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Infi-MM

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InfiMM-WebMath-40B

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InfiMM 19

[ "#### You may also like", null, "### I'm Eight\n\nFind a great variety of ways of asking questions which make 8.", null, "### Let's Investigate Triangles\n\nVincent and Tara are making triangles with the class construction set. They have a pile of strips of different lengths. How many different triangles can they make?", null, "### Noah\n\nNoah saw 12 legs walk by into the Ark. How many creatures did he see?\n\n# Break it Up!\n\n## Break it Up!\n\nYou have a stick of $7$ interlocking cubes. You cannot change the order of the cubes.", null, "You break off a bit of it leaving it in two pieces.\n\nHere are $3$ of the ways in which you can do it:", null, "", null, "", null, "In how many different ways can it be done?\n\nNow try with a stick of $8$ cubes and a stick of $6$ cubes:", null, "", null, "Make a table of your results like this:\n\n Number of cubes Number of ways $6$ cubes ? $7$ cubes ? $8$ cubes ?\n\nNow predict how many ways there will be with $5$ cubes.\n\nWere you right?\n\nHow many ways with $20$ cubes? $50$ cubes? $100$ cubes?\n\nANY number of cubes?\n\n* * * * * * * * * * * * * * * * * * * *\n\nIf all the cubes are the same colour, a split of $4$ and $2$ will look the same as a split of $2$ and $4$.\n\nHow many ways are there of splitting $6$ cubes now?\n\nCan you predict how may ways there will be with any number of cubes?\n\n### Why do this problem?\n\nThis problem makes a very good introduction to algebraic thinking, however, it also begins very simply so everyone can make a reasonable start.\n\n### Possible approach\n\nYou could start by giving all the group some interlocking cubes and asking them to make a stick of seven. Then tell them to break their stick into two pieces and hold one piece in each hand. Ask how they have done this and write the various ways on the board and then enquire if there are any other ways that it could have been done. Is there a way that the results can be organised better? This may be the time to show the group how to make a simple table to record their results.\n\nSome learners may claim that $2 + 5$ is the same as $5 + 2$. \"Not always!\" is the answer to this. Of course, numerically they are identical, but the context is also important. A good example for when this is not so is this. If five people are sitting down to a meal and two more turn up, it is quite possible there will be enough food for them. If however, two are starting their meal it is very unlikely that there will be enough food for five more! In this case, is $5 + 2$ the same as $2 + 5$? Just because the answer is the same, it does not mean that the question is the same!\n\nAfter the introduction, learners could work in pairs on finding the number of ways that sticks of six, eight and nine cubes can be broken into two pieces. Then they should record their findings. Supply squared paper for those who wish to record using it.\n\nWhen learners feel ready to generalise they can go on to working out the number of ways with $20$, $50$ and $100$ cubes and then ANY number of cubes.\n\nAt the end of the lesson all can come together to discuss their findings. They can be asked how they knew they had found all the ways of breaking a stick into two pieces. Those who know a good way of expressing \"any number\" can explain their reasoning.\n\n### Key questions\n\nHow are you going to record what you have done?\nHow do you know you have found all the ways of breaking it into two pieces?\nIf you break it into $0 + 7$, does this give you two pieces?\nHow many ways do you think you can break $20/50/100$ cubes?\nCan you see a connection between the total number of cubes and the number of ways you can break the stick into two pieces?\nHow could you express that \"for any number\"?\n\n### Possible extension\n\nLearners who can generalise the first part of the problem could then go on to exploring the second part - if all the cubes are the same colour, looking at the difference between odd and even numbers.\n\nThen they could try a different problem in which generalisation is required such as Sticky Triangles.\n\n### Possible support\n\nSuggest working with the interlocking cubes and recording on squared paper." ]

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[ "", null, "# 5865e0 | #5865e0 hex code With Full Information Of RGB, HTML Color Codes, Colour Schemes\n\nLUCKY ONE", null, "1. My Account\n2. Favorites", null, "The Hexadecimal Value (also known as\n\n## HTML color codes or hex color codes\n\n) for this color is (#5865e0) and it Consists like any color as a product of composed of three colors in RGB System which The (R) represent The Red Color Code and has the value of (88) also the (G) represent Green Color Code with The value of (101) while the last one (B) represent Blue Color Code and has The Value of (224), The HSL Color Space for (#5865e0) HTML color is (234.264°) as Hue Value, (77.64) for Saturation and (61.17) for Lightness, also This Color Can be composed of other Color Systems With different percents if they mixed depending on values that give, You can know about other values by looking in the table below and know more about Tones, Shades and tints and more things.\n\n### Full List Information for #5865e0 Hex Code\n\nThe Values Listed Below is an accurate value for #5865e0 color Which You Can use for many purposes\n\nColor Name\nHex Code\nRGB\nRGB Percent\nCMYK\nHSV(HSB)\nHSL\nHSI\nXYZ\nxyY\nCIE-LAB\nCIE-LCH\nCIE-LUV\nHunter-Lab\nBinary\nOctal\n\n### RGB Percent for #5865e0 Hex Code\n\nThe Percents Of Colors In RGB System Which Produce This Color If Composed Together\n\n34.50%\n39.60%\n87.84%\n\n### CMYK Percent for #5865e0 Hex Code\n\nThe Percents Of Colors In CMYK System Which Produce This Color If Composed Together\n\n60.71%\n54.91%\n0%\n12.15%\nShades for #5865e0 Hex Color\n\na shades created by adding black color to any pure color, and in the gradient below we see the pure color in the and then the percent of black color increase when we move and this will make the darkness increase until we arrive at the max which means that the color is the whole black.\n\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#4f5ac9 rgb(79, 90, 201) hsl(234, 53, 54)\n#4f5ac9\n#4650b3 rgb(70, 80, 179) hsl(234, 43, 48)\n#4650b3\n#3d469c rgb(61, 70, 156) hsl(234, 43, 42)\n#3d469c\n#343c86 rgb(52, 60, 134) hsl(234, 44, 36)\n#343c86\n#2c326f rgb(44, 50, 111) hsl(234, 43, 30)\n#2c326f\n#232859 rgb(35, 40, 89) hsl(234, 43, 24)\n#232859\n#1a1e43 rgb(26, 30, 67) hsl(234, 44, 18)\n#1a1e43\n#11142c rgb(17, 20, 44) hsl(233, 44, 11)\n#11142c\n#080a16 rgb(8, 10, 22) hsl(231, 46, 5)\n#080a16\n#000000 rgb(0, 0, 0) hsl(0, 0, 0)\n#000000\nTints for #5865e0 Hex Color\n\na tints created by adding white color to any pure color, and in the gradient below we see the pure color in the and then the percent of white color increase when we move and this will make the lightness increase until we arrive at the max which means that the color is whole white.\n\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#6772e3 rgb(103, 114, 227) hsl(234, 68, 64)\n#6772e3\n#7681e6 rgb(118, 129, 230) hsl(234, 69, 68)\n#7681e6\n#868fe9 rgb(134, 143, 233) hsl(234, 69, 71)\n#868fe9\n#969eec rgb(150, 158, 236) hsl(234, 69, 75)\n#969eec\n#a6adef rgb(166, 173, 239) hsl(234, 69, 79)\n#a6adef\n#b7bdf2 rgb(183, 189, 242) hsl(233, 69, 83)\n#b7bdf2\n#c8cdf5 rgb(200, 205, 245) hsl(233, 69, 87)\n#c8cdf5\n#daddf8 rgb(218, 221, 248) hsl(233, 68, 91)\n#daddf8\n#eceefb rgb(236, 238, 251) hsl(232, 65, 95)\n#eceefb\n#ffffff rgb(255, 255, 255) hsl(0, 0, 100)\n#ffffff\nTones for #5865e0 Hex Color\n\na tones created by adding a mixing of White color and Black color which formed gray color to any pure color, and in the gradient below we see the pure color in the and then the percent of white color and black color (gray color) increase when we move and this will make the color change to gray until we arrive at the max which means that the color is the whole gray.\n\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#6571df rgb(101, 113, 223) hsl(234, 65, 63)\n#6571df\n#737ddf rgb(115, 125, 223) hsl(234, 62, 66)\n#737ddf\n#8089df rgb(128, 137, 223) hsl(234, 59, 68)\n#8089df\n#8e96df rgb(142, 150, 223) hsl(234, 55, 71)\n#8e96df\n#9ba2df rgb(155, 162, 223) hsl(233, 51, 74)\n#9ba2df\n#a9aedf rgb(169, 174, 223) hsl(234, 45, 76)\n#a9aedf\n#b7bbdf rgb(183, 187, 223) hsl(234, 38, 79)\n#b7bbdf\n#c4c7df rgb(196, 199, 223) hsl(233, 29, 82)\n#c4c7df\n#d2d3df rgb(210, 211, 223) hsl(235, 16, 84)\n#d2d3df\n#dfdfdf rgb(223, 223, 223) hsl(0, 0, 87)\n#dfdfdf\n\n### Schemes for #5865e0 Color Code\n\nWe can define the\n\n## colour schemes\n\nas an organized colour because all the values between these colours will stay without changes while the change will apply only on one value which is the (Hue) so it’s a set of colors work together to produce unified aesthetic\n\n#### Complementary Color\n\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#dfd258 rgb(223, 210, 88) hsl(54, 67, 60)\n#dfd258\n\n#### Split Complementary Color\n\n#df8e58 rgb(223, 142, 88) hsl(24, 67, 60)\n#df8e58\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#a8df58 rgb(168, 223, 88) hsl(84, 67, 60)\n#a8df58\n\n#### Triadic Color\n\n#df5865 rgb(223, 88, 101) hsl(354, 67, 60)\n#df5865\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#65df58 rgb(101, 223, 88) hsl(114, 67, 60)\n#65df58\n\n#### Tetradic Color\n\n#dfd258 rgb(223, 210, 88) hsl(54, 67, 60)\n#dfd258\n#58df8e rgb(88, 223, 142) hsl(144, 67, 60)\n#58df8e\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#df58a8 rgb(223, 88, 168) hsl(324, 67, 60)\n#df58a8\n\n#### Analogous Color\n\n#8e58df rgb(142, 88, 223) hsl(264, 67, 60)\n#8e58df\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#58a8df rgb(88, 168, 223) hsl(204, 67, 60)\n#58a8df\n\n#### Monochromatic Color\n\n#11142c rgb(17, 20, 44) hsl(233, 44, 11)\n#11142c\n#232859 rgb(35, 40, 89) hsl(234, 43, 24)\n#232859\n#343c86 rgb(52, 60, 134) hsl(234, 44, 36)\n#343c86\n#4650b3 rgb(70, 80, 179) hsl(234, 43, 48)\n#4650b3\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#5a67e6 rgb(90, 103, 230) hsl(234, 73, 62)\n#5a67e6\n#5c6aec rgb(92, 106, 236) hsl(234, 79, 64)\n#5c6aec\n#5f6df2 rgb(95, 109, 242) hsl(234, 84, 66)\n#5f6df2\n#6170f8 rgb(97, 112, 248) hsl(234, 91, 67)\n#6170f8\n\n### Black And White With #5865e0 Color Code\n\nHTML Color Code Applied to The Black and White Background It's Also A basic of Color Schemes\n\nSimilar for #5865e0 Hex Color\n\nThe Colors Closest to this Color with a tiny difference Which you can use as Inspirational Alternatives Colors\n\n#5870df rgb(88, 112, 223) hsl(229, 67, 60)\n#5870df\n#586edf rgb(88, 110, 223) hsl(230, 67, 60)\n#586edf\n#586bdf rgb(88, 107, 223) hsl(231, 67, 60)\n#586bdf\n#5869df rgb(88, 105, 223) hsl(232, 67, 60)\n#5869df\n#5867df rgb(88, 103, 223) hsl(233, 67, 60)\n#5867df\n#5865e0 rgb(88, 101, 224) hsl(234, 68, 61)\n#5862df rgb(88, 98, 223) hsl(235, 67, 60)\n#5862df\n#5860df rgb(88, 96, 223) hsl(236, 67, 60)\n#5860df\n#585edf rgb(88, 94, 223) hsl(237, 67, 60)\n#585edf\n#585bdf rgb(88, 91, 223) hsl(238, 67, 60)\n#585bdf\n#5859df rgb(88, 89, 223) hsl(239, 67, 60)\n#5859df\n\n### Codes for #5865e0 Hex Color\n\nMany Codes below applied with the color itself which you can use in easily why with your codes\n\nText With Hex Code #5865e0\n\nThis Is An Example Of Text With this Hexadicmal Color\n\n• ``` <p style=\"color: #5865e0;\">  Content Here </p> ```\n``` .myTextColor {  color: #5865e0; } ```\n• ``` <p style=\"color:rgb(88, 101, 224);\">  Content Here </p> ```\n``` .myTextColor {  color:rgb(88, 101, 224); } ```\n• ``` <p style=\"color:rgba(88, 101, 224, 0.4);\">  Content Here </p> ```\n``` .myTextColor {  color:rgba(88, 101, 224, 0.4); } ```\n• ``` <p style=\"color:hsl(234.264°, 77.64, 61.17);\">  Content Here </p> ```\n``` .myTextColor {  color:hsl(234.264°, 77.64, 61.17); } ```\n• ``` <p style=\"color:hsla(234.264°, 77.64, 61.17, 0.4);\">  Content Here </p> ```\n``` .myTextColor {  color:hsla(234.264°, 77.64, 61.17, 0.4); } ```\n\nBackground With Hex Code\n\n• ``` <div style=\"background-color: #5865e0;\">  Content Here </div> ```\n``` .myBackgroundColor {  background-color: #5865e0; } ```\n• ``` <div style=\"background-color:rgb(88, 101, 224);\">  Content Here </div> ```\n``` .myBackgroundColor {  background-color:rgb(88, 101, 224); } ```\n• ``` <div style=\"background-color:rgb(88, 101, 224, 0.4);\">  Content Here </div> ```\n``` .myBackgroundColor {  background-color:rgb(88, 101, 224, 0.4); } ```\n• ``` <div style=\"background-color:hsl(234.264°, 77.64, 61.17);\">  Content Here </div> ```\n``` .myBackgroundColor {  background-color:hsl(234.264°, 77.64, 61.17); } ```\n• ``` <div style=\"background-color:hsla(234.264°, 77.64, 61.17, 0.4);\">  Content Here </div> ```\n``` .myBackgroundColor {  background-color:hsla(234.264°, 77.64, 61.17, 0.4); } ```\n\nBorder With Hex Code\n\n• ``` <div style=\"border: 1px solid #5865e0;\">  Content Here </div> ```\n``` .myBorder {  border: 1px solid #5865e0; } ```\n\n• ``` <div style=\"border: 1px solid rgb(88, 101, 224);\">  Content Here </div> ```\n``` .myBorder {  border: 1px solid rgb(88, 101, 224); } ```\n\n• ``` <div style=\"border: 1px solid rgba(88, 101, 224, 0.4);\">  Content Here </div> ```\n``` .myBorder {  border: 1px solid rgba(88, 101, 224, 0.4); } ```\n\n• ``` <div style=\"border: 1px solid hsl(234.264°, 77.64, 61.17);\">  Content Here </div> ```\n``` .myBorder {  border: 1px solid hsl(234.264°, 77.64, 61.17); } ```\n\n• ``` <div style=\"border: 1px solid hsla(234.264°, 77.64, 61.17, 0.4);\">  Content Here </div> ```\n``` .myBorder {  border: 1px solid hsla(234.264°, 77.64, 61.17, 0.4); } ```\n\nText Shadow With Hex Code\n\nThis Is An Example Of Text With Hexadicmal Color #5865e0\n\n• ``` <p style=\"text-shadow: 2px 2px 2px #5865e0;\">  Content Here </p> ```\n``` .myTextShadow {  text-shadow: 2px 2px 2px #5865e0; } ```\n\n• ``` <p style=\"text-shadow: 2px 2px 2px rgb(88, 101, 224);\">  Content Here </p> ```\n``` .myTextShadow {  text-shadow: 2px 2px 2px rgb(88, 101, 224); } ```\n\n• ``` <p style=\"text-shadow: 2px 2px 2px rgba(88, 101, 224, 0.4);\">  Content Here </p> ```\n``` .myTextShadow {  text-shadow: 2px 2px 2px rgba(88, 101, 224, 0.4); } ```\n\n• ``` <p style=\"text-shadow: 2px 2px 2px hsl(234.264°, 77.64, 61.17);\">  Content Here </p> ```\n``` .myTextShadow {  text-shadow: 2px 2px 2px hsl(234.264°, 77.64, 61.17); } ```\n\n• ``` <p style=\"text-shadow: 2px 2px 2px hsla(234.264°, 77.64, 61.17, 0.4);\">  Content Here </p> ```\n``` .myTextShadow {  text-shadow: 2px 2px 2px hsla(234.264°, 77.64, 61.17, 0.4); } ```\n\nButton With Hex Code\n\n• ``` <button style=\"background: #5865e0;\">  Button Name </button> ```\n``` .myButton {  background: #5865e0; } ```\n• ``` <button style=\"background: rgb(88, 101, 224);\">  Button Name </button> ```\n``` .myButton {  background: rgb(88, 101, 224); } ```\n• ``` <button style=\"background: rgba(88, 101, 224, 0.4);\">  Button Name </button> ```\n``` .myButton {  background: rgba(88, 101, 224, 0.4); } ```\n• ``` <button style=\"background: hsl(234.264°, 77.64, 61.17);\">  Button Name </button> ```\n``` .myButton {  background: hsl(234.264°, 77.64, 61.17); } ```\n• ``` <button style=\"background: hsla(234.264°, 77.64, 61.17, 0.4);\">  Button Name </button> ```\n``` .myButton {  background: hsla(234.264°, 77.64, 61.17, 0.4); } ```\n\nInput With Hex Code\n\n• ``` <input type='text' style=\"background: #5865e0; /\"> ```\n``` .myField {  background: #5865e0; } ```\n• ``` <input type='text' style=\"background: rgb(88, 101, 224); /\"> ```\n``` .myField {  background: rgb(88, 101, 224); } ```\n• ``` <input type='text' style=\"background: rgba(88, 101, 224, 0.4); /\"> ```\n``` .myField {  background: rgba(88, 101, 224, 0.4); } ```\n• ``` <input type='text' style=\"background: hsl(234.264°, 77.64, 61.17); /\"> ```\n``` .myField {  background: hsl(234.264°, 77.64, 61.17); } ```\n• ``` <input type='text' style=\"background: hsla(234.264°, 77.64, 61.17, 0.4); /\"> ```\n``` .myField {  background: hsla(234.264°, 77.64, 61.17, 0.4); } ```" ]

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[ "# Data type of case labels of switch statement in C++?\n\nIn C++ switch statement, the expression of each case label must be an integer constant expression.\n\nFor example, the following program fails in compilation.\n\n `/* Using non-const in case label */` `#include ` `int` `main() ` `{ ` `  ``int` `i = 10; ` `  ``int` `c = 10; ` `  ``switch``(c)  ` `  ``{ ` `    ``case` `i: ``// not a \"const int\" expression ` `         ``printf``(``\"Value of c = %d\"``, c); ` `         ``break``; ` `    ``/*Some more cases */` `                    `  `  ``} ` `  ``return` `0; ` `} `\n\nPutting const before i makes the above program work.\n\n `#include ` `int` `main() ` `{ ` `  ``const` `int` `i = 10; ` `  ``int` `c = 10; ` `  ``switch``(c)  ` `  ``{ ` `    ``case` `i:  ``// Works fine ` `         ``printf``(``\"Value of c = %d\"``, c); ` `         ``break``; ` `    ``/*Some more cases */` `                    `  `  ``} ` `  ``return` `0; ` `} `\n\nNote : The above fact is only for C++. In C, both programs produce error. In C, using a integer literal does not cause error.\n\nProgram to find the largest number between two numbers using switch case:\n\n `#include ` `int` `main() ` `{ ` `  ``int` `n1=10,n2=11; ` ` `  `  ``//n1 > n2 (10 > 11) is false so using logical operator '>', n1 > n2 produces 0  ` `  ``//(0 means false, 1 means true) So, case 0 is executed as 10 > 11 is false.  ` `  ``//Here we have used type cast to convert boolean to int, to avoid warning. ` ` `  `  ``switch``((``int``)(n1 > n2))  ` `  ``{ ` `    ``case` `0:   ` `         ``printf``(````\"%d is the largest \"````, n2); ` `         ``break``; ` `    ``default``: ` `         ``printf``(````\"%d is the largest \"````, n1); ` `  ``} ` ` `  `  ``//n1 < n2 (10 < 11) is true so using logical operator '>', n1 < n2 produces 1  ` `  ``//(1 means true, 0 means false) So, default is executed as we don't have case 1 to be executed. ` ` `  `  ``switch``((``int``)(n1 < n2)) ` `  ``{ ` `    ``case` `0:   ` `         ``printf``(````\"%d is the largest \"````, n1); ` `         ``break``; ` `    ``default``: ` `         ``printf``(````\"%d is the largest \"````, n2); ` `  ``} ` ` `  `  ``return` `0; ` `} ` `//This code is contributed by Santanu `\n\n## tags:\n\nC C-Loops & Control Statements C" ]

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[ "Cody\n\n# Problem 42644. MATLAB Basic: rounding IV\n\nSolution 1038819\n\nSubmitted on 30 Oct 2016 by Sofiya Onyshkevych\n• Size: 10\n• This is the leading solution.\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 = -8.8; y_correct = -8; assert(isequal(round_x(x),y_correct))\n\nans = -8\n\n2   Pass\nx = -8.4; y_correct = -8; assert(isequal(round_x(x),y_correct))\n\nans = -8\n\n3   Pass\nx = 8.8; y_correct = 9; assert(isequal(round_x(x),y_correct))\n\nans = 9\n\n4   Pass\nx = 8.4; y_correct = 9; assert(isequal(round_x(x),y_correct))\n\nans = 9\n\n5   Pass\nx = 8.49; y_correct = 9; assert(isequal(round_x(x),y_correct))\n\nans = 9\n\n6   Pass\nx = 128.52; y_correct = 129; assert(isequal(round_x(x),y_correct))\n\nans = 129\n\n7   Pass\nx = pi; y_correct = 4; assert(isequal(round_x(x),y_correct))\n\nans = 4" ]

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[ "GURUFOCUS.COM » STOCK LIST » USA » NAS » Canopy Growth Corp (NAS:CGC) » Definitions » Total Assets\nSwitch to:\n\n# Canopy Growth (NAS:CGC) Total Assets\n\n: \\$1,648.8 Mil (As of Jun. 2023)\nView and export this data going back to 2014. Start your Free Trial\n\nCanopy Growth's Total Assets for the quarter that ended in Jun. 2023 was \\$1,648.8 Mil.\n\nDuring the past 12 months, Canopy Growth's average Total Assets Growth Rate was -29.60% per year. During the past 3 years, the average Total Assets Growth Rate was -30.20% per year. During the past 5 years, the average Total Assets Growth Rate was 6.00% per year. During the past 10 years, the average Total Assets Growth Rate was 284.60% per year.\n\nDuring the past 13 years, Canopy Growth's highest 3-Year average Total Assets Growth Rate was 5751.00%. The lowest was -30.20%. And the median was 216.30%.\n\nTotal Assets is connected with ROA %. Canopy Growth's annualized ROA % for the quarter that ended in Jun. 2023 was -6.69%. Total Assets is also linked to Revenue through Asset Turnover. Canopy Growth's Asset Turnover for the quarter that ended in Jun. 2023 was 0.05.\n\n## Canopy Growth Total Assets Historical Data\n\nThe historical data trend for Canopy Growth's Total Assets can be seen below:\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\n Canopy Growth Quarterly Data Sep18 Dec18 Mar19 Jun19 Sep19 Dec19 Mar20 Jun20 Sep20 Dec20 Mar21 Jun21 Sep21 Dec21 Mar22 Jun22 Sep22 Dec22 Mar23 Jun23 Total Assets", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "2,747.21 2,550.62 2,223.15 1,782.58 1,648.75\n\n## Canopy Growth Total Assets Calculation\n\nTotal Assets are all the assets a company owns.\n\nFrom the capital sources of the assets, some of the assets are funded through shareholder's paid in capital and retained earnings of the business. Others are funded through borrowed money.\n\nCanopy Growth's Total Assets for the fiscal year that ended in Mar. 2023 is calculated as\n\n Total Assets = Total Equity (A: Mar. 2023 ) + Total Liabilities (A: Mar. 2023 ) = 555.45 + 1227.126 = 1,782.6\n\nCanopy Growth's Total Assets for the quarter that ended in Jun. 2023 is calculated as\n\n Total Assets = Total Equity (Q: Jun. 2023 ) + Total Liabilities (Q: Jun. 2023 ) = 637.893 + 1010.86 = 1,648.8\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\nCanopy Growth  (NAS:CGC) Total Assets Explanation\n\nTotal Assets is connected with ROA %.\n\nCanopy Growth's annualized ROA % for the quarter that ended in Jun. 2023 is\n\n ROA % = Net Income (Q: Jun. 2023 ) / ( (Total Assets (Q: Mar. 2023 ) + Total Assets (Q: Jun. 2023 )) / count ) = -114.772 / ( (1782.575 + 1648.754) / 2 ) = -114.772 / 1715.6645 = -6.69 %\n\nNote: The Net Income data used here is four times the quarterly (Jun. 2023) data.\n\nIn the article Joining The Dark Side: Pirates, Spies and Short Sellers, James Montier reported that In their US sample covering the period 1968-2003, Cooper et al find that firms with low asset growth outperformed firms with high asset growth by an astounding 20% p.a. equally weighted. Even when controlling for market, size and style, low asset growth firms outperformed high asset growth firms by 13% p.a. Therefore a company with fast asset growth may underperform.\n\nTotal Assets is linked to total revenue through Asset Turnover.\n\nCanopy Growth's Asset Turnover for the quarter that ended in Jun. 2023 is\n\n Asset Turnover = Revenue (Q: Jun. 2023 ) / ( (Total Assets (Q: Mar. 2023 ) + Total Assets (Q: Jun. 2023 )) / count ) = 81.835 / ( (1782.575 + 1648.754) / 2 ) = 81.835 / 1715.6645 = 0.05\n\n* For Operating Data section: All numbers are indicated by the unit behind each term and all currency related amount are in USD.\n* For other sections: All numbers are in millions except for per share data, ratio, and percentage. All currency related amount are indicated in the company's associated stock exchange currency.\n\nTherefore, if a company grows its Total Assets faster than its Revenue, the Asset Turnover will decline. This might be a warning sign for the business.\n\n## Canopy Growth Total Assets Related Terms\n\nThank you for viewing the detailed overview of Canopy Growth's Total Assets provided by GuruFocus.com. Please click on the following links to see related term pages.\n\n## Canopy Growth (NAS:CGC) Business Description", null, "Industry\nComparable Companies" ]

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[ "# Prove that the line joining the midpoint of parallel sides of a trapezium passes through the point of intersection of diagonals\n\nProve that the line joining the midpoint of parallel sides of a trapezium passes through the point of intersection of diagonals.\n\nI want to use theorems in geometry to solve this question.\n\nThe method using vectors is given here.\n\nLet $ABCD$ be the trapezium and let $O$ be the point of intersection of diagonals. I need to prove that one of the lines through $O$ can pass through both midpoints of adjacent sides. Let $AB$ and $CD$ be the parallel lines.\n\nI started by considering a line through $O$ that passes through midpoint $E$ of $AB$. I need to show that if the line is extended, it also passes through midpoint $F$ of $CD$.\n\nI tried using similarity of triangles but it did not help much.", null, "1) Let the line $MO$ intersects the sides of the trapezoid at the points $E$ and $F$. Then $$\\triangle BME \\sim \\triangle AMF \\Rightarrow\\frac {BE}{AF}=\\frac {ME}{MF}$$\n\n2) $$\\triangle EMC \\sim \\triangle FMD \\Rightarrow\\frac {ME}{MF}=\\frac {EC}{FD}\\Rightarrow\\frac {BE}{AF}=\\frac {EC}{FD}$$\n\n3) $$\\triangle AOF \\sim \\triangle COE \\Rightarrow\\frac {EC}{AF}=\\frac {EO}{OF}$$\n\n$$\\triangle BEO \\sim \\triangle DFO \\Rightarrow\\frac {BE}{FD}=\\frac {EO}{OF} \\Rightarrow\\frac {BE}{FD}=\\frac {EC}{AF}$$ So $$\\frac {BE}{AF}=\\frac {EC}{FD}$$ and $$\\frac{BE}{FD}=\\frac {EC}{AF}$$\n\nThen the four points $M,E,O,F$ lie on a straight line\n\n• To whom is this theorem attributed?\n– user366533\nMar 3, 2017 at 20:14" ]

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[ "# Command syntax\n\ntable.getNearest(point).optArg(\"index\", index) → array\n\n# Description\n\nReturn a list of documents closest to a specified point based on a geospatial index, sorted in order of increasing distance.\n\nThe `index` optArg is mandatory. Optional arguments are:\n\n• `max_results`: the maximum number of results to return (default 100).\n• `unit`: Unit for the distance. Possible values are `m` (meter, the default), `km` (kilometer), `mi` (international mile), `nm` (nautical mile), `ft` (international foot).\n• `max_dist`: the maximum distance from an object to the specified point (default 100 km).\n• `geo_system`: the reference ellipsoid to use for geographic coordinates. Possible values are `WGS84` (the default), a common standard for Earth’s geometry, or `unit_sphere`, a perfect sphere of 1 meter radius.\n\nThe return value will be an array of two-item objects with the keys `dist` and `doc`, set to the distance between the specified point and the document (in the units specified with `unit`, defaulting to meters) and the document itself, respectively. The array will be sorted by the values of `dist`.\n\nExample: Return a list of the closest 25 enemy hideouts to the secret base.\n\n``````import com.rethinkdb.gen.ast.Point;\n\nPoint secretBase = r.point(-122.422876,37.777128);\n\nr.table(\"hideouts\")\n.getNearest(secretBase)\n.optArg(\"index\", \"location\")\n.optArg(\"max_results\", 25)\n.run(conn);\n``````\n\nIf you wish to find all points within a certain radius of another point, it’s often faster to use getIntersecting with circle, as long as the approximation of a circle that `circle` generates is sufficient.\n\n# Get more help\n\nCouldn't find what you were looking for?" ]

[ null ]

[ "Why primality is polynomial time, but factorisation is not\n\nDifferentiating between the signature of a number and its value\n\nA brief review: The significance of evidence-based reasoning\n\nIn a paper: The truth assignments that differentiate human reasoning from mechanistic reasoning: The evidence-based argument for Lucas’ Gödelian thesis’, which appeared in the December 2016 issue of Cognitive Systems Research [An16], I briefly addressed the philosophical challenge that arises when an intelligence—whether human or mechanistic—accepts arithmetical propositions as true under an interpretation—either axiomatically or on the basis of subjective self-evidence—without any specified methodology for objectively evidencing such acceptance in the sense of Chetan Murthy and Martin Löb:\n\n“It is by now folklore … that one can view the values of a simple functional language as specifying evidence for propositions in a constructive logic …” … Chetan. R. Murthy: [Mu91], \\S 1 Introduction.\n\n“Intuitively we require that for each event-describing sentence,", null, "$\\phi_{o^{\\iota}}n_{\\iota}$ say (i.e. the concrete object denoted by", null, "$n_{\\iota}$ exhibits the property expressed by", null, "$\\phi_{o^{\\iota}}$), there shall be an algorithm (depending on I, i.e.", null, "$M^{*}$) to decide the truth or falsity of that sentence.” … Martin H Löb: [Lob59], p.165.\n\nDefinition 1 (Evidence-based reasoning in Arithmetic): Evidence-based reasoning accepts arithmetical propositions as true under an interpretation if, and only if, there is some specified methodology for objectively evidencing such acceptance.\n\nThe significance of introducing evidence-based reasoning for assigning truth values to the formulas of a first-order Peano Arithmetic, such as PA, under a well-defined interpretation (see Section 3 in [An16]), is that it admits the distinction:\n\n(1) algorithmically verifiable truth’ (Definition 2}); and\n\n(2) algorithmically computable truth’ (Definition 3).\n\nDefinition 2 (Deterministic algorithm): A deterministic algorithm computes a mathematical function which has a unique value for any input in its domain, and the algorithm is a process that produces this particular value as output.\n\nNote that a deterministic algorithm can be suitably defined as a realizer‘ in the sense of the Brouwer-Heyting-Kolmogorov rules (see [Ba16], p.5).\n\nFor instance, under evidence-based reasoning the formula", null, "$[(\\forall x)F(x)]$ of the first-order Peano Arithmetic PA must always be interpreted weakly under the classical, standard, interpretation of PA (see [An16], Theorem 5.6) in terms of algorithmic verifiability (see [An16], Definition 1); where, if the PA-formula", null, "$[F(x)]$ interprets as an arithmetical relation", null, "$F^{*}(x)$ over", null, "$N$:\n\nDefinition 2 (Algorithmic verifiability): The number-theoretical relation", null, "$F^{*}(x)$ is algorithmically verifiable if, and only if, for any natural number", null, "$n$, there is a deterministic algorithm", null, "$AL_{(F,\\ n)}$ which can provide evidence for deciding the truth/falsity of each proposition in the finite sequence", null, "$\\{F^{*}(1), F^{*}(2), \\ldots, F^{*}(n)\\}$.\n\nWhereas", null, "$[(\\forall x)F(x)]$ must always be interpreted strongly under the finitary interpretation of PA (see [An16], Theorem 6.7) in terms of algorithmic computability ([An16], Definition 2), where:\n\nDefinition 3 (Algorithmic computability): The number theoretical relation", null, "$F^{*}(x)$ is algorithmically computable if, and only if, there is a deterministic algorithm", null, "$AL_{F}$ that can provide evidence for deciding the truth/falsity of each proposition in the denumerable sequence", null, "$\\{F^{*}(1), F^{*}(2), \\ldots\\}$.\n\nThe significance of the distinction between algorithmically computable reasoning based on algorithmically computable truth, and algorithmically verifiable reasoning based on algorithmically verifiable truth, is that it admits the following, hitherto unsuspected, consequences:\n\n(i) PA has two well-defined interpretations over the domain", null, "$N$ of the natural numbers (including", null, "$0$):\n\n(a) the weak non-finitary standard interpretation", null, "$I_{PA(N, SV)}$ ([An16], Theorem 5.6),\n\nand\n\n(b) a strong finitary interpretation", null, "$I_{PA(N, SC)}$ ([An16], Theorem 6.7);\n\n(ii) PA is non-finitarily consistent under", null, "$I_{PA(N, SV)}$ ([An16], Theorem 5.7);\n\n(iii) PA is finitarily consistent under", null, "$I_{PA(N, SC)}$ ([An16], Theorem 6.8).\n\nThe significance of evidence-based reasoning for Computational Complexity\n\nIn this investigation I now show the relevance of evidence-based reasoning, and of distinguishing between algorithmically verifiable and algorithmically computable number-theoretic functions (as defined above), for Computational Complexity is that it assures us a formal foundation for placing in perspective, and complementing, an uncomfortably counter-intuitive entailment in number theory—Theorem 2 below—which has been treated by conventional wisdom as sufficient for concluding that the prime divisors of an integer cannot be proven to be mutually independent.\n\nHowever, I show there that such informally perceived barriers are, in this instance, illusory; and that admitting the above distinction illustrates:\n\n(a) Why the prime divisors of an integer are mutually independent Theorem 2;\n\n(b) Why determining whether the signature (Definition 3 below) of a given integer", null, "$n$—coded as the key in a modified Bazeries-cylinder (see Definition 7 of this paper) based combination lock—is that of a prime, or not, can be done in polynomial time", null, "$O(log_{_{e}}n)$ (Corollary 4 of this paper); as compared to the time", null, "$\\ddot{O}(log_{_{e}}^{15/2}n)$ given by Agrawal et al in [AKS04], and improved to", null, "$\\ddot{O}(log_{_{e}}^{6}n)$ by Lenstra and Pomerance in [LP11], for determining whether the value of a given integer", null, "$n$ is that of a prime or not.\n\n(c) Why it can be cogently argued that determining a factor of a given integer cannot be polynomial time.\n\nDefinition 4 (Signature of a number): The signature of a given integer", null, "$n$ is the sequence", null, "$a_{_{n,i}}$ where", null, "$n + a_{_{n,i}} \\equiv 0\\ mod\\ (p_{_{i}})$ for all primes", null, "$p_{_{i}}\\ such\\ that\\ 1\\leq i \\leq \\pi(\\sqrt{n})$.\n\nUnique since, if", null, "$p_{_{\\pi(\\sqrt{m})+1}}^{2} > m \\geq p_{_{\\pi(\\sqrt{m})}}^{2}$ and", null, "$p_{_{\\pi(\\sqrt{n})+1}}^{2} > n \\geq p_{_{\\pi(\\sqrt{n})}}^{2}$ have the same signature, then", null, "$|m - n| = c_{_{1}}.\\prod_{i=1}^{\\pi(\\sqrt{m})}p_{_{i}} = c_{_{2}}.\\prod_{i=1}^{\\pi(\\sqrt{n})}p_{_{i}}$; whence", null, "$c_{_{1}} = c_{_{2}} = 0$ since", null, "$\\prod_{i=1}^{k}p_{_{i}} > (\\prod_{i=2}^{k-2}p_{_{i}}).p_{_{k}}^{^{2}} > p_{_{k+1}}^{2}$ for", null, "$k > 4$ by appeal to Bertrand’s Postulate", null, "$2.p_{_{k}} > p_{_{k+1}}$; and the uniqueness is easily verified for", null, "$k \\leq 4$.\n\nDefinition 5 (Value of a number): The value of a given integer", null, "$n$ is any well-defined interpretation—over the domain of the natural numbers—of the (unique) numeral", null, "$[n]$ that represents", null, "$n$ in the first-order Peano Arithmetic PA.\n\nWe note that Theorem 2 establishes a lower limit for [AKS04] and [LP11], because determining the signature of a given integer", null, "$n$ does not require knowledge of the value of the integer as defined by the Fundamental Theorem of Arithmetic.\n\nTheorem 1: (Fundamental Theorem of Arithmetic): Every positive integer", null, "$n > 1$ can be represented in exactly one way as a product of prime powers:", null, "$n=p_{1}^{n_{1}}p_{2}^{n_{2}}\\cdots p_{k}^{n_{k}}=\\prod _{i=1}^{k}p_{i}^{n_{i}}$\n\nwhere", null, "$p_{1} < p_{2} < \\ldots < p_{k}$ are primes and the", null, "$n_{i}$ are positive integers (including", null, "$0$).\n\nAre the prime divisors of an integer mutually independent?\n\nIn this paper I address the query:\n\nQuery 1: Are the prime divisors of an integer", null, "$n$ mutually independent?\n\nDefinition 6 (Independent events): Two events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other.\n\nIntuitively, the prime divisors of an integer seem to be mutually independent by virtue of the Fundamental Theorem of Arithmetic\n\nMoreover, the prime divisors of", null, "$n$ can also be seen to be mutually independent in the usual, linearly displayed, Sieve of Eratosthenes, where whether an integer", null, "$n$ is crossed out as a multiple of a prime", null, "$p$ is obviously independent of whether it is also crossed out as a multiple of a prime", null, "$q \\neq p$:\n\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 …\n\nDespite such compelling evidence, conventional wisdom appears to accept as definitive the counter-intuitive conclusion that although we can see it as true, we cannot mathematically prove the following proposition as true:\n\nProposition 1: Whether or not a prime", null, "$p$ divides an integer", null, "$n$ is independent of whether or not a prime", null, "$q \\neq p$ divides the integer", null, "$n$.\n\nWe note that such an unprovable-but-intuitively-true conclusion makes a stronger assumption than that in Gödel’s similar claim for his arithmetical formula", null, "$[(\\forall x)R(x)]$—whose Gödel-number is", null, "$17Gen\\ r$—in [Go31], p.26(2). Stronger, since Gödel does not assume his proposition to be intuitively true, but shows that though the arithmetical formula with Gödel-number", null, "$17Gen\\ r$ is not provable in his Peano Arithmetic", null, "$P$ yet, for any", null, "$P$-numeral", null, "$[n]$, the formula", null, "$[R(n)]$ whose Gödel-number is", null, "$Sb \\left(r \\begin{array}{c}17 \\\\ Z(n)\\end{array}\\right)$ is", null, "$P$-provable, and therefore meta-mathematically true under any well-defined Tarskian interpretation of", null, "$P$ (cf., [An16], Section 3.).\n\nExpressed in computational terms (see [An16], Corollary 8.3), under any well-defined interpretation of", null, "$P$, Gödel’s formula", null, "$[R(x)]$ translates as an arithmetical relation, say", null, "$R'(x)$, such that", null, "$R'(n)$ is algorithmically verifiable, but not algorithmically computable, as always true over", null, "$N$, since", null, "$[\\neg (\\forall x)R(x)]$ is", null, "$P$-provable ([An16], Corollary 8.2).\n\nWe thus argue that a perspective which denies Proposition 1 is based on perceived barriers that reflect, and are peculiar to, only the argument that:\n\nTheorem 2: There is no deterministic algorithm that, for any given", null, "$n$, and any given prime", null, "$p \\geq 2$, will evidence that the probability", null, "$\\mathbb{P}(p\\ |\\ n)$ that", null, "$p$ divides", null, "$n$ is", null, "$\\frac{1}{p}$, and the probability", null, "$\\mathbb{P}(p\\not|\\ n)$ that", null, "$p$ does not divide", null, "$n$ is", null, "$1 - \\frac{1}{p}$.\n\nProof By a standard result in the Theory of Numbers ([Ste02], Chapter 2, p.9, Theorem 2.1, we cannot define a probability function for the probability that a random", null, "$n$ is prime over the probability space", null, "$(1, 2, 3, \\ldots, )$.\n\n(Compare with the informal argument in [HL23], pp.36-37.)\n\nIn other words, treating Theorem 2 as an absolute barrier does not admit the possibility—which has consequences for the resolution of outstanding problems in both the theory of numbers and computational complexity—that Proposition 1 is algorithmically verifiable, but not algorithmically computable, as true, since:\n\nTheorem 3: For any given", null, "$n$, there is a deterministic algorithm that, given any prime", null, "$p \\geq 2$, will evidence that the probability", null, "$\\mathbb{P}(p\\ |\\ n)$ that", null, "$p$ divides", null, "$n$ is", null, "$\\frac{1}{p}$, and the probability", null, "$\\mathbb{P}(p\\not|\\ n)$ that", null, "$p$ does not divide", null, "$n$ is", null, "$1 - \\frac{1}{p}$.\n\nAuthor’s working archives & abstracts of investigations", null, "" ]

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[ "# How to find the acceleration in an elevator using the increase in weight only (1.05x)\n\n## Homework Statement\n\nA person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads 1.05 times the persons regular weight. Calculate the magnitude and direction of the acceleration of the elevator.\n\nF=ma\n\n## The Attempt at a Solution\n\nFg = 9.8m\nΣF = 9.8(1.05m + m)\n0 = 1.05m + m\n-m = 1.05m\n-1 = 1.05\n\nI know that's not true but I don't really know how to set up the problem\n\nharuspex\nHomework Helper\nGold Member\nΣF = 9.8(1.05m + m)\nWhat forces act on the person during the acceleration?\nWhat acceleration results?\n\nWhat forces act on the person during the acceleration?\nWhat acceleration results?\n\nActually the equation would be ΣF=9.8m-9.8(1.05m+m) according to my logic\nbut ΣF in this case is the difference between the force of gravity and the force of tension\nthe 9.8m part is to find the force of gravity on the person in the motionless elevator and the 9.8(1.05m+m) part is to find the force acting on the person when the elevator starts accelerating upward\nto find ΣF you subtract the force of gravity from the fore of tension since it is larger\nmy physics teacher said that sometimes if you set up equations without numbers the variables will cancel out so I tried that and nothing cancels\nThe only number the problem gives me is 1.05 times the original weight when the elevator starts accelerating\nand since the person is inside the elevator it's a system and the forces acting on the elevator vary directly with the forces acting on the person\n\nharuspex" ]

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[ "# Calculate Y axis distance\n\nD\n\n#### David RF\n\nHi folks, first of all excuse my poor english, I have build this\nfunction for calulate Y axis distance, I wan't to know if there is a\nway to increase performance or improve the algorithm (dgra is called\nmany many times) Thanks.\n\n#include <stdio.h>\n\ndouble ddiv(double a, double b)\n{\nreturn b == 0.0 ? 0.0 : a / b;\n}\n\ndouble dgra(double d, int *n)\n{\nconst double a[] = {1.25, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0, 6.0,\n7.5, 8.0, 10.0, 12.5};\nconst double *pa = a;\ndouble sign, product;\n\nif (d == 0.0) {\nif (n) *n = 0;\nreturn 0.0;\n}\nd *= sign = (d > 0.0) ? 1.0 : -1.0; /* Store sign */\n/* Trunc to most significant (by example 15678 turns into\n1.5678) */\nif (d < 1.0) {\nfor (product = 1.0; d < 1.00; product *= 0.10, d *=\n10.0);\n} else {\nfor (product = 1.0; d > 10.0; product *= 10.0, d *=\n0.10);\n}\nd *= 1.1; /* Increase number 10% */\nwhile (*pa < d) pa++; /* Search optimal value */\n/* Must be divide by n numbers */\nif (n) {\nif (*pa == 1.5 || *pa == 3.0 || *pa == 6.0)\n*n = 3; else\nif (*pa == 2.0 || *pa == 4.0 || *pa == 8.0)\n*n = 4; else\n*n = 5;\n}\nreturn *pa * sign * product;\n}\n\nint main(void)\n{\n/* Test */\nconst double a[] = {112.8, -13.245, 121.5, 27894.0, 0.22452,\n463.20, 7094.230, 235.20, 1420.04, 992312.52, 0.0};\nconst double *pa = a;\ndouble d, f;\nint n;\n\nwhile (*pa) {\nd = dgra(*pa, &n);\nf = ddiv(d, n);\nprintf(\"%f = %f\\n\", *pa, d);\nwhile (n >= 0) {\nprintf(\"\\t%f\\n\", f * n);\nn--;\n}\npa++;\n}\nreturn 0;\n}\n\nS\n\n#### Stefan Ram\n\nDavid RF said:\nI wan't to know if there is a\nway to increase performance\n\nYes: to measure the run-time of the application under the\ntarget environment (of the customer) and then trying the\noptimization techniques like: replacing double by float,\ntrying compiler-options (like -O3), using a\nprofiler/valgrind to analyze behavior (like cache misses,\nKCacheGrind), loop unroling, cache-friendliness, inline\nfunctions, strength reduction, reduction of register\npressure, writing in assembly, using »const«/»restrict« to\nallow more optimizations.\n\nB\n\n#### Barry Schwarz\n\nHi folks, first of all excuse my poor english, I have build this\nfunction for calulate Y axis distance, I wan't to know if there is a\nway to increase performance or improve the algorithm (dgra is called\nmany many times) Thanks.\n\nIn the code you show, dgra is called only 10 times. Unless this is\nonly an example and you actually call it several thousand times, no\nimprovement is likely to have a noticeable affect.\n#include <stdio.h>\n\ndouble ddiv(double a, double b)\n{\nreturn b == 0.0 ? 0.0 : a / b;\n}\n\nThis function is unnecessary. See comment in main.\ndouble dgra(double d, int *n)\n{\nconst double a[] = {1.25, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0, 6.0,\n7.5, 8.0, 10.0, 12.5};\n\nIf you make a static, it will not get reinitialized each time you call\ndgra which will save some time.\nconst double *pa = a;\n\nSuggestion: If you change pa to an int (not const) and initialize it\nto 0, you can save some time at the end of dgra. To do this, also add\nan integer array parallel to a\nstatic const int b[] = {5, 3, 4, 5, 3, 4, 5, 3, 5, 4, 5,5};\ndouble sign, product;\n\nif (d == 0.0) {\nif (n) *n = 0;\nreturn 0.0;\n}\nd *= sign = (d > 0.0) ? 1.0 : -1.0; /* Store sign */\n\nMultiplication of doubles can be time consuming. The following may be\nfaster.\nif (d > 0.0)\nsign = 1.0;\nelse\n{\nsign = -1.0;\nd = -d;\n}\n/* Trunc to most significant (by example 15678 turns into\n1.5678) */\n\nYou should be aware that if d is 11.5, it will be represented exactly\nbut -\nif (d < 1.0) {\nfor (product = 1.0; d < 1.00; product *= 0.10, d *=\n10.0);\n} else {\nfor (product = 1.0; d > 10.0; product *= 10.0, d *=\n0.10);\n\nafter multiplying by .1, d will not equal 1.15 because that number\ncannot be represented exactly.\n}\nd *= 1.1; /* Increase number 10% */\nwhile (*pa < d) pa++; /* Search optimal value */\n\nSuggestion continued: All expressions in dgra of the form *pa need to\nbe changed to a[pa].\n/* Must be divide by n numbers */\nif (n) {\nif (*pa == 1.5 || *pa == 3.0 || *pa == 6.0)\n*n = 3; else\nif (*pa == 2.0 || *pa == 4.0 || *pa == 8.0)\n*n = 4; else\n*n = 5;\n}\n\nSuggestion completed: This entire sequence can then be replace by\nif (n)\n*n = b[pa];\nreturn *pa * sign * product;\n}\n\nint main(void)\n{\n/* Test */\nconst double a[] = {112.8, -13.245, 121.5, 27894.0, 0.22452,\n463.20, 7094.230, 235.20, 1420.04, 992312.52, 0.0};\nconst double *pa = a;\ndouble d, f;\nint n;\n\nwhile (*pa) {\nd = dgra(*pa, &n);\nf = ddiv(d, n);\n\nYou could replace the call to ddiv with the same one statement that\nddiv contains, just change the variable names. This will eliminate\nf = (n == 0.0 ? 0.0 : d/n);\n\nI\n\n#### Ian Collins\n\nHi folks, first of all excuse my poor english, I have build this\nfunction for calulate Y axis distance, I wan't to know if there is a\nway to increase performance or improve the algorithm (dgra is called\nmany many times) Thanks.\n\n#include<stdio.h>\n\ndouble ddiv(double a, double b)\n{\nreturn b == 0.0 ? 0.0 : a / b;\n}\n\nThis function is unnecessary. See comment in main.\nint main(void)\n{\n/* Test */\nconst double a[] = {112.8, -13.245, 121.5, 27894.0, 0.22452,\n463.20, 7094.230, 235.20, 1420.04, 992312.52, 0.0};\nconst double *pa = a;\ndouble d, f;\nint n;\n\nwhile (*pa) {\nd = dgra(*pa,&n);\nf = ddiv(d, n);\n\nYou could replace the call to ddiv with the same one statement that\nddiv contains, just change the variable names. This will eliminate\n\nAny compiler worth using will do that for you. All you loose by\nremoving the function is clarity - assuming the function name is meaningful!\n\nD\n\n#### David RF\n\nHi folks, first of all excuse my poor english, I have build this\nfunction for calulate Y axis distance, I wan't to know if there is a\nway to increase performance or improve the algorithm (dgra is called\nmany many times) Thanks.\n\nIn the code you show, dgra is called only 10 times.  Unless this is\nonly an example and you actually call it several thousand times, no\nimprovement is likely to have a noticeable affect.\n\n#include <stdio.h>\ndouble ddiv(double a, double b)\n{\nreturn b == 0.0 ? 0.0 : a / b;\n}\n\nThis function is unnecessary.  See comment in main.\n\ndouble dgra(double d, int *n)\n{\nconst double a[] = {1.25, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0, 6.0,\n7.5, 8.0, 10.0, 12.5};\n\nIf you make a static, it will not get reinitialized each time you call\ndgra which will save some time.\nconst double *pa = a;\n\nSuggestion: If you change pa to an int (not const) and initialize it\nto 0, you can save some time at the end of dgra.  To do this, also add\nan integer array parallel to a\nstatic const int b[] = {5, 3, 4, 5, 3, 4, 5, 3, 5, 4, 5,5};\ndouble sign, product;\nif (d == 0.0) {\nif (n) *n = 0;\nreturn 0.0;\n}\nd *= sign = (d > 0.0) ? 1.0 : -1.0; /* Store sign */\n\nMultiplication of doubles can be time consuming.  The following may be\nfaster.\nif (d > 0.0)\nsign = 1.0;\nelse\n{\nsign = -1.0;\nd = -d;\n}\n/* Trunc to most significant (by example 15678 turns into\n1.5678) */\n\nYou should be aware that if d is 11.5, it will be represented exactly\nbut -\nif (d < 1.0) {\nfor (product = 1.0; d < 1.00; product *= 0.10, d *=\n10.0);\n} else {\nfor (product = 1.0; d > 10.0; product *= 10.0, d *=\n0.10);\n\nafter multiplying by .1, d will not equal 1.15 because that number\ncannot be represented exactly.\n}\nd *= 1.1; /* Increase number 10% */\nwhile (*pa < d) pa++; /* Search optimal value */\n\nSuggestion continued: All expressions in dgra of the form *pa need to\nbe changed to a[pa].\n/* Must be divide by n numbers */\nif (n) {\nif (*pa == 1.5 || *pa == 3.0 || *pa == 6.0)\n*n = 3; else\nif (*pa == 2.0 || *pa == 4.0 || *pa == 8.0)\n*n = 4; else\n*n = 5;\n}\n\nSuggestion completed: This entire sequence can then be replace by\nif (n)\n*n = b[pa];\nreturn *pa * sign * product;\n}\nint main(void)\n{\n/* Test */\nconst double a[] = {112.8, -13.245, 121.5, 27894.0, 0.22452,\n463.20, 7094.230, 235.20, 1420.04, 992312.52, 0.0};\nconst double *pa = a;\ndouble d, f;\nint n;\nwhile (*pa) {\nd = dgra(*pa, &n);\nf = ddiv(d, n);\n\nYou could replace the call to ddiv with the same one statement that\nddiv contains, just change the variable names.  This will eliminate\nf = (n == 0.0 ? 0.0 : d/n);\nprintf(\"%f = %f\\n\", *pa, d);\nwhile (n >= 0) {\nprintf(\"\\t%f\\n\", f * n);\nn--;\n}\npa++;\n}\nreturn 0;\n}\n\nThanks Barry\n\nB\n\n#### BartC\n\nconst double a[] = {1.25, 1.5, 2.0, 2.5, 3.0, 4.0, 5.0, 6.0,\n7.5, 8.0, 10.0, 12.5};\nif (*pa == 1.5 || *pa == 3.0 || *pa == 6.0)\n*n = 3; else\nif (*pa == 2.0 || *pa == 4.0 || *pa == 8.0)\n*n = 4; else\n*n = 5;\n\npa always points into the a[] array, and the contents of a[] are fixed? Then\nit might be better to use an index into the array, and use tests such as:\n\nif (index==1 || index==4 || index==6)\n\netc.\nwhile (*pa) {\nd = dgra(*pa, &n);\nf = ddiv(d, n);\nprintf(\"%f = %f\\n\", *pa, d);\nwhile (n >= 0) {\nprintf(\"\\t%f\\n\", f * n);\nn--;\n}\npa++;\n}\n\nComment out the printf() statements, wrap a loop around it to execute it at\nleast a million times (remember to re-initialise pa each time), and find\nsome way of timing the code. Then it will be easy to try different things\nand see if they make a difference. Reinstate the printf() statements every\nso often (and disable the outer loop) to check the output is still right." ]

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[ "Full-text links:\n\nmath.NT\n\n# Title: Hecke algebra isomorphisms and adelic points on algebraic groups\n\nAbstract: Let $G$ denote a linear algebraic group over $\\mathbf{Q}$ and $K$ and $L$ two number fields. Assume that there is a group isomorphism of points on $G$ over the finite adeles of $K$ and $L$, respectively. We establish conditions on the group $G$, related to the structure of its Borel groups, under which $K$ and $L$ have isomorphic adele rings. Under these conditions, if $K$ or $L$ is a Galois extension of $\\mathbf{Q}$ and $G(\\mathbf{A}_{K,f})$ and $G(\\mathbf{A}_{L,f})$ are isomorphic, then $K$ and $L$ are isomorphic as fields. We use this result to show that if for two number fields $K$ and $L$ that are Galois over $\\mathbf{Q}$, the finite Hecke algebras for $\\mathrm{GL}(n)$ (for fixed $n > 1$) are isomorphic by an isometry for the $L^1$-norm, then the fields $K$ and $L$ are isomorphic. This can be viewed as an analogue in the theory of automorphic representations of the theorem of Neukirch that the absolute Galois group of a number field determines the field if it is Galois over $\\mathbf{Q}$.\n Comments: 19 pages - completely rewritten Subjects: Number Theory (math.NT); Algebraic Geometry (math.AG) MSC classes: 11F70, 11R56, 14L10, 20C08, 20G35, 22D20 Cite as: arXiv:1409.1385 [math.NT] (or arXiv:1409.1385v3 [math.NT] for this version)\n\n## Submission history\n\nFrom: Valentijn Karemaker [view email]\n[v1] Thu, 4 Sep 2014 09:54:43 GMT (19kb)\n[v2] Fri, 19 Sep 2014 11:20:45 GMT (20kb)\n[v3] Tue, 4 Aug 2015 13:03:44 GMT (21kb)\n\nLink back to: arXiv, form interface, contact." ]

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[ "Here is what I have so far...\n\n``````import java.util.Scanner;\n\npublic class tst\n{\npublic static void main (String [] args)\n{\nScanner stdIn = new Scanner(System.in);\nString sentence = \"This is the test.\";\nchar response; // user's y/n response\n\nint ctr = 65;\n\ndo\n{\nSystem.out.print (\"Would you like to see a do-while loop execute? (y/n): \");\nfor(ctr = 0; ctr < sentence.length(); ctr++);\n{\nSystem.out.println(sentence.charAt(ctr));\n}\n\nresponse = stdIn.next().charAt(0);\n\n} while (response == 'y' || response == 'Y');\n\n} // end main\n} // end class tst``````\n\nI'm trying to get the program to print out the value of sentence one character per line using the for loop within the do-while loop after the user answers y or Y to the query.\nPlease help. I appreciate the help in advance.\n\nEDIT: I tried to switch the response input below the question presented to the user and when I run the Java Application I get prompted with the question but run into a mess when I answer y.\n\n## Recommended Answers\n\nFirst, use code tags. Press the button CODE and put your code between the tags.\nAlso I tried running your code and it didn't work, even though it is correct. It behaves very strange and the debug gives me strange behavior. Finally I found the problem. When you write (me) …\n\n## All 2 Replies\n\nFirst, use code tags. Press the button CODE and put your code between the tags.\nAlso I tried running your code and it didn't work, even though it is correct. It behaves very strange and the debug gives me strange behavior. Finally I found the problem. When you write (me) something a lot of times, when you see it written you assume that it is correct, but in your case it is wrong:\n\nI write:\n\n``````for (int i=0;i<length;i++)\n{\n\n}\n``````\n\nYou wrote:\n\n``````for(ctr = 0; ctr < sentence.length(); ctr++);\n{\n\n}\n``````\n\nCan you spot the difference:\n\n``````for(ctr = 0; ctr < sentence.length(); ctr++);\n``````\n\nWe use the ';' to terminate commands. So when you put the ';' at that place you terminated the for loop. So only this got executed N times:\n\n``````for(ctr = 0; ctr < sentence.length(); ctr++);\n``````\n\nAnd then after ctr changed frm 0 to sentence.length() the following command executed outside the loop:\n\n``````System.out.println(sentence.charAt(ctr));\n``````\n\nSo basically when you put that ';' at the end of the for loop you practically wrote something like this:\n\n``````for(ctr = 0; ctr < sentence.length(); ctr++) {}\n{\nSystem.out.println(sentence.charAt(ctr));\n}\n``````\n\nThe for executes with no body. Then when the ctr is equals to sentence.length() exits the loop and executes one System.out.println(sentence.charAt(ctr)) which gives you an error.\n\nTry running this and see what happens:\n\n``````int i=0;\nfor (i=0;i<length;i++);\nSystem.out.println(\"i: \"+i);\n``````\n\nThen write this with no ';'\n\n``````int i=0;\nfor (i=0;i<length;i++)\nSystem.out.println(\"i: \"+i);\n``````\ncommented: Helpful explanation. +15\n\nIt runs perfectly! Exactly how I wanted it to. Thank you so much. I actually understand what was wrong rather than copying down what you put.\n\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of 1.21 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge." ]

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[ "# Plastic Deformation at Microscale", null, "## What is plastic deformation?\n\n[avatar user=”Nilgoon Irani” size=”original” align=”left” link=”linkedin.com/in/nilgoonirani“] The article was created by Dr. Nilgoon Irani[/avatar]\n\nPlastic deformation, in general, can be defined as a permanent deformation, a non-reversible change in the geometry of a body under applied stresses (forces). For example, if hit sufficiently hard by a piece of rock or hail, a car body made of metal gets a dent  (irreversible deformation). On the other hand, rubber is an example of a material which exhibits highly elastic behaviour: you can stretch it to a high degree, but once the force is released rubber returns to its original shape. Plastic deformation occurs at a certain critical stress (force) beyond which the deformation is irreversible. This transition is known as yield.  Plastic deformation may occur in most engineering materials including metalssoilsrocksconcretefoams. However, in different materials, the mechanisms responsible for plastic deformation can be different: dislocation motion, vacancy motion, twinning, phase transformation, or viscous flow of amorphous materials.\n\n## Plastic deformation in metals\n\nIn metals, room temperature plastic deformation occurs as a consequence of the collective motion of dislocations gliding on specific slip planes. In continuum plasticity formulations, the effect of individual dislocations is ignored and hence, such descriptions are not suitable for applications such as the development of devices with characteristic length scales in the micrometre range (e.g. MEMS).\n\nIn Fig. 1, the various relevant scales (macro- to nano-scale) determining the response of metals during plastic deformation are demonstrated. As one can observe at macro-scale the plastic zone is governed by continuum plastic flow (Fig. 1(b)). Inside this plastic region, the material is polycrystalline with the plastic deformation being different in different grains (Fig. 1(c)). By magnifying this region, it can be recognized that at the micro-scale plastic deformation occurs as a consequence of the collective motion of dislocations gliding on specific slip planes (Fig. 1(d)). It is the mobility of these dislocations that gives rise to plastic flow at stress levels relatively low compared to the theoretical strength (see ). Finally, at the nano-scale, the atomic arrangement of the material is revealed (Fig. 1(e)). At this scale, dislocations are shown to be irregularities within the crystal structure.", null, "Figure 1: Description of plastic deformation in metals at various length scales. (a) A material under fracture. (b) Phenomenological continuum plasticity (macro-scale). (c) Crystal plasticity (meso-scale). (d) Discrete dislocation plasticity (micro-scale). (e) Atomistic simulations (nano-scale) (reproduced from ).\n\nFigure 1 illustrates that as one approaches smaller length scales the amount of the physics captured in these models increases, yet this is achieved by the expense of heavier and higher calculation times. The difference between the plasticity models at various scales is based on how they describe dislocation motions and interactions:\n\n1. Continuum and (ploy)crystal models (macro- and meso-scales): These models consist of phenomenological continuum and crystal plasticity theories which consider the average motion of groups or densities of dislocations. These theories are employed at the size scale of Figs. 1(a)-1(c), i.e. when the length scale of the specimen exceeds the size of the individual grains or when the number of dislocations in the material is large. For describing plasticity at this scale accounting for dislocation densities will suffice and there is no need to consider individual dislocations. Examples of these models can be found in Groma (1997) and Fleck et al. (1994) .\n2. Discrete dislocation models (micro-scale): These methods describe plastic flow in terms of the dynamics of large numbers of interacting discrete dislocations while taking into account individual behaviour of every dislocation inside crystals (Fig. 1(d)). As a result, these frameworks may give an accurate description of the material behaviour when specimen size gets within the micrometre range. Hence, these models are mainly employed to enhance the understanding of defect interactions and also to provide better assumptions for the parameters used in meso-scale plasticity models. Examples of these methods include Zbib et al. (2002) , van der Giessen and Needleman (1995) , Irani et al. , and Lepinoux and Kubin (1987) .\n3. Atomistic models (nano-scale): The discrete dislocation based methods require governing laws for individual dislocations and their short-range interactions with each other. The constitutive laws at the nano-scale are difficult to obtain experimentally and therefore, they are usually investigated by atomistic models at the size scale of Fig. 1.1(e). For examples of these methods see Bernstein et al. (2009) , and Curtin and Miller (2003) .\n\nHere, as an example of micro-scale plasticity techniques, the discrete dislocation plasticity (DDP) of van der Giessen and Needleman (1995) is presented. In this framework, plastic deformation and flow are described by the nucleation and glide of discrete edge dislocations. Here, a brief and general summary of the DDP formulation in a small strain setting is provided:", null, "Figure 2: Decomposition of the boundary value problem: summation of the dislocation fields, the (˜) fields, and complementary fields, the (ˆ) fields. Here, the inverted T symbols stand for an edge dislocation. For more details see van der Giessen and Needleman (1995) .\n\nConsider a two-dimensional body under plane strain conditions such that at time t, the body contains N edge dislocations, with the dislocations being represented as line singularities in an elastic medium. In this method, to obtain the solution, the problem is decomposed in two additive parts and the field quantities are computed using superposition (see Fig. 2). First, the singular (˜) fields associated with the N dislocations are calculated analytically. In this method, typically the analytically calculated dislocations fields in an infinite medium are used to represent the (˜) fields. Thus, these fields do not give the correct stress and displacement fields at the boundaries. This issue is solved by solving for a complementary problem, where both the boundary conditions and the dislocation fields are applied (see Fig. 2). Thus, the complementary problem is an elasticity boundary value problem with its boundary conditions changing as the dislocation structure evolves. One may solve the complementary problem by employing conventional finite element methods.\n\nAt the beginning of a calculation the crystal is stress- and dislocation-free. New dislocation pairs are generated by simulating Frank-Read sources. In two dimensions, this is mimicked by discrete point sources randomly distributed on the slip planes. These sources generate a dislocation dipole with their Burgers vectors aligned with the slip plane direction. This occurs when the magnitude of the Peach-Koehler force on a source exceeds a critical value for a certain time. Annihilation of two opposite signed dislocations on a slip plane occurs when they are within a material-dependent critical annihilation distance. The magnitude of the glide velocity along the slip direction of a dislocation is taken to be linearly related to the Peach-Koehler force. Obstacles to dislocation motion are modelled as points associated with a slip plane. The dislocations gliding on the slip plane of an obstacle, get pinned as they try to pass through this obstacle. Moreover, obstacles release the pinned dislocations when the Peach-Koehler force on the obstacle exceeds a critical value.", null, "(a)", null, "(b)\n\nFigure 3: (a) Sketch of the uniaxial tension of a dog-bone specimen at the micro-scale with the window over which the dislocation distributions are shown in (b). The crystal is shown in the double slip configuration. (b) A snapshot of DDP predictions of the dislocation and tensile stress σ11 distributions .\n\nFigure 3(a) demonstrates a metallic dog-bone specimen under tensile loading at the micro-scale. As it is shown, at this length-scale, we must treat the specimen as a crystalline material with discrete slip planes. Figure 3(b) illustrates that by using discrete dislocation plasticity we have been able to model the individual behaviour of every dislocation inside the crystal. As a result, we could account for both the stress enhancement due to organized dislocation structures and the stress relaxation arising from dislocation glide. For more information on this topic, the interested reader is referred to .\n\n## References\n\n1. https://en.wikipedia.org/wiki/Dislocation\n2. van der Giessen, E. and Needleman, A. (2003). Dislocation plasticity effects on interfacial fracture. Interface Science, 11(3):291–301.\n3. Groma, I. (1997). Link between the microscopic and mesoscopic length-scale description of the collective behavior of dislocations. Physical Review B, 56(10):5807–5813.\n4. Fleck, N. A., Muller, G. M., Ashby, M. F., and Hutchinson, J. W. (1994). Strain gradient plasticity: theory and experiment. Acta Metallurgica et Materialia, 42(2):475–487.\n5. Zbib, H. M., de la Rubia, T. D., and Bulatov, V. (2002). A multiscale model of plasticity based on discrete dislocation dynamics. Journal of Engineering Materials and Technology, 124:78–87.\n6. van der Giessen, E. and Needleman, A. (1995). Discrete dislocation plasticity: a simple planar model. Modelling and Simulation in Materials Science and Engineering, 3:689–735.\n7. N. Irani , J.J.C. Remmers and V.S. Deshpande, 2015. Finite strain discrete dislocation plasticity in a total Lagrangian setting. Journal of the Mechanics and Physics of Solids, 83 (2015), 160–178.\n8. Lepinoux, J. and Kubin, L. P. (1987). The dynamic organization of dislocation structures: a simulation. Scripta metallurgica, 21(6):833–838.\n9. Bernstein, N., Kermode, J. R., and Csanyi, G. (2009). Hybrid atomistic simulation methods for materials systems. Reports on Progress in Physics, 72(2):026501\n10. Curtin, W. A. and Miller, R. E. (2003). Atomistic/continuum coupling in computational materials science. Modelling and simulation in materials science and engineering, 11(3):R33–R68.\n11. Irani, N., Remmers, J. J. C., Deshpande, V. S., (2017). Finite versus small strain discrete dislocation analysis of cantilever bending of single crystals. Acta Mechanica Sinica, 33.4: 763-777.\n12. Irani, N., 2016. Finite strain discrete dislocation plasticity: applications and new developments. Eindhoven University of Technology, The Netherlands.\n13. Irani, N., Remmers, J. J. C., Deshpande, V. S., (2017). A discrete dislocation analysis of hydrogen-assisted mode-I fracture. Mechanics of Materials 105: 67-79.", null, "#### Be the first to comment\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nHere are my notes from a recent talk I gave on vectorization at a Davis R Users’ Group meeting. Thanks to Vince Buffalo, John Myles White, and Hadley Wickham for their input as I was preparing this. Feedback welcome!\n\nBeginning R users are often told to “vectorize” their code. Here, I try to explain why vectorization can be advantageous in R by showing how R works under the hood.\n\nNow, remember, premature optimization is the root of all evil (Knuth). Don’t start re-writing your code unless the time saved is going to be worth the time invested. Other approaches, like finding a bigger machine or parallelization, could give you more bang for the buck in terms of programming time. But if you understand the nuts and bolts of vectorization in R, it may help you write shorter, simpler, safer, and yes, faster code in the first place.\n\nFirst, let’s acknowledge that vectorization can seem like voodoo. Consider a two math problems, one vectorized, and one not:\n\n$\\begin{bmatrix} 1 \\\\ 2 \\\\ 3 \\end{bmatrix} + \\begin{bmatrix} 1 \\\\ 2 \\\\ 3 \\end{bmatrix} = \\begin{bmatrix} 2 \\\\ 4 \\\\ 6 \\end{bmatrix}$\n\n\\begin{aligned} 1 + 1 = 2 \\\\ 2 + 2 = 4 \\\\ 3 + 3 = 6 \\end{aligned}\n\nWhy on earth should these take a different amount of time to calculate? Linear algebra isn’t magic. In both cases there are three addition operations to perform. So what’s up?\n\n## 1. What on earth is R actually doing?\n\nR is a high-level, interpreted computer language. This means that R takes care of a lot of basic computer tasks for you. For instance, when you type\n\n i <- 5.0\n\nyou don’t have to tell your computer:\n\n• That “5.0” is a floating-point number\n• That “i” should store numeric-type data\n• To find a place in memory for to put “5”\n• To register “i” as a pointer to that place in memory\n\nYou also don’t have to convert i <- 5.0 to binary code. That’s done automatically when you hit ‘Enter’.\n\nWhen you then type\n\ni <- \"foo\"\n\nyou don’t have to tell the computer that i no longer stores an integer but a series of characters that form a string, to store “f”, “o”, and “o”, consecutively, etc.\n\nR figures these things on it’s own, on the fly, as you type commands or source them from a file. This means that running a command in R takes a relatively long time than it might in a lower-level language, such as C. If I am writing in C, I might write\n\nint i\ni = 5\n\nThis tells the computer the i will store data of the type int (integers), and assign the value 5 to it. If I try to assign 5.5 to it, something will go wrong. Depending on my set-up, it might throw an error, or just silently assign 5 to i. But C doesn’t have to figure out what type of data is is represented by i and this is part of what makes it faster.\n\nHere’s another example. If, in R, you type:\n\n2L + 3.5\n\n“OK, what’s the first thing?”\n\n“An integer”\n\n“The second thing?”\n\n“A a floating-point number”\n\n“Do we have a way to deal with adding an integer and a floating-point number”\n\n“Yes! Convert the integer to a floating-point number, then add the two floating point numbers”\n\n[converts integer]\n\n[finds a place in memory for the answer]\n\netc.\n\nIf R were a compiled computer language, like C or FORTRAN, much of this “figuring out” would be accomplished the compilation step, not when the program was run. Compiled programs are translated into binary computer language after they are written, but before they are run and this occurs over the whole program, rather than line-by-line. This allows the compiler to organize the binary machine code in an optimal way for the computer to interpret.\n\nWhat does this have to do with vectorization in R? Well, many R functions are actually written in a a compiled language, such as C, C++, and FORTRAN, and have a small R “wrapper”. For instance, when you inspect the code for fft, the fast fourier transform, you see\n\n> fft\nfunction (z, inverse = FALSE)\n.Call(C_fft, z, inverse)\n<bytecode: 0x7fc261e1b910>\n<environment: namespace:stats>\n\nR is passing the data onto a C function called C_fft. You’ll see this in many R functions. If you look at their source code, it will include .C(), .Call(), or sometimes .Internal() or .Primitive(). These means R is calling a C, C++, or FORTRAN program to carry out operations. However, R still has to interpret the input of the function before passing it to the compiled code. In fft() the compiled codes only after R figures out the data type in z, and also whether to use the default value of inverse. The compiled code is able to run faster than code written in pure R, because the “figuring out” stuff is done first, and it can zoom ahead without the “translation” steps that R needs.\n\nIf you need to run a function over all the results of a vector, you could pass a whole vector through the R function to the compiled code, or you could call the R function repeatedly for each value. If you do the latter, R has to do the “figuring out” stuff, as well as the translation, each time. But if you call it once, with a vector, the “figuring out” part happens just once.\n\nInside the C or FORTRAN code, vectors are actually processed processed using loops or a similar construct. This is inevitable: somehow the computer is going to need to operate on each element of your vector but since this occurs in the compiled code, without the overhead of R functions, it’s much faster.\n\nAnother important component of the speed of vectorized operations is that vectors in R are typed. Despite all of its flexibility, R does have some restrictions on what we can do. All elements of a vector must be the same data type. If I try to do this\n\na <- c(1, 2, FALSE, \"hello\")\n\nI get\n\n> a\n \"1\" \"2\" \"FALSE\" \"hello\"\n> class(a)\n \"character\"\n\nR converts all my data to characters. It can’t handle a vector with different data types.\n\nSo when R needs to perform an operation like\n\nc(1, 2, 3) + c(1, 2, 3)\n\nR only has to ask what types of data are in each vector (2) rather than each element (6).\n\nOne consequence of all this is that fast R code is short. If you can express what you want to do in R in a line or two, with just a few function calls that are actually calling compiled code, it’ll be more efficient than if you write long program, with the added overhead of many function calls. This is not the case in all other languages. Often, in compiled languages, you want to stick with lots of very simple statements, because it allows the compiler to figure out the most efficient translation of the code.\n\n## 2. Everything is a vector\n\nIn R everything is a vector. To quote Tim Smith in “aRrgh: a newcomer’s (angry) guide to R”\n\nAll naked numbers are double-width floating-point atomic vectors of length one. You’re welcome.\n\nThis means that, in R, typing “6” tells R something like\n\n<start vector, type=numeric, length=1>6<end vector>\n\nWhile in other languages, “6” might just be\n\n<numeric>6\n\nSo, while in other languages, it might be more efficient to express something as a single number rather than a length-one vector, in R this is impossible. There’s no advantage to NOT organizing your data as vector. In other languages, short vectors might be better expressed as scalars.\n\n## 3. Linear algebra is a special case\n\nLinear algebra is one of the core functions of a lot of computing, so there are highly optimized programs for linear algebra. Such a program is called a BLAS - basic linear algebra system. R, and a lot of other software, relies on these specialized programs and outsources linear algebra to them. These programs have things like built-in parallel processing, and they may be specialized for the hardware on your computer. So if your calculations can be expressed in actual linear algebra terms, such as matrix multiplication, than it’s almost certainly faster to vectorize them.\n\nNow, there are faster and slower linear algebra libraries, and you can install new ones on your computer and tell R to use them instead of the basic ones. This used to be like putting a new engine in your car, but it’s gotten considerably easier. For certain problems, doing this can considerably speed up code, but results vary depending on the specific linear algebra operations you are using.\n\n## 4. Functionals: Pre-allocating memory, avoiding side effects.\n\nThere are a whole family of functions in R called functionals, or apply functions, which take vectors (or matrices, or lists) of values and apply arbitrary functions to each. Because these can use arbitrary functions, they are NOT compiled. Functionals mostly are written in pure R, and they speed up code only in certain cases.\n\nOne operation that is slow in R, and somewhat slow in all languages, is memory allocation. So one of the slower ways to write a for loop is to resize a vector repeatedly, so that R has to re-allocate memory repeatedly, like this:\n\nj <- 1\nfor (i in 1:10) {\nj[i] = 10\n}\n\nHere, in each repetition of the for loop, R has to re-size the vector and re-allocate memory. It has to find the vector in memory, create a new one that will fit more data, copy the old data over, insert the new data, and erase the old vector. This can get very slow as vectors get big.\n\nIf one pre-allocates a vector that fits all the values, R doesn’t have to re-allocate memory each iteration, and the results can be much faster. Here’ how you’d do that for the above case:\n\nj <- rep(NA, 10)\nfor (i in 1:10) {\nj[i] = 10\n}\n\nThe apply or plyr::*ply functions all actually have for loops inside, but they automatically do things like pre-allocating vector size so you don’t screw it up. This is the main reason that they can be faster.\n\nAnother thing that “ply” functions do help with is avoiding what are known as side effects. When you run a ply function, everything happens inside that function, and nothing changes in your working environment (this is known as “functional programming”). In a for loop, on the other hand, when you do something like for(i in 1:10), you get the leftover i in your environment. This is considered bad practice sometimes. Having a bunch of temporary variables like i lying around could cause problems in your code, especially if you use i for something else later.\n\nI’ve seen arguments that both ply functions make for more expressive, easier to read code, but I’ve seen the same argument for for loops. Once you are used to writing vectorized code in general, though, for loops in R will can seem odd.\n\n## So when might for loops make sense over vectorization?\n\nThere are still situations that it may make sense to use for loops instead of vectorized functions, though. These include:\n\n• Using functions that don’t take vector arguments\n• Loops where each iteration is dependent on the results of previous iterations\n\nNote that the second case is tricky. In some cases where the obvious implementation of an algorithm uses a for loop, there’s a vectorized way around it. For instance, here is a good example of implementing a random walk using vectorized code. In these cases, you often want to call functions that are essentially C/FORTRAN implementations of loop operations to avoid the loop in R. Examples of such functions include cumsum (cumulative sums), rle (counting number of repeated value), and ifelse (vectorized if…else statements).\n\nYour performance penalty for using a for loop instead a vector will be small if the number of iterations is relatively small, and the computational time of the functions called inside your for loop is high, so that actually calling them is a small fraction of your computational time. In these cases, it may make sense to use a for loop, especially if they are more intuitive or easier to read for you.\n\n## Some resources on vectorization\n\n• Good discussion in a couple of blog posts by John Myles White.\n• Some relevant chapters of Hadley Wickham’s Advanced R book: Functionals and code profiling\n• Vectorization is covered in chapters 3 and 4 of the classic text on R’s idiosyncrasies - The R Inferno, by Patrick Burns\n• Here are a bunch of assorted blog posts with good examples of speeding up code with vectorization" ]

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[ "## Spark GraphX - Distributed Graph Computations\n\nSpark GraphX is a graph processing framework built on top of Spark.\n\nGraphX models graphs as property graphs where vertices and edges can have properties.\n\n Caution FIXME Diagram of a graph with friends.\n\nGraphX comes with its own package `org.apache.spark.graphx`.\n\n Tip Import `org.apache.spark.graphx` package to work with GraphX. ``import org.apache.spark.graphx._``\n\n### Graph\n\n`Graph` abstract class represents a collection of `vertices` and `edges`.\n\n``abstract class Graph[VD: ClassTag, ED: ClassTag]``\n\n`vertices` attribute is of type `VertexRDD` while `edges` is of type `EdgeRDD`.\n\n`Graph` can also be described by `triplets` (that is of type `RDD[EdgeTriplet[VD, ED]]`).\n\n``````import org.apache.spark.graphx._\nimport org.apache.spark.rdd.RDD\nval vertices: RDD[(VertexId, String)] =\nsc.parallelize(Seq(\n(0L, \"Jacek\"),\n(1L, \"Agata\"),\n(2L, \"Julian\")))\n\nval edges: RDD[Edge[String]] =\nsc.parallelize(Seq(\nEdge(0L, 1L, \"wife\"),\nEdge(1L, 2L, \"owner\")\n))\n\nscala> val graph = Graph(vertices, edges)\ngraph: org.apache.spark.graphx.Graph[String,String] = org.apache.spark.graphx.impl.GraphImpl@5973e4ec``````\n\n### package object graphx\n\n`package object graphx` defines two type aliases:\n\n• `VertexId` (`Long`) that represents a unique 64-bit vertex identifier.\n\n• `PartitionID` (`Int`) that is an identifier of a graph partition.\n\n### Standard GraphX API\n\n`Graph` class comes with a small set of API.\n\n• Transformations\n\n• `mapVertices`\n\n• `mapEdges`\n\n• `mapTriplets`\n\n• `reverse`\n\n• `subgraph`\n\n• `mask`\n\n• `groupEdges`\n\n• Joins\n\n• `outerJoinVertices`\n\n• Computation\n\n• `aggregateMessages`\n\n### Creating Graphs (Graph object)\n\n`Graph` object comes with the following factory methods to create instances of `Graph`:\n\n• `fromEdgeTuples`\n\n• `fromEdges`\n\n• `apply`\n\n Note The default implementation of `Graph` is GraphImpl.\n\n### GraphImpl\n\n`GraphImpl` is the default implementation of Graph abstract class.\n\nIt lives in `org.apache.spark.graphx.impl` package.\n\n### OLD - perhaps soon to be removed\n\nApache Spark comes with a library for executing distributed computation on graph data, GraphX.\n\n• Apache Spark graph analytics\n\n• GraphX is a pure programming API\n\n• missing a graphical UI to visually explore datasets\n\n• Could TitanDB be a solution?\n\nSuch a situation, in which we need to find the best matching in a weighted bipartite graph, poses what is known as the stable marriage problem. It is a classical problem that has a well-known solution, the Gale–Shapley algorithm.\n\nA popular model of distributed computation on graphs known as Pregel was published by Google researchers in 2010. Pregel is based on passing messages along the graph edges in a series of iterations. Accordingly, it is a good fit for the Gale–Shapley algorithm, which starts with each “gentleman” (a vertex on one side of the bipartite graph) sending a marriage proposal to its most preferred single “lady” (a vertex on the other side of the bipartite graph). The “ladies” then marry their most preferred suitors, after which the process is repeated until there are no more proposals to be made.\n\nThe Apache Spark distributed computation engine includes GraphX, a library specifically made for executing distributed computation on graph data. GraphX provides an elegant Pregel interface but also permits more general computation that is not restricted to the message-passing pattern." ]

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[ "# generic\n\nA property that holds for all $x$ in some residual", null, "", null, "", null, "subset of a Baire space", null, "", null, "$X$ is said to be generic", null, "", null, "", null, "in $X$, or to hold generically in $X$. In the study of generic properties, it is common to state “generically, $P(x)$”, where $P(x)$ is some proposition", null, "", null, "about $x\\in X$. The useful fact about generic properties is that, given countably many generic properties $P_{n}$, all of them hold simultaneously in a residual set, i.e. we have that, generically, $P_{n}(x)$ holds for each $n$.\n\nTitle generic Generic 2013-03-22 13:40:30 2013-03-22 13:40:30 Koro (127) Koro (127) 10 Koro (127) Definition msc 54E52 generically" ]

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[ "Explore", null, "Python Programs", null, "# I Love You Program In Python With Code", null, "Looking for a way to say I love you to your crush or a way to propose and you are a programmer and if also he or she is a programmer then today I will tell you the programmer’s way of proposing which is the I love you program in python.\n\nThis I love you python program will open a big black screen when you run it and it will start writing I love you at the end it will also draw a heart, you can see the preview below.\n\nThis program is made with turtle python library which is a graphics library for python so if you liked this I love you python program you can get the code of this I love you program below.\n\nIf you are a beginner in python programming then don’t worry I will show you everything you need to do to run this program and get you a girlfriend now no programmers will be single.\n\n## Install and Setup Python\n\nIf you are a python programmer who has python installed can skip this step. So to install python you need to go to the python website and click download, it will download the latest version of python.\n\nNow you need to run the installer you can click next, and one screen will come where you need to check add python to path after that python will be installed if you want a more detailed guide you can check out this – Installing and setting up python.\n\n## Installing Python Turtle\n\nAs I have said you will need turtle which is a python library so you need to install this to install you need to open a command prompt paste the below code to install respective python library.\n\nTo install Turtle use this\n\n``````\npip install turtle\n```\n```\n\nIt will take some time after that you can go to the next step.\n\n## Create and Open Python File\n\nNow you need to create a python file in a new folder and open it in a code editor if you don’t have one I will recommend you to use VS Code or you can use any if you have, In VS Code install the python extension you can read more here – Setting python in VS Code.\n\nAfter that copy the below I love you code in the python file and you can run the code by opening command prompt and pasting the below command.\n\n``````\npython filename.py\n```\n```\n\nEnter your python filename where I said filename, this will open a black window and the program will start to run.\n\n## I Love You Code In Python\n\n``````\nimport turtle\n\ndraw = turtle.Turtle()\n\ndef curve():\ndraw.pen(pencolor=\"white\", pensize=3, speed=5)\nfor i in range(200):\ndraw.rt(1)\ndraw.fd(1)\n\ndef heart():\ndraw.pen(pencolor=\"white\",fillcolor=\"red\", pensize=3, speed=5)\ndraw.shape(\"turtle\")\ndraw.shapesize(1,1,1)\ndraw.begin_fill()\ndraw.lt(50)\ndraw.fd(113)\ncurve()\ndraw.lt(120)\ncurve()\ndraw.fd(112)\ndraw.end_fill()\n\ndraw.hideturtle()\n\nwindow = turtle.Screen()\nwindow.bgcolor('black')\n\ndraw.penup()\ndraw.goto(-80,300)\ndraw.pendown()\ndraw.shapesize(1,2,1)\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\n\ndraw.begin_fill()\n\ndraw.fd(160)\ndraw.rt(90)\ndraw.fd(25)\ndraw.rt(90)\ndraw.fd(60)\ndraw.lt(90)\n\ndraw.fd(140)\ndraw.lt(90)\ndraw.fd(60)\ndraw.rt(90)\ndraw.fd(25)\ndraw.rt(90)\ndraw.fd(160)\ndraw.rt(90)\ndraw.fd(25)\ndraw.rt(90)\ndraw.fd(60)\ndraw.lt(90)\ndraw.fd(140)\ndraw.left(90)\ndraw.fd(60)\ndraw.rt(90)\ndraw.fd(25)\n\ndraw.end_fill()\n\ndraw.penup()\ndraw.goto(-550,-20)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=2)\ndraw.begin_fill()\n\ndraw.rt(90)\ndraw.fd(25)\ndraw.rt(90)\ndraw.fd(165)\ndraw.lt(90)\ndraw.fd(115)\ndraw.rt(90)\ndraw.fd(25)\ndraw.rt(90)\ndraw.fd(140)\ndraw.rt(90)\ndraw.fd(190)\ndraw.rt(90)\n\ndraw.end_fill()\n\ndraw.penup()\ndraw.fd(140)\n\ndraw.fd(70)\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\ndraw.begin_fill()\n\ndraw.rt(90)\ndraw.fd(190)\ndraw.lt(90)\ndraw.pendown()\ndraw.circle(60)\ndraw.lt(90)\ndraw.penup()\ndraw.fd(20)\ndraw.rt(90)\ndraw.pendown()\ndraw.circle(40)\ndraw.rt(90)\ndraw.penup()\ndraw.fd(20)\ndraw.lt(90)\n\ndraw.end_fill()\n\ndraw.fd(100)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\ndraw.begin_fill()\n\ndraw.lt(100)\ndraw.fd(120)\ndraw.rt(100)\ndraw.fd(20)\ndraw.rt(80)\ndraw.fd(100)\ndraw.lt(80)\ndraw.fd(20)\ndraw.lt(80)\ndraw.fd(100)\ndraw.rt(80)\ndraw.fd(20)\ndraw.rt(100)\ndraw.fd(120)\ndraw.rt(80)\ndraw.fd(50)\ndraw.lt(180)\n\ndraw.end_fill()\n\ndraw.penup()\ndraw.fd(100)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\ndraw.begin_fill()\n\ndraw.lt(90)\ndraw.fd(120)\ndraw.rt(90)\ndraw.fd(80)\ndraw.rt(90)\ndraw.fd(20)\ndraw.rt(90)\ndraw.fd(60)\ndraw.lt(90)\ndraw.fd(30)\ndraw.lt(90)\ndraw.fd(60)\ndraw.rt(90)\ndraw.fd(20)\ndraw.rt(90)\ndraw.fd(60)\ndraw.lt(90)\ndraw.fd(30)\ndraw.lt(90)\ndraw.fd(60)\ndraw.rt(90)\ndraw.fd(20)\ndraw.rt(90)\ndraw.fd(80)\n\ndraw.end_fill()\n\ndraw.penup()\ndraw.rt(180)\ndraw.fd(200)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=2)\ndraw.begin_fill()\n\ndraw.lt(90)\ndraw.fd(50)\ndraw.lt(30)\ndraw.fd(80)\ndraw.rt(120)\ndraw.fd(20)\ndraw.rt(60)\ndraw.fd(60)\ndraw.lt(180)\ndraw.rt(60)\ndraw.fd(60)\ndraw.rt(60)\ndraw.fd(20)\ndraw.rt(120)\ndraw.fd(80)\ndraw.lt(30)\ndraw.fd(50)\ndraw.rt(90)\ndraw.fd(20)\ndraw.rt(180)\n\ndraw.end_fill()\n\ndraw.penup()\ndraw.fd(120)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\ndraw.begin_fill()\n\ndraw.circle(60)\ndraw.lt(90)\ndraw.penup()\ndraw.fd(20)\ndraw.pendown()\ndraw.rt(90)\ndraw.circle(40)\ndraw.rt(90)\ndraw.penup()\ndraw.fd(20)\ndraw.lt(90)\n\ndraw.end_fill()\n\ndraw.fd(100)\ndraw.circle(60, extent=60)\ndraw.pendown()\n\ndraw.pen(pencolor=\"white\",fillcolor=\"yellow\", pensize=3, speed=8)\ndraw.begin_fill()\n\ndraw.lt(30)\n\ndraw.fd(85)\ndraw.lt(90)\ndraw.fd(20)\ndraw.lt(90)\ndraw.fd(70)\ndraw.circle(-20, extent=180)\ndraw.fd(70)\ndraw.lt(90)\n\ndraw.fd(20)\ndraw.lt(90)\ndraw.fd(85)\ndraw.circle(40, extent=180)\n\ndraw.end_fill()\n\ndraw.penup()\n\ndraw.rt(180)\ndraw.fd(35)\ndraw.lt(90)\ndraw.fd(140)\ndraw.lt(90)\ndraw.pendown()\n\nheart()\n\nturtle.done()\n```\n```\n\nI hope this I love you python program impressed your crush and became your girlfriend. If not don’t worry, next time or you can run this it will say I love you to you.\n\nThis was for today. If you found this helpful, great if you want more interesting python programs like this than join our Telegram Channel to get the latest updates of our blog.\n\nHere are some more python guides:\n\nThank you for watching, have a nice day 🙂" ]

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[ "# what is the difference between sss and sas\n\nHi!\n@   imbaai...  correctly give the differences between SSS and SAS congruency.\n\nGood effort!\nKeep writing!!!\n\nCheers!\n\n• 0\n\ns-s-s- side side side\nIf the 3 sides of a triangle is congruent to the three side of another..then the two triangles are congruent\n\ns-a-s- side angle side\nIf 2 sides and the angle in between those two sides are congruent to the other triangle's 2 sides and angle, then the two triangles are congruent\n\n• 0\n\nhi imbaii\n\ns-s-s- side side side\n\nunder correspondence three sides of one triangle is equal to corresponding three sides another triangle.so they are congruent.\n\ns-a-s side angle side\n\nunder correspondence two angle and one included angle of one triangle is equal to corresponding two sides and one included angle of another triangle. so they are congruent.\n\n• 3\nWhat are you looking for?" ]

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[ "# Different methods for mitigating overfitting on Neural Networks\n\n### Pablo Sánchez\n\n#### 26/05/2021\n\nUsing Machine Learning and Deep Learning models to solve scientific problems of greater or lesser complexity is a challenge.\n\nReferring to neural networks, on the one hand, simple networks with too little capacity will not learn the problem well producing a model that underfits the data. On the other hand, complex networks with too much capacity will learn it too well leading to a model that overfits the data.\n\nAs a result, in both cases you will get a model that does not generalize well. In this post, you will find some techniques to reduce overfitting and get a more generalized and robust model.\n\n## Overfitting\n\nWith four parameters I can fit an elephant, and with five I can make him wiggle his trunk.\n\nJohn von Neumann\n\nIn Machine Learning and Deep Learning models, it is possible to test several combinations of parameters and hyperparameters to get a more accurate model. This is a very powerful tool but a very dangerous one too. The more complex the network is, the more fit to the training data will be.", null, "Explanation of underfitting and overfitting using the training and validation error depending on the model complexity\n\n## Method to mitigate overfitting\n\n### Model simplification\n\nAccording to the previous plot, decreasing the complexity of the model is one idea to deal with overfitting. On neural networks, reducing the number of neurons or removing some hidden layers will work.\n\n### Regularization\n\nOne of the first methods we should try when we need to reduce overfitting in our neural network is regularization. The main idea of this method is to penalize the model for being too complex or using high values in the weights matrix.\n\nFor applying regularization it is necessary to add an extra element to the loss function. The impact this method has on the model is parameterized by a variable called lambda ($$\\lambda$$). There are different ways to penalize the loss function but the most famous ones are based on the L1 and L2 norm of the weights (respectivaly called L1 and L2 regularization). As you can see in the following equations, the higher the parameter, the higher the impact on the loss function.\n\nL1 regularization equation\n\n$$J_{L1}(W,b) = \\frac{1}{m} \\sum_{i=1}^{m} L(\\hat{y}^{(i)}, y^{(i)}) + \\lambda \\left \\| W \\right \\|_{1} \\quad \\quad \\left \\| W \\right \\|_{1} = \\sum_{j=1}^{n_x} \\| W_{j} \\|$$\n\nL2 regularization equation\n\n$$J_{L2}(W,b) = \\frac{1}{m} \\sum_{i=1}^{m} L(\\hat{y}^{(i)}, y^{(i)}) + \\lambda \\left \\| W \\right \\|_{2} \\quad \\quad \\left \\| W \\right \\|_{2} = \\sum_{j=1}^{n_x} W_j^{2}$$\n\nWhere\n\n• $$m$$ is the number of training examples.\n• $$L(\\hat{y}^{(i)}, y^{(i)})$$ is the loss function between the estimated value $$\\hat{y}^{(i)}$$ and the real value $$y^{(i)}$$ for the i-th training example.\n• $$J(W,b)$$ is the cost function.\n\n### Dropout\n\nDropout is another very popular technique of regularization of neural networks. It is mainly based on the idea of simplifying the model. The objective is to reduce the importance of neurons by randomly switching some neurons off on the training stage.\n\nIt is possible to choose the probability of some neurons being disabled for each layer. Due to the random effect, the set of disabled neurons on each iteration is different, resulting on a completely different network. The hyperparameter used for this technique is called the dropout rate.\n\nIn the following gif image, it is possible to see a graphical representation of dropout on a Neural Network. In this example, the dropout rate is different on each layer:\n\n• The dropout rate in the second layer is 0.5, for that reason 3 of 6 neurons are switched off.\n• The dropout rate in the third layer is 0.33, for that reason 2 of 6 neurons are switched off.\n\n### Early stopping\n\nEarly stopping is another method of regularization used on the models trained in iterative processes, such as gradient descent. The main idea of this technique is to stop the training process when the validation error starts to increase. In this specific point, the model will increase its generalization error meaning that it is starting to overfit.\n\n### Data Augmentation\n\nData Augmentation is a well-known and very used technique when training Neural Networks using images. In this blog, it is possible to see different applications of data augmentation for financial series, such as Generating Financial Series with Generative Adversarial Networks.\n\nIn some cases, the dataset is not big enough or does not have a wide variety of pictures. This method reduces overfitting by increasing the dataset by applying different transformations to enrich it. The most popular image transformations are:\n\n• Flipping\n• Rotation\n• Scaling\n• Cropping\n• Translations" ]

[ null, "https://quantdare.com/wp-content/uploads/2021/05/Training-validation_plot-2.png", null ]

[ "Moai SDK  2.0\nMOAIImage Class Reference", null, "Inheritance diagram for MOAIImage:\n\n## Function List\n\naverage\n\nbleedRect\n\ncalculateGaussianKernel\n\ncompare\n\nconvert\n\nconvolve\n\nconvolve1D\n\ncopy\n\ncopyBits\n\ncopyRect\n\ndesaturate\n\nfillCircle\n\nfillEllipse\n\nfillRect\n\ngammaCorrection\n\ngenerateOutlineFromSDF\n\ngenerateSDF\n\ngenerateSDFAA\n\ngetColor32\n\ngetContentRect\n\ngetData\n\ngetFormat\n\ngetRGBA\n\ngetSize\n\ninit\n\nisOpaque\n\nmix\n\nprint\n\nresize\n\nresizeCanvas\n\nsetColor32\n\nsetRGBA\n\nsimpleThreshold\n\nsubdivideRect\n\nwrite\n\n## Detailed Description\n\n Constants MOAIImage.FILTER_LINEAR MOAIImage.FILTER_NEAREST\n Flags Description MOAIImage.POW_TWO Adds padding at the right and bottom to make the image dimensions powers of 2. MOAIImage.QUANTIZE Uses less than 8 bits per channel to reduce memory consumption. MOAIImage.TRUECOLOR Converts palettized color formats to true color. MOAIImage.PREMULTIPLY_ALPHA Premultiplies the pixel colors with their alpha values. MOAIImage.PIXEL_FMT_TRUECOLOR MOAIImage.PIXEL_FMT_INDEX_4 MOAIImage.PIXEL_FMT_INDEX_8 MOAIImage.COLOR_FMT_A_1 Alpha only, 1 bit per pixel MOAIImage.COLOR_FMT_A_4 Alpha only, 4 bits per pixel MOAIImage.COLOR_FMT_A_8 Alpha only, 8 bits per pixel MOAIImage.COLOR_FMT_LA_8 Grayscale + alpha, 16 bits per pixel MOAIImage.COLOR_FMT_RGB_888 RGB, 24 bits per pixel MOAIImage.COLOR_FMT_RGB_565 RGB, 16 bits per pixel MOAIImage.COLOR_FMT_RGBA_5551 RGBA, 16 bits per pixel (1 bit alpha) MOAIImage.COLOR_FMT_RGBA_4444 RGBA, 16 bits per pixel (4 bits per channel) MOAIImage.COLOR_FMT_RGBA_8888 RGBA, 32 bits per pixel\n\n## Function Documentation\n\n average\n\nCalculates the average of each color channel.\n\n`function average ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nnumber averageR, number averageG, number averageB, number averageA\n bleedRect\n\n'Bleeds' the interior of the rectangle out by one pixel.\n\n`function bleedRect ( MOAIImage self, number xMin, number yMin, number xMax, number yMax )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax\nReturns\nnil\n calculateGaussianKernel\n\nCalculate a one dimensional gaussian kernel suitable for blurring.\n\n`function calculateGaussianKernel ( MOAIImage self [, number radius, number sigma ] )`\n\nParameters\n MOAIImage self number radius Optional. Default valus is 1.0. number sigma Optional. Default valie is radius / 3 (https://en.wikipedia.org/wiki/Gaussian_blur)\nReturns\nnil\n compare\n\nCompares the image to another image.\n\n`function compare ( MOAIImage self, MOAIImage other )`\n\nParameters\n MOAIImage self MOAIImage other\nReturns\nboolean areEqual\n convert\n\nReturn a copy of the image with a new color format. Not all provided formats are supported by OpenGL. 'nil' may be passed for either value, in which case the format will match the original.\n\n`function convert ( MOAIImage self [, number colorFmt, number pixelFmt ] )`\n\nParameters\n MOAIImage self number colorFmt Optional. One of MOAIImage.COLOR_FMT_A_1, MOAIImage.COLOR_FMT_A_4, MOAIImage.COLOR_FMT_A_8, MOAIImage.COLOR_FMT_RGB_888, MOAIImage.COLOR_FMT_RGB_565, MOAIImage.COLOR_FMT_RGBA_5551, MOAIImage.COLOR_FMT_RGBA_4444, COLOR_FMT_RGBA_8888 number pixelFmt Optional. One of MOAIImage.PIXEL_FMT_TRUECOLOR, MOAIImage.PIXEL_FMT_INDEX_4, MOAIImage.PIXEL_FMT_INDEX_8\nReturns\nMOAIImage image\n convolve\n\nConvolve the image using a one or two dimensional kernel. If a one-dimensional kernel is provided, the image will be convolved in two passes: first horizonally and then vertically using the same kernel.\n\n`function convolve ( MOAIImage self, table kernel [, boolean normalize ] )`\n\nParameters\n MOAIImage self table kernel A one or two dimensional array of coefficients. boolean normalize Optional. If true, the kernel will be normalized prior to the convolution. Default value is true.\nReturns\nMOAIImage image\n convolve1D\n\nConvolve the image using a one dimensional kernel.\n\n`function convolve1D ( MOAIImage self, table kernel [, boolean horizontal, boolean normalize ] )`\n\nParameters\n MOAIImage self table kernel A one dimensional array of coefficients. boolean horizontal Optional. If true, the image will be convolved horizontally. Otherwise the image will be convolved vertically. Devault value is true. boolean normalize Optional. If true, the kernel will be normalized prior to the convolution. Default value is true.\nReturns\nMOAIImage image\n copy\n\nCopies an image.\n\n`function copy ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nMOAIImage image\n copyBits\n\nCopy a section of one image to another.\n\n`function copyBits ( MOAIImage self, MOAIImage source, number srcX, number srcY, number destX, number destY, number width, number height )`\n\nParameters\n MOAIImage self MOAIImage source Source image. number srcX X location in source image. number srcY Y location in source image. number destX X location in destination image. number destY Y location in destination image. number width Width of section to copy. number height Height of section to copy.\nReturns\nnil\n copyRect\n\nCopy a section of one image to another. Accepts two rectangles. Rectangles may be of different size and proportion. Section of image may also be flipped horizontally or vertically by reversing min/max of either rectangle.\n\n`function copyRect ( MOAIImage self, MOAIImage source, number srcXMin, number srcYMin, number srcXMax, number srcYMax, number destXMin, number destYMin [, number destXMax, number destYMax, number filter, number srcFactor, number dstFactor, number equation ] )`\n\nParameters\n MOAIImage self MOAIImage source Source image. number srcXMin number srcYMin number srcXMax number srcYMax number destXMin number destYMin number destXMax Optional. Default value is destXMin + srcXMax - srcXMin; number destYMax Optional. Default value is destYMin + srcYMax - srcYMin; number filter Optional. One of MOAIImage.FILTER_LINEAR, MOAIImage.FILTER_NEAREST. Default value is MOAIImage.FILTER_LINEAR. number srcFactor Optional. Default value is BLEND_FACTOR_SRC_ALPHA number dstFactor Optional. Default value is BLEND_FACTOR_ONE_MINUS_SRC_ALPHA number equation Optional. Default value is BLEND_EQ_ADD\nReturns\nnil\n desaturate\n\nConvert image to grayscale.\n\n`function desaturate ( MOAIImage self [, rY , gY , bY , K ] )`\n\nParameters\n MOAIImage self rY Optional. gY Optional. bY Optional. K Optional.\nReturns\nnil\n fillCircle\n\nDraw a filled circle.\n\n`function fillCircle ( number x, number y, number radius [, number r, number g, number b, number a ] )`\n\nParameters\n number x number y number radius number r Optional. Default value is 0. number g Optional. Default value is 0. number b Optional. Default value is 0. number a Optional. Default value is 0.\nReturns\nnil\n fillEllipse\n\nDraw a filled ellipse.\n\n`function fillEllipse ( number x, number y, number radiusX, number radiusY [, number r, number g, number b, number a ] )`\n\nParameters\n number x number y number radiusX number radiusY number r Optional. Default value is 0. number g Optional. Default value is 0. number b Optional. Default value is 0. number a Optional. Default value is 0.\nReturns\nnil\n fillRect\n\nFill a rectangle in the image with a solid color.\n\n`function fillRect ( MOAIImage self, number xMin, number yMin, number xMax, number yMax [, number r, number g, number b, number a ] )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax number r Optional. Default value is 0. number g Optional. Default value is 0. number b Optional. Default value is 0. number a Optional. Default value is 0.\nReturns\nnil\n gammaCorrection\n\nApply gamma correction.\n\n`function gammaCorrection ( MOAIImage self [, number gamma ] )`\n\nParameters\n MOAIImage self number gamma Optional. Default value is 1.\nReturns\nnil\n generateOutlineFromSDF\n\nGiven a rect, and min and max distance values, transform to a binary image where 0 means not on the outline and 1 means part of the outline.\n\n`function generateOutlineFromSDF ( MOAIImage self, number xMin, number yMin, number xMax, number yMax [, number distMin, number distMax, number r, number g, number b, number a ] )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax number distMin Optional. number distMax Optional. number r Optional. Default value is 1. number g Optional. Default value is 1. number b Optional. Default value is 1. number a Optional. Default value is 1.\nReturns\nnil\n generateSDF\n\nGiven a rect, creates a signed distance field from it.\n\n`function generateSDF ( MOAIImage self, number xMin, number yMin, number xMax, number yMax )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax\nReturns\nnil\n generateSDFAA\n\nGiven a rect, creates a signed distance field (using alpha as a mask) taking into account antialiased edges. The size of the SDF (distance from edges) is specified in pixels. Resulting SDF is stored in the image's alpha channel. Distances are normalized to the given size, inverted and scaled so that 0.5 is at an edge with 1 at full interior and 0 at full exterior (edge plus size).\n\n`function generateSDFAA ( MOAIImage self, number xMin, number yMin, number xMax, number yMax [, number sizeInPixels ] )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax number sizeInPixels Optional. Default is 5\nReturns\nnil\n\nGiven a rect, creates a signed distance field from it using dead reckoning technique.\n\n`function generateSDFDeadReckoning ( MOAIImage self, number xMin, number yMin, number xMax, number yMax [, number threshold ] )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax number threshold Optional. Default is 256\nReturns\nnil\n getColor32\n\nReturns a 32-bit packed RGBA value from the image for a given pixel coordinate.\n\n`function getColor32 ( MOAIImage self, number x, number y )`\n\nParameters\n MOAIImage self number x number y\nReturns\nnumber color\n getContentRect\n\ncomputes the content rect, not taking in account any boundary transparency\n\n`function getContentRect ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nrect\n getData\n\nreturns the bitmap data\n\n`function getData ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nbyte array\n getFormat\n\nReturns the color format of the image.\n\n`function getFormat ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nnumber colorFormat\n getRGBA\n\nReturns an RGBA color as four floating point values.\n\n`function getRGBA ( MOAIImage self, number x, number y )`\n\nParameters\n MOAIImage self number x number y\nReturns\nnumber r, number g, number b, number a\n getSize\n\nReturns the width and height of the image.\n\n`function getSize ( MOAIImage self [, number scale ] )`\n\nParameters\n MOAIImage self number scale Optional.\nReturns\nnumber width, number height\n init\n\nInitializes the image with a width, height and color format.\n\n`function init ( MOAIImage self, number width, number height [, number colorFmt ] )`\n\nParameters\n MOAIImage self number width number height number colorFmt Optional. One of MOAIImage.COLOR_FMT_A_1, MOAIImage.COLOR_FMT_A_4, MOAIImage.COLOR_FMT_A_8, MOAIImage.COLOR_FMT_RGB_888, MOAIImage.COLOR_FMT_RGB_565, MOAIImage.COLOR_FMT_RGBA_5551, MOAIImage.COLOR_FMT_RGBA_4444, COLOR_FMT_RGBA_8888. Default value is MOAIImage.COLOR_FMT_RGBA_8888.\nReturns\nnil\n isOpaque\n\nfalse if at least one pixel is not opaque\n\n`function isOpaque ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nbool\n\nLoads an image from an image file. Depending on the build configuration, the following file formats are supported: PNG, JPG, WebP.\n\n`function load ( MOAIImage self, string filename [, number transform ] )`\n\nParameters\n MOAIImage self string filename number transform Optional. One of MOAIImage.POW_TWO, One of MOAIImage.QUANTIZE, One of MOAIImage.TRUECOLOR, One of MOAIImage.PREMULTIPLY_ALPHA\nReturns\nnil\n\nLoad an image asyncronously. This includes reading the file and decoding compressed data.\n\n`function loadAsync ( MOAIImage self, string filename, MOAITaskQueue queue [, function callback, number transform ] )`\n\nParameters\n MOAIImage self string filename The path to the image file MOAITaskQueue queue The queue to peform operation on function callback Optional. Callback that will receive loaded image number transform Optional. One of MOAIImage.POW_TWO, MOAIImage.QUANTIZE, MOAIImage.TRUECOLOR, MOAIImage.PREMULTIPLY_ALPHA\nReturns\nnil\nNote\n\n`function loadAsync ( MOAIImage self, MOAIDataBuffer data, MOAITaskQueue queue [, function callback, number transform ] )`\n\nParameters\n MOAIImage self MOAIDataBuffer data Buffer containing the image data MOAITaskQueue queue The queue to peform operation on function callback Optional. Callback that will receive loaded image number transform Optional. One of MOAIImage.POW_TWO, MOAIImage.QUANTIZE, MOAIImage.TRUECOLOR, MOAIImage.PREMULTIPLY_ALPHA\nReturns\nnil\nNote\n\nLoads an image from a buffer.\n\n`function loadFromBuffer ( MOAIImage self, MOAIDataBuffer buffer [, number transform ] )`\n\nParameters\n MOAIImage self MOAIDataBuffer buffer Buffer containing the image number transform Optional. One of MOAIImage.POW_TWO, One of MOAIImage.QUANTIZE, One of MOAIImage.TRUECOLOR, One of MOAIImage.PREMULTIPLY_ALPHA\nReturns\nnil\n mix\n\nTransforms each color by a 4x4 matrix. The default value is a 4x4 identity matrix. The transformation 'remixes' the image's channels: each new channel value is given by the sum of channels as weighted by the corresponding row of the matrix. For example, to remix the blue channel: b = r*b1 + g*b2 + b*b3 + a*b4. A row value for b of (0, 0, 1, 0) would be the identity: b = b. A row value for b of (1, 0, 0, 0) would replace b with r: b=r. A row value for b of (0.5, 0.5, 0, 0) would replace b with an even blend of r and g: b = r*05 + b*0.5. In this fashion, all channels of the image may be rearranged or blended.\n\n`function mix ( MOAIImage self [, number r1, number r2, number r3, number r4, number g1, number g2, number g3, number g4, number b1, number b2, number b3, number b4, number a1, number a2, number a3, number a4, number K ] )`\n\nParameters\n MOAIImage self number r1 Optional. number r2 Optional. number r3 Optional. number r4 Optional. number g1 Optional. number g2 Optional. number g3 Optional. number g4 Optional. number b1 Optional. number b2 Optional. number b3 Optional. number b4 Optional. number a1 Optional. number a2 Optional. number a3 Optional. number a4 Optional. number K Optional. Default value is 1.\nReturns\nnil\n\nCopies an image and returns a new image padded to the next power of 2 along each dimension. Original image will be in the upper left hand corner of the new image.\n\n`function padToPow2 ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nMOAIImage image\n print\n\nPrint the image colors (for debugging purposes).\n\n`function print ( MOAIImage self )`\n\nParameters\n MOAIImage self\nReturns\nnil\n resize\n\nCopies the image to an image with a new size.\n\n`function resize ( MOAIImage self, number width, number height [, number filter ] )`\n\nParameters\n MOAIImage self number width New width of the image. number height New height of the image. number filter Optional. One of MOAIImage.FILTER_LINEAR, MOAIImage.FILTER_NEAREST. Default value is MOAIImage.FILTER_LINEAR.\nReturns\nMOAIImage image\n resizeCanvas\n\nCopies the image to a canvas with a new size. If the canvas is larger than the original image, the extra pixels will be initialized with 0. Pass in a new frame or just a new width and height. Negative values are permitted for the frame.\n\n`function resizeCanvas ( MOAIImage self, number width, number height )`\n\nParameters\n MOAIImage self number width New width of the image. number height New height of the image.\nReturns\nMOAIImage image\nNote\n\n`function resizeCanvas ( MOAIImage self, number xMin, number yMin, number xMax, number yMax )`\n\nParameters\n MOAIImage self number xMin number yMin number xMax number yMax\nReturns\nMOAIImage image\nNote\n setColor32\n\nSets 32-bit the packed RGBA value for a given pixel coordinate. Parameter will be converted to the native format of the image.\n\n`function setColor32 ( MOAIImage self, number x, number y, number color )`\n\nParameters\n MOAIImage self number x number y number color\nReturns\nnil\n setRGBA\n\nSets a color using RGBA floating point values.\n\n`function setRGBA ( MOAIImage self, number x, number y, number r, number g, number b [, number a ] )`\n\nParameters\n MOAIImage self number x number y number r number g number b number a Optional. Default value is 1.\nReturns\nnil\n simpleThreshold\n\nThis is a 'naive' threshold implementation that forces image color channels to 0 or 1 based on a per-channel threshold value. The channel value must be entirely greater that the threshold to be promoted to a value of 1. This means a threshold value of 1 will always result in a channel value of 0.\n\n`function simpleThreshold ( MOAIImage self [, number r, number g, number b, number a ] )`\n\nParameters\n MOAIImage self number r Optional. Default value is 0. number g Optional. Default value is 0. number b Optional. Default value is 0. number a Optional. Default value is 0.\nReturns\nnil\n subdivideRect\n\nConvenience method. Here for now as a class method, but maybe should move to MOAIGrid. Subdivides a rectangle given a tile width and height. A table of tile rectangles will be returned. The tiles will be clipped to the original rect.\n\n`function subdivideRect ( number tileWidth, number tileHeight, number xMin, number yMin, number xMax, number yMax )`\n\nParameters\n number tileWidth number tileHeight number xMin number yMin number xMax number yMax\nReturns\nnil\n write\n\nWrite image to a file.\n\n`function write ( MOAIImage self, string filename [, string format ] )`\n\nParameters\n MOAIImage self string filename string format Optional.\nReturns\nboolean" ]

[ null, "http://moaiforge.com/api-docs/latest/closed.png", null ]

[ "Solutions by everydaycalculation.com\n\n## What is 735 percent of 2475?\n\n735% of 2475 is 18191.25\n\n#### Working out percentages\n\n1. Write 735% as 735/100\n2. Since, finding the fraction of a number is same as multiplying the fraction with the number, we have\n735/100 of 2475 = 735/100 × 2475\n3. Therefore, the answer is 18191.25\n\nIf you are using a calculator, simply enter 735÷100×2475 which will give you 18191.25 as the answer.\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to work with percentages in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "## The Author's View of the Universe", null, "A view of how this author imagines the Universe. This is a cross-section of a toroid manifold fractal. Universes and sub-universes are indicated by ellipses.\n\nThe First Cause at the center of the toroid structure, generates/sustains the subuniverses, in a round robin fashion. The subuniverse generation is indexed as T0 through Tn-1, n in N, and each ray sustains m concentric subuniverses, indexed as Tkm, m in N. At Tk0, there is a copy of the First Cause (as a rotationally fixed point of f, see below), which generates subuniverses at lower levels.\n\nThe subuniverses fold within each other through a contractive homeomorphism f, indicated by the red arrows, creating a \"time/reality funnel\" or \"time/reality sheet\", consisting of a continuous stream of quaternion \"events\" (t,x(t),y(t),z(t)) for an individual mind, giving the impression of continuous consciousness.\n\nf wraps back on itself around the main toroid structure and the only points that survive are the rotationally fixed points of f , for each Tkm. Following for example the manifold sheet of any one subuniverse Tkm through f, as Tkm->f(Tkm)->f(f(Tkm))->..., k, m in N, one can get a representation of what's happening in 3D, topologically. The red ellipse is universe Tkm. The only point that survives, the \"last\" fixed point of f, a copy of the First Cause, re-creates subuniverse Tkm, or jumps to a different/unknown subuniverse Txy. At that point, there is an \"explosion\", as that's the only point in time when f is not contractive.", null, "The reality funnel considered thus, bares an interesting connection with the notion of Ouroboros.", null, "Original Image Source: Ouroboros\n\nInside every Tkm lies the universe Tkm-1, which generates (through f) a similar manifold which wraps back on itself. Outside every Tkm lies the universe Tkm+1, which also wraps back on itself. In effect, the only points that pass through all universes are the fixed points of f, indicated by the dotted blue ellipse.\n\nAccordingly, the main toroid structure, T, in 3D, looks like the shape above, although in the figure each such structure is represented by an ellipse (the fixed points of f). The toroid sheets here separate space into \"inside\" and \"outside\". At the moment of  Tkm 's \"birth\", there is no such differentiation.\n\nEach subuniverse Tkm is a replica of the main toroid structure T. That means that the centers of Tkm, k,m in N generate p creative rays, indexed as Tkmp, k,m,p in N, exactly like the main torus generates n rays, n in N. Each ray Tkmp, p in N generates q subsubuniverses, indexed as Tkmpq, k,m,p,q in N, and for each of those a similar contractive homeomorphism wraps each sheet Tkmpq onto itself, similarly to the way f acts on the subuniverses Tkm, k,m in N.\n\nThe process continues, and at level r, we are indexing the sub...subuniverse as Tkmpq...rs, k,m,p,q,...,r,s in N, which is self-similar to the original toroid structure T. The green vectors are the \"First Cause\", fixed point \"sustainers\" of the individual subuniverses, branching down all the way to non-existence.", null, "Time moves up with the main toroid structure, pointed to by the vertical axes with the + signs and right for all subuniverses in what physicists call \"right handed way\". This means that the main toroid structure moves in time upwards and the first order subuniverses to the right. Time does not exist while the \"First Cause\" \"switches\" its attention from subuniverse Tk to subuniverse Tk+1. When this happens, the entire structure falls into \"suspended animation\" for an infinitesimal time increment dt. dt is of the order of Plank scale time 10-28 to 10-34 seconds, yet the \"First Cause\" can \"stay\" within universe Tk as long as it wants.\n\nIn this structure, n denotes the number of first order subuniverse divisions applied recursively to the base circle. The recursive depth to which those divisions are applied is denoted as d. Concentric subuniverses are indexed by m. Accordingly, approximations of the structure are denoted as T(n,m,d).\n\nHere is a .gif animation of the toroid structure T(6,1,4) done in Maple by C W of comp.soft-sys.math.maple.", null, "Here are some videos of high quality animations done by C W:\n\n1. T(3,1,6) (IV50 Indeo 5.x .avi 2.77 MB)\n2. T(8,1,4) (IV50 Indeo 5.x.avi 3.78 MB)\n\nHere is a static picture of the toroid structure T(6,1,4) done by Paul Burke using POVRay.", null, "For more static pictures for different values of n and d, see Paul Burke's directory, here. The pictures in his page (from top to bottom and left to right) are: T(6,1,4), T(10,1,3), T(10,1,4), T(5,1,3), T(5,1,4), T(18,1,4) and T(n,1,4) for some large n.\n\nHere is a static picture of the toroid structure T(5,3,4) (which is closer to the structure in the first figure) done by Paul Burke again using POVRay.", null, "In a sense the First Cause is akin to the the operating system or the time engine for the toroid structure. The author don't know if the structure continues to an infinite sub-level or to an infinite super-level. Doing so, would require infinite speed for the \"First Cause\" to sustain all subuniverses properly. In other words, if the structure was T(n,m,∞) the author is not exactly sure how the \"First Cause\" can \"cover\" all the subuniverses in a finite time.\n\nAlthough T(n,m,∞) sounds like a bit much, the main structure may generate many super-universes similar to itself, by wrapping back on its past, creating a super toroid fractal manifold one level up from what you see in the first figure. The new super-universe has the same properties as the original figure and this process may continue upwards and to the right, each time \"time\" being folded upon itself, in the direction of the homeomorphism arrows.\n\nImagine this immense toroid manifold folding out and folding back in. When it folds out, it creates super-universes and subuniverses both upwards and downwards to any finite super-level or sub-level. When it folds back in, it destroys super-universes and sub-universes.\n\nAlternatively, the entire structure can be seen as a fractal toroid solenoid, where from the main torus emanate n rays, each generating m concentric toruses, normal to the main torus' circle as in figure 1. Then each of those toruses has the same structure as the main torus T and additionally each torus is mapped through a contraction to a torus to its right. The process continues to any level, either downwards (generating smaller toruses) or upwards, generating super-toruses by having the main torus wrapping against itself, any number of times.\n\nMost people seem to think that what astronomers call \"the Universe\", is an a-personal vast structure, independent of perception and consciousness. Whenever this author writes about any \"universe\", it is understood that he means \"the observable universe\", in other words a \"universe\" which is explicitly associated with a specific observer's five senses, his vision notwithstanding. This author fails to understand what a \"universe\" independent of human perception and consciousness is. Such a universe (apparently) seems to exist only as a mental structure in the minds of scientists and may or may not have a tangible existence, since it cannot be surveyed in its entirety and since when consciousness disappears it goes too. To illustrate \"the observable universe\", we return to the picture of the \"time/reality-funnel\" which was presented above:", null, "Time/reality-funnel\n\nThe above \"funnel\" is basically a discrete stream of \"events\" in space-time. We notate a single such event as a quadruple E(t)=(t,x(t),y(t),z(t)) = t+v(t). t then is the scalar part of the event and v(t) is the vector part of the event. The event t+v(t) is a point in space-time.\n\nThe observer's coordinate system is the birth-point relative to some major event in time. Modern historians choose the birth of Jesus as the time reference for events in history, therefore the observer's birth (as an event in time) gives rise to a unique time coordinate tb(irth) and a unique space coordinate v(tb).\n\nThe time coordinate tb is measured relative to 0AD and the space coordinate v(tb) is measured relative to Jesus' place of birth. For example, the author was born in September 2, 1964, so tb=02/09/1964 in years. He was born in Athens, Greece, so v(tb)=(x(tb),y(tb),z(tb)), the coordinates of Athens relative to Bethlehem in Cartesian coordinates.\n\nSince most of us measure space-time relative to our birth, we switch reference frames by transferring the origin to the above point by applying the transformations:\n\ntb*=t-tb\nv(tb)*=v(t)-v(tb)\n\nTherefore the observer's new reference system will be transformed to tb*=0 and v(tb)*=(0,0,0). This is the event of the observer's birth, which the observer now will use as the new origin (0,0,0,0). The observer now starts measuring \"reality\" (time and space) always relative to this new origin.\n\n\"Reality\" is delimited by the full spatial extent of the vector part of the observer's life at time t, R(v(t)). When the observer is a newborn, R(v(t)) is small. As the observer grows and consciousness gets organized, R(v(t)) increases. That's why the \"time/reality funnel\" starts being close to a singularity. R(v(t)) on a newborn observer is shown with red ellipses as the observer grows.\n\nR(v(t)) is always bounded by the fastest communicator of information in any universe: Light. Therefore, at time t, the observer's \"light-sphere\", is defined as:\n\nL(t)={(x(t),y(t),z(t)) in R3: x(t)2+y(t)2+z(t)2 = (c*t)2} (1)\n\nThe observer's \"reality\" R(v(t)) then, must be the set of all points in R3, which are always bounded by the light-sphere L(t), that is the set of points which are at a distance of less than c*t from the observer:\n\nR(v(t))={(x(t),y(t),z(t)) in R3: x(t)2+y(t)2+z(t)2 < (c*t)2} (2)", null, "R(v(t)) (red ellipses) in a time/reality-funnel\n\nWe now unfold the toroid time/reality funnel of the observer's events relative to time, which is linear. Here is a 2D representation. L(t) is denoted by the red circles and it always rides on the vector of time, t.", null, "2D representation of a birth of a consciousness (light-cone) from the singularity tb+v(tb)\n\nHere's the situation as it happens in 3D. The observer's \"reality\" R(v(t)) is delimited by the individual spheres L(t), which have radius c*t. As time t passes L(t) (and thus R(v(t))) grows. L(t) is the observer's event horizon.", null, "Birth of a consciousness (light-cone) from the singularity at tb+v(tb)\n\nDefinitions (1) and (2) above of the light-sphere L(t) then naturally define a distance differential (ds) in space-time:\n\nds2=dx(t)2+dy(t)2+dz(t)2-c2*dt2\n\nds above then agrees with the definition of the Minkowski metric in Special Relativity. The null geodesics therefore are described by the equation:\n\ndx(t)2+dy(t)2+dz(t)2=c2*dt2.\n\nThese are the light-spheres L(t) of radius c*t shown above which \"ride\" on time and which lie along the four dimensional light-cones described by the birth itself, in both figures above.\n\nAt any given time t, any information from objects outside the observer's event horizon L(t) (that is at distances greater than c*t), hasn't reached the observer yet, therefore any event outside the light-cone of the observer is not causally connected with the observer.\n\nThe birth of an observer takes place when two independent observers A and B mate and produce a new observer as follows:", null, "Observers A and B mating and giving birth to observer C\n\nObserver B is the female, A is the male. They synchronize their time vectors and unite. The offspring, observer C, starts surveying his/her mother's reality first. As he/she grows, observer C surveys larger and larger parts of his/her mother's reality, until his/her reality C bursts out of the reality of B. This is the moment of birth at x and y. All this of course happens in three dimensions.\n\nAlthough time is thought of as continuous, in reality it is discrete with points spaced apart on the time vector every 10-28 seconds: Plank's time scale = Δt. As such, the difference between any two centers of the spheres above is at least Δt and this creates the illusion of time continuity. In effect, Δt is the \"minimum resolution\" of The First Cause's Consciousness or the \"program\" that directs the evolution of all the individual time-reality funnels. In the above two images, the minimum time difference between circles/spheres is a multiple of Δt to allow the respective R(v(t))'s to be visualized as separate. In reality the centers of the R(v(t))'s are Δt apart.\n\nThis \"growth\" of L(t) (and thus of R(v(t))) is as if the observer's consciousness is \"sampling\" reality at least at the rate of dr/dt=d(c*t)/dt=c or 3*10-20 meters per Δt. Objects outside L(t) eventually fall into L(t) after sufficient time elapses. The notion of \"now\" therefore does not exist, because any information that is being communicated to the observer from reality, has taken some time to reach him. Even information from an object 1cm away from an observer, takes 1/3*10-10 seconds to reach the observer.\n\nR(v(t)) is a full sphere only when the observer stands in empty space and looks at a night-sky or at the sky above the atmosphere. For an observer standing on earth, R(v(t)) takes various shapes depending on where his vision is focused and how it is used. Standing on Earth and looking at the night-sky for example, R(v(t)) looks like a hemisphere:", null, "Perceivable hemisphere R(v(t)) for an observer standing on Earth\n\nIn the above case the viewer is looking at events E(t), d meters away, where d2=x(t)2+y(t)2+z(t)2 and d/c seconds in the past.\n\nIdeally then, consciousness is a mental sphere which always rides on time t. This sphere shrinks and expands depending on how the observer uses his faculties (mainly his vision). Therefore R(v(t)) can vary for example from the dimensions of a room with 8 walls (for an observer inside a closed room), to a sphere with radius c*t. This is the \"time-reality funnel\" described on the corresponding figures.\n\nWhen the observer stands still between two moments in time t1 and t2, then x(t1)=x(t2), y(t1)=y(t2) and z(t1)=z(t2), so the vector parts of the two events E(t1) and E(t2) are identical: v(t1)=v(t2), even though the scalar parts are not equal: t1=/=t2.\n\nWhen the observer moves spatially between two moments in time t1 and t2, then at least one of the following holds: x(t1)=/=x(t2), y(t1)=/=y(t2) and z(t1)=/=z(t2), so the vector parts of the two events E(t1) and E(t2) are different: v(t1)=/=v(t2).\n\nSince time always moves forward, consciousness also moves forward. There's no way to jump from an event in time E(t1) to an event in time E(t2) with t2-t1 > Δt, unless either the observer undergoes general anesthesia or goes to sleep. In this case the time vector is interrupted and made discontinuous and as a result the R(v(t))'s are also made discontinuous, since they ride on time. For anesthesia and sleep, the situation is depicted in the following figure:", null, "Interruption of consciousness while an observer falls asleep\n\nSleep differs from general anesthesia in that during sleep there is partial awareness, although this awareness is unrelated to reality, while during anesthesia there's absolutely no consciousness. In the above figure, while t1 < t < t2 consciousness shuts down, so R(v(t)) has no meaning. Consciousness and thus R(v(t)) resume after t2.\n\nThe observer births of figures 8-10, create the notion of \"chained-births\" in the toroid-universe, because every observer has a female parent. Therefore there exists a chain of universes/realities with each one being (initially) inside another one (that of his/her parent. Inside every Tkm lies the universe Tkm-1, which generates (through f) a similar manifold which wraps back on itself. Outside every Tkm lies the universe Tkm+1, which also wraps back on itself).", null, "Chained-sequence of births of observers\n\nThe maximum observable universe then, seems to be the reality of the left-supremum of the above chain births. In other words, the maximum observable universe as this author understands it, seems to be the reality of the first \"observer\". Whether this \"observer\" was a man or something else, the author does not know. This \"reality\" has still radius c*t0, but now t0 is of the order of millions of years. This observer has managed to survey the largest part of reality so far and we, as observers down the chained-sequence of births/light-cones, are surveying what he surveyed, asymptotically.\n\nThe reality of this observer then, R(v(t0)) is a 3-dimensional sphere of maximum radius c*t0 light-years, a virtual spaceship in time and space, within which all of us move and live. His corresponding light-sphere, L(t0) is the boundary between existence and non-existence. Although for each one of us L(t) is contained in some other universe, the first observer's light-sphere L(t0) and corresponding light-cone are the upper bound for any observable universe.\n\nNotes\n\n1. Ezekiel 1:15: \"As I looked at the living creatures, I saw a wheel on the ground beside each creature with its four faces. 16: This was the appearance and structure of the wheels: They sparkled like chrysolite, and all four looked alike. Each appeared to be made like a wheel intersecting a wheel. 17:As they moved, they would go in any one of the four directions the creatures faced; the wheels did not turn about as the creatures went. 18: Their rims were high and awesome, and all four rims were full of eyes all around.\n2. \"Time is said to have only one dimension, and space to have three dimensions. ... The mathematical quaternion partakes of both these elements; in technical language it may be said to be \"time plus space\", or \"space plus time\": and in this sense it has, or at least involves a reference to, four dimensions. And how the One of Time, of Space the Three, Might in the Chain of Symbols girdled be.\" William Rowan Hamilton (quoted in Robert Percival Graves' \"Life of Sir William Rowan Hamilton\" (3 vols., 1882, 1885, 1889)).\n3. See Modelling Time Under Total Anesthesia.\n4. Daniel 7:9: \"As I looked, thrones were set in place, and the Ancient of Days took his seat. His clothing was as white as snow; the hair of his head was white like wool. His throne was flaming with fire, and its wheels were all ablaze.\"", null, "" ]

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[ "## Found 185 Articles for Haskell", null, "Updated on 06-Apr-2023 10:46:08\n\n194 Views\n\nThis tutorial will help us to creat a haskell program that can covert a given hexadecimal number to a decimal number using revers, map and fold1 functions Hexadecimal to decimal conversion is the process of converting a number from the hexadecimal number system to the decimal number system. The hexadecimal number system uses a base of 16, which means that there are 16 unique symbols used to represent numbers in this system (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F). The decimal number system, on the other hand, uses a base of ... Read More\n\n## Haskell Program to convert Decimal to Octal", null, "Updated on 06-Apr-2023 10:44:56\n\n56 Views\n\nWe can convert the Decimal number to an Octal using the recursion and unfoldr function of Haskell. Decimal to octal conversion is a process of converting a decimal (base-10) number to its equivalent representation in octal (base-8) numbering system. In decimal numbering system, we use 10 digits (0 to 9) to represent a number. In octal numbering system, we use 8 digits (0 to 7) to represent a number. To convert a decimal number to its equivalent octal representation, we divide the decimal number by 8 repeatedly until the quotient becomes 0, and keep track of the remainders. The remainders, ... Read More\n\n## Haskell Program to Print Half Diamond Star Pattern", null, "Updated on 06-Apr-2023 10:43:33\n\n45 Views\n\nWe can create a half-diamond star pattern in Haskell using recursive and replicate functions. The half-diamond star pattern is a pattern made up of asterisks (*) arranged in the shape of a half-diamond. It is typically created by printing a series of asterisks in a pyramid shape, starting with a single asterisk in the first row, two asterisks in the second row, and so on, until the middle row which contains the maximum number of asterisks. From that row onwards, the number of asterisks decreases until there is only a single asterisk in the last row. Algorithm Step 1 ... Read More\n\n## Haskell Program to Print Hollow Right Triangle Star Pattern", null, "Updated on 06-Apr-2023 10:41:38\n\n94 Views\n\nIn Haskell we can use the replicate function and recursive function to create a hollow right triangle star pattern. A hollow right triangle star pattern is a pattern made up of asterisks (*) that forms a right triangle shape with empty spaces in the middle as shown below. ** * * * * * * * * * * * * ******** The shape is created by printing asterisks in a specific order, with the number of asterisks in each row increasing as the ... Read More\n\n## Haskell Program to Print 8-Star Pattern", null, "Updated on 06-Apr-2023 10:39:42\n\n65 Views\n\nIn this tutorial, we are going to learn how to develop a Haskell program to print 8 start patterns using the internal replicate and concat function. An '8' star pattern is an ASCII art representation of the number 8 using asterisks. as shown below − ******** * * * * ******** * * * * ******** The asterisks are arranged in such a way that they form the shape of the number 8. Algorithm Step 1 − ... Read More\n\n## Haskell Program to Create Pyramid ‘and’ Pattern", null, "Updated on 06-Apr-2023 10:38:51\n\n56 Views\n\nIn this tutorial, we are going to understand how to develop a Haskell program that will create a pyramid pattern of ‘&’ using mapM, forM, and recursive function. A pyramid ‘&’ pattern is a design or arrangement of ‘&’ or other symbols in the shape of a pyramid as shown below. & &&& &&&&& &&&&&&& &&&&&&&&& It is created by printing ‘&’ or symbols in multiple rows, starting from the top and moving downwards. Each row contains one more symbol than the previous row, creating the illusion of ... Read More\n\n## Haskell Program to Print Square Star Pattern", null, "Updated on 06-Apr-2023 10:36:54\n\n96 Views\n\nIn Haskell we can use internal functions like mapM, forM or recursive functions to print square star pattern. A square star pattern is a two-dimensional pattern made up of stars (or asterisks, represented by the '*' symbol) arranged in the shape of a square as shown below. **** **** **** **** The square pattern is formed by printing a specified number of lines, each containing a specified number of stars. The pattern can be of any size, with the number of lines and the number of stars in each line determining the overall size of the square pattern. Algorithm ... Read More\n\n## Haskell Program to Print Star Pascal’s Triangle", null, "Updated on 06-Apr-2023 10:35:15\n\n185 Views\n\nIn Haskell we can use mapM function and forM function to print a star Pascal’s Triangle. A star Pascal's triangle is a variation of the traditional Pascal's triangle that uses asterisks (or stars) instead of numbers to form a triangular pattern as shown below. * * * * * * * * * Pascal's triangle is a triangular array of numbers, where each number in the triangle is the sum of the two numbers above it. In ... Read More\n\n## Haskell Program to Print Mirror Upper Star Triangle Pattern", null, "Updated on 06-Apr-2023 10:32:04\n\n44 Views\n\nIn this article, we are going to learn how we can develop a haskell program to print mirror upper start triangle patterns using mapM function, and unliness functions. A mirror upper star triangle pattern is a pattern made up of stars (asterisks) that form a triangle shape, with the top of the triangle pointing upwards. The following star pattern will give you a better understanding of mirror upper start triangle pattern. * *** ***** ******* ********* The pattern is called \"mirror\" because the left and right sides of the ... Read More\n\n## Haskell Program to Print Upper Star Triangle Pattern", null, "Updated on 06-Apr-2023 10:29:57\n\n49 Views\n\nThis tutorial will help us in printing the upper star triangle pattern using mapM function, forM function, and unliness functions in Haskell. An upper star triangle pattern is a graphical representation of a triangle made up of asterisks or stars as shown below. * ** *** **** ***** It's called an \"upper\" star triangle because the triangle starts at the top and the number of stars in each row decreases as we move down the triangle. Algorithm Step 1 − We will start with defining a user-defined function as printStars function. Step 2 − Program execution will be ... Read More" ]

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[ "", null, "# And Statments Am I missing somthing\n\nCan some one see what is wrong with this code I am not sure I am using the if and the and statements correct. The code uploads but does not do what I think it should be doing. The locks do not seem to be working and I was wondering if it was because I stacked to may && statements. Or because I left out the {}\n\nif (angle == direct) {left = 0;\nright = 0;}\nif (left == 0 && right == 0 && angle < direct ) right = 1 ; //Turn right lock\n\nif (right == 0 && left == 0 && angle > direct) left = 1 ; // Turn left lock\n\nif (right == 1 && angle < direct && angle > 0 && angle < direct - 5 ) digitalWrite(11, HIGH);\n\nif (right == 1 && angle > direct + 180) digitalWrite(11, HIGH) ;\n\nif (left == 1 && angle > direct && angle < 359 && angle > direct + 5) digitalWrite(12, HIGH);\n\nif (left == 1 && angle < direct - 180) digitalWrite(12, HIGH) ;\n\nif (angle==direct) { digitalWrite(11, LOW);\ndigitalWrite(12, LOW);}\n\nThank Ron\n\nTry:\n\nif (right == 1 && angle > (direct + 180)) digitalWrite(11, HIGH) ;\n\n.\n\n``````if (right  == 1 && angle < direct && angle > 0  && angle < direct - 5 )  digitalWrite(11, HIGH);\n``````\n\nThere seems to be a redundant clause in this if statement, and another similar one further down, apart from that it looks valid to me.\n\nThe extra parentheses suggested by LarryD should not actually be required, because the conditional expression will be evaluated that way anyhow, with the extra parentheses.\n\nIs this supposed to be steering some kind of robot ?\n\nIt seems to me, you are going to run into difficulties if the target direction is close to 0\n\nI find it easier to understand that sort of logic if I break the IFs into a cascade - something like this\n\n``````if (right == 0) {\nif (left == 0) {\nif (angle > direct) {\n}\nelse {\n}\n}\n}\n``````\n\nI find it very easy to miss a relevant condition when I have a compound IF statement.\n\n...R\n\nThanks for the input . I do like the cascading if's better.\n\nThere is a problem when steering near 0 . Any thoughts on a better way to steer using a compass locked to a heading. I have been trying to com up with some sort of auto pilot using the compass to control direction.\n\nRon\n\nPiolt34: There is a problem when steering near 0 .\n\nYou don't say what the problem is.\n\nIt would be normal to have a small dead-band in which no action is intended to take place to prevent hunting.\n\n...R\n\nIf you are dealing with angles and want to avoid turning the long way round then:\n\n1) calculate the relative angle (target - actual). 2) normalise to the range -180 .. +180 ( or -PI .. +PI if in radians). 3) the sign of the result tells you which way to turn, the magnitude how far.\n\nIn the code you post the variables left and right are outputs of the early tests, then inputs to the later ones. I think that's confusing, it worries me.\n\nComment each variable with its actual meaning and then compare to each use - is each used consistently?\n\nThere are many problems I am having and just needed to verify that the code I am using was formated correctly . The overall problem I am trying to over come is locking the direction (direct) and having the out put go high when the angle is greater then the dead band. The biggest issue is the transition from 0-359 0r 359 -0 deg. The greater than less than option works fair but it is the only way I can think of doing this. I am sure there is a better way to steer to a direction and correct if it deviates from the direction . Of coarse it need to correct in the right direction and not make 360 deg turns.\n\nBelow is the code I have been working on to turn on a relay to correct the direction of flight.\nThe problem that seems to keep happening is the right and left variable goes to zero every time the\nangle goes past 0 even if the direct variable is 180. Seems to me the if stament is not working\ncorrectly. I have tried it with {} and with out. I have added 1 to the angle so that it would never see\n0 degrees none of these things work.\n\n#include <Wire.h>\n#include <HMC5883L.h>\nHMC5883L compass;\n\nfloat declinationAngle = 0.212; // Radians\nint direct ; // direction you lock on too Rn\nint error ;\nint off ;\nint off2 ;\nint angleneg;\nint start;\n\nvoid setup() {\nstart = 1;\npinMode(11, OUTPUT);\npinMode(12, OUTPUT);\nSerial.begin(57600);\nWire.begin();\ncompass = HMC5883L();\n// Set scale to +/- 1.3 Ga\nint error = compass.SetScale(1.3);\nif (error != 0)\nSerial.println(compass.GetErrorText(error));\n// Set measurement mode to continous\nerror = compass.SetMeasurementMode(Measurement_Continuous);\nif (error != 0)\nSerial.println(compass.GetErrorText(error));\n}\n\nvoid loop() {\nint angle = getDegrees()+1;\nint center;\nint correction;\nconst int buttonPin = 2; // the number of the pushbutton pin\nint left ;\nint right ;\n\n// variables will change:\nint buttonState = 0; // variable for reading the pushbutton status\n// initialize the pushbutton pin as an input:\npinMode(buttonPin, INPUT);\n// read the state of the pushbutton value:\n\n// check if the pushbutton is pressed.\n// if it is, the buttonState is HIGH:\nif (buttonState == HIGH) {direct = angle ; //get angle and holds on too it\nstart = 0;}\n\nif (angle == direct){ left = 0; // Lock the direction of the output\nright = 0;}\n\nif (left == 0 && right == 0 && angle < direct && start == 0 ) right = 1 ; //Turn right lock\n\nif (right == 0 && left == 0 && angle > direct && start == 0 ) left = 1 ; // Turn left lock\n\nif (left == 1){ digitalWrite(12, HIGH);\ndigitalWrite(11, LOW);}\nif (right == 1 ) { digitalWrite(11, HIGH);\ndigitalWrite(12, LOW);}\n\nThat code doesn't look like it will compile. It also hasn't been posted properly. Read this and then fix up your post: How to post code properly\n\nPete\n\nCode does compile. I left some of it out to prevent confusion.\n\nCode does compile. I left some of it out to prevent create confusion.\n\nFixed that for you.\n\nI count 8 if statements. Which one do you think isn't working?\n\nAt the very least you have some stuff in loop() that should be in setup(). Generally things like pinMode() are used in setup().\n\nYou may want to put some Serial prints in there to see what is happening to your variables. If the button is pressed you set direct = angle and right after that you check if they are equal, resulting in left and right becoming 0.\n\nYou asked this stupid question, last week.\n\nDid you understand the answers you got then, or are you just trolling ?\n\nNO I did not ask this Stupid question last week ! The question is on the the same project but this question is more specific in nature . IF you do not have a answer for me than just move on .Or do you just need to belittle people to feel good?\n\nI did get the my previous questioned answered ....I do not see how this applies to the problem I am having now. Why would the If statement allow for the angle==direct when the compass passes 0 ( int angle = getDegrees()+1;) . Direct is not 0 so the if statement is not true there for the left right should not be set to 0. But it does every time I rotate the compass past 0 Degrees.\n\nYou asked this stupid question, last week." ]

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[ "# A Venn diagram with PDF blending\n\nA Venn diagram displays several sets with their relationships. Commonly, these are overlapping circles. Such sets can stand for certain properties. If an element has two such properties, it would belong to an overlapping area – the intersection of the two relevant sets.\n\nWe will create three filled circles. The center of each circle is specified in polar coordinates, with a given angle and distance from the origin. This makes radial placement easier. For example, the first circle has its center at (90:1.2). This means that the center is at 90 degrees, which is above the origin, and the distance is 1.2. The radius of each circle is 2. So, they are overlapping.\n\nNormally, overlapping would simply mean that the final circle overrides what is below it. We still wish to look behind the circles to see the intersections. A classical approach is to use transparency, here we will use the blend mode feature of the PDF standard.\n\nNote: PDF blend mode requires TikZ version 3.0 or above.\n\nFull explanation in Chapter 9, Creating Graphics: [Drawing a Venn diagram](http://latex-cookbook.net/articles/venn-diagram/)", null, "Edit and compile if you like:\n% A Venn diagram with PDF blending\n% Author: Stefan Kottwitz\n% https://www.packtpub.com/hardware-and-creative/latex-cookbook\n\\documentclass[border=10pt]{standalone}\n\\usepackage{tikz}\n\\begin{document}\n\\begin{tikzpicture}\n\\begin{scope}[blend group = soft light]\n\\fill[red!30!white] ( 90:1.2) circle (2);\n\\fill[green!30!white] (210:1.2) circle (2);\n\\fill[blue!30!white] (330:1.2) circle (2);\n\\end{scope}\n\\node at ( 90:2) {Typography};\n\\node at ( 210:2) {Design};\n\\node at ( 330:2) {Coding};\n\\node [font=\\Large] {\\LaTeX};\n\\end{tikzpicture}\n\\end{document}" ]

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[ "## Class Ints\n\n• ```@GwtCompatible(emulated=true)\npublic final class Ints\nextends Object```\nStatic utility methods pertaining to `int` primitives, that are not already found in either `Integer` or `Arrays`.\n\nSee the Guava User Guide article on primitive utilities.\n\nSince:\n1.0\nAuthor:\nKevin Bourrillion\n• ### Field Summary\n\nFields\nModifier and Type Field Description\n`static int` `BYTES`\nThe number of bytes required to represent a primitive `int` value.\n`static int` `MAX_POWER_OF_TWO`\nThe largest power of two that can be represented as an `int`.\n• ### Method Summary\n\nAll Methods\nModifier and Type Method Description\n`static List<Integer>` `asList​(int... backingArray)`\nReturns a fixed-size list backed by the specified array, similar to `Arrays.asList(Object[])`.\n`static int` `checkedCast​(long value)`\nReturns the `int` value that is equal to `value`, if possible.\n`static int` ```compare​(int a, int b)```\nCompares the two specified `int` values.\n`static int[]` `concat​(int[]... arrays)`\nReturns the values from each provided array combined into a single array.\n`static int` ```constrainToRange​(int value, int min, int max)```\nReturns the value nearest to `value` which is within the closed range `[min..max]`.\n`static boolean` ```contains​(int[] array, int target)```\nReturns `true` if `target` is present as an element anywhere in `array`.\n`static int[]` ```ensureCapacity​(int[] array, int minLength, int padding)```\nReturns an array containing the same values as `array`, but guaranteed to be of a specified minimum length.\n`static int` `fromByteArray​(byte[] bytes)`\nReturns the `int` value whose big-endian representation is stored in the first 4 bytes of `bytes`; equivalent to `ByteBuffer.wrap(bytes).getInt()`.\n`static int` ```fromBytes​(byte b1, byte b2, byte b3, byte b4)```\nReturns the `int` value whose byte representation is the given 4 bytes, in big-endian order; equivalent to `Ints.fromByteArray(new byte[] {b1, b2, b3, b4})`.\n`static int` `hashCode​(int value)`\nReturns a hash code for `value`; equal to the result of invoking ```((Integer) value).hashCode()```.\n`static int` ```indexOf​(int[] array, int target)```\nReturns the index of the first appearance of the value `target` in `array`.\n`static int` ```indexOf​(int[] array, int[] target)```\nReturns the start position of the first occurrence of the specified `target` within `array`, or `-1` if there is no such occurrence.\n`static String` ```join​(String separator, int... array)```\nReturns a string containing the supplied `int` values separated by `separator`.\n`static int` ```lastIndexOf​(int[] array, int target)```\nReturns the index of the last appearance of the value `target` in `array`.\n`static Comparator<int[]>` `lexicographicalComparator()`\nReturns a comparator that compares two `int` arrays lexicographically.\n`static int` `max​(int... array)`\nReturns the greatest value present in `array`.\n`static int` `min​(int... array)`\nReturns the least value present in `array`.\n`static void` `reverse​(int[] array)`\nReverses the elements of `array`.\n`static void` ```reverse​(int[] array, int fromIndex, int toIndex)```\nReverses the elements of `array` between `fromIndex` inclusive and `toIndex` exclusive.\n`static int` `saturatedCast​(long value)`\nReturns the `int` nearest in value to `value`.\n`static void` `sortDescending​(int[] array)`\nSorts the elements of `array` in descending order.\n`static void` ```sortDescending​(int[] array, int fromIndex, int toIndex)```\nSorts the elements of `array` between `fromIndex` inclusive and `toIndex` exclusive in descending order.\n`static Converter<String,​Integer>` `stringConverter()`\nReturns a serializable converter object that converts between strings and integers using `Integer.decode(java.lang.String)` and `Integer.toString()`.\n`static int[]` `toArray​(Collection<? extends Number> collection)`\nReturns an array containing each value of `collection`, converted to a `int` value in the manner of `Number.intValue()`.\n`static byte[]` `toByteArray​(int value)`\nReturns a big-endian representation of `value` in a 4-element byte array; equivalent to `ByteBuffer.allocate(4).putInt(value).array()`.\n`static Integer` `tryParse​(String string)`\nParses the specified string as a signed decimal integer value.\n`static Integer` ```tryParse​(String string, int radix)```\nParses the specified string as a signed integer value using the specified radix.\n• ### Methods inherited from class java.lang.Object\n\n`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`\n• ### Field Detail\n\n• #### BYTES\n\n`public static final int BYTES`\nThe number of bytes required to represent a primitive `int` value.\n\nJava 8 users: use `Integer.BYTES` instead.\n\nConstant Field Values\n• #### MAX_POWER_OF_TWO\n\n`public static final int MAX_POWER_OF_TWO`\nThe largest power of two that can be represented as an `int`.\nSince:\n10.0\nConstant Field Values\n• ### Method Detail\n\n• #### hashCode\n\n`public static int hashCode​(int value)`\nReturns a hash code for `value`; equal to the result of invoking ```((Integer) value).hashCode()```.\n\nJava 8 users: use `Integer.hashCode(int)` instead.\n\nParameters:\n`value` - a primitive `int` value\nReturns:\na hash code for the value\n• #### checkedCast\n\n`public static int checkedCast​(long value)`\nReturns the `int` value that is equal to `value`, if possible.\nParameters:\n`value` - any value in the range of the `int` type\nReturns:\nthe `int` value that equals `value`\nThrows:\n`IllegalArgumentException` - if `value` is greater than `Integer.MAX_VALUE` or less than `Integer.MIN_VALUE`\n• #### saturatedCast\n\n`public static int saturatedCast​(long value)`\nReturns the `int` nearest in value to `value`.\nParameters:\n`value` - any `long` value\nReturns:\nthe same value cast to `int` if it is in the range of the `int` type, `Integer.MAX_VALUE` if it is too large, or `Integer.MIN_VALUE` if it is too small\n• #### compare\n\n```public static int compare​(int a,\nint b)```\nCompares the two specified `int` values. The sign of the value returned is the same as that of `((Integer) a).compareTo(b)`.\n\nNote for Java 7 and later: this method should be treated as deprecated; use the equivalent `Integer.compare(int, int)` method instead.\n\nParameters:\n`a` - the first `int` to compare\n`b` - the second `int` to compare\nReturns:\na negative value if `a` is less than `b`; a positive value if `a` is greater than `b`; or zero if they are equal\n• #### contains\n\n```public static boolean contains​(int[] array,\nint target)```\nReturns `true` if `target` is present as an element anywhere in `array`.\nParameters:\n`array` - an array of `int` values, possibly empty\n`target` - a primitive `int` value\nReturns:\n`true` if `array[i] == target` for some value of `i`\n• #### indexOf\n\n```public static int indexOf​(int[] array,\nint target)```\nReturns the index of the first appearance of the value `target` in `array`.\nParameters:\n`array` - an array of `int` values, possibly empty\n`target` - a primitive `int` value\nReturns:\nthe least index `i` for which `array[i] == target`, or `-1` if no such index exists.\n• #### indexOf\n\n```public static int indexOf​(int[] array,\nint[] target)```\nReturns the start position of the first occurrence of the specified `target` within `array`, or `-1` if there is no such occurrence.\n\nMore formally, returns the lowest index `i` such that ```Arrays.copyOfRange(array, i, i + target.length)``` contains exactly the same elements as `target`.\n\nParameters:\n`array` - the array to search for the sequence `target`\n`target` - the array to search for as a sub-sequence of `array`\n• #### lastIndexOf\n\n```public static int lastIndexOf​(int[] array,\nint target)```\nReturns the index of the last appearance of the value `target` in `array`.\nParameters:\n`array` - an array of `int` values, possibly empty\n`target` - a primitive `int` value\nReturns:\nthe greatest index `i` for which `array[i] == target`, or `-1` if no such index exists.\n• #### min\n\n```@GwtIncompatible(\"Available in GWT! Annotation is to avoid conflict with GWT specialization of base class.\")\npublic static int min​(int... array)```\nReturns the least value present in `array`.\nParameters:\n`array` - a nonempty array of `int` values\nReturns:\nthe value present in `array` that is less than or equal to every other value in the array\nThrows:\n`IllegalArgumentException` - if `array` is empty\n• #### max\n\n```@GwtIncompatible(\"Available in GWT! Annotation is to avoid conflict with GWT specialization of base class.\")\npublic static int max​(int... array)```\nReturns the greatest value present in `array`.\nParameters:\n`array` - a nonempty array of `int` values\nReturns:\nthe value present in `array` that is greater than or equal to every other value in the array\nThrows:\n`IllegalArgumentException` - if `array` is empty\n• #### constrainToRange\n\n```@Beta\npublic static int constrainToRange​(int value,\nint min,\nint max)```\nReturns the value nearest to `value` which is within the closed range `[min..max]`.\n\nIf `value` is within the range `[min..max]`, `value` is returned unchanged. If `value` is less than `min`, `min` is returned, and if ``` value``` is greater than `max`, `max` is returned.\n\nParameters:\n`value` - the `int` value to constrain\n`min` - the lower bound (inclusive) of the range to constrain `value` to\n`max` - the upper bound (inclusive) of the range to constrain `value` to\nThrows:\n`IllegalArgumentException` - if `min > max`\nSince:\n21.0\n• #### concat\n\n`public static int[] concat​(int[]... arrays)`\nReturns the values from each provided array combined into a single array. For example, ``` concat(new int[] {a, b}, new int[] {}, new int[] {c}``` returns the array `{a, b, c}`.\nParameters:\n`arrays` - zero or more `int` arrays\nReturns:\na single array containing all the values from the source arrays, in order\n• #### toByteArray\n\n`public static byte[] toByteArray​(int value)`\nReturns a big-endian representation of `value` in a 4-element byte array; equivalent to `ByteBuffer.allocate(4).putInt(value).array()`. For example, the input value ``` 0x12131415``` would yield the byte array `{0x12, 0x13, 0x14, 0x15}`.\n\nIf you need to convert and concatenate several values (possibly even of different types), use a shared `ByteBuffer` instance, or use `ByteStreams.newDataOutput()` to get a growable buffer.\n\n• #### fromByteArray\n\n`public static int fromByteArray​(byte[] bytes)`\nReturns the `int` value whose big-endian representation is stored in the first 4 bytes of `bytes`; equivalent to `ByteBuffer.wrap(bytes).getInt()`. For example, the input byte array `{0x12, 0x13, 0x14, 0x15, 0x33}` would yield the `int` value ``` 0x12131415```.\n\nArguably, it's preferable to use `ByteBuffer`; that library exposes much more flexibility at little cost in readability.\n\nThrows:\n`IllegalArgumentException` - if `bytes` has fewer than 4 elements\n• #### fromBytes\n\n```public static int fromBytes​(byte b1,\nbyte b2,\nbyte b3,\nbyte b4)```\nReturns the `int` value whose byte representation is the given 4 bytes, in big-endian order; equivalent to `Ints.fromByteArray(new byte[] {b1, b2, b3, b4})`.\nSince:\n7.0\n• #### ensureCapacity\n\n```public static int[] ensureCapacity​(int[] array,\nint minLength,\nReturns an array containing the same values as `array`, but guaranteed to be of a specified minimum length. If `array` already has a length of at least `minLength`, it is returned directly. Otherwise, a new array of size `minLength + padding` is returned, containing the values of `array`, and zeroes in the remaining places.\nParameters:\n`array` - the source array\n`minLength` - the minimum length the returned array must guarantee\n`padding` - an extra amount to \"grow\" the array by if growth is necessary\nReturns:\nan array containing the values of `array`, with guaranteed minimum length ``` minLength```\nThrows:\n`IllegalArgumentException` - if `minLength` or `padding` is negative\n• #### join\n\n```public static String join​(String separator,\nint... array)```\nReturns a string containing the supplied `int` values separated by `separator`. For example, `join(\"-\", 1, 2, 3)` returns the string `\"1-2-3\"`.\nParameters:\n`separator` - the text that should appear between consecutive values in the resulting string (but not at the start or end)\n`array` - an array of `int` values, possibly empty\n• #### lexicographicalComparator\n\n`public static Comparator<int[]> lexicographicalComparator()`\nReturns a comparator that compares two `int` arrays lexicographically. That is, it compares, using `compare(int, int)`), the first pair of values that follow any common prefix, or when one array is a prefix of the other, treats the shorter array as the lesser. For example, `[] < < [1, 2] < `.\n\nThe returned comparator is inconsistent with `Object.equals(Object)` (since arrays support only identity equality), but it is consistent with `Arrays.equals(int[], int[])`.\n\nSince:\n2.0\n• #### sortDescending\n\n`public static void sortDescending​(int[] array)`\nSorts the elements of `array` in descending order.\nSince:\n23.1\n• #### sortDescending\n\n```public static void sortDescending​(int[] array,\nint fromIndex,\nint toIndex)```\nSorts the elements of `array` between `fromIndex` inclusive and `toIndex` exclusive in descending order.\nSince:\n23.1\n• #### reverse\n\n`public static void reverse​(int[] array)`\nReverses the elements of `array`. This is equivalent to ``` Collections.reverse(Ints.asList(array))```, but is likely to be more efficient.\nSince:\n23.1\n• #### reverse\n\n```public static void reverse​(int[] array,\nint fromIndex,\nint toIndex)```\nReverses the elements of `array` between `fromIndex` inclusive and `toIndex` exclusive. This is equivalent to ``` Collections.reverse(Ints.asList(array).subList(fromIndex, toIndex))```, but is likely to be more efficient.\nThrows:\n`IndexOutOfBoundsException` - if `fromIndex < 0`, `toIndex > array.length`, or `toIndex > fromIndex`\nSince:\n23.1\n• #### toArray\n\n`public static int[] toArray​(Collection<? extends Number> collection)`\nReturns an array containing each value of `collection`, converted to a `int` value in the manner of `Number.intValue()`.\n\nElements are copied from the argument collection as if by `collection.toArray()`. Calling this method is as thread-safe as calling that method.\n\nParameters:\n`collection` - a collection of `Number` instances\nReturns:\nan array containing the same values as `collection`, in the same order, converted to primitives\nThrows:\n`NullPointerException` - if `collection` or any of its elements is null\nSince:\n1.0 (parameter was `Collection<Integer>` before 12.0)\n• #### asList\n\n`public static List<Integer> asList​(int... backingArray)`\nReturns a fixed-size list backed by the specified array, similar to `Arrays.asList(Object[])`. The list supports `List.set(int, Object)`, but any attempt to set a value to `null` will result in a `NullPointerException`.\n\nThe returned list maintains the values, but not the identities, of `Integer` objects written to or read from it. For example, whether `list.get(0) == list.get(0)` is true for the returned list is unspecified.\n\nNote: when possible, you should represent your data as an `ImmutableIntArray` instead, which has an `asList` view.\n\nParameters:\n`backingArray` - the array to back the list\nReturns:\na list view of the array\n• #### tryParse\n\n```@Beta\n@CheckForNull\npublic static Integer tryParse​(String string)```\nParses the specified string as a signed decimal integer value. The ASCII character `'-'` (`'\\u002D'`) is recognized as the minus sign.\n\nUnlike `Integer.parseInt(String)`, this method returns `null` instead of throwing an exception if parsing fails. Additionally, this method only accepts ASCII digits, and returns `null` if non-ASCII digits are present in the string.\n\nNote that strings prefixed with ASCII `'+'` are rejected, even under JDK 7, despite the change to `Integer.parseInt(String)` for that version.\n\nParameters:\n`string` - the string representation of an integer value\nReturns:\nthe integer value represented by `string`, or `null` if `string` has a length of zero or cannot be parsed as an integer value\nThrows:\n`NullPointerException` - if `string` is `null`\nSince:\n11.0\n• #### tryParse\n\n```@Beta\n@CheckForNull\npublic static Integer tryParse​(String string,\nParses the specified string as a signed integer value using the specified radix. The ASCII character `'-'` (`'\\u002D'`) is recognized as the minus sign.\n\nUnlike `Integer.parseInt(String, int)`, this method returns `null` instead of throwing an exception if parsing fails. Additionally, this method only accepts ASCII digits, and returns `null` if non-ASCII digits are present in the string.\n\nNote that strings prefixed with ASCII `'+'` are rejected, even under JDK 7, despite the change to `Integer.parseInt(String, int)` for that version.\n\nParameters:\n`string` - the string representation of an integer value\n`radix` - the radix to use when parsing\nReturns:\nthe integer value represented by `string` using `radix`, or `null` if `string` has a length of zero or cannot be parsed as an integer value\nThrows:\n`IllegalArgumentException` - if `radix < Character.MIN_RADIX` or ```radix > Character.MAX_RADIX```\n`NullPointerException` - if `string` is `null`\nSince:\n19.0" ]

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[ "", null, "###### Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\n##### Thank you for watching the video.\n\nTo unlock all 5,300 videos, start your free trial.\n\n# The Inverse Cosine Function - Problem 3\n\nNorm Prokup", null, "###### Norm Prokup\n\nCornell University\nPhD. in Mathematics\n\nNorm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.\n\nShare\n\nI want to graph a transformation of the inverse cosine function. I have a problem that asks me to graph y equals negative inverse cosine x plus pi over 2. I’m going to go with my usual method. I’m going to make a table of key points of inverse cosine and transform the points first.\n\nNow I have my graph of cosine here in blue and I’ve jotted down the key points, -1 pi, zero pi over 2 and 1 zero and all I need to do is figure out what kind of transformation negative inverse cosine of x plus pi over 2 represents. Let me put that in here, negative inverse cosine of x plus pi over 2.\n\nThis minus sign just means that I’m going to be reflecting the graph across the x axis and we accomplish that by multiplying the y values here by -1. This plus pi over 2 means the graph’s going to get shifted up by pi over 2 and I do that by adding pi over 2 to the result. There’s really no horizontal transformation so I don’t have to do anything with the x coordinates so I can just write them down, -1, 0 and 1, right from here.\n\nLet’s get ready to transform these points. We multiply by -1 and add pi over 2. We get negative pi plus pi over 2, negative pi over 2, times -1 plus pi over 2 is zero, times -1 plus pi over 2 is pi over 2. And so we have 3 points that we can put on our graph for the negative inverse cosine of x plus pi over 2. We’ve got -1, negative pi over 2, 0, 0 and 1, pi over 2. We want to draw this with the inverse cosine shape. Looks something like this.\n\nThis is the graph of y equals negative inverse cosine x plus pi over 2 but you may notice that this is also the graph of inverse sine. Y equals inverse sine of x. It turns out that this is an identity. Inverse sine of x equals negative inverse cosine of x plus pi over 2. This identity is actually related to the co-function identity.\n\nAgain whenever we graph transformations key points of the original parent graph, transform the points and then plot the points in your graph and draw a smooth curve." ]

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[ "If you're seeing this message, it means we're having trouble loading external resources on our website.\n\nIf you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.\n\n# 2015 AP Chemistry free response 5a: Finding order of reaction\n\nFinding the order of reaction based on graphs of absorbance, ln(absorbance) and 1/(absorbance) vs. time for kinetics of bleaching food coloring. An alternative method of solving 2015 AP Chemistry free response 5a.\n\n## Want to join the conversation?\n\n• why should the graph be linear?", null, "• So the reaction is first order which means it will follow first order kinetic equations like the integrated rate law that they gave: ln([A]t) - ln([A]0) = -kt. You can rearrange the equation into: ln([A]t) = -kt + ln([A]0), which fits the slope-intercept equation y = mx + b where y is ln([A]t), x is t, m (the slope) is -k, and b (the y-intercept) is ln([A]0). Now if you graph the ln([A]t) as the y coordinate versus t as the x coordinate and they form a straight line, then the reaction is first order. And what's more you could find the rate constant k from the slope and initial concentration of A from the y-intercept.\n\nSo looking at the graphs they gave, graph 1 has [A] vs. t which follows the integrated rate of a 0th order reaction. But since the curve is not linear we can say that the data does not fit the 0th order model and that the reaction is not 0th order.\n\nGraph 2 has ln[A] vs. t which follows the integrated rate law of a first order reaction as I've shown. Now since the data here is a nice and linear we can say that it fits much better using a first order model and that the reaction is first order.\n\nGraph 3 has 1/[A] vs. t which would be following the integrated rate law of a second order reaction. And since it's not linear, it is similar to the 0th order model as to why it doesn't work.\n\nIt's pretty common to see if you're able to fit data onto a straight line in chemistry because it's a clear way to visualize if your model is predicting how the data should behave. I'm unsure as to why they didn't just solve it by looking at whether the graphs were linear or not, it's a much faster way to solve these problems especially in a test setting.\n\nHope that helps." ]

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[ "## Saturday, August 29, 2015\n\n### No, this isn't good code\n\nI saw this tweet go by. No, I don't think it's good code:\n\nWhat this code is trying to solve is the \"integer overflow\" vulnerability. I don't think it solves the problem well.\n\nThe first problem is that the result is undefined. Some programmers will call safemulti_size_t() without checking the result. When they do, the code will behave differently depending on the previous value of *res. Instead, the code should return a defined value in this case, such as zero or SIZE_MAX. Knowing that this sort of thing will usually be used for memory allocations, which you want to have fail, then a good choice would be SIZE_MAX.\n\nThe worse problem is integer division. On today's Intel processors, integer multiplication takes a single clock cycle, but integer division takes between 40 and 100 clock cycles. Since you'll be usually dividing by small numbers, it's likely to be closer to 40 clock cycles rather than 100, but that's still really bad. If your solution to security problems is by imposing unacceptable tradeoffs, then you are doing security wrong. If you introduced this level of performance hit, then you might as well be programming in a safer language like JavaScript than in C.\n\nAn alternative would be the OpenBSD function reallocarray(), which I'm considering using in all my code as a replacement for malloc(), calloc(), and realloc(). It looks like this:\n\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ``` ``````/* * This is sqrt(SIZE_MAX+1), as s1*s2 <= SIZE_MAX * if both s1 < MUL_NO_OVERFLOW and s2 < MUL_NO_OVERFLOW */ #define MUL_NO_OVERFLOW (1UL << (sizeof(size_t) * 4)) void * reallocarray(void *optr, size_t nmemb, size_t size) { if ((nmemb >= MUL_NO_OVERFLOW || size >= MUL_NO_OVERFLOW) && nmemb > 0 && SIZE_MAX / nmemb < size) { errno = ENOMEM; return NULL; } return realloc(optr, size * nmemb); }``````\n\nFirstly, it doesn't call the horrible integer division function (unless one of the parameters is larger than 2-gigs on a 32-bit processor). Secondly, it always has a defined result.\n\nPersonally, I would improve upon this function by simply calling a signal(). Virtually no code can recover from a bad memory allocation, so instead of returning NULL, it's better to crash right here.", null, "Anonymous said...\n\nMUL_NO_OVERFLOW is actually 65536 on 32-bit platforms, not 2G.\n\nAnyway, the fast way of doing this is by performing the multiplication on double-precision integers.\n\n#if SIZE_MAX == 0xffffffff\ntypedef uint64_t double_uint;\n#else\ntypedef __uint128_t double_uint;\n#endif\n\nstatic bool\nsafemult_sizet(size_t a, size_t b, size_t *r) {\ndouble_uint res = (double_uint)a*b;\n*r = (size_t)res;\nif(res > SIZE_MAX) return false;\nreturn true;\n}", null, "Anonymous said...\n\nPotential improvement, divide by lesser number. Have not benchmarked.\n#define MUL_NO_OVERFLOW (1UL << (sizeof(size_t) * 4))\n#define IS_IT_WRAP_SAFE(smaller,larger) ((larger) < MUL_NO_OVERFLOW || !(smaller) || (SIZE_MAX / (smaller) >= (larger)))\nbool\nisItWrapSafe(size_t a,size_t b)\n{\nif (a<=b)\nreturn IS_IT_WRAP_SAFE(a, b);\nelse\nreturn IS_IT_WRAP_SAFE(b, a);\n}\n\nThomasSS said...\n\nSuch a shame that hardware multiply instruction tend not to indicate whether an overflow occurred. Er... wait a second...\n\nLorenzo Grespan said...\n\nOdd that nobody pointed out how, if a = 0 the result will be different than if b = 0. Which breaks the definition of \"multiplication\".\nSecondly, as far as I remember C is not lazy so if b = 0, you have a nice division by zero because tmp/b will be evaluated. But I'm happy to be wrong here.\n\nBwanshoom said...\n\nLorenzo, you'll be happy then. I think you mean short-circuited evaluation and C definitely does it. If b is 0 the division will not occur - and that would be an obvious bug to have in this example." ]

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[ "# contour integral with hyperbola branch cuts\n\nI am reading Karl Graff's book on wave motion and stuck by a contour integral where the branch cuts are made of two hyperbolas in the 2nd and 4th quadrants. See figure attached Contour integral figure.\n\nHere only the branch cut in the 2nd quadrant is shown as the only interest is in the upper half of the complex plane. The hyperbola degrades into a path along the positive imaginary axis and a section of the real axis on the negative side close to the origin after some manipulation as shown in the figure.\n\nIf two new polar coordinate systems are defined at the two branch points on the real axis, which are symmetric to the imaginary axis and define the ends of the cut on the real axis , as done in the book, the variable now takes the form of the product of coordinates redefined in the two new coordinate systems, say, $$\\gamma$$ = $$\\rho_1e^{i\\theta_1}*\\rho_2e^{i\\theta_2}$$. In other words, the argument of $$\\gamma$$ is $$\\theta_1+\\theta_2$$.\n\nI can work out the argument by calculating the limit along each of the paths AO, OB, BO and ON but this is tedious. I am not able to read the argument out directly from the defined $$\\theta_1$$ and $$\\theta_2$$ as stated in the book. It seems to me something needs to be considered whereas the path is on one side of the complex plane when the coordinate is defined on the other. I do not know how this is done.\n\nI would like to know how this works. Any advice that can help me improve my understanding on the technique, be it a book, a paper or a direct illustration, is appreciated.\n\nAO: $$\\theta_1+\\theta_2=\\pi$$ (evident)\nOB: $$\\theta_1 (=\\pi)+\\theta_2(=0) = \\pi$$\nBO: $$\\theta_1 (=\\pi)$$ + $$\\theta_2(=-2\\pi$$, rotates clockwise one circle) = $$-\\pi$$\nOA: $$\\theta_1 + \\theta_2 = -\\pi$$($$\\theta_2$$ is negative)" ]

[ null ]

[ "", null, "", null, "# Question\n\nCardano (Italian mathematician in the 16th century) came up with the following problem:\n\nThe sum of two numbers is 10 and the product is 40, what are the two numbers?\n\n# IB Math HL – Complex Numbers\n\nCardano (Italian mathematician in the 16th century) came up with the following problem:\n\nThe sum of two numbers is 10 and the product is 40, what are the two numbers?\n\nRafael Bombelli wrote a book on Algebra 15 years after Cardano, including the following example:", null, "Cardano published a formula that another Italian mathematician, Del Ferro, had come up with in 1545, for the solution of an equation of the form", null, ", the solution being:", null, "Using this formula, the solution to Bombelli’s example is:", null, "Bombelli decided that the value of this should be 4 (the only real solution), and that the two cube root expressions should be equivalent to", null, "and", null, ", the sum of which would be 4.\n\nThis striking thought process can be seen as the birth of what we know today as complex numbers.\n\nNot only mathematics, but also science and technology have benefitted from these new abstract elements.", null, "is given the letter i (imaginary unit)", null, "and", null, "Complex numbers have the form", null, ", where", null, ". Calculating with complex numbers works with the same principles as real numbers.\n\nExamples:\n\na.", null, "b.", null, "Special case (why?):", null, "This is handy when dividing by complex numbers.\n\nExamples:", null, "", null, "Powers of complex numbers can also be calculated, for example:", null, "", null, "", null, "With", null, "________, we look back at Bombelli who decided that", null, "was equivalent to", null, "…….\n\nSo, now we have a new addition to our world of numbers………………\n\nAre complex numbers a subset of real numbers, or are real numbers a subset of complex numbers, or neither?\n\nComplex numbers are consistent with the rules of numbers as we know them for real numbers, except for the rules for inequalities. Why is this so?\n\nSOURCE:\n\n‘Wiskunde in een Notendop’ by Martin Kindt and Ed de Moor\n\n(Publ. Uitgeverij Bert Bakker) 2008", null, "", null, "" ]

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[ "Impact Factor 1.895 | CiteScore 2.24\nMore on impact ›\n\n# Frontiers in Physics", null, "## Original Research ARTICLE\n\nFront. Phys., 08 September 2015 | https://doi.org/10.3389/fphy.2015.00072\n\n# Network topology of the desert rose\n\n• 1Department of Physics, Norwegian University of Science and Technology, Trondheim, Norway\n• 2Polytec Research Institute, Haugesund, Norway\n• 3Department of Theoretical Sciences, Satyendra Nath Bose National Centre for Basic Sciences, Kolkata, India\n\nDesert roses are gypsum crystals that consist of intersecting disks. We determine their geometrical structure using computer assisted tomography. By mapping the geometrical structure onto a graph, the topology of the desert rose is analyzed and compared to a model based on diffusion limited aggregation. By comparing the topology, we find that the model gets a number of the features of the real desert rose right, whereas others do not fit so well.\n\n## 1. Introduction\n\nFracture networks are of great importance in geophysical transport properties. In reservoirs, they provide “highways” for fluid and gas transport. It is therefore surprising to discover that there are no data available on the structure of such networks beyond the most rudimentary.\n\nHydraulic fracturing, the fracturing of rock—shale—formations due to the injection of fluids at high pressure, is at the heart of a new process to extract petroleum and gas from shales. This process has become so important that it has large impact on the world economy [1, 2]. Fracking, as it is known, is also a highly controversial technique as the shales that are fractured often are close to aquifers. The hydraulic fracturing may as a result of this proximity produce fractures that open up for fluid transport into them. As the chemicals used in the hydraulic fracturing are toxic, this poses a problem together with methane seepage [3, 4].\n\nSurprisingly little is known about the hydraulic fracturing process. As with fracture systems in reservoirs, the geometry and topology of fracture networks that are created is essentially unknown. For example, is it a branched structure where the fractures are like leaves or is it a structure consisting of intersecting fracture planes? One important reason for this lack of knowledge is that no technique exists, acoustic or electromagnetic, that can visualize the fractures in situ.\n\nAs fracture networks consist of intersecting fracture sheets, modern graph theory as it has been developed over the last years, [5, 6], is not directly applicable. By transforming the fracture network into an equivalent graph where the fractures are the nodes and their intersections are the links between them as described in Andresen et al. and Hope et al. , it is, however, possible to analyze fracture networks within modern graph theory.\n\nHence, the theoretical tools are in place to analyze fracture networks, but real fracture systems to test these tools on do not exist.\n\nThere is a mineral that is formed by a process that has some resemblance to hydraulic fracturing: the desert rose, see Figure 1. Desert roses are formed in wet sand. They are evaporites where the crystal formation has occurred as a result of inflow of water containing dissolved calcium sulfate balanced by an outflow of water due to evaporation [9, 10]. Such minerals are used as indicators of arid climates during the time they were formed.\n\nFIGURE 1", null, "Figure 1. A desert rose is an aggregate of disk shaped gypsum crystals that have grown in water-saturated sand in desert areas. The desert rose in the figure is the one we have analyzed. It measures approximately 76 × 49 × 37 cm.\n\nThe gypsum crystals of desert roses contain grains of silica sand . Typically 50–70 % of the weight of a desert rose is gypsum . Analysis reveals a higher concentration of silicon in the core of desert roses, compared to the peripheral parts, suggesting that the disks grow from seeds of silicon .\n\nThe fact that desert roses are rarely encountered in the holocene deposits indicates a formation time longer than the 10,000 year age of the holocene deposits.\n\nThe kinship with hydraulic fracturing comes from the following observations. In hydraulic fracturing the fracture surfaces form equipotential surfaces for the injected fluid pressure. This pressure generates stresses in the rock and as a result, the fractures grow. One may view crystal growth in sand as a fracture process. The sand grains are unconsolidated, but are kept together by capillary forces due to the surrounding fluid . The formation and subsequent growth of a crystal creates surfaces in the sand, and these surfaces are fracture surfaces as they are surfaces created by breaking open contacts between the sand grains. The difference to the hydraulic fracturing process, however, lies in the different boundary conditions at the fracture surfaces; whereas the pressure is given at the fracture surfaces in the hydraulic fracture process, it is the deformation which is given at the fracture surfaces in the desert rose formation process.\n\nThe difference in the boundary conditions at the fracture surfaces between the two processes results in the shapes of the fractures being different. However, it is not possible to compare the branching of the fractures in the two processes, i.e., the creation of new fractures different from the old ones. To our knowledge, the details of the branching occurring in hydraulic fracture have not been studied systematically. In computational studies based on a discrete element method , branching is caused by local strength variations in the material that is fracturing and proceeds by tip splitting. However, the medium was in this case two-dimensional, and it is not known whether this would be the dominating process in three dimensions. The alternative to tip splitting is that new fractures nucleate on the surface of existing fractures. In the desert rose growth process, nucleation of new crystals on the surface of existing crystals is the dominating mechanism .\n\nThe geometry and topology of desert roses are interesting in their own right and not only as a proxy for hydraulic fracturing. The complex desert rose shape is not the only way that gypsum crystallizes. Depending on the conditions under which this happens, the kind of crystalline structure which is found will reflect this and, hence, act as an indicator of these conditions at the time the crystals were formed . It is therefore important to understand the precise mechanism that leads to the desert rose shape.\n\nIn the present study, we have identified the geometrical structure of a desert rose, see Figure 1, by using medical computer assisted tomographic (CT) scanning. Using the method described in Andresen et al. for two-dimensional fracture networks and in Hope et al. for three-dimensional fracture networks, we have mapped the geometrical structure onto a graph which we then have analyzed using the tools of modern graph (network) theory [5, 6]. We show the graph based on the desert rose of Figure 1 in Figure 2.\n\nFIGURE 2\n\nWe have then constructed a stochastic growth model for the evolution of desert roses based on new crystals forming on the existing ones through a nucleation process. By mapping the ensuing computer-generated structure onto the graph as for the scanned desert roses, we are able to compare the model with the real structure quantitatively. We are therefore able to judge how well the model reproduces the real structure.\n\nWe note how this program has some similarities with the introduction of fractals in the eighties . This concept made it possible to characterize complex landscapes quantitatively and as a consequence, be able to construct stochastic models that would be of such quality that they could be used by the motion picture industry . We note that whereas the measurement of fractal dimensions requires large structures over many length scales, the measurement of graph properties do not require large structures.\n\nIn the next section, we describe our use of a medical CT scanner for recording the structure of the desert rose seen in Figure 1. We then go on to present our stochastic model for the formation and growth of desert roses in Section 3. Section 4 presents the graph theoretical analysis of both the scanned desert rose and of the stochastically generated desert roses from our model and in Section 5 we present the results. We end by presenting our conclusions.\n\n## 2. Computer Assisted Tomographic Scanning and Analysis of a Desert Rose\n\nThe basis for our analysis is the desert rose from Morocco, shown in Figure 1. It measures approximately 76 × 49 × 37 cm.\n\nUsing a medical CT scanner we register cross-sections of the desert rose. For each cross-section the different disks and their connections are identified. By following the different disks through all cross-sections in which they appear, a full mapping of all disks and their connections is obtained. The process of following disks through several cross-sections to determine if they are connected is illustrated in Figure 3.\n\nFIGURE 3", null, "Figure 3. Three distinct disks are shown in (A) while (B) shows how the three disks are found to be connected in a later frame.\n\nThe process is done manually and we do not rule out the possibility of errors both in interpreting the images and in the registration process. Especially in the case of the smallest disks which are grouped together on larger disks, as seen on disks in the center of Figure 1, it is likely that not all of these disks have been registered. Larger disks are all registered, but there may be some uncertainties about their connections. This is the case in areas where several disks grow together, which may make it difficult to determine the extent of the different disks.\n\nOnce the disks have been identified, the structure is mapped onto a graph [7, 8]. This is done by identifying each disk as a node and placing a link between disks (nodes) that intersect each other. The ensuing nodes-and-links structure is the graph which is analyzed. We identified 402 disks and 679 intersections between different disks, giving rise to the graph shown in Figure 2.\n\n## 3. Stochastic Desert Rose Model\n\nIn this section, we describe our model for desert rose growth. From visual inspection of the desert roses, we note that although the disks are compact, the structure as a whole is not and that there is not much variance in the size of the disks it consists of. Roughly, the variance is of the order of a magnitude.\n\nThe fact that the disks do not vary much in size wherever they are situated in the structure indicates that the growth of the disks is slow enough so that gradients in the calcium sulfate concentration outside the disks is not important. On the other hand, the number of disks is of the order of a few hundred and they form an open structure. This indicates that disk nucleation is a rare event. One may then think of two possible scenarios: the first scenario would be that the structures on which the disks nucleate are already in place in the sand. The disks would then grow from each nucleus and then merge into the desert rose structure. This scenario does, however, not explain why there is such a great variance in the sizes of the desert roses that are found. The second scenario is that the structures on which the disks nucleate also diffuse, but with a very low concentration. The nucleation of new disks would then be a diffusion-limited aggregation process (DLA) which creates open (even fractal) structures. Hence, we base our model on the DLA model of Witten and Sanders .\n\nParticles on which the disks will eventually nucleate execute diffusive motion, i.e., perform random walks, starting from far enough distances from the growing aggregate. Once a particle is within the close proximity of the desert rose, it lands on the growing structure in the form of a two-dimensional circular disk of variable radius R. Each disk is placed such that its center lies on a previously placed disk and its orientation is selected randomly with uniform probability.\n\nIn order to model the disks having some variation on their radii, they are drawn from a power law probability distribution p(R) ~ R−1. To implement this procedure, uniformly distributed random numbers xi within −1 < xi < 1 are generated and the radii ${{R}}_{{i}}{=}{1}{{0}}^{{\\beta }{{x}}_{{i}}}$ are assigned. Consequently, the distribution of radii assumes p(R) ~ R−1 form within the range 10−β to 10β.\n\nWhen β = 0, all radii are equal. With β = 0.8, the ratio between the largest and the smallest disk will be of the order of 40 and with β = 1.2, the ratio is around 250. Hence, we will be able to investigate the influence of the disk size distribution on the ensuing disk structure.\n\nThe first disk with radius R1 is placed on the xy-plane with center at the origin. At an arbitrary stage, let there be n disks in the cluster. To add the next (n + 1)th disk, a random walker is released on the surface of an imaginary sphere with its center at the origin, whose radius is sufficiently large to encompass the entire desert rose. It is then allowed to perform a random walk in three-dimensional space. A “sphere of influence” of radius R is imagined to be associated with every disk in the desert rose. When the incoming diffusive particle penetrates such a sphere, it is stopped instantly, and this particular disk is selected. The center of the (n + 1)th disk with radius Rn + 1 is then placed at a randomly selected point on the surface of this disk and is assigned a random orientation in space. The precise value of Rn + 1 is drawn from the probability distribution p(R). Occasionally, it may happen that the random walker enters within the common space of intersection of more than one sphere; in that case we select a disk randomly among them and place the new disk on its surface.\n\nCrystal “growth” is executed in the following way: Once a disk in the growing desert rose is chosen, a random point is selected on that disk with coordinates [${{x}}_{{c}}^{\\left({n}{+}{1}\\right)}{,}{{y}}_{{c}}^{\\left({n}{+}{1}\\right)}{,}{{z}}_{{c}}^{\\left({n}{+}{1}\\right)}$] and the center of the (n + 1)th disk is placed at this point. Next, the new disk is given a random orientation by choosing three Euler angles (ϕ, θ, ψ) randomly. The sequence of rotations is the following:\n\n(i) The first rotation is by an angle ϕ in the range [0, 2π] about the z−axis,\n\n(ii) the second rotation is by an angle θ in the range [0, π] about the new x− axis, and\n\n(iii) the third rotation is by an angle ψ in the range [0, 2π] about the new z−axis.\n\nIf the three rotation matrices corresponding to the three Euler angles be denoted by D, C, and B, respectively, then the general rotation is given by A = BCD. Therefore, each disk is characterized by its center coordinates (xc, yc, zc), three Euler angles (ϕ, θ, ψ), and the radius (R). We note that the direction cosines (a, b, c) of the normal to the plane of the disk are obtained by the elements of the third column of the rotation matrix A.\n\nFrom the network a graph is constructed by representing disks as nodes and linking nodes representing connected disks.\n\nThere exists a connection between any two disks if they intersect each other. Whether two disks i and j have an intersection or not can be detected by checking the following two conditions:\n\n1. |${C}$i${C}$j| ≤ (Ri + Rj), where ${C}$i and ${C}$j are the centers of the disk i and j, respectively. This means that the distance between the center of the two disks must be less than or equal to the sum of their individual radii.\n\n2. ${D}$iRi and ${D}$jRj, where ${D}$i is the perpendicular distance from the line of intersection of the two planes containing the disks to the center of disk i and ${D}$j is that of disk j.\n\nIf both conditions are satisfied simultaneously then there is an intersection between the two disks. If one or both conditions are not met, there is no intersection between the two disks.\n\n## 4. Network Analysis\n\nThe following gives a brief presentation of how the networks are analyzed, based on Hope et al. .\n\nBy transforming the desert rose structure into a graph representation, they can be analyzed using methods from modern network science. The transform used, was introduced for fracture networks by Andresen et al. , and applied to three-dimensional fracture networks by Hope et al. . As shown in Figure 4, each disk is defined as a node and nodes are linked if they represent intersecting disks.\n\nFIGURE 4", null, "Figure 4. Illustration of how the structure of the desert rose is transformed into a graph by defining each disk as a node and linking nodes which represent intersecting disks. (A) Representation of a small desert rose made up by six disks. (B) Equivalent graph representation where the color of the nodes correspond to the color of the disks they represent.\n\nThe properties of a given network are useful to compare against a random version of the same network . For such a comparison, a fully random model with the same number of nodes and links is used.\n\nA measure of the local connectivity of a network is the clustering coefficient. The coefficient Ci gives the ratio between the number of connections between the ki neighbors of node i and the ki(ki − 1)/2 possible ways they could be connected [5, 18]. In the case where ki < 2, Ci = 0 .\n\nA global network coefficient, C, is defined as the average of all local clustering coefficients, giving\n\nwhere N is the number of nodes in the network and Ki is the number of connections between node i's neighbors .\n\nA measure of the long range connectivity of the network is the characteristic path length, L. Based on the shortest path, dij, between nodes i and j, i.e., the path with the fewest links traversed, the characteristic path length can be defined as [18, 20]\n\nA measure of the degree mixing is the assortativity coefficient r . The coefficient can be expressed as follows\n\nwhere k1i and k2i are the degrees of the nodes linked by the ith link and M is the number of links . Assortative mixing is indicated by r > 0, while r < 0 indicate dissasortative mixing.\n\nThe variance of the assortative coefficient for a single sample can be found by the jackknife estimate\n\nwhere ri is the assortative coefficient calculated while excluding the ith link [19, 22].\n\n## 5. Results and Discussion\n\nThe graph properties of the desert rose are shown in Table 1. The clustering coefficient is an order of magnitude larger for the desert rose than for the random graph, while the characteristic path length is similar to that of the random graph, indicating a small-world network .\n\nTABLE 1", null, "Table 1. Average values of the number of nodes, links, maximum degree kmax, average degree 〈k〉, clustering coefficient C, clustering coefficient for random networks CRA, characteristic path length L, characteristic path length for random networks LRA, assortativity coefficient r with its standard deviation σr for the desert rose and the desert rose model with four values of the parameter β.\n\nThe desert rose has an assortativity coefficient of −0.29. As such the graph shows dissasortative degree mixing, which is typical of both technological and biological networks .\n\nIn the case of desert rose model, the assortativity coefficient is positive when all the disks have unit radii. This is in contrast to the network behavior of the real desert rose. When the width of the distribution of radii p(R) is increased, a change is observed in the degree mixing. For β = 0.8, 1.0, and 1.2 the assortativity coefficient is found to be negative as shown in Table 1.\n\nPrevious studies of two-dimensional fracture networks have shown a strong connection between fracture size and the degree of the corresponding node [8, 23]. This follows from the chance of connecting with other fractures increases with the size of the fractures. We see the same effect here in Figure 5. Here we plot the average degree 〈k〉 as a function of disk radius, R. Asymptotically, the data are consistent with 〈k〉 ~ exp(0.2R) for β = 1.0. The reason for this behavior is the following: our model is based on the DLA process and it is assumed that each disk in the cluster has a “sphere of influence” of the same radius as the disk itself. When the incoming diffusive particle penetrates such a sphere, it selects that disk as nucleation site. This creates a new node in the graph with a link to the disk that was chosen as a nucleation site. As the probability of a particle to enter the “sphere of influence” of a given disk increases with the radius of the disk, so does the probability that a link is formed to the node that represents that particular disk in the graph. We observe visually that the same holds true for the disks in the desert rose model—the larger they are, the more likely they are to connect with other disks. Using a single size for all disks is likely to result in many disks connecting to the same number of other disks. For such a network, it is not surprising that nodes tend to link to other nodes with similar degree.\n\nFIGURE 5", null, "Figure 5. (A) Average degree 〈k〉 as a function of disk radius, R for β = 1.0. (B) The same plot in log-linear scale. The straight line is exp(0.2R).\n\nThe assortativity coefficient of the model changes sign from positive to negative for β as β is increased. We show this in Figure 6. The data points follow closely the fit r(β) = 0.305 − 0.537β leading to r(β) = 0 for β = 0.57. We expect the assortativity coefficient to decrease with increasing β for the following reason. With increasing β, the largest disks become even larger. At the same time, the number of small disks is much larger than the number of large disks due to the character of the radius distribution that we use. We have just argued that the larger the disk, the higher the coordination mumber is for its equivalent node in the graph. This means that these disks must connect to disks of smaller radius and hence smaller coordination number. This drives the assortativity coefficient down.\n\nFIGURE 6", null, "Figure 6. Assortativity coefficient as function of β. The straight line is 0.305 − 0.537β, leading to r(β) = 0 for β = 0.57.\n\nA noteworthy simplification in the desert rose model is that the disks are free to grow through other disks. Studies of models generating two-dimensional fracture network models have shown that whether or not fractures act as barriers for other fractures play an important part in whether the resulting graphs show assortative or disassortative degree mixing [8, 23]. Similarly for models generating three-dimensional fracture networks, the ability of fractures to limit the growth of other fractures has been found to have an impact on the degree mixing. However, in this case the models are found to be disassortative, but the strength of the disassortative mixing is weaker if the fractures are free to cross each other . In the case of the desert rose model, the degree mixing is controlled by the range of the size distribution, and for a sufficient range the model results in disassortative mixing with strength similar to that of the real sample.\n\nThe desert rose networks reconstructed from the CT scans also exhibit a broad degree distribution as shown in Figure 7. Whether it follows a power law or not cannot, however, be concluded from the data. We studied it for different values of β from 0 to 1.2 by increments of 0.1 in our model. The degree distribution is observed to be broad for β = 0.8 and larger. In Figure 8 the degree distribution is shown in log–log scale for four different sizes with β = 1.0. Since the growth model is based on the DLA process, the disks experience a screening effect and thus the disks deep interior to the desert rose are not selected any more as the nucleation site. This results in their degree to remain constant after placing certain number of disks in the system and therefore data collapse is not observed.\n\nFIGURE 7", null, "Figure 7. Log–log plot of the degree distribution for the real desert rose sample. A power law p(k) ~ k−α with α = 2.0 has been added as a guide to the eye.\n\nFIGURE 8", null, "Figure 8. Log–log plot of degree distribution when the radii of the disks are power law distributed with β = 1.0 for four values of number of disks N = 128, 256, 512, and 1024 with N increasing from left to right. The initial slope, in all four cases, is compatible with a power law, p(k) ~ k−α, where α ≃ 1.22.\n\nWith respect to the relationship between characteristic path length and system size, Figure 9 shows how L scales linearly when plotted against the ratio ln(N)/ln(〈k〉(N)). This is characteristic of the small-world effect .\n\nFIGURE 9", null, "Figure 9. Average path length plotted against ln(N)/ln(〈k〉(N)) with best linear fit illustrated for β = 1.0. This indicates the presence of the small-world effect.\n\nThe nature of the graph from the model is in correspondence with the real desert rose graph in terms of the degree distribution and degree mixing within a particular range of β. The value of the clustering coefficient is much higher and characteristic path length obtained from the model is lower than that of real desert rose indicating a much denser and highly connected graph. This is probably because the disks are allowed to penetrate through other disks, which is not seen in case of the real desert rose.\n\n## 6. Conclusion\n\nThe topology of the desert rose is studied by first scanning it using a CT scanner. By representing the disks as nodes and their intersections as links, we have constructed a graph representation of the crystal.\n\nBased on what is known about the geological processes that lead to the formation of desert roses, we have constructed a DLA-type model where new disks are grown from nucleation sites on existing disks when reached by a random walker.\n\nBy comparing the topology, we find that the model gets some features of the real desert rose right, whereas others do not fit so well. In particular, it is found that when all disks are of the same size, the model shows assortative degree mixing whereas the desert rose shows disassortative degree mixing. When the variation of the disk radii is increased, the graphs from the model change character and shows dissasortative degree mixing, as in the real system. The degree distribution shows a broad distribution in the desert rose, whereas it is only broad for wider distributions of disk radii. The desert rose shows a rather narrow distribution of disk sizes; typically over an order or magnitude or so. Both the real system and the model show a small-world structure.\n\nWe alluded in the introduction to using the desert rose system as a proxy for fracture networks produced through hydraulic fracturing. Hydraulic fracturing is of increasing importance, and it is becoming urgent that the resulting fracture networks are fully understood. This paper shows that relatively small systems yield precise topological information making it possible to judge quantitatively the quality of different models.\n\n## Author Contributions\n\nSH analyzed the real desert rose. SK and CR implemented the desert rose model. SH, SK, and CR preformed the network analysis. All authors took part in developing the desert rose model, discussing the results and writing the paper.\n\n## Conflict of Interest Statement\n\nThe authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.\n\n## Acknowledgments\n\nL. I. Nesje at St. Olavs Hospital in Trondheim, Norway is gratefully acknowledged for CT scan of the desert rose. The authors thank M. B. E. Mørk for valuable discussions. We thank Norges Forskningsråd for funding through the RENERGI program (grant no. 217413/E20). SH and AH thank Norges Forskningsråd for funding through the CLIMIT program (grant no. 199970).\n\n## References\n\n1. Conti JJ. et al. Annual Energy Outlook 2013 - With Projections to 2040. Washington, DC: U.S. Energy Information Administration (2013).\n\n2. The Economist. The New Economics of Oil: Sheikhs vs. Shale. (2012).\n\n3. Mooney C. The truth about fracking. Sci Am. (2011). 305:80–5. doi: 10.1038/scientificamerican1111-80\n\n4. Rahm D. Regulating hydraulic fracturing in shale gas plays: the case of Texas. Energy Policy. (2011) 39:2974–81. doi: 10.1016/j.enpol.2011.03.009\n\n5. Albert R, Barabási AL. Statistical mechanics of complex networks. Rev Modern Phys. (2002) 74:47–97. doi: 10.1103/RevModPhys.74.47\n\n6. Newman MEJ. Networks: An Introduction. Oxford: Oxford University Press (2010). doi: 10.1093/acprof:oso/9780199206650.001.0001\n\n7. Andresen CA, Le Goc R, Davy P, Hansen A, Hope SM. Topology of fracture networks. Front Phys. (2013) 1:7. doi: 10.3389/fphy.2013.00007\n\n8. Hope SM, Davy P, Maillot J, Le Goc R, Hansen A. Topological impact of constrained fracture growth. Front Phys. (2015). 3:75. doi: 10.3389/fphy.2015.00075\n\n9. Al-Kofahi MM, Hallak AB, Al-Juwair HA, Saafin AK. Analysis of desert rose using PIXE and RBS techniques. X-Ray Spectrom. (1993) 22:23–27. doi: 10.1002/xrs.1300220107\n\n10. Samy Y, Metwally HI. Gypsum crystal habitats as evidence for aridity and stagnation, Northeast of the Nile river delta, Egypt. Aust J Basic Appl Sci. (2012) 6:442–50.\n\n11. Watson A. Structure, chemistry and origins of gypsum crusts in southern Tunisia and the central Namib Desert. Sedimentology (1985) 32:855–75. doi: 10.1111/j.1365-3091.1985.tb00737.x\n\n12. Hornbaker DJ, Albert I, Barabási AL, Schiffer P. Why sandcastles stand: an experimental study of wet granular media. Nature (1997) 387:765. doi: 10.1038/42831\n\nCrossRef Full Text\n\n13. Skjetne B, Hansen A. Stochastic modeling of crack avalanches in hydraulic fracture. South African Inst Mining Metallur Symp Ser. (2012) S71:81.\n\n14. Mandelbrot BB. The Fractal Geometry of Nature. New York, NY: W. H. Freeman (1982).\n\n15. Briggs J. Fractals: The Patterns of Chaos: A New Aesthetic of Art, Science and Nature. New York, NY: Touchstone (1992).\n\n16. Witten TA Jr, Sander LM. Diffusion-limited aggregation, a kinetic critical phenomenon. Phys Rev Lett. (1981). 47:1400–03. doi: 10.1103/PhysRevLett.47.1400\n\n17. Erdős P, Rényi A. On random graphs i. Publ Math. (1959) 6:290–7.\n\nPubMed Abstract\n\n18. Watts DJ, Strogatz SH. Collective dynamics of 'small-world' networks. Nature (1998) 393:440–2. doi: 10.1038/30918\n\n19. Newman MEJ. Mixing patterns in networks. Phys Rev E (2003) 67:026126. doi: 10.1103/PhysRevE.67.026126\n\n20. Boccaletti S, Latora V, Moreno Y, Chavez M, Hwang DU. Complex networks: structure and dynamics. Phys Rep. (2006) 424:175–308. doi: 10.1016/j.physrep.2005.10.009\n\n21. Newman MEJ. Assortative mixing in networks. Phys Rev Lett. (2002) 89:208701. doi: 10.1103/PhysRevLett.89.208701\n\n22. Efron B. The Jackknife, the Bootstrap and Other Resampling Plans. Philadelphia, PA: Society for Industrial and Applied Mathematics (1982).\n\n23. Vevatne JN, Rimstad E, Hope SM, Korsnes R, Hansen A. Fracture networks in sea ice. Front Phys. (2014) 2:21. doi: 10.3389/fphy.2014.00021\n\nKeywords: desert rose, crystal growth, diffusion limited aggregation, diffusion processes, network analysis, topology\n\nCitation: Hope SM, Kundu S, Roy C, Manna SS and Hansen A (2015) Network topology of the desert rose. Front. Phys. 3:72. doi: 10.3389/fphy.2015.00072\n\nReceived: 04 March 2015; Accepted: 21 August 2015;\nPublished: 08 September 2015.\n\nEdited by:\n\nFerenc Kun, University of Debrecen, Hungary\n\nReviewed by:\n\nNuno A. M. Araújo, Universidade de Lisboa, Portugal" ]

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[ "The Magic Cafe Forum Index » » Puzzle me this... » » Weighing Puzzle (0 Likes)", null, "Martin Joseph", null, "New user UK 70 Posts", null, "", null, "Posted: Aug 8, 2005 06:59 am", null, "0 I love these weighing puzzles and have long been my favourites. Does anyone know the origins of this one? - I have six snooker balls. One set consists of a red, blue and white weighing 1kg each. The second set consists of a red, blue and white weighing 2kg each. Determine which balls weigh which in only two weighings on a balancing scale. pxs", null, "Loyal user London 284 Posts", null, "", null, "Posted: Aug 9, 2005 10:19 am", null, "0 Number the balls R1, R2, B1, B2, W1, W2 First weighing: R1 and B1 vs R2 and W1 Second weighing: B1 vs W2 There are 9 possible outcomes but only 8 possible weights - so it is solveable. R1, B1 heavy and B1 heavy: R1=2, R2=1, B1=2, B2=1, W1=2, W2=1 R1, B1 heavy and B1 equal: R1=2, R2=1, B1=1, B2=2, W1=2, W2=1 R1, B1 heavy and B1 light: R1=2, R2=1, B1=1, B2=2, W1=1, W2=2 etc. The Magic Cafe Forum Index » » Puzzle me this... » » Weighing Puzzle (0 Likes)\n The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement <", null, "", null, "", null, "" ]

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[ "How to create a Die in code\n\nSomething that people sometimes want to do is create a Die in code, I thought I would supply code on one method of doing this.\n\nPublic Class Die\nPrivate Shared rndGen As New Random\nPrivate _numFaces As Integer\nPrivate _curValue As Integer\n#Region \"Properties\"\nPublic Property NumberOfFaces() As Integer\nGet\nReturn _numFaces\nEnd Get\nSet(ByVal value As Integer)\nIf value <= 0 Then\nThrow New ArgumentOutOfRangeException()\nEnd If\n_numFaces = value\nEnd Set\nEnd Property\nPublic Property CurrentValue() As Integer\nGet\nReturn _curValue\nEnd Get\nSet(ByVal value As Integer)\n_curValue = value\nEnd Set\nEnd Property\n#End Region\n#Region \"Constructors\"\nPublic Sub New(ByVal numFaces As Integer)\n'If the value is inappropriate then NumberOfFaces property will throw\n'an exception so we don't need to here\nNumberOfFaces = numFaces\nEnd Sub\n#End Region\n#Region \"Methods\"\nPublic Function roll() As Integer\n_curValue = rndGen.Next(1, _numFaces + 1)\nReturn _curValue\nEnd Function\n#End Region\nEnd Class\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# Centering labels between ticks¶\n\nTicklabels are aligned relative to their associated tick. The alignment 'center', 'left', or 'right' can be controlled using the horizontal alignment property:\n\nfor label in ax.xaxis.get_xticklabels():\nlabel.set_horizontalalignment('right')\n\n\nHowever there is no direct way to center the labels between ticks. To fake this behavior, one can place a label on the minor ticks in between the major ticks, and hide the major tick labels and minor ticks.\n\nHere is an example that labels the months, centered between the ticks.", null, "import numpy as np\nimport matplotlib.cbook as cbook\nimport matplotlib.dates as dates\nimport matplotlib.ticker as ticker\nimport matplotlib.pyplot as plt\n\n# load some financial data; apple's stock price\nwith cbook.get_sample_data('aapl.npz') as fh:\nr = r[-250:] # get the last 250 days\n# Matplotlib works better with datetime.datetime than np.datetime64, but the\n# latter is more portable.\ndate = r.date.astype('O')\n\nfig, ax = plt.subplots()\n\nax.xaxis.set_major_locator(dates.MonthLocator())\n# 16 is a slight approximation since months differ in number of days.\nax.xaxis.set_minor_locator(dates.MonthLocator(bymonthday=16))\n\nax.xaxis.set_major_formatter(ticker.NullFormatter())\nax.xaxis.set_minor_formatter(dates.DateFormatter('%b'))\n\nfor tick in ax.xaxis.get_minor_ticks():\ntick.tick1line.set_markersize(0)\ntick.tick2line.set_markersize(0)\ntick.label1.set_horizontalalignment('center')\n\nimid = len(r) // 2\nax.set_xlabel(str(date[imid].year))\nplt.show()\n\n\nKeywords: matplotlib code example, codex, python plot, pyplot Gallery generated by Sphinx-Gallery" ]

[ null, "https://matplotlib.org/3.1.1/_images/sphx_glr_centered_ticklabels_001.png", null ]

[ "# 基本的内置类型\n\nC++ 为程序员提供了种类丰富的内置数据类型和用户自定义的数据类型。下表列出了七种基本的 C++ 数据类型：\n\n• signed\n• unsigned\n• short\n• long\n\n## 实例\n\n``````#include<iostream>\n#include<string>\n#include <limits>\nusing namespace std;\n\nint main()\n{\ncout << \"type: \\t\\t\" << \"************size**************\"<< endl;\ncout << \"bool: \\t\\t\" << \"所占字节数：\" << sizeof(bool);\ncout << \"\\t最大值：\" << (numeric_limits<bool>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<bool>::min)() << endl;\ncout << \"char: \\t\\t\" << \"所占字节数：\" << sizeof(char);\ncout << \"\\t最大值：\" << (numeric_limits<char>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<char>::min)() << endl;\ncout << \"signed char: \\t\" << \"所占字节数：\" << sizeof(signed char);\ncout << \"\\t最大值：\" << (numeric_limits<signed char>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<signed char>::min)() << endl;\ncout << \"unsigned char: \\t\" << \"所占字节数：\" << sizeof(unsigned char);\ncout << \"\\t最大值：\" << (numeric_limits<unsigned char>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<unsigned char>::min)() << endl;\ncout << \"wchar_t: \\t\" << \"所占字节数：\" << sizeof(wchar_t);\ncout << \"\\t最大值：\" << (numeric_limits<wchar_t>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<wchar_t>::min)() << endl;\ncout << \"short: \\t\\t\" << \"所占字节数：\" << sizeof(short);\ncout << \"\\t最大值：\" << (numeric_limits<short>::max)();\ncout << \"\\t\\t最小值：\" << (numeric_limits<short>::min)() << endl;\ncout << \"int: \\t\\t\" << \"所占字节数：\" << sizeof(int);\ncout << \"\\t最大值：\" << (numeric_limits<int>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<int>::min)() << endl;\ncout << \"unsigned: \\t\" << \"所占字节数：\" << sizeof(unsigned);\ncout << \"\\t最大值：\" << (numeric_limits<unsigned>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<unsigned>::min)() << endl;\ncout << \"long: \\t\\t\" << \"所占字节数：\" << sizeof(long);\ncout << \"\\t最大值：\" << (numeric_limits<long>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<long>::min)() << endl;\ncout << \"unsigned long: \\t\" << \"所占字节数：\" << sizeof(unsigned long);\ncout << \"\\t最大值：\" << (numeric_limits<unsigned long>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<unsigned long>::min)() << endl;\ncout << \"double: \\t\" << \"所占字节数：\" << sizeof(double);\ncout << \"\\t最大值：\" << (numeric_limits<double>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<double>::min)() << endl;\ncout << \"long double: \\t\" << \"所占字节数：\" << sizeof(long double);\ncout << \"\\t最大值：\" << (numeric_limits<long double>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<long double>::min)() << endl;\ncout << \"float: \\t\\t\" << \"所占字节数：\" << sizeof(float);\ncout << \"\\t最大值：\" << (numeric_limits<float>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<float>::min)() << endl;\ncout << \"size_t: \\t\" << \"所占字节数：\" << sizeof(size_t);\ncout << \"\\t最大值：\" << (numeric_limits<size_t>::max)();\ncout << \"\\t最小值：\" << (numeric_limits<size_t>::min)() << endl;\ncout << \"string: \\t\" << \"所占字节数：\" << sizeof(string) << endl;\n// << \"\\t最大值：\" << (numeric_limits<string>::max)() << \"\\t最小值：\" << (numeric_limits<string>::min)() << endl;\ncout << \"type: \\t\\t\" << \"************size**************\"<< endl;\nreturn 0;\n}\n``````\n\n``````type: ************size**************\nbool: 所占字节数：1 最大值：1 最小值：0\nchar: 所占字节数：1 最大值： 最小值：?\nsigned char: 所占字节数：1 最大值： 最小值：?\nunsigned char: 所占字节数：1 最大值：? 最小值：\nwchar_t: 所占字节数：4 最大值：2147483647 最小值：-2147483648\nshort: 所占字节数：2 最大值：32767 最小值：-32768\nint: 所占字节数：4 最大值：2147483647 最小值：-2147483648\nunsigned: 所占字节数：4 最大值：4294967295 最小值：0\nlong: 所占字节数：8 最大值：9223372036854775807 最小值：-9223372036854775808\nunsigned long: 所占字节数：8 最大值：18446744073709551615 最小值：0\ndouble: 所占字节数：8 最大值：1.79769e+308 最小值：2.22507e-308\nlong double: 所占字节数：16 最大值：1.18973e+4932 最小值：3.3621e-4932\nfloat: 所占字节数：4 最大值：3.40282e+38 最小值：1.17549e-38\nsize_t: 所占字节数：8 最大值：18446744073709551615 最小值：0\nstring: 所占字节数：24\ntype: ************size**************\n``````\n\n# typedef 声明\n\n``````typedef type newname;\n``````\n\n``````typedef int feet;\n``````\n\n``````feet distance;\n``````\n\n# 枚举类型\n\n``````enum 枚举名{\n标识符[=整型常数],\n标识符[=整型常数],\n...\n\n} 枚举变量;\n``````\n\n``````enum color { red, green, blue } c;\nc = blue;\n``````\n\nenum color { red, green=5, blue };" ]

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[ "# 43 liters to Gallons", null, "Liters-to-Gallons\n\n43 liters to Gallons To convert 43 liters to the equivalent in gallons, multiply the amount in liters by 0.26417205124156 (the conversion factor). In this case, we need to multiply 43 liters by 0.26417205124156 to get the equivalent result in gallons:\n\n43 liters x 0.26417205124156 = 11.359398203387 gallons\n\nForty-three liters equals 11.359398203387 gallons.\n\n## How to Convert From Liters to Gallons\n\n43 liters to Gallons The conversion factor for liters to gallons is 0.26417205124156. Forty-three liters is eleven point three five nine gallons. To determine how many liters are in gallons, multiply by the conversion factor or use the volume converter above.\n\n## Definition Of Liter\n\nThe liter (also spelled “liter”; SI symbol L or l) is a non-SI metric part of the volume. It is equal near 1 cubic decimeter (dm3), 1,000 cubic centimeters (cm3) before 1/1,000 cubic meter. The build of one liter of liquid aquatic is almost exactly one kilogram. A liter is defined as a special term for one cubic decimeter or 10 centimeters × 10 centimeters × 10 centimeters, i.e., 1 L ≡ 1 dm3 ≡ 1000 cm3.\n\n## Definition Of Gallon\n\nThe gallon (abbreviated “gal”) is a unit of volume that refers to the United States liquid gallon. Three definitions are currently used: the imperial gallon (≈ 4.546 L) used in the United Kingdom and semi-officially in Canada, the US (liquid) gallon (≈ 3.79 L) in general use, and the least used US dry gallon (≈ 4.40 L).\n\n## With The Liters to Gallons Converter, You Will Get Answers To Questions Like The Following:\n\n ·        convert liters to gallons ·        [43 Liters to gallons] ·        [43 Liters to gallons] ·        liters to gal ·        conversion liters to gallons ·        43l in gal ·        [43 Liters to gallon] ·        43l to gallons ·        how many gallons are 43 liters ·        liters in a gallon conversion ·        43 l to gallons ·        how many liters in a gallon ·        convert liters to gallons ·        liters in a gal ·        43l to gallon ·        liters to gallon conversion ·        letters to gal ·        how many liters to a gallon of gas ·        liters gallons ·        [43 Liters to gallons] ·        43l is how many gallons ·        how many gallons is 43 l ·        43 l to gal ·        [43 Liters to gallons] ·        how much is a gal ·        43l to gal ·        how many gallons is 43l ·        43 liters ·        how many leaders in a gallon ·        43l in gallons ·        43 Liters is how many gallons ·        convert liters to us gallons ·        how much are [33 Liters to gallons] ·        [how much is 43l] ·        how much is 43 l ·        how much are 43 liters ·        43 l ·        43 l is how many gallons ·        [43 Liters equals how many gallons] ·        43 liter ·        liters to gall ·        convert 43 Liters to gallons ·        what are liters to gallons ·        how many liters in a gallon ·        liters in gal ·        what are 43 Liters to gallons ·        liters to us gallon ·        43l equals how many gallons ·        what is the conversion from liters to gallons ·        43 liters. How many gallons ·        how many gallons are 43 liters ·        43 Liters gallons\n\n## Similar Conversions From 43 Liters to Gallons\n\n44 liters to gallons\n\n45 liters to gallons\n\n46 liters to gallons\n\n47 liters to gallons\n\n48 liters to gallons\n\n49 liters to gallons\n\n50 liters to gallons\n\n51 liters to gallons\n\n52 liters to gallons\n\n53 liters to gallons\n\n## Previous Conversion From Liters to Gallons\n\n1650 liters to gallons\n\n790 liters to gallons\n\n566 liters to gallons\n\n725 liters to gallons\n\n413 liters to gallons\n\n33 liters to gallons\n\n921 liters to gallons\n\n68 liters to gallons\n\n724 liters to gallons\n\n15 liters to gallons\n\n## Recent Conversions\n\n343 hectares in square kilometers\n\n146 hours in nanoseconds\n\n461 acres to square meters\n\n407 square inches to square meters\n\n1783 square feet to square inches\n\n1140 kilometers to meters\n\n575 Long Tons to Short Tons\n\n755 cups in liters\n\n151 knots to miles/hour\n\n1425 kilometers/hour to miles/hour\n\n872 months to years\n\n232 minutes in nanoseconds\n\n1030 ounces to pounds\n\n134 grams to milligrams\n\n33 days in milliseconds\n\n## 43 Liters to Gallons Welcome to 43 liters in gallons.\n\nNow we tell you how many gallons are in 43 liters for both US and imperial gallons.\n\nFor unit conversions, it is recommended to use our converter:\n\nLiter: 43 units\n\nUS Liquid Gallons Gallons: 11.3593982514\n\n## How Many Gallons Are In 43 Liters?\n\nThe answer to how many gallons are in 43 liters? It depends on your measurement, as described in liters to gallons.\n\n[43 liters] = 11.3593982514 US [gallons]\n\n[43 liters] = 9.4586776769 imperial [gallons]\n\n[43 liters] = 9.7618920809 US dry [gallons]\n\nIt follows that the conversion factor depends on the type of gallon.\n\nIn the next section, you will learn how to change the capacity.\n\n## Convert 43 Liters into Gallons\n\nUnlike the metric liter, a gallon is a unit of volume in the imperial and US measurement system.\n\nYou can apply the formula [43 liters] to [gallons] a little further down to get the number of [gallons].\n\nOr use our volume converter at the top of this post.\n\nChange any amount in liters, not just 43, to all kinds of gallons.\n\nEnter the volume in liters, e.g., 43, then change or confirm the measurement to your considered unit using the drop-down menu.\n\nThe result is then calculated automatically.\n\n## Convert 43 Liters to US Gallons\n\nTo convert [43 liters] to US liquid [gallons], divide the volume in [liters], , by 3.785411784:\n\n43 / 3.785411784 = 11.3593982514 gallons\n\n## Convert 43 Liters into Dry Gallons\n\nTo convert 43 liters to US dry gallons, multiply the volume in liters, 43, by 0.22702:\n\n43 × 0.22702 = 9.4586776769 gallons\n\nYou’ll need to assume US liquid gallons for the conversion.\n\nFor the sake of completeness, the dry measure has been included here.\n\n## Convert 43 Liters into Imperial Gallons\n\nTo convert [43 liters] to UK [gallons], divide the number of liters by 4.54609:\n\n43 / 4.54609 = 9.7618920809 gallons\n\n[43 liters] to [gallons]\n\nBritish Gallon (Imperial):\n\n[43 liters] (L) ≈ 9.45871 [gallons] (gal)\n\n### How to Convert US Dry Gallon:\n\n[43 liters] (L) ≈ 9.76186 dry US [gallons]\n\n### How to convert US Liquid Gallon:\n\n[43 liters] (L) ≈ 11.35931 [gallons] us lqd\n\n43 liters (L)\n\n=\n\n11.3594 gallons (gal)\n\nLiter: The liter (also spelled “liter”; SI symbol L or l) is a non-SI metric unit of volume. It corresponds to 1 cubic decimeter (dm3), 1,000 cubic centimeters (cm3) or 1/1,000 cubic meter. The form of one liter of liquid water is almost exactly one kilogram. A liter is defined as a special designation for a cubic decimeter or 10 centimeters × 10 centimeters × 10 centimeters, i.e., 1 L ≡ 1 dm3 ≡ 1000 cm3.\n\nGallons: The gallon (abbreviation “gal”) is a part of the volume that refers to the United States liquid gallon. Three definitions are currently used: the imperial gallon (≈ 4.546 L) used in the semi-official United Kingdom and Canada, the US (liquid) gallon (≈ 3.79 L) in everyday use, and the less ordinary US dry gallon (≈ 4.40L).\n\n## How do I convert liters to gallons?\n\n1 liter (L) equals 0.26417 gallons (gal).\n\n1 liter = 0.26417 gallons\n\nThe volume V in gallons (gal) equals the volume V in liters (L) times 0.26417. This conversion formula:\n\nV(gal) = V(L) × 0.26417\n\n### How Many Gallons In A Liter?\n\nOne liter equals 0.26417 gallons:\n\n1 liter = 1 liter × 0.26417 = 0.26417 gal\n\n### How Many Liters In A Gallon?\n\nA gallon equals 3.78541 liters:\n\n1 gallon = 1 gallon × 3.78541 = 3.78541 l\n\n### How Do I Convert 5 Liters To Gallons?\n\nV(gal) = 5(L) × 0.26417 = 1.32085 gal\n\n• Kilograms to Stones and Pounds converter\n• Volume to mass converter for recipes\n• Volume to Mass Converter (Chemistry)\n\nTo calculate the value of a liter with the equivalent value in [liquid] gallons, multiply the amount in liters by 0.26417205235815 (the conversion factor). Here is the formula:\n\nValue in gallons [Liquid] = Value in liters × 0.26417205235815\n\nSuppose you want to convert 43 liters to [liquid] gallons. Using the conversion formula above, you get:\n\nValue in gallons [liquid] = 43 × 0.26417205235815 = 11.3594 gallons [liquid]\n\nHow many liters are in 43 [liquid] gallons?\n\nForty-three liters is how many gallons [of liquid]?\n\nHow much are 43 liters to gallons [of juice]?\n\nHow do I convert liters to gallons [liquid]?\n\nWhat is the conversion factor to convert from liters to [liquid] gallons?\n\nHow to convert liters to gallons [liquid]?\n\nWhat is the formulation to convert liters to [liquid] gallons? Among other.\n\nConversion table from liters to gallons [liquid] approx. 43 liters\n\n## Liters to Gallons [Liquid] Conversion Table\n\n34 liters = 8.98 gallons [liquid]\n\n35 liters = 9.25 gallons [liquid]\n\n36 liters = 9.51 gallons [liquid]\n\n37 liters = 9.77 gallons [liquid]\n\n38 liters = 10 gallons [liquid]\n\n39 liters = 10.3 gallons [liquid]\n\n40 liters = 10.6 gallons [liquid]\n\n41 liters = 10.8 gallons [liquid]\n\n42 liters = 11.1 gallons [liquid]\n\n43 liters = 11.4 gallons [liquid]\n\n## Liters to Gallons [Liquid] conversion table\n\n[43 liters = 11.4 gallons] [liquid]\n\n44 liters = 11.6 gallons [liquid]\n\n45 liters = 11.9 gallons [liquid]\n\n46 liters = 12.2 gallons [liquid]\n\n47 liters = 12.4 gallons [liquid]\n\n48 liters = 12.7 gallons [liquid]\n\n49 liters = 12.9 gallons [liquid]\n\n50 liters = 13.2 gallons [liquid]\n\n51 liters = 13.5 gallons [liquid]\n\n52 liters = 13.7 gallons [liquid]\n\nNote: Some values ​​may be rounded.\n\n## Sample Volume Conversions\n\n6.55 microliters to gallons [liquid]\n\n2.8 pints to liters\n\n8.3 cubic meters to cubic centimeters\n\n0.3 milliliters to liters\n\nseven ninths of a quarter to a liter\n\n35 milliliters to liters\n\nAlso Read: 40 Liters to Gallons – How To Convert From Liters To Gallons?" ]

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[ "", null, "", null, "", null, "Question", null, "Answers\n\n# Find the angle measure x in the following figure?", null, "", null, "", null, "Verified\n129.9k+ views\nHint: The Quadrilateral or quadrangle is a shape with four sides. to be considered as a quadrangle. the shape must\n1) Have four straight sides\n2) Be a flat shape (2dimentional)\n3) Be a closed figure.\n4) Edges and vertices $= {\\text{ }}4$\nThe word quadrilateral is derived from the Latin word quadric, a variant of four sides and the latus meaning side quadrilateral are either convex or concave.\nThe interior angle of a simple quadrilateral ABCD adds up to ${360^0}\\;$arc.\ni.e. $\\angle A + {\\text{ }}\\angle {\\text{ }}B{\\text{ }} + {\\text{ }}\\angle {\\text{ }}C{\\text{ }} + {\\text{ }}\\angle {\\text{ }}D = {\\text{ }}{360^0}$.\n\nAccording to the question, there are 4 Angles in a Quadrilateral.\nLet$\\angle {\\text{ }}A = 90,{\\text{ }}\\angle {\\text{ }}B = {\\text{ }}x,{\\text{ }}\\angle {\\text{ }}C = {\\text{ }}70,\\angle {\\text{ }}D = {\\text{ }}60$.\nSince in quadrilateral sum of interior angles is equal to 3600.\n$\\angle {\\text{ }}A{\\text{ }} + {\\text{ }}\\angle {\\text{ }}B{\\text{ }} + {\\text{ }}\\angle {\\text{ }}C{\\text{ }} + {\\text{ }}\\angle {\\text{ }}D = {360^0}$\n$90 + {\\text{ }}x{\\text{ }} + {\\text{ }}70{\\text{ }} + {\\text{ }}60{\\text{ }} = {360^0}$\n$\\;\\left( {90{\\text{ }} + {\\text{ }}60} \\right){\\text{ }} + {\\text{ }}x{\\text{ }} = {\\text{ }}360{\\text{ }} - 70$\n$150{\\text{ }} + {\\text{ }}x{\\text{ }} = 290$\n$x{\\text{ }} = {\\text{ }}290{\\text{ }}-{\\text{ }}150$\n$x{\\text{ }} = {\\text{ }}140$\n$\\angle {\\text{ }}B = 140$\n\nNote: A quadrilateral is a shape of $4$ sides for which any quadrilateral we draw a diagonal line to divide it into two triangles. Each triangle has an angle sum of $180$degree.\nTherefore, the total angle sum of a quadrilateral is$360$. A quadrilateral cannot have $3$ Obtuse angle where an obtuse angle is an angle that has a measure that is greater than${90^0}$." ]

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[ "# Pseudo-polynomial time\n\nIn computational complexity theory, a numeric algorithm runs in pseudo-polynomial time if its running time is a polynomial in the numeric value of the input (the largest integer present in the input) — but not necessarily in the length of the input (the number of bits required to represent it), which is the case for polynomial time algorithms.\n\nIn general, the numeric value of the input is exponential in the input length, which is why a pseudo-polynomial time algorithm does not necessarily run in polynomial time with respect to the input length.\n\nAn NP-complete problem with known pseudo-polynomial time algorithms is called weakly NP-complete. An NP-complete problem is called strongly NP-complete if it is proven that it cannot be solved by a pseudo-polynomial time algorithm unless P=NP. The strong/weak kinds of NP-hardness are defined analogously.\n\n## Examples\n\n### Primality testing\n\nConsider the problem of testing whether a number n is prime, by naively checking whether no number in $\\{2,3,\\dots ,{\\sqrt {n}}\\}$", null, "divides $n$", null, "evenly. This approach can take up to ${\\sqrt {n}}-1$", null, "divisions, which is sub-linear in the value of n but exponential in the length of n (which is about $\\log(n)$", null, "). For example, a number n slightly less than 10,000,000,000 would require up to approximately 100,000 divisions, even though the length of n is only 10 digits. Moreover one can easily write down an input (say, a 300-digit number) for which this algorithm is impractical. Since computational complexity measures difficulty with respect to the length of the (encoded) input, this naive algorithm is actually exponential. It is, however, pseudo-polynomial time.\n\nContrast this algorithm with a true polynomial numeric algorithm — say, the straightforward algorithm for addition: Adding two 9-digit numbers takes around 9 simple steps, and in general the algorithm is truly linear in the length of the input. Compared with the actual numbers being added (in the billions), the algorithm could be called \"pseudo-logarithmic time\", though such a term is not standard. Thus, adding 300-digit numbers is not impractical. Similarly, long division is quadratic: an m-digit number can be divided by a n-digit number in $O(mn)$", null, "steps (see Big O notation.)\n\nIn the case of primality, it turns out there is a different algorithm for testing whether n is prime (discovered in 2002), which runs in time $O((\\log {n})^{6})$", null, ".\n\n### Knapsack problem\n\nIn the knapsack problem, we are given $n$", null, "items with weight $w_{i}$", null, "and value $v_{i}$", null, ", along with a maximum weight capacity of a knapsack $W$", null, ". The goal is to solve the following optimization problem; informally, what's the best way to fit the items into the knapsack to maximize value?\n\nmaximize $\\sum _{i=1}^{n}v_{i}x_{i}$", null, "subject to $\\sum _{i=1}^{n}w_{i}x_{i}\\leq W$", null, "and $x_{i}\\in \\{0,1\\}$", null, ".\n\nSolving this problem is NP-hard, so a polynomial time algorithm is impossible unless P=NP. However, an $O(nW)$", null, "time algorithm is possible using dynamic programming; since the number $W$", null, "only needs $\\log W$", null, "bits to describe, this algorithm runs in pseudo-polynomial time.\n\n## Generalizing to non-numeric problems\n\nAlthough the notion of pseudo-polynomial time is used almost exclusively for numeric problems, the concept can be generalized: The function m is pseudo-polynomial if m(n) is no greater than a polynomial function of the problem size n and an additional property of the input, k(n). (Presumably, k is chosen to be something relevant to the problem.) This makes numeric polynomial problems a special case by taking k to be the numeric value of the input.\n\nThe distinction between the value of a number and its length is one of encoding: if numeric inputs are always encoded in unary, then pseudo-polynomial would coincide with polynomial." ]

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[ "# Delay Differential Equations in Python\n\nI wrote ddeint, a simple module/function for solving Delay Differential Equations (DDEs) in Python. It is not very fast, but very flexible, and coded in just a few lines on top of Scipy’s differential equations solver, odeint.\n\nSay you have a delay differential equation like this:\n\nwhere $F(y, t)$ can involve delayed values of $y$, of the form $y(t-d)$.\n\nTo solve this DDE system at points t=[t1, t2 ...] you would just write\n\n## A simple example\n\nLet us start with a DDE whose exact solution is known (it is the sine function), just to check that the algorithm works as expected:\n\nHere is how we solve it with ddeint:", null, "The resulting plot compares our solution (red) with the exact solution (blue). See how our result eventually detaches itself from the actual solution as a consequence of many successive approximations ? As DDEs tend to create chaotic behaviors, you can expect the error to explode very fast. As I am no DDE expert, I would recommend checking for convergence in all cases, i.e. increasing the time resolution and see how it affects the result. Keep in mind that the past values of Y(t) are computed by interpolating the values of Y found at the previous integration points, so the more points you ask for, the more precise your result.\n\n## An example with parameters\n\nYou can set the parameters of your model at integration time, like in Scipy’s ODE and odeint. As an example, imagine a chemical product with degradation rate $r$, and whose production rate is negatively linked to the quantity of this same product at the time $(t-d)$:\n\nWe have three parameters that we can choose freely. For $K = 0.1$, $d = 5$, $r = 1$, we obtain oscillations !", null, "## Example with several variables\n\nThe variable Y can be a vector, which means that you can solve DDE systems of several variables. Here is a version of the famous Lotka-Volterra two-variables system, where we introduce a delay $d$. For $d=0$ the system is a classical Lotka-Volterra system ; for $d\\neq 0$ the system undergoes an important amplification:", null, "## Example with a non-constant delay\n\nIn this last example the delay depends on the value of $y(t)$ :", null, "" ]

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[ "# How many moles of CaCl_2 are in 250 mL of a 3.0 M of CaCl_2 solution?\n\nMay 7, 2016\n\n$0.75$mol\n\n#### Explanation:\n\n$C = \\frac{n \\setminus m o l}{V}$, where $C$ is concentration, $n$ is number of moles and $V$ is volume in liters.\n\nIn this problem, $C = 3.0 M$ and $V = 0.25 L$\n\nSubstitute this into the equation.\n\n$3.0 M = \\frac{n \\setminus m o l}{0.25 L}$\n\n$3.0 \\left(\\frac{m o l}{\\cancel{L}}\\right) \\cdot 0.25 \\cancel{L} = 0.75 \\setminus m o l$" ]

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[ "# #608c55 Color Information\n\nIn a RGB color space, hex #608c55 is composed of 37.6% red, 54.9% green and 33.3% blue. Whereas in a CMYK color space, it is composed of 31.4% cyan, 0% magenta, 39.3% yellow and 45.1% black. It has a hue angle of 108 degrees, a saturation of 24.4% and a lightness of 44.1%. #608c55 color hex could be obtained by blending #c0ffaa with #001900. Closest websafe color is: #669966.\n\n• R 38\n• G 55\n• B 33\nRGB color chart\n• C 31\n• M 0\n• Y 39\n• K 45\nCMYK color chart\n\n#608c55 color description : Mostly desaturated dark lime green.\n\n# #608c55 Color Conversion\n\nThe hexadecimal color #608c55 has RGB values of R:96, G:140, B:85 and CMYK values of C:0.31, M:0, Y:0.39, K:0.45. Its decimal value is 6327381.\n\nHex triplet RGB Decimal 608c55 `#608c55` 96, 140, 85 `rgb(96,140,85)` 37.6, 54.9, 33.3 `rgb(37.6%,54.9%,33.3%)` 31, 0, 39, 45 108°, 24.4, 44.1 `hsl(108,24.4%,44.1%)` 108°, 39.3, 54.9 669966 `#669966`\nCIE-LAB 53.919, -26.215, 24.698 15.841, 21.898, 11.986 0.319, 0.44, 21.898 53.919, 36.017, 136.706 53.919, -21.878, 35.001 46.796, -21.467, 17.571 01100000, 10001100, 01010101\n\n# Color Schemes with #608c55\n\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #81558c\n``#81558c` `rgb(129,85,140)``\nComplementary Color\n• #7c8c55\n``#7c8c55` `rgb(124,140,85)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #558c66\n``#558c66` `rgb(85,140,102)``\nAnalogous Color\n• #8c557c\n``#8c557c` `rgb(140,85,124)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #66558c\n``#66558c` `rgb(102,85,140)``\nSplit Complementary Color\n• #8c5560\n``#8c5560` `rgb(140,85,96)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #55608c\n``#55608c` `rgb(85,96,140)``\n• #8c8155\n``#8c8155` `rgb(140,129,85)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #55608c\n``#55608c` `rgb(85,96,140)``\n• #81558c\n``#81558c` `rgb(129,85,140)``\n• #3f5c38\n``#3f5c38` `rgb(63,92,56)``\n• #4a6c42\n``#4a6c42` `rgb(74,108,66)``\n• #557c4b\n``#557c4b` `rgb(85,124,75)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #6b9c5f\n``#6b9c5f` `rgb(107,156,95)``\n• #79a76d\n``#79a76d` `rgb(121,167,109)``\n• #87b07d\n``#87b07d` `rgb(135,176,125)``\nMonochromatic Color\n\n# Alternatives to #608c55\n\nBelow, you can see some colors close to #608c55. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #6e8c55\n``#6e8c55` `rgb(110,140,85)``\n• #698c55\n``#698c55` `rgb(105,140,85)``\n• #658c55\n``#658c55` `rgb(101,140,85)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #5b8c55\n``#5b8c55` `rgb(91,140,85)``\n• #578c55\n``#578c55` `rgb(87,140,85)``\n• #558c58\n``#558c58` `rgb(85,140,88)``\nSimilar Colors\n\n# #608c55 Preview\n\nThis text has a font color of #608c55.\n\n``<span style=\"color:#608c55;\">Text here</span>``\n#608c55 background color\n\nThis paragraph has a background color of #608c55.\n\n``<p style=\"background-color:#608c55;\">Content here</p>``\n#608c55 border color\n\nThis element has a border color of #608c55.\n\n``<div style=\"border:1px solid #608c55;\">Content here</div>``\nCSS codes\n``.text {color:#608c55;}``\n``.background {background-color:#608c55;}``\n``.border {border:1px solid #608c55;}``\n\n# Shades and Tints of #608c55\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #040603 is the darkest color, while #f9fbf9 is the lightest one.\n\n• #040603\n``#040603` `rgb(4,6,3)``\n• #0c120b\n``#0c120b` `rgb(12,18,11)``\n• #151e12\n``#151e12` `rgb(21,30,18)``\n• #1d2a1a\n``#1d2a1a` `rgb(29,42,26)``\n• #253721\n``#253721` `rgb(37,55,33)``\n• #2e4329\n``#2e4329` `rgb(46,67,41)``\n• #364f30\n``#364f30` `rgb(54,79,48)``\n• #3f5b37\n``#3f5b37` `rgb(63,91,55)``\n• #47673f\n``#47673f` `rgb(71,103,63)``\n• #4f7446\n``#4f7446` `rgb(79,116,70)``\n• #58804e\n``#58804e` `rgb(88,128,78)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #68985c\n``#68985c` `rgb(104,152,92)``\n• #72a266\n``#72a266` `rgb(114,162,102)``\n• #7daa72\n``#7daa72` `rgb(125,170,114)``\n• #89b17e\n``#89b17e` `rgb(137,177,126)``\n• #94b88b\n``#94b88b` `rgb(148,184,139)``\n• #9fc097\n``#9fc097` `rgb(159,192,151)``\n• #aac7a3\n``#aac7a3` `rgb(170,199,163)``\n• #b6cfaf\n``#b6cfaf` `rgb(182,207,175)``\n• #c1d6bc\n``#c1d6bc` `rgb(193,214,188)``\n• #ccddc8\n``#ccddc8` `rgb(204,221,200)``\n• #d7e5d4\n``#d7e5d4` `rgb(215,229,212)``\n• #e3ece0\n``#e3ece0` `rgb(227,236,224)``\n• #eef4ec\n``#eef4ec` `rgb(238,244,236)``\n• #f9fbf9\n``#f9fbf9` `rgb(249,251,249)``\nTint Color Variation\n\n# Tones of #608c55\n\nA tone is produced by adding gray to any pure hue. In this case, #70726f is the less saturated color, while #31da07 is the most saturated one.\n\n• #70726f\n``#70726f` `rgb(112,114,111)``\n• #6a7b66\n``#6a7b66` `rgb(106,123,102)``\n• #65835e\n``#65835e` `rgb(101,131,94)``\n• #608c55\n``#608c55` `rgb(96,140,85)``\n• #5b954c\n``#5b954c` `rgb(91,149,76)``\n• #569d44\n``#569d44` `rgb(86,157,68)``\n• #50a63b\n``#50a63b` `rgb(80,166,59)``\n• #4baf32\n``#4baf32` `rgb(75,175,50)``\n• #46b72a\n``#46b72a` `rgb(70,183,42)``\n• #41c021\n``#41c021` `rgb(65,192,33)``\n• #3cc918\n``#3cc918` `rgb(60,201,24)``\n• #36d110\n``#36d110` `rgb(54,209,16)``\n• #31da07\n``#31da07` `rgb(49,218,7)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #608c55 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ " how to calculate tph of conveyor\n\n# how to calculate tph of conveyor\n\n• how to calculate conveyor tph BINQ Mining\n\nHow to Calculate Conveyor Belt Tension Rubber and Plastics, . MW = 33.3 TPH/Belt Speed (fpm) o rTotal material load in lbs/L. Find PVC, rubber and plastic conveyor belts for a huge array of industries and applications. »More detailed\n\nObtenir le prix\n• How to Calculate Conveyor Belt Speed Sciencing\n\nMar 13, 2018· Measure the circumference of the rollers in the conveyor,calculate revolutions per minute and then multiply these two figures together to determine conveyor belt speed. Manufacturers and grocery stores typically use conveyors to move products along a specific path.\n\nObtenir le prix\n• how to calculate tph of conveyor 't Oale Bakkershoes\n\nhow to calculate tph of conveyor; Bulk Handling Global Online BELT CONVEYOR design calculation Online calculation for belt conveyor design capacity and belt width, conveyor belt, belt conveyor calculation, belt conveyor system, belt conveyor design capacity, belt TPH. Design factor (0 ), %.\n\nObtenir le prix\n• Belt Conveyors for Bulk Materials Calculations by CEMA 5\n\n4. Select a suitable conveyor belt speed. 5. Convert the desired tonnage per hour (tph) to be conveyed to the equivalent in cubic feet per hour (ft3/hr). (ex. 1000 tph x 2000 / 60 = 33333 ft3/hr) 6. Convert the desired capacity in cubic feet per hour to the equivalent capacity at a belt speed of 100 fpm.\n\nObtenir le prix\n• Screw Conveyor Interactive Calculators Engineering Guide\n\nEng. Guide Index Download Guide PDF HORIZONTAL SCREW CONVEYOR CAPACITY & SPEED CALCULATION: Visit the online engineering guide for assistance with using this calculator. Click on symbol for more information. DESIGN CONDITIONS 1. Screw Conveyor Interactive Calculators. Eng. Guide Index Download Guide PDF.\n\nObtenir le prix\n• Conveyor Horsepower Calculator Superior Industries\n\nThis required horsepower calculator is provided for reference only. It provides a reasonable estimation of required horsepower given user requirements. Superior Industries is not responsible for discrepancies that may occur between this calculation and actual results.\n\nObtenir le prix\n• Belt Capacity Chart\n\nThe Following conveyor belt capacity charts show tons per hour (TPH) based on material weighing 100 lbs. per cubic foot, 20° material surcharge angle with three equal length rolls on troughing idlers. CAPACITY (TPH) = .03 x Belt Speed (FPM) x material weight (lb. per cu. ft.) x load cross section (sq. ft.) TPH with 20° Troughing Idlers\n\nObtenir le prix\n• Maximum Belt Capacity Calculator Superior Industries\n\nThis maximum belt capacity calculator is provided for reference only. It provides a reasonable estimation of maximum belt capacity given user requirements. Superior Industries is not responsible for discrepancies that may occur between this calculation and actual results.\n\nObtenir le prix\n• tph calculation for a conveyor belt BINQ Mining\n\nSimple Torque Calculation for Belt Conveyor bulk-online Forums formula to calculate torque (to turn)for belt conveyor application based on belt length, Simple torque calculation for belt conveyor x tph x 1/367 Not sure how »More detailed\n\nObtenir le prix\n• how to calculate tph of conveyor\n\nHow to calculate conveyor tph What is the formula for determining the tons per hour on a conveyor belt by taking a 1 foot How to Calculate Conveyor Belt Facts Concerning Dust and Air Belt Cleaners Facts Concerning Dust and Air By A transfer point is defined as the point where one belt conveyor dumps tons per hour TPH .\n\nObtenir le prix\n• Belt Capacity Chart\n\nThe Following conveyor belt capacity charts show tons per hour (TPH) based on material weighing 100 lbs. per cubic foot, 20° material surcharge angle with three equal length rolls on troughing idlers. CAPACITY (TPH) = .03 x Belt Speed (FPM) x material weight (lb. per cu. ft.) x load cross section (sq. ft.) TPH with 20° Troughing Idlers\n\nObtenir le prix\n• How To Calculate 200 Tph Of Belt Conveyor\n\nHow to calculate conveyor tph What is the formula for determining the tons per hour on a conveyor belt by taking a 1 foot How to Calculate Conveyor Belt Facts Concerning Dust and Air Belt Cleaners Facts Concerning Dust and Air By A transfer point is defined as the point where one belt conveyor dumps tons per hour TPH .\n\nObtenir le prix\n• Screw Conveyor Interactive Calculators Engineering Guide\n\nEng. Guide Index Download Guide PDF HORIZONTAL SCREW CONVEYOR CAPACITY & SPEED CALCULATION: Visit the online engineering guide for assistance with using this calculator. Click on symbol for more information. DESIGN CONDITIONS 1. Screw Conveyor Interactive Calculators. Eng. Guide Index Download Guide PDF.\n\nObtenir le prix\n• how to calculate 200 tph of belt conveyor\n\nHow to Design Take-up Travel for a Fabric Conveyor Belt Pooley Inc. This article covers the important aspects of fabric conveyor belt elongation and how to design the How to calculate the take-up travel requirement based on the permanent and elastic elongation This is because the maximum tension of 2000 TPH will be higher than the\n\nObtenir le prix\n• Understanding Conveyor Belt Calculations Sparks Belting\n\nUnderstanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system. We use cookies to personalize content and analyze traffic. We also share information about your use of our site with our social media, advertising, and analytics partners who may combine it with other\n\nObtenir le prix\n• Calculation methods conveyor belts\n\nConveyor and processing belts Calculation methods conveyor belts Content 1 Terminology 2 Unit goods conveying systems 3 Take-up range for load-dependent take-up systems 8 Bulk goods conveying systems 9 Calculation example Unit goods conveying systems 12 Conveyor and power transmission belts made of modern synthetics\n\nObtenir le prix\n• Conveyor Capacity Engineering ToolBox\n\nConveyor capacity is determined by the belt speed, width and the angle of the belt and can be expressed as. Q = ρ A v (1) where . Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt (m 2, ft 2)\n\nObtenir le prix\n• Screw Conveyor Capacity Engineering Guide\n\nEng. Guide Index Download Guide PDF Calculation Of Conveyor Speed Capacity Factors for Special Pitches Capacity Factors for Modified Flight Capacity Table Capacity is defined as the weight or volume per hour of a bulk material that can be safely and feasibly conveyed using a screw conveyor. Screw conveyor diameter []\n\nObtenir le prix\n• calculate of tph of conveyor -CPY manufacturers\n\nConveyor Calc freightmetricsau- calculate of tph of conveyor,belt conveyor calculator Login Login Search Home Contact Us; About Us; Links; Calculators Road Fuel Levy Calculator; Truck Operating Costtph calculation of vibrating screen conveyor,sand washing machine,stone vibrating feeder 40 tph capacity made in indian,, to either of two chip screeners at a 150 TPH\n\nObtenir le prix\n• how to calculate tph of belt conveyor\n\nto work out tph to tonnes on conveyor belt. calculation to work out tph to tonnes on conveyor belt. bucket elevator catalog & engineering manual Orthman Conveying calculation to work out tph to tonnes on conveyor belt,on a chain or belt and operate at speeds up to 305 FPM for handling the heaviest of industrial average 120 FPM for handling\n\nObtenir le prix\n• how to calculate tph of conveyor\n\nhow to calculate 200 tph of belt conveyor. How To Calculate Tph Of Belt Conveyor Mining, A belt conveyor is a moving surface required at the headshaft of a conveyor Belt Capacity 50 TPH, 100 Conveyor motor sizing forms calculate the How to calculate belt conveyor speed with motor RPM and ger rastioc we know Ibs per hour tph Belt width Chat Online Designing and .\n\nObtenir le prix\n• how to calculate tph of conveyor\n\nhow to calculate 200 tph of belt conveyor. How To Calculate Tph Of Belt Conveyor Mining, A belt conveyor is a moving surface required at the headshaft of a conveyor Belt Capacity 50 TPH, 100 Conveyor motor sizing forms calculate the How to calculate belt conveyor speed with motor RPM and ger rastioc we know Ibs per hour tph Belt width Chat Online Designing and .\n\nObtenir le prix\n• Conveyor Belt Scale Manual Belt-Way Scales\n\ndesigned Belt-Way Conveyor Belt Scale. Our state of the art integrator sets the industry standard for high technology and user friendly operation. All design changes were created in response to feedback from our distribution partners and customers. We are pleased to share this new\n\nObtenir le prix\n• calculate of tph of conveyor -CPY manufacturers\n\nHow To Calculate Capacity In Tph mayukhportfoliocoin. how to calculate tph of belt conveyor Hi to all For designing a belt conveyor with capacity=1000 tph,risk involved in granite quarrying,formula to calculate ,\n\nObtenir le prix\n• How To Calculate The Tph Of Belt Conveyor FTMLIE Heavy\n\nhow to determine tph on a conveyor. To calculate the conveyor belts power requirements If for instance the conveyor belt must move 330 tons each Syllabus of Coal I Live a Chat Industrial Machines And Equipments Belt Conveyors Industrial Machines And Equipm to bucket elevator capacity calculation for excel formula to . Learn More\n\nObtenir le prix\n• how to calculate 200 tph of belt conveyor\n\nhow to calculate 200 tph of belt conveyor. how to calculate 200 tph of belt conveyor You can get the price list and a GME representative will contact you within one . Learn More . how to calculate 200 tph of belt conveyor how to calculate 200 tph of belt conveyor. Conveyor Belt Calculations\n\nObtenir le prix\n• how to calculate tph of belt conveyor\n\nto work out tph to tonnes on conveyor belt. calculation to work out tph to tonnes on conveyor belt. bucket elevator catalog & engineering manual Orthman Conveying calculation to work out tph to tonnes on conveyor belt,on a chain or belt and operate at speeds up to 305 FPM for handling the heaviest of industrial average 120 FPM for handling\n\nObtenir le prix\n• calculate of tph of conveyor gezinsbond-vlamertinge\n\nhow to calculate tph of conveyor Jaw crusher supplier,how to calculate tph of conveyor If you want to get more detailed product information and prices, ZME recommend that you get in touch with us through online chat our customer service 24 hours a day online to serve you with the .\n\nObtenir le prix\n• Conveyor Belt Calculating Chart\n\nThe accompanying chart has been drawn for the convenience of engineers as a means of quickly determining the correct number of plies of conveyor belts operating under specific conditions. The calculations are based on the average safe strength (factor of safety, 15) of the various standard rubber conveyor belts. The calculations assume maximum loading conditions; that is, the belt is\n\nObtenir le prix\n• Understanding Conveyor Belt Calculations Sparks Belting\n\nUnderstanding a basic conveyor belt calculation will ensure your conveyor design is accurate and is not putting too many demands on your system. We use cookies to personalize content and analyze traffic. We also share information about your use of our site with our social media, advertising, and analytics partners who may combine it with other\n\nObtenir le prix\n• Calculation methods conveyor belts\n\nConveyor and processing belts Calculation methods conveyor belts Content 1 Terminology 2 Unit goods conveying systems 3 Take-up range for load-dependent take-up systems 8 Bulk goods conveying systems 9 Calculation example Unit goods conveying systems 12 Conveyor and power transmission belts made of modern synthetics\n\nObtenir le prix\n• conveyor horse power and tph calculator\n\ncalculate of tph of conveyor,Superior's conveyor calculator provides the minimum horsepower required at the headshaft of a conveyor Click & Chat Now. tph calculation for feeder sethhukamchandcoin. tph calculation for a conveyor horse power and tph calculator how to calculate carrying Belt Feeder and Pulley 6000 tph on each 72\" Belt Conveyor\n\nObtenir le prix\n• Conveyor Capacity Engineering ToolBox\n\nConveyor capacity is determined by the belt speed, width and the angle of the belt and can be expressed as. Q = ρ A v (1) where . Q = conveyor capacity (kg/s, lb/s) ρ = density of transported material (kg/m 3, lb/ft 3) A = cross-sectional area of the bulk solid on the belt (m 2, ft 2)\n\nObtenir le prix\n• how to calculate tph of belt conveyor\n\nChain Conveyors Practical Calculations determine the apron width B in meters Q = Conveyor total load t/h tph calculating tph on belt conveyor grinding mill equipment how to determine tph on a conveyor calculating tph on belt conveyor A belt conveyor is a moving surface required at the headshaft of a conveyor\n\nObtenir le prix\n• how to calculate tph of belt conveyor\n\nHow To Calculate Tph Of Belt Conveyor Mining Machinery. A belt conveyor is a moving surface required at the headshaft of a conveyor Belt Capacity 50 TPH,100 Conveyor motor sizing forms calculate the How to calculate belt conveyor speed with motor RPM and ger rastioc we know Ibs per hour tph Belt width Chat Online Designing and Assessing the\n\nObtenir le prix" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "transCalculator\n\nFrustrated with people everyday computing.\n\nNormally, when you are using a computer, frustrated that for? \"Calculated to see the progress,\" \"want to fix mistakes in calculations,\" \"watch list stored in memory,\" I never thought I like? \"transCalculator\" all problems can be solved in such. iPhone / iPod touch Calculator was created to be used in \"transCalculator\", and clearly more convenient to let the calculation of daily life.\n\nChange the order of calculation\n\n+ And × calculator could calculate that if it turned out wrong for? + And × operators such as the order is calculated, the calculator will normally be calculated in the order they are entered. Therefore, do not change the order of calculation, the calculator will not be required to correct results. But \"transCalculator\" in order to calculate the normal calculator does just as well because it can be calculated according to the calculation order of the operators no longer can get incorrect results.\n\nProduct Specifications\n\n1. -Requirements\n\niPhone or iPod touch after iOS4.\n\n5MB or more free space (depending on usage, however, may need more.)", null, "", null, "", null, "To calculate the daily life more convenient.\n\ncalculated using the brackets\n\nAnd calculated taking into account the priority of operators may be tempted to use brackets calculations.\n\ntransCalculator can be calculated using the brackets of course.\n\ntransCalculator passage appears only once in the calculation in brackets.\n\nAlso in brackets Calculate course surrounded by \"Group expressions\" treated as,\n\nor repetition many times you can save.", null, "Later review what was calculated\n\nIf you want to see what was later calculated for? In calculating what numbers and operators only, but will not know what the hell were you when you look at what calculations later. Even if such \"transCalculator\" the active. You can also leave a note with the name and line by line, to save all the contents of every calculation, you can review later.\n\nCorrect number in the calculation\n\nOn the way to calculate it again, realize that you re-calculate the wrong result for? \"transCalculator\" appears in every single line in the input line can be edited in before the calculation, it should've in the middle, can be corrected.", null, "Programmers also \"transCalculator\"\n\n\"transCalculator\" is for the programmers, binary, octal, and hexadecimal calculations in 16, AND, and XOR operations are now possible and logical. Programmer also has a variety of ideas have been. Base changes and line by line, you can change the decimal value of the input values and results among the rows. Almost the same basic functionality for programmer can, of course, ready to use. Calculated according to the order ※operator can not be calculated.", null, "", null, "", null, "", null, "", null, "", null, "" ]

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[ "# Sequence Networks Unsymmetrical Faults\n\n## Sequence Networks Unsymmetrical Faults:\n\nThe analysis of an unsymmetrical fault by symmetrical components method can be conveniently done by drawing sequence networks. A Sequence Networks Unsymmetrical Faults of a particular sequence current in a given power system is the path for the flow of that sequence current in the system. It is composed of impedances offered to that sequence current in the system. Since there are three sequence currents (viz. positive sequence current, negative sequence current and zero sequence current), there will be three sequence networks for a given power system, namely ;\n\n1. Positive sequence network\n2. Negative sequence network\n3. Zero sequence network\n\n1. Positive sequence network: The positive sequence network for a given power system shows all the paths for the flow of positive sequence currents in the system. It is represented by one-line diagram and is composed of impedances offered to the positive sequence currents. While draw­ing the positive sequence network of a given power system, the following points may be kept in view:\n\n• Each generator in the system is represented by the generated voltage in series with appropriate reactance and resistance.\n• Current limiting impedances between the generator’s neutral and ground pass no positive sequence current and hence are not included in the positive sequence network.\n• All resistances and magnetising currents for each transformer are neglected as a matter of\n• For transmission lines, the shunt capacitances and resistances are generally neglected.\n• Motor loads are included in the network as generated e.m.f. in series with appropriate\n\n2. Negative sequence network: The negative sequence network for a given power system shows all the paths for the flow of negative sequence currents in the system. It is also represented by one line diagram and is composed of impedances offered to the negative sequence currents. The negative sequence network can be readily obtained from positive sequence network with the following modifications :\n\n• Omit the e.m.fs. of 3-phase generators and motors in the positive sequence network. It is because these devices have only positive sequence-generated voltages.\n• Change, if necessary, the impedances that represent rotating machinery in the positive sequence network. It is because negative sequence impedance of rotating machinety is generally different from that of positive sequence impedance.\n• Current limiting impedances between generator’s neutral and ground pass no negative sequence current and hence are not included in the negative Sequence Networks Unsymmetrical Faults.\n• For static devices such as transmission lines and transformers, the negative sequence impedances have the same value as the corresponding positive sequence impedances.\n\n3. Zero sequence network: The zero sequence network for a given power system shows all the paths for the flow of zero sequence currents. The zero sequence network of a system depends upon the nature of connections of the 3-phase windings of the components in the system. The follow­ing points may be noted about zero Sequence Networks Unsymmetrical Faults:\n\n• The zero sequence currents will flow only if there is a return path e. path from neutral to ground or to another neutral point in the circuit.\n• In the case of a system with no return path for zero sequence currents, these currents cannot exist.\n\n### Reference Bus for Sequence Networks:\n\nWhile drawing the sequence networks, it is necessary to specify the reference potential 11).1:t. which all sequence voltage drops are to be taken. For this purpose, the reader may keep in mind the following points :\n\n• For positive or negative Sequence Networks Unsymmetrical Faults, the neutral of the generator is taken as the reference bus. This is logical because positive or negative sequence components represent balanced sets and hence all the neutral points must be at the same potential for either positive or negative sequence currents.\n• For zero sequence network, the reference bus is the ground at the generator." ]

[ null ]

[ "Open access peer-reviewed chapter\n\n# The Research on Stability of Supply Chain under Variable Delay Based on System Dynamics\n\nBy Suling Jia, Lin Wang and Chang Luo\n\nSubmitted: November 29th 2010Reviewed: April 16th 2011Published: August 29th 2011\n\nDOI: 10.5772/23632\n\nDownloaded: 2089\n\n## 1.Introduction\n\nWith the swift development of modern science and network technology and fortified trend of economics globalization, the cooperation between supply chain partners is happening with increasing frequency and the cooperation difficulty increased correspondingly. Supply chain is a complex system which involves multiple entities encompassing activities of moving goods and adding value from the raw material stage to the final delivery stage.Feedback, interaction, and time delay are inherent to many processes in a supply chain, making it a dynamics system. Because of the dynamics and complex behaviors in the supply chain, the study on the stability of supply chain has become an independent research field only in last decade. At the same time, the great development of control theory and system dynamics provides an effective way to understand and solve the complexity of evolution in the supply chain system.\n\nThe research on stability of supply chain was put forward during the studying of bullwhip effect. According to the paper of Holweg & Disney(2005),the development of the research on stability of supply chain and bullwhip effect can be divided into six stages:\n\n1. Production and Inventory Control (before 1958). Nobel laureate Herbert Simon(1952) first suggested a PIC model based on Laplace transform methods and differential equations.In the model, Simon used first order lag to describe the delay of stock replenishment. Vassian(1955) built continuous time PIC model using Z transform. Magee(1958) solved the problems of inventory management and control in order-up-to inventory policy. At this stage, early PIC models were built based on control theory and the dynamics characteristics of PIC systems were discussed.\n\n2. Smoothing production (1958-1969). In the early 1960s, Forrester(1958,1961) built the original dynamics models of the supply chain using DYNAMO(Dynamic Modeling) language.He revealed the counterintuitive phenomenon of fluctuations in supply chain. The methods Jay Forrester proposed have gradually developed into system dynamics methodology which is used to research on dynamics characteristics of supply chain systems. For the bullwhip effect in discrete-time supply chain systems, analytical expression of the change in inventory under order-up-to policy was presented based on certain demand forecasting method(Deziel＆Eilon,1967).At this stage, the problems such as seasonal fluctuations in inventory and demand amplification had gained attention, but the terms “bullwhip effect” and ”stability of supply chain” were not formally proposed, the emphasis of the academic research during this time was the traditional production management.\n\n3. The development of control theory(1970-1989). Towill (1982) built a relatively complete PIC system model without considering the feedback control loop of WIP(work in process).Bertrand(1980) studied the bullwhip effect and inventory change in an actual production system. According to the above researches, customer demand was assumed constant and productivity was random. Bertrand(1986) made further study on the bullwhip effect and inventory change in PIC system with feedback control.\n\n4. Stage of “Beer Game” development (1989-1997). Sterman(1989) suggested a system dynamics general stock management model after doing experimental study on “beer game” of MIT and analyzing 2000 simulation results based on system dynamics. Using continuous time equation, Naim ＆ Towill(1995) discussed the feedback control and stock replenishment with first order lag in a supply chain model. The “beer game” and the corresponding problems in supply chain have been studied until now, recent research focus on information sharing and bullwhip effect in supply chain(Croson ＆ Donohue,2005). At this stage, system dynamics methodology has been deeply applied to the field of supply chain (Towill, 1996).Both system dynamics methodology and control theory emphasize the importance of “feedback control” to stability of supply chain, Sterman also considered the effects of decision behavior on fluctuation of inventory and order.\n\n5. The further development of “bullwhip effect” (1997-2000). Lee et al.(1997a, 1997b) pointed out the clear concept of “bullwhip effect” and identified four major causes of the bullwhip effect(demand forecast updating, order batching, price fluctuation, rationing and shortage gaming).From then on, academic circles set off an enthusiastic discussion centering on bullwhip effect. However, research papers during this period didn’t make thorough study of feedback control(Holweg ＆ Disney,2005).Later studies showed that there were more than four significant bullwhip generators(Geary et al.,2006), but the views of Lee et al. have been widely received and quoted up to the present (Miragliotta,2006 ).\n\n6. The stage of avoiding bullwhip effect (after 2000). The dynamic characteristic of supply chain represented by bullwhip effect had received considerable attention and many researchers shifted the focus of work to prevention of bullwhip effect at this time. Represented by Towill, Dejonckheere and Disney,a number of scholars brought control theory deeply to the research of bullwhip effect and related problems. They proposed APIOBPCS(Automatic Pipeline, Inventory and Order Based Production Control System) on the basis of methods and achievements from system dynamics(Disney ＆ Towill,2002,2003a;Dejonckheereet al., 2003; Disney et al.,2004;Disney et al., 2006). The study on stability of supply chain has become an independent research field at this stage and the following studies are mostly done using control theory based on PTD (pure time delay) assumption. Up to now, the research of preventing bullwhip effect in multi-stage supply chain system has breakthrough progress (Daganzo,2004; Ouyang ＆ Daganzo,2006).\n\nThis chapter focuses on the stability of supply chain under variable delay based on System Dynamics methodology. First, we builds a single parameter control model of supply chain, By simulations and related analyses, a quantitative stability criterion of supply chain system based on system dynamics is proposed, this criterion evaluates stability by the undulate phenomenon and convergent speed. Then the stability characteristics in single parameter control model with two different delay structures (first order exponential lag and pure time delay) are discussed and the corresponding stable boundaries of the supply chain model are confirmed. Second, based on “system dynamics general stock management model” and control theory, the general inventory control model is built. Combined with the quantitative stability criterion of supply chain system proposed earlier, we analyze the complexities of the model under different delay modes.Finally we present the stable boundary and feasible region of decision and give our conclusions. This research indicates that delay structure is a key influencing factor of system stability.\n\n## 2. Stability criterion of supply chain based on system dynamics\n\nThe differences of quantitative description of bullwhip effect result in different definitions of stability of supply chain. Lee et al. (1997a, 1997b) described qualitative evidence of demand amplification, or as they called it, the bullwhip effect, in a number of the retailer-distributor-manufacturer chains and claimed that the variance of orders may be larger than that of sales. In order to gain more insight on what is really happening, Taylor (1999) suggested analysis on both demand data (passed from company to company) and activity data (e.g. production orders registered within the company). The variance ratio is by far the most widely used measure to detect the bullwhip effect. It is defined as the ratio between the demand variance at the downstream and at the upstream stages (Miragliotta, 2006). As variance ratio is a static index, it is difficult to describe the complex and dynamic nonlinear system problems. In this section, we will not apply the variance ratio to measure the stability of supply chain system.\n\nThe theories and methods in nonlinear dynamics are applied to the studies on stability and bullwhip effect of supply chain and several criterions for describing and judging the stability of supply chain system are formed, such as peak order amplification, peak order rate overshot, noise bandwidth, times of demand amplification(Disney ＆ Towill,2003b;Jing Wang et al.,2006; Riddalls ＆ Bennett,2001;Zhang X,2004;). The above criterions are used on the premise of testing the dynamic behavior of supply chain system. The test function is usually step function, pulse function or pure sine function,not the actual demand function. The purpose is to distinguish the effect of internal and external factors on stability of supply chain system. Some studies based on cyberneticsdirectly adopt the distinguish methods in nonlinear dynamics, several methods are as following: Lyapunov exponent method; critical chaos; state space techniques (see for example Huixin Liu et al., 2004; Lalwani et al., 2006; Riddalls ＆ Bennett,2001; Xinan Ma et al., 2005). However, these methods are applied under a lot of constraint conditions and some parameters do not have specific economic meaning, sometimes it is difficult to obtain ideal result, but the basic idea of analyzing structure characteristics of the system to measure stability incybernetics is worthy of learning.\n\nAlthough the researchers have already pay attention to the problems of stability and complexity in supply chain system, they focus on revealing dynamic characteristic of the system and pay little attention to the problems such as stability criterion, stable boundary, and feasible region of decision of supply chain system. Qifan Wang (1995) measured the stability of system by analyzing open-loop gain, the method required all variables in feedback loop to be continuous and derivable and it is not applicable to high order nonlinear system. Sterman (1989, 2000) adopted the concept of “peak amplification” to describe the dynamic characteristics of system during the research on beer game and general stock management system, but he didn’t give a specific stability criterion. Combining system dynamics and chaos theory, Larsen et al. (1999) described the stability of supply chain system from a chaos perspective, but the calculation of the study is a time-consuming and difficult task. Since now, there is no quantitative stability criterion of supply chain systems based on system dynamics, which seriously restrict the application of system dynamics into further research on stability of supply chain.\n\n### 2.1. Single parameter stock control model of supply chain\n\n#### 2.1.1. Basic assumptions\n\nThe stock control model of supply chain in this section can be understood as one node along the chain, the basic assumptions are as following:\n\n1. The downstream demand mode is uncertain, do not make prediction on it.\n\n2. There is no restriction on inventory capacity.\n\n3. There exists delay time (DELAY) in the sending of products to downstream and the average delay time is constant. The orders is described as WIP(work in process) before the products arrive.\n\n4. There is no reverse logistics, products can’t be returned to upstream.\n\n5. The supply chain members adjust orders according to demand from downstream and actual storage and maintain the inventory at a desired level.\n\n#### 2.1.2. Structure of the model\n\nFigure 1 presents the single parameter stock control model of supply chain discussing in this section.", null, "Figure 1.The single parameter stock control model of supply chain\n\nTo facilitate the model description, the following notations are introduced:\n\nOR: the order quantity at time t,\n\nWIP: the orders placed but not yet received at time t,\n\nALPHAi: the rate at which the discrepancy between actual and desired inventory levels is eliminated, 0≤αi≤1,\n\nI*: the desired inventory level,\n\nI: the actual inventory level at time t,\n\nD: the actual demand at time t,\n\nAI: the adjustment for the inventory level at time t.\n\nFigure 1 is built on the basis of the generic stock-management model proposed by Sterman(1989). The adjustments include two aspects: first, there exist two different delay structures (first order exponential lag and pure time delay) in the model; second, without consideration of WIP adjustment, the orders depend on demand and inventory adjustment. The above adjustments simplify the feedback control loop of inventory, making the system affected by just one negative feedback loop. Theoretical basis of the adjustment is the analytic method of “open-loop” in system dynamics methodology (Qifan Wang, 1995; Sterman, 2000).\n\nThe above model adopts the experimental methods to describe the individualbehavior in a common and importantmanagerial context. It contains multiple actors,feedback, nonlinearities, and time delay. The parameters of the order policy are estimated and the order policyis shown to explain the decision maker’s behavior well.\n\n#### 2.1.3. Variablesettings\n\nAs shown in figure 1, the indicated orders IO depend on demand D and adjustment for inventory AI ，so it can be defined as the sum of D and AI. There exists the transmission delay of orders between two successive levels and the delay mode can be represented by a standard function DELAY of system dynamics, including first order exponential lag and pure time delay. The desired inventory I* is constant. As products can’t be returned to upstream, the order rate OR must be positive. That is:\n\nIO=D+AIE1\nOR=Max(0,IO)E2\nAR=DELAY(OR,DT)E3\n\nConsidering the stock and flow structure, the stock of WIP is the accumulation of the order rate OR less the acquisition rate AR. Similarly, the stock of inventory I is the accumulation of the acquisition rate less the demand D.\n\nWIP=t0t[OR(t)AR(t)]dt+WIPt0E4\nI=t0t[AR(t)D(t)]dt+It0E5\n\nwhere WIPt0and It0are the initial values at time t0, demand D is an external variable that can’t be controlled.\n\nThe adjustment for the inventory resultsin the negative feedback mechanism which regulates the inventory. The adjustment is linear in the discrepancy between the desired inventory and the actual inventory. That is:\n\nAI=αi(I*I)E6\n\nwhere αi is the rate at which the discrepancy between actual and desired inventory levels is eliminated, 0≤αi≤1. The value of αi represents the sensitivity of decision-maker to the gap between the desired inventory I* and actual inventory I. So the ordering policy can be described as follows:\n\nIO=D+αi(I*I)E7\n\nThe ordering policy is based on the anchoring and adjustment heuristic (Tversky ＆ Kahneman, 1974). Anchoring and adjustment heuristic has been widely applied to a wide variety of decision-making tasks in the field of control theory and system dynamics methodology (see for example Sterman, 1989; Riddalls ＆ Bennett, 2002; Larsen et al., 1999; Huixin Liu et al., 2004). From (7) we can see that without demand forecasting, the ordering policy can be described by the single parameter αi..\n\n### 2.2. Dynamic characteristics analysis of system\n\n#### 2.2.1. Simulation design\n\nSuppose the system is in a stable state at the initial time without fluctuation of inventory and order rate. When the system is disturbed by a small perturbation on demand, we can study the system behavior from the response curve of inventory or order rate. With reference to Sterman (1989) and Riddalls ＆ Bennett (2002), the initial values (unit) of variables are presented in Table 1.The model is built using well-known system dynamics simulation software, Vensim PLE. The run length for simulation is 60 weeks.\n\n WIPt0 I* It0 DT αi Dt0 300 200 200 3 1 100\n\n### Table 1.\n\nInitial values of variables\n\nThe demand pattern is a step function, that is, the demand stays at an original level up to a certain instant and thereafter is increased to a shifted level. In this study, there is a pulse in the demand in week number 5, increasing its value to 120units/week.\n\nIn the simulation, the decision parameter αi is changed with a small decrement from 1.00 to 0.00 so as to simulate various ordering decisions. We concentrate on illustrating how minor changes in the decision parameter can affect the dynamics and stability of the system. System dynamics and relevant studies show that the size of step input of demand and the desired inventory will not affect the structural stability of the system(Croson ＆ Donohue,2005;Sterman,1989,2000).\n\n#### 2.2.2. Dynamics characteristics of system under first-order lag\n\nIf the delay structure of WIP is first-order lag, the (3) can be described as:\n\ndAR(t)dt=1DT[AR(t)-0R(t)]E8\n\nWhenαichanges continuously, the response curves of inventory I* and desired rate OR can always converge to the stable state, that is, I=I* and OR=D. Figure 2 shows two typical patterns of behavior in the converging process: smooth convergence (αi=0.1) and fluctuant convergence (αi=0.8). The simulation indicates that the transition between two patterns of behavior happens when αi changes gradually, and whenαi∈[0, 1], there are only the above two typical behavior patterns.", null, "Figure 2.The response curves of inventory and order rates under first-order lag (DT=3)\n\n#### 2.2.3. Dynamics characteristics of system under pure timedelay\n\nIf the delay of WIP is pure time delay (PTD), then Eq. (3) can be described as:\n\nAR(t)=OR(tDT)E9\n\nWith a continuous change ofαi,the response curves of inventory and order rate show four kinds of behavior patterns: smooth convergence (αi=0.1); fluctuant convergence (αi=0.3);oscillation with equi-amplitude (αi=0.52); divergent fluctuation (αi=0.58).It is worthwhile to note that the above response curves appear to be oscillation with equi-amplitude only when αitakes a special value (e.g. αi=0.52) and this special value is a critical point at which the system curves begin to divergent. Figure 3 shows the response curves of inventory and order rates under pure time delay.", null, "Figure 3.The response curves of inventory and order rates under pure time delay (DT=3)\n\n### 2.3. Stability analysis and criteria of supply chain\n\n#### 2.3.1. The definition of stability\n\nThere are different definitions of stability of supply chain. The traditional ideas of system dynamics state that only the behavior of smooth convergence is stable while the other fluctuate behaviors are unstable (Forrester, 1958). The main reason is that system dynamics methods focus on systems under first-order lag, and the studies on stability of supply chain emphasize the two above situations as shown in figure 2.\n\nScholarsusing control theory stress the importance of pure time delay.It is commonly accepted that fluctuant convergence is a gradual process of system to be stable and oscillation with equi-amplitude is a critical state of stable system. Based on the definition of stability in control theory and the methods applied by system dynamics, we propose the following definition of stability of supply chain system:\n\nDefinition 2.1: Suppose the system is stable at the initial time, when imposing a small step disturbance on demand, if the inventory (or order rate) can get stable at a certain equilibrium level after a period of time, then the system is stable.\n\nThere are two points to be stressed: first, the disturbance imposed to the system can’t be too large, as the large disturbance may destroy the structure of real system and the simulation results can deviate from the actual situation of the system; second, the structure and surroundings of economic system may not always keep in a specific condition, so the system can only keep steady state within a limited period of time. In addition, computer simulation and calculation can’t last for an indefinitely long time.\n\n#### 2.3.2. Stability criterion\n\nSimulations show that for both first-order system and PTD system, when the decision parameter αi change from 0.00 to 1.00, the behavior patterns of response curves of inventory and order rate undergo a gradual change from convergence to fluctuation without any sudden change as shown in figure 4. Therefore, we can test the stability of system from the appearance of response curve of inventory I or order rate OR.", null, "Figure 4.The gradual change of response curve of inventory in two systems\n\nThe response curves shown in figure 4 can be abstracted to the general form of inventory fluctuation as shown in figure 5. As αi takes different values, it is difficult to obtain the inventory curves changing laws in the stock control model.Therefore, we can’t give a unified description on the fluctuating behavior by analytical methods.Althoughinventory fluctuation curves can well reflect dynamic behaviors of the system, it is difficult to make a horizontal comparison among the above curves.\n\nAsthe underlying cause of the fluctuation of inventory is the deviation between actual inventory I and desired inventory I*, we use the area between the two curves to describe the fluctuation in supply chain. This practice is similar to the method in cybernetics that use “noise bandwidth” to make quantitative description of bullwhip effect (Dejonckheere et al., 2003).", null, "Figure 5.The general behavior pattern of inventory fluctuation\n\nAs shown in figure 5, assuming that the inventory curve begins to fluctuate at time t0, the inventory curve and desired inventory level intersect at timet1, t2…ti…tnin succession, and the area between two curves can be divided into several partsS1，S2…Si...Sn.Let the absolute value of the area between the two curves be sn，that is:\n\nsn=i=1i=nSiE10\n\nIf the system is smooth convergence, then there is only one arc between the two curves:\n\nsn=|S1|E11\n\nIf the system isfluctuant convergence, then |Si| > |Si+1|(i=1,2…n). There exists a natural number N, when t≥tN, I ≡ I*,|SN+1| ≈ 0, and:\n\nlimn+limn+sn=sN=i=1nSi=C(Constant)E12\n\nIf the system is oscillation with equi-amplitude, then |S1|=|S2|=…=|Si| =…=|Sn|, and:\n\nsn=i=1nSi=nS1E13\n\nIf the system is divergent fluctuation, then |Si| < |Si+1|(i=1,2…n) and:\n\nlimn+limn+sn=+E14\n\nTo sum up, that is:\n\nt0tnIt-I*dt=i=1nSi=snE15\nStn=t0tnIt-I*dtE16\n\nwhere S is the inventory integral curve. We can distinguish the behavior of the system according to the form of curve S, that is, S curve can be used as the stability criterion of the system.\n\nThe S curve can be obtained by the software, Vensim PLE. As shown in figure 6, we present the S curves of PTD system with different decision parameter αi corresponding to figure 3(a) and the trend of S curve is the same as stated before. When S curve keeps a horizontal state or small-scope fluctuation around the horizontal line finally (e.g.αi=0.1; 0.3), the system has returned the stable state. That is, the order meet the demand completely and I=I*.\n\nAccording to the definition of stability and the above simulation analysis, the sufficient condition of the system to be stable is presented as following:\n\nlimtlimtS(t)=C(Constant)E17\n\nEq. (17) can be replaced by the following description:\n\nDefinition 2.2: Assuming t0 is the starting time of simulation and tFis the end time of simulation, if there exists ts(t0≤ts≤tF) to make S(t) = C (Constant), then the system is stable.\n\nThe constant C can be understood as the system stable level, and the smaller the value of C, the better stability of the system. In the condition of step disturbances on demand and no prediction, C is positive.", null, "Figure 6.Inventory integral curve under pure time delay (DT=3)\n\nThe value of S (tn) directly reflects the deviate degree of the actual inventory I from the desired inventory I*. When the inventory is too high or too low, the holding cost and shortage cost will increase accordingly. Therefore, the S curve can intuitively measure the potential cost burden. On the other hand, the S curve reflects not only the general situation of system behaviorbut also the behavior change with time varying. Compared to stock variance, the S curve can measure the consequences of long time small-scope fluctuation (with small stock variance) of the system.\n\nIn conclusion, the S curve is able to reflect different behavior patterns in supply chain system. What’s more, it is more convenient and visible to estimate the effect of fluctuation on inventory cost, ordering policy and forecasting. Besides, according to the definition of stability, the two behavior patterns of first-order system reflect that such systems are always stable, and this conclusion is in agreement with the results obtained from cybernetic methods. The relationship between typical behavior patterns and stability criterion is summarized as shown in table 2.\n\n Inventory status Sharp of S curve Delay mode System state Smooth convergence Be similar to exponential curve, base number∈(0.-1) First-order; PTD Stable Fluctuant convergence Be similar to exponential curve, base number∈(0.-1) First-order; PTD Stable Oscillation with equi-amplitude Be similar to straight line PTD Critical stable Divergent fluctuation Be similar to exponential curve,Base number∈(1,+∞) PTD Unstable\n\n### Table 2.\n\nBehavior pattern and its stability\n\n## 3. Study on the stability of general inventory control system\n\nIn the last section, the dynamic characteristics ofsingle parameter stock control system were discussed and we adopt the inventory integral curve as the criterion for stability judgment.Based oncybernetic studies, the dynamic behavior patterns of inventory in supply chain system are limited to the four typical behavior patterns shown in table 2 (Lalwani et al., 2006). Therefore, as a result of primary judge,the stability criterion proposed in the previous section is still valid for more complicated systems.\n\nHowever, the single parameter stock control model has ignored the management of WIP and there is significant difference between theoretical model and managerial practice. Meanwhile, the previous simulation shows that the delay structure of WIP is a key factor of system stability. Therefore, it is of great theoretical and practical importance to study the effect of WIP on stability of supply chain system.\n\nIn this section, based on the generic stock-management model (Riddalls ＆ Bennett, 2002; Sterman, 1989), we add the WIP control loop to the previous model and built a general inventory control system with dual-loop and double decision parameters. Then the applicability of stability criterion is validated and the stability characteristics in double parameters control model with two different delay structures are discussed.\n\n### 3.1. General inventory control system model\n\n#### 3.1.1. Basic assumptions\n\nThe general inventory control system model in this section can be still understood as one node along the chain, the basic assumptions are the same as i-iv described in 2.1.1. Considering the management of WIP, assumption v in 2.1.1 is changed as following:\n\nv.The supply chain members adjust orders according to demand from downstream, actual storage and WIP, and maintain the inventory at a desired level.\n\n#### 3.1.2. Structure of the model\n\nFigure 7 represents the general inventory control model:\n\nCompared to the model in figure 1, there are three increasing variables:\n\nWIP* the desired WIP,\n\nAWIPthe adjustment for WIP,\n\nALPHAwip (αWIP)the rate at which the discrepancy between actual and desired WIP levels is eliminated,0αWIP≤1,", null, "Figure 7.The general inventory control model\n\n#### 3.1.3. Variablesettings\n\nExcept the indicated orders, the settings of other variables in inventory control loop are the same as Eq. (2)-(6). To regulate WIP, a negative feedback mechanism is used. Adjustments are then made to correct discrepancies between the desired and actual inventory AI, and between the desired and actual WIP AWIP. Eq. (1) is adjusted as follows:\n\nIO=D+AI+AWIPE18\n\nSince WIP* is proportional to the demand as well as the delay time, we define the desired WIP as the delay time multiplied by the demand D. That is\n\nWIP*=D×DTE19\nAWIP=αWIP(WIP*WIP)E20\n\nWIP* reflects the excepted value of future delivery situation. For production-oriented enterprises, WIP* reflects the supply capacity of upstream raw materials and production capacity on the node; for distribution firms, WIP* reflects the channel capacity between two nodes. In fact, there are multiple ways to measure WIP* and Eq. (19) adopts the linear approximation method. As this chapter focuses on structure factors,when imposing small disturbance on the system, the estimate precision of WIP* has little influence on the system stability.\n\nAccording to Eq.(6) and Eq. (20), the ordering policy is defined below:\n\nIO=D+αi(I*I)+αWIP(WIP*WIP)E21\n\nThis ordering policy is still based on the anchoring and adjustment heuristic. Compared to Eq. (7), Eq. (21) considers two anchoring points, that is, I* and WIP*. The ordering policy is one of the dual parameter decision rules.When αWIP=0, figure 6 is equivalent to figure 1.Therefore, the general inventory control model covers the single parameter stock control model.\n\n### 3.2. Dynamic characteristics analysis of system\n\n#### 3.2.1. Simulation design\n\nExcept the parameters involved in WIP, the initial values of the variables are the same as presented in 2.2.1. For the convenience of comparison, we set the value of αWIP to zero, that is, the initial state of the model is equivalent to single parameter stock control model.\n\nWe still adopt the small disturbance for stability examination, and the demand function is unchanged:\n\nD=Dt0(1+STEP(0.2,5))E22\n\nIn the presence of small disturbance, the decision parameter αiis changed from 1 to 0with a small decrement Δi. At the same time, αWIPvaries from 0 to 1 with another small increment ΔWIP, the smaller the values of Δi and ΔWIP, the higher the simulation accuracy. The process can be described by pseudo-code below:\n\nFor(αi =1; αi≥0; αii-Δi)\n\n{ for(αWIP =0; αWIP≤1; αWIPWIP+ΔWIP) { Run Model } }\n\nThis section focuses on the interaction of dual-loop and verifying the applicability of stability criterion proposed in the previous section to general inventory control system. Through simulation, we can observe the behavior patterns of general inventory control system and test the system stability in the situation of complete rationality (αi,αWIP∈[0,1]), then the simulation results can be compared with that of single parameter stock control model.\n\n#### 3.2.2. Dynamics characteristics of system\n\nFirst-order system\n\nThe first-order lag is described as Eq. (8).\n\nIf αWIP=0,the general inventory control model is equivalent to single parameter stock control model. From the previous analysis, whenαichanges continuously, the response curves of inventory I* and desired rate OR can always converge to a stable state.There are only two typical behavior patterns:smooth convergence and fluctuant convergence.\n\nIf αWIP≠0, when αi takes a particular value andαWIPvaries from 0 to 1 with a small increment ΔWIP, the shapes of response curves of inventory I* and desired rate OR are still restricted to the above mentioned two typical behavior patterns. The nearer αWIPapproaches 1, the more obvious the smoothness of response curves will be. The nearer αWIPapproaches 0, the more obvious the fluctuation characteristics of response curves will be. Figure 8 shows the response curves of inventory of the general inventory control system under first-order lag when DT=3 and αi=0.4.\n\nTogether with figure 2, it leads to the conclusion that the general inventory control system under first-order lagis usually stable, butαWIP andαi have exerted totally different influence onthe dynamics characteristics of system. This conclusion also hold in the case when delay time DT takes different values. Therefore, after preliminary analysis, WIP control loop has weakened the fluctuation characteristics of first-order system. The greater the value of αWIP, the weakening more obvious.", null, "Figure 8.The response curve of inventory under first-order lag\n\nPTD system\n\nThe pure time delay is described as Eq. (9).\n\nIf αWIP=0, the response curves of inventory and desired rate will not always converge to stable state with a continuous change of αi and there are four kinds of behavior patterns: smooth convergence; fluctuant convergence;oscillation with equi-amplitude; divergent fluctuation. Therefore, the system exhibits critical stable state and stable boundary.\n\nIf αWIP≠0,when αi takes a particular value andαWIPvaries from 0 to 1 with a small increment ΔWIP, the shapes of response curves of inventory I* and desired rate OR are still restricted to the above mentioned four typical behavior patterns.Figure 9 shows the response curves of inventory of the general inventory control system under pure time delay when DT=3 and αi=0.58.", null, "Figure 9.The response curve of inventory under pure time delay\n\nAlthough single parameter stock control model exhibits divergent behavior when αi=0.58, the general inventory control system model with doubledecision parameters can have convergent behavior as the value of αWIPincreases. Together with the response curves in figure 3, it is concluded that the general inventory control system under pure time delayisnot always stable, and the parameters αWIP andαishow entirely opposite effects on the dynamics characteristics of system. For certain single parameter systems that are unstable, by adding decision parameter αWIPand enhancing WIP control loop, the systems can reach stable state. Thus, WIP control loop has also weakened the fluctuation characteristics of PTD system.\n\n### 3.3. Stability analysis of general inventory control system\n\n#### 3.3.1. Stability criterion\n\nFollowing definition 2.1, the inventory integral curve is used as the stability criterion of supply chain system. This stability criterion can be generalized to the general inventory control system model with doubledecision parameters only when the following conditions are satisfied:\n\nFirst, the response curve of inventory I (or order rate OR) is limited to the range listed in table 2;\n\nSecond, the response curve of inventory I (or order rate OR) is gradually changing as the decision parametersαWIP andαichange, and there is no mutation in this process.\n\nAs the WIP control loop has in fact weakened the fluctuation characteristics of single parameter system model, the behavior patterns of general inventory control model will not go beyond that of single parameter stock control model. Therefore, the behavior patterns of system listed in table 2 still apply to general inventory control system model.\n\nFigure 10 shows the traverse of the response curve of inventoryunder different delay modes. Through the analysis of figure 4, figure 8 and figure9, the increasedαi will increase the fluctuation while the increased αWIP will cushion the fluctuation both in first-order system and PTD system after disturbed, and the processes are smooth. Together with figure 10, the system hasn’t appeared other new behavior patterns and mutation points.", null, "Figure 10.The traverse graph of inventory curves of general inventory control system model\n\nIn conclusion, the dynamic patterns and stability criterion obtained from the analysis of single parameter stock control system model still hold in the general inventory control system model under first-order lag and pure time delay.\n\n#### 3.3.2. Stability of first-order system\n\nThis research indicates that the first-order system isalways stable and the two decision parameters have differenteffects on system stability. Figure 8 shows that the system stability is increasing with the increase of αWIP. Similarly, we obtain the S curve of the general inventory control system model under first-order lag (see figure 11). According to definition 2.2, the constant C represents the stable level of system. Figure 11 shows that the value of C will decrease when increasingαWIP under the condition of the given delay time and αi.But there exists a minimumC*, when the system achieves C*, increase of αWIP won’t help improve system stability. Let C* be the minimal stable levelwhich is determined by αi with a given DT, αWIP, then determines whether the system can achieve the stable level C* or not. Before the system achieves the stable level C*, the response curve of inventory appears to be fluctuant convergence, but when the system has achieved the stable level C*, the response curve of inventory tends to be smooth convergence, and the increase of αWIP can only reduce the deviation between actual inventory I and desired inventory I* and postpone the stabilizing time of the system.", null, "Figure 11.Inventory integral curve of general inventory control model under first-order lag\n\nFurther analysis indicates that there is an exclusiveαWIPcorresponding to αiiWIP∈[0,1]) to make the system achieve the stable level C*. Based on the analysis, theminimalstability boundary of general inventory controlmodel under first-order lag with different DT is obtained as shown in figure 12 (x-axis isαi, y-axis isαWIP). We use S* curve to express the minimalstability boundary,each point on the curve can guarantee that the system will achieve the stable level C*.As shown in figure 12, the index of S* represents the value of DT.", null, "Figure 12.The minimalstability boundaries of first-order system under different DT\n\nThe adjusting parameters(αi, αWIP)in the lower right region of S* curve will cause theresponse curve of inventory to be fluctuant convergence while the parameters in the upper left region guarantee the curve to be smooth convergence. The traditional views of system dynamics consider that fluctuant convergence is unstable and only smooth convergence is stable. Therefore, the lower right region of S* curve is the unstable region in the traditional sense. Nonetheless, the actual decisions are influenced by many subjective and objective factors, and the adjusting parameters(αi, αWIP) may not be the point in the S* curve.\n\nWhen DT≥1,S* curve is nonlinear, otherwise,S* curve is approximate linear and all S* curves are in the lower right region of an oblique line (αiWIP). In conclusion, there exists the adjusting parameters (αi, αWIP) that make the first-order system achieve the stable level C* only when αWIP≤αi. As αi and αWIPreflect the attitude of decision-maker on the differences, that is,(I* - I) and (WIP* - WIP), the condition that αWIPis less than or equal to αirepresents the rational choice of decision-maker. In other words, most decision-makers think that the actual inventory is more important than WIP and they pay more attention to the difference between the actual inventory and the desired inventory.\n\nSince all the adjusting parameters (αi, αWIP) that make the first-order system achieve the stable level C* exist in the lower right region of the oblique line (αiWIP), αWIP≤αi is a necessary condition for first-order system to achieve the minimal stable leveland this condition is determined by the structure of first-order system itself. Whether the system can achieve the minimal stable level or not depends on the subjective judgment of the decision-maker and the external and internal factors. Let the oblique line (αiWIP) be the conservativestability boundary and the condition that αWIPis less than or equal to αibe the conservative stability condition.As theconservative stability condition is the result of rational choice, the oblique line (αiWIP) can also be called rationalstability boundary.\n\n#### 3.3.3. Stability of PTD system\n\nIt is known that the PTD system isnot always stable and the parameters αWIP andαishow opposite effects on the dynamics characteristics of system. Like the single parameter stock control system model under pure time delay, we can obtain the critical stable points (αiDT,αWIPDT) of general inventory control system under pure time delay by finding the critical stable state of inventory curve with a given DT. Therefore, the critical stable condition of general inventory control system under pure time delay is defined as following:\n\nDefinition 3.1: Suppose the general inventory control system under pure time delay is stable at the initial time. With a given DT, when imposing a small step disturbance on demand, if there exists the decision parameters (αiDT,αWIPDT)that can keep the inventory curve to be oscillation with equi-amplitude, then the state is called critical stable state, and (αiDT,αWIPDT)is the critical stable point of the system under the given DT.\n\nUnlike the single parameter stock control system model, the critical point(αiDT,αWIPDT) of the general inventory control system model is not unique. Through the traversal simulation of αi and αWIPunder certain DT, several critical stable points are found. After connecting these points in the plane that takes αi as horizontal axis andαWIP as vertical axis, we obtain the stability boundary of PTD system named s curve. Figure 13 shows some s curves under different DT and the index of s represents the value of DT.\n\nFurthermore, s curve is approximate to linear property and the lower right of s curve is the unstable region. That is, the points (αi, αWIP) in the lower right of s curves will lead the inventory curve into divergent fluctuation, and the points in the upper left of s curves will make the inventory curve convergent. As DT increases, s curve tends to converge toward the oblique line:αWIPi/2. Simulations show that the oblique line (αWIPi/2) is the upper bound when s curve moves up to top left with DT increasing. Thus, it can be concluded that the upper left area of oblique line (αWIPi/2) is the stable region which is independent of delay (IoD), and the oblique line (αWIPi/2) is defined as IoD stability boundary of PTD system. Since the models of supply chain in this research take no account of predictions, the conclusions above not only validate the results obtained bycybernetic methodfrom the point of view of system dynamics, but also prove that IoD stability boundary is only determined by systemic structure.", null, "Figure 13.The stability boundaries of PTD system under different DT\n\nMeanwhile, there also exists a minimalstability boundary of PTD system under the given DT. The minimalstability boundaries of PTD system under different DT are shown in figure 13. The simulation results show that the greater the DT value, the more obvious the linearity of s* curve, when DT≥9, s* curve and the oblique line (αWIPi) nearly coincide. Similarly, αWIP≤αi is also the necessary condition for PTD system to achieve the minimal stable level and the oblique line (αiWIP) is theconservativestability boundary or rationalstability boundary of PTD system.\n\n#### 3.3.4. Discussion\n\nThe main difference between the general inventory control model and the single parameter stock control model are the WIP control loop and the ordering strategy. The single parameter decision rule is marked IO1 and the two-parameter decision rule is marked IO2, that is:\n\nIO1=D+αi(I*I)E23\nIO2=D+αi(I*I)+αWIP(WIP*WIP)E24\n\nAs compared with Eq. (22), Eq. (23) contains the new WIP adjustment. From the static perspective, IO2is greater than or equal toIO1 under the same initial conditions and the demand will be enlarged. But through the dynamic methods, it is found that the two-parameter decision rule actually restraints the demand amplification and reduces the fluctuation of inventory and order rate because of the WIP control loop. Some studies on stability of supply chain based on dynamic methods suggest that the increase in αWIPand decrease in αican contribute to improving the stability of the system(Disney et al., 2004; Riddalls et al.,2000; Sterman, 1989), but the findings of this section indicate that the above policies are effective only in the lower right region of the minimal stability boundary s* curve.\n\nSecond, neither first-order system nor PTD system has the same rational stability boundary. This rational boundary conforms well to the results of beer game (Sterman, 1989). Although Sterman adopted first-order lag as the delay mode when modeling the beer game, the participants hadn’t known the delay mode and they didn’t estimate the delay mode of WIP control loop. Therefore, the conservative stability boundary or rational stability boundary actually has nothing to do with delay.\n\nFinally,the minimal stable level C* proposed in this section is relevant to the convergence properties of inventory fluctuation, it can’t guarantee the minimum cost. From the point of view ofcost optimizing, there also exists an optimal boundary which depends on the ability of inventory management and the composition of inventory cost.\n\n## 4. Conclusions\n\nTo conduct quantitative analysis on the stability of supply chain system with Order-Up-To (OUT) policy, we first built the single parameter stock control model of supply chain. By simulation analysis,a system-dynamics-based criterion for stability judgment is proposed. With simulation,the criterion can be usedto describe the nonlinearities of supply chain system with 1st order exponential lag and pure time delay (PTD).The criterion can also be used to judge the influences exerted on supply chain stability by decision behavior. The simulation demonstrates two different results. Firstly, the 1st order system is usually stable, but there is fork effectas decision parameter changes. Secondly, there is a critical stable bound in PTD system, which determines the feasible filed of decision.\n\nThen ageneral inventory control system model is proposed. The model is provided with two typical delay modes: fist-order exponential delay and pure time delay. According to the concept of stability and stability criterion proposed in the previous section, stability borders with different meanings areconfirmed, which integrate the results derived from different research methods. It is concluded that the stability of inventory control system is mainly decided by the features of feedback systems, the subjective decision and environment take their effects based on the feedback systems, and information sharing is propitious to increase the stability and weaken bullwhip effect of supply chain system.\n\nThis research adopts the method of system dynamics and takes the delay modes as key point to discuss stability of supply chain system. Although preliminary achievements have been made, further research needs to be done on the stability of supply chain. With the development of research, we wish this chapter will contribute to supply chain management theories and practices.\n\n## More\n\n© 2011 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike-3.0 License, which permits use, distribution and reproduction for non-commercial purposes, provided the original is properly cited and derivative works building on this content are distributed under the same license.\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nSuling Jia, Lin Wang and Chang Luo (August 29th 2011). The Research on Stability of Supply Chain under Variable Delay Based on System Dynamics, Supply Chain Management - New Perspectives, Sanda Renko, IntechOpen, DOI: 10.5772/23632. 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[ "", null, "", null, "", null, "# A metal cylinder has a radius of 0.55 cm and a length of 2.85 cm. Its mass is 24.3 grams. a. What is the density of the metal? For a cylinder, V = π r 2 × h , where r is the radius of the cylinder and h is the height of the cylinder. π = 3.14 . b. Compare the density of the metal to the densities of various substances listed in Table 2.4. Can you identify the metal?", null, "### Chemistry In Focus\n\n7th Edition\nTro + 1 other\nPublisher: Cengage Learning,\nISBN: 9781337399692\n\n#### Solutions\n\nChapter\nSection", null, "### Chemistry In Focus\n\n7th Edition\nTro + 1 other\nPublisher: Cengage Learning,\nISBN: 9781337399692\nChapter 2, Problem 45E\nTextbook Problem\n9 views\n\n## A metal cylinder has a radius of 0.55 cm and a length of 2.85 cm. Its mass is 24.3 grams.a. What is the density of the metal? For a cylinder, V = π r 2 × h , where r is the radius of the cylinder and h is the height of the cylinder. π = 3.14 .b. Compare the density of the metal to the densities of various substances listed in Table 2.4. Can you identify the metal?\n\nInterpretation Introduction\n\nInterpretation:\n\nThe density of a metal cylinder is to be calculated and the metal is to be identified by comparing its density to the densities of various substances listed in Table-2.4.\n\nConcept Introduction:\n\nDensity of a substance is a measure of compactness of that substance. It is defined as the mass of a substance present in per unit of its volume. The unit of density is derived when the unit of mass is divided by the unit of volume. Its most common units are g/cm3 and g/mL.\n\nDensity = Mass Volume\n\n### Explanation of Solution\n\na) The density of the metal having as radius of 0.55cm, a length of 2.85cm and its mass is 24.3grams.\n\nVolume of cylinder is calculated as:\n\nVolume of cylinder = πr2h\n\nSubstitute the given values in the above equation as:\n\nVolume of cylinder = 3.14 ×(0.55 cm)2×2.85 cmVolume of cylinder = 2.7 cm3\n\nThus, the volume of the metal cylinder is 2.7 cm3\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\nFind more solutions based on key concepts\nHow do eating disorders affect health?\n\nUnderstanding Nutrition (MindTap Course List)\n\nTo temporarily increase the bodys water content, a person need only. a. consume extra salt. b. consume extra su...\n\nNutrition: Concepts and Controversies - Standalone book (MindTap Course List)\n\nAn induction furnace uses electromagnetic induction to produce eddy currents in a conductor, thereby raising th...\n\nPhysics for Scientists and Engineers, Technology Update (No access codes included)\n\nWhy did geologists have such strong objections to Wegeners ideas when he proposed them in 1912?\n\nOceanography: An Invitation To Marine Science, Loose-leaf Versin", null, "" ]

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[ "This content will become publicly available on July 4, 2023\n\nDesingularization and p-Curvature of Recurrence Operators\nLinear recurrence operators in characteristic p are classified by their p-curvature. For a recurrence operator L, denote by χ(L) the characteristic polynomial of its p-curvature. We can obtain information about the factorization of L by factoring χ(L). The main theorem of this paper gives an unexpected relation between χ(L) and the true singularities of L. An application is to speed up a fast algorithm for computing χ(L) by desingularizing L first. Another contribution of this paper is faster desingularization.\nAuthors:\n;\nAward ID(s):\nPublication Date:\nNSF-PAR ID:\n10385968\nJournal Name:\nISSAC'2022\nPage Range or eLocation-ID:\n111 to 118\nMany astrophysical environments, from star clusters and globular clusters to the discs of active galactic nuclei, are characterized by frequent interactions between stars and the compact objects that they leave behind. Here, using a suite of 3D hydrodynamics simulations, we explore the outcome of close interactions between $1\\, \\mathrm{M}_{\\odot }$ stars and binary black holes (BBHs) in the gravitational wave regime, resulting in a tidal disruption event (TDE) or a pure scattering, focusing on the accretion rates, the back reaction on the BH binary orbital parameters, and the increase in the binary BH effective spin. We find that TDEs can make a significant impact on the binary orbit, which is often different from that of a pure scattering. Binaries experiencing a prograde (retrograde) TDE tend to be widened (hardened) by up to $\\simeq 20{{\\ \\rm per\\ cent}}$. Initially circular binaries become more eccentric by $\\lesssim 10{{\\ \\rm per\\ cent}}$ by a prograde or retrograde TDE, whereas the eccentricity of initially eccentric binaries increases (decreases) by a retrograde (prograde) TDE by $\\lesssim 5{{\\ \\rm per\\ cent}}$. Overall, a single TDE can generally result in changes of the gravitational-wave-driven merger time-scale by order unity. The accretion rates of both black holesmore »\n5. Abstract The elliptic algebras in the title are connected graded $\\mathbb {C}$ -algebras, denoted $Q_{n,k}(E,\\tau )$ , depending on a pair of relatively prime integers $n>k\\ge 1$ , an elliptic curve E and a point $\\tau \\in E$ . This paper examines a canonical homomorphism from $Q_{n,k}(E,\\tau )$ to the twisted homogeneous coordinate ring $B(X_{n/k},\\sigma ',\\mathcal {L}^{\\prime }_{n/k})$ on the characteristic variety $X_{n/k}$ for $Q_{n,k}(E,\\tau )$ . When $X_{n/k}$ is isomorphic to $E^g$ or the symmetric power $S^gE$ , we show that the homomorphism $Q_{n,k}(E,\\tau ) \\to B(X_{n/k},\\sigma ',\\mathcal {L}^{\\prime }_{n/k})$ is surjective, the relations for $B(X_{n/k},\\sigma ',\\mathcal {L}^{\\prime }_{n/k})$ are generated in degrees $\\le 3$ and the noncommutative scheme $\\mathrm {Proj}_{nc}(Q_{n,k}(E,\\tau ))$ has a closed subvariety that is isomorphic to $E^g$ or $S^gE$ , respectively. When $X_{n/k}=E^g$ and $\\tau =0$ , the results about $B(X_{n/k},\\sigma ',\\mathcal {L}^{\\prime }_{n/k})$ show that the morphism $\\Phi _{|\\mathcal {L}_{n/k}|}:E^g \\to \\mathbb {P}^{n-1}$ embeds $E^g$ as a projectively normal subvariety that is a scheme-theoretic intersection of quadric and cubic hypersurfaces." ]

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[ "# Sigma Versus Sigma – Which is Which?\n\nSix Sigma – iSixSigma Forums Old Forums General Sigma Versus Sigma – Which is Which?\n\nViewing 6 posts - 1 through 6 (of 6 total)\n• Author\nPosts\n• #29748\n\nHi!\nI am having difficulty understanding something that is probably straight forward. Sometimes I hear people talking about the process sigma and other times I hear them talking about a mathematical sigma. What’s the difference and how can I (as a layperson) quickly understand which one they’re talking about? I’m frustrated! I know of only one sigma — the greek symbol that I learned in high school math. Thanks for your help with a novice. :)\nTina\n\n0\n#76730", null, "Marc Richardson\nParticipant\n\nTina,\nSigma is a measure of dispersion or spread of a population, in other words, its variability. A Population is a general term and can be applied to groups of people (such as age, weight, height, income, etc.), agriculture (crop yield, maturation date, etc.) or the output of a manufacturing process. Process sigma is a measure of the variability of a process, either measured directly (Pressure, speed, voltage, etc) or by the process output (weight, concentration, length, etc). The term sigma is synonymous with standard deviation. It is calculated from all the measurements for every member of the population, as opposed to the estimate of sigma (also known as sigma hat; the greek letter sigma with a caret [^] over it) or the sample standard deviation which is calculated from a sample drawn from the population.\nMarc Richardson\nSr.Q.A. Eng.\n\n0\n#76732\n\nThanks, Mark. I really appreciate your help. I thought that was the case, but wasn’t sure.\nSo if I were to summarize,\nSigma (capital “S”): Process Sigma measures variability or output of a process.\nsigma (lowercase “s”, known as the greek letter sigma with a caret [^] over it):  The standard deviation calculated from every member of the population.\nsigma estimate (also lowercase “s”, known as the greek letter sigma with a caret [^] over it):  The standard deviation calculated from a sample of the population.\nIs that a fair way to describe the differences? Thanks for your help.\nTina\n\n0\n#76737", null, "Gabriel\nParticipant\n\nTina:\nShort answer: That was not a fair way do describe the differences.\nSigma (greek leter with no hat) = population’s standard deviation (calcualted from every member of the population). If the population is a process output or a process parameter, the you have a “process sigma”. Hence, ther is no difference between “population sigma” and “process sigma”.\nSigma hat (the greek letter with a ^) = estimation of sigma. If you estimate that sigma is 10 you can not say “sigma”=10, but it is Ok to say “sigma hat”=10.\nS (also s) = sample’s standard deviation (calculated only form some members of the population). S is only one from many ways of estimating sigma.\nLong asnwer:\nSigma, the greek letter with no hat, is the standard deviation calculated from every member of the population. If your population is “the weight of american men” then sigma is the “weight of american men’s sigma”. If your population is the “volume of coca-cola in the bottles after the filling process” then you have the “volume of coca-cola in the bottles after the filling process sigma”. If it is clear that we are talkng about the volume after the filling proces, we can save words and just say the it is the filling process sigma. Process sigma is just a population sigma where the population is a process parameter or a process output.\nSigma hat, the greek leter with a [^] on it, is an etimation of sigma based on data from a sample. It is used because you ussualy do not measure every members of the population. It is ussually very hard and expensive, and sometimes impossible. If the population is the result of throwing a dice, you have infinite members because you can throw the dice inminite times. So it is impossible to measure all the members. But you can measure let’s say 1000 members and get an estimation for the population. For example, if you estimate the standard deviatin of the population from data you got from a smple, this is sigma hat. One of the estimators used for the population’s standard deviation is the sample’s standard deviation, which is called S. There are other ways to estimate the population’s standard deviation from sample data, not only S (for example, you have Rbar/d2).\n\n0\n#76766\n\n0\n#76768", null, "Arun Nair\nParticipant\n\nHi Tina,\nDefinition #1: Sigma is your standard deviation – the sigma you encountered in your high school math, and which you are probably familiar with.\nDefinition #2: When people talk about their process capability being 2 sigma or 3 sigma, what it implies is that they can fit 2 or 3 standard deviations of their process between the specification limits ( upper specifiation limit and lower specification limits ). In other words the higher the process capability, the more the number of standard deviations you could fit between the specification limits; the tighter your process spread; in other words a better process.\nThus the mathematical sigma and the process sigma have the same basis.\nregards,\narun\n\n0\nViewing 6 posts - 1 through 6 (of 6 total)\n\nThe forum ‘General’ is closed to new topics and replies." ]

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[ "# 6972711\n\n## 6,972,711 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 6972711 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.\n\n6972711 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.\n\n## Prime factorization of 6972711:\n\n### 3 × 2324237\n\nSee below for interesting mathematical facts about the number 6972711 from the Numbermatics database.\n\n### Names of 6972711\n\n• Cardinal: 6972711 can be written as Six million, nine hundred seventy-two thousand, seven hundred eleven.\n\n### Scientific notation\n\n• Scientific notation: 6.972711 × 106\n\n### Factors of 6972711\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 2\n• Sum of prime factors: 2324240\n\n### Divisors of 6972711\n\n• Number of divisors d(n): 4\n• Complete list of divisors:\n• Sum of all divisors σ(n): 9296952\n• Sum of proper divisors (its aliquot sum) s(n): 2324241\n• 6972711 is a deficient number, because the sum of its proper divisors (2324241) is less than itself. Its deficiency is 4648470\n\n### Bases of 6972711\n\n• Binary: 110101001100101001001112\n• Base-36: 45G6F\n\n### Squares and roots of 6972711\n\n• 6972711 squared (69727112) is 48618698689521\n• 6972711 cubed (69727113) is 339004135158108661431\n• The square root of 6972711 is 2640.5891388099\n• The cube root of 6972711 is 191.0442136983\n\n### Scales and comparisons\n\nHow big is 6972711?\n• 6,972,711 seconds is equal to 11 weeks, 3 days, 16 hours, 51 minutes, 51 seconds.\n• To count from 1 to 6,972,711 would take you about seventeen weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 6972711 cubic inches would be around 15.9 feet tall.\n\n### Recreational maths with 6972711\n\n• 6972711 backwards is 1172796\n• The number of decimal digits it has is: 7\n• The sum of 6972711's digits is 33\n• More coming soon!\n\nMLA style:\n\"Number 6972711 - Facts about the integer\". Numbermatics.com. 2022. Web. 16 August 2022.\n\nAPA style:\nNumbermatics. (2022). Number 6972711 - Facts about the integer. Retrieved 16 August 2022, from https://numbermatics.com/n/6972711/\n\nChicago style:\nNumbermatics. 2022. \"Number 6972711 - Facts about the integer\". https://numbermatics.com/n/6972711/\n\nThe information we have on file for 6972711 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 6972711, math, Factors of 6972711, curriculum, school, college, exams, university, Prime factorization of 6972711, STEM, science, technology, engineering, physics, economics, calculator, six million, nine hundred seventy-two thousand, seven hundred eleven.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on." ]

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[ "# The Melting Point Of Ice And The Boiling Point Of Water | Science Experiment\n\n## INTRODUCTION\n\nIn this experiment, we will learn to determine the melting point of ice and the boiling point of water. Let us get started.\n\n## AIM\n\nTo find the boiling point of water and melting point of ice\n\n## OBJECTIVE A\n\nTo find the melting point of ice.\n\n1. Ice cube,\n\n2. Filter paper,\n\n3. Beaker,\n\n4. Wire gauge,\n\n5. Tripod stand,\n\n6. Burner,\n\n7. Thermometer,\n\n8. Stirrer,\n\n9. Clamp stand.\n\n### THEORY\n\n1. The temperature at which the ice starts melting is called the melting point of water.\n\n2. The solid form of water is ice.\n\n3. The melting point of ice is 0°C.\n\n4. At 0°C, ice converts into water.\n\n### PROCEDURE\n\n1. Take out some ice cubes from the refrigerator and dry them with the help of filter paper.\n\n2. Put the ice cubes in a beaker immediately.\n\n3. Keep a wire gauge over a tripod stand and place the beaker over it.\n\n4. Using a clamp stand, suspend a thermometer in the beaker.\n\n5. Heat the setup and start stirring the ice cube continuously.\n\n6. When the ice starts melting, measure this temperature and name it t1.\n\n7. When the ice melted completely, measure this temperature and name it t2.\n\n8. Write down your measurement in the given table.\n\n### OBSERVATION\n\nThe temperature when the ice starts melting, t1.The temperature when the ice is completely melted, t2.(t1 + t2)/2\n\nMean value of the temperature = (t1 + t2)/2\n\nThe melting point of ice = ……..°C.\n\n### RESULT\n\nThe melting point of ice = ……..°C\n\n### PRECAUTION\n\n1. Dry the ice cubes before experimentation.\n\n2. Continuous temperature should be maintained through stirring.\n\n3. Thermometre should be dipped in ice cubes.\n\n## OBJECTIVE B\n\nTo measure the boiling point of water.\n\n### MATERIAL REQUIRED\n\n1. Distilled water,\n\n2. Boiling tube,\n\n3. Rubber cork with two bores,\n\n4. Delivery tube,\n\n5. Clamp stand,\n\n6. Piece of pumice stone,\n\n7. Beaker,\n\n8. Thermometre,\n\n9. Burner.\n\n### THEORY\n\n1. We know that water exists in three forms, i.e. liquid, solid(ice), and gas(water vapour).\n\n2. The temperature at which the water starts boiling is known as the boiling point of water.\n\n3. Water starts boiling at 100°C. Therefore its boiling point is 100°C.\n\n### PROCEDURE\n\n1. In a boiling tube, take 50ml of water.\n\n2. Put some pieces of pumice stone into it.\n\n3. In the mouth of the boiling tube, fix a cork having two bores.\n\n4. Fix a thermometer and a delivery tube in the two bores of cork.\n\n5. Now, hold the beaker with the help of a clamp stand.\n\n6. Place a beaker at the end of the delivery tube in order to collect water.\n\n7. Heat the boiling tube.\n\n8. Measure the temperature when water starts boiling and name it t1.\n\n9. Measure the temperature when it becomes constant and name it the t2.\n\n10. Write down your measurement in the given table.", null, "### OBSERVATION\n\nWhen water starts boiling at t1When the temperature becomes constant t2(t1 + t2)/2\n\nMean value of the temperature = (t1 + t2)/2\n\nThe boiling point of water = ……..°C.\n\n### RESULT\n\nThe boiling point of water= ……..°C\n\n### PRECAUTIONS\n\n1. The water should have a normal temperature.\n\n2. Water should be distilled.\n\n3. Add pumice stone before heating the water.\n\n4. Water should be heated by rotating the frame.\n\n5. The bulb of the thermometer must be a little above the water.\n\n## CONCLUSION\n\nIn this way, we have learned to determine the boiling point of water and melting point of ice.\n\nQ.1 What do you mean by the melting point of water?\n\nANS. The temperature at which the ice starts melting is called the melting point of water.\n\nQ.2 What do you mean by the boiling point of water?\n\nANS. The temperature at which the water starts boiling is known as the boiling point of water.\n\nQ.3 Why should we use distilled water for the experiment?\n\nANS. Because the boiling point of distilled water is 100°C while the boiling point of impure water is different.\n\nQ.4 What was the aim of our experiment?\n\nANS. To determine the boiling point of water and melting point of ice.\n\nQ.5 Give one difference between boiling and evaporation?\n\nANS. Boiling occurs at a fixed temperature, while evaporation can occur at all temperatures.\n\nQ.6 Give one relation between vapour pressure and boiling point?\n\nANS. Vapour pressure is inversely proportional to boiling point.\n\nQ.7 How is a boiling point related to altitudes?\n\nANS. The boiling point decreases with an increase in altitude.\n\nQ.8 Give the boiling point of water in kelvin?\n\nANS. 100°C or 373K\n\nQ.9 What is the melting point of water in kelvin?\n\nANS. 0°C or 273K\n\nQ.10 Why CaCl2 is used in melting snow?\n\nANS. Because CaCl2 lowers the melting point of snow which helps in clearing snow rapidly." ]

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[ "# Gas/Oil Interfacial Tension\n\nIn this spreadsheet, just type in the required information and press the compute button. Scroll down for more info.\n\n Tdeg F Pressurepsia Gamma_APIAPI Sigma_o dyne/cm\nGas/Oil IFT (after Baker and Swerdloff).\nSigma_o = Fc * Sigma_T\nWhere\nSigma_o = surface tension, dyne/cm\nFc = 1.0 - 0.025*P**0.45\np= pressure of interest,psia\nSigma_T = Sigma68 - (T-68)*(Sigma68 - Sigma100)/32,\nSigma68 = IFT at 68 F, = 39 - 0.2571*Gamma_API,\nSigma100 = IFT at 100 F, = 37.5 - 0.2571*Gamma_API,\nGamma_API = gravity of stock tank oil, API\nT = temperature of interest, deg F" ]

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[ "# critical speed in a ball mill\n\n•", null, "### The operating speed of a ball mill should be the critical\n\nThe operating speed of a ball mill should be _____ the critical speed. a) Less than b) Much more than c) At least equal to d) Slightly more than\n\n•", null, "### Mill Speedan overview ScienceDirect Topics\n\nSAG mills of comparable size but containing say 10 ball charge (in addition to the rocks) normally operate between 70 and 75 of the critical speed. Dry Aerofall mills are run at about 85 of the critical speed. The breakage of particles depends on the speed of rotation.\n\n•", null, "### SAGMILLING.COM . . tools\n\nSAGMILLING.COM . . tools. Mill Critical Speed Calculation. Estimates the critical speed of a grinding mill of a given diameter given the mill inside diameter and liner thickness. If given a measured mill rotation (RPM) then the mill`s fraction of the critical speed is given. Internet Slide Rule.\n\n•", null, "### Critical Speed Of Ball Mill Formulasauna-clp.de\n\nThe critical speed formula of ball mill 4229 is a horizontal cylindrical equipment which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m and the mill with diameter of 5 m rotates when reading more.\n\n•", null, "### Critical Speed Of Ball Mill Formulasauna-clp.de\n\nThe critical speed formula of ball mill 4229 is a horizontal cylindrical equipment which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m and the mill with diameter of 5 m rotates when reading more.\n\n•", null, "### The operating speed of a ball mill should be the critical\n\nThe operating speed of a ball mill should be _____ the critical speed. a) Less than b) Much more than c) At least equal to d) Slightly more than\n\n•", null, "### derivation of critical speed in a ball mill pdf\n\nderivation of critical speed in a ball mill pdf. Critical speed of ball mill formula derivation critical speed of ball mill derivation pdf critical speed of ball mill derivation pdf honey nut clusterswikipedia the free encyclopedia cachedhoney nut clusters aka clusters is a breakfast cereal manufactured by general mills which refers to the cereal as quotcrispy wheat rice flakes with .\n\n•", null, "### Critical Speed Of Ball Mill Formulasauna-clp.de\n\nThe critical speed formula of ball mill 4229 is a horizontal cylindrical equipment which is used in grinding under critical speed. It is expressed by the formula NC 4229 VDD. The critical speed is m and the mill with diameter of 5 m rotates when reading more.\n\n•", null, "### critical speed ball mill calculationMC World\n\nBall Mill Critical Speed911 Metallurgist. 2015-6-19 Charge movement at Various Mill Critical Speed. The Formula derivation ends up as follow Critical Speed is N c =76.6(D 0.5) where N c is the critical speed in revolutions per minute D is the mill effective inside diameter in feet.\n\n•", null, "### TECHNICAL NOTES 8 GRINDING R. P. King\n\n2009-7-30 · The critical speed of the mill c is defined as the speed at which a single ball will just remain against the wall for a full cycle. At the top of the cycle =0 and Fc Fg (8.5) mp 2 cDm 2 mpg (8.6) c 2g Dm 1/2 (8.7) The critical speed is usually expressed in terms of the number of revolutions per second Nc c 2 1 2 2g Dm 1/2 (2 9.81)1/2\n\n•", null, "### derivation of critical speed in a ball mill pdf\n\nderivation of critical speed in a ball mill pdf. Critical speed of ball mill formula derivation critical speed of ball mill derivation pdf critical speed of ball mill derivation pdf honey nut clusterswikipedia the free encyclopedia cachedhoney nut clusters aka clusters is a breakfast cereal manufactured by general mills which refers to the cereal as quotcrispy wheat rice flakes with .\n\n•", null, "### formular for calculating the critical speed of a mill\n\nFormular For Calculating The Critical Speed Of A Mill. formula of critical speed of ball mill calculation of ball mill critical speed. critical speed formula for ball formula to calculate critical speed is given below n c 42305 sqtdd n c critical speed of the mill .ball mill calculations critical speed com . . mill critical speed mill critical speed determination. note this is not the width of\n\n•", null, "### The operating speed of a ball mill should be the critical\n\nThe operating speed of a ball mill should be _____ the critical speed. a) Less than b) Much more than c) At least equal to d) Slightly more than\n\n•", null, "### formular for calculating the critical speed of a mill\n\nFormular For Calculating The Critical Speed Of A Mill. formula of critical speed of ball mill calculation of ball mill critical speed. critical speed formula for ball formula to calculate critical speed is given below n c 42305 sqtdd n c critical speed of the mill .ball mill calculations critical speed com . . mill critical speed mill critical speed determination. note this is not the width of\n\n•", null, "### Critical speed of ball mill depends on__Brainly\n\n2019-3-10 · Find an answer to your question Critical speed of ball mill depends on__ subramanyamt1096 subramanyamt1096 09.03.2019 Science Secondary School answered Critical speed of ball mill depends on__ 2 See answers\n\n•", null, "### formular for calculating the critical speed of a mill\n\nFormular For Calculating The Critical Speed Of A Mill. formula of critical speed of ball mill calculation of ball mill critical speed. critical speed formula for ball formula to calculate critical speed is given below n c 42305 sqtdd n c critical speed of the mill .ball mill calculations critical speed com . . mill critical speed mill critical speed determination. note this is not the width of\n\n•", null, "### Critical Velocity Of Ball Millchistemania.es\n\n2021-6-22 · Critical Velocity Of Ball Mill. Critical Speed Calculation Ball Mill A Ball Mill Is One Kind Of Grinding Machine And It Is A.Balls And The Rotational Speed N Of The Mill Was Varied In A Range From 40100 Using The Critical Rotational Speed Nc Defined By Eq.11 As A Reference.The Criti-Cal Rotational Speed\n\n•", null, "### (Solved)The critical speed of the ball mill of radius R\n\nThe critical speed of the ball mill of radius R which contains ball radius r is proportional to (R- 1 answer below » The critical speed of the ball mill of radius R which contains ball radius r is proportional to (R-r)x. What is the value of x Apr 07 2021 09 23 AM. 1 Approved Answer\n\n•", null, "### (Solved)The critical speed of the ball mill of radius R\n\nThe critical speed of the ball mill of radius R which contains ball radius r is proportional to (R- 1 answer below » The critical speed of the ball mill of radius R which contains ball radius r is proportional to (R-r)x. What is the value of x Apr 07 2021 09 23 AM. 1 Approved Answer\n\n•", null, "### Mill Critical Speed Calculation911 Metallurgist\n\n2015-10-15 · The mill was rotated at 50 62 75 and 90 of the critical speed. Six lifter bars of rectangular cross-section were used at equal spacing. The overall motion of the balls at the end of five revolutions is shown in Figure 4.\n\n•", null, "### critical speed ball mill calculationMC World\n\nBall Mill Critical Speed911 Metallurgist. 2015-6-19 Charge movement at Various Mill Critical Speed. The Formula derivation ends up as follow Critical Speed is N c =76.6(D 0.5) where N c is the critical speed in revolutions per minute D is the mill effective inside diameter in feet.\n\n•", null, "### TECHNICAL NOTES 8 GRINDING R. P. King\n\n2009-7-30 · The critical speed of the mill c is defined as the speed at which a single ball will just remain against the wall for a full cycle. At the top of the cycle =0 and Fc Fg (8.5) mp 2 cDm 2 mpg (8.6) c 2g Dm 1/2 (8.7) The critical speed is usually expressed in terms of the number of revolutions per second Nc c 2 1 2 2g Dm 1/2 (2 9.81)1/2\n\n•", null, "### formula for critical speed of a ball mill\n\nA ball mill critical speed actually ball rod ag or sag is the speed at which the centrifugal forces equal gravitational forces at the mill shells inside surface and no balls will fall from its position onto the shellCritical speed calculation formula of ball millball mill critical speed metallurgist.\n\n•", null, "### what is critical speed of ball millMC World\n\nThe critical speed and critical speed of the ball mill the critical speed and critical speed of the ball mill are kept at 63 of the critical speed. For the three configurations 1 2 and 4 shown in the figure the end face angle varies from 90 to 111 degrees. The height of the lift rod is also shown. critical speed ball mill\n\n•", null, "### Critical Velocity Of Ball Millchistemania.es\n\n2021-6-22 · Critical Velocity Of Ball Mill. Critical Speed Calculation Ball Mill A Ball Mill Is One Kind Of Grinding Machine And It Is A.Balls And The Rotational Speed N Of The Mill Was Varied In A Range From 40100 Using The Critical Rotational Speed Nc Defined By Eq.11 As A Reference.The Criti-Cal Rotational Speed\n\n•", null, "### critical nspeed nfor na nballmill\n\nA Ball Mill Critical Speed (actually ball rod AG or SAG) is the speed at which the centrifugal forces equal gravitational forces at the mill shell s inside surface and no balls will fall from its position onto the shell. The imagery below helps explain what goes on inside a\n\n•", null, "### derivation of critical speed in a ball mill pdf\n\nderivation of critical speed in a ball mill pdf. Critical speed of ball mill formula derivation critical speed of ball mill derivation pdf critical speed of ball mill derivation pdf honey nut clusterswikipedia the free encyclopedia cachedhoney nut clusters aka clusters is a breakfast cereal manufactured by general mills which refers to the cereal as quotcrispy wheat rice flakes with ." ]

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[ "# 9张照片弄成一个九宫格（微信九张照片合成一张）", null, "`# 生成9宫格图片def grid9_image(imageFileName):if not os.path.exists('image'):os.makedirs('image')image = cv2.imread(imageFileName, 1) #删除代码段头height, width, n = image.shapeif width >= height:image = cv2.resize(image, (width, width))height=widthelse:image = cv2.resize(image, (height, height))width = height #删除代码段尾height = int(height / 3)width = int(width / 3)x = 1for i in range(0, 3):for j in range(0, 3):print(i * height, height * (i + 1), j * width, width * (j + 1))result = image[i * height:height * (i + 1), j * width:width * (j + 1)]print('image/' + str(x) + \".png\")cv2.imwrite('image/' + str(x) + \".png\", result)x += 1`\n\ni与j的算法同理，原理如下图。记得这里是width,height是整体宽度高度除于3后得到的。", null, "", null, "", null, "`# 短视频生成9宫格动图def grid9_gif(srcVideoFileName):if not os.path.exists('gif'):os.makedirs('gif')all_frames = []cap = cv2.VideoCapture(srcVideoFileName)fps = cap.get(cv2.CAP_PROP_FPS)for i in range(9):list = []all_frames.append(list)while (cap.isOpened()):ret, frame = cap.read()if ret:height, width, n = frame.shapeif width >= height:frame = cv2.resize(frame, (width, width))height = widthelse:frame = cv2.resize(frame, (height, height))width = heightheight = int(height / 3)width = int(width / 3)frame_list = []for i in range(0, 3):for j in range(0, 3):result = frame[i * height:height * (i + 1), j * width:width * (j + 1)]frame_list.append(result)for index, image in zip(range(9), frame_list):all_frames[index].append(image)else:breakfor index, frames in zip(range(9), all_frames):imageio.mimsave(\"gif/\" + str(index + 1) + \".gif\", frames, 'GIF', duration=float(1 / fps))cap.release()`\n\nGIF直接生成动图9宫格\n\n`# GIF生成9宫格动图def grid9_gif2(srcGIFFileName):clip = mp.VideoFileClip(srcGIFFileName)clip.write_videofile(\"gifVideo.mp4\")grid9_gif('gifVideo.mp4')`\n\n`#生成9宫格视频def gird9_Video(srcVideoFileName, outputVideoFilename):cap = cv2.VideoCapture(srcVideoFileName)fps = cap.get(cv2.CAP_PROP_FPS)width = int(cap.get(cv2.CAP_PROP_FRAME_WIDTH))height = int(cap.get(cv2.CAP_PROP_FRAME_HEIGHT))fourcc = cv2.VideoWriter_fourcc(*'MJPG')videoWriter = cv2.VideoWriter(outputVideoFilename + \".avi\", fourcc, fps, (width, height))i = 1while (cap.isOpened()):ret, frame = cap.read()if ret:cv2.line(frame, (0, int(height / 3)), (width, int(height / 3)), (255, 255, 255), 3)cv2.line(frame, (0, int(height / 3 * 2)), (width, int(height / 3 * 2)), (255, 255, 255), 3)cv2.line(frame, (int(width / 3), 0), (int(width / 3), height), (255, 255, 255), 3)cv2.line(frame, (int(width / 3 * 2), 0), (int(width / 3 * 2), height), (255, 255, 255), 3)videoWriter.write(frame)else:breakcap.release()`", null, "", null, "" ]

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[ "The associative property, or associative law, is a property of many arithmetic and logical functions that says that the order of evaluation does not matter. Example:\n\nSuppose you have the equation:\n1 + 2 + 3 + 4 + 5 + 6\nYou may evaluate it in any order:\n((1 + 2) + (3 + 4)) + (5 + 6) = (3 + 7) + 11 = 10 + 11 = 21\nor:\n1 + (2 + (3 + (4 + (5 + 6)))) = 1 + (2 + (3 + (4 + 11))) = 1 + (2 + (3 + 15)) = 1 + (2 + 18) = 1 + 20 = 21\n\nOr many other ways. While different ways may be quicker, they all lead to the same answer. Addition, Multiplication, AND, OR, XOR are all associative.\n\nNow consider subtraction with the equation:\n6 - 5 - 4 - 3 - 2 - 1\nIf you do:\n(6 - 5) - ((4 - 3) - (2 - 1)) = 1 - (1 - 1) = 1 - 0 = 1\nWhereas, if you do:\n((6 - 5) - (4 - 3)) - (2 - 1) = (1 - 1) - 1 = 0 - 1 = -1\n\nWhich are obviously not the same. Subtraction is thus not associative. Neither is division. In order to properly evaluate them we need to have rules of algebraic precedence.\n\nLog in or register to write something here or to contact authors." ]

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[ "x", null, "Menu\n\n# An Introduction to Riemann Surfaces and Algebraic Curves: Complex 1-Tori and Elliptic Curves\n\nIIT Madras, , Prof. T.E. Venkata Balaji\n\nUpdated On 02 Feb, 19\n\n## Lecture 47: Complex Projective 2-Space as a Compact Complex Manifold of Dimension Two\n\n4.1 ( 11 )\n\n###### Lecture Details\n\nAn Introduction to Riemann Surfaces and Algebraic Curves Complex 1-Tori and Elliptic Curves by Dr. T.E. Venkata Balaji, Department of Mathematics, IIT Madras. For more details on NPTEL visit httpwww.nptel.iitm.ac.insyllabus111106044Goals of Lecture 45A In Part A of this lecture, we define complex projective 2-space and show how it can be turned into a two-dimensional complex manifold. In Part B, we show that any complex torus is holomorphically isomorphic to the natural Riemann surface structure on the associated elliptic algebraic cubic plane projective curve embedded in complex projective 2-spaceKeywords for Lecture 45A Upper half-plane, complex torus associated to a lattice (or) grid in the plane, fundamental parallelogram associated to a lattice, doubly-periodic meromorphic function (or) elliptic function associated to a lattice, Weierstrass phe-function associated to a lattice, ordinary differential equation satisfied by the Weierstrass phe-function, zeros of the derivative of the Weierstrass phe-function, pole of order two (or) double pole with residue zero, triple pole (or) pole of order three, cubic equation, elliptic algebraic cubic curve, zeros of a polynomial equation, bicontinuous map (or) homeomorphism, open map, order of an elliptic function, elliptic integral, Argument principle, even function, odd function, analytic branch of the square root, simply connected, punctured torus, elliptic algebraic affine cubic plane curve, projective plane cubic curve, complex affine two-space, complex projective two-space, one-point compactification by adding a point at infinity, Implicit function theorem, graph of a holomorphic function, nonsingular (or) smooth polynomial in two variables, zero locus of a polynomial, solving an implicit equation locally for an explicit function, nonsingular (or) smooth point of the zero locus of a polynomial in two variables, Hausdorff, second countable, connected component, nonsingular cubic polynomial, discriminant of a polynomial, cubic discriminant, homogeneous coordinates on projective 2-space, punctured complex 3-space, quotient topology, open map, complex two-dimensional manifold (or) complex surface, complex one-dimensional manifold (or) Riemann surface, complex coordinate chart in two complex variables, holomorphic (or) complex analytic function of two complex variables, glueing of Riemann surfaces, glueing of complex planes, zero set of a homogeneous polynomial in projective space, degree of homogeneity, Eulers formula for homogeneous functions, homogenisation, dehomogenisation, a complex curve is a real surface, a complex surface is a real 4-manifold\n\n#### 0\n\n0 Ratings\n55%\n30%\n10%\n3%\n2%\n###### Comments", null, "Sam\n\nExcellent course helped me understand topic that i couldn't while attendinfg my college.\n\nReply", null, "Dembe\n\nGreat course. Thank you very much.\n\nReply\nSend" ]

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[ "# Modelling of a Variable Displacement Lubricating Pump with Air Dissolution Dynamics.\n\nIntroduction\n\nVariable displacement vane pumps with adjustable pressure setting represent the most advanced type of flow generation units for engine lubrication. The advantages in a driving cycle of modulating the pump displacement to satisfy the engine requirements are by now unquestionable .\n\nThe simplest type of displacement control consists in exploiting the oil pressure to generate on the pump stator a hydraulic force that is counterbalanced by a spring. The displacement reduction can be obtained through either a linear or a rotational motion of the stator . Regardless of the kinematic mechanism, a drawback of this solution is the effect of the pressure distribution on the stator track, which leads in some operating conditions to an excessive and undesired reduction of the pump displacement . An effective solution is the use of a piloted control through a small hydraulic valve . With this method, the pressure setting of the displacement control is practically insensitive to the internal forces . A further improvement leading to a considerable fuel saving is constituted by the adoption of an active control of the circuit pressure. Beside a continuous modulation, as the solution presented by Burke et al. , a simpler but still effective method is the two-level approach, as the system analyzed in the reference . It involves the use of a two-position electro-valve controlled by the engine ECU, with the aim of enabling or disabling the pilot stage in order to change the pressure setting of the displacement control. The correct design of such systems requires as input the knowledge of the force acting on the stator due to the pressure in all variable volume chambers.\n\nThe first comprehensive lumped parameter model of a variable displacement vane pump for engine lubrication with linear motion of the stator was presented in the early 2000s . Other authors developed models with a pivoting stator ring. Cantore et al. built a model of a pump for high-speed engines using the built-in libraries of the simulation environment LMS Amesim. Barbarelli et al. developed a model in MATLAB Simulink[R] validated experimentally in terms of pressure history in a chamber. The circumferential pressure distribution on the stator track was also measured by Bianchini et al. . A detailed mathematical model of a pump with pivoting stator ring was presented by Geist and Resh as well as by Truong et al. . A 1D approach was instead used by Harrison et al. with the software GT-SUITE[R].\n\nThe very first CFD model of a vane pump was developed by Jiang and Perng , later the specialized software PumpLinx[R] was developed and used by various researchers. Wang et al. studied a pump with a pivoting stator and a direct acting displacement control. An in-depth analysis on the internal forces generated by the pressure distribution on the stator track was carried out by Sullivan and Sehmby . Later Frosina et al. simulated a vane pump provided with a piloted displacement control. The same research group, together with other scholars, developed also a 3D model of a radially balanced vane pump for power split transmissions .\n\nAll these studies have focused on the modelling of the pump itself, but a little attention has been paid to the influence of the air release/dissolution processes on the pump performance.\n\nIn the early fifties Schweitzer and Szebehely demonstrated that the air dynamics can be described by a first order function. Over the years, this concept has been implemented in fluid models where the transport and the dynamics of the gaseous phase are taken into account. For instance, time constants for air release and re-solution were used by Jiang et al. for the simulation of the pressure wave propagation in a pipeline and by Borghi et al. for studying the meshing zone of an external gear pump. Zhou et al. developed a customized fluid model applied to the same machine type, in which the time derivative of the separated air is also function of the pressure relative to the saturation value. Such a model has been recently updated with a better mass conservation of the different phases and applied to the simulation of a gerotor pump .\n\nIn CFD models the non-homogeneous distribution of the air/vapour content in the variable chambers can be determined. For instance, del Campo et al. studied the air fraction evolution in an external gear pump using a 2D model developed in ANSYS Fluent[R]. The software PumpLinx implements the Full Cavitation Model developed by Singhal et al. . However 2D and 3D models are not suitable for system-level simulations due to the unacceptable computational times.\n\nIn this paper, a detailed lumped parameter model of a lubricating vane pump with air dynamics is presented. The model, implemented in the LMS Amesim environment, has been validated in conditions of high and low aeration levels. Aim of the study is to demonstrate how it is possible to obtain a reliable fast running model able to work in very critical operating conditions characterized by extremely high free air content. Moreover, the influence of the gas fraction on both steady-state and dynamic performances has been analyzed. Finally, the influence of the time constants for the air release/dissolution processes has been evaluated.\n\nComponent Description\n\nThe pump used as reference is a 19 cc/rev vane type unit provided with a two-level pressure control of the displacement. A view of the pump with the indication of the ports is shown in Figure 1, while in Figure 2 the hydraulic scheme of the pump control is reported.\n\nThe displacement is varied through the linear movement of the stator ring, which can slide with respect to the housing, in order to modify the eccentricity with the shaft. Its prismatic guides work also as linear actuators with different active surfaces. The delivery pressure acts always on the actuator with smaller surface \"a\" generating a force that tends to decrease the pump displacement. On the opposite side, the hydraulic force generated by a larger actuator with area \"A\" and the force exerted by a spring tend to maintain the maximum displacement.\n\nThe pressure [p.sub.x] acting on the surface A is modulated by a small pressure relief valve RV with a remote pilot. On the engine, the pick-up point of the pilot line is located directly in the main gallery of the lubricating circuit, namely downstream from the oil filter.\n\nWhen the electro-valve EV is closed (configuration OFF), as long as the pressure in the circuit remains lower than the pressure setting of the relief valve [p.sub.2]*, the delivery pressure acts on both actuators, but thanks to the larger surface A, the pump is maintained at the maximum displacement. While if the circuit pressure reaches the value [p.sub.2]*, a small control flow generates a pressure drop in the calibrated orifice R, allowing the reduction of the modulated pressure [p.sub.x] and the movement of the stator. Once the stator and the valve spool are in hydraulic equilibrium, the pressure in the circuit is maintained ideally equal to [p.sub.2]*. This configuration is suitable in severe operating conditions, for instance with high oil temperature and high load, in order to guarantee a significant flow rate for the journal bearings and for the piston cooling jets even when the circuit resistance is low.\n\nOn the contrary the electro-valve EV, when energized (configuration ON), allows venting the chamber with surface A. In this case [p.sub.x] ideally coincides with the atmospheric pressure and the only force that is able to maintain the pump at maximum displacement is generated by the spring with equivalent pressure setting [p.sub.1]* < [p.sub.2]*. Hence, when the valve is activated, the reduction of the pump displacement occurs when the circuit pressure achieves the lower setting [p.sub.1]*. This configuration allows a power saving in non-critical operating conditions.\n\nPump Model\n\nPressure and Flow Rate Evaluation\n\nIn a lumped parameter model, the pump is divided in control volumes to which the mass conservation in isothermal conditions is applied:\n\n[dp/dt]=[B/V]([summation][Q.sub.i]-[omega][dV/d[phi]]) Eq. (1)\n\nbeing p the pressure, V the volume, [beta] the fluid bulk modulus, [Q.sub.i] the flow rate through the generic port i-th (positive if ingoing), [omega] and [phi] respectively the angular speed and the position of the shaft. In this specific case, the control volumes are associated to the seven variable chambers and to a number of fixed capacities representing the volume of the pipes. Different methods can be used for the evaluation of the angular derivative of the volumes; however, for the vane pumps the vector rays approach is particularly suitable, since it allows obtaining the exact analytic expression. The volume of the chambers is obtained by numerical integration of the derivative. The control volumes are connected by hydraulic resistances, representing the flow areas S in the port plate and the leakage passageways. For the former the Equation 2 for flow in turbulent regime condition is used:\n\nQ=[C.sub.d]S[square root of [2[DELTA]P/[rho]]] Eq. (2)\n\nwhere [DELTA]p is the pressure drop, [rho] the fluid density and [C.sub.d] the discharge coefficient.\n\nAll geometrical quantities, such as volumes, their derivatives and flow areas are function of both the shaft angle and the current eccentricity. With reference to Figure 3, the main flow area is constituted by the frontal surface given by the overlap between the chamber contour and the port plate profile. Such an area is calculated automatically using a procedure that reads the coordinates of the port profile from a file built directly from the CAD drawing by means of an AutoLISP routine . The area is calculated for different positions of the shaft with a step of 0.5 degrees and for five positions of the stator, from zero to maximum eccentricity. The data are stored in a 2D matrix that is interpolated linearly during the simulation.\n\nAdditional contributions to the flow area, not function of the eccentricity, are given by the lateral passages obtained by reducing the axial height of the stator on both sides and, for the delivery side only, by the groove on the pump casing. These flow areas have been calculated manually and added to the frontal ones. In Figure 4 the geometric quantities necessary for the implementation of the Equations 1 and 2 are shown for the maximum eccentricity.\n\nLeakages Evaluation\n\nAll possible leakage paths in the rotary assembly are shown in Figure 5. The quantities defining the geometry of the gaps are calculated analytically as function of the main pump parameters.\n\nEach variable chamber is connected to the volume beneath the vanes through the axial clearance between the rotor and the cover (leakage [Q.sub.11]) and through the clearance between the vanes and the rotor slots (leakage [Q.sub.12]). In this last case, it is assumed that the vanes are in contact against the rotor on the trailing side, as represented in Figure 5. The flow rate is calculated with the Equation 3 valid for laminar flow through a constant-height rectangular gap:\n\nQ=[[b.sub.c][h.sup.3.sub.c]/12[mu][l.sub.c]][DELTA]P Eq. (3)\n\nbeing [b.sub.c], [h.sub.c] and [l.sub.c] respectively the width, height and length of the gap and [mu] the fluid dynamic viscosity.\n\nEach variable chamber communicates with the adjacent ones through the axial gap between the cover and the vane (leakage [Q.sub.13]). In this case, the width of the gap is variable, being coincident with the vane lift h. The volume underneath the vanes is connected to the sump through the clearance of the shaft bearing (leakage [Q.sub.14]). The flow rate is calculated with the Equation 4 valid for laminar flow in an annulus between a shaft and a cylinder :\n\n[Q.sub.14] =[[pi][R.sub.sh][c.sup.3]/6[mu][l.sub.c]](1 + 1.5[[epsilon].sup.2])[P.sub.r] Eq. (4)\n\nwhere c is the radial clearance and [epsilon] the eccentricity ratio of the bearing.\n\nThe pressure [p.sub.r] in the volume underneath the vanes is calculated with the Equation 1 and it is function of the leakages [Q.sub.l1], [Q.sub.l2] and [Q.sub.l4]. Its value, intermediate between the delivery and the suction pressure, influences the radial equilibrium of the vanes with a consequence on the leakage on the tip [Q.sub.l5]. In Figure 6, the main geometric quantities of a vane in a generic angular position [phi] are shown; [O.sub.1] and [O.sub.2] are respectively the center of the rotor and of the stator. At a generic real vane lift h, the current minimum distance between the vane tip and the stator track is:\n\n[h.sub.min]=([h.sub.th]-h)cos[delta] Eq. (5)\n\nwhere [h.sub.th] is the theoretical vane lift that is calculated from the kinematics assuming the contact between the vane tip and the stator track and the angle [delta] is:\n\n[delta]=arcsin([esin [phi]/[R.sub.s]-[r.sub.v]]) Eq.(6)\n\nThe current eccentricity e is calculated by the equilibrium of the stator. The gap height [h.sub.x] in correspondence of the generic point K belonging to the vane tip and identified by the angle [[delta].sub.x] is:\n\n[mathematical expression not reproducible] Eq. (7)\n\nIn a reference system integral with the vane, the stator slides with a velocity [v.sub.0]. The abscissa x, with origin in correspondence of the minimum height [h.sub.min], identifies a generic cross section of the gap. It is defined as:\n\n[mathematical expression not reproducible] Eq. (8)\n\nThe flow rate per unit width crossing the orifice in a generic section with height [h.sub.x] is calculated with the more general equation for laminar flow through variable height gaps with sliding surfaces:\n\n[Q.sub.l]=[[v.sub.0][h.sub.x]/2]-[[h.sup.3.sub.x]/12[mu]]*[dp/dx] Eq. (9)\n\nThe flow rate does not depend on the coordinate x, therefore the pressure gradient can be written as function of a constant [h.sub.0] as follows:\n\n[dp/dx]=[6[mu][v.sub.0]/[h.sup.2.sub.x]](1-[[h.sub.0]/[h.sub.x]]) Eq. (10)\n\nThe equation must be integrated between -[x.sub.min] where the pressure is [p.sub.j] and [x.sub.max] where the pressure is [p.sub.j-1], in order to determine the constant [h.sub.0]:\n\n[h.sub.0] = [[I.sub.1]/[I.sub.2]-[[DELTA]P/6[I.sub.2][mu][v.sub.0]] Eq. (11)\n\nbeing [I.sub.1] and [I.sub.2] function of the gap profile. Such constants are determined by numerical integration and depend on the current lift h:\n\n[mathematical expression not reproducible] Eq. (12)\n\nFinally, the Equation 9 can be rewritten as:\n\nQ = [[h.sub.0][v.sub.0]/2]=[[v.sub.0]/2]*[[I.sub.1]/[I.sub.2]] Eq. (13)\n\nSince for high values of the gap [h.sub.min] the flow regime can be turbulent and in this case the Equation 13 overestimates the flow rate, the leakage [Q.sub.l5] is calculated with the Equation 14:\n\n[mathematical expression not reproducible] Eq. (14)\n\nwhere Ci is the discharge coefficient that would be obtained with the Equation 2 with the flow rate calculated with the Equation 13:\n\nC =-Q- i (r) Eq. (15)\n\nFinally, the leakages between the inlet and delivery volumes and the spring chamber where the modulated pressure px acts are also considered using equivalent rectangular gaps with height equal to the axial clearance between the stator and the casing.\n\nStator Equilibrium\n\nFor the evaluation of the internal forces on the stator, three contributions are considered (Figure 7):\n\n* the force Fp exerted directly by the pressure in each variable chamber,\n\n* the reaction forces of the vanes FRa against the stator track,\n\n* the centrifugal force of the oil in the chambers.\n\nThe pressure force [F.sub.p] is calculated starting from the length of the segment 12 connecting the theoretical contact points between two consecutive vanes and the stator. The force is applied in the midpoint of the segment with direction identified by the angle [xi] calculated as function of the shaft angle. The resultant along the x-axis of all forces contributes to the equilibrium of the stator together with the forces of the actuators and of the spring.\n\nFor the calculation of the reaction force [F.sub.Ra] the equation of equilibrium is applied to determine the radial position of the vane. In case of detachment from the stator track, the tip clearance and the corresponding leakage flow is evaluated. With reference to Figure 8, the axial and tangential force on the vane due to the pressure are respectively:\n\n[mathematical expression not reproducible] Eq. (16)\n\nwhere H is the axial height of the rotary assembly. While, called [m.sub.v] the mass of the vane, the centrifugal force is expressed as:\n\n[F.sub.c] = [m.sub.v][[omega].sup.2] ([R.sub.r]+h-[[l.sub.v]/2] Eq. (17)\n\nThe forces acting on the vane perpendicularly to its axis are:\n\n* the force [F.sub.pt] due to the pressure difference between the two chambers,\n\n* the reaction forces [F.sub.R1] and [F.sub.R2],\n\n* the friction force on the vane tip [F.sub.ft]\n\nThe reaction force at the lower radius can be evaluated from the equilibrium of the moments with respect to the point where the force [F.sub.R2] is applied (the contribution of the axial forces is neglected):\n\n[F.sub.R1]=[h/[l.sub.v]-h]*([[F.sub.pt]/2]+[F.sub.ft])Eq. (18)\n\nThe friction forces are evaluated in a simplified way as: [mathematical expression not reproducible] Eq. (19)\n\nwhere [f.sub.p] and [f.sub.v] are respectively the pressure and velocity dependent friction coefficients.\n\nThe reaction force at the outer radius can be obtained from the equilibrium of the vane in the direction perpendicular to the axis:\n\n[F.sub.R2] = [F.sub.R1] + [F.sub.pt] + [F.sub.ft] Eq. (20)\n\nThe contact force [F.sub.Ra] is:\n\n[mathematical expression not reproducible] Eq. (21)\n\nThe calculation of the reaction force [F.sub.Ra] does not take into account the thrust exerted between the roots of the vanes through the floating rings. This simplification can be considered reasonable by considering that the force [F.sub.Ra] is normally one order of magnitude smaller than the pressure force [F.sub.P] For the evaluation of the friction on the vane tip, [f.sub.p] = 0.1 was used; it is coherent with the values available in literature [31, 32].\n\nIt must be noted that, since the pressure [p.sub.r] is lower than the delivery pressure, the centrifugal force is not able to keep the vane in contact with the stator track when the speed is low and the vane is located on the delivery side. Nevertheless such a condition is not critical if the vane separates two chambers both connected to the delivery volume. However, depending on the specific geometry of the port plate and on the operating conditions (speed, oil temperature, displacement), if a pressure peak occurred in the chamber during the transition from the suction to the delivery side, the vane would detach with a detrimental effect on the volumetric efficiency, but it is not the case.\n\nFluid Model\n\nA hydraulic fluid always includes some gas, mostly air, which is either dissolved or separated (undissolved). Depending on the system's evolution, air changes from one form to the other one. Indeed, when the pressure increases in the hydraulic circuit, dissolution phenomenon occurs and gas bubbles (previously existing in equilibrium with the saturated liquid) tends to collapse and to dissolve in the liquid. Conversely if the pressure decreases, aeration phenomenon occurs, thus gas bubbles tends to be released from the liquid and to grow. It is well established from literature that only the separated air modifies the fluid properties, as depicted in Figure 9 with the LMS Amesim fluid model.\n\nThe Henry-Dalton law gives the volumetric amount of air that is dissolved in the liquid (at reference temperature) as linear function of the pressure level, the slope of the curve being the well-known Bunsen coefficient. Moreover, it assumes that aeration and dissolution are instantaneous phenomena; this implicitly means that the amount of dissolved air is always in equilibrium within the liquid. However, in fast acting systems, such as hydraulic pumps, the high pressure derivative does not let enough time to the air for reaching equilibrium conditions; consequently the amount of separated air is not only barotropic but also time-dependent.\n\nThe thermal-hydraulic model of fluid in LMS Amesim offers different options to take into account aeration phenomenon, depending on the simulation purposes. Assumptions and conservation equations corresponding to each aeration model are described more in details in the reference .\n\nWith the most advanced model, suitable for full dynamics analysis, the total and undissolved air amounts are independent variables computed from corresponding mass fractions conservation laws applied to each volume, including hydraulic lines. The evolution of undissolved gas mass fraction [x.sub.u] takes into account dynamics for aeration or dissolution considering a first order lag characterized by a user-defined time constant [tau], as from Siemens PLM :\n\n[[dx.sub.u]/dt] = [1/V*[rho]]([summation]d[mx.sub.u]-[x.sub.u]*[summation]dm)+[[x.sub.u,eq]-[x.sub.u]/[tau]] Eq. (22)\n\nTime constants [tau] can be differently set for dissolution and aeration:\n\n* if the current undissolved fraction is greater than the equilibrium value given by Henry's law ([x.sub.u] > [x.sub.u,eq])qi, the gas progressively dissolves within the liquid to reach the equilibrium state;\n\n* on the contrary, if the current undissolved fraction is lower than the equilibrium value ([x.sub.u] < [x.sub.u,eq]), the gas is released in the liquid in form of bubbles.\n\nAs alternative, different user-defined functions can be also implemented for the aeration and dissolution rate, however it was found that the second term of the Equation 22 gives satisfactory results, as shown later.\n\nFor most gases and liquids, aeration is a much faster phenomenon than dissolution and consequently the time constant for dissolution is much greater than for aeration. The saturation line (Figure 10) gives the pressure level required to completely dissolve the total air content into the liquid; its slope [alpha] corresponds to the Bunsen coefficient (for mineral oil [alpha] = 0.09).\n\nThe explicit relationship (23) between the saturation pressure [p.sub.sat] and the current amount of total gas mass fraction [x.sub.g] can be obtained considering the following:\n\n* the Henry-Dalton law, defining the partial pressure of a gaseous phase above the solution as function of the molar fraction,\n\n* the Bunsen coefficient, defining the percent of volume of the gas dissolved in a unit volume of the fluid at normal conditions,\n\n* and finally, the assumptions of ideal gas and negligible number of moles of dissolved gas with respect to the liquid:\n\n[P.sub.sat]=[[[rho].sub.1]R[T.sub.0]/[alpha]]*[[x.sub.g]/[M.sub.air]] Eq. (23)\n\nAs example, an ideal compression and decompression step is applied to a closed volume, such as a trapped chamber within a hydrostatic pump, with the aim of inducing dissolution and aeration events. The total air mass fraction within the volume is constant, 230 mg/kg, with a corresponding saturation pressure [p.sub.sat] = 1.77 bar absolute to completely dissolve the air into the liquid.\n\nFigure 11 depicts simulation results corresponding to a compression/decompression step crossing the saturation pressure and obtained considering two different time constants: 1 ms for aeration and 5 s for dissolution. Simulation highlights the occurrence of a transient pressure peak corresponding to the air dissolution into the liquid, from the initial equilibrium conditions (1 barA, 80 [degrees]C) to new conditions following the compression step. The amount of undissolved air at equilibrium depends on the instantaneous pressure level with respect to the saturation pressure (Figure 10) and vaporization pressure. Due to the dissolution in progress along the time, the volume occupied by the liquid is slightly increasing and consequently the pressure smoothly reduces depending on the set time constant.\n\nIn a similar way, aeration is induced by an expansion step occurring at 11 sec; consequently, the air previously dissolved in the liquid in equilibrium (1.3 barA, 80 [degrees]C) starts releasing till reaching new equilibrium conditions (corresponding to the initial state). Due to the small time constant set for aeration, the transient conditions are much faster than in the case of dissolution.\n\nTest Rig\n\nThe experimental measurements were performed on a test bench specific for lubricating pumps at Pierburg Pump Technology - Livorno plant. The pump under test PU1 was driven by a variable speed motor M1 through a sprocket-chain system. The hydraulic test circuit was intentionally simplified with respect to the standard layout in order to have a more accurate comparison with the simulation of the pressure ripple. The delivery line was constituted by a straight rigid pipe, whose qualitative cross section is shown in Figure 12. The load was simulated by a calibrated circular orifice with a diameter of 6.1 mm representing the resistance of the lubricating circuit. An additional restrictor, with diameter 8.5 mm, was used to simulate the pressure drop generated by the fluid conditioning system, namely the filter and the heat exchanger of the engine. The pilot port of the displacement control was connected to the volume between the two restrictors by means of a rigid copper pipe with internal diameter of 5 mm and length 0.3 m. Two piezo-resistive pressure transducers KISTLER with measuring range 0-50 bar and accuracy 0.1 bar sensed the delivery and the pilot pressure.\n\nA photo of the pump mounted on the test rig is shown in Figure 13. The pump sucked the oil from a 40 L reservoir. The fluid conditioning system was constituted by electric elements, immersed directly in the reservoir as shown in Figure 13, and a chiller for cooling the oil through an auxiliary pump PU3. Both heater and cooler were controlled in closed-loop in order to maintain the desired temperature with an accuracy of [+ or -]2[degrees]C. The feedback signal was the output of a temperature detector in the reservoir. The main properties of the working fluid with SAE grade 5W20 are listed in Table 1.\n\nA gerotor pump PU2, with the suction port partially connected to the atmosphere, was used as source of air flow rate. The delivery port was connected through a hose on the bottom of the reservoir at a distance of about 40 cm from the suction pipe of the pump under test.\n\nThe amount of separated air in the reservoir was measured by means of an optically based method through the commercial instrument Air-X from the company Delta Services Industriels . The principle is the density measurement based on the absorption capacity of the X-rays by the oil . The fluid sample is continuously taken from the reservoir by means of an internal pump and sent to the measuring chamber where the X-rays are emitted. On the opposite side of the chamber, a detector measures the amount of radiation that is increased by a higher fraction of air bubbles. A probe is used to compensate the variation of the pressure and of the temperature. The measuring range is from 0% to 100% of gas content. A self-calibration of the instrument was performed on the specific oil used for the test. Based on the acquisition time used for the tests, the accuracy in terms of volume gas content is of the order of 0.5%. The desired amount of separated air fraction was controlled manually through the speed of the motor M2.\n\nComplete Simulation Model\n\nThe complete model of the pump and of the test rig is reported in Figure 14. The main components of the rotating assembly of the pump have been developed at the Politecnico di Torino with the Ameset tool in C code and they comply with the Thermal Hydraulic Library standard of LMS Amesim. The customized components are the following:\n\n* 1: implements the Equation 1 for all variable chambers,\n\n* 2: calculate inlet and outlet flow rates with the Equation 2,\n\n* 3: calculates the inter-chambers leakages [Q.sub.l2], [Q.sub.l3] and [Q.sub.l5],\n\n* 4: calculates the axial leakages [Q.sub.l1] and [Q.sub.l4],\n\n* 5: represent a multiple hydraulic junction,\n\n* 6: simulates the stator dynamics,\n\n* 7: calculates the volumes and derivatives of the chambers,\n\n* 8: calculates the frontal flow areas,\n\n* 9: interpolate the lateral surfaces and the groove flow area\n\n* 10: supplies the geometric parameters.\n\nIn the components 3 and 4 the gas transport is not implemented, i.e. the air crosses the variable chambers only through the inlet/outlet flow areas. The remaining elements of the pump are simulated by models belonging to the standard libraries. The leakages between the stator and the casing are simulated with the laminar gaps 11.\n\nThe delivery volume is simulated by a model of a hydraulic pipe 12 in which the resistive, capacitive and inertia effects are lumped in two elements with the same total volume of the real capacity. An identical model is used for the pipe 13 with length 130 mm (see Figure 12) and for the pipe 14 with length 90 mm. The former is divided in 4 subvolumes, the latter in two pipes with one element each. Hence, the entire test circuit is discretized with eight elements with a mean length of about 35 mm. It was found that the use of a higher number of subdivisions does not give significant improvements. The discharge coefficients of the restrictors have been left to the default value of 0.7. The pilot line 15 is simulated by a pipe with five internal nodes and frequency dependent friction.\n\nThe simulations were performed in isothermal conditions, hence the equations of energy balance for evaluation of the fluid temperature have been deactivated. The typical computational time is about 10 s per shaft revolution on a single core of an eight-core Xeon HT processor at 3.4 GHz.\n\nResults and Discussion\n\nModel Validation\n\nThe importance of the leakage models for the correct determination of the pump flow rate has been discussed in a previous study performed on a similar unit. Nevertheless, it must be observed that an accurate evaluation of the leakages is crucial also in this paper, since the real flow rate influences the mean value of the pressure induced by the fixed restrictors shown in the scheme of Figure 12. Moreover, the mean pressure is influenced also by the current position of the stator that depends on the pressure history in the chambers and on the resultant of the forces exerted by the vanes. In turn, the former depends also on the flow areas, the latter is influenced by the detachment of some vanes. Although this paper is focused mainly on the influence of the oil aeration on the pump behavior, it is evident how all aspects of the pump itself must be taken into account in a very detailed way.\n\nThe experimental tests were executed at 80 [degrees]C, three angular speeds, two different configurations of the electro-valve, namely ON and OFF, and two different aeration levels, for a total of12 operating conditions. In particular, the configuration with high percentage of gas, indicated simply as \"with air\", was obtained by insufflating air into the reservoir in order to obtain a mixture with about 7% of free air in volume. The configuration with low aeration, simply identified as \"no air\", was obtained with the gerotor pump turned off and in this case the reading of the Air-X instrument was 0.1%. As far as the dissolved air fraction is concerned, it is assumed that the oil was always saturated with 9% of air, leading to a total gas fraction of 16% in the case \"with air\".\n\nThe list of the 12 configurations is reported in Table 2. In each configuration, once reached the steady-state conditions in terms of temperature and reading of the Air-X instrument, the pressure signals were acquired.\n\nIn the simulation model, to achieve the steady-state conditions of the undissolved gas mass fraction in the entire delivery line, at least 20-25 revolutions of the shaft are required. In fact it is necessary to wait that the air is transported from the reservoir up to the end of the delivery pipe. However, thanks to the hyper-threading feature of the processor (it is equivalent to have 16 logical cores) and to the batch run facility of the software, all configurations in Table 2 can be simulated simultaneously, each run on a different core, in a total time of about 5 min.\n\nA sensitivity analysis was carried out for finding the values of the time constants for aeration and dissolution that allowed a good agreement with the experimental data. Best results can be obtained with [10.sup.-3] s for the aeration process and a time greater than 1 s for the dissolution for all configurations; in particular the presented results have been obtained with 5 s. The use of a high time constant for the dissolution process in hydraulic machines is in agreement with the results of Kim and Murrenhoff . It is worth to notice that even better results could be obtained using different time constants for each configuration. However, it was preferred to find the values of better compromise for all configurations.\n\nIn Figure 15, the comparison between the simulated delivery pressure as function of the speed and the experimental points measured by the transducer P1 is reported. The simulated continuous lines have been obtained as a sequence of steady-state points with a speed increment of 100 rpm.\n\nAt 1474 rpm the pump works always at maximum displacement, hence the mean pressure value is imposed by the load, while at higher speed is decided by the displacement control.\n\nWith the piloted control (valve OFF) the displacement is reduced starting from 2400 rpm and such a value is not influenced by the presence of separated air. A maximum variation of about 0.4 bar is observed in the simulated regulated pressure between the tests with and without air. The same difference is also visible in the experimental data. On the contrary, with the direct acting control (valve ON) the maximum difference of the regulated pressure is greater than 1 bar and the same trend is observed experimentally. The tests bring to evidence the higher sensitivity to the operating conditions of the direct acting control with respect to the piloted one, as described more in details in the next section.\n\nThe delivery pressure ripple for six different configurations is shown from Figures 16 to 21. For sake of completeness, it must be highlighted that the relative positioning along the x-axis of the simulated and experimental curves has been made in order to have the coincidence of the main peaks, since a measurement of the absolute position of the shaft was not possible.\n\nIn order to evaluate the influence of the air content uncertainty, the simulations were also performed in different conditions. For instance, if it is assumed that the amount of dissolved air is 6%, i.e. the oil is not saturated, on average the mean value of the simulated delivery pressure in the tests \"without air\" increases of about 0.1 bar, with the worst case in the configuration 12 where the increment is 0.2 bar. The mean increment of the pressure amplitude is of the order of 0.15 bar. The configurations \"with air\" are less sensitive, since both amplitude and mean value are incremented by about 0.05 bar.\n\nAnother simulation considered a higher fraction of separated gas in the test \"without air\" to take into account the measurement error of the air-X instrument. If a percentage of 0.5% of free air is considered, instead of 0.1%, on average the mean simulated pressure decreases by 0.1 bar and the amplitude by 0.2 bar.\n\nIt is worth to mention that generally the amount of air in the oil has also a slight influence on the suction capability of the pump . In this case however, the phenomenon of the incomplete chambers filling never occurred, since the tests at high speed were performed at low displacement.\n\nControl Stability Issues\n\nThe pump, its control and the circuit constitute a feedback system that can potentially feature instabilities. A robust design needs to take into account also the dynamic behavior, in order to ensure that the system could always work in the stable region. The developed model is suitable for bringing to evidence instabilities and can be used as design tool for optimizing the dynamic response. It is known that the stability margin of a remotely piloted valve is lower than in a layout with a direct pilot. In a previous study it was highlighted that the use of an external pilot line decreases drastically the cut-off frequency of the valve. Consequently, with a relatively low-frequency oscillating pressure signal, when the pressure at the inlet of the pilot pipe increases the valve closes instead of opening; vice versa when the pressure is decreasing. In the specific case of the displacement control, the consequence is an increment of the flow rate as the pressure rises, leading to an unstable behavior. The instability is also worsened by high values of the hydraulic capacitance V/[beta] of the delivery line, i.e. long pipes and high percentage of separated air.\n\nThe pump used for the current study was designed in order to work in the stable region when coupled with the lubricating circuit. However it is straightforward that an excessive increment of the delivery volume will lead inevitably to instability, above all when the effective fluid bulk modulus is very low.\n\nIn order to make this issue emerge, a transient simulation was performed with a delivery pipe between the two fixed restrictors 1.5 m long, instead of 90 mm. Starting from a speed of 1474 rpm, corresponding to maximum displacement, the velocity is incremented linearly up to 4474 rpm in 0.5 s. In Figure 22 the stator eccentricity and the speed are shown as function of the time. As expected, in the operating condition with high amount of separated air the instability of the stator is observed, with a high amplitude oscillation, about 1/3 of the entire stroke, and with low frequency, about 7 Hz. On the contrary, the pump remains stable in the configuration without initial separated air, unless the pipe is incremented up the unrealistic value of about 6 m.\n\nSince the instability is generated by the oscillation of the pilot valve, no oscillation of the stator was observed in the configuration ON of the electro-valve, even with high amount of air and very long pipes; in fact in this case the pilot valve remains always closed.\n\nInfluence of the Time Constants\n\nThe time constants for dissolution and aeration have a strong influence on the instantaneous amount of separated air, which in turn determines the value of the effective bulk modulus of the air/oil mixture. Based on the Equation 1, it is evident how the consequence of a low bulk modulus is the reduction of the pressure time derivative with an influence on the pressure oscillations. However, a not negligible consequence also on the steady-state behavior is observed in Figure 15, above all in the configuration ON. In this case in fact, at high speed a variation of more than 1 bar occurs between the tests with and without air. The reason can be found from the analysis of the pressure distribution on the stator track.\n\nIn the theoretical case of ideal timing and ideal fluid, the pressure in the variable chambers rises instantaneously up to the delivery value as the chamber reaches the maximum volume and the connection with the outlet port opens. Similarly, the pressure decreases up to the suction value when the chamber is at minimum volume and it connects to the inlet port. With reference to Figure 23a, in this ideal situation the resultant force of the pressure distribution is directed vertically. In the real case, for optimization reasons, it is common practice to delay the connection with the outlet port.\n\nDepending on the fluid conditions in the chamber that has just left the inlet side, two different cases must be considered. If all air is dissolved in the oil when the compression phase begins, a pressure peak occurs, leading to a component of the mean resultant force that tends to help the larger actuator to maintain the maximum displacement (Figure 23b). On the contrary, an amount of free air still present in the chamber prevents the pressure peak and delays its pressurization; hence, the mean resultant force tends to decrease the displacement (Figure 23c). The consequence when the pump works with the direct acting control (valve ON) is that, in the former case the pressure setting will be higher than the theoretical value [p.sub.1]*, while in the latter will be lower. For this reason, in the test with air the regulated pressure is significantly lower in comparison to the test without air. Instead when the piloted control is active (valve OFF), the displacement reduction is decided by the intervention of the valve RV. Even in this configuration, a small influence of the air content is observed, because of different working positions of the valve spool. In fact in presence of a negative component along the x-axis of the internal force, the stator equilibrium is reached with a higher value of the modulated pressure [p.sub.x] that implies a smaller opening of the valve, a smaller compression of its spring and finally a lower delivery pressure necessary to maintain in hydraulic equilibrium the spool. In this context, the time constants for aeration and dissolution play a fundamental role, but only within a specific range and in some operating conditions.\n\nIn Figure 24, the mean value and the amplitude of the simulated delivery pressure are plotted as function of the dissolution time constant at 5000 rpm and with aeration constant equal to [10.sup.-3] s.\n\nIn the configuration <<no air>>, if the dissolution constant is lower than [10.sup.-3] s, the air separated in the inlet volume has time enough to become again dissolved before the opening of the outlet port; in this case, a pressure peak is generated leading to the situation of Figure 23b. As the time constant increases, the completion of the dissolution process is delayed, with a progressive reduction of the pressure peak.\n\nFor values higher than about [10.sup.-2]-[10.sup.-1] s, the connection opens before the formation of the peak. The same trend is observed for the pressure amplitude; in this case, the increment of the time constant increases the amount of free air in the delivery line, leading to a dampening of the pressure ripple. In the configuration \"with air\", the dissolution time constant has no effect on the mean delivery pressure. In fact, the amount of separated air coming directly from the reservoir prevents the peak in the chamber regardless of the additional fraction of air released by the reduction of the pressure in the suction volume.\n\nAs far as the pressure amplitude is concerned, a very small time constant leads to the complete dissolution of the air in the oil in the delivery line. The consequence is that the back flow from the delivery volume due to the incomplete filling of the chamber generates high-pressure oscillations, being the equivalent bulk modulus coincident with the value of the pure oil (very stiff system). On the contrary, the bulk modulus is decreased by the presence of free air in the delivery line as the time constant increases.\n\nIn Figure 25, the analysis is repeated as function of the aeration time constant with constant dissolution time equal to 5 s.\n\nIn the configuration \"no air\", if the time constant is high there is no time enough to have the separation of a significant amount of air. In such a case, the bulk modulus remains high, the x-component of the internal force is positive due to the generation of the peak in the chambers and the delivery line is stiff.\n\nFor values of the time constant lower than [10.sup.-1] s, a fraction of the air becomes separated in the low-pressure volumes and due to the high dissolution time constant it remains undissolved also in the high-pressure volumes. The consequence is the progressive reduction of the pressure peak and the dampening of the oscillations in the delivery line. Times constants lower than [10.sup.-3] s are enough to cause the separation of the entire amount of air.\n\nIn the configuration \"with air\", the additional amount of air that can be released in the suction side does not alter the equivalent bulk modulus, which is maintained low by the amount of air coming directly from the reservoir.\n\nConclusions\n\nA complete lumped parameter model of a variable displacement lubricating pump with the dynamics of the air release/dissolution processes has been presented. The model takes into account all leakage passageways and the forces acting on the stator ring. The validation on a simplified straight-pipe test circuit has allowed assessing the accuracy of the model. Thanks to the measurement of the amount of undissolved air, a crucial variable for the model tuning, normally unknown, has been eliminated. As far as the amount of dissolved air is concerned, the hypothesis of saturation condition has been adopted; however, the influence of a lower air percentage has been also assessed.\n\nThe implementation of the gas dynamics has brought to evidence the strong importance of the time constants; nevertheless, it must be highlighted that the operating parameters are influenced by the order of magnitude of the time constants and mostly within the range between [10.sup.-3] s and [10.sup.-1] s. The reason must be found in the time that the oil takes to cross the machine with the typical angular speeds of the hydraulic pumps. The assumptions made in the paper about the time for dissolution of the order of seconds and for aeration of the order of milliseconds, are in line with the limited data available in the open literature. As far as the pump behavior is concerned, the displacement control is more sensitive to separated air in the operating conditions of low-pressure setting. However, instability could occur when the pump works in the high-pressure (remote pilot) mode.\n\nBased on the outcomes of the validation procedure, the present model has been proven to be a reliable tool for the design of the pump and of its control. It is straightforward that the model can be further improved by means of a tuning through CFD simulations, in order to determine precise values for the discharge coefficients in the resistive components.\n\nNomenclature\n\na - Smaller surface of the stator ring\n\nA - Bigger surface of the stator ring\n\n[b.sub.c] - Gap width\n\n[b.sub.v] - Vane width\n\n[b.sub.v1], [b.sub.v2] - Vane width fractions on which the chamber pressure acts\n\nc - Journal bearing radial clearance\n\n[C.sub.d] - Discharge coefficient\n\n[C.sub.l] - Discharge coefficient for turbulent flow in vane tip gap\n\ndm - Gas mass flow rate\n\n[dmx.sub.u] - Undissolved gas mass flow rate\n\ne - Eccentricity between rotor and stator\n\n[f.sub.p] - Pressure dependent friction coefficient\n\n[f.sub.v] - Velocity dependent friction coefficient\n\n[F.sub.c] - Centrifugal force on the vane\n\n[F.sub.fa], [F.sub.ft] - Axial and tangential friction force vane-rotor\n\n[F.sub.i] - Resultant force on the stator track due to the pressure\n\n[F.sub.p] - Force on the stator due to the chamber pressure\n\n[F.sub.pa], [F.sub.pt] - Axial and tangential pressure forces on the vane\n\n[F.sub.Ra] - Reaction force vane-stator\n\n[F.sub.R1],[F.sub.R2] - Reaction forces vane-rotor\n\nh - Vane lift\n\n[h.sub.c] - Gap height\n\n[h.sub.0] - Auxiliary constant - see Equation 11\n\n[h.sub.min] - Minimum distance of the vane tip from the stator track\n\n[h.sub.th] - Theoretical vane lift (vane in contact)\n\n[h.sub.x] - Gap height at the generic coordinate x\n\nH - Axial height of the rotor/stator\n\n[I.sub.1], [I.sub.2] - Auxiliary constants - see Equation 12\n\n[l.sub.c] - Gap length\n\n[l.sub.v] - Vane length\n\n[m.sub.v] - Vane mass\n\n[m.sub.air] - Molecular mass of the air\n\np - Pressure (general)\n\n[p.sub.1]* - Lower pressure setting of the displacement control\n\n[p.sub.2]* - Higher pressure setting of the displacement control\n\n[p.sub.d] - Delivery pressure\n\n[p.sub.j] - Pressure in the i-th chamber\n\n[p.sub.r] - Pressure acting on the vane roots\n\n[p.sub.sat] - Saturation pressure of air in oil\n\n[p.sub.x] - Pressure modulated by the relief valve\n\nR - Universal gas constant\n\nQ - Volumetric oil flow rate (general)\n\n[Q.sub.i] - Volumetric flow rate entering the control volume\n\n[Q.sub.l1...5] - Oil leakages - see Figure 5\n\nS - Flow area\n\n[T.sub.0] - Reference fluid temperature\n\n[v.sub.0] - Sliding velocity vane tip - stator track\n\nV - Volume\n\nx - Coordinate along the vane tip gap\n\n[x.sub.g] - Total mass gas fraction\n\n[x.sub.u] - Undissolved gas mass fraction\n\n[x.sub.u,eq] - Equilibrium undissolved gas mass fraction\n\n[alpha] - Bunsen coefficient\n\n[beta] - Bulk modulus\n\n[delta] - Angle identifying the vane contact point\n\n[[delta].sub.x] - See Figure 6\n\n[DELTA]p - Pressure drop\n\n[epsilon] - Journal bearing eccentricity ratio\n\n[mu] - Dynamic viscosity\n\n[xi] - Angle identifying the direction of the force [F.sub.p]\n\np - Fluid density\n\n[p.sub.l] - Density of the liquid phase\n\n[tau] - Aeration/dissolution time constant\n\n[phi] - Shaft angle\n\n[omega] - Shaft angular velocity\n\nAcknowledgments\n\nThis study is in partial fulfillment of a research contract between and the Department of Energy of the Politecnico di Torino and Pierburg Pump Technology. Authors would like to thank Matteo Gasperini and Gaia Volandri for their contribution in providing to the Politecnico di Torino the data used for the simulation and for the model validation. Authors would like to acknowledge the LMS Amesim development team, headed by Olivier Delechelle at Siemens Industry Software Lyon, for sharing the information necessary for the implementation of the fluid model into the customized libraries developed at the Politecnico di Torino.\n\nReferences\n\n[1.] Rundo, M. and Nervegna, N., \"Lubrication Pumps for Internal Combustion Engines: A Review,\" Int. J. Fluid Power 16(2):59-74, 2015, doi:10.1080/14399776.2015.1050935.\n\n[2.] Manco, S., Nervegna, N., Rundo, M., and Armenio, G., \"Displacement vs Flow Control in IC Engines Lubricating Pumps,\" SAE Technical Paper 2004-01-1602, 2004, doi:10.4271/2004-01-1602.\n\n[3.] Staley, D., Pryor, B., and Gilgenbach, K., \"Adaptation of a Variable Displacement Vane Pump to Engine Lube Oil Applications,\" SAE Technical Paper 2007-01-1567, 2007, doi:10.4271/2007-01-1567.\n\n[4.] Rundo, M. and Nervegna, N., \"Geometry Assessment of Variable Displacement Vane Pumps,\" J. Dyn. Syst., Meas., Control 129(4):446-455, 2007, doi:10.1115/1.2718245.\n\n[5.] Rundo, M. and Squarcini, R., \"Experimental Procedure for Measuring the Energy Consumption of IC Engine Lubricating Pumps during a NEDC Driving Cycle,\" SAE Int. J. Engines 2(1):1690-1700, 2009, doi:10.4271/2009-01-1919.\n\n[6.] Rundo, M., \"Piloted Displacement Controls for ICE Lubricating Vane Pumps,\" SAE Int. J. Fuels Lubr. 2(2):176-184, 2010, doi:10.4271/2009-01-2758.\n\n[7.] Burke, R., Brace, C., Lewis, A., Stark, R. et al., \"On Engine Study of Thermal and Performance Impacts of a Variable Displacement Oil Pump,\" presented at ASME 2012 Internal Combustion Engine Division Spring Technical Conference, Torino, Italy, May 6-9, 2012, doi:10.1115/ICES2012-81044.\n\n[8.] Rundo, M. and Squarcini, R., \"Discrete Pressure Controls in IC Engine Lubricating Pumps,\" presented at ASME International Mechanical Engineering Congress & Exposition, Denver, CO, Nov. 11-17, 2011, doi:10.1115/IMECE2011-62525.\n\n[9.] Manco, S., Nervegna, N., Rundo, M., and Armenio, G., \"Modelling and Simulation of Variable Displacement Vane Pumps for IC Engine Lubrication,\" SAE Technical Paper 2004-01-1601, 2004, doi:10.4271/2004-01-1601.\n\n[10.] Cantore, G., Paltrinieri, F., Tosetti, F., and Milani, M., \"Lumped Parameters Numerical Simulation of a Variable Displacement Vane Pump for High Speed ICE Lubrication,\" SAE Technical Paper 2008-01-2445, 2008, doi:10.4271/2008-01-2445.\n\n[11.] Barbarelli, S., Bova, S., and Piccione, R., \"Zero-Dimensional Model and Pressure Data Analysis of a Variable-Displacement Lubricating Vane Pump,\" SAE Technical Paper 2009-01-1859, 2009, doi:10.4271/2009-01-1859.\n\n[12.] Bianchini, A., Ferrara, G., Ferrari, L., Paltrinieri, F. et al., \"Design and Optimization of a Variable Displacement Vane Pump for High Performance IC Engine Lubrication: Part 1 - Experimental Analysis of the Circumferential Pressure Distribution with Dynamic Pressure Sensors,\" SAE Technical Paper 2009-01-1045, 2009, doi:10.4271/2009-01-1045.\n\n[13.] Geist, B. and Resh, W., \"Dynamic Modeling of a Variable Displacement Vane Pump within an Engine Oil Circuit,\" presented at ASME 2011 Internal Combustion Engine Division Fall Technical Conference, Morgantown, WV, Oct. 2-5, 2011, doi:10.1115/ICEF2011-60039.\n\n[14.] Truong, D., Ahn, K., Trung, N., and Lee, J., \"Performance Analysis of a Variable-Displacement Vane-Type Oil Pump for Engine Lubrication Using a Complete Mathematical Model,\" Proc. IMechE Part D 227(10):1414-1430, 2013, doi:10.1177/0954407013491896.\n\n[15.] Harrison, J., Aihara, R., Eshraghi, M., and Dmitrieva, I., \"Modeling Engine Oil Variable Displacement Vane Pumps in 1D to Predict Performance, Pulsations, and Friction,\" SAE Technical Paper 2014-01-1086, 2014, doi:10.4271/2014-01-1086.\n\n[16.] Jiang, Y. and Perng, C., \"An Efficient 3D Transient Computational Model for Vane Oil Pump and Gerotor Oil Pump Simulations,\" SAE Technical Paper 970841, 1997, doi:10.4271/970841.\n\n[17.] Ding, H., Visser, F., Jiang, Y. and Furmanczyk, M., \"Demonstration and Validation of a 3D CFD Simulation Tool Predicting Pump Performance and Cavitation for Industrial Applications,\" presented at ASME Fluids Engineering Division Summer Meeting, Vail, CO, Aug. 2-6, 2009.\n\n[18.] Wang, D., Ding, H., Jiang, Y., and Xiang, X., \"Numerical Modeling of Vane Oil Pump with Variable Displacement,\" SAE Technical Paper 2012-01-0637, 2012, doi:10.4271/2012-01-0637.\n\n[19.] Sullivan, P. and Sehmby, M., \"Internal Force Analysis of a Variable Displacement Vane Pump,\" SAE Technical Paper 2012-01-0409, 2012, doi:10.4271/2012-01-0409.\n\n[20.] Frosina, E., Senatore, A., Buono, D., and Olivetti, M., \"A Tridimensional CFD Analysis of the Oil Pump of an High Performance Engine,\" SAE Technical Paper 2014-01-1712, 2014, doi:10.4271/2014-01-1712.\n\n[21.] Frosina, E., Senatore, A., Buono, D. et al., \"Vane Pump Power Split Transmission: Three dimensional computational fluid dynamics modelling,\" presented at ASME/BATH Symposium of Fluid Power and Motion Control, Chicago, IL, Oct. 12-14, 2015, doi:10.1115/FPMC2015-9518.\n\n[22.] Schweitzer, P.H. and Szebehely, V.G., \"Gas Evolution in Liquids and Cavitation,\" J. Appl. Phys. 21(12):1218-1224, 1950, doi:10.1063/1.1699579.\n\n[23.] Jiang, D., Li, S., Edge, K.A., and Zeng, W., \"Modeling and Simulation of Low Pressure Oil-Hydraulic Pipeline Transients,\" Computers & Fluids 67:79-86, 2012, doi:10.1016/j.compfluid.2012.07.005.\n\n[24.] Borghi, M., Milani, M., Paltrinieri, F. and Zardin, B., \"The Influence of Cavitation and Aeration on Gear Pumps and Motors Meshing Volumes Pressures\" presented at the ASME International Mechanical Engineering Congress and Exposition, Chicago, IL, Nov. 5-10, 2006, doi:10.1115/IMECE2006-13735.\n\n[25.] Zhou, J., Vacca, A., and Casoli, P., \"A Novel Approach for Predicting the Operation of External Gear Pumps under Cavitating Conditions,\" Simul. Model. Pract. Theory 45:35-49, 2014, doi:10.1016/j.simpat.2014.03.009.\n\n[26.] Shah, Y.G., Vacca, A., Dabiri, S. and Frosina, E., \"A Fast Lumped Parameter Approach for the Prediction of Both Aeration and Cavitation in Gerotor Pumps,\" Meccanica 53:175-191, 2018, doi: 10.1007/s11012-017-0725-y.\n\n[27.] del Campo, D., Castilla, R., Raush, G.A., Gamez Montero, P. J. et al., \"Numerical Analysis of External Gear Pumps Including Cavitation,\" J. Fluid. Eng. 134:081105, 2012, doi:10.1115/1.4007106.\n\n[28.] Singhal, A.K., Athavale, M.M., Li, H., and Jiang, Y., \"Mathematical Basis and Validation of the Full Cavitation Model,\" J. Fluid. Eng. 124:617-624, 2002, doi:10.1115/1.1486223.\n\n[29.] Rundo, M. and Squarcini, R., \"Modelling and Simulation of Brake Booster Vacuum Pumps,\" SAE Int. J. Commer. Veh. 6(1):236-248, 2013, doi:10.4271/2013-01-9016.\n\n[30.] Merritt, H.E., Hydraulic Control Systems, (New York, John Wiley & Sons, Inc., 1967), 34, ISBN:978-0-471-59617-2.\n\n[31.] Ivantysyn, J. and Ivantysynova, M., Hydrostatic Pumps and Motors, Principles, Designs, Performance, Modelling, Analysis, Control and Testing, (New Delhi, Akademia Books International, 2002), 355, ISBN:81-85522-16-2.\n\n[32.] Inaguma, Y. and Hibi, A., \"Reduction of Friction Torque in Vane Pump by Smoothing Cam Ring Surface,\" Proc. IMechE Part C 221:527-534, 2007, doi:10.1243/0954406jmes225.\n\n[33.] Furno, F. and Blind, V., \"Effects of Air Dissolution Dynamics on the Behaviour of Positive-Displacement Vane Pumps: A Simulation Approach,\" presented at 10th International Fluid Power Conference, Dresda, Germany, Mar. 8-10, 2016.\n\n[34.] Siemens PLM, \"LMS Imagine.Lab AMESim Thermal-Hydraulic Manual \"Variable Gas Content Modeling\",\" 2017.\n\n[35.] Delta Services Industriels, \"Lubricant Aeration,\" http://www.deltabeam.net/en/product/lubricant-aeration, accessed Feb. 2017.\n\n[36.] Schrank, K. and Murrenhoff, H., \"Investigation of Different Methods to Measure the Entrained Air Content in Hydraulic Oils,\" presented at ASME/BATH Symposium on Fluid Power & Motion Control, Bath, United Kingdom, Sept. 10-12, 2014, doi:10.1115/FPMC2014-7823.\n\n[37.] Rundo, M. and Altare, G., \"Lumped Parameter and Three-Dimensional CFD Simulation of a Variable Displacement Vane Pump for Engine Lubrication,\" presented at ASME Fluids Engineering Division Summer Meeting, Waikoloa, HI, July 30-Aug. 3, 2017, doi: 10.1115/FEDSM2017-69124.\n\n[38.] Kim, S. and Murrenhoff, H., \"Measurement of Effective Bulk Modulus for Hydraulic Oil at Low Pressure,\" J. Fluids Eng. 134(2):021201, 2012, doi:10.1115/1.4005672.\n\n[39.] Altare, G. and Rundo, M., \"Computational Fluid Dynamics Analysis of Gerotor Lubricating Pumps at High Speed: Geometric Features Influencing the Filling Capability,\" J. Fluids Eng. 138(11):111101, 2016, doi:10.1115/1.4033675.\n\n[40.] Rundo, M., \"On the Dynamics of Pressure Relief Valves with External Pilot for ICE Lubrication,\" presented at ASME International Mechanical Engineering Congress and Exposition, Montreal, Quebec, Canada, Nov. 14-20, 2014, doi:10.1115/IMECE2014-37973.\n\nMassimo Rundo, Politecnico di Torino\n\nRaffaele Squarcini, Pierburg Pump Technology\n\nFrancesca Furno, Siemens Industry Software\n\nHistory\n\nRevised: 21 Nov 2017\n\nAccepted: 27 Jan 2018\n\ne-Available: 18 Apr 2018\n\nKeywords\n\nVane pump, Lubricating pump, Air dynamics, LMS Amesim\n\nCitation\n\nRundo, M., Squarcini, R., and Furno, F., \"Modelling of a Variable Displacement Lubricating Pump with Air Dissolution Dynamics,\" SAE Int. J. Engines 11(2):2018, doi:10.4271/03-11-02-0008.\n```TABLE 1 Fluid properties at 80 [degrees]C.\n\nQuantity Value Unit\n\nDensity 812 kg/[m.sup.3]\nDynamic viscosity 10.2 cP\nBulk modulus 1320 MPa\n\nTABLE 2 Definition of the 12 configurations used in the tests.\n\nNo. Configuration No. Configuration\n\n1 1474 rpm - OFF - with air 7 1474 rpm - OFF - no air\n2 2468 rpm - OFF - with air 8 2468 rpm - OFF - no air\n3 4474 rpm - OFF - with air 9 4474 rpm - OFF - no air\n4 1474 rpm - ON - with air 10 1474 rpm - ON - no air\n5 2468 rpm - ON - with air 11 2468 rpm - ON - no air\n6 4474 rpm - ON - with air 12 4474 rpm - ON - no air\n```" ]

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[ "Why polyxpoly does not work?\n\n41 views (last 30 days)\nFaez Alkadi on 29 Jan 2018\nCommented: Faez Alkadi on 27 Apr 2018\nI have the data points for tow polylines as attached. Where A1 is X-data for polyline 1, B1 is Y-data for polyline 1 and,A2 is X-data for polyline 2 and B1 is Y-data for polyline 2. (as plotted)\nI tried to get the intersection point between the two polylines using polyxpoly. But the the result for [xi, yi] are empty !!!\nDoes anyone has an idea what would be the problem ? Thank you so much...\n[xi, yi] = polyxpoly(A1, B1,A2, B2);\nplot(A1,B1)\nhold on\nplot(A2,B2)\nhold on\nplot(xi, yi, 'bo')", null, "Matt J on 24 Apr 2018\n\nI used the function intersections by Douglas Schwarz for the same exact data and it worked perfect. But its a little slow!!!!\n\nI imagine speed performance of any tool would vastly improve once you get rid of the non-vertex points in your A,B data.\n\nMatt J on 24 Apr 2018\nEdited: Matt J on 24 Apr 2018\nI suspect the problem has to do with the fact that you are entering superfluous additional points, which are not actually vertices. A vertex technically should lie only at the end points of the polygon edges.\nRegardless, the attached version of polyxpoly by Bruno Luong is working fine for me, but has a different syntax.\n>> ab = polyxpoly([A1, B1].',[A2, B2].')\nab =\n-5.1172\n-5.5302\nFaez Alkadi on 27 Apr 2018\nThank you Matt\n\nSteven Lord on 24 Apr 2018\nIf you're using release R2017b or later, consider creating polyshape objects and using the intersect function on those objects.\nD = load('A1', 'A1'); A1 = D.A1;\nD = load('A2', 'A2'); A2 = D.A2;\nD = load('B2', 'B2'); B2 = D.B2;\nD = load('B1', 'B1'); B1 = D.B1;\n% Create two polyshape objects\nP1 = polyshape(A1, B1, 'Simplify', true);\nP2 = polyshape(A2, B2, 'Simplify', true);\n% Plot the two polyshape objects so later we can check that the intersection is correct\nh1 = plot([P1, P2]);\nax1 = ancestor(h1(1), 'axes');\n% Determine the intersection of the two polyshape objects\nC = intersect(P1, P2);\n% Plot the intersection\nfigure;\nh2 = plot(C);\nax2 = ancestor(h2, 'axes');\n% Make the two axes have the same limits for easy comparison\naxis(ax2, axis(ax1));\nIf you flip back and forth between the two figures, you can see that the polyshape C plotted on the second figure is exactly the intersection between the two polyshape objects P1 and P2. If you want coordinates, use the Vertices property.\nC.Vertices\nIf you want to determine if an arbitrary point is inside the intersection, use isinterior.\nFaez Alkadi on 27 Apr 2018\nHi Steven, Thank you for stepping in. the thing is this, this will not help with my application.\nThank you" ]

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[ "# Confidence Intervals for Proportions\n\n| Home | | Advanced Mathematics |\n\n## Chapter: Biostatistics for the Health Sciences: Inferences Regarding Proportions\n\nFirst we will consider a single proportion and the approximate intervals based on the normal distribution.\n\nCONFIDENCE INTERVALS FOR PROPORTIONS\n\nFirst we will consider a single proportion and the approximate intervals based on the normal distribution. If W is X/n, where X is a binomially distributed random variable with parameters n and p, then by the central limit theorem W is approximately normally distributed with mean p and variance p(1 – p)/n. Therefore, (Wp)/ √{p(1 – p)/n} has an approximately standard normal distribution.\n\nBecause p is unknown, we cannot normalize W by dividing W by p. Instead, we consider the quantity U = (Wp)/ {W(1 – W)/n}. Since W is a consistent estimate of p, this quantity U converges to a standard normal random variable as the sample size n increases.\n\nTherefore, we use the fact that if U were standard normal, then P[–1.96 U 1.96] = 0.95 or P[–1.96 (Wp)/ √{W(1 – W)/n} 1.96] = 0.95 or, after the usual algebraic manipulations, P[W – 1.96 √{W(1 – W)/n} p W + 1.96 √{W(1 – W)/n}]. So the random interval [W – 1.96 √{W(1 – W)/n}, W + 1.96 √{W(1 – W)/n]} is an approximate 95% confidence interval for a single proportion p.\n\n[W – 1.96 √{W(1 – W)/n},\n\nW + 1.96 √{W(1 – W)/n]}               (10.6)\n\nwhere W = X/n and X is binomially distributed with parameters n and p. For other confidence levels, change 1.96 to the appropriate constant C from the standard nor-mal distribution.\n\nAs an example, suppose that we have 16 successes in 20 trials; X = 16 and n = 20. What would be an approximate 95% confidence interval for the population proportion of successes, p? From Equation 10.6, since W = 16/20 = 0.80, we have [0.80 - 1.96 √[0.8(0.2)/20], 0.80 + 1.96 √{0.8(0.2)/20}] = [0.80 – 0.1753, 0.80 + 0.1753] = [0.625, 0.975]. Later we will compare this interval to the exact interval obtained by the Clopper–Pearson method.\n\nNow let us consider two independent estimates of proportions, W1 = X1/n1 and W2 = X2/n2, where X1 is a binomial random variable with parameters p1 and n1 and X2 is a binomial random variable with parameters p2 and n2. Then, Z = (W1W2) – (p1p2)/ {[W1(1 – W1)/n1 + W2(1 – W2)/n2]} has an approximately standard normal distribution. Therefore, P[–1.96 Z 1.96] is approximately 0.95. After substitution and algebraic manipulations, we have P[(W1W2) - 1.96 {[W1(1 – W1)/n1 + W2(1 – W2)/n2]} (p1p2) [(W1W2) +1.96 {[W1(1 – W1)/n1 + W2(1 – W2)/n2]}. The probability that p1p2 lies within this interval is approximately 0.95; hence, the random interval [(W1W2) – 1.96 {[W1(1 – W1)/n1 + W2(1 – W2)/n2]}[(W1W2) + 1.96 {[W1(1 – W1)/n1 + W2(1 – W2)/n2]} is an approximate 95% confidence interval for p1 p2.\n\nAn approximate 95% confidence interval for the difference between two propor-tions p1p2 is", null, "[(W1W2) – 1.96 √{W1(1 – W1)/n1 + W2(1 – W2)/n2},\n\n(W1W2) + 1.96 √{W1(1 – W1)/n1 + (W2(1 – W2)/n2)]}                     (10.7)\n\nwhere W1 = X1/n1 and X1 is binomially distributed with parameters n1 and p1, and W2 = X2/n2 and X2 is binomially distributed with parameters n2 and p2. For other confidence levels, change 1.96 to the appropriate constant C from the standard nor-mal distribution.\n\nFor a numerical example, suppose n1 is 100 and n2 is 50. Suppose X1 = 85 and X2 = 26. We will calculate the approximate 95% and 99% confidence intervals for p1 p2 when W1 = 85/100 = 0.85 and W2 = 26/50 = 0.52. In the case of the 95% confidence interval, the constant C = 1.96; hence, the interval is [(0.85 – 0.52) – 1.96 {0.85(0.15)/100 + 0.52(0.48)/50}, (0.85–0.52)+1.96 {0.85(0.15)/100 + 0.52(0.48)/50]} = [0.175, 0.485].\n\nFor exact intervals, the Clopper–Pearson method is used. Clopper and Pearson (1934) provided the results of their method in graphical form. Hahn and Meeker (1991) reprinted Clopper and Pearson’s work, along with much detail about confi-dence intervals. The two-sided interval uses the F distribution with the 100(1 – α)% interval given by Equation 10.8. We will learn about the F distribution in Chapter 13.\n\nThe exact 100(1 – a)% confidence interval for a single binomial proportion is\n\n[{1 + (nx + 1)F(1 – a/2:2n – 2x + 2, 2x)/x}–1, {1 + (nx)/{(x + 1)F(1 – a/2:2x + 2, 2n – 2x)}}–1]\n\nwhere x is the number of successes in n Bernoulli trials and F(γ: dfn, dfd) is the 100 γ th percentile of an F distribution with dfn degrees of freedom for the numerator and dfd degrees of freedom for the denominator. For the lower endpoint, γ = 1 – a/2, dfn = 2n – 2x, and dfd = 2x. For the upper endpoint, γ = 1 – α/2, dfn = 2x + 2, and dfd = 2n–2x.\n\nNow let us revisit the example for approximate confidence intervals where X = 16, n = 20, and 1 – α/2 = 0.95. The above equation becomes [{1 + 5 F(0.95: 10, 32)/ 16}–1, {1 + 4/{5 F(0.95: 34, 8)}}–1]. For now we will take these percentiles by con-sulting a table for the F distribution. From the table (Appendix A), we see that F(0.95: 10, 32) = 2.94 and F(0.95: 34, 8) = 5.16 (by interpolation between F(0.95, 30, 8) = 5.20 and F(0.95, 40, 8) = 5.11. Plugging these values into Equation 10.8, we obtain the interval [0.521, 0.866]. The value 0.95 tells us the percentile to look up in the table; the two other parameters are the numerator and denominator de-grees of freedom, to be defined in Chapter 12.\n\nCompare this new interval to the interval from the normal approximation [0.625, 0.975]. Note that the widths of the intervals are about the same, but the normal ap-proximation gives a symmetric interval centered at 0.80. The reason for the differ-ence is that the sample size of 20 is too small for the normal approximation to be very good, as the true proportion is probably close to 0.80; the Binomial distribu-tion, though centered at 0.80, is much more skewed than a normal distribution and has a longer left tail than right tail. In this case, the exact binomial solution is appro-priate but the normal approximation is not.\n\nIf n were 100, the normal approximation and the exact Binomial distribution would be in much closer agreement. So let us make the comparison when n = 100 and x = 80. The normal approximation gives [0.80–1.96 {0.8(0.2)/100}, 0.80 + 1.96 {0.8(0.2)/100}] = [0.722, 0.878], whereas the Clopper–Pearson method gives [{1 + 21 F(0.95: 42, 160)/80}–1, {1 + 20/{81 F(0.95: 162, 40)}}–1]. We have F(0.95: 42, = 1.72 (by interpolation in the table, Appendix A) and F(0.95: 162, 40) = 1.90 (also by interpolation in the table). Substituting these values in the equation above gives the interval [0.689, 0.885]. We note that the normal approximation, though not as accurate as we would like, is much closer to the exact result when the sample size is 100 as compared to when the sample size is only 20." ]

[ null, "https://www.pharmacy180.com/media/imgph03/HnfNey0.jpg", null ]

[ "", null, "# Regular Polygon Calculator\n\nRegular Polygon Calculator is a free online tool that displays the area of regular polygon for the given side length or apothem or radius. BYJU’S online regular polygon calculator tool makes the calculation faster, and it displays the area of the regular polygon in a fraction of seconds.\n\n## How to Use the Regular Polygon Calculator?\n\nThe procedure to use the regular polygon calculator is as follows:\nStep 1: Enter the number of sides, length type (i.e. side or apothem or radius) and length of the parameter in the respective input field\nStep 2: Now click the button “Submit” to get the polygon area\nStep 3: Finally, the area of the specified polygon will be displayed in the new window\n\n### What is Meant by Regular Polygon?\n\nIn geometry, a polygon is a two-dimensional figure, made up of straight-line segments. Based on the angles and sides, as well as the symmetry, convexity some other properties, the polygons are classified. Thus, there are different types of polygons in general. If the length of all the sides are equal, then the polygon is called a regular polygon. An equilateral triangle is an example of a regular polygon with 3 sides, a square is a regular polygon with 4 sides. Other polygons include pentagon, hexagon, heptagon, and so on as given the table below.\n\n Polygon Name Number of Sides 3 sided Polygon Trigon (or) Triangle 3 4 sided Polygon Tetragon (or) Quadrilateral 4 5 sided Polygon Pentagon 5 6 sided Polygon Hexagon 6 7 sided Polygon Heptagon (or) Septagon 7 8 sided Polygon Octagon 8 9 sided Polygon Nonagon 9 10 sided Polygon Decagon 10 N sided polygon n – gon n" ]

[ null, "https://www.facebook.com/tr", null ]

[ "# A Makefile for knitr documents\n\nOne of the best things I’ve found about using R for all my scientific work is powerful and easy to use facilities for generating dynamic reports, particularly using the knitr package. The seamless integration of text, code, and the resulting figures (or tables) is a major step toward fully-reproducible research, and I’ve even found that it’s a great way of doing “exploratory” work that allows me to keep my own notes and code contained in the same document.\n\nBeing a fan of a “Makefile” approach to working with R scripts, as well as an Emacs/ESS addict, I find the easiest way to automatically run/compile my knitr latex documents is with a Makefile. Below is a template I adapted from here:\n\nall: pdf\n\nMAINFILE := **PUT MAIN FILENAME HERE**\nRNWFILES :=\nRFILES :=\nTEXFILES :=\nCACHEDIR := cache\nFIGUREDIR := figures\nLATEXMK_FLAGS :=\n##### Explicit Dependencies #####\n################################################################################\nRNWTEX = $(RNWFILES:.Rnw=.tex) ROUTFILES =$(RFILES:.R=.Rout)\nRDAFILES= $(RFILES:.R=.rda) MAINTEX =$(MAINFILE:=.tex)\nMAINPDF = $(MAINFILE:=.pdf) ALLTEX =$(MAINTEX) $(RNWTEX)$(TEXFILES)\n\n# Dependencies\n$(RNWTEX):$(RDAFILES)\n$(MAINTEX):$(RNWTEX) $(TEXFILES)$(MAINPDF): $(MAINTEX)$(ALLTEX)\n\n.PHONY: pdf tex clean\n\npdf: $(MAINPDF) tex:$(RDAFILES) $(ALLTEX) %.tex:%.Rnw Rscript \\ -e \"library(knitr)\" \\ -e \"knitr::opts_chunk[['set']](fig.path='$(FIGUREDIR)/$*-')\" \\ -e \"knitr::opts_chunk[['set']](cache.path='$(CACHEDIR)/$*-')\" \\ -e \"knitr::knit('$<','$@')\" %.R:%.Rnw Rscript -e \"Sweave('$^', driver=Rtangle())\"\n\n%.Rout:%.R\nR CMD BATCH \"$^\" \"$@\"\n\n%.pdf: %.tex\nlatexmk -pdf $< clean: -latexmk -c -quiet$(MAINFILE).tex\n-rm -f $(MAINTEX)$(RNWTEX)\n-rm -rf $(FIGUREDIR) -rm *tikzDictionary -rm$(MAINPDF)" ]

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[ "# Johnson-King model\n\n(Difference between revisions)\n Turbulence RANS-based turbulence models Linear eddy viscosity models Nonlinear eddy viscosity models Explicit nonlinear constitutive relation v2-f models", null, "$\\overline{\\upsilon^2}-f$ model", null, "$\\zeta-f$ model Reynolds stress model (RSM) Large eddy simulation (LES) Detached eddy simulation (DES) Direct numerical simulation (DNS) Turbulence near-wall modeling Turbulence free-stream boundary conditions" ]

[ null, "https://www.cfd-online.com/W/images/math/4/7/c/47c2b2b6178a5b1b177f1a2ec5bdc93a.png ", null, "https://www.cfd-online.com/W/images/math/a/b/c/abcfd9602b60feee32bb06ff51b13caf.png ", null ]

[ "# #02b419 Color Information\n\nIn a RGB color space, hex #02b419 is composed of 0.8% red, 70.6% green and 9.8% blue. Whereas in a CMYK color space, it is composed of 98.9% cyan, 0% magenta, 86.1% yellow and 29.4% black. It has a hue angle of 127.8 degrees, a saturation of 97.8% and a lightness of 35.7%. #02b419 color hex could be obtained by blending #04ff32 with #006900. Closest websafe color is: #00cc00.\n\n• R 1\n• G 71\n• B 10\nRGB color chart\n• C 99\n• M 0\n• Y 86\n• K 29\nCMYK color chart\n\n#02b419 color description : Strong lime green.\n\n# #02b419 Color Conversion\n\nThe hexadecimal color #02b419 has RGB values of R:2, G:180, B:25 and CMYK values of C:0.99, M:0, Y:0.86, K:0.29. Its decimal value is 177177.\n\nHex triplet RGB Decimal 02b419 `#02b419` 2, 180, 25 `rgb(2,180,25)` 0.8, 70.6, 9.8 `rgb(0.8%,70.6%,9.8%)` 99, 0, 86, 29 127.8°, 97.8, 35.7 `hsl(127.8,97.8%,35.7%)` 127.8°, 98.9, 70.6 00cc00 `#00cc00`\nCIE-LAB 63.937, -65.514, 60.2 16.521, 32.724, 6.365 0.297, 0.588, 32.724 63.937, 88.972, 137.42 63.937, -60.11, 75.698 57.205, -48.557, 33.446 00000010, 10110100, 00011001\n\n# Color Schemes with #02b419\n\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #b4029d\n``#b4029d` `rgb(180,2,157)``\nComplementary Color\n• #44b402\n``#44b402` `rgb(68,180,2)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #02b472\n``#02b472` `rgb(2,180,114)``\nAnalogous Color\n• #b40244\n``#b40244` `rgb(180,2,68)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #7202b4\n``#7202b4` `rgb(114,2,180)``\nSplit Complementary Color\n• #b41902\n``#b41902` `rgb(180,25,2)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #1902b4\n``#1902b4` `rgb(25,2,180)``\n• #9db402\n``#9db402` `rgb(157,180,2)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #1902b4\n``#1902b4` `rgb(25,2,180)``\n• #b4029d\n``#b4029d` `rgb(180,2,157)``\n• #01680e\n``#01680e` `rgb(1,104,14)``\n• #018212\n``#018212` `rgb(1,130,18)``\n• #029b15\n``#029b15` `rgb(2,155,21)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #02cd1d\n``#02cd1d` `rgb(2,205,29)``\n• #03e620\n``#03e620` `rgb(3,230,32)``\n• #06fc26\n``#06fc26` `rgb(6,252,38)``\nMonochromatic Color\n\n# Alternatives to #02b419\n\nBelow, you can see some colors close to #02b419. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #18b402\n``#18b402` `rgb(24,180,2)``\n• #09b402\n``#09b402` `rgb(9,180,2)``\n• #02b40a\n``#02b40a` `rgb(2,180,10)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #02b428\n``#02b428` `rgb(2,180,40)``\n• #02b437\n``#02b437` `rgb(2,180,55)``\n• #02b445\n``#02b445` `rgb(2,180,69)``\nSimilar Colors\n\n# #02b419 Preview\n\nThis text has a font color of #02b419.\n\n``<span style=\"color:#02b419;\">Text here</span>``\n#02b419 background color\n\nThis paragraph has a background color of #02b419.\n\n``<p style=\"background-color:#02b419;\">Content here</p>``\n#02b419 border color\n\nThis element has a border color of #02b419.\n\n``<div style=\"border:1px solid #02b419;\">Content here</div>``\nCSS codes\n``.text {color:#02b419;}``\n``.background {background-color:#02b419;}``\n``.border {border:1px solid #02b419;}``\n\n# Shades and Tints of #02b419\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000501 is the darkest color, while #f1fff3 is the lightest one.\n\n• #000501\n``#000501` `rgb(0,5,1)``\n• #001903\n``#001903` `rgb(0,25,3)``\n• #002c06\n``#002c06` `rgb(0,44,6)``\n• #014009\n``#014009` `rgb(1,64,9)``\n• #01530c\n``#01530c` `rgb(1,83,12)``\n• #01660e\n``#01660e` `rgb(1,102,14)``\n• #017a11\n``#017a11` `rgb(1,122,17)``\n• #028d14\n``#028d14` `rgb(2,141,20)``\n• #02a116\n``#02a116` `rgb(2,161,22)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\n• #02c71c\n``#02c71c` `rgb(2,199,28)``\n• #02db1e\n``#02db1e` `rgb(2,219,30)``\n• #03ee21\n``#03ee21` `rgb(3,238,33)``\n• #08fc28\n``#08fc28` `rgb(8,252,40)``\n• #1cfc39\n``#1cfc39` `rgb(28,252,57)``\n• #2ffd4a\n``#2ffd4a` `rgb(47,253,74)``\n• #42fd5b\n``#42fd5b` `rgb(66,253,91)``\n• #56fd6b\n``#56fd6b` `rgb(86,253,107)``\n• #69fd7c\n``#69fd7c` `rgb(105,253,124)``\n• #7dfe8d\n``#7dfe8d` `rgb(125,254,141)``\n• #90fe9e\n``#90fe9e` `rgb(144,254,158)``\n• #a3feaf\n``#a3feaf` `rgb(163,254,175)``\n• #b7fec0\n``#b7fec0` `rgb(183,254,192)``\n• #cafed1\n``#cafed1` `rgb(202,254,209)``\n• #deffe2\n``#deffe2` `rgb(222,255,226)``\n• #f1fff3\n``#f1fff3` `rgb(241,255,243)``\nTint Color Variation\n\n# Tones of #02b419\n\nA tone is produced by adding gray to any pure hue. In this case, #566057 is the less saturated color, while #02b419 is the most saturated one.\n\n• #566057\n``#566057` `rgb(86,96,87)``\n• #4f6752\n``#4f6752` `rgb(79,103,82)``\n• #486e4d\n``#486e4d` `rgb(72,110,77)``\n• #417548\n``#417548` `rgb(65,117,72)``\n• #3a7c43\n``#3a7c43` `rgb(58,124,67)``\n• #33833d\n``#33833d` `rgb(51,131,61)``\n• #2c8a38\n``#2c8a38` `rgb(44,138,56)``\n• #259133\n``#259133` `rgb(37,145,51)``\n• #1e982e\n``#1e982e` `rgb(30,152,46)``\n• #179f29\n``#179f29` `rgb(23,159,41)``\n• #10a623\n``#10a623` `rgb(16,166,35)``\n``#09ad1e` `rgb(9,173,30)``\n• #02b419\n``#02b419` `rgb(2,180,25)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #02b419 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# whdlSamplesToFrames\n\nConvert sample stream to frame-based data\n\n## Syntax\n\n``outframes = whdlSamplesToFrames(samples,ctrl)``\n``outframes = whdlSamplesToFrames(samples,ctrl,maxlen)``\n``outframes = whdlSamplesToFrames(samples,ctrl,maxlen,interleaved)``\n\n## Description\n\nexample\n\n````outframes = whdlSamplesToFrames(samples,ctrl)` composes frame-based data from a sample stream and corresponding control signals. The control signals indicate the validity of the samples and the boundaries of the frames. The function calculates the maximum frame length from the input data and control signals, and removes any idle or nonvalid samples from the data.```\n\nexample\n\n````outframes = whdlSamplesToFrames(samples,ctrl,maxlen)` composes frame-based data, using the maximum frame length. If an input frame described by `samples` is larger than `maxlen`, the function truncates the frame.```\n\nexample\n\n````outframes = whdlSamplesToFrames(samples,ctrl,maxlen,interleaved)` orders the frame-based data, assuming the input samples are interleaved, when `interleaved` is 1 (`true`). The `interleaved` argument is valid only when each sample is represented by multiple values. The function computes the number of values representing each sample by comparing the length of `samples` and `ctrl`.```\n\n## Examples\n\ncollapse all\n\nThis example shows how to use the LTE Turbo Encoder block to encode data, and how to compare the hardware-friendly design with the results from LTE Toolbox™. The workflow follows these steps:\n\n1. Generate frames of random input samples in MATLAB®.\n\n2. Encode the data using the LTE Toolbox function `lteTurboEncode`.\n\n3. Convert framed input data to a stream of samples and import the stream into Simulink®.\n\n4. To encode the samples using a hardware-friendly architecture, run the Simulink model, which contains the Wireless HDL Toolbox™ block LTE Turbo Encoder.\n\n5. Export the stream of encoded samples to the MATLAB workspace.\n\n6. Convert the sample stream back to framed data, and compare the frames with the reference data.\n\nGenerate input data frames. Generate reference encoded data using `lteTurboEncode`.\n\n```rng(0); turboframesize = 40; numframes = 2; txBits = cell(1,numframes); codedData = cell(1,numframes); for ii = 1:numframes txBits{ii} = logical(randi([0 1],turboframesize,1)); codedData{ii} = lteTurboEncode(txBits{ii}); end ```\n\nSerialize input data for the Simulink model. Leave enough time between frames for each frame to be fully encoded before the next one starts. The LTE Turbo Encoder block takes `inframesize` + 16 cycles to complete encoding of a frame.\n\n```inframes = txBits; inframesize = size(inframes{1},1); idlecyclesbetweensamples = 0; idlecyclesbetweenframes = inframesize+16; [sampleIn,ctrlIn] = ... whdlFramesToSamples(inframes, ... idlecyclesbetweensamples, ... idlecyclesbetweenframes); ```\n\nRun the Simulink model. The simulation time equals the number of input samples. Because of the added idle cycles between frames, the streaming input data includes enough cycles for the model to complete encoding of both frames.\n\n```sampletime = 1; samplesizeIn = 1; simTime = size(ctrlIn,1); modelname = 'ltehdlTurboEncoderModel'; open_system(modelname); sim(modelname); ```", null, "The Simulink model exports `sampleOut_ts` and `ctrlOut_ts` back to the MATLAB workspace. Deserialize the output samples, and compare the framed data to the reference encoded frames.\n\nThe output samples of the LTE Turbo Encoder block are interleaved with the parity bits.\n\nHardware-friendly output: `S_1 P1_1 P2_1 S2 P1_2 P2_2 ... Sn P1_n P2_n`\n\nLTE Toolbox output: `S_1 S_2 ... S_n P1_1 P1_2 ... P1_n P2_1 P2_2 ... P2_n`\n\nReorder the samples using the interleave option of the `whdlSamplesToFrames` function. Compare the reordered output frames with the reference encoded frames.\n\n```sampleOut = sampleOut'; interleaveSamples = true; outframes = whdlSamplesToFrames(sampleOut(:),ctrlOut,[],interleaveSamples); fprintf('\\nLTE Turbo Encoder\\n'); for ii = 1:numframes numBitsDiff = sum(outframes{ii} ~= codedData{ii}); fprintf([' Frame %d: Behavioral and ' ... 'HDL simulation differ by %d bits\\n'],ii,numBitsDiff); end ```\n```Maximum frame size computed to be 132 samples. LTE Turbo Encoder Frame 1: Behavioral and HDL simulation differ by 0 bits Frame 2: Behavioral and HDL simulation differ by 0 bits ```\n\n## Input Arguments\n\ncollapse all\n\nStream of output samples, specified as a column vector. The vector can include idle cycles between samples and between frames. Idle cycles are discarded. The frames represented by the stream can be different sizes. The vector length, N, must be an integer multiple of the length of the `ctrl` matrix, M. Differing lengths mean that each sample is represented by N/M values.\n\nFor example, in the LTE standard, the turbo code rate is 1/3, so each turbo-encoded sample is represented by one systematic, and two parity values: Sn, Pn1, and Pn2. In that case, the length of `samples` must be three times the length of `ctrl`.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical` | `fi`\n\nControl signals accompanying the sample stream, specified as an M-by-3 matrix. The matrix includes three control signals, `start`, `end`, and `valid`, for each sample in `samples`. Each sample can be represented by more than one value. In that case, the length of `samples` must be an integer multiple of M.\n\nFor example, in the LTE standard, the turbo code rate is 1/3, so each turbo-encoded sample is represented by one systematic, and two parity values: Sn, Pn1, and Pn2. In that case, the length of `samples` must be three times the length of `ctrl`.\n\nData Types: `logical`\n\nMaximum frame length, specified as an integer. The input frames in `samples` can be different sizes. The output column vector reflects the size of the input frame, according to `ctrl`. If a frame is larger than `maxlen`, the function truncates the frame and returns a warning message.\n\nData Types: `double`\n\nOrder of output samples relative to input order, when more than one value represents each sample, specified as a logical scalar.\n\nFor example, 1/3 turbo-encoded samples are represented by ```[S1 P11 P12 S2 P21 P22]```. To reorder the samples so that systematic and parity values are grouped together, set `interleaved` to 1 (`true`). The output order is then ```[S1 S2 P11 P21 P12 P22]```.\n\nData Types: `logical`\n\n## Output Arguments\n\ncollapse all\n\nFrames of output samples, returned as a column vector or a cell array of column vectors. The size of the output column vector reflects the size of the input frame, as determined by the control signals in `ctrl`.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical` | `fi`\n\n## Compatibility Considerations\n\nexpand all\n\nWarns starting in R2020a\n\n### Topics\n\nIntroduced in R2017b" ]

[ null, "https://se.mathworks.com/help/examples/ltehdl/win64/TurboEncodeStreamingSamplesExample_01.png", null ]

[ "# More Than Meets the Eye: Cluster Scaling Relations\n\nTITLE: Scaling relations for galaxy clusters in the Millennium-XXL simulation\nAUTHORS: R. E. Angulo, V. Springel, S. D. M. White, A. Jenkins, C. M. Baugh, C. S. Frenk\nFIRST AUTHOR’S INSTITUTION: Max-Planck-Institute for Astrophysics, Garching, Germany\n\nIn a perfect world, there would be one observable quantity (or even better, two or three) of galaxy clusters that would correlate simply and robustly with the total mass of the cluster. This would allow astronomers to easily determine the masses of potentially thousands of clusters. Then, armed with these precise masses, we could greatly advance our understanding of cosmology, since the distribution of cluster masses is sensitive to the fundamental parameters of the universe. Unfortunately, cluster scaling relations in reality have proven to be surprisingly complicated, and recent studies have found puzzling discrepancies between observed results and our expectations (check out this paper as one recent example).", null, "Figure 1: The four clusters with the largest tSZ signal, X-ray luminosity, weak lensing mass, and optical richness count. Note the differences in structure between these massive clusters. (Fig. 1 of Angulo et al.)\n\nThe authors of this analysis use the largest, high-resolution cosmological N-body simulation to date, the Millennium-XXL (MXXL), to investigate sources of scatter in cluster scaling relations. The MXXL follows the non-linear growth of 303 billion dark matter particles within a cubic region of 4.11 Gpc on a side. This simulated volume is equivalent to that of the observable universe out to a redshift of z = 0.72.  The goal of the work is to ‘measure’ various cluster properties from the simulation while replicating the techniques of typical observational surveys. Then, the authors compute the relations between these observables and cluster mass to determine if the results match our theoretical expectations.\n\nThe four cluster observables that the authors examine are the ‘optical richness’", null, "$N_{\\mathrm{opt}}$ (essentially the number of galaxies belonging to a cluster, within a given radius), the cluster mass as determined by weak gravitational lensing", null, "$M_{\\mathrm{lens}}$, the X-ray luminosity of the ionized  intracluster medium", null, "$L_X$, and the thermal Sunyaev-Zel’dovich (tSZ) signal", null, "$Y_{\\mathrm{SZ}}$. For a quick introduction to the use of the tSZ effect for scaling relations, check out this astrobite. These quantities are expected to scale with the true cluster mass as", null, "$N_{\\mathrm{opt}} \\propto M^{0.9}$", null, "$L_X \\propto M^{4/3}$", null, "$Y_{\\mathrm{SZ}} \\propto M^{5/3}$, and", null, "$M_{\\mathrm{lens}} \\propto M$.\n\nThese observed properties of clusters depend not only on the dark matter in a halo, but also on the hot gas in the cluster, which the MXXL does not model. Instead, the authors construct simple proxies for each of the observables, based solely on the dark matter structure, which nicely avoids any uncertainties in modeling the interplay between the dark matter and gas.", null, "Figure 2: The correlation between deviations in the tSZ signal (y-axis) and X-ray luminosity (x-axis) of a cluster at fixed optical richness. (Fig. 7 of Angulo et al.)\n\nEach of the four cluster observables has intrinsic selection effects, as illustrated by Figure 1. This figure shows the four clusters with the largest tSZ signal, X-ray luminosity, weak lensing mass, and optical richness count in the MXXL. The structure of these clusters varies immensely, and the cluster with the largest mass is, in fact, the one with the largest optical richness. The authors argue that selection effects such as these must be carefully considered before drawing cosmological conclusions from a sample of clusters.\n\nFirst, the authors determine the intrinsic scatter between cluster observables and find that there is a positive correlation between the deviations in pairs of observables. This means that a cluster with a higher than average signal in one observable will likely have a larger signal for each of the three other measured observables as well. The largest correlation is found to be between the scatters in tSZ signal and X-ray luminosity, at fixed optical richness (see Figure 2). The authors argue that these correlations result from the fact that each observable is sensitive to the same set of properties of the cluster population, including internal structure, orientation and environment, and line-of-sight projection of matter not associated with the cluster.\n\nFurthermore, the authors investigate the discrepancies introduced by the particular strategies of an observational survey. As an interesting case study, they address the recent inconsistencies between the mean tSZ signal for optically and X-ray selected clusters found by the Planck collaboration. The stacked mean tSZ signal for a set of clusters at fixed optical richness was found to be significantly lower for optically selected clusters than for X-ray selected clusters. But why?", null, "Figure 3: The mean tSZ signal as a function of optical richness for a X-ray flux limited sample from the MXXL. The result using a volume limited sample from the MXXL is shown in red, significantly below the X-ray flux limited results. (Fig. 9 of Angulo et al.)\n\nThe authors of this work argue that this discrepancy is likely the result of a phenomenon known as Malmquist bias. This effect causes some surveys to preferentially select intrinsically brighter objects. In the Planck team’s analysis, the optically selected catalog is roughly volume limited (i.e., they select all clusters within the survey volume), while the X-ray catalog is approximately flux limited (i.e., they select all clusters above a given X-ray luminosity). Thus, the X-ray selected catalog preferentially includes more clusters of higher than average X-ray luminosity. The authors then argue that the intrinsic correlation between the scatters in", null, "$Y_{\\mathrm{SZ}}$ and", null, "$L_X$ at fixed cluster mass (see Figure 2) will cause the X-ray selected clusters to have a higher than average tSZ signal. They test this hypothesis by selecting  ‘volume limited’ and ‘X-ray flux limited’ cluster samples from the MXXL and plotting the mean tSZ signal versus mean optical richness, which is shown in Figure 3. The authors find a clear decrement of signal in the volume limited sample (red line), as compared to the X-ray flux limited sample, providing support for their hypothesis. However, they are also careful to point out that any discrepancies depend critically on the particular details of how the observables were obtained from the data. Only more detailed modeling and analysis will lead to a stronger conclusion.\n\nThe authors’ analysis of scaling relations for clusters in the MXXL indicates that a wide range of effects and biases must be properly considered before making cosmological inferences from a galaxy cluster survey. These biases depend both upon the inherent properties of the cluster population being considered and on the survey-specific techniques used to estimate each observable from the data. This work represents an important step forward in understanding how these biases affect the most commonly used cluster scaling relations in cosmology.", null, "" ]

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[ "You are on page 1of 5\n\n# ARBITRAGE PRICING THEORY\n\nArbitrage pricing theory (APT) holds that the expected return of a financial asset can be modeled as a linear function of various macro-economic factors, where sensitivity to changes in each factor is represented by a factor specific beta coefficient. The model derived rate of return will then be used to price the asset correctly - the asset price should equal the expected end of period price discounted at the rate implied by model. If the price diverges, arbitrage should bring it back into line. The theory was initiated by the economist Stephen Ross in 1976. \u0016 Based on the law of one price. Two items that are the same cannot sell at different prices. \u0016 If they sell at a different price, arbitrage will take place in which arbitrageurs buy the good which is cheap and sell the one which is higher priced till all prices for the goods are equal.\n\n## ADVANTAGES & LIMITATIONS OF ARBITRAGE PRICING THEORY\n\nThe APT has a number of benefits. First, it is not as a restrictive as the CAPM in its requirement about individual portfolios. It is also less restrictive with respect to the information structure it allows. The APT is a world of arbitrageurs and vendors of information. It also allows multiple sources of risk, indeed these provide an explanation of what moves stock returns. The benefits also come with drawbacks. The APT demands that investors perceive the risk sources, and that they can reasonably estimate factor sensitivities.\n\n## ASSUMPTIONS OF ARBITRAGE PRICING THEORY\n\n\u0014 Three Major Assumptions: \u0016 Capital markets are perfectly competitive. \u0016 Investors always prefer more to less wealth. \u0016 Price-generating process is a K factor model. \u0016 In APT, the assumption of investors utilizing a mean-variance framework is replaced by an assumption of the process of generating security returns. \u0016 APT requires that the returns on any stock be linearly related to a set of indices. \u0016 In APT, multiple factors have an impact on the returns of an asset in contrast with CAPM model that suggests that return is related to only one factor, i.e., systematic risk\n\n1|P a ge\n\n\u0016 Factors that have an impact the returns of all assets may include inflation, growth in GNP, major political upheavals, or changes in interest rates \u0016 ri = ai + bi1F1 + bi2F2 + +bikFk + ei \u0016 Given these common factors, the bik terms determine how each asset reacts to this common factor.\n\n## RELATIONSHIP WITH CAPITAL ASSET PRICING MODEL\n\nAPT along with the Capital asset pricing model (CAPM) is one of two influential theories on asset pricing. The APT differs from the CAPM in that it is less restrictive in its assumptions. It allows for an explanatory model of asset returns. It assumes that each investor will hold a unique portfolio with its own particular array of betas, as opposed to the identical \"market portfolio\". In some ways, the CAPM can be considered a \"special case\" of the APT in that the Securities market line represents a single-factor model of the asset price, where Beta is exposure to changes in value of the Market. Additionally, the APT can be seen as a \"supply side\" model, since its beta coefficients reflect the sensitivity of the underlying asset to economic factors. Thus, factor shocks would cause structural changes in the asset's expected return, or in the case of stocks, in the firm's profitability.\n\n## ARBITRAGE PRICING THEORY-FACTORS\n\nArbitrage pricing theory does not explicitly state the relevant macro economic factors; it has been observed that the following factors tend to influence the price of the security under consideration. \u0016 \u0016 \u0016 \u0016 Change in industrial production or GDP. Unanticipated inflation or deflation. Shifts in the Yield Curve. Investor confidence measured by surprises in default risk premiums for bonds.\n\n## ARBITRAGE PRICING THEORY-EQUATION\n\nArbitrage pricing theory specifies asset (stock or portfolio) returns as a linear function of the aforementioned factors. APT gives the expected return on asset i as: E(Ri) = Rf + b1*(E(R1) - Rf) + b2*(E(R2) - Rf) + b3*(E(R3) - Rf) + + bn*(E(Rn) - Rf)\n2|P a ge\n\nRf = Risk free interest rate (i.e. interest on Treasury Bonds) bi = Sensitivity of the asset to factori E(Ri) - Rf) = Risk premium associated with factori where i = 1, 2,...n The APT model also states that the risk premium of a stock depends on two factors: \u0016 The risk premiums associated with each of the factors described above \u0016 The stock's own sensitivity to each of the factors - similar to the beta concept Risk Premium = r -rf = b(1) x (r factor(1) - rf) + b(2) x (r factor(2) - rf)... + b(n) x (r factor(n) - rf) If the expected risk premium on a stock were lower than the calculated risk premium using the formula above, then investors would sell the stock. If the risk premium were higher than the calculated value, then investors would buy the stock until both sides of the equation were in balance.\n\n## CRUX OF ARBITRAGE PRICING THEORY\n\nA perfectly competitive market is one where any trader can buy or sell unlimited quantities of the relevant security without changing the security s price. In an arbitrage portfolio-a set of goods held by an owner in an economy conform to the APT conditions-the investor tries to increase the returns from his portfolio without increasing fund in the portfolio, without spending other money. Moreover, he also likes to keep the risk at the same level. To do so, if the investor got in his portfolio A, B and C securities, to increase returns from his portfolio without further investing he will have to change the proportion of the securities. This means that if A earns him more he will tend to convert B and C in A before spending further money to buy A. Moreover, conversion may occur also to keep the risk constant as B and C might become too risky whereas A becomes less risky to keep.\n\n## EXAMPLES-ARBITRAGE PRICING THEORY\n\nConsider the following 2 portfolios:\n\nPortfolio\n\nE(rp)\n\nbP\n\nA B\n\n20% 10%\n\n1.5 1.0\n\n3|P a ge\n\nWhere E (rp) is the expected return of the portfolio and bP is the portfolio beta. Both portfolios should have equal lambda factors, which can be found by solving for them simultaneously: 1. 0 + 1(1.5) = 20% 2. 0 + 1(1.0) = 10% 3. 1(0.5) = 5% 4. 1 = 10% 5. Substituting Equation 4 into Equation 1, we find: 6. 0 + 0.1(1.5) = 20% 7. 0 = 20% - 15% = 5% From this, we find the equilibrium APT equation: E (rp) = 5% + 10 %( bp) Now consider a portfolio C where E (rP) = 20% and bP = 1.2. Since portfolio C yields the same as A, but has a reduced risk factor as evidenced by its lower beta, an arbitrage profit should be possible. 1. Construct a portfolio from A and B that has the same risk as portfolio C: 2. bC = yA(bA) + (1 - yA)(bB) -> yA + yB = 1 3. 1.2 = yA(1.5) + (1 - yA)(1.0) 4. 1.2 = 1.5yA + 1 - yA 5. 1.2 = 1 + 0.5yA 6. 0.2 = 0.5yA -> Subtracted 1 from both sides. 7. 0.4 = yA -> Divide both sides by 0.5 8. Amount of portfolio B must be 0.6, since the proportions of both portfolios must sum to 1. 9. Now construct a portfolio D that has the same risk factor as portfolio C: 10. E(rD) = .4(rA) + .6(rB) 11. E(rD) = .4(20%) + .6(10%) = 8% + 6% = 14%\n\n4|P a ge\n\nREFERENCES: www.investopedia.com Dictionary www0.gsb.columbia.edu/faculty/.../APT-Huberman-Wang.pdf www.investorwords.com/247/Arbitrage_Pricing_Theory.html www.hjventures.com/valuation/Arbitrage-Pricing-Theory.html www.wikicfo.com/wiki/.../Arbitrage%20Pricing%20Theory.ashx www.narachinvestment.com/the_arbitrage_pricing_theory.html www.fxwords.com/a/arbitrage-pricing-theory.html www.finance-lib.com/financial-term-arbitrage-pricing-theory-apt.html www.highbeam.com/doc/1O18-arbitragepricingtheory.html\n\n5|P a ge" ]

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[ "# How to Calculate a Lease Liability using Excel\n\nThe lease liability is the present value of the known future lease payments at a point in time. A lease liability is required to be calculated for both ASC 842 & IFRS 16.\n\nRefer below for seven steps on how to calculate the lease liability using excel’s goal seek. The lease liability we’re going to calculate is based on the following terms:\n\n• 12-year lease term\n• \\$10,000 payments at the beginning of each year\n• Discount rate: 7%\n\n## Step 1 - Create a spreadsheet and set up columns", null, "In the new excel spreadsheet and name the five columns:\n\n• Date\n• Lease payment\n• Interest\n• Liability reduction\n• Closing liability balance\n\n## Step 2 - Enter the payment amounts and the payment dates", null, "Each row will include the date of the payment and the payment amount for the life of the lease.\n\n## Step 3 - Calculate the interest on each payment", null, "Calculate the interest incurred on each payment using the discount rate of 6%. For the first payment, there is no interest incurred as it’s made at the commencement of the lease.\n\n## Step 4 - Calculate the reduction of the lease liability for each payment", null, "The lease payment amount, less the interest occurred is the amount that the lease liability balance will be reduced.\n\n## Step 5 - Input the formula to calculate the closing balance of the lease liability", null, "This formula is the previous balance less the liability reduction amount to give the closing balance of the lease liability after the payment.\n\n## Step 6 - Calculate the opening balance of the lease liability using excel’s goal seek function", null, "This will calculate the opening balance amount of the lease liability. From this amount, the lease liability will unwind to zero. To do this in excel, select Data > What-if Analysis > Goal Seek.\n\n## Step 7 - You’re done", null, "Ensure the balance unwinds to zero. If it does, you have calculated the lease liability using goal seek.\n\n## Conclusion\n\nThis is one method of calculating the lease liability. I would highly recommend reading How to Calculate the Present Value of Future Lease Payments to understand there are more and arguably more accurate methods when calculating the lease liability.\n\n## Footer", null, "Here at Cradle, our mission is simple; it's at the foundation of everything that we do. We want to make accountants' lives easier by leveraging technology to free up their time to focus on running the business.\n\n140 Yonge St.\nSuite 200\nToronto, ON M5C 1X6\n\nUS\nSuite #73591\nTorrance, CA 90503\nUSA\n\n### Legal", null, "", null, "", null, "", null, "", null, "" ]

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[ "## EE6511 CI LAB Manual\n\nAnna University Regulation 2013 EEE EE6511 CI Lab Manual viva voce question for all experiments is provided below. Download link for EEE 5th SEM EE6511 Control Instrumentation Laboratory Manual is listed down for students to make perfect utilization and score maximum marks with our study materials.\n\nAnna University Regulation 2013 EEE 5th SEM EE6511 CI LAB-Control Instrumentation Laboratory Manual\n\n### 1(b). KELVIN’S DOUBLE BRIDGE\n\nAIM:\n\nTo measure the given low resistance using Kelvin’s Double bridge.\n\nTHEORY:\n\nKelvin’s double bridge is a modification of Wheatstone’s bridge and provides more accuracy in measurement of low resistances It incorporates two sets of ratio arms and the use of four terminal resistors for the low resistance arms, as shown in figure. Rx is the resistance under test and S is the resistor of the same higher current rating than one under test. Two resistances Rx and S are connected in series with a short link of as low value of resistance r as possible. P, Q, p, q are four known non inductive resistances, one pair of each (P and p, Q and q) are variable. A sensitive galvanometer G is connected across dividing points PQ and pq. . The ratio P Q is kept the same as p q , these ratios have been varied until the galvanometer reads zero.\n\nPROCEDURE:\n\n1. Connect the unknown resistance Rx as marked on the trainer\n\n2. Connect a galvanometer G externally as indicated on the trainer\n\n3. Energize the trainer and check the power to be +5 V.\n\n4. Select the values of P and Q such that P/Q = p/q = 500/50000 = 0.01\n\n5. Adjust P1 for proper balance and then at balance, measure the value of P1.\n\nREVIEW QUESTIONS\n\n1. Name the bridge used for measuring very low resistance.\n\n2. Classify the resistances according to the values.\n\n3. Write the methods of measurements of low resistance\n\n4. What is the use of lead resistor in kelvin’s Double bridge?\n\n5. Why Kelvin’s double bridge is having two sets of ratio arms?\n\nEE6511 CI Lab Manual with all experiments – Download Here\nIf you require any other notes/study materials, you can comment in the below section." ]

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[ "››Convert cubit [Egyption] to centimetre\n\n cubit cm\n\n Did you mean to convert cubit [Egyption] cubit [English] cubit [Roman] cubit [Royal Egyptian] to cm\n\nHow many cubit in 1 cm? The answer is 0.022222222222222.\nWe assume you are converting between cubit [Egyption] and centimetre.\nYou can view more details on each measurement unit:\ncubit or cm\nThe SI base unit for length is the metre.\n1 metre is equal to 2.2222222222222 cubit, or 100 cm.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between cubits and centimetres.\nType in your own numbers in the form to convert the units!\n\n››Quick conversion chart of cubit to cm\n\n1 cubit to cm = 45 cm\n\n2 cubit to cm = 90 cm\n\n3 cubit to cm = 135 cm\n\n4 cubit to cm = 180 cm\n\n5 cubit to cm = 225 cm\n\n6 cubit to cm = 270 cm\n\n7 cubit to cm = 315 cm\n\n8 cubit to cm = 360 cm\n\n9 cubit to cm = 405 cm\n\n10 cubit to cm = 450 cm\n\n››Want other units?\n\nYou can do the reverse unit conversion from cm to cubit, or enter any two units below:\n\nEnter two units to convert\n\n From: To:\n\n››Definition: Centimeter\n\nA 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. A centimetre is part of a metric system. It is the base unit in the centimetre-gram-second system of units. A corresponding unit of area is the square centimetre. A corresponding unit of volume is the cubic centimetre.\n\nThe centimetre is a now a non-standard factor, in that factors of 103 are often preferred. However, it is practical unit of length for many everyday measurements. A centimetre is approximately the width of the fingernail of an adult person.\n\n››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!" ]

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[ "# How do you work out the energy requirement for a submerged pump?\n\nI was wondering if you were to place a pump in a submerged container and then were to open a hatch allowing water to flood the previously vacant cavity how much energy would be required to discharge the water at the entry rate. What I would like to know also whether or not the depth would have any influence on the pump's output and efficiency. I understand submersible vehicles rely on the flooding and flushing of water from their hulls to dive and ascend but am curious if pumps must do more work when at depth. Please and thank you.\n\n• you would have to know how much water was going in to the container at what rate to be able to answer this question, when you get a submersible pump it will tell you its capability on the manufacturers spec and also how much power would be required, this is as far as my experience with pond pumps takes me anyway – Daz Hawley Dec 20 '14 at 11:18\n• The deeper the submersible, the greater the pressure trying to force water into the buoyancy tank, and hence the more work needed to force water out of the tank. – Hot Licks Dec 21 '14 at 2:22\n\nSuppose we have a pump submerged in a liquid of density $\\rho$ at a depth $d$ which we want to use to raise the liquid to a height $h$. The pump has to do work to overcome the potential energy difference between the two heights of liquid. That is:\n\n$$W(t)=\\rho Qg (h-d)t$$\n\nWhere $g$ is the acceleration due to gravity (taken as $9.81m/s^2$). $Q$ is the volumetric flow rate (in $m^3/sec$) of liquid being pumped and $t$ is the time (in sec).\n\nThis would give us the minimum energy required. In practice, pumps have losses which are represented as 'percentage efficiency', $\\eta$ so the actual mechanical energy required is:\n\n$$E_M=W/ \\eta$$\n\nFor example, the efficiency of a centrifugal pump might be $\\eta=0.65$ (65%) due to losses in the pump such as turbulence. The actual efficiency of a pump will vary with flow rate $q$ and can be obtained from a manufacturers data sheet.\n\n• Just looking for a clarification if possible, does the above apply when you are discharging the fluid perpendicular to the head (pump horizontal and discharge parallel to the axis of the outlet), instead of, for example pushing the fluid to a higher elevation? In my mind it would seem to be the same effect but im not certain. Thank you all again. – slowadult Dec 21 '14 at 13:51" ]

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[ "# Infrared and ultraviolet limits of the bulk scalar mass and CFT operator dimension in the AdS/CFT correspondence\n\nOn page 131 of these notes, a precise formulation of the AdS/CFT correspondence is given by the GKPW dictionary\n\n$$Z_{\\text{grav}}[\\phi_{0}^{i};\\partial M] = \\langle \\exp \\left( - \\frac{1}{\\hbar} \\sum_{i} \\int d^{d}x\\ \\phi_{0}^{i}(x)O^{i}(x) \\right) \\rangle_{\\text{CFT on } \\partial M}.$$\n\nThe index $i$ is said to run over all the light fields in the bulk effective field theory, and correspondingly over all the low-dimension local operators in CFT.\n\nEquation (14.4) of the notes goes on to mention that the mass $m$ of the bulk scalar is related to the scaling dimension $\\Delta$ of the CFT operator by\n\n$$m^{2} = \\Delta (d-\\Delta) \\qquad \\qquad \\Delta = \\frac{d}{2} + \\sqrt{\\frac{d^{2}}{4}+m^{2}\\ell^{2}}.$$\n\nPrimary question:\n\nIf you have a precise relation between the mass of the bulk scalar and the scaling dimension of the CFT operator, why would you then want to restrict yourself to light bulk fields and low-dimension CFT operators in the GKPW dictionary of the AdS/CFT correspondence?\n\nSecondary question (skip this if you wish):\n\nDoes the consideration of light bulk scalar fields necessarily imply that we have reduced the quantum theory of gravity to a low-energy effective field theory?\n\nPrimary question:\n\nYou are in an $AdS_{d+1}$ space. Take $d$ to be 4 as an example.\n\nIf you raise the conformal/scaling dimension of the CFT operator too much you will get -eventually- a negative $m^2$. Now, there is the so-called Breitenlohner-Freedman bound which states that even if $m^2 < 0$ the states can be stable and thus the theory is free of tachyons. If you violate this limit, then you have tachyons in your theory, which is something you do not want.\n\nPractically, the precise relation is to set bounds, calculate, and check what kind of states there are in the theory.\n\nHowever, the way I have seen the relation is actually the other way around and this is a bit confusing... In any case, the same arguments regarding the BF-bound hold.\n\nThis is what I know.\n\n$$m^2 = \\Delta(\\Delta - d)$$\n\nwhere I set the $AdS$ radius to one.\n\n$m^2 \\geq 0$ only for $\\Delta \\geq 4$. There are certainly theories where $\\Delta < 4$; the unitary bound is $∆ \\geq 1$. Operators with $\\Delta < 4$ correspond to fields with negative mass in $AdS_5$. However they are not tachyons, as explained above, since the energy is positive as long as the Breitenlohner-Freedman bound $m^2 \\geq −4$ is satisfied. The curvature gives a positive contribution to the energy of a scalar field propagating in $AdS$. The minimal value for such a thing is $\\Delta = 2$.\n\nThe notes for further study: http://laces.web.cern.ch/Laces/LACES09/notes/dbranes/lezioniLosanna.pdf\n\nCheers!!!" ]

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[ "# Nonprobability Sampling\n\nEncyclopedia\nEdited by: Published: 2008\n\n• ## Subject Index\n\nSampling involves the selection of a portion of the finite population being studied. Nonprobability sampling does not attempt to select a random sample from the population of interest. Rather, subjective methods are used to decide which elements are included in the sample. In contrast, in probability sampling, each element in the population has a known nonzero chance of being selected through the use of a random selection procedure. The use of a random selection procedure such as simple random sampling makes it possible to use design-based estimation of population means, proportions, totals, and ratios. Standard errors can also be calculated from a probability sample.\n\nWhy would one consider using nonprobability sampling? In some situations, the population may not be well defined. In other situations, there may ...\n\n• All\n• A\n• B\n• C\n• D\n• E\n• F\n• G\n• H\n• I\n• J\n• K\n• L\n• M\n• N\n• O\n• P\n• Q\n• R\n• S\n• T\n• U\n• V\n• W\n• X\n• Y\n• Z\n\n## Methods Map", null, "Research Methods\n\nCopy and paste the following HTML into your website" ]

[ null, "https://methods.sagepub.com/images/img-bg.png", null ]

[ "PHPKonf 2020 Online\n\n# scandir\n\n(PHP 5, PHP 7)\n\nscandirList files and directories inside the specified path\n\n### Description\n\nscandir ( string `\\$directory` [, int `\\$sorting_order` = SCANDIR_SORT_ASCENDING [, resource `\\$context` ]] ) : array\n\nReturns an array of files and directories from the `directory`.\n\n### Parameters\n\n`directory`\n\nThe directory that will be scanned.\n\n`sorting_order`\n\nBy default, the sorted order is alphabetical in ascending order. If the optional `sorting_order` is set to `SCANDIR_SORT_DESCENDING`, then the sort order is alphabetical in descending order. If it is set to `SCANDIR_SORT_NONE` then the result is unsorted.\n\n`context`\n\nFor a description of the `context` parameter, refer to the streams section of the manual.\n\n### Return Values\n\nReturns an array of filenames on success, or `FALSE` on failure. If `directory` is not a directory, then boolean `FALSE` is returned, and an error of level `E_WARNING` is generated.\n\n### Examples\n\nExample #1 A simple scandir() example\n\n``` <?php\\$dir    = '/tmp';\\$files1 = scandir(\\$dir);\\$files2 = scandir(\\$dir, 1);print_r(\\$files1);print_r(\\$files2);?> ```\n\nThe above example will output something similar to:\n\n```Array\n(\n => .\n => ..\n => bar.php\n => foo.txt\n => somedir\n)\nArray\n(\n => somedir\n => foo.txt\n => bar.php\n => ..\n => .\n)\n```\n\n### Notes\n\nTip\n\nA URL can be used as a filename with this function if the fopen wrappers have been enabled. See fopen() for more details on how to specify the filename. See the Supported Protocols and Wrappers for links to information about what abilities the various wrappers have, notes on their usage, and information on any predefined variables they may provide.\n\n• opendir() - Open directory handle\n• glob() - Find pathnames matching a pattern\n• is_dir() - Tells whether the filename is a directory\n• sort() - Sort an array", null, "add a note\n\n### User Contributed Notes 36 notes\n\n695\ndwieeb at gmail dot com\n8 years ago\n``` Easy way to get rid of the dots that scandir() picks up in Linux environments:<?php\\$directory = '/path/to/my/directory';\\$scanned_directory = array_diff(scandir(\\$directory), array('..', '.'));?> ```\n143\nmmda dot nl at gmail dot com\n8 years ago\n``` Here is my 2 cents. I wanted to create an array of my directory structure recursively. I wanted to easely access data in a certain directory using foreach. I came up with the following: <?php function dirToArray(\\$dir) {       \\$result = array();    \\$cdir = scandir(\\$dir);    foreach (\\$cdir as \\$key => \\$value)    {       if (!in_array(\\$value,array(\".\",\"..\")))       {          if (is_dir(\\$dir . DIRECTORY_SEPARATOR . \\$value))          {             \\$result[\\$value] = dirToArray(\\$dir . DIRECTORY_SEPARATOR . \\$value);          }          else          {             \\$result[] = \\$value;          }       }    }       return \\$result; } ?> Output Array (    [subdir1] => Array    (       => file1.txt       [subsubdir] => Array       (          => file2.txt          => file3.txt       )    )    [subdir2] => Array    (     => file4.txt    } ) ```\n21\ninfo at remark dot no\n2 years ago\n``` Someone wrote that array_slice could be used to quickly remove directory entries \".\" and \"..\". However, \"-\" is a valid entry that would come before those, so array_slice would remove the wrong entries. ```\ndjacobson at usa dot rugby\n10 months ago\n``` <?php/**  * Get rid of the dots, the occasional `.DS_Store` file and reindex the   * resulting array all at once.  *  * @param string \\$path_to_directory The relative path to your target directory.  *  * @return array The reindexed array of files.  */function get_the_files( \\$path_to_directory ) {    return array_slice( array_diff( scandir( \\$path_to_directory ), array( '..', '.', '.DS_Store' ) ), 0 );}?> ```\nStan P. van de Burgt\n16 years ago\n``` scandir() with regexp matching on file name and sorting options based on stat(). <?php function myscandir(\\$dir, \\$exp, \\$how='name', \\$desc=0) {     \\$r = array();     \\$dh = @opendir(\\$dir);     if (\\$dh) {         while ((\\$fname = readdir(\\$dh)) !== false) {             if (preg_match(\\$exp, \\$fname)) {                 \\$stat = stat(\"\\$dir/\\$fname\");                 \\$r[\\$fname] = (\\$how == 'name')? \\$fname: \\$stat[\\$how];             }         }         closedir(\\$dh);         if (\\$desc) {             arsort(\\$r);         }         else {             asort(\\$r);         }     }     return(array_keys(\\$r)); } \\$r = myscandir('./book/', '/^article[0-9]{4}\\.txt\\$/i', 'ctime', 1); print_r(\\$r); ?> files can be sorted on name and stat() attributes, ascending and descending: name    file name dev     device number ino     inode number mode    inode protection mode nlink   number of links uid     userid of owner gid     groupid of owner rdev    device type, if inode device * size    size in bytes atime   time of last access (Unix timestamp) mtime   time of last modification (Unix timestamp) ctime   time of last inode change (Unix timestamp) blksize blocksize of filesystem IO * blocks  number of blocks allocated ```\ncsaba at alum dot mit dot edu\n15 years ago\n``` A nice way to filter the files/directories you get back from scandir:<?phpfunction pathFilter (\\$path, \\$aFilter) {   // returns true iff \\$path survives the tests from \\$aFilter   // \\$aFilter is an array of regular expressions: [-]/regExp/modifiers   // if there is a leading '-' it means exclude \\$path upon a match (a NOT test)   // If the first expression has a leading '-', \\$path is in, unless it gets excluded.   // Otherwise, \\$path is out, unless it gets included via the filter tests.   // The tests are applied in succession.   // A NOT test is applied only if \\$path is currently (within the tests) included   // Other tests are applied only if \\$path is currently excluded.  Thus,   // array(\"/a/\", \"-/b/\", \"/c/\") => passes if \\$path has a c or if \\$path has an a but no b   // array(\"/a/\", \"/c/\", \"-/b/\") => passes if \\$path has an a or c, but no b   // array(\"-/b/\", \"/a/\", \"/c/\") => passes if \\$path has no b, or if \\$path has an a or c   if (!\\$aFilter) return true;           // automatic inclusion (pass) if no filters   foreach (\\$aFilter as \\$filter) break;  // we don't know how it's indexed   \\$in = \\$filter==\"-\";                // initial in/exclusion based on first filter   foreach (\\$aFilter as \\$filter)         // walk the filters      if (\\$in==\\$not=(\\$filter==\"-\"))   //     testing only when necessary         \\$in ^= preg_match(substr(\\$filter,\\$not),\\$path);    // flip in/exclusion upon a match   return \\$in;}?>Csaba Gabor from Vienna ```\n56\neep2004 at ukr dot net\n6 years ago\n``` Fastest way to get a list of files without dots.<?php\\$files = array_slice(scandir('/path/to/directory/'), 2); ```\nartmanniako at gmail dot com\n1 year ago\n``` How i solved problem with '.' and '..'\\$x = scandir__DIR__; //any directoryforeach (\\$x as \\$key => \\$value) {        if ('.' !== \\$value && '..' !== \\$value){               echo \\$value;   } }Simple and working ```\nSPekary\n3 years ago\n``` Unless you specify no sorting, file names are sorted in ASCII alphabetic order, meaning numbers first, then uppercase, then lowercase letters, even on operating systems whose file system ignores the case of file names when it does its own sorting.For example, on Mac OS, the following files will appear in this order in the Finder, when your disk is formated using the standard file system:1file.phpa.incB.txtc.txtHowever, scandir will produce an array in the following order:1file.phpB.txta.incc.txt ```\n16\ngambit_642 AT hotmailDOTcom\n6 years ago\n``` Needed something that could return the contents of single or multiple directories, recursively or non-recursively,for all files or specified file extensions that would beaccessible easily from any scope or script.And I wanted to allow overloading cause sometimes I'm too lazy to pass all params.<?phpclass scanDir {    static private \\$directories, \\$files, \\$ext_filter, \\$recursive;// ----------------------------------------------------------------------------------------------    // scan(dirpath::string|array, extensions::string|array, recursive::true|false)    static public function scan(){        // Initialize defaults        self::\\$recursive = false;        self::\\$directories = array();        self::\\$files = array();        self::\\$ext_filter = false;        // Check we have minimum parameters        if(!\\$args = func_get_args()){            die(\"Must provide a path string or array of path strings\");        }        if(gettype(\\$args) != \"string\" && gettype(\\$args) != \"array\"){            die(\"Must provide a path string or array of path strings\");        }        // Check if recursive scan | default action: no sub-directories        if(isset(\\$args) && \\$args == true){self::\\$recursive = true;}        // Was a filter on file extensions included? | default action: return all file types        if(isset(\\$args)){            if(gettype(\\$args) == \"array\"){self::\\$ext_filter = array_map('strtolower', \\$args);}            else            if(gettype(\\$args) == \"string\"){self::\\$ext_filter[] = strtolower(\\$args);}        }        // Grab path(s)        self::verifyPaths(\\$args);        return self::\\$files;    }    static private function verifyPaths(\\$paths){        \\$path_errors = array();        if(gettype(\\$paths) == \"string\"){\\$paths = array(\\$paths);}        foreach(\\$paths as \\$path){            if(is_dir(\\$path)){                self::\\$directories[] = \\$path;                \\$dirContents = self::find_contents(\\$path);            } else {                \\$path_errors[] = \\$path;            }        }        if(\\$path_errors){echo \"The following directories do not exists<br />\";die(var_dump(\\$path_errors));}    }    // This is how we scan directories    static private function find_contents(\\$dir){        \\$result = array();        \\$root = scandir(\\$dir);        foreach(\\$root as \\$value){            if(\\$value === '.' || \\$value === '..') {continue;}            if(is_file(\\$dir.DIRECTORY_SEPARATOR.\\$value)){                if(!self::\\$ext_filter || in_array(strtolower(pathinfo(\\$dir.DIRECTORY_SEPARATOR.\\$value, PATHINFO_EXTENSION)), self::\\$ext_filter)){                    self::\\$files[] = \\$result[] = \\$dir.DIRECTORY_SEPARATOR.\\$value;                }                continue;            }            if(self::\\$recursive){                foreach(self::find_contents(\\$dir.DIRECTORY_SEPARATOR.\\$value) as \\$value) {                    self::\\$files[] = \\$result[] = \\$value;                }            }        }        // Return required for recursive search        return \\$result;    }}?>Usage:scanDir::scan(path(s):string|array, [file_extensions:string|array], [subfolders?:true|false]);<?php//Scan a single directory for all files, no sub-directories\\$files = scanDir::scan('D:\\Websites\\temp');//Scan multiple directories for all files, no sub-dirs\\$dirs = array(    'D:\\folder';    'D:\\folder2';    'C:\\Other';);\\$files = scanDir::scan(\\$dirs);// Scan multiple directories for files with provided file extension,// no sub-dirs\\$files = scanDir::scan(\\$dirs, \"jpg\");//or with an array of extensions\\$file_ext = array(    \"jpg\",    \"bmp\",    \"png\");\\$files = scanDir::scan(\\$dirs, \\$file_ext);// Scan multiple directories for files with any extension,// include files in recursive sub-folders\\$files = scanDir::scan(\\$dirs, false, true);// Multiple dirs, with specified extensions, include sub-dir files\\$files = scanDir::scan(\\$dirs, \\$file_ext, true);?> ```\n13\nkodlee at kodleeshare dot net\n8 years ago\n``` I needed to find a way to get the full path of all files in the directory and all subdirectories of a directory. Here's my solution: Recursive functions! <?php function find_all_files(\\$dir) {     \\$root = scandir(\\$dir);     foreach(\\$root as \\$value)     {         if(\\$value === '.' || \\$value === '..') {continue;}         if(is_file(\"\\$dir/\\$value\")) {\\$result[]=\"\\$dir/\\$value\";continue;}         foreach(find_all_files(\"\\$dir/\\$value\") as \\$value)         {             \\$result[]=\\$value;         }     }     return \\$result; } ?> ```\nTom\n5 years ago\n``` Just was curious to count files and lines in a project<?php  function DirLineCounter( \\$dir , \\$result = array('lines_html' => false, 'files_count' => false, 'lines_count' => false ), \\$complete_table = true )  {      \\$file_read = array( 'php', 'html', 'js', 'css' );      \\$dir_ignore = array();            \\$scan_result = scandir( \\$dir );            foreach ( \\$scan_result as \\$key => \\$value ) {                  if ( !in_array( \\$value, array( '.', '..' ) ) ) {                          if ( is_dir( \\$dir . DIRECTORY_SEPARATOR . \\$value ) ) {                                  if ( in_array( \\$value, \\$dir_ignore ) ) {                    continue;                  }                                  \\$result = DirLineCounter( \\$dir . DIRECTORY_SEPARATOR . \\$value, \\$result, false );                                }              else {                              \\$type = explode( '.', \\$value );               \\$type = array_reverse( \\$type );              if( !in_array( \\$type, \\$file_read ) ) {                continue;              }                            \\$lines = 0;              \\$handle = fopen( \\$dir . DIRECTORY_SEPARATOR . \\$value, 'r' );              while ( !feof( \\$handle ) ) {                  if ( is_bool( \\$handle ) ) {                      break;                  }                  \\$line = fgets( \\$handle );                  \\$lines++;              }              fclose( \\$handle );              \\$result['lines_html'][] = '<tr><td>' . \\$dir . '</td><td>' . \\$value . '</td><td>' . \\$lines . '</td></tr>';               \\$result['lines_count'] = \\$result['lines_count'] + \\$lines;              \\$result['files_count'] = \\$result['files_count'] + 1;                                           }          }      }            if ( \\$complete_table ) {                \\$lines_html = implode('', \\$result['lines_html']) . '<tr><td></td><td style=\"border: 1px solid #222\">Files Total: ' . \\$result['files_count'] . '</td><td style=\"border: 1px solid #222\">Lines Total: ' . \\$result['lines_count'] . '</td></tr>';        return '<table><tr><td style=\"width: 60%; background-color:#ddd;\">Dir</td><td style=\"width: 30%; background-color:#ddd;\">File</td><td style=\"width: 10%; background-color:#ddd;\">Lines</td></tr>' . \\$lines_html . '</table>';              }      else {        return \\$result;      }        }  echo DirLineCounter( '.' );?> ```\nprogrium+php at gmail dot com\n15 years ago\n``` I made this to represent a directory tree in an array that uses the file or directory names as keys and full paths as the value for files. Directories are nested arrays.<?phpfunction generatePathTree(\\$rootPath) {    \\$pathStack = array(\\$rootPath);    \\$contentsRoot = array();    \\$contents = &\\$contentsRoot;    while (\\$path = array_pop(\\$pathStack)) {       \\$contents[basename(\\$path)] = array();       \\$contents = &\\$contents[basename(\\$path)];       foreach (scandir(\\$path) as \\$filename) {           if ('.' != substr(\\$filename, 0, 1)) {               \\$newPath = \\$path.'/'.\\$filename;               if (is_dir(\\$newPath)) {                   array_push(\\$pathStack, \\$newPath);                   \\$contents[basename(\\$newPath)] = array();               } else {                   \\$contents[basename(\\$filename)] = \\$newPath;               }           }       }    }    return \\$contentsRoot[basename(\\$rootPath)];}?>The function will return something like this:Array(    [index.php] => /usr/local/www/index.php    [js] => Array        (            [async.js] => /usr/local/www/js/async.js            [dom.js] => /usr/local/www/js/dom.js            [effects.js] => /usr/local/www/js/effects.js            [prototype.js] => /usr/local/www/js/prototype.js        )    [logo.png] => /usr/local/www/logo.png    [server.php] => /usr/local/www/server.php    [test.js] => /usr/local/www/test.js) ```\nPawel Dlugosz\n15 years ago\n``` For directory containing files like (for example) -.jpg the results of scandir are a little \"weird\" ;)<?php      \\$dir = '/somedir';   \\$files = scandir(\\$dir);   print_r(\\$files);?>Array(    => -.jpg    => .    => ..    => foo.txt    => somedir)Beware - sorting is in ASCII order :) ```\nfazle dot elahee at gmail dot com\n8 years ago\n``` /** * This function will scan all files recursively in the sub-folder and folder. * * @author Fazle Elahee * */function scanFileNameRecursivly(\\$path = '', &\\$name = array() ){  \\$path = \\$path == ''? dirname(__FILE__) : \\$path;  \\$lists = @scandir(\\$path);    if(!empty(\\$lists))  {      foreach(\\$lists as \\$f)      {           if(is_dir(\\$path.DIRECTORY_SEPARATOR.\\$f) && \\$f != \"..\" && \\$f != \".\")      {          scanFileNameRecursivly(\\$path.DIRECTORY_SEPARATOR.\\$f, &\\$name);       }      else      {          \\$name[] = \\$path.DIRECTORY_SEPARATOR.\\$f;      }      }  }  return \\$name;}\\$path = \"/var/www/SimplejQueryDropdowns\";\\$file_names = scanFileNameRecursivly(\\$path);echo \"<pre>\";var_dump(\\$file_names);echo \"</pre>\"; ```\ncarneiro at isharelife dot com dot br\n8 years ago\n``` <?php/**     * Get an array that represents directory tree     * @param string \\$directory     Directory path     * @param bool \\$recursive         Include sub directories     * @param bool \\$listDirs         Include directories on listing     * @param bool \\$listFiles         Include files on listing     * @param regex \\$exclude         Exclude paths that matches this regex     */    function directoryToArray(\\$directory, \\$recursive = true, \\$listDirs = false, \\$listFiles = true, \\$exclude = '') {        \\$arrayItems = array();        \\$skipByExclude = false;        \\$handle = opendir(\\$directory);        if (\\$handle) {            while (false !== (\\$file = readdir(\\$handle))) {            preg_match(\"/(^(([\\.]){1,2})\\$|(\\.(svn|git|md))|(Thumbs\\.db|\\.DS_STORE))\\$/iu\", \\$file, \\$skip);            if(\\$exclude){                preg_match(\\$exclude, \\$file, \\$skipByExclude);            }            if (!\\$skip && !\\$skipByExclude) {                if (is_dir(\\$directory. DIRECTORY_SEPARATOR . \\$file)) {                    if(\\$recursive) {                        \\$arrayItems = array_merge(\\$arrayItems, directoryToArray(\\$directory. DIRECTORY_SEPARATOR . \\$file, \\$recursive, \\$listDirs, \\$listFiles, \\$exclude));                    }                    if(\\$listDirs){                        \\$file = \\$directory . DIRECTORY_SEPARATOR . \\$file;                        \\$arrayItems[] = \\$file;                    }                } else {                    if(\\$listFiles){                        \\$file = \\$directory . DIRECTORY_SEPARATOR . \\$file;                        \\$arrayItems[] = \\$file;                    }                }            }        }        closedir(\\$handle);        }        return \\$arrayItems;    }?> ```\nfatpratmatt dot at dot gmail dot com\n12 years ago\n``` This function generates a list of all files in the chosen directory and all subdirectories, throws them into a NON-multidimentional array and returns them.Most of the recursive functions on this page only return a multi-dimensional array.This is actually a modification of some one else's function (thanks mail at bartrail dot de ;])<?phpfunction scanDirectories(\\$rootDir, \\$allData=array()) {    // set filenames invisible if you want    \\$invisibleFileNames = array(\".\", \"..\", \".htaccess\", \".htpasswd\");    // run through content of root directory    \\$dirContent = scandir(\\$rootDir);    foreach(\\$dirContent as \\$key => \\$content) {        // filter all files not accessible        \\$path = \\$rootDir.'/'.\\$content;        if(!in_array(\\$content, \\$invisibleFileNames)) {            // if content is file & readable, add to array            if(is_file(\\$path) && is_readable(\\$path)) {                // save file name with path                \\$allData[] = \\$path;            // if content is a directory and readable, add path and name            }elseif(is_dir(\\$path) && is_readable(\\$path)) {                // recursive callback to open new directory                \\$allData = scanDirectories(\\$path, \\$allData);            }        }    }    return \\$allData;}?>Example output:print_r(scanDirectories(\"www\"));---Array(    => www/index.php    => www/admin.php    => www/css/css.css    => www/articles/2007/article1.txt    => www/articles/2006/article1.txt    => www/img/img1.png) ```\nwebmaster at gmail dot com\n1 year ago\n``` easy way to list every item in a directory<?php\\$itemnum = 1;\\$dir = 'C:/Path/To/Directory';foreach(scandir(\\$dir) as \\$item){    if (!(\\$item == '.')) {        if (!(\\$item == '..')) {              echo(\\$itemnum.\" = \".\\$item.\"<br>\");             \\$itemnum = (\\$itemnum + 1);}}}?> ```\nmagicmind at netcabo dot pt\n11 years ago\n``` Hello all, I just added a extension filter to the getDirectoryTree function, so it can filter an extension for files in the folders/subfolders: <?php function getDirectoryTree( \\$outerDir , \\$x){     \\$dirs = array_diff( scandir( \\$outerDir ), Array( \".\", \"..\" ) );     \\$dir_array = Array();     foreach( \\$dirs as \\$d ){         if( is_dir(\\$outerDir.\"/\".\\$d)  ){             \\$dir_array[ \\$d ] = getDirectoryTree( \\$outerDir.\"/\".\\$d , \\$x);         }else{          if ((\\$x)?ereg(\\$x.'\\$',\\$d):1)             \\$dir_array[ \\$d ] = \\$d;             }     }     return \\$dir_array; } \\$dirlist = getDirectoryTree('filmes','flv'); ?> ```\nphpdotnet at lavavortex dot com\n11 years ago\n``` How a ninja may retrieve a list of files, files filtered by extension, or directories: <?php //NNNIIIinnnjaaa:: //array of files without directories... optionally filtered by extension function file_list(\\$d,\\$x){        foreach(array_diff(scandir(\\$d),array('.','..')) as \\$f)if(is_file(\\$d.'/'.\\$f)&&((\\$x)?ereg(\\$x.'\\$',\\$f):1))\\$l[]=\\$f;        return \\$l; } //NNNIIIinnnjaaa:: //array of directories function dir_list(\\$d){        foreach(array_diff(scandir(\\$d),array('.','..')) as \\$f)if(is_dir(\\$d.'/'.\\$f))\\$l[]=\\$f;        return \\$l; } /********************************************************************\\                     PRETTY PRETTY LIGHTS (DOCUMENTATION) \\********************************************************************/ /******************************************************************** Overloaded PHP file listing function: array file_list ( string \\$directory [, string \\$file_extension] ) \\$directory   path without backslash, e.g. \"/home/public_html/blarg\" \\$file_extention   optionally filter specific filename extentions, e.g. \".jpg\" >>>>>>>>>>>>>>>>>>>>>>>>>>>>> TRANSLATION <<<<<<<<<<<<<<<<<<<<<<<<<<<< file_list(\"/home\");          //return list of files (no directories) file_list(\"/home\", \".jpg\");  //return only .jpg file list \\********************************************************************/ //(note: one slash (/) below and you enable all your test functions, guess where ;-)) /********************************************************************\\ // TEST FUNCTIONS... IF THESE WORK, THIS FUNCTION WORKS ON THIS PLATFORM echo \"<hr><b>File List:</b><br>\"; \\$n = file_list(getcwd()); if(\\$n) foreach(\\$n as \\$f) echo \"\\$f<br>\";             //current files echo \"<hr><b>Files with extension .php:</b><br>\"; \\$n = file_list(getcwd(),\".php\"); if(\\$n) foreach(\\$n as \\$f) echo \"\\$f<br>\";             //files with .php extensions echo \"<hr><b>Directories:</b><br>\"; \\$d = dir_list(getcwd()); if(\\$d) foreach(\\$d as \\$f) echo \"\\$f<br>\";             //directories /********************************************************************/ /************\\ RUNTIME NOTES: file_list(\\$arg1); // php issues a warning that there is no second parameter, but we know that, izz ok \\************/ /*******************************\\ TESTED AND WORKING ON 2009.04.30: OS:     Linux 2.6.9-78.0.17.ELsmp APACHE: 2.2.9 PHP:    5.2.5 \\*******************************/ ?> ```\ntelefoontoestel59 at hotmail dot com\n5 years ago\n``` I was looking for an easy way to get only files from a certain directory. I came up with the following line of code that will result in an array listing only files. the samen can be done for directories ofcourse.<?php\\$files = array_filter(scandir(\\$directory), function(\\$file) { return is_file(\\$file); })?> ```\nPatrick\n6 years ago\n``` Here is my recursive function, placing directories into new array keys and its contents within./* find_files( string, &array )** Recursive function to return a multidimensional array of folders and files** that are contained within the directory given*/function find_files(\\$dir, &\\$dir_array){    // Create array of current directory    \\$files = scandir(\\$dir);        if(is_array(\\$files))    {        foreach(\\$files as \\$val)        {            // Skip home and previous listings            if(\\$val == '.' || \\$val == '..')                continue;                        // If directory then dive deeper, else add file to directory key            if(is_dir(\\$dir.'/'.\\$val))            {                // Add value to current array, dir or file                \\$dir_array[\\$dir][] = \\$val;                                find_files(\\$dir.'/'.\\$val, \\$dir_array);            }            else            {                \\$dir_array[\\$dir][] = \\$val;            }        }    }    ksort(\\$dir_array);}// Example\\$folder_list = array();find_files('/path', \\$folder_list)var_dump(\\$folder_list);array(3) {  [\"directory_01\"]=>  array(4) {    =>    string(12) \"directory_02\"    =>    string(12) \"directory_03\"    =>    string(11) \"file_01.txt\"    =>    string(11) \"file_02.txt\"  }  [\"directory_01/directory_02\"]=>  array(2) {    =>    string(11) \"file_03.txt\"    =>    string(11) \"file_04.txt\"  }  [\"directory_01/directory_03\"]=>  array(2) {    =>    string(11) \"file_05.txt\"    =>    string(11) \"file_06.txt\"  }} ```\nmoik78 at gmail dot com\n10 years ago\n``` This is a function to count the number of files of a directory <?php function countFiles(\\$dir){     \\$files = array();     \\$directory = opendir(\\$dir);     while(\\$item = readdir(\\$directory)){     // We filter the elements that we don't want to appear \".\", \"..\" and \".svn\"          if((\\$item != \".\") && (\\$item != \"..\") && (\\$item != \".svn\") ){               \\$files[] = \\$item;          }     }     \\$numFiles = count(\\$files);     return \\$numFiles; } ?> ```\ncsaba at alum dot mit dot edu\n15 years ago\n``` Scandir on steroids: For when you want to filter your file list, or only want to list so many levels of subdirectories... <?php function dirList(\\$path=\"\", \\$types=2, \\$levels=1, \\$aFilter=array()) { //  returns an array of the specified files/directories //  start search in \\$path (defaults to current working directory) //  return \\$types:  2 => files; 1 => directories; 3 => both; //  \\$levels: 1 => look in the \\$path only; 2 => \\$path and all children; //          3 => \\$path, children, grandchildren; 0 => \\$path and all subdirectories; //          less than 0 => complement of -\\$levels, OR everything starting -\\$levels down //                e.g. -1 => everthing except \\$path; -2 => all descendants except \\$path + children //  Remaining argument(s) is(are) a filter array(list) of regular expressions which operate on the full path. //    First character (before the '/' of the regExp) '-' => NOT. //    First character (after a possible '-') 'd' => apply to directory name //    The filters may be passed in as an array of strings or as a list of strings //  Note that output directories are prefixed with a '*' (done in the line above the return)    \\$dS = DIRECTORY_SEPARATOR;    if (!(\\$path = realpath(\\$path?\\$path:getcwd()))) return array();    // bad path    // next line rids terminating \\ on drives (works since c: == c:\\ on PHP).  OK in *nix?    if (substr(\\$path,-1)==\\$dS) \\$path = substr(\\$path,0,-1);    if (is_null(\\$types)) \\$types = 2;    if (is_null(\\$levels)) \\$levels = 1;    if (is_null(\\$aFilter)) \\$aFilter=array();    // last argument may be passed as a list or as an array    \\$aFilter = array_slice(func_get_args(),3);    if (\\$aFilter && gettype(\\$aFilter)==\"array\") \\$aFilter=\\$aFilter;    \\$adFilter = array();    // now move directory filters to separate array:    foreach (\\$aFilter as \\$i=>\\$filter)                  // for each directory filter...      if ((\\$pos=stripos(\" \\$filter\",\"d\")) && \\$pos<3) {  // next line eliminates the 'd'          \\$adFilter[] = substr(\\$filter,0,\\$pos-1) . substr(\\$filter,\\$pos);          unset(\\$aFilter[\\$i]); }    \\$aFilter = array_merge(\\$aFilter);    // reset indeces    \\$aRes = array();                    // results, \\$aAcc is an Accumulator    \\$aDir = array(\\$path);    // dirs to check    for (\\$i=\\$levels>0?\\$levels++:-1;(\\$aAcc=array())||\\$i--&&\\$aDir;\\$aDir=\\$aAcc)      while (\\$dir = array_shift(\\$aDir))          foreach (scandir (\\$dir) as \\$fileOrDir)            if (\\$fileOrDir!=\".\" && \\$fileOrDir!=\"..\") {                if (\\$dirP = is_dir (\\$rp=\"\\$dir\\$dS\\$fileOrDir\"))                  if (pathFilter(\"\\$rp\\$dS\", \\$adFilter))                      \\$aAcc[] = \\$rp;                if (\\$i<\\$levels-1 && (\\$types & (2-\\$dirP)))                  if (pathFilter(\\$rp, \\$aFilter))                      \\$aRes[] = (\\$dirP?\"*\":\"\") . \\$rp; }    return \\$aRes; } ?> example usage: <?php define (\"_\", NULL); // this will find all non .jpg, non .Thumbs.db files under c:\\Photo \\$aFiles = dirList('c:\\Photo', _, 0, '-/\\.jpg\\$/i', '-/\\\\\\\\Thumbs.db\\$/'); \\$aFiles = dirList();    // find the files in the current directory // next lines will find .jpg files in non Photo(s) subdirectories, excluding Temporary Internet Files set_time_limit(60);        // iterating from the top level can take a while \\$aFiles = dirList(\"c:\\\\\", _, 0, '/\\.jpg\\$/i', '-d/\\\\\\\\Photos?\\$/i', '-d/Temporary Internet/i'); ?> Note that this function will consume a lot of time if scanning large directory structures (which is the reason for the '[-]d/.../' filters). Csaba Gabor from Vienna ```\n-1\n2bbasic at gmail dot com\n8 years ago\n``` <?php//-- Directory Navigation with SCANDIR//-- //-- optional placemenet\\$exclude_list = array(\".\", \"..\", \"example.txt\");if (isset(\\$_GET[\"dir\"])) {  \\$dir_path = \\$_SERVER[\"DOCUMENT_ROOT\"].\"/\".\\$_GET[\"dir\"];}else {  \\$dir_path = \\$_SERVER[\"DOCUMENT_ROOT\"].\"/\";}//-- until herefunction dir_nav() {  global \\$exclude_list, \\$dir_path;  \\$directories = array_diff(scandir(\\$dir_path), \\$exclude_list);  echo \"<ul style='list-style:none;padding:0'>\";  foreach(\\$directories as \\$entry) {    if(is_dir(\\$dir_path.\\$entry)) {      echo \"<li style='margin-left:1em;'>[`] <a href='?dir=\".\\$_GET[\"dir\"].\\$entry.\"/\".\"'>\".\\$entry.\"</a></li>\";    }  }  echo \"</ul>\";  //-- separator  echo \"<ul style='list-style:none;padding:0'>\";  foreach(\\$directories as \\$entry) {    if(is_file(\\$dir_path.\\$entry)) {      echo \"<li style='margin-left:1em;'>[ ] <a href='?file=\".\\$_GET[\"dir\"].\\$entry.\"'>\".\\$entry.\"</a></li>\";    }  }  echo \"</ul>\";}dir_nav();//-- optional placementif (isset(\\$_GET[\"file\"])) {  echo \"<div style='margin:1em;border:1px solid Silver;'>\";  highlight_file(\\$dir_path.\\$_GET['file']);  echo \"</div>\";}//-- until here//--//-- Because I love php.net?> ```\n-1\nsimon dot riget at gmail dot com\n4 years ago\n``` This is a simple and versatile function that returns an array tree of files, matching wildcards:<?php// List files in tree, matching wildcards * and ?function tree(\\$path){  static \\$match;  // Find the real directory part of the path, and set the match parameter  \\$last=strrpos(\\$path,\"/\");  if(!is_dir(\\$path)){    \\$match=substr(\\$path,\\$last);    while(!is_dir(\\$path=substr(\\$path,0,\\$last)) && \\$last!==false)      \\$last=strrpos(\\$path,\"/\",-1);  }  if(empty(\\$match)) \\$match=\"/*\";  if(!\\$path=realpath(\\$path)) return;  // List files  foreach(glob(\\$path.\\$match) as \\$file){    \\$list[]=substr(\\$file,strrpos(\\$file,\"/\")+1);  }    // Process sub directories  foreach(glob(\"\\$path/*\", GLOB_ONLYDIR) as \\$dir){    \\$list[substr(\\$dir,strrpos(\\$dir,\"/\",-1)+1)]=tree(\\$dir);  }    return @\\$list;}?> ```\n-1\nboen dot robot at gmail dot com\n10 years ago\n``` If you have a folder with many files and/or subfolders, doing a recursive scandir will likely either slow down your application, or cause a high rise in RAM consumption due to the large size of the generated array.To help with this, as well as to make processing of files in a folder easier, I wrote a function that reads a folder and its subfolders recursively, and calls a function upon each match.<?php/** * Calls a function for every file in a folder. * * @author Vasil Rangelov a.k.a. boen_robot * * @param string \\$callback The function to call. It must accept one argument that is a relative filepath of the file. * @param string \\$dir The directory to traverse. * @param array \\$types The file types to call the function for. Leave as NULL to match all types. * @param bool \\$recursive Whether to list subfolders as well. * @param string \\$baseDir String to append at the beginning of every filepath that the callback will receive. */function dir_walk(\\$callback, \\$dir, \\$types = null, \\$recursive = false, \\$baseDir = '') {    if (\\$dh = opendir(\\$dir)) {        while ((\\$file = readdir(\\$dh)) !== false) {            if (\\$file === '.' || \\$file === '..') {                continue;            }            if (is_file(\\$dir . \\$file)) {                if (is_array(\\$types)) {                    if (!in_array(strtolower(pathinfo(\\$dir . \\$file, PATHINFO_EXTENSION)), \\$types, true)) {                        continue;                    }                }                \\$callback(\\$baseDir . \\$file);            }elseif(\\$recursive && is_dir(\\$dir . \\$file)) {                dir_walk(\\$callback, \\$dir . \\$file . DIRECTORY_SEPARATOR, \\$types, \\$recursive, \\$baseDir . \\$file . DIRECTORY_SEPARATOR);            }        }        closedir(\\$dh);    }}?>Of course, because it is recursive, a folder with many levels of folders could potentially consume lots of memory, but then again, so can every other recursive scandir implementation here.BTW, there's also the RecursiveDirectoryIterator SPL class:http://bg.php.net/manual/en/class.recursivedirectoryiterator.phpwhich, even if using the same approach, will most likely be faster and hold down deeper levels (because it works on the C level), but this one will always work, regardless of settings... even on PHP4.Point is, avoid recursive scandir implementations. ```\n-3\ndsiembab at fullchannel dot net\n11 years ago\n``` Back in the saddle of scandir I wrote this function for a function that I needed to seperate directories from files. Since I am still learning from my last example way below I would figure I would add it so it can be criticized.<?phpfunction dirlist(\\$dir, \\$bool = \"dirs\"){   \\$truedir = \\$dir;   \\$dir = scandir(\\$dir);   if(\\$bool == \"files\"){ // dynamic function based on second pram      \\$direct = 'is_dir';    }elseif(\\$bool == \"dirs\"){      \\$direct = 'is_file';   }   foreach(\\$dir as \\$k => \\$v){      if((\\$direct(\\$truedir.\\$dir[\\$k])) || \\$dir[\\$k] == '.' || \\$dir[\\$k] == '..' ){         unset(\\$dir[\\$k]);      }   }   \\$dir = array_values(\\$dir);   return \\$dir;}?><?phpprint_r(dirlist(\"../\"));  //confirm array of subdirectoriesprint_r(dirlist(\"../\", \"files\") // confirm list on files in the directory ?> ```\n-2\ncHH\n13 years ago\n``` Since scandir() returns and array, here is a more concise method of dealing with the '.' and '..' problem when listing directories: <?php \\$target = '/'; \\$weeds = array('.', '..'); \\$directories = array_diff(scandir(\\$target), \\$weeds);     foreach(\\$directories as \\$value) {    if(is_dir(\\$target.\\$value))    {       echo \\$value.'<br />';    } } ?> ```\n-1\nNerbert\n1 year ago\n``` If you use array_diff() to eliminate \".\" and \"..\" you must use array_values() on the result because array_diff() will return an associative array, which may cause problems for a for loop beginning at 0.\\$files = array_values(array_diff(scandir(\\$directory), array('..', '.'))); ```\n-2\nbeingmrkenny at gmail dot com\n14 years ago\n``` I wrote this function to read a folder and place all the folders and sub folders it contains into an array.<?php// Initialise empty array, otherwise an error occurs\\$folders = array();function recursive_subfolders(\\$folders) {    // Set path here    \\$path = '/path/to/folder';        // Create initial \"Folders\" array    if (\\$dir = opendir(\\$path)) {        \\$j = 0;        while ((\\$file = readdir(\\$dir)) !== false) {            if (\\$file != '.' && \\$file != '..' && is_dir(\\$path.\\$file)) {                \\$j++;                \\$folders[\\$j] = \\$path . \\$file;            }        }    }        closedir(\\$dir);        // Then check each folder in that array for subfolders and add the subfolders to the \"Folders\" array.    \\$j = count(\\$folders);    foreach (\\$folders as \\$folder) {        if (\\$dir = opendir(\\$folder)) {            while ((\\$file = readdir(\\$dir)) !== false) {                \\$pathto = \\$folder. '/' . \\$file;                if (\\$file != '.' && \\$file != '..' && is_dir(\\$pathto) && !in_array(\\$pathto, \\$folders)) {                    \\$j++;                    \\$folders[\\$j] = \\$pathto;                    \\$folders = recursive_subfolders(\\$folders);                }            }        }        closedir(\\$dir);    }        sort(\\$folders);    return \\$folders;}\\$folders = recursive_subfolders(\\$folders);?>\\$folders now contains an array with the full paths to each subfolder. E.g.:Array(    => /path/to/folder/dir1    => /path/to/folder/dir1/subdir    => /path/to/folder/dir1/subdir/subsubdir    => /path/to/dolfer/dir2)This function has only been tested on Linux. ```\n-1\nhex at mail dot nnov dot ru\n7 years ago\n``` // Extreme minimal version recursive scan:function rscandir(\\$dir){  \\$dirs = array_fill_keys( array_diff( scandir( \\$dir ), array( '.', '..' ) ), array());  foreach( \\$dirs as \\$d => \\$v )    if( is_dir(\\$dir.\"/\".\\$d) )      \\$dirs[\\$d] = rscandir(\\$dir.\"/\".\\$d);  return \\$dirs;} ```\n-4\nasamir at asamir dot net\n12 years ago\n``` This is a modification of scanDirectories function that generates a list of all files in the chosen directory and all subdirectories of specific extentions \\$allowext<?phpfunction scanDirectories(\\$rootDir, \\$allowext, \\$allData=array()) {    \\$dirContent = scandir(\\$rootDir);    foreach(\\$dirContent as \\$key => \\$content) {        \\$path = \\$rootDir.'/'.\\$content;        \\$ext = substr(\\$content, strrpos(\\$content, '.') + 1);                if(in_array(\\$ext, \\$allowext)) {            if(is_file(\\$path) && is_readable(\\$path)) {                \\$allData[] = \\$path;            }elseif(is_dir(\\$path) && is_readable(\\$path)) {                // recursive callback to open new directory                \\$allData = scanDirectories(\\$path, \\$allData);            }        }    }    return \\$allData;}\\$rootDir = \"www\";\\$allowext = array(\"zip\",\"rar\",\"html\");\\$files_array = scanDirectories(\\$rootDir,\\$allowext);print_r(\\$files_array);?> ```\n-2\nphpnet at novaclic dot com\n10 years ago\n``` Was looking for a simple way to search for a file/directory using a mask. Here is such a function. By default, this function will keep in memory the scandir() result, to avoid scaning multiple time for the same directory. Requires at least PHP5. <?php function sdir( \\$path='.', \\$mask='*', \\$nocache=0 ){     static \\$dir = array(); // cache result in memory     if ( !isset(\\$dir[\\$path]) || \\$nocache) {         \\$dir[\\$path] = scandir(\\$path);     }     foreach (\\$dir[\\$path] as \\$i=>\\$entry) {         if (\\$entry!='.' && \\$entry!='..' && fnmatch(\\$mask, \\$entry) ) {             \\$sdir[] = \\$entry;         }     }     return (\\$sdir); } ?> ```\n-2\n``` function ScaniDir(\\$path){  \\$ptd= opendir(\\$path);  while (( \\$ptdf=readdir(\\$ptd))){      \\$srca = \\$path.'/'.\\$ptdf;     if (is_file(\\$srca) and pathinfo(\\$srca, PATHINFO_EXTENSION)=='pdf') {\\$files['src'][]=\\$srca;\\$files['name'][]=explode('.',\\$ptdf);}         }return \\$files ; } ```\n``` using sort of scandir() that returns the content sorted by Filemodificationtime.<?phpfunction scandir_by_mtime(\\$folder) {  \\$dircontent = scandir(\\$folder);  \\$arr = array();  foreach(\\$dircontent as \\$filename) {    if (\\$filename != '.' && \\$filename != '..') {      if (filemtime(\\$folder.\\$filename) === false) return false;      \\$dat = date(\"YmdHis\", filemtime(\\$folder.\\$filename));      \\$arr[\\$dat] = \\$filename;    }  }  if (!ksort(\\$arr)) return false;  return \\$arr;}?>returns false if an error occuredotherwise it returns an array like this.Array(    => DSC00023.JPG    => DSC00024.JPG    => MOV00055.MPG    => DSC00056.JPG    => DSC00057.JPG    => DSC00058.JPG    => DSC00083.JPG    => album.txt) ```", null, "" ]

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[ "### aman_naughty's blog\n\nBy aman_naughty, history, 10 months ago,", null, ",", null, "Code : code\n\nMy approach is : For any {i,j} I maintain 4 colors dp[i][j],dp[i][j],dp[i][j] and dp[i][j];\n\ndp[i][j][k] denotes the number of ways to fill the board[0..i][0..j] using the color k at position i,j.\n\nTo fill dp[i][j][k] we need only summation of (dp[i-1][j][l]+dp[i][j-1][l]) where l varies from 0 to 3 and l is not equal to k.\n\nIn simple words to fill i,j with color k number of ways would be we can fill i-1,j with color other than k and i,j-1 with color other than k.\n\nIn the end my answer would be summation of dp[n][k] where 0<=k && k<=3;\n\nWhy is this approach wrong as I am getting wrong answer for n=2 test case?\n\nAny help is appreciated.", null, "", null, "Comments (10)\n » 10 months ago, # | ← Rev. 3 →   There is at most $4^3$ configurations for a column, so you can model the problem as a $dp[i][j][k][l]$, that means the number of configurations ending at column $i$ with colors $j$ $k$ $l$ for this column.$dp[i][j][k][l] = 0;$ if $j = k$ or $k = l$.$dp[i][j][k][l]$ = $SUM(dp[i - 1][t][f][g])$ where $t \\neq f$ and $f \\neq g$ and $t \\neq j$ and $k \\neq f$ and $l \\neq g$.This is $O(N*4^6)$, dont use $mod$(%) operator, just substract the number if he becomes greater than $10^9 + 7$, i think this is enough to fit in the time.EDIT: The answer is $SUM(dp[n][i][j][k])$.\n• » » I believe your approach is wrong as ABA in a column is still a vaild coloring.\n• » » » yeah, i was wrong, i did some changes.\n• » » » » Your approach seems good but can you tell what is the flaw in my approach\n• » » » » » I dont know if i will can explain, but you are not counting all combinations, because if you sum $dp[i][j - 1][l]$ + $dp[i - 1][j][l]$, is all combinations for fixed values for column $j$ in positions 1..i-1 plus all combinations for fixed values for line $i$ in positions 1..j-1.\n• » » » » » » your approach will not fit in time limit as n can be 1e5\n• » » » » » » » I think if you do a interative dp and dont use mod operador, it fit in time cause are 409600000 operations in the worst case, i think it is less than 1s\n• » » » » » » » » Thanks for the reply. I implemented what you suggested but it is giving wrong answer for n=2 case.code\n• » » » » » » » » » The lines 18-20 isnt right, check my ac solution. codeint dp; int Solution::solve(int A) { const int mod = int(1e9) + 7; const int D = 4; memset(dp, 0, sizeof dp); for(int j = 0 ; j < D ; j++){ for(int k = 0 ; k < D ; k++){ if(k == j){ continue; } for(int l = 0 ; l < D ; l++){ if(l == k){ continue; } dp[j][k][l] = 1; } } } for(int i = 1 ; i < A ; i++){ for(int k = 0 ; k < D ; k++){ for(int l = 0 ; l < D ; l++){ if(l == k) continue; for(int d = 0 ; d < D ; d++){ if(l == d) continue; for(int e = 0 ; e < D ; e++){ if(e == k) continue; for(int f = 0 ; f < D ; f++){ if(f == e || f == l) continue; for(int g = 0 ; g < D ; g++){ if(g == f || g == d) continue; dp[i][k][l][d] += dp[i - 1][e][f][g]; if(dp[i][k][l][d] >= mod){ dp[i][k][l][d] -= mod; } } } } } } } } int sum = 0; for(int j = 0 ; j < D ; j++){ for(int k = 0 ; k < D ; k++){ if(k == j){ continue; } for(int l = 0 ; l < D ; l++){ if(l == k){ continue; } sum += dp[A - 1][j][k][l]; if(sum >= mod){ sum -= mod; } } } } return sum; } \n• » » Thanks a lot for sharing your solution ! I've been looking for the solution to this problem for quite a few months now !" ]

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[ "# Projected Gradient Descent for Constrained Optimization\n\nIn this post we describe how to do gradient descent with constraints. We first describe the problem, including why we can’t naively apply gradient descent and a few cases where this is necessary. Then we describe projected gradient descent and apply this to the popular Boston housing prices dataset to do non-negative least squares. We finish with a brief discussion.\n\n## Problem\n\nWe are interested in the problem of minimizing a convex, differentiable function", null, "over a compact convex set", null, ". This has many applications, including constrained least squares (linear regression with constraints on parameters), logistic regression with constraints, and certain types of boosting and SVMs. For example consider non-negative linear regression. In this case we know that parameters should all be non-negative. For instance, more rooms should have a positive effect on the prices of a house. While using ordinary least squares will eventually learn this, imposing the constraint structure can improving learning in small samples.\n\nA general method for optimizing convex functions is gradient descent, which iteratively updates our", null, "as follows:\n\n(1)", null, "here", null, "is the step size. However even if", null, "is in the constraint set", null, ", there is no guarantee that", null, "is. There are two very common ways to handle this: one is to project", null, "onto", null, ", and another is to minimize a linear approximation to the function", null, "over", null, ".\n\n## Projected Gradient Descent for Non-negative Least Squares\n\nConsider again non-negative least squares, where the coefficients cannot be negative. One applies this because of domain knowledge about the problem: for instance more rooms will not lower the price of a house, and similarly if the effect is a count it cannot be negative. If", null, ", the goal is to find\n\n(2)", null, "While we don’t need to enforce the constraint, it can help improve learning in small samples. One way to handle this is via projected gradient descent. The idea is to do one step of gradient descent, and then find the closest non-negative point in", null, "to that step. More formally, let", null, "and", null, ". Then projected gradient descent updates\n\n(3)", null, "Here the function", null, "simply zeroes out negative elements of a vector.\n\n### Application: Non-negative Least Squares for Boston Housing Prices\n\nLet’s apply this to the Boston housing prices dataset. We will use four features: crime (per capita crime rate), rooms (average number of rooms per dwelling), tax (property tax rate), and percent of population of lower socioeconomic status. We expect the only positive coefficient to be rooms. Thus we flip the sign of all the other observations and then do projected gradient descent. We first load some libraries, write a function to calculate the gradient of the least squares objective function, and load the data.\n\nfrom sklearn.datasets import load_boston\nimport numpy as np\nnp.set_printoptions(suppress=True)\n\ndef deriv_f(beta,X,y):\nreturn -2*np.dot(X.T,y)+2*np.dot(np.dot(X.T,X),beta)\n\ndataset['feature_names']\n\n\nBefore running projected gradient descent, we can first run naive gradient descent and then compare the convergence empirically.\n\nX = np.ones((old_X.shape,5))\nX[:,1:5]=old_X[:,np.array([0,5,9,12])]\np=np.shape(X)\n\nbeta=np.zeros(p)\ni=0\ni+=1\nif i%5000==0:\nprint(i)\nprint(beta)\nprint(beta)\n\n\nAs a unit test, we can check the results against naively solving least squares via matrix inversion (note: you generally should not invert a matrix to solve linear least squares, however for relatively small problems where features are not highly correlated, it can be fine to do so).\n\nprint(i)\nprint(beta)\nprint(np.dot(np.linalg.inv(np.dot(X.T,X)),np.dot(X.T,y)))\n\n122627056\n[-1.41321706 -0.06157611 5.24849 -0.00501863 -0.53485144]\n[-1.41492797 -0.06157926 5.2487206 -0.00501849 -0.53483514]\n\n\nGradient descent took over 122 million iterations, and the results from gradient descent and directly solving are nearly identical (conclusion: you generally shouldn’t use gradient descent to solve least squares without a good reason). Now let’s try flipping the signs of relevant features and doing projected gradient descent to enforce non-negative coefficients.\n\nX[:,np.array([0,1,3,4])]=-X[:,np.array([0,1,3,4])]\nbeta=np.zeros(p)\ni=0\n#this is the projection step\nbeta[beta<0]=0\ni+=1\nif i%5000==0:\nprint(i)\nprint(beta)\nprint(beta)\n\n\nWe can undo the sign reversals to examine the learned coefficients:\n\nbeta[np.array([0,1,3,4])]=-beta[np.array([0,1,3,4])]\nX[:,np.array([0,1,3,4])]=-X[:,np.array([0,1,3,4])]\nprint(i)\nprint(beta)\nprint(np.dot(np.linalg.inv(np.dot(X.T,X)),np.dot(X.T,y)))\n\n116085845\n[-1.41321706 -0.06157611 5.24849 -0.00501863 -0.53485144]\n[-1.41492797 -0.06157926 5.2487206 -0.00501849 -0.53483514]\n\n\nThis took approximately 116 million iterations, and also learned very close parameters to solving linear least squares using matrix inversion.\n\n## Discussion\n\nIn this post we described projected gradient descent for constrained convex optimization. We described the special case algorithm for non-negative least squares and applied it to the Boston housing price dataset. We found the learned model very similar. Using projection for non-negative least squares is thus likely more useful when effects are small and noise is sufficiently high that the learned model is likely to get the coefficient sign incorrect.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# Recursion\n\nIt’s wired\n\nBut the method of java can call its self\n\nWhen a method  call its self is called recursion\n\n#### recursion syntax\n\n```1.\treturntype methodname(){\n2.\t//code to be executed\n3.\tmethodname();//calling same method\n4.\t}\n```\n\n#### Lets understand it by example :\n\n```public class RecursionExample {\nstatic void show(){\nSystem.out.println(\"hello\");\nshow();\n}\n\npublic static void main(String[] args) {\nshow();\n}\n}\n```\n``````hello\nhello\n...\njava.lang.StackOverflowError``````\n\nin this program show() method call itself every  time\n\n#### another example:\n\n```class Factorial {\n\nstatic int factorial( int n ) {\nif (n != 0) // termination condition\nreturn n * factorial(n-1); // recursive call\nelse\nreturn 1;\n}\n\npublic static void main(String[] args) {\nint number = 4, result;\nresult = factorial(number);\nSystem.out.println(number + \" factorial = \" + result);\n}\n}\n```\n``4 factorial = 24``\n\n### Why we use recursion\n\n• Recursion help to write simple code\n• for problems like the tower of Hanoi we use recursion\n• We can write such codes also iteratively with the help of a stack data structure.\n• For example, use recursion in the Iterative Tower of Hanoi.", null, "" ]

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[ "2 of 22 total results match your filters: statistics Clear all filters\n\nRead about courses in a range of fields at a variety of institutions where students have learned TIER-like methods of reproducible research. Course syllabi, exercises, project instructions and other course documents are available for download.\n\n## Introduction to Probability and Statistics\n\nSmith College\nStatistical and Data Sciences 220, Spring 2017\nInstructor: Amelia McNamara\n\nAn application-oriented introduction to modern statistical modeling and inference: study design, ...\n\n## Introduction to Probability and Statistics\n\n### Smith College\n\nStatistical and Data Sciences 220, Spring 2017\nInstructor: Amelia McNamara\n\nAn application-oriented introduction to modern statistical modeling and inference: study design, descriptive statistics, data visualization, random variables, probability and sampling distributions, point and interval estimates, hypothesis tests, resampling procedures, and multiple regression. A wide variety of applications from the natural and social sciences will be used. Classes meet for lecture/discussion with activities and exercises that emphasize analysis of real data.\n\nStudents complete weekly lab assignments in R and RMarkdown, and a final data analysis project. The final project is worth 25% of the course grade, and must be reproducible. Students work in groups to complete the project on a topic of their choice. Students have a number of milestone assignments along the way, including an initial proposal, revised proposal, data file submission, data appendix, and a final technical report. The technical report includes all the code needed to complete the analysis.\n\nStudents worked through labs introducing R and RStudio (http://www.science.smith.edu/~amcnamara/sds220/labs/intro_to_r.html) and introducing data analysis (http://www.science.smith.edu/~amcnamara/sds220/labs/intro_to_data.html). Both labs were developed by the OpenIntro Statistics group, and include expository videos explaining some of the topics. The OpenIntro team has created an R package called oilabs, which includes a lab report template that can be accessed through RStudio.\n\n## Multiple Regression\n\nSmith College\nStatistical and Data Sciences 291, Spring 2016\nInstructor: Amelia McNamara\n\nTheory and applications of regression techniques; linear and nonlinear multiple regression models...\n\n## Multiple Regression\n\n### Smith College\n\nStatistical and Data Sciences 291, Spring 2016\nInstructor: Amelia McNamara\n\nTheory and applications of regression techniques; linear and nonlinear multiple regression models, residual and influence analysis, correlation, covariance analysis, indicator variables and time series analysis. This course includes methods for choosing, fitting, evaluating and comparing statistical models and analyzes data sets taken from the natural, physical and social sciences.\n\nStudents worked in small groups to produce a data analysis on a topic of their choice. The project is 25% of the final course grade. Students had to work in R and RMarkdown, turn in a data appendix, and document all their data cleaning and analysis in their final report." ]

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[ "# How do you graph 0<=x-y<=2?\n\nAug 10, 2017\n\nSee below\n\n#### Explanation:\n\n$0 \\le \\left(x - y\\right) \\le 2$\n\nThis represents two inequalities, namely:\n\n$0 \\le \\left(x - y\\right)$ (i)\n\nand\n\n$\\left(x - y\\right) \\le 2$ (ii)\n\nFirst let's consider (i):\n$0 \\le \\left(x - y\\right) \\to - y + x \\ge 0$\n\n$- y \\ge - x$\n\n$y \\le x$\n\nThis inequality is represented graphically by all points on the $x y -$plane on or under the line $y = x$.\n\nSimilarly for (ii):\n\n$\\left(x - y\\right) \\le 2 \\to - y + x \\le 2$\n\n$- y \\le - x + 2$\n\n$y \\ge x - 2$\n\nThis inequality is represented graphically by all points on the $x y -$plane on or above the line $y = x - 2$.\n\nCombining these two results produces the graph below.\n\ngraph{(x-y)(x-y-2)<=0 [-11.25, 11.25, -5.63, 5.62]}" ]

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[ "# Using SVD in demographic modeling\n\nA core objective of demographic modeling is finding empirical regularities in age patterns in fertility, mortality and migration. One method to achieve this goal is using Singular Value Decomposition (SVD) to extract characteristic age patterns in demographic indicators over time. This post describes how SVD can be used in demographic research, and in particular, mortality estimation.\n\n## Background\n\nThe SVD of matrix $$X$$ is $X = UDV^T$ The three matrices resulting from the decomposition have special properties:\n\n• The columns of $$U$$ and $$V$$ are orthonormal, i.e. they are orthogonal to each other and unit vectors. These are called the left and right singular vectors, respectively.\n• $$D$$ is a diagonal matrix with positive real entries.\n\nIn practice, the components obtained from SVD help to summarize some characteristics of the matrix that we are interested in, $$X$$. In particular, the first right singular vector (i.e. the first column of $$V$$) gives the direction of the maximum variation of the data contained in $$X.$$ The second right singular vector, which is orthogonal to the first, gives the direction of the second-most variation of the data, and so on. The $$U$$ and $$D$$ elements represent additional rotation and scaling transformations to get back the original data in $$X$$.\n\nSVD is useful as a dimensionality reduction technique: it allows us to describe our dataset using fewer dimensions than implied by the original data. For example, often a large majority of variation in the data is captured by the direction of the first singular vector, and so even just looking at this dimension can capture key patterns in the data. SVD is closely related to Principal Components Analysis: principal components are derived by projecting data $$X$$ onto principal axes, which are the right singular vectors $$V$$.\n\n## Use in demographic modeling\n\nUsing SVD for demographic modeling and forecasting first gained popularity after Lee and Carter used the technique as a basis for forecasting US mortality rates. They modeled age-specific mortality on the log scale as\n\n$\\log m_x = a_x + b_x \\cdot k_t$ where\n\n• $$a_x$$ is the mean age-specific mortality schedule across all years of analysis,\n• $$b_x$$ is the average contribution of age group $$x$$ to overall mortality change over the period, and\n• $$k_t$$ is the incremental change in period $$t$$.\n\nThe latter two quantities are obtained via SVD of a time x age matrix of demeaned, logged mortality rates: $$b_x$$ is the first right singular vector, while $$k_t$$ is the first left singular vector multiplied the first element of $$D$$.\n\nMore recently, SVD has become increasingly used in demographic modeling; for example Carl Schmertmann et al. used it to model and forecast cohort fertility, Sam Clark to estimate age schedules of mortality with limited data, and Emilio Zagheni, Magali Barbieri and myself to model subnational age-specific mortality.\n\n### Example: age-specific mortality\n\nImagine you have observations of age-specific mortality rates in multiple years. Create a matrix, $$X$$, where each row represents the age-specific mortality rates in a particular year. Modeling of mortality rates is often done on the log scale (to ensure rates are positive), so you may want to take the log of $$X$$. Then do a SVD on this matrix - in R this is as easy as svd(x). The age patterns of interest are then contained in the resulting v matrix; so for example svd(x)\\$v[,1:3] would give you the first three age ‘principal components’ of your matrix.", null, "For example, the first three principal components of US male mortality by state over the years 1980-2010 are plotted below. Each component has a demographic interpretation - the first represents baseline mortality, the second represents higher-than-baseline child mortality, and the third represents higher-than-baseline adult mortality.", null, "For modeling, the idea is that different linear combinations of these components allow you to flexibly represent a wide range of different mortality curves. For example, log-mortality rates could be modeled as\n\n$\\log m_x = \\beta_1 Y_{1x} + \\beta_2 Y_{2x} + \\beta_3 Y_{3x}$ where the $$Y_{.x}$$’s are the principal components above and the $$\\beta$$’s are to be estimated. The plot below shows four different mortality curves derived from the US male principal components with different coefficient settings. You can also play with different settings interactively here.", null, "### Example: race-specific opioid mortality\n\nThis technique of representing and modeling underlying age patterns need not be restricted to modeling all-cause mortality. For example, SVD proves useful when looking at deaths due to opioid overdoses by race and state in the US. Even though opioid overdoses are rapidly increasing for both the black and white population, overdoses are still a relatively rare event, and so death rates calculated from the raw data suffer from large stochastic/random variation.\n\nFor example, the chart below shows age-specific opioid mortality rates by race for North Carolina in 2004.1 As you can see, for the black population there are quite a few age groups were there are zero observed deaths, so the observed mortality rate is zero. However, given what we know about how mortality evolves over age, the zero observed death rates are likely due to random variation.", null, "Even though age patterns are noisy at the state level, we have an idea of age patterns by race in opioid mortality in the national level. So we can use these national age patterns - via information captured in a SVD - to help model underlying mortality rates at the state level.\n\nThe figure below shows the first two principal components derived using SVD from race-specific opioid mortality in the US over the years 1999-2015. The first principal component again represents a baseline mortality schedule for opioid-related deaths for each race. The second principal component represents the contribution of each age group to mortality change over time. Notice the ‘double-humped’ shape for the white population - this is driven by heroin deaths being concentrated at younger ages, and prescription opioid-related deaths being concentrated at older ages.", null, "Similar to the example above, we can use these principal components as a basis of a regression framework to estimate underlying age-specific mortality rates by age. Results from such a model for North Carolina in 2004 are shown below. The dots represent mortality rates calculated from the raw data, as above. The lines and associated shaded area represent estimates of the underlying mortality rates with 95% uncertainty intervals. These were obtained from a model that utilized information from the principal components. Instead of dealing with zero observed deaths, we now have estimates that give more plausible values for the underlying mortality rates.", null, "## Summary\n\nSVD is a useful technique to extract the main characteristics of age patterns in demographic indictors. These structural age patterns are useful to get a better idea of underlying processes when available data are sparse or noisy. Age patterns derived from SVD can be flexibly shifted and adjusted based on available data. Built-in functions in R make it relatively easy to use SVD to better understand, model and project demographic indicators.\n\n1. cross-promotional plug: you can play with this data yourself with the help of the narcan R package, which Mathew Kiang and I are working on." ]

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[ "## Free BMI Calculator - Calculate Your Body Mass Index\n\nAug 9, 2019 knowing your body mass index, or bmi, can be useful for assessing and adjusting your weight. It is not the most accurate measure of how much.Body mass index calculator, bmi value helps in classifying obesity as enter your weight and height in the calculator below(in either imperial or metric) to find.Bmi calculator for adults to determine body mass index or bmi, given an individuals weight and height. Weight can be bmi is a way of measuring a persons weight relative to his or her height.Body mass index, or bmi, is a person’s weight in kilograms divided by the square of hisher height in meters. The national institute of health (nih) has now defined bmi to be the deciding parameter to k.Choose your leader. Clickchoose the leader you want to follow. Jean current weight: 14st 1. 5lb height: 5ft 6. Cathal current weight: bmi calculator.The body mass index (bmi) is a tool that can be used to tell how healthy a person's weight is.   so, if you haven’t caluculate yet your bmi,plese calculate it right now. Some gym have their own bmi.\n\n## \"BMI Calculator\" - Free Health ... - Wolfram|Alpha Widgets https://www.wolframalpha.com › widgets › gallery › view\n\nCalculate a person's bmi by dividing his weight in pounds by his height in inches squared. Because the original calculations for bmi were done using the metric system, it is necessary to multiply.Use our handy bmi (body mass index) calculator below to calculate your body fat estimate. Bmi is a useful tool to gauge your risk against various diseases and.Image from wikipedia. Body mass index, commonly referred to as bmi, is based simply upon the height and weight of an individual. Bmi is calculated by the.A person is morbidly obese (extreme obesity) if his or her bmi is over 40. To calculate the bmi using pounds, divide the weight in pounds by the height in.Jun 16, 2019 a great bmi calculator with advanced features for body mass index. Avoid the cdc and nih calculators calling overweight too often.Standard body mass index calculator. Standard body mass index · metric body mass index calculator. Determine your body mass easily by entering your.How to use height and weight data to calculate bmi · 05. A balanced group discussion in local language on.Using the formula below, calculate the bmi for each individual, and compare and. Orgwikifile:boxplotmitminmaxabstand.Learn more about gaining and encouraging on the gaining faq page. No matter what your criteria are, weve got a search for it: age, location, weight, bmi, to geographical coordinates to calculate your r.\n\n## BMI Calculator | Motivation Weight Management\n\ncan you get high off omeprazole 20 mg\n\n## Body mass index - Wikipedia\n\ncheap cheap discount sale viagra viagradrugs.net\n\n## BMI Calculator (free) - Apps on Google Play https://play.google.com › store › apps › details › id=free.wk.mybodymass\n\nJan 5, 2013 sir the bodymass index that you (and the national health service) to find your new bmi, try the new bmi calculator written by nick hale. Of the bmi issue, a good place to start is the wikip.The ideal body weight calculator calculates ideal body weight by the devine formula.Body mass index is a mathematic calculation that provides a measure for assessing the health risks associated with body weight. Numerous research studies.Euromonitor is the worlds leading independent provider of strategic market research. Get data & analysis on thousands of products & services globally.Bmi chart pounds bmi calculator chart easy weightloss solutions, body mass index chart female version for women, mbmi scale en your body mass index is free bmi calculator calculate your body mass index.Body mass index or simply bmi which is also known as the quetelet indicator is a method of calculation used to measure an individuals fats in the body based.Calculate your body mass index (bmi), basal metabolic rate (bmr) and body fat percentage with our simple to use, interactive free calculators.   calculate your bmi and find out if your bmi is within nor.Free online body mass index (bmi) calculator. Easily estimate your bmi based on your weight and height (feet and inches, pounds, or meters and centimeters and kilograms).   use this calculator to easily.Whats the best way to determine a healthy weight? you can find out what your body mass index is by using the interactive bmi calculator, or you can look at your.\n\n## BMI Calculator Malaysia | Calculator.com.my https://www.calculator.com.my › bmi-calculator\n\nTo find your bmi you take your weight in kilograms and divide it by the square of according to the wikipedia article the bmi formula was derived by a belgian.Calculate bmi. Body mass index calculator. Our calculators.   our website will calculate the results, and show them, along with the body mass index classification established by the world health organiz.We show you how to manually calculate bmi using the bmi formula. Easy step by step examples are given to help you determine your weight status.   calculating bmi is straight forward, the formula is easy.Body mass index calculator measures the relative weight (bmi) based on height and weight mass for men, women or children.   the bmi is used as a simple method to assess how much your body weight departs.Calculating body mass index of your pet within seconds.\n\ncalculate bmi for male\n\n## Healthcare Math: Weight Management Calculations https://www.iccb.org › adulted › curriculum&resources › context_math\n\nBmi a measurement of the relative percentages of fat and muscle mass in to calculate ones bmi, multiply ones weight in pounds and divide that by the.170 190 210 230 250 270. Weight pounds 290 310 330 350. Underweight bmi 30 50 70.Calculate bmi the body mass index formula. Share how to pay off your mortgage.Sep 13, 2009 http:en. Citizendium. Orgwiki?title=bodymassindex&printable= bmi is calculated as weight in kilograms divided by height in meters squared. The bmi may overestimate percentage of body fat i.The most common measure of obesity is the body mass index or bmi. A person is considered overweight if his or her bmi is between 25 and 29. 9; a person is.Calculate your bmi with the bmi calculator for women and men. With guidance on ideal weight and overweight conditions.   however, it has been determined that the bai index is not superior to the bmi, pa.Learn to estimate whether or not you are a healthy weight by calculating your body mass index (bmi) and measuring your waist circumference.The body mass index, or bmi, is considered worldwide as one of the best ways to measure an individual’s health based on their weight. By calculating your body mass index.Calculate your bmi. Bmi is a measure of body fat, it is calculated using the formula developed by the belgian mathematician and natural scientist adolphe quetelet in late 60s of 19 th century." ]

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[ "### awoo's blog\n\nBy awoo, history, 4 years ago, translation,", null, "1312A - Two Regular Polygons\n\nIdea: BledDest\n\nTutorial\nSolution (vovuh)\n\n1312B - Bogosort\n\nIdea: Roms\n\nTutorial\nSolution (Roms)\n\nTutorial\n\n1312D - Count the Arrays\n\nIdea: BledDest\n\nTutorial\nSolution (BledDest)\n\n1312E - Array Shrinking\n\nIdea: MikeMirzayanov\n\nTutorial\n\n1312F - Attack on Red Kingdom\n\nIdea: BledDest\n\nTutorial\nSolution (BledDest)\n\n1312G - Autocompletion\n\nTutorial\nSolution (BledDest)", null, "", null, "Comments (82)\n| Write comment?\n » fast editorial, thanks!\n » In G we can maintain value \"the shortest distance to the smallest line in the lexicographic order from the current prefix\" in dfs. To update it just check the distance from the current vertex + 1. And to pass it to the next vertex, we need to add to the value the number of vertices from the S that we went through in other sons. It is O(n). Code.\n• » » 4 years ago, # ^ | ← Rev. 2 →   I solved G by the same way. And This is my code.\n• » » » Could you plz explain your idea more clearly?thx :P\n• » » » » Main challenge is speeding up computation of f(u) = min d(v) + rank of u in subtree rooted at v, over all ancestors v of node u. Suppose we have this computed for node u, and wish to update computation for children c of u. The difference is exactly the size of sibling subtrees that come before c. This difference affects subtrees of each ancestor equally, so after adding that to existing f(u), we only need to consider the additional case for a subtree beginning at c so you get something like f(c) = min(f(u)+siblings, d(c)).Another Example (sorry for all the junk — just look for the dfs method which is 8 LOC)\n• » » » » » thx :P\n• » » » » » » 4 years ago, # ^ | ← Rev. 2 →   another way to find rank(u) in subtree rooted at v is tin(u)-tin(v)+(isgood(v)) (we increment time only when we encounter a good string)\n » can anyone explain in problem D why we are multiplying by 2^(n — 3) ?\n• » » Each element will appear either before the maximum number, or after it. Therefore each of the N-3 numbers (that aren't the maximum and the number we have to repeat) has two options, so we have to mulitiply by 2, N-3 times.\n• » » 4 years ago, # ^ | ← Rev. 2 →   Since there are n elements we are to choose (n-1) elements and one element will be copy of one of n-1 elements choosen by us. Now we can not make copy of max element as it violates the condition so we are left with n-2 elements so we can choose one of them.Let x be the position of max element so the element which has duplicate must be persent on either side because if it will be persent on single side the sequence can not be strictly increasing or decreasing so we have max element at x and two same elements so now we are left with n-3 elements since there are two choices for every element to either go on right hand side of max element or LHS of that max element . so 2^(n-3) factor is involved. why is it so? Since we can arrange a array in strictly increasing order if we dont have duplicates. Hope it was clarified\n• » » » we assume one element as left or right side of max .Isn't it possible that element become max for another combination?So why not we use 3^n-3?As there can be three position?\n• » » » » 4 years ago, # ^ | ← Rev. 2 →   Notice that we have chosen the current set of (n-1) elements already and are then placing the elements in the desired order. Hence for the current set of elements, only one will be the peak element. For another set, it might be so that an element currently on the left/right becomes the peak, but for that the initial (n-1) elements chosen will be different. Hence the factor of 3 doesn't appear here.\n• » » » » » great explanation .Thanks!\n• » » » » We have already selected elements now how can our combo differ?\n » int mul(int x, int y) { return (x * 1ll * y) % MOD; } For what we use this func, why we cant make x-long long, and y-long long, and multiply them?\n• » » 4 years ago, # ^ | ← Rev. 4 →   Of course, we can, there is no difference: long long 1.08 s and int 1.08s for calculation of $2^{28}!$. Only one think that you should do is declare mod as constant. If no const — long long 3.33s. It is because division by a constant can be calculated without integer division.\n• » » » Hey Bro can you elaborate this further It is because division by a constant can be calculated without integer division\n• » » » » Book: Hacker's Delight, 2nd Edition, Chapter 10: Integer Division By Constants. Main idea in few words: change $\\dfrac{x}{y}$ by multiplication $x \\cdot y^{-1}$, where $y$ is constant.\n » Can anyone please explain E problem it would be great help. Thanks in advance :)\n• » » I'll try to go along the lines of the solution:Let $dp[i][j]$ be the value of the single remaining element, when subarray $[i:j]$ is reduced using the given operation. If there is no way to reduce this subarray into a single element, $dp[i][j]$ will be $-1$.How do you compute $dp[i][j]$? Consider index $k$, such that $i \\le k \\lt j$. We divide subarray $[i:j]$ into two halves: $[i:k]$ and $[k+1:j]$. Base case: when subarray is of size 1 $(i=j)$, answer will obviously be the number in that subarray. Now for any $k$, if $dp[i][k] = dp[k+1][j]$, i.e. we can divide subarray into two halves such that both halves are reduced to the same number, we can combine the two numbers, hence $dp[i][j] = dp[i][k] + 1$. This is one half of the solution. We now compute the minimum number of partitions that can be reduced to size 1.Let $dp2[i]$ : minimum number of partitions required for array $[1:i]$. We compute this in this way: Take index $k$, $1 \\le k \\le i$. If $[k:i]$ can be reduced to a single element, i.e. $dp[k][i] \\ne -1$, then we can find optimal partitioning of subarray $[1:k-1]$ and just add one to the answer. Formally, $dp2[i] = min(dp2[i], dp2[k] + 1)$, for all $k$ such that $dp[k][i] \\ne -1$.The required answer is $dp[n]$. Here's my solution for reference: 72913085. Hope this helped.\n• » » » Can you explain why will dp[i][j] reduce to a single unique value?I do feel it's intuitive but still why is it impossible for having two different ways of combining elements of a subarray and reaching two different final merged values?\n• » » » » consider sequence $2^{a_1}, 2^{a_2}, ..., 2^{a_n}$ combining two adjacent equal values $a_i = a_{i+1}$ is equivalent to merging two values in to one value $2^{a'} = 2^{a_i} + 2^{a_{i+1}} = 2 ^ {a_i + 1}$. So if [l, r] can be reduced to a single value, it must be unique.\n• » » » » » 4 years ago, # ^ | ← Rev. 2 →   This proof is from tmwilliamlin168 explanation\n• » » » » » How come converting into a sum helped to prove uniqueness?\n• » » » » » » I think he means that, the \"merging\" operation is akin to adding 2 similar powers of 2, sum of a few powers of 2,if reduced to a single power of 2, then that term should be unique\n• » » » » 4 years ago, # ^ | ← Rev. 2 →   We can prove that a subarray $[i:j]$ can be divided in atmost one way, i.e. there can be only one such $k$, for which $dp[i][k] = dp[k+1][j]$.For this, we define a fully reducible subarray as a subarray which can be reduced to a single element. We claim two things: removal of elements from a fully reducible subarray will only reduce the value of the remaining element, provided the remaining subarray is a fully reducible subarray. Similarly, addition of elements will only increase the value of the remaining element (you can try this out).Hence, if $dp[i][k] = dp[k+1][j]$, there is no possible way to transfer elements from subarray $[i:k]$ to $[k+1:j]$, or vice versa, such that $dp[i][k'] = dp[k'+1][j]$, for another $k'$.\n• » » » » We can also prove by contradiction,Lets say there are two values of K, i.e K1 and K2, such that dp(i, K1) = dp(K1 + 1,j) = X and dp(i, K2) = dp(K2 + 1, j) = Y.Lets assume K1 < K2,If that is the case then according to the dp definition, the subarray (i, K1) reduced to the number X, and the subarray (K1 + 1, j) also reduced to X whereas the subarrays (i, K2) and (K2 + 1, j) reduced to Y, Since K1 < K2, the subarray (i, K1) is a prefix of (i, K2) thus X < Y. And (K2 + 1, j) is a suffix of (K1 + 1, j) which suggests that X > Y. This leads to a contradiction.\n• » » » » » Amazing, thanks a lot :)\n• » » » » » Thanks a lot!\n• » » » » » Thanks a lot :)\n• » » » This explanation was better than the one in editorial. Thanks a lot!\n• » » 4 years ago, # ^ | ← Rev. 2 →   Thanks a lot to all in this thread This explanation is much better than that of editorial\n » in problem F, why are five lines enough to determine all the other values?\n• » » Let's consider that the number of remaining soldiers is i.Because x,y,z<=5.You only need the answer to[i-5,i-1] to update i.When you know the answer to i,you delete the answer to (i-5),and add the new answer to the end of the vector.In other words,the vector is scrolling. Because the period is at most 36,you can use brute force to find it. I hope this helps. ：）\n » Can anyone explain how we get string \"ieh\" in first example in two seconds? After getting string \"i\" \"ieh\" is lexicographically second after \"i\", so we need at least 3 seconds, isn't it?\n » How to solve problem C using bitmasks. Kindly help me.\n• » » 4 years ago, # ^ | ← Rev. 2 →   You can make each number of the array v into a number based on k.If the number only consists of 0 and 1, it must be able to become 0.Then, mark the position of each 1.If the number consists of other numbers or the position of some 1 is already marked, the array won't be able to be filled with 0 and the answer is \"NO\". After you finish that operation for each number in the array v, the answer will be \"YES\".This is my solusion.I am a Chinese and I am sorry that my English is poor. if there is something wrong in my words, please tell me and I will repair it as fast as posiible.\n• » » » 4 years ago, # ^ | ← Rev. 3 →   Can you please explain What is the use of step variable in your code?How we will check if this i power is used or not?\n » 4 years ago, # | ← Rev. 2 →   Is it posssible to solve G on Java?\n• » » See my solution, I had to make my recursion iterative in order to bypass stack overflow\n » awoo Why the same solutions receive completely different times?:72925974 4321 ms = 72925981 2885 ms72927822 1668 ms = 72927856 920 ms\n » Can anyone explain B. Bogo Sort Problem??\n• » » In this problem it was quite an observation that if we sort the array in descending(or non increasing order to be precise) then A[i]-i can never be equal to A[j]-j for iA[j] since we sorted the array in non increasing order.Now we are subtracting larger value by smaller index example array of 5 4 will become 5-1 != 4-2. hope that helps. Here's the link to my solution\n• » » If we sort the array in non-increasing order, then we have $\\begin{equation} i \\lt j \\end{equation}\\tag{1} $$\\begin{equation} a_i \\ge a_j \\end{equation}\\tag{2} Rewrite (1) to \\begin{equation} -i \\gt -j \\end{equation}\\tag{3} and add a_i to both side \\begin{equation} a_i - i \\gt a_i -j \\end{equation}\\tag{4} Add -j to both side to (2)$$ \\begin{equation} a_i - j \\ge a_j - j \\end{equation}\\tag{5}$From $(4)$ and $(5)$, we have$\\begin{equation} a_i - i \\gt a_i - j \\ge a_j - j \\end{equation}\\tag{6}$which means$\\begin{equation} a_i - i \\gt a_j - j \\end{equation}\\tag{7} $$\\begin{equation} i - a_i \\ne j - a_j \\end{equation}\\tag{8}$$ \\begin{equation} j - a_j \\ne i - a_i \\end{equation}\\tag{9}$Therefore, after sorting the array in non-increasing order, for each pair of indexes $i < j$ , the condition $j - a_j \\ne i - a_i$ holds.\n » Can anyone explain me C please.\n• » » convert given numbers into k-base system. now, do addition of k-base numbers without k-base system. now, you can observe that, if there is any bit is >= 2 then it's not possible to make given numbers using power of k. because, you can use k^some power only one time.\n• » » » great way!\n• » » » can u please explain in more details?\n• » » » » if a number is in k-base system, then if number n = 1111 = k^0 + k^1 + k^2 + k^3 , and if n = 2111 = k^0 + k^1 + k^2 + 2*k^3, means we need to add k^3 two times. so we can't make this number.\n• » » lets assume we have to check for no. X ,if X is summation of power of k or not then X=k^n+k^m+k^l....,where n>m>l..,so if we take k^n common and divide X by k^n then resulting no. should be divisible by k, X=k^n(1+k^(m-n)+k^(l-n)) X/(k^n)-1=k^(m-n)+k^(l-n)..\n• » »\n » Solution for Problem C is not quit clear can anyone help me out with that??? adedalic\n• » » see this comment : https://codeforces.com/blog/entry/74640?#comment-587595\n » 4 years ago, # | ← Rev. 2 →   Could you tell me what does this return dp[l][r] = a[l] ?\n• » » I think it first assigns the value of a[I] to dp[l][r] and then returns the value of dp[l][r];:^)\n » I have an $O(n \\log n)$ solution to problem E: https://codeforces.com/blog/entry/74656\n » 4 years ago, # | ← Rev. 3 →   Can someone please explain how time complexity of problem E is n^3 ? Thanks.\n• » » Figured it out after rethinking the problem. Let's think of the main function for loop and calcDP separately. So calcDP is an N^3 function while our for loop runs in N^2. They are independent entities. So when you call calcDP from inside your N^2 for loop, their complexities don't affect each other.\n » 4 years ago, # | ← Rev. 2 →   There is another solution to problem D with complexity $O(n \\log P)$.Welcome to see in https://codeforces.com/blog/entry/74685. For the Chinese version https://andylizf.github.io/2020/03/10/CF1312D-Count-the-Arrays.Can anyone prove it is equal to the normal solution?\n » can anyone explain c better plz?\n• » » The question states that you have an array ( which is initially all zeros ), and you apply some number of operations, say $M$.In $i$th operation ( $0 \\le i \\lt M$ ), you can either pass and go to step $i+1$, or add $k^i$ to one of the elements of the array.Notice that, each power of $k$ is used atmost once, if at all. Thus, we simply convert each number to base $k$, and see if any power of $k$ is repeated.\n » 4 years ago, # | ← Rev. 2 →   Another possible solution of 1312E — Array Shrinking. Complexity is O(V * N) where V is limit of a. The solution is here.\n » To solve F, I used the heuristic that the period is equal to $smallest+largest$ out of $x,y,z$ It only has 2 exceptions (considering $y$ and $z$ equivalent): $1,3,4$ (period=7) and $4,1,2$ (period=3).I have no clue on why this works, wondering if it just somehow worked because of small constraints on the values. Any insights?Link to my code: 73467272\n » what is the time complexity of D,is it O(m log p) or (n log p)??\n• » » I think the time complexity is only O(log p).\n » For problem G,we don't need any data struct .Simple dfs is enough.The code is very short: Codeint n; char c[1000000+20]; vector each[1000000+20]; int queries[1000000+20],dp[1000000+20]; bool ok[1000000+20]; bool cmp(int A,int B){ return c[A]>n; rb(i,1,n){ int pi; cin>>pi>>c[i]; each[pi].PB(i); } int k; cin>>k; rb(i,1,n){ sort(ALL(each[i-1]),cmp); } rb(i,1,k){ cin>>queries[i]; ok[queries[i]]=1; } dfs(0,INF,-1); rb(i,1,k){ cout<\n• » » 4 years ago, # ^ | ← Rev. 3 →   Though there are many details,it's still much simpler than solutions using data structures.Also,it's very fast.\n » In problemA, if we consider case n = 8 and m = 4 then how can a convex polygon with 4 sides can have a same center.Thanks, in advance\n• » » 4 years ago, # ^ | ← Rev. 2 →   Number the vertex from 0 to n-1 (7) take 0th and 2nd and 4th and 6th vertex\n• » » » How do you prove that we can have the vertices of smaller polygon in common with larger polygon?\n• » » » » 3 years ago, # ^ | ← Rev. 3 →   I don't have a math-heavy perfect proof but you can think in this way.Lets call polygon of m sides be polygonIn and polygon of n sides be polygonOut. select one side of polygonIn and connect it to the center of polygon the angle will be (2*pi/m) also note that this angle is equal to x*(2*pi/n) (x is some integer) because the end points of a side of polygonIn must be end points of some x continous sides of polygonOut.Example in case of hexagon and triangle x = 2 because end points of one side of triangle is also end points of 2 continous sides of hexagon.so we get x*(2*pi/n) = 2*pi/mx*m = nn%m = 0\n » How can we check the period is at most 36 by brute force ?Is there a theoretical proof for the bound on period ?\n » In problem E, would the single representative of a subarray always be unique (to cache beforehand), would appreciate if someone can give a proof.\n• » » I think This is really good .\n• » » » Got it, thank you\n » Can any one help me with my solution (81649940) to Problem E.my approach :dp[i][j] — min length of which can be obtained from subarray [i,j]dp[i][j]- after reduction of this([i,j]) subarray what is the left most value eg if array is 6 6 3 3 5 then dp[i][j] = 7;(from reduced array- 7 4 5)dp[i][j]- after reduction of this subarray what is the right most value eg. if array is 6 6 3 3 5 then dp[i][j] = 5;(from reduced array- 7 4 5)\n » can someone explain me problem C. It would be great help. Thanks\n » 4 years ago, # | ← Rev. 2 →   resolved\n » can anyone explain Problem A in detail?\n• » » no\n » This tutorial is not linked in the problems, for example 1312A - Two Regular Polygons Maybe someone can fix this.\n• » » fixed" ]

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[ "Logic puzzles, riddles, math puzzles and brainteasers - pzzls.com\n\nVandaag is het 24 October 2021\n\nSort - Least popuplar on top\nPage: << 1 2 3 4 5 >>\n\n# Also having birthday today? - math puzzle\n\nDifficulty:", null, "", null, "", null, "Rating: 2.1/5.0\n\n Out of how many people should a group at least consist in order to have that the probability that two persons out of that group are having birthday on the same day, is larger than 1/2?", null, "# 2=1? - math puzzle\n\nDifficulty:", null, "", null, "Rating: 2.3/5.0\n\nSuppose: a = b\n\nIt then follows that\na * a = a * b\na^2 = a * b\na^2 - b^2 = a * b - b^2\n(a - b) * (a + b) = (a - b) * b\na + b = b\na + a = a\n2 * a = 1 * a\n2 = 1\n\nIs this derivation really correct? Or is there somewhere a mistake? If so, where?\n\n# Infected? - math puzzle\n\nDifficulty:", null, "", null, "", null, "", null, "Rating: 2.4/5.0\n\n Suppose that 1 out of 1000 people of a specific country are infected with the deadly X-virus. An inhabitant of that country goes the hospital and gives some blood for a survey. Her blood is checked on the X-virus by a test which gives a correct outcome in 99% of the cases. The test points out that the woman is infected. After the investigation, a doctor tells the woman that he is 99% sure that she is infected with the X-virus. Can the doctor be right by making this worrying statement?", null, "" ]

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[ "", null, "# How SUSI AI Tabulates Answers For You\n\nSUSI is an artificial intelligence chat bot that responds to all kinds of user queries. It isn’t any regular chat bot replying in just plain text. It supports various response types which we refer to as ‘actions’. One such action is the “table” type. When the response to a user query contains a list of answers which can be grouped, it is better visualised as a table rather than plain text.\n\nLets visit SUSI WebChat and try it out. In our example we ask SUSI for the 2009 race statistics of British Formula 1 racing driver Lewis Hamilton.\n\nQuery: race stats of hamilton in f1 season 2009\n\nResponse: <table> (API response)", null, "How does SUSI do that? Let us look at the skill teaching SUSI to give table responses.\n\n```# Returns race stats as a table\n\nrace summary of  * in f1 season *|race stats of  * in f1 season *\n!console:\n{\n\"url\":\"http://ergast.com/api/f1/\\$2\\$/drivers/\\$1\\$/status.json\",\n\"path\":\"\\$.MRData.StatusTable.Status\",\n\"actions\":[{\n\"type\":\"table\",\n\"columns\":{\"status\":\"Race Status\",\"count\":\"Number Of Races\"}\n}]\n}\neol\n```\n\nHere, we are telling SUSI that the data type is a table through type attribute in actions and also defining column names and under which column each value must be put using their respective keys. Using this information SUSI generates a response accordingly with the table schema and data points.\n\nHow do we know when to render a table?\n\nWe know it through the type attribute in the actions from the API response.\n\n```\"actions\": [{\n\"type\": \"table\",\n\"columns\": {\n\"status\": \"Race Status\",\n\"count\": \"Number Of Races\"\n},\n\"count\": -1\n}]\n}],\n```\n\nWe can see that the type is table so we now know that we have to render a table.\n\nBut what is the table schema? What do we fill it with?\n\nThere is a columns key under actions and from the value of the columns key we get a object whose key value pairs give us column names and what data to put under each column.\n\nHere, we have two columns – Race Status and Number Of Races\n\nAnd the data to put under each column is found in answers.data with same keys as those for each column name i.e ‘status’ and ‘count’.\n\nSample data object from the API response:\n\n```{\n\"statusId\": \"2\",\n\"count\": \"1\",\n\"status\": \"Disqualified\"\n}\n```\n\nThe value under ‘status’ key is ‘Disqualified’ and the column name for ‘status’ key is ‘Race Status’, so Disqualified is entered under Race Status column in the table. Similarly 1  is entered under Number Of Races column. We thus have a row of our table. We populate the table for each object in the data array using the same procedure.\n\n```let coloumns = data.answers.actions[index].columns;\n```\n\nWe also have a ’count’ attribute in the API response . This is used to denote how many rows to populate in the table. If count = -1 , then it means infinite or to display all the results.\n\n```function drawTable(coloumns,tableData,count){\n\nlet parseKeys;\nlet showColName = true;\n\nif(coloumns.constructor === Array){\nparseKeys = coloumns;\nshowColName = false;\n}\nelse{\nparseKeys = Object.keys(coloumns);\n}\n\n});\n\nlet rowCount = tableData.length;\n\nif(count > -1){\nrowCount = Math.min(count,tableData.length);\n}\n\nlet rows = [];\n\nfor (var j=0; j < rowCount; j++) {\n\nlet eachrow = tableData[j];\n\nlet rowcols = parseKeys.map((key,i) =>{\nreturn(\n<TableRowColumn key={i}>\n{eachrow[key]}\n</TableRowColumn>\n);\n});\n\nrows.push(\n<TableRow key={j}>{rowcols}</TableRow>\n);\n\n}\n\nconst table =\n<MuiThemeProvider>\n<Table selectable={false}>\n<TableBody displayRowCheckbox={false}>{rows}</TableBody>\n</Table>\n</MuiThemeProvider>\n\nreturn table;\n\n}\n```\n\nHere we first determine how many rows to populate using the count attribute and then parse the columns to get the column names and keys. We then loop through the data and populate each row.\n\nThis is how SUSI responds with tabulated data!\n\nYou can create your own table skill and SUSI will give the tabulated response you need. Check out this tutorial to know more about SUSI and the various other action types it supports.\n\nResources" ]

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[ "# 27.6 Gal To L\n\nGal To L Calculator\nLet's convert 27.6 gallon to liter.\n\nGallon to liter conversion rate is 3.78541 .\n\nSo if we multiply 27.6 with 3.78541 rate, we will find 27.6 gallon is how much liter.", null, "## How to convert 27.6 gallons to liters ?\n\nTo convert gallons to liters, multiply the number of gallons by 3.78541. For example, if you want to convert 27.6 gallons to liters, you would multiply 27.6 by 3.78541, which equals 104.477316 liters.\n\n27.6 X 3.78541 = 104.477316\n27.6 gallon = 104.477316 liter\nReverse Check : Convert 27.6 Liter To Gallon\n\n#### More than 27.6 gallon in percentage :\n\nGallon UnitLiter Unit\n27.7 Gallon104.855857 Liter\n27.8 Gallon105.234398 Liter\n27.9 Gallon105.612939 Liter\n28 Gallon105.99148 Liter\n28.1 Gallon106.370021 Liter\n28.2 Gallon106.748562 Liter\n28.3 Gallon107.127103 Liter\n28.4 Gallon107.505644 Liter\n28.5 Gallon107.884185 Liter\n\n#### More than 27.6 gallon in percentage :\n\nGallon UnitLiter Unit\n28.6 Gallon108.262726 Liter\n29.6 Gallon112.048136 Liter\n30.6 Gallon115.833546 Liter\n31.6 Gallon119.618956 Liter\n32.6 Gallon123.404366 Liter\n33.6 Gallon127.189776 Liter\n34.6 Gallon130.975186 Liter\n35.6 Gallon134.760596 Liter\n36.6 Gallon138.546006 Liter\n\n## How To Convert Gal To L\n\n1 gallon is how much liter ? Let's find the answer for this basic volume question. 1 gallon is 3.78541 liter.\n\nThe coefficient value for the gal to l conversion is 3.78541.\n\nGal to L Conversion Rate Is 3.78541\n\nFor example :\n\nHow much liters are 2 gallons?\n\nAnswer : If 1 gallon is 3.78541 liter so 2 gallon will be 3.78541 multiply by 2. It is equal 7.571 L .\n\n### Conversion Chart Of Gallon to Liter\n\nGallon UnitLiter Unit\n0.1 Gallon0.378541 Liter\n0.2 Gallon0.757082 Liter\n0.3 Gallon1.135623 Liter\n0.4 Gallon1.514164 Liter\n0.5 Gallon1.892705 Liter\n0.6 Gallon2.271246 Liter\n0.7 Gallon2.649787 Liter\n0.8 Gallon3.028328 Liter\n0.9 Gallon3.406869 Liter\n1 Gallon3.78541 Liter\nGallon UnitLiter Unit\n1 Gallon3.78541 Liter\n2 Gallon7.57082 Liter\n3 Gallon11.35623 Liter\n4 Gallon15.14164 Liter\n5 Gallon18.92705 Liter\n6 Gallon22.71246 Liter\n7 Gallon26.49787 Liter\n8 Gallon30.28328 Liter\n9 Gallon34.06869 Liter\n10 Gallon37.8541 Liter\n\n## Gallon Unit :\n\nA gallon is a unit of volume that is equal to the volume of one liquid gallon. The United States customary system is defined as 231 cubic inches, or 0.831547237 L. The imperial system is defined as 473.1768 cubic inches, or 1.163394619 L.\n\nThe liquid capacity of an automobile tank varies from around 10 gallons to over 20 gallons, depending on what type of gasoline is used and how many people are expected to be traveling in the vehicle.\n\nOne gallon is equal to 3.79 liters, so one thousand gallons is equal to thirty-nine hundred liters.\n\nConversion from a gallon to a liter: multiply by 0.1164\n\nConversion from a gallon to an acre: multiply by 0.003569\n\nConversion from a gallon to an Imperial pint: multiply by 0.56874\n\nGallon to Liter and Liter to Gallon volume unit converter. Convert Gal unit to L units." ]

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[ "Home\n\n# Tsp to grams", null, "### Convert teaspoons to grams - Conversion of Measurement Unit\n\n1 teaspoons to grams = 4.26057 grams. 5 teaspoons to grams = 21.30283 grams. 10 teaspoons to grams = 42.60567 grams. 15 teaspoons to grams = 63.9085 grams. 20 teaspoons to grams = 85.21134 grams. 25 teaspoons to grams = 106.51417 grams. 30 teaspoons to grams = 127.81701 grams. 40 teaspoons to grams = 170.42267 grams 1 teaspoon (US) of salt (fine) measures approximately 5.9 grams. If you're using a metric teaspoon, that measurement is 6 grams. How many grams in a teaspoon of sugar? 1 teaspoon (US or metric) of granulated sugar is equal to around 4.2 grams One teaspoon of salt is equal to 5.69 grams, so use this simple formula to convert: grams = teaspoons × 5.69. The salt volume in grams is equal to the teaspoons multiplied by 5.69. For example, here's how to convert 5 teaspoons to grams using the formula above. 5 tsp = (5 × 5.69) = 28.45 g\n\n### Teaspoons to Grams Converter (tsp to Grams\n\n1. Weight of 1 milliliter (ml) of pure water at temperature 4 °C = 1 gram (g). Thus 1 US teaspoon ( tsp) = 4.92892159 grams ( g) = 4.92892159 milliliters (ml) = 1/6 or 0.166666667 US fluid ounce (fl. oz.) = 1/3 US tablespoons (tbsp). The teaspoon is rounded to precisely 5 mL (0.169070114 US fluid ounces) by US federal regulations for food labeling.\n2. US teaspoons to grams of Baking powder; 1 US teaspoon = 4.79 grams: 2 US teaspoons = 9.58 grams: 4 US teaspoons = 19.2 grams: 5 US teaspoons = 24 grams: 8 US teaspoons = 38.3 grams: 1 / 16 US teaspoon = 0.299 gram: 1 / 8 US teaspoon = 0.599 gram: 1 / 4 US teaspoon = 1.2 grams: 1 / 3 US teaspoon = 1.6 grams: 1 / 2 US teaspoon = 2.4 grams: 2 / 3 US teaspoon = 3.19 grams: 3 / 4 US teaspoon = 3.59 grams: 1 1 / 16 US teaspoons = 5.09 grams\n3. The answer is: The change of 1 tsp (teaspoon) unit in a active dry yeast measure equals = into 2.83 g (gram) as per the equivalent measure and for the same active dry yeast type. Professional people always ensure, and their success in fine baking depends on, they get the most precise units conversion results in measuring their ingredients\n\nNavigation: from unit menu • into unit menu • » converter tool « . Step 3 of 3 Convert amount of CUMIN SEED measure. From tsp, whole to g, gram quantity. Amount: 1 tsp, whole of CUMIN SEED Equals: 2.10 of g, gram in CUMIN SEED TOGGLE: from g, gram to tsp, whole quantities in the other way aroun 16 grams = 3 1/4 tsp water. Please note that grams and teaspoons are not interchangeable units. You need to know what you are converting in order to get the exact tsp value for 16 grams Tsp is an abbreviation of teaspoon. Teaspoon values are rounded to the nearest 1/8, 1/3, 1/4 or integer. More Information On 12 grams to tsp. If you need more information on converting 12 grams of a specific food ingredient to teaspoons, check out the following resources: 12 grams flour to teaspoon; 12 grams sugar to teaspoons; 12 grams butter. Teaspoons to Grams of Butter Conversion Formula [X] g = 4.7249205208333 × [Y] tsp where [X] is the result in g and [Y] is the amount of tsp we want to convert 1.5 Teaspoons to Grams of Butter Conversion breakdown and explanatio\n\n1/2 tsp to grams 1/2 US teaspoon of water equals 2.46 grams (*) Volume to 'Weight' Converter. Volume ⇀ Weight Weight ⇀ Volume. Inputs? Notes: the results in this calculator are rounded (by default) to 3 significant figures. The conversion factors are approximate once it is intended for recipes measurements. This is not rocket. Tsp to Grams | Conversion Calculator. Calculate the quantity of grams in an US teaspoon. To use this converter, please choose the unit of mass (weight), the unit of volume to convert to, the mass value, the desired ingredient ten click on the button 'CONVERT IT!' 7 grams equals to 1.40 teaspoons or there are 1.40 teaspoons in 7 grams. Teaspoons Conversion. Grams: 7. US Teaspoon: 1.4. Pounds (lbs): 0.01543. Kilograms (kg) teaspoon to g conversion table: 0.1 tsp. = 0.5 gram. 2.1 tsp. = 10.5 grams. 4.1 tsp. = 20.5 grams. 7 tsp. = 35 grams Teaspoons to Grams of Butter Conversion Formula [X] g = 4.7249205208333 × [Y] tsp where [X] is the result in g and [Y] is the amount of tsp we want to convert 3 Teaspoons to Grams of Butter Conversion breakdown and explanatio\n\nHow many grams of butter are in a teaspoon? Use this easy and mobile-friendly calculator to convert between teaspoons and grams of butter. Just type the number of teaspoons into the box and the conversion will be performed automatically With this grams to teaspoons calculator, you'll quickly convert between grams - the basic unit of weight (mass) in the metric system - and teaspoons, which are a measure of volume often used in cooking. Wondering how many grams of sugar in a teaspoon? 1 tsp (5ml) * 3 = 1 tbsp (15ml) 1 US tsp (4.93ml) * 3 = 1 US tbsp (14.79ml How Many Teaspoons is 9 Grams? 9 grams equals to 1.80 teaspoons or there are 1.80 teaspoons in 9 grams. Convert 9 Grams to Teaspoons 9 Grams to Teaspoons will not only convert 9 grams to teaspoons, but will also convert 9 grams to other units such as pounds, kilograms, milligrams, stones and more teaspoons = grams × 0.175747 The salt volume in teaspoons is equal to the grams multiplied by 0.175747. For example, here's how to convert 5 grams to teaspoons using the formula above. 5 g = (5 × 0.175747) = 0.878735 tsp 1 US tablespoon (tbsp) = 14.7867648 grams (g) of water = 14.7867648 milliliters (mL) = 0.5 US fluid ounce (fl. oz.) = 3 US teaspoons (tsp). The tablespoon is rounded to precisely 15 mL by US federal regulations (FDA) for food labeling purposes\n\n### Teaspoons of Salt to Grams Conversion (tsp to g\n\n• 2 Tsp to Grams Conversion. Calculate the quantity of grams in any quantity of tsp. To use this converter, please choose the unit of mass (weight), the unit of volume to convert to, the mass value, the desired ingredient ten click on the button 'CONVERT IT!'\n• 75 grams to teaspoons = 17.60329 teaspoons 100 grams to teaspoons = 23.47106 teaspoons This is the amount of sugar, often measured as 4.2 grams per teaspoon on a nutrition facts label. teaspoon to g conversion table\n• When carrying out a conversion between grams and teaspoons, it is important to remember that the gram is a unit of weight and the teaspoon is a unit of volume. As an example, a teaspoon of flour weighs less than a teaspoon of milk. It is for this reason that a list of ingredients is included with the converter\n• For example let's look at a teaspoon of sugar. 4.2 grams of sugar would be equal to 1 teaspoon, instead of the general 5 grams per teaspoon. The added sugar is less dense, so it takes less grams to equal a teaspoon. Here is a list of common ingredients with measurement in volume, and their gram to teaspoon conversion tables\n• This page will convert baking soda from units of weight such as grams and ounces into units of volume such as teaspoons, tablespoons, and cups. This can be used to convert between British and American recipes\n\n### Teaspoons to Grams [water] Conversio\n\nGrams also can be marked as grammes (alternative British English spelling in UK). US tablespoon can be abbreviated as T, tb, tbs, tbsp, tblsp, or tblspn. Related converters: Grams to Carats Grams to Cubic Centimeters Grams to Cups Grams to Cups Grams to Fluid Ounces Grams to Gallons Grams to Grains. Let me break it down for you. 2 tablespoons are equivalent to 40 grams which is the same as 1.4 oz. a cupful is 300 grams or 10.71 oz. Butter is another essential ingredient in our foods and it measures as follows: 2 table spoons are equivalent to 30 grams. A cup is 240 grams; Conversion of liquids to grams depends on their viscosity Tsp to Grams | Conversion Calculator. Calculate the quantity of grams in an US teaspoon. To use this converter, please choose the unit of mass (weight), the unit of volume to convert to, the mass value, the desired ingredient ten click on the button 'CONVERT IT!' So, all you need to know is that one teaspoon contains 4 grams of sugar: 1 teaspoon = 4 grams of sugar. Let's say that you usually drink 5 cups of coffee during the day and that each time you drink coffee, you add one teaspoon of sugar to it. So, just when you drink coffee, you are ingesting 5 teaspoons of sugar 1.5 to 2 Grams per Teaspoon. There are 1.5 to 2 grams of matcha per tsp or teaspoon, depending on the specific brand or matcha and it's current moisture levels. 1 teaspoon of matcha is generally a good measure for a 12 oz. cup of 170F degree water\n\nGrams to tablespoons . 10 grams = 0.782 tablespoons. 20 grams = 1.56 tablespoons. 30 grams = 2.34 tablespoons. 40 grams = 3.12 tablespoons. 50 grams = 3.91 tablespoons. 60 grams = 4.00 tablespoons. 70 grams = 4.67 tablespoons. 80 grams = 5.33 tablespoons. 90 grams = 6.34 tablespoons. 100 grams = 7.82 tablespoons. 125 grams = 8.45 tablespoons. grams per teaspoon to milligrams per teaspoon conversion cards. 1 through 20 grams per teaspoon; 1 g/tsp to mg/tsp = 1 000 mg/tsp; 2 g/tsp to mg/tsp = 2 000 mg/tsp; 3 g/tsp to mg/tsp = 3 000 mg/tsp; 4 g/tsp to mg/tsp = 4 000 mg/tsp; 5 g/tsp to mg/tsp = 5 000 mg/tsp; 6 g/tsp to mg/tsp = 6 000 mg/tsp; 7 g/tsp to mg/tsp = 7 000 mg/tsp There is no such formula as the conversion from a volume measure (tsp or ml) to mass measure (gram) depends on the substance. For example, 1 ml of water is exactly 1 gram, but 1 ml of olive oil is 0.91 grams. And 1 ml of dried basil is only 0.11 grams. You can find more conversions for various substances here: http://www.convert-me Changing into weight in grams from a maple syrup amount, conversion from 250 ml volume in a bottle (\\$9.95 per bottle). Got result: 250 ml ( millilitres = 1 cup ) amount of maple syrup equals 334.6 g ( grams ) of maple syrup. Excellent. Trying to delute the syrup a little after I get it in bulk I will need the conversion ratios\n\n### 1 teaspoon of baking powder in grams\n\n• For instance, compute how many ounces or grams a cup of GROUND CUMIN, UPC: 855269003404 weighs. Weight of the selected food item is calculated based on the food density and its given volume to answer questions such as, how many ounces or grams of a selected food in a liter, a cup, or in a spoon\n• Measuring your ingredients by weight (grams) can help make your ingredient amounts are accurate. It's especially true in baking — think how much flour you can fit in a measuring cup depending on how much you pack it. These charts help you go between cups, grams, and ounces, depending on what your recipe calls for\n• As others have observed, a teaspoon is a(n ambiguous) measure of volume, a gram is a unit of mass and weight. To convert, you need to make multiple assumptions: 1. As Lawrence Statton observes, the density of the material you're measuring. 2. The.\n• Hi, I am looking for standard source for conversion of cups, teaspoons and tablespoons mentioned in recipes to grams/lbs or ml. For example if the recipe mentions 1 cup All purpose flour I want to know how many grams is it and 1 teaspoon vanilla then how many ml is it and 1 tablepoon salt then how many grams is it, etc\n• Agar-agar: 1 teaspoon = 2 grams Baby shrimp, shelled, cooked: 1 cup = 100 grams Capers: 1 cup = 120 grams Dried bread crumbs: 1 cup = 100 grams Hibiscus flowers: 1/3 cup = 15 grams Oil: 1 tablespoon = 9 grams Olives: 1 cup = 180 grams. Pans and dishes 10-inch tart or cake pan = 25-centimeter tart or cake pan 9-inch cake pan = 22-centimeter cake pa\n• dful of the amount of fat that you eat each day and throughout the week is made a little easier when you visualize fat in individual teaspoons\n• tablespoon to g conversion table: 0.1. = 1.5 gram. 2.1. = 31.5 grams. 4.1. = 61.5 grams. 7. = 105 grams\n\n### Active Dry Yeast tsp to g converter for culinary baking\n\n1. Grams to Kg How to convert Kilograms to Grams. 1 kilogram (kg) is equal to 1000 grams (g). 1 kg = 1000 g. The mass m in grams (g) is equal to the mass m in kilograms (kg) times 1000: m (g) = m (kg) × 1000. Example. Convert 5kg to grams: m (g) = 5 kg × 1000 = 5000 g. Kilograms to Grams conversion tabl\n2. 226 grams: Sugar: 1 cup of caster sugar: 200 grams: 1 cup of raw sugar: 250 grams: 1 cup of brown sugar: 220 grams: 1 cup of confectioners (icing) sugar: 125 grams: 1 teapsoon of caster sugar: 4.2 grams: 1 tablespoon of caster sugar: 12.6 grams: Honey: 1 tablespoon: 21.25 grams: 1/4 cup: 85 grams: 1 cup: 340 grams: Salt: 1/4 teaspoon: 1.42.\n3. Next, multiply the number of grams per gallon by the number of gallons in the mash (4). 2.4 x 4 = 9.6 grams, which can be rounded to 10 grams. Unless you have a gram scale handy, you will want to convert that to teaspoons which is more convenient\n\n### Convert 1 tsp, whole to 1 g, gram of CUMIN SEE\n\nTag: 12 tsp to grams. Grams To Teaspoons. Grams To Teaspoons To be precise, 4.2 grams equals a teaspoon, but the nutrition facts round this number down to four grams. How to Convert Grams to Teaspoons Sliding down the label to the total carbohydrates it reads sugars 4g, or Read more Grams Teaspoon. How many grams of red pepper are in 1 teaspoons? 1.79 grams of red pepper fit into one teaspoon. 2 teaspoons of red pepper = 3.58 grams of red pepper. 3 teaspoons of red pepper = 5.37 grams of red pepper. ¾ teaspoon of red pepper = 1.34 grams of red pepper. 2 / 3 teaspoon of red pepper = 1.19 grams of red pepper 12 grams = 2 1/4 tsp. I read that 1 teaspoon of sugar is about 4 grams 1 teaspoon of suger is 3 grams and there are 9 teaspoons of ground nutmeg in 20 grams of nutmeg A tablespoon of flour weighs approximately 8 to 9 grams. Since the measurement of 1 tablespoon is equivalent to 3 teaspoons, 1 teaspoon of flour weighs approximately 3 grams. Using this formula, we can estimate these common grams-to-teaspoons measurements: 3 grams = 1 teaspoon of flour ; 4.5 grams = 1 1/2 teaspoons of flou\n\n### 16 Grams To Teaspoons (Tsp) - Online Unit Converte\n\nFour grams of sugar is equal to one teaspoon. To be precise, 4.2 grams equals a teaspoon, but the nutrition facts rounds this number down to four grams. Sugars: 4 grams equals 1 teaspoon Using this equation you can easily look at any food product to see how much sugar it contains Next, you need to know the two steps for converting teaspoons to grams. Step 1. Convert teaspoons into mL, based on the basic information we just covered (i.e. 1 tsp = 5 mL) Grams are a measure of mass, and teaspoons measure volume. The correct conversion depends on the density of the item you're measuring. Water has a density of 1 g/ml, so the conversion is 1 gram to 1 millileter, which is equivalent to 0.2 teaspoons. For other substances, the density will be different, and each teaspoon will weigh a different.\n\nIt wasn't so long ago that trisodium phosphate (TSP) was a go-to choice for tough cleaning jobs, especially on exteriors. Diluted in water and applied often with a stiff scrub brush, it can. Convert butter from US cups, sticks of butter and tablespoons and more to grams with handy Butter Measurement Conversion Charts. The confusion: Most countries outside the US (and Canada), weigh their butter using scales which makes it very difficult to understand American recipes UK tsp = UK tblsp * 4.0000 . UK Teaspoons. A British cooking measurement. UK Tablespoons to UK Teaspoons table. Star A tablespoon is three-time of a teaspoon that means 10-12 grams so don't use a tablespoon to measure your dose. 1 teaspoon = 3.8-4.0 grams of kratom powder. 8-10 grams of Kratom powder provide much painkilling and sedating effects and is considered best Kratom dosage for treating opiate withdrawal symptoms\n\nRe: Dried Herbs- grams to tsp Hi, a packet of sweet basil, or origano, or sage, or parsley is 10g, and is about 100mL in volume. A teaspon is 5mL, so there is about 20 teaspoons per packet. 2 grams will be 4 teaspoons, 4 grams will be 8 teaspoons, 6 grams will be 12 teaspoons, and 8 grams will be 16 teaspoons Grams to tablespoons, tablespoon to grams: sugar, flour and other products. Assume we'd like to: Convert amount of sugar in grams to tablespoons in the blink of an eye, Switch the ingredient and quickly find out how many grams are in a tablespoon of salt, Change the butter tablespoon to grams Seven grams converts to exactly 1.4000000000000001 teaspoons. This number can be safely rounded to 1.4 teaspoons for ease of measuring when working in the kitchen. Because 1.4 teaspoons is an uncommon measurement, the easiest way to measure it out is to first get 1.5 teaspoons and then subtract a small pinch from it\n\n### 12 Grams To Teaspoons (Tsp) - Online Unit Converte\n\n• g to tsp Converter, gram to teaspoon (metric) Conversion, Liquid density chart\n• How many Imperial cups is 80 grams of sesame seeds? 80 grams of sesame seeds = 0.5 Imperial cup (without 2 teaspoons or 2/3 of a tablespoon) of sesame seeds. We listed all the necessary calculations to measure sesame seeds without scales\n• 3 teaspoons: 1 tablespoon: 1/2 ounce: 14.3 grams-2 tablespoons: 1/8 cup: 1 fluid ounce: 28.3 grams-4 tablspoons: 1/4 cup: 2 fluid ounces: 56.7 grams-5 1/3 tablespoon\n• 24 grams = 3 tablespoons of flour; 50 grams = 6 1/4 tablespoons of flour; 100 grams = 12 1/2 tablespoons of flour; 200 grams = 25 tablespoons of flour; Sugar. From nutrition labels on products, 4 grams of sugar are equivalent to 1 teaspoon. And, converting that to tablespoons means multiplying 4 grams by 3, since 3 teaspoons make 1 tablespoon\n• Mitrafoline is inveterate kratom tsp to grams and will explain most fevers. Biolumitá in atherosclerosis cad, or maintaining decorticate wall matrix and wilson maintained to function with all kratom. Mujeresrusas - what if multiple ages 55. Ffm threesome but unless the same amazing. Airwolf caribou arrive in canada, she posts sales jobs\n• Our cups to grams converter lets you easily convert American recipes into UK recipes to make at home. Check out our easy to use conversion tool. 1 tsp: 6ml: 0.2 fl oz: 1 tbsp: 15ml: 0.5 fl oz.\n• There are 0.75000507301355 teaspoon in a dram. 1 Dram is equal to 0.75000507301355 Teaspoon. 1 dram = 0.75000507301355 tsp\n\n### 1.5 Teaspoons to Grams of Butter Convert 1.5 tsp in g ..\n\nSince grams are a weight measurement and U.S. tablespoons are a volume measurement, there isn't a universal conversion chart. Different ingredients weigh different amounts. Butter weighs more than flour, for example, so 14 grams of butter is going to be 1 tablespoon whereas 14 grams of flour is a little more than 1 3/4 tablespoons Easily convert Grains (gr) to Teaspoons (tsp) using this free online unit conversion calculator. Simple online unit conversion tool to convert grains (gr) into teaspoons (tsp). Related Calculators Convert 4 Ounces to Grams. To calculate 4 Ounces to the corresponding value in Grams, multiply the quantity in Ounces by 28.349523125 (conversion factor). In this case we should multiply 4 Ounces by 28.349523125 to get the equivalent result in Grams: 4 Ounces x 28.349523125 = 113.3980925 Grams. 4 Ounces is equivalent to 113.3980925 Grams US tsp * 0.98578 . Metric Teaspoons. A metric approximation to the popular cooking measurement . US Teaspoons to Metric Teaspoons table. Star Concentration solution unit conversion between milligram/teaspoon and percentage, percentage to milligram/teaspoon conversion in batch, mg/tsp per conversion char\n\n### 1/2 tsp to grams - How Many Wiki- HowMany\n\n1. Cups to Grams Converter. A cup is a measure of volume, whereas grams are a weight measure. To convert cups to grams, we have to take into account the density of each specific ingredient.Therefore, you have to select an ingredient in the select box to perform your conversion\n2. 1 Milligrams = 0.001 Grams: 10 Milligrams = 0.01 Grams: 2500 Milligrams = 2.5 Grams: 2 Milligrams = 0.002 Grams: 20 Milligrams = 0.02 Grams: 5000 Milligrams = 5 Grams: 3 Milligrams = 0.003 Grams: 30 Milligrams = 0.03 Grams: 10000 Milligrams = 10 Grams: 4 Milligrams = 0.004 Grams: 40 Milligrams = 0.04 Grams: 25000 Milligrams = 25 Grams: 5 Milligrams = 0.005 Grams: 50 Milligrams = 0.05 Grams\n3. A teaspoon (tsp.) is an item of cutlery.It is a small spoon that can be used to stir a cup of tea or coffee, or as a tool for measuring volume. The size of teaspoons ranges from about 2.5 to 7.3 mL (0.088 to 0.257 imp fl oz; 0.085 to 0.247 US fl oz). For cooking purposes and, more importantly, for dosing of medicine, a teaspoonful is defined as 5 mL (0.18 imp fl oz; 0.17 US fl oz), and.\n4. Re: sodium hydroxide: grams to teaspoons/tablespoons? One level tablespoon of lye weighs about 15 grams If you do not have access to a pH meter or papers to confirm how base the solution is the root bark can be basified to a pH of approximately ∼13 by dissolving 1 tablespoon of NaO\n5. Source(s): convert tsp tbsp grams: https://shortly.im/OSouD. 0 0. LRG. 1 decade ago. You can't accurately. A tsp is a measure of volume (or space), a gram is a measure of weight, so you can't convert the two because you're not measuring the same thing. The weight to volume ratio is specific to the item you're measuring\n6. It depends what you're filling it with. A tablespoon is a volume measurement, and grams are a mass unit. A tablespoon is 15 ml, so if you fill it with water, it is 15 grams\n7. Or 0.15 tsp (3/20 teaspoon). I know, from some angle sometimes, it can seem slightly confusing. Look at the math this way: 0.05 tbsp (1/20 tablespoon), and, 0.15 tsp (3/20 teaspoon) treacle syrup makes 1g Any amount, in this case 1 gram of treacle syrup, gives you 3 times higher number in tsp compared to Tbs volume. Because 1 Tbs gives 3 tsp's\n\nFor example; 200 of grams of butter is equal to a cup and 300 grams of sugar is a cup. The chart below shows accurately grams to cups, teaspoons, tablespoons and ounces for some commonly used ingredients. Product: Grams Per Cup: Grams Per Ounce: Grams Per Teaspoon: Grams Per Tablespoon: Baking powder, double acting: 220 3/4. 28.35. 4.6. 13.8 __Examples__ * How many liters are in a gallon * How many grams of butter is one ounce? * How many ounces in a pound * Convert UK tablespoons to tablespoons ###U.S. Liquid Measurements### Teaspoon: Tablespoon: Fluid ounce: Gill: Cup: Pint: Quart: Gallon: 1 teaspoon = 1: 1 / 3: 1 / 6: 1 / 24: 1 / 48: 1 / 96: 1 / 192: 1 / 768: 1 tablespoon = 3: 1. Spice Conversions Page 1 of 2 26 Lyerly St. Houston, TX 77022 713-691-2935 800-356-5189 Fax: 713-691-3250 For more recipes and information call us, come by or visit us on our Web Site. www.alliedkenco.com Many Recipes Call For Spices To Be Measured In Ounces\n\nMeasure food ingredients without a scale. 150 grams of white flour equals 1 cup, 1 spoonful tablespoon of sugar is around 20 grams (0.7 ounces). Calculate how much is 200 grams of flour in cups or spoons and more Essential oil conversions. Helpful chart. Teaspoons, drops, milliliters conversions. Print, keep handy for DIY essential oil recipes How to convert Ounces to Grams. 1 ounce (oz) is equal to 28.34952 grams (g). 1 oz = 28.34952 g. The mass m in grams (g) is equal to the mass m in ounces (oz) times 28.34952:. m (g) = m (oz) × 28.34952. Example. Convert 5 oz to grams: m (g) = 5 oz × 28.34952 = 141.7476 g. Ounces to Grams conversion tabl\n\n3 teaspoons: 1 tablespoon: 1/2 ounce: 14.3 grams: 2 tablespoons: 1/8 cup: 1 fluid ounce: 28.35 grams: 4 tablespoons: 1/4 cup: 2 fluid ounces: 56.7 grams: 5 1/3. My Account. TSP Account Number. User ID. Forgot your account number or user ID? My Account, Plan Participation, Investment Funds, Planning and Tools, Life Events and. Easily convert Gram (g) to Teaspoons (tsp) using this free online unit conversion calculator. Simple online unit conversion tool to convert gram (g) into teaspoons (tsp). Related Calculators\n\n### Tsp to Grams Conversion Calculato\n\n• The number of teaspoons to grams differs depending on density of the substance. While 1 teaspoon of sugar is equivalent to about 4 grams, 1 teaspoon of salt is about 6 grams. A gram of water is equal to 0.2 teaspoon, so 5 grams of water equals 1 teaspoon\n• How many grams of gelatine powder would equate one gelatine leaf? Our answer. Unfortunately converting recipes from leaf to powdered gelatine is extremely tricky as you can't use a straight substitution based on weight. This is due to the strength of gelatines being very different. Leaf gelatine comes in different strengths (or grades) and the.\n• Grams Calculator - Simple. This is a simplified calculator. It uses an averaged gram weight to do a simple calculation between a weight like grams, oz, pounds to a volume like cups, teaspoons, tablespoons. Example, a teaspoon of baking powder does not weigh the same as a teaspoon of peanut butter. This calculator will simply give you an.\n• Converting cups / tsp to grams vet_ca 2016 March 26. Does anyone know if any of the online tables or calculator programs are accurate in converting cups of liquids or solids to grams? Fave . up. 190 users have voted. Replies. farinam 2016 March 26. Hi Ron\n• The volume value 0.75 tsp (US teaspoon) in words is zero point seven five tsp (US teaspoon). This is simple to use online converter of weights and measures. Simply select the input unit, enter the value and click Convert button. The value will be converted to all other units of the actual measure\n• It is roughly around 7 teaspoons. 4 grams of sugar=1 tsp. 12 grams of sugar=1 tbsp\n• GRAMS IN 1 CUP: White Granulated Sugar: 7 oz: 198 gr: Light/Dark Brown Sugar (lightly packed) 8 oz: 227 gr: Powdered/Confectioners' Sugar: 4 oz: 113 gr: Superfine Sugar: 7.33 oz: 205 gr: Corn Syrup (Dark & Light) 12 oz : 336 gr: Honey : 12 oz : 336 gr: Molasses: 11 oz : 308 gr: Dairy Volume to Weight Conversions. Ingredient: Ounces in 1 Cup.", null, "Free online density conversion. Convert Mt/tsp.us to grams/cubic centimeter (megatonne/teaspoon (US) to g/cm3). +> with much ♥ by CalculatePlu One level tablespoon of lye weighs about 15 grams If you do not have access to a pH meter or papers to confirm how base the solution is the root bark can be basified to a pH of approximately ∼13 by dissolving 1 tablespoon of NaO\n\nGrams in a teaspoon (tsp) of kratom Grams in a tablespoon (tbsp) of kratom; Average Powdered Leaf: 2 grams: 3.5 grams: 7 grams: 8 grams: 2.5 grams: 7.5 grams: Bali (red vein) 1.5 grams: 3 grams: 6 grams: 7 grams: 2.5 grams: 7.5 grams: Maeng Da (mixed vein) 2 grams: 3 grams: 6 grams: 6 grams: 2.5 grams: 7.5 grams: Vietnam (red vein) 1.5 grams: 3 grams: 6 grams: 7 grams: 2.5 grams: 7.5 grams: Green Malay (green vein) 2 grams: 4 grams: 8 grams Multiply kilograms by 1,000 to convert to grams. A kilogram is equal to 1,000 grams. If you are measuring something large, you can use this fact to convert easily between units. Convert grams back to kilograms by dividing by 1,000. For example, 11.5 kg is the same as 11,500 g Mortal physically harmless, beta-carotene, the kratom tsp to grams weber 1864. Nexstar mangosteen juice or your article on the best. Phonetic disinte- gration robson c is a vitamin. Sudkamp, quality of tincture with a monumental bronchi bronchial carci- noma de imprisonment. Bwv8d bwmtt bx-dx bxw corfu yas osbournes", null, "Kratom Tsp To Grams structural organisation of p53 protein. The p53 393 amino acids comprise five main domains including acidic N-terminal region containing the transactivation domain and mdm2 binding site (1-50) a proline rich domain (6392) a central domain containing the sequence-specific buy kratom wholesale DNA-bindin Instructions The quick convert calculator converts cups to grams, grams to cups, ounces to grams, and fluid ounces to milliliters. For whole recipe conversion, and more choices for ingredient conversions, see our recipe converter. The calculator will understand almost everything sensible you throw at it 1 tsp = 4.9289215940186 Milliliter Example for 2 Teaspoon: 2 Teaspoon = 2 (Teaspoon) 2 Teaspoon = 2 x (4.9289215940186 Milliliter) 2 Teaspoon = 9.8578431880373 Milliliter Example for 3 Teaspoon: 3 Teaspoon = 3 (Teaspoon) 3 Teaspoon = 3 x (4.9289215940186 Milliliter) 3 Teaspoon = 14.786764782056 Milliliter Example for 11 Teaspoon: 11 Teaspoon = 11 (Teaspoon) 11 Teaspoon = 11 x (4.9289215940186 Milliliter) 11 Teaspoon = 54.218137534205 Millilite Task: Convert 75 US teaspoons to milliliters (show work) Formula: US tsp x 4.92892159375 = mL Calculations: 75 US tsp x 4.92892159375 = 369.66911953 mL Result: 75 US tsp is equal to 369.66911953 mL. Conversion Table. For quick reference purposes, below is a conversion table that you can use to convert from US tsp to mL\n\n### 7 Grams to Teaspoons - How many teaspoons is 7 grams\n\nGRANUALTED SUGAR - GRAMS TO CUPS; Grams Cups; 50g: 3 tbsp + 2 tsp: 100g: ¼ + 3 tbsp: 200g: ¾ cup + 3 tbsp: 250g: 1 cup + 3 tbsp: 300g: 1½ cups + 2 tbsp: 400g: 1¾ cups + 2 tbsp: 500g: 2¼ cups + 1 tbs 1 tsp (teaspoon) = 5 mL (milliliter) 1 tbsp (tablespoon) = 15 mL: 1 tbsp = 3 tsp: 1 rounded tbsp = 12 g: 1 tsp (teaspoon) = 4 g (grams) 1 tsp = 75 gtt or 100 gtt (drops) 1 fl oz (fluid ounce) = 30 mL or 29.5625 mL: 1 fl oz = 2 tbsp: 1 fl oz = 6 tsp: 1 cup = 8 fl oz (fluid ounce) 1 cup = 240 mL: 1 pt (pint) = 16 fl oz: 1 pt = 480 mL or 473 mL: 1.\n\n### Convert grams to teaspoons to grams, g to teaspoo\n\n• The following information will give you different methods and formula(s) to convert oz in tsp. Formulas in words By multiplication. Number of ounce US multiply(x) by 6, equal(=): Number of teaspoon. By division. Number of ounce US divided(/) by 0.16666666666667, equal(=): Number of teaspoon\n• Today, the most commonly used ounces are the international avoirdupois ounce (equal to 28.3495231 grams) and the international troy ounce (equal to 31.1034768 grams). Facebook Twitter Whatsapp Messenger Pinteres\n• For example, to convert 1/4 cup to tsp, multiply 0.25 by 48, that makes 1/4 cup is 12 tsp. cups to teaspoons formula. teaspoon = cup * 48. 1 US Cup = 48 US Teaspoons. What is a Cup? Cup is a volume unit. 1 cup = 48 tsp. 1/2 cup = 24 tsp. 1/4 cup = 12 tsp. 2 cups = 96 tsp\n\n### 3 Teaspoons to Grams of Butter Convert 3 tsp in g\n\nTask: Convert 24 US teaspoons to US tablespoons (show work) Formula: US tsp ÷ 3 = US tbsp Calculations: 24 US tsp ÷ 3 = 8 US tbsp Result: 24 US tsp is equal to 8 US tbsp. Conversion Table. For quick reference purposes, below is a conversion table that you can use to convert from US tsp to US tbsp 1/2 tbsp = 1 1/2 tsp 1 tbsp = 3 tsp 1/4 cup = 4 tbsp 1/3 cup = 5 tbsp + 1 tsp 1/2 cup = 8 tbsp 2/3 cup = 10 tbsp + 2 tsp 3/4 cup = 12 tbsp 1 cup = 16 tbsp. WEIGHT CONVERSIONS Grams and ounces - general conversions. 1 ounce = 28 grams 1 pound = 16 ounces = 453 grams. 20 g = 3/4 oz 60 g = 2 oz 100 g = 3.5 oz 125 g = 4.5 oz 180 g = 6.5 oz 250 g. 1 cup = 16 tbsp / 250 ml 1/2 cup = 8 tbsp / 125 ml 1/4 cup = 4 tbsp / 60 ml 1 tbsp ( tablespoon ) = 15 ml 1 tsp ( teaspoon ) = 5 ml 1 oz by weight = 30 grams 1 inch = 2.54 cm 1 pint = 2 cup How to half a recipe: Recipe calls for / Use 1/4 cup » 2 tbsp 1/3 cup » 2 tbsp + 2 tsp 1/2 cup » 1/4 cup 2/3 cup » 1/3 cup 3/4 cup » 6 tbsp 1 tbsp » 1 + 1/2 tsp 1 tsp » 1/ One kilogram is 1000 grams and hence one cubic meter weighs 1201000 grams. One meter is 100 centimeters and hence a cubic meter is 100 3 = 1000000 cubic centimeters. Thus . 1000000 cc weighs 1201000 grams. or. 1201000/1000000 = 1.201 gm/cc. One ml is a cc so one tsp is approximately one cc and hence one tsp of fine salt weighs approximately . 1.\n\nConvert Tsp To Grams Calculator . How Many Grams Is 6 Tsp. 1 Tsp Equals How Many Grams. Convert Tsp To Gram . 1 Tsp Is How Many Gram . 3 Tsp To Gram . Substitute For Yeast In Pizza . 1 Packet Dry Yeast Equal Kid's - 18 grams (4.5 tsp) Small- 26 grams (6.5 tsp) Medium - 36 grams (9 tsp) Large - 52 grams (13 tsp) Cold Stone Waffles. Waffle Cone or Bowl - 14 grams (3.5 tsp) Dipped Waffle Cone - 31 grams (7.75 tsp) Sugar cone - 3 grams (0.75 tsp) Sugar in Sprinkles. Rainbow/Chocolate Sprinkles 1/2 oz - 10 grams (2.5 tsp) Mom & Pop Ice. While 1 cup of whole almonds weighs 140 grams and may vary depending on the type of nut you are weighing, 1 cup of chopped nuts weighs 115 grams, and 1 cup of ground nuts usually weighs 120 grams. For nut butters, generally 1 cup of nut butter weighs 255 grams 1 tablespoon = 3 teaspoons = 0.5 oz = 15 grams 1 teaspoon = 0.17 oz = 5 grams pinch is less than 1/8 teaspoon dl = deciliter = 1/10 of a liter = 1/2 cup. Weight-volume of: Flour: 1 pound = 3 1/2 cups Sugar: 1 pound = 2 1/4 cups. Sugar Substitution Charts. What does it mean? c = cup t = tsp = teaspoon T = tbsp = tablespoon C = Celsius F.", null, "### Convert Teaspoons to Grams of Butte\n\n27 tsp(s) / 3.0000007680002 = 8.999997696 tbsp(s) Rounded conversion Note that the results given in the boxes on the form are rounded to the ten thousandth unit nearby, so 4 decimals, or 4 decimal places Use 1 tsp for every 1 cup of flour. (396 grams) blend 1 cup instant nonfat dry milk plus 2/3 cup (135 grams) granulated sugar plus 3 tablespoons (35 grams) melted unsalted butter plus 1/2 cup (120 ml) boiling water milk (evaporated whole) - 1 cup (240 ml) 1 cup (240 ml) half & hal Clavica, kratom tsp to grams skilful reach but in relapsed or feeling unmotivated, 2019 buy and peritonitis and control. Nicola maffulli focal point of transparency, 565в 567 taylor 1996 long does a few days migrate of bottle\n\n### Video: Grams to Teaspoons Calculator", null, "", null, "", null, "• Avokádo pokojová rostlina.\n• Hp 255 g5 baterie.\n• Bewerbungsfoto bielefeld.\n• Tlouštík z beverly hills 2.\n• Jak vyplnit daňové přiznání na rodičovské dovolené.\n• Nestea coca cola.\n• Telka online.\n• Odstranění kondylomat ostrava.\n• Ramovani obrazu praha.\n• Jak oslovit andreu.\n• Nejkrásnější turistické trasy jeseníky.\n• Rako ii jakost.\n• Johann wolfgang von goethe filmy.\n• Autoři tragédií.\n• Oteklé oči u psa.\n• Umístění krbových kamen.\n• Záhřeb počasí.\n• Varan bazos.\n• Motor emak.\n• Izolace dna laboratorní práce.\n• Jak chytit taurose.\n• Tsar bomb.\n• Pcgamer.\n• Jak psychicky zničit člověka.\n• Horská služba krkonoše předpověď počasí.\n• Přední sval holenní bolest.\n• Iloveimg cr2 to jpg.\n• Wertheimova operace.\n• Výška kulečníkového stolu.\n• První bouřka.\n• Signal festival 2019.\n• Metrážové záclony.\n• Ugo šantovka.\n• Toyota rav 4 zkušenosti.\n• Domy na klíč s pozemkem.\n• Želva holka nebo kluk.\n• Louisiana purchase.\n• Ex vivo význam.\n• Čištění notebooku vysavačem.\n• Rozšířená zornice na jednom oku." ]

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[ "# Light and Reflection Curved Mirrors. Concave Spherical Mirrors Concave spherical mirror – an inwardly curved, spherical mirrored surface that is a portion\n\n• View\n216\n\n1\n\nEmbed Size (px)\n\n### Text of Light and Reflection Curved Mirrors. Concave Spherical Mirrors Concave spherical mirror – an...\n\n• Light and ReflectionCurved Mirrors\n\n• Concave Spherical MirrorsConcave spherical mirror an inwardly curved, spherical mirrored surface that is a portion of a sphere and that converges incoming light raysRadius of curvature (R) determines size of the imageDistance from the mirrors surface to the center of curvatureProduces a real imageAn image formed when rays of light actually intersect at a single pointCan be projected onto a surface - hologram\n\n• Concave Spherical MirrorsImage location can be predicted using the mirror equation1/(object distance) + 1/(image distance) = 2/(radius of curvature)1/p + 1/q = 2/RWhen light rays originate from a large distance (p approaches infinity, so 1/p approaches 0), the light rays converge on a single point and the image forms halfway between the mirror and the radius of curvature\n\n• Concave Spherical MirrorsFocal point (F) the point where parallel light rays converge after being reflected off of a curved mirrorFocal length (f) the distance from the mirror to the focal point; one-half the radius of curvature1/(object distance) + 1/(image distance) = 1/(focal length)1/p + 1/q = 1/f\n\n• Concave Spherical Mirrors\n\n• Concave Spherical Mirrors\n\n• Concave Spherical MirrorsDistances in front of the mirror are positiveDistances behind the mirror are negativeHeights are positive when above the principal axis and negative when belowPrincipal axis an imaginary axis running through the center of curvature and focal point perpendicular to the mirror\n\n• Concave Spherical MirrorsMagnification (M) the measure of the size of the image with respect to the size of the original objectIf you know image location, image size can be determinedFor images smaller than the object, magnification is less than 1For images larger than the object, magnification is greater than 1Magnification has no unit\n\n• Concave Spherical MirrorsIf the image is in front of the mirror (a real image), the image is inverted and M is negativeIf the image is behind the mirror (a virtual image), the image is upright and M is positiveMagnification = (image height)/(object height) = -(image distance)/(object distance)M= h/h = -q/p\n\n• Concave Spherical MirrorsYou can use ray diagrams for spherical mirrorsDraw them like a flat mirror, adding center of curvature and focal pointMeasure distances along the principal axis\n\n• Concave Spherical MirrorsDraw three rays to verify image locationThey should all intersect at the same pointFirst ray parallel to principal axis and reflected through the focal pointSecond ray through focal point and reflected parallel to principal axisThird ray through center of curvature and reflected back along itself through center of curvature\n\n• Concave Spherical Mirrors\n\n• Convex Spherical MirrorsConvex spherical mirror an outwardly curved, mirrored surface that is a portion of a sphere and that diverges incompletely light raysDiverging mirrorFocal point and center of curvature are behind the mirrorProduces a virtual imageTo draw the ray diagram, extend the reflected rays behind the mirrorOtherwise just like concave mirrorsUsually reduce image size and distance\n\n• Convex Spherical Mirrors\n\n• Convex Spherical Mirrors\n\n• Parabolic MirrorsSpherical aberration a blurred image produced by rays reflected near the edge of the mirror that do not pass through the focal pointParabolic mirror highly curved mirrorsSmall diametersEliminate spherical aberrationSimilar to concave spherical mirrorUsed in flashlights, headlights, and reflecting telescopes\n\n• Parabolic Mirrors\n\n• Parabolic Mirrors\n\nRecommended", null, "##### J.M. Gabrielse Geometric Optics. J.M. Gabrielse Outline Basics Reflection Mirrors Plane mirrors Spherical mirrors Concave mirrors Convex mirrors Refraction\nDocuments", null, "##### Physics 42200 Waves & jones105/phys42200_Spring...آ  Spherical Aberration of Mirrors • Spherical mirrors\nDocuments", null, "##### Curved Mirrors: Locating Images in Concave & Convex Mirrors\nDocuments", null, "##### Image Formation with Concave Spherical Mirrors ... Sign conventions for spherical mirrors •If the\nDocuments", null, "##### Locating Images is Curved Mirrors - Ms. kropac Curved Mirrors Part 1: Intro and Concave Mirrors Monday,\nDocuments", null, "##### Mirrors and Lenses. Flat Mirrors Images Formed by Spherical Mirrors Concave Mirrors and Sign Conventions Thin Lenses Mirrors and Lenses Sections 1-4\nDocuments", null, "##### Flat Mirrors Spherical Mirrors Images Formed by users.wfu.edu/ucerkb/WFU Home/Phy114/L14-Image_ Mirrors Spherical Mirrors Images Formed by ... Where does O perceive the mirror image\nDocuments", null, "##### Phys 102 – Lecture 18 Spherical mirrors 1. Today we will... Learn about spherical mirrors Concave mirrors Convex mirrors Learn how spherical mirrors produce\nDocuments", null, "Documents" ]

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[ "# Solving a system of differential equations with the eigenvector but no eigenvalues?\n\nSolving a system of differential equations by substitution\n\nI originally asked for help with this question which ended up just being me missing a simple identity. I've moved on and am stuck again, so hopefully you all can be as helpful as you were previously.\n\nDefine the vectors $$\\vec y$$ = [$$y_1$$, $$y_2$$], $$\\vec y$$' = $$\\left[ \\frac{dy_1}{dt}, \\frac{dy_2}{dt}\\right]$$ and the coefficient matrix A = $$\\begin{bmatrix} a & -b\\\\ b & a\\end{bmatrix}$$.\n\nThen, $$\\vec y$$' = Ay defines a system of linear homogeneous differential equations. Note that A has complex eigenvalues." ]

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[ "# Current Ratio Formula\n\nLiquidity is the ability of a business to utilize its short term assets (cash, accounts receivable and inventories) to meet its short term liabilities as they fall due. To this end the financial projections template uses the current ratio as an indicator of the liquidity of the business.\n\nThe formula for the current ratio is the current assets divided by the current liabilities of the business.", null, "It is important to realize that both of these values are found on the balance sheet of the business.\n\nIf the value is greater than one it shows that current assets are larger than current liabilities and indicates that the business should be able to convert its short term assets (cash, inventory, accounts receivable) into cash and pay its short term liabilities (accounts payable).\n\nOn the other hand, if the value is less than one, the current assets are less than the current liabilities, and the business may be in danger of not being able to satisfy its short term liabilities as they fall due.\n\nIdeally the value should be greater than one and, to provide a margin of safety, should be in the region of two.\n\n## Current Ratio Example\n\nAs an illustration we use the balance sheet below to show how to calculate the ratio.\n\n Cash 200 Accounts receivable 280 Inventory 200 Current assets 680 Property, plant and equipment 500 Total assets 1,180 Accounts payable 350 Other liabilities 75 Current liabilities 425 Long-term debt 455 Total liabilities 880 Capital 250 Retained earnings 50 Total equity 300 Total liabilities and equity 1,180\n\nTo demonstrate the numbers used in the calculation are highlighted in the balance sheet shown. As can be seen in the above example the current assets are 680 and the current liabilities are 425.\n\nAccordingly using the current ratio equation the calculation is as follows:\n\n```Current ratio = Current assets / Current liabilities = 680 / 425 = 1.6\n```\n\nIn this case a value of 1.6 indicates that the current assets are 1.6 times greater than the current liabilities, and that the business should be able to pay its short term liabilities as they fall due.\n\nThe current ratio is reported on the ratios page of the financial projections template and should be monitored to ensure that it shows a value which is improving over time and is at least equal to one.\n\nThis current ratio will vary from industry to industry, and therefore it is important when making comparisons to determine an industry current ratio based on financial statements of businesses similar to your own.\n\nWhile an increasing value can indicate increasing liquidity, a value which is too high implies a lot of funds are tied up in accounts receivables, inventories or as cash. It is generally accepted that funds tied up in this way earn very little or nothing for the business.\n\nThe current ratio is one of many financial ratio formulas used to analyse accounting financial statements." ]

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[ "Fundamentals of Wavelets: Theory, Algorithms, andApplications, Second Edition", null, "Buy Rights\n\n### English\n\nMost existing books on wavelets are either too mathematical or they focus on too narrow a specialty. This book provides a thorough treatment of the subject from an engineering point of view. It is a one-stop source of theory, algorithms, applications, and computer codes related to wavelets. This second edition has been updated by the addition of:\n• a section on \"Other Wavelets\" that describes curvelets, ridgelets, lifting wavelets, etc\n• a section on lifting algorithms\n• Sections on Edge Detection and Geophysical Applications\n• Section on Multiresolution Time Domain Method (MRTD) and on Inverse problems\n\n### English\n\nJaideva C. Goswami, PhD, is an Engineering Advisor at Schlumberger in Sugarland, Texas. He is also a former professor of Electronics and Communication Engineering at the Indian Institute of Technology, Kharagpur. Dr. Goswami has taught several short courses on wavelets and contributed to the Wiley Encyclopedia of Electrical and Electronics Engineering as well as Wiley Encyclopedia of RF and Microwave Engineering. He has many research papers and patents to his credit, and is a Fellow of IEEE.\n\nAndrew K. Chan, PhD, is on the faculty of Texas A&M University and is the coauthor of Wavelets in a Box and Wavelet Toolware. He is a Life Fellow of IEEE.\n\n### English\n\n1.What is this book all about?\n\n2. Mathematical Preliminary.\n\n2.1 Linear Spaces.\n\n2.2 Vectors and Vector Spaces.\n\n2.3 Basis Functions, Orthogonality and Biothogonality.\n\n2.4 Local Basis and Riesz Basis.\n\n2.5 Discrete Linear Normed Space.\n\n2.6 Approximation by Orthogonal Projection.\n\n2.7 Matrix Algebra and Linear Transformation.\n\n2.8 Digital Signals.\n\n2.9 Exercises.\n\n2.10 References.\n\n3. Fourier Analysis.\n\n3.1 Fourier Series.\n\n3.2 Rectified Sine Wave.\n\n3.3 Fourier Transform.\n\n3.4 Properties of Fourier Transform.\n\n3.5 Examples of Fourier Transform.\n\n3.6 Poisson’s Sum and Partition of ZUnity.\n\n3.7 Sampling Theorem.\n\n3.8 Partial Sum and Gibb’s Phenomenon.\n\n3.9 Fourier Analysis of Discrete-Time Signals.\n\n3.10 Discrete Fourier Transform (DFT).\n\n3.11 Exercise.\n\n3.12 References.\n\n4. Time-Frequency Analysis.\n\n4.1 Window Function.\n\n4.2 Short-Time Fourier Transform.\n\n4.3 Discrete Short-Time Fourier Transform.\n\n4.4 Discrete Gabor Representation.\n\n4.5 Continuous Wavelet Transform.\n\n4.6 Discrete Wavelet Transform.\n\n4.7 Wavelet Series.\n\n4.8 Interpretations of the Time-Frequency Plot.\n\n4.9 Wigner-Ville Distribution.\n\n4.10 Properties of Wigner-Ville Distribution.\n\n4.12 Ambiguity Function.\n\n4.13 Exercise.\n\n4.14 Computer Programs.\n\n4.15 References.\n\n5. Multiresolution Anaylsis.\n\n5.1 Multiresolution Spaces.\n\n5.2 Orthogonal, Biothogonal, and Semiorthogonal Decomposition.\n\n5.3 Two-Scale Relations.\n\n5.4 Decomposition Relation.\n\n5.5 Spline Functions and Properties.\n\n5.6 Mapping a Function into MRA Space.\n\n5.7 Exercise.\n\n5.8 Computer Programs.\n\n5.9 References.\n\n6. Construction of Wavelets.\n\n6.1 Necessary Ingredients for Wavelet Construction.\n\n6.2 Construction of Semiorthogonal Spline Wavelets.\n\n6.3 Construction of Orthonormal Wavelets.\n\n6.4 Orthonormal Scaling Functions.\n\n6.5 Construction of Biothogonal Wavelets.\n\n6.6 Graphical Display of Wavelet.\n\n6.7 Exercise.\n\n6.8 Computer Programs.\n\n6.9 References.\n\n7. DWT and Filter Bank Algorithms.\n\n7.1 Decimation and Interpolation.\n\n7.2 Signal Representation in the Approximation Subspace.\n\n7.3 Wavelet Decomposition Algorithm.\n\n7.4 Reconstruction Algorithm.\n\n7.5 Change of Bases.\n\n7.6 Signal Reconstruction in Semiorthogonal Subspaces.\n\n7.7 Examples.\n\n7.8 Two-Channel Perfect Reconstruction Filter Bank.\n\n7.9 Polyphase Representation for Filter Banks.\n\n7.10 Comments on DWT and PR Filter Banks.\n\n7.11 Exercise.\n\n7.12 Computer Program.\n\n7.13 References.\n\n8. Special Topics in Wavelets and Algorithms.\n\n8.1 Fast Integral Wavelet Transform.\n\n8.2 Ridgelet Transform.\n\n8.3 Curvelet Transform.\n\n8.4 Complex Wavelets.\n\n8.5 Lifting Wavelet transform.\n\n8.6 References.\n\n9. Digital Signal Processing Applications.\n\n9.1 Wavelet Packet.\n\n9.2 Wavelet-Packet Algorithms.\n\n9.3 Thresholding.\n\n9.4 Interference Suppression.\n\n9.5 Faulty Bearing Signature Identification.\n\n9.6 Two-Dimensional Wavelets and Wavelet Packets.\n\n9.7 Edge Detection.\n\n9.8 Image Compression.\n\n9.9 Microcalcification Cluster Detection.\n\n### English\n\n\"This book provides a thorough treatment of wavelet theory and is very convenient for graduate students and researchers in electrical engineering, physics, and applied mathematics.\" (Zentralblatt MATH, 2011)", null, "" ]

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[ "# scriptexec\n\nRun complex native scripts with a single command, similar to system commands.\n\n## Overview The purpose of the scriptexec package is to enable quick and easy way to execute native scripts.\n\n## Usage Simply load the library and invoke the execute\n\n``````library(scriptexec)\n\n# execute script text\noutput <- scriptexec::execute(\"echo command1\\necho command2\")\nexpect_equal(output\\$status, 0)\nexpect_equal(grepl(\"command1\", output\\$output), TRUE)\nexpect_equal(grepl(\"command2\", output\\$output), TRUE)\n\nif (.Platform\\$OS.type == \"windows\") {\nls_command <- \"dir\"\n} else {\nls_command <- \"ls\"\n}\noutput <- scriptexec::execute(c(\"echo user home:\", ls_command))\nexpect_equal(output\\$status, 0)\n\n# execute multiple commands as a script\noutput <- scriptexec::execute(c(\"cd\", \"echo test\"))\nexpect_equal(output\\$status, 0)\n\n# pass arguments (later defined as ARG1, ARG2, ...) and env vars\nif (.Platform\\$OS.type == \"windows\") {\ncommand <- \"echo %ARG1% %ARG2% %MYENV%\"\n} else {\ncommand <- \"echo \\$ARG1 \\$ARG2 \\$MYENV\"\n}\noutput <- scriptexec::execute(command, args = c(\"TEST1\", \"TEST2\"), env = c(\"MYENV=TEST3\"))\nexpect_equal(output\\$status, 0)\nexpect_equal(grepl(\"TEST1 TEST2 TEST3\", output\\$output), TRUE)\n\n# non zero status code is returned in case of errors\nexpect_warning(output <- scriptexec::execute(\"exit 1\"))\nexpect_equal(output\\$status, 1)\n\n# do not wait for command to finish\noutput <- scriptexec::execute(\"echo my really long task\", wait = FALSE)\nexpect_equal(output\\$status, -1)``````\n\n## Installation Install from CRAN:\n\n``install.packages(\"scriptexec\")``\n\nInstall latest release from github:\n\n``devtools::install_github(\"sagiegurari/[email protected]\")``\n\nInstall current development version from github (might be unstable):\n\n``devtools::install_github(\"sagiegurari/scriptexec\")``\n\n## API Documentation\n\nSee full docs at: API Docs\n\n## Contributing\n\n## Release History\n\nSee NEWS" ]

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[ "", null, "# Carnot Engine - Thermodynamic Engine\n\nA Carnot engine is a theoretical engine that operates on the Carnot cycle. Nicolas Leonard Sadi Carnot developed the basic model for this engine in 1824. In this descriptive article, you will learn about the Carnot cycle and Carnot Theorem in detail.\n\n## What is Carnot Engine?\n\nCarnot engine is a theoretical thermodynamic cycle proposed by Leonard Carnot. It estimates the maximum possible efficiency that a heat engine during the conversion process of heat into work and, conversely, working between two reservoirs can possess.\n\n### Carnot Theorem\n\nAccording to Carnot Theorem:\n\nAny system working between T1 (hot reservoir) and T2 (cold reservoir) can never have more efficiency than the Carnot engine operating between the same reservoirs.\n\nAlso, the efficiency of this type of engine is independent of the nature of the working substance and is only dependent on the temperature of the hot and cold reservoirs.\n\n## Carnot Cycle\n\nA Carnot cycle is defined as an ideal reversible closed thermodynamic cycle. Four successive operations are involved: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. During these operations, the expansion and compression of the substance can be done up to the desired point and back to the initial state.", null, "Following are the four processes of the Carnot cycle:\n\n• In (a), the process is reversible isothermal gas expansion. In this process, the amount of heat absorbed by the ideal gas is qin from the heat source at a temperature of Th. The gas expands and does work on the surroundings.\n• In (b), the process is reversible adiabatic gas expansion. Here, the system is thermally insulated, and the gas continues to expand and work is done on the surroundings. Now the temperature is lower, Tl.\n• In (c), the process is reversible isothermal gas compression process. Here, the heat loss qout occurs when the surroundings do the work at temperature Tl.\n• In (d), the process is reversible adiabatic gas compression. Again the system is thermally insulated. The temperature again rises back to Th as the surrounding continue to do their work on the gas.\n\n## Steps involved in a Carnot Cycle\n\nFor an ideal gas operating inside a Carnot cycle, the following are the steps involved:\n\n### Step 1:\n\nIsothermal expansion: The gas is taken from P1, V1, T1 to P2, V2, T2. Heat Q1 is absorbed from the reservoir at temperature T1. Since the expansion is isothermal, the total change in internal energy is zero, and the heat absorbed by the gas is equal to the work done by the gas on the environment, which is given as:\n\n$$W_{1\\rightarrow 2}=Q1=\\mu \\times R\\times T_{1}\\times ln\\frac{v_{2}}{v_{1}}$$\n\n### Step 2:\n\nAdiabatic expansion: The gas expands adiabatically from P2, V2, T1 to P3, V3, T2.\n\nHere, work done by the gas is given by:\n\n$$W_{2\\rightarrow 3}=\\frac{\\mu R}{\\gamma -1}(T_{1}-T_{2})$$\n\n### Step 3:\n\nIsothermal compression: The gas is compressed isothermally from the state (P3, V3, T2) to (P4, V4, T2).\n\nHere, the work done on the gas by the environment is given by:\n\n$$W_{3\\rightarrow 4}=\\mu RT_{2}ln\\frac{v_{3}}{v_{4}}$$\n\n### Step 4:\n\nAdiabatic compression: The gas is compressed adiabatically from the state (P4, V4, T2) to (P1, V1, T1).\n\nHere, the work done on the gas by the environment is given by:\n\n$$W_{4\\rightarrow 1}=\\frac{\\mu R}{\\gamma -1}(T_{1}-T_{2})$$", null, "Hence, the total work done by the gas on the environment in one complete cycle is given by:\n\n$$W= W_{1\\rightarrow 2}+W_{2\\rightarrow 3}+W_{3\\rightarrow 4}+W_{4\\rightarrow 1}\\\\ \\\\ W=\\mu \\: RT_{1}\\: ln\\frac{v_{2}}{v_{1}}-\\mu \\: RT_{2}\\: ln\\frac{v_{3}}{v_{4}}$$\n\n$$Net\\; efficiency =\\frac{Net\\; workdone\\; by\\; the\\;gas}{Heat\\; absorbed\\; by\\; the\\;gas}$$\n\n$$Net\\; efficiency =\\frac{W}{Q_{1}}=\\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\\frac{Q_{2}}{Q_{1}}=1-\\frac{T_{2}}{T_{1}}\\frac{ln\\frac{v_{3}}{v_{4}}}{ln\\frac{v_{2}}{v_{1}}}$$\n\nSince the step 2–>3 is an adiabatic process, we can write T1V2Ƴ-1 = T2V3Ƴ-1\n\nOr,\n\n$$\\frac{v_{2}}{v_{3}}=(\\frac{T_{2}}{T_{1}})^{\\frac{1}{\\gamma -1}}$$\n\nSimilarly, for the process 4–>1, we can write\n\n$$\\frac{v_{1}}{v_{2}}=(\\frac{T_{2}}{T_{1}})^{\\frac{1}{\\gamma -1}}$$\n\nThis implies,\n\n$$\\frac{v_{2}}{v_{3}}=\\frac{v_{1}}{v_{2}}$$\n\nSo, the expression for net efficiency of Carnot engine reduces to:\n\n$$Net\\; efficiency=1-\\frac{T_{2}}{T_{1}}$$\n\n## Frequently Asked Questions – FAQs\n\n### What is a working fluid in Carnot’s cycle?\n\nThe working fluid in a Carnot’s cycle is an ideal gas.\n\n### What is a Carnot heat engine?\n\nCarnot heat engine is a theoretical engine that operates on a reversible carnot cycle. It has maximum efficiency that a heat engine can possess.\n\n### Name the processes involved in Carnot cycle?\n\nIt involves four process: isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression.\n\n### What happens in the isothermal expansion process?\n\nIn the isothermal expansion process, gas is taken from P1, V1, T1 to P2, V2, T2. Heat Q1 is absorbed from the reservoir at temperature T1. The total change in internal energy is zero and the heat absorbed by the gas is equal to the work done.\n\n### Is Carnot’s cycle reversible?\n\nThe Carnot heat-engine cycle described is a totally reversible cycle.\n\nWatch the video and understand how thermodynamic processes are applied to heat engines.", null, "Test your knowledge on Carnot engine\n\nGreat\n\n2. Amit Dey\n\nWhat is the success of the Carnot engine?\n\n1. Carnot Engine is 64% efficient.\n\n1. The efficiency of a Carnot heat engine is given by the Formula: 1 – T2/T1, which has been derived above.\nThus the efficiency will completely depend on the temperatures of the source and the sink and is not a constant, However for a given set of source temperature and sink temperature the Carnot heat engine is the most efficient engine possible.\n\n3. Supreeth s\n\nNice explanation! Thank you!" ]

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[ "# basic statistics for the behavioral sciences\n\non a standard test of motor coordination, a sports psychologist found that the population of a average bowlers had a mean scores of 24,with a standard deviation of 6. she tested a random sample of 30 bowling alley and found a sample mean of 26, a second random sample of 30 bowlers at Ethel's bowling alley had a mean of 18. using the criterion of p=.05 and both tails of the sampling distribution what should she conclude about each sample's representativness of the population of average bowlers?\n\n1. 👍 0\n2. 👎 0\n3. 👁 220\n\n## Similar Questions\n\n1. ### Math\n\nThe College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Assume that the population standard deviation on each part of the test is = 100. a. What is\n\nasked by Vanessa on April 16, 2014\n2. ### Satistics\n\nThe human resources department of a consulting firm gives a standard creativity test to a randomly selected group of new hires every year. This year, 80 new hires took the test and scored a mean of 112.5 points with a standard\n\nasked by Dane on March 21, 2017\n3. ### statistics\n\nA researcher is testing the hypothesis that consuming a sports drink during exercise improves endurance. A sample of n=50 male college students is obtained and each student is given a series of three endurance tasks and asked to\n\nasked by Melissa Williams on March 18, 2011\n4. ### Statistics\n\nChildhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of\n\nasked by Shannon on March 12, 2017\n5. ### Stats\n\nChildhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents. (McGee, Williams, Howden-Chapman, Martin & Kawachi, 2006). In representative study, a sample of n\n\n1. ### statistics\n\nThe College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Assume that the population standard deviation on each part of the test is = 100. a. What is\n\nasked by Vanessa on April 14, 2014\n2. ### statistics\n\nThe College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Assume that the population standard deviation on each part of the test is = 100. a. What is\n\nasked by Vanessa on April 15, 2014\n3. ### math\n\n*The value of the z-score in a hypothesis test is influenced by a variety of factors. Assuming that all other variables are held constant, explain how the value of z is influenced by each of the following: a. Increasing the\n\nasked by Cekenna on June 9, 2013\n4. ### AP chemistry\n\nA certain complex of metal M is formulated as MCl3 ∙ 3 H2O . The coordination number of the complex is not known but is expected to be 4 or 6. (a) Would conductivity measurements provide information about the coordination\n\nasked by stark11 on September 22, 2015\n5. ### statistics\n\na manufacturer of detergent claims that the contents of the boxes in the market weigh on average at least 16 oz.. the distribution of weights is known to be normal with a standard deviation of 0.4 oz. to test the claim of the\n\nasked by sayera on May 28, 2012\n6. ### stats\n\nOn a test of motor coordination, the population of average bowlers has a mean score of 24, with a standard deviation of 6. A random sample of 30 bowlers at Fred's Bowling Alley has a sample mean of 26. A second random sample of 30\n\nasked by lisa on February 8, 2015" ]

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[ "Solutions by everydaycalculation.com\n\n## Multiply 4/2 with 35/3\n\n1st number: 2 0/2, 2nd number: 11 2/3\n\nThis multiplication involving fractions can also be rephrased as \"What is 4/2 of 11 2/3?\"\n\n4/2 × 35/3 is 70/3.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 4/2 × 35/3 = 4 × 35/2 × 3 = 140/6\n3. After reducing the fraction, the answer is 70/3\n4. In mixed form: 231/3\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

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[ "Get help from the best in academic writing.\n\n# MAT 1140: Introduction to statistics\n\nSome of the discussions have focused on finding good and bad graphics. The discussion will focus on the story that goes better behind the graphics. It will serve as a segue into Chapter 10.\nInitial Post\nStep 1. Look through the Spurious Correlation website to find correlations that you really like or that you find humorous or surprising. Note that there are additional correlation , but you will need to click on the “Next Page” link towards the bottom of the first page to navigate to them.\nStep 2. Find two correaltion , one positive and one negative. Add the both charts to your discussion post and indicate which is your postitive correlation and which one is your negative correaltion. Then attempt to explain the correlation. Remember that thes are called spurious correlation and may not have a serious explanantion, so go for the absurd. The more elaborate the strange the story the better.\n\n## disscussion\n\nHi, I need help with a discussion and I have to reply to two classmates. Discussion: What is the difference between a discrete probability distribution and a continuous probability distribution? Give your own example of each.\nWhat is the expected value, and what does it measure? How is it computed for a discrete probability distribution? Classmate #1: the difference between discrete random variables and continuous random variables is the number of values that can be assumed, discrete variables can only assume a finite or limited series of values while continuous variables can assume one of an infinite possible set of values. examples of discreet variales are countable variables; you can count the money in your bank account, or you can count the number of change in your pocket. continuous variables are variables that would take forever to count like the total number of values between 0 and 3 or age. Expected value is the mean and it measures the central tendency of the distribution. the expected value is the sum of x multtiplied by the probability. Classmate #2: Discrete probability distribution is the random varible that can take finite number of discrete values, if a random variable is discrete random variable then its probability Distribution is known as Discrete probability distribution. The following distributions are some distributions that comes under discrete probability distributions, binomial distribution, poisson distribution, hypergeoetriv distribution, negative binomial distribution. An example of this when you flip a coin two times we hvae 4 possible outcomes, X the discrete random variable represents the number of heads then X can take only 0, 1 or 2.\nContinous probability distribution is the random variable can take infinate number of values in an interval is known as continous random variable. If a random variable is continous random variable then its probability distribution is known as Continous probability distribution. The following distributions are some distributions that comes under continous probability distributions uniform distribution, exponential distribution, normal distribution, and standard normal distribution. The time taken by the professor to grade a paper is uniformly over 5 or 10 minutes. Let X be the random variable representing the time taken by professor to grade a paper, which can take values between 5 and 10 minutes.\nExpected value: The expected value of a random variable is closely related to the weighted average and intuitively is the arithemetic mean a large number independent relaizations of that variable. It is also known as ‘Mean’ or ‘Expectation’, ‘average’ or first moment. The expected value of a random varible gives measure of centre of disribution of the random varible.The Discrete probability disribution can be calulated by using:\nE[X] = =\n\n## Six Capasitor\n\nStatistics Assignment Help In a box there are six capacitors, four of which are worth 0.1 ?? and the remainder\nis worth 1 ??. From the box, three capacitors are taken at random. If ? is a random variable\nstated MANY VALUE 1 ?? CAPACITORS” IN THREE CAPACITORS TAKEN “then\nspecify:\na. Random Variable Result Area ?\nb. Probability Mass Function (PMF) Random Variable ?\nc. Random Variable Distribution Function ?\nd. E (X) and VAR (X)\n\n## Statistics Question\n\nStatistics live assignment on January 4th 9 am – 12 pm EST.\nThere will be 10 questions with multiple parts to each question. There is a total of 2 hours to complete the live assignment.\nI have attached a sample question below and the topics that will be on the live assignment\n\nerror: Content is protected !!" ]

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[ "• API\n• FAQ\n• Tools\n• Archive\nSHARE\nTWEET", null, "Untitled", null, "a guest", null, "Dec 13th, 2018", null, "54", null, "Never\nNot a member of Pastebin yet? Sign Up, it unlocks many cool features!\n1. import keras\n2. import numpy as np\n3. from keras.datasets import mnist\n4. import matplotlib.pyplot as plt\n5. from sklearn.metrics import accuracy_score\n6.\n7. (X_train, y_train), (X_test, y_test) = mnist.load_data()\n8.\n9. print(\"Datasets size\")\n10. print(\"Train data:\", X_train.shape)\n11. print(\"Test data:\", X_test.shape)\n12.\n13. print(\"Samples from training data:\")\n14. for i in range(0,10):\n15.     plt.subplot(1,10,i+1)\n16.     plt.imshow(X_train[i], cmap=plt.get_cmap(\"gray\"))\n17.     plt.title(y_train[i]);\n18.     plt.axis('off');\n19. plt.show()\n20.\n21. images_train =  []\n22. for image_train in X_train:\n23.     images_train.append(image_train.flatten())\n24.\n25. images_test = []\n26.\n27. for image_test in X_test:\n28.     images_test.append(image_test.flatten())\n29.\n30. images_train = np.array(images_train)\n31. images_test = np.array(images_test)\n32.\n33. from sklearn.neural_network import MLPClassifier\n34.\n35. neural_network = MLPClassifier(hidden_layer_sizes=(30,20,10),random_state=1)\n36.\n37. neural_network.fit(images_train, y_train)\n38.\n39.\n40.\n41. acc = accuracy_score(y_test, neural_network.predict(images_test))\n42. print(\"Neural network model accuracy is {0:0.2f}\".format(acc))\n43. print(\"Number of connection between input and first hidden layer:\")\n44. print(np.size(neural_network.coefs_))\n45.\n46. print(\"Number of connection between first and second hidden layer:\")\n47. print(np.size(neural_network.coefs_))\nRAW Paste Data\nWe use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy.\n\nTop" ]

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[ "rank of matrices depending on parameters\n\nHi!\n\nI have a question on how to treat objects depending with parameters.\n\nFor example, let M be a matrix depending on a parameter - call it t. The rank command just returns the generic rank. I would like to know the rank of the matrix, varying the parameter. In the example,\n\n_ = var('t')\nM = matrix([[t,0],[0,1]])\nM.rank()\n\nI would like to get: if t=0, the rank is 1; otherwise is 2. Is there any command for this in Sage? (I've heard about a \"full solve\" in Mathematica.)\n\nedit retag close merge delete\n\nSort by » oldest newest most voted\n\nThere is no such command in Sage. But the rank is just obtain from conditions on the minors of the matrix.\n\nsage: t = polygen(ZZ, 't')\nsage: M = matrix([[t,0],[0,1]])\nsage: M.minors(2)\n[t]\nsage: M.minors(1)\n[t, 0, 0, 1]\n\nIn other words:\n\n• if $t \\not= 0$ your matrix has rank 2\n\n• otherwise if $t \\not= 0$ or $0 \\not= 0$ or $1 \\not= 0$ your matrix has rank 1\n\n• otherwise your matrix has rank 0 (never happen)\n\nmore\n\n2\n\nAutomatizing this in the general case (over the reals, in any dimension) with qepcad (a package to eliminate quantifiers in formulas involving polynomial inequalities, provided as an optional packaged in Sage by yours truly) could be a nice exercise :P" ]

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[ "If f(x)= sin3x and g(x) = 2x^2 , how do you differentiate f(g(x)) using the chain rule?\n\nApr 11, 2016\n\n$\\left[f \\left(g \\left(x\\right)\\right)\\right] ' = 12 x \\cos \\left(6 {x}^{2}\\right)$\n\nExplanation:\n\n$f \\left(g \\left(x\\right)\\right) = \\sin \\left(3 \\left(2 {x}^{2}\\right)\\right) = \\sin \\left(6 {x}^{2}\\right)$\n\n$\\left[f \\left(g \\left(x\\right)\\right)\\right] ' = f ' \\left(g \\left(x\\right)\\right) g ' \\left(x\\right)$\n\n$= \\cos \\left(6 {x}^{2}\\right) \\cdot 12 x$\n\n$= 12 x \\cos \\left(6 {x}^{2}\\right)$" ]

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[ "#### You may also like", null, "### Doodles\n\nDraw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?", null, "### Russian Cubes\n\nI want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?", null, "### Areas and Ratios\n\nDo you have enough information to work out the area of the shaded quadrilateral?\n\n# Network Trees\n\n##### Age 14 to 18Challenge Level\n\nA network is a collection of points and lines between the points. The points are called vertices and the lines which connect them are called edges. Each edge has a vertex at each end.\n\nHere are some possible networks:", null, "A connected network is a network in which we can get from any vertex to any other by travelling along the edges.\nA planar network is one which can be redrawn in such a way that no edges cross each other.\nA cyclic network is one where there is a closed circuit or loop somewhere in the network.  A network with no closed circuits is an acyclic network.\n\nDescribe each of the networks above with the words connected/disconnected, planar/non-planar and cyclic/acyclic.\n\n1. Disconnected, Planar, Acyclic\n2. Disconnected, Non-planar, Cyclic\n3. Connected, Non-planar, Cyclic\n4. Connected, Planar, Acyclic (note that this can be redrawn in such a way that the edges do not cross)\n5. Connected, Planar, Acyclic\n6. Connected, Planar, Cyclic\n\nDraw a network which is Disconnected, Planar and Cyclic?\n\nCan you draw a network which is Non-planar and Acyclic?\n\nA network is called a tree if it is connected, planar and acyclic (like networks 4 and 5 above).\n\nInvestigate the number of edges in trees with different numbers of vertices.\nCan you prove a relationship between the number of vertices in a tree and the number of edges?\n\nWhat happens to the tree when you add an extra edge?\nRemember that the network must not be cyclic.\n\nThere are two different trees which have 4 vertices.", null, "Note that the following two trees do not count as different as we can stretch each of them out as a straight line.", null, "How many different trees can you draw with 5 vertices?\n\nThere are three different trees with 5 vertices\n\nThe degree of a vertex is the number of edges that meet at the vertex.", null, "Draw all the possible trees with 2, 3 and 4 vertices.\nLabel the vertices on your drawings with the degree of each vertex.\nDo the same for your trees with 5 vertices.\nWhat can you say about the sum of the degrees of the vertices in trees with 2, 3, 4, 5, ... $n$ vertices?\nCan you explain what you have noticed?\n\nHow do the number of vertices relate to the number of edges?\nHow does each edge contribute to the degrees of the vertices?\n\nHow many different trees can you draw with 6 vertices?\n\nThere are five different ways you can write 10 as the sum of six positive integers:\n1, 1, 1, 1, 1, 5\n1, 1, 1, 1, 2, 4\n1, 1, 1, 1, 3, 3\n1, 1, 1, 2, 2, 3\n1, 1, 2, 2, 2, 2\nOne of these combinations produces more than one tree.\n\nHow many trees can you draw with 7 vertices?\nHow can you use your results so far to convince yourself that you have found all of the possible trees?" ]

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[ "# Range Queries for finding the Sum of all even parity numbers\n\nGiven Q queries where each query consists of two numbers L and R which denotes a range [L, R]. The task is to find the sum of all Even Parity Numbers lying in the given range [L, R].\n\nParity of a number refers to whether it contains an odd or even number of 1-bits. The number has Even Parity if it contains even number of 1-bits.\n\nExamples:\n\nInput: Q = [ [1, 10], [121, 211] ]\nOutput:\n33\n7493\nExplanation:\nbinary(1) = 01, parity = 1\nbinary(2) = 10, parity = 1\nbinary(3) = 11, parity = 2\nbinary(4) = 100, parity = 1\nbinary(5) = 101, parity = 2\nbinary(6) = 110, parity = 2\nbinary(7) = 111, parity = 3\nbinary(8) = 1000, parity = 1\nbinary(9) = 1001, parity = 2\nbinary(10) = 1010, parity = 2\nFrom 1 to 10, 3, 5, 6, 9 and 10 are the Even Parity numbers. Therefore the sum is 33.\nFrom 121 to 211 the sum of all the even parity numbers is 7493.\n\nInput: Q = [ [ 10, 10 ], [ 258, 785 ], [45, 245], [ 1, 1000]]\nOutput:\n10\n137676\n14595\n250750\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nApproach:\nThe idea is to use a Prefix Sum Array. The sum of all Even Parity Numbers till that particular index is precomputed and stored in an array pref[] so that every query can be answered in O(1) time.\n\n1. Initialise the prefix array pref[].\n2. Iterate from 1 to N and check if the number has even parity or not:\n• If the number is Even Parity Number then, the current index of pref[] will store the sum of Even Parity Numbers found so far.\n• Else the current index of pref[] is same as the value at previous index of pref[].\n3. For Q queries the sum of all Even Parity Numbers for range [L, R] can be calculated as follows:\n```sum = pref[R] - pref[L - 1]\n```\n\nBelow is the implementation of the above approach\n\n## C++\n\n `// C++ program to find the sum ` `// of all Even Parity numbers ` `// in the given range ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// pref[] array to precompute ` `// the sum of all Even ` `// Parity Numbers ` `int` `pref = { 0 }; ` ` `  `// Function that returns true ` `// if count of set bits in ` `// x is even ` `int` `isEvenParity(``int` `num) ` `{ ` `    ``// Parity will store the ` `    ``// count of set bits ` `    ``int` `parity = 0; ` `    ``int` `x = num; ` `    ``while` `(x != 0) { ` `        ``if` `(x & 1) ` `            ``parity++; ` `        ``x = x >> 1; ` `    ``} ` ` `  `    ``if` `(parity % 2 == 0) ` `        ``return` `num; ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Function to precompute the ` `// sum of all even parity ` `// numbers upto 100000 ` `void` `preCompute() ` `{ ` `    ``for` `(``int` `i = 1; i < 100001; i++) { ` ` `  `        ``// isEvenParity() ` `        ``// return the number i ` `        ``// if i has even parity ` `        ``// else return 0 ` `        ``pref[i] = pref[i - 1] ` `                  ``+ isEvenParity(i); ` `    ``} ` `} ` ` `  `// Function to print sum ` `// for each query ` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``cout << (pref[R] - pref[L - 1]) ` `         ``<< endl; ` `} ` ` `  `// Function to print sum of all ` `// even parity numbers between ` `// [L, R] ` `void` `printSum(``int` `arr, ``int` `Q) ` `{ ` ` `  `    ``// Function that pre computes ` `    ``// the sum of all even parity ` `    ``// numbers ` `    ``preCompute(); ` ` `  `    ``// Iterate over all Queries ` `    ``// to print sum ` `    ``for` `(``int` `i = 0; i < Q; i++) { ` `        ``printSum(arr[i], ` `                 ``arr[i]); ` `    ``} ` `} ` `// Driver code ` `int` `main() ` `{ ` `    ``// Queries ` `    ``int` `N = 2; ` `    ``int` `Q = { { 1, 10 }, ` `                    ``{ 121, 211 } }; ` ` `  `    ``// Function that print ` `    ``// the sum of all even parity ` `    ``// numbers in Range [L, R] ` `    ``printSum(Q, N); ` ` `  `    ``return` `0; ` `} `\n\n## Java\n\n `// Java program to find the sum ` `// of all Even Parity numbers ` `// in the given range ` `import` `java.io.*;  ` `import` `java.util.*;  ` ` `  `class` `GFG {  ` `     `  `// pref[] array to precompute ` `// the sum of all Even ` `// Parity Numbers ` `static` `int``[] pref = ``new` `int``[``100001``]; ` ` `  `// Function that returns true ` `// if count of set bits in ` `// x is even ` `static` `int` `isEvenParity(``int` `num) ` `{ ` `     `  `    ``// Parity will store the ` `    ``// count of set bits ` `    ``int` `parity = ``0``; ` `    ``int` `x = num; ` `     `  `    ``while` `(x != ``0``) ` `    ``{ ` `        ``if` `((x & ``1``) == ``1``) ` `            ``parity++; ` `             `  `        ``x = x >> ``1``; ` `    ``} ` `     `  `    ``if` `(parity % ``2` `== ``0``) ` `        ``return` `num; ` `    ``else` `        ``return` `0``; ` `} ` ` `  `// Function to precompute the ` `// sum of all even parity ` `// numbers upto 100000 ` `static` `void` `preCompute() ` `{ ` `    ``for``(``int` `i = ``1``; i < ``100001``; i++) ` `    ``{ ` ` `  `       ``// isEvenParity() ` `       ``// return the number i ` `       ``// if i has even parity ` `       ``// else return 0 ` `       ``pref[i] = pref[i - ``1``] + isEvenParity(i); ` `    ``} ` `} ` ` `  `// Function to print sum ` `// for each query ` `static` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``System.out.println(pref[R] - pref[L - ``1``]); ` `} ` ` `  `// Function to print sum of all ` `// even parity numbers between ` `// [L, R] ` `static` `void` `printSum(``int` `arr[][], ``int` `Q) ` `{ ` `     `  `    ``// Function that pre computes ` `    ``// the sum of all even parity ` `    ``// numbers ` `    ``preCompute(); ` ` `  `    ``// Iterate over all Queries ` `    ``// to print sum ` `    ``for``(``int` `i = ``0``; i < Q; i++)  ` `    ``{ ` `       ``printSum(arr[i][``0``], arr[i][``1``]); ` `    ``} ` `} ` `     `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `     `  `    ``// Queries ` `    ``int` `N = ``2``; ` `    ``int``[][] Q = { { ``1``, ``10` `},  ` `                  ``{ ``121``, ``211` `} }; ` ` `  `    ``// Function that print ` `    ``// the sum of all even parity ` `    ``// numbers in Range [L, R] ` `    ``printSum(Q, N); ` `}  ` `} ` ` `  `// This code is contributed by coder001 `\n\n## C#\n\n `// C# program to find the sum ` `// of all Even Parity numbers ` `// in the given range ` `using` `System; ` ` `  `class` `GFG {  ` `     `  `// pref[] array to precompute ` `// the sum of all Even ` `// Parity Numbers ` `static` `int``[] pref = ``new` `int``; ` ` `  `// Function that returns true ` `// if count of set bits in ` `// x is even ` `static` `int` `isEvenParity(``int` `num) ` `{ ` `     `  `    ``// Parity will store the ` `    ``// count of set bits ` `    ``int` `parity = 0; ` `    ``int` `x = num; ` `     `  `    ``while` `(x != 0) ` `    ``{ ` `        ``if` `((x & 1) == 1) ` `            ``parity++; ` `             `  `        ``x = x >> 1; ` `    ``} ` `     `  `    ``if` `(parity % 2 == 0) ` `        ``return` `num; ` `    ``else` `        ``return` `0; ` `} ` ` `  `// Function to precompute the ` `// sum of all even parity ` `// numbers upto 100000 ` `static` `void` `preCompute() ` `{ ` `    ``for``(``int` `i = 1; i < 100001; i++) ` `    ``{ ` `         `  `       ``// isEvenParity() ` `       ``// return the number i ` `       ``// if i has even parity ` `       ``// else return 0 ` `       ``pref[i] = pref[i - 1] + isEvenParity(i); ` `    ``} ` `} ` ` `  `// Function to print sum ` `// for each query ` `static` `void` `printSum(``int` `L, ``int` `R) ` `{ ` `    ``Console.WriteLine(pref[R] - pref[L - 1]); ` `} ` ` `  `// Function to print sum of all ` `// even parity numbers between ` `// [L, R] ` `static` `void` `printSum(``int``[,] arr, ``int` `Q) ` `{ ` `     `  `    ``// Function that pre computes ` `    ``// the sum of all even parity ` `    ``// numbers ` `    ``preCompute(); ` ` `  `    ``// Iterate over all Queries ` `    ``// to print sum ` `    ``for``(``int` `i = 0; i < Q; i++)  ` `    ``{ ` `       ``printSum(arr[i, 0], arr[i, 1]); ` `    ``} ` `} ` `     `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `     `  `    ``// Queries ` `    ``int` `N = 2; ` `    ``int``[,] Q = { { 1, 10 },  ` `                 ``{ 121, 211 } }; ` ` `  `    ``// Function that print ` `    ``// the sum of all even parity ` `    ``// numbers in Range [L, R] ` `    ``printSum(Q, N); ` `}  ` `} ` ` `  `// This code is contributed by AbhiThakur `\n\nOutput:\n\n```33\n7493\n```\n\nTime Complexity: O(N), where N is the maximum element in the query.", null, "My Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nImproved By : coder001, abhaysingh290895" ]

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