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2021-03-30T04:53:59.830 | <p>Is there a way to quantify the amount of radiofrequency interference? Do any units exist that quantifies how much radiofrequency interference is there?</p>
| |electromagnetism|radio| | <p>The severity of radio frequency interference is described by comparing the power content of the interfering noise to that of the desired signal. The power content is measured in units of <em>decibels</em> and the ratio of signal power to noise power gives the <em>signal to noise ratio</em>.</p>
<p>For a signal to be <em>readable</em> it must be stronger than the underlying noise interference (this is called the <em>noise floor</em>). Decibel math states that for two signals that differ by 3 decibels, the power content of the stronger signal is <em>twice</em> that of the weaker signal. So to lift a signal above the noise floor by +3dB requires the signal to be twice the strength of the noise floor.</p>
| 42298 | Is there a way to quantify the amount of radiofrequency interference? |
2021-03-30T14:56:43.840 | <p>I am unable to find it again but I once heard of a type of air pump that involved a pair of offset archimedes screws inside a cylinder. The point of the device was that as the screws would turn against eachother it encapsulated air into pockets (like how a centrifugal fan would) but then these pockets would compress their contents as the screws turned against eachother. What was this pump called?</p>
| |mechanical-engineering|pumps|mechanical| | <p>Rotary screw and scroll , neither uses centrifugal force. Sullair Co. makes both and other types of compressors . Their website should have more information.</p>
| 42307 | What is this mechanical pump called? |
2021-03-31T09:48:44.450 | <p>I want to separate into two matrices in Matlab like this:</p>
<p><span class="math-container">$$
Q=Sq^{T}Sq
$$</span>
for example this matrix:
<span class="math-container">$$
Q=\Biggm[\matrix{92.316 &31.78&240.417\cr
31.78 &194.66 &275.47\cr
240.417 &275.47 &938.99}\Biggm]
$$</span></p>
| |matlab| | <p>You can the Cholesky factorization.</p>
<p>In matlab its very easy.</p>
<pre><code>A= [92.316 31.78 240.417; 31.78 194.66 275.47; 240.417 275.47 938.99]
B=chol(A)
B'*B
</code></pre>
| 42314 | How I can separate a matrix into two matrices in Matlab? |
2021-03-31T10:43:18.187 | <p>Below is the equation of convective flux of an arbitrary property that is carried by particles. When particles move they carry their property along with them. Given a fixed surface in space, in order to calculate the amount of property crossing a surface S we just use the formula. Note that <span class="math-container">$\Psi$</span> is the amount of the property per unit mass, <span class="math-container">$\rho$</span> is the density of particles, <span class="math-container">$v$</span> is the velocity and <span class="math-container">$n$</span> is the normal vector of the surface.</p>
<p><span class="math-container">$$\Phi =\int \rho\ \Psi\ v.n\ dS\ $$</span></p>
<p>Derivation of the formula is taken <a href="https://www.youtube.com/watch?v=YZJF7pNlig8&list=PLp-8j6_XqvKQcI3w1QXKaxM6whld1O_yy&index=2" rel="nofollow noreferrer">from</a></p>
<p>I don't understand why do we have to include the normal component of the velocity vector of the particles (<span class="math-container">$v.n$</span>). I understand that the particles are crossing the surface, but why can't they just cross the surface while keeping their original direction?</p>
<p>By the way, the surface is not meant to be real, it is just virtual... that's why the particles are able to cross it after all.</p>
| |convection|mass-transfer| | <p>The flux of a quantity is by definition the rate per time per area of a given quantity normal to the area in question. If you use the original direction of the particles and this direction is not normal to the surface in question, then you are not evaluating the flux through the surface. To use the original particle direction, you would need to use a surface orthogonal to this direction. This thread puts it well: <a href="https://physics.stackexchange.com/questions/346144/why-surface-normal-is-used-while-defining-flux-through-an-open-surface">https://physics.stackexchange.com/questions/346144/why-surface-normal-is-used-while-defining-flux-through-an-open-surface</a>
The normal component of the quantity is the only component that pushes quantity through the surface. The other components play no role in transporting particles through the normal surface. Mathematically, I believe this stems back to rules of the dot product.</p>
| 42316 | Why do we consider the normal component of velocity in a flux and not the original direction of velocity? |
2021-03-31T20:33:07.517 | <p>I have a flat aluminium plate and I need to mount a digital indicator[1] on it.
The digital indicator doesn't seem to have any mounting holes itself so I
think I need to use a clamp to hold it.</p>
<p>I don't have much experience with industry standard parts in this area and I'd
rather not make something myself if I can buy it. So is there any standard
part or way I can mount this digital indicator on a flat aluminium plate? I can
drill holes in this plate no problem.</p>
<p>Any help is appreciated!</p>
<p>[1] <a href="https://shop.mitutoyo.eu/web/mitutoyo/en/mitutoyo/01.04.04A/Digital%20Indicator%20ID-H%2C%20CEE%20AC-Adapter/$catalogue/mitutoyoData/PR/543-563D/index.xhtml" rel="nofollow noreferrer">https://shop.mitutoyo.eu/web/mitutoyo/en/mitutoyo/01.04.04A/Digital%20Indicator%20ID-H%2C%20CEE%20AC-Adapter/$catalogue/mitutoyoData/PR/543-563D/index.xhtml</a></p>
| |mechanical-engineering|materials|fasteners| | <p>You could have a 8mm hole drilled and reamed slightly undersize for a press fit 8mm dowel pin and use a 8mm/8mm swivel clamp which would let you set the angle and height as desired (photo from McMaster).</p>
<p><a href="https://i.stack.imgur.com/merie.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/merie.png" alt="enter image description here" /></a></p>
| 42328 | Mounting a digital indicator on a plate |
2021-04-01T07:34:49.017 | <p>In the realm of structural engineering, construction, foundation design, what is "<strong>crs</strong>" when talking about piles/posts? Is it an abbreviation? Does it refer to some sort of size?</p>
<p>Some examples:</p>
<p><a href="https://i.stack.imgur.com/YCIgj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YCIgj.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/7JW36.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7JW36.png" alt="enter image description here" /></a></p>
| |structural-engineering|foundations|pile-foundations| | <p>Does not apply to this example, but just for completeness: I've also seen CRS as an abbreviation for cold-rolled steel in a material callout in machine drawings, e.g. "1018 CRS".</p>
| 42336 | What does "crs" mean in structural engineering / construction? |
2021-04-02T17:37:11.653 | <p>I am a physics teacher, but have found myself teaching engineering. I have come across the following pulley question, and am starting to doubt my own understanding of pulleys !</p>
<p>Until now the rule that <strong>mechanical advantage equals 2*no. of moveable pulleys</strong> seems to have hold fast, but I am not sure that it works in this case. I am wondering if instead a better rule is that <strong>mechanical advantage equals no. of supporting ropes</strong> ?</p>
<p>For this particular question, would the no. of ropes supporting the load be three or four ?</p>
<p>I am not sure whether or not to count the far left section where the effort is applied.</p>
<p>Any thoughts on which is the most reliable rule, and how to apply it to this question, would be most gratefully received please. Thank you.</p>
<p><a href="https://i.stack.imgur.com/gL4l5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gL4l5.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|pulleys| | <p>The tension on the rope is everywhere the same and its equal to F.</p>
<p>So if you did a free body diagram on the following system by <em>sectioning along the ropes</em>:</p>
<p><a href="https://i.stack.imgur.com/eUV2F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eUV2F.png" alt="enter image description here" /></a></p>
<p>what you get from the equilibrium is <span class="math-container">$4F = 48[N]$</span>.</p>
<p>I hope that is sufficient as an explanation, I tend to find that <em>problems with pulleys can have different configurations and as such it is always better to turn to the basics</em>.</p>
| 42367 | Mechanical Advantage of 3 Pulley System |
2021-04-02T20:04:14.713 | <p>Some slides associated to the book <em>Handbook of Marine Craft Hydrodynamics and Motion Control</em> (2021 edition) are provided by the author of the book <a href="https://www.fossen.biz/wiley/" rel="nofollow noreferrer">here</a>.</p>
<p>At page 12 of the slides for Chapter 4, the moment due to the restoration forces (gravity, buoyancy) of a floating vessel is computed, expressed in the body frame of the vessel. According to page 14, such moment is computed in the Center of Flotation; however, the lever arm computed at page 12 is the following:</p>
<p><span class="math-container">$$r_{\text{GM}}^b = \begin{bmatrix}
-\text{GM}_L \sin\theta \\
\text{GM}_T \sin\phi \\
0
\end{bmatrix}$$</span></p>
<p>where <span class="math-container">$GM_L$</span> is the longitudinal metacentric height, <span class="math-container">$GM_T$</span> is the transverse metacentric height, <span class="math-container">$\theta$</span> is the pitch angle of the vessel, <span class="math-container">$\phi$</span> is its roll angle, and the superscript <span class="math-container">$b$</span> indicates that the arm is represented in the vessel body frame.</p>
<p>The geometrical meaning of the second component of the arm is shown in a figure on page 10, as follows:</p>
<p><a href="https://i.stack.imgur.com/gI4Ch.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gI4Ch.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$CB$</span> stands for Center of Buoyancy, and <span class="math-container">$CG$</span> is the Center of Gravity.</p>
<p>Then the moment is computed as follows:
<span class="math-container">$$\mathbf{m}_r^b=\mathbf{r}_{GM}^b\times \mathbf{f}_b^b$$</span></p>
<p>where <span class="math-container">$\it{f_b^b}$</span> is the buoyancy force expressed in body frame.</p>
<p>By looking at this, it seems to me that the moment was computed about the Center of Gravity, not about the Center of Flotation, as claimed later.</p>
<p>Is this right, or is there any implicit assumption here that makes the assertion on the slides correct (for example the vessel is boxed-shaped)?</p>
| |torque|marine-engineering|ships|hydrostatics| | <p>The gravitational force and the buoyancy force create what is known as a <strong>moment couple</strong> (or sometimes just <a href="https://en.wikipedia.org/wiki/Couple_(mechanics)" rel="nofollow noreferrer">couple</a>).</p>
<p>Essentially they are two equal (because buoyancy and gravity should be equal in a boat), parallel and opposite facing forces, when have a zero translational effect and produce only rotation. See the following example</p>
<p><a href="https://i.stack.imgur.com/bVuVs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bVuVs.png" alt="enter image description here" /></a></p>
<p>For the above example it doesn't matter which point in space you calculate the moment of the two forces. will always get <span class="math-container">$F\cdot d$</span>.</p>
<p>So as pointed by TigerGuy, it doesn't matter which point you take as reference for the moment.</p>
| 42373 | Computation of the moment of restoring forces for a floating vessel |
2021-04-03T12:01:33.387 | <p>I have an assignment for my Mobile Robotics course to design the drive mechanism for a legged hexapod and demonstrate its gait in motion.</p>
<p>The idea is to design the hexapod as per the following schematic:</p>
<p><a href="https://i.stack.imgur.com/PY0Mb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PY0Mb.jpg" alt="Schematic of the Hexapod" /></a></p>
<p>There are 2 motors in the mechanism, each driving 3 sets of wheels. Wheels 1,3 & 5 are connected to one motor and wheels 2,4 & 6 are connected to the other. The wheels work as sort of a crank to produce motion in the legs and both the motors are 180 degrees out of phase to enable a working gait.</p>
<p>The part where I am having trouble is designing a gear mechanism that connects the 3 wheels in this triangular configuration to a single motor. If anyone has ideas for that, or further suggestions, it will indeed be very helpful.</p>
| |design|gears|robotics|machine-design| | <p>Ok I think I understand the intention now. Here is a thought of how to do it with timing belts. 6 half axles and 2 common shafts M1 and M2.</p>
<p>All 6 half axles would all need to be horizontally adjustable to tension the 6 belts. Everything could mount to "2D" vertically oriented sheet metal or laser cut parts, and the same shafts, bearings, and sprockets could perhaps be used throughout. The trick is getting the tensioning process to be reasonable. Bit of a mess as a concept, to be honest, compared to 6 individual motors.</p>
<p><img src="https://i.stack.imgur.com/56fNB.jpg" alt="enter image description here" /></p>
<p>PS- if the range of angular motion is finite, like 120 deg or less, then replace the belts and sprockets with linkages to drastically simplify</p>
| 42380 | Gear configuration for a legged hexapod |
2021-04-05T11:25:11.227 | <p>How can I construct the free body diagram of this figure? I am in a topic of equilibrium of particles so the total force will be equal to zero. The two springs are indicated to give equal forces.<a href="https://i.stack.imgur.com/Kvt1r.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kvt1r.jpg" alt="enter image description here" /></a></p>
<p>Am I correct that I need to add the masses of the blocks and then I will make an FBD together with two spring forces?</p>
| |statics| | <p>What you need to do is you need to isolate the masses from the environment. I.e. create a section.</p>
<p><a href="https://i.stack.imgur.com/58DUK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/58DUK.png" alt="enter image description here" /></a></p>
<p>At the points that section intersects with other objects (i.e. spring) you should place forces there with the correct direction.</p>
<p>Additionally you need to add any external forces like weight on the masses.</p>
<p><a href="https://i.stack.imgur.com/QwptR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QwptR.png" alt="enter image description here" /></a></p>
| 42405 | Constructing a Free Body Diagram (Stacked Blocks) |
2021-04-06T02:52:37.513 | <p>Can anyone help with this Impact problem? It does not seem very difficult but I am having trouble putting all the parts together.</p>
<p>Thanks,
<a href="https://i.stack.imgur.com/WntnA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WntnA.png" alt="Impact Problem" /></a></p>
| |mechanical-engineering|structural-engineering|structural-analysis|impact| | <p>Its just a matter of using the right formulas and algebraically finding an equation that will leave you with 1 unknown. In this case it is h.
<a href="https://i.stack.imgur.com/CG3Xq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CG3Xq.png" alt="Impact Height Answer" /></a></p>
| 42419 | How to calculate beam impact problem |
2021-04-06T15:53:38.723 | <p>So I'm trying to come up with a simple hack to heat up food in a water bath to hopefully +-1ºC of a target temperature, but I really don't understand the physics of it all so much. I'm pretty sure I can keep a hot plate at a somewhat steady temperature by cycling it on an off at regular intervals, but I'm not sure what that means for the food itself.</p>
<p>Here's what I'm thinking: If I can get the hot plate to stay steady at a temperature of my choice, when will place a pot of water on it and the water will stop heating up after it hits some lower temperature point, say at 75ºC. That means it has reached thermal equilibrium and is heating up at the same rate as it loses heat to the environment - and as long as the heat output of the plate is the same and the environment is stable, the water shouldn't drastically change temperature.</p>
<p>What I want to know is: will the final equilibrium temperature will be significantly different after putting some food in it?</p>
<p>I did some googling and managed to come with the following, probably wildly incorrect reasoning:</p>
<p>Most foods' specific heat is around 3 kJ/KgºC, and waters' is around 4 kJ/KgºC. I suppose that means given the same heat input, food will heat up about 1.33x more than water?</p>
<p>And thus a pot with, say, 3kg water and 1kg of meat (assuming a specific heat of 3 kJ/KgºC for the meat), all the stuff in the pot will have an average specific heat of 3.25 kJ/KgºC (3+3+3+4=13, divided by 4kg = 3.25), so it will all heat up 1.08x more?</p>
<p>What exactly does heating up 1.08x more mean, though? Do I just multiply 75ºC by 1.08? Surely not, right? Do I need to stick 1.08 in some formula or convert it to kelvin or something? What even are numbers, really? Thanks!</p>
<p>Just to clarify, what I'm concerned here is the final temperature once the pot reaches equilibrium again, not how much the temperature will drop when the food is added.</p>
<p>EDIT: To further clarify - what I want to calculate is what the final equilibrium temperature of a pot of water will be after placing some food in it, knowing only the final equilibrium temperature of the water before placing the food.</p>
| |heat-transfer|thermal-conduction| | <p><strong>TL;DR: The temperature in pot of water when food is placed in it is not only a matter of heat capacity. Other parameters like heat conductivity and convection have a significant effect on the transient temperature response.</strong></p>
<p>I will try to answer by taking a different simpler example, before tackling your example. Before I do that, I will point out that water has probably the highest heat capacity out of common substances (at around 4 [kJ/kgK]). The reason that meat has approximately 3 [kJ/kgK] is because meat is about 70-75% water.</p>
<h2>A short explanation of heat capacity</h2>
<p>Heat capacity <span class="math-container">$c_p$</span> is the amount of heat energy in J you have to give to a kg of substance in order for it to raise 1 degree K. So in order to raise about 10 degrees Kelvin 1 kg of water you'd have to give it about 40 kJ. Equivalently, if you add 40 kJ of energy to 10 kg of water the temperature will increase only by 1 degree.</p>
<p>(Another quantity you might want to learn about is enthalpy, but for simplicity's sake we'll not be discussing about it).</p>
<h2>Example:</h2>
<p>Let assume that you have 1 kg of water at 20 <span class="math-container">$^oC$</span>, and 1 kg of water at 60 <span class="math-container">$^oC$</span>. If you put the two together in a pot (and assume that heat capacity remains constant - although it changes with temperature), then the temperature will be approximately 40 degrees.</p>
<p>So what happens is that the hotter water loses about 80 [kJ] <span class="math-container">$= (1[kg]\cdot 4\left [\frac{kJ}{kg K}\right]\cdot (-20) [K] $</span>, while the cooler water gains those 80 [kJ] and raises its temperature by 20 [K]. (This is actually clearer when talking in terms of enthalpy but for this purpose this should be sufficient).</p>
<p>In a similar example, if there were 3 kg of 20<span class="math-container">$^o$</span>C, and 1 kg at 60<span class="math-container">$^o$</span>C, then the hotter will loose 120 kJ, which will be gained by the cooler water.</p>
<p>However, all the above are due to the fact that <strong>the quantities discussed above are liquids, and their instant mixture allows quick exchange of heat.</strong> Things are different for solids. The reason is that they are at a different phase.</p>
<h2>different phase</h2>
<p>Assume now that you have a ice cube of 1 kg, which is at -10 degrees, and 1kg water that is at approximately 20 degrees.</p>
<p>The heat energy in the system is equal to 2kg of water at 5<span class="math-container">$^oC$</span>. However if you place the icecube in the water it will take some time until ice changes to water (if you could somehow put this into a thermally insulated container (with minimal heat capacity), the resulting temperature after a while will be 5<span class="math-container">$^oC$</span>. However, for a long time you will have an icecube of -10<span class="math-container">$^oC$</span> with an outer layer which is close to zero.</p>
<p>Now, if your take an icepick and break up the ice cube and then place it into a blender, and then join them together, then almost instantaneously the result will be 2 kg of water at 5<span class="math-container">$^oC$</span>.</p>
<h2>heat conductivity k</h2>
<p>So the idea that creeps up now is <strong>heat conductivity</strong>. This is the rate by which heat flows through a material. (Actually, its also convection, but this post is starting to look like a few weeks worth of lecture notes for a physics class, so I won't delve into that).</p>
<p>So heat conductivity is one of the main factors that affect the heat flux. That is why when you hold a metal spoon (high conductivity) over a fire, eventually you feel the temperature rise, while, if you hold a wooden spoon (low conductivity) you don't feel any change in temperature unless your hand is on fire. (Ice by the way has a surprisingly low heat conductivity coefficient which is one of the reason's why igloos are made of ice).</p>
<h2>heat transfer mechanisms</h2>
<p>The following image shows the different mechanisms of heat transfer. That might serve for the following (short) discussion of your example.</p>
<p><a href="https://i.stack.imgur.com/ooNOB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ooNOB.jpg" alt="enter image description here" /></a></p>
<h2>your example</h2>
<p>Putting a piece of meat into water is subject to all the above (and more). When you put a piece of meat at 5<span class="math-container">$^oC$</span>, into water of 45<span class="math-container">$^oC$</span>, the heat isn't immediately exchanged. It takes some time for the heat to propagate into the meat. this is why you get the different zones in the cooked meat below (although this is not boiled).</p>
<p><a href="https://i.stack.imgur.com/7b9ah.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7b9ah.jpg" alt="enter image description here" /></a></p>
<p><strong>Bottom line</strong>: The way heat transfers in a pot of water when food is added, is affected by many parameters (conductivity, convection and heat capacity). Maintaining a constant temperature in the water between <span class="math-container">$\pm 1^oC $</span> will be a challenge.</p>
| 42438 | How do I calculate final equilibrium temperature of a pot of water on a stovetop vs. after I add food to it? |
2021-04-07T07:51:16.863 | <p>I drew a puzzle like joint sketch and split a body with it for 3d printing.</p>
<p>But I have to add a small gap between them for the tolerance of the machine.</p>
<p>Any way on how to do it not manually?</p>
<p><a href="https://i.stack.imgur.com/KuAby.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KuAby.png" alt="enter image description here" /></a></p>
| |solidworks|cutting|joining| | <p>Use a thin cut extrude, instead of the "Split" tool.</p>
<p>Set this to be midplane, and add the total tolerance gap that you require depending on your printer, and how loose you want the puzzle pieces to be.</p>
<p><a href="https://i.imgur.com/CVLXooP.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/CVLXooP.gif" alt="Thin cut extrude" /></a></p>
| 42454 | How to add a gap when splitting for joint in Solidworks |
2021-04-08T02:45:04.840 | <p>At datum height I have a pipe A (ID = 25mm) that discharges water at a volumetric flow rate of 10l/min, and unknown pressure. When this pipe is connected to a polyethylene pipe B (ID = 12mm), 50m long running to a height of 6m, then the end of this pipe B discharges water at 3l/min</p>
<p>Do I have enough information to calculate what the change in discharge rate at the end of pipe B would be if pipe B was of diameter 25mm instead of 12mm, and if so how could I? The equations I'm seeing seem to require a pressure measurement.</p>
<p>I've seen <a href="https://engineering.stackexchange.com/questions/35087/relative-improvement-in-flow-from-diameter-change-with-sub-milliimeter-tubing">this answer</a>, but it appears to ignore friction losses.</p>
<p>I'm guessing minor losses from the transition at the join of unlike pipe sizes can probably be ignored for an approximate answer.</p>
| |fluid-mechanics| | <p>Yes, you do have enough information. The equations you've seen that require a pressure measurement are the Colebrook-White equation (or its embodiment in the Moody chart) and the Darcy-Weisbach equation (as a definition of Darcy friction factor), right? Applying those equations three times to the situations described in your first paragraph (once to pipe A in isolation; once to pipe A when connected to pipe B; and once to pipe B when connected to pipe A) will give you three simultaneous equations, which you can use to compute the values of three unknowns: the pressure difference between entrance and discharge, the length of pipe A, and the pressure difference between the pipe junction and the discharge. The first two of those three unknowns can be expected to stay the same in the situation described in your second paragraph, so armed with their values, you can apply the equations to each pipe in that situation to get two simultaneous equations, from which you can compute values of two unknowns: the flow rate and the new value of the pressure difference between the pipe junction and the discharge.</p>
<p>(In case you're worried about the appearance of relative roughness in the Colebrook-White equation, I think "polyethylene" is intended to imply that the roughness can be safely treated as zero.)</p>
| 42466 | Calculating the change in flow rate with a change in pipe diameter |
2021-04-08T10:23:33.110 | <p>Is a concrete screw like the <a href="https://www.hilti.co.uk/c/CLS_FASTENER_7135/CLS_SCREW_ANCHORS_7135/r2523586?CHA_GLOBAL_ANC_SIZE=10%20mm&CHD_ANCHOR_LENGTH=150%20mm&combo_content=fdbaf0c8059ef27f1cb0778dcd54b011&salespackquantity=50%20pc&itemCode=2079918" rel="nofollow noreferrer">HILTI HUS3</a> removable?</p>
<p>I.e. can it be repeatedly unscrewed and reinstalled as required? Will this damage the fixing or the hole in the concrete?</p>
<p>We're thinking about using it for a bund that occasionally will need to be removed for access.</p>
| |concrete|fasteners| | <p>No, they are usually fitted once and left.</p>
<p>If you need to remove them then you need to drill new holes.</p>
<p>Or you should consider fitting a wooden framework to the concrete and attaching to that so it can be easily removed and refitted, without disturbing the concrete fixings.</p>
| 42469 | Are concrete screws suitable for repeated removal and reinstallation? |
2021-04-09T15:19:19.660 | <p>I have commonly seen that the work for an isentropic compression is written as:
<span class="math-container">$$
W_{in} = c_{v}(T_1 - T_2)
$$</span>
What confuses me is that since it is a compression (volume is changing) why do we use the constant volume specific heat?</p>
<p>Thank you kindly for your time and help.</p>
| |thermodynamics|compressed-air|compressors| | <p>Your problem is that much of the terminology of thermodynamics was defined before the underlying physical concepts were properly understood, which leaves us stuck with a lot of confusing names for things now.</p>
<p>In particular, you're worried about the applicability of the quantity commonly known as the "specific heat capacity at constant volume". The first thing you need to know is that that's a misleading name for the quantity, for two reasons. The first reason is that "heat capacity" is an oxymoron: "capacity" signifies storage, but we now know (originally from Joule, 1845, <em>Lond. Edinb. Dublin Philos. Mag. J. Sci.</em> <strong>27</strong>(179):205-207) that, when it's being stored, energy doesn't have an identity as "heat" or "work", it's just energy; it only develops an identity as "heat" or "work" when it's being transferred. The second reason is that, as you've noticed, it's important even when the volume is not constant.</p>
<p>Therefore, a better name for the quantity commonly known as the "specific heat capacity at constant volume" would, following Stedman (1963, <em>Educ. Train.</em> <strong>5</strong>(3): 127-128), be "specific internal energy capacity". It is, as that name suggests, the quotient of the change in specific internal energy of a body by the change in the temperature of that body.</p>
<p>Now let's think about the situation you're trying to analyse: an isentropic non-flow process, in which some material is compressed. In a non-flow process (unless something exotic is going on), internal energy is the only form in which the material can store energy; and "isentropic" implies that no heat is transferred, so the only way in energy can get in or out is as work. Hence, the specific work done is equal to the change in specific internal energy, which is (by definition) equal to the product of the specific internal energy capacity and the change in temperature, as your first equation says.</p>
<p>Now you may be wondering how the specific internal energy capacity came by the misleading name "specific heat capacity at constant volume" in the first place. Well, think about an isochoric (i.e. constant volume) non-flow process. As before, in a non-flow process (unless something exotic is going on), internal energy is the only form in which the material can store energy; and if the volume is constant in a non-flow process (again, unless something exotic is going on), then no work is done, so the only way energy can get in or out is as heat. Therefore, the specific input of heat is equal to the change in specific internal energy, which is (by definition) equal to the product of the specific internal energy capacity and the change in temperature, i.e. <em>in an isochoric non-flow process</em>, the quotient of the specific input of heat by the change in temperature is equal to the specific internal energy capacity. The quantity that Stedman and I suggest should be called "specific internal energy capacity" was first discussed in the context of measurements of the specific input of heat required to bring about a change in temperature in an isochoric process, before anyone knew about the interchangeability of heat and work or the idea of internal energy, and hence it got the name "specific heat capacity at constant volume" with which we seem to be stuck today.</p>
| 42491 | Specific heat at constant volume used in non constant volume processes |
2021-04-09T21:49:46.867 | <p>I've been going thru the book by Hodges and Roithmayr regarding dynamics theory and I can't seem to prove that:
<a href="https://i.stack.imgur.com/1a2cQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1a2cQ.png" alt="enter image description here" /></a></p>
<p>I have tried using multiple substitutions, including the fact that every component of the direction cosine matrix is equal to its cofactor, in multiple, different ways. But I feel like I've been walking in circles. The last attempt I made was, when using the definition of angular velocity:</p>
<p><a href="https://i.stack.imgur.com/8mVTP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8mVTP.png" alt="enter image description here" /></a></p>
<p>And then writing every derivative of the coordinate system by substituting its direction cosines and substituting the cofactors inside. This led to another interesting thing:</p>
<p><a href="https://i.stack.imgur.com/M1QLW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M1QLW.png" alt="enter image description here" /></a></p>
<p>Which may or may not be related to the proof I'm looking for, but I cannot work out how. My guess is that this would be a pretty standard equation to use, since angular velocity is such a big thing in dynamics, but I could not find it anywhere.</p>
| |dynamics| | <p>Knowing that the derivative of a vector in frame A can be written as:</p>
<p><a href="https://i.stack.imgur.com/AckJm.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AckJm.gif" alt="enter image description here" /></a></p>
<p>And similarly, the derivative of the same vector written in the B frame can be written as:</p>
<p><a href="https://i.stack.imgur.com/6eNn3.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6eNn3.gif" alt="enter image description here" /></a></p>
<p>Substituting the second equation in the first one we have:</p>
<p><a href="https://i.stack.imgur.com/ESkSz.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ESkSz.gif" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/sVzYu.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sVzYu.gif" alt="enter image description here" /></a></p>
<p>For the above to be true, either the vector being differentiated is a null vector (which is a trivial solution) or:</p>
<p><a href="https://i.stack.imgur.com/JSnLR.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JSnLR.gif" alt="enter image description here" /></a></p>
| 42500 | Proof of angular velocity identity |
2021-04-10T08:38:11.533 | <p>I am studying rock caverns lined with concrete for compressed air storage. I would like to know if it's possible to convert the permeability coefficient from m/s to m^2 or Darcy.
Thank you.</p>
| |airflow|compressed-air| | <p>The word <strong>"convert"</strong> is reserved for units that measure the same quantity (like distance in inches or cm). In this case however you need to use the word <strong>"calculate"</strong> because the units are dissimilar. Yes, the <a href="https://en.wikipedia.org/wiki/Darcy%27s_law" rel="nofollow noreferrer">Darcy's Law</a> equation is what you are looking for.</p>
<p><a href="https://i.stack.imgur.com/O6aiJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O6aiJ.png" alt="enter image description here" /></a></p>
<p>In SI units, <a href="https://en.wikipedia.org/wiki/Permeability_(Earth_sciences)#Units" rel="nofollow noreferrer">permeability</a> "k" is measured in <strong>m<sup>2</sup></strong>.<br>
In SI units, <a href="https://en.wikipedia.org/wiki/Darcy%27s_law#Description" rel="nofollow noreferrer">porous medium gas flux</a> "q" or "Q/A" is measured in (m<sup>3</sup>/s) / (m<sup>2</sup>) which reduces to <strong>m/s</strong>. Q is volumetric flowrate (m<sup>3</sup>/s) and A is area (m<sup>2</sup>). Algebraically rearrange q= Q/A as necessary such as Q= q × A.</p>
<p>So to calculate you also need the dynamic viscosity of air "µ" in kg/m-s. Note that this value is temperature and pressure dependent as you can see in this <a href="https://www.engineersedge.com/physics/viscosity_of_air_dynamic_and_kinematic_14483.htm" rel="nofollow noreferrer">table</a>.</p>
<p>You will also need the differential pressure in <a href="https://en.wikipedia.org/wiki/Pascal_(unit)" rel="nofollow noreferrer">Pascals</a> "Δp" in Pa or kg/ms<sup>2</sup>. The difference in pressure between the inside of our cave and atmospheric pressure.</p>
<p>And finally you will need the length "L" in meters. This is distance the gas must travel through the porous medium or in your case the thickness of your concrete layer.</p>
<p>When you solve the equation, you can confirm you used the correct units by canceling them out.</p>
| 42504 | How to calculate rock permeability in m^2 from gas flux in m/s? |
2021-04-10T21:11:54.117 | <p>An electrical engineer here. Using a stepper motor for a robotic arm. A stepper motor at the base of the arm has a max holding torque of 6kgcm. The arm is around 30cm in length from this base motor meaning that the max weight that can be supported at the end of the arm is 200 grams. There are other motors along this length that the base motor has to support too. How can i mechanically increase the torque of the motor so it can support 800g to 1kg ? The max torque (6kgcm) is provided at the highest rated voltage on the electric side of things, Is there any mechanical methods to improve torque of a stepper motor without wrecking its precision/step accuracy? I've heard gearboxes increase torque but I want to know all options and find the best one.</p>
| |motors|torque|robotics|stepper-motor| | <p>Gearboxes are very convenient to integrate with stepper motors, and sounds like the most likely solution by far. Just for completeness, some other options are:</p>
<ul>
<li>Don't know if this counts, but something fairly common that is a gear, but not a "standard NEMA-xx gearbox", would be a worm gear.</li>
<li>Timing belt with the appropriate reduction. The belts have some layout possibilities that complement those you can get from gears, and (IMO) would usually be done for reasons of mechanical layout rather than cost. Usually gears are the first try.</li>
<li>A different motor and/or driver. In particular larger diameter will help torque, but same diameter but longer length, may also help. If the existing motor is fine-pitch, like 400 steps/rev, then a standard 200 steps/rev motor might give more torque also. Other than increasing stepper motor diameter, these are all relatively small improvements, like +50% can be expected if it was not already optimized.</li>
<li>If range of angular motion is small, a lever mechanism, combined with cams, or linkages, or some other kind of simple high ratio gear (like a worm gear again), could be used. There is an endless variety of "old-school" mechanisms with usually very narrow/specialized applications, so would need to know about the situation.</li>
<li>Adding an external force to support part of your load, e.g. a spring of some kind</li>
</ul>
| 42512 | Mechanically increase motor torque |
2021-04-11T08:30:38.817 | <p>I'm breaking my brain for a two weeks trying to find a way to build a fireproof box 100 cm × 100 cm × 100 cm with 1 door at the cheapest price.</p>
<ul>
<li>The immediate candidate is sheet metal, which needs to be at least 1.5 mm thick in order to be welded, but metal sheet is expensive (at least locally, $600 for the box).</li>
<li>Rockwool is going to be extremely heavy for this size</li>
<li>Polycarbonate is not fireproof, and also costly</li>
<li>drywall not fireproof, costly</li>
</ul>
<p>I'm looking for an idea.</p>
| |metals|metal-folding|fire| | <p>If the problem were to just make a box that itself did not burn in an housefire, that seems trivial and somewhat useless - 100 cm cube of brick that holds 0 (pick your volumetric units) contents</p>
<p>If the goal is to protect the contents from a fire you have to understand that many items one wants to protect would deterioriate with heat.</p>
<p>Even the best insulation (probably aerogel) can only buy you time. If it takes half an hour for the inside of my box to reach 100 degrees in a 500 degree fire, I just need to keep the fire on it longer and eventually I can get the contents to hit pretty close to the 500 degrees.</p>
<p>If you use a more active solution such as a compound that will absorb heat, possibly releasing a relatively inert gas that will pressurize the box so that flames flow away rather than in, cost will go up and it will still only last as long as your chemical anti-fuel. On top of it, you will have to maintain the compound, and prevent it from exploding from thermal shocks of the fire (adding another layer to the problem).</p>
<p>Also fyi, it is possible to spot and seam weld extremely thin metal sheets if the process conditions are appropriately controlled. Anyway, I suspect you will end up with a 22ga spot welded box, 'sealed' with ceramic adhesive or braze and lined with insulation. "Keeps your stuff at Xdegrees or below for a whole Y hours in a Zdegree fire. (Only tested for fires from A - Trash, wood, paper. C - Electrical equipment)"</p>
| 42517 | What is the cheapest way to build a fireproof box? |
2021-04-11T13:33:00.920 | <p>I am working out the temperature for an Ideal diesel cycle:</p>
<p>Based on the isentropic properties for an ideal gas.
<span class="math-container">$$T_f = T_i * r^{k-1}$$</span>
Where <span class="math-container">$T_i = 50 ºF = 510ºR$</span></p>
<p>If I use ºF, then <span class="math-container">$T_f = 172.9ºF = 632.9 ºR$</span></p>
<p>If I use ºR, then <span class="math-container">$T_f = 1763 ºR = 13030 º F$</span></p>
<p>It gives me completely different temperatures; how am I supposed to determine which one is correct?</p>
| |mechanical-engineering|thermodynamics|unit| | <p>We cannot obtain negative temperatures in thermodynamics calculations. Only the absolute scales prevent this. The two absolute scales are Kelvin and degrees Rankin.</p>
<p>Here are the two calculations to show that both give the same result. I take that <span class="math-container">$r^{k-1} = 3.45$</span> based on using your Rankin values.</p>
<p><span class="math-container">$$T_f = (50 + 459.67)(3.45) = 1763\ ^o\mathrm{R}$$</span></p>
<p><span class="math-container">$$T_i = ((50 - 32)/1.8) + 273.15 = 283.15\ \mathrm{K}$$</span></p>
<p><span class="math-container">$$T_f = (283.15)(3.45) = 979.44\ \mathrm{K}$$</span></p>
<p><span class="math-container">$$T_f = 1.8(979.44 -273.15) + 32 + 459.67 = 1763\ ^o\mathrm{R}$$</span></p>
<p>In summary, the K and <span class="math-container">$^o$</span>R scales are interchangeable in thermodynamic calculations because they are absolute scales. The <span class="math-container">$^o$</span>C and <span class="math-container">$^o$</span>F scales are not thermodynamic temperature scales.</p>
<p>This of course presumes that, for whichever absolute temperature scale you use, you also use the appropriately scaled constants. Your equation is unitless when divided by <span class="math-container">$T_i$</span>, so no conversion is needed. But, for a simple yet effective counter, consider that the value of the gas law constant <span class="math-container">$R$</span> is not the same in the Kelvin unit scale as it is in the Rankin unit scale.</p>
<p>So, always use absolute temperature scales in thermodynamics calculations. Convert back to either of the other two "colloquial" (common usage) scales when desired. And always check that units on all other factors are consistent with the absolute temperature scale that you choose.</p>
<p>As a side note, some move is afoot to use degrees Kelvin interchangeably with Kelvin. This would make all temperature scales prefaced by the word "degrees". I find the practice a bit hard to accept. Cementing the notion that the former is an absolute scale while the latter is a difference helps distinguish these two confusions.</p>
<p><span class="math-container">$$ 273.15\ K = 0\ ^oC \ \ \mathrm{but}\ \ 273.15\ ^oK = 273.15\ ^oC$$</span></p>
| 42520 | Inconsistencies with units for thermodynamic identities |
2021-04-11T14:40:56.220 | <p>I am working with an arbitrarily complicated and long-winded transfer function that looks something like this:</p>
<p><span class="math-container">$$
H(s) = \frac{as^4+\cdots+b}{cs^6+\cdots+ds^2}
$$</span>
where letters represent constants. First I have the bod plot which has a consistently decreasing magnitude which indicates that t acts similar to an integrator for magnitude, but I am unsure about what the almost constant -180 degree phase margin indicates and why there could be this trough and peak around 10 rad/s.</p>
<p><a href="https://i.stack.imgur.com/1WGuo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1WGuo.jpg" alt="bode plot" /></a></p>
<p>This Nyquist diagram is below as well. I think this is showing that the system is stable as expected as no open-loop poles appear on the right-hand real axis.</p>
<p><a href="https://i.stack.imgur.com/rwa0G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rwa0G.jpg" alt="nyquist" /></a>
I zoomed in on the origin and this is what I was able to see. It seems like it does not ever cross into the positive real axis.</p>
<p><a href="https://i.stack.imgur.com/n7OZw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n7OZw.jpg" alt="enter image description here" /></a></p>
<p>Nichols diagrams are something I have not worked with before and so am trying to understand their significance. This certainly mirrors the bode plot where a -180 degree phase dominates but is this saying that the phase is about -180 degrees for pretty much all open-loop gain values?</p>
<p><a href="https://i.stack.imgur.com/kiHy9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kiHy9.jpg" alt="nichols" /></a></p>
| |control-engineering|control-theory|pid-control|transfer-function| | <p>Your systems shows extremely close pole-zero cancellation. So much even that it nearly removes 4 poles and zeros. Lets look at why, starting with the Bode plot:</p>
<p>The magnitude plot is constantly decreasing with a slope of -40dB/decade. Following basic rules this already implies that at the given frequencies, the system can be approximated using a double integrator. The rule of thumb for this is: -20dB/decade for each pure integrator. The phase plot backs this statement with a nearly constant -180 degrees phase angle. Same to the magnitude plot, the rule of thumb is -90 degrees for each pure integrator. (pure integrator means <span class="math-container">$\alpha\frac{1}{s}$</span>). However, the interesting part here is the little bump in this phase plot. This behaviour can be explained using the following example:</p>
<p><span class="math-container">$$H(s)|_{s\ll\text{min}(\alpha, \beta)} = \frac{s+\alpha}{s(s+\beta)}$$</span></p>
<p>Let imagine drawing a bode plot for this. At low frequencies, this can be approximated as <span class="math-container">$$\frac{\alpha}{\beta s}$$</span> so a pure integrator. For very high frequencies this approximates to:</p>
<p><span class="math-container">$$H(s)|_{s\gg\text{max}(\alpha, \beta)} = \frac{s}{s^2} = \frac{1}{s}$$</span></p>
<p>So as you see, the low frequency approximations behaves equal to the high frequency approximation. What happens in between depend on which value is smaller. is <span class="math-container">$\alpha < \beta$</span> then the magnitude plot approaches a horizontal slope (and a 0-degree phase angle) for values larger than <span class="math-container">$\alpha$</span>, but again a -1 slope for values larger than <span class="math-container">$\beta$</span>. However, if <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> are very close, this effect is practically not visible in the magnitude plot. The phase plot does show a bump as the effect in phase changes much faster. As such, to return to your case, the bode plot shows almost the perfect behaviour of a double integrator system. Given your system representation, it is safe to assume 4 poles are closely cancelled by 4 zeros.</p>
<p>One thing that remains unknown is whether these cancelled poles and zeros might be in the right half plane. This is where the nyquist plot play a role. The nyquist stability criterion uses the amount of encirclements of the point -1 to determine stability of the system. This criterion roughly shows that the amount of encirclements of the -1 point is equal to the amount of unstable poles in the closed loop system and the amount of open loop unstable poles. If the net result is > 0 (so clock-wise encirclements) the closed-loop response is unstable, if it is < 0 (so counter-clockwise encirclements) the open loop is unstable (I really hope I am right here).</p>
<p>Your graph does not show any encirclements. However, due to the presence of 2 pure integrators, the graph shown is not complete. This should include a full rotation in the clockwise direction with an infinite radius. As such, the contour does actually encircle the -1 point more than once in the clockwise direction. This means that the closed-loop system has unstable poles, but if the open loop system has them as well is sadly not known (many sources state that you should know this).</p>
<p>And as for your third plot, I havent seen that one either so cannot give a proper interpretation. But despite that, I hope some of this is usefull!</p>
| 42523 | How do I interpret the following Bode, Nyquist and Nichols diagrams? |
2021-04-11T18:48:54.057 | <p>In my project I have to choose a form tolerances ( parallelism ) between the handle ( purple part ) and the part that sits inside it ( the grey part ), where the two parts slide on each other.
But I don't know how to choose the right reference line on my technical drawing ( drafting ). <a href="https://i.stack.imgur.com/KoaOB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KoaOB.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/bav9E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bav9E.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|technical-drawing|machine-design|machine-elements| | <p>You have to show two reference lines for tolerance measurements. One is the longitudinal axis of the linked parts. The other one is the line through the center of the pin as shown below.</p>
<p><a href="https://i.stack.imgur.com/h1wyg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h1wyg.png" alt="enter image description here" /></a></p>
<p>Does this answer your question?</p>
| 42531 | How to find a reference line for " Form Tolerances"? |
2021-04-12T02:58:23.217 | <p>I am in the process of building an antenna mast that can tilt from horizontal (down) to vertical (up). The mast (arm) is 360" and weighs a total of 105 lbs. I need to calculate the amount of force placed on a winch in order to lift the mast (arm) from the down to up position. All the formulas I am finding are for calculating the force required to lift a weight at the end of a leaver. It seems to me that this will not work in my case since the weight is distributed evenly over the entire length of the arm.</p>
<p>I have two options (plans) on how to place my winch. I need to determine which option will require the least effort but also the most stable.</p>
<p>Can anyone shed some light on this? I have attached a diagram.</p>
<p><a href="https://i.stack.imgur.com/VU218.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VU218.png" alt="Diagram of Plans" /></a></p>
| |mechanical-engineering|structural-engineering|torque|stresses|pulleys| | <p>If you are looking for just which option would require the least effort then the answer is the <strong>second case (PLAN 2)</strong>.</p>
<p><a href="https://i.stack.imgur.com/GrGBQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GrGBQ.png" alt="enter image description here" /></a></p>
<p>The help you to understand why is that, do the following mental experiment. Imagine that the support is so high that the fulcrum coincides with the center of the arm. For plan 2 in this scenario, the winch effort to rotate the arm would be almost zero.</p>
<p>(This can be easily proven mathematically, but there is no point because of issues that need to be addressed and are presented on the following section).</p>
<h2>Issues that need to be addressed</h2>
<p>Having said the above, there are a lot of issues that you need to consider, before attempting this in your current configuration:</p>
<ul>
<li>the vibrational effects</li>
<li>the strength of the arm</li>
</ul>
<p>To make an estimate you need to provide:</p>
<ul>
<li>the moment of area of the arm (affects strength and vibrational behaviour)</li>
<li>cross-sectional area of the arm (affects strength and vibrational behaviour)</li>
<li>rate at which you intend to raise the arm (this will have an effect on the forces)</li>
</ul>
| 42542 | Calculating Load On Winch From a Distributed Load? |
2021-04-12T10:58:38.037 | <p>I have the <strong>Fanuc robotic M-2000iA/2300</strong> data sheet in which it is being told that its maximum reach is <strong>3734 mm</strong> and its payload capacity is <strong>2300 kg</strong>, so how is its maximum reach being calculated and in respect to which axis?</p>
<p>Also, what will the load capacity be at that point?</p>
<p><a href="https://i.stack.imgur.com/9ODY1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ODY1.png" alt="Fanuc Robotic M-2000iA/2300" /></a></p>
<p><a href="https://www.fanuc.eu/%7E/media/files/pdf/products/robots/robots-datasheets-en/m-2000ia/datasheet%20m-2000ia-2300.pdf" rel="nofollow noreferrer">Datasheet</a></p>
| |robotics| | <p>If you notice in the datasheet, for the weight it says that it is at the wrist.</p>
<p>Having said that, usually there isn't any standard calculation method for robotic arms. It is basically striking a balance between:</p>
<ul>
<li>maximizing the load (to satisfy marketing)</li>
<li>minimizing the load (to make the legal team stop complaining about possible lawsuits).</li>
</ul>
<p>There was one notable exception I have had the pleasure to encounter (it was not on a robotic arm, but on a <a href="https://en.wikipedia.org/wiki/Stewart_platform" rel="nofollow noreferrer">Stewart platform</a>) that takes the time/effort to provide software that calculates the allowable load at any given position in space. Probably there are more like that.</p>
| 42547 | How can I calculate a robotic arm's maximum reach? |
2021-04-12T18:52:24.907 | <p>If I put a Peltier device between a heat source and a heat sink (Peltier turned off),
How well will it conduct heat? as opposed to lets say a copper block?</p>
<p>Thank you</p>
| |heat-transfer|thermocouple| | <p>Most thermometric cooling modules (Peltier devices) are made out of <a href="https://en.wikipedia.org/wiki/Bismuth_telluride" rel="nofollow noreferrer">Bismuth Telluride</a> which has a thermal conductivity of 1.20 W/(m·K), similar to ordinary glass.</p>
<p>From this wiki <a href="https://en.wikipedia.org/wiki/List_of_thermal_conductivities" rel="nofollow noreferrer">list of thermal conductivities</a>: <br>
Copper, 401 W/(m·K)<br>
Aluminum, 237 W/(m·K)</p>
<p>So for a given temperature differential, a thermometric cooling module in the off state would conduct 334 times less heat than copper and 197 times less heat than aluminum.</p>
| 42554 | Does a Peltier device conduct heat when turned off? |
2021-04-12T20:21:06.533 | <p>I'm currently investigating fault recovery in a UAV using MATLAB.</p>
<p><em>I've been given several variables:</em></p>
<ul>
<li><strong>phi</strong> = the roll angle</li>
<li><strong>psi</strong> = the yaw/heading angle</li>
<li><strong>beta</strong> = the side slip</li>
<li><strong>p</strong> = roll rate</li>
<li><strong>r</strong> = yaw/heading rate</li>
<li><strong>delta_r</strong> = angle of the rudder</li>
<li><strong>delta_a</strong> = angle of the aileron (the aileron is currently inactive).</li>
</ul>
<p><strong>My input is the rudder angle.</strong></p>
<p><strong>I've simulated a step fault in the yaw/heading sensor and I'm trying to recover from it.</strong></p>
<p><em>To detect that there was a fault in the heading sensors:</em> I compared the actual side-slip with the modeled side slip. If there is a difference between the "real' system and the model in heading angle, but no difference in side-slip, then the heading sensor is faulty.</p>
<p><em>To recover from this fault:</em> I want to reconfigure the side-slip so I can use it to determine the heading/yaw angle. In order to do this I need to multiply the side-slip by the transfer function between side-slip and heading/yaw angle.</p>
<p><strong>We've been given the following equations:</strong></p>
<p><a href="https://i.stack.imgur.com/U0FaJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U0FaJ.jpg" alt="enter image description here" /></a></p>
<p><strong>I'm having some trouble coming up with a transfer function between side slip and heading angle.</strong></p>
<p><em>I figured the best thing to do, as I am using a simulation filled with arrays, is to turn all the derivatives into their discrete equivalent</em> - of course this is means I'm not using a transfer function, but an equation. Taking the equation for the derivative of beta and discretizing it...</p>
<p><a href="https://i.stack.imgur.com/F9tDN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F9tDN.jpg" alt="enter image description here" /></a></p>
<p><strong>When I plug this in, the system doesn't recover.</strong></p>
<p><strong>I even tried doing Laplace transforms</strong>, <em>but I get stuck here:</em></p>
<p><a href="https://i.stack.imgur.com/pL3li.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pL3li.jpg" alt="enter image description here" /></a></p>
<p><strong>And I'm not sure how I could apply that in MATLAB</strong></p>
<p><em>Is there a mistake in my logic or approach?</em> <em>How would I turn the discrete system into a transfer function?</em></p>
| |control-engineering|dynamics|matlab|transfer-function| | <p>Find the transfer function <span class="math-container">$\delta_r(t)\to \psi (t)$</span> and <span class="math-container">$\delta _r(t)\to \beta (t)$</span>. Divide the first transfer function by the second to get the transfer function <span class="math-container">$\beta (t)\to \psi (t)$</span>.</p>
<p>These calculations can be tedious. I used Mathematica to get the following result.</p>
<pre><code>ssm = StateSpaceModel[{p'[t] == -11.450 p[t] + 2.7185 r[t] -
19.4399 \[Beta][t] + 78.4002 Subscript[\[Delta], a][t] -
2.7282 Subscript[\[Delta], r][t],
r'[t] ==
0.5068 p[t] - 2.9875 r[t] + 23.3434 \[Beta][t] -
3.469 Subscript[\[Delta], a][t] +
13.9685 Subscript[\[Delta], r][t], \[Beta]'[t] ==
0.0922 p[t] - 0.9957 r[t] - 0.4680 \[Beta][t] +
0.3256 \[Phi][t], \[Phi]'[t] ==
p[t] + 0.0926 r[t], \[Psi]'[t] == r[t]}, {p[t],
r[t], \[Beta][t], \[Phi][t], \[Psi][t]}, {Subscript[\[Delta], r][
t]}, {\[Psi][t], \[Beta][t]}, t];
tfm = TransferFunctionModel[
Simplify[Divide @@ Flatten[TransferFunctionModel[ssm, s][s]]], s]
</code></pre>
<p><span class="math-container">$$\frac{0.986478 s^3+11.6592 s^2+6.59389 s+4.77963}{-1. s^3-10.9882 s^2+1.02337 s}$$</span></p>
<p>The transfer function is unstable (assuming I did not make any typos).</p>
<pre><code>TransferFunctionPoles[tfm]
</code></pre>
<blockquote>
<p>{{{-11.0805, 0, 0.0923575}}}</p>
</blockquote>
<p>However, the transfer function from <span class="math-container">$\psi (t)\to \beta (t)$</span> is stable</p>
<p><span class="math-container">$$ \frac{-1.01371 s^3-11.1388 s^2+1.0374 s}{s^3+11.819 s^2+6.68427 s+4.84515}$$</span></p>
<p>and we can plot its step response.</p>
<pre><code>Plot[Evaluate@OutputResponse[tfm, UnitStep[t], {t, 0, 20}], {t, 0, 20}, PlotRange -> All]
</code></pre>
<p><a href="https://i.stack.imgur.com/xHiQH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xHiQH.png" alt="enter image description here" /></a></p>
| 42555 | Fault Recovery of a UAV using MATLAB |
2021-04-13T16:56:07.880 | <p>I have two 3 by 4 foot roof windows facing south and want to keep the heat out in summer.
What I did on the hottest days was I opened them a bit and strapped cardboard on top. That way a bit (hot) wind came in but the sunlight stayed out.
Of course there are electric shutters, but I guess they will not last ten years, if at all, in these harsh conditions. In summer you can fry eggs on the roof and in winter it's covered with snow and ice.</p>
<p>My best bet was to permanently glue aluminum foil from emergency rescue blankets onto the glass but I do not know what the best removable glue would for that, since these things will also not last forever, but are cheap....</p>
<p>What is your suggestion here? Thanks in advance :-)</p>
| |thermal-insulation| | <p>Home improvement centers have R10 insulating foam boards the come in 2' x 4' (feet) and some are self adhesive.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/qyb7v.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qyb7v.jpg" alt="R10 foam board" /></a></p>
<p>.</p>
<p>If you have certain species of birds you need to cover them with plastic sheets tho.</p>
<p>The other alternative is to cover the roof windows with a suspended roof of plywood raised 8 inches from the skylights to allow breeze from the sides wash the heat from the windows.</p>
| 42569 | Thermally insulate roof windows |
2021-04-13T18:11:48.783 | <p>Found some printable M8 screws on Thingiverse (<a href="https://www.thingiverse.com/thing:4805480" rel="nofollow noreferrer">https://www.thingiverse.com/thing:4805480</a>), which made me wonder if I can get M4 screws, just by resizing them factor 0.5.</p>
| |mechanical-engineering|bolting| | <p>The ISO metric bolt has its diameter as the numeral part of the name.</p>
<p>So M8 has a diameter of 8mm compared to that of the M4 bolt. So far scaling would make sense.</p>
<p>However, the pitch of them is not proportional and would be wrong if you scale them.</p>
<p>her is a figure from Wikipedia. <a href="https://www.wikiwand.com/en/ISO_metric_screw_thread" rel="noreferrer">Wiki</a></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/XBoxu.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XBoxu.png" alt="enter image description here" /></a></p>
| 42573 | Are M8 screws double the size of M4 screws? |
2021-04-14T00:39:12.650 | <p>In the example problem below, how were the signs of the torques at x = 0 and x = L found? At x = 0, the direction of pulley rotation is unspecified but the direction is assumed to be positive. And at x = L, if you use the right hand rule the torque is applied in the positive x direction but the author states that torque is negative by the right hand rule. Is this because the shaft is stationary and must supply the same torque in the opposite direction at x = 0 and x = L?</p>
<p><a href="https://i.stack.imgur.com/gvmCq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gvmCq.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/fDeFd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fDeFd.png" alt="enter image description here" /></a></p>
| |vibration| | <p>Yes, you are right.</p>
<p>If it helps try to imagine a horizontal spring with two equal masses at the two ends on a frictionless surface, vibrating about the center of mass of the system, which is at the middle of the length of the spring. They always have opposite and equal magnitude acceleration and velocity and displacement.</p>
| 42579 | Sign of torque in torsional vibration of a shaft |
2021-04-14T22:57:19.160 | <p>Recently I was thinking about home improvement, and one idea that is haunting me is that I could use some kind of refrigerator for gaming consoles and laptops.</p>
<p>The obvious benefit is that it maintains a constant low temperature. Also, the doors usually can be closed firmly and it won't let dust come inside, so there isn't any need to disassemble and clean hardware. That is a good thing in long term. Despite it makes airflow less intensive, according to my understanding it should not be a problem, as all that is needed for hardware is cooled air, not new air.</p>
<p>And whenever I get what I consider a good idea, I usually stop for a minute and ask myself a question "why doesn't this exist yet?". So there are definitely big-scale expensive enterprise servers which could possibly use the same approach to reduce maintenance effort, but that is not the case. So the disadvantages I can think of is that refrigerators are just not designed to cool a source of constant heat for a long period of time and (probably) will break quickly. Another problem that came to mind is that refrigerator boxes tend to generate condensate and it might be harmful for hardware.</p>
<p>Am I correct with my assumptions, or is it not that crucial and are there in fact similar solutions?</p>
| |airflow|cooling|computer-engineering| | <p>Its called a chiller rack - fridges are designed to be <em>insulated</em> (keep out heat), while a chiller rack pumps in coolant (often cooled water from a refrigeration unit, either on the rack itself or elsewhere.</p>
<p>Its not enough to keep pumping in cold air, or to <em>keep</em> heat out (when you're cooling heat generating devices), you need to vent it somewhere (outside? into the server room hot isles?). You can't really have a heat generating server in an <em>insulated</em> box - Cooling is significantly about air flow, and most sensible solutions use some form of forced conduction.</p>
<p>There's also rack mount air conditioning units that pump air out via something like a dryer hose. <a href="https://www.youtube.com/watch?v=SZOZTRpQI4Q" rel="nofollow noreferrer">Craft computing has a video of one</a> installed in a rack - however unlike a refrigerator - it has air vents and fans to <em>move</em> heated air out.</p>
<p>That said - there's another principle of cooling - thermal mass, and <em>rooms</em> have a lot of thermal mass. By pumping out your heat into the room, and <em>cooling the room</em>, except in the most extreme cases (for example high density blade servers), its just simpler and more maintenance free than chilling per rack, or even plumbing in a cooling system per server.</p>
| 42598 | Why not put servers in a refrigerator? |
2021-04-15T03:13:36.103 | <p>Is there a convention to use (0, 0, 0) as the center of a CAD drawing? If so, why? Wouldn't it be easier to use (0, 0, 0) as the bottom left corner and start drawing on the positive coordinates?</p>
| |cad| | <p>While there are no conventions for the cad file itself, most folks will look to use the coordinate system datums to their advantage. This involves considering the fit and function of the part.</p>
<p>For example if a block is bolted, consider using the bolt hole rather than the edge. However if the two edges of a block have a bigger impact on how it fits, consider using their connecting corner.</p>
<p>This can also reduce time required to recalculate constrained positions in an assembly when parametric parts are modified. Since the constrained features will not move as much, smaller transformations are performed on the part's coordinates as a whole and it tends to make convergence to meeting constraints easier.</p>
| 42600 | What is the convention for the origin coordinates of a CAD drawing? |
2021-04-15T07:32:38.207 | <p>A composite action beam may have a concrete section the top, carrying the compression load, and steel at the bottom, carrying tension.</p>
<p>But I have always thought that concrete had lower strength for a given mass then steel, compressive or otherwise. So why do engineers build steel concrete composite action floor systems at all? How could it possibly save weight?</p>
<p>To clarify; <strong>What are the practically significant advantages of s composite beam of steel and concrete over one made of purr steel?</strong></p>
| |beam|steel|concrete|composite| | <p>I think I have figured it out.</p>
<p>The transfer of loads in a floor system has 2 main stages.</p>
<ol>
<li>Transfer of load from any random point on the slab to the beams. This has to be accompished by bending and shear in the slab.</li>
<li>Transfer of loads down the beams to the columns. This has to be accomplished through bending and shear in the beams.</li>
</ol>
<ul>
<li><p>Concrete, due to it's low density, creates a thick slab for a given weight, making the slab more efficient at absorbing the bending moments at stage 1.</p>
</li>
<li><p>Rigidly attaching the concrete to the beams with shear studs synergetically works to</p>
<ul>
<li>absorb some of the compression load due to beam bending, that would otherwise have to be carried by the beam's top flange and</li>
<li>prevent any part of the concrete slab from being in tension due to bending, thus increasing the bending strength of the slab.</li>
</ul>
</li>
</ul>
<p>The above combines with the advantages stated in other answers to make concrete-steel composite floors a cheap, practical, lightweight solution.</p>
| 42603 | Exactly what makes steel-concrete composite beam better than pure steel? |
2021-04-16T11:06:47.223 | <p><a href="https://i.stack.imgur.com/O0tln.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O0tln.png" alt="enter image description here" /></a></p>
<p>The above image is part of table 3.1 of Eurocode steel design standard. Looking for example at the first row (steel grade S 235), we see that there are two values for yield strength <span class="math-container">$f_y$</span>, one for element thickness <span class="math-container">$t < 40mm$</span> and another for thickness <span class="math-container">$t > 40mm$</span>. <strong>But why is the yield strength <em>less</em> for the thicker element?</strong> I am aware that the yield strength depends on the manufacturing process of the element. For example, in hot rolled steel there are residual stresses due to uneven cooling of the steel element after hot rolling process. But I would assume that thicker elements require less rolling, and therefore less residual stresses. So why do we assume the value to be <em>lower</em> for these elements instead? Am I misunderstanding something? Thank you!</p>
| |structural-engineering|stresses|steel| | <p>For hot rolled , the thicker section cools slower so has coarser pearlite and coarser grain size because the grains have more time to grow larger during the slow cool. A secondary factor is the thicker section is worked less / less strain to refine grains. The lower strength affect of the slower cooling thick sections also occurs during normalize and quench and temper heat-treatments.</p>
| 42622 | Why is the yield strength of steel assumed to be less for a thicker element? |
2021-04-16T11:13:36.167 | <p>So I understand that shear stress can be either in the the form of transverse or torsional or both at the same time. And normal stress is due to the moment or a force normal to the surface. My question is why is it that sometimes when it asks for the max shear stress we use the <span class="math-container">$\tau_{max}= \sqrt{\left( \frac{\sigma}{2}\right)^2 + \tau^2}$</span> and sometimes we use <span class="math-container">$\tau_{max} = \tau_v + \tau_T$</span> to find the max shear stress?</p>
<p>Why sometimes do we consider the normal stress in calculating the max shear stress and sometimes we do not?</p>
<p><a href="https://i.stack.imgur.com/Litwk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Litwk.png" alt="This is an example where it says find the max shear stress and we use <span class="math-container">$\tau_{max}= \tau_v + \tau_T$</span>" /></a></p>
| |mechanical-engineering|materials|structural-analysis|torque|shear| | <p>In the example you are presenting the shear stresses on cross-section of beam AB are as in the image below.</p>
<p><a href="https://i.stack.imgur.com/xnIOb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xnIOb.png" alt="enter image description here" /></a></p>
<ul>
<li>with <span class="math-container">$\color{green}{green}$</span> is the shear stress <span class="math-container">$\tau_v$</span> which has a single direction (parallel to the shear force)</li>
<li>with <span class="math-container">$\color{red}{red}$</span> is the torsional stress <span class="math-container">$\tau_T$</span>, which has a changing direction (see the following graph)</li>
</ul>
<p><a href="https://i.stack.imgur.com/j4OA7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j4OA7.png" alt="enter image description here" /></a></p>
<p>if you notice at points</p>
<ul>
<li>B and D the stresses are perpendicular. In that case <span class="math-container">$\tau=\sqrt{\tau_v^2 +\tau_t^2}$</span></li>
<li>C the stresses are cancelling each other. In that case <span class="math-container">$\tau_C=\tau_v - \tau_t$</span></li>
<li>A the stresses are contributing to each other. In that case <span class="math-container">$\tau_C=\tau_v + \tau_t$</span></li>
</ul>
<p>Now, regarding the other question of your post (i.e. <span class="math-container">$\tau_{max}= \sqrt{\left( \frac{\sigma}{2}\right)^2 + \tau^2}$</span>), again its best if you post an example you have in mind.</p>
| 42623 | Determining max shear stress |
2021-04-16T14:23:11.353 | <p>I'm having difficulties with calculating the distribution of forces and moments in a statically determinate structure shown below.</p>
<p><a href="https://i.stack.imgur.com/Rn9L9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rn9L9.png" alt="enter image description here" /></a></p>
<p>I'm especially confused about the arc part. Will there be only bending moments, or should we expect torque as well? Any help with approaching this problem is appreciated!</p>
| |statics|beam|moments| | <p>When you want to determine the internal reactions to a structure (in a example such as this), then the best course of action is</p>
<ol>
<li>to isolate the structure of interest by creating section around the part.</li>
</ol>
<p><a href="https://i.stack.imgur.com/pxmeN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pxmeN.png" alt="enter image description here" /></a></p>
<ol start="2">
<li>The next step is to draw the internal forces on the points where the section (blue bubble) intersects with the structure.
Please notice, that in this example (by habit more than anything else), I am using a convenience reference system which uses direction</li>
</ol>
<ul>
<li>dir 1: normal to the section (for tensile/compressive forces)</li>
<li>dir 2: perpendicular to the axial direction on the plane of the image</li>
<li>dir 3: perpendicular to the axial direction and to the plane of the image (Z-axis).</li>
</ul>
<p><a href="https://i.stack.imgur.com/Y2eew.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y2eew.png" alt="enter image description here" /></a></p>
<p>In this part of the example I am only plotting (for simplicity):</p>
<ul>
<li><span class="math-container">$\color{blue}{N}$</span>: the axial force (which is normal to the section)</li>
<li><span class="math-container">$\color{green}{Q_2}$</span>: the shear force perpendicular to the axial force</li>
<li><span class="math-container">$\color{red}{M_3}$</span>: the bending moment around the z-axis (perpendicular to the image plane).</li>
</ul>
<p><em>In the section presented above</em>, the last two (namely <span class="math-container">$\color{green}{Q_2}$</span> and <span class="math-container">$\color{red}{M_3}$</span>) are zero.</p>
<p>However, as you gathered there are 3 other internal forces (for the 3D case), namely:</p>
<ul>
<li><span class="math-container">$\color{green}{Q_3}$</span>: the shear force with direction (parallel to the Z -axis)</li>
<li><span class="math-container">$\color{red}{M_t}=\color{red}{M_1}$</span>: the torsional moment</li>
<li><span class="math-container">$\color{red}{M_2}$</span>: a bending moment in the direction of <span class="math-container">$Q_2$</span>.</li>
</ul>
<p><a href="https://i.stack.imgur.com/OxMnq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OxMnq.png" alt="enter image description here" /></a></p>
<h2>are there torsional moments?</h2>
<p>Yes in this example, at different points you will have different moments and forces.</p>
<p>Apart from bending moments you will also find torsional moment on the straight beam on the support and partially on the curved part of the structure.</p>
| 42630 | Diagrams of forces and moments of a statically determinate structure |
2021-04-17T12:53:00.367 | <p><a href="https://i.stack.imgur.com/YIo0b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YIo0b.png" alt="enter image description here" /></a></p>
<p>Above is the definition of fundamental basic wind velocity from Eurocode 1. According to its instructions, wind loads on building are supposed to be calculated based on this velocity. It says that this velocity needs to have an annual risk of being exceeded of 2 %. <strong>But isn't this quite a risk?</strong> If we design our building's extreme wind loads according to this value, doesn't that mean our building has 2 % chance of failing? To me that seems quite high.</p>
<p>Also according to Eurocode, in reliability class RC2, we should have a reliability index of at least 4.7, which approximately corresponds to probability of failure less than <span class="math-container">$10^{-6}$</span>. This is much lower than 2%.</p>
| |structural-engineering|civil-engineering|structural-analysis|structures|wind| | <p>Buildings are typically designed with a 50-year lifespan. So a 2% yearly chance of passing the limit makes intuitive sense.</p>
<p>But you're right: that'd basically mean we expect the structure to collapse sometime during those 50 years.</p>
<p>So that load with a 2% yearly chance of being surpassed makes sense... but only as a starting point.</p>
<p>The thing to remember is that you aren't getting that wind load, calculating its internal stresses and then checking to see if your members resist that load.</p>
<p>After all, there's a bunch of other safety factors involved in that calculation. If you're using LRFD design procedures, then there are two factors: one for the applied load and one for the structure's strength. The factors themselves depend on the code you're using.</p>
<p>In Brazil, the relevant code is usually NBR 8681, which gives a safety factor of 1.4 for wind loads (when they're the primary load in a combination). For the structural strength, the code depends on the material. For steel, it's 1.35.</p>
<p>So, using these coefficients, you start out with a wind load <span class="math-container">$W$</span> representing a 2% chance. But you then effectively design your structure for a load of <span class="math-container">$1.4 \cdot 1.35 W = 1.89W$</span>, almost double the initial value.</p>
<p>And given that the distribution of wind speeds is roughly normal (well, more Weibull, but close enough) and that we start at the tail of the distribution (2%), getting a load twice as large as that throws us deep into the tail, with a much lower chance of happening, basically ever.</p>
| 42651 | How is any kind of structural reliability achieved if the probability of annual wind speed exceedance is 0.02? |
2021-04-18T10:16:56.463 | <p>I asking for a building that contains a desalination plant and a storage of water bottles (in carton boxes).</p>
<p>Should we split the factory and the storage or not (according to NFPA 13)?</p>
<p>And which section in the code that deals with this matter?</p>
| |structural-engineering|safety| | <p>NFPA 13 is the industry <strong>standard</strong> on all matters related with sprinkler systems in the USA. However, as a standard, it needs to be adopted by each of the States Building Department to incorporate into the State Building Code. Also, the State may impose additional requirements, as Amendments to NFPA 13, which are also to be met. So, in your case, you shall contact the local building department as the first stop.</p>
<p>To your specific question, the paragraphs below explains well:</p>
<p>" Since fire sprinkler design involves plenty of technical concepts, the NFPA 13 dedicates its Chapter 3 to defining key terms that are used throughout the document. An automatic sprinkler is defined as a “a fire suppression or control device that operates automatically when its heat-activated element is heated to its thermal rating or above, allowing water to discharge over a specified area.” There is a common misconception that automatic sprinklers are activated by smoke, but actually they respond to heat.</p>
<p><strong>General requirements apply for all sprinkler systems regardless of the type of building or specific configuration, unless there is a direct exception in the code. If a building uses fire sprinklers, NFPA 13 demands full coverage for the entire property unless the standard indicates clearly that a specific building area is optional."</strong></p>
<p>The NFPA 13 standard is nearly 500 pages long, and very detailed in requirements and specifications. You can get it from Amazon for 79 dollars, a money well spent, I think.</p>
<p>Add - On top of the jurisdiction of your facility, the NFPA standard, you should find out from the EPA what substances/materials are permitted to be stored alongside of the manufacturing area, and limit on the quantity.</p>
| 42660 | Should we split a building containing a factory and a storage according to (NFPA 13) code? |
2021-04-18T15:07:36.560 | <p><a href="https://i.stack.imgur.com/YKYw2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YKYw2.png" alt="enter image description here" /></a></p>
<p>Above is the definition of the probability factor from Eurocode 1-4. It is used in context of calculating the extreme wind load on a building. Ignoring some other factors, it is used as follows:</p>
<p><span class="math-container">$$v_b = v_{b,0} * c_{prob}$$</span></p>
<p>where <span class="math-container">$v_{b,0}$</span> is the extreme wind velocity with less than 0.02 probability of being exceeded. The factor <span class="math-container">$c_{prob}$</span> is used to reduce the extreme wind velocity in cases when considering time periods shorter than the 50 year-return period (0.02 probability).</p>
<p>I assume the formula comes from Gumbel distribution:</p>
<p><span class="math-container">$$p = e^{-e^{-\frac{v-\mu}{\beta}}}$$</span></p>
<p>which can be solved for <span class="math-container">$v$</span>:</p>
<p><span class="math-container">$$v = \mu-\beta \ln(-\ln(p))$$</span></p>
<p>where <span class="math-container">$p$</span> is the probability of encountering extreme wind <span class="math-container">$v$</span>.</p>
<p>Taking the ratio <span class="math-container">$\frac{v}{v_{b,0}}$</span> and considering that <span class="math-container">$v_{b,0}$</span> is defined to have to probability of being exceeded of 0.02, we get:</p>
<p><span class="math-container">$$\frac{v}{v_{b,0}}=\frac{\mu-\beta \ln(-\ln(1-p))}{\mu-\beta \ln(-\ln(0.98))}$$</span></p>
<p>Defining <span class="math-container">$K = \frac{\beta}{\mu}$</span>, we get:</p>
<p><span class="math-container">$$\frac{v}{v_{b,0}} = c_{prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}$$</span></p>
<p>Finally to my question: <strong>Where does the exponent <em>n</em> come from in the Eurocode formula?</strong> As I have derived to formula in the picture, there is no need for the exponent. Is my derivation flawed? Is there something I don't know?</p>
<p>EDIT: Only reason I assumed Gumbel distribution was that I was able to get a similar formula using it. There seems to be nothing in Eurocodes that goes into detail as to what distribution is actually used.</p>
| |structural-engineering|mathematics| | <p>Your derivation seems absolutely reasonable, as does your assumption about the Gumbel Distribution, with one minor detail.</p>
<p>IMHO, the reason why the exponent <span class="math-container">$n$</span> exists, is that the Eurocode does not really care about the velocity but about wind pressures (ultimately about loads on the structure). So the idea, is that you are trying to capture the extreme value of the wind pressure (though the wind speed).</p>
<p>So all the analysis you carried out is not about <span class="math-container">$\frac{v}{v_{b,0}}$</span> but about wind pressures <span class="math-container">$\frac{q}{q_{b,0}}$</span>. So, instead its:</p>
<p><span class="math-container">$$\frac{q}{q_{b,0}}= c_{\color{red}{q},prob} = \frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))}
$$</span></p>
<p>However, because the wind pressure <span class="math-container">$q$</span> is proportional to the square of the wind velocity so:</p>
<p><span class="math-container">$$\frac{q}{q_{b,0}} = \frac{v^2}{v_{b,0}^2}=\left(\frac{v}{v_{b,0}}\right)^2 $$</span></p>
<p><span class="math-container">$$\left(\frac{q}{q_{b,0}} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span></p>
<p><span class="math-container">$$\left(c_{q,prob} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span>
<span class="math-container">$$\left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5} = \frac{v}{v_{b,0}}$$</span></p>
<p>However, because <span class="math-container">$c_{prob}$</span> is a factor that is applied to velocity:</p>
<p><span class="math-container">$$ \frac{v}{v_{b,0}} = \left(\frac{1-K \ln(-\ln(1-p))}{1-K \ln(-\ln(0.98))} \right)^{0.5}$$</span></p>
| 42666 | Question regarding the extreme wind velocity distribution in Eurocode |
2021-04-18T21:15:27.433 | <p>Im working on a project that has gearing system integrated. From my simulation results im dealing with a relatively small amount of input force at the gearbox and would like to know if the force would be transmitted through the gears which i chose at all. I cant seem to find any consistent or not so complicated information about how to calculate the static friction to overcome before motion. Can anyone point me in the right direction? A simplified estimate or model will do. I would like information on spur and bevel gears. Thanks</p>
<p>Edit.
I added a pic as per request.
<a href="https://i.stack.imgur.com/VICZd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VICZd.jpg" alt="enter image description here" /></a></p>
| |gears|mechanisms|machine-design| | <p>As others have pointed out there are many other factors that affect the net efficiency of a gear mesh, but a good rule of thumb is that any decent spur gearbox is 95% efficient at each mesh. <a href="http://maxpowergears.com/know-gear-efficient/#:%7E:text=Its%20efficiency%20varies%20from%2094,98%25%20with%20lower%20gears%20ratios.&text=Straight%20bevel%20gearing%20is%20similar,(93%25%20to%2097%25)." rel="nofollow noreferrer">Approximate gear efficiencies by type</a></p>
| 42669 | static friction of gear pair |
2021-04-19T14:06:34.837 | <p>I have had some trouble determining the best way to measure a dovetail slot and it's mating component. The parts have some difficulty mating occasionally, and our QC department is having trouble getting accurate measurements of the angles using a CMM and other basic measuring devices. I was thinking we might be able to make a go/no-go gauges, but I'm not convinced this is the best route. And ideas would be helpful. Thanks!</p>
<p><a href="https://i.stack.imgur.com/gGqrB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gGqrB.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|machine-design|quality-engineering| | <p>I assume measuring the depth of the dovetail is not an issue.</p>
<p>The classic manual technique is what jko describes, using two pairs of gage pins (or just one pair if the angle is known to be good a-priori). When doing this, you should also account for any difference in depth on the two sides, by the way.</p>
<p>If the CMM machine has some kind of spherical probe you might be able to use that in an equivalent way to get angles and widths.</p>
| 42686 | How to measure a dovetail slot and it's mating component? |
2021-04-20T19:07:39.780 | <p>A cyclist will have the pedal spindle somewhere under the front part of their foot, usually around or just behind the ball of the foot. This means that the ankle is part of the driving movement - at a minimum, holding itself in a single position while the other muscles apply torque to the pedals, so the 'ankle muscles' are acting as a static stabilising force; or in some riders the ankle position changes throughout the pedal stroke, possibly helping to apply force around more of the pedal circle, possibly helping around the 'dead spots' where the pedals are at the 12 and 6 o'clock positions. This video illustrates it very well, from 3m08s: <a href="https://www.youtube.com/watch?v=E7K8VcBXUM4#t=3m08s" rel="nofollow noreferrer">https://www.youtube.com/watch?v=E7K8VcBXUM4#t=3m08s</a></p>
<p><strong>Are there any machines designed to utilise a similar mechanism?</strong> Off the top of my head something like a steam train doesn't, it uses the fewest, simplest parts to enable the linear movement of a piston to be translated to the circular movement of the wheels.</p>
<p><strong>What would be the advantages/disadvantages of such a mechanical setup?</strong></p>
| |mechanical-engineering|torque|pistons| | <h2>Ankling is to make up for the inefficiencies of turning a crank with a device designed to walk/run.</h2>
<p>So unless you have a stupid designer who make the input power inefficient (thanks, evolution), there is no need to create complex linkages that help turn a crank that helps account for muscle contraction and extra movement. A crank is used to turn linear motion into rotary. The most efficient is turning rotary motion in to rotary, so if you really wanted to make it more efficient, you'd lose the linear motion entirely. Steam turbines versus old time piston steam engines.</p>
<p>Everything else is wasted motion and friction.</p>
| 42699 | Is there any mechanical engineering solution that employs 'ankling', as per a cyclist's ankle/foot/pedal setup? |
2021-04-21T13:11:00.660 | <p>Given two pumps that are similar :</p>
<p><span class="math-container">$H_1=70-1.41x10^{-3}Q_1^2$</span></p>
<p><span class="math-container">$H_2=100-1.1x10^{-3}Q_2^2$</span></p>
<p>Find the geometric scale :</p>
<p><span class="math-container">$\frac{D1}{D2}= ?$</span></p>
<p>I am also given maximum efficiency = 1.</p>
<p>I tried solving but I got stuck I don’t know which pi terms should I use for instance:</p>
<p>I tried
<span class="math-container">$ \pi_{gh},\pi_Q,\pi_{\nu} $</span>
and got stuck with hard complicated algebra.</p>
<p>Here is the result in the denominator it should be <span class="math-container">$Q_2^{2}$</span>.
Left side of the lower (2nd) equation.</p>
<p><img src="https://i.stack.imgur.com/n1vz7m.jpg" alt="enter image description here" /></p>
| |fluid-mechanics|pumps|hydraulics|pipelines|prototyping| | <p>The key principle is that, when expressed in terms of the non-dimensional head coefficient <span class="math-container">$K_H := gH/\left(D\omega\right)^2$</span> and the non-dimensional flow coefficient <span class="math-container">$K_Q := Q/\left(D^3\omega\right)$</span>, there is a single head-flow characteristic for the whole family of geometrically similar pumps. That is to say, when expressed in terms of those quantities, the given dimensional head-flow characteristics of the two pumps will both collapse onto the same equation.</p>
<p>So the first given head-flow characteristic
<span class="math-container">$$H_1 = A_1-B_1Q_1^2$$</span>
<span class="math-container">$$\Rightarrow \frac{g_1H_1}{\left(D_1\omega_1\right)^2} = \frac{g_1A_1}{\left(D_1\omega_1\right)^2}-\frac{g_1B_1Q_1^2}{\left(D_1\omega_1\right)^2} = \left(\frac{g_1A_1}{\left(D_1\omega_1\right)^2}\right)-\left(g_1D_1^4B_1\right)\left(\frac{Q_1}{D_1^3\omega_1}\right)^2\textrm{,}$$</span>
i.e.
<span class="math-container">$$K_H = \left(\frac{g_1A_1}{\left(D_1\omega_1\right)^2}\right)-\left(g_1D_1^4B_1\right)K_Q^2\textrm{.}$$</span>
Similarly, the second given head-flow characteristic
<span class="math-container">$$H_2 = A_2-B_2Q_2^2$$</span>
leads to
<span class="math-container">$$K_H = \left(\frac{g_2A_2}{\left(D_2\omega_2\right)^2}\right)-\left(g_2D_2^4B_2\right)K_Q^2\textrm{.}$$</span></p>
<p>In order for both head-flow characteristics to have collapsed to the same equation as required, we must have
<span class="math-container">$$\frac{g_1A_1}{\left(D_1\omega_1\right)^2} = \frac{g_2A_2}{\left(D_2\omega_2\right)^2}$$</span>
and
<span class="math-container">$$g_1D_1^4B_1 = g_2D_2^4B_2\textrm{.}$$</span></p>
<p>The second of those two equations rearranges to
<span class="math-container">$$\frac{D_1}{D_2} = \left(\frac{g_2}{g_1}\frac{B_2}{B_1}\right)^{1/4}\frac{\omega_2}{\omega_1}\textrm{;}$$</span>
If we assume both pumps are on the same planet <span class="math-container">$g_1 = g_2$</span>, then
<span class="math-container">$$\frac{D_1}{D_2} = \left(\frac{B_2}{B_1}\right)^{1/4} = \left(\frac{1.1}{1.41}\right)^{1/4} = 0.94\textrm{.}$$</span></p>
<p>I observe in passing that, in the dimensional head-flow characteristic equations as they appear at the start of the question, someone was very naughty in forgetting to provide units on the numerical <span class="math-container">$A_i$</span> and <span class="math-container">$B_i$</span> values, thus making the equations dimensionally inconsistent; this probably makes it harder than it needs to be to see the way forward with the problem.</p>
| 42714 | Pumps' geometric similarity |
2021-04-21T22:46:21.190 | <p>I'm installing solar panels (not connected to the grid) in units out in a field for research we're doing that looks like this (this is an outdated image, but I can't find the updated one unfortunately):</p>
<p><a href="https://i.stack.imgur.com/PCbxI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PCbxI.png" alt="Outdated model of the system" /></a></p>
<p>I have to safely measure power output from each unit, and each unit has 9 panels with the following specs:</p>
<ul>
<li>Maximum Power: 100W</li>
<li>Maximum System Voltage: 600V DC (UL) Optimum</li>
<li>Operating Voltage (Vmp): 18.6V</li>
<li>Open-Circuit Voltage (Voc): 22.3V</li>
<li>Optimum Operating Current (Imp): 5.38A</li>
<li>Short-Circuit Current (Isc): 5.86A</li>
<li>Operating Temperature:-40°F to 176°F</li>
<li>Output Cables: 14 AWG (2 ft long)</li>
<li>Maximum Series Fuse Rating: 15A</li>
</ul>
<p>(The panels are coming from Amazon, so I'm not sure how accurate the specs are, and the link to reference them is at the bottom)</p>
<p>These panels are going into an agricultural field, so my primary concerns are: 1) Not electrocuting anyone, 2) not starting a fire, 3) being able to accurately measure the power coming from each unit.</p>
<p>This is my main plan so far, which I'd like feedback on:</p>
<ol>
<li>Wire all 9 panels in series to minimize amperage</li>
<li>Use 14 AWG wire to take the power out of the field to some sort of resistive load to use the power</li>
<li>The unit will be grounded using grounding rods</li>
<li>On the circuit, there will be a 10A circuit breaker, a manual on/off switch</li>
<li>The resistive load will be connected to a GFCI</li>
<li>Power will be measured using some sort of Arduino based module</li>
</ol>
<p>Two notes:</p>
<ol>
<li>I'm not sure what the best, safest (and if possible, cheapest) way to burn the power from the solar panels is. I was thinking using a space heater since each unit is allegedly only putting out a maximum of 900W and commercial space heaters are rated for 1875W. I know I could ground each unit and not have a load and the power will just discharge, but my understanding is I can't measure that.</li>
<li>On the subject of measurements. Are any of the micro controller compatible power meters I find on Amazon going to be sufficient for this task, or do I need to look for something specific?</li>
</ol>
<p>This has to be done ideally asap, but definitely before the end of May. Although, again, my main concern is doing this safely first and foremost. I am willing to hold off on measuring the power as long as there's as the primary concern is discharging power safely then figuring out how to measure power. I imagine that would involve routing power to a grounding rod, but I'm not sure.</p>
<p>Any advice and feedback is greatly appreciated, and I am happy to answer any questions.</p>
<p>Solar panel link: <a href="https://rads.stackoverflow.com/amzn/click/com/B07JXYTFF7" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Renogy-Monocrystalline-Solar-Compact-Design/dp/B07JXYTFF7/ref=sr_1_1?dchild=1&keywords=Renogy+Solar+Panel+2pcs+100+Watt+12+Volt+Monocrystalline%2C+2-Pack+Compact+Design&qid=1619041063&sr=8-1</a></p>
| |electrical-engineering|measurements|safety|photovoltaics| | <p>Long story short, probably the safest and cheapest way to measure while maximizing the output of the panels is to buy a small inverter which is connected to the internet and logs the data at a cloud, and connect the inverter to the grid.</p>
<p>On the grid almost nobody will notice the excess energy (probably they won't mind either, unless you are on some net metering scheme).</p>
<p>The other option is to connect - to the output of the inverter - some sort electrical resistance heating unit - preferably with twice the required input (e.g. for the 900W install a 2kW), which is properly cooled (maybe with a fan or otherwise)</p>
| 42720 | Safely Measuring Voltage from Solar Panels in an Experimental Research Field |
2021-04-22T11:05:44.880 | <p>I'm trying to gauge the efficiency of a night-cooled stone to condense water from the humid air at seashores at noon. I'm assuming the stone is thermally insulated once it's in thermal equilibrium with the lowest possible temperature at night. Then at noon, the humid air is passed through the stone and the (assume thermally isolated) system now consists of hot, humid air and the cool stone.</p>
<p>I understand that at lower temperatures, air holds lesser vapor, so if hot humid air is condensed, some vapor will condense.</p>
<p>I'm using the specific heat capacity of vapor, air and stone to try and come up with the final temperature, so that I can see how much vapor will air condense at that temperature by subtracting the after-cooling holding capacity per meter-cubed from the initial one.</p>
<p>The problem here that I must use the latent heat of vaporisation for the energy lost to the stone for the water condensed, but then the vapor is at a temperature lower than 100 Celsius so should I use the specific heat capacity of vapor or latent heat or specific heat capacity of water?! I just got confused.</p>
<p>As <span class="math-container">$q_{vapor}+q_{air}=-q_{stone}$</span>, I came up with this <span class="math-container">$$m_{vapor}c_{Pvapor}(T_{1}-T_{F})+m_{air}c_{Pair}(T_{1}-T_{F})=-m_{stone}c_{Pstone}(T_{1}-T_{F})$$</span></p>
<p>Something just seems wrong to me here because of the vapor temperature being below boiling point of water. I know that temperature is AVERAGE internal energy of a mass of substance and all. However it still confuses me. Someone clear it up for me.</p>
| |fluid-mechanics|thermodynamics|heat-transfer| | <p>You confused me too, lol. Lets back up.</p>
<p>Water will only condense when moist air is cooled below its dew point. The amount of water the air can store is not important to us in this case; just when it is at the dew point.</p>
<p>This means that in order for this system to produce any water, the night temperature must be less than the noon dew point. The final temp will be the noon dew point because any warmer and water will not condense, it will evaporate. As such, if it continues past that point it will evaporate all the water it gained.</p>
<p>To calculate how much water will be produced you will use something like:</p>
<pre><code>Rock_Specific_Heat * Rock Mass * ΔT(night_temp to noon_dewpoint) =
Water_Latent_Heat_of_Vaporization * Water Mass
+ Air_Specific_Heat(with humidity) * Air Mass * ΔT(noon_temp to noon_dewpoint)
</code></pre>
<p>Also note that the <a href="https://en.wikipedia.org/wiki/Latent_heat#Specific_latent_heat_for_condensation_of_water_in_clouds" rel="nofollow noreferrer">latent heat equation for condensation</a> in atmospheric conditions is different than pure steam.</p>
<p>Then you will have to add in the humidity. Use a <a href="https://en.wikipedia.org/wiki/Psychrometrics#/media/File:PsychrometricChart.SeaLevel.SI.svg" rel="nofollow noreferrer">Psychometric Chart</a> convert relative humidity to a humidity mass ratio kg/kg.</p>
<p>Still half baked, but I think that gets you on the right track. Good Luck!</p>
| 42727 | Calculating final temperature of humid air and cool stone |
2021-04-23T06:03:55.990 | <p>I'm currently working through some rather old documentation which gives the thermal conductivity of a material as 9.8 x 10⁻⁶ Btu/s-in.²-°F/in. There are many typical ways to give a thermal conductivity in imperial units, but this does not appear to be one of them - in fact, I cannot find any other instance of thermal conductivity being expressed this way, and even the document I'm working with only uses it a handful of times, none in specifying a known value or constant I could check against another unit of thermal conductivity.</p>
<p><a href="https://i.stack.imgur.com/5WxXq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5WxXq.png" alt="The unit as it appears in the original document" /></a></p>
<p>I've worked through the units, and it seems most likely to me, though I'm far from confident in this, that this unit is equivalent to 2943641 watts per meter kelvin, making the given value equal to 28.85 W/mK (which seems like a plausible value).</p>
<p>However, I'm far from confident in my calculation here, in particular about the meaning of the dashes and period in the unit.</p>
<p>Is 1 Btu/s-in.²-°F/in = 2943641 W/mK the correct conversion factor? How would one read this unit aloud - "BTU per second-inch squared-degree farenheit per inch"?</p>
| |thermal-conduction|unit| | <p>Thermal conductivity is commonly denoted as <span class="math-container">$k$</span> or <span class="math-container">$\lambda$</span>. The units <span class="math-container">$\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}}}$</span> you state are perfectly OK, although awkward. I would have preferred to—at least—simplify the inches:</p>
<p><span class="math-container">$$\mathrm{\frac{BTU}{s\cdot in \cdot ^\circ\mkern-5mu F}}$$</span></p>
<p>or, even better, write the expression as:</p>
<p><span class="math-container">$$\mathrm{\frac{\frac{BTU}{s}}{ in \cdot ^\circ\mkern-5muF} = \frac{1055\; W}{ in \cdot ^\circ\mkern-5muF}}$$</span></p>
<p>The conversion to SI is as follows:</p>
<p><span class="math-container">$$1\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} = 74767.7 \frac{W}{m\cdot K}}$$</span></p>
<p>So the value you've got of <span class="math-container">$9.8 \times 10^{-6}\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} }$</span> is equal to approximately <span class="math-container">$0.7327\mathrm{\frac{W}{m\cdot K}}$</span>.</p>
| 42738 | Thermal conductivity listed in the rather complicated Btu/s-in.²-°F/in - am I converting this right? |
2021-04-23T07:27:58.640 | <p>I am working with a public dataset (mentioned in few papers) of vibration measurements of broken and healthy gearbox under various load. Most of the measured channels posses the same parameters no matter whether the gearbox is broken or not - almost same FFT, same histogram of the raw data, same cepstrum, same IMF (obtained by EMD).</p>
<p>However one channel carry a significant difference in vibrations from the broken and the healthy gearbox (see picture). I checked some papers and books about this topic (fault detection in gearboxes, FFT analysis of gearboxes etc.), but they seems to represent way to nice and clean results far from the dataset observations I have. So I am still quite unable to understand why the FFTs looks different in this way. Few notes about the dataset:</p>
<ul>
<li>frequency of measurement / or speed and size of gears are unknown (I guess something for the FFT x axis - it has no real meaning)</li>
<li>the broken gearbox suppose to suffer with a "broken tooth"</li>
</ul>
<p><strong>My question:</strong></p>
<p>Can you form any hypothesis, why the frequency spectrum is changed in the way displayed in Figure in relation to "broken tooth" fault?</p>
<p><a href="https://i.stack.imgur.com/3XaNx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3XaNx.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|gears|vibration| | <p>Vibrations are the manifestation of things occurring at some frequency. This can be one per rev from imbalance (nothing is ever perfectly balanced) in gears spinning, one per tooth from each tooth changing where it presses against another tooth, one per specific tooth if a tooth exhibits an anomaly differentiating it from other teeth, specific interactions between two specific teeth, frequencies of components as they vibrate and turn energy into noise and heat. Overlay all these and you get a plot like that.</p>
<p>Note that the per-rev frequencies depend on how fast you turn shafts. Modes, etc can vary depending on how you affix and agitate something.</p>
<p>While it is possible to design models for a specific gearbox (you'd have to precisely measure every part), it is not cost-effective. Much of the vibrations and interactions result from deviations from the nominal values. What you can however do is track a specific gearbox's vibrations over its lifetime. You look at it when it is new (and good) while spun at a specific rpm. Then you look at it again when spun at the same rpm.</p>
<p>You asked for a hypothesis on a broken tooth so here goes:</p>
<p>Assuming your gearbox still functions with the break (Depending on the specifics of the break and gearbox, you could expect the output to stop altogether or turn the output differently. Such cases are better detected with other sensors):</p>
<p>What you might pick up include broken bits that are now vibrating on their own, a possible reduction in your one per tooth (this should be more apparent in an fft of the phase-aligned difference between good and bad rather than a difference of fft), and an increase in your one per rev due to imbalance (probably the main thing you can look for in data like what you showed).</p>
| 42740 | Wierd faulty gearbox FFT |
2021-04-23T17:38:10.430 | <p><a href="https://i.stack.imgur.com/buEMt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/buEMt.jpg" alt="enter image description here" /></a></p>
<p>I am not properly understanding this question. please help!</p>
<p>Most of the process on PV diagram have this kind of curvature</p>
<p><a href="https://i.stack.imgur.com/in7wu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/in7wu.png" alt="enter image description here" /></a></p>
<p>But in the question above, its bit weird.</p>
<p>My thoughts:</p>
<p>Pressure, temperature and volume all increasing.</p>
<p>So It looks like combustion process than expansion.</p>
<p>I am unable to figure out answer for it. They all seems correct.</p>
<p>There is definitely net work (-ve may be but there is net work)</p>
<p>Not sure about Net heat output.</p>
<p>As expanding density might decrease but pressure is increasing so it might increase.</p>
<p>And enthalpy is more likely to increase as all the quantities are increasing.</p>
<p>Please clarify the above thoughts and explain the answer.</p>
<p>I searched a lot all over the internet. Did not found any process look similar to this!</p>
| |thermodynamics|temperature| | <p>The process is general in a closed system. The gas is ideal. Adding isotherms on the <span class="math-container">$p,V$</span> axes will show whether <span class="math-container">$T$</span> increases / decreases as <span class="math-container">$p$</span> and <span class="math-container">$V$</span> increase. Work is <span class="math-container">$-\int p_{ext}\ dV$</span>. An assumption of reversibility or constant external pressure will be needed to determine whether 1) is / is not true. The internal energy of an ideal gas depends only on the temperature of the ideal gas. The use of <span class="math-container">$\Delta U = C_V \Delta T = q + w$</span> will tell whether 2) is / is not true. The system is closed, so the amount (moles) of gas is constant. Volume changes and density is the inverse of molar volume. The will address whether 3) is / is not true. The enthalpy of an ideal gas also depends only on its temperature. This will address whether 4) is / is not true.</p>
| 42750 | PV Diagram Pressure temperature volume increasing what is net effect on output? |
2021-04-23T22:13:37.973 | <p>Is there something like a consumable cleaning tape for automatically cleaning parts during manufacturing? Some parts need to be cleaned before automatic assembly, e.g. magnets before adhesive being applied to one of their surfaces. Tape may be soaked in cleaning agent like isopropyl alcohol. I'm looking for something ready to use, before trying to make it on my own.</p>
<p><a href="https://i.stack.imgur.com/YtRBE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YtRBE.png" alt="enter image description here" /></a></p>
| |manufacturing-engineering|production-technology| | <p>I have never seen a cleaning system like this; but it may exist. It may have some merits for having good control and very high purity cleaning; but probably overly complicated for just an adhesive bonding application. Usually manufacturing tries to avoid batch operations when possible to avoid down time. Industry also avoids things that are disposable or have to be batch reconditioned to avoid the additional cost.</p>
<p>A simpler system for your application may be to just use compressed air and a nozzle angled to the surface. Biggest benefit here is that the nozzle does not have to be lined up perfectly with the part geometry to operate correctly. If there is contamination like oil on the surface that air alone will not remove, you could add some soapy water or alcohol via venturi into the cleaning air stream. Alcohol will work great, but soapy water is probably safer for personnel (no chemical exposure or fire risk) and doesn't have volatile organic compound VOC environmental restrictions. This system can also be combined with a vacuum system to minimize contamination released outside of the cleaning machine.</p>
<p>If your application does require the precision of a tape like system, maybe consider a spinning disk cleaner that could be self cleaned between each part. Similar functionality, but mechanically simpler and no disposal/reconditioning of the cleaning tape.</p>
| 42757 | Cleaning tape for automatic part cleaning during manufacturing |
2021-04-24T12:31:41.143 | <p><a href="https://i.stack.imgur.com/6Imth.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Imth.png" alt="enter image description here" /></a>
In this diagram, assuming there is no friction between the pin joint and the rod, I understand that the rod will rotate about <span class="math-container">$O$</span>. However, let's say that the rod is released from <span class="math-container">$ \theta = 0$</span> and that when <span class="math-container">$ \theta = 90$</span> the rod is vertical at that instant in time. If we drew the forces on the rod at this instant, there would be a weight force and a horizontal and vertical reaction force acting at <span class="math-container">$O$</span>. What is confusing me is that surely the horizontal reaction force at this instant must be 0 since there is no other force to balance it and no acceleration occurs in the x direction (the angular acceleration at 90 degrees will be 0 and so it follows that the linear acceleration is also 0 from <span class="math-container">$ a = r \alpha$</span>).</p>
<p>However, in order for the rod to keep rotating doesn't a moment need to be applied about <span class="math-container">$O$</span>? And in the situation where <span class="math-container">$ \theta = 90$</span> the horizontal reaction force is the only force that can provide this moment, leading me to believe that this force is indeed non zero. Therefore, at <span class="math-container">$ \theta = 90$</span> will the horizontal reaction force be 0 N or non zero and if it is non zero how would you calculate it?</p>
| |dynamics| | <p>Since the center of mass offsets the center of support with a distance <span class="math-container">$"r"$</span>, The rod will rotate (with point <span class="math-container">$"B"$</span> heads down) about the frictionless pin <span class="math-container">$(point "O")$</span> in a circular motion and convert the potential energy into kinetic energy. Along the path, the gravity force <span class="math-container">$(W = m*g)$</span> will have a radial force component and a tangential force component. The magnitudes of the component forces vary with position, but the net horizontal force remains at zero at all time, so there is no horizontal reaction at the support pin as <span class="math-container">$\sum Fx = 0$</span> remains valid.</p>
<p>Due to angular velocity and acceleration, the rod will continue to move past the vertical position <span class="math-container">$(\theta = 90^o)$</span> until the energy has exhausted by the drag (air friction) and the reversed kinetic motion. From here on, it swings back and forth similar to a pendulum. that will swing for a long time without interference from an external force.</p>
<p>Note, without externally applied force (torque, moment), the rod will not make a full circle rotation.</p>
<p><a href="https://i.stack.imgur.com/Ta1x1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ta1x1.png" alt="enter image description here" /></a></p>
| 42764 | Reaction force at a pin joint for a rod that is rotating about a fixed axis |
2021-04-24T19:36:34.210 | <p>I was watching a video that derives the forced vibration response of a beam. The link to the video is <a href="https://www.youtube.com/watch?v=6VuLZDgRhZ4&list=PLMXj6GKKnHI6Lftj7CXr9WusMkXi5s9yH&index=56" rel="nofollow noreferrer">https://www.youtube.com/watch?v=6VuLZDgRhZ4&list=PLMXj6GKKnHI6Lftj7CXr9WusMkXi5s9yH&index=56</a></p>
<p><a href="https://i.stack.imgur.com/58pvh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/58pvh.png" alt="enter image description here" /></a></p>
<p>In the derivation, why is it correct to use the expressions for <span class="math-container">$w_n$</span> and <span class="math-container">$U_n$</span> (natural frequencies and mode shapes) that were derived based on free vibration of a beam?</p>
<p>These expressions were derived using the method of separation of variables by equating the first two terms in the equation of motion (EOM in the image above, with f equal to 0 for free vibration) and then equating them to a constant. The two separate equations, one that depends on time and another that depends on position, are set equal to 0 and boundary conditions are used to find the natural frequencies and mode shapes.</p>
<blockquote>
<p>But why is valid to perform these steps when the right side of the EOM is not zero?</p>
</blockquote>
| |vibration| | <p>If I get your question correctly what you are asking essentially is, why are we using the formulas for the free vibration to predict the forced behaviour.</p>
<p><strong>TL;DR: The general case of the forced excitation can be modelled as many cases of constant forces acting on a very small time interval</strong></p>
<h2>Impulsive excitation</h2>
<p>I will start from this because its the basis for what follows. Basically, if you have an impulsive excitation you have a force, that acts for a very small time. The impulse <span class="math-container">$I$</span> is then <span class="math-container">$I= F\cdot \Delta t$</span>, where F is the force (assumed constant) and <span class="math-container">$\Delta t$</span> is the time duration that the force acts.</p>
<p>So, on first glance, you could say that the equation of motion is:</p>
<p><span class="math-container">$$ \ddot{x}+ 2\zeta\omega_n\dot{x}+\omega_n^2 x = \frac{F(t)}{m_{eq}} $$</span></p>
<p>However, if you think about it after the force stops acting, then essentially you have a <strong>free vibration</strong>. So, if the <span class="math-container">$\Delta t$</span> is sufficiently small and you know the velocity before, then you can obtain the velocity just after the force stopped acting, through the principle of impulse momentum. Assuming the mass was initally at rest (to simplify the example), the velocity of the mass (<span class="math-container">$v_0$</span>)just after the force stopped acting will be:</p>
<p><span class="math-container">$$ v_0 = \frac{I}{m}$$</span></p>
<p>So now, you have a problem, with no forced (thus a free vibration) with non zero initial conditions. <strong>So the forced problem devolves to a problem of free vibration with initial conditions.</strong></p>
<p><span class="math-container">$$ \ddot{x}+ 2\zeta\omega_n\dot{x}+\omega_n^2 x = 0 , \text{ with } x(0)=0, \;\;\dot{x}=\frac{I}{m_{eq}} $$</span></p>
<h2>General Excitation</h2>
<p>If you get the general case of the forced excitation it may look like</p>
<p><a href="https://i.stack.imgur.com/QKX4U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QKX4U.png" alt="enter image description here" /></a></p>
<p>However that can be partitioned wrt to time to look like the following image. In that case the force can be thought as constant (if not, then you can take a much smaller interval)</p>
<p><a href="https://i.stack.imgur.com/DLikY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DLikY.png" alt="enter image description here" /></a></p>
<p>Now the area on each of those intervals is <span class="math-container">$\int F\cdot dt$</span>, therefore its impulse of the force <span class="math-container">$F_i$</span></p>
<p><a href="https://i.stack.imgur.com/dd9Kd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dd9Kd.png" alt="enter image description here" /></a></p>
<h2>Putting things together</h2>
<p>So now, what you can do, is take a general excitation (any in fact), break it up into smaller pieces. Then for each piece calculate the impulse, and then the response x(t) as a free vibration with initial conditions. Finally you take all the responses and stitch them together.</p>
<p>The stitching part is not exactly what you'd call a walk in the park. It is actually called "<a href="https://en.wikipedia.org/wiki/Convolution" rel="nofollow noreferrer">Convolution integral/Duhamel integral</a>", and unfortunately is beyond my capacity to explain in a way that its comprehensive in this forum, but if you are interested you can read about it.</p>
| 42771 | Forced vibration of beam |
2021-04-25T02:55:26.083 | <p>How would you add a fillet between two parts in a Solidworks assembly, and propagating the fillet body to one of the parts?</p>
<p>In the example below I have two parts, yellow and gray, and when adding an assembly fillet feature, Solidworks adds it to the gray part. I'd like to treat the grey cylinder as if it were part of the yellow part for the purposes of creating a fillet between them, and then propagate that fillet to the gray part. The second image shows what I'd like to achieve, but between two assembly parts instead of two part features.</p>
<p><a href="https://i.stack.imgur.com/1fCpr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1fCpr.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/H6LOf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H6LOf.png" alt="enter image description here" /></a></p>
<p>I found an example of a convoluted way of getting something like this to work, but I was hoping things have improved since the video was published in 2013.
<a href="https://www.youtube.com/watch?v=6IAGTlvgBgU" rel="nofollow noreferrer">https://www.youtube.com/watch?v=6IAGTlvgBgU</a></p>
| |solidworks| | <p>This is similar to Johnathan's answer, but I would add some dummy extrudes to the grey part, apply the fillet, and then cut off the extrudes.</p>
<p>That's assuming the fillet you're generating is going to be complicated. In most practical cases I would probably just extrude or revolve a simplified fillet onto the gray part by manually drawing it.</p>
| 42775 | adding fillet between parts in an assembly |
2021-04-25T10:45:01.583 | <p>Is this a picture of a plate girder train bridge or a deck truss train bridge?</p>
<p><a href="https://i.stack.imgur.com/8Xzr0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8Xzr0.jpg" alt="enter image description here" /></a></p>
| |rail|bridges| | <p>It's a truss bridge.</p>
<p>A quick Google Image search will make the difference between the bridge types clear.</p>
<p>This is a <a href="https://www.google.com.br/search?q=plate+girder+train+bridge&tbm=isch" rel="nofollow noreferrer">plate girder bridge</a>:</p>
<p><a href="https://i.stack.imgur.com/eUCIX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eUCIX.png" alt="enter image description here" /></a></p>
<p>And this is a <a href="https://www.google.com.br/search?q=deck+truss+train+bridge&tbm=isch" rel="nofollow noreferrer">truss bridge</a>:</p>
<p><a href="https://i.stack.imgur.com/oL9Ls.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oL9Ls.png" alt="enter image description here" /></a></p>
<p>Basically, a girder bridge uses a few massive longitudinal beams while a truss instead uses a criss-cross of smaller beams instead.</p>
| 42779 | is this a picture of a plate girder train bridge or a deck truss train bridge? |
2021-04-25T14:30:52.487 | <p>So, let us say a body moves with 10 m/s, mass = 10 kg.
At time t=0 a force of 30 N is acting on it. Gradually (linearly) the force increases and finally it becomes 130 N after a period of 2 second .</p>
<p>Now, we can say that change in force every second = <span class="math-container">$30+50t.$</span></p>
<p>Now, <span class="math-container">$dp = dt (30+50t)$</span></p>
<p>Now, integrating and putting limits as t = 0 to t= 2 seconds. We get final momentum = <span class="math-container">$160\ kg/m s^2$</span>. Now, we cannot say that F=160/2 since value of instantaneous force is not constant.</p>
<p>Part 1: Finding final velocity and momentum:</p>
<p>If I say <span class="math-container">$130N=10*a $</span>, then a=<span class="math-container">$13\ m/s^2.$</span></p>
<p>Final velocity = u+at = 10 + (13)(2) = 36m/s.</p>
<p>Now, Final momentum becomes <span class="math-container">$m*v =10*36$</span> = <span class="math-container">$360\ kgm/s^2. $</span></p>
<p>Q1 Now, why is 160 not equal to 360 as the final momentum?</p>
<p>Q2 are final velocity and acceleration correct?</p>
| |applied-mechanics|kinematics| | <p>I see this problem differently.</p>
<p>The 10kg mass was moving at a constant speed at 10 m/s, so <span class="math-container">$V_0 = 10 m/s$</span>. 2 seconds After a 30N force was applied, the force increased to 130N due to acceleration, this can be expressed as <span class="math-container">$a = (F_2 - F_1)/m = (130 - 30)/10 = 10 m/s^2$</span>. Now we can find the final velocity, <span class="math-container">$V_F = V_0 + a*t = 10 + 10*2 = 30 m/s^2$</span>.</p>
<p>For <span class="math-container">$V_0 = 10 m/s$</span> and <span class="math-container">$V_F = 30 m/s$</span></p>
<p><span class="math-container">$p = m*(V_0 + V_F) = 10*(10 + 30) = 400 kg*m/s$</span></p>
<p><em><strong>Edit: Adding figure below.</strong></em>
<a href="https://i.stack.imgur.com/ta280.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ta280.png" alt="enter image description here" /></a></p>
| 42782 | Why is final momentum found by two formulae in case of increasing force not equal? |
2021-04-25T16:32:37.897 | <p>For a Q in which we are firing a bullet of 0.04kg onto a wooden block which is stopped after a distance of 60cm after penetrating inside it. It’s u=90m/s.</p>
<p>For this Q , we can find a=-6750m/s^2. Using<span class="math-container">$ v^2 = u^2 +2as$</span></p>
<p>Now, this a can also be compared with a body moving with<span class="math-container">$ -6750 m/s^2 $</span>.Not necessarily a wooden block needs to be there.</p>
<p>Then , second point. My book provides average resistive force by the block on bullet = <span class="math-container">$0.04*6750=270N.$</span></p>
<p>Now , I am confused since the mass of block is not given . How can find the average resistive force exerted by it.</p>
<p>One way I though of is this but I am not sure if it correct or wrong.</p>
<p>So , using Newton’s third law here.</p>
<ol>
<li><p>Force by the bullet on block = -( Force by block on bullet). OR Force by bullet on block = -(force by bullet on block ). Both are not similar here, why is that?</p>
</li>
<li><p><span class="math-container">$0.04*(-6750) $</span>= -(mass of wooden block * -6750). So , in the end we force has to be equal. Therefore , we considered F=270N since finding force by bullet on block is not what Q asked. It asked is the average resistive force by block on bullet.</p>
</li>
<li><p>Is writing average here really necessary ?</p>
</li>
</ol>
| |applied-mechanics|kinematics| | <p>There is no need to know the mass of the block. If it were movable yes, but in this case you can assume that is a fixed wall.</p>
<p>So the whole idea is that you are assuming a constant (average) deceleration of the bullet. Since you already know the deceleration on the bullet, then from the equation</p>
<p><span class="math-container">$$\sum F = m\cdot a$$</span></p>
<p>you know that the only force on the X axis is the deceleration force (most likely through friction).</p>
<p>With that you obtain <span class="math-container">$F = 270 [N]$</span></p>
<p><strong>Another way/route</strong> would be to calculate the kinetic energy of the bullet, and then (Eventually) equate that with the work of the friction on the system until the bullet stops.</p>
<p>So you would get</p>
<p><span class="math-container">$$\frac{1}{2} m u^2 = F\cdot s$$</span>
<span class="math-container">$$ F=\frac{1}{2s} m u^2 $$</span></p>
<p>Which (surprise surpise) yields
<span class="math-container">$$ F=270 [N]$$</span></p>
| 42785 | Firing a bullet on wooden block and then finding the average resistive force by block? |
2021-04-25T20:18:05.480 | <p>I am here regarding a personal project but I am not a mech engineer. The shaft is a watch stem that is displaced from the crown stem. The two shafts are positioned like in the photo below but are 0.9mm in diameter and about 2mm to 4mm apart (it can vary). The shafts can be threaded too. One of the shafts rotates and drives the other and it also can be pushed and pulled along it's axis a distance of 1mm and the other shaft needs to follow that.</p>
<p>I have considered a universal joint with a third shaft/rod between the two, but they cannot be sourced at this size and are hard to manufacture and I am not not certain that the linear force will transmit well with a steep angle.</p>
<p>The other option is use an ordinary gear on each shaft for the rotation. And then couple the linear motion of the shafts with a metal bracket sitting around both shafts that is allows free rotation of the shafts but secures the linear motion of the shafts.</p>
<p>Another similar option would be to use 4 bevel gears. They would transmit the rotational power between shafts easily and the angled surfaces would restrict both shafts to the same linear movement.</p>
<p>Could someone help confirm this for me, I have no knowledge if this will work.</p>
<p><img src="https://i.stack.imgur.com/6ckk6.png" alt="From" />
<img src="https://i.stack.imgur.com/7k3J4.png" alt="second" /></p>
| |mechanical-engineering|gears|coupling| | <p>How about an arrangement like this?</p>
<p><a href="https://i.stack.imgur.com/JSMdl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JSMdl.png" alt="enter image description here" /></a></p>
<p>It might be easier to make than herringbone gears depending on you equipment.</p>
| 42789 | What joint can I use for two thin parallel shafts to transmit both rotational and and linear power (in the axis of the shaft)? |
2021-04-26T13:02:49.967 | <p>I want to make a structure like the following</p>
<p><a href="https://i.stack.imgur.com/1T42y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1T42y.png" alt="[1]: https://i.stack.imgur.com/npCdL.png" /></a></p>
<p><a href="https://i.stack.imgur.com/R7aXs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R7aXs.png" alt="enter image description here" /></a></p>
<p>where:</p>
<ul>
<li>L is the total length and its fixed</li>
<li><span class="math-container">$\lambda$</span> is the distance between each arc</li>
<li>n: is the number of repeated structures in the center</li>
</ul>
<p>I have tried several ways to do that. The best two so far are both using a sweep along a line. One is using a <em>Equation driven line</em> and the other one is using arcs and a <strong>linear pattern</strong>, however in both cases I am missing something.</p>
<p>In general my specs are -ideally- the following. I want to :</p>
<ul>
<li>manage all parameters as <em>global variables</em> from the "Manage Equations"</li>
<li>center the non straight structure</li>
<li>define different <span class="math-container">$\lambda$</span>, <span class="math-container">$n$</span></li>
<li>have a fixed length L</li>
<li>(if at all possible) not be limited to arcs but also more generic shapes, eg. triangles or squares.</li>
</ul>
| |design|solidworks| | <p>I have a partial solution for this. By doing two opposing patterns of sketch blocks, rotated by 180 degrees to each other, and dimensioning the other sketch entities to fit around these, I can make a sketch which will update as you desire, including the more complex geometry, but unfortunately when you do an extrude, it creates disjoint bodies.</p>
<p>This is easily fixed by creating a new blank sketch, selecting the original sketch, and then pressing convert entities.</p>
<p>This can then handle parametric changes in the Lambda and the Length, but if you change the Number, then it will 'fall over'. It's quick to manually fix, just edit the sketch, CTRL+A to delete, select the origial, convert entities, rebuild. This could perhaps be assigned to a macro, set to run automatically on rebuild.</p>
<p>I have uploaded the file that I created for this demo here: <a href="http://www.filedropper.com/parameteriseddemo" rel="nofollow noreferrer">http://www.filedropper.com/parameteriseddemo</a></p>
<p><a href="https://i.imgur.com/TdjfD1c.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/TdjfD1c.gif" alt="Thin cut extrude" /></a></p>
| 42799 | What is the best way to build this parametrically in solidworks? |
2021-04-26T17:03:22.613 | <p><a href="https://i.stack.imgur.com/LWjJL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LWjJL.jpg" alt="enter image description here" /></a></p>
<p>These are suction-mounting hooks. I want to visualize and understand how do they stick on walls.</p>
<p>After these hooks stick on wall. Like we compressed them. How? Since gases are compressible up to a good finite limit. What makes it difficult for them to come off if I pull them. It should be easily able come off also. Online , it says that the air gets enclosed inside . But I am not getting exactly what it means by that. How does that air enclosed creates a way in which it becomes difficult for us to pull the hook off.</p>
| |materials|pressure|applied-mechanics|air| | <p><strong>TL;DR: Its not the air that's enclosed inside that creates the force, but the lack of air - or more precisely pressure.</strong></p>
<p>First of all some nomenclature (this is for the closed type, there is also the open type):</p>
<p><a href="https://i.stack.imgur.com/qI6nD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qI6nD.png" alt="enter image description here" /></a></p>
<p>When you press a suction cup on a surface what happens is that the air inside it is pushed out. During the compression phase of the cup the air comes out of the lip to the atmospheric pressure. So if you have an original volume <span class="math-container">$V_0$</span> of the suction cup, and then the volume reduces to <span class="math-container">$V_2 = \frac{V_0}{2}$</span> (usually there is even less volume), then the mass of the air trapped inside the suction cup is about half the original (keep this in mind for later on).</p>
<p><a href="https://i.stack.imgur.com/MxJ9b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MxJ9b.png" alt="enter image description here" /></a>
<em>Figure 2: Air exiting . (source: <a href="http://ijiset.com/vol4/v4s6/IJISET_V4_I06_23.pdf" rel="nofollow noreferrer">ijiset.com</a>)</em></p>
<p>The dome is flexible, so while pushing it is easy to change its shape (thus force the the air out). Additionally the shape of the lip (at a tangent with the object makes it easy for the air to be driven out). However, after the external force is removed then because of the material elasticity, the suction cup is trying to return to the original shape.</p>
<p><a href="https://i.stack.imgur.com/SY503.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SY503.png" alt="enter image description here" /></a></p>
<p><em>Figure 3: Different stages. (source: <a href="http://abetterchemtext.com/gases/images/s_cup3.png" rel="nofollow noreferrer">Abetterchemtext</a>)</em></p>
<p>Following on with the example, because the suction cup tries to recover (partially), as the volume of the suction cup tries to increase the air density reduces, and so does the pressure. So what you end up is a low pressure underneath the suction cup. If the suction cup were allowed to recover its original shape (it doesn't) then the pressure in the example would be half the pressure.</p>
<p><a href="https://i.stack.imgur.com/1tpfu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1tpfu.png" alt="enter image description here" /></a></p>
<p><em>Figure 4: Low and high pressure regions. (source: <a href="http://www.wired.com/2013/08/how-strong-are-the-suction-wheels-in-bad-piggies/" rel="nofollow noreferrer">wired</a>)</em></p>
<p>Then the "suction" force is equal to :</p>
<p><span class="math-container">$$F = A\cdot dP $$</span></p>
<p>where :</p>
<ul>
<li>A is the area of the suction cup (its projection)</li>
<li>dP is the difference in pressure (<span class="math-container">$P_{High}- P_{low}$</span>)</li>
</ul>
<hr />
<h2>numerical example</h2>
<p>Assuming a diameter of 50 mm, and a pressure difference at about half the pressure of the atmospheric pressure, then you can estimate that the pull out force would be approximately:</p>
<p><span class="math-container">$$F = \frac{\pi d^2}{4} \Delta P $$</span>
<span class="math-container">$$F \approx 2 \; [N]$$</span></p>
<p>Notice that if you double the diameter the force quadruples, so you can get significantly higher forces.</p>
<h2>Why some surfaces are better</h2>
<p>The reason why smooth non porous surfaces tend to behave better with suction cups is that there is not much leakage (from the outside in).</p>
<p><a href="https://i.stack.imgur.com/EjH60.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EjH60.png" alt="enter image description here" /></a></p>
<p>Air coming in the suction chamber would increase the pressure inside the chamber, and therefore reduce the pressure differential <span class="math-container">$\Delta P$</span>.</p>
| 42803 | How do suction cups work? |
2021-04-27T12:37:22.577 | <p>I am looking for a way to generate a linear motion with a triangle motion profile for my hobby project. The travel distance should be 6cm (so 12 cm for round-trip) with constant speed 1.2 m/s or higher. The solution should be able to oscillate at 10 Hz for minutes at time.</p>
<p>The travel distance basically rules out the use solenoids, and the triangle motion profile requirement rules out basic slider-crank mechanism where the linear motion speed changes over time. I am hoping that this is a trivial problem for proper engineers, but as a hobbyist I am struggling to come up with a work-able solution.</p>
<p>Any suggestion will be appreciated.</p>
<p>Thanks</p>
| |mechanical-engineering|linear-motion|linear-motors| | <p>Probably many possibilities. Here's what comes to my mind:</p>
<p>If the load is light, you could mount it on a timing belt. It could still be on a linear guide, to keep its motion perfectly straight. One end of the timing belt is an idler pulley (probably with a belt tension adjustment feature), and the other end is the motor. Drive the timing belt with any motor that can produce constant speed with the precision you need.</p>
<p>Conventionally, electromechanical / optical / magnetic limit switches would sense the end positions and trigger motor to reverse direction.</p>
<p>A stepper motor with a small gear reduction may work out too, though the step resolution** may be an issue at 10Hz. In that case, a single limit switch might be sufficient. If the stepper driver electronics can sense stalling (from hitting a hard stop), you can even go with no switch at all. Some fancy industrial ones from FESTO I've used do this, but that is massively overpriced and overkill.</p>
<p>Other thoughts:</p>
<ul>
<li>If the stepper motor step-resolution is not a limitation, you could also use any rotary-to-linear mechanism, and program the stepper motor's motion profile to compensate for the non-linearity.</li>
<li>Also take a look at windshield-wiper mechanisms, they have a clever mechanical auto-reverse gear shift arrangement that activates when the motion reaches the ends. That could perhaps be used here.</li>
<li>With any electronically-controlled motor situation, some controlled acceleration at the ends of motion will probably be necessary, as a theoretical triangle profile has infinite acceleration at the ends.</li>
</ul>
<hr />
<p>** For stepper motor resolution and gear reduction considerations, use typical values of 200steps/rev, prior to any gear reduction, and a max step rate of 1000 steps/sec</p>
| 42825 | Linear motion with triangle profile |
2021-04-27T12:50:15.377 | <p>The formula is:
<span class="math-container">$$T= \dfrac{\left(\sum (\text{coefficient of tension})_i\right) \cdot g_{effective}}{\sum \frac{(\text{coefficient of tension})_i^2}{m_i}}$$</span>
I tried to prove this formula but not getting it. An example for this formula is below:</p>
<p><a href="https://i.stack.imgur.com/wuhHj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wuhHj.jpg" alt="enter image description here" /></a></p>
<p>Explaining the Q:</p>
<p>1g + 2g means correction of tension is 1 for 4kg block and 2 for 2kg block. In the end, g is always the same. Then, in the denominator it is the addition of <span class="math-container">$\text{coefficient of }T^2 / \text{mass}$</span>.</p>
<p>About <span class="math-container">$g$</span> effective on top, in case of inclined plane.its value would be <span class="math-container">$g \sin \theta$</span> if it is like this:</p>
<p><a href="https://i.stack.imgur.com/XwdmA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XwdmA.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|applied-mechanics| | <p>I'll be using subscript notation <span class="math-container">$_2$</span>and <span class="math-container">$_4$</span> to denote the mass with 2 kg and 4 kg respectively. Also note that <span class="math-container">$a_2$</span> and <span class="math-container">$a_4$</span> both point upwards which I consider as the positive direction.</p>
<p><a href="https://i.stack.imgur.com/gcyYs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gcyYs.png" alt="enter image description here" /></a></p>
<p>There are 2 equations of motion</p>
<ul>
<li>for mass 2:
<span class="math-container">$$m_2\cdot a_2 = -m_2 \cdot g +2T \tag{eoq:m2}$$</span></li>
<li>for mass 4kg
<span class="math-container">$$m_4\cdot a_4 = -m_4 \cdot g +T \tag{eoq:m4}$$</span></li>
</ul>
<p>Additionally you get the kinematic constraint (assuming the wire is not deformable as)</p>
<p><span class="math-container">$$a_4 = -2a_2 \tag{KC}$$</span></p>
<p>you can rewrite <span class="math-container">$eoq:m2,\ eoq:m4$</span> as:</p>
<p><span class="math-container">$$a_2 = -g +\color{red}{2} \frac{T}{m_2} $$</span>
<span class="math-container">$$a_4 = -g +\color{red}{1}\frac{T}{m_4} $$</span>
<span class="math-container">$$\color{red}{1} a_4 = -\color{red}{2}a_2 $$</span></p>
<p>With the <span class="math-container">$\color{red}{red}$</span> are the correction coefficients g, you are mentioning. Therefore:
<span class="math-container">$$a_2 = -g +\color{red}{g_2} \frac{T}{m_2} $$</span>
<span class="math-container">$$a_4 = -g +\color{red}{g_4}\frac{T}{m_4} $$</span>
<span class="math-container">$$\color{red}{g_4} a_4 = -\color{red}{g_2}a_2 $$</span></p>
<p>Then you solve the system of the equations of motion with the kinematic constraint:
<span class="math-container">$$
\begin{cases}
a_2 = -g +g_2 \frac{T}{m_2} \\
\color{red}{a_4} = -g + g_4\frac{T}{m_4} \\
\color{red}{g_4 a_4 = -g_2 a_2}
\end{cases} \Rightarrow
\begin{cases}
a_2 = -g +g_2 \frac{T}{m_2} \\
-\frac{g_2}{g_4}a_2 = -g + g_4\frac{T}{m_4}
\end{cases}
$$</span></p>
<p><span class="math-container">$$
\begin{cases}
a_2 = -g +g_2 \frac{T}{m_2} \\
a_2 = \frac{{g_4}}{{g_2}}\left(g - g_4\frac{T}{m_4}\right)
\end{cases} \Rightarrow
-g +g_2 \frac{T}{m_2} = \frac{{g_4}}{{g_2}}\left(g - g_4\frac{T}{m_4}\right)
$$</span></p>
<p><span class="math-container">$$
+g_2 \frac{T}{m_2} +\frac{g_4}{g_2} g_4\frac{T}{m_4} = +g +\frac{g_4}{g_2} g
$$</span>
<span class="math-container">$$
\left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)T = \left(1 +\frac{g_4}{g_2}\right) g
$$</span>
<span class="math-container">$$
T = \frac{ \left(1 +\frac{g_4}{g_2}\right) g }{\left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)}$$</span></p>
<p>To simplify this we multiply by <span class="math-container">$g_2 $</span> mnominator and denominator:
<span class="math-container">$$
T = \frac{g_2}{g_2} \frac{ \left(1 +\frac{g_4}{g_2}\right) g }{\left(\frac{g_2}{m_2} + \frac{1}{g_2} \frac{g_4^2}{m_4}\right)}\rightarrow
T = \frac{ \left(g_2 +g_4\right) g }{\left(\frac{g_2^2}{m_2} + \frac{g_4^2}{m_4}\right)}$$</span></p>
<p>So finally you get:</p>
<p><span class="math-container">$$
T = \frac{ \left(\sum g_i\right) g }{\sum \frac{g_i^2}{m_i}}
$$</span></p>
<hr />
<p><strong>POINT OF CAUTION</strong> Although this has been proven for this particular set of pulley, it is something I've never come across. And despite, solving it myself (and marveling at the elegance that someone actually contrived this), I would be very cautious in using this equation in another problem without cross-checking it.</p>
| 42826 | Any idea on how to prove this formula of tension? |
2021-04-28T12:46:48.230 | <p>I am designing a joint which I want to break at a specific force. I want to design a 2 parts connected by a joint that breaks when a certain amount of force is applied and thus prevent any damage to those individual components themselves. I also want the joint to be easily replaceable without replacing the individual parts themselves. I'm hoping to 3D Print the joint.</p>
<p>I have seen some petrol stations that have pipes attached such that they will break and cut-off fuel supply if they are pulled harshly(i.e. By someone who forgot to take the nozzle off their car and tried to drive away). I don't know the name of the mechanical linkage they use. But I'm hoping to achieve a similar affect.</p>
<p>I would appreciate if you could point me to relevant information sources and design of such type of linkages.</p>
| |mechanical-engineering|structural-engineering|design|3d-printing| | <p>Another option is the shear pin. It was first thing that came to mind (Pete W suggested it also already).</p>
<p>A joint involving a shear pin (in tension and compression) would look something like the following (here its called a <em>clevis pin</em>, and sometimes you find it as <em>hinge pin</em>).</p>
<p><a href="https://i.stack.imgur.com/r9mU2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r9mU2.png" alt="enter image description here" /></a></p>
<h2>force calculation</h2>
<p>The force required for the shear pin to fail is basically determined by the smallest cross-section subjected to shear. Although theoretically you can have any cross-section, usually , (to allow for rotation) the shear pin is cylindrical. So the Force for a cylindrical shear pin will be calculated by:</p>
<p><span class="math-container">$$F_{max} = \sigma_s \cdot A$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\sigma_s$</span> is the shear stress at failure</li>
<li><span class="math-container">$A$</span>: is the smallest cross-section under stress.
<ul>
<li>For a filled cylinder <span class="math-container">$A=\pi r^2$</span></li>
<li>for a hollow cylinder like the following</li>
</ul>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/lOtMr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lOtMr.png" alt="enter image description here" /></a></p>
<h2>3d Printing relevance</h2>
<p>The really good thing about this and 3D printing is that you can fine tune the exact force that you want. With 3D printing you can design a hollow shear pin and adjust :</p>
<ul>
<li>thickness of walls</li>
<li>printing orientation</li>
<li>generate different cross-section geometries like</li>
</ul>
<p><a href="https://i.stack.imgur.com/fRZzR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fRZzR.png" alt="Shear pin with modifie geometry " /></a></p>
| 42851 | Mechanical joint that break at specific force |
2021-04-28T15:44:08.963 | <p><a href="https://i.stack.imgur.com/yakyY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yakyY.jpg" alt="enter image description here" /></a>
I have named the strings as S1,S2 and S3 on diagram. T is the tension in the string.</p>
<p>If I draw the Free body diagram for m1, I think it should be 2T force on it but on internet , it is just T. So , why do I think it is 2T?</p>
<p>Just imagine that when m2 is pulled , the pulley connecting mass m1 also gets pulled with 2T tension. Then , We have a fixed wall. So , the fixed wall can’t and would never move. So , it applies a T force on the string which travel to mass m1 and pulls it towards m2 direction. Now , till here the force is applied by string S1.</p>
<p>Then , when the pulley gets pulled. The string S2 also applied another T on the mass m1 which pulls it too. So , in conclusion . Total tension on the mass m1 is 2T.</p>
<p>So , where is it that I am going wrong ?</p>
<p>EDIT: Example of double tension or 2T .</p>
<p><a href="https://i.stack.imgur.com/phqSi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/phqSi.jpg" alt="enter image description here" /></a></p>
<p>So ,here. There is a 2T force applied on 6kg mass.</p>
| |mechanical-engineering|applied-mechanics| | <p>Something that might help you clear this is the following sketch.</p>
<p><a href="https://i.stack.imgur.com/1GJdq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1GJdq.png" alt="enter image description here" /></a></p>
<p>I just drew the forces in the direction that they will allow the rope/wire to be taut.</p>
<p>In your original sketch the directions of the wire forces from the wall and m1 and the wall were the opposite way. The problem there is that if the forces are drawn in your original way then the rope would collapse. Wires can transfer <strong>tension</strong> forces but not compressive. So ropes can only transfer tensile forces.</p>
<p><a href="https://i.stack.imgur.com/eK6hb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eK6hb.png" alt="enter image description here" /></a></p>
<p>In the above example the tension force that you have is at every point of the rope exactly the same. Unless you provide an axial force (so unless another person steps in there is no reason for the force to increase. And you will notice (in the image below) that the tension forces drop at the right hand side (the rough sketch underneath is the magnitude of the axial/tensile force)</p>
<p><a href="https://i.stack.imgur.com/4MvhB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4MvhB.png" alt="enter image description here" /></a></p>
<p>To come back to your original example, I've highlighted the rope from <span class="math-container">$m_1$</span> with a green thick line. You see that it (ultimately) attaches to the wall at the other end. On that continuous piece of rope the force should be always equal.</p>
<p><a href="https://i.stack.imgur.com/rCWGa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rCWGa.png" alt="enter image description here" /></a></p>
<p>Now the reason that the force on the right is 2T, is because the pulley is in equilibrium, so the algebraic sum of the forces should be zero. Since, you know the force on the left hand of the image, then you can easily derive that the force to <strong><span class="math-container">$m_2$</span></strong> is <span class="math-container">$2T$</span></p>
| 42856 | Why is the tension on mass m1 not 2T here? |
2021-04-28T17:14:13.237 | <p><a href="https://i.stack.imgur.com/BwN8y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BwN8y.jpg" alt="enter image description here" /></a></p>
<p>Assumption :All frictionless surfaces.</p>
<p>EDIT: In the book , it doesn’t say friction has to be present but as PeteW has mentioned , i think it should be necessary too. So , in that case let me take two cases. One when friction is present between 5Kg and 1kg block and other when it is not. So , depending upon it. Check it.</p>
<p>The triangle wedge (white color) is always fixed. The blue color wedge of 5kg moves down and so does 1kg is given in Q.</p>
<p>So , the main question to find the time when 1kg block touches the inclined plane. Assume in one case a inertial frame and in another , any non inertial frame of your own choice.</p>
<p>My main confusion is that how do we know if the 1kg block will move towards right direction. It can be left or no direction also.</p>
<p>If I draw the FBD for 1kg block , it has N from 5kg which cancels out due to its mg=1g. So , there is no external force to make it move right.</p>
| |mechanical-engineering|applied-mechanics|kinematics| | <p>If you are asking why does the 1 kg mass (I'll denote it as <span class="math-container">$m_1$</span>), moves towards the inclined plane, then IMHO the exercise you shouldn't have any friction.</p>
<ul>
<li><strong>If there is no friction</strong>, then the <span class="math-container">$m_1$</span> will move downwards, because there will be no horizontal force to keep it attached to the Wedge (I suspect its mass is 5 kg, so I will denote it as <span class="math-container">$m_5$</span>. Below is the FBD and the kinetic diagrams of the two moving masses <span class="math-container">$m_1, m_5$</span>.</li>
</ul>
<p><a href="https://i.stack.imgur.com/nrWmc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nrWmc.png" alt="enter image description here" /></a></p>
<p>The acceleration of mass 1 will be downward, while the wedge will accelerate in a direction along the plane.</p>
<ul>
<li><strong>If there is friction</strong>, then the material will probably never meet the inclined face. The reason is that the friction will be the only force on the horizontal plane, so the <span class="math-container">$m_1$</span> will move with <span class="math-container">$m_5$</span>. (Of course, if you change the slope you might get a reaction so great from the wall, that the friction cannot overcome).</li>
</ul>
<hr />
<p>If you are asking about the forces, then in order to calculate the reactions and the acceleration you'd have to solve the following system</p>
<p><span class="math-container">$$\begin{cases}
- m_1 \cdot g + N_1 = -m_1\cdot a_1\\
- N_{2x} = - m_5\cdot a_{5x}\\
-m_5 g + N_{2y} - N_1 = - m_5\cdot a_{5y}\\
a_1 = a_{5y}\\
\frac{A_{5x}}{\cos(\phi)} = \frac{a_{5y}}{\sin(\phi)} = a_5\\
\frac{N_{2x}}{\sin(\phi)} = \frac{N_{2y}}{\cos(\phi)} = N_2
\end{cases}
$$</span></p>
<p>If you do the substitutions you obtain the following system with 4 equations and 4 unknowns <span class="math-container">$a_1, a_5, N_1, N_2 $</span>:</p>
<p><span class="math-container">$$\begin{cases}
- m_1 \cdot g + \color{red}{N_1} = -m_1\cdot \color{red}{a_1}\\
- \color{red}{N_{2}}\sin(\phi) = - m_5\cdot \color{red}{a_{5}}\cos(\phi)\\
-m_5 g + \color{red}{N_2} \cos(\phi) - N_1= - m_5\cdot\color{red}{ a_{5}}\sin(\phi)\\
\color{red}{a_1} = \color{red}{a_{5}}\sin(\phi)\\
\end{cases}
$$</span></p>
<p>The results (using g= 9.81 <span class="math-container">$m/s^2$</span>)should be:</p>
<ul>
<li><span class="math-container">$N_1$</span>: 5.834 N</li>
<li><span class="math-container">$N_2$</span>: 43.82 N,</li>
<li><span class="math-container">$a_1$</span>: 3.975 <span class="math-container">$m/s^2$</span></li>
<li><span class="math-container">$a_5$</span>: 6.6 <span class="math-container">$m/s^2$</span></li>
</ul>
<p>I'll leave the numerical solution to any interested party.</p>
| 42857 | Why does the 1kg mass move towards inclined plane here? |
2021-04-28T21:19:34.183 | <p>I was looking at a spec sheet for a photopolymer and saw these two different subsections under the "Thermal" section. I tried Googling around as I usually do, but no site could give me the clear difference between the two and some simply defined the two as the same. I have a hard time believing that since there are two distinct test methods indicated by the ASTM. Could someone please clarify what the difference is?</p>
<p>PS - To further clarify, I mean the difference between the two results, not the ASTM test methods.</p>
<p><a href="https://i.stack.imgur.com/NrkNo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NrkNo.png" alt="Spec Sheet" /></a></p>
| |materials|material-science| | <ul>
<li>CTE characterizes dimensional change, with no load. It corresponds to a relatively fundamental physical principle, making it easy to use in a design calculation</li>
<li>Heat Deflection test characterizes the deformation under heat AND load, which includes a variety of phenomena happening at the same time. The result is dependent on the particular geometry and loading conditions specified by the test, in a way that makes it hard to extrapolate for design purposes. On the other hand, it is more useful when "shopping" for a material, by making it possible to compare them in standard conditions for applications that are limited by both heat and load at the same time.</li>
<li>For the two different lines of the heat deflection test, they correspond to different standard loads.</li>
<li>Annoyingly, the actual standards are not made freely available, but a general description can usually be found with a little searching. (e.g. <a href="https://www.intertek.com/polymers/testlopedia/heat-deflection-temperature-astm-d648/" rel="nofollow noreferrer">here is one for ASTM-D648</a> )</li>
</ul>
| 42864 | Heat Deflection VS Coefficient of Thermal Expansion? |
2021-04-29T08:12:35.830 | <p>A block can slide on a smooth inclined plane of inclination <span class="math-container">$\theta$</span> kept on the floor of a lift. When the lift is descending with retardation i.e. <span class="math-container">$ a \; m/s^2 $</span>. Find the acceleration of the block relative to the in inclined plane.</p>
<p>How I solved till now:</p>
<p><a href="https://i.stack.imgur.com/rlFVm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rlFVm.jpg" alt="enter image description here" /></a>
Above is the FBD I drew.</p>
<p>I have written some points and want to confirm them:</p>
<ol>
<li>From an inertial frame. The acceleration of the block should be
mg+ma?</li>
<li>Since it asks for acceleration of <span class="math-container">$m$</span> relative to the incline, I can
take a non inertial frame which is sitting on the inclined plane at
the top. So, I have marked a person there. For that person, the lift
will be at rest. So, then the acceleration of the block would be <span class="math-container">$mg
sin \ \theta$</span>. Since, I view it from NIF. Then, I add a pseudo force
which will be <span class="math-container">$-ma$</span> (a is acceleration of lift). I have assumed
that the person also has an acceleration <span class="math-container">$a m/s^2$</span> as given in Q
(then only the person sees the lift at rest). Then, the pseudo force
is always negative. So, it acts in the direction towards the NIF on
the mass <span class="math-container">$m$</span>. So, we get:</li>
</ol>
<p><span class="math-container">$mg sin\ \theta$</span> - <span class="math-container">$ma sin \ theta$</span> = m*(relative acc of block wrt inclined plane).</p>
<p>Therefore, answer is <span class="math-container">$(g-a)sin \ \theta$</span>.</p>
<p>I have solved this much. But my answer is wrong. The correct answer is <span class="math-container">$g+a= sin \ \theta$</span></p>
<p>Also, if there is a way to solve it in inertial frame, it would be great to know, since most answers solve it with non inertial. Is there a way to solve with inertial frame also?</p>
<p>EDIT:</p>
<p><a href="https://i.stack.imgur.com/O86vx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O86vx.jpg" alt="enter image description here" /></a>I solved it this way now. (Different way).It is a photo. So , I hope it is clear to just read. Only thing I got wrong is that when I added a pseudo force.</p>
<p>My equation becomes :</p>
<p>N(normal by the wedge in the lift)-<span class="math-container">$m(g+a)$</span>sin <span class="math-container">$\theta$</span> - ma = <span class="math-container">$m*(velocity of block wrt inclined plane = $</span>a_v$).</p>
<p><span class="math-container">$m(g+a) =N.$</span></p>
<p><span class="math-container">$m(g+a) $</span>- <span class="math-container">$m(g+a)$</span>sin <span class="math-container">$\theta$</span> - <span class="math-container">$ma$</span>= <span class="math-container">$m* a_0. $</span></p>
<p>This is what I got.</p>
<p>So , if I don’t include the N and -ma. Then , my answer comes to be right.</p>
| |mechanical-engineering|applied-mechanics|kinematics| | <p>As shown in this <a href="https://engineering.stackexchange.com/questions/42878/why-is-the-acceleration-of-wedge-and-block-equal-like-this/42880?noredirect=1#comment76843_42880">thread</a>, the acceleration of a mass on an incline equals <span class="math-container">$g*sin\theta$</span>. When the inclination is lowered with a deacceleration of -<span class="math-container">$a_{dec}$</span>,
from <strong>relativity</strong>, it means the mass is moving backward from the original frame, thus, <span class="math-container">$a = (g - (-a_{dec}))sin\theta = (g + a_{dec})sin\theta$</span>.</p>
<p><a href="https://i.stack.imgur.com/ARPI0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ARPI0.png" alt="enter image description here" /></a></p>
| 42875 | Find the acceleration of a block relative to incline plane in non and inertial frames? |
2021-04-29T11:19:24.803 | <p><a href="https://i.stack.imgur.com/g2EA9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g2EA9.jpg" alt="enter image description here" /></a></p>
<p>No friction at all anywhere. Let me explain the diagram in steps and the Q.</p>
<ol>
<li><p>green colour let us start with. It represent a wedge on which there is mass of 10kg (red colour). That’s it the first picture.</p>
</li>
<li><p>dotted line in white represent when the block touches the bottom of the wedge.</p>
</li>
<li><p>represent the dotted yellow line which represents the direction of acceleration for 10 kg mass. Which is kind of like unknown or not in a symmetry.</p>
</li>
<li><p>Where the Q lies</p>
</li>
</ol>
<p>The Dark yellow line represent displacements. Below one is AC I.e displacement of the wedge. Now , only thing in this Q what I don’t understand is that why is Displacement of mass 10kg BC even though the direction of acc for 10 kg is the dotted line.</p>
<p>If you wish to know why the dotted line , there is an example in my textbook which tells us that. Also , tells us that because of the dotted line kind of acceleration, we use pseudo force for solving such questions.</p>
<p>EDIT: Where a different concept and answer is also used.
<a href="https://i.stack.imgur.com/nOj2Z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nOj2Z.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|applied-mechanics| | <p>In order to find the acceleration the process is similar to this <a href="https://engineering.stackexchange.com/questions/42875/find-the-acceleration-of-a-block-relative-to-incline-plane-in-non-and-inertial-f">question</a>.</p>
<p><strong>For the inertial observer the angle of the acceleration (<span class="math-container">$\phi$</span>) with the horizontal axis will be steeper than the angle of the inclined plane (<span class="math-container">$\theta$</span>) (See red line in the kinetic diagram).</strong></p>
<p><a href="https://i.stack.imgur.com/Dh3EI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dh3EI.png" alt="enter image description here" /></a></p>
<p>As you can see on the inertial Frame you have:</p>
<ul>
<li><span class="math-container">$a_A$</span> the acceleration of A with respect to the inertial frame. (the direction is not known)</li>
<li><span class="math-container">$a_W$</span> is the acceleration of the wedge with respect to the inertial frame. Since the wedge will be accelerating then the acceleration vector is pointing to the right.</li>
</ul>
<p>The remaining vector <span class="math-container">$a_{A|W}$</span> is the acceleration vector with respect to the frame of reference of the Wedge. For a person standing on the Wedge as it is moving, then the <strong>acceleration vector <span class="math-container">$a_{A|W}$</span> will be parallel to the inclined plane (it will form an angle <span class="math-container">$\theta$</span> with the horizontal plane)</strong>.</p>
<p>So you have two vectors <span class="math-container">$a_W$</span> and <span class="math-container">$a_{A|W}$</span> , that you know their directions. Therefore because of the equation:</p>
<p><span class="math-container">$$\vec{a}_A = \vec{a}_W + \vec{a}_{A|W}$$</span></p>
<p>you can add up the accelerations and obtain <span class="math-container">$a_A$</span>. (<em>apology for the similarities with the other question but the process is exactly the same</em>)</p>
<p>The main difference know is that the motion of the wedge is dependent on the mass A, and on the angle <span class="math-container">$\theta$</span>. So the diagrams for the wedge are</p>
<p><a href="https://i.stack.imgur.com/9yy1C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9yy1C.png" alt="enter image description here" /></a></p>
<p>The resulting equations are the following
<span class="math-container">$$\begin{cases}
- N_{1x} = -m a_{Ax}\\
- mg + N_{1y} = -m a_{Ay}\\
N_{1x} = m a_{W}\\
- N_{1y} + N_2 - mg = 0\\
a_{Ax} = a_{Wx} + a_{(A|W)x}
a_{Ay} = 0 + a_{(A|W)y}
\end{cases}$$</span></p>
<p>which you can solve for the equations.</p>
| 42878 | Why is the acceleration of wedge and block equal like this |
2021-04-29T11:22:34.973 | <p>in short, I want to implement a PID controller that takes in distance and returns velocity.</p>
<p>Details: I am simulating robots following each other. I wish to maintain a constant distance, <code>d</code> between these robots. I can obtain the distance between the robots at any time and then I can adjust the velocity of the following robot accordingly. My problem is the formulation that achieves the transformation of the information provided by the PID to velocity so as to achieve this constant distance. My current attempt is to use the error provided by the PID as the velocity for the following robot, that is;</p>
<p><code>v2 = e(t)</code>
where <code>v2</code> is the velocity of the second/following robot and <code>e(t)</code> is the error yielded by the PID at time <code>t</code>.</p>
<p>The problem with this approach is that it vehicle velocity is only as big as the error, so that way velocity only increases if the error increases, so if the constant distance I want is relatively small. I can never achieve this.</p>
<p>I also tried using the error as the acceleration, with the idea that the velocity should only grow as error grows and decrease in a similar fashion according to:</p>
<p><code>v2 = v2 + e(t) * dt</code></p>
<p>where 'dt' is the change in time.</p>
<p>But then when I get negative errors. I wish to avoid reversing or backward motion. So we can call it a stop-go implementation if possible. But ideally when the error in the desired distance reduces ultimately both velocities will be moving with the same velocity. Any approaches and suggestions will be highly appreciated. Thanks very much. I am doing all implementation tests in python. If more clarification is needed please let me know.</p>
| |pid-control|distance-measurement| | <p>Forget PID at first. If we assume PID, we're skipping steps.</p>
<p>Write down what we know. From your description:</p>
<ul>
<li>process variable : position</li>
<li>input : (other robot's position - following distance)</li>
<li>loop output : this robot's position</li>
<li>plant input : velocity</li>
<li>plant output : this robot's position</li>
<li>plant dynamics : simple integrator</li>
</ul>
<p>Now we can draw a system diagram:</p>
<p><a href="https://i.stack.imgur.com/W2ffS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W2ffS.jpg" alt="diagram1" /></a></p>
<p>Note that we left out two very important things: limits on velocity and acceleration. But for now, let's neglect these so we can solve the inside case, when those limits are not reached. In reality, the performance experienced by the user will very much be determined by the edge cases where those limits are in effect.</p>
<p>In a typical PID, G=1 and F = the PID expression.</p>
<p>We can get both stability and zero steady state error for this system with just PI.</p>
<p>A PI controller has a form</p>
<p>F = Kp ( 1 + 1/((Ti)s) ) and G=1</p>
<p>For convenience of analysis, and let's use an alternate form. (The Kp will be different, but no loss of generality. This is still a PI controller)</p>
<p>F = (Kp/s)(1+(Tz)s) and G=1</p>
<p>This form makes the zero in the forward path clearly visible, at s=-1/Tz. The zero in the forward path will remain in the closed loop transfer function, and thus produce step response overshoot, which is not desired per the problem statement.</p>
<p>We can get the exact same loop stability by keeping the product FG constant, while eliminating the step response overshoot, (although slower response). Thus we refactor FG into:</p>
<p>F = (Kp/s) and G = (1+(Tz)s)</p>
<p>Now we have deviated from the PI controller, this is slightly different, even though it results in the same stability expression for the complete loop (denominator of closed loop transfer function).</p>
<p>As an optional last step, we can also rearrange the loop into the so-called pseudo-derivative form to get rid of the possibly-troublesome derivative, this rearrangement is shown below.</p>
<p><a href="https://i.stack.imgur.com/GLPaT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GLPaT.jpg" alt="enter image description here" /></a></p>
<p>The rest of the work is about implementing decent saturation and anti-windup logic for the large signal issues. This isn't entirely trivial, and is the real "secret sauce" in some control situations.</p>
| 42879 | object following using pid |
2021-04-30T05:45:30.020 | <p>I am an Indian, ravaged by the Covid-19 pandemic in my country. I come from a non engineering background (Physics). I was thinking about the following.</p>
<p>If one has an oxygen concentrator, i.e. a device which removes impurities such as N2 and CO2 from air, to release pure oxygen (-93%) using electricity.</p>
<p><a href="https://en.wikipedia.org/wiki/Oxygen_concentrator" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Oxygen_concentrator</a></p>
<p>It supplies oxygen at the rate of 5 liters / min. Can it be used to fill multiple empty oxygen cylinder, which are basically pressurized oxygen kept in cans. This way, if there are multiple patients suffering from Covid-19, they can be supplied oxygen using the multiple cylinders. Is there any simple and efficient way for one to fill the cylinders with oxygen, bear in mind the cylinders are having oxygen at very high pressure.</p>
| |mechanical-engineering|fluid-mechanics| | <p>Pressurizing oxygen, especially for medical use requires a purpose built compressor. Other compressors add oil that can damage the lungs and catch fire. Oil-less diaphragm compressors are better, but may be damaged by pure oxygen and are also not rated for medical use. Pressurizing oxygen is not a good DIY project. A different solution to consider is to produce oxygen on-site at low pressure.</p>
<p>A year ago when the US was being hit with the first wave of Covid, I joined the Apollo Oxygen Concentrator project of the <a href="https://helpfulengineering.org/projects/" rel="nofollow noreferrer">Helpful Engineering group</a> (lots of other good Covid projects on there too). They had designed a point-of-use oxygen concentrator. Noise and reliability are additional hurdles, but it avoids having to store the oxygen.</p>
<p><a href="https://drive.google.com/drive/folders/1AWvZsK41v0T8z20f_-QFAzTxI99lZbBv?ths=true" rel="nofollow noreferrer">Apollo Oxygen Concentrator Google Docs</a><br>
<a href="https://github.com/oxycon/ProjectApollo" rel="nofollow noreferrer">Apollo Oxygen Concentrator Software</a><br>
<a href="https://helpfulengineering.slack.com/archives/C0105E2T0UB" rel="nofollow noreferrer">Apollo Oxygen Concentrator Slack Channel</a></p>
<p>This <a href="https://oxikit.com/" rel="nofollow noreferrer">oxikit company</a> is a little more commercial, but also worth having a look at.</p>
| 42889 | Pressurizing Oxygen simply and efficiently |
2021-04-30T08:53:40.217 | <p><a href="https://i.stack.imgur.com/mw9LY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mw9LY.png" alt="enter image description here" /></a></p>
<p>What I did was I found the <span class="math-container">$I_{short circuit(AB)}$</span> = 10/5 = 2A by shorting AB.</p>
<p>How am I supposed to find <span class="math-container">$R_{AB}$</span> in order to solve for <span class="math-container">$V_{Thevenin}$</span>? It can't be 5//10 right?</p>
| |electrical-engineering| | <p>For an open circuit,Thevenin Equivalent Voltage is the Open circuit voltage.I use Node Voltage method to find the Thevenin Equivalent Voltage.
Voc = Va - 0
[(Vs - Va)/R1] + Is = (Va/R2)
Va = [Vs+(R1<em>Is)] / [1+(R1/R2)]
Va = [10+(5</em>1)] / [1+(5/10)]
Va = (10+5)/[(10+5)/10]
Va = (15/15)(10/1)
Voc = Va = 10V
Therefore, Thevenin’s Equivalent Voltage is 10V.</p>
| 42890 | Finding Thevenin Equivalent Voltage |
2021-04-30T10:38:46.197 | <p>In a robot or spacecraft attitude representation, using quaternion is relevant.
The kinematic equation is as below</p>
<p><span class="math-container">$$
\begin{bmatrix}
\dot{q}_0 \\ \dot{q}
\end{bmatrix} =
\frac{1}{2}
\begin{bmatrix}
-q^\top \\ q_0\,I_3 + q^\times
\end{bmatrix} \omega.
$$</span></p>
<p>In this equation quaternions derivative are calculated based on angular velocity <span class="math-container">$\omega$</span>. I want to calculate <span class="math-container">$\omega$</span> based on quaternions derivatives. So I should inverse the matrix multiplied to <span class="math-container">$\omega$</span> in above equation. Is there any straight forward way to inverse it analytically?</p>
| |kinematics| | <p>In order to make the notation shorter for the remaining calculations in my answer I will denote the matrix with</p>
<p><span class="math-container">$$
H = \frac{1}{2}
\begin{bmatrix}
-q^\top \\ q_0\,I_3 + q^\times
\end{bmatrix}.
$$</span></p>
<p>In this case you have four equations but only three unknowns. So there might not be a solution that exactly matches all equations, especially when the time derivative of the quaternion is perturbed, i.e. by numerical rounding errors. However, there is an elegant least squares solution that minimizes <span class="math-container">$\|\dot{\textbf{q}} - H\,\omega\|$</span>. Here I use <span class="math-container">$\textbf{q}$</span> to denote the entire quaternion <span class="math-container">$\begin{bmatrix}q_0 & q^\top\end{bmatrix}^\top$</span>. This least squares solution can shown to be</p>
<p><span class="math-container">$$
\omega = \left(H^\top H\right)^{-1} H^\top \dot{\textbf{q}}.
$$</span></p>
<p>It can be noted that if the quaternion is of unit length then the matrix, whose inverse is taken, simplifies to</p>
<p><span class="math-container">$$
\left(H^\top H\right)^{-1} = 4\,I_3.
$$</span></p>
<p>Thus the solution for the angular velocity in a reduced form can also be written as</p>
<p><span class="math-container">$$
\omega = 4\,H^\top \dot{\textbf{q}} = 2
\begin{bmatrix}
-q & q_0\,I_3 \!-\! q^\times
\end{bmatrix}
\begin{bmatrix} \dot{q}_0 \\ \dot{q}
\end{bmatrix}.
$$</span></p>
| 42892 | Inverse of quaternion kinematic matrix |
2021-04-30T14:53:39.050 | <p>I have a linear solenoid with a tapped M3 hole at the end of the plunger, like this:</p>
<p><a href="https://i.stack.imgur.com/rOkkz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rOkkz.png" alt="Linear solenoid with tapped M3 hole in plunger" /></a></p>
<p>I need to connect this plunger to a lever, so I want to add a mount which has a connection something like this:</p>
<p><a href="https://i.stack.imgur.com/YEu1v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YEu1v.png" alt="Linear solenoid with desired mount on plunger" /></a></p>
<p>Does anyone have any suggestions as to where I could find an M3 mount like this which I could attach to my solenoid?</p>
| |electromagnetism|linear-motion| | <p><a href="https://i.stack.imgur.com/0aR00.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0aR00.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. An image search for "M3 clevis male" throws up a wheelbarrow-full of images.</em></p>
<p>You could use a female thread version but would need to connect the them with a male stud and maybe some thread locking compound.</p>
| 42895 | Linear solenoid mount |
2021-05-01T06:51:28.260 | <p>I'm creating a report on the possible implementation of a sub hydro system and need some help. I've attached the image of what I'm proposing, where I've essentially created a head via a riser B with a water inlet just below sea level, which will then flow down into the hydraulic turbine and fill up the pipeline and Riser A to create electrical energy.</p>
<p>In it's initial state, the entire system is full of water. When power isn't in demand I've incorporated an ESP to then evacuate the pipeline and Riser A back out of Riser B (Riser B is still filled after). Initially I calculated the required energy needed to evacuate Riser A and the pipeline which would act as potential energy using the simple energy equation E=mgh. That potential energy would then be converted into electrical energy upon subsequent flooding of the system again.</p>
<p>I'm having trouble in determining the ideal size of the turbine though because I'm not sure how to calculate the flow rate that will be coming into Riser B and subsequently into the turbine itself since Riser B is never completely empty. I assume that I won't need that large a one due to the the large head, but I'm not sure of the appropriate governing equations or if they're even needed given my potential energy calculations.</p>
<p>I'll be testing multiple values of pipe diameter and pipeline length but original values were:</p>
<p>Riser A = 500m</p>
<p>Riser B = 50m</p>
<p>Pipeline L = 10000m</p>
<p>Pipeline, Riser A/B Diameter = 500mm</p>
<p>Any help would be appreciated, because I'm stuck in a bit of a rut finding the connection between the two as I'm using two different ideas and joining them.</p>
<p>Also for reference the ESP will be powered using excess energy from a wind turbine (not shown).</p>
<p><a href="https://i.stack.imgur.com/jGikK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jGikK.png" alt="enter image description here" /></a></p>
| |turbines|renewable-energy|flow-control|hydroelectric-dams|offshore| | <p>I can't easily help you with the flow rate calculations but I offer the following observations:</p>
<ul>
<li>This is an undersea pumped storage system.</li>
<li>The capacity of the sloped pipe is <span class="math-container">$\pi r^2 l = \pi\cdot 0.25 \cdot 10000 = 1963 \ \mathrm {m^3} $</span>. It will hold 2001 T of sea water.</li>
<li>The empty pipe anchors will have to counter a buoyancy of 2000 T.</li>
<li>The pressure on the empty pipe will be 50 bar at 500 m.</li>
<li>Riser A has a capacity of 500/10000 of that, so 0.5% and I'll ignore it in the calculations.</li>
<li>You've put the hydraulic turbine at the top of the pipe. Your working head, therefore is 50 m no matter how deep the second riser is. The 10 km of pipe and 500 m depth buys you nothing other than storage and pain. You would get the same benefit from a 2000 m<sub>3</sub> low-profile storage at the bottom of your 50 m Riser B. These would have to withstand 5 bar external pressure when empty so I'm picturing cylindrical tanks with domed tops and bottoms - maybe 1000 L (1 T) capacity each. Each would need a water inlet/outlet at the bottom manifolded together and an air inlet/outlet at the top also manifolded and connected to the surface by a 50 m Riser A.</li>
<li>Most land-based pumped storage systems use the turbine as a pump to return the water to the reservoir. Your system has a second pump for this. This is duplication of equipment with associated cost.</li>
<li>Your evacuation pump is only 50 m deep. You can't evacuate the sloped pipe. To evacuate the sloped pipe the ESP would have to be at the bottom of Riser A.</li>
<li>The energy available to your turbine is <span class="math-container">$mgh = 2001000 \times 9.81 \times 50 = 491 \ \mathrm{MJ} = \frac {491\text{M}}{3600} = 136 \ \mathrm {kWh}$</span>. If you can get 10c per kWh that's worth 13.60 in your favorite currency.</li>
<li>To pump the water out from 500 m will also be determined by <span class="math-container">$mgh$</span> and this will work out to ten times the 50 m drop, <span class="math-container">$1360 \ \text{kWh}$</span>. Cost 136.00 of your favorite currency - and that assumes that you can purchase off-peak at the same cost as your feed-in tarriff.</li>
</ul>
<p>I'm not investing in your company for now!</p>
| 42906 | I want to determine the flow rate and required turbine size for an offshore hydroelectric scheme and need help |
2021-05-02T02:27:25.323 | <p>I'm making metal frame to support a wooden deck.</p>
<p>The yellow object is the concrete base where there is a 2" trough where my metal frame needs to support a 300-500lbs deck load.</p>
<p>The span between the unsupported trough is 15'. The length of the cross brace is 22"</p>
<p>The red frame is 2" welded steel tubing. Wall thickness 1/8".</p>
<p>I understand 2" will have huge deflection at 15' unsupported span but will adding cross beam like this reduce this deflection to be acceptable to support a ground-level deck?</p>
<p><a href="https://i.stack.imgur.com/FCFGX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FCFGX.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Pjt0p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pjt0p.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/51EtM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/51EtM.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/bxSc2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bxSc2.png" alt="enter image description here" /></a></p>
| |steel| | <p>For HSS <span class="math-container">$2"$</span> x <span class="math-container">$2"$</span> x <span class="math-container">$1/8"$</span>, <span class="math-container">$E = 29000 ksi, Ix = 0.486 in^4$</span></p>
<p>Assume uniformly distributed load. Let <span class="math-container">$w = 500 lbs/15 ft = 33.33 plf$</span>, round up to <span class="math-container">$40 plf$</span> to include misc. weight.</p>
<p><span class="math-container">$\Delta = 5*w*L^4/384*E*I = (5*40*15^4*12^3)/(2*384*29*10^6*0.486) = 1.616"$</span></p>
<p><span class="math-container">$\Delta \approx L/111, NG(*)$</span></p>
<p><strong>(*) Range of usual deflection limits:</strong></p>
<ul>
<li><p>Live Load: <span class="math-container">$L/180 - L/360$</span></p>
</li>
<li><p>Total Load: <span class="math-container">$L/120 - L/240$</span></p>
</li>
</ul>
<p>As noted above, depending on the nature of the given load, the deflection limit varies.
Since the deflection of your beam has exceeded the lowest limit, you need to determine a deflection limit according to the type of load, and that is suitable for your application.</p>
<p>(Note: Adding the cross bar does not add beam stiffness. You need a stronger beam)</p>
<p>As an example, assume <span class="math-container">$L/180$</span> governs, let's see what size of beam will satisfy it:</p>
<p><span class="math-container">$I_{new} = I_{old}*(L/111)/(L/180) = 0.486*(180/111) = 0.788 in^4$</span></p>
<ul>
<li>HSS <span class="math-container">$2"$</span> x <span class="math-container">$2"$</span> x <span class="math-container">$5/16", I = 0.815in^4, OK$</span></li>
</ul>
<p>However, as the new tube is almost twice as heavy as the original tube, you need to re-check the deflection with the modified load.</p>
<p>Also, don't forget to check the flexural stress: <span class="math-container">$f_b = M/S <= 0.66fy$</span>,</p>
<p>where <span class="math-container">$M = w*L^2/8$</span>.</p>
| 42922 | Deflection of 2" steel tubing with braces across 15' unsupported span? |
2021-05-02T16:08:17.810 | <p>I recently noticed that the cylindrical shape of a railroad tank car is not completely straight but has a bend in the middle. The entire tank is a bit lower there. What's the reason for this bend?</p>
<p>This is also visible on the drawing of a <a href="https://en.wikipedia.org/wiki/DOT-117_tank_car" rel="noreferrer">DOT 117</a> tank car on Wikipedia (the red nearly-horizontal lines were added by myself).</p>
<p>I first thought it had something to do with pressure containment, but the model 117 is apparently used for non-pressured goods only.</p>
<p><a href="https://i.stack.imgur.com/cvhkP.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/cvhkP.jpg" alt="drawing of a dot 117 railroad car" /></a></p>
| |mechanical-engineering|design|rail| | <p>Same reason the Corsair fighter has bent wings. Makes clearances easier.</p>
<p>Look at the lumpy, sticky-outy things on a tank car. You have the <strong>trucks/bogeys</strong> at the end bottom... and the massive <strong>fill valve and work safety area</strong> at center top. Lowering the center is advantageous to the car's clearance envelope, allowing it to have more tank volume in the same clearance plate (loading gauge for Brits, not that it would fit in any British tunnel save for the Eurostar).</p>
<p>For the last 100 years, railroads have been careening toward heavier and heavier cars, while many older system tunnels do not have high clearances. (Lines commercially viable for double-stack container service are being notched for double-stacks).</p>
<hr />
<p>Assures head space too. While the bend eases unloading, as NMech covers amply, <strong>it also prohibits entirely filling</strong>; there is an airspace above the load that simply cannot be filled, short of drawing a vacuum on the tank (which the Mythbusters showed is a bad idea).</p>
<p>Loads have different volumes at different temperatures. Notice the steam-heating apparatus in the cutaway on the drawing. That means the car is intended for loads that may be too thick to flow without heating. That material's volume will increase while you are heating it, or simply the inevitable effect of 16 hours in the Arizona sun on a car painted flat-black. You do not want the car to "hydraulic lock"; that would send pressure up very rapidly and crack the tank.</p>
<p>That "unfillable" volume assures there is head space for the liquid to expand into; of course this compresses the gas there somewhat, increasing pressure on the whole vessel; but nothing it can't handle.</p>
| 42931 | Why are railroad tank cars bent in the middle? |
2021-05-03T00:01:15.527 | <p>If I have an object of 90kg, moving along the ground with motors attached to wheels, how would I calculate how much torque, rpm and power those motors would need to have to accelerate the object to 20 mph in 10 seconds? If you have any resources like videos or readings, or just an explanation of how to do it, I would appreciate it. I realize this may be a duplicate question, but I cannot find something that explains this simply. I am a novice to this sort of thing and through all of my googling I just can't figure it out.</p>
| |automotive-engineering| | <p>Assuming that:</p>
<ul>
<li>there is no rolling resistance</li>
<li>there is no air resistance</li>
<li>the acceleration is constant during the 10[s]</li>
</ul>
<p>Then the total force between the wheels the ground would be equal to :</p>
<p><span class="math-container">$$\sum F = m \cdot a $$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$a=\frac{20 [mph]}{10 [s]}= 0.891 [m/s^2]$</span></li>
<li><span class="math-container">$m=90 [kg]$</span>
Therefore, <span class="math-container">$F= 80.47 [N]$</span></li>
</ul>
<p>Then you can calculate the properties you are after can be easily calculated as:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">quantity</th>
<th style="text-align: center;">formula</th>
<th style="text-align: center;">Comments</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">rpm</td>
<td style="text-align: center;"><span class="math-container">$n = \frac{60}{2\cdot \pi \cdot r_{wheel} }u $</span></td>
<td style="text-align: center;"></td>
</tr>
<tr>
<td style="text-align: center;">Torque</td>
<td style="text-align: center;"><span class="math-container">$F\cdot r_{wheel}$</span></td>
<td style="text-align: center;"></td>
</tr>
<tr>
<td style="text-align: center;">Power</td>
<td style="text-align: center;"><span class="math-container">$P = F\cdot u $</span></td>
<td style="text-align: center;">~720[W]</td>
</tr>
</tbody>
</table>
</div>
<p>The rpm and the torque are depended by the Radius of the wheel (<span class="math-container">$r_{wheel}$</span></p>
| 42943 | Finding the Torque, RPM and power required to move an object |
2021-05-03T00:30:00.660 | <p>I am having difficulty with a CSWA question that asks to find the mass of the following part. Here is a link to my attempt: <a href="https://drive.google.com/file/d/162No5ugwQ-9cRkdcp-78Ohq08T8UngBd/view?usp=sharing" rel="nofollow noreferrer">https://drive.google.com/file/d/162No5ugwQ-9cRkdcp-78Ohq08T8UngBd/view?usp=sharing</a></p>
<p><a href="https://i.stack.imgur.com/C03F4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C03F4.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/vD6Q1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vD6Q1.png" alt="enter image description here" /></a></p>
<p>Can anyone identify what I did wrong? I selected answer choice A which is closest to the mass of my model but that is incorrect. Any help would be much appreciated.</p>
| |solidworks| | <p>OK. So, there's actually a few problems.</p>
<ol>
<li>Per my comment, you selected the wrong material, which has the wrong density.</li>
<li>I quickly modelled the part myself so I could do a part compare to see how your geometry differed. The "Red bit" is your part sticking out of the correct geometry. You can see that the position of the top isn't in the correct place.</li>
</ol>
<p><a href="https://i.stack.imgur.com/YzjBQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YzjBQ.png" alt="enter image description here" /></a></p>
<ol start="3">
<li>I then opened your sketch to see how you had got this wrong - and was met with a fully 'blue' sketch. Your sketches should <em>always</em> be fully defiled (black). You simply had not dimensioned the position of the upper circle in the horizontal direction at all - the orange dimension shown on the image is one <em>I added</em> to show that it's not 30mm per the drawing.</li>
</ol>
<p><a href="https://i.stack.imgur.com/Ay8gd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ay8gd.png" alt="enter image description here" /></a></p>
<ol start="4">
<li>When I added this missing dimension, and set it to 30mm, the model fell over/broke, - your whole model has been built in an unstable way using odd tools (boss extrude instead of fillet on the R100s, for example?</li>
</ol>
<ol start="5">
<li>I've attached an example model here: <a href="http://www.filedropper.com/cswaexample" rel="nofollow noreferrer">http://www.filedropper.com/cswaexample</a> I'm happy to help explain some best practice and tutor you through this model to give some transferrable skills for future CSWA tests, but, StackExchange isn't the best forum for this. Try the SOLIDWORKS Discord?</li>
</ol>
| 42945 | Solidworks CSWA part modelling question |
2021-05-03T09:05:39.073 | <p>Picture of a High Voltage Direct Curent powerline (HVDC line) on the Læsø Island, Denmark (from Google Street View):
<a href="https://i.stack.imgur.com/57uOQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/57uOQ.jpg" alt="enter image description here" /></a></p>
<p>We can observe that there are 2 conductor cables, and that both are insulated from the tower (which is grounded). Because the line is in DC it makes sense to have 2 cables.</p>
<p>However, I see no reason why one of those conductors would not be grounded. For example on railway electrification, one of the conductors (the overhead line or 3rd rail) is always isolated and the other of the conductors is the rail track itself and is grounded.</p>
<p>Grounding one of the conductors on the HVDC line might save money on isolators, not to mention electric and safety issues if the HVDC line as a whole recieves a too large voltage offset compared to ground.</p>
| |electrical-engineering|power-transmission|electrical-grid| | <h2>Because they're both live</h2>
<p>Just because you see multiple wires doesn't mean they are carrying different things. They are frequently paralleled for more ampacity.</p>
<p>It's perfectly common in HVDC lines to use earth as the return path. For instance, the <a href="https://en.wikipedia.org/wiki/Pacific_DC_Intertie" rel="nofollow noreferrer">Pacific Intertie</a> does just that, with absolutely massive grounding electrodes at each end.</p>
<blockquote>
<p>The grounding system at Celilo consists of 1,067 cast iron anodes buried in a two-foot (60 cm) trench of petroleum coke, which behaves as an electrode, arranged in a ring of 2.0 miles (3,255 m) circumference at Rice Flats (near Rice, Oregon).</p>
<p>The Sylmar grounding system is a line of 24 silicon-iron alloy electrodes submerged in the Pacific Ocean at Will Rogers State Beach[5] suspended in concrete enclosures about 2 to 3 feet (0.5 to 1 m) above the ocean floor.</p>
</blockquote>
<p>In that case, the multiple wires on the pole are all live, typically configured as redundant paths so the route is not entirely downed if one wire takes a hit.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Konti%E2%80%93Skan" rel="nofollow noreferrer">Konti-Skan HVDC line</a> you are talking about does exactly the same thing, although it varies, using a number of methods. In some places it does carry the ground wire on the towers, but on top and insulated from the towers.</p>
<blockquote>
<p>Before the implementation of Konti–Skan 2 both conductors were switched parallel, today one conductor is used for the high voltage pole of Konti–Skan 1, while the other is used by that of Konti–Skan 2.</p>
</blockquote>
<p>The multiple conductors are either redundant or have different destinations.</p>
| 42951 | Why are both conductors in a High Voltage Direct Curent powerline isolated from ground? |
2021-05-03T11:36:14.627 | <p><a href="https://i.stack.imgur.com/hRdHZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hRdHZ.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/KBfKJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KBfKJ.jpg" alt="enter image description here" /></a>
Using any method find the constraint relation between mass 1 and mass 2.</p>
<p>Assumptions: No friction , massless string and pulley.</p>
<p>Q: Find constraint relation between 1kg and 2kg block.</p>
<hr />
<h2>what I tried so far</h2>
<p>So, I used a method in which I consider the length of string. Length of red string is <span class="math-container">$2x_1$</span>. Now , we can notice that one string is long joining 1kg and other joining the pulley is short. But since the assumptions are there. We can consider it like this. If you need a proof , then let me know. I shall give an example too in edit afterwards.</p>
<p>So , then length of string for 2nd pulley is <span class="math-container">$2x_p$</span>.</p>
<p>I assume that the Length of string for <span class="math-container">$2kg $</span>block is <span class="math-container">$2x_2$</span>. However, I feel this is wrong since the same string is connected to 1kg block as well.</p>
<p>So, I get is <span class="math-container">$2x_1+2x_2+2x_p $</span>but my answer is coming wrong.</p>
<p>So, it would be helpful if you can help me just correct where my mistake is or share another method if you’d like that.</p>
<h2>another simpler example</h2>
<p>EDIT: an example <a href="https://i.stack.imgur.com/J35pL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J35pL.jpg" alt="enter image description here" /></a></p>
<p>Here , <span class="math-container">$a_1 , a_2 , a_3 $</span> are accelerations of <span class="math-container">$1kg , 2kg , 3kg $</span>block respectively.</p>
| |mechanical-engineering|applied-mechanics|dynamics| | <p>Using the following coordinates:</p>
<p><a href="https://i.stack.imgur.com/qzjGu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qzjGu.png" alt="enter image description here" /></a></p>
<p>i.e.: using pulley one (top left as the reference point:</p>
<ul>
<li><span class="math-container">$x_1$</span> the distance of mass 1 from pulley 1</li>
<li><span class="math-container">$x_2$</span> the distance of mass 2 from pulley 1</li>
<li><span class="math-container">$P_2$</span> the distance of pulley 2 from pulley 1</li>
<li><span class="math-container">$P_3$</span> the distance of pulley 3 from pulley 1</li>
</ul>
<p>then the equations are the following:</p>
<ul>
<li>for rope wrapped around pulley 1
<span class="math-container">$$L_1 = x_1 + P_2$$</span></li>
<li>for rope wrapped around pulley 2
<span class="math-container">$$L_2 = (x_1 - P_2) + (P_3-P_2)$$</span></li>
<li>for rope wrapped around pulley 3
<span class="math-container">$$ L_3 = (x_1 - P_3) + (X_2 - P_3)$$</span></li>
</ul>
<p>From those relationships you get:</p>
<p><span class="math-container">$$\begin{cases}
P_2 = L_1 - x_1\\
P_3 = L_2 - x_1 +2 P_2 \\
L_3 = x_1 + x_2 - 2 P_3
\end{cases} \rightarrow
\begin{cases}
P_2 = L_1 - x_1\\
P_3 = L_2 - x_1 +2 (L_1 - x_1) \\
L_3 = x_1 + x_2 - 2 P_3
\end{cases} \Rightarrow
\begin{cases}
P_2 = L_1 - x_1\\
P_3 = L_2 +2 L_1 - 3 x_1 \\
L_3 = x_1 + x_2 - 2 (L_2 +2 L_1 - 3 x_1)
\end{cases}
$$</span>
The final equation is what you want:</p>
<p><span class="math-container">$$L_3 = x_1 + x_2 - 2 (L_2 +2 L_1 - 3 x_1)$$</span>
<span class="math-container">$$L_3 = x_1 + x_2 - 2 L_2 -4 L_1 + 6 x_1$$</span>
<span class="math-container">$$L_3 + 2 L_2 + 4 L_1 = 7x_1 + x_2 $$</span></p>
<p>By differentiating with respect to time twice you get
<span class="math-container">$$0 = 7\ddot{x}_1 + \ddot{x}_2 $$</span>
<span class="math-container">$$0 = 7a_1 + a_2 $$</span></p>
<hr />
<p>Additionally there should be a way to obtain this by the energies. Probably by obtaining the equations of motion through langrange's method.</p>
| 42954 | Find a constraint relation for this Q of 3 pulleys and 2 blocks attached with string |
2021-05-04T04:46:33.447 | <p>I am welding up a rack to support a distributed load of up to 1000kg, what is the process to work out what steel section size I should use? Many Thanks.</p>
| |structural-engineering|steel| | <p>Because the weight is quite high, and the use is for supports, I will assume that you will use the same cross-section for the shelve columns (if there are any). In you use the same cross-section, you probably will not have a problem with buckling</p>
<p>I am assuming RHS refers to Rectangular Hollow Cross-section.</p>
<p>Regarding the process that you need to follow IMHO for the shelves is:</p>
<ol>
<li><p>Define the allowable deflection <span class="math-container">$\delta_{all}$</span> (usually its about <span class="math-container">$\frac{1}{200}L_{\text{shelves span}}$</span> and the allowable stress (usually its safety factor times the yield stress <span class="math-container">$N\cdot \sigma_y$</span>)</p>
</li>
<li><p>Determine the support type. Are the shelves simply supported or are they welded on the frame. In any case I'll be using the formulas for simply supported beams.</p>
</li>
</ol>
<ul>
<li>max deflection: <span class="math-container">$\delta_{max} = \frac{5 wl^4}{384EI}$</span></li>
<li>max operating stress: <span class="math-container">$\sigma = \frac{wl^2}{8I}\cdot \frac{y_{max}}{2}$</span></li>
</ul>
<p>where:</p>
<ul>
<li><span class="math-container">$E$</span> is the young's Modulus</li>
<li><span class="math-container">$I$</span> is the second moment of area which you can find in tables or calculate</li>
<li><span class="math-container">$y_{max}$</span> is the height of the RHS.</li>
</ul>
<ol start="3">
<li>you can solve for I (which is depended on the geometric properties of the cross-section), for both constraints (Deflection and stress) and you will obtain two parameters:</li>
</ol>
<p><span class="math-container">$$I_{Deflection} = \frac{5 wl^4}{384E\delta_{max}}, \qquad \frac{2 I}{y_{max}}= \frac{wl^2}{8\sigma_{Allowed}}
$$</span></p>
<hr />
<p><strong>UPDATE:</strong></p>
<p>The equations for the fixed ends (see welded shelves on the frame) are:</p>
<ul>
<li>max deflection: <span class="math-container">$\delta_{max} = \frac{wl^4}{384EI}$</span></li>
<li>max operating stress: <span class="math-container">$\sigma = \frac{wl^2}{12 I}\cdot \frac{y_{max}}{2}$</span></li>
</ul>
<p>As you can see they are less stringent (that is why I opted for the simple beam), because that would built in a safety factor.</p>
| 42968 | How to calculate what size steel RHS to use? |
2021-05-04T10:49:58.083 | <p><a href="https://i.stack.imgur.com/Mm69Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mm69Y.png" alt="enter image description here" /></a></p>
<ol>
<li>The input signal is limited and lies in range <span class="math-container">$[-U;+U]$</span>. <span class="math-container">$U$</span> - unknown.</li>
<li>The output signal must be rescaled to be in the range <span class="math-container">$[-1;+1]$</span>, i.e. is <span class="math-container">$+1$</span> if the input is at steady-state <span class="math-container">$+U$</span>, and <span class="math-container">$-1$</span> if the input at steady-state <span class="math-container">$-U$</span>.</li>
<li>Between <span class="math-container">$-1$</span> and <span class="math-container">$+1$</span> the output signal must also occupy some rescaled value.</li>
</ol>
<p>What can be used as a block <span class="math-container">$ ??? $</span> for such a conversion? Is there a linear / non-linear filter that does this, and such and which is described by the differential equation?</p>
<p><em>Remarks:</em>
<em>Red line, just my fantasy about how the input signal changes</em></p>
| |control-engineering|control-theory|dynamics|nonlinear-control| | <ul>
<li>Input signal: <em>±U</em> V.</li>
<li>Scaled output signal: <em>±1</em> V.</li>
<li>Required gain: <span class="math-container">$ \frac 1 U $</span>.</li>
</ul>
<p><img src="https://i.stack.imgur.com/v9xJs.png" alt="schematic" /></p>
<p><em>Figure 1. Possible solutions.</em></p>
<ul>
<li><p>If <em>U</em> > 1 then a simple potential divider may suffice. <span class="math-container">$ V_{OUT} = \frac {R_2}{R_1+R_2} $</span>. Bear in mind that whatever follows this circuit may load it somewhat so keep the parallel combination value of R1 || R2 < 10% of the value of the following stage.</p>
</li>
<li><p>If <em>U</em> < 1 then a non-inverting amplifier is required. Gain is given by <span class="math-container">$ A = 1 + \frac {R_4}{R_3 + R_4} $</span>.</p>
</li>
</ul>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Question</th>
<th>Response</th>
</tr>
</thead>
<tbody>
<tr>
<td>Is there a linear / non-linear filter that does this, ...</td>
<td>A filter is for modifying a signal's frequency content. This is not what you are trying to do so a filter is not an appropriate solution.</td>
</tr>
<tr>
<td>... and such and which is described by the differential equation?</td>
<td>You are looking for a simple attenuation or amplification function. There is no need for differential equations.</td>
</tr>
</tbody>
</table>
</div> | 42976 | Input signal rescaling block |
2021-05-05T06:17:11.577 | <p>Ever since I was a little boy, I've always had this idea that electricity can only go through metal materials. For example, if I put a 100% metal screwdriver into the socket in the wall, I'm going to get electrocuted. Logical. Or if I take a long wire of copper (or even gold) and hold it with my hands and then stick the far end into a powered-on toaster, I'm going to get zapped.</p>
<p>Also ever since I was a child, I've had this screwdriver which (of course) has a metal "main part" (for the actual screwing), but whose "hand part" is made of what looks like glass or plastic. It's transparent. And there's some kind of little "spring thingie" inside it near the top. This has always confused me. I think at some point, an adult told me that it's a special screwdriver which is safe to use when there's live power. But is it?</p>
<p>Why is that "spring thingie" necessary at all? Isn't simply the glass/plastic cover around the metal enough to insulate my hands from the electricity that only travels through metal? I don't understand why there would be a need for some "special" kind of screwdriver with a spring inside it to protect me from electricity when it doesn't travel through non-metal materials anyway.</p>
<p>And even if this is such a special screwdriver, how can I know that it actually still works after all these years? How could I trust it even if it were brand new? What is it about that spring that somehow makes it safe to use it on live power?</p>
<p>I vaguely remember that classic story about the famous scientist who went out with a kite in a thunderstorm and got zapped by lightning hitting the kite, then traveling through the <em><strong>non-metal string</strong></em> to his hands. But did that really happen? Or is that a major misunderstanding on my part? Or did he somehow use some kind of metal wire (unlikely)? This story seems to go completely against my understanding of electricity, in that it apparently <em>does</em> travel through non-metal materials.</p>
| |electrical-engineering|power|power-electronics|electrical|safety| | <p>Your question has several sub-questions. I'll try to address them to the best of my knowledge (there are other people more knowledgeable than me on the subject).</p>
<hr />
<h2>Spring Thingie</h2>
<p>Regarding the spring thingie, I think you are referring to this:</p>
<p><a href="https://i.stack.imgur.com/RIiXE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RIiXE.png" alt="enter image description here" /></a>
<em>Figure 1: mains Tester screwdriver (source: <a href="https://gr.rsdelivers.com/product/rs-pro/kers-mainst/rs-pro-mains-tester-screwdriver-100-250-v-ac/6226781?cm_mmc=GR-PLA-DS3A-_-google-_-PLA_GR_EN_Hand_Tools_Whoop-_-(GR:Whoop!)+Mains+Test+Screwdrivers-_-&matchtype=&pla-327948832682&s_kwcid=AL!7457!3!513863661756!!!g!327948832682!&gclid=Cj0KCQjw4cOEBhDMARIsAA3XDRieZaA0Eey1j6dKujoohQ0HyvbnbTOSH7n-DdLMee6i8O7OklZ3KRMaArOIEALw_wcB&gclsrc=aw.ds" rel="nofollow noreferrer">RS website</a></em>)</p>
<p>The idea is that you can use this to test if a socket has a live wire in it (or which is the live wire). If you press this on a live wire, the idea is that the lamp (usually a neon) will illuminate and you will know that there is voltage difference.</p>
<p>By itself, the spring (or in general this type of screwdriver) is not safer than the other types of screwdrivers with a solid insulator handle. More specifically, the mains tester screwdrivers, will be able to conduct electricity at lower voltages, compared to the other type (see below )</p>
<p><a href="https://i.stack.imgur.com/t6odG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t6odG.png" alt="enter image description here" /></a></p>
<p><em>Figure 2: electrician screwdriver with solid insulator handle (source: <a href="https://rads.stackoverflow.com/amzn/click/com/B001A2T6ES" rel="nofollow noreferrer" rel="nofollow noreferrer">amazon.com</a>)</em></p>
<hr />
<p>Regarding electricity travelling through non plastic materials.</p>
<p>First of all, there can be voltage difference between any material. However, each material has a different resistance to allowing electrons travelling through it.</p>
<p>That property is <em><strong>Dielectric strength</strong></em> and its measured in <span class="math-container">$\dfrac{V}{m}$</span>.
The higher the value of Dielectric strength, the higher is the required voltage across two surfaces to allow electrons to travel (essentially create an arc). So for conductors the value is really small, while for insulators it increases. So (at least to my knowledge, I am not an expert like Transistor):</p>
<blockquote>
<p><strong>current can travel through any material, if the voltage differential is high enough</strong>.</p>
</blockquote>
<p>So coming back to the example of Benjamin Franklin with the thunderstorm, a typical lightning flash is in the order of about a few hundred million Volts. This is a very high voltage differential, and as a result, you'd either require a thin but very good insulator, or a thicker but not as good insulator.</p>
| 42988 | Does electricity go through non-metal materials or not? |
2021-05-06T06:58:20.970 | <p>Some weeks ago I downloaded a couple of 3d models from Thingieverse in STL format and I wanted to make some minor modifications before printing. The modification were:</p>
<ul>
<li>I wanted to split a multi part STL files into separate models (I had trouble with printing all of them together)</li>
<li>I wanted to move a few millimeters a feature to improve (IMHO) the load carrying capacity of the model.</li>
</ul>
<p>However, upon importing in in Solidworks I realized that I was unable to edit by applying new features based on the existing STL. (From what I recall I was not even able to remove some of the parts with a cut extrude feature). I did a little search and realised that generally STL files are not editable. (In the end, I used blender to edit the part, and I made the other part from scratch).</p>
<p>So my (main) question is:</p>
<blockquote>
<p>Is there a procedure I can follow so that I can convert an STL file to a native sldprt file (FeatureWorks has a similar functionality but I couldn't apply it)</p>
</blockquote>
<p>Additionally, I have the following bonus questions ( I can create separate post if you feel the question needs more focus) :</p>
<ol>
<li>Is the non editability of the STL a feature meant to protect intellectual property (similar to a PDF)?</li>
<li>What is the concept of STL in layman's terms? Does it store, vertices/edges/faces/volumes? (I got the impression that the STL build the model by creating basic tetrahedra)</li>
</ol>
| |design|solidworks| | <p>Let's look at your bonus questions first:</p>
<ol>
<li><p>No. Intellectual property isn't relevant, and STLs are definitely "editable" - I use Meshmixer to edit them directly. It <em>is</em> meant to make a file safe to open on pretty much any device with predictable results, similar to a PDF.</p>
</li>
<li><p>STL's store the vertices.</p>
</li>
</ol>
<p>SOLIDWORKS creates models using what's known as "Boundary Representation" or BREP, and creates a 'mathematically perfect' shape (to within floating point calculation tolerances). You will notice that a cylinder created in SOLIDWORKS is a single smooth surface, where a cylinder saved as an STL is made up of numerous flat polygons, which bridge the gaps between the vertices. The size of these facets (and so the tradeoff between STL complexity and accuracy to the ideal geometry) is determined during export from BREP to Mesh.</p>
<p>Mesh support is pretty new in SOLIDWORKS, it's improved hugely over the more recent software versions, but still has a way to go. The information below is based on SW2020.</p>
<p>When you drag/drop an STL into SOLIDWORKS, by default, it imports as a graphic body. Your first step should be: INSERT -> FEATURES -> CONVERT TO MESH BODY. Select everything, hit OK, and then delete all graphic bodies from the feature tree, leaving only "Imported" bodies.</p>
<ul>
<li>Splitting multi-part STLs is usually handled directly in the slicing software. You can simply delete the "Imported" bodies that you don't want, or use a "Delete/Keep Bodies" feature to do so non-destructively if you prefer.</li>
<li>Moving a feature is a little trickier, and may be best handled, again, with mesh-specific software such as Meshmixer or Blender, depending on your specific model. That said, it <em>is</em> possible to edit Mesh bodies with SOLIDWORKS. There are two main ways you can interact with Mesh bodies. These are with <em>surfaces</em> (e.g. Surface Cut, Split, etc.), or <em>with another mesh body</em>, using the "Combine" tool.</li>
</ul>
<p>So, to Add a feature to your mesh, you can model said lump as normal, use "Convert to Mesh Body" to create a mesh, with whatever resolution you require, and then use "Combine" and "Add" to join these together.</p>
<p>To Cut a feature from your mesh, it's best to use the same process with the "Subtract" option, as this gives you control over the resolution. Alternatively, if the cutting tool sticks out of the original mesh, you can delete one face of the solid to make a surface, and use surface cut.</p>
<p>To <em>move</em> a feature, therefore, your would need use a surface or plane with the "split" tool to separate the feature from the main body, use "move/copy bodies" to place it in your desired location, and then use "combine" to reattach it when you are done.</p>
<p>With a native SOLIDWORKS file, you can use features like "move face", and the other faces around the feature being moved are able to 'grow' to fill in any gaps. STL files simply do not contain the data of where their faces are in the same way, and cannot afford this functionality.</p>
<p>In a Mesh editing specific software, you would be able to select the STL faces/vertices directly and move these around, but this would likely leave deformed polygons either side of the moving feature - which requires additional steps to fix anyway!</p>
<p>EDIT:
I never answered your <em>actual</em> main question - can you convert an .STL to a native .SLDPRT, allowing you to interact using all tools?
<strong>Yes and no.</strong></p>
<p>You can convert the mesh to a solid BREP made of lots of flat surfaces. This takes a <em>very</em> long time, and is not recommended except on the simplest of models.</p>
<p>You can also use the Scan-To-3D Add in to assist on re-modelling a mesh part. You still need to model the part 'from scratch', but sketch generation is considerably sped up. This is a whole 'nother can of worms, and not really intended for 'Thingiverse modifications" <a href="http://help.solidworks.com/2020/english/SolidWorks/scanto3d/c_Scanto3d_overview.htm" rel="nofollow noreferrer">http://help.solidworks.com/2020/english/SolidWorks/scanto3d/c_Scanto3d_overview.htm</a></p>
| 43014 | On modifying imported STL files in Solidworks |
2021-05-06T08:35:30.210 | <p>I am being asked to compare a fan which is going obsolete with its proposed replacement. From a mechanical and electrical point of view, they are almost completely identical so I have no concerns from that point of view. However, when I look at their respective performance curves, they are somewhat different, especially in the low air flow region. This is outside of my area of competence, and although I have done some reading about it, I am not sure what to make of it. Can somebody explain to me which one is best and why (I have purposefully not put any legend)?</p>
<p><a href="https://i.stack.imgur.com/Gw6Tn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gw6Tn.png" alt="enter image description here" /></a></p>
| |pressure|airflow|fan| | <p>The yellow curve corresponds to a fan that will provide mostly a set pressure differential, say, keeping fumes from escaping a tank through elsewhere than dedicated vents. If there is a circumstance that reduces the pressure differential, it will speed up and increase the flow, to compensate.</p>
<p>The blue curve is a fan that doesn't expect any significant pressure differential and will provide a constant air flow, e.g. for cooling a machine or providing air circulation through a room, maintaining near-constant flow regardless of factors that may restrict the pressure - likely higher RPM would result in extra noise, so it peaks at a certain threshold and doesn't attempt to pump more air than necessary.</p>
| 43015 | Fan performance curve |
2021-05-06T12:59:43.333 | <p>Say you are trying to measure the liquid flow rate (mass or volume) through a pipe and you want to get the best measurement possible, and you have two flow meters. In one scenario, you put one flow meter and leave the other out as a back up. In the other scenario, you split the flow up into two pipes, put a flow meter on each, and then join them back up later. In the final scenario, you hook them up in series on one pipe only.</p>
<p>My instinct is that using two measuring devices instead of one is always better, and I'm fairly sure that the series case is the best scenario. However, what happens if you can't put them in series? Is it better to just have one, or to split the flow up if you can?</p>
<p>For the purposes of these scenarios, I'm always calibrating the whole unit end to end, so I would always calibrate both meters as one.</p>
| |fluid-mechanics|pipelines|fluid|instrumentation| | <p>In series would usually be best, with two caveats:</p>
<ul>
<li>the setup has to be such that the upstream meter does not affect the downstream meter, either systematically or by adding noise.</li>
<li>if the total flow is near the full scale of the sensor, and the sensor type is such that there is a "sweet spot" in accuracy or noise over the range</li>
</ul>
<p>The first of the above is a particular issue I've seen with thermal mass flow meters at low flow rates. I.e. the upstream sensor heats the flow a little, throwing off the downstream sensor. I suspect coriolis and ultrasonic technologies can also be affected.</p>
<p>To get best measurement, the thing to do is make sure flow for all sensors is fully developed, free from vorticity, free from temperature gradients (both axial and radial), and everything is at thermal equilibrium. Probably these are the most important.</p>
<p>For flows that are rising and falling in time, different sensor technologies can produce distortions in the frequency domain. A combination of technologies (e.g. differential pressure and thermal) could improve overall performance in such dynamic situations (e.g. dispensing), but it needs careful setup again. The idea is one of the technologies provides low frequency accuracy, and the other can inherently capture the mid band dynamics with less distortion.</p>
<p>For calibration, I would prefer a mass measurement.</p>
| 43019 | Question on fluid flow measurement uncertainty |
2021-05-07T06:34:11.743 | <p>Can anyone help inform a discussion I'm having with fellow modellers on a model railway forum?</p>
<p>I have a locomotive powered by an electric motor turning a 20mm diameter worm gear meshing with a spur gear. The loco performs badly, it lacks power. I am planning to exchange the worm gear for one which is 6.5mm diameter as I'm convinced this will increase the amount of force the motor can exert on the spur gear.</p>
<p>Some on the forum say it will make no difference - one rotation of the worm will move the spur gear forward by one tooth and the effort needed is the same regardless of the diameter of the worm gear.</p>
<p>I've tried supporting my argument with rudimentary physics but I'm confused as to whether i use formulae associated with rotational inertia, torque or centrifugal force. None seems to be directly applicable.</p>
<p>Assuming all other factors are constant, how do I calculate the force needed to turn the two sizes of worm gear?</p>
| |mechanical-engineering|gears|torque|moments| | <p>Even though the spur gear tooth is moved by one rotation, there is a force associated with it which is related to the friction (or in general the load on the spur gear).</p>
<p>To simplify things, let's assume that the system is in steady state (i.e. no angular acceleration). In that case you don't need to worry about inertial effects.</p>
<p>So, lets assume that the torque needed on the spur gear axis is <span class="math-container">$M_s$</span>. Then the tangential force on the spur gear (which is causing rotation) will be equal to:</p>
<p><span class="math-container">$$F_{s,t} = \frac{M_s}{R_s}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$F_{s,t}$</span> is the Tangential component on the spur gear</li>
<li><span class="math-container">$M_{s}$</span> is the required torque</li>
<li><span class="math-container">$R_{s}$</span> is radius of the spur gear</li>
</ul>
<p>The tangential component of the force on the worm gear <span class="math-container">$F_{w,t}$</span> will be equal to the spur gear <span class="math-container">$F_{s,t}$</span>.</p>
<p>The following image shows the forces that are developed on the worm gear.</p>
<p><a href="https://i.stack.imgur.com/lSYMm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lSYMm.png" alt="enter image description here" /></a>
<em>Figure 1: Forces of worm gear (source: <a href="https://khkgears.net/new/gear_knowledge/gear_technical_reference/gear_forces.html" rel="nofollow noreferrer">khdgears</a></em></p>
<p>It is impossible to have a worm gear pair mesh without radial and axial forces. However,<br />
the other forces (radial <span class="math-container">$r$</span> and axial <span class="math-container">$x$</span>) will <strong>be proportional</strong> and will be depended on the angles of the worm gear.</p>
<p>Now the force that <em>dominates</em> the torque on the worm gear is transverse to the plane of the spur gear. Notice that <span class="math-container">$F_{t,1}$</span> and <span class="math-container">$F_{x_2}$</span> are collinear, and <span class="math-container">$F_{t,2}$</span> and <span class="math-container">$F_{x_1}$</span> are collinear. To put things into perspective, in this situation, the worm gear is gear 1, and the spur gear is no 2.</p>
<p>So, although, the tangential force of the spur gear <span class="math-container">$F_{s,t}$</span> becomes axial on the worm gear, the worm gears tangential force is still proportional to <span class="math-container">$F_{s,t}$</span>. Therefore:</p>
<p><span class="math-container">$$F_{w,t}\propto F_{s,t}$$</span></p>
<p>However, now the torque required to keep the system moving would need to be equal to (actually its higher due to losses but to simplify things let's assume no losses):</p>
<p><span class="math-container">$$M_w = F_{w,t}\cdot R_w$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$R_w$</span> is the radius of the worm gear.</li>
</ul>
<p>So in the end:</p>
<p><span class="math-container">$$M_w \propto \frac{M_s}{R_s}\cdot R_w$$</span></p>
<p>So as you can see, increasing the diameter will result in higher torque requirements.</p>
| 43031 | How does the diameter of a worm gear affect the amount of force needed to turn it |
2021-05-07T11:02:19.547 | <p>okay so I am really struggling to see how to resolve the Normal reaction force and Weight into the r and theta direction please? Why is it that the angle is taken to the left of N and right of mg please?<a href="https://i.stack.imgur.com/zjkei.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zjkei.jpg" alt="enter image description here" /></a></p>
<p>Can someone give me some tips on this please? And why isn’t the N force resolved with the angle to the right of it attached to the theta line please?</p>
| |mathematics|kinematics| | <p>You need to review the trigonometric functions. In your question, assume the forces are to be resolved into component forces along the orthogonal axes r & q,</p>
<p><span class="math-container">$\sum F_r = Rcos\theta - Ncos\theta = (R-N)cos\theta$</span></p>
<p><span class="math-container">$\sum F_q = F + Rsin\theta - Nsin\theta = F + (R-N)sin\theta$</span></p>
<p><a href="https://i.stack.imgur.com/aukHW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aukHW.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/Tfruc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tfruc.png" alt="enter image description here" /></a></p>
<p>If you have any doubt, you can prove the validity of the component forces by the "rule of the right triangle":</p>
<p><span class="math-container">$c^2 = a^2 + b^2$</span>, thus</p>
<p><span class="math-container">$R = \sqrt{R^2sin^2\theta + R^2cos^2\theta} = \sqrt{R^2(sin^2\theta + cos^2\theta)}= \sqrt{R^2} = R$</span>, <strong>CHECK!</strong></p>
| 43035 | Resolving the forces into components |
2021-05-08T15:17:03.537 | <p>I have a Philips trimmer which when fully charged feels much heavier than when it is not charged. Also , the blades move much faster when it is fully charged I.e it’s weight , speed of blades kind of light increases as it more charged. What is it that can contribute to make it feel more heavy ?</p>
<p>Also , please let me know if you have any confusion in my Q or need more info.</p>
| |mechanical-engineering|electrical-engineering|dynamics|vibration| | <p>When you are operating a trimmer you have a mass of about 10 gr moving side to side.</p>
<p>Let's assume that the amplitude of the motion is about 0.1mm, and that the frequency is f= 500Hz. So the blade will be vibrating:</p>
<p><span class="math-container">$$x(t) = X\cdot \sin(2\pi f t)$$</span></p>
<p>That means that is velocity will be:
<span class="math-container">$$\dot{x}(t) = (2\pi f)X\cdot \cos(2\pi f t)$$</span></p>
<p>and the acceleration will be:</p>
<p><span class="math-container">$$\ddot{x}(t) = (2\pi f)^2X\cdot \sin(2\pi f t)$$</span></p>
<p>If you put the number in you get that the maximum acceleration is <span class="math-container">$a= 986 \left[\frac{m}{s^2}\right]$</span>.</p>
<p>The RMS value would be <span class="math-container">$a\approx 700\left[\frac{m}{s^2}\right]$</span>.</p>
<p>Therefore, the inertial force would be equal to:</p>
<p><span class="math-container">$$F_{rms}= m\cdot a_{rms}= 7 [N]$$</span></p>
<p>That force would need to be counteracted by your hands (partly) and that is the force you perceive as the added weight of the trimmer.</p>
<hr />
<p><strong>Intentional Shortcuts in the analysis</strong></p>
<p>The force calculated above would not actually be transmitted as it (I used approximate number to demonstrate the mechanism). The vibrating force of the blade is significantly attenuated by transmission through the (intentionally) heavy trimmer. If the trimmer were lighter at the same oscillating frequency, the force would be higher.</p>
<p>In real life, the frequencies are higher, and the forces. However, I intentionally did not use typical values, because I do not have damping ratios or stiffness matrix constants for the trimmer structure (which would actually be very much depended upon implementation).</p>
<hr />
<p><strong>Regarding the fully charged</strong>: I suspect your trimmer does not control the vibrating frequency actively. I am guessing it powers directly the (DC?) motor. That has the result, that the higher the frequency the higher the torque and the final velocity of the motor.</p>
| 43050 | Why does my trimmer feels heavier when it is operating and its fully charged? |
2021-05-08T23:17:01.470 | <p>I have a 600cc 4 cylinder Honda engine that my race car team uses.</p>
<p>The rules of the race mandate that we install an intake that is substantially smaller than the stock factory intake size to limit power.</p>
<p>This year, we want to optimize the stock camshaft profile to increase power across the whole curve (or at least the low end).</p>
<p>The engine is running on stock factory cams but its air intake is much more restricted that stock so the current camshaft profile is not optimal. So this means we need to reduce duration/overlap on intake and exhaust camshafts. I am looking for help on how to find the new profile.</p>
<p>(I'm guessing it will be something like this but I need to calculate actual numbers).
<a href="https://i.stack.imgur.com/fQA2a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fQA2a.png" alt="camshaft profiles" /></a></p>
<p>How do I go about doing this? (Using spreadsheets)</p>
<p>Thank you</p>
| |fluid-mechanics|automotive-engineering|airflow| | <p>First of all, do NOT USE EXCEL. Sooner or later it will lead to undetected errors and great sadness.</p>
<p>Next, whether you use "Excel formulas" or proper code in tools like MATLAB or python, you need to start with the equations and formulas and boundary conditions you want to work with. Only after you have this can you start to make software do the optimization you desire. Sadly, we are nowhere near the STTNG - level of software yet ("Computer, design an optimum cam shape for me, and while you're at it pop a couple out of the Replicator")</p>
| 43056 | How to use excel spreadsheet formulas to find optimal camshaft profile |
2021-05-09T13:09:25.130 | <p>In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is μ. Find the minimum horizontal force F required to keep the block stationary with respect to wedge.<a href="https://i.stack.imgur.com/JDW61.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JDW61.jpg" alt="enter image description here" /></a></p>
<ol>
<li><p>For this Q , total acc of <span class="math-container">$M+m$</span>= <span class="math-container">$F/M+m. $</span></p>
</li>
<li><p>Considering inertial frame of reference , for mass M and m.FBD is as follows :<a href="https://i.stack.imgur.com/UDgNi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UDgNi.jpg" alt="enter image description here" /></a></p>
</li>
</ol>
<p>I haven’t marked N1 or N2 but can assume them according to FBD.</p>
<p>For mass M ,</p>
<ol>
<li><p>μN1 on left means the friction from surface.</p>
</li>
<li><p><span class="math-container">$N_m$</span> means normal force by the mass m which is equal to <span class="math-container">$N_M$</span>.</p>
</li>
</ol>
<p>For mass m ,</p>
<p>1 )μN2 on left is because frictional force is always in direction opposite to motion. Since the N2=0(no surface from ground for mass m) (as per Q), So <span class="math-container">$μN2=0$</span>.</p>
<p>Q 1 Why did we not take <span class="math-container">$N_M$</span> as the normal force in μN2 ?</p>
<p>2 )<span class="math-container">$mg=0 $</span></p>
<p>Total F= μN1 + <span class="math-container">$N_m$</span> where <span class="math-container">$N_m$</span> = μ*N = m * acc of mass m. Since ,</p>
<p>μN=0. Therefore , either m or a = 0 it has to be .</p>
<p>So , total <span class="math-container">$F= μN$</span> only.</p>
<p>But correct answer is <span class="math-container">$(M+m)*a$</span> where <span class="math-container">$a=g/μ$</span></p>
<p>So , I just wish to know where am I wrong in my calculation.</p>
| |mechanical-engineering|applied-mechanics|friction| | <p>What you are learning here is the effect of the "inertia force" due to motion.</p>
<p>In order for both blocks to move together, we can simply write the equation, <span class="math-container">$F = (M + m)*a$</span>, by the law of motion.</p>
<p>Then for the small block not to fall during the motion, there must have a normal force exerted from the smaller block on the larger block to produce friction force required to maintain the small block in place. In here, the friction force is <span class="math-container">$m*g$</span>, and the inertia force is <span class="math-container">$m*a$</span>, thus <span class="math-container">$N = m*g/\mu = m*a$</span>, <span class="math-container">$a = g/\mu$</span>. (The normal force <span class="math-container">$N$</span> is the effect of the "Inertia Force", which has the same intensity as the mass in motion and always in the reversed direction of the motion.)</p>
<p><a href="https://i.stack.imgur.com/3iz4A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3iz4A.png" alt="enter image description here" /></a></p>
| 43062 | Find minimum horizontal force required to keep the block and wedge stationary? |
2021-05-09T19:18:46.573 | <p>Peace and blessings beloved. I come from a family of tradesmen (plumbers, electricians, welders, carpenters, painters, etc) but never took a real interest in learning about said trades (or anything trade-related). I seen people work on and extensively repair households but never from the ground up so that particular part of the process is new to me. As these jobs relate to Engineering I had no clue or understanding of where each would fall, align with, etc.</p>
<p>As I began to research what specifically goes into building a home (from the womb to conception) I came across <a href="https://en.wikipedia.org/wiki/Engineering" rel="nofollow noreferrer">Engineering</a> and began to also research what type of schooling they go through to attain that position, noticing the <a href="https://typesofengineeringdegrees.org/" rel="nofollow noreferrer">plethora of Engineers</a>. Sifting through I came across various Engineering disciples that are stated to be involved with the process that included (not limited to) <a href="https://qr.ae/pGnrls" rel="nofollow noreferrer">Structural, Mechanical, Electrical</a>, <a href="https://designeverest.com/blog/why-hire-an-engineer-or-architect-for-residential-project/" rel="nofollow noreferrer">Architects</a>, <a href="https://ashleyvance.com/blog/planning-build-home" rel="nofollow noreferrer">Civil</a>, and maybe even a <a href="https://www.peforhire.com/blog/home-building-basics-what-engineers-are-needed-to-build-a-home/" rel="nofollow noreferrer">Sustainable Engineer</a> (if one wants to build a Green Home). Seeing that there are a lot of nuances and technicalities involved I came across some articles and places that argued one more over the other, which to a novice can be very confusing.</p>
<p>Knowing that size, region, and environment all play a role in the process, generally speaking, which type of engineers are involved in the process (from start to finish) of designing and constructing residential homes? Whether the house is an urban, suburban, or rural setting. Thank you.</p>
| |building-design|construction-management|home-improvement| | <p>As an engineer (with partner also an engineer) who designed a 4 bed house to exceed the energy requirements of the building regs, and by a huge amount that some new builds are just beginning to match its performance some 15years later I have to point out that a good engineer or two, and a good architect are all that is needed.</p>
<p>So, had to brush up on some engineering theory to work out if the 20cm thickness of insulation would support the total mass of the house as no other support went through that slab except sewage, water and electricity pipes.</p>
<p>What was paramount was to be able to make use of the sun - passive solar design. Ie avoid summer sun by the use of shading, get free heat in winter by letting sunlight into the house etc</p>
<p>So, being able to use degree-days tables and interpolate data to approximate the weather to run performance simulations was the crux to the design - and the architect had no idea...</p>
<p>Main thrust of our engineering degrees was thermal and power plant analysis along with experience with renewables : solar pv and thermal etc.</p>
| 43072 | What type of engineer/profession specialize in the design and building of residential homes/houses? |
2021-05-11T20:16:02.073 | <p>Recently I was weighting stuff with my dad and as I was playing with one of the old 2 kg weights I felt with my fingers number "5". After cleaning and upon further inspection it turned out the writing on the weight says "5Φ". We checked and it indeed weights 2 kg as my dad remembered. It was done with a simple balance scale by comparing with another weight of similar age that is properly marked as "2 kg". There are no other marks on that weight. On the opposite side is a bulge but it's an irregular blob - I think it's remainder of sprue.</p>
<p>So, question that arised in my head is - <strong>what does that Φ mean? Is it some rare unit of mass? Some sort of preferred series or standard for weights?</strong><br />
I was looking for answers and found nothing. I ruled out pounds that was first thing that came to my mind because the difference was to big (5 pounds is approx. 2.26 kg) - though there are some signs of wear and corrosion it's in rather good shape and I don't believe it weighted 0.26 kg more originally. Also, considering I'm from Poland and the weight is really old it probably was made during the soviet era (maybe even in USSR) - so that <em>phi</em> could be actually cyrilic <em>fe</em> - but that also gave me no results.</p>
<p>As a last resort I decided to ask here - maybe some member of that community will know the answer. Below is a photo of the weight in question.</p>
<p><a href="https://i.stack.imgur.com/Ypb5F.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Ypb5F.jpg" alt="photo of the weight in question" /></a></p>
| |measurements|unit| | <p>You are probably right (it being of Russian origin).</p>
<p>The weight seems to be 5 funt.</p>
<p>1 funt is the Russian equivalent of a pound. It is written as Фунт, funt, and around 1900 it was the basic unit of weight measurement in Russia (so it survived in the USSR days), but now its obsolete.</p>
<p>1 Φ is about 409.5 grams.</p>
| 43112 | What unit of weight is that? |
2021-05-12T18:13:02.377 | <p>I am trying to implement my own FE code for shell structures. Most literature provide formulations and strategies on how to compute shell element stiffness matrix. I can already compute deformations using my FE code, but I don't know how to get the stress field. Can somebody direct me to the right paper/textbook?</p>
| |structural-analysis|stresses|finite-element-method| | <p>Here be dragons. Or, to paraphrase a hadith of the Prophet Mohammed, "There are seventy ways to compute element stresses, and all of them are wrong."</p>
<p>The naive approach is to differentiate the element shape functions to find the strain at a point, and then use the stress-strain relationship for the material.</p>
<p>The problem is that since the shape functions are (usually) low order polynomials, differentiating them reduces the order by one. Therefore the calculated strain and stress distribution in the element is not physically realistic, and is discontinuous across element boundaries. For example in a 3-node triangular membrane element, this method implies the stress and strain are constant over the each element.</p>
<p>Back in the 1970s, Barlow published a paper which pointed out that these "low order" stresses and strains have a higher-order accuracy at a few well-defined points in each element, which were subsequently named "Barlow points".</p>
<p>If you use numerical integration to formulate the stiffness matrix, you are effectively "sampling" the element strain energy (i.e. stress <span class="math-container">$\times$</span> strain) at the numerical integration points and minimizing it by differentiating it to form the stiffness matrix. Therefore it is not surprising that the Barlow points coincide with the numerical integration points, for many element formulations.</p>
<p>However, "these are not the <strike>droids</strike> stresses you are looking for", because the critical stress locations are often on the boundary of the structure (and hence on the boundary of an element) and the integration points are at internal locations in the element.</p>
<p>Therefore, the final step is to somehow interpolate or extrapolate the Barlow point stresses to create a stress distribution for the entire structure. A naive way to do this is to interpolate within each element to calculate nodal stresses at the element, and then find the "average nodal stress" from all the elements which connect to each node.</p>
<p>That is a simple way to produce pretty computer graphics (and some commercial FE post processing software uses this algorithm), but a bit of experimentation (e.g. modelling a cantilever beam with shell elements) will show that the maximum errors (which are large) are on the boundaries where an engineer wants to know the stress, because (obviously) on a boundary there are fewer points to average.</p>
<p>A more sophisticated idea is to define a continuous stress function for the structure based on element shape functions, and do a least-squares fit to match that with the Barlow point stresses.</p>
<p>But those are still not always the stresses you are looking for, because a shell element model may contain assumptions (e.g. step changes in thickness or different materials in different parts of the model) which mean that assuming the stresses and strains are <em>continuous</em> everywhere is also physically wrong...</p>
<p>... and all of the above has ignored the complications of formulating curved shell elements, or using flat elements to model curved real-world shells. For example, it might make sense to average the stresses from flat elements modelling a spherical or cylindrical structure, but it doesn't make any sense at all to average the stress at the edges or corners of a box, where the sides, top and bottom are modelled as flat shells.</p>
<p>None of the above really answers the question, but at least you now know why a full answer is more likely to be a book (or a PhD thesis) not a post on Engineering.SE.</p>
<p>All this is still a research topic. Barlow's original 1976 paper is online here (paywall), plus a (free) list of more than 300 citations: <a href="https://onlinelibrary.wiley.com/doi/abs/10.1002/nme.1620100202" rel="nofollow noreferrer">https://onlinelibrary.wiley.com/doi/abs/10.1002/nme.1620100202</a></p>
| 43121 | Computing the stress field of a structure made of 4-nodes shell elements (accounting for membrane action, bending and shear) |
2021-05-12T22:21:23.840 | <p>In the design of boat hulls (and aircraft) reduction in material thickness is important and the hull is usually made as an assembly (welded or riveted) with stringers and frames. The design of stiffened panels is a complex problem. Are there a set of simplified rules for the design of stringers and frames for boat hulls?</p>
| |frame| | <p>They are called scantling rules. Herreshoff and Nevin had scantling rules for wooden yachts. Classification Societies such as Lloyds and BV have scantling rules for ships. Gerr's <em>Boat Strength</em> has rules for small boats covering several hull materials.</p>
| 43128 | Rules of thumb in the design of stringers and frames for hull design |
2021-05-13T03:05:28.250 | <p>Consider this question:
<a href="https://i.stack.imgur.com/37RJ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/37RJ9.png" alt="Question" /></a></p>
<p>I solve this eventual question and my answer is <code>7.29667</code>:</p>
<p><a href="https://i.stack.imgur.com/B2sVD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B2sVD.png" alt="Calculation" /></a></p>
<p>Among the question choices are <code>7 x 7 ft</code> and <code>7.5 x 7.5 ft</code>. Which one should I choose? I guess if a safety margin is intended, the larger choice i.e. <code>7.5 x 7.5 ft</code> should be selected. Am I right?</p>
<p>There are situations like this in which two question choices are close to my final answer and I'm not quite sure which one to pick. Is considering safety margin a good criterion to pick the choice? Thanks for your help!</p>
| |structural-engineering|civil-engineering|structural-analysis|foundations|soil| | <p>You are worrying about the numbers and forgetting what the numbers mean. This is for the PE exam, so this is a very important topic to make clear in your mind.</p>
<p>You solved the problem. You came up with an answer for what is required. You now have two options:</p>
<ol>
<li>Choose a small footing that your own solution just proved is too small for the requirements.</li>
<li>Choose a large footing that your own solution just proved is larger than is required. This will be conservative, but it is the only option that meets the requirements.</li>
</ol>
<p>If you look at the two options above, there is only one solution. This is a question for the PE exam. One of the primary purposes of professional licensure is to promote the <strong>safety</strong> of the public. When in doubt, always choose the safer option.</p>
| 43131 | PE exam: situations in which two question choices are close to my final answer: how to pick one? |
2021-05-13T05:59:40.290 | <p>For the problem below, I tried solving it, but I don't know where I should get the value of normal acceleration, guess I was lost.<a href="https://i.stack.imgur.com/pEYgQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pEYgQ.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/cR1AZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cR1AZ.jpg" alt="enter image description here" /></a>
And here's what I solved so far
<a href="https://i.stack.imgur.com/SgbYe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SgbYe.jpg" alt="enter image description here" /></a></p>
| |dynamics|acceleration| | <p>You are almost there.</p>
<p>In order to calculate the tangential and the normal acceleration you need to take them as vectors.</p>
<p><a href="https://i.stack.imgur.com/WQJRA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WQJRA.png" alt="enter image description here" /></a></p>
<p>The angles are</p>
<ul>
<li><span class="math-container">$\theta= \arctan(16/12)=-53.13 deg$</span> for velocity</li>
<li><span class="math-container">$\phi= \arctan(16/12)=-26.51 deg$</span> for accelaration</li>
</ul>
<p>Therefore the vectors for acceleration and velocity would be</p>
<p><a href="https://i.stack.imgur.com/qsOwt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qsOwt.png" alt="enter image description here" /></a></p>
<p>And then you can find the component of tangential and normal acceleration as:</p>
<ul>
<li><strong>tangential:</strong> <span class="math-container">$a_t =a \cos(\theta-\phi)$</span></li>
<li><strong>normal:</strong> <span class="math-container">$a_n =a \sin(\theta-\phi) = 5\frac{m}{s^2}$</span></li>
</ul>
<p>Then you can just substitute:</p>
<p><span class="math-container">$$R = \frac{v^2}{a_n}= \frac{20^2}{5}= \frac{400}{5}= 80 in$$</span></p>
| 43132 | How to get the normal acceleration |
2021-05-14T08:49:04.410 | <p>first of i know it seems like a homework assignment but it does not, it is calculations i need to do at work and it has been a long time since i studied this material. so I need help to calculate if a given clutch can handle a given torque but I'm not sure if my calculation are right.</p>
<p>what I did so far</p>
<p><a href="https://i.stack.imgur.com/NPMT9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NPMT9.png" alt="enter image description here" /></a></p>
<p>on the left there is a motor on the right is the cluch with a connector.</p>
<p>ive build a FBD:
<a href="https://i.stack.imgur.com/nPshu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nPshu.png" alt="enter image description here" /></a></p>
<p>and ended up with this eq:</p>
<p><span class="math-container">$$\tau_{max} = \frac{T*r_1}{J} = \frac{2T*r_1}{\pi*r^3}=\frac{2*2.5}{\pi*(6.05*10^{-3})^3} = 7.2 [MPa]$$</span></p>
<p>but I don't know if this is the way to calculate and I don't know how to calculate if the point connecting the two diameters is strong enough.</p>
| |mechanical-engineering|motors|torque| | <p>Your calculation is generally correct .</p>
<p>The basic formula is:</p>
<p><span class="math-container">$$\tau_{max} = \frac{T*r_1}{J}$$</span></p>
<p>Although the Torque is constant (<span class="math-container">$2.5 [Nm]$</span>), what changes is the radii and the second moment of area.</p>
<ul>
<li><p>For solid sections
<span class="math-container">$$J_{solid} = \frac{\pi r^4}{4}$$</span></p>
</li>
<li><p>For hollow sections
<span class="math-container">$$J_{hollow} = \frac{\pi \left(r_o^4 - r_i^4\right)}{4}$$</span></p>
</li>
</ul>
<p>where:</p>
<ul>
<li><span class="math-container">$r_o$</span>: the external radius</li>
<li><span class="math-container">$r_i$</span>: the internal radius</li>
</ul>
<hr />
<p>if you are worried for increases in stress due to the change in diameter, you can look at stress concentrations.</p>
| 43155 | calculate torsion in a diffrent sized shaft |
2021-05-14T10:36:43.487 | <p>A UK subterranean reinforced concrete structure built in the late 1960s has experienced sustained water ingress over time as shown by the deposition of minerals of the walls:</p>
<p><a href="https://i.stack.imgur.com/7H8cO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7H8cO.jpg" alt="enter image description here" /></a></p>
<p>According to the original drawings the construction joints were detailed with 190 mm PVC eyeleted watafoil waterbar.</p>
<p><a href="https://i.stack.imgur.com/8WqK8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8WqK8.png" alt="enter image description here" /></a></p>
<p>My question: Has the PVC water bar degraded over the 50+ years since installation? Or was it never installed correctly and ground water has been leaking in ever since?</p>
| |civil-engineering|reinforced-concrete|waterproofing| | <p>Based on limited studies, PVC can degrade underwater due to chemical and bio-microbe attacks, it can lose weight and become more fragile after an extended time. But under the same attacks, the cement can also have degraded and lost the bond with the PVC water stop. It looks like this is the case as the photo is shown - a stretch of watermarks tracing the construction joint.</p>
<p>Both the design and construction methods may share the blame for water intrusion.</p>
<ul>
<li><p>Design didn't incorporate water-tight requirement which was developed during the 80s and 90s. The requirement includes an increased amount of reinforcement to reduce concrete stress, and an increased thickness of the concrete clear cover to the reinforcement, both are in efforts to minimize the occurrence of crack, the spacing, and the width of the crack. Also, the resulting thicker wall lengthens the path for water migration.</p>
</li>
<li><p>Construction care. Since the water stop is much lighter than the concrete in the fluid state, it is difficult to maintain the water stop in the proper position, and resulting in locations with an inadequate cover thickness that shorten the water flow path. Also, under-vibration and over-vibration both can cause dislocation and poor bonding.</p>
</li>
</ul>
<p>However, after the structure has stood for 50 years, I wouldn't put too much blames on the design and construction. For leakage problem caused by which tends to occur shortly after the construction, and the problem tends to be widely spread, and can only get worse with time.</p>
<p>IMO, any localized repair will not be durable. Water will find its way to get in if the repair does not correct the problems that may be present somewhere away from the leakage. If accessible, waterproofing the exterior wall face is the most effective, but costly, method. Otherwise, a false wall with a proper drainage/dewatering/ventilation system may be considered.</p>
| 43157 | Do PVC concrete joint water stops degrade over time? Or were they never installed correctly? |
2021-05-15T14:54:06.093 | <p>Here , we have a car on a banking road. Considering two cases :</p>
<p><a href="https://i.stack.imgur.com/MXarU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MXarU.jpg" alt="enter image description here" /></a>
Frictionless and friction surface of the road.</p>
<p>I give an initial velocity to the car by a push for a second. Then , the cases discuss about the direction of friction and motion of the car.</p>
<p>Case 1: Frictionless road surface</p>
<p>We will have an <span class="math-container">$N*sin$$\theta$</span> force acting towards the circle. Initial direction of motion was along the road but isn’t it that due to the banking of road and frictionless road. The car would slowly start to slip down <span class="math-container">$? $</span>Is this case also uniform circular motion <span class="math-container">$? $</span></p>
<p>non uniform circular motion is not possible here since friction is not present. Therefore , no acceleration.</p>
<p>Also , can we say no centripetal and tangential acceleration is present<span class="math-container">$ ?$</span></p>
<p>_____ 1)N cos <span class="math-container">$\theta$</span> = mg</p>
<p>_____ 2)N sin <span class="math-container">$\theta$</span> = <span class="math-container">$\frac{Mv^2}{r}$</span> . From this , we can say the car moves down <span class="math-container">$?$</span>.</p>
<p>Case 2: Friction is present between road surface and car.</p>
<p>All directions are same as the above except for friction. So , can I say that friction also acts in the same direction as <span class="math-container">$Nsin$$\theta$</span> ? in case of uniform circular motion.</p>
<p>In non uniform , friction force will be at an angle alpha to the centripetal acceleration <span class="math-container">$?$</span></p>
<p>By the <span class="math-container">$?$</span>, I wish to understand what would happen.</p>
| |mechanical-engineering|applied-mechanics|kinematics| | <p>If the road width is wide enough:</p>
<ul>
<li><p>No friction and no air friction: the car will seek a wider curve up the bank and circle around or go around an ellipse if not initially pointed straight, if the initial velocity, Vi is greater than,<span class="math-container">$$V_i> \sqrt {rgtan\theta} $$</span>
Or else a lower orbit.</p>
</li>
<li><p>If there is friction:</p>
</li>
<li><p>If <span class="math-container">$V_i> \sqrt{\frac{rg (sin\theta+\mu cos\theta)}{cos\theta-\mu sin\theta}} $</span> Then the car will skid up and rotate in a higher orbit likely an ellipse similar to a rocket changing orbit.</p>
</li>
<li><p>If <span class="math-container">$Vi< \sqrt{\frac{rg (sin\theta-\mu cos\theta)}{cos\theta+\mu sin\theta}}$</span></p>
</li>
</ul>
<p>the car will skid down and rotate in a lower orbit.</p>
<p>Otherwise, it will stay the course.</p>
<h1>Edit</h1>
<p>Just for the record the accepted answer is wrong.</p>
<p>For example in frictionless situation as <span class="math-container">$\theta$</span> approaches 90 the speed, V, must approach infinity, otherwise the car will slide down due to lack of friction. but <span class="math-container">$sin90=1 $</span> which implies <span class="math-container">$ V = \sqrt{Rg}.$</span></p>
| 43174 | Would a car slip down on a frictionless banked road and situation for friction road |
2021-05-15T19:40:00.720 | <p>I am designing my own optical rotary encoder with a non-standard number of encoder positions for a specialty application. I have gotten a 0.35mm acrylic with the pattern printed on it but now I need to punch out the center hole.</p>
<p>A smaller hole could be much more easily punched precisely by hand or machine.</p>
<p>If a punch could be designed with a small rod in the center, it would align just right with the center hole.</p>
<p>This method could also be used to punch out the whole disk. First a big punch with rod could be used to punch the disk, and then a smaller one to punch out the center hole.</p>
<p>The biggest problem that arises is where/how to get a bit that has an alignment rod in the center. Perhaps they already exist, but I just don't know how to search. I don't trust my lathe to drill out the center of a punch correctly, as the punch will likely be made of the same material as my drill bits. How can I precisely punch these holes in thin acrylic sheet?</p>
| |tools| | <p>The fact that it isn’t Gray code is a reason to start over. When you do, add a tooling hole for the center. With careful use of a punch properly sized for the tooling hole you should get good enough alignment for this large of a pattern.</p>
<p>PCB manufacture uses tooling holes and fiducial marks to get good alignment between layers. If you want to make more of these that’s a good way to make high precision encoder wheels.</p>
| 43183 | Punch Hole Exactly in Center of Acrylic |
2021-05-18T16:58:07.050 | <p>What amount of energy would it take to replace the entirety of America's water consumption with desalinated water? And how does it compare with the country's current energy generation capacity?</p>
<p>By "water consumption" I generally mean things that one usually thinks of when thinking of using water, including but not limited to household, commercial, agricultural, mining, and industrial uses. I'm not talking about filling reservoirs for recreation, hydroelectricity generation, or outflows meant to preserve fisheries. If there is another case worth considering, feel free to ask, or simply note it in the answer.</p>
<p>By "amount of energy", I mean to take seawater and turn it into usable water, not to pump such desalinated water from, say, San Diego to Denver. This would clearly be a non-trivial amount, but I think it would be too hard to calculate with much precision. (I guess the lower bound would be the current hydroelectric generation capacity of the USA, and a reasonable upper bound would be 10x that amount.)</p>
| |civil-engineering|water-resources|environmental-engineering|hydrology| | <h2>Amount of Water Used</h2>
<p>The amount of water used in the USA in 2015 for all uses other than thermoelectric power (1) (read: to cool power plants) was <a href="https://pubs.usgs.gov/fs/2018/3035/fs20183035.pdf" rel="nofollow noreferrer">189B gal/day</a>, or 715M m^3/day.</p>
<h2>Amount of Energy Needed</h2>
<p>The amount of electrical energy it takes to desalinate ocean water is in the range of <a href="https://en.wikipedia.org/wiki/Desalination#Energy_consumption" rel="nofollow noreferrer">3-5.5 kWh/m^3</a>.</p>
<p>Therefore the amount of electrical energy it takes to desalinate a day's worth of water in the USA is at least 2.15B kWh/day.</p>
<h2>Comparison with Amount of Energy Produced</h2>
<p>The amount of electrical energy produced in the USA in 2020 was <a href="https://www.eia.gov/tools/faqs/faq.php?id=427&t=3" rel="nofollow noreferrer">4.01T kWh</a>, which works out to 10.93B kWh/day. This means the USA would need to increase electrical energy production by 19.7% in order to derive the fresh water for the USA's daily water needs from ocean water (2).</p>
<p>Electrical energy production has historically increased <a href="https://www.youtube.com/watch?v=7dfyG6FXsUU&t=184s" rel="nofollow noreferrer">at the rate of 4%</a> per year (4), meaning that this energy need could be met in about five years, not accounting for changes in energy and water use.</p>
<p>Suppose we wanted to derive the energy solely from renewables. The amount of electrical energy produced from renewable sources (not including hydropower) in the USA for 2020 was <a href="https://www.eia.gov/tools/faqs/faq.php?id=427&t=3" rel="nofollow noreferrer">501B kWh/year</a> or 1.37B kWh/day. By increasing the electrical energy production of non-hydro renewables by 157%, America could completely replace its water needs via desalination.</p>
<h3>Notes</h3>
<ol>
<li>I omit power plant cooling because either this often already comes from seawater, or the use of this water can be sharply reduced by converting to closed-loop systems. (<a href="https://pubs.usgs.gov/fs/2018/3035/fs20183035.pdf" rel="nofollow noreferrer">"Powerplants that used once-through cooling systems accounted for 96 percent of all thermoelectric-power withdrawals."</a>)</li>
<li>Deriving this fresh water via waste water processing would take <a href="https://en.wikipedia.org/wiki/Reverse_osmosis#Desalination" rel="nofollow noreferrer">much less energy</a>.</li>
<li>This compares 2020 energy production against 2015 water consumption, but this shouldn't keep us from getting to a pretty close solution.</li>
<li>Yes I referenced Engineering Explained directly rather than derive the growth rate directly. Fwiw we use the <a href="https://www.eia.gov/tools/faqs/faq.php?id=427&t=3" rel="nofollow noreferrer">same source</a> for USA electrical energy production.</li>
</ol>
| 43239 | How much energy is needed for the USA to only use desalinated water? |
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