"increment of aggregated process) rises. For instance, for aggregation level equals 4096 tail of the distribution converges to Gaussian distribution.\n\nVolatility clustering, that one can observe in Fig. 3 is a result of the presence of long memory in volatility. In order to quantify it we have considered four different measures of the volatility. The first one is the simplest squared values of returns",
null,
"",
null,
"",
null,
"(increments). Second is the definition of volatility as a standard deviation of returns in a rolling window of size nt:\n\nThird is the widely-used volatility estimator as a Exponentially-Weighted Moving Average (EWMA), which can be defined as\n\noi = ^ Астг2_! + (1 — A) rf2_ і, (18)\n\nwhere A є [0,1] is the rate of decay of the exponential weight within time window. Finally, we have also considered Muller estimator of the volatility (Muller 2000) which is similar to the EWMA, but involves recursion both of lagged and current squared returns:\n\nOn M = f 1 M + (u — M) rl_і + (1 — u) rl; (19)\n\nwhere a = (ln—ln_i)/r is the rate of decay of the exponential weight; m = e_a is an exponential weight itself and u = (1 — M)/a. Autocorrelation functions computed for above estimators of volatility are presented in Fig. 6.",
null,
"As one can see from the Fig. 6, autocorrelation of all proxies of volatility is significantly non zero in a very wide range (of the lags up to 1000 and more). Compared to",
null,
"Fig. 7 Log-log plot of moments of increments (4) calculated using the MRW sample of length 217 for A2 = 0.06, ct = 7.5 • 10_5, L = 2048 and q = 1,2, 3, 4, 5. Dashed lines correspond to linear fit of dependency (4)\n\nautocorrelation for squared returns, the rate of decay of autocorrelations for standard deviation, EWMA and Muller estimator decay much slower due to the fact, that above the three estimators perform recursive procedures for volatility computation. These recursion-based estimators capture the features of volatility behavior better than squared returns.\n\nIn order to illustrate the scale invariance in simulated MRW sample, one have to consider moments of increments of the realization (4). As described above, the presence of scale invariance is qualified with the power law behavior of the the moments (4). As one can see from the Fig. 7 this holds for the analyzed MRW process, as the absolute moments Mq (l) for all q has linear or close to linear (for q = 5) form in log-log scale, which tells about the presence of power law dependency in the ordinary scale.\n\nДобавить комментарий\n\n## Financial Econometrics and Empirical Market Microstructure\n\n### Modeling Financial Market Using Percolation Theory\n\nAnastasiya Byachkova and Artem Simonov Abstract Econophysics is a relatively new discipline. It is one of the most interesting and promising trends in modeling complex economic systems such as financial …\n\n### Multifractal Formalism for Stochastic Processes\n\nOriginal definition of fractal was proposed by Mandelbrot with respect to sets. He defined fractal as a mathematical set with fractal dimension is strictly larger than its topological dimension (Mandelbrot …\n\nRisk management is a core discipline in a rapidly changing world. From finance to ecology, we face unprecedented systemic risks from increasingly coupled global systems. Non-linearities render long term predictions …\n\n## Как с нами связаться:\n\nУкраина:\nг.Александрия\nтел./факс +38 05235 77193 Бухгалтерия\n+38 050 512 11 94 — гл. инженер-менеджер (продажи всего оборудования)\n\n+38 050 457 13 30 — Рашид - продажи новинок\ne-mail: [email protected]\nСхема проезда к производственному офису:\nСхема проезда к МСД\n\nПартнеры МСД\n\n## Контакты для заказов шлакоблочного оборудования:\n\n+38 096 992 9559 Инна (вайбер, вацап, телеграм)\nЭл. почта: [email protected]"
] | [
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"https://mc.yandex.ru/watch/86890482",
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"https://msd.com.ua/img/1140/image179.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90419793,"math_prob":0.96598065,"size":4354,"snap":"2022-27-2022-33","text_gpt3_token_len":1062,"char_repetition_ratio":0.12298851,"word_repetition_ratio":0.026560426,"special_character_ratio":0.23954985,"punctuation_ratio":0.10024155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98975927,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T00:50:29Z\",\"WARC-Record-ID\":\"<urn:uuid:adac3877-5c7b-41fe-ac6f-f4641c15e5b8>\",\"Content-Length\":\"41477\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ead9c12f-bbd7-40f2-8e3b-e0921bfbca2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6ec803e3-43a3-4142-812e-3886142d7336>\",\"WARC-IP-Address\":\"87.118.82.20\",\"WARC-Target-URI\":\"https://msd.com.ua/financial-econometrics-and-empirical-market-microstructure/statistical-properties-of-mrw-process/\",\"WARC-Payload-Digest\":\"sha1:AOZDY3MYQ54ZZPTWSFEMXGNS6N6IHFZQ\",\"WARC-Block-Digest\":\"sha1:ZCWEPIV3XYHTFI3YUX3EO2DRL4LUXL5P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104506762.79_warc_CC-MAIN-20220704232527-20220705022527-00583.warc.gz\"}"} |
https://www.stat.math.ethz.ch/pipermail/r-help/2017-May/446941.html | [
"# [R] Odd results from rpart classification tree\n\nTherneau, Terry M., Ph.D. therneau at mayo.edu\nMon May 15 14:43:11 CEST 2017\n\n```You are mixing up two of the steps in rpart. 1: how to find the best candidate split and\n2: evaluation of that split.\n\nWith the \"class\" method we use the information or Gini criteria for step 1. The code\nfinds a worthwhile candidate split at 0.5 using exactly the calculations you outline. For\nstep 2 the criteria is the \"decision theory\" loss. In your data the estimated rate is 0\nfor the left node and 15/45 = .333 for the right node. As a decision rule both predict\ny=0 (since both are < 1/2). The split predicts 0 on the left and 0 on the right, so does\nnothing.\n\nThe CART book (Brieman, Freidman, Olshen and Stone) on which rpart is based highlights the\ndifference between odds-regression (for which the final prediction is a percent, and error\nis Gini) and classification. For the former treat y as continuous.\n\nTerry T.\n\nOn 05/15/2017 05:00 AM, r-help-request at r-project.org wrote:\n> The following code produces a tree with only a root. However, clearly the tree with a split at x=0.5 is better. rpart doesn't seem to want to produce it.\n>\n> Running the following produces a tree with only root.\n>\n> y <- c(rep(0,65),rep(1,15),rep(0,20))\n> x <- c(rep(0,70),rep(1,30))\n> f <- rpart(y ~ x, method='class', minsplit=1, cp=0.0001, parms=list(split='gini'))\n>\n> Computing the improvement for a split at x=0.5 manually:\n>\n> obs_L <- y[x<.5]\n> obs_R <- y[x>.5]\n> n_L <- sum(x<.5)\n> n_R <- sum(x>.5)\n> gini <- function(p) {sum(p*(1-p))}\n> impurity_root <- gini(prop.table(table(y)))\n> impurity_L <- gini(prop.table(table(obs_L)))\n> impurity_R <- gini(prop.table(table(obs_R)))\n> impurity <- impurity_root * n - (n_L*impurity_L + n_R*impurity_R) # 2.880952\n>\n> Thus, an improvement of 2.88 should result in a split. It does not.\n>\n> Why?\n>\n> Jonathan\n>\n>\n\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81326807,"math_prob":0.96628207,"size":1921,"snap":"2022-05-2022-21","text_gpt3_token_len":585,"char_repetition_ratio":0.10015649,"word_repetition_ratio":0.0062111802,"special_character_ratio":0.33368036,"punctuation_ratio":0.1558753,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99609035,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T18:43:06Z\",\"WARC-Record-ID\":\"<urn:uuid:b2f05558-50be-4271-a2f9-4b5ffb1c799d>\",\"Content-Length\":\"5040\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0c7f0bf4-c4dc-4caa-b353-e835c6b64f33>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a67ecf4-3e42-4755-877a-864a687af352>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://www.stat.math.ethz.ch/pipermail/r-help/2017-May/446941.html\",\"WARC-Payload-Digest\":\"sha1:G4WHH2MOATI32YTZBIB24AQKMASUJDJG\",\"WARC-Block-Digest\":\"sha1:J4S33BYSYOILJ7QXNMTGEWRLRWPSTXH7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662675072.99_warc_CC-MAIN-20220527174336-20220527204336-00789.warc.gz\"}"} |
https://math.stackexchange.com/questions/1985951/show-that-any-uncountable-set-a-of-mathbbr-can-be-divided-into-two-uncoun | [
"# Show that any uncountable set $A$ of $\\mathbb{R}$ can be divided into two uncountably infinite disjoint subsets\n\nHow would one prove using the Axiom of Choice that any uncountable set $A$ of $\\mathbb{R}$ can be divided into two uncountably infinite disjoint subsets? I haven't seen this kind of proof exactly, because I'm not looking for a partition.\n\nFormally, let $D$ be the set with the following property: $x$ is a member of $D$ just in case the set of all elements of $A$ strictly greater than $x$ is uncountable and the set of all elements of $A$ strictly less than $x$ is uncountable. How would one show that $D$ is nonempty, preferably without directly resorting to cardinal arithmetic?\n\nUpdate: I appreciate all the feedback, including a solution. Although a nice proof has been given, I'm interested in seeing how it can be done by using the rationals in particular to split the set.\n\n• On a related note: math.stackexchange.com/questions/853113/… – lulu Oct 26 '16 at 13:26\n• I've added a second argument to show that we can indeed use a rational in the splitting. – Andrés E. Caicedo Oct 26 '16 at 16:07\n• @AndrésE.Caicedo I saw, thank you very much. But it seems as though Asaf's suggestion was a bit different and perhaps a bit more straightforward. I'm curious as to what he meant exactly. – CuriousKid7 Oct 26 '16 at 16:43\n\n## 2 Answers\n\nFor every rational $q$ let $A_q^+=\\{x\\in A\\mid x\\geq q\\}$ and $A_q^-=\\{x\\in A\\mid x<q\\}$. If for some $q$ both are uncountable, super. If not, what can you say about $A$ itself?\n\nNow, you might wonder as to the use of the axiom of choice here, and it hides beneath the surface: in order to prove that every countable union of countable sets is countable, we have to use the axiom of choice.\n\nEdit: Let $r=\\sup\\{q\\mid A^-_q\\text{ is countable}\\}$ and $s=\\inf\\{q\\mid A^+_q\\text{ is countable}\\}$.\n\n1. If either $r$ or $s$ is not a real number (i.e., $\\pm\\infty$) then we get that $A$ is countable.\n\n2. If $r>s$, take a rational $q$ witnessing that, then $A^-_q$ and $A^+_q$ are both countable, what do we get?\n\n3. If $r<s$, take a rational $q$ witnessing that, then $A^-_q$ and $A^+_q$ are both uncountable.\n\n4. Finally, if $r=s=q$, take an increasing sequence of rationals $r_n$ and a decreasing sequence of rationals $s_n$, both converging to $q$. Then $A\\cap\\bigcup_{p<q}A_p^+=A\\cap\\bigcup_{n\\in\\Bbb N}A^+_{r_n}$ is countable; and $A\\cap\\bigcup_{p>r} A^-_p$ is countable. Again, we get that $A$ is countable.\n\nSo if $A$ is uncountable, there is a rational $q$ such that $r<q<s$ and $A_q^-$ and $A^+_q$ are both uncountable.\n\n• Good answer, and also good for anticipating the question of where AC is invoked. – Matthew Leingang Oct 26 '16 at 13:05\n• It ain't my first rodeo, when it comes to the axiom of choice. ;-) – Asaf Karagila Oct 26 '16 at 13:05\n• @CuriousKid7: I've added stuff to the answer. – Asaf Karagila Oct 26 '16 at 16:57\n• Ah, this is the right way. Cool! – Andrés E. Caicedo Oct 26 '16 at 17:08\n• @Andres: Well, in general? As far as I know, from the work of Eilon Bilinsky, a student of Moti Gitik, the answer is indeed positive. It was an odd problem, posed in the early 1920s that wasn't given its due attention over the years. (I helped Eilon with the history of the problem, I can try and retrace my steps if you want, but only on Monday.) – Asaf Karagila Oct 26 '16 at 19:25\n\nThe question immediately reminded me of this. Here is an argument following the same basic idea at the beginning of that argument:\n\nFirst, consider $B=\\{x\\in A\\mid A\\cap(-\\infty,x]$ is countable$\\}$, and note that $B$ itself is countable: The point is that if $x\\in B$ then $A\\cap(-\\infty,x]\\subseteq B$. Now, if $B\\ne\\emptyset$, let $t=\\sup B$, fix an increasing sequence of elements of $B$ converging to $t$, $t_0\\le t_1\\le t_2\\le\\dots\\to t$, and note that $$B\\subseteq A\\cap(-\\infty,t]\\subseteq \\{t\\}\\cup\\bigcup_n A\\cap(-\\infty,t_n],$$ and the expression on the right is a countable union of countable sets, and therefore countable (by the axiom of choice).\n\nNow let $C=A\\setminus B$. Note that $C$ is uncountable and that if $x\\in C$ then $C\\cap(-\\infty,x]$ is itself uncountable. Otherwise, $x\\in B$ (a contradiction), since $A\\cap(-\\infty,x]\\subseteq B\\cup(C\\cap(-\\infty,x])$. Forget about $A$ itself now and work with $C$. The same argument as above shows that $D=\\{x\\in C\\mid C\\cap[x,\\infty)$ is countable$\\}$ is countable as well.\n\nFinally, let $E=C\\setminus D$, and note that for any $x\\in E$, both $E\\cap(-\\infty,x)$ and $E\\cap(x,\\infty)$ are uncountable.\n\nLet me present another argument, showing that we can actually find a rational $q$ such that both $A\\cap(-\\infty,q)$ and $A\\cap(q,\\infty)$ are uncountable:\n\nSince $$A\\subseteq\\mathbb Z\\cup\\bigcup_{l\\in\\mathbb Z}A\\cap(l,l+1),$$ at least one of the intervals $A\\cap(l,l+1)$ is uncountable (by the axiom of choice). We may as well assume that, for such an $l$, $A\\setminus(l,l+1)$ is countable, or we are done.\n\nLet $I_0=(l,l+1)$. Suppose we have found $I_n=(a,b)$ of length at most $1/2^n$ with rational endpoints such that $A\\cap I_n$ is uncountable but $A\\setminus I_n$ is countable. Find rational points $q_m$ for $m\\in\\mathbb Z$ such that $a<q_k<q_m<b$ whenever $k<m$, $\\lim_{k\\to-\\infty} q_k=a$, $\\lim_{m\\to\\infty}q_m=b$, and $q_{m+1}-q_m\\le 1/2^{n+1}$ for all $m\\in\\mathbb Z$. As before, $$A\\cap I_n\\subseteq\\{q_m\\}_{m\\in\\mathbb Z}\\cup\\bigcup_{m\\in\\mathbb Z}A\\cap(q_m,q_{m+1}),$$ so for some $m$, $A\\cap(q_m,q_{m+1})$ is uncountable. As before, we may assume that $A\\setminus (q_m,q_{m+1})$ is countable, or else we are done. Now set $I_{n+1}=(q_m,q_{m+1})$.\n\nThe claim is that at some stage $n$ we must actually be done, that is, $A\\setminus(q_m,q_{m+1})$ must also be uncountable and therefore at least one of $q_m,q_{m+1}$ works as the rational $q$ we were after.\n\nThe proof is by contradiction: Otherwise, the recursion produces an infinite, shrinking sequence of intervals $I_n=(a_n,b_n)$ with $A\\cap I_n$ uncountable and $A\\setminus I_n$ countable for all $n\\in\\mathbb N$. Notice that $\\bigcap_n I_n$ is either empty or a singleton $\\{t\\}$, and set $t=0$ if $\\bigcap_n I_n=\\emptyset$. We have $$A\\subseteq \\{t\\}\\cup\\bigcup_{n\\in\\mathbb N}\\{a_n,b_n\\}\\cup\\bigcup_{n\\in\\mathbb N}A\\setminus I_n,$$ so $A$ is countable."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7724227,"math_prob":0.9997057,"size":2883,"snap":"2021-21-2021-25","text_gpt3_token_len":1030,"char_repetition_ratio":0.1622091,"word_repetition_ratio":0.005167959,"special_character_ratio":0.32604927,"punctuation_ratio":0.1263318,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-20T18:53:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9e3906c1-d1ec-46b4-9d3b-d6fb174e62bb>\",\"Content-Length\":\"187120\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea066a0d-8db3-4676-ac46-ac608b8bba18>\",\"WARC-Concurrent-To\":\"<urn:uuid:41677c97-e69c-46d9-81cc-5bd2f47704c1>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1985951/show-that-any-uncountable-set-a-of-mathbbr-can-be-divided-into-two-uncoun\",\"WARC-Payload-Digest\":\"sha1:BAAHQAQO3AI5TWVIXALAIZ6SX7URDMXK\",\"WARC-Block-Digest\":\"sha1:B4FYL5QKFHAWRPKMRSKQATNRTGDC3ZQH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488253106.51_warc_CC-MAIN-20210620175043-20210620205043-00323.warc.gz\"}"} |
https://www.netlib.org/lapack/explore-html/d2/d36/a18672_ga3d55441f4aa7be616d35cea3f4a95ac7.html | [
"",
null,
"LAPACK 3.8.0 LAPACK: Linear Algebra PACKage\n\n## ◆ zspt03()\n\n subroutine zspt03 ( character UPLO, integer N, complex*16, dimension( * ) A, complex*16, dimension( * ) AINV, complex*16, dimension( ldw, * ) WORK, integer LDW, double precision, dimension( * ) RWORK, double precision RCOND, double precision RESID )\n\nZSPT03\n\nPurpose:\n``` ZSPT03 computes the residual for a complex symmetric packed matrix\ntimes its inverse:\nnorm( I - A*AINV ) / ( N * norm(A) * norm(AINV) * EPS ),\nwhere EPS is the machine epsilon.```\nParameters\n [in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the upper or lower triangular part of the complex symmetric matrix A is stored: = 'U': Upper triangular = 'L': Lower triangular``` [in] N ``` N is INTEGER The number of rows and columns of the matrix A. N >= 0.``` [in] A ``` A is COMPLEX*16 array, dimension (N*(N+1)/2) The original complex symmetric matrix A, stored as a packed triangular matrix.``` [in] AINV ``` AINV is COMPLEX*16 array, dimension (N*(N+1)/2) The (symmetric) inverse of the matrix A, stored as a packed triangular matrix.``` [out] WORK ` WORK is COMPLEX*16 array, dimension (LDW,N)` [in] LDW ``` LDW is INTEGER The leading dimension of the array WORK. LDW >= max(1,N).``` [out] RWORK ` RWORK is DOUBLE PRECISION array, dimension (N)` [out] RCOND ``` RCOND is DOUBLE PRECISION The reciprocal of the condition number of A, computed as ( 1/norm(A) ) / norm(AINV).``` [out] RESID ``` RESID is DOUBLE PRECISION norm(I - A*AINV) / ( N * norm(A) * norm(AINV) * EPS )```\nDate\nDecember 2016\n\nDefinition at line 112 of file zspt03.f.\n\n112 *\n113 * -- LAPACK test routine (version 3.7.0) --\n114 * -- LAPACK is a software package provided by Univ. of Tennessee, --\n115 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--\n116 * December 2016\n117 *\n118 * .. Scalar Arguments ..\n119 CHARACTER uplo\n120 INTEGER ldw, n\n121 DOUBLE PRECISION rcond, resid\n122 * ..\n123 * .. Array Arguments ..\n124 DOUBLE PRECISION rwork( * )\n125 COMPLEX*16 a( * ), ainv( * ), work( ldw, * )\n126 * ..\n127 *\n128 * =====================================================================\n129 *\n130 * .. Parameters ..\n131 DOUBLE PRECISION zero, one\n132 parameter( zero = 0.0d+0, one = 1.0d+0 )\n133 * ..\n134 * .. Local Scalars ..\n135 INTEGER i, icol, j, jcol, k, kcol, nall\n136 DOUBLE PRECISION ainvnm, anorm, eps\n137 COMPLEX*16 t\n138 * ..\n139 * .. External Functions ..\n140 LOGICAL lsame\n141 DOUBLE PRECISION dlamch, zlange, zlansp\n142 COMPLEX*16 zdotu\n143 EXTERNAL lsame, dlamch, zlange, zlansp, zdotu\n144 * ..\n145 * .. Intrinsic Functions ..\n146 INTRINSIC dble\n147 * ..\n148 * .. Executable Statements ..\n149 *\n150 * Quick exit if N = 0.\n151 *\n152 IF( n.LE.0 ) THEN\n153 rcond = one\n154 resid = zero\n155 RETURN\n156 END IF\n157 *\n158 * Exit with RESID = 1/EPS if ANORM = 0 or AINVNM = 0.\n159 *\n160 eps = dlamch( 'Epsilon' )\n161 anorm = zlansp( '1', uplo, n, a, rwork )\n162 ainvnm = zlansp( '1', uplo, n, ainv, rwork )\n163 IF( anorm.LE.zero .OR. ainvnm.LE.zero ) THEN\n164 rcond = zero\n165 resid = one / eps\n166 RETURN\n167 END IF\n168 rcond = ( one / anorm ) / ainvnm\n169 *\n170 * Case where both A and AINV are upper triangular:\n171 * Each element of - A * AINV is computed by taking the dot product\n172 * of a row of A with a column of AINV.\n173 *\n174 IF( lsame( uplo, 'U' ) ) THEN\n175 DO 70 i = 1, n\n176 icol = ( ( i-1 )*i ) / 2 + 1\n177 *\n178 * Code when J <= I\n179 *\n180 DO 30 j = 1, i\n181 jcol = ( ( j-1 )*j ) / 2 + 1\n182 t = zdotu( j, a( icol ), 1, ainv( jcol ), 1 )\n183 jcol = jcol + 2*j - 1\n184 kcol = icol - 1\n185 DO 10 k = j + 1, i\n186 t = t + a( kcol+k )*ainv( jcol )\n187 jcol = jcol + k\n188 10 CONTINUE\n189 kcol = kcol + 2*i\n190 DO 20 k = i + 1, n\n191 t = t + a( kcol )*ainv( jcol )\n192 kcol = kcol + k\n193 jcol = jcol + k\n194 20 CONTINUE\n195 work( i, j ) = -t\n196 30 CONTINUE\n197 *\n198 * Code when J > I\n199 *\n200 DO 60 j = i + 1, n\n201 jcol = ( ( j-1 )*j ) / 2 + 1\n202 t = zdotu( i, a( icol ), 1, ainv( jcol ), 1 )\n203 jcol = jcol - 1\n204 kcol = icol + 2*i - 1\n205 DO 40 k = i + 1, j\n206 t = t + a( kcol )*ainv( jcol+k )\n207 kcol = kcol + k\n208 40 CONTINUE\n209 jcol = jcol + 2*j\n210 DO 50 k = j + 1, n\n211 t = t + a( kcol )*ainv( jcol )\n212 kcol = kcol + k\n213 jcol = jcol + k\n214 50 CONTINUE\n215 work( i, j ) = -t\n216 60 CONTINUE\n217 70 CONTINUE\n218 ELSE\n219 *\n220 * Case where both A and AINV are lower triangular\n221 *\n222 nall = ( n*( n+1 ) ) / 2\n223 DO 140 i = 1, n\n224 *\n225 * Code when J <= I\n226 *\n227 icol = nall - ( ( n-i+1 )*( n-i+2 ) ) / 2 + 1\n228 DO 100 j = 1, i\n229 jcol = nall - ( ( n-j )*( n-j+1 ) ) / 2 - ( n-i )\n230 t = zdotu( n-i+1, a( icol ), 1, ainv( jcol ), 1 )\n231 kcol = i\n232 jcol = j\n233 DO 80 k = 1, j - 1\n234 t = t + a( kcol )*ainv( jcol )\n235 jcol = jcol + n - k\n236 kcol = kcol + n - k\n237 80 CONTINUE\n238 jcol = jcol - j\n239 DO 90 k = j, i - 1\n240 t = t + a( kcol )*ainv( jcol+k )\n241 kcol = kcol + n - k\n242 90 CONTINUE\n243 work( i, j ) = -t\n244 100 CONTINUE\n245 *\n246 * Code when J > I\n247 *\n248 icol = nall - ( ( n-i )*( n-i+1 ) ) / 2\n249 DO 130 j = i + 1, n\n250 jcol = nall - ( ( n-j+1 )*( n-j+2 ) ) / 2 + 1\n251 t = zdotu( n-j+1, a( icol-n+j ), 1, ainv( jcol ), 1 )\n252 kcol = i\n253 jcol = j\n254 DO 110 k = 1, i - 1\n255 t = t + a( kcol )*ainv( jcol )\n256 jcol = jcol + n - k\n257 kcol = kcol + n - k\n258 110 CONTINUE\n259 kcol = kcol - i\n260 DO 120 k = i, j - 1\n261 t = t + a( kcol+k )*ainv( jcol )\n262 jcol = jcol + n - k\n263 120 CONTINUE\n264 work( i, j ) = -t\n265 130 CONTINUE\n266 140 CONTINUE\n267 END IF\n268 *\n269 * Add the identity matrix to WORK .\n270 *\n271 DO 150 i = 1, n\n272 work( i, i ) = work( i, i ) + one\n273 150 CONTINUE\n274 *\n275 * Compute norm(I - A*AINV) / (N * norm(A) * norm(AINV) * EPS)\n276 *\n277 resid = zlange( '1', n, n, work, ldw, rwork )\n278 *\n279 resid = ( ( resid*rcond ) / eps ) / dble( n )\n280 *\n281 RETURN\n282 *\n283 * End of ZSPT03\n284 *\ndouble precision function zlange(NORM, M, N, A, LDA, WORK)\nZLANGE returns the value of the 1-norm, Frobenius norm, infinity-norm, or the largest absolute value ...\nDefinition: zlange.f:117\ndouble precision function zlansp(NORM, UPLO, N, AP, WORK)\nZLANSP returns the value of the 1-norm, or the Frobenius norm, or the infinity norm, or the element of largest absolute value of a symmetric matrix supplied in packed form.\nDefinition: zlansp.f:117\nlogical function lsame(CA, CB)\nLSAME\nDefinition: lsame.f:55\ndouble precision function dlamch(CMACH)\nDLAMCH\nDefinition: dlamch.f:65\ncomplex *16 function zdotu(N, ZX, INCX, ZY, INCY)\nZDOTU\nDefinition: zdotu.f:85\nHere is the caller graph for this function:"
] | [
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"https://www.netlib.org/lapack/explore-html/lapack.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6037414,"math_prob":0.99573576,"size":1490,"snap":"2021-21-2021-25","text_gpt3_token_len":457,"char_repetition_ratio":0.12786002,"word_repetition_ratio":0.09053498,"special_character_ratio":0.31342283,"punctuation_ratio":0.14134276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99559605,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-15T06:35:16Z\",\"WARC-Record-ID\":\"<urn:uuid:35e19f16-aa87-4427-88bd-acb4e897a86a>\",\"Content-Length\":\"37777\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4775680e-2238-44ab-a26b-89b50bbd1b16>\",\"WARC-Concurrent-To\":\"<urn:uuid:44eff0b8-eb04-4e36-aceb-8172cce043cb>\",\"WARC-IP-Address\":\"160.36.131.221\",\"WARC-Target-URI\":\"https://www.netlib.org/lapack/explore-html/d2/d36/a18672_ga3d55441f4aa7be616d35cea3f4a95ac7.html\",\"WARC-Payload-Digest\":\"sha1:W7MFG7I2BET5NRXAKE5YOXVMRBV4KY3P\",\"WARC-Block-Digest\":\"sha1:RH62GTT26YVB4U2SN6X5YV7KQIGTNCEY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487617599.15_warc_CC-MAIN-20210615053457-20210615083457-00238.warc.gz\"}"} |
http://forums.wolfram.com/mathgroup/archive/2007/Jun/msg00911.html | [
"",
null,
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"question\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg77935] question\n• From: dimitris <dimmechan at yahoo.com>\n• Date: Wed, 20 Jun 2007 05:28:38 -0400 (EDT)\n\n```Hi.\n\nSay\n\no=(16-x^2)^(1/2)-(4-x)^(1/2)*(4+x)^(1/2);\n\nIn another CAS one must use assumptions\nin order to simplify\n\n>o:=sqrt(16-x^2)-sqrt(4-x)*sqrt(4+x);\n> combine(u) assuming x>-4,x<4 ;\n> simplify(%);\n\n2 1/2 1/2 1/2\no := (16 - x ) - (4 - x) (4 + x)\n\n2 1/2 1/2\n(16 - x ) - ((4 - x) (4 + x))\n\n0\n\nIn Mathematica one simply gets\n\nIn:=\nSimplify[o]\n\nOut=\n0\n\nI would like to hear your comments on this issue.\n\nDimitris\n\n```\n\n• Follow-Ups:\n• Prev by Date: Re: v6 GraphicsGrid with PlotLabel\n• Next by Date: NDSolve with (compiled) numerical functions\n• Previous by thread: Re: question\n• Next by thread: Re: question"
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https://wiki.gnuradio.org/index.php?title=FFT&printable=yes | [
"# FFT\n\nJump to: navigation, search\n\nThis block takes in a vector of floats or complex values and calculates the FFT. If all you do is want to see the frequency domain of a signal, the QT GUI Frequency Sink is more user friendly.\n\nNote that even if the input signal is real, the output will be complex, so you must use a Complex to Mag or similar block if you want to see magnitude.\n\n## Parameters\n\nFFT Size\nNumber of samples used in each FFT calculation, which also determines how many points are in the output.\nForward/Reverse\nWhether to do an FFT or IFFT.\nWindow\nType of window to apply to each set of samples before the FFT is taken, default is a blackmanharris window. The argument of the window.blackmanharris() function is simply how many points in the window, which must match the FFT size. If an empty window \"[]\" is supplied then no windowing math is performed.\nShift\nWhether or not to do an \"fft shift\" which puts DC (0 Hz) in the center. If you are not sure what this means, leave it set to Yes. Only active when input type is complex.\nNum Threads\nNumber of threads to assign to perform the FFTs.\n\n## Example Flowgraph\n\nThis flowgraph shows how the FFT block can be used to reproduce the behavior of the QT GUI Frequency Sink block. Both outputs match (besides an arbitrary scaling factor), but using the FFT block directly requires converting from a stream to vector, and performing the magnitude and log manually.\n\n## Source Files\n\nC++ files\nComplex input\nReal input\nCore algorithms\nHeader files\nComplex input\nReal input\nPublic header files\nComplex input\nReal input\nBlock definition\nGRC yaml"
] | [
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https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=18&t=48239&p=171601 | [
"## 1B.21\n\nAliya Jain 2B\nPosts: 86\nJoined: Wed Sep 11, 2019 12:16 am\n\n### 1B.21\n\n1B.21 reads \"A baseball must weigh between 5.00 and 5.25 ounces (1 ounce 5 28.3 g). What is the wavelength of a 5.15-ounce baseball thrown at 92 mph?\" I understand the basic premise of the question, but I'm kind of confused about the conversations? Can someone explain the initial steps\n\nSVajragiri_1C\nPosts: 91\nJoined: Thu Jul 11, 2019 12:16 am\n\n### Re: 1B.21\n\nFirst convert the 5.15 ounces to grams using the conversion factor that is provided. Then you have to convert the 92 mph to meters per second by using the miles to meters conversion factor and the hours to minutes to seconds conversion factors (you might have to look this up online). Then plug the data into De Broglie's equation to solve for the wavelength.\n\nCharisse Vu 1H\nPosts: 86\nJoined: Thu Jul 25, 2019 12:17 am\n\n### Re: 1B.21\n\nThis question is dealing with deBroglie's equation, as it is asking for the wavelength of something that has mass and velocity (momentum). Since the units are not what we want, we must use dimensional analysis to convert ounces to kilograms and miles per hour to meters per second. It is given that one ounce is equivalent to 28.3 grams. You would use dimensional analysis to convert the units from ounces to grams, then from grams to kilograms. There are 1,690.3 meters in one mile, and 3600 seconds in one hour. Using this, you would again use dimensional analysis to convert the units to meters per second. Then, you would substitute these values into deBroglie's equation to find the wavelength.\n\nCamille 4I\nPosts: 57\nJoined: Sat Aug 24, 2019 12:18 am\n\n### Re: 1B.21\n\nWould I be able to convert to kilograms after proceeding with the equation? I already did the De Broglie equation and am wondering if it is possible to convert these units after doing the calculations."
] | [
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https://www.myelectrical2015.com/2020/03/transistor-biasing-interview-short.html | [
"## 01/03/2020\n\n### Transistor Biasing Interview Short Question Answer - 1\n\n 1 Explain the term : Biasing of the transistor Biasing of the transistor The proper flow of zero signal collector current, maintain input circuit always forward biased and output circuit always reversed biased in all the conditions is known as the biasing of the transistor. 2 Explain the term : Faithful amplification of the transistor signal Faithful amplification When a transistor is used as an amplifier, a weak signal is given to the base of the transistor and amplified output is obtained from collector circuit. The process of obtaining amplified output signal without change of its shape is called as faithful amplification. 3 What is meaning of the ‘zero signal collector current’? Zero signal collector current The zero signals means that the ac signal is not given to the transistor. The flow of collector current when only DC biasing signal is given to the transistor is called as zero signal collector current. 4 What are the conditions to be satisfied for faithful amplification of the transistor? OR What are the essential conditions for biasing of the transistor? Conditions for faithful amplification of the transistor Proper flow of zero signal collector current Emitter – base circuit remains in forward biased Collector – base circuit remains in reverse biased 5 What is the DC voltage level to be maintaining at the input and output circuit of the transistor for faithful amplification? In order to achieve faithful amplification in CE transistor Emitter – base voltage (VBE) > 0.5 V Germanium transistor 0.7 V Silicon transistor Collector – base voltage (VCE) > 0.5 V Germanium transistor 1.0 V Silicon transistor At the emitter – base circuit once the potential barrier ( Si - 0.7 V and Ge - 0.5 V ) overcomes, the input circuit should remain in forward biased. The collector – emitter voltage always greater than the knee voltage ( Ge - 0.5 V and Si - 1.0 V ) otherwise the output circuit should not be properly reverse biased. 6 Explain the term : Stabilization of the transistor Stabilization It is process of making the transistor operating point independent of temperature changes and variation of the transistor inherent parameters. 7 Describe the different method of Transistor biasing Transistor biasing method Base resistor Emitter bias Collector feedback resistor Voltage divider ( Universal biasing ) 8 Explain the term : Stability factor Stability factor It is defined as the rate of change of collector current with respect to collector leakage current at constant base current and current amplification factor β in the common emitter transistor. 9 Describe the disadvantages of base resistor bias. Disadvantages of Base resistor bias Poor stabilization and Possibility of thermal run away due to high stability factor 10 Which biasing method is also known as fixed bias method? Why? The base resistor method is known as fixed bias method because the base current does not depends upon the collector current. 11 Why the fixed bias method has poor thermal run away? The fixed bias method has poor thermal run away due to large value of stability factor. The stability factor in this method is S = β + 1 12 What should be ideal value of stability factor? One 13 Give reason : The base bias resistor method of transistor is unstable. Base bias method The Q point of the transistor in the base bias method depends upon current amplification factor β therefore it is unstable. . 14 Give reason : The emitter bias method of the transistor is stable. Emitter bias method The Q point in the emitter bias method does not depend upon current amplification factor β and variation of base – emitter voltage. Therefore the operating point Q is stable in this method. 15 Describe the disadvantages of collector feedback resistor biasing method. Disadvantages of collector feedback resistor biasing Gain of the amplifier reduces due to negative feedback circuit. High stability factor The operating point some extent depend upon temperature. 16 Describe the effect of temperature on current amplification factor β and base emitter voltage. As the temperature increases the current amplification factor β increase whereas the base emitter voltage decreases. 17 Which biasing method is known as universal biasing method? Voltage divider method 18 What is reason behind the name voltage divider bias? Voltage divider bias There are two resistances R1 and R2 are connected across the supply voltage and the voltage divider is done by these resistors R1 and R2. These resistances provide biasing therefore it is called as voltage divider bias. 19 Which biasing method provides highest thermal stability? Voltage divider method 20 Explain the term : Mid - point biasing Mid – point biasing The transistor amplifier circuit is designed such that its operating point lies at the center of the DC load line. As the operating point lies on the center of load line, it is called as mid-point biasing. 21 Describe the importance of the midpoint biasing. Importance of midpoint biasing When the AC signal is given to the base of the transistor, the magnitude of the output current and output voltage of the transistor varies according to the position of the operating point. If the operating point lies above the midpoint on the DC load line, the transistor is operated in the saturation region and part of the output signal will be clipped off. If the operating point lies below the midpoint on the DC load line, the transistor is operated in the cur-off region resulting portion of the output signal will be clipped. The mid – point biasing is best amplification method because there is no cut of output signal and transistor is operated in the active region. 22 Explain the term : Thermal runaway Thermal runaway The collector leakage current in the transistor greatly depends upon the temperature. The flow of the collector current produces heat at the collector junction. This will increase the temperature of the transistor and if stabilization is not done properly, collector leakage current is further increased. IC = βIB + ( β +1 )ICBO Where ICBO = Collector leakage current If the ICBO increases, the IC also increases and this process is cumulative resulting collector current increases rapidly to burn out transistor. The burn out of the sterilized transistor is known as thermal runaway.\n\nYou may also like :"
] | [
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https://openreview.net/forum?id=Hg-IBdIo5e9 | [
"## An Explore-then-Commit Algorithm for Submodular Maximization Under Full-bandit Feedback",
null,
"Abstract: We investigate the problem of combinatorial multi-armed bandits with stochastic submodular (in expectation) rewards and full-bandit feedback, where no extra information other than the reward of selected action at each time step $t$ is observed. We propose a simple algorithm, Explore-Then-Commit Greedy (ETCG) and prove that it achieves a $(1-1/e)$-regret upper bound of $\\mathcal{O}(n^\\frac{1}{3}k^\\frac{4}{3}T^\\frac{2}{3}\\log(T)^\\frac{1}{2})$ for a horizon $T$, number of base elements $n$, and cardinality constraint $k$. We also show in experiments with synthetic and real-world data that the ETCG empirically outperforms other full-bandit methods."
] | [
null,
"https://openreview.net/images/pdf_icon_blue.svg",
null
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https://docs_92.abinit.org/topics/EffectiveMass/ | [
"# EffectiveMass\n\nThis page gives hints on how to perform an effective mass calculation with the ABINIT package.\n\n## Introduction¶\n\nThe direct estimation of effective masses from DFT band curvature using topic_DFPT has been implemented within the linear response part of ABINIT [Laflamme2016]. This method avoids the use of finite differences to estimate these masses, which eliminates the associated numerical noise and convergence study. To compute the effective masses, one has to set the keyword efmas to 1 within a calculation of the derivative of ground-state wavefunctions with respect to wavevector (rfelfd = 1 or 2). The effective masses will then be computed for all k-points and bands present in the calculation. One can optionally specify the range of bands to be treated for each k-point with the keyword efmas_bands.\n\nAn additional feature of the effective mass implementation is the correct treatment of degenerate bands. Indeed, the concept of effective mass breaks down at degenerate band extrema since it is no longer possible to describe band curvature using a tensor [Luttinger1955], [Mecholsky2014]. However, using the concept of transport equivalent effective mass’‘ [Mecholsky2014] and its adaptation to the k.p framework, the implementation is able to provide the user with effective mass tensors which, while not describing the band curvature, describe accurately the contribution of the individual bands to transport properties.\n\nThe implementation supports both NCPP and PAW schemes.\n\nSpin-polarized systems (nspden = 2) as well as spinors (nspinor = 2) can be treated, although the spin-orbit interaction can only be treated in the PAW case.\n\nThe treatment of degeneracies is limited to the extremal points of the band structure (which are the most relevant in any case).\n\nBy the way, the first derivative of the eigenenergies is also computed and printed during a d/dk calculation, and corresponds to the electronic velocity.\n\ncompulsory:\n\n• efmas EFfective MASs\n• rfelfd Response Function with respect to the ELectric FielD\n\nbasic:\n\n• efmas_dirs EFfective MASs, DIRectionS to be calculated\n• efmas_n_dirs EFfective MASs, Number of DIRectionS\n• efmas_ntheta EFfective MASs, Number of points for integration w/r to THETA\n\nuseful:\n\nexpert:\n\n• efmas_deg EFfective MASs, activate DEGenerate formalism\n\nv7:\n\nv8:"
] | [
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https://www.onlinemath4all.com/real-world-problems-on-pythagorean-theorem.html | [
"# REAL WORLD PROBLEMS ON PYTHAGOREAN THEOREM\n\nProblem 1 :\n\nA man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?\n\nSolution :",
null,
"the distance of his current position from the starting point = √182 + 242\n\n= √(324 + 576)\n\n= √900\n\n= 30 m\n\nSo, the required distance is 30 m.\n\nProblem 2 :\n\nThere are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C street? (Using figure).",
null,
"Solution :\n\nBy choosing the C street, he has to cover the distance,\n\n= √22 + 1.52\n\n= √(4 + 2.25)\n\n= √6.25\n\n= 2.5 miles\n\nBy choosing the alternative way, he has to cover the distance = 2 + 1.5\n\n= 3.5 miles\n\nThe difference between these two paths = 3.5 - 2.5\n\n= 1 mile\n\nSo, by choosing the direct path, he may save 1 miles faster than other way.\n\nProblem 3 :\n\nTo get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41 m east. To the nearest meter, how many meters would be saved if it were possible to make a way through the pond?\n\nSolution :\n\nBy drawing the rough picture using the given information, we get",
null,
"AC = √342 + 412\n\n= √1156 + 1681\n\n= √2837\n\n= 53.26\n\nMiles saved = (34 + 41) - 53.26\n\n= 75 - 53.26\n\n= 21.74 m\n\nProblem 4 :\n\nIn the rectangle WXYZ, XY + YZ = 17 cm, and XZ + YW = 26 cm. Calculate the length and breadth of the rectangle?",
null,
"Solution :\n\nXY + YZ = 17 cm\n\nXZ + YW = 26 cm\n\nTo calculate : - Length and breadth of the rectangle.\n\nWe know that,\n\nDiagonals of a rectangle are equal.\n\nSo,\n\nXZ = YW\n\nThen,\n\nXZ = YW = 26/2 = 13 cm\n\nIn ∆XYZ, let YZ = P. Then\n\nXY = 17 - P\n\nThen, by Pythagoras theorem,\n\n(P)2 + (17 - P)2 = (13)2\n\nP2 + 289 - 34P + P= 169\n\n2P2 - 34P = 169 - 289\n\n2(P2 - 17P) = - 120\n\nP2 - 17P = - 120/2\n\nP2 - 17P = - 60\n\nP2 - 17P + 60 = 0\n\nP2 - 12P - 5P + 60 = 0\n\nP(P - 12) - 5(P - 12) = 0\n\n(P - 12)(P - 5) = 0\n\nP - 12 = 0 or P = 12\n\nP = 12 cm or P = 5 cm\n\nNow,\n\nYZ = P = 12 cm [Because , YZ is the length of the rectangle ,so we will assign it the greatest value of P]\n\nAgain, XY = (17 - P) = (17 - 12) cm = 5 cm\n\n[Because , XY is thee breadth]\n\nKindly mail your feedback to [email protected]\n\nWe always appreciate your feedback.\n\n## Recent Articles",
null,
"1. ### Power Rule of Logarithms\n\nOct 04, 22 11:08 PM\n\nPower Rule of Logarithms - Concept - Solved Problems\n\n2. ### Product Rule of Logarithms\n\nOct 04, 22 11:07 PM\n\nProduct Rule of Logarithms - Concept - Solved Problems"
] | [
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"https://www.onlinemath4all.com/images/x10thnewsylabusex4.3q1.png.pagespeed.ic.8Ppmhu1CDL.png",
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"https://www.onlinemath4all.com/images/x10thnewsylabusex4.3q3.png.pagespeed.ic.81FixlfKN4.png",
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https://cs.nyu.edu/pipermail/fom/2009-September/014007.html | [
"[FOM] 361: Finite Promise Games\n\nHarvey Friedman friedman at math.ohio-state.edu\nWed Sep 2 13:02:05 EDT 2009\n\nWe have had the idea that perhaps it would be even more elementary to\nhave Bob make PROMISES during the game. Then Bob wins the game if and\nonly if all of Bob's PROMISES are kept.\n\nI had thought till now that this was purely cosmetic, and might even\nobscure the mathematical content - i.e., possible connections with\ncore mathematics.\n\nHowever, I now realize that bringing in promises during the game\nserves to open up new possibilities for further simplifications of the\ngame - especially from the point of view of the nonprofessional.\n\nFINITE INTEGER GAMES WITH PROMISES\nby\nHarvey M. Friedman\nSeptember 2, 2009\n\nAs usual, SMAH+ = ZFC + \"for all k there is a strongly k-Mahlo\ncardinal\". SMAH = ZFC + {there is a strongly k-Mahlo cardinal}_k. That\nvarious two person games can be won, is provable in SMAH+ but not in\nZFC.\n\n1. Finite PL Copy/Inversion game.\n2. Finite Polynomial PL Copy/Inversion game.\n3. Finite Order Invariant Copy/Inversion game.\n4. Finite Linear Copy/Inversion game.\n\n1. FINITE PL COPY/INVERT GAMES.\n\nZ is the set of all integers. N is the set of all nonnegative\nintegers. All letters represent integers.\n\nWe say that T:N^k into N is PL if and only if it is piecewise linear\nwith integer coefficients. I.e., T can be defined by various affine\nfunctions with integer coefficients on each of finitely many pieces,\nwhere each piece is defined by a finite set of linear inequalities\nwith integer coefficients.\n\nLet T:N^k into N be PL. A T inversion of x is some y_1,...,y_k < x\nsuch that T(y_1,...,y_k) = x.\n\nWe now define the game G(T,n,s), n,s >= 1. G(T,n,s) consists of n\nalternating plays by Alice and Bob.\n\nAt every stage of the game, Alice is required to play x in [0,s] of\nthe form y + z or w!, where y,z are integers previously played by Bob\n(Alice's offering). Bob is then required to\n\n(accept x) play x and PROMISE that there will be no T inversion of x\namong the integers ever played by Bob; or\n(reject x) PROMISE that x is never played by Bob, and play a T\ninversion of x.\n\nAS IN ALL GAMES THROUGHOUT THIS ABSTRACT, Bob wins if and only if Bob\nhas kept all of his PROMISES. We will not mention this obvious winning\ncondition again.\n\nTHEOREM 1.1. Let T be in PL and n,s >= 1. Bob wins G(T,n,s).\n\nPROPOSITION 1.2. Let T be in PL and n >= 1. If m,s are sufficiently\nlarge then Bob wins G(T,n,s) even if Bob accepts all factorials > m\noffered by Alice.\n\nTHEOREM 1.3. Theorem A1.1 is provable in RCA_0. Proposition 1.2 is\nprovable in SMAH+ but not in any consistent fragment of SMAH.\nProposition 1.2 is provably equivalent, in ACA, to 1-Con(SMAH). The\n\"sufficiently large\" as a function of T,n is a provably recursive\nfunction of SMAH+ that eventually dominates all provably recursive\nfunctions of SMAH. For each n, Proposition 1.2 is provable in RCA_0,\nbut the proofs grow in length as n increases, according to a provably\nrecursive function of SMAH+ that eventually dominates all provably\nrecursive functions of SMAH.\n\n2. FINITE POLYNOMIAL COPY/INVERT GAMES.\n\nLet P:Z^k into Z be a polynomial with integer coefficients. A special\nP inversion at x in Z consists of 0 < y_1,...,y_n < x/2 such that\nP(y_1,...,y_n) = x.\n\nWe now define the game G(P,Q,n,s), n,s >= 1, where P,Q:Z^k into Z are\npolynomials with integer coefficients. G(P,Q,n,s) consists of n\nalternating plays by Alice and Bob.\n\nAt every stage of the game, Alice is required to play x in [-s,s] of\nthe form P(y), Q(y), or z!!, where y is a k tuple of integers\npreviously played by Bob (Alice's offering). Bob is then required to\n\n(accept x) play x and PROMISE that there will be no special P or Q\ninversion of x among the integers ever played by Bob; or\n(reject x) PROMISE that x is never played by Bob, and play a special P\nor Q inversion of x.\n\nTHEOREM 2.1. Let P,Q:Z^k into Z be polynomials with integer\ncoefficients, and n,s >= 1. Bob wins G(P,Q,n,s).\n\nPROPOSITION 2.2. Let P,Q:Z^k into Z be polynomials with integer\ncoefficients, and n >= 1. If m,s are sufficiently large then Bob wins\nG(P,Q,n,s) even if Bob accepts all double factorials > m offered by\nAlice.\n\nTHEOREM 2.3. Theorem 2.1 is provable in RCA_0. Proposition 2.2 is\nprovable in SMAH+ but not in any consistent fragment of SMAH.\nProposition 2.2 is provably equivalent, in ACA, to 1-Con(SMAH). The\n\"sufficiently large\" as a function of P,Q,n is a provably recursive\nfunction of SMAH+ that eventually dominates all provably recursive\nfunctions of SMAH. For each n, Proposition 2.2 is provable in RCA_0,\nbut the proofs grow in length as n increases, according to a provably\nrecursive function of SMAH+ that eventually dominates all provably\nrecursive functions of SMAH.\n\n3. FINITE ORDER INVARIANT COPY/INVERT GAMES.\n\nLet R be contained in N^2k be order invariant. An R inversion of\nx_1,...,x_k consists of some y_1,...,y_k such that max(y_1,...,y_k) <\nmax(x_1,...,x_k) and R(x_1,...,x_k,y_1,...,y_k).\n\nWe now define the game G(R,n,s), n,s >= 1. G(R,n,s) consists of n\nalternating plays by Alice and Bob.\n\nAt every stage of the game, Alice is required to play k integers\nx_1,...,x_k in [-s,s], each of the form z or z+1 or w!, where z is an\ninteger previously played by Bob (Alice's offering). Bob is then\nrequired to\n\n(accept x) play x_1,...,x_k and PROMISE that no R inversion of\nx_1,...,x_k is the k tuple ever simultaneously played by Bob; or\n(reject x) PROMISE that x_1,...,x_k is not the k tuple ever\nsimultaneously played by Bob, and play an R inversion of x_1,...,x_k.\n\nTHEOREM 3.1. Let R contained in N^2k be order invariant and n,s >= 1.\nBob wins G(R,n,s).\n\nPROPOSITION 3.2. Let R contained in N^2k be order invariant and n,s >=\n1. Bob wins G(R,n,s) even if Bob always plays outside the interval\n((8kn)!/2,(8kn)!).\nNote that Proposition 3.2 is explicitly Pi01.\n\nTHEOREM 3.3. Theorem 3.1 is provable in RCA_0. Proposition 3.2 is\nprovable in SMAH+ but not in any consistent fragment of SMAH.\nProposition 3.2 is provably equivalent, in ACA, to Con(SMAH).\n\nNote how large factorials are being protected from below. They are\n\"limit points\".\n\n4. FINITE LINEAR COPY/INVERT GAMES.\n\nDEFINITION. We say that x,y in N^k are additively equivalent if and\nonly if the following holds. Suppose some given length <= k sum of\ncoordinates of x is less than another given length <= k sum of\ncoordinates of x. Then the corresponding length <= k sum of\ncoordinates of y is less than the corresponding length <= k sum of\ncoordinates of y.\n\nWe now define the game G(v_1,...,v_p;n,s), p,n,s >= 1, v_1,...,v_p in\nN^k. G(v_1,...,v_p;n,s) consists of n alternating plays by Alice and\nBob.\n\nAt every stage of the game, Alice is required to play an integer x in\n[0,x] of the form y + z or w!, where y,z are integers previously\nplayed by Bob (Alice's offering). Bob is then required to\n\n(accept x) play x and PROMISE that x cannot be written as y_1 + ... +\ny_k, where (y_1,...,y_k) is additively equivalent to some v_j, and\ny_1,...,y_k are integers played by Bob at various times: or\n(reject x) PROMISE that x is never played by Bob, and play\ny_1,...,y_k, where x = y_1 + ... + y_k and (y_1,...,y_k) is additively\nequivalent to some v_j.\n\nTHEOREM 4.1. Let v_1,...,v_p be in N^k and n,s >= 1. Bob wins\nG(v_1,...,v_p;n,s).\n\nPROPOSITION 4.2. Let v_1,...,v_p be in N^k and n,s >= 1. If m is\nsufficiently large then Bob wins G(v_1,...,v_p;n,s) even if Bob\naccepts all factorials > m offered by Alice.\n\nTHEOREM 4.3. Theorem A4.1 is provable in RCA_0. Proposition A4.2 is\nprovable in SMAH+ but not in any consistent fragment of SMAH.\nProposition A4.2 is provably equivalent, in ACA, to 1-Con(SMAH). The\n\"sufficiently large\" as a function of v_1,...,v_p is a provably\nrecursive function of SMAH+ that eventually dominates all provably\nrecursive functions of SMAH. For each n, Proposition 4.2 is provable\nin RCA_0, but the proofs grow in length as n increases, according to a\nprovably recursive function of SMAH+ that eventually dominates all\nprovably recursive functions of SMAH.\n\n*************************************\n\nmanuscripts. This is the 361s6 in a series of self contained numbered\npostings to FOM covering a wide range of topics in f.o.m. The list of\nprevious numbered postings #1-349 can be found at n/a in the FOM\narchives.\n\n350: one dimensional set series 7/23/09 12:11AM\n351: Mapping Theorems/Mahlo/Subtle 8/6/09 10:59PM\n352: Mapping Theorems/simpler 8/7/09 10:06PM\n353: Function Generation 1 8/9/09 12:09PM\n354: Mahlo Cardinals in HIGH SCHOOL 1 8/9/09 6:37PM\n355: Mahlo Cardinals in HIGH SCHOOL 2 8/10/09 6:18PM\n356: Simplified HIGH SCHOOL and Mapping Theorem 8/14/09 9:31AM\n357: HIGH SCHOOL Games/Update 8/20/09 10:42AM\n358: clearer statements of HIGH SCHOOL Games 8/23/09 2:42AM\n359: finite two person HIGH SCHOOL games 8/24/09 1:28PM\n360: Finite Linear/LImited Memory Games 8/31/09 6:14PM\n\nHarvey Friedman"
] | [
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https://math.stackexchange.com/questions/2085047/complementary-convex-sets-in-bbb-rn/2085073 | [
"# Complementary, convex sets in $\\Bbb R^n$\n\nIs it true or false?\n\nIf $A\\subset \\Bbb R^n$ is convex and $\\Bbb R^n\\setminus A$ is convex then $A=\\emptyset$, $A=\\Bbb R^n$ or $\\partial A$ is a hyperplane.\n\nI remember vaguely this fact from university, but I'm not sure if it is true or not. (My guts say to me that it is true, but...)\n\nHints or references are welcome.\n\nTo try it myself, I have considered two points $P\\in A$ and $Q\\in\\Bbb R^n\\setminus A$. Then there is a \"first\" point $X$ ($\\sup$ can be used to formalize this idea) in the segment $PQ$ such that $PX$ is not in $A$. (Or a \"last\" point $X$ such that $PX$ is in $A$). But I still have to prove that all these points $X$ are in a hyperplane.\n\nIs it true for infinite dimensional vector spaces?\n\n• The closure of a convex set is convex. The boundary of a set $A$ equals to $\\overline{A} \\cap \\overline{A^c}$. So such a set has convex boundary. – Crostul Jan 5 '17 at 17:47\n\nLet $A\\subset\\mathbb{R}^n$ be a convex set and $x\\in \\partial A$. Then there exists a hyperplane $H$ such that $x\\in H$ and $A$ lies in one of the half spaces for $H$."
] | [
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