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2018-10-07T14:43:35.513 | <p>I have observed that when a 4-wheeled vehicle takes a turn, one of it rear tire moves with more speed as compared to the other. For example A car is taking right turn, its right rear tire will be slower as compared to left rear tire.</p>
<p>How is that done?<br>
Without this system, the vehicle wont take a turn and collapse.</p>
| |car| | <p>It depends. </p>
<p>The driven wheels need to be connected by a <a href="https://en.wikipedia.org/wiki/Differential_(mechanical_device)" rel="nofollow noreferrer">differential</a> whcih allows power to be transferred to both wheels but still allowing them to turn at different speeds. </p>
<p>It's not that easy to comprehends exactly how this works mechanically without having a physical model to play with but it does work , at least within certain limits. </p>
<p>The problem with a gear based differential is that if one wheel loses traction the opposite wheel doesn't get any power either whcih is a big issue on slippery surfaces like snow, ice, mud or gravel. </p>
<p>A simple fix is to lock the diff which improves traction but means that the wheels need to slip on cornering on slipper surfaces this may be acceptable but is a problem on dry pavement. </p>
<p>Many 4x4 vehicles have the ability to lock and unlock their differentials according to the conditions. </p>
<p>A more sophisticated solution is a <a href="https://en.wikipedia.org/wiki/Limited-slip_differential" rel="nofollow noreferrer">limited slip differential</a> whcih allows the drive wheels to rotate at different speeds but ensures that both get at least some torque in all conditions. </p>
<p>There are various ways to implement LSDs but some sort of viscous coupling (a bit like the torque converter in an automatic gearbox) is common). </p>
<p>These are often used in conditions where vehicle performance is traction limited ie some wheel slip is frequent but not consistent eg in rallying and racing. </p>
<p>Modern systems allow the degree of slip to be controlled either by directly by the driver or by a computer with reference to some pre-set parameters. Indeed diff settings are one of the important driver-controlled parameters in a modern F1 car. </p>
<p>In a 'loose' setting any loss of traction on a driven wheel will reduce overall torque so you want ta tight diff for exiting a low speed corner conversely a tight diff will tend to push the car in more of a straight line and induce under-steer (at least in general). </p>
<p>This is further complicated by that fact that cornering at any speed tends to induce roll which unloads the inside wheel to the inside driven wheel will tend to lose traction before the outside one. </p>
<p>It isn't <em>impossible</em> to turn without a differential most go-karts don't have diffs and they are more essential on fwd than rwd cars. A fws car with a simple Limited Slip Diff tends to have fairly epic torque steer. But in rwd you can use engine torque to deliberately slide the rear and break traction to assist turn-in so the tyres provide the slip rather than the transmission. </p>
<p>This is very obvious in 'drift' racing, whcih tends to use locked or relatively tight LSD. </p>
| 24073 | Proccess of a 4-wheeled vehicle taking turn |
2018-10-08T03:09:49.643 | <p>Take a loom at this diagram
<a href="https://i.stack.imgur.com/ANglD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ANglD.jpg" alt="enter image description here" /></a></p>
<p>You have probably heard of this before but im just curious what laser they are using to do this. They do t appear to be ionizing air. It seems the one laser is just for color on the particle but what is the other? I would like to try this at home and make a dot for fun but i dont quite understand whats going on here. Any ideas?</p>
| |lasers| | <p>The full letter is here:</p>
<p><a href="https://www.nature.com/articles/nature25176" rel="nofollow noreferrer">https://www.nature.com/articles/nature25176</a></p>
<p>They have 4 lasers: an RGB beam (3 lasers) that provides colour and an invisible 405nm beam that traps the particle. The RGB beam is pretty self-explanatory: you shine a beam of the colour you want the line to be. </p>
<p>The trapping beam is more complicated, by using lenses with particular imperfections you can focus the beam so that there is a dark volume near the focal point. Particles in this volume will be kept there by the photophoretic force. There is more information in this paper: </p>
<p><a href="https://www.osapublishing.org/oe/abstract.cfm?uri=oe-19-18-17350" rel="nofollow noreferrer">https://www.osapublishing.org/oe/abstract.cfm?uri=oe-19-18-17350</a></p>
<p>Playing with invisible lasers can be pretty dangerous </p>
| 24079 | How 2 lasers interacting causes a dot of light |
2018-10-08T14:51:39.253 | <p>I have to do a durability approval of feather key (shaft-hub connection) and the formulas are dependent if the shaft respectively the hub is ductile or brittle. </p>
<p>But what is the decision about to say it is ductile or brittle.</p>
<ul>
<li>My norm for feather key isn’t saying anything how to distinguish if the metal is ductile or brittle.</li>
<li>The norm for pressfits uses the parameter fracture elongation and reduction of area.</li>
<li>And the literature is talking about brittle metal when the Maximum Stress Theory is valid <span class="math-container">$(\sigma_{limit}/\tau_{limit}=1)$</span> and ductile metal when the Shear Stress Hypothesis is valid <span class="math-container">$(\sigma_{limit}/ \tau_{limit}=2)$</span>.
<ul>
<li><span class="math-container">$\sigma_{limit}$</span> is tensile strength</li>
<li><span class="math-container">$\tau_{limit}$</span> is the strength when plasticity begins</li>
</ul></li>
</ul>
<p>Okay, there are many possibilities… But which one is the right possibility? Or is there a connection between my findings that I’m not aware of?</p>
<p>Thanks in advance for your help.</p>
| |materials|steel|metals|strength| | <p>The simplest way to tell if a metal is ductile or brittle is to look at a fracture surface (e.g. from a tensile or impact test). To have either a ductile or brittle metal defines the failure mechanism which you can tell by having either a <a href="http://home.jtan.com/~joe/bicycle/images/spoke_2.jpg" rel="nofollow noreferrer">ductile fracture</a> or a <a href="http://vgoinc.com/wp-content/uploads/2015/11/SEM-Image3.jpg" rel="nofollow noreferrer">brittle fracture</a>.</p>
<p>Elongation and reduction of area as well as notch toughness (impact test) and hardness testing can give a quantitative value to ductility, by measuring the amount of energy the material absorbs and converts to mechanical deformation. These values are normally used to allow 'design ductility' which is less interested in the failure mechanism and a lot of the time only cares if a material can absorb enough energy not to fail as soon as it is over stressed (or for deliberate deformation). But there isn't a fixed limit for defining a material as ductile or brittle as it depends on the material and is usually used to define a design criteria that must be met. </p>
<p>However if you have an elongation greater than 20% you are probably dealing with a ductile metal, but this can depend on the alloy type/condition and is not always the case for all materials. </p>
| 24088 | With which parameter can you specify whether a metal is ductile or brittle |
2018-10-08T21:28:07.780 | <p>So upon watching videos and reading articles i understand few things. For one when the laser beam is focused to a tight enough point when placing a material in there (people use crayola markers) it will trap and burn that particle for a little while.
<a href="https://i.stack.imgur.com/RjgWn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RjgWn.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/hSkqC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hSkqC.png" alt="enter image description here" /></a>
My question is how can i do this without having to touch a marker to it? Can I achieve a particle effect through something else or atleast something that wont burn off the particle right away. Also is this effect as easy as talking a 405nm laser, focusing it with some sharp lenses and then having that particle get trapped. I have seen a few diagrams but still it is unclear all the equipment they use and then in a video it is just a laser focused to a point.
Such as these:
<a href="https://i.stack.imgur.com/TIaBg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TIaBg.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/KB65D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KB65D.png" alt="enter image description here" /></a></p>
<p>UPDATE: Upon reading more i see the mW required is only around 5mW which isnt bad at all. I also see people use graphite but im not sure how ling that particle lasts? Forever until the beam is broken or only for a certain amount of time. To use graphite could i simple take a block of it and wave it in front of the laser at its lowest density point to capture the particle? My other confusion is the lens required what is so special? Is there maybe an equation that includes a class of particles, wavelength, power, and focal length of lens to see where i stand with this. I know this isnt what stack overflow is for but i just want to set this up and see it and understand it rather then just read so maybe a good diagram that i or any member coming here could understand in order to setup this experiment.</p>
| |lasers| | <p>I was able to order a 25mW laser in the 532nm wavelength and trap a particle using the following optics: a 50mm focal lens which after it was focused I allowed it to expand to a 1cm diameter beam width. Then I used another focusing lens of fl=25mm and that focused beam is where the trapping begins.
<a href="https://i.stack.imgur.com/VHFEx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VHFEx.jpg" alt="enter image description here"></a>
By touching a black marker to it (black because any light colors will simply be reflected and not absorbed) and when the black pieces of sharpie marker would burn off they would be trapped into the laser beam at its lowest density point.
<a href="https://i.stack.imgur.com/uROTa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uROTa.jpg" alt="enter image description here"></a>
As for how long it lasts without burning, I don't think it is burning that is the problem, its wind and air. The slightest breath can cause the particle to move away. Sometimes you can have a particle stay for minuets and some just a second.</p>
| 24093 | Trapping a particle in a laser beam (optical tweezers) |
2018-10-09T14:15:11.647 | <p>Some days ago, I took a plane (Airbus A330) for a pretty long flight - around 8hrs. During the flight, we experienced some turbulences and I thought : what if the turbulences last the whole flight?</p>
<p>How would a plane react to 8+ hrs under mid to heavy turbulences during the whole flight?</p>
| |aircraft-design| | <p>Duration of turbulence is not a factor, intensity is!</p>
<p>All modern airplanes have been designed with a flexible structure ready to take punishing weather and vibrate with the impact of the shock wave to dissipate the energy of turbulence.</p>
<p>They have been designed to vibrate longitudinally along the fuselage and rotationally to take in the torque induced by wing shock. The wings are designed to flex and vibrate to behave in a ductile mode.</p>
<p>Even the empennage, the end of the fuselage including the rudder and elevators, is designed to twist and shake safely.</p>
<p>However the pilot during an intense turbulence or even in the absence of it can subject the plane to stresses above the safe levels, do maneuvers that are not safe for that weather and cause damage to structure.</p>
<p>Fatigue which is the first culprit to look into happens mostly by the routine stresses of flight and engine vibration and has to be checked at routine service intervals.</p>
<p>FAA crash reports are a good source to start researching about effect of bad weather on flights, planes, and pilots' decision making. </p>
<p>I did some search on Youtube and found this on flutter of the airplanes in calm weather. <a href="https://youtu.be/egDWh7jnNic" rel="nofollow noreferrer">Flutter</a></p>
| 24102 | For how long could a plane handle turbulences? |
2018-10-10T01:56:17.813 | <p>I'm prototyping a hardware that involves many thin walls to bend around curves and such, so it needs to be flexible, cheap, and also transparent. It may also be the same plastic used in a production version.</p>
<p>The only example I can think of is the plastic you find around the inside of a new dress shirt collar. (McMaster Carr doesn't seem to have anything like it, except polycarbonate, which seems expensive and overkill; surely I'm thinking of a much cheaper plastic. Maybe PP, but not finding enough info to confirm, or find out where to get some in that form.) I almost went with transparency rolls for transparency projectors, but it's too thin/flexible (5 mils).</p>
| |materials|plastic|prototyping| | <p>Polyester film / PET / PETE, according to McMaster-Carr support, is a clear, flexible, inexpensive plastic, and is the type found around the inside of a new dress shirt collar. </p>
<p>It's minimum cold bend radius (no heating necessary), at least of certain PET materials, can be approximated from its thickness as R=150T where T is the thickness.</p>
| 24109 | What type of plastic is a good clear, flexible plastic for prototyping? (~0.5mm - 1.5mm thick) |
2018-10-11T01:20:29.757 | <p><a href="https://i.stack.imgur.com/QLYjT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QLYjT.png" alt="\\this is the qeustion i am facing"></a></p>
<p>First, I thought the neutral axis would just be the y bar which can be obtained from the centroid of the cross-section but is different. I have tried letting the net moment at the wall = 0 and the reaction force I got is 1kn * 1m. But I am kinda lost on how to continue.</p>
| |structural-analysis|beam| | <p>The stresses from bending and axial should be added up by superposition. This moves the location of zero stress (neutral axis). I have included a figure to demonstrate the principle, and provided the calculations again, please note the axial stress is 40 kN/mm^2</p>
<p><span class="math-container">$$ \sigma_{bending}= m/S =\frac{1kN*1m}{(50\times 100^2)/6} =6000N/500mm^2 = 12N/mm^2$$</span> </p>
<p><span class="math-container">$$ \sigma_{axial}= 200kN/(50*100)mm^2 = 40Nmm^2 $$</span></p>
<p><span class="math-container">$$ \sigma_{max} = \sigma_{bending} + \sigma_{axial} = 12N/mm^2 + 40N/mm^2 = 52 N/mm^2$$</span>
<span class="math-container">$$ \sigma_{min} = -\sigma_{bending} + \sigma_{axial} = -12N/mm^2 + 40N/mm^2 = 28 N/mm^2$$</span></p>
<p>We can see from the scaled image, as well as the math, that the axial stress overcomes the compression stress due to bending. There is no location on the cross section where the stress is zero. Therefore the neutral axis is below the beam. We can determine this distance by similar triangles (or equation of a line, or whatever you prefer).</p>
<p><span class="math-container">$$\frac{52-28}{100} =\frac{52}{D_{NA}+D} -> (D_{NA} + D) = \frac{52*100}{52-28} = 216.67mm $$</span></p>
<p>So the total depth from the bottom surface of the beam to the neutral axis is the difference of the depth and the calculated dimension.
<span class="math-container">$D_{NA} = (216.67 - 100)mm = 116.67mm$</span>
<a href="https://i.stack.imgur.com/xFXm2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xFXm2.png" alt="scaled view showing the stress diagram"></a></p>
| 24120 | Neutral axis in canteliver beam at support |
2018-10-12T06:48:32.800 | <p>Consider 2 oddly shaped surfaces and pick a point on each of the surfaces. At the point considered we can always draw a normal to the surface. Now if the two surfaces are in contact at this same point will the normal of one surface always be coincident to the normal of the other surface?. </p>
<p>I am asking this because, when studying about gears, there is always talk of the common normal and I am trying to understand if the presence of common normal is due to the design of the gear tooth or there is always a common normal between 2 surfaces in contact.</p>
| |gears| | <p>If two surfaces are tangentially in contact meaning they have the same partial derivatives at that point; then they do have a common normal.</p>
<p>And it is the cross product of the two tangent vectors, </p>
<p><span class="math-container">$ \frac{\partial f}{\partial x} \space and\space \frac{\partial f}{ \partial y}.$</span></p>
<p>With f being the surface function.</p>
<p>However two surfaces can be in contact intersecting each other or touching at an edge, where they won't have a common normal!</p>
| 24129 | Intuitive understanding of the common normal at the point of contact of 2 gear teeth |
2018-10-12T11:08:22.810 | <p>I'm using the Eurocodes for some structural analysis and am looking for the limiting value of deflection of a plate.
EN 1993-1-7, Section 8.2 states:
<em>For limiting values of out of plane deflection w see application standard.</em></p>
<p>I'm not sure to which application standard this is referring.</p>
<p>Can anyone point me in the right direction?</p>
<p>Thanks</p>
| |structural-analysis|structures|standards| | <p>In case anyone else is ever looking for a limiting case for deflection of plates, in relation to the Eurocodes, you can find this in:</p>
<p>EN 1993-4-1 Section 9.8.2 (2)</p>
<p>Hope this helps someone!</p>
| 24131 | What is the 'application standard' referred to in EN 1993-1-7 |
2018-10-13T00:16:18.747 | <p>First post here, trying to get a little help with Autodesk Inventor.</p>
<p>I am trying to make a part that is basically a wheel with little nubs on the end (think of a lego with tabs on the outer circumference). </p>
<p>What I tried so far is making a rod and using the Constraint function to make the parts one piece. Is there a more direct way to do this? I can't seem to find a way to just add the tabs/nubs on the edge of the circle. </p>
<p><a href="https://i.stack.imgur.com/YihlA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YihlA.png" alt="enter image description here"></a></p>
| |cad|autodesk-inventor| | <p>The way I would model the component shown on the LHS of your image is like this:</p>
<p><a href="https://media.giphy.com/media/1yTg8TIe7D2Nfx5oze/giphy.gif" rel="nofollow noreferrer"><img src="https://media.giphy.com/media/1yTg8TIe7D2Nfx5oze/giphy.gif" alt="enter image description here"></a></p>
<ol>
<li>Sketch the outer diameter of the circle on one of the home planes, say XZ, with the centre of the circle at the origin.</li>
<li>Extrude the circle an equal distance in each direction, such that the XZ plane runs through it's centre</li>
<li>Sketch half of a 'nub' on the same plane, with a centreline along the X axis</li>
<li>Revolve the 'master nub' by 360degrees around the centreline in the sketch (coincident with the X axis)</li>
<li>Use the circular pattern tool to make 4 copies of this nub, equally spaced, around the Y Axis.</li>
</ol>
<p>The reason I suggest this approach is that it ensure that the nubs are all identical to one another, and all equally spaced and pointing radially out from the centre of the circle. IF you want to edit the form of the nubs, there's a single sketch to edit, and all other nubs will auto-update.</p>
<p>Using relations to control the position of things is dangerous, as they rely on the internal ID's of selected faces etc, and these are notorious for getting 'lost' as a result of model changes elsewhere in your workflow.</p>
| 24141 | How to add radial protrusions to a disc in Autodesk Inventor |
2018-10-14T09:07:20.637 | <p>The surface area of the head of a tunnel boring machine is usually flat. A cone-shaped head would increase the surface area. The question is if it could speed up the boring process.</p>
<p>I know that with current machines the boring process is also limited by other factors. But that is not part of the question.</p>
| |tunnels| | <p>Because with a flat head the waste material can be (relatively easily) collected and removed from the cutting face.</p>
<p>With a cone-shaped bit that removal is not so easily accomplished.</p>
<p>Cone-shaped bits tend to be used to push or compress the material out of the way, which is fine for "softer" materials ...</p>
| 24157 | Why are tunnel boring machines not using cone-shaped drills? |
2018-10-14T18:45:07.577 | <p>I am looking to solve the 1D Unsteady Homentropic Flow equations using the Method of Characteristics (MOC). I am using Zucrow and Hoffman's Gas Dynamics Volume 1 as a reference and I am currently stuck.</p>
<p>In chapter 13, the authors begin describing the use of MOC to solve the 1D unsteady flow equations. I understand their methodology till I get to the step that states to substitute equation (13.16) back to equation (13.15).</p>
<p>For this example:
<span class="math-container">$$
u = u(x,t)
$$</span></p>
<p>and
<span class="math-container">$$
p = p(x,t)
$$</span></p>
<p>For homentropic flow, <span class="math-container">$$ \rho = \rho(p) $$</span> </p>
<p>Equation (13.15) is the following:</p>
<p><span class="math-container">$$
(\rho u \sigma_1 + \rho a^2 \sigma_2) ( u_x + \frac{\rho \sigma_1}{\rho u \sigma_1 + \rho a^2 \sigma_2}u_t) + (\sigma_1 + u\sigma_2 )(p_x + \frac{\sigma_2}{\sigma_1 + u\sigma_2} p_t) = 0
$$</span></p>
<p>Equation (13.16) is the following:
<span class="math-container">$$
\frac{du}{dx} = u_x + \lambda u_t
$$</span></p>
<p><span class="math-container">$$
\frac{dp}{dx} = p_x + \lambda p_t
$$</span></p>
<p>The resulting equation (13.18) when you substitute (13.16) into (13.15) is:</p>
<p><span class="math-container">$$
(\rho u \sigma_1 + \rho a^2 \sigma_2) du + (\sigma_1 + u\sigma_2 ) dp = 0
$$</span></p>
<p>I understand that
<span class="math-container">$$
\lambda = \frac{\sigma_1}{u\sigma_1 + a^2\sigma_2} = \frac{\sigma_2}{\sigma_1+u\sigma_2}
$$</span></p>
<p>However, I have not been able to get (13.18) from back-substituting (13.16) into (13.15).</p>
<p>Does anyone have any suggestions for me to consider, look into, etc.?</p>
<p>Thank you!</p>
| |fluid-mechanics|compressible-flow| | <p>In (13.15) use the first equation for <span class="math-container">$\lambda$</span> in the first term, and the second equation in the second term. That gives
<span class="math-container">$$(\rho u \sigma_1 + \rho a^2 \sigma_2) \frac{du}{dx} + (\sigma_1 + u\sigma_2 )\frac{dp}{dx} = 0.$$</span></p>
<p>Then use the fact that <span class="math-container">$$\frac{du}{dx} \Bigm / \frac{dp}{dx} = \frac{du}{dp}.$$</span></p>
| 24162 | Solving 1D Unsteady Homentropic Flow using Method of Characteristics |
2018-10-15T19:29:08.827 | <p>Actually, I see some papers use viscoelasticity or hyperelastic model for the same materials. It seems to me that those models were invented in different eras, and the hyperelastic is a newer model than viscoelasticity. Am I right?</p>
<p>What's the difference?</p>
<p>What is the best way to decide which model to use when we want to simulate a material?</p>
| |materials|finite-element-method|modeling|elastic-modulus| | <p>Superelasticity becomes linear at small deformations, but becomes nonlinear at medium or large deformations; But nonlinear elasticity is nonlinear when subjected to small deformations.</p>
| 24180 | Viscoelasticity and Hyperelastic model, history and difference |
2018-10-16T12:20:34.677 | <p>I'm currently trying to run my solutions in batch mode because in the future I'll be running simulations which I want to be done in the background.</p>
<p>I've tried to adapt the code from here: <a href="https://www.sharcnet.ca/Software/Fluent6/html/ug/node22.htm" rel="nofollow noreferrer">https://www.sharcnet.ca/Software/Fluent6/html/ug/node22.htm</a></p>
<p>Using some additional commands from here: <a href="https://www.sharcnet.ca/Software/Ansys/16.2.3/en-us/help/flu_gs/flu_ug_sec_startup_option.html" rel="nofollow noreferrer">https://www.sharcnet.ca/Software/Ansys/16.2.3/en-us/help/flu_gs/flu_ug_sec_startup_option.html</a></p>
<p>What I tried to do was:</p>
<pre><code>fluent 3d -gr -t36 < inputfile >& outputfile &
</code></pre>
<p>Where the inputfile has no extension and has the following content:</p>
<pre><code>; Read case file
rc CoolingChannelNoBatWall3D.cas
; Initialize the solution
/solve/initialize/initialize-flow
; Calculate 500 iterations
it 500
; Write case & data file
wcd CoolingChannelNoBatWall3DRes.cas
; Exit FLUENT
exit
yes
</code></pre>
<p>This starts the GUI, even though I tried to suppress it. After starting the GUI it does nothing. What am I doing wrong here?</p>
<p>EDIT:</p>
<pre><code>fluent 3d -t 36 -g -i < inputfile.jou >& outputfile.jou &
</code></pre>
<p>Does not start the gui, but it's also quite hard to see if anything's really happening.</p>
| |fluent| | <p>Solution was easier than expected, I didn't give the full file path to the fluent command: </p>
<p><code>\ansys_inc\v192\fluent\bin\fluent 3d -t 36 -g -i < inputfile.jou >& outputfile.jou &</code></p>
| 24186 | Solving parallel without GUI in batch mode not working as expected (Linux) |
2018-10-17T05:58:56.123 | <p>Is resolution of forces similar to expressing a vector as a linear combination of some other vectors in linear algebra?
Similarly, Is the state of stress at a point in 3D similar to expressing the applied load as a linear combination of 9 other vectors(the shear and normal components along the direction of the three axes)?</p>
| |stresses|strength| | <p>Sort of. </p>
<p>When you resolve a vector into its component, you have to choose a coordinate system, resolving the vectors into components is the same as projecting the original vector on the coordinate axis. So yes the resolution is equivalent of linear combination. </p>
<p>If the stress tensor is symmetric then you need only six basis matrix to construct the linear space. </p>
<p>Otherwise you need nine component to describe the linear space. Notice, you cannot construct a <strong>general linear matrix space</strong> with vectors, you need matrixes. So the state of stress is generally the linear combination of nine other matrixes. </p>
| 24199 | Resolution of forces and state of stress at a point |
2018-10-17T20:27:39.210 | <p>I'm building a 4' x 8' gantry-style CNC router table. The spindle motor uses pneumatics to power its draw bar (for automatic tool changes). There are three separate pneumatic supply lines that go to the spindle motor.</p>
<p>One is a constant air supply of low pressure that serves to blow debris out of the tool mount during tool changes. The other two are the "open" and "close" lines.</p>
<p>I can choose to place the electrically-actuated pneumatic valves and various pneumatic components on the base of the machine, on the gantry, or on the spindle mount itself.</p>
<p>In the first case, three pneumatic lines must run several meters to reach the spindle, but the bulk of the pneumatic hardware is kept off the moving parts of the machine (reducing vibration and moving mass).</p>
<p>In the last case, the electrical control lines have to be run, but the pneumatic control lines are shortened (except for the main supply line), and a number of valves and regulators have to be placed on the gantry carriage. The second case is similar, but the equipment is mounted to the gantry itself.</p>
<p>I'm leaning toward just running the three pneumatic lines the full distance and keeping as much equipment as possible off the moving parts of the machine. But I don't know if the long pneumatic lines will cause other problems, perhaps sluggishness due to the elasticity of the tubing (they're typical poly or PVC pneumatic tubing). Is this even a concern?</p>
| |pneumatic| | <p>I would keep all the control hardware/components off of the moving members as much as possible and in the same general area for ease of wiring and trouble shooting. Use nylon tubing, its more "rigid" such that the tubing diameter will not flex or change. </p>
<p>In electro-pneumatic machines, I usually build a pneumatic "panel/enclosure" with all my air prep/sensors/valves/etc, just as you would your electrical control, and keep it close to the electrical panel/enclosure. </p>
<p>In this application, I wouldn't consider any time delays due to the tubing lengths to be significant. </p>
| 24207 | Best balance of electrical and pneumatic control lines in CNC router table? |
2018-10-18T08:40:32.807 | <p>I want to build my own off-grid system but confused with what sizes (V, I) to get.</p>
<p>I want to run a 24v system powering 4.4kWh/day. I have 4 hours of usable sunlight a day and space for panels and equipment is not an issue (within reason).</p>
<p>Looking at the calculations:</p>
<p>Electrics = 220v @ 60Hz<br>
Load = 4.4kWh/day<br>
Sunlight = 4 hrs </p>
<p>The PV array I then calculate at:</p>
<p>PV Array = 4400Wh / 4hrs = 1100W (10 x 100w panels)</p>
<p>The batteries I then work out at:</p>
<p>4400W / 24v = 184Ah / 80% discharge = 220Ah</p>
<p>A battery example [1] - 24v 220Ah deep-cycle li-ion (2000times(80%DOD))<br>
* or if you want to go down building your own battery pack with 18650's - 7 in series (to get 24v) x 55 packs in parallel (to get 220Ah).</p>
<p>The calculation problems I cant find anyone explaining on Google is:</p>
<pre><code>a) What size charge controller do I need? I see options from 10A to 100A.
b) What size inverter do I need? I see options up from 100W to 3000W and more.
c) Does the PV array not need to be x2 the size so it can power the house and charge the batteries at the same time during the day?
</code></pre>
<p>[1] <a href="https://www.alibaba.com/product-detail/Custom-24-Volt-Solar-24V-200Ah_60670679703.html?spm=a2700.7724857.normalList.25.15c91a82Albyfj" rel="nofollow noreferrer">https://www.alibaba.com/product-detail/Custom-24-Volt-Solar-24V-200Ah_60670679703.html?spm=a2700.7724857.normalList.25.15c91a82Albyfj</a></p>
| |solar-energy| | <p>The answers to (a) and (b) depend on the <em>peak</em> load you want to use - i.e. the highest powered devices you want to run. Note that these might be things that you only run for a few minutes (e.g. an electric kettle), so they don't eat up the whole of your daily 4.4kWh energy budget.</p>
<p>The answer to (c) is "no". For example, suppose your appliances used 1100W for the entire four hours of daylight. The panels would be powering the appliances but not charging the battery at all. But <em>that doesn't matter</em>, because you have used all your daily energy budget in those four hours, so <em>to stay within the design limits of your system</em> you will not be using any electricity at all for the remaining 20 hours of the day, and it doesn't matter that the battery isn't charged!</p>
<p>Of course that is an extreme example to make the point easily - but the same thing will apply for any usage pattern over 24 hours.</p>
<p>Note, you need to include the efficiency factors of the components in your system design. The efficiency of the battery charging system won't be 100%. The inverter should have a high efficiency, but it won't be 100% either. Its efficiency might vary depending on the load (and it might be worse for light loads than for heavy ones).</p>
<p>Also, ask yourself whether "4.4kWh per day" is a <em>long term</em> average (over a week, or a month, or a year) rather than the daily limit for every day. You might have some high-power devices (e.g. a washing machine) that you don't use every day, so you need to store energy for a longer period of time - i.e. you need a bigger battery capacity.</p>
| 24210 | Electronics confusion |
2018-10-19T03:46:00.880 | <p><a href="https://i.stack.imgur.com/bkgKJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bkgKJ.png" alt="enter image description here"></a></p>
<p>I solved this problem by doing a mass balance for the mixing chamber and an energy balance for the water heater and the mixing chamber as follows:</p>
<p><span class="math-container">$\dot{m}_{\text{mix}}=\dot{m}_{\text{Tank}}+\dot{m}_{\text{Cold}}\\
\\
\dot{m}_{\text{mix}}\hat{H}_{\text{mix}}=\dot{m}_{\text{Tank}}\hat{H}_{\text{tank}}+\dot{m}_{\text{Cold}}\hat{H}_{\text{cold}}\\
\\
\frac{m_{\text{water in tank}}C_{\text{v}}\left ( T_2-T_1 \right )}{\Delta t}=\dot{W}_{\text{e}}+\dot{m}_{\text{Cold}}\hat{H}_{\text{cold}}-\dot{m}_{\text{Tank}}\hat{H}_{\text{tank}}$</span></p>
<p>resulting in:</p>
<p><span class="math-container">$\dot{m}_{\text{Tank}}=0.05838 \text{ kg/s}\\
{T}_{\text{shower}}=44.67 \text{ °C}$</span></p>
<p>and my results are in agreement with the solutions manual. But, if I do a mass balance in the water heater I obtain</p>
<p><span class="math-container">$\dot{m}_{\text{water heater}}=\dot{m}_{\text{Cold}}-\dot{m}_{\text{TanK}}$</span> </p>
<p>and if I replace the value that I have obtained previously, I do not obtain a zero mass flow for the tank. So, ¿Is my answer or mass balance wrong, or I misunderstand something in the problem statement?</p>
<p>Regards</p>
| |mechanical-engineering|thermodynamics|chemical-engineering| | <p>The results are correct and not contradictory at all, but the way you interpret the problems is not totally correct. </p>
<p>For mass or impulse balance, you have to first define a <strong>control volume</strong>. I define it as the volume of the tank, then the continuity equation: </p>
<p><span class="math-container">$$\iint_S \rho(\vec{v}_{in}.\vec{n}) \,ds+\iint_S \rho(\vec{v}_{out}.\vec{n}) \,ds+\frac{\partial }{\partial t}\iiint_V \rho \,dV = 0$$</span></p>
<p>The first two terms are mass flux into and out of the control volume respectively, and the third term represents the mass accumulation, here the last term is zero.
<span class="math-container">$\rho$</span> is the density and <span class="math-container">$\vec{v}_{in}$</span>, <span class="math-container">$\vec{v}_{out}$</span> the velocity of the fluid. Notice the terms in the above equation have the dimension of <span class="math-container">$\frac{kg}{s}$</span>. </p>
<p>By defining the tank as the control volume, the mass per second that exits the tank is indeed equal to <span class="math-container">$0.05838\frac{kg}{s} $</span> (assuming your calculations are correct). By assuming no accumulation in the tank, this mass should be replaced by the cold water, however this is not necessarily equal to the mass of cold water mixing in the mixing chamber. You confuse the control volume of the tank with the control volume of the mixing chamber. Another way to say it, is there are two channels that provide the cold water, one goes into the tank with the same mass rate as the hot water comes out of the other side, and the one goes inside the mixing chamber.
Now if you define the control volume as the volume of the tank and the mixing chamber then you'll see, the <span class="math-container">$m_{mix}$</span> is equal to <span class="math-container">$m_{tank} + m_{cold, chamber} +m_{cold, tank} $</span></p>
| 24231 | Contradictory result for mass balance |
2018-10-19T06:53:57.363 | <p>I would like to know what kind of glass is used in biometric attendance machines over which we press our thumb? Is there something special about these glasses or their manufacturing?</p>
| |glass| | <p>According to <a href="https://www.us.schott.com/innovation/security-at-your-fingertips/" rel="nofollow noreferrer">this glass manufacturer</a> it is more about the optical properties than mechanical ones. They are looking for:</p>
<ul>
<li>Homogeneous illumination of the touchsurface</li>
<li>Transparent and durable in UV and IR light radiation</li>
<li>Temperature-resistant to heat development in LEDs</li>
</ul>
<p>They use <strong>borosilicate glass</strong> and <strong>aluminosilicate glass</strong>, which are by themselves scratch-resistant.</p>
| 24232 | Glass used in biometric machines |
2018-10-19T15:20:25.460 | <p>I am trying to bend this sheet metal concave lip inside the electrical box. I also need the top of the bent part below the top edge of the outside of the box.
<a href="https://i.stack.imgur.com/pQ671.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pQ671.png" alt=""></a></p>
<p><a href="https://i.stack.imgur.com/4QCy2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4QCy2.png" alt="enter image description here"></a></p>
| |cad|autodesk-inventor|metal-folding|surface-modelling| | <p>I've had a quick go - it's hard to be more precise without measurements, and it should be stressed that the actual bent shape of the metal is <em>very</em> weird, and difficult to model. Hopefully this will give you a method to get started with, however, and you can adjust spline handle lengths and dimensions etc. as you go.</p>
<p><a href="https://i.stack.imgur.com/AEhTY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AEhTY.png" alt="First Attempt"></a></p>
<p>The steps go as follows:</p>
<ol>
<li>Model the 'non-weird' bits, and float them in space.</li>
<li>Create a 3D sketch with splines connecting the two bodies, set to be tangent to the edges. Right click the spline, and click "activate handle" to give yourself more control over its form. Also include lines connecting these splines to create a closed loop.</li>
<li>Use "Boundary Patch" to generate a surface body inside the 3D sketch you just created</li>
<li>Mirror the Surface Body to create a symmetrical copy of it.</li>
<li>Sketch on the surface that will have the bend attached, and draw two lines. Use the split tool to split the surface accordingly.</li>
<li>Use the Delete Face tool to remove the faces of the solid body which will join the bent part. It has now also become a surface body.</li>
<li>Use Boundary patch again to create the outer face of the bend this time with automatic edge chain deselected, and selecting the edges of the surface bodies, rather than anything in a sketch. Set a Tangent Condition on the edges that are 'bent', and leave it as Contact for the 'cut' edges. of the tab.</li>
<li>Repeat for the inner face of the bend</li>
<li>Use the 'Stitch Surface' Tool, and uncheck "Maintain as Surface" to merge all the surface bodies and create a solid.</li>
<li>Create a new work plane at a slight angle, and sketch a circle on this for use as an extrude cut for the cutout hole. This is an approximation, of course. You could use the draft tools to create a taper in this hole closer to real life, but without the object for reference, I've decided this is good enough - you can experiment with other methods!</li>
<li>All Done</li>
<li>Mirror/Rotate/Whichever to create the opposite side of the box. Also, to create a 12th point on this list and make my image nicer :) <a href="https://i.stack.imgur.com/EMSrO.png" rel="nofollow noreferrer">High resolution version available here</a> </li>
</ol>
<p><a href="https://i.stack.imgur.com/EMSrO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EMSrO.png" alt="enter image description here"></a></p>
| 24242 | Bending a concave sheet metal lip in AutoDesk Inventor |
2018-10-19T20:37:39.173 | <p>I was wondering, if there is a shock load resistant electromechanical actuator that can withstand powerful impacts?
Is there an electromechanical counterpart for pneumatic cylinders that is as shockresistant as hydraulic or pneumatic actuators?</p>
<p>And if not, is there some kind of way to make them withstand impacts via external parts?</p>
| |robotics|actuator|linear-motors| | <p>EDrive Corporation makes a linear actuator called the Eliminator SS rated at 65in/sec and a thrust of 4000lbf. If you want to increase its shock resistance you can take the actuator and mount it in a ruggedized housing of some sort with an opening for actuator arm to do its work. You can call the company and find out if it has a shock rating.</p>
| 24246 | shock-resistant electromechanical linear actuator? |
2018-10-20T03:41:59.073 | <p>I have read about a device called drumometer, which measures the number of strokes of hitting drum stick.</p>
<p>I think this is almost the same, but instead, I would like to measure the number of strokes/impacts made when I shake a very small object inside a very small container.</p>
<p>To give an example, imagine a cube where one face is removed, hence making a hole. Then, I put the end (say X) of a stick inside that cube and hold the other end (say Y) from outside. I then start vibrating (I mean like stirring movement) the stick so that the X end hits all faces of the cube. Then I want to measure how many times the X end of that stick hits not necessarily all faces but at least one face (since I will shake it randomly anyway).</p>
<p>The cube could be as small as 1 cm^3 (it might not even be a cube, just something with a hole that small where a stick can be put inside and shook from outside), and the stick is just long enough for me to hold it and small enough to get through the hole.</p>
<p>Thus, in relation with that drumometer, it kinda looks like an even smaller "drumometer" is needed. What do you think?</p>
<p>I don't have good background in measurement, only have some elementary freshmen undergraduate physics background instead. This is gonna be very helpful in my experiment once I know how to measure the number of strokes.
*feel free to edit my tags if not related</p>
| |measurements|vibration|instrumentation| | <p>Piezoelectric thin film sensors can be made as small as you want (well maybe not Planck length) :-)</p>
<p>All you have to do is pre-amp the signal from the sensor and send the output to a scope to see the waveform caused by the impact, or any other recording device you want.</p>
| 24250 | How to measure number of strokes in a small space |
2018-10-20T17:52:32.217 | <p>I'd like to use a titanium wire to make some jewelry. I ordered two kinds of titanium wire, 0.3mm thick and 0.5mm thick. Both of them are grade 1 titanium.</p>
<p>How do I calculate the strength they'll have, i.e. the amount of weight they could lift before snapping? </p>
<p>I know that titanium grade 1 has an ultimate tensile strength of 240 MPa, but I don't know how to use that information to get the amount of force needed to snap the wire.</p>
| |materials| | <p>Stress = Force / Area, and "tensile strength" really means "tensile stress".</p>
<p>You need to be careful with units (and some people would recommend always convert everything into "basic SI units" of meters and Pascals), but using MPa and mm conveniently gives the force in Newtons.</p>
<p>The cross section area of your thin wire is <span class="math-container">$3.14 \times 0.3 \times 0.3 \,/\, 4 = 0.07065$</span> square mm, so the force to snap the wire is <span class="math-container">$0.07065 \times 240 = 16.956$</span> Newtons, or a weight of <span class="math-container">$16.956 / 9.81 = 1.73$</span> Kg.</p>
<p>For the thicker wire the force is <span class="math-container">$4.8$</span> Kg.</p>
<p>Note these are only approximate. You want to have a safety factor of at least 2, and maybe as high as 5, so the <em>safe</em> weights might be more like 350 grams and 960 grams.</p>
<p>The best thing to do would be attach something weighing that amount to a bit of wire, handle it roughly (for example jerk the wire rather than picking it up carefully) and convince yourself it isn't going to break before you start creating your jewellery.</p>
| 24260 | How to calculate strength for titanium wire |
2018-10-20T20:22:24.083 | <p>I am doing research into lunar regolith's hardness in relation to abrasion of lunar tools and need to choose a hardness scale (either Mohs, or Rockwell A, B, or C). Is either of these metrics more legitimate/accurate than the other?</p>
| |mechanical-engineering|materials| | <p>Rockwell,Brinell, Knoop, Vickers, etc are for materials like metals that have at least a little ductility as the rely on plastic deformation. Mohs is for brittle minerals although it is somewhat qualitative. Abrasion of tools is not that simple anyway. For example : aluminum is softer than many metals but aluminum castings are very abrasive to cutting tools as the high silicon content combines with the aluminum to make hard abrasive particles in the soft matrix. </p>
| 24263 | Mineral Hardness Scales |
2018-10-21T12:16:01.673 | <p>If we give a sinusoidal <span class="math-container">$\cos(\omega t)$</span> to a 1/s, it means that you just integrate the given input. </p>
<p>What if we have the same input but Transfer Funtion with a pole, let's say 1/(s+1), what will be the output of the system ?</p>
<p>Edit #1</p>
<p>According to my hand calculations, I guess there no way rather than dealing with Laplace Transform.</p>
<p>Hint: As you may know,</p>
<p><span class="math-container">$$\begin{gather}
\cos(\omega t) \rightarrow s/(s^2+w^2) \\
h(s) \rightarrow 1/(s+1)
\end{gather}$$</span></p>
<p>output --> multiply them, and then apply partial fraction expansion, and then Laplace^-1</p>
<p>P.S.(Took approx. 30 min)</p>
| |transfer-function| | <p>For a sinusoidal input signal at frequency <span class="math-container">$\omega$</span> in rad/s you can use that the steady state response of the system would be another sinusoidal signal whose amplitude is scaled by <span class="math-container">$|H(j\,\omega)|$</span> and its phase is shifted by <span class="math-container">$\angle H(j\,\omega)$</span>. The total output of the system would be the sum of the steady state response and the transient response. The transient response only depends on the initial conditions and <span class="math-container">$H(s)$</span>.</p>
<p>For example when the input is <span class="math-container">$\cos (\omega\,t)$</span> and <span class="math-container">$H(s)=1/(1+s)$</span> you get that the steady state response is equal to <span class="math-container">$|H(j\,\omega)|\cos(\omega\,t+\angle H(j\,\omega))$</span>. The values for <span class="math-container">$|H(j\,\omega)|$</span> and <span class="math-container">$\angle H(j\,\omega)$</span> can be found by multiplying and dividing <span class="math-container">$H(j\,\omega)$</span> by the complex conjugate of its denominator</p>
<p><span class="math-container">$$
H(j\,\omega) = \frac{1}{1+j\,\omega}\frac{1-j\,\omega}{1-j\,\omega} = \frac{1-j\,\omega}{1+\omega^2}
$$</span></p>
<p>such that</p>
<p><span class="math-container">$$
|H(j\,\omega)| = \frac{\sqrt{1 + \omega^2}}{1+\omega^2}
$$</span>
<span class="math-container">$$
\angle H(j\,\omega) = atan2(-\omega, 1)
$$</span></p>
<p>The transient response of <span class="math-container">$H(s)$</span> are a scaled summation of <span class="math-container">$e^{\lambda\,t}$</span>, where <span class="math-container">$\lambda$</span> are the poles of <span class="math-container">$H(s)$</span>. So for this example the transient would be <span class="math-container">$C\,e^{-t}$</span> with <span class="math-container">$C$</span> depending on the initial conditions.</p>
| 24275 | Input-Output Relation |
2018-10-22T06:09:46.147 | <p>I'm new to solidworks, and I built my part and would like to put it into a drawing. Its a small square box-like base with one side of the base having holes in it. When I put my parts into the drawing, it shows 2 sides of the base, a bottom view, and an isometric view, but it never shows the side/plane with the holes on it. My question is this, how do I have a solidworks drawing show a different plane or side of the base?</p>
| |solidworks|drawings| | <p>You can use the "Projected View" button to add another view, by selecting an existing view. You should bear in mind, however, that you should avoid adding unnecessary views if you can fully define the geometry of the object with only a few. I can't comment on this without an image of your object, however.</p>
<p><a href="https://i.stack.imgur.com/sUdNJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sUdNJ.png" alt="Projected View Button"></a></p>
| 24291 | Show all sides of a solidworks part in a solidworks drawing |
2018-10-22T19:25:17.913 | <p>I have seen anti-condensation coatings for glass, but as far as I know they're designed to cause any moisture that does form on them to sheet rather than bead. They do not, however, actually prevent condensation, and I don't see how any product could (other than extreme insulation, of course). The formation of condensation is an issue of physics; warm moist air meets cold surface and since cold air holds less moisture than warm, the moisture forms on the cold surface. On the other hand, engineers are pretty clever, so if there's a way to prevent this, it may be possible, even if I can't figure out how.</p>
<p>This question is actually prompted by some reviews of some <a href="https://rads.stackoverflow.com/amzn/click/B07CT98WJM" rel="nofollow noreferrer">ice packs</a> I recently purchased from Amazon, in which reviewers complained about condensation, which seems like a nonsense complaint.</p>
<p>Sorry about the tag, but I tried "condensation," "dew," "moisture," "water," and maybe a couple of others and got no match.</p>
| |surface-preparation| | <p>The condensation mechanism of water vapor onto a chilled surface requires not only a chilled surface but also the presence of nucleation seed sites in the surface to kickstart the condensation kinetics.</p>
<p>Those nucleation seeds are sites on the surface (usually microscopic pores, crevices or specks of dust) which have a little water already trapped in them. The supersaturated vapor nearby "sees" that water and comes out of solution right there, adding to the amount of water originally present. Growing globs of condensed water then result, which can then freeze solid if conditions are right.</p>
<p>By coating a chilled surface with an extremely low-energy substance like a fluorosilicon oil or wax or a thin layer of a solid fluoropolymer, those surface sites are poisoned and no longer present themselves to the nearby water vapor as pre-existing bodies of condensate- and condensation is delayed or possibly eliminated altogether. </p>
<p>The problem with inhibiting condensation with low-energy coatings is that they eventually degrade and lose their effectiveness by accumulating dirt and grime. So a brand-new jacket made of Gore-Tex(tm) inhibits conndensation for a few months of wear but then supports condensation like anything else after it gets dusty & dirty. </p>
| 24303 | Would it be possible to put an anti-condensation coating on, say, an ice bag such that it will not get wet in use? |
2018-10-22T23:55:12.653 | <p><a href="https://i.stack.imgur.com/Eh80H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Eh80H.png" alt="enter image description here"></a></p>
<p>I'm an industrial engineering student taking engineering mechanics, and I'm quite confused by this problem our proffessor gave us, the chair is supposely in equilibrium, yet in order to counter the moment created by the weight pressed on the chair upward y vector vertical force on leg be would need to be eade, and cos(b) is greater than cos(a) - and she says there's no friction to insure horizntal equilibrium through resistance, could someone please show me where I'm screwing up on this, - please help! I don't want the answer I just want to see what I 'm doing wrong and fi< it, I just want to know how to solve this problem... </p>
| |statics| | <p>Using the principle of virtual work...</p>
<p>If the chair is rotated infinite infinitesimally about the point A, each point will move an amount vertically that is proportional to the horizontal distance from A. The only points that we are interested are B where the reaction force acts upwards and where W acts downwards.</p>
<p><span class="math-container">$$ (200+200+100) \ R_B + (-65/2) \ (60+200+100)=0$$</span>
<span class="math-container">$$500 \ R_B = 11700$$</span>
<span class="math-container">$$R_B = 117/5= 23.4$$</span></p>
<p>The same procedure can how be applied by rotating the chair an infinitesimal amount about the point B.</p>
<p><span class="math-container">$$(200+200+100) \ R_A +(-65/2)\ (200-60)=0$$</span>
<span class="math-container">$$ 500 \ R_A = 4550 $$</span>
<span class="math-container">$$R_A = 45.5/5 = 9.1$$</span></p>
| 24312 | Statics problem ignoring horizontal reactions seems to require horizontal forces |
2018-10-23T10:07:32.560 | <p>I'm trying to wrap my head around the question: "what is the optimal slip ratio and how do I go about controlling it?". I'm working on a control strategy for an electrical vehicle. The issue I'm facing is that of maximizing traction. The way I can measure slip ratio is by measuring difference between wheel velocity and vehicle velocity (both of which let's assume are measured separately).</p>
<p>Now, what I need to do (I think) is to figure out a way to maximize traction by always keeping the slip ratio at some specific value. But what value? Also it seems wrong to always try to keep wheel speeds at an offset from vehicle speed. I suppose I only should be doing this during acceleration/deceleration? How can I do this in a generic way without having to implement different strategies for accel/decel? </p>
| |automotive-engineering| | <p>Ended up not worrying about slip and instead just making sure the tires don't spin out by synchronizing their velocity but without forcing them to spin at the same velocity (ie I implemented coupling). Now slip is held at optimal value by the very nature of this control strategy. </p>
| 24317 | How to actually control slip? |
2018-10-23T14:45:10.660 | <p><a href="https://en.wikipedia.org/wiki/Thermostat#Pneumatic_thermostats" rel="nofollow noreferrer">According to Wikipedia</a>, pneumatic thermostats were invented after electric thermostats.</p>
<p>Yet <a href="http://members.questline.com/Article.aspx?articleID=26707&accountID=4536&nl=15031" rel="nofollow noreferrer">this article</a> claims that "at least 25 percent of U.S. commercial buildings are served by pneumatic controls, according to the U.S. Department of Energy". I would guess that they were even more popular in the past.</p>
<p>This seems surprising. Why set up an air compressor system and run pneumatic tubes through the walls? Wouldn't an electric system be easier to set up and maintain?</p>
<p>What were the factors that made pneumatic thermostats so popular?</p>
| |building-design|pneumatic|engineering-history| | <p>In the 1940's and 50's, dynamic system controllers were invented which used air pressure to work. It was possible in fact to design and build not only analog control mechanisms but also binary logic devices that ran on compressed air and which could perform simple switching functions. These devices were less expensive and more durable than electronic logic based on vacuum tubes and relays and were popular all the way up into the early 1970's. They were called "pneumatic logic" or "air control" and they disappeared quickly when solid state logic became cheap. </p>
| 24322 | Why were pneumatic thermostats commonly installed in buildings? |
2018-10-25T13:57:48.440 | <p>Studying Fluid Mechanics, I started to notice that almost every textbook/website uses a specific point to make calculations about the pressure in a <strong>liquid</strong> at a given depth (hydrostatic pressure): <strong>the geometric center</strong> (<em>as shown in the images below</em>), when presenting pressure gauges/manometers/piezometers.</p>
<p><strong>Note:</strong> This happens regardless of the field to which the book is directed (I looked in textbooks of Fluid Mechanics for Civil, for Electrical, for Mechanical...).
<img src="https://i.stack.imgur.com/qX4KO.png" alt="Pressure gauges/Manometers/Piezometers" />
<sub>Images sources: MATHalino/PennState College of Engineering (MNE)/The SensorsGuide/University of Sydney (MDP)/ScienceStruck/Chegg </sub></p>
<p><img src="https://i.stack.imgur.com/V33Io.jpg" alt="Pressure gauges/Manometers/Piezometers" />
<sub>Sources: Introduction to Fluid Mechanics - Nakayama & Boucher/Mecânica dos Fluidos - Noções e Aplicações - Sylvio R. Bistafa/Chegg</sub></p>
<hr />
<p>One of the textbooks I looked at even draws attention to this fact, but it doesn't explain the reason for the choice:</p>
<blockquote>
<p>Note the origin of the measurement of h, in the center of the tube</p>
<p> <img src="https://i.stack.imgur.com/q3uxu.png" width="400" /><br />
<sub>Source: Mecânica dos Fluidos - Franco Brunetti</sub></p>
</blockquote>
<p>A similar behavior can be identified when textbooks present liquids in motion: they use the centerline of the pipe to make calculations/measurements. Here's an example:</p>
<p> <img src="https://i.stack.imgur.com/Ejj2a.jpg" width="400" /><br />
<sub>Source: Fluid Mechanics for Civil Engineers - N.B. Webber</sub></p>
<hr />
<p>So why is the choice of geometric center/centerline of the pipe so common when measuring/calculating pressure? Some hypotheses:</p>
<ul>
<li>Maybe all the textbooks/websites are unconsciously copying each other?</li>
<li>Maybe is this some kind of "convention"?</li>
<li>Maybe it is because a point in the horizontal plane of the geometric center gives the average pressure of a tank?</li>
</ul>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>I'll divide my answer in two cases. First, I'll talk about <strong>liquids in motion</strong> (assuming incompressible flow). Then, I'll talk about <strong>liquids at rest</strong>.</p>
<hr>
<h2>Liquid Flow:</h2>
<p>Reading the comments of <a href="https://youtu.be/_VDVi_zq3nk" rel="nofollow noreferrer">this</a> YouTube video about piezometers made by Donald Elger, I found the answer for this case:</p>
<blockquote>
<p>Why is it [the pressure measurement with piezometer] taken from the middle of the pipe?</p>
<p>Elger's answer: The pressure variation across a section of a pipe is hydrostatic; thus, the pressure will vary linearly with radius and the pressure at the center of the pipe is the average pressure. If you use this value of pressure in your calculations, this will be give you the <strong>most accurate results</strong>. Thus, engineers nearly always apply or measure the pressure at the center of the pipe. </p>
</blockquote>
<p>The question that came to me as soon as I read this was: <a href="https://engineering.stackexchange.com/questions/24425">"Why using average pressure in calculations gives the most accurate results?"</a>. </p>
<p>(I recommend reading my answer to this question before proceeding)</p>
<p>Briefly, in general, the average pressure gives the most accurate results if used in calculations because there are many applications/cases in which the locations with <span class="math-container">$P=P_{average}$</span> are the best places for experimental data collection.</p>
<p>In the case of a pipe, this location is its centerline. So, I believe that this is why textbooks generally choose this location in case of liquids in motion: the centerline is associated with <span class="math-container">$P_{average}$</span> that, in its turn, is associated with the best places for experimental data collection for many applications.</p>
<hr>
<h2>Liquids at rest:</h2>
<p>For this case, firstly I would like to quote part of the answer written by David White to my question on Physics.SE <a href="https://physics.stackexchange.com/a/436482/102936">"Where is the right place to put the pressure gauge to measure the pressure of a tank?"</a>:</p>
<blockquote>
<p>The location depends on why you are measuring the pressure. There will be a process reason for the pressure measurement, and that will determine the location of the pressure measuring device.</p>
</blockquote>
<p>When textbooks present pressure gauges/manometers/piezometers for the first time, the presentation is usually "application neutral" (i.e., there's no process reason), the diagrams/sketches/figures are only to illustrate the concepts/formulas. Therefore there are no best points as in the liquid flow case, for two reasons: </p>
<ul>
<li>There is no process reason that determines the location of measurement;</li>
<li>Since the liquid is at rest, there are no points that lead to most accurate results, they all provide the same accuracy.</li>
</ul>
<p><strong>But the authors need to choose a point to do the pressure-related calculations...</strong></p>
<p>After everything I've researched, my hypothesis is that the "point choice" of hydrostatics was imported from hydrodynamics. So, instead of choosing a random point to pressure-related calculations, they choose one that at least has importance/meaning for other areas of Fluid Mechanics.</p>
| 24357 | Why textbooks use geometric center/centerline of the pipe when calculating/measuring pressure? |
2018-10-27T18:28:19.237 | <p>I am having a bar and surface (made of aluminum). I need to attach some material to the end of the bar so that, if I apply any force (25 N-sideways) to bar it has to stick with the surface. </p>
<p>Force is applied normally to the flatter surface of the bar. I need to select a material at the end of bar so that the force applied on the bar is minimal.</p>
<p>What methodology should I choose to find an answer? </p>
<p><a href="https://i.stack.imgur.com/ncIZY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ncIZY.png" alt="enter image description here"></a></p>
<p>Aim is lowest necessary force to keep the rod intact (with a material added to rod end)</p>
| |mechanical-engineering|materials|applied-mechanics|tribology| | <p>I'm trying to approach the problem at the high school level. </p>
<p>The vertical force is proportional to the tangential force acting at the other end of the rod, the ratio of these two is called the static coefficient of friction. </p>
<p>Here the applied forces are due to robotic arm and the weight of the rod:
<span class="math-container">$$(F+mg)\mu_s=25\ [N]$$</span> or:
<span class="math-container">$$F\mu_s=25\ [N] - mg\mu_s$$</span></p>
<p>As you can see we have an equation with two unknowns, <span class="math-container">$F$</span> the vertical force of robots arm and the material dependent parameter <span class="math-container">$\mu_s$</span>. </p>
<p>If you look at the plot: <a href="https://i.stack.imgur.com/eR8Bs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eR8Bs.png" alt="enter image description here"></a> </p>
<p>you observe if you increase the <span class="math-container">$\mu_s$</span>, the applied force will decrease, but it makes it difficult for the rod to move freely, if you increase the force the coefficient of friction will decrease but you want to keep the applied force minimum so this is not an option. So we need to define a range for the force <span class="math-container">$F$</span> to solve the problem with bounded extrema, or simply choose a heavier rod, but make sure the force <span class="math-container">$25\ [N]$</span> can pull it back. </p>
| 24394 | Approach to solve a frictional problem? |
2018-10-28T07:51:13.117 | <p>I've worked with <a href="http://www.supraalloys.com/titanium-grades.php" rel="noreferrer">different grades of titanium</a>. Mostly with grade 5, and also grade 1 which is ductile.</p>
<p>I know that some glasses are made of titanium and are called "memory titanium", I guess because they bend back to their original shape?</p>
<p>My question is, which grade of titanium would that be?</p>
| |materials|metals| | <p>Those glasses are made out of Shape Memory Alloys (SMA), from which the most common ones include nickel, titanium, copper, zinc, or iron in their composition. I don't think the tag: <strong>grade of titanium</strong>, is applicable for SMAs even it they contain a high percentage of titanium. </p>
<p>As a functional materials, their properties and applications are so different that they have their own system. For example, one of the most recognizable brands of SMAs, Nitinol (nickel-titanium):</p>
<blockquote>
<p>...offers different <strong>grades of Nitinol</strong>, which are distinguished by the
austenite start temperature of the ingot: </p>
<ul>
<li><p>Nitinol #1 (superelastic) -35 to -10 °C</p></li>
<li><p>Nitinol #2 (superelastic) -45 to -15 °C</p></li>
<li><p>Nitinol #4 (superelastic) -10 to +10 °C</p></li>
<li><p>Nitinol #5 (shape memory) ≥ +85 °C</p></li>
<li><p>Nitinol #6 (shape memory) +35 to +85 °C</p></li>
<li><p>Nitinol #8 (shape memory) +10 to + 35 °C</p></li>
<li><p>Nitinol #9 (superelastic) ≥ +35 °C</p></li>
</ul>
</blockquote>
<p>From: <a href="https://cdn.thomasnet.com/ccp/00956637/229312.pdf" rel="nofollow noreferrer">https://cdn.thomasnet.com/ccp/00956637/229312.pdf</a></p>
<p>It is possible to tune the shape-memory and superelasticity properties of the alloy by modifying its composition. And speaking about the glasses (spectacles), you are more interested in the <a href="https://www.youtube.com/watch?v=m4fQHgjgc3Y" rel="nofollow noreferrer">superelasticity behavior</a> of the material rather than in its <a href="https://www.youtube.com/watch?v=XPrg8EZlD1E" rel="nofollow noreferrer">shape memory effect</a>.</p>
| 24401 | What grade of titanium are "memory titanium" glasses made of? |
2018-10-28T14:17:05.697 | <p>I am wondering how the screw top of a container made of either aluminium or brass is manufactured at scale. A picture of the sort of thing I am talking about is below. Currently the only method I can think of is hydroforming. Can anyone enlighten me?</p>
<p><a href="https://i.stack.imgur.com/oJBou.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oJBou.jpg" alt="Example Image"></a></p>
| |mechanical-engineering|manufacturing-engineering|industrial-engineering| | <p>These threads are rolled. The blank is automatically fed to sit around a central wheel which has the form of the thread. An external arm with the opposing half of the form then lowers to apply pressure and form the threads, before being automatically ejected from the central wheel, ready for the next part. </p>
<p>This video <a href="https://www.youtube.com/watch?v=HykVRyTsV20" rel="nofollow noreferrer">https://www.youtube.com/watch?v=HykVRyTsV20</a> shows a machine forming the threads on a cap, but the process is identical for the container itself.</p>
| 24404 | Manufacturing screw thread onto lid |
2018-10-29T00:19:56.063 | <p>According to <a href="https://en.wikipedia.org/wiki/Sulfur_vulcanization" rel="nofollow noreferrer">this Wikipedia page</a>, vulcanization is an important process to make rubber harder and more durable. However, <a href="https://en.wikipedia.org/wiki/Plasticizer" rel="nofollow noreferrer">another Wikipedia page</a> says that plasticizers play a significant role in rubber manufacturing, acting as a softener.</p>
<p>Is there a reason why both are needed when processing rubber? It seems to me that they exhibit an exact opposite effect on the rubber being processed and hence only one is needed.</p>
| |materials|manufacturing-engineering| | <p>yes, there is, as follows. </p>
<p>Unvulcanized rubber will soften and turn into taffy when it gets warm. Vulcanization cross-links the polymer strands and prevents the rubber from turning into taffy at high temperatures. </p>
<p>This also necessarily makes the rubber stiffer and more resistant to deformation at lower temperatures, which can yield a rubber compound which is not soft enough for a particular application. This tendency can be counteracted by milling mineral oils (plasticizers) into the rubber to get the rubber molecules to slip more easily past one another (within the limits set by the cross-linking). </p>
<p>This means that the process of engineering a rubber batch to have the right combination of flexibility and durability involves balancing the benefit of cross-linking against the need for rubber-like behavior, and this involves careful blending of plasticizers into the rubber resin. In so doing, the engineer can get anything from supersoft to almost rock-hard rubber, dialed-in for a broad range of special applications. </p>
| 24412 | Is there a reason why both plasticizer and vulcanizing agent are needed during processing of rubber? |
2018-10-29T18:56:54.193 | <p>I am not asking for specific numbers of any production plant. I want to know the general direct worker/employee ratio of a plant.</p>
<p>Example: if there are 1000 employees working usually 700 of them are direct workers etc.</p>
<p>I am looking for estimates backed up with sources or personal experience.
Also there may be multiple shifts in a production plant so I am looking for the ratio in a single shift since all the jobs may not get a night shift. The plants I am interested in are the ones that mass produce for international brands. Including the country the plant is in would also be helpful for categorizing the information.</p>
<p>Direct worker means a worker directly working on the production line while employee means all the the people getting paid in that plant.</p>
<p><strong>Edit:</strong> Added more information.</p>
| |automotive-engineering|manufacturing-engineering|car| | <p>Below is a breakdown for the Fiat Chrysler Automobile Transmission plant in Kokomo, Indiana USA. (Note: Hourly = Direct, salaried = Indirect)</p>
<p><strong>Employment</strong>: 3,830 (3,411 hourly; 419 salaried)</p>
<p><strong>Floor Space</strong>: 3.1 million square feet </p>
<p><strong>Acreage</strong>: 110 acre site with casting plant</p>
<p>Transmission Plant I</p>
<p><strong>Employment</strong>: 2,230 (1,950 hourly, 280 salaried)</p>
<p><strong>Floor Space</strong>: 1.2 million square feet</p>
<p><strong>Acreage</strong>: 233</p>
<p>The plant is fairly automated. Below are some video links of the plant. </p>
<p><a href="https://i.stack.imgur.com/NqNWR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zO1Rz.jpg" alt="Fiat Chrysler Transmission Plant 1"></a></p>
<p><sub><em>Click on image for a larger version of the image.</em></sub></p>
<p><a href="https://i.stack.imgur.com/ASwwU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/alf0B.jpg" alt="Fiat Chrysler Transmission Plant 2"></a></p>
<p><sub><em>Click on image for a larger version of the image.</em></sub></p>
<p>Additional details and video can be found <a href="http://media.fcanorthamerica.com/newsrelease.do?id=322&mid=" rel="nofollow noreferrer">here</a> and <a href="http://media.fcanorthamerica.com/newsrelease.do?id=349&mid=" rel="nofollow noreferrer">here</a></p>
| 24421 | What is the percentage of direct workers in an automotive production plant compared to all the employees? |
2018-10-30T03:04:51.987 | <p>In the saga of trying to answer the question that came into my mind while studying the basic concepts of Fluid Mechanics, "Why textbooks use geometric center to calculate hydrostatic pressure when presenting pressure gauges?", after asking it to my professor of Fluid Mechanics, consulting ~20 textbooks, asking it as a part of <a href="https://physics.stackexchange.com/questions/436475/">this</a> question on Physics.SE, <a href="https://engineering.stackexchange.com/questions/24357/">asking it here</a> by the recommendation of a Physics.SE user, I didn't gave up and finally found an answer while reading the comments on this YouTube video:</p>
<p><a href="https://youtu.be/_VDVi_zq3nk" rel="nofollow noreferrer">How a Piezometer Works by Donald Elger</a></p>
<p><img src="https://i.stack.imgur.com/zBmDJ.jpg" alt="Piezometer diagram"></p>
<blockquote>
<p>Why is it [the pressure measurement with piezometer] taken from the middle of the pipe?</p>
<p>Elger's answer: The pressure variation across a section of a pipe is hydrostatic; thus, the pressure will vary linearly with radius and the pressure at the center of the pipe is the average pressure. If you use this value of pressure in your calculations, this will be give you the <strong>most accurate results</strong>. Thus, engineers nearly always apply or measure the pressure at the center of the pipe. </p>
</blockquote>
<hr>
<p>With this new information, a new question arose: <strong>Why average pressure gives the most accurate results if used in calculations? What "calculations" is he referring to?</strong></p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>I also asked this question on <a href="https://www.quora.com/Why-does-using-the-average-pressure-in-calculations-give-the-most-accurate-results" rel="nofollow noreferrer">Quora</a> and started sending requests. Someone answered it. I'll post the answer.</p>
<blockquote>
<p>On reading the context for this question, i.e the best location for measurement of pressure along a pipe and why it is the center, it helps to revisit the fundamentals of pipe flow. Essentially the center of the pipe has zero shear stress since the velocity profile is typically symmetric and almost no turbulent shear. If you trace the centerline of the pipe, you would see that the total pressure at the inlet is converted to a mix of static pressure and kinematic pressure, with almost no losses. This is not true near the wall, where there are viscous losses in the boundary layer region and there might be significant turbulence or reverse flow. So the center of the pipe is a cleaner place to read the total pressure or static pressure. Of course, the sensor will cause disturbance in the flow which needs to be accounted for.
<strong>Author: Roopesh Mathur</strong></p>
</blockquote>
<hr>
<p>I built an example to complement Roopesh's answer and give an example of the "calculations" that Elger's answer mentions. </p>
<p>Consider an experiment in which a Pitot tube is used and there is a flow with velocity profile given by:
<span class="math-container">$$v(h)=V_{max}\cdot\left(1-\frac{\left|h-R\right|}{R}\right)^{1/7} ,\space 0\leq h\leq2R \space\space$$</span>
Note that <span class="math-container">$v(h)=v(2R-h)$</span>, so the velocity profile is symmetric, with axis of symmetry passing through <span class="math-container">$h=R$</span>. Our goal is to <strong>determine <span class="math-container">$V_{max}$</span></strong>. Below is an image that illustrates the experiment:</p>
<p> <img src="https://i.stack.imgur.com/rN0qj.png" width="500" />
<sub>(Adapted from Fluid Mechanics - Yunus A. Çengel & Cimbala) </sub></p>
<p>The Pitot tube can measure the stagnation pressure at a point, where <span class="math-container">$P_{stag}=P+\rho\frac{v^2}{2}$</span>. If a piezometer is used in conjunction with a Pitot tube, it's possible to calculate the fluid velocity at a specific location, using the static pressure <span class="math-container">$P$</span> at of this location, measured with the piezometer, and the stagnation pressure at that location, measured with the Pitot tube:</p>
<p><span class="math-container">$$v=\sqrt{\frac{2(P_{stag}-P)}{\rho}}$$</span>
Since <span class="math-container">$v=v(h)$</span> , by the velocity profile formula, we have:</p>
<p><span class="math-container">$$\left. \begin{array}{r}
v=v(h)\\
P=P(h)=\gamma\cdot (h+k)\\
P_{stag}=P_{stag}(h)=P(h)+\rho\frac{v^2(h)}{2}
\end{array}\right\} v(h)=\sqrt{\frac{2\left[P_{stag}(h)-P(h)\right]}{\rho}}
$$</span>
To determine <span class="math-container">$V_{max}$</span> it's necessary to obtain the velocity at a specific height – using the Pitot tube, the piezometer and the Pitot velocity formula – and then replace the experimental value found in the velocity profile formula. At first, we can choose any height to do the measurements! </p>
<p>Roopesh's answer tell us the best height to be chosen in order to get the most accurate result: the height of the pipe centerline
(<span class="math-container">$h=R\space$</span> in my example), because there we have "zero shear stress" and "almost no turbulent shear". Furthermore in this height there are "almost no losses" in the total pressure. Then, we have:</p>
<p><span class="math-container">$$v(R)=\sqrt{\frac{2\left[P_{stag}(R)-P(R)\right]}{\rho}}=\sqrt{\frac{2\left[P_{stag}(R)-P_{average}\right]}{\rho}}$$</span></p>
<p>And this confirms what Elger said:"If you use this value [average pressure] in your calculations, this will be give you the most accurate results". </p>
<hr>
<p>So, in general, the average pressure gives the most accurate results if used in calculations because there are many applications/cases in which the locations with <span class="math-container">$P=P_{average}$</span> are the best places for experimental data collection.</p>
| 24425 | Why using average pressure in calculations gives the most accurate results? |
2018-10-30T17:48:08.813 | <p>Looking to add a "high oil temp" warning on a machine tool by adding a temperature switch into the hydraulic circuit, right after the radiator before returning to tank. I know monitoring from the tank would be better, but tapping into the tank is not possible.</p>
<p>The switch I would like to use....</p>
<p><a href="https://www.mcmaster.com/7079k95" rel="nofollow noreferrer">https://www.mcmaster.com/7079k95</a></p>
<p>....is rated for 120/240 AC. </p>
<p>I would like to use the switch to trigger an input on my PLC. The switch is normally open and would close at over temperature, allowing a +24vDC to connect to the PLC input(PNP). </p>
<p>Can this switch, as only rated for AC, be used to carry the low-amperage +24vDC signal? </p>
| |mechanical-engineering|electrical-engineering|control-engineering| | <p>There are important differences between switches designed for AC and those designed for DC, the main distinction being "breaking speed", i.e. how fast the contacts are separated when the switch is turned off.</p>
<p>However, that is unlikely to matter at the current levels you describe. The problem is greatest when there is an inductive load like a motor involved, which leads to arcing across the contacts when they are separated. Briefly, energy is stored in an inductor in the form of magnetic energy and it needs to go somewhere when the circuit is broken causing the magnetic field to collapse. The voltage rises until the energy can be dissipated and this causes arcing across the contacts that can degrade them.</p>
<p>I expect your PLC load is low inductance and low current, so it shouldn't be any problem in your specific application. Other readers might want to look further and pick a suitably rated switch for motor, transformer, or other inductive loads.</p>
<p><a href="https://www.mouser.com/blog/which-switch-who-cares-if-its-ac-or-dc" rel="nofollow noreferrer">https://www.mouser.com/blog/which-switch-who-cares-if-its-ac-or-dc</a></p>
| 24430 | Can an AC rated temperture switch be used in a DC control circuit? |
2018-10-31T10:08:13.263 | <p>I saw this meme on Facebook showing a coiled railway overbridge and an alternative plan. I replied that straightening that overbridge will force it to start at the other side of the road, and you need to build a u-turn and there is a valuable building off the frame. Is the coil rational, and why?</p>
<p><a href="https://i.stack.imgur.com/9abjQ.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/9abjQ.jpg" alt="enter image description here"></a></p>
| |bridges|highway-engineering| | <p>In 1909, the same thing was done <em>twice</em> on the Canadian Pacific Railway rail line through Kicking Horse Pass in the Rocky Mountains: <a href="https://i.stack.imgur.com/EwtL3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EwtL3.jpg" alt="Spiral tunnels at Kicking Horse Pass"></a></p>
<p>Two circular tunnels (dashed white lines) were dug into the mountains, lengthening the route in order to decrease its steepness.</p>
<p>This is exactly the same situation and solution as in the original question.</p>
| 24438 | Why was a coil chosen over a straight for this overbridge? |
2018-10-31T17:30:26.737 | <p>I'm designing a funnel/bottle for a powder. My goal is to have all of the bottle's powder contents come out when inverted. However, the powder has a tendency to stick to itself (like flour or brown sugar) and so I will run into a chance that the powder will not fall into the bottom vessel by gravity alone. </p>
<p>To make matters more complicated, this is not an open funnel on the top. It will be more like a plastic water bottle in that when it's rotated to dumping position, the top isn't open to the air and venting doesn't currently appear to be an option.</p>
<p><a href="https://i.stack.imgur.com/Ch81f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ch81f.png" alt="enter image description here"></a></p>
<p>My question is, what funnel surface type will increase the flow to help dispel the powder contents? Some ideas that come to mind are smooth, vertical guides, or stepped stairs. Note my funnel outlet is fixed in diameter.</p>
<p><strong>Smooth</strong></p>
<p>This one makes the most sense as it's a "smooth" flow. However, I've been surprised by how air and water behave in relationship to fluid dynamics (ex: golf ball dimples or pickup trucks being more aerodynamic with the tailgate up). And although fluid dynamics doesn't cover powders, I would imagine there's some similarities.</p>
<p><a href="https://i.stack.imgur.com/nixh7.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nixh7.gif" alt="enter image description here"></a></p>
<p><strong>Vertical Guides</strong></p>
<p>In theory, these ridges would break up the powder clumping on the way down.</p>
<p><a href="https://i.stack.imgur.com/vwzkP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vwzkP.png" alt="enter image description here"></a></p>
<p><strong>Stepped</strong></p>
<p>In theory, these steps would add turbulence on the way down and help break up the powder on each "step down."</p>
<p><a href="https://i.stack.imgur.com/XLpKf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XLpKf.png" alt="enter image description here"></a></p>
<p><strong>Extra question:</strong> Regardless of the features in the funnel (or lack thereof), what's the minimum angle from vertical that would be ideal for a funnel to dispense powders by gravity alone?</p>
| |fluid-mechanics|friction|flow-control|turbulence|open-channel-flow| | <p>The problem you have is more complicated than you think. Any industry that deals with the storage and movement of granulated/powdered materials has to deal with this and most have unique solutions.</p>
<p>You need to look at the <a href="https://www.chemengonline.com/hopper-design-principles/" rel="nofollow noreferrer">principles of hopper design</a>.</p>
<p>Some of the things that affect any design will be:</p>
<ul>
<li>The <a href="https://en.wikipedia.org/wiki/Angle_of_repose" rel="nofollow noreferrer">angle of repose</a>
of the material you intend to use.</li>
<li>The moisture content of the material.</li>
<li>The angle of the cone in the funnel/hopper.</li>
<li>The material used in the cone and wall of the bottle/hopper/funnel
and its friction factor.</li>
<li>The ability of the powdered material to create a near vertical wall
and stand up, despite its angle of repose.</li>
</ul>
<p>A stepped cone for the funnel is a bad idea because it creates hang-ups for the material to get caught on. Also the steps reduce the overall angle of the cone. Generally, the steeper the angle of the cone, the better. </p>
| 24442 | Ideal funnel design for powders |
2018-10-31T18:30:08.960 | <p>I am trying to select a BLE module for low power application. I have looked into different BLE modules from different chip manufacturers. For example:</p>
<ol>
<li><a href="https://www.inventeksys.com/products-page__trashed/ble/ble/" rel="nofollow noreferrer">ISM78G1D Bluetooth Low Energy SIP from Inventek</a> </li>
<li><a href="https://www.silabs.com/products/wireless/bluetooth/bluetooth-low-energy-modules/bgm13s-bluetooth-sip-module" rel="nofollow noreferrer">Blue Gecko BGM13S Bluetooth® Module - SiP from Silicon Labs</a></li>
<li><a href="http://www.ti.com/product/CC2652R" rel="nofollow noreferrer">SimpleLink Multi-Standard CC2652R Wireless MCU from TI</a></li>
<li><a href="http://www.cypress.com/products/ez-ble-and-ez-bt-bluetooth-modules" rel="nofollow noreferrer">EZ-BLE and EZ-BT Bluetooth Modules from Cypress</a></li>
<li><a href="https://www.nordicsemi.com/eng/Products/nRF52840" rel="nofollow noreferrer">nRF52840 from Nordic Semiconductor</a></li>
<li><a href="https://www.artik.io/modules/artik-020/" rel="nofollow noreferrer">Samsung ARTIK™ 020</a></li>
</ol>
<p>I am in the process of learning strategies on selecting a BLE module given a few choices. The application will be used in a low power application which includes a sensor to detect objects. We plan to connect the BLE module to the host micro-controller controlling the sensor. The BLE module will communicate to a Gateway (currently being defined) that will transmit the data to the cloud.</p>
<p>Appreciated if someone can provide some guidance on what criteria to use when selecting a BLE module. Thanks for your help!</p>
| |electrical-engineering|wireless-communication| | <p>Looking at if the device supports BLE 4.0, 4.1, 4.2 or 5.0 is a good start. Not all BLE modules are 5.0 capable. What type is distance range might be a parameter to consider some BLE module have line of sight range of 1km or more at a low data rates. Below is a short side by side comparison to help get started. </p>
<p><a href="https://i.stack.imgur.com/jZsxQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jZsxQ.jpg" alt="enter image description here"></a></p>
| 24443 | What criteria to focus on when selecting a BLE module? |
2018-10-31T20:20:04.153 | <p><a href="https://i.stack.imgur.com/gVjzl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gVjzl.png" alt="enter image description here"></a>I would prefer an equation.
The premise is that I'm trying to calculate how thick an iron cylinder using a screw jack would need to be to support 20kg of weight for engineering class. We haven't learned anything along these lines and I'd really like to know.</p>
<p>Thanks!</p>
| |mechanical-engineering|structural-engineering| | <p>There are stablished criteria by codes to design columns, but we ignore these for the benefit of clarity here.</p>
<p>Basically we want to design an Iron column strong enough to support a 20 kg load with a length of <span class="math-container">$ L= 2\times 3ft = 6ft$</span></p>
<p>We multiplied by 2 because the column is a cantilever column, it is not restrained laterally on top.</p>
<p>First we check for the slenderness.</p>
<p>Usually the columns with a radius of gyration (radius of gyration is the radius at which you can assume all the mass is located), greater then L/20 are considered slender and need to be checked for buckling load.
Radius of gyration of a cylinder is :</p>
<p><span class="math-container">$ R_{gyration} = D/4$</span></p>
<p>So we know any diameter less than 4.5 cm should be considered a slender column and checked for buckling.</p>
<p>Say for trial and error we pick as a first guess a 1.5 cm diameter post.</p>
<p><span class="math-container">$ p_{critical} = \frac {\pi^2EI}{L^2}
\\\ and \ I = \frac {pi D^4}{64} $</span></p>
<p>This is the Maximum buckling load.</p>
<p>So I = 0.24cm^4 and </p>
<p><span class="math-container">$P = \pi^2(0.24E)/ 183^2 = \text{ approx} \ 76kg$</span> Assuming Iron's E at 100 GPa. This strength is reasonable and has a factor of safety of approx. 4, I would stop here.</p>
<p>But through the same steps picking a size smaller we gat closer to the optimal size.</p>
<p>Finally we check the result Against the compression strength with a factor of safety, and check for likelihood of overturning of the base block, which is missing the width. </p>
| 24446 | How do I calculate how much weight an iron cylinder can support? |
2018-11-01T20:05:18.790 | <p><a href="https://i.stack.imgur.com/WVcgG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WVcgG.png" alt="enter image description here"></a></p>
<p>I'm an engineering student, our professor assigned us this truss problem I found <span class="math-container">$A_y$</span> and <span class="math-container">$L_y$</span> by using the moment about l, and found ab, now what? It seems like I have too many unknowns to solve for after that...</p>
| |statics| | <p>You say you've already identified the zero-force members, so I'll skip part (a) of the question. The structure then becomes (deleting all the zero-force members except for DE which is needed for structural stability):</p>
<p><a href="https://i.stack.imgur.com/MiIlP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MiIlP.png" alt="enter image description here"></a></p>
<p>In this case, I prefer to work from the cantilever and then move in.</p>
<p>Since we know DE is zero-force, we therefore know <span class="math-container">$EG = 20\text{ kN (compression)}$</span>.</p>
<p>Since the only other member with a vertical component on <span class="math-container">$G$</span> is <span class="math-container">$DG$</span>, we can obtain that <span class="math-container">$$\begin{align}
DG &= 20\times\dfrac{\sqrt{7^2+DG_v^2}}{DG_v} \\
\text{where }DG_v &= \dfrac{5}{3\times7}\times2\times7 = 3.33\text{ m} \\
\therefore DG &= 46.5\text{ kN (tension)} \\
\therefore GH &= 46.5\times\dfrac{7}{\sqrt{7^2+3.33^2}} = 42.0\text{ kN (compression)}
\end{align}$$</span></p>
<p>Looking at <span class="math-container">$D$</span>, we have the applied load, <span class="math-container">$CD$</span> (with horizontal and diagonal components), <span class="math-container">$DE$</span> (which we know is zero-force), <span class="math-container">$DG$</span> and <span class="math-container">$DH$</span>. <span class="math-container">$CD$</span> is the only one which can absorb <span class="math-container">$DG$</span>'s horizontal component (which is equal to the result found for <span class="math-container">$GH$</span>, so we can find that</p>
<p><span class="math-container">$$\begin{align}
CD &= 42.0\times\dfrac{\sqrt{7^2+CD_v^2}}{7} \\
\text{where }CD_v &= 5 - 3.33 = 1.67\text{ m} \\
\therefore CD &= 43.2\text{ kN (tension)} \\
\therefore DH &= 43.2\times\dfrac{1.67}{\sqrt{7^2+1.67^2}} - 46.5\times\dfrac{3.33}{\sqrt{7^2+3.33^2}} + 40 = 50\text{ kN (compression)}
\end{align}$$</span></p>
<p>Likewise, looking at <span class="math-container">$H$</span>, only <span class="math-container">$CH$</span> can absorb <span class="math-container">$DH$</span>'s vertical load, so by repeating the calculations above, we get that:</p>
<p><span class="math-container">$$\begin{align}
CH &= 86.0\text{ kN (tension)} \\
HI &= 112.0\text{ kN (compression)}
\end{align}$$</span></p>
<p>And then looking at <span class="math-container">$C$</span>, only <span class="math-container">$BC$</span> can absorb <span class="math-container">$CD$</span>'s and <span class="math-container">$CH$</span>'s horizontal components, so we then calculate that:</p>
<p><span class="math-container">$$\begin{align}
BC &= 169.3\text{ kN (tension)} \\
CI &= 195.3\text{ kN (compression)}
\end{align}$$</span></p>
<p>Obviously <span class="math-container">$BI = HI = 112.0\text{ kN (compression)}$</span> and <span class="math-container">$IL = CI = 195.3\text{ kN (compression)}$</span>.</p>
<p>Then looking at <span class="math-container">$A$</span>, we use your previously calculated reactions to obtain</p>
<p><span class="math-container">$$\begin{align}
AB &= 84.8\text{ kN (tension)} \\
AL &= 10.1\text{ kN (compression)}
\end{align}$$</span></p>
<p>And likewise with <span class="math-container">$L$</span> (or from <span class="math-container">$B$</span>), we obtain the last remaining <span class="math-container">$BL = 23.8\text{ kN (tension)}$</span>.</p>
<p>And we're done.</p>
<p><a href="https://i.stack.imgur.com/gZqIi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gZqIi.png" alt="enter image description here"></a></p>
| 24455 | I'm taking statics, and I'm stuck on this truss problem, how do do I solve this? |
2018-11-02T15:15:26.443 | <p>When attempting to trap a particle in a laser beam i was able to achieve the effect.
<a href="https://i.stack.imgur.com/AvAOd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AvAOd.jpg" alt="enter image description here"></a>
Pretty cool but now when i put of piece of plexiglass that is 1/8” thick i can no longer achieve the effect. The beam hits a focal lens when its diameter is around 6mm and focuses 50mm away. I put the piece if plexiglass 15mm after the focal lens. Now it is obvious that my 25mW laser will lose power but i wonder how, where, and how much? I imagine reflection and scattering accounts for most of it but there is more that plays a part. I want to know if theres a equation that takes a material, thickness, laser wavelength and power and will maybe give me my power on the other side of the material.</p>
| |lasers| | <p>The equation you are looking for is</p>
<p><span class="math-container">$$I(z) = (1-R) \cdot I_0 \cdot e^{-\alpha z}$$</span></p>
<p>where</p>
<p><span class="math-container">$I(z)$</span>: intensity of the laser beam after the distance <span class="math-container">$z$</span></p>
<p><span class="math-container">$R$</span>: reflectance of material</p>
<p><span class="math-container">$I_0$</span>: intensity of laser beam on the surface</p>
<p><span class="math-container">$\alpha$</span>: absorption coefficient (NOT the same as absorptance)</p>
<p><span class="math-container">$z$</span>: thickness of the material</p>
<p>When you plug the thickness of the plexiglas plate in as <span class="math-container">$z$</span>, you get the intensity of the laser beam after it passes through the plate. </p>
<p>Note that the equation considers intensity instead of power, so you need to find the beam diameter on the surface of your plexiglas plate. </p>
<p>Further, you need to find the absorption coefficient of the material for the given wavelength. The absorption coefficient (or <a href="https://en.wikipedia.org/wiki/Attenuation_coefficient" rel="nofollow noreferrer">attenuation coefficient</a>) is not a dimensionless number like the absorptance, but has the dimension <span class="math-container">$m^{-1}$</span>, so they should not be mixed up.</p>
| 24470 | Calculating energy loss from laser when shining through a material |
2018-11-02T16:05:16.320 | <p>I have a little creek in my back yard. It's slow moving and safe (as safe as a slow moving creek can be.) I'd like to build a little set of stone steps into the creek and am not sure of the best structural/engineering/least impactful way of do it.</p>
<p>So here's the main question:</p>
<p>Should the stairs be pointing up river or down river? </p>
<p>I've attached an image (very rough) of what I'm thinking. i'm not sure if the top version or the bottom would be better all else considered equal.</p>
<p><a href="https://i.stack.imgur.com/GLLdi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GLLdi.jpg" alt="enter image description here"></a></p>
| |fluid-mechanics| | <p>From a fluid mechanics perspective, the lower the fluid interaction the better.<br>
As you have seen, a lot depends on how the original creek is shaped. </p>
<p>If you face the stairs into the fluid flow, you will impact the most physical stress on the rocks. When you move water, it is going to go somewhere else. So, it will erode the banks. This approach will also result in a vortex being generated behind the stairs that will erode the ground behind your stairs.</p>
<p>The pocketed entry drawn in your first example will last longer. Just be careful to reinforce the banks.</p>
| 24473 | Which way should I build stairs into a creek? |
2018-11-03T06:46:42.717 | <p>Object <strong>A</strong> has mass <span class="math-container">$m_A$</span>, and is travelling west with velocity <span class="math-container">$v_A$</span>. Object <strong>B</strong> has mass <span class="math-container">$m_B$</span> and is travelling east with velocity <span class="math-container">$-v_B$</span>. The objects collide in a perfectly inelastic collision so that the velocities of the two objects after the collision are the same. Prior to the collision, the objects are made of uniform, identical materials and shapes.</p>
<p><em>How much of the kinetic energy is absorbed by deformation of object <strong>A</strong>, and how much of the kinetic energy is absorbed by deformation of object <strong>B</strong>?</em></p>
<p>If there is not a unique solution to this question because of a lack of necessary constraints, please answer the question with some examples for the unspecified constraints.</p>
<p>EDIT: when I said same shapes, I meant they are both cylinders or prisms or spheres or something. The dimensions and masses of the shapes may be different.</p>
| |mechanical-engineering|structural-engineering|materials| | <p>For Simplicity we assume the objects are moving on the same path but opposite direction, however it would be only a bit complex if they were not moving colinearly if one introduces vectors.</p>
<p>the momentum of particles before the collision and after are preserved.</p>
<p><span class="math-container">$m_Av_A + m_Bv_B = (m_A +m_b)v_{(A+B)} $</span></p>
<p>So let's consider object A. It has changed it's velocity by the amount <span class="math-container">$v_A -v_{(A+B)}$</span></p>
<p>So the kinetic energy it has lost is <span class="math-container">$ E_k= \frac{ m \space[v_A -v_{(A+B)}]^2}{2}$</span></p>
<p>And same way we can calculate the loss for B.</p>
| 24485 | Distribution of energy absorption in inelastic collision |
2018-11-03T10:40:46.360 | <p>I'm building a CNC router around a HSD ES330 spindle motor that has pneumatic tool-changing ability. The manual recommends the following pneumatic circuit:</p>
<p><a href="https://i.stack.imgur.com/ghIvy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ghIvy.png" alt="enter image description here"></a></p>
<p>I'm not sure I understand the purpose of the "quick bleeder valve" (circuit element #5 in the diagram). Searching for "quick bleeder valve" doesn't turn up anything with that symbol, or a part I'm confident buying on McMaster-Carr. I know the one-way flow restrictor (element #4) is supposed to slow the tool ejection speed; I guess during tool unlocking, air returns via #2, but I'm not sure if that exhaust air escapes through the bleeder valve, or the 5-2 valve #6.</p>
<p>I do have a one-way flow <a href="https://www.mcmaster.com/6692K11" rel="nofollow noreferrer">restrictor</a> (although I'm not entirely sure how to adjust it).</p>
<ul>
<li>What is the "quick bleeder valve?"</li>
<li>Do I really need it?</li>
<li>Where is this air exhausted?</li>
</ul>
| |pneumatic| | <p>It's common for CNC equipment to use air pressure to manipulate tools and collets and mandrels and other "non-precision" mechanical components. In the case of tooling, the engagement sockets provide the precision, while the air pressure locks the collars to the mechanism.</p>
<p>Air pressure can be rapidly applied in a closed system, but not so rapidly reversed when using a pump, for example. Considering the cost of air (as in as free as air), simply releasing the pressure to the atmosphere resolves the question of what to do with it.</p>
<p>In your question, one can expect that the circuit will use air pressure to lock a tool in place, and when the sequence requires that the tool be released, the dump valve/quick bleeder valve/quick exhaust valve will be actuated.</p>
<p>I use the last term, because my search resulted in better returns for various valve suppliers with "quick exhaust valve" as opposed to any other terms.</p>
<p>There are mechanical exhaust valves (not applicable here) and solenoid exhaust valves, which become part of an electrical circuit to seal the valve and to open the valve. I found prices in the triple digit range, but also some lower cost items. <a href="https://www.valvesonline.com.au/pneumatic-quick-exhaust-valve" rel="nofollow noreferrer">This one</a> is sourced from Australia, but included as a representative example rather than a solid suggestion.</p>
<p><a href="https://i.stack.imgur.com/faryL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/faryL.png" alt="exhaust valve"></a></p>
<p>Depending on your system air pressure, it may be necessary to install an exhaust line or a deflector of some form. When the valve dumps, it will push on anything in line, which may be insignificant, or may be undesirable.</p>
<p>The link to the inline flow restrictor shows a reference to a dial and one of the photos shows gradations along the circumference of the fitting. It's difficult to see if there is a screwdriver slot for adjustment or merely something turned with a wrench.</p>
| 24488 | What is the "quick bleeder valve" in this tool changer pneumatic schematic (and other questions)? |
2018-11-03T19:28:16.327 | <p>I am trying to design a pipe which takes water from a river and transports it to a reservoir.</p>
<p>However, I do not want my pipe to take all of the water from stream, because that will create all sorts of environmental problems.</p>
<p>How do I make sure that this does not happen?</p>
<p>The flow rate in the stream is <span class="math-container">$0.02 m^{3}/s$</span> and the stream is <span class="math-container">$1m$</span> wide and <span class="math-container">$1.25m$</span> deep. </p>
<p>Do I just need to make sure that the flow rate in my pipe is not greater than <span class="math-container">$0.02 m^{3}/s$</span>?</p>
| |hydraulics|pipelines|open-channel-flow| | <p>As long as the pipe intake is higher than the stream bed, it cannot drain it dry.
The bottom is sized to maintain a minimum safe flow rate for the stream ecology.
The middle is sized for the maximum flow rate of your application. A dry pipe signals that the stream ecology is in danger.
The top is for overflow when your intake pipe is full.</p>
| 24500 | How to make sure a pipe does not take all of the water from a stream? |
2018-11-03T20:48:31.870 | <p>What should I choose from today. </p>
<p>I am looking at washing machines, and I see that LG has this Inverter Direct Drive (6 motion cleaning) technology in their machines, I see that its not like a year old one, but I am also wondering how long do these last?</p>
<p>So can someone tell me how does direct drive perform nowadays, especially in LG products;</p>
<ul>
<li><p>If its more reliable then how cheaper is belt drive fixing?</p></li>
<li><p>Are there more stuff to go wrong with that technology?</p></li>
<li><p>I see some manufacturers use only inverter technology, what does that
mean to the average consumer, should I choose that over non-inverter?</p></li>
</ul>
| |mechanical-engineering|electrical-engineering|motors| | <p>There are tons of advantages and disadvantages, but I will mention only the relevant cons and pros.</p>
<p>Indirect (belt) drive usually produces less noise in comparison to direct drive (continue power transmission vs discrete power transmission like gears and chains), but you know engineers had done a lot to reduce the noise in household equipment, unfortunately not yet in industrial sector.</p>
<p>Belts are far more prone to relaxation and then they can transmit less torque. As you might have noticed after some years after you replace the belt. Again this is less observable for home and kitchen equipment. </p>
<p>Belt drives dampen vibrations, direct drives produce more vibration specially when the rpm goes up. But in washing machines, the direction of rotation changes time to time and direct drive offers more stability. </p>
<p>The maintenance cost of belt drives is lower than that for direct drive, they are easier to disengage for repairing.</p>
<p>Belt drives apply greater load on the main shaft of the machine, it has a direct effect on the life time of bearings, shaft misalignment and as a result, oil leakage and more vibrations. </p>
<p>Direct drives requires less space in comparison to belt drives, so smaller machines.</p>
<p>Belts are sensitive to the surrounding temperature, humidity, oil ..., so the probability of failure is higher plus the micro slip.</p>
<p>Inverters are a sort of speed and power control over electric motors, long story short, they make the rotation of induction motors smoother in comparison to classic motors without inverters, of course they cost the clients more, but it also regulates power consumption. </p>
| 24504 | Inverter + Direct drive vs. Belt drive motor in washing machines? |
2018-11-04T22:14:06.183 | <p>I'm familiar with left hand threads on bicycle pedals. The action of pedaling for that side of the bike would un-thread a right-hand threaded pedal, although some people have hacked a solution.</p>
<p>I have a chain saw that uses a left hand threaded axle to tighten the bar which tensions the chain.</p>
<p>The image below shows the cover over the bar, and also the legend embossed in the plastic.</p>
<p><a href="https://i.stack.imgur.com/JjeCF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JjeCF.jpg" alt="chain saw cover"></a></p>
<p>The right hand "wing nut" is an ordinary thread and serves to secure the cover and the bar, once the tension is set with the left hand knob. Clockwise rotation for the locking nut works as expected. The left hand knob turns a gear in a pinion arrangement, clockwise pushes the bar away from the motor, as per the legend.</p>
<p><a href="https://i.stack.imgur.com/vz2XU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vz2XU.png" alt="chain saw inside with text"></a></p>
<p>The left hand knob has unrestricted rotation (within reason) and has gear teeth as part of the construction. The pinion gear resides under the knob, engaging the teeth and rotating the axle, which resides in the groove shown in the photo. Near the end of the axle, the threads push or pull the rectangular tab (pin) which in turn pushes the chain bar.</p>
<p><a href="https://i.stack.imgur.com/O3hQM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O3hQM.png" alt="chain saw with additional parts"></a></p>
<p>As can be seen in this photo, the end of the axle has a severe bend, preventing movement of the tab/pin, removing the saw from service.</p>
<p>Unfortunately, all the sources I've been able to find for the replacement part have indications of "no longer available." Pricing indicates a cost in the US$3.00 range, but that's not helpful without product.</p>
<p>I have a Harbor Freight mini-lathe (toy lathe) that can handle replication of this part, although I'd like to avoid the left-hand thread operation.</p>
<p>I see no practical reason to not use a right hand thread. There are no rotational forces as in the previous pedal example. Once the bar is properly tensioned, the locking nut secures the bar, rendering the axle and pin immobile.</p>
<p>I can easily cut a piece of steel to replace the tab/pin, cut and tap the hole to match the rod which would be cut in the lathe, threaded with an appropriately sized die and mill a flat on the gear drive end.</p>
<p>What technical/engineering aspects would prevent me from making this work?</p>
<p>I do not consider the human aspect of a reversed legend on the knob to be a factor.</p>
| |threads| | <p>This is a user experience/design consideration. Users are simply conditioned to expect that rotating clockwise causes a movement to the right, and anticlockwise to the left - this is how a car steering wheel works, or bike handlebars etc. etc.</p>
<p>The mark of good design is where you are able to intuitively guess what the operation of an actuator does, before you use it. If there were to be a RH thread, then you would have to rotate the knob in the un-intuitive direction.</p>
<p>As @Solar Mike explained, once this constraint is understood, avoiding the LH thread would cause other undesirable mechanical features such as an idler, or extended shaft. The LH thread is the simplest of these options, so it's the one that was manufactured.</p>
<p>You could replace this component with a RH threaded variant, but make sure to mark the outer cover to make the new operation direction very, very clear, particularly if it's possible you'd lend this out to a friend!</p>
| 24526 | What is the justification for left-hand threads? |
2018-11-04T23:51:17.083 | <p>I am trying to heat the oil in a 1/2 inch thick rectangular steel box by using an Insulated heated blanket covered on all 6 faces of the box, the working conditions of the operation are at -50 Celsius, the temperature of the oil is required to be at 15 Celcius during the operation, so I need to determine how much heat/power is required by the blanket and how much time will it take for the temperature of the oil to change from -50 Celsius to 15 Celsius?</p>
<p>one face resistive diagram,
Rb = resistance by the blanket thickness of 1 inch,
Rw = rectangular steel box wall</p>
<p><a href="https://i.stack.imgur.com/p7MOx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p7MOx.png" alt="enter image description here"></a></p>
| |thermodynamics|heat-transfer|energy|thermal-conduction| | <h1>Foundations</h1>
<p>The governing equation for the heat flow to the tank is</p>
<p><span class="math-container">$$
k\nabla^2 T + \rho \tilde{C}_p \frac{\partial T}{\partial t} = \hat{\dot{q}}_o
$$</span></p>
<p>where temperature, thermal conductivity, density, and specific heat refer to the oil. The heat flow is per unit volume.</p>
<p>--> Assume the oil is well-mixed. The internal gradients disappear, giving</p>
<p><span class="math-container">$$
\rho \tilde{C}_p V \frac{d T}{d t} = \dot{q}_o
$$</span></p>
<p>The governing equation for the heat flow to the air is</p>
<p><span class="math-container">$$ (T_h - T_a) = \dot{q}_a R_a $$</span></p>
<p>using the temperatures of the heater and air as well as the thermal resistance to the air. Allow that <span class="math-container">$\dot{q}_o$</span> is also defined by a thermal resistance equation as <span class="math-container">$(T_h - T) = \dot{q}_o R_o$</span>. The total power taken is <span class="math-container">$\dot{q}_T = \dot{q}_o + \dot{q}_a$</span>.</p>
<p>The system has three unknowns, <span class="math-container">$T$</span>, <span class="math-container">$T_h$</span>, and <span class="math-container">$\dot{q}_T$</span>.</p>
<h1>Solutions</h1>
<h2>General</h2>
<p>Define <span class="math-container">$\Theta = (T_h - T) / (T_a - T_o) = (T_h - T) / \Delta T_{ao}$</span>, where <span class="math-container">$T_o$</span> is the initial temperature of the oil. At time <span class="math-container">$t = 0$</span>, we can set a boundary condition <span class="math-container">$\Theta = 1$</span>. This means, the heater as the same temperature as the air to start. The option to set <span class="math-container">$\Theta(0) = 0$</span> (the heater the same as the oil) is more difficult to resolve. We can combine and rewrite the first governing equations as</p>
<p><span class="math-container">$$
-\rho \tilde{C}_p V \frac{d \Theta}{d t} = \frac{\Theta}{R_o} \Rightarrow - \frac{d \Theta}{d \tau} = \Theta
$$</span></p>
<p>with <span class="math-container">$\tau = t / \rho \tilde{C}_p V R_o$</span>. We find the solution as</p>
<p><span class="math-container">$$ \Theta = \exp(-\tau)$$</span></p>
<p>This gives an equation for the temperature difference between the heater and the oil as a function of time.</p>
<p>Write the thermal resistance equations as</p>
<p><span class="math-container">$$\dot{q}_o R_o = \Theta \Delta T_{ao}$$</span>
<span class="math-container">$$\dot{q}_a R_a = - \Phi \Delta T_{ao}$$</span></p>
<p>with <span class="math-container">$\Phi = (T_a - T_h) / \Delta T_{ao}$</span>. The boundary condition is that <span class="math-container">$\Phi = 0$</span> at <span class="math-container">$t = 0$</span>. We end with one equation.</p>
<p><span class="math-container">$$\dot{q}_T = \left(\frac{\Theta}{R_o} - \frac{\Phi}{R_a}\right)\Delta T_{ao}$$</span></p>
<h1>Solutions</h1>
<p>Solve the equation for <span class="math-container">$\Theta = \exp(-\tau)$</span> to obtain <span class="math-container">$T_h - T$</span> versus <span class="math-container">$t$</span>.</p>
<h2>Constant Total Power</h2>
<p>Solve the equation for <span class="math-container">$\dot{q}_T$</span> to obtain <span class="math-container">$\Phi(\tau)$</span>.</p>
<p><span class="math-container">$$\Phi(\tau) = \frac{R_a}{R_o}\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}} $$</span></p>
<p>Solve this for <span class="math-container">$T_h(\tau)$</span> from <span class="math-container">$T_h(\tau) = T_a - \Phi(\tau) \Delta T_{ao}$</span>. Plug the equation for <span class="math-container">$T_h(\tau)$</span> into the equation <span class="math-container">$T = T_h(\tau) - \exp(-\tau)\Delta T_{ao}$</span> to find the temperature change of the oil with time.</p>
<p><span class="math-container">$$ \frac{T_a - T}{\Delta T_{ao}} = \left(1 + \frac{R_a}{R_o}\right)\exp(-\tau) - \frac{R_a \dot{q}_T}{\Delta T_{ao}}$$</span></p>
<h2>Constant Heater Temperature</h2>
<p>Solve the equation <span class="math-container">$\Theta(\tau)$</span> directly for <span class="math-container">$T$</span> versus <span class="math-container">$t$</span>. The value of <span class="math-container">$\Phi$</span> is a constant. Plug both <span class="math-container">$\Theta$</span> and <span class="math-container">$\Phi$</span> into the equation for <span class="math-container">$\dot{q}_T$</span> to find <span class="math-container">$\dot{q}_T$</span> versus <span class="math-container">$t$</span>.</p>
<h2>Translation to Picture</h2>
<p>The translation to the picture occurs as</p>
<p><span class="math-container">$$R_o = R_{CL} + R_w + R_{B1}$$</span>
<span class="math-container">$$R_a = R_{CA} + R_{B2}$$</span>
<span class="math-container">$$T = T_{Lubricant} \ \ \ T_h = T_s \ \ \ T_a = T_\infty$$</span></p>
<h2>Example</h2>
<p>Here is an example of a plot of <span class="math-container">$\frac{T_a - T}{\Delta T_{ao}}$</span> versus <span class="math-container">$\tau$</span> generated using a python/Jupyter code.</p>
<p><a href="https://i.stack.imgur.com/B5Q7O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B5Q7O.png" alt="temperature profile plot"></a></p>
<h1>Summary</h1>
<p>Take the starting point that you regulate the temperature of the heater jacket to obtain a desired heating time. Allow the power to the heater to vary. The power will start high and fall off over time (as the oil heats). Estimate the initial power using values at <span class="math-container">$t = 0$</span>. This will give you an easy estimate to the starting point for the level power that you need when time is a constraint.</p>
<p>When you start with the initial power and hold it constant, you will heat the oil faster. The temperature of the jacket will increase with time. Find the time when you have to cutoff or reduce the power to avoid burning out the heater.</p>
<p>When time is critical, you will do better to insulate (increase <span class="math-container">$R_a/R_o$</span>) than to increase power. Notice that <span class="math-container">$(1 + R_a/R_o)$</span> and <span class="math-container">$R_a$</span> are multipliers, the former especially to an exponential of time, while power is just an additive term.</p>
| 24531 | Conductive Heat Transfer calculation over time |
2018-11-05T16:47:47.563 | <p>ICE Engines are designed to burn the fuel and air mixtures at high pressures to increase the combustion efficiency. But hydrogen combustion efficiency is already high at 1 atm and at constant Volume combustion, peak pressure may reach around 8 atm. So why compressing the hydrogen and air mixture in Hydrogen ICE engines? If we could avoid compression of fuel and air, we can design much simpler and cheaper Hydrogen engines.</p>
| |mechanical-engineering|automotive-engineering|combustion| | <p>SolarMike's comment basically answers the question, but let me flesh it out a little.</p>
<p>Let's take a typical engine, say it has a compression ratio of 12:1, and weighs 500 pounds, and has a displacement of 4 liters. Now let's reduce the compression ratio to only 2:1. To get the same amount of power output, the displacement now how to be 24 liters. (can you see why? burning one-sixth as many fuel molecules per liter, so need six times more liters). So now the cylinders and pistons have to be about six times bigger as well, so now the car/truck has to be a lot bigger overall to fit this giant engine in there. Also those bigger cylinders are going to weigh a lot more, potentially as much as the entire car/truck, which hurts the performance of the vehicle. Overall it just doesn't make any sense to use this kind of engine. The power density is too low. </p>
| 24542 | Why to compress the hydrogen + air mixture in Hydrogen ICE engines when combustion efficiency of hydrogen at 1 atm is already high? |
2018-11-07T21:19:42.567 | <p>I have been searching for historical structures in the province of Van, Turkey, by examining the publicly available satellite imagery like Bing Maps. There is a historic bridge near Kaçit, Çatak. Its name is Zeril Bridge. I tried to figure out the places which are connected by the bridge. The two sides of the stream seem to be very steep and it seems to me that the bridge is situated in a strange location.</p>
<p>From the viewpoint of civil engineering, if you build this bridge today, what would your purpose for locating the bridge at that point be?</p>
<p>Coordinates of the Zeril Bridge: 37.915422 , 42.984053</p>
<p><a href="https://i.stack.imgur.com/5SzUj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5SzUj.jpg" alt="enter image description here"></a></p>
| |civil-engineering|bridges|architecture| | <p>The location makes sense from a civil engineering perspective. The bridge has to span the river channel without being washed away during flooding events. A narrow, steep canyon works well for this because the bridge can be built high enough over the water without it having to be too long. The height gives it protection from floods, the steepness of the canyon walls keeps the span as short as possible. </p>
<p>Additionally, the steepness of the canyon walls gives an indication of the strength of the supporting bedrock for the abutments. The steeper the canyon sides, the more an indication of strong, hard rock. If the side slopes were flatter, this would indicate looser, more erodible soils under the ends of the bridge (the abutments). Obviously, stronger, harder rock is superior to easily eroded soils for the purposes of holding up the ends of the bridge. </p>
<p>As long as the bridge is built during low water times of the year, access to the site isn't too difficult. The bridge is close enough to the main river channel to be convenient enough for travelers while still taking advantage of the favorable characteristics listed above. </p>
| 24583 | From the viewpoint of a civil engineer, what was the purpose of this bridge? |
2018-11-09T02:46:40.663 | <p>I find it difficult to understand the concept of flux and field.I have searched the internet and refered a few books.From what I understand, generally flux can be defined as "some quantity per unit of some other quantity" E.g: "the force of attraction at a point in space per unit charge placed at that point is called electric flux" , "the rate of heat transfer per unit area is called heat flux in conduction".
Please suggest whether I am right or wrong.</p>
| |electrical-engineering|heat-transfer| | <blockquote>
<p>From what I understand, generally flux can be defined as "some quantity per unit of some other quantity".</p>
</blockquote>
<p>That's about right. The <em>some other quantity</em> is often one of particular interest Typically, it is a surface, or a facet of a surface, used in control volume analysis. And there is a flux operator. It returns a scalar quantity from an vector operation. In the case of fluids, the fluid advects the quantity of interest. So you have a description of the density of some thing of interest at time t (a scalar field) and a velocity field of the fluid (a vector field), and a reference surface S. The Flux operator tallies the rate at which stuff is advecting through the surface.</p>
<p>Of course, when dealing with pure field equations as in Electromagnetism, there is no need for an advective ether. Flux pertains to the vector field itself with respect to some surface of interest.</p>
<p>We often have a system of equations that all involve this same control volume, and which have multiple conservation laws, which relate to a potential field. The flux idea is very useful in writing these equations in a compact form. For example, the equation below can represent a system of three conservation equations. <span class="math-container">$F$</span> being the flux function.</p>
<p><span class="math-container">$$\frac{d}{dt} \iint_{\Omega}\mathbf q\ dx\ dy\ + \int_{\partial \Omega} F(\mathbf q)\cdot\mathbf n\ ds = \iint_{\Omega} {\psi}\ dx\ dy $$</span></p>
<p>The equations would be conservation of mass, conservation of x momentum, and conservation of y momentum.</p>
<p>See page 4 of this document to see this in action. - <a href="https://pdfs.semanticscholar.org/5844/0936fd380a3e0806cbb7b2507f05d5e8868b.pdf" rel="nofollow noreferrer">Numerical Approximation of the Nonlinear Shallow Water Equations
with Topography and Dry Beds:
A Godunov-Type Scheme, by
David L. George
</a></p>
<p>There is a nice series of three videos from the Kahn Academy that deals with the flux operator. They play in sequence. The first one is <a href="https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/3d-flux/v/conceputal-understanding-of-flux-in-three-dimensions" rel="nofollow noreferrer">here</a>.</p>
<p>(There should be a badge for learning mathjax. That formula took me 45 minutes.)</p>
| 24593 | The concept of flux |
2018-11-09T03:41:45.270 | <p>I am looking for answers to the following questions concerning FreeCAD from someone who is familiar with implementation of FreeCAD :</p>
<ol>
<li><p>Which library FreeCAD uses to parse STEP file format?</p></li>
<li><p>Which library FreeCAD uses to convert file of STEP format to Obj format?</p></li>
<li><p>Is it possible (or how hard is it) to use the above mentioned libraries as standalone libraries in other projects. I have checked some options, but they are all proprietary software.</p></li>
<li><p>If one has many files of STEP format that need to be converted to Obj format, is it possible to write a script to call FreeCAD to do it in a batch. </p></li>
</ol>
| |cad| | <ol>
<li>Open Cascade.</li>
<li>Open Cascade provides triangulation. Parsing step is pretty easy, hell the standard documents a parser. But having a triangulation and underlying data manipulation package makes the difference. This is why its called a CAD kernel.</li>
<li>Yes. Its nontrivial to use cascade directly. That said it does include examples.</li>
<li>Yes.</li>
</ol>
| 24594 | The underlying library FreeCAD uses to manipulate STEP file format and format conversion |
2018-11-09T17:30:28.440 | <p>I know that a hydronic system forms a closed loop. When the boiler is turned on, the water temperature will increase and expand. Since this system forms a closed loop, there is no volume for the hot water to expand. Therefore, the pressure inside the system will increase and explosion might occur if the pressure increases significantly. To maintain a constant pressure, an expansion vessel is installed to contain the increase in the water volume as it expands as it heats up.</p>
<p><img src="https://i.stack.imgur.com/hcdnw.gif" alt="enter image description here"></p>
<p>The expansion vessel is divided into two compartments separated by a diaphragm. There is water on one side and air on the other side. Initially the air pressure equals the system pressure. When the water heats up its pressure increases. Eventually, the pressure exceeds that of the air causing the diaphragm to bend back in the air space, compressing the air. Since the water is allowed to expand, this means the system pressure is maintained constant. On the other hand, the air pressure has been increased because its volume decreases. Now, more increase in the water temperature will increase its pressure. The hot water will not be able to push the diaphragm more unless the pressure water becomes greater than the air pressure, which means the system pressure is not maintained constant. Am I wrong? </p>
| |mechanical-engineering|fluid-mechanics|hvac| | <p>A device used to heat domestic potable water to about 120 <sup><span class="math-container">$\circ$</span></sup>F (50 <sup><span class="math-container">$\circ$</span></sup>C) is usually called a hot water heater. To heat water above that temperature the device used is called a boiler.</p>
<p>Hot water boilers are usually required to heat hot water systems and can operate from low pressures to 160 psig (1100 kPa) max.</p>
<p>Steam boilers provide steam to a steam system that is used for heating only.</p>
<p>Low pressure steam boiler or heating steam boiler can operate to a maximum pressure of 15 psig (103 kPa). High pressure steam boiler (as per ASME) operates at pressures over 15 psig (103 kPa).</p>
<p>High pressure and high temperature hot water boilers, the ones exceeding the above, are found in Manufacturing plants, hospitals etc. These types of boilers will need heavy duty non pressurized expansion tanks. </p>
| 24601 | How does an expansion tank work? |
2018-11-09T20:48:29.433 | <p>For my 4th year dissertation in Civil Engineering I've got to investigate the difference between BS5950, Eurocode 3 and ABAQUS failures. The only thing is, in the BS5950 and Eurocode 3 the failure can only include Major OR Minor axis which is what I need. However, in ABAQUS, the software itself 'decides' in which way it will fail. It starts with Major axis but at the end it takes into account the Minor axis too which gives me an error in comparing these results with the BS and Eurocode. </p>
<p>My question is, is there any way you can tell ABAQUS that the only way the failure can occur is in Major axis? without taking into account the Minor axis. </p>
<p>I'm using BBC Basic to create the ABAQUS input files and then run them through the Command. I have already tried to put 'boundary conditions' (a maximum length it can move along) but it gave me a too high of an error between with and without the boundary conditions so I can't do that. </p>
<p>Has anyone ever do that kind of investigation? Any advice/ideas?</p>
<p>Thanks in advance.</p>
| |structural-engineering|civil-engineering|beam|abaqus|software| | <p>Standards use simplified method of defining failures most times based on experimental data. On that other hand, FEA Software try to simulate exactly what happens in your material.</p>
<p>I am not familiar with abacus but I am familiar with FE and other programs. In order to trigger failure, you have to reach the stress of your yield criterion. I am not really sure if you can change your criterion for each direction thought.</p>
<p>An other way is to cheat by assuming an anisotropic material, rasing enough your material's stiffness of one of the axis will eventually trigger failure in the other axis.</p>
<p>If I were you, I would first try to connect the failure in the other axis with a different part of the standard (that failure should be checked somewhere). Then setup the problem in a way that your main axis will be the crucial.</p>
| 24605 | Abaqus Failure for I-Beam Section in Major direction only |
2018-11-09T22:51:54.787 | <p>I have a mass - spring - damper system with external force and I am trying to simulate it using Matlab. I want to have a linearly parameterized form and use the least squares method to find the estimators. I have reached the stage where I have the form:</p>
<p><span class="math-container">$$ y = \theta^T z $$</span>
<span class="math-container">$$\theta = [\theta_1 \ \ \theta_2 \ \ \theta_3]^T$$</span>
<span class="math-container">$$z = [-s\frac{y}{Λ(s)} \ \ -\frac{y}{Λ(s)} \ \ \frac{u}{Λ(s)}]^T$$</span></p>
<p>where <span class="math-container">$\ Λ(s) $</span> is a stable filter <span class="math-container">$\ Λ(s) = s^2 + 3s + 2 $</span> and <span class="math-container">$\ y $</span> is the output of the system (the mass relocation) and <span class="math-container">$\ u $</span> is the input of the system (the external force).</p>
<p>I am trying to calculate the elements of <span class="math-container">$\ z $</span> vector and for <span class="math-container">$\ z_1 = -s\frac{y}{Λ(s)} $</span> for example I have done the following:</p>
<p><span class="math-container">$$ z_1 = -s\frac{y}{Λ(s)} \Rightarrow \ddot{z_1}+3\dot{z_1}+2z_1 = -\dot{y} $$</span></p>
<p><span class="math-container">$$x_1 = z_1 \Rightarrow \dot{{x_1}} = x_2 - y$$</span></p>
<p><span class="math-container">$$x_2 = \dot{z_1}+y \Rightarrow \dot{{x_2}} = -3x_2 - 2x_1 + 3y$$</span></p>
<p>From now on I want to use the <code>ode45(...)</code> function but I am not able to figure out how to finally calculate <span class="math-container">$\ z_1 $</span> and then put it in the <span class="math-container">$\ z $</span> vector.</p>
| |control-engineering|control-theory|matlab| | <p>This is how the whole backward vector <span class="math-container">$\ z $</span> is computed using matlab <code>ode45(...)</code> solver:</p>
<pre><code>tspan - 0:0.01:10;
eqtn = @(t,x,y) [x(2)-interp1(tspan(:),y(:),t) ; -3*x(2)-2*x(1)+3*interp1(tspan,y(:),t)];
x = [0;0];
[~,phi_1] = ode45(@(t,x)eqtn(t,x,y),tspan, x);
eqtn = @(t,x,y) [x(2) ; -3*x(2)-2*x(1)-interp1(tspan,y(:),t)];
x = [0;0];
[~,phi_2] = ode45(@(t,x)eqtn(t,x,y),tspan, x);
eqtn = @(t,x,u) [x(2) ; -3*x(2)-2*x(1)+interp1(tspan,u(:),t)];
x = [0;0];
[~,phi_3] = ode45(@(t,x)eqtn(t,x,u),tspan, x);
z = [phi_1(:,1) phi_2(:,1) phi_3(:,1)];
</code></pre>
<p>It is assumed that the vectors <span class="math-container">$\ y \ \& \ u $</span> are known and represent the output and the input of the original system correspondingly. </p>
| 24608 | Simulating a mass - spring - damper system |
2018-11-10T21:33:52.277 | <p>So, I have a situation where we are putting load on a plate from above, and the plate has a step in it, which was made by milling a bigger block of steel (i.e. there are no layers and everything is homogeneous). Conventional engineering knowledge tells us that if we increase the fillet radii, we will get a better fatigue performance for this abrupt change in EI within this plate, which is great. But a bigger fillet will also mean the item that this plate is supporting (which in this case fits into the milled area) becomes a limiting factor with how sharp their corners are. I would like to know - is there a fillet design that I can use in this situation that will allow me to improve fatigue performance without necessarily compromising the way my object fits into this milled area? I have proposed some different versions in the sketch attached, and I would like someone well versed in this area to help me understand which one is best, and for what reasons. Thanks!<img src="https://i.stack.imgur.com/9nBT9.png" alt="![enter image description here">]<a href="https://i.stack.imgur.com/9nBT9.png" rel="noreferrer">1</a></p>
| |fatigue| | <p>I have seen 'lollipop cutters' used for this purpose, just a single pass into the side (closest to option B). The thickness of the shank means it's not a perfectly square bottomed slot, but was sufficient in the instance in question. </p>
<p>The image below shows the cutter type:</p>
<p><img src="https://i.stack.imgur.com/xtlZi.jpg" alt="enter image description here"></p>
<p>And the profile in question:</p>
<p><a href="https://i.stack.imgur.com/dg8EPm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dg8EPm.png" alt="corner undercut radius"></a></p>
<p>The method of adding an undercut radius will certainly improve the fatigue life of the part by drastically reducing the stress concentration factor at that corner.</p>
<p>The only situation in which I can imagine this having a detrimental effect is where the wall section is sufficiently thin that the removal of extra material makes it too thin for the forces involved (as illustrated below), but from the sketch in your original question, I doubt very much that this will be the case.</p>
<p><a href="https://i.stack.imgur.com/hCGLSm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hCGLSm.png" alt="thin wall undercut to leave not enough material"></a></p>
<p>It's worth noting that this does leave a very small remaining interference from the 'perfectly square corner', but in my application the mating part was not so perfectly sized to cause a problem here. Also, it was fully enclosed on all sides (such that these illustrations are a section view). Lollipop cutters do not plunge well (if at all), so cutting downwards as shown on the right is only possible if you can enter the workpiece from the side (rather than above, in my case), or perhaps if a pilot slot is milled downwards first with a small endmill first.</p>
<p><a href="https://i.stack.imgur.com/dxHYLm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dxHYLm.png" alt="interference issue"></a></p>
| 24615 | Fatigue resistance at filleted corners - Design detail |
2018-11-11T02:15:28.913 | <p>I need to construct a table that will hold a water tank. The filled tank will weigh up to 1300 pounds. I have the table leg frames worked out to hold 2500 pounds so no worries there. (<a href="https://www.benchdepot.com/cleveland-series-workbenches/1553/159/cleveland-series-frame-with-legs-industrial-work-benches.html" rel="nofollow noreferrer">Here is a link to details.</a>) </p>
<p><a href="https://i.stack.imgur.com/6iwCt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6iwCt.jpg" alt="enter image description here"></a></p>
<p>They are just a 28" x 34" rectangle. The tank itself has a footprint of 30" x 38" and is made of molded polyethylene, so don't count on its rigidity for support. The table top will be 36" x 48". (It is bigger than the tank solely because I can buy it off the shelf at that size without any fabrication cost.) There will be no motion to account for nor any rapid changes in the weight. So my question is how thick does the stainless steel table top need to be to safely bear 1300 pounds?</p>
| |mechanical-engineering|structural-engineering|steel| | <p>The Stress formula for a rectangular plate is:<a href="https://www.engineersedge.com/material_science/rectangular_plate_uniform_load_13643.htm" rel="nofollow noreferrer">source,</a></p>
<p>Rectangular plate, uniform load, simply supported (Empirical) equations.
Since comers tend to rise off the supports, vertical movement must be prevented without restricting rotation.
Symbols used:</p>
<ul>
<li><span class="math-container">$a$</span> = major length of rectangular plate, (in)</li>
<li><span class="math-container">$b$</span> = minor length of rectangular plate, (in)</li>
<li><span class="math-container">$p$</span> = uniform pressure loading, ( Ibs/in')</li>
<li><span class="math-container">$\nu$</span> = Poisson's ratio</li>
<li><span class="math-container">$E$</span> = Young's modulus, ( lbs/in2)= 29,000,000 psi for a 36 steel, with 3 as factor of safety we get 18000 psi for allowable stress.</li>
<li><span class="math-container">$t$</span> = plate thickness, (in)</li>
<li><span class="math-container">$S_m$</span> = maximum stress, (lbs/in2)</li>
<li><span class="math-container">$y_m$</span> = maximum deflection, (m, in)</li>
</ul>
<p><span class="math-container">$$\begin{align}
\sigma_{max} &= \frac {0.75 pb^2}{t^2[1.61(b/a)^3 + 1 ] } \\
y_m &= \frac {0.142pb^4}{Et^3[2.21(b/a )^3+1]}
\end{align}$$</span></p>
<p>This is for edge simply supported, but you have cantilever edges. to make the task simpler we just reduce the p by a factor of 0.75, so 0.75(1300)/ (36)(48) = 0.564 psi</p>
<p><span class="math-container">$$\begin{align}
t^2 &= \frac{(0.564(28^2)0.75)}{ 18000[1.61(28/34)^3 +1]} \\
&= .0987 \\
\therefore t &= 0.1\text{ in}
\end{align}$$</span></p>
<p>Say pick 1/8 thick sheet for practicality.</p>
<p>Check my arithmetic please.</p>
| 24617 | How thick does my stainless steel table top need to be? |
2018-11-11T13:11:10.027 | <p>I am simulating a mass - damper - system in Matlab and I have the following vector (1x100) as input to my system:</p>
<pre><code>ut = linspace(0, 10);
u = 5 * sin(2 * ut) + 10.5; % input of our system - external force
</code></pre>
<p>Now, I want to take the value of the output for each value of the input vector using this differential equation:</p>
<pre><code>function dx = odefun_4(t,x)
m = 15;
b = 0.2;
k = 2;
dx = [x(2); u/m - (b/m)*x(2) - (k/m)*x(1)];
end
</code></pre>
<p>which is called from my main <code>.m</code> file like this:</p>
<pre><code>[t,X] = ode45(@odefun_4, [0 10], [0;0])
</code></pre>
<p>I tried to pass the <code>u</code> vector by making it global variable but I get an error stating: <code>Dimensions of arrays being concatenated not consistent.</code>
I also tried to set the <code>Refine</code> parameter using <code>odeset</code> and the <code>options</code> argument of <code>ode45</code> solver but still don't get the desired results. I could really use some help.</p>
| |mathematics|matlab| | <p>If you want <span class="math-container">$u$</span> to be considered as a smooth continuous function, then you should define it as such since you also have <span class="math-container">$t$</span> available as well. So you could use:</p>
<pre><code>function dx = odefun_4(t,x)
m = 15;
b = 0.2;
k = 2;
u = 5 * sin(2 * t) + 10.5;
dx = [x(2); u/m - (b/m)*x(2) - (k/m)*x(1)];
end
</code></pre>
<p>I often also find it more convenient to use anonymous functions in Matlab, for example:</p>
<pre><code>m = 15;
b = 0.2;
k = 2;
u = @(t) 5 * sin(2 * t) + 10.5;
odefun_4 = @(t,x) [x(2); u(t)/m - (b/m)*x(2) - (k/m)*x(1)];
</code></pre>
| 24620 | Get results of output for each one value of the input - MATLAB ODE |
2018-11-11T15:36:11.660 | <p>My professor, when discussing about Random Walk model during diffusion in metals said that "Vacancies are equilibrium defects but dislocations are not". I could not understand why. Why is it so?</p>
| |materials|metallurgy| | <p>Nature will always try to minimize the number of dislocations and grain-boundaries, whereas in the case of vacancies, there is an optimum number that will lead to the lowest Gibbs energy.</p>
| 24623 | Why are vacancies in solids called as equilibrium defects? |
2018-11-13T08:50:51.537 | <p>In my engineering textbook, I didnt understand the example given to explain the magnitude of the moment. It asks to calculate the magnitude of the moment about the base point O of the 600-N force in five different ways. Now, I dont understand the third way it calculated the moment. What I don't understand is why d(1) = 4 + 2 tan(40) and so the moment is 460(d(1)). </p>
<p>It used the principle of transmissibliity to move the vector force to point B and so we dont need one of the component as it is parallel with the distance. And to calculate the distance we have 4 + 2 tan(40). I understand upto 4 but I dont understand why we add 2tan(40). How is 2 vectored with tan(40)? </p>
<p>Use picture I have uploaded. <img src="https://i.stack.imgur.com/WFbZP.jpg" width="300" /></p>
| |civil-engineering|applied-mechanics|moments| | <p>This is a trigonometry question. </p>
<p>Point B is a certain distance upwards from Point A. How much? You know the distance that it is to the left from Point A (<span class="math-container">$2\text{m}$</span>) and the angle that it is from point A (<span class="math-container">$40°$</span>), so using standard trigonometry <span class="math-container">$\tan(\theta)=\frac{opposite}{adjacent}$</span>, you can calculate the vertical distance by rearranging to give: <span class="math-container">$opposite=adjacent\tan(\theta)$</span>, or in this specific case, <span class="math-container">$x=2\tan(40)$</span>, such that <span class="math-container">$d_1 = 4+2\tan(40)$</span>.</p>
<p><img src="https://i.stack.imgur.com/ib8xSh.jpg" width="400" /></p>
| 24645 | A way to calculate the magnitude of the moment |
2018-11-13T12:36:28.213 | <p>Zinc Plating is known for its ability to prevent most corrosion, and it is done by just dipping a metal in hot pool of zinc and letting it cooldown. Meanwhile, stainless steel, an alloy that is capable of resisting corrosion.</p>
<p>Zinc Plating Advantages:</p>
<ul>
<li>Cheap</li>
</ul>
<p>Stainless Steel Advantages:</p>
<ul>
<li>Is not coating (hence, scratching will not expose for corrosion)</li>
<li>Hardnesss of Material (<span class="math-container">$UTS_{Stainless} \approx 860 \frac {N}{mm^2}$</span>) comparatively to Mild Steel (<span class="math-container">$UTS_{Mild Steel} \approx 470 \frac {N}{mm^2}$</span>)</li>
</ul>
<p><strong>Other than what is said, in what circumstance would you use stainless steel over galvanized steel?</strong></p>
| |metallurgy|metals| | <p>Stainless steel can be used up to temperatures of about 1000C. The corrosion resistance of zinc plating decreases rapidly above 100C, and embrittlement can occur above 500C.</p>
<p>Zinc plating has lower resistance to chemical corrosion from acids and alkalis than stainless steel.</p>
<p>Aside from mechanical damage caused by scratching, the rate of corrosion may not be uniform over the whole structure, leading to local failures of the protective layer.</p>
<p>Zinc plating will fail if it is in contact with some common construction materials, including copper, brass, chrome plated materials, unplated iron and steel, bitumen, some species of timber (including oak, cedar, and some fir species) etc.</p>
<p>Fabricating a structure from plated raw materials may damage the plating layer and reduce the corrosion resistance (e.g. assembly using bolts and screws, bending plated sheet metal, etc). </p>
<p>Welding plated steel requires special precautions to make good quality joints and to protect the operator from fumes containing zinc compounds.</p>
| 24647 | Galvanized Steel (Zinc Plated Steel) vs Stainless Steel |
2018-11-13T16:23:45.170 | <p>I am actuating a small ceramic rotary valve with a cheap mg996 servo motor. The valve has a 90 degree max rotation from fully closed to fully open. Currently I have the servo shaft directly coupled to the valve shaft, and simply limit maximum rotation of the servo to 90 degrees in software.</p>
<p>It would be nice to have the extra resolution afforded by utilizing the servo's full range.</p>
<p>Is there a simple way to reduce the 180 degree rotation to 90 degrees without gears? Space is tight and complexity needs to be minimized.</p>
<p>A push/pull rod on an arm may be the best option, but I'm not aware of other ways to do this.</p>
<p>Note: I have access to a 3d printer.</p>
<p>Thanks!</p>
| |valves|servo| | <p>If space is tight and complexity must be minimized, why not just buy a servo that has 90 degree travel to begin with. There are a number of these available for cheap.</p>
| 24650 | Reduce 180 degree rotation to 90 degree rotation without gears |
2018-11-13T22:29:35.697 | <p>This is a gear that transfers force from a motorcycle kickstarter to the transmission. What method should I use to repair the missing teeth.</p>
<p><img src="https://i.stack.imgur.com/KIBbJ.jpg" alt="enter image description here"><img src="https://i.stack.imgur.com/eyQP2.jpg" alt="enter image description here"></p>
| |motors|gears|transmission| | <p>Oddly enough, this problem can be fixed using a set of car keys and your wallet. Here are the steps to follow:</p>
<p>1) put the keys in the ignition of your car, start it, and drive to the motorcycle scrapyard.</p>
<p>2) once there, exchange several pieces of local currency for a gear pulled out of another identical motorcycle. </p>
<p>3) place the gear in your pocket and drive your car home (optional step: purchase a six of your favorite dry-hopped IPA before arrival).</p>
<p>4) installation procedure is the reverse of the disassembly procedure. </p>
<p>5) after washing hands, open the first bottle of IPA. </p>
<p>6) share and enjoy.</p>
| 24651 | How to fix this gear |
2018-11-14T03:46:23.920 | <p><strong>Fastest plane</strong> ever built: SR-71 ,Max Speed: 3.3 Mach , Development - <strong>1950s</strong></p>
<p><strong>Fastest drone</strong> ever built: Lockheed-D21 ,Max Speed: 3.3 Mach, Development -<strong>1960s</strong></p>
<p><strong>Fastest cruise missile</strong> ever built: Brahmos , Max Speed: 3.3 Mach, Development - <strong>2000s</strong></p>
<p>Why Ram/turbo-Ram jets got stuck around 3.3 Mach? Is there any technological challenge beyond 3.3 Mach like heating or combustion instability or low thrust? </p>
| |mechanical-engineering|aerospace-engineering|aerodynamics|propulsion| | <p>At those speeds, several effects become important. We consider air-breathing engines in what follows.</p>
<p>First of all, aerodynamic heating of the airframe becomes important. In broad terms, to prevent structural failure, the maximum temperature of the hottest portions of the airframe must be held below red heat. Going faster than mach 3.3 pushes this limit.</p>
<p>Second, for an air-breathing engine that propels the airframe from sea level and zero airspeed to mach 3.3 at high altitude, the temperature at the face of the compressor stage must be below the maximum temperature limit of the first stage compressor blades, which is roughly 800F. Going faster than this means the temperature at the face of the engine will exceed the temperature rating of the blades and the engine will self-destruct. </p>
| 24656 | Why there are no jets (missile or aircraft) faster than Mach 3.3? |
2018-11-15T03:01:38.327 | <p>I have both car and motorbike. Just common car and motorbike. My car is Toyota 2005, and my motorbike is Honda 2016, I bough both brand new. Both them are fuel injection. Before them, I owned car and motorbike, but both them were carburetor. With those old automotive, I have to warm the engine up until they reach their stationary. The requirement time will be around 5-10 minutes, depend on the weather and time. Morning in rainy season will be different to day time in dry season, however.</p>
<p>With my new automotive that are fuel injection, my habits still there, I still need to take that time to warm up the engine. But I read in many occasions in automotive news said that for fuel injection just required half and maximum one minute to warm up. For my car, since I start the engine up, it will go to around 1,300 RPM, and required around 7 minutes to become stationary, around 700 RPM. The stationary's RPM is clearly stated at the engine cap, with a yellow poster posted there. For the motorbike, I have no idea as no stationary indicator. But another car, Honda, I saw that it takes shorter time, and one Mazda car I saw just required around half to one minute to warm up, as said in the automotive news.</p>
<p>In my understanding, warm up is intended to warm the engine up slowly to avoid the engine's temperature changes so quickly that of course is not good for the engine itself. So my question is, how that warming the engine could be so different? And what is actually that warming up intended? Several time I tested my car without proper warming up, the result is very bad, something like knocking in the engine.</p>
| |automotive-engineering| | <p>The specific issue with carburettors are that they are purely mechanical devices and can't directly measure the mass flow rate of fuel they are delivering and as such are quite sensitive to temperature. </p>
<p>Carburettor engines often have a 'choke', either automatic or manual which overrides the normal operation of the carburettor, causing the engine to run rich. This improves starting and warm up but isn't good under load so you need to wait untill the choke turns itself off before you use the engine hard. </p>
<p>With fuel injection the ecu has direct control over the air/fuel mix and can adjust it on the fly according to conditions so all the warm up need to do is get the oil warm enough to work and it doesn't need to worry about getting the flue system in thermal equilibrium as it can adjust air fuel ratio directly as required. </p>
| 24667 | What is the Meaning of To Warm Up an Engine? |
2018-11-16T12:11:30.283 | <p>So this is very general, but... How are simulations for vehicles done? I know that computational fluid dynamic simulations are used, but why are repeated tests of actual hardware done too? Obviously this is because there are some things that are found through actual physical testing. So this leads into my question. Are simulations refined from the actual test data? how are phenomena from the actual test translated into simulation? Eyeballing things? just gathering data, making a hypothesis, and then running the simulation again from the top?
I guess what my question ultimately boils down to is: What does a simulation cycle look like?</p>
| |aerospace-engineering|simulation|aerodynamics| | <p>That is a very broad question. It can be as "simple" as visually comparing the simulation and test results, and deciding on the basis of experience how the tweak the simulation if it isn't accurate enough (whatever "accurate" means, in a particular situation). After the model has been changed to agree with the test data, you can then use it to simulate other conditions which were not tested, with more confidence that the results will be meaningful.</p>
<p>At the other extreme, this may be treated as a mathematical optimisation problem. You set up a model with some parameters (geometry, material properties, etc) that can be varied, tell the software what the "answers" should be (as measured in the tests) and the software then runs the model repeatedly, changing the parameters to get the "best" agreement.</p>
<p>The problem with a fully automatic approach is that the initial model may be completely unrealistic (for example because some important feature was left out altogether, either by mistake, or because nobody thought it was important enough to be worth including!) but the software managed to make the answers line up with the test data by using unrealistic values for the parameters, and getting to a situation where several "large errors" cancelled out but the final results "looked about right".</p>
<p>It is fairly routine to use "automatic model generation" in a more limited way. For example in finite element analysis, you might define the geometry of the component, and the software automatically generates the finite element mesh, and then refines it by subdividing some of the elements in regions where the numerical solution hasn't converged, according to some predefined measure of "converged". This is usually much quicker (and more accurate) than generating the complete mesh by hand.</p>
<p>The amount of work involved in validating models can cover a huge range, from one engineer working for part of a day up to several person-years of work for a large, complex, and critically important model.</p>
| 24683 | What is the cycle of simulation editing? |
2018-11-16T17:51:06.060 | <p>We shared <strong>Mechanics of materials</strong> with structural engineers back in college. I learned some basic concept about structural engineering, but besides the basic knowledge i don't know anything about structural engineering. </p>
<p>If i recall correctly, an engineer found a way to distribute the weight of buildings over a wide area to reduce the stress, i think he managed to solve the problem of sinking buildings for the first time in Chicago, and up to this moment, his solution has been popular. </p>
<p>If the area has a strong bedrock, manhattan for instance then, the buildings are much more stable, but my question targets the areas with very weak bedrock, The Hague for instance. </p>
<p>I can say, there is channel is every street, when i walk in streets i can actually hear the vibrations of water beneath my feet, if you dig the ground four or five meters, in city centre you'll reach water. </p>
<p>I wonder how engineers manage to build skyscrapers in this area?
The concentration of tall buildings is not as same as in manhattan, Chicago or ... . But still, how those few tall buildings are still on the ground? </p>
| |structural-engineering| | <p>Usually using piles in multiple vertical arrays is the answer. The friction between the soil and rock against a long vertical concrete column can be substantial enough to provide more than enough reaction force to the compressive load applied by structures on it. A good example is the Shard in London which was built this way. The clay soils in London are notoriously unstable and the 97 floor tower rests on these piles. <a href="https://youtu.be/PIqTIB7R5zc" rel="nofollow noreferrer">Shard construction</a>
<a href="https://youtu.be/1DN5dTwbWUk" rel="nofollow noreferrer">Impossible Engineering Documentary</a></p>
| 24689 | Why skyscrapers don't sink into ground? |
2018-11-16T19:55:46.267 | <p>My cousin is a mechanical engineer, got contacted today from someone who claimed to be working for the supreme board for international arbitrators (<a href="http://sbia.org.uk" rel="nofollow noreferrer">sbia</a>) and <a href="http://www.icac-lebanon.com/lb/" rel="nofollow noreferrer">ICAC</a> in the Middle East and north Africa region.</p>
<p>She told him he would have to pay $300 in order to get a 1 day course, or workshop, afterwords he would become an international arbitrator for engineering issues related to his field. Full sized images <a href="https://i.stack.imgur.com/Oa3JJ.jpg" rel="nofollow noreferrer">here</a> and <a href="https://i.stack.imgur.com/atM1d.jpg" rel="nofollow noreferrer">here</a></p>
<p><a href="https://i.stack.imgur.com/Oa3JJm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Oa3JJm.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/atM1dm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/atM1dm.jpg" alt="enter image description here"></a></p>
<p>My cousin and I think this is a scam, and their website is bad and there's no mention about SBIA anywhere, no results on google, no social media presence. I would like to ask you about it though, have you ever heard about it? Is this a scam?</p>
<p><strong>Edit</strong></p>
<p>I think they are copycatting this <a href="http://www.incodir.org/" rel="nofollow noreferrer">http://www.incodir.org/</a>, not only that, I hope you could read arabic, on the first image, it says in parentheses </p>
<p><a href="https://people.bayt.com/reem-alfanek/?lang=ar" rel="nofollow noreferrer">ريم الفانك</a></p>
<p>which also works in the organization I mentioned, I'm going to try to contact her somehow, either she's being impersonated in which case I'll see if we can press charges, otherwise she's part of the scam, I hope not.</p>
<p><strong>Edit 2</strong></p>
<p>I tried contacting [email protected], the email didn't get delivered to report the case, well then I must conclude that they too are a part of the scam sadly, the rabbit hole goes deep.</p>
| |mechanical-engineering|electrical-engineering|civil-engineering|international| | <p>They can't even write English on their certificate. </p>
<blockquote>
<p>... training of good stand, and is in accord with American and
European standers...</p>
</blockquote>
<p>That's an obvious typo. Of course it should say</p>
<blockquote>
<p>... training of good sit, and is in accord with American and
European sitters...</p>
</blockquote>
<p>Is it worth paying $300 for a joke certificate? That depends on the buyer's sense of humour, I suppose.</p>
| 24691 | Is there such thing as supreme board for international arbitrators? |
2018-11-17T13:44:53.293 | <p>I have shared the question statement, its diagram and my solution but my solution is wrong and I have no idea why is it wrong, please help me out.</p>
<p><a href="https://i.stack.imgur.com/E5fCl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E5fCl.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/0dd2C.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0dd2C.jpg" alt="enter image description here"></a></p>
| |statics| | <p>Using method of joints we can calculate the maximum compression according to the values given being 6kN. As that strain is greater than 8kN we should assume 6kN to be achieved first.</p>
<p>Force P should not exceed the maximum compression value of 6kN at the vertical component of the truss CD.</p>
<p><span class="math-container">$$6 \sin 60=3 \sqrt {3}=5.196kN$$</span></p>
<p>There is no way that tension can even come close to 8kN in the truss without exceeding the 6kN compression.</p>
| 24702 | Statics Question |
2018-11-18T11:32:43.850 | <p>I am curious about the design considerations for a hydraulic system.</p>
<p>Let's say you have a 3HP 1200RPM motor
you also have a hydraulic pump that's rated 4LPM and pressure of 40MPA rated for 40RPM MAX</p>
<p>You can put many different types of gear box on the 3HP motor to get it to operate in the N-40RPM range</p>
<p>Let's forget about the power loss through heat when using the gearbox.</p>
<p>I am trying to understand the relationship between HP and RPM of the electric motor, Pressure of the system and output Torque.</p>
<p>Let's say I am working at 200Bar, electric motor doing 40RPM and the hydraulic motor running at 4LPM.</p>
<p>How would that compare to say going to 20RPM and keeping 200Bar and 4LPM.</p>
<p>I do understand that raising the pressure would increase the heat and possibly require cooling.</p>
<p>Any suggestions on how to calculate these relationships?</p>
| |mechanical-engineering|fluid-mechanics|torque|hydraulics| | <p>To summarize your problem description, you have ...</p>
<ul>
<li>3 HP, 1200 RPM electric motor</li>
<li>40 MPa (400 bar) pump rated 4 LPM @ 40 RPM</li>
<li>Gearbox ratio TBD between motor and pump</li>
<li>Desired operation of 20MPa (200bar) at 4 LPM</li>
</ul>
<p>First, we can make a few observations from the problem data. </p>
<ol>
<li>The full-load torque of your electric motor. Note that this number is a nominal rating. Typical general-purpose induction motors can run at 10-15% higher torque than the full-load rating without damage, as long as cooling air flow is not obstructed.
<span class="math-container">$$P_{motor} = 3 \ hp * {1 \ kW \over 1.341 \ hp} = 2.24 \ kW$$</span></li>
</ol>
<p><span class="math-container">$$M_{motor} = {P_{motor}\over2*\pi*{n/60}} = 17.8 \ Nm$$</span></p>
<ol start="2">
<li><p>Where did you find a pump with those specs? It's possible in theory but I've never seen anything close. A typical 4 LPM positive displacement pump would have a max speed over 1800 rpm. This avoids the need for a gearbox. Rexroth, Parker, Hydac, Eaton... many options available for pumps. Gear-type (200 bar) or radial piston (400 bar) pumps are typically the most cost-effective at your flow rate. Here is one catalog for example -
<a href="https://dc-us.resource.bosch.com/media/us/products_13/product_groups_1/industrial_hydraulics_5/pdfs_4/re11263.pdf" rel="nofollow noreferrer">https://dc-us.resource.bosch.com/media/us/products_13/product_groups_1/industrial_hydraulics_5/pdfs_4/re11263.pdf</a> </p></li>
<li><p>In case you really do have a pump along those lines... here is the power transmitted through the hydraulic fluid:
<span class="math-container">$$P_{flow} = p*Q*{100000 \ Pa \over 1 \ bar}*{1 \ m^3/s \over 60000 \ lpm} = 1.33 \ kW$$</span></p></li>
<li><p>Choose a typical pump efficiency of 85% and you have the power required to spin the pump:
<span class="math-container">$$P_{pump} = {P_{flow} \over 0.85} = 1.57 kW$$</span>
This is safely within the nominal motor power rating of 2.24 kW, including whatever gearbox efficiency you want to assume. </p></li>
<li><p>Lastly we can verify that motor torque is also sufficient to drive the pump at 40 rpm.
<span class="math-container">$$M_{pump} = {P_{pump}\over2*\pi*{n/60}} = 374.8 \ Nm$$</span>
So your gearbox would need to amplify torque by
<span class="math-container">$${M_{pump} \over M_{motor}} = {374.8 \ Nm \over 17.8 \ Nm} = 21.1 \ min \ gear \ ratio$$</span>
Which is accomplished with plenty of overhead thanks to the speed ratio:
<span class="math-container">$${n_{motor} \over n_{pump}} ={1200 \ rpm \over 40 \ rpm} = 30.0 \ actual \ gear \ ratio$$</span></p></li>
</ol>
<p>Units: <span class="math-container">$$P[kW], \ \ p[bar], \ \ n[rpm], \ \ Q[lpm], \ \ M[Nm]$$</span></p>
| 24717 | Calculate hydraulic torque from motor HP, hydraulic pump and gearbox |
2018-11-18T14:46:47.213 | <p>"In a sling psychrometer(with dry and wet bulbs), if the moist air is not saturated(with water vapour) then it evaporates some water on the wet bulb and in turn cools the wet bulb.So the wet bulb temperature falls below the dry bulb temperature" .This is the explanation given in the book Iam using(Fundamentals of thermodynamics by Moran and Shapiro).
1)Does this mean, the wet bulb and dry bulb temperatures are initialy same(before exposing to air)?
2)If evaporation occurs continuously as air is blown over the bulbs, why isnt the temperature reducing continuously(it stops at a point)?</p>
| |thermodynamics|hvac| | <p>1) yes,</p>
<p>2) evaporation continues which reduces the wet bulb temperature until the temperature stays constant because the rate of evaporative heat loss is balanced by the heat absorbed due to the temperature difference between the wet bulb temperature and ambient temperature is in equilibrium. The humdity can then be obtained from a Mollier or psychrometric chart.</p>
| 24720 | Wet bulb temperature in psychrometry |
2018-11-19T20:51:51.750 | <p>If the diameter before knurling is 80. Is the knurled diameter also 80?</p>
| |technical-drawing| | <p>Knurling affects the roughness of a surface, but does not affect its diameter much. No material is removed, so for every indentation, a ridge is formed. This has the effect of making the measured diameter using calipers, for example, slightly larger*. If the knurled diameter is an inspection dimension, then you may wish to place a note on the drawing, after experimental verification, to show the QA team what to look for.</p>
<p>Here is a manufacturer's table of expected increase in diameter for various knurls: <a href="http://accu-trak.com/technicalinfo/table1.html" rel="nofollow noreferrer">http://accu-trak.com/technicalinfo/table1.html</a> (thanks to @D Duck)</p>
<p>Remember the primary purpose of the drawing is to give information to ensure it will be manufactured correctly. Both the person machining the turned 'blank', and the person adding the knurl to that will be primarily concerned with the 'smooth' diameter.</p>
<p>*The average diameter actually goes down, since volume is related to radius cubed, meaning the indentations are deeper than the ridges are tall, relative to the original diameter. In reality this is hard to measure, and I've never seen an engineering drawing where the knurled dimension was different to the turned dimension. </p>
| 24736 | Is the diameter before knurling assumed to be the same as diameter after knurling? |
2018-11-20T06:57:00.397 | <p><a href="https://www.hindawi.com/journals/bmri/2017/4701481/" rel="nofollow noreferrer">Research shows that applying 1 MHz non-focused ultrasound at 3 W/cm2 may reduce subcutaneous fat thickness.</a> In fact, high-end and expensive (costs $5000 >) <a href="https://en.wikipedia.org/wiki/High-intensity_focused_ultrasound" rel="nofollow noreferrer">HIFU</a> ultrasonic transducers are now being used in cosmetic clinics that allow the user to regulate ultrasound frequency, pulsing, power and time in order to target body fat. </p>
<p>To my surprise, I noticed that one can also buy inexpensive (costs only $50), portable 1Mhz Ultrasound cavitation device from <a href="https://www.ebay.ca/itm/1MHz-Ultrasonic-Cavitation-Ultrasound-Face-Body-Massage-Skin-Tightening-Anti-Age/222369021374?hash=item33c639fdbe:g:--wAAOSwWMFa2uxK:rk:5:pf:0" rel="nofollow noreferrer">Ebay</a> or <a href="https://www.aliexpress.com/item/1MHz-Ultrasound-Cavitation-EMS-Body-Slimming-Massager-Loss-Weight-Anti-Cellulite-Fat-Burner-Galvanic-Infrared-Ultrasonic/32884572943.html?src=google&albslr=200431112&src=google&albch=shopping&acnt=494-037-6276&isdl=y&slnk=&plac=&mtctp=&albbt=Google_7_shopping&aff_platform=google&aff_short_key=UneMJZVf&&albcp=1633541485&albag=62156539197&trgt=296904914040&crea=en32884572943&netw=u&device=c&gclid=Cj0KCQiA28nfBRCDARIsANc5BFBp1inb-DaYexIKqC9yO8fc9QJJIJS8ERwr6Rt7CZKim8q2bmIsegsaAgDgEALw_wcB&gclsrc=aw.ds" rel="nofollow noreferrer">Aliexpress</a> that also claims to reduce body fat. Scam or not, one way to justify its claim will be to measure and verify the emitting frequency and intensity.</p>
<p><a href="https://www.aliexpress.com/item/foot-massager-ultrasonic-slimming-body-firming-lifting-face-Health-Care-Products-lose-weight-burn-fat-electronic/32239480611.html?spm=a2g0s.9042311.0.0.3da24c4dN2ciMo" rel="nofollow noreferrer">I bought such a device from Aliexpress (costed $130)</a> to play around with it. My skin can feel a definite vibration when I turn on the device, however, it is very hard to tell if ultrasound is actually exciting my tissue lesions. </p>
<p>I want to know how to set up an inexpensive testing setup and:</p>
<p>1) Find out if the device is really emitting ultrasound. If yes, how would I measure its frequency? Device manual says 1MHz, but I suspecting it might be a bs.</p>
<p>2) How to measure the intensity and penetration depth. Finding one value should lead me to calculate another.</p>
<p>3) Find out if the device's ultrasound is focused or non-focused. My gut feeling is the ultrasound is non-focused or weakly focused. </p>
<p>One way to find more clues would be to rip through the device and look at the transducer circuit. The transducer hardware might have a model number that can be used to find it's spec sheet online. However, I haven't unscrewed the device yet and I doubt I would find anything helpful on the circuit board as it probably came directly from a questionable Chinese factory.</p>
<p>Further reading:
There has been plenty of research conducted so far regarding effects of non-focused ultrasound on animal contouring:</p>
<p>1) <a href="https://www.ncbi.nlm.nih.gov/pubmed/23220874" rel="nofollow noreferrer">The effects of nonfocused external ultrasound on tissue temperature and adipocyte morphology.</a>
2) <a href="http://www.electrotherapy.org/modality/ultrasound-dose-calculation" rel="nofollow noreferrer">Ultrasound dose calculation</a></p>
| |experimental-physics|biomedical-engineering|ultrasound|waves| | <p>You can measure this with a piezoelectric transducer and an oscilliscope. Several companies sell thin film piezo transducers that you can just stick onto your device under test. The transducer has leads that you would connect to a piezo pre-amplifier which will amplify and buffer the signal coming from your DUT. Then attach a scope lead to the output of the pre-amp and you can see the signal on your scope.</p>
| 24747 | How to measure and verify ultrasound frequency and intensity from a cheap ultrasonic body contouring device? |
2018-11-21T01:36:51.527 | <p>This is the problem I'm trying to solve:</p>
<p><img src="https://d2vlcm61l7u1fs.cloudfront.net/media%2F4a5%2F4a5383da-7b6d-4ec7-baf6-ca99ebf309da%2FphpQwuYae.png" alt="figure"></p>
<p>I'm looking at the worked-out solution from my professor and it shows that the sum of the forces in the x-direction is equal to:</p>
<p>[d/dt(momentum) in the x-direction, out] - [d/dt(momentum) in the x-direction, in]</p>
<p>I don't understand what is meant by momentum out vs. momentum in and why one is positive and the other is negative. </p>
<p>Also, I apologize if this is a dumb question.</p>
| |mechanical-engineering|fluid-mechanics| | <p>If it is a solid body, the force applied on the body is the rate of change of momentum of the body in the direction of force(here we talk about the mass of the body and its velocity).In case of a fluid, we talk about the mass flow rate and velocity.So the initial momentum of the flow before the bend(and nozzle) is the momentum in and after interacting with the nozzle, the momentum of the flow leaving out is the momentum out.The difference will give the force acting on the fluid by the bend.I think you should include pressure force in x direction.</p>
| 24770 | How to use the linear momentum equation to determine the forces required to hold a nozzle stationary? |
2018-11-21T21:04:04.813 | <p>I'm trying to make some code to predict the performance of a sounding rocket I want to build. To do this, I need the atmospheric density at any point in the flight for the time-step simulation I made.</p>
<p>I got my atmospheric density in slug/ft^3 from <a href="https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/standard-atmosphere-d_604.html</a> and put all that data in this python script, under the assumption that the function that would best describe this data would be a negative exponential.</p>
<pre><code>from scipy import optimize
import numpy as np
import matplotlib.pyplot as plt
def fxn(x, a, b, c):
return a*np.exp(-b*x) + c
altitudes = np.array([0,5000,10000,15000,20000,25000,30000,35000,40000,45000,50000,60000,70000,80000,90000,100000,150000,200000,250000])
density1 = np.array([23.8,20.48,17.56,14.96,12.67,10.66,8.91,7.38,5.87,4.62,3.64,2.26,1.39,.86,.56,.33,.037,.0053,.00065])
altitudeLinSpace = np.linspace(0,500000,500000)
#Popt returns a array containing constants a,b,c... etc for function 'fxn'
popt, pcov = optimize.curve_fit(fxn, altitudes, density1, p0=(1, 1e-6, 1))
print(popt)
#Plots data
plt.grid(True)
plt.ylim((0,25))
plt.plot(altitudes, density1)
plt.plot(altitudeLinSpace, fxn(altitudeLinSpace, popt[0], popt[1], popt[2]))
plt.show()
</code></pre>
<p>Which gives me this graph when I plot the solution ( Altitude ft (x), density in slug/ft^3 * 10^-4 (y)):
<a href="https://i.stack.imgur.com/jhSkF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jhSkF.png" alt="Altitude(x), density in slug/ft^3 * 10^-4 (y)"></a></p>
<p>The orange curve is what python found as a solution, blue is the plot of the data I used to curve fit. Because of the poor fit, my guess is that the data is actually not a negative exponential, but I'm not really sure how to determine the kind of function this might be...</p>
<p>I'm not looking for perfect, but using a polynomial is a less than ideal solution here as the 'wiggles' give too much error at higher altitudes.</p>
<p>Any suggestions on how I could fit this data better would be greatly appreciated!</p>
<p>Solution from Math SE: <a href="https://math.stackexchange.com/questions/3009560/curve-fitting-atmosphere">https://math.stackexchange.com/questions/3009560/curve-fitting-atmosphere</a></p>
| |fluid-mechanics| | <p>The density varies exponentially - but at rate dependent not just on altitude but also temperature.</p>
<p><a href="https://i.stack.imgur.com/Z1zyz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z1zyz.jpg" alt="enter image description here"></a>
<a href="https://scied.ucar.edu/atmosphere-layers" rel="nofollow noreferrer">source</a></p>
<p><a href="https://i.stack.imgur.com/uXy5c.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uXy5c.gif" alt="enter image description here"></a>
<a href="http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap2.html" rel="nofollow noreferrer">source</a></p>
<p>If you want a more precise fit, you'll have to split it up into atmospheric layers: </p>
<ul>
<li>Troposphere: 0 to 12 km (0 to 7 miles)</li>
<li>Stratosphere: 12 to 50 km (7 to 31 miles)</li>
<li>Mesosphere: 50 to 80 km (31 to 50 miles)</li>
<li>Thermosphere: 80 to 700 km (50 to 440 miles)</li>
<li>Exosphere: 700 to 10,000 km (440 to 6,200 miles)</li>
</ul>
<p>and fit the exponent into each of these - possibly introducing additional thresholds for stratosphere (at 20km and 35km) - a spline created of exponential curves. Trying to find a single simple function to fit reality with this sort of complexity is a misguided attempt.</p>
| 24784 | Curve Fitting Atmospheric Density |
2018-11-22T21:14:02.670 | <p>I have a question about the characteristic dimension. </p>
<p>Q1
The drag on an airship is known to depend on the air density, ρ , dynamic viscosity,
μ , speed relative to the air , V , and a characteristic dimension, L . An airship is to
operate at 20 m/s in air at standard conditions. A model is constructed at 1/20 scale
and tested in a wind tunnel at the same air temperature to determine drag.
(a) Perform a dimensional analysis to determine the criterion which should be
considered to obtain dynamic similarity. </p>
<p>like the question said for the characteristic dimension L, How can i know what dimension is L.</p>
| |fluid-mechanics| | <p>The drag will actually depend on the frontal area. This can be calculated as <span class="math-container">$constant*L^2$</span>. What the constant is depends on what you <em>select</em> L to be. </p>
<p>The question is concerned specifically with the dimensionality, i.e. how does the drag force scale as you vary each of the components in the equation?</p>
<p>In this example, you're looking at a 1/20th scale model. A full scale model will be 20 times as long, 20 times as wide, have wheels 20 times as big, and consequently, the frontal area will be 400 times larger.</p>
<p>It doesnt matter what you select as “L” since the constant is determined experimentally. I’d suggest something easy to measure, like overall length of the ship.</p>
<p>If, in this particular model, the length is 4 times the diameter, and you define L=diameter, then the constant that is measured experimentally would simply be 1/4 of the value it would have been if you defined L=length.</p>
| 24804 | Dimension Analysis For fluid mechanic |
2018-11-22T22:29:55.403 | <p>I'm trying to design a plumbing system for a school project, and I want to verify that my initial piping layout from water source to outlets is logical before I start making calculations and programs for the system. Here is my drawing, thanks to Rhodie:</p>
<p><a href="https://i.stack.imgur.com/PIOxI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PIOxI.png" alt="LAyout"></a>
The green curve represents a hill that the building is on. My logic for this design was that by minimizing the number of fittings I have in my system I can reduce the required head I would need to generate by the pump and save some money.</p>
<p>Any advice or comments would be greatly appreciated!</p>
| |fluid-mechanics|civil-engineering|building-design| | <p>My first question is why would you want that length of pipe? The total length could be substantially reduced if the vertical pipe was placed close to point 1. Your materials cost would be halved. The way to calculate the best vertical point would be to spur outlet 1 straight from the riser and the other 2 distant outlets from that same single riser.</p>
<p>Pressure head has nothing to do with number of fittings. It is calculated according to gravity×vertical volume of water above inlet: your pump. </p>
| 24805 | Optimal plumbing layout for 3 storey building? |
2018-11-24T01:03:31.390 | <p>It is my understanding that of all three states of matter, gases have the lowest thermal conductivity. </p>
<p>If this is the case, why are <em>solid</em> materials, such as fiberglass and polyurethane, used for this purpose? </p>
<p>Would it not make the most sense to simply fill the space with air?</p>
| |heat-transfer| | <h1>Background</h1>
<p>The trade off is between conduction and convection. Heat conduction <span class="math-container">$\dot{q}_k$</span> (W) is set as</p>
<p><span class="math-container">$$ \dot{q}_k = k A \frac{\Delta T}{w} $$</span></p>
<p>where <span class="math-container">$k$</span> is thermal conductivity (W/m K), <span class="math-container">$A$</span> is area (m<span class="math-container">$^2$</span>), <span class="math-container">$\Delta T$</span> is temperature difference (<span class="math-container">$^o$</span>K), and <span class="math-container">$w$</span> is the width (m). Heat convection <span class="math-container">$\dot{q}_h$</span> is set as</p>
<p><span class="math-container">$$ \dot{q}_h = h A \Delta T $$</span></p>
<p>where <span class="math-container">$h$</span> is convection coefficient (W/m<span class="math-container">$^2$</span> K). As the thickness of the gap <span class="math-container">$w$</span> increases, the rate of conduction decreases. At zero thickness, the convection coefficient <span class="math-container">$h$</span> is zero. As <span class="math-container">$w$</span> increases, <span class="math-container">$h$</span> increases to a value indicative of a natural convection system.</p>
<p>Armed with this information, you can look up values of <span class="math-container">$k$</span> for air and <span class="math-container">$h$</span> for natural convection in air. Derive a ratio as</p>
<p><span class="math-container">$$\frac{\dot{q}_h}{\dot{q}_k} = \frac{hw}{k}$$</span></p>
<p>Use this to find the value of <span class="math-container">$w$</span> where convection dominates conduction. Any air gap thinner than this will be OK because convection will be lower than conduction, and conduction is low in any case. Any air gap thicker will be controlled by convection.</p>
<p>This presumes that air is not "blowing" in the gap. At that point, <span class="math-container">$h$</span> can increase by one or two orders of magnitude. By example, what was acceptable for natural convection as a 1cm air gap becomes acceptable perhaps only below 0.1mm as an air gap.</p>
<h1>Examples</h1>
<p>By reference with <span class="math-container">$k_{air} \approx$</span> 0.02 W/m K as per <a href="https://en.wikipedia.org/wiki/List_of_thermal_conductivities" rel="nofollow noreferrer">this reference</a>, we can find as per <a href="https://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html" rel="nofollow noreferrer">this reference</a></p>
<p>Free Convection (low): <span class="math-container">$h \approx 0.5$</span> W/m<span class="math-container">$^2$</span> K <span class="math-container">$\Rightarrow w \approx 2$</span> cm</p>
<p>Free Convection (high): <span class="math-container">$h \approx 500$</span> W/m<span class="math-container">$^2$</span> K <span class="math-container">$\Rightarrow w \approx 40~\mu$</span>m</p>
<p>Maintaining a low free convection state in a vertical gap that is 2 cm wide is difficult. Once the air at the top of the gap becomes even slightly hotter than the air at the bottom, a convection cell is established. The air rotates from bottom to top of the cell, and the convection coefficient goes up by an order of magnitude already.</p>
<h1>A Case for an Exception?</h1>
<p>Modern homes with brick exteriors have an air gap between the brick and the panel insulation.</p>
<p><a href="https://www.carsondunlop.com/inspection/blog/brick-houses-solid-masonry-vs-brick-veneer/" rel="nofollow noreferrer">https://www.carsondunlop.com/inspection/blog/brick-houses-solid-masonry-vs-brick-veneer/</a></p>
<p><a href="https://www.finehomebuilding.com/2013/03/06/best-practices-methods-for-installing-brick-or-stone-veneer" rel="nofollow noreferrer">https://www.finehomebuilding.com/2013/03/06/best-practices-methods-for-installing-brick-or-stone-veneer</a></p>
<p>This is to avoid the case where moisture will collect at the junction of brick-touching-wall. It also allows the bricks to dry properly when they become wet.</p>
<p>The air gap is a modest thermal insulator. However, when the sun shines on the brick, free convection cells can form in the gap. This is one reason why thermal insulation is still used between the brick+air gap and the interior house walls.</p>
| 24826 | Wouldn't it be most effective to insulate a house with just an air space? |
2018-11-24T22:27:31.413 | <p>Example: let's say that I have a rectangular part that I would like machined. I'd like to cut it to change the length, but the width and thickness are fine as is. The width and thickness are not nice round numbers. Is there any convention on how to communicate this through a technical drawing? Something like "AS IS"?</p>
<p>Related to <a href="https://engineering.stackexchange.com/questions/11130/do-i-need-to-mention-every-dimension-on-a-to-scale-drawing]">Do I need to mention every dimension on a to-scale drawing?</a></p>
| |drafting|technical-drawing|drawings| | <p>The way to indicate dimensions that do not have to be machined is with a so-called <a href="https://en.wikipedia.org/wiki/Reference_dimension" rel="nofollow noreferrer">reference dimension</a>. The dimension used to be followed by 'REF', but nowadays the dimension is simply placed in between parentheses, e.g. (1 mm).</p>
| 24840 | How to communicate to keep original dimensions of a part in a technical drawing? |
2018-11-25T04:31:10.227 | <p><a href="https://i.stack.imgur.com/UYVuT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYVuT.png" alt="question"></a>
<a href="https://i.stack.imgur.com/ca1Ot.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ca1Ot.png" alt="solution"></a></p>
<p>I am having a lot trouble understanding this problem, I don't even know where to begin or how to articulate my difficulties, but I will try my best.</p>
<ol>
<li>Is the question asking to find the torque developed at the part in the picture where it says "Pipe supported by wall"? </li>
<li>What is meant by the "Horizontal section" of the pipe? In the solution it says that it's going 24 in. into the paper. Okay, so what? Is water flowing through there? I thought it was flowing through the pipe supported by the wall. Or is it flowing through both the pipe support by the wall AND the pipe section that's going 24 in. into the paper?</li>
<li>In the solution, I don't understand how the moment arm for the inlet flow is zero and the moment arm for the outlet flow is 24 in. Where is this being measured from? I thought it was being measured from the "Pipe supported by wall"? Where is the force being applied?</li>
</ol>
<p>I apologize for my extreme confusion. I will be extremely thankful for any help that I can get. </p>
| |mechanical-engineering|fluid-mechanics|torque| | <p>This pipe's bends and the flow of water in it create two reactions:</p>
<p>1- momentum on the wall support created by the first 90 degree elbow, which is not asked for in the question.</p>
<p>2- torque created by the out flow downward which is being asked and answered by the question.</p>
<p><a href="https://i.stack.imgur.com/Mswya.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mswya.jpg" alt="This is a 3d sketch of the pipe."></a></p>
| 24846 | How to use the moment-of-momentum equation for water flowing through a pipe? |
2018-11-26T09:33:58.443 | <p>I'm trying to get the compressive strength (and Young's modulus) of a clay sample. </p>
<p>The method I found is called 'Unconfined Compressive Strength of Cohesive Soil'. According to ASTM D2166M, I have to set the loading rate to 0.5-2%/min (axial strain).</p>
<p>The machine I have though, is stress-controlled. It measures the load in Pa/s. </p>
<p>Since concrete samples can be tested through a similar method; is there a method that I can use to measure the elastic modulus and/or compressive strength of the clay using something with a rate of Pa/s?</p>
<p>For example, if I set the loading to something very low (e.g 1Pa/s) will I get acceptable results?
(Note: for concrete it is acceptable to use 0.15MPa/s)</p>
| |stresses|geotechnical-engineering|elastic-modulus|soil| | <p>I would use a Shear Box test for this. Stress testing is for maximum load-bearing capacity...</p>
<p>Shearing is what gives you the modulated graphic results you seek. Have a look at Smith's Soil Mechanics 9th ed for more details. </p>
| 24862 | Is it possible to calculate Young's Modulus of Clay from load-controlled tests? |
2018-11-26T22:08:07.900 | <p>Why is it that, at least in Ontario Canada, camber diagrams are required for detailed design drawings (Not shop drawings) for bridges with steel girders, but are not required for CPCI Girders?</p>
<p>My understanding is that the camber drawing is required for the steel manufacture to put in a vertical curved to the beam so that when the dead load of the superstructure is added on top of the girders the whole thing bends to meet the road profile.</p>
<p>For concrete girder this is not done, though a similar process seems to be followed. The screed elevation is adjusted to account for deflection of the concrete, including long term deflection from shrinkage and creep.</p>
<p>I would have thought a screed elevation table would be required for both steel and concrete girder designs, so why have to two different approaches for dealing with deflection to match the road profile?</p>
| |structural-engineering|bridges| | <p>CPCI girders are prestressed precast concrete girders, right? (Every region seems to have their own acronym for these things so it's not the name I'm familiar with but the mechanics should be the same.)</p>
<p>The short version of the answer is that the prestressing will usually give you the appropriate amount of precamber even if the girder is cast in a form with a flat bottom.</p>
<p>Roughly speaking, the amount of prestressing is chosen to give a bending moment that counters the permanent load and a small percentage of the variable load (by placing most of the prestressing in the bottom of the beam). This gives the girder an upwards precamber straight out of the form, but once it's installed and the bridge completed, the total deflection will be around zero.</p>
<p>Obviously, it will not always be possible to make it end up perfectly horizontal. And the resulting deflection is nearly impossible to calculate with millimeter accuracy due to uncertainties in creep both before and after installation and the elasticity modulus of the concrete when the prestressing is applied. But in most cases it will even out nicely enough for a precamber diagram not to be worth the trouble.</p>
| 24870 | Camber Diagram for Steel Girders but not for Concrete Girders |
2018-11-26T23:19:01.200 | <p>I am new to vibration analysis/simulation etc. so I may be missing something very basic here so please forgive me if I'm explaining things poorly or not making sense.</p>
<p>To familiarize myself with vibration simulation I am trying to simulate the frequency response spectrum for a tuning fork. I am using Solidworks.
I have </p>
<ul>
<li>created the geometry for a tuning fork </li>
<li>fixed the base and calculated the eigenfrequencies eigenmode shapes</li>
<li>Applied a uniform base displacement excitation (same magnitude for all frequencies)</li>
<li>Plotted the lateral displacement of one of the tines as a function of excitation frequency</li>
<li>The part is made from copper (young modulus supplied by software) and I am using a modal damping ratio of 0.02 for all modes.</li>
<li>The excitation amplitude is 1mm</li>
</ul>
<p>Here is an image of the main "tuning fork" mode showing the fixed base (green arrows) and excitation vector (purple arrow).<a href="https://i.stack.imgur.com/vDrbA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vDrbA.png" alt="enter image description here"></a></p>
<p>Here is a plot of the magnitude lateral displacement (x direction) of a node at the tip of one of tines of the tuning force versus excitation frequency.</p>
<p><a href="https://i.stack.imgur.com/wEA0b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wEA0b.png" alt="enter image description here"></a></p>
<p>The peak at ~500 Hz corresponds to the mode shown above. The peak at ~2.5 kHz corresponds to a higher frequency vibrational mode.</p>
<p><strong>Question</strong></p>
<p>I am confused about the high frequency response of the tuning fork. At low frequency I see that the response is falling off from the two resonances towards lower frequencies. This makes sense. If you slowly move the tuning fork up and down (from the base) there will be no stress or deflection since everything is slow so there will be no lateral displacement.</p>
<p>However, I would ALSO expect the response to drop off at high frequencies as well.. I guess my intuition here is motivated by the following. Consider a mass attached to a base by a spring and a damper. If the base is shaken then at high frequencies the response of the mass falls off, it is as if there is nothing connecting the mass and the base at high frequency. I guess I expect something similar for the tuning fork. If I drive at high enough frequency the tuning forks can't even tell there is a drive..</p>
<p>Can someone please explain to me:</p>
<p>1) If it is correct that this response function should be constant versus frequency above the tuning fork resonance frequencies and</p>
<p>2) Whether it is correct or incorrect can you give me any intuition for why we should expect whatever is the correct behavior?</p>
| |vibration|simulation|frequency-response| | <p>You said you applied a constant <em>displacement</em> to the base. In that situation, if the structure doesn't have any modes that can are excited at a particular frequency, the response is going to be approximately the same as a rigid body being moved with a fixed amplitude, which is what your plot shows.</p>
<p>Note that if you apply a fixed displacement, the <em>excitation force</em> will increase for higher frequencies (and except where there is a resonance in the displacement response) it will be approximately proportional to the frequency squared.</p>
<p>If you applied a constant <em>force</em>, you would see the displacement falling as the frequency increased, but unless the structure is constrained in some way, the displacement amplitude will increase to infinity as the frequency gets close to zero.</p>
<p>Another way to understand this is think abut a 2 DOF system being excited at a frequency much higher than its two resonant frequencies. In that situation the elastic forces are small compared with the inertia forces, so we can ignore the stiffness term in the equation of motion and approximate it as <span class="math-container">$$\begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} = \begin{bmatrix}F \\ 0 \end{bmatrix} $$</span> where F is the excitation force applied at point 1. </p>
<p>Now if we apply a constant amplitude <em>displacement</em> <span class="math-container">$X$</span> at point 1, we have <span class="math-container">$\ddot x_1 = -\omega^2X$</span> and the two unknowns in the equations are <span class="math-container">$\ddot x_2$</span> and <span class="math-container">$F$</span>. </p>
<p>From the second equation we have <span class="math-container">$$\ddot x_2 = - \frac{m_{21}}{m_{22}} \ddot x_1$$</span> which means that <span class="math-container">$$x_2 = - \frac{m_{21}}{m_{22}}X.$$</span></p>
<p>In other words, for a constant base amplitude, the tip amplitude of your tuning fork will also be approximately constant at high frequencies, except when there is a resonance that affects the motion of the system.</p>
| 24873 | Base Driven Tuning Fork Mechanical Response Spectrum |
2018-11-27T18:13:16.360 | <p>Is there any way a portal frame with pin joints at the top of the columns can be stable? Are portal frames ever done with pin joints at the bases? (Aside from those with raft foundations/combined footings)?</p>
| |structural-engineering|structural-analysis| | <p>Yes a portal frame with at least one fixed (moment bearing) joint on top can be stable.</p>
<p>Portal frames with pin joints at the base could theoretically bear moment, provided they have fixed joints on top of the columns, but they are not advisable.</p>
<p>In fixed base and fixed top, a frame shares the moment at all joints and vibrates under lateral loads more in uniformity with the rest of the building.
Also it lends itself to smaller cross sections, hence more ductile behavior and less stiffness. Too much stiffness in a portal frame invites all the shear concentrated over its length and is detrimental to the overall lateral performance of the building. </p>
| 24886 | Portal frame stability |
2018-11-28T10:45:52.623 | <p>I'm trying to design a measurement system, the primary objective of which is to reliably detect a 10% change in the the volume of fluid flowing through a pipe.</p>
<p>I'm struggling to come up with a required accuracy/precision for the system based on something on something other than "This is simply as accurate as I could make it".</p>
<p>Note I am only really interested in relative change, i'm not sure how this effects the system requirements in terms of required precision/accuracy. For example, there two primary sources of error: </p>
<ul>
<li>Measurement error (accuracy) </li>
<li>Estimated internal diameter of the pipe (precision)</li>
</ul>
<p>I'm assuming I can completely negate the error introduced from the estimated internal diameter of the pipe and place the requirements solely on the "measurement error"</p>
<p>Is there a statistical technique to determine the accuracy required to reliably detect X change in some variable?</p>
| |measurements|statistics| | <h1>Direct Measurement of Volumetric Flow</h1>
<p>Let's handle the (easy) statement of accuracy first. Validating the accuracy of the measurement of volume flow requires that you calibrate your chosen measuring device with a second approach. That second approach must be at least as precise if not better than when you put the device in to practice. A best method will use a standards test, for example as defined by NIST or other agencies, as the calibration. Alternatively, when you intend to buy a device to measure volume flow and just use it directly without in-house calibration, you should insist on getting a calibration test report.</p>
<p>Now, let's handle the issue of sensitivity. You want to measure a change in volumetric flow <span class="math-container">$\Delta \dot{V}$</span> within a standard of <span class="math-container">$\Delta \dot{V} / \dot{V} \geq 0.1$</span>. This is akin to asking to what you need in order to measure a signal <span class="math-container">$S$</span> over a noise level <span class="math-container">$N$</span>. A guideline to reliable measurement in this case is that <span class="math-container">$S/N \geq 3$</span> to state that you have a signal above the noise. For a hypothetic system with no inherent flow noise (the volumetric flow is perfectly stable), the "noise" is the precision of the device itself. At a minimum, you must therefore assure that <span class="math-container">$\Delta \dot{V} / \dot{V} \leq 0.1/3 = 0.033$</span>. You will want a device with a relative measurement precision of at least 3.3%, preferably (much) better. When you intend to buy a device, you should confirm that its measurement precision fits your needs.</p>
<h1>Indirect Measurement of Volumetric Flow</h1>
<p>Suppose you are measuring mass flow of an incompressible fluid and you want to correlate this with volume flow in a pipe. To first order, this is an application of linear propagation of uncertainties. The starting equation is</p>
<p><span class="math-container">$$ \dot{V} = \frac{\dot{m}}{\rho} $$</span></p>
<p>You know that you can measure <span class="math-container">$\dot{m}$</span> with a device that has a relative measurement precision <span class="math-container">$\delta \dot{m}$</span>. You know that your fluid has a density that is reported with a precision of <span class="math-container">$\delta \rho$</span>. To first order, the uncertainty propagation equation is as follows:</p>
<p><span class="math-container">$$\left(\frac{\Delta\dot{V}}{\dot{V}}\right)^2 = \left(\frac{\delta\dot{m}}{\dot{m}}\right)^2 + \left(\frac{\delta\rho}{\rho}\right)^2 $$</span></p>
<p>To <em>reliably</em> measure a 10% change in volumetric flow, you should have a total uncertainty in your measurement system <span class="math-container">$\frac{\Delta\dot{V}}{\dot{V}}$</span> that is <em>significantly</em> below 10%. This is a statement of the precision of your system.</p>
<p>As a second example, when you are measuring velocity to obtain volumetric flow, the equation for velocity and volumetric flow</p>
<p><span class="math-container">$$\dot{V} = v \pi d^2 / 4$$</span></p>
<p>gives this</p>
<p><span class="math-container">$$\left(\frac{\Delta\dot{V}}{\dot{V}}\right)^2 = \left(\frac{\delta v}{v}\right)^2 + 4\left(\frac{\delta d}{d}\right)^2 $$</span></p>
<p>The precision of your measurement depends on the relative precision of the velocity flow device and the relative uncertainty of the tube diameter.</p>
<p>As with direct measurements, the accuracy of your system depends on how well it is calibrated. Calibration might in this case be done by using a separate device that measures volumetric flow and making a plot of volumetric flow versus mass flow. A perfect linear correlation is expected to validate the accuracy of your correlation equation.</p>
| 24898 | What measurement accuracy/precision is required to detect a 10% relative change in the measured quantity |
2018-11-28T15:52:15.353 | <p>I want to measure the vibrations of a solid object (a tuning fork, a bone conducting headphone, the chest during singing) and to isolate them from the vibrations of the air. So I am looking for a meter that will pick those vibrations only when it's in contact when the vibrating object (hence a meter that just reads frequency probably won't work for me)</p>
<p>I've tried to use a smartphone's accelerometer but it didn't pick anything. I've also to use a phone's sound meter in order to see whether it's picks more sound when in contact with such object but the reading was the same.</p>
| |vibration| | <p>You might want to look at contact pickups designed for acoustic musical instruments. </p>
<p>Try searching for contact microphones of contact transducers. These tend to be based on piezo electric devices and sense vibrations through direct contact as opposed to acoustic microphones whcih generally use some sort of diaphragm. </p>
<p>Edit </p>
<p>The instrument pickups area available as a complete package with an analogue audio output, usually via a 1/4" jack socket. Most PCs have an audio input although you should check that the impedance of the pickup matches that of the PCs input. You can also get various <a href="https://en.wikipedia.org/wiki/DI_unit" rel="nofollow noreferrer">balancing devices</a> to facilitate this although it may not be essential. </p>
<p>As mentioned in the comments instrument pickups are designed for an audible frequency range which 200Hz is well within albeit at the low end but individual device specs should tell you their response range. </p>
<p>You can also get the sensors individually but you will at least need to fit an output connector and possibly also a pre-amp or passive filter. </p>
| 24904 | How can I measure the vibrations of a solid object? |
2018-11-28T17:32:37.327 | <p>This stitch is composed of a single string/thread and is common on dogfood and grain sacks. It can be removed by just pulling on the end. Here is an example I found on <a href="https://www.youtube.com/watch?v=Q-NHxRwds4g" rel="nofollow noreferrer">youtube</a>.</p>
<p>Thanks!</p>
| |manufacturing-engineering| | <p>It is called "chain stitch" and was the stitch used in the first designs of sewing machines, because it is easy to produce mechanically. The fact that it can be pulled out unless the end of the string is knotted (as explained in the OP's video link) is a disadvantage for most purposes. For example, if the string breaks one side of the break will often quickly unravel.</p>
<p>The standard sewing machine stitch is now "lock stitch" which uses two threads, not one, and can't be pulled undone in the same way. Chain stitch is only used for special applications, or as decoration.</p>
<p>See <a href="https://home.howstuffworks.com/sewing-machine1.htm" rel="noreferrer">https://home.howstuffworks.com/sewing-machine1.htm</a> for an animation of how both stitches are made by machine. </p>
| 24907 | What is the name for a removable sewn bag stitch? |
2018-11-30T21:46:55.663 | <p>I need help figuring the distribution of a horizontal shearing force in an elastic solid trapezoid. When force <span class="math-container">$F_1$</span> is applied to all the top surface of the trapezoid, at equilibrium, is the force in all horizontal layers of the solid the same? That is <span class="math-container">$F_1=F_2=F_3=F_4$</span>? Or is the shear stress the same and the force varies between layers?</p>
<p><a href="https://i.stack.imgur.com/IkUNH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IkUNH.png" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>Consider the main trapezoid, a cantilever column( assuming it is thin enough to be subject of plane shear, not a solid) with load F applied to the top.</p>
<p>We know that <span class="math-container">$ \ \Sigma F_x =0 $</span></p>
<p>This means if we consider the free body diagram of the column at any horizontal level including the base and the level shown in your diagram, shear, V is constant and is always:</p>
<p>V = -F1. </p>
<p>However the shear flow, q is not the same and shear stress increases as the length of the base of trapezoid in the free body diagram decrease. e.g. if the length of the base is half of the length on top the shear flow q is two times the shear flow of the top. </p>
| 24936 | shearing force distribution in a trapezoid |
2018-12-02T08:35:52.440 | <p>All I know about welding is it is a technique in which electrode is connected to one terminal and the gripper connected to workpiece is in connection with the other terminal. As current passes and when the electrode is touched to the workpiece,an arc is generated which melts electrode and helps in the formation of a joint. As the circuit is completed with a conductor in-between,doesn't it result in short circuit?</p>
| |welding|current|shock| | <p>No, this is not classed as a "short-circuit". This is a circuit with a designed current flow - ie a 60 to 150A range is common...</p>
<p>A "short-circuit" is a circuit where the current flow is not following the designed current path ie it is taking a "shorter" route to ground.</p>
| 24952 | Short circuit in welding |
2018-12-02T21:07:56.203 | <h1>Problem</h1>
<p>I have two USRP antennas close to each other. I know that signal strength on the receiving end has to be at least -80 dBm. Both antennas have 2 dBi of gain. I want to compute what is maximum propagation loss so that this system still works. Signal is being transmitted with power of 10 mW.</p>
<h1>Attempt to solve</h1>
<p>Since i want to know the maximum amount of power loss during transmission to one antenna to another. I can form equation as. <span class="math-container">$p_0=$</span> power of transmitted signal, <span class="math-container">$p_{m}=$</span> minimum acceptable signal strength, <span class="math-container">$x=$</span> maximum amount of signal loss during transmission.
<span class="math-container">$$ p_0 \cdot x = p_{\text{ml}} $$</span>
<span class="math-container">$$ x = \frac{p_0}{p_{ml}} $$</span></p>
<p>My solution is ratio of powers so I can use dB units, by converting them with base-10 logarithm.</p>
<p><span class="math-container">$$ x_{dB}=10\cdot \log_{10}\left(\frac{p_0}{p_{ml}}\right)$$</span></p>
<p>i need to change -80 dBm to millliWatts with <span class="math-container">$-80\text{ dBm}=10^{-80/10}\text{ mW}$</span>. After this I can sum antenna gain to my answer since we are using logarithms.</p>
<p><span class="math-container">$$ x_{dB}=10\log_{10}\left(\frac{10^{-80/10}\text{ mW}}{10\text{ mW}}\right)+2\text{ dBi} +2\text{ dBi}$$</span></p>
<p><span class="math-container">$$ x_{dB}=-86\text{ dB} $$</span></p>
<p>However my answer seems to be incorrect ?</p>
| |telecommunication|radio| | <p>There is simple error on second line: it should be like this
<span class="math-container">$$ p_0\cdot x = p_{ml} $$</span>
<span class="math-container">$$ x = \frac{p_{ml}}{p_0} $$</span>
the fraction was wrong.
Then we have:
<span class="math-container">$$ x_{dB}= 10 \log_{10}(\frac{10\text{mW}}{10^{-80/10}\text{mW}})+2\text{dBi}+2 \text{dBi}= 94 \text{dB} $$</span></p>
<p>This answer seems to be correct since it corresponds to physical measurements. </p>
| 24962 | Propagation loss in USRP antenna |
2018-12-03T03:59:38.890 | <p><img src="https://i.stack.imgur.com/cHZIS.jpg" alt=""></p>
<p>I know the moment of inertia about X and Y axes. The area has been rotated through an angle theta. How to calculate the moment of inertia about x1 and y1 axes?</p>
| |structural-engineering|mechanical| | <p>Say your Iy is the max and Ix is minimum I before rotation of the axis.</p>
<p>We build the Mohr circle plotting Ix and Iy on its x axis. And we draw a vertical along Y axis and call it xy axis, somewhere on the left outside of the circle.</p>
<p>So the center of circle is at <span class="math-container">$ (Ix+ Iy)/2 \ $</span> and its radius is <span class="math-container">$ R = (I_y - I_x)/2 \ $</span> </p>
<p>now say you want to calculate the Ix' at 30 degrees anticlockwise rotation wrt x axis. So you cut an angle 60 degrees, twice the 30, and mark the coordinates of where it intersects the circle. </p>
<p>The Y is the Ixy max of the new rotated axis and if you draw a line from this point passing through the center of the Mohr circle it intersect the circle on the other side at Ixy min. If you want you could also wright these as equations.</p>
| 24968 | Moment of inertia of rotated area |
2018-12-05T03:26:58.340 | <p>If I have a single Boolean value that I wish to transmit wirelessly over a relatively short distance (assume <2m) then what of the many possible options would be the most <em>practical</em>.</p>
<p>To define what I mean my <em>practical</em>, it is some combination of:</p>
<ul>
<li>low cost</li>
<li>low power</li>
<li>environmentally robust</li>
<li>high data integrity</li>
<li>high reliability</li>
<li>low bulk/weight</li>
</ul>
<p>All of which apply equally to both the transmitter and receiver. I assume there will be some tradeoff between these </p>
<p>Other concerns that might impact the answers, but are of less interest to me:</p>
<ul>
<li>The transmission may be repeated, but not at a high frequency (assume <2Hz), although there is no penalty for higher frequency capability.</li>
<li>The transmission need only be unidirectional, although bi-directionality might be a bonus.</li>
<li>The transmission does not have any security concerns, although some level of security might be a bonus.</li>
<li>There is only one transmitter/receiver pair in the local area, although the ability to operate with multiple unique transmissions in the same area might be a bonus.</li>
<li>The transmission is one-to-one between transmitter and receiver, although the ability to transmit to many receivers, or from many transmitters, might be a bonus.</li>
<li>The solution should be off-the-shelf to a certain extent, although if bespoke (or high volume) options significantly improve the solution in some way then I would be interested to hear about them.</li>
<li>The solution should ideally be legal/unlicensed in most countries, but I'd be interested in what illegal or licensed options could be possible too.</li>
</ul>
<p>I don't have a specific application in mind, I'm more interested as a theoretical exercise.</p>
<p>Note that there is a related question <a href="https://engineering.stackexchange.com/questions/23209/best-way-to-communicate-data-wirelessly-with-specific-conditions">here</a>, but currently unanswered and with slightly different criteria.</p>
| |wireless-communication| | <p>Few of the technologies that fit well are Bluetooth Low Energy and Thread. Both of them operate in the 2.4 GHZ ISM band. Also you might want take a look at Zigbee. These all offer basic level of security features. </p>
<p>Lora and Sigfox are two other technologies that might work well. I have little knowledge on these technologies </p>
<p>As the other post indicate IR also can be considered. But I think you will have to create the security layer. </p>
| 24992 | Most practical way to wirelessly transmit a single bit of information? |
2018-12-05T04:13:08.507 | <p>you might find my question stupid. I am unfortunately not an engineer and didn't find the answer on google.
I want to calculate the <strong>braking energy</strong> of a vehicle <strong>decelerating from v2 to v1</strong> (km/h). Given are the <strong>Wheel inertia mass moment</strong> "I" and the <strong>wheel diameter</strong> "d" as well as the <strong>vehicle mass</strong> "M".</p>
<p>Could you give me the equation needed for that? </p>
<p>Thank you ! </p>
| |mechanical-engineering|automotive-engineering|energy|regenerative-braking| | <p>there are two kinetic types of energy that are involved in a moving car: </p>
<p>Linear and rotational kinetic energy, <span class="math-container">$$ K_{e \ linear} = \frac {1}{2}mv^2 \ ; K_{e \ rotatinal} = \frac {1}{2}I \omega^2 $$</span></p>
<p>The deceleration from V1 to V2 will reduce the energy. This energy is being saved in hybrid and electrical cars as regenerative energy.</p>
<p>The linear energy change is equal to: </p>
<p><span class="math-container">$ K_{e \ linear \ chang} = \frac {1}{2}mv_{1}^2 - \frac {1}{2}mv_{2}^2 $</span></p>
<p>And rotational energy change is equal to: </p>
<p><span class="math-container">$ K_{e \ rotatinal \ change} = \frac {1}{2}I_{1} \omega^2 - \frac {1}{2}I_{2} \omega^2 $</span></p>
<p>And <span class="math-container">$ \ V = r \omega = d \omega/2 \ and\ \omega= V/r $</span> with r being the radius of the wheel, r = d/2.</p>
<p>However, because of small "I" of the wheels compared to the mass of the car, the contribution of the wheel's change of energy is not significant. </p>
| 24993 | Calculating the Braking Energy of a vehicle |
2018-12-06T13:10:10.290 | <p>I have been trying to understand a particular aspect of the Chernobyl accident - the role of the graphite displacers on the end of the control rods. The basic story is that the graphite displacers caused the reactivity of the core to increase when the rods moved down, a non-intuitive effect that the operators were not expecting/sufficiently aware of.</p>
<p>All literature I can find tells basically the same story, but I shall refer here to <a href="http://www.rri.kyoto-u.ac.jp/NSRG/reports/kr79/kr79pdf/Malko1.pdf" rel="nofollow noreferrer">this PDF</a>.</p>
<p>On page 17 of that PDF it describes the graphite displacers and shows a schematic of a control rod in the raised position and in the "initial insertion" position.</p>
<p>The text indicates that:</p>
<blockquote>
<p>On moving down of absorbers into the core, their displacers displace water columns from the lower part of the core. Thus, inserting of absorbers from their extreme top position introduces a positive reactivity into the core because graphite absorbs neutrons much less than water</p>
</blockquote>
<p>The graphic shows a control rod "extreme top" and "initial insertion" positions. In the second case the graphite has moved in the channel, displacing water there and increasing reactivity in the lower part of the core. The image has a "+" symbol showing this.</p>
<p>Also, there is reduced reactivity above the displacer as a result of water replacing the graphite in that position, and reduced reactivity at the very top of the core due to the presence of the boron control rod. There are "-" symbols showing these effects.</p>
<p><strong>My question is, if the graphite displacer was already in the water column, how could moving it downwards change the overall reactivity of the core?</strong></p>
<p>Obviously the reactivity distribution changes, but it seems to me looking at the diagram that the (+) and (-) from the graphite movement would cancel out overall, leaving the (-) from the boron rod: an overall reduction in reactivity.</p>
<p>Clearly I'm misunderstanding something.</p>
| |nuclear-engineering| | <p>I was looking at the same thing. I think this answers it better: <a href="http://accidont.ru/ENG/rodes.html" rel="nofollow noreferrer">http://accidont.ru/ENG/rodes.html</a></p>
<p>Looking at the diagram from the accepted answer:
<a href="https://i.stack.imgur.com/Ai7Yp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ai7Yp.jpg" alt="enter image description here"></a>
I think that they did not pull the control roads all the way up as in position II as that would make the system slower in getting more power.</p>
<p>given the following:</p>
<ul>
<li>the up and down parts of the fuel rods are made of graphite (marked with purple in the diagram)</li>
<li>there is a space between of 1.25m from the bottom of the graphite part of the control rod and the bottom of the graphite part of the fuel rod</li>
<li>the water in that space acts as an inhibitor slowing down the reaction in the bottom part of the core and absorbing neutrons.</li>
</ul>
<p>when:</p>
<ul>
<li>the control rod is lowered to the point where the previous gap that was filled with water is now displaced by graphite </li>
</ul>
<p>then:</p>
<ul>
<li>you have a zone of only graphite at the bottom of the core, a hotspot of graphite</li>
<li>graphite from the control rod and from the core itself will accelerate the fission</li>
<li>as you continue lowering the control rod, the bottom of the core you will continue to accelerate the fission at the bottom of the core as there will always be graphite at the bottom until the control rods get in position IV</li>
<li>it takes 18 seconds for the control rods to get in the IV position</li>
</ul>
<p>So i think that is why there is a power surge at the bottom of the core, but as i have no taring on this subject, and this only a theory i could be wrong.</p>
| 25006 | How did the RBMK control rod design cause an increase in reactivity when moved downwards? |
2018-12-06T13:48:22.280 | <p>I have a very sensitive scale used for dynamic weighing (by dynamic I mean that the scale and load cells are actually beneath a belt conveyor that actively weight product when it´s being transported along the belt).</p>
<p>There´s a lot of variation in the measurements (+- 8g, while the manufacturer says it should be max +-2g).</p>
<p>I want to rule out vibration as a source of enviromental influence, so I gathered some data with a pair of 3 axis accelerometer dataloggers mounted on the structure beneath the load cells.</p>
<p>I analyzed the data in Matlab and found that the maximum vibration frequency when the equipment is being static is 1.066 Hz (EDIT 3); peaks of 1.33 g in the z axis (perpendicular to the floor).</p>
<p>It´s being a while since I´ve done anything with vibrations and I´m a little rusty on the theory and practical experiments.</p>
<p>I´m not sure what I should be looking for and I don´t have (didn´t find) anything about what the normal structural vibrations are for a lab room or specifications from the scale manufacturer on how much vibrations in can withstand or filter out.</p>
<p>I´d appreciate any suggestions.</p>
<p>EDIT: </p>
<p>The Scale is installed on top of a mezanine 5m tall.
The Accelerometer specifications are:
Sensor Type MEMS semiconductor
Acceleration Sampling Rate (Datalogger) 200Hz
Acceleration Range ±18g
Acceleration Resolution 0.00625g
Acceleration Accuracy ±0.5g
Bandwidth 0 to 60Hz
Sampling rate (Software) 500 ms to 24 hours
Memory 4Mbit Flash; 112028 Motion Detection samples per axis,
or 168042 Normal samples
Data Format Time stamped peak acceleration, average and peak vector sum
Dimensions 3.7 x 1.1 x 0.8" (95 x 28 x 21mm)
Weight 1oz (20g)</p>
<p>The FFT results in the X,Y and Z axis were:</p>
<p><a href="https://i.stack.imgur.com/Kp4AB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kp4AB.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/5tLZQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5tLZQ.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/BTEV5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BTEV5.png" alt="enter image description here"></a></p>
<p>EDIT 3: Here´s both time and frequency plots:
<a href="https://i.stack.imgur.com/rOgCq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rOgCq.png" alt="enter image description here"></a></p>
| |structural-analysis|vibration|matlab|industrial-engineering|modal-analysis| | <p>So, look at the time history that you added in edit 3. Notice how it is never ever ever below 1. That's very telling. Let us assume that you have a MEMS accelerometer is able to read down to 0 Hertz. So in the vertical direction, we expect to see 1g due to gravity PLUS the vibration. In other words, we expect to see <span class="math-container">$y(t) = 1 + A cos(\omega t)$</span>, where A is the amplitude of vibration and <span class="math-container">$\omega$</span> is the frequency. Now A may be a function of time, and there might be multiple frequencies, but this is the general idea. So, we expect to see something that is symmetric about 1. Sometimes greater than 1 and sometimes less than 1. E.g. if A = 0.3g, then you'd expect to see between 0.7 and 1.3g. Given that signal, we could do a fourier transform or other analysis to determine the values of A and <span class="math-container">$\omega$</span>.</p>
<p>But that is NOT what we see here. The value is always positive. So, based on this, and the fact that this is billed as <em>datalogger</em>, and not just a raw accelerometer, I very strongly suspect that what you are getting is not actually the variable y(t). Instead, I think it is an already processed value. i.e. the datalogger is probably sampling a block of time, perhaps 1 second, perhaps 100 ms, could be anything, and then doing some kind of processing, perhaps RMS over that time period. i.e. you aren't getting y(t), but instead something more akin to <span class="math-container">$\sqrt{\frac{1}{N} \sum y_i ^2} $</span>. Because this is an already processed value, taking a fourier transform of it does not give any meaningful result. </p>
<p>If you work out the math, you'll see that if we take that <span class="math-container">$1 + A cos(\omega t)$</span>, the RMS value will never be less than 1, which is consistent with the data you posted. </p>
<p>Put another way, when you datalogger spits out a number like "1.3g" at a given time, it is not telling you that the instaneous value of acceleration is 1.3g. It is saying that the average value, over a range of frequencies, and over a certain time range, is 1.3g. There may be various settings to your datalogger, and it might be doing a different type of processing, perhaps it's a peak over a frequency range, or something else. But I don't think you are getting raw acceleration. </p>
| 25007 | Analizing Vibrations Influence on Dynamic Weighing |
2018-12-07T09:50:14.303 | <p>In a grid-connected (utility-interactive) battery-based PV system, what is the purpose of the 'Isolated Subpanel' terminal of the inverter?</p>
<p><strong>EDIT</strong>: Found this descriptive image and hence answered the question myelf.</p>
<p><a href="https://i.stack.imgur.com/3PDbQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3PDbQ.jpg" alt="enter image description here"></a></p>
| |electrical-engineering|solar-energy|photovoltaics| | <p>I found out the answer. The isolated subpanel serves a load-center or distribution panel for backup loads (incase of power outage, etc).</p>
| 25020 | What is the purpose of the isolated subpanel terminal of a solar inverter? |
2018-12-08T05:16:49.007 | <p>I am interested in discovering a method to relate air flow to temperature reduction efficiency.</p>
<p>The purpose of this is in application to atmospheric water generation.</p>
<p>The apparatus will involve a fan (with a CFM rating), a long coil that will be externally cooled, and the temperature it is cooled to with perhaps another post processing stage. That would involve the narrow diameter pipe opening into a larger diameter "dew point chamber" or a way to cool the air further by narrow to wider expansion.</p>
<p>All of the condensed water will be collected into a collection reservoir.</p>
<p>So, the question is: How can I relate the CFM (cubic feet per minute) of air flow, the diameter of the pipe, the length of the pipe, the temperature of the pipe, the diameter of the expansion chamber --> the amount of water collected?</p>
<p>I am not an engineer. I am a simple person with aspirations to design a viable atmospheric water generator. I do not want to discuss energy requirements. I only want to discuss the variables I have outlined.</p>
<p>EDIT: I understand the idea of relative humidity vs. water available to "mine" from the air via condensation. My question pertains to the apparatus itself and how the variables relate. My concern is a high air flow will only have a certain rate of cooling based upon the length, diameter, and temperature of the pipe. My goal is to develop, with someone's kind help, a sort of formula for determining the temperature of the starting air (ambient) and the output temperature based on the variables.</p>
<p>Thank you, my fellow smart humans!</p>
| |airflow|cooling| | <p>This is a straightforward calculation for an engineer trained in the art of what is called HVAC (heating/ventilation/air conditioning) engineering but a full outlining of the solution process would require you to grasp many new physics concepts (most of which are not simple), an unfamiliar dictionary of engineering jargon, and the mastery of coupled algebraic equations. As such it is beyond the scope of what can be dealt with here. I would recommend that you partner up with an HVAC engineer to help you with this project.</p>
| 25026 | Variables Relating to Atmospheric Water Generation |
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