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2018-12-08T17:18:13.297 | <p>Recently I've been scratching my head over an assignment, and I have trouble understanding a couple of the parts.
<a href="https://i.stack.imgur.com/fHTpU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fHTpU.png" alt="Sketch"></a></p>
<p>The issue I have is with the overall calculations, since we've just started on mechanical issues and I have no experience except easy transmissions. Therefor, I am wondering how I calculate the ratio of the compound gear between A and B, and how many turns the worm gear have to turn to make the C spur gear travel a specified distance of 100mm. </p>
<p>The question goes as follows;</p>
<blockquote>
<p>How many revolutions does the worm gear have to complete, for the
spring to compress entirely?</p>
</blockquote>
<p>The C gear is connected to a rack that then compresses a spring, that'll have to travel 100mm. </p>
<p>Thanks!</p>
| |mechanical-engineering|transmission| | <p>In order for a tooth on C to travel 100mm, the teeth on B also need to travel 100mm. Let's not worry about how many revolutions that is. </p>
<p>A has 2.6 times as many teeth as B,and travels the same number of rotations, so, that means that a tooth on A must travel 260mm.</p>
<p>The pitch of the worm is 4mm, so, it needs to rotate 260/4=65 times in order to cause the teeth if A to move 260mm, and consequently the teeth of the rack meshed with C 100mm</p>
| 25031 | Calculations of compound gears? |
2018-12-08T23:29:06.387 | <p>In an experiment used to find the Young's modulus of a steel bar, the formula for change in height is given as follows:
<span class="math-container">$$\delta_x=\frac{Fx^2}{6EI}(3L-x)$$</span>
<span class="math-container">$$I=\frac{bh^3}{12}$$</span>
where:
<span class="math-container">$δ$</span>- the change in height of the steel bar under load at a distance L</p>
<p><span class="math-container">$δ_x$</span>- the change in height of the steel bar under load at the point where deflection is measured</p>
<p><span class="math-container">$L$</span>- the distance between the point of application of the load and the end support</p>
<p><span class="math-container">$x$</span>- the distance between the point at which deflection is measured and the end support</p>
<p><span class="math-container">$E$</span>- Young’s modulus</p>
<p><span class="math-container">$F$</span>- applied load</p>
<p><span class="math-container">$I$</span>- second moment of area of the cross- section</p>
<p><span class="math-container">$b$</span>- the width of the beam</p>
<p><span class="math-container">$h$</span>- the height of the beam</p>
<p><span class="math-container">$w$</span>- the weight of each value in the weighted mean</p>
<p><span class="math-container">$m$</span>- the mass hung from the bar</p>
<p>The problem is outlined below:
<a href="https://i.stack.imgur.com/LsnEF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LsnEF.png" alt="formula background"></a></p>
<p>The problem is that I do not understand how they have derived this top formula. Thanks in advance.</p>
| |materials| | <p>I try to tackel the problem in two different ways, as suggested in the comments. </p>
<p>The first approach is based on Bernoulli beam theory, a set of differential equations, (if you want to know how drive them, then you can ask it here or in physics SE):</p>
<p><span class="math-container">$$q(x)=-\frac{dv}{dx}=\frac{d^2M}{dx^2}=-EI_{yy}\frac{d^3\alpha}{dx^3}=EI_{yy}\frac{d^4u}{dx^4}$$</span></p>
<p>I respect the <strong>right handed coordinate system</strong>, the <span class="math-container">$x$</span> axis lies along the beam, <span class="math-container">$q$</span>, <span class="math-container">$v$</span>, <span class="math-container">$M$</span>, <span class="math-container">$\alpha$</span> and <span class="math-container">$u$</span> represent the vertical force, shear force, moment, curvature and displacement respectively.</p>
<p>I assume you know how to drive the moment, if you don't, then take a look at the second method: </p>
<p><span class="math-container">$$M=F(L-x)$$</span></p>
<p>Integrating the equation twice results in: </p>
<p><span class="math-container">$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})-C_1x+C_2$$</span>. </p>
<p>The boundary conditions in this case are: </p>
<p><span class="math-container">$$EI_{yy}\alpha(x=0)=0 \rightarrow C_1=0$$</span>
<span class="math-container">$$EI_{yy}u(x=0)=0 \rightarrow C_2=0$$</span></p>
<p>So:</p>
<p><span class="math-container">$$EI_{yy}u(x)=-F(L\frac{x^2}{2}-\frac{x^3}{6})$$</span>.</p>
<p>The second way is a bit more lengthy but it helps you to develop your insight:</p>
<p><a href="https://i.stack.imgur.com/gMiWZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gMiWZ.png" alt="enter image description here"></a></p>
<p>Image: Mechanics of materials Prof. Wim VAN PAEPEGEM</p>
<p>First we find the reaction forces: </p>
<p><span class="math-container">$$R_A+F=0 \rightarrow R_A=F$$</span></p>
<p><span class="math-container">$$RM_A=-F.L$$</span></p>
<p>Now we wish to find the shear forces and the moment, you can choose between the right hand side of the beam or the left hand side, as you can see in the photo, i find it easier to write the equilibrium equations for the right hand side (BC).</p>
<p><span class="math-container">$$-V+F=0 \rightarrow V=-F $$</span></p>
<p><span class="math-container">$$M+F(L-x)=0 \rightarrow M=F(L-x)$$</span></p>
<p>you can now integrate the moment as i did before. </p>
| 25034 | derivation of Young's modulus experiment formula |
2018-12-09T18:56:47.760 | <p>Referring to this diagram:</p>
<p><a href="https://i.stack.imgur.com/CZju9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CZju9.png" alt="Peltier Element"></a></p>
<p>How does the height of the P and N components affect heat or cold generated on this thermoelectric element?</p>
<p>The idea I keep having is that it seems like the hot and cold sides are too close to each other. Would increasing the distance, via lengthening the height of the P and N components help to segregate the differential more or would this decrease the differential produced?</p>
| |heat-transfer|cooling| | <p>Here is the problem: the electrical current has to flow through the chunks of junction material. That material has electrical resistance and will therefore heat up as the current flows, negating the cooling effect. So you do not want the P & N chunks to be too thick.</p>
<p>On the other hand, if you make then too thin in response, then it's too easy for heat to flow across the chunks from the hot side to the cold side, short-circuiting the cooling effect. </p>
<p>This means two things: 1) the thickness of the chunks is a compromise between these two competing influences, and 2) the holy grail of this business is a material that simultaneously possesses high electrical conductivity and low thermal conductivity, which are to a great extent contradictory properties. </p>
| 25040 | Peltier Element - Segregation of the Hot/Cold |
2018-12-10T03:36:19.297 | <p><strong>Answer briefly how the object-oriented methodology transforms an information system.</strong></p>
<hr>
<p>I really do not know how to answer that question. I know what is an information system ("System capable of collecting, storing, processing and distributing information in time and form for decision making within an organization") and also what is object-oriented methodology ("discipline related to programming") but I do not know how the latter models the first one.</p>
| |systems-engineering| | <p>Object-oriented methodology allows representing the system as a hierarchical structure of self-contained subsystems and entities comprising it. This allows separate development and testing of the subsystems apart from each other, good code reuse (through inheritance - similar objects; and through use of object libraries - use of pre-made objects by 3rd parties) and abstracting internals of subsystems apart from the systems using them, creating consistent interfaces between them, easy to extend, debug and modify.</p>
<p>For an easy analogy: You build the big system out of big bricks, each of which you build of smaller bricks, and these - of yet smaller, each with own quirks and special purposes, some being just special variants of others, others being pre-packaged ready-to-use bigger bricks. They all need to fit just right, but if you make a small brick of the tiny bricks and it fits just right with the bigger brick, you can just reuse it everywhere it's needed (and if it's faulty, you can pinpoint the fault and fix it easily.) If you started building the entire system out of the tiniest bricks from start, you'd end up with a mess where every segment, no matter how common, needs to be made from scratch all over again, and if something's broken, you don't follow the simple chain big brick; smaller brick within it, even smaller within, the bad tiny one - you'd need to track all the hundreds of tiny bricks from top to the faulty one through that mess.</p>
| 25046 | In what way the object-oriented methodology model an information system? |
2018-12-10T05:20:43.253 | <p>This is from Stephen Hawking's latest book; <em>Brief Answers to the Big Questions</em>. </p>
<blockquote>
<p>The speed at which we can send a rocket is governed by two things, the speed of the exhaust and the fraction of its mass that the rocket loses as it accelerates.</p>
</blockquote>
| |aerospace-engineering|rocketry| | <p>He refers to <a href="https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation" rel="noreferrer">Tsiolkovsky's Rocket Equation</a>:</p>
<p><span class="math-container">$$ \Delta v=v_e \ln {\frac {m_0}{m_f}} $$</span></p>
<p>where: </p>
<p><span class="math-container">$v_e$</span> is the exhaust velocity;</p>
<p><span class="math-container">${\frac {m_0}{m_f}}$</span> is the fraction of mass; <span class="math-container">$m_f$</span> - the "dry mass"/"final mass" (rocket without fuel) and <span class="math-container">$m_0$</span> - "wet mass"/"launch mass" (rocket fully fueled up.) </p>
<p>This equation is one of the most important in rocket science - describing the change of velocity a rocket can achieve. The implications are that the larger the difference between the mass of fuel and the mass of the craft, the larger the velocity achievable (but the <span class="math-container">$\ln$</span> results in diminishing returns as mass of fuel is increased) and that engines that impart the exhaust with most velocity provide most performance - linearly, without that pesky <span class="math-container">$\ln$</span> - but then... with square root of energy needed; <span class="math-container">$E={1\over 2} mv^2; v = \sqrt{2E \over m}$</span>. So, increasing power of the engine - chemical energy of fuel, amount of electrical energy imparted by ion drive - results in diminishing returns again.</p>
<p>One of Hawking's last ideas - "<a href="https://en.wikipedia.org/wiki/Breakthrough_Starshot" rel="noreferrer">Breakthrough Starshot</a>" nicely sidesteps both problems. The propellant is photons, moving at speed of light, and the craft doesn't carry any fuel - the propellant is beamed from a ground-based station through a powerful laser. </p>
| 25047 | the speed of the exhaust |
2018-12-10T05:37:21.947 | <p>Are solar panels placed horizontally at the equators since the solar altitude (solar window) changes direction from north to south?</p>
<p><a href="https://i.stack.imgur.com/Qq7AU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qq7AU.jpg" alt="enter image description here"></a></p>
| |solar-energy| | <p>Basically, yes, but any shading caused by buildings / other obstructions may make one direction preferential.</p>
<p>You can find sites that will do simulations online...</p>
| 25048 | How are solar panels at equator placed? |
2018-12-10T08:09:08.770 | <p>What is the solar altitude at Bangalore at solar noon during summer solstice, March equinox, Winter solstice and September equinox? In other words, What is the solar window (solar altitude angle range) of Bangalore?</p>
<p>PS: Consider latitude of Bangalore to be 11 deg, North.</p>
| |solar-energy| | <p>Earth axial tilt is 23.5°. For 11°N the solar altitude at noon of the equinoxes will be 90-11=79°; 90-11-23.5=55.5° at winter solstice and - it would be 90-11+23.5=102.5° at the summer solstice, but we normalize it to angle between nearest horizon and the Sun, so for values >90 we take 180-(value): 180-102.5=77.5° at the summer solstice. </p>
<p>The actual solar window though will be the range between minimum and maximum altitude: 55.5°-90° as the Sun passes through Zenith sometime in the late spring and early summer.</p>
| 25051 | What is the solar window (solar altitude angle range) of Bangalore? |
2018-12-11T18:23:06.213 | <p>Let <span class="math-container">$\omega_0 \in \mathbb{R}$</span> and <span class="math-container">$\omega_0 > 0$</span>. Be <span class="math-container">$G(s)$</span> a transfer function defined as:</p>
<p><span class="math-container">$$G(s)=\frac{1-\frac{s}{\omega_0}}{1+\frac{s}{\omega_0}}$$</span></p>
<p>We're interested into evaluating its phase response in a continuous LTI system. First, we split <span class="math-container">$G(j\omega)$</span> into its real and imaginary parts (assuming <span class="math-container">$\tau=\frac{\omega}{\omega_0}$</span>):</p>
<p><span class="math-container">$$G(j\omega) = \frac{1}{1+\tau^2}\left[1-\tau^2+j\left(-2\tau\right)\right]$$</span></p>
<p>Then using the definition of <em>phase</em> of a transfer function we have:</p>
<p><span class="math-container">$$\angle G(j\omega)=\arctan\left(\frac{\mathbb{Im}\{G(j\omega)\}}
{\mathbb{Re}\{G(j\omega)\}}\right)=\arctan\left(\frac{-2\tau}{1-\tau^2}\right)$$</span></p>
<p>So, for any <span class="math-container">$\tau \gg 1$</span> (which implies that <span class="math-container">$\omega \gg 1$</span>) then the following should hold for Taylor's series:</p>
<p><span class="math-container">$$\angle G(j\omega) \simeq \arctan\left(\frac{2}{\tau}\right) \simeq \frac{2}{\tau}$$</span>
which approaches <span class="math-container">$0$</span> as <span class="math-container">$\omega$</span> gets bigger.</p>
<p>Here comes the discrepancy. Let <span class="math-container">$\omega_0 = 1$</span> (for plotting reasons), why does <a href="https://www.wolframalpha.com/input/?i=bode%20(1-s)%2F(1%2Bs)" rel="nofollow noreferrer">Wolfram Alpha</a> give me the following Bode plot?</p>
<p><a href="https://i.stack.imgur.com/Mxxzu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mxxzu.png" alt="Wolfram Alpha Bode plot"></a></p>
<p>Also, I expected one discontinuity point at <span class="math-container">$\omega = \omega_0 = 1$</span> (and another one for <span class="math-container">$\omega = -\omega_0$</span> not shown in the plot), but I guess I messed up something since the beginning.</p>
<p><strong>2018-12-12</strong> (in response to Sam F.) Here's how I've separated the real from the imaginary part of <span class="math-container">$G(j\omega)$</span>.</p>
<p><span class="math-container">$$
\begin{align}
G(j\omega)
&=\frac{1-j\frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}} \\
&=\frac{1-j\tau}{1+j\tau} \\
&=\frac{1-j\tau}{1+j\tau}\cdot\frac{1-j\tau}{1-j\tau} \\
&=\frac{(1-j\tau)^2}{1^2-(j\tau)^2} \\
&=\frac{1^2+(j\tau)^2-2\cdot 1 \cdot j\tau }{1+\tau^2} \\
&=\frac{1- \tau^2+j(-2\tau)}{1+\tau^2} \\
\end{align}
$$</span></p>
| |transfer-function| | <p>The plots are correct.</p>
<p>This is how we do it on our heads in about a minute after years of practice.</p>
<p>The magnitude:</p>
<p>Consider the numerator and denominator separately. Numerator rise with a slope of <span class="math-container">$20$</span> <span class="math-container">$dB/dec$</span>, but at the same time denominator falls with the exact same slope, now it's clear that these two tend to cancel each other out, the results is a constant number. As you know the <span class="math-container">$20log1 = 0$</span> so it explains the magnitude plot. Notice if you look at the magnitude of the transfer function you'll see no real pole.</p>
<p>The phase </p>
<p><span class="math-container">$arctan(*)$</span> is a continue function, at the break frequency it is equal to <span class="math-container">$-\frac{\pi}{2}$</span>, as we plot the phase in logarithmic scale we have to add the phases. Both denominator and numerator drop <span class="math-container">$-\frac{\pi}{2}$</span> over the whole frequency band. So <span class="math-container">$-\pi$</span> or 180°. </p>
| 25070 | Phase response of a continuous transfer function |
2018-12-12T09:39:52.277 | <p>What is the degree of cooling caused by air flowing from smaller diameter into a large diameter? How much does air cool as it expands?</p>
<p><a href="https://i.stack.imgur.com/YhlXF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YhlXF.jpg" alt="enter image description here"></a></p>
<p>If the air inlet was the biggest hole on the right and the outlets were the full smaller holes, would the air cool as it left each hole? Let's say that the surface area of the inlet equals the outlets.</p>
<p>Could these types of apparatus be linked in series to produce a staged cooling effect? </p>
<p>Does cooling only occur in change of pressure?</p>
| |heat-transfer|airflow|cooling| | <p>The principal you are referring to is a fairly fundamental law. The Charles law. V/T = k. This assumes that there is no compression, and constant ambient temperature.</p>
<p>To answer your question without measurements is difficult, however it would appear from inspection that the unit surface area of the output is greater than the unit surface area of the input, so I would imagine temperature would drop proportionately. </p>
| 25075 | Air Cools As It Expands |
2018-12-12T20:35:34.760 | <p>Referring to the pasted in ductwork diagram, if an inline ventilation fan with a high CFM airflow was used to circulate air within a sealed ductwork system, would the fast moving airflow result in a significant pressure drop within the ductwork and cause a section of it to implode?</p>
<p>Or since it is a sealed system and air can neither leave nor enter it, would the result be that the fan would essentially just be circulating pressurized air around the ductwork and there would thus be no pressure drop within the ductwork? Referring to the ductwork diagram again, would static and dynamic pressure (in PSI) at points A, B, C, and D be approximately the same if measured first with the fan turned off and then measured again with the fan turned on?</p>
<p>Although this appears to be a homework question, it is not. I am contemplating running ductwork from my house to my garage and I want to make sure that what I construct will not implode.</p>
<p><a href="https://i.stack.imgur.com/1CJFG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1CJFG.png" alt="Ductwork diagram"></a></p>
| |fluid-mechanics|airflow|heating-systems| | <p>Closed or not, a fan will always raise the pressure in front of it, and lower the pressure behind it. In order for the ducting to implode you would need a relatively strong vacuum. What is needed would be based on the rigidity of the ducting. However, a couple points that are important to remember:</p>
<ul>
<li><p>Its a closed system, so mass is conserved. Regardless of where the air density is at any given point, the air is available.</p></li>
<li><p>While a fan will slightly raise the pressure on the outlet side, even a powerful fan, is not a compressor. It may move the air but it will not compress the air significantly, so the "make up air" will be readily available on the intake side.</p></li>
<li><p>Natural convection dictates that an increase pressure gradient will increase air flow. Even if there was compression, in a closed system, the rate of compression would be compensated by the rate of airflow.</p></li>
</ul>
| 25079 | Will circulating air within a sealed ductwork system cause it to implode? |
2018-12-14T20:11:40.530 | <p>I'm currently researching the strength of domes underwater and I went down the rabithole of material science. The basics of this research (or though experiment) is to bring a dome filled with air to the seafloor and see if it implodes or not. The specifics are as follows. </p>
<pre><code> The dome is 3D-printed
Outer Diameter: 10 cm
Inner Diameter: 8 cm (so the thickness of the dome is 1 cm) (this can vary)
Material: PLA
</code></pre>
<p><em>QUESTION: What is the maximum amount of pressure this dome can take before it breaks?</em></p>
<p>I've already found the following data from this <a href="https://downloads.makerbot.com/legal/MakerBot_R__PLA_and_ABS_Strength_Data.pdf" rel="nofollow noreferrer">source</a></p>
<pre><code>Properties PLA:
- Impact strength:
- STD: 1.8 ft-lb/in
- MAX: 4.1 ft-lb/in
- Compressive Strength
- STD: 2600 PSI
- MAX: 13600 PSI
- Tensile Strength
- STD: 6783 PSI
- MAX: 9531 PSI
- Flexural Strength
- STD: 8970 PSI
- MAX: 13731 PSI
</code></pre>
<p>It seems likely that I'll be needing the compressive strength here, because the theoretical dome is being put at the bottom of the theoretical sea. It's going to be compressed. But these numbers have very little context (at least for me, I don't have an engineering degree). I mainly worry about thickness. In my mind, when a wall is thicker, it means it can hold more weight, can withstand heavier punches and generally breaks less easily. I don't see this reflected in this number. It also doesn't take the shape of the object in mind. A sphere is supposedly stronger then any other shape (I found this <a href="https://www.monolithic.org/benefits/benefits-strength" rel="nofollow noreferrer">cool site</a> selling dome bunkers). But this too doesn't seem reflected in the compressive strength value.</p>
<p>It might also have to do with <a href="https://en.wikipedia.org/wiki/Size_effect_on_structural_strength" rel="nofollow noreferrer">the size effect</a>, but that seems to weaken - not strengthen - structures and would only be applicable if I made the outer diameter variable.</p>
<p>If anyone can help me any further it would be very much appreciated. I really want to make clear that I'm not asking you to do my homework (this isn't really homework, but still). Pointing me vaguely in the right direction is also extremely helpful. Thanks in advance!</p>
| |materials|pressure|strength| | <p>A shell structure under external pressure is likely to fail in buckling before it reaches the compressive strength of the material.</p>
<p>Buckling of curved shells is a complex problem in general, but for a complete sphere a simple formula (source <a href="http://eprints.keele.ac.uk/1563/1/Stability.pdf" rel="nofollow noreferrer">here</a>) is <span class="math-container">$$P = \frac{2E}{\sqrt{3(1-\nu^2)}}\frac{t^2}{r^2}$$</span> </p>
<p>Where <span class="math-container">$E$</span> and <span class="math-container">$\nu$</span> are Young's modulus and Poisson's ratio, <span class="math-container">$t$</span> is the shell thickness and <span class="math-container">$r$</span> is the mean radius. </p>
<p>Taking <span class="math-container">$E$</span> = 510,000 psi and <span class="math-container">$\nu$</span> = 0.33 (sources, <a href="https://www.makeitfrom.com/material-properties/Polylactic-Acid-PLA-Polylactide" rel="nofollow noreferrer">here</a> and <a href="https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20160010284.pdf" rel="nofollow noreferrer">here</a>) This gives about <span class="math-container">$31,000$</span> psi for your dimensions.</p>
<p>These buckling formulas are very imprecise and I would take a safety factor of at least 10 times, which would give a safe buckling load of about 3000 psi.</p>
<p>Since the maximum pressure using the "STD" compressive stress is about 1200 psi (using the formula in mg4w's answer), in this case buckling is not the limiting factor, but the buckling load is proportional to <span class="math-container">$t^2$</span> while the compressive load is proportional to <span class="math-container">$t$</span>, so for a thinner shell the buckling load decreases faster than the compressive load.</p>
<p>So taking the pressure gradient in water as about 0.5 psi per foot of depth, (there's no point using more than one significant figure here!) a sphere with your dimensions should be good for a depth of about 2,500 feet.</p>
<p>But a dome with a flat base would fail before that, since the base would break at the edges where it joined the sphere. Assuming the base is rigidly supported by the dome around its edges, the maximum stress (source <a href="http://www.roymech.co.uk/Useful_Tables/Mechanics/Plates.html" rel="nofollow noreferrer">here</a>) is <span class="math-container">$$\frac{3Pr^2}{4t^2}$$</span>. </p>
<p>Taking the maximum tensile strength as 6783 psi, that gives a pressure of about 450 psi at failure - much smaller than the other failure modes considered above, and limiting the depth to about 900 feet (with no safety factor applied!)</p>
| 25096 | Strength of Plastic Underwater Domes with various thicknesses |
2018-12-16T09:21:26.677 | <p>Please bear with me - I'm not an engineer, I'm a software developer so lack the language / knowledge to describe what I want precisely.</p>
<p>I'm controlling a servo to rotate /pan a platform to hold a camera (think pan/tilt). This is fine.</p>
<p>However, I'd like to house the servo/electronics in a waterproof box, and have a platform on top of the box that Pans.</p>
<p>How can I transfer the rotation through a waterproof enclosure? Is there such a thing as a waterproof bearing, that the servo can rotate one side within the enclosure, and have the rotation exposed on the outside, so I can attach the Panning platform to it?</p>
<p>Does such a thing exist? If so, what is it called?</p>
<p>If not, surely this is a problems that has already been solved
Can anyone direct me to a solution for this?</p>
<p>Thank you!</p>
| |bearings|servo|waterproofing| | <p>I needed a "shaft gland" / "gland seal" used in boats (and for my application, radio controlled boats" to solve the problem.</p>
| 25113 | Waterproof Pan control/bearing/mechanism |
2018-12-17T17:02:11.403 | <p>I am trying to construct a lift handle in the CAD Sketchup. The circled part of the image shows the arch I'm referring to.</p>
<p><img src="https://i.stack.imgur.com/sUgRv.jpg" alt=""></p>
<p>The program, Sketchup, has a 'tool' called push/pull, and another tool called rotate. I have a video uploaded on streamable demonstrating this, but I'm not sure if it's within the site's rule to do so. Here's a brief clip partially demonstrating it in action: <a href="https://i.imgur.com/pgVRw9d.mp4" rel="nofollow noreferrer">https://i.imgur.com/pgVRw9d.mp4</a> </p>
<p>I use the push/pull tool, to raise the face of the object by 250 mm, and the rotate tool, to rotate it by 15 degrees. <span class="math-container">$90/15 = 6$</span>. The arch ends up with an overall height of 1074 mm.</p>
<p>I repeat the task 2 more times, but with the angle of 7.5 and 30 degrees.
With 7.5 degrees, the arch looks much curvier, but the height is 2032 mm. With 30 degrees, the arc sides are noticeably pointy, but its height is more acceptable, 591 mm.</p>
<p>Here is an image of all 3 arcs together: </p>
<p><img src="https://i.stack.imgur.com/Vw9DT.jpg" alt=""></p>
<p>The dimension of each side on the inner part of the arc, are all 250 mm.</p>
<p>My question is, what equation may I use, so I end up with an arch that looks curvy enough, in this case 15 degrees or lower, and has a height of 508 mm from the 'ground' (or x-axis)?</p>
<p>I tried the following, with no success:</p>
<p>Circumference of a circle <span class="math-container">$C=2\pi r$</span></p>
<p>The arc in this case represents 1/4th of a circle; so <span class="math-container">$C_{arc}=2\pi r/4$</span></p>
<p><span class="math-container">$$\begin{align}
2\pi508/4 &= 797.96 \\
\dfrac{797.96}{90/15} &= 132.99
\end{align}$$</span></p>
<p>I push/pull the face of the object by 132.99, and rotate its top face by 15 degrees. I repeat this process 5 more times. The 6th arch rotated, is perfectly perpendicular to the x-axis. Unfortunately, the height is 571 mm, instead of 508 mm.</p>
| |cad|geometry| | <p>It looks like you need a = 118.1978.</p>
<p>I came up with this using the formulas on <a href="https://www.calculatorsoup.com/calculators/geometry-plane/polygon.php" rel="nofollow noreferrer">this site</a>. A section through your handle looks like part of a regular polygon with 12, 24, or 48 sides. </p>
<p>You can use the calculator on that site, but you have to make a minor adjustment because your first segment is perpendicular to the surface (so if you placed a mirror image below it you would have two adjacent sides with no angle in between). You can account for this by rotating their illustration by 1/2 the angle of each segment, and using r instead of R in their computation, and then adding half the side length to the answer. In other words, their calculated r + a/2 = height.</p>
<p>Using this approach, I perfectly reproduced all four of your results.</p>
<p>To calculate a, given the desired height is a little more complicated because of the modification I described above. In general, you can do it by solving two equations simultaneously: a = 2 r tan(pi/n) and r + a/2 = height.</p>
<p>Plugging in your desired height = 508, and n = 24, I get the answer a = 118.1978.</p>
<p>EDIT: Here's a sketch that explains why r + a/2 = 508 mm:
<a href="https://i.stack.imgur.com/p6nzN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p6nzN.png" alt="sketch"></a></p>
| 25128 | Building a half-arch of a specific height, while consistently using an angle divisible by 90 |
2018-12-17T17:11:48.860 | <p>Does notching or cutting out the corner of an angle brace distribute stress better along the flange of a member with shear lag or reduce the stressed edge of the flange's ability to resist deformation?</p>
<p>Previously, user CableStay <a href="https://engineering.stackexchange.com/a/4196/33">had illustrated Shear Lag</a> concept in a bolted angle iron member very well.</p>
<p><a href="https://i.stack.imgur.com/niLm3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/niLm3.png" alt="enter image description here"></a></p>
<p>My question then asks if it is possible to better distribute the stress of the uniform stress closer to the end of the member by notching or cutting the corner of it.</p>
<p><a href="https://i.stack.imgur.com/aVMdj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aVMdj.png" alt="enter image description here"></a> </p>
| |structural-engineering|shear| | <p>It makes sense to stream line the shape of connection at the end so as to eliminate the part that is not in the path of stress for two reasons:</p>
<p>A - In a situation of dynamic loading such as a truss in a bridge under impact of vibration of passing traffic the part which is not sharing the stress, does not strain with the rest of the structure either, and in the long run this sea-saw effect causes fatigue in the connection.</p>
<p>B - In a large structure with many complex joints removing excessive material leads to easier inspection and maintenance of the structure and lighter weight, think about power towers, bridges, stadium roofs, etc.</p>
| 25129 | Does notching the corner of an angle brace member help or hurt when shear lag is present? |
2018-12-17T18:28:02.557 | <p>I am trying to find out if the 12 V DC electric motors that I have are powerful enough to launch a 100 g tennis ball 30 m away. The tennis ball is put down a pipe and ends up wedged in between 2 wheels (radius = 0.05 m) driven by the two 12 V DC electric motors. The picture below should hopefully clarify the situation:</p>
<p><a href="https://i.stack.imgur.com/6aljX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6aljX.png" alt="enter image description here"></a></p>
<ul>
<li>The electric motors are spinning at ~7000 RPM (no load), the motor specs are here: <a href="http://www.banebots.com/product/M5-RS555-12.html" rel="nofollow noreferrer">http://www.banebots.com/product/M5-RS555-12.html</a></li>
<li>A wheel (with mass of 0.05 kg and radius of 50 mm) is connected to each motor</li>
<li>The tennis ball is dropped down the pipe and ends up in between the two rotating disks</li>
<li>The launching angle is 20 degrees</li>
</ul>
<p>I calculated how far the ball flies if I keep the RPM constant based on the horizontal component of the linear velocity in combination with the flying time and launching angle. However, this does not account for the mass of the tennis ball. </p>
<p>When the ball passes through the two wheels, the electric motors will see an opposite moment. I guess I can calculate the moment by multiplying the mass * g * radius. However, it is not clear to me how I can then accurately calculate the effect on the RPM of the motors. I can optimize the location of wheels so that the ball isn't squeezed too much, so I was hoping to ignore this in the calculation and build in some margin instead. In the end, I would like the ball to fly at least 30 m.</p>
<p>Does anybody have an idea on how I can calculate if the motors are strong enough to launch the ball 30 m away?</p>
| |mechanical-engineering|torque| | <p>I'm using another type of 12 V motor (here's the <a href="https://www.ebay.com/itm/DIY-DC-12V-24V-3000-6000-RPM-High-Speed-Adjustable-Marshmallows-Machine-Motor/183249689117?_trkparms=aid%3D111001%26algo%3DREC.SEED%26ao%3D1%26asc%3D20160908105057%26meid%3D596a41df3df24f59ba9119e416bcd7ca%26pid%3D100675%26rk%3D5%26rkt%3D15%26sd%3D302910615730%26itm%3D183249689117&_trksid=p2481888.c100675.m4236&_trkparms=pageci%3Aa57edf9d-28ce-11e9-bbe4-74dbd1809234%7Cparentrq%3Abab096751680aa16af9741a9ffd48612%7Ciid%3A1" rel="nofollow noreferrer">link</a> to it on eBay) that has 1.2 Kg of torque and 3000 rpm speed. I used two 5 inch diameter plastic wheels and tested how far the two motors can launch the ball. At about 45 degrees of angle, it can shoot the ball ~22-23 meters. I hope this info helps! </p>
| 25130 | Torque required to launch a tennis ball |
2018-12-17T23:01:15.587 | <p>For building something like a big building or a parking garage, there is a need for a large flat surface of concrete. Wondering how they make the dirt and the concrete flat. It's hard to make a tiny flat surface at the beach in sand let along thousands of square feet. Wondering what sorts of machines tools or techniques are used at a high level.</p>
| |surface-preparation| | <p>Easier done than thought. </p>
<p>"Grading" and "levelling" are carried out in stages of approximation. </p>
<p>A grading machine or a common dozer can work up an even surface (horizontal or uniform gradient). </p>
<p>As for the good question, "It's hard to make a tiny flat surface at the beach in sand let alone thousands of square feet. Wondering what sorts of machines tools or techniques are used at a high level.": Remember that the surface of a liquid is <strong>always</strong> horizontal. A layer of concrete usually evens out the minor variations (bumps and pits), giving you a horizontal "flat" surface. </p>
| 25138 | How land is made flat at scale |
2018-12-18T08:29:00.177 | <p>Anyone tried mounting thermal imaging camera on ordinary optical microscope?</p>
<p>I know it is unlikely to work. Because of different refraction index and possible absorption of IR radiation by the lens.</p>
| |thermal-radiation| | <p>The glass used in visible light microscopes does not pass much IR. For strong signals you need calcium fluorite lenses, which are expensive and rare. </p>
| 25148 | Mount thermal imaging camera on ordinary optical microscope? |
2018-12-18T23:34:52.410 | <p>I'm looking at googles mocktails maker. And it puts 2 holes in a container. 1 hole is connected to a pump that will take water out and the other connects to a backflow prevention valve that only allows air in and not liquid.</p>
<p>What is the point of having that valve? It doesn't connect to anything. Also when I test it with and without having that valve the water does the exact same thing.</p>
<p>Reference:
<a href="https://github.com/deeplocal/mocktailsmixer#step-19-attach-tubing-and-check-valve" rel="nofollow noreferrer">https://github.com/deeplocal/mocktailsmixer#step-19-attach-tubing-and-check-valve</a></p>
<p><a href="https://i.stack.imgur.com/reknM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/reknM.jpg" alt="enter image description here"></a></p>
<p>EDIT: Does it have to do with keeping the container pressurized? Why would I need to have a constant pressure if the pump doesn't require it.</p>
| |mechanical-engineering|pumps|airflow|valves| | <p>yes it is just to maintain the pressure in the container at the same level as atmosphere.</p>
<p>Many small pumps malfunction when the source reservoir has vacuum built up in it.</p>
<p>Even simple bottle water dispensers have check valves that lets air replace used water and break the vacuum.</p>
<p>Especially unable to handle the negative head are peristaltic pumps, which function by a rotary triangle of wheels squeezing the water in a tube forward, because of a random bubble of air trapped in the tube between the wheels. </p>
| 25160 | What is the point of this valve that only allows air in and not liquid? |
2018-12-19T00:21:14.080 | <p>I'm trying to make a quick release male and female connector similar to the one shown here.<a href="https://i.stack.imgur.com/JjbiS.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JjbiS.gif" alt="enter image description here"></a></p>
<p>One connector is attached to the bottom of a container filled with liquid. When it is not connected it is closed and none of the liquid can flow out of it. However when it is in contact with another connector it then opens and allows the liquid to flow out of it. Any ideas where these kind of connectors can be found or made?</p>
| |mechanical-engineering|valves|liquid|connections| | <p>There are many, many different commercially-available designs of fluid connectors that shut off automatically when disconnected and open up when connected. They are available in plastic and metal, permanent and disposable, in all sorts of sizes. Try a search on "make-and-break fluid connectors".</p>
| 25162 | Is there a male and female valve that is closed until connected together? |
2018-12-20T04:36:25.170 | <p>Suppose you were to fasten a string to a ping pong ball, then fasten the string to a stationary object, and then use a vacuum cleaner with a strong vacuum to suck the ball partially into the vacuum hose. Due to the strong vacuum's pull on the ball, the string should be very taut and the ball should also be centered inside the vacuum hose.</p>
<p>Based on this setup and referring to the two drawings below, if you were to then push down on the string with your finger at the point indicated by the arrow in Figure A, would the tautness of the string move the ball from a centered position in the hose down until it comes to rest on the bottom of the vacuum hose?</p>
<p>Or, would pushing down on the string at that point cause a bend in the string, as shown in Figure B, with the result being that the ball would be pulled back towards the opening of the vacuum hose with the ball remaining centered inside the hose due to the strong airflow flowing around it?</p>
<p>The reason I'm asking this question is because I have been unsuccessful with keeping the string fastened to the ping pong ball, even with super glue, so I am unable to see what the ball will do.</p>
<p><a href="https://i.stack.imgur.com/Z450S.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Z450S.png" alt="enter image description here"></a></p>
| |fluid-mechanics|airflow|aerodynamics| | <p>No - the sphere would likely stick to one wall, or bounce between them.</p>
<p>A sphere is not a very aerodynamic shape, having both a relatively blunt front, and sudden cut off at the back. This causes a lot of turbulence, such as the familiar vortex shedding shown here:</p>
<p><a href="https://www.grc.nasa.gov/www/k-12/airplane/Images/mix.gif" rel="nofollow noreferrer"><img src="https://www.grc.nasa.gov/www/k-12/airplane/Images/mix.gif" alt="Vortex Shedding around a sphere"></a></p>
<p>This shows how the air flows around a fixed sphere in free space - in your hose, the vortex forming on one side of the sphere will push it towards the wall, whereupon the flow will change as a result of the wall being there.</p>
<p>The key thing to note, is that the highly turbulent flow is not a stable equilibrium state - think of trying to balance a pencil vertically on your finger.</p>
<p>If you replaced the sphere with a different shape, imagine a Nerf football-type situation, that is designed to create a stable flow, then you would be able to get the behaviour that you're after - more like trying to keep the pencil vertically by holding the top, and dangling it downwards.</p>
| 25189 | Will the airflow inside a vacuum hose keep a sphere centered inside of it? |
2018-12-20T12:53:45.567 | <p>I have checked the definition of tonal masking on Wikipedia and google but I saw the definition of sound masking and auditory masking.</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Auditory_masking" rel="nofollow noreferrer">Auditory masking</a> </li>
<li><a href="https://en.wikipedia.org/wiki/Sound_masking" rel="nofollow noreferrer">Sound masking</a> </li>
</ul>
<p>But I want to know what exactly is tonal masking?</p>
| |electrical-engineering|telecommunication|acoustics|sound-isolation|electronic-filters| | <p>Tonal masking is using tones at specific frequencies for sound masking, as compared with using noise containing a continuous band of frequencies.</p>
<p>In most practical situations noise is more effective for masking than tones, but experiments with pure tones are a way to try to understand how masking actually "works" in human hearing.</p>
| 25192 | What is tonal masking? |
2018-12-20T21:01:09.910 | <p>My in-view sketch toolbar has disappeared recently -</p>
<p><a href="https://i.stack.imgur.com/6Q8Jr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Q8Jr.jpg" alt="enter image description here"></a></p>
<p>How can you make this reappear? So confused, don't even know what it's name is!</p>
| |solidworks| | <p>To reset the default toolbar to its original tool set, select Toolbar Options > Add or Remove Buttons > Main Toolbar > Reset Toolbar, or on the Customize dialog box'sToolbars tab, select Main Toolbar and click the Reset mbutton. From the menu bar, select View > Customize Toolbar.</p>
| 25199 | How to restore In-view Sketch Toolbar? |
2018-12-22T05:03:42.363 | <p>Geothermal systems are used in regions across the U.S. for home use, but industrial geothermal plants are only used in areas where subteranian temperatures are very high. Why is this?</p>
<p><a href="https://en.wikipedia.org/wiki/Geothermal_energy_in_the_United_States" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Geothermal_energy_in_the_United_States</a> </p>
<p><a href="https://www.energy.gov/eere/geothermal/geothermal-maps" rel="nofollow noreferrer">https://www.energy.gov/eere/geothermal/geothermal-maps</a></p>
| |heat-transfer|geotechnical-engineering|heating-systems| | <p>Because they're two completely different systems that have, in some circles, accidentally been given the same names.</p>
<p>The things that are sometimes (particularly in the USA) called home geothermal systems are ground-source heat pumps. The energy for these comes from sunlight on the ground over the year. Electricity is used to move heat from a source temperature that is broadly equal to the average ambient air temperature over the year (let's say 10-20 degrees C), to 30-90 degrees C to provide space- and water- heating.</p>
<p>Industrial geothermal systems are heat exchangers. The energy for these comes from the much hotter temperatures underground, that arises from the heat of magma, heating rocks above it. So in this case, the source temperature is 300 degrees celsius or higher. That's high enough to provide space and water heating without a heat-pump. It's also high enough (relative to ambient temperatures) to drive a heat-engine, which is pretty much the opposite of a heat pump. This heat engine is then used to generate electricity.</p>
<p>(A heat pump uses work to generate a temperature difference. A heat engine is the inverse of this: it uses a temperature difference to generate work.)</p>
| 25208 | Why are home geothermal systems usable anywhere in the U.S. but industrial geothermal plants limited in geographical use? |
2018-12-26T22:08:17.980 | <p>This is a screenshot of a lab where we were experimenting Vapor-Compression Refrig. Systems.</p>
<p>I am trying to understand what is happening between points 1-2 because I thought that because of max isothermal efficiency the gradient of the line shouldn't be more than the constant entropy line between the two points. </p>
<p>Also I am not sure how to explain the dip between points 5-1.</p>
<p><a href="https://i.stack.imgur.com/YwAJE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YwAJE.png" alt="Refrigerant: R134a"></a>
<a href="https://i.stack.imgur.com/VyyaJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VyyaJ.png" alt="Schematics"></a>
<a href="https://i.stack.imgur.com/Lb3pf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lb3pf.png" alt="System schematics"></a></p>
| |mechanical-engineering|thermodynamics| | <blockquote>
<p>I am trying to understand what is happening between points 1-2 because
I thought that because of max isothermal efficiency the gradient of
the line shouldn't be more than the constant entropy line between the
two points.</p>
</blockquote>
<h1>Short Answer</h1>
<p>Some cooling took place in the compression step between points <span class="math-container">$1$</span> and <span class="math-container">$2$</span> that isn't shown because lines are easier for a program to draw.</p>
<h1>Longer Answer</h1>
<p>I think the graph as shown suffers from the fact that the cycle is drawn with straight lines from point-to-point instead of curves that would reflect rational design.</p>
<p>That's not to say that the curve shown is impossible. Any movement on a P-H diagram is possible so long as your machine has a cold sink to dump its waste heat and the machine has energy to perform work. However, in refrigeration examples, practicality means you are limited in your movement on a P-H diagram such that <strong>you shouldn't cross to lower temperature isotherms below that of your cold sink's temperature unless you are in an evaporation step</strong>. Otherwise you're wasting energy in some way or another. For example, the isotherm barrier you can't cross until the evaporation step is the outlet temperature of your condenser. In the drawing that temperature appears to be <span class="math-container">$36^{\circ}\text{C}$</span>-ish.</p>
<p>I think you're right to be suspicious of the <span class="math-container">$1\rightarrow2$</span> step because if you interpret its path literally, its initial course shows a decrease in entropy (it's P-H path is steeper than the isentropic line) while near <span class="math-container">$20^{\circ}\text{C}$</span>. In other words, it is initially decreasing in entropy while below the isotherm of the cold sink, <span class="math-container">$36^{\circ}\text{C}$</span>. In order to decrease in entropy during compression, <a href="https://engineering.stackexchange.com/questions/18885/what-is-the-reason-for-a-reduction-in-entropy-in-a-compressor">heat transfer from gas to the environment must simultaneously occur for each incremental increase in pressure during the compression</a> (as mentioned by <a href="https://engineering.stackexchange.com/users/12754/phil-sweet">Phil Sweet</a>). <strong>The initial path of the <span class="math-container">$1\rightarrow2$</span> line only makes sense if there is a <span class="math-container">$20^{\circ}\text{C}$</span> cold sink available to the refrigeration machine which clearly isn't the case</strong>. If this were true, then the <span class="math-container">$2\rightarrow3$</span> line would ride the <span class="math-container">$20^{\circ}\text{C}$</span> isotherm, instead of the <span class="math-container">$36^{\circ}\text{C}$</span> isotherm.</p>
<p>That said, it is possible to shift the compression line to the left while never falling in temperature below the <span class="math-container">$36^{\circ}\text{C}$</span> isotherm. For example, the ideal refrigeration cycle, a portion of a <a href="https://en.wikipedia.org/wiki/Carnot_cycle" rel="nofollow noreferrer">Carnot cycle</a>, starting from point <span class="math-container">$1$</span> would follow an isentropic line (somewhere between the <span class="math-container">$1.8$</span> and <span class="math-container">$1.9$</span> blue isentropic lines) up until it intersects the cold sink isotherm (isentropic compression), <span class="math-container">$36^{\circ}\text{C}$</span>. Then, it would ride this isotherm (somewhere between the red <span class="math-container">$20^{\circ}\text{C}$</span> and <span class="math-container">$40^{\circ}\text{C}$</span> isotherm lines ) up to the black dew point line (note: isothermal compression = compression + cooling) at a point to the left of point <span class="math-container">$2$</span>. I drew in magenta what this ideal portion of a carnot cycle would look like:</p>
<p><a href="https://i.stack.imgur.com/7B22L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7B22L.png" alt="annotation depicting idealized compression step in refrigeration cycle"></a></p>
<p>Real world compressor equipment probably would <strong>not</strong> be able to follow this ideal curve since it is cheaper to simply have one single-stage compressor instead of many compression stages interlaced with many heat exchangers. Also, riding close to the dew point line is dangerous since formation of incompressible fluids damage most compressor types.</p>
<p>My guess for why the <span class="math-container">$1\rightarrow2$</span> line is so wrong is because the software generating point <span class="math-container">$2$</span> assumed that some cooling took place during or after the compression step and simply connected point <span class="math-container">$1$</span> with point <span class="math-container">$2$</span> with a straight line because it's simpler to program drawing a straight line.</p>
<blockquote>
<p>Also I am not sure how to explain the dip between points 5-1.</p>
</blockquote>
<p>The drop in pressure for the <span class="math-container">$5\rightarrow1$</span> step is probably due to friction in the piping between the evaporator and the compressor suction inlet. The rise in enthalpy is probably due to imperfect insulation of said piping (the environment warming the cold refrigerant vapor).</p>
| 25258 | Is this refrigeration cycle thermodynamically possible? |
2018-12-27T13:17:30.937 | <p>I need help calculating required torque and power for an elevator to select appropriate motor and gearbox. </p>
<p>The mass (weight of elevator plus max load) is 2000kg</p>
<p>The speed of the elevator is 10 m/s. </p>
<p>The radius of the pulley is 0,5 m</p>
<p>I want to use a counterweight - mass of counterweight is not specified, what is appropriate here?
I dont get it how to count when I have a counterweight in the system. </p>
| |mechanical-engineering|electrical-engineering| | <p>Let's say we ignore the friction for simplicity now, and assume mass of counter weight 600kg, this should be the weight of elevator plus cable and depending on the traffic pattern some extra weight.</p>
<p>More important than the speed is what acceleration you need. Say we choose 4m/s otherwise at constant speed of 10m/s we just need much smaller tension of holding the difference weight of 2000- 600=1400kg. So the tension would be only 13729N and torque would be 6860N.m, but we will see the torque needed to accelerate the elevator is much larger.</p>
<p>The tension force in the cable to accelerate the elevator and its counter weight is
<span class="math-container">$$ F_{net} = T_{cable} - mg = m\alpha $$</span> <span class="math-container">$$ and \ F_{net}= \ m\alpha = 2600* \alpha = 2600*4 = 10400N $$</span>
So:</p>
<p><span class="math-container">$$ T_{cable}= 10400 + mg =10400 + 1400 *9.8 = 24120N $$</span></p>
<p>now we can calculate the torque
<span class="math-container">$$ \tau = f*r = 24120 * 0.5 = 12060 N.m $$</span>
So we see the torque needed to accelerate the elevator is nearly two times that which is needed to move it at 10m/s. The real scenario is each manufacturer uses their own proprietary pattern of relays and controls that adjust the acceleration in a curve , ramping up, for comfort and minimizing wear and tear.</p>
| 25263 | Torque and power for elevator with counterweight |
2018-12-27T19:00:49.153 | <p>I have a Delta ASB0912L DC Brushless fan. PDF <a href="https://www.tedss.com/DataSheets/2031/2031000026.pdf" rel="nofollow noreferrer">here</a>. </p>
<p>In the datasheet it says, </p>
<pre><code>RPM 3800
CFM 67.80
IN H2O 0.302
</code></pre>
<p>On the blower, squirrel cage fan, Dayton from Grainger, found <a href="https://www.grainger.com/product/DAYTON-Rectangular-Permanent-Split-1TDP1" rel="nofollow noreferrer">here (Rectangular Permanent Split Capacitor OEM Specialty Blower, Flange: Yes, Wheel Dia: 3", 115VAC)</a></p>
<p>It says,</p>
<pre><code>RPM 3010
CFM @ 0.200-In. SP78
CFM @ 0.300-In. SP74
</code></pre>
<p>I want to exhaust fumes from a spray booth. I know there are many articles and plans online. I am trying to understand the math here.</p>
<p>It seems intuitive to me the Grainger fan is more powerful and will remove more air than the CPU fan. But the math makes no sense to me.</p>
<p>On the CPU fan it has .302 for Static Pressure and .3 for the Grainger and the CFM is pretty close.</p>
<p>I am moving air through 6' of ductwork, 4" in diameter, and neither tell me what CFM means in this context.</p>
<p>Given a 2'Lx2'Wx2'H spray box. Again, I'm not asking for plans. I can get that already. I only want to understand what these numbers mean and the spray box application gives the question context.</p>
<p>Why doesn't the CPU fan move almost as much as the Blower based on the specifications? Also, is the Delta IN H20 given static pressure or is that something else?</p>
| |pressure|statics|airflow|hvac| | <ol>
<li>Why doesn't the CPU fan move almost as much as the Blower based on the specifications? </li>
</ol>
<p>It is because of the working principle of fan and blower. The fan takes the air axially(axis of rotation) and pushing it in the axial direction. But the Blower works like a centrifugal compressor in which axial inlet and radial/tangential outlet are present. Fundamentally, due to the pressure reactions, the centrifugal pump/turbines are more efficient than the axial pump/turbines. That is what you are seeing as the data in the sheets.</p>
<ol start="2">
<li>Also, is the Delta IN H20 given static pressure or is that something else? </li>
</ol>
<p>Both are in the units of inches of water column.</p>
| 25268 | What is the difference in CFM, at a given static pressure, between a CPU fan and a Squirrel Cage Blower fan when moving air through ductwork? |
2018-12-28T05:52:29.033 | <p>When we increase the temperature of water in an open container, the vapour pressure increases.But boiling (phase transition at constant temperature and pressure) occurs only when the vapour pressure equals atmospheric pressure.My understanding is, till that point the atmosphere can hold the water vapour like a cover or lid but when equilibrium partial pressure of water vapour(vapour pressure) tries to increase beyond atmospheric pressure, the atmosphere cant hold the vapour and boiling starts.Is this right?</p>
| |fluid-mechanics|thermodynamics| | <p>Boiling only occurs when vapour pressure becomes equal to atmospheric pressure. The reason behind it is that, atmospheric pressure resists the phase change i. e
Liquid to gas. But when we equalize this pressure with atmospheric pressure the affect of atmospheric is neutrized and then energy which is absorbed by molecule is used to change the phase of liquid to solid. In this case temperature will not change because energy that is absorbed is used to break intermolecular forces and hence it is not contributing in increasing kienetic energy</p>
| 25274 | Why boiling occurs when vapour pressure equals the local atmospheric pressure |
2018-12-28T11:58:17.907 | <p>Please could someone explain the reason for using negative superscripts in SI units, and more precisely when to use negative versus positive?</p>
<p>I do not understand, for example, why the volume of a fuel would be given in m<sup>3</sup> but heat of combustion would be given in J m<sup>-3</sup>.</p>
| |mechanical-engineering| | <p>Negative exponents means <strong>per</strong>. In countries that were heavily influenced by the British, units such as meters per second would most likely be written <span class="math-container">$m/s$</span>. In Europe and elsewhere it could be written with negative exponents as <span class="math-container">$m.s^{-1}$</span>.</p>
<p>In your example <span class="math-container">$J m^{-3}$</span> means joules per cubic meter.</p>
| 25276 | What is the meaning of negative exponents in SI units? |
2018-12-28T16:14:51.590 | <p>During a lecture my teacher told me that the UPSTREAM in ADSL is very much affected from near end crosstalk due to DOWNSTREAM, while DOWNSTREAM is not affected by this phenomena. This lead to the decision to dedicate lower frequency to the UPSTREAM and higher frequency to the DOWNSTREAM. Is this correct? In case it is, how is the DOWNSTREAM not affected by near end crosstalk effect?</p>
| |telecommunication| | <p>Most of the noise is physically generated where the ISP's multiplexors split up the signal onto individual wires to each consumer. </p>
<p>The upstream data signal is generated by the consumer and has been attenuated by the cable when it reaches that point in the network. </p>
<p>For downstream data, the cable attenuates both the downstream signal and the noise which reaches the consumer.</p>
<p>So the signal to noise ratio is better for downstream data and the bandwidth can be higher.</p>
| 25279 | ADSL near end crosstalk (NEXT) |
2018-12-28T19:38:19.627 | <p>I'm building a motorized standing desk and am trying to calculate what linear actuators I need. If I have 4 linear actuators (one on each corner) each able to lift 25kg does that mean that the desk would be able to lift a full 100kg?</p>
<p>In other words, when linear actuators are sharing a load is the total drive force equal to the sum of the actuators individual drive force? Or is it more complicated than that?</p>
| |linear-motion| | <p>Yes, the four will share the 100kg, each carrying 25kg. </p>
<p>Only thing is you want to connect them to same rocker switch and install the linear actuators symmetrically with respect to the center of your desk.</p>
<p><strong>Edit</strong></p>
<p>After some comments I modify my answer this way. Lets say there is a heavy vise you need to have on the desk and it throws the center of gravity of the desk from the center to x and y centimeters off the center. Lets call the distance between the base of the legs X on the long side and Y on the short side.</p>
<p>Then the loads distributed to the four legs are going to change as follows.</p>
<p>The load to the right hand side pair of legs <span class="math-container">$$ is = l00*x/X = R_{load}$$</span>
<span class="math-container">$$ R_{load}*y/Y =load\ on\ the\ upper\ right\ leg. $$</span>
We repeat this process and find the individual loads for each leg.</p>
| 25284 | Sharing load between linear actuators |
2018-12-29T05:41:42.103 | <p>I am interested in building a very basic water jet propulsion system for a toy boat. Before I build it, I would like to know from a conceptual standpoint if this design of a water jet propulsion system will actually produce forward thrust.</p>
<p>Please refer to the conceptual drawing below of this water jet design. This drawing shows a top-down view of the toy boat.</p>
<p>An embedded motor-propeller will pull water into the pipe section coming from the stern of the toy boat and will at the same time be forcing water out of the pipe section going back to the stern of the toy boat. The force which should propel the toy boat forward should come from the dynamic pressure of the rushing water pushing against the inner walls of the two 90 degree elbow sections of the pipe. Will this particular design of a water jet produce forward thrust as expected?</p>
<p>Although I know that a traditional inline water jet system would be the more ideal thing to construct, I am very interested in finding out if a boat can be propelled using just the dynamic pressure(s) generated within a pipe.</p>
<p><a href="https://i.stack.imgur.com/RhbnN.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RhbnN.png" alt="enter image description here"></a></p>
| |mechanical-engineering|fluid-mechanics|propulsion|marine-engineering| | <p>Based on the other answers to this question, this simple water jet design should produce forward thrust. Yet for the toy boat to travel forward in a straight line it needs an additional part added to it. To cancel out the anticlockwise rotation which will be caused by the combined thrust and torque, I believe adding a small rudder into the inlet pipe section will provide the means of generating a clockwise torque to cancel out the anticlockwise torque. See modified picture below. I will probably need to keep readjusting the rudder towards the Port side and fixing it in place until I find the ideal position to make the toy boat travel in a straight line. I will then use the toy boat's main rudder to steer the boat.</p>
<p><a href="https://i.stack.imgur.com/POHaB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/POHaB.png" alt="enter image description here"></a></p>
| 25292 | Will this water jet propulsion system design produce any forward thrust? |
2018-12-29T18:53:18.250 | <p>Here is a picture of a caliper that I found:</p>
<p><a href="https://i.stack.imgur.com/mo0uR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mo0uR.jpg" alt="enter image description here"></a></p>
<p>Calipers are such basic tools, but I have trouble understanding how they work. Here the main scale says "mm" at the end, and the Vernier scale says "0.05mm". I've read some explanations saying that if the Vernier constant is, say, 0.1, on the Vernier scale the "1" is at a distance of 9/10 from 0, so that if an object that is 0.1mm thick is measured, the "1" on the Vernier scale is moved 1/10 to the right and the "1" is aligned with the "1" on the main scale, and therefore the measurement is taken to be 0.1mm.</p>
<p>But here we have a Vernier scale with 0.05mm as the constant. Shouldn't this mean that the "1" of the Vernier scale should be 0.05mm short of the "1" on the main scale? But this is clearly not the case, the "1" spacing is clearly much larger than the "1" spacing on the main scale. <strong>So how does this caliper function?</strong></p>
| |measurements| | <p>The problem with most analog meters and measuring devices is you have no feedback on your reading to confirm you are correct.</p>
<p><a href="https://www.stefanelli.eng.br/en/virtual-vernier-caliper-simulator-05-millimeter/" rel="nofollow noreferrer">Metric vernier caliper, (calliper or pachymeter), read in millimeter and vernier scale 0.05 mm</a> </p>
<p>This site explains how the vernier caliper works in detail, but more importantly has an interactive applet at the bottom. You can move the caliper and it provides the reading. The magnifying glass allows you to zoom in.</p>
<p>The 0 marker gives you the first two digits. Where the movable arm aligns exactly with the fixed caliper gives you the digits.</p>
<p><a href="https://i.stack.imgur.com/cSICw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cSICw.png" alt="Vernier Calipers"></a></p>
<p>In the image shown, the 0 on the movable arm is after 27 and checking the 0 to 10 scale on the movable arm aligns with the fixed arm at 7.5. This makes the reading 27.75mm. </p>
<p>Notice there are 20 divisions on the movable arm, which corresponds to the nearest 0.5mm.</p>
<p>Make a few measurements and this will get you over the learning curve. Put paper over the bottom of the screen to help you learn.</p>
| 25299 | Confused as to how to read this caliper |
2018-12-30T00:56:48.123 | <p>So I’m driving down the highway in my sedan on a speed of lets say 60 mph. Why is it that my car sways just before it passes or it passes me? Almost always where the semis first tire is (closest to the hood)? What forces are acting upon it? What could be a possible explanation to why it happens? Why is it only at that spot that the car sways? </p>
| |automotive-engineering|airflow|car| | <p>As far as I understand a semi track travelling on highway has at least three different zones of interest. They are:</p>
<ol>
<li>Front </li>
<li>Middle </li>
<li>Back </li>
</ol>
<p>Below are the three regions pictorially. </p>
<p><a href="https://i.stack.imgur.com/3pZAj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3pZAj.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/kOyZH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kOyZH.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/lKyrK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lKyrK.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/n5c3w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n5c3w.jpg" alt="enter image description here"></a></p>
<p>In order for the truck to pass through it diverts outwards air in front of the semi truck. The empty area middle (between the tractor and rear wheels) it creates vacuum which is reason sometime small car like the sedan get dragged toward the semi. The back of the semi also creates air void thus and air gets pulled in thus a small sedan could swayed into the lane the semi is travelling. </p>
| 25305 | Passing by a semi truck on a highway |
2018-12-30T12:31:42.407 | <p>I had an appliance (wireless audio receiver), which has operational range 662-698 mhz (and had same ranged antenna exactly from factory), like this:</p>
<p><a href="https://i.stack.imgur.com/U9Jtw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U9Jtw.jpg" alt="enter image description here"></a></p>
<p>However, that antenna was broken accidentally and I want to replace the antenna, but can't find exactly that model. However, I've found a bit wider ranged antenna: 632-698 mhz</p>
<p>Can that work without problems? may i replace with that?</p>
<p>EDIT: exact model is: shure <code>ULXP4 (662-698 mhz)</code>: <a href="https://web.archive.org/web/20140113201654/http://cdn.shure.com/brochure/upload/38/pdf_2010_ulx_brochure.pdf" rel="nofollow noreferrer">Sophisticated, scalable
wireless solutions</a> (also here specs: <a href="http://alss1.com/wp-content/uploads/2013/07/ULX4PK.pdf" rel="nofollow noreferrer">ULX Wireless System Specification Sheet</a>) and replacement antenna I want is: <a href="http://bit.do/fgergsadf" rel="nofollow noreferrer">http://bit.do/fgergsadf</a></p>
| |wireless-communication| | <p>The operational frequency operational range of the proposed antenna is within the frequency range of the old antenna. It should not be an issue from frequency standpoint. If the gains are similar chances are that the replacement antenna might work.</p>
<p>I also noticed the following statement for your alternative antenna. </p>
<p><a href="https://i.stack.imgur.com/bcrem.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bcrem.png" alt="enter image description here"></a></p>
<p>So the alternative antenna is not design your UXLP4 wireless receiver. Wanted to make you aware of the differences. </p>
<p>I did get chance to review <a href="https://www.shureasia.com/productdocumentsfiles/default/discontinued/wireless/ulxwirelessuserguide-36fc66950413b32b410df0a3e73af99c.pdf" rel="nofollow noreferrer">ULX Wireless System User Guide</a> Looks like you might have other antenna options for your receiver. Some those options are listed below. </p>
<p><a href="https://i.stack.imgur.com/9eHRC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9eHRC.png" alt="enter image description here"></a></p>
<p>Depending on region you might have to select the appropriate regulatory transmit and receive frequency / channel</p>
<p><a href="https://i.stack.imgur.com/UYKm9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYKm9.png" alt="transmit and receive Frequency / Channel "></a></p>
<p>On a side note United States of America, Federal Communication Commission (FCC) has reassigned the 600MHz frequency spectrum for cellular communication as licenses frequency band. Therefore as of October 2018 it is illegal to use unless authorized by the FCC. That most likely explains the absence of 600 MHz frequency spectrum antenna for wireless receivers. </p>
<hr>
<p><strong>References:</strong> </p>
<ul>
<li><a href="https://nofilmschool.com/2017/11/how-600-mhz-wireless-auction-will-impact-filmmaking" rel="nofollow noreferrer">Everything You Need to Know About How the 600 MHz Wireless Auction Will Impact Filmmaking</a></li>
<li><a href="https://en-us.sennheiser.com/spectrum" rel="nofollow noreferrer">Guide to Wireless Microphone Operation</a></li>
</ul>
| 25313 | Can wider range (i.e. 600-700 MHz) antenna replace lower range (650-700 MHz) antenna? |
2018-12-31T11:49:09.797 | <p>I am trying to calculate some rather long and complex drive trains, this is a project that I have worked up to, but its the first problem of its kind I have come across, to try and explain the position I am in please follow my example of how I am trying to calculate the final ratios and if I am going wrong, please can you point out where I have gone wrong and if you have the time, why.</p>
<p>Example:
In the image below is a simple drive train, it consists of three shafts with four gears turning.( I know it doesn't look it, but please assume all teeth are of the correct size and mesh with no problems)</p>
<ul>
<li>Shaft 1</li>
<li><ul>
<li>Input Gear 1 (Purple) has 24 teeth</li>
</ul></li>
<li>Shaft 2</li>
<li><ul>
<li>Driven 2 (Red) has 12 teeth</li>
</ul></li>
<li><ul>
<li>Driven 3 (Blue) has 15 teeth</li>
</ul></li>
<li>Shaft 3 </li>
<li><ul>
<li>Output Gear(Green) has 10 teeth.</li>
</ul></li>
</ul>
<p>So the power travels through Purple,Red, Blue Green.</p>
<p>Now for my Math.</p>
<p>though a 2:1 is connected to a 3:2 I have been calculating the ratio like this, but I am sometimes getting different answers with long (up 12 gears) trains in my project.</p>
<p>I would calculate the final ratio like this:</p>
<p>(12/24)x(15/12)x(10/15) = 0.416(RECUR.)</p>
<p>This is where I am going wrong
Is the Final ratio 100:416 (1:0.416)
or is it
1/0.416 = 1.666 = ~16:10 or is it something else?
<a href="https://i.stack.imgur.com/P57ZR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P57ZR.png" alt="Image shows example drive train as described in the example above"></a></p>
<p>Any help is greatly appreciated, im stuck at the moment and need to get on!</p>
| |mechanical-engineering| | <p>According to the <a href="https://woodgears.ca/gear/ratio.html" rel="nofollow noreferrer">Woodgears</a> web site, one takes the gear ratios independently and "adds them together" as shown in the video within the linked site.</p>
<p>In your example, the progression is as follows:</p>
<p>24:12, 15:10 which reduces to 2:1 and 3:2</p>
<p>Two (first ratio) times one and one-half (second ratio) is 3, which provides for a 3:1 increase or a 1:3 reduction if the reverse is created. There is no math between the red and blue gears as one rotation of the red is one rotation of the blue.</p>
| 25318 | How can I calculate final gear ratio when the drive train contains multiple gears with some reductions on the same shaft? |
2018-12-31T12:40:27.913 | <p>For example we have an inverted pendulum attached to a cart, with the cart having a speed limitation. Having derived the linearized model, how can I calculate the maximum possible deviation on the angle, and angular rate from the linearized model and the speed limitation ?</p>
| |control-engineering|control-theory| | <p>Lets set the following variables,</p>
<p>1- Length of the pendulum = r</p>
<p>2- Mass of pendulum = m</p>
<p>3- Small angle from vertical = a</p>
<p>So the horizontal acceleration of you pendulum, "A" is</p>
<p><span class="math-container">$A = \frac{mg *a}{r} $</span></p>
<p>Your control system has to be able to compensate for at least this acceleration A just to keep the tilted pendulum in check, a bit more to bring it back to stable, vertical position.</p>
| 25320 | How can I calculate the maximum possible deviation from equilibrium, that can still be compensated, given a limitation on the control input? |
2018-12-31T14:00:58.020 | <p>I can roughly understand the working of an axial flow gas turbine i.e. the fluid moves over the blade and gets redirected, its whirl velocity component changes and the blade is pushed in the opposite direction.But I couldnt understand the working of a table fan or an aircraft propeller.I know that its the opposite of an axial flow gas turbine but couldnt really imagine it.Please explain the principle?</p>
| |thermodynamics|turbomachinery| | <p>A table fan or a plane propeller have blades, pitched into an angle, which cuts into stationary air and acelerates it forward.</p>
<p>They are basically rotating wings, but working in reverse sense. Wings move up and create lift when wind goes through them. Fans and proppellers go through the air and cause it to move. </p>
<p>In a moving car if you hold your hand out of the window and try to keep your open hand like a row but gently angled you feel the pressure of the air trying to lift your arm up, and if you resist it you feel you are pushing the air down. And you may wish you had bigger hands to fly with.</p>
<p>A fan does the exact thing, by moving through the air in a screwdriver motion, it pushes it forward.</p>
<p>Or you could imagine tiny stationary air bubbles being plowed through by the blade and proppeled by the slant on the surface of it.</p>
| 25321 | Working of a fan or propeller |
2018-12-31T18:36:14.217 | <p>According to Elon in <a href="https://youtu.be/nSIzsMlwMUY?t=786" rel="nofollow noreferrer">this video (@13:06)</a>, he states that the reinforced segments are 70% just dirt, but in normal tunneling, they actually truck the dirt out and bring new dirt in which seems very inefficient. </p>
<p>Is it really that simple or is Elon oversimplifying the challenge?</p>
| |civil-engineering|tunnels| | <p>Regarding the word <strong>dirt</strong>. It depends on what a person means by dirt. Generally, it is understood that dirt means loose material such as soil or sand on or near the surface of the earth.</p>
<p>In mining however, and this may apply to tunneling, the word dirt can sometimes be used used to describe the broken rock, or mined material, that is excavated from the mine or tunnel in the process of mining or tunneling. In this situation it's a form of jargon or colloquialism used by some in the English speaking mining community and possibly the tunneling community.</p>
<p>One reason why locally excavated "dirt" may not be suitable as a ground reinforcing or ground supporting material is that it's strength properties are inadequate for it to be used. Another reason is that it could be a material that expands and contracts when subject to changes in moisture content. This could cause the supporting medium to buckle and decouple from the ground mass. Another reason is the material being removed might be soft.</p>
<p>Materials that are unsuitable include: clay; chalk; talc; limestone; ultramafic rocks such as komatiites ; most sedimentary rocks such a sandstone and mud stone; platey material such as mica; and oxidized or weathered rocks.</p>
<p>Materials that are better suited to be used for ground support purposes are hard, competent, non-friable materials with high strength characteristics, such as basalt and dolerite. Which is why these types or rock are used as aggregate materials for concrete. Generally, these materials need to be brought in sites.</p>
<p>In countries like Bangladesh, situated in a massive river delta, where there are no hard rock sources, broken baked clay is sometimes used as aggregate for road construction purposes.</p>
| 25323 | Why do normal tunnel boring operations not use the dirt (from the boring itself) for reinforcement rather than bringing it from other sources? |
2019-01-02T19:31:17.567 | <p>Is there any kind of board or mat that when we place it on the floor and wire it, a self-made connected device can read these data when it is being pushed and holding onto it, at multiple points?</p>
<ul>
<li>Location where the force is applied onto. Something working similar to a touchscreen, or a trackpad, or at least a keyboard. Except it's not any of those :-D</li>
<li>Magnitude of the force</li>
</ul>
<p>I need to build a footstep detecting device consisting a board that can be wired and detect point of pressure and how much amount it is. It is something working like a capacitive sensing device. But able to handle greater pressure.</p>
| |electrical-engineering|power-electronics| | <p>You are looking for Pressure sensing mats.</p>
<p>They are used for exactly what you want to use them for, getting stepped on for footprint and gait analysis. </p>
| 25349 | A board or mat that we can detect a push at an exact point |
2019-01-03T00:54:19.653 | <p>I have an RV garage but no RV. I'm considering building a platform 10ft above the floor. The garage is 14 feet tall and the area that I want the platform is 16ft a 16ft.</p>
<p>The details are stopping me though.</p>
<p>I'm thinking that I'll use a 2x6 every 2 feet as the joists. (see my picture attached.) Would a,b,c,or d be better?</p>
<p><strong>a:</strong> each joist runs the full length and is attached to a 2x6 that has already been attached to the wall</p>
<p><strong>b:</strong> i split the joists in half and have a middle joist running perpendicular.</p>
<p><strong>c:</strong> similar to a but the joists are connected directly to the wall studs rather than an intermediary 2x6</p>
<p><strong>d:</strong> combination of b and c.</p>
<p>Maybe there's a better way, so I'm open to suggestions.</p>
<p>Are 2x6s strong even strong enough for this?</p>
<p>Photo attached to show what I mean. The perspective view is to show how it fits in my garage. There are 3 solid walls and 1 open. I'm thinking that the open side will be where the joists are parallel to, (so each end of the main 8 joists will be connected to a solid wall)</p>
<p><a href="https://i.stack.imgur.com/tickP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tickP.png" alt="enter image description here"></a></p>
| |structural-engineering| | <p>Whatever joist plan you use , you can make it substantially stronger (and neater appearance) by attaching plywood to the bottom. I used glue and screws because the connection must be strong . This more or less makes an "H" beam , the plywood becomes the primary tension member; chose 1/2 , 3/4 , etc. depending on how stiff and strong you want. With your relatively long span , you could consider steel strapping on the bottom of the joists , they would be covered by the plywood . A little more hassle drilling screw holes but serious increase in strength.</p>
| 25350 | Structurally sound option for overhead storage platform in garage |
2019-01-03T12:01:12.097 | <p>I have a cold body of constant temperature A and a hot surface several times it's area. What is the best way I can absorb the heat from that surface using A. Will sheets of metal like alluminium or copper help?</p>
| |heat-transfer|cooling| | <p>A shape such as the one below will help. Make the piece from metal (copper). Insulate the edges.</p>
<p><a href="https://i.stack.imgur.com/Z81Wq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z81Wq.png" alt="metal adapter piece"></a></p>
<p>As the differences in area get larger, this will eventually have diminishing returns in improvement. When the goal is high heat transfer, thinner is better. Of course, at some point, the thin piece will be the same as just putting the smaller cold piece directly on the larger hot piece. The trade off is between the heat transfer perpendicular versus lateral in the piece. When the goal is uniform temperature on the (larger) hot surface, thicker may be better. In the latter case, an optimal shape (curvature of the sidewalls) may be possible instead to minimize lateral temperature differences versus the perpendicular temperature difference.</p>
| 25355 | Absorbing heat from a surface |
2019-01-03T17:17:34.500 | <p><a href="https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html" rel="nofollow noreferrer">The Engineering Toolbox</a> presents a chart of "common" materials and their thermal conductivity coefficients. I'd read or learned some time ago that diamond was superior in this respect. The chart shows diamond at 1000 W/m K compared to the next closest, silver at 429 W/m K. </p>
<p>My objective is to reduce the grip of a thermal pad between a heated bed and a glass surface. The heat transfer from the bed to the glass was improved by the addition of this thermal transfer pad, but the pad is very sticky, one millimeter thick and relatively fragile. I've only now discovered that <a href="https://www.sparkfun.com/products/retired/14283" rel="nofollow noreferrer">this product</a> is discontinued!</p>
<p>Pursuant to this objective is the ability to lift the glass, with the thermal pad (dusted?) remaining on the heater, without the pad adhering to the glass. It's not adhesive, but it sticks with the power of gekko.</p>
<p>I'm considering to powder one surface of the pad, but do not wish to reduce the thermal conductivity. I'm aware that diamond powder exists for lapping/grinding purposes, but have also found that most of the product listings provide the expression "synthetic diamond" in the description.</p>
<p>Even though SE discourages more than one question per post, these are all related to my objective.</p>
<p>Is there a better choice than diamond powder to provide a thermally conductive release mechanism between the thermal gap filler pad and the glass?</p>
<p>Related:
Will synthetic diamond powder display the same thermal transfer characteristics?</p>
<p>Edit Added from comments:
The bed is a flat electrical heater (3D printer) with a range from ambient to about 100°C. There is normally a 1mm air gap to the glass bed. As such, the air gap has to be heated before the glass reaches the desired temperature. Having installed the 1mm transfer pad reduced the elapsed time involved.</p>
| |thermal-conduction| | <p>Based on the question and comments, it seems you want something akin to this.</p>
<p><a href="https://i.stack.imgur.com/jEoEC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jEoEC.png" alt="powder tray"></a></p>
<p>This is a tray that holds a conductive powder. The powder is used to allow the glass bed to "self-level" independently of the hot plate.</p>
<p>As for the conductive powder, any metal powder would do. The thermal conductivity increase by using a powder in the gap versus having an air gap is significant enough, let alone that you want to use the most conductive material possible.</p>
<p>This would have two further issues. First, the powder should not stick to the glass bed. The choice may be have to be made through experiments. Second, the gap of 1 mm is small, and this may make such a tray design nearly impossible from a mechanical perspective. Indeed, even when you can create a 1 mm tray, the depth of the powder may be too thin to allow the self-leveling of the glass independent of the hot plate. In other words, when the powder is too thin, any thermal movement of the hot plate will be transmitted to the glass because the powder is essentially compacted.</p>
<p>One method may avoid the second problem. Increase the gap between the hot plate and the glass. This will allow the (thicker) powder to act both as a thermal conductor as well as an expansion/compression "blanket" between the hot plate and the glass. The downside is, as the gap increases, the temperature difference between the hot plate and the glass will increase proportionally. The end point will be that what you gain in mechanical stability by increasing the gap will eventually put you back to the same situation as having a 1 mm air gap. This will happen when <span class="math-container">$(k/d) \approx h$</span> (<span class="math-container">$k$</span> is conductivity, <span class="math-container">$d$</span> is gap width, and <span class="math-container">$h$</span> is the convection coefficient of air).</p>
<p>A second approach to this problem is to replace the air gap with a liquid gap. Flow water or oil between the hot plate and the glass. This will increase the convection coefficient in the gap by a factor of 10 - 100.</p>
| 25360 | Powder form of material with high thermal transfer |
2019-01-03T18:15:36.543 | <p>There is an interesting and potentially useful property to do with the procession of Gyroscopes, as well as devices like the "Powerball" toy.</p>
<p><a href="https://i.stack.imgur.com/NXH2h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NXH2h.png" alt="Gyroscope"></a></p>
<p>When the gyroscope is rotating (W<sub>s</sub>) and then force is applied to increase the procession (W<sub>p</sub>) the W<sub>s</sub> rotation increases.</p>
<p>Experiments with the Powerball show that a small increase in W<sub>p</sub> results in a very large increase in both W<sub>s</sub> and the resistance to an increase of rotational velocity in W<sub>p</sub>.</p>
<p>More intuitively, imagine the dynamics here as being like a variable gear that increases in gear ratio at an accelerating rate as the speed of the input increases.</p>
<p>Would it be practical to design a vehicle such as a human powered bike or a motorised car that used this property of a gyroscope instead of a conventional gearing system.</p>
<p><a href="https://i.stack.imgur.com/me9RW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/me9RW.png" alt="Bike Gears and Chain"></a></p>
| |mechanical-engineering|mathematics|mechanisms|mechanical| | <p>Naively, this looks as if it would be rather efficient, as there are few obvious mechanisms for heat loss. However, I'm unsure as to how make such a mechanism reliable. What if the wheel starts turning in the wrong direction?</p>
<p>Gearing would probably still be required, given the rotational speeds required for the exploited gyroscopic effects, so heat would be lost there. But the lack of chain makes for one fewer mode of failure.</p>
<p>If one make the gyroscopic effect reliable, such an engine might be reliable, perhaps more so than the traditional cog & chain, but given the need to step down the rotation speed to drive the wheel, there's not likely to be a noticeable energy saving compared to your traditional bike.</p>
<p>Overall, a nice idea. It would be worth building such a thing to investigate further.</p>
| 25361 | Gyroscope/Powerball gearing system for bikes? |
2019-01-04T13:24:47.050 | <p>I need a long arm which will transmit motion from a motor to a pen. It needs to be about 500mm long.</p>
<p>What I want to try and do is reduce the bend (i.e make it stiff) but also make it fairly light. </p>
<p>Question A)
In this case, am I better off going for a tube rather than a solid rod?</p>
<p>Question B)
Our of the materials available to me (which is carbon fibre, aluminium, steel or wood) - is there any of this which would have an obvious benefit?</p>
| |materials| | <p>A compromise between efficiency and practicality:</p>
<p>Use a 1/4" type L copper tubing with brass fitting adaptor (1/4 to 1/2) or similar as base connection.</p>
<p>Then solder 2 gauge 16 electrical copper wires from the sides of brass fitting to the tip of tubing at the 50cm end so it forms a rigid horizontal long triangle.</p>
<p>Readily available materials, total cost under $30.00 including the soldering gun, stiff enough vertically to support its own weight. And reinforced laterally by the two wires to take the load of pen.</p>
| 25370 | Stiff rod / tube - What material and diameter |
2019-01-05T16:14:35.803 | <p>I think I'm missing something important here.</p>
<p>For a fluid with a velocity <span class="math-container">$u$</span> in a pipe with cross section area <span class="math-container">$S$</span>, the volumetric flow rate is equal to:</p>
<p><span class="math-container">$$Q_v = uS$$</span></p>
<p>If we now consider vectors, the volumetric flow rate is:</p>
<p><span class="math-container">$$ Qv = \int \vec{u} ·\vec{n} dS$$</span></p>
<p>Where <span class="math-container">$\vec{n}$</span> is the normal vector to the cross sectional area. For example putting numbers, considering only one axis <span class="math-container">$x$</span>:</p>
<p><span class="math-container">$$ \vec{u} = 10 \vec{i} \qquad \vec{n} = \vec{i} \qquad S = 10 \\ Qv = 100$$</span></p>
<p>However from the divergence theorem we know that:</p>
<p><span class="math-container">$$ \int \vec{u} · \vec{n} dS = \int \nabla · \vec{u} dV$$</span></p>
<p>And for incompressible fluids <span class="math-container">$\nabla · \vec{u} = 0$</span>, so it would give <span class="math-container">$Q_v = 0$</span>.</p>
<p>Obviously i'm doing something very wrong here, but I cannot see what. Am I considering something wrong in these equations? Is the normal vector in the divergence theorem refering to another surface in this case, or to both surfaces, these being, both <span class="math-container">$\vec{n} = \vec{i}$</span> and <span class="math-container">$\vec{n} = -\vec{i}$</span>? </p>
| |pipelines|fluid| | <p>The divergence theorem only applies to a <strong>closed</strong> surface that encloses the volume <span class="math-container">$V$</span>. </p>
<p>If you construct a closed surface by "shrink wrapping" a section of pipe and closing off <em>both</em> ends, enclosing a volume <span class="math-container">$V$</span>, the volumetric flow out the <em>complete</em> surface is zero, because the flow into one end is equal and opposite to the flow into of the other end - i.e. whatever goes into one end must come out of the other end.</p>
| 25389 | Volumetric flow in a pipe from velocity |
2019-01-07T15:23:19.850 | <p>In the term of <strong>machine design</strong> and <strong>mechanical engineering</strong>.</p>
<p>If same spring is to be used in both smooth and rough road what will be the effect in suspension of the vehicle and if instead of using same spring for smooth and rough road how about using spring with <strong>variable spring rate</strong>.</p>
<p>I am new to this and sorry for my bad English.
But i am stocked in finding the comparative answer of this question.
Can anyone help me out here????
Please....</p>
| |mechanical-engineering|design| | <p>It is a compromise between comfort and practicality. Yes the off-road cars would have a smother ride if they have constant rate springs, but then they would need a much larger suspension to accommodate all the play.</p>
<p>Say two cars with similar mass on two surfaces with same speed: </p>
<p>1-a well leveled, smooth track.</p>
<p>2- a coarse bumpy off-road trail.</p>
<p>The first car is expected to hit obstacles of say 1 inch height and the second 3 inches.</p>
<p>In a spring with constant ratio k, F= kx and the jerk or acceleration the bump will give to the car will be F = m*a therefore a = f/m.</p>
<p>but if the spring in the second car is to absorb a 3inch obstacle with the same constant k it needs to recoil 3-5 times x or even more depending on the speed, and this will call for a very deep spring box.</p>
<p>So to optimise the performance of the spring they design the spring with a gradually increasing stiffness, e.g. <span class="math-container">$F = k*x^2$</span></p>
<p>This way the suspension can take a much larger range of bumps and at the same time is comfortable when meeting small obstacles and tough enough to take the big ones and still not bottom out.</p>
<p>The way the accomplish this is to build the spring with gradually thicker coil diameter, or use spring leaves, or design the geometry in a way that the play of spring becomes stiff as it recoils.</p>
| 25414 | Racing cars have spring with constant rate whereas Off road vehicles have variable spring rate |
2019-01-07T16:00:16.703 | <p>I designed a LQR controller and want to implement it on a microcontroller. I don't know about digital control and don't know when to apply it.</p>
<p>After a brief research, I found out that I need to apply the z-Transform, instead of Laplace-Transform. However the gain matrix is derived from time-domain.</p>
<p>The microcontroller I use is a OpenCR1.0.</p>
| |control-engineering|control-theory|optimal-control| | <p>You can also solve the LQR problem for discrete systems instead of continuous. One does needs to solve the <a href="https://en.wikipedia.org/wiki/Algebraic_Riccati_equation" rel="nofollow noreferrer">DARE instead of CARE</a>. But this does require having a discretized state space model. A simple but bad discetization method would be <a href="https://en.wikipedia.org/wiki/Euler_method" rel="nofollow noreferrer">forward Euler</a>. But there are a lot of other methods out there, such as zero-order-hold, first-order-hold and bilinear-transform/Tustin. In certain situations one method would be preferable over another and a decent overview is given in <a href="https://www.youtube.com/playlist?list=PLUMWjy5jgHK0MLv6Ksf-NHi7Ur8NRNU4Z" rel="nofollow noreferrer">this</a> YouTube series by Brian Douglas. But at low frequencies they all yield the same response, so if the sample rate is significantly faster than the dynamics of the system then it won't matter much which method you pick.</p>
<p>It can also be noted that it is very likely that the continuous LQR gain you calculated for the continuous system will also work on a digital controller if the absolute values of the closed loop poles lie significantly below the sample frequency. Namely most often zero-order-hold would be a good representation of what the digital controller does. Namely it calculates the controller output and holds it for one sample time. This roughly adds half a sample time delay, which might make the system unstable if the closed loop poles lie close to the sample frequency. But a delay only reduces the phase margin of the controller, but LQR has a guaranteed phase margin of 60°, so should always be able to handle some decent amount of delay. For example when use would use poles placement you would not have this 60° phase margin guarantee. But with LQR the performance might not be as optimal as the LQR designed for the discretized system.</p>
| 25415 | When is it necessary to consider digial control rather than directly coding in the gain matrix? |
2019-01-08T05:41:14.463 | <p>I read an article about Alberta, Canada losing millions per day from air barrels and am now doing some research, but having a hard time finding information. </p>
<p>I am curious to know more about air barrels and why they occur. Are there better systems out there that prevent air barrels or is it a constant, inevitable issue for pipelines? In particular, I am interested in the economics that result in air barrels, as well as the infrastructural issues or limitations that companies use to justify the purchase of air barrels. If you happen to know anything else about them, please do tell.</p>
| |petroleum-engineering| | <p>Air barrel means oil pipeline capacity which is booked or contracted by an oil company while they already know they can not deliver that volume of oil. Basically they over book the line.</p>
<p>Alberta pipeline in order to maintain the line or repair accidents has to cut down the flow of oil, so the oil companies overbook to be able to have guaranteed delivery of what they can produce. </p>
<p>However some companies game this known loophole to leverage the markets.</p>
<p><a href="https://thetyee.ca/Opinion/2018/11/28/Air-Barrels-Pipeline-System/" rel="nofollow noreferrer">Here is an article.</a></p>
| 25430 | What are air barrels (with regards to oil pipelines) and why are they a problem? |
2019-01-08T08:10:48.060 | <p>I'm currently trying to do an initial design for a propeller. In order to do this I'm trying to use <a href="http://web.mit.edu/drela/Public/web/xrotor/" rel="nofollow noreferrer">Xrotor</a>. Xrotor allows the user to enter certain information about both the propeller geometry, flight conditions and the airfoil lift and drag data after which it performs its calculations. Xrotor requires the following values for the lift and drag data (example values are given):</p>
<pre><code>========================================================================
1) Zero-lift alpha (deg): 0.00 7) Minimum Cd : 0.0070
2) d(Cl)/d(alpha) : 6.280 8) Cl at minimum Cd : 0.150
3) d(Cl)/d(alpha)@stall : 0.100 9) d(Cd)/d(Cl**2) : 0.0040
4) Maximum Cl : 2.00 10) Reference Re number : 2000000.
5) Minimum Cl : -1.50 11) Re scaling exponent : -0.2000
6) Cl increment to stall: 0.200 12) Cm : -0.100
13) Mcrit : 0.620
========================================================================
</code></pre>
<p>This information can be obtained from the drag polars of the airfoil.</p>
<p><a href="https://i.stack.imgur.com/HS92z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HS92z.png" alt="<span class="math-container">$\alpha,C_l$</span> plot"></a></p>
<p><a href="https://i.stack.imgur.com/Utyzq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Utyzq.png" alt="<span class="math-container">$C_d,C_l$</span> plot"></a></p>
<p>[<img src="https://i.stack.imgur.com/MV3lx.png" alt="360 degree extrapolated polar using JBlade"></p>
<p>The 360 degree polar was made using <a href="https://sites.google.com/site/joaomorgado23/Home" rel="nofollow noreferrer">JBlade</a>. The only value that Xrotor requests that I am not sure on how to calculate is the <code>d(Cd)/d(Cl^2)</code> figure. I'm fairly confident that this can be obtained using the <span class="math-container">$C_d,C_l$</span> <a href="https://en.wikipedia.org/wiki/Drag_polar" rel="nofollow noreferrer">drag polar</a>, but I'm not completely confident on how I should go about it. What is the correct way of determining this value?</p>
| |mechanical-engineering|fluid-mechanics|drag| | <p><span class="math-container">$d C_d/dC_l^2 = dC_d/(2 \cdot C_l \cdot dC_l) = (1/(2\cdot C_l)) \cdot(1/$</span>Slope of <span class="math-container">$C_d/C_l$</span> curve<span class="math-container">$)$</span>.</p>
<p>Use a few <span class="math-container">$\alpha$</span>s of interest and read the <span class="math-container">$C_l$</span>s. Find the slope at those <span class="math-container">$C_l$</span>s. The ratio should be consistant.</p>
| 25436 | How to obtain the $dC_d/dC_l^2$ value from the drag polar of an airfoil for Xrotor? |
2019-01-08T10:07:31.500 | <p>We are heating up our houses in winter to ex.: 21C to then have an insulated box inside the house, which we cold down to 5 and -15C, and we call it a refrigerator.. :)</p>
<p>The situation: </p>
<ul>
<li>we would buy a new refrigerator and put it in a house having 21C constantly</li>
<li>we would also have a "heat-pump boiler" for heating water. </li>
<li>the "heat-pump boiler" outputs colder air than what is there in the house normally, that is how "heat-pump boilers" work, ok. </li>
</ul>
<p>Like if I would put out the whole refrigerator to outside, where there is ex.: day: 5C night: -5C, I would be sure that the electricity consumption (operation cost) would go down, because it wouldn't have to battle with the indoor 21C. </p>
<p><strong>The question</strong>: would blowing cold air from the "heat-pump boiler" to the back (to the bottom or top of the back?) of a refrigerator help reduce the operational cost for the refrigerator? </p>
| |thermodynamics|heat-transfer|temperature|refrigeration| | <p>I would design the fridge's heat exchanger in such a way that the exothermic side is kept wetted and exposed to a natural air stream. The resulting evaporation will have a greater cooling effect than just cool air. </p>
| 25437 | Save money with using a refrigerator correctly - blowing cold air to the back of it? |
2019-01-08T18:40:25.797 | <p>The setup: I'm trying to mount a leadscrew powered by this motor <a href="https://www.pitsco.com/TETRIX-MAX-TorqueNADO-Motor-with-Encoder" rel="nofollow noreferrer">Tetrix Max Torquendado Motor</a> on a 40ish pound robot, in order to lift it 4" off the ground. I already have the linear motion, latching mechanism, etc. I'd like it to go as fast as possible, but for now, let's just say that it can take 4 seconds to lift the robot. There will be 40ish pounds of tension acting on the screw when the robot is hanging, and minimal compression. Also, resolution is unimportant. It seems like there's a simple way to calculate the pitch/diameter leadscrew needed based on the torque, motor, and time requirements. What leadscrew should be used, and generally how could I calculate this sort of thing?</p>
| |robotics|linear-motion| | <p>The relationship between torque and force for a frictionless single-start leadscrew is [1]:</p>
<p><span class="math-container">$$
\frac{F}{\tau} = \frac{2 \pi}{p}
$$</span></p>
<p>where <span class="math-container">$F$</span> is linear lifting force, <span class="math-container">$\tau$</span> is torque, and <span class="math-container">$p$</span> is pitch.</p>
<p>You can choose a suitable torque which that motor is best operated at, you know the force is the weight of the robot, so you can calculate pitch.</p>
<p>The speed of lifting is determined by the power delivered by the motor. So you can get maximum speed by operating the motor at a torque that maximizes its power output. That's typically half its stall torque (700 oz-in) so choose <span class="math-container">$\tau$</span> = 350 oz-in.</p>
<p>[1] <a href="https://en.wikipedia.org/wiki/Screw_(simple_machine)#Torque_form" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Screw_(simple_machine)#Torque_form</a></p>
| 25448 | What lead screw to use for specific torque and speed requirements? |
2019-01-10T16:46:50.570 | <p>I am wondering if a spinning permanent magnet's interaction with the Earth's magnetic field will produce a propelling force strong enough to propel a toy boat through the water. This propelling force would be most effective when the toy boat travels due North and stays on that course.</p>
<p>Please reference the drawing below. A permanent magnet would be fastened to a non-metallic bar that is perpendicular to an electric motor's shaft and on the other end of this bar would be a non-metallic counterweight. Say that the toy boat is 1 foot long, the magnet is a .5 inch cube N52 Neodymium magnet, the motor is spinning at 3000 rpm, and boat is in an indoor swimming pool. </p>
<p>The basic concept is that as the magnet is rotated from the position of facing due North to facing due South (180 degree rotation), the motor will have to push the magnet through the Earth's magnetic field and the reaction to this is that the toy boat will be pushed through the water. As the magnet is rotated from the position of facing due South to facing due North (180 degree rotation), the motor will not have to push the magnet through the Earth's magnetic field. So, during each full/360 degree rotation of the motor's shaft, the magnet's interaction with the Earth's magnetic field should create a paddle wheel effect that should propel the toy boat forward.</p>
<p>Could this magnetic paddle wheel effect propel this boat through the water?</p>
<p>Also, I am wondering if the rotational speed of the motor has a direct effect on the amount of propelling force that is generated. For example, will the motor have a harder time pushing the magnet through the Earth's magnetic field when it is spinning at 3000 rpm as compared to spinning at only 500 rpm? </p>
<p>EDIT</p>
<p>I had to insert a revised drawing because the original drawing had the magnet positioned in the wrong orientation. Also added text to help convey concept.</p>
<p><a href="https://i.stack.imgur.com/GvUhS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GvUhS.png" alt="enter image description here"></a></p>
| |mechanical-engineering|design|applied-mechanics|experimental-physics| | <p>Like anyone who has used a compass knows, earth's magnetic field does exert a force on magnetic objects. In theory this force could be harnessed to do work, but in practice it is too small to be a concern. </p>
<p>For example, a 1 meter long wire carrying 1 amp of current sitting perpendicular to earth's magnetic field will be subject to about 0.0001 Newtons of force. That's not going to be enough to overcome friction let alone build up any noticeable amount of speed. For comparison, dropping a paper clip off the side of a table exerts 100 times that force. </p>
<p>As to your "paddle wheel": The magnet is always in earth's field and is the equivalent of a paddle wheel that is fully submerged in the water. That means the force exerted on your spinning magnet will follow a typical sinusoid pattern, resulting in net zero work. You would be better off just leaving the magnet stationary, but don't expect your boat to move a noticeable distance unless you see all the iron objects in the world flying towards the poles. </p>
| 25470 | Could a spinning permanent magnet's interaction with the Earth's magnetic field propel a boat through the water? |
2019-01-11T10:03:36.823 | <p>I have a physics background so apologize my ignorance.</p>
<p>I'm confused by the constants <span class="math-container">$D$</span>, <span class="math-container">$\kappa$</span>, and <span class="math-container">$J$</span> when it comes to torsion of beams. From what I understand <span class="math-container">$\kappa$</span> and <span class="math-container">$J$</span> are the same thing, <span class="math-container">$J$</span> is (usually) for round beams with isotropic materials while <span class="math-container">$\kappa$</span> is for beams with unusual shapes and/or material properties.</p>
<p>What I really don't understand is how you get from <span class="math-container">$D$</span> to <span class="math-container">$\kappa$</span> though. In a <a href="https://www.sciencedirect.com/science/article/pii/S0263822309004693" rel="nofollow noreferrer">paper</a>, call it paper 1, I found, in the case of an orthotropic beam with rectangular cross section, they derive equations for torsional rigidity <span class="math-container">$D$</span> then mention equations such as</p>
<p><span class="math-container">$$
\kappa=\frac{D}{2ba^3G^0_x}
$$</span></p>
<p>where <span class="math-container">$b$</span> and <span class="math-container">$a$</span> are the height and width of the beam while <span class="math-container">$G^0_x$</span> is the shear modulus. Where did this equation come from? Is it a standard result? Did they derive the equation for <span class="math-container">$\kappa$</span> from FEM simulations?</p>
<p>Then in <a href="https://link.springer.com/article/10.1007%2Fs00707-017-2067-1" rel="nofollow noreferrer">this</a> paper (call it 2), which deals with torsion of arbitrarily shaped orthotropic composites, they talk about the torsional rigidity <span class="math-container">$GJ$</span>, which makes me doubt my previous statement about <span class="math-container">$J$</span>. One of the equations that they give is that the torsional rigidity factor <span class="math-container">$\beta$</span> (which is <span class="math-container">$\kappa$</span>, right?) is</p>
<p><span class="math-container">$$
\beta=\frac{GJ}{ab^3G_{zx}}
$$</span></p>
<p>From what I can tell <span class="math-container">$GJ=D$</span> from paper 1 and like I said <span class="math-container">$\beta=\kappa$</span>. So once again I wonder how <span class="math-container">$D$</span> (or <span class="math-container">$GJ$</span>) and <span class="math-container">$\kappa$</span> (or <span class="math-container">$\beta$</span>) are related.</p>
<p>I should explain that the reason why I'm looking to understand better these relationships is because I'm 3D printing springs. I'm looking to find a relationship between the shear moduli of an orthotropic material and the spring constant of a coil spring.</p>
<p>One method that I found is to relate the energy stored in a spring to the energy stored by a beam under torsion <span class="math-container">$\frac{1}{2}kx^2=\frac{1}{2}\kappa \theta^2$</span>, where <span class="math-container">$k$</span> and <span class="math-container">$x$</span> are the spring constant and deflection, and <span class="math-container">$\kappa$</span> and <span class="math-container">$\theta$</span> are the torsional constant and angle of twist of the material.</p>
<p>The two big problems that I have are that my springs have square cross sections (because they're easier to print) and are orthotropic. Hence why I'm having so much trouble.</p>
| |materials|beam|modeling|springs| | <p>Different texts use different notation.</p>
<p><span class="math-container">$K$</span> is sometimes called <span class="math-container">$J$</span>, torsional stiffness constant, units of <span class="math-container">$L^4$</span>.</p>
<p><span class="math-container">$G$</span> is shear modulus of rigidity which is sometimes assumed to be 40% of <span class="math-container">$E$</span>.</p>
<p><span class="math-container">$GJ$</span> or <span class="math-container">$JK$</span> is called section modulus of torsion rigidity as <span class="math-container">$EI$</span> is in a beam bending moment.</p>
<p>The frequently used equations can be found in Roark's Formulas for Stress and Strain, 7th edition, chapter 10, pp 404.</p>
<p>I am using my phone app, but if you need more detail I responded on my computer using <span class="math-container">$\LaTeX$</span>.</p>
<p>The equations for torsion in a rectangular solid section:</p>
<p><span class="math-container">$$\theta =\frac { TL }{ KG }$$</span></p>
<p><span class="math-container">$${ \tau }_{ max }=\frac { 3T }{ 8a{ b }^{ 2 } } \left[ 1+0.6095\left(\frac { b }{ a }\right) + 0.8865{ \left(\frac { b }{ a } \right) }^{ 2 }-1.8023{ \left(\frac { b }{ a } \right) }^{ 3 }+0.9100{ \left(\frac { b }{ a } \right) }^{ 4 } \right]$$</span></p>
<p><span class="math-container">$$K=a{ b }^{ 3 }\left[ \frac { 16 }{ 3 } -3.36\frac { b }{ a } \left(1-\frac { { b
}^{ 4 } }{ { 12a }^{ 4 } } \right) \right] \text{ for } a\ge b $$</span></p>
<p><span class="math-container">$\theta = \text{angle of rotation [rad]} \\
\tau = \text{torsion stress} \\
K = \text{torsional stiffness constant}$</span></p>
| 25474 | Torsional rigidity D vs torsional rigidity factor k vs torsion constant J |
2019-01-11T16:02:41.070 | <p>Does surface finish affect steels strength properties? If so, how does strength increase or decrease with the type of surface finish applied?</p>
| |strength|surface-preparation| | <p>There are a lot of different types of strength to describe steel (Compressive, tensile, yield, ultimate, fatigue, hardness, toughness, etc). Surface finish affects some of these strength parameters, while some others are independent of surface finish. </p>
<p>Surface finish generally cannot affect the intrinsic property of the underlying steel, which dictates strength parameters that are inherent to the steel material in question. That will include strength parameters such as compressive strength, tensile strength, yield strength, ultimate strength, etc. </p>
<p>Surface finish has significant impacts however to fatigue strength, since different surface finishes will generally have a large impact on how the S-N curve will look like for that particular piece of steel. Generally, smoother surfaces that has been machined or work hardened will have an improved fatigue strength over rougher surfaces. </p>
<p>Hardness, as indicated by @kamran, will generally be impacted by compressive surface finishing techniques applied, making it harder in the process. Stated in @kamran's answer as well, roughening will increase hardness since most roughening methods generally apply some compressive force to steel surfaces. I do want to note that other finishing operations can affect hardness indirectly as well. For example, while milling does not really change hardness at all of the underlying material, you can change the hardness of the surface that you are testing by milling, because hardness of steel do not necessarily stay consistent throughout the entire piece.</p>
| 25478 | Surface Finish and Strength |
2019-01-12T05:40:43.470 | <p><a href="https://i.stack.imgur.com/qMspd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qMspd.jpg" alt="enter image description here"></a>
I would be appreciate if anyone can solve problems below for this beam. <br></p>
<p>・Find the all sectional forces <span class="math-container">$F_x,F_y,M_x,M_z$</span> for both elements① and element②. <br>
・Find elastic strain energy of element① and element②. <br>
Flexural rigidities of two elements are both <span class="math-container">$EI$</span> and torsional rigidities of them are <span class="math-container">$GJ$</span>.</p>
| |mechanical-engineering|materials| | <p>Lets start from the point A and name the other end of the beam B and the end at support, where element 1 is connected c, and radius of the beams, R.</p>
<p>Loads are P concentrated at A and q distributed between points A and B over L= a. </p>
<p>For the circular beam ,</p>
<p><span class="math-container">$I = \frac{\pi R^4}{4} ,\ \quad J = \frac{π R^4}{ 2} ,\quad Tmax = (π / 2) τ_{max} R^3 \quad τ = T r / J $</span></p>
<p>τ = shear stress (Pa, psi)</p>
<p>T = twisting moment (Nm, in lb)</p>
<p>r = distance from center to stressed surface in the given position (m, in)</p>
<p>J = Polar Moment of Inertia of Area (m4, in4)</p>
<p>The deflection at the point A is sum of two deflections due to loads P and q,</p>
<p><span class="math-container">$ \delta = P a^3 / (3 E I) \quad + q a^4 / (8 E I) $</span></p>
<p>The deflection on point B is rotation under torsion due to beam AB torque and deflection under P an q, which is </p>
<p><span class="math-container">$(P + q*a)a^3/3EI $</span></p>
<p><span class="math-container">$ \theta_{rotation} = L T / (J G)= a(P*a + q*a^2/2 )/GJ $</span></p>
<p><span class="math-container">$and\quad \delta = (P +qa )a^3 / (3 E I) $</span></p>
<p>Elastic, strain, energy of beam AB , U is the work done by the forces which caused deflection. </p>
<p><span class="math-container">$U_{AB} = m^2L/2EI= (a*P+a^2q/2)^2/2EI$</span></p>
<p><span class="math-container">$ U_{C B}= \theta *T/2 = T^2*a/2GJ = (P*a + q*a^2/2)^2/2GJ
$</span></p>
<p><span class="math-container">$ F_{x, ab} = 0 \quad F_{y, ab} = P +qa\quad \\ M_{z,ab} =aP +qa^2/2$</span></p>
<p><span class="math-container">$F{x,bc }=0\quad F{y,bc} =P+aq \\ M{z,bc}= a( P+aq )$</span></p>
<p>Note: all coordinate are assumed local WRT the beam.</p>
| 25485 | A problem on mechanics of materials |
2019-01-12T20:08:00.660 | <p>I am trying to construct a torsion pendulum/balance for a physics experiment but the research papers are always missing specifics on how the certain connections are made.</p>
<p>I am looking for recommendations on rigid connections for:</p>
<ol>
<li>How to attach a glass mirror to tungsten wire.</li>
<li>How to attach tungsten wire to torsion pendulum/balance beam.</li>
<li>How to attach PLA to glass.</li>
<li>How to attach glass to glass (I do not have the capability to melt them).</li>
</ol>
<p><a href="http://www.leydenscience.org/physics/gravitation/cavend.htm" rel="nofollow noreferrer">Here is a link</a> to an example experimental setup. The balance beam I am using is actually quartz glass tube. The wire is tungsten. The plastic components are 3d printed and are PLA. The total weight is about 22lbs suspended from the tungsten wire.</p>
<p>(I posted this in the Physics exchange and they closed it claiming it is an engineering problem)</p>
| |mechanical-engineering|structural-engineering|materials|welding|joining| | <p>Keeping things simple...</p>
<ol>
<li>epoxy</li>
<li>design a fitting suitable for 3D printing to mechanically attach the parts </li>
<li>if the part needs to detach, design a fitting, otherwise epoxy</li>
<li>epoxy</li>
</ol>
| 25494 | Recommendations on a rigid connection between materials |
2019-01-14T17:21:35.630 | <p>I'm modifying a lawn tractor to pull single axle trailers in a woodland environment. They are rear wheel drive, have enough power, but traction is our biggest issue even when using tractor/ATV tyres.</p>
<p>My question is - what height (in relation to the rear axle) should the tow hitch be at and why?</p>
<p>I have looked at tractor pulling for comparisons and </p>
<ul>
<li><p>they seem to go as high as they can within the rules but I understand this is due to benefits of lifting the front of the sled off the ground.</p></li>
<li><p>they extend the hitch to influence front wheel lift - I guess this is to put more weight on the back axle.</p></li>
</ul>
<p>Front wheels lifting on woodland slopes isn't ideal so I have always opted to go low (below axle line) as this pulls the front down and I don't see how it could reduce the weight on the back wheels.</p>
<p>But a physics explanation would be really helpful.</p>
<p><strong>EDIT:</strong> I also found these two links which undertake a complex analysis of sled pulling with trucks under the title 'Physics of a truck pull, using a 2WD pickup truck'. </p>
<ul>
<li><a href="http://jimhawley.ca/downloads/Truck_Pull/Single_equation_model.pdf" rel="nofollow noreferrer">Part I - A simplified physical model captured by a single equation</a></li>
<li><a href="http://jimhawley.ca/downloads/Truck_Pull/Enhancements_to_the_dynamics.pdf" rel="nofollow noreferrer">Part II - Enhancements to the dynamics of the simplified model</a></li>
</ul>
<p>Thanks.</p>
| |car| | <p>Let's say you need to tow 1000kg weight. And you have your hitch at 10 cm, 01meters above the axel of your tractor.</p>
<p>If you pull it with an acceleration of 2 meters per second or 1/5 g. You have an overturning torque trying to lift up the front wheels of </p>
<p><span class="math-container">$1000*2*0.1= 200 *9.8 Nm.$</span></p>
<p>This is enough torque to lift the front wheel of a small garden tractor. If you put the connection lower than axel you loose part of traction due to the redistribution of part of the mass to front wheels.</p>
<p>So may be best is to hook the trailer hitch lined up with axel or barely above it.</p>
| 25519 | Physics of Trailer Hitch Height |
2019-01-14T21:37:32.373 | <p>If you draw a diagram with a spring, is there any convention about how it should behave with lateral loads? I guess it might be either:</p>
<ul>
<li>Rigid </li>
<li>Free</li>
<li>Elastic with bending and/or shear stiffness</li>
<li>Pin jointed - Initially free but as the spring rotates, it increases
stiffness due to its orientation changing.</li>
</ul>
<p><a href="https://i.stack.imgur.com/BIECP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BIECP.png" alt="spring with lateral force diagram"></a></p>
<p>Here's an example where it looks like we're supposed to neglect any bending or shearing of the springs. For instance, if the object moves in the y direction, only the 3 bottom springs would deflect and we wouldn't consider the bending stiffness of the other 3 springs. Is that right and common?
<a href="https://i.stack.imgur.com/WJaf9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WJaf9.jpg" alt="enter image description here"></a>
From <a href="http://ecoursesonline.iasri.res.in/mod/page/view.php?id=126222" rel="nofollow noreferrer">http://ecoursesonline.iasri.res.in/mod/page/view.php?id=126222</a></p>
| |springs|diagram| | <p>The convention for an idealized spring in such a schematic is that the lateral stiffness is zero but also that it does not move out of its original orientation with respect to its base (i.e., it does not tilt). The idealized spring is characterized by only one parameter: the axial stiffness. (It is thus assumed that the axial deflections are sufficiently small that the stiffness is constant and displacement independent. This assumption holds for small perturbations of any stable solid.)</p>
<p>Since the lateral stiffness is zero, other constraints must limit a body's motion in the directions orthogonal to the spring axis. This can be seen in your second schematic, in which each degree of freedom is constrained by the axial stiffness of one or more springs. In contrast, your first diagram is not well defined: the block should accelerate to the right for any positive <span class="math-container">$F$</span>, which then results in a discrepancy because a real spring would naturally resist such acceleration. To indicate this resistance, you'd either add a pseudospring with horizontal orientation or move away from lumped components altogether, perhaps modeling the spring using the method of compliant joints or by finite element analysis.</p>
<p>A good source of sample schematics for springs and rigid bodies is vibration textbooks, e.g., Tongue's <em>Principles of Vibrations</em>. Thorough and rigorous textbooks will emphasize that the block in your first diagram must be constrained (e.g., by walls) from moving left or right or in or out of the screen/page.</p>
| 25522 | Convention for spring symbols with lateral force |
2019-01-15T13:57:22.197 | <p>Some sources point to difficulties which may destroy the bridge, so the Wikipedia article on the Crimean Bridge e.g. has: </p>
<p>"The geology of the Kerch Strait is difficult: it has a tectonic fault, and the bedrock is covered by a 60 m (197 ft) layer of silt.[44] About 70 mud volcanoes have been found in the area of the strait.[44] More than 7,000 piles support the bridges; these piles have been driven up to 91 m (300 ft) beneath the water surface.[44] Some of the piles are at an angle to make the structure more stable during earthquakes.[44]</p>
<p>Some experts have expressed doubts that the construction is durable, given the tectonic and sea current conditions in the strait.[44][45]</p>
<ol start="44">
<li><p>Pollock, Emily (6 July 2018), <a href="http://web.archive.org/web/20181013090024/https://www.engineering.com/BIM/ArticleID/17007/Europes-Longest-Bridge-Spans-Troubled-Waters.aspx" rel="nofollow noreferrer">Europe’s Longest Bridge Spans Troubled Waters</a>, Engineering.com, archived from the original on 13 October 2018 </p></li>
<li><p><a href="https://economics.unian.info/10296951-kerch-strait-bridge-may-collapse-at-any-time-expert.html" rel="nofollow noreferrer">Kerch Strait Bridge may collapse at any time – expert</a>, UNIAN, 12 October 2018</p></li>
</ol>
| |bridges| | <p>Evaluating safety and risks involved in a project of this magnitude and complexity need a multifaceted scientific research, composed of teams representing different areas of technology, with experts to sample the soils, water current; chemical engineers, meteorologist, seismologists, etc, expensive imaging, testing machinery. It could cost millions of dollars and many years.</p>
<p>I had two classmates in college many years ago who spent their entire Ph.D. time doing an assessment of the safety of a small single span 110-foot bridge in Massachusets, with unlimited help from undergrad students at the lab.</p>
<p>Your question is way too ambitious for a site like this.</p>
| 25531 | Is the Crimean Bridge in danger of collapsing? |
2019-01-16T08:56:41.913 | <p>I have been doing a bunch of research online into how to build a basement that is below the water table. The common wisdom is to "not waste the money doing that" but as someone who doesn't like to be told no and always likes to fully understand their options, I am curious how I could go about building a basement that is below the water table.</p>
<p>I know it is <em>possible</em> to build a waterproof basement because submarines exist, marinas often build underwater sections of buildings, pools are a thing, etc. The ideal would be to have the primary structure for the basement be poured concrete with no windows or other holes in the walls (plumbing/electricity/air would come in through the ceiling from the higher floors).</p>
<p>For simplicity, lets assume that I don't need sewer out of the basement (perhaps by using an up-flush system to pump sewage/water out). Lets also imagine that the basement is on a flood plain so I can build it while it is dry, but it <em>will</em> be completely submersed in water later in the year.</p>
<p>Finally, again for simplicity lets assume that I am building in a location that has no building codes I need to abide by (so any option is on the table), but I do care about the structure being sound for a long time (e.g., 100 years).</p>
<p>What are the techniques I could use to water <strong>PROOF</strong> the basement so that it can be fully submerged under water for extended periods of time without leaking?</p>
| |structural-engineering|civil-engineering|structures| | <p>you need to build your basement with concrete, on deep foundation that penetrates into competent layer of soil and is below the level of under ground water.</p>
<p>Assuming your structure is built strong enough to withstand all the hydrostatic and soil pressure, we need to insulate and water proof it.</p>
<p>There are waterproof plastic sheets that are applied over a waterproof rolled over polyurethane film.</p>
<p>This insulation covers all around and from top to the bottom of the exterior of concrete wall, leading on the bottom to a trench sloped to carry any water to a sump pump. </p>
<p>The wall is then covered with 12 inches of gravel built over a 4 inch diameter perforated PVC pipe running on the trench behind the wall, to collect the water and take it to the said sump pump. then backfilled with the dirt from excavation and compacted. </p>
<p>This is standard practice in areas with aquifer. The packages are readily available at the big box building material stores.</p>
<p>They come with manuals and have link to instruction youtube videos. here is a sketch,<a href="https://i.stack.imgur.com/W9doa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W9doa.jpg" alt="wall detail"></a></p>
| 25541 | How does one build a waterproof basement? |
2019-01-16T19:31:34.313 | <p>Background:
My professor for system dynamics at my university gave us an engineering dynamics concept quiz. A particular question I am interested in presented two figures I have re-drawn below. The pulleys are stated to have no mass or friction. No gravitational accelerations, or mass was given for the left figure.</p>
<p>My professor said (without proof) that the figure with the idealized force (right figure) will accelerate faster. I would like to know why.</p>
<p>I am posting here because my professor could not understand where my confusion was and thus could not answer my question to my satisfaction.</p>
<p>I thought that regardless of the source of force, the accelerations should be the same. How is this possible? <a href="https://i.stack.imgur.com/kKuWO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kKuWO.png" alt="Left figure has a mass weighting 50N. Right figure is an idealized force of the same magnitude of 50N"></a></p>
| |mechanical-engineering|dynamics| | <p>The accelerations will not be same. What you do is you reduce the masses as rotational inertia then apply the force as moment. </p>
<p>Since the fïrst drawing has more rotational inerta it will accelerate slower.</p>
| 25559 | Pulley question: Which 10N block will accelerate faster? |
2019-01-16T20:15:40.710 | <p>My question is about a tank filled with some liquid (properties of the liquid are known and can be considered constant. Specific heat equals 1.5 kJ/(kg.K)). The initial temperature of the liquid in the tank is 90 ºC. Now imagine that 2000 L/h of this liquid are recirculating, but before returning to the tank they are heated somehow (doesn't matter how) until they reach 240 ºC (assume it remains a liquid at this temperature).</p>
<p>How can I estimate <strong>how much time</strong> it would take before the tank temperature reaches 230 ºC. The liquid volume on the tank is constant and equal to 10000 L.</p>
<p>It's kind of an open problem (the values are only hypothetical so that the problem can be implement on Excel). How should I start analyzing it? What equations should I be looking for?</p>
| |heat-transfer| | <p>The mass balance gives mass flow in = mass flow out = <span class="math-container">$\dot{m}$</span>. The energy balance on an insulated, constant volume tank gives</p>
<p><span class="math-container">$$\dot{m}\tilde{C}_p(T - T_h) + m_T\tilde{C}_V\frac{dT}{dt} = 0 $$</span></p>
<p>where the variables are tank temperature <span class="math-container">$T$</span>, hot temperature <span class="math-container">$T_h$</span>, specific heat capacities <span class="math-container">$\tilde{C}_p$</span> and <span class="math-container">$\tilde{C}_V$</span>, and tank fluid mass <span class="math-container">$m_T$</span>. The ratio <span class="math-container">$m_T/\dot{m}$</span> is the residence time in the tank <span class="math-container">$t_R$</span>. The residence time is also a ratio of tank volume to volumetric flow. Allow <span class="math-container">$\tau = t/t_R$</span> and <span class="math-container">$\theta = (T_h - T)/(T_h - T_o)$</span> where <span class="math-container">$T_o$</span> is the initial temperature in the tank. The differential equation becomes</p>
<p><span class="math-container">$$\theta = \gamma \frac{d\theta}{d\tau} $$</span></p>
<p>with <span class="math-container">$\gamma = \tilde{C}_V / \tilde{C}_p$</span>. When <span class="math-container">$\gamma = 1$</span>, the answer follows with the boundary condition that <span class="math-container">$\theta = 1$</span> at <span class="math-container">$\tau = 0$</span>.</p>
<p><span class="math-container">$$\theta = \exp(-\tau) $$</span></p>
<p>An example plot is below.</p>
<p><a href="https://i.stack.imgur.com/j9Eqe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j9Eqe.png" alt="plot of dimensionless temperature profile"></a></p>
<p>Modifications for <span class="math-container">$\gamma \neq 1$</span> are left as an exercise.</p>
| 25562 | How to calculate the time to heat up a tank using recirculating liquid |
2019-01-17T19:36:15.963 | <p>I'm trying to do an experiment on high pressure's effect on the electrical impedance of a really corrosive solution. I'm aiming for at least 2,000 psi with a pressure chamber but the only material I found that's chemically inert is PTFE and it's really limiting. One of my design is to use a PTFE diaphragm at the opening of the tube and press it inside the chamber to create pressure, and the other one is to push a PTFE piston into a tube directly but I don't know if it will hold much pressure. </p>
<p>The problem is that I can't really find a pre-made diaphragm that's under $500 that suits my chamber. Also, I can't find a way to fit a pressure gauge into the chamber and seal it properly. If you have an idea where to find these parts or any suggestions on how I should improve my design to better accommodate these challenges, that would be a great help. Thanks in advance! </p>
| |design|pressure|chemical-engineering|experimental-physics|seals| | <p>High Pressure corrosion testing is usually done in alloys like Hastelloy C ( or similar). Autoclave Engineers corp makes a lot of this equipment. Of course in this area it is usually done on a "money is no factor" basis.</p>
| 25586 | Issues creating a high pressure chamber for corrosive solution |
2019-01-20T11:51:09.253 | <p>I have shared a question from mechanics of solids and my solution with it, as you can see I am trying to take the derivative of the stress equations to find the min, max points but its going wrong. I need to know why my solution is wrong.<a href="https://i.stack.imgur.com/B0SWq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B0SWq.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/fVygq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fVygq.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/tYRob.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tYRob.jpg" alt="enter image description here"></a></p>
| |stresses| | <p>Let us assume that the greatest stress in any segment is at its top. (You can verify this on your own.) </p>
<p>Let us further assume that if the maximum stresses in the segments are not all equal, then we can adjust the geometry (here, <span class="math-container">$b$</span>) to reduce the larger value. (You can verify this on your own.) </p>
<p>In other words, we've optimized the geometry when the maximum stresses in all segments are equal. (The limit for a continuously changing cross section, i.e., an infinite number of segments, would be a constant stress throughout the hanging rod. This problem comes up in the context of <a href="https://en.wikipedia.org/wiki/Space_elevator#Cable_section" rel="nofollow noreferrer">space elevator tethers</a>.)</p>
<p>The maximum stresses in the upper and lower section are therefore </p>
<p><span class="math-container">$$\sigma_\mathrm{max,\,upper}=\frac{\mathrm{load}}{\mathrm{cross}\mbox{-}\mathrm{sectional\,area}}=\rho\left(\frac{A_bb+A_a(100\,\mathrm{m}-b)}{A_a}\right)=\rho\left[\left(\frac{A_b}{A_a}-1\right)b+100\,\mathrm{m}\right]$$</span></p>
<p>and</p>
<p><span class="math-container">$$\sigma_\mathrm{max,\,lower}=\frac{\mathrm{load}}{\mathrm{cross}\mbox{-}\mathrm{sectional\,area}}=\frac{\rho A_bb}{A_b}=\rho b,$$</span></p>
<p>respectively, where <span class="math-container">$A_i$</span> is the cross-sectional area of segment <span class="math-container">$i$</span>, and we wish to equate these stresses to determine the optimum geometry. The result is </p>
<p><span class="math-container">$$b_\mathrm{optimum}=\frac{100\,\mathrm{m}}{2-A_b/A_a},$$</span></p>
<p>and the minimum maximum stress is <span class="math-container">$\rho_\mathrm{optimum} b$</span>.</p>
<p>Does this result in the answer you were looking for?</p>
| 25620 | Mechanics of Solids / Strength of Materials |
2019-01-21T04:25:16.773 | <p>I am using strut-and-tie method to design a corbel. I noticed that Eurocode mentions strut-and-tie only in section 6, which is ultimate limit state design. How do I perform serviceability limit state design for the corbel using strut-and-tie method?</p>
| |structural-engineering|reinforced-concrete|eurocodes| | <p>One important thing to remember about the strut-and-tie method is that as a lower-bound method it is based on the plasticity theory. That is, the principle is to find a safe and statically admissible stress distribution, and if any such distribution is possible the structure will not collapse at this load. This principle only makes sense for the ultimate limit state. However, if the assumed load path is close to what linear elasticity theory would give, the stresses will be a decent approximation to a detailed finite element calculation and may very well be sufficiently accurate. Therefore, the difficult part of the calculation is knowing how well your strut-and-tie model aligns with elasticity theory.</p>
<p>There is a bit more on this in EN 1992-1-1 section 5.6.4.</p>
<p>But if you do your SLS check with a strut-and-tie model, the actual calculations are quite straightforward. Calculate the stress in concrete and reinforcement and check them against the SLS limits in section 7.2. Use the reinforcement stress to calculate a crack width and check that against the limit from section 7.3. And you're done. The strut-and-tie approach will not give you a way to calculate a deflection but for a typical corbel that wouldn't be very interesting anyway.</p>
| 25623 | How to perform SLS check with strut-and-tie method? |
2019-01-21T14:14:58.943 | <p>I apologize for the vague title. I know my problem, but I am having problems describing it in concise words, which is making it difficult for me to google.</p>
<p>I am building a CNC machine that is similar to a pick and place machine. I used extruded aluminum for the rails, and I am certain that everything is squared up. I had a local machine shop create a plate that has a grid of holes, so that I can attach "parts" that can be used by the CNC machine. However, this grid of holes is not machined at the tolerance that I need (budget constraints). I need a system for making sure that the "parts" are square to the cnc frame. One of my ideas was to use leveling springs similar to how you adjust a the bed on a 3D printer to be square with the frame. However, I am unsure if this is the best idea, and I would like a second opinion. </p>
<p>It is important that the machine has a resolution of 200um (if you were to move in a straight line across the machine ~500mm all parts that should be on that line should not be more than 200um above or below it). I know that some solutions maybe more expensive than having the plate machined to lower tolerance, but if I were to mass produce this machine I would like a way to adjust the alignment of parts.</p>
<p>Thank you for taking the time to read and/or answer my question!</p>
<p><a href="https://i.stack.imgur.com/NVsonl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NVsonl.png" alt="enter image description here"></a></p>
| |machining|cnc| | <p>At a bare minimum,I would accurately mount a "fence" in one of the directions, X or Y. pretty much a piece of tool steel stock, or anything precision ground say, 1/4" x 1" wide x however long you can fit, mounted to the working table at one of its ends. The fence should be indicated to within .001" or so, alignment with whatever axis you pick. Now when you go to put parts on the work table, you can use squares and parallels, referenced off your fence which you know is whiten .001" alignment of a particular axis(square to the other), to accurately align whatever feature on the part your using as the parts reference. Then your parts will be accurately located withe respect to the fence, which is accurately located with respect to the axis' of travel. This would be the most "universal" way to set up your machine. If your working on the same parts constantly, I would design fixtures for holding them. All depends how fast you want to work. Dowel pin the fence to the table so it can be removed and put back without indicating again.</p>
| 25630 | System for aligning/truing objects to each other in a plane for CNC machine |
2019-01-23T01:20:41.303 | <p>I am trying to figure out what the efficiency loss of a leaking hydraulic cylinder is. This <a href="https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwjm5pPi3YLgAhUQCKwKHfM1BX8QFjAAegQICRAC&url=http%3A%2F%2Fwww.mate.tue.nl%2Fmate%2Fpdfs%2F10299.pdf&usg=AOvVaw0UI7OlgfDT9rIm3kF0_Dc8" rel="nofollow noreferrer">master's thesis</a> found that a controllable pitch propeller (essentially a hydraulic cylinder inside a rotating shaft to control the pitch of the propeller) had external leakage of 0.5 ltr/min at 40 bar and internal leakage of 3 ltr/min at 100 bar. I was wondering if it would be possible to convert these numbers into a rough efficiency loss?</p>
<p>I have been google searching for days to try to find a reference to indicate how to make this calculation. I also went all out and went to a local university library to use Engineering Village -- hoping to find some paper that did an analogous study but couldn't find much to help.</p>
<p>This seems relevant but not sure how to use it: <a href="https://engineering.stackexchange.com/questions/19228/losses-through-holes-drilled-in-a-pipe">Losses through holes drilled in a pipe</a></p>
| |hydraulics|energy-efficiency| | <p>The efficiency is a dimensionless quantity and is calculated by power output devided by power input.</p>
<p>I hope you have the value of your power input? But you can also calculate it with the following formula. The left one is for your (electro-)motor which is convertible in your hydraulic pump.</p>
<p><span class="math-container">$P_{mechanical}=M[Nm]*2*n[1/sec]*\pi=p[bar]*Q[l/min]=P_{hydraulic}$</span></p>
<p>There are three (main-)possibilities to get some energy loss in your system:</p>
<ul>
<li>the mechanical one caused by friction in your pump, motor, valves … - <span class="math-container">$P_{loss-mechanical} [W]$</span></li>
<li>the pressure one caused by perfusion-resistance of the elements in your hydraulic system, e.g. in your piping – <span class="math-container">$P_{loss-\Delta p} [W]$</span></li>
<li>the volumetric one caused by leakage in your hydraulic system – <span class="math-container">$P_{loss-volumetric} [W]$</span></li>
</ul>
<p>The volumetric loss is your issue - mentioned by the two leakage (internal and external). </p>
<p><span class="math-container">$P_{l- vol} [W]= \Sigma (p[bar]*Q[l/min])=10^5*40bar*1/60000*0,5l/min + 10^5*100bar*1/60000*3l/min=533,33 W$</span></p>
<p>Note the <span class="math-container">$10^5$</span> and the <span class="math-container">$1/60000$</span> are values because of the different units…</p>
<p>Now the equation for efficiency:</p>
<p><span class="math-container">$n= P_{output}/P_{input}=(P_{input}-\Sigma(P_{loss}))/P_{input}=(P_{input}-P_{l- vol})/P_{input}=(P_{input}-533,33 W)/P_{input}$</span></p>
| 25653 | Calculate Efficiency Loss of a Leak in a Hydraulic Cylinder |
2019-01-23T20:05:12.633 | <p>After reading about submarines in World War II, I was curious about their battery capacity, specifically in comparison to modern Battery Electric Vehicles (ie Tesla, Bolt, Leaf, i3, etc).</p>
<p>I haven't been able to find a source that either answers the question in kWh or gives me enough information to calculate it myself.</p>
<p><strong>My current guess is about 30 kWh</strong>, based on a figure of 12000 Ah in a 120 cell system, and a voltage of 2.75 V - 1.05 V</p>
<p>I would accept an answer for any class of submarine in WWII, but I was looking at the US Balao class.</p>
| |electrical-engineering|battery| | <p>Collecting bits from sites,</p>
<p><a href="https://fleetsubmarine.com/battery.html" rel="nofollow noreferrer">fleetsubmarines</a></p>
<blockquote>
<p>World War II American fleet submarines had two batteries, each
composed of 126 cells. By comparison, a 12-volt car battery contains
only 6 cells, each producing about 2.25 volts when fully charged, with
a maximum power output of about 45-50 amps. Each cell in a submarine
battery produces from 1.06 volts when fully discharged, to 2.75 volts
at the optimum output, so connecting the 126 cells in each battery in
series gives a usable output of from about 210 to 350 volts, and a
power output of as much as 15,000 amps with both batteries connected
in parallel.
(no mention of total amp-hrs) </p>
</blockquote>
<p><a href="https://www.quora.com/How-long-does-a-diesel-submarine-have-to-run-its-engines-to-fully-charge-its-batteries" rel="nofollow noreferrer">quora</a></p>
<blockquote>
<p>My submarines (Oberon class of the 1960s-1990s) had two lead acid
batteries containing 224 cells each with a nominal voltage of 440
volts.. The cells were rated 74.20 ampere-hours at a 5 hour rate
(nominal voltage of each cell was 2.2 V)</p>
</blockquote>
<p>448*74*2.2 = 73kWh </p>
<p><a href="https://uboat.net/articles/id/54" rel="nofollow noreferrer">uboat.net</a></p>
<blockquote>
<p>The US Navy "Balao" type submarine (1944/45) was fitted with 4 four
Elliot Main (Electric Motors) two on each shaft, with a total
horsepower of 2,740. While submerged, these motors were powered by two
massive (each cell weighing 1650#) 126-cell batteries (in series)
capable of delivering 5,320 Amp/Hrs each.</p>
</blockquote>
<p>Assuming they meant Amp-hrs, and guessing 2.2 V per cell, 2.2 * 2*5320 = 23.4 kWh</p>
| 25659 | What is the capacity of a WWII Submarine battery in kWh? |
2019-01-23T23:24:22.030 | <p>Let's say I have a 300 rpm motor and a roller connected to it's shaft and I would like to use it for pulling fabric from a printer-kind machine. But think that printer's printing speed is changing according to printed surface and I want this motor's shaft to spin freely when opposite load occurs on this roller by fabric.</p>
<p>So basically, if printer prints 100rpm long fabric, 200rpm of motor will spin freely as printer pull other side of fabric. When printer prints another length of fabric, motor will roll again. Roller (or idler?) should be able to spin asynchronously.</p>
<p>I don't know much about bearings and mechanical parts so my question might be weird, but I heard some kind of bearing can solve my problem. I tried to explain but sorry if it's nonsense. Thank you for your help in advance.</p>
| |mechanical-engineering|motors|bearings| | <p>Ok, so what you are searching for is a torque limiter (slip type) which only transmits torques below a certain threshold (be careful here, since you need to allow some torques to accelerate your shaft). Slip type means that it automatically continues transmitting torque if the torque is below a certain threshold.</p>
<p>The most simple type are so called friction plates which are literally just two plates which are preloaded to each other and attached to the shaft on both sides. If too much torque is acting on it, frictional forces will be lower than transmitted force and the rotate relatively to each other. </p>
<p>Beside, keep in mind that the mass of the roller is quite important!</p>
<p>See:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Torque_limiter" rel="nofollow noreferrer">Wikipedia</a></li>
<li><a href="https://www.motioncontroltips.com/torque-limiter/" rel="nofollow noreferrer">Motioncontroltips</a></li>
</ul>
| 25661 | What kind of bearing should be used for free rotation under load? |
2019-01-24T15:45:22.050 | <p>I'm really confused. I've only ever used metric in my life. Recently I'm trying to build a IOT device and there is plumbing and water involved.</p>
<p>I was looking for a 19mm fitting, but on Amazon and other sites it's always advertised in inches. <a href="https://www.amazon.co.uk/gp/product/B0087OSUY4/ref=ppx_yo_dt_b_asin_title_o00__o00_s00?ie=UTF8&psc=1" rel="nofollow noreferrer">John Guest 3/4" BSP x 1/4" Push Fit Tap Adapter</a> </p>
<p>I did the conversion and 3/4 inch, according to my calculation, should be around 19mm. But whenever I order these fittings they always send me a larger size. The resulting size is always around 25mm when I measure it.</p>
<p>To me it seems like 3/4 inch doesn't actually mean 3 quarters of an inch?</p>
| |measurements| | <p>No, as noted by@ user 1683793 , 3/4 inch is the NAME of a size of ASTM pipe ( not the size) , very close to API pipe. Any 3/4 pipe will thread to any other whether is is Schedule 10 or Sch 120 , stainless or copper, etc. ; only the wall thickness and inner diameter change. Most often Sch 40 is used for domestic service. And tubing has nothing to do with pipe other than it may look similar. I am not familiar with pipe threaded to ISO,BS, DIN, JIS, GOST, etc. Although I suspect CSA ( Canadian) is very close to ASTM. </p>
| 25669 | Plumbing Imperial units confusion - 3/4" is supposed to be around 19mm, right? |
2019-01-25T14:53:14.510 | <p>In this flow control simple experiment in the lab, i am made to control the flow through the pipe and at the same time record down the output given by the sensor in voltage. I am told to record down values for increasing output and decreasing output.
The lowest flow is 0cc/min, and highest flow is 3000cc/min. Just wondering why must i take datas for both increasing and decreasing output. Is this so that i can determine the recommended operating range of the sensor? which will be the min. value to the max value of the both sets of outputs (increasing and decreasing) ? </p>
<p>The Objective of this experiment is by the way to investigate the linearity and hysteresis of the flow sensor. </p>
| |flow-control| | <blockquote>
<p>The Objective of this experiment is by the way to investigate the
linearity and hysteresis of the flow sensor.</p>
</blockquote>
<p>The key word here is <strong>hysteresis</strong>: the output of the sensor at the same flow rate might not the same depending on whether you are in the increasing or decreasing output direction. This is why you are being asked to test in both directions. Check out <a href="https://en.wikipedia.org/wiki/Hysteresis" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hysteresis</a> for more details on hysteresis.</p>
<p>Here's a typical characteristic curve with hysteresis (albeit from a different domain):</p>
<p><a href="https://i.stack.imgur.com/GnKvh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GnKvh.jpg" alt="enter image description here"></a></p>
<p>You also ask about <strong>linearity</strong>: it's pretty basic but it means that if you plot your sensor output vs. flow rate, you basically get a straight line if the sensor is linear, or a curved line if not:</p>
<p><a href="https://i.stack.imgur.com/Z5eEa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z5eEa.png" alt="enter image description here"></a></p>
<p>If the sensor is linear but has hysteresis, then you'll essentially get 2 parallel lines. If it's linear but without any hysteresis, both lines will be on top of each other and you'll get a unique characteristic regardless of which direction you go.</p>
| 25681 | Flow control experiment |
2019-01-27T01:20:35.037 | <p>Would a passive and controlled amount of electricity fed through post-tension cables, heat up a slab, without ruining the dependability and health of the post-tension cables, and the concrete itself?</p>
<p>I am wondering because I coat concrete slabs with different types of paint, and in the winter, it is super slow to add layers of coats on top of one another because of the waiting time for drying and curing of the paint. If I could heat up the slab through the post tension, then re-grout the post tension ends, it would totally change the game for me. I am just afraid of possible snapping the cables. In the winter I can get away with only one coat for the whole day. In the summer I can put down 5 coats of paint over a 7,200 square foot area with ease.</p>
| |reinforced-concrete|prestressed-concrete| | <p>Nobody else applied any numbers to this so here is my take.</p>
<p>The smallest post tension strand I found on the web was 1/2 inch in diameter or (0.25^2*PI) about 0.2 square inches.
We want to heat up a 7,200 square foot area slightly to assist in drying paint.
A square that size would be about 85 feet on a side or 1020 inches.</p>
<p>So assume we have an iron conductor 1020 inches long with an area of 0.2 square inches. </p>
<p>The resistance of iron is about 4E-6 Ohm-Inch so doing the math:</p>
<pre><code>ohms=(4e-6 ohm inch)*(1020 inch)/(0.2 inch^2)=0.02 ohms
</code></pre>
<p>Remembering P=IE and E=IR we have P=I^2R so if you put your welder on the strand and pump some current through, I get:</p>
<ul>
<li>20A 8W</li>
<li>40A 32W</li>
<li>60A 72W</li>
<li>100A 200W</li>
<li>200A 800W</li>
</ul>
<p>Now,this is inside the concrete. If you put in 800W without being sure what is happening inside, you are ... not going to be pleased with the results. A little reading shows that if you keep the temperature of the strand below 200°C, you won't compromise the strength... of the strand but I would not want to bet on the concrete if it was expected to set at that temperature.</p>
<p>What I would do is connect a welder at some minimum current and feel around and feel around to see if the strand is warm before entering the slab. I expect that the heat transfer inside with contact with the concrete will be greater than outside to air so if it is not too hot to touch, you are OK. Let it sit for a while and turn up the current a little, keeping in mind that the power goes up as square of the current.</p>
| 25692 | Using electricity to heat up a slab through post-tension cables? |
2019-01-28T19:45:39.140 | <p><strong><em>Don't try this at home. I am taking personal responsibility for what happens in my car.</em></strong></p>
<p>Hi all,
I recently purchased a fire extinguisher for my car (it weighs 1.5 lbs or 0.68 kg). I browsed the internet for a decent bracket to mount it inside my car, under the front passenger seat. Everything was fairly expensive which made me want to make a DIY version that would be cheaper.</p>
<p>I have been trying to find what sort of materials I can/should use. The last thing I want is the bracket to break and have a 1.5 lbs metallic projectile flying inside the car in the event of a crash.</p>
<p>The bracket should be 16" or ~40 cm wide so that I can use the bolts on which the passenger seat is mounted to attach the bracket. (See picture below)</p>
<p><a href="https://i.stack.imgur.com/rPUZ6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rPUZ6.jpg" alt="enter image description here"></a></p>
<p>Obviously I could use some thick steel and be safe but I don't really have easy access to tools to cut metals so I would like to try something else.</p>
<p>My requirement for the bracket is that it should be able to hold on to the fire extinguisher in the event of a frontal collision on the freeway. Which would be an impact of, if my memory serves me right, of </p>
<code>65 mph + 65 mph = 130 mph or ~210 kmh</code></p>
If my math is correct this would be equal to </p>
<code>210000 * 0.68 = 3966 N</code></p>
<p>After that I'm not sure where to go...
If I was considering using a PVC pipe, would I be able to relate those Newtons to the tensile strength of the pipe to figure out if it would hold up?</p>
<p>I had also thought about 3D printing the bracket but a friend pointed out that it's hard to make measurements with PLA because the printer layers the material so the "resistance" or "strength" is different in different directions.</p>
<p>I am also open to other material suggestions!
Thank you.</p>
| |structural-engineering|materials|automotive-engineering|structural-analysis|strength| | <p>I don't know what you thought you were doing multiplying <span class="math-container">$210000 \times 0.68$</span> but it certainly doesn't give a force in Newtons.</p>
<p><a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3503406/" rel="nofollow noreferrer">Measured crash data</a> shows the maximum deceleration is of the order of <span class="math-container">$50g$</span>. In other words your 1.5lb object has a force of <span class="math-container">$1.5\times50 = 75$</span> lb acting on it.</p>
<p>The simplest way to restrain it is to realise that in a frontal crash it will be trying to accelerate forwards, relative to the vehicle. Therefore if you mount it on something where it can't move forwards, such as the front of the footwell, the rear of a seat, or even inside the glove box, you don't have much to worry about.</p>
<p>Note: the reference was based on crashes at lower speeds, but that is not a critical issue. If the crash is enough to crumple the front of the car by a significant amount, the maximum <em>force and acceleration</em> will be more or less independent of the speed, but the <em>duration</em> of the impact and the <em>total amount of deformation</em> will increase. There is not much point worrying about the state of the fire extinguisher, if a high speed frontal crash deforms the car so much that you are trapped in the seat with the steering wheel pushed against your chest, and you can't reach the extinguisher to operate it even if you are still conscious!</p>
| 25723 | Material for car mounted fire extinguisher bracket |
2019-01-30T18:08:58.927 | <p>A uniform block is in the form of a square ABCD of sides 2 m and a unit thickness. E is a point on AD such that ED = xm. The portion EDC is then removed from the block. What is the maximum value of x such that the block will not topple?</p>
| |mechanical-engineering|civil-engineering|statics|dynamics| | <p>The CG of the cut shape should fall inside the base footprint which is, 2- xm.</p>
<p>The CG is <span class="math-container">$ \dfrac{ \sum A_n* d\bar{x}}{ \sum A} $</span></p>
<p>Area of square is 2*2= 4 and its dx=2</p>
<p>Area of the triangle cut off is xm*2/2= xm and its dx = 2-xm/2 </p>
<p><span class="math-container">$ CG_x = \dfrac {4 *2- ( 2-xm/2 ))*xm }{4-xm } \quad$</span>, note we factored the E, thickness, out. </p>
<p>And this cg must be smaller than the base, <span class="math-container">$2-xm$</span></p>
| 25743 | Centroid and toppling |
2019-01-31T08:46:48.003 | <p>After testing a power transformer(pt) and closing its wiring terminals I've encountered that phase S 63 kV correct voltage and phases R and T shows incorrect voltage 32 kV.</p>
<p>Can any body tell why?</p>
| |electrical-engineering|control-engineering|circuits| | <p>As I Know when you close the breaker you have seen the voltages.
You should check the secondary circuits fuses.
This voltages shows open circuits of one poles.
If the polarity of poles were incorrect you shoud have seen 37Kv on other poles.
If fuses were healthy you shoud check the open circuits of PT secondery circuits.</p>
| 25757 | Unbalance voltage mesurements from 63 kV power transformer |
2019-01-31T10:37:39.183 | <p>This question is about a technical aspect of naval warfare, and I think that this site is the closest fit. <em>History</em> would have been my next guess.</p>
<p>In the TV show “The Last Ship” the protagonists have an Arleigh-Burke class guided missile destroyer. It seems to engage targets mainly with rockets of different sizes. Incoming rockets are shot down either with the point defense turrets or with other rockets. It is probably skewed for the plot of the show, but they seem to be able to shoot down most incoming missiles, making the ship appear near invincible.</p>
<p>In the fifth season they are up against a battleship and are struck with incoming shells from canons in two different episodes. The first time the destroyer gets rendered disabled, the second time it starts to sink.</p>
<p>Reading up on this I noticed that modern fleets do not have battleships with huge cannons any more, guided missile destroyers are the largest warships that are built now. Given that in the show the rockets appear somewhat ineffective and the shells seem to be unblockable, I wonder why modern ships use mostly rockets.</p>
<p>Judging by their presentation of cyber warfare, the military accuracy is probably not good enough to extrapolate this. But if guided missiles are supposed to be so much better, did the destroyer have a serious problem with the battleship just for plot reasons or is that somewhat realistic?</p>
| |ships| | <p>Ship-to-ship combat is now restricted to the internet's various fandoms.</p>
<p>If you are close enough to launch a massive projectile to a nearby ship, you are close enough to get hit by one of it's lighter guided missiles. </p>
<p>Those guided missiles tend to be more accurate over distance and much lighter to carry. Missiles can also do the penetrate and then explode for more damage where the big-hunk-o-metal can only do penetration. </p>
<p>A well designed ship can still sail with a few holes in it.</p>
<p>Shooting a massive shell requires a lot more propellant than a self propelled projectile. And the launching rig needs to be very sturdy to catch the recoil.</p>
| 25758 | Why don't modern warships still use cannons? |
2019-01-31T20:47:18.667 | <p>From my office on the 21st floor of a building in Vancouver, BC two parking garages in the process of being demolished are visible. A third is just out of view a few more blocks away.</p>
<p>One of these structures is being taken down by a team consisting of an excavator with a jackhammer, an excavator with a claw and a man holding a hose. The jackhammer punches holes in the cement, and the claw breaks it up and allows it to fall to the floor below.</p>
<p>A couple of blocks away a different approach has been taken where it seems they are cutting the concrete into chunks and then lowering it down with a crane.</p>
<p>We've all seen the news clips and YouTube videos of demolition via pyrotechnics.</p>
<p>Are there other approaches?</p>
<p>Presumably, demolition by explosion requires a large amount of upfront time, effort and engineering skill as <a href="https://engineering.stackexchange.com/questions/16490/controlled-demolition-of-a-sky-scraper/16495#16495">detailed in this answer</a>. All of this culminates in one significant event, the detonation, and then substantial cleanup.</p>
<p>The jackhammer and claw approach seems like it is at the opposite end of the spectrum and requires relatively little skill and lots of labor to proceed at an excruciatingly slow and noisy pace.</p>
<p>Cutting and lowering seems to be in the middle with more expensive machinery to cut and skilled operators required. It also seems to move very slowly but anecdotally seems to have none of the noise associated with jackhammering.</p>
<p>What does the decision tree/matrix look like for picking an approach? Surely cost plays heavily into it but watching these sites move forward day by day I'm curious as to how two very similar sites in the same city (so permitting is presumably similar) chose different approaches and neither chose to use explosives.</p>
| |structural-engineering|civil-engineering|building-physics| | <p>According to university of Wolverhampton demolition masters degree literature, explosives are used where the shock is not likely to have detrimental effects on surrounding structures. Consideration of dust is important due to silicosis. The cost is high but relatively speaking not much different than other methods. Material from this method is usually a large mound which requires removal by the other two methods and also requires professionally qualified personnel to employ but they are very scarce and extremely well paid.</p>
<p>Cut and remove requires machine tools of higher rental costs but less expertise. Material may be removed to a specific site for disposal or reuse. There may be some hazardous material residue that should be dealt with outside the demolition area.</p>
<p>The crush and shift method is similar but dust and debris is less of a consideration than what lies underneath the foundations. With explosives there is need to consider the ground impact on subterranean services such as gas, water electricity and transportation tunnels. The other methods avoid this issue. </p>
| 25769 | What approaches are there for demolishing a structure and how is an approach chosen? |
2019-02-01T01:02:58.973 | <p>If I am carrying out a reaction by flowing 200 SCCM of a gas in a 20mm diameter pipe at 100mm length, if I change the pipe diameter to 10mm:</p>
<ol>
<li>Do I adjust the mass flow rate of the gases?</li>
<li>If so, how do you calculate this?</li>
</ol>
<p>Thank you for helping. Looking to learn.
Cheers</p>
| |flow-control|compressed-gases| | <p>It depends on if you have a smooth laminar flow with little friction and steady state flow, or transient, and or turbulent flow.</p>
<p>Assuming you pipe lining is smooth, by changing the inside diameter of the pipe to 10mm, you get the same mass flow rate but at four times the speed.
Bernoulli equation applies, conservation of kineticand potential energy.</p>
<p>But if your pipe lining is rough, then you may need to use Poiseuille's Law if the flow is laminar:</p>
<p><span class="math-container">$ Flow = \frac {P_1 -P_2}{R} \quad, \ R = \frac{8\eta*L}{\pi*r^4} \quad \eta = viscosity $</span></p>
<p>In many practical cases both rules apply, meaning you have pressure loss due to viscosity and you have pressure loss to constrictions, however for a short length as 100mm the friction loss is just nominal and the flow rate is constant.</p>
| 25771 | Do you adjust gas flow rate (SCCM) to volume of chamber? |
2019-02-01T11:10:22.763 | <p>In Mechanisms and Machines Theory we were introduced to Assur groups and structural analysis where a mechanism is subdivised into elementary Assur groups of different classes.
But what is really the practical use of it? </p>
| |mechanisms|robotics| | <p>Things have to be assembled. Assur group gives you a catalog of possible ways to make a rigid assembly out of joints. Each joint in this case represents a connection that will be assembled later, or locked down in case of foldable structures meant for reuse.</p>
<p>Being 0 degrees of freedom means that analysis of these is straight forward. Recognizing them will also make kinematic and dynamic analysis easier since you can replace them with a single rigid body.</p>
<p>Since assur groups can be predicted in a organized manner nothing stops you from generating a very big catalogue of these. Allowing you to systematically search for new solutions for the task at hand.</p>
| 25777 | What is the practical use of Assur Groups and structural analysis? |
2019-02-01T15:01:29.627 | <p>When performing a pavement design using AASHTOWare Pavement ME for AC on AC Overlay, the pavement distress prediction for bottom-up (alligator) cracking and AC thermal cracking seem to give incorrect results. </p>
<p>First, the Reliability for both these distresses seems fixed at 50%. It does not allow me to increase the value.</p>
<p>Secondly, if I go ahead and run the program for 10 years the two distresses almost always remain zero (or close to zero).</p>
<p>Any ideas on this?</p>
| |pavement| | <p>Note that you can change the reliability for total fatigue cracking and total transverse cracking which includes the cracking in the existing AC layer that reflects through the AC overlay and new cracks that develop in the overlay. </p>
<p>The reliability is set at 50 percent for new fatigue cracking in the AC overlay and new transverse cracks in the AC overlay. The reason the reliability is set to 50 percent is because it is impossible to separate new crack development in the AC overlay from the reflective fatigue and transverse cracks in the existing AC layer that reflect through the overlay. As such, the reliability is applied to the total fatigue and total transverse cracks which combines new crack development with reflective cracks.</p>
<p>To provide an answer to the second part of your question, more information is required. AC overlay thickness and the location of the site in terms of climate is important relative to version 2.3.1. If the site is located in the south or in a mild climate, version 2.3.1 will almost never calculate transverse cracks – they will be close to “0” when using the global calibration coefficient for transverse cracks. </p>
<p>If the AC overlay thickness is relatively thick, then it will take a long time for additional fatigue cracks to develop in the existing AC layer because the tensile strains are significantly reduced. The version to be released in July of this year has a coefficient for transverse cracks that is climate dependent.</p>
| 25778 | Why do the distress values for Pavement ME Design for alligator cracking and thermal cracking for AC pavements seem to give incorrect results? |
2019-02-01T18:47:03.187 | <p>Which is more economical to improve or change the soil that is used to build a structure?</p>
<p>my answer is that it depends on each case, but I would like to get more information, or where to search about it.</p>
<p>Thank you.</p>
| |fluid-mechanics|civil-engineering|geotechnical-engineering|soil|soil-mechanics| | <p>There is many factors leading to the decision of, either improving the existing soil by adding the missing aggregate and mixing and compacting, bringing it to optimal moisture, density and strength. Or it is better to remove it entirely and import good soil and compact it.</p>
<p>These are decisions that can only be answered by a licenced soils engineer, or if the remedy is known to your local building department they usually do a jobsite observation and give you their recommendation.</p>
<p>If they request a soils report then a soils engineer will drill exploratory holes into the ground and take enough samples as he deems needed to take to his lab and run the whole battery of tests on the samples. </p>
<p>Then he will compile a report and recommends on what needs to be done. Sometimes they recommend stabilizing the soils by adding the certain sizes of sand and gravel, sometimes they recommend replacing the bad strata. And sometimes they recommend a mix of both and also drilling deep pylons and building certain retaining walls and concrete or steel soldier piles. </p>
<p>They give detailed specific instructions to modify the soil and design the foundation. and how to maintain the site in the future as far as drainage and surface water control.</p>
| 25782 | Soil mechanics-Geotechnical engineering |
2019-02-02T18:14:59.667 | <p>I'm a neuroscientist interested in building a device to replicate the following study.</p>
<p><a href="https://www.nature.com/articles/s41598-018-24924-9" rel="nofollow noreferrer">https://www.nature.com/articles/s41598-018-24924-9</a></p>
<p>Essentially, they sent air through a nose catheter to stimulate mechanically sensitive cells of the nose. They had 8 seconds of continuous air, followed by 12 seconds off (0.05 Hz). This was to mimic the breathing associated with deep meditation, and they report it induced alterations to consciousness.</p>
<p>I'm interested in building a device that can replicate their paradigm, as well as send air bursts at an arbitrary frequency of my choosing. Ideally the range should be between very slow (~0.05 Hz) and higher frequencies (~100 Hz). </p>
<p>I understand pulsing most standard pumps won't work for those higher frequencies. My mechE friend proposed some two way valve that would be able to pulse between two directions, the nose catheter, or just out to the air. I could then just use a standard continuous pump or fan.</p>
<p>I'm unfamiliar with the terminology and haven't been able to find such an electrical valve. Does anyone know of any such valves, or perhaps has a better design in mind?</p>
<p>Thanks everyone!</p>
| |mechanical-engineering|electrical-engineering|valves| | <p>The RDWorks forum has a discussion which covers adding an <a href="https://rads.stackoverflow.com/amzn/click/com/B019PTHS7Y" rel="nofollow noreferrer" rel="nofollow noreferrer">air diverter valve</a> that can be operated by the Ruida class of laser cutter controllers. It is a simple matter of directing a relay to the line that trips the solenoid of this valve.</p>
<p>Priced at US$22.49 currently, it won't break the bank. Combined with an arduino and a MOSFET to handle the 24v load current of the coil, you should be able to regulate the airflow in the manner you describe.</p>
<p><a href="https://i.stack.imgur.com/62AJW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/62AJW.jpg" alt="air valve"></a></p>
| 25792 | Build Device for Emitting Weak Air Bursts at Specified Frequency |
2019-02-04T12:25:10.307 | <p>I am working on a telemetry system where a liquid level sensor (Hydrostatic pressure sensor) is used to monitor the volume of fuel (gasoline / diesel) in an underground storage tank.</p>
<p>The sensor has to immersed in the fuel and has a cable that is to exit the tank and then be connected to an acquisition unit housed in a panel.</p>
<p>The main problem i will like ideas on is that, the sensor cable needs to form a liquid-tight seal at the point it exits the tank.</p>
<p>I have come across a component called a CordGrip / Cable gland but will like to know if any other methods exist</p>
<p>The final solution should take into account that the exit point of the cable is exposed to the outside environment of sunshine and rain</p>
| |sensors|process-engineering|instrumentation| | <p>As this is a fuel tank, you need to consider explosion hazards. If I understood you correctly, you plan to place a pressure sensor low in the (underground) tank and have the cable enter the tank near the top. This means the seal will not only need to be tight against fuels, but also against fuel vapor. I don't know where you are, so I don't know which ex-protection codes apply if any. Make sure YOU know which codes are relevant and follow them.</p>
<p>If you use cable ducts, make sure they are compliant with the relevant codes and stand up to a the fuel-ladden atmosphere (organic solvents attack some plastics).</p>
<p>Other ways to solve your problem:</p>
<ul>
<li>screw the pressure sensor directly into a socket on the tank (<a href="https://www.vega.com/en/home_us/products/product-catalog/pressure/process-pressure/vegabar-14" rel="nofollow noreferrer">possible product</a>), this is likely impossible in your case as the tank is underground.</li>
<li>screw a pressure transmitter into a socket at the top of the tank, have the actual sensor hanging at a steel rope into the fluid (<a href="https://www.vega.com/en/home_us/products/product-catalog/level/hydrostatic/vegabar-87" rel="nofollow noreferrer">possible product</a>)</li>
<li>Use a different level mesurement technology like radar or ultrasonic that measuers from the top (though that is likely more expensive, but you can maintain the sensor more easily). The beauty of radar is that with some tank materials (not metals obvs.), you can measure through the tank so the whole sensor does not contact your medium, there's one less seal you need to worry about re. explosive vapors and so on.</li>
</ul>
<p>I would go with option two as a compromise between ease of service and cost. Find a reputable vendor of measuring devices near you and talk to their application engineer. </p>
| 25823 | How to create a Liquid tight cable seal in a fuel storage tank |
2019-02-04T16:16:59.107 | <p>I'm designing a precision adjustment bracket, and I'd like to have a component that allows the field tech to make small adjustments to push and pull on each axis of adjustment. </p>
<p>I went looking to see how others have solved this problem and found an assembly that seems common to several precision adjustment assemblies. I'm a bit embarrassed to say that, as a degreed mechanical engineer.... I don't know what these are called. </p>
<p><a href="https://i.stack.imgur.com/9ZZn2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ZZn2.png" alt="Adjustment assembly 1"></a></p>
<p>I would say that this is a captive screw that threads into a barrel nut, but that doesn't seem to be it. At least, I can't turn up anything suitable on <a href="https://www.mcmaster.com" rel="nofollow noreferrer">McMaster's site</a> when I try it. Pro tip - Probably don't Google "captive <em>bolt</em>." Apparently <em>that</em> name is more commonly used to refer to the device used to slaughter animals.</p>
<p>I also can't seem to find barrel nuts that have a threaded end as pictured in the assembly; <a href="https://en.wikipedia.org/wiki/Barrel_nut" rel="nofollow noreferrer">barrel nut</a> seems to be exclusively the drop-in fastener used for a lot of flat-pack furniture. </p>
<p>Here's another view of the same item from above:</p>
<p><a href="https://i.stack.imgur.com/eS51O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eS51O.png" alt="Adjustment assembly 2"></a></p>
<p>And then what I believe is the same style assembly in a different product:</p>
<p><a href="https://i.stack.imgur.com/jl60g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jl60g.png" alt="Adjustment assembly 3"></a></p>
<p>In all the brackets where it's not just "loosen the nut and tap with a wrench until it's generally correct," the assemblies I've shown seem to be the standard solution. I'd like to incorporate it into my solution, but <em>I need to know what it's called</em> to be able to source the parts. </p>
| |mechanisms|terminology| | <p>I never quite found exactly what I'm looking for, but I did come close, so I'll post what I've got. If anyone ever does have a better answer, please post it and I'll accept that one as correct. </p>
<p>The parts seem to go by a lot of names, but <a href="https://engineering.stackexchange.com/questions/25826/what-is-this-captive-screw-and-barrel-nut-adjustment-assembly-called-picture-at?noredirect=1#comment47559_25826">Phil Sweet's comment</a> led me down the track to get where I am. An IPB, "illustrated parts breakdown," of a reel mower showed a similar assembly where they called one part a <strong>pivot pin</strong>. I had tried "swivel pin" to no avail, but <strong>pivot pin</strong> started to turn up some results. </p>
<p>I'm not affiliated with the following site, but they seemed to be the only people I could find that had a McMaster-style site that let me configure the parts by material grade, thread spec, length, etc., so I'll mention them here for that reason: <a href="https://us.misumi-ec.com" rel="noreferrer">https://us.misumi-ec.com</a></p>
<p>They sell the components for an assembly that they depict in their catalog as follows:</p>
<p><a href="https://i.stack.imgur.com/llDS9.png" rel="noreferrer"><img src="https://i.stack.imgur.com/llDS9.png" alt="Adjustment Assy"></a></p>
<p>The assembly is made up of an "adjusting bolt":</p>
<p><a href="https://i.stack.imgur.com/9R6WP.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/9R6WP.jpg" alt="Adjusting bolt1"></a> </p>
<p><a href="https://i.stack.imgur.com/ZtW2T.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/ZtW2T.gif" alt="enter image description here"></a></p>
<p>The "neck" on the adjusting bolt, noted with dimension "T" above, sits in a "block for adjusting bolts":</p>
<p><a href="https://i.stack.imgur.com/wvzd4.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/wvzd4.jpg" alt="Blocks for adjusting bolt"></a></p>
<p>The block wraps around the "neck" of the adjusting bolt, such that the bottom of the hex head cap is on one side of it and the top of the shoulder is on the other side. Once you put the adjusting bolt into the slot in the block it can't move axially. </p>
<p>Then you thread the adjusting bolt into the "Brackets for Adjustment Screws Bolt Type":</p>
<p><a href="https://i.stack.imgur.com/KGQYS.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/KGQYS.jpg" alt="Bracket for adjustment screws bolt type"></a></p>
<p>My issue with this setup is that the block can't swivel/pivot, so it wasn't of any use for me. I continued to look for a while and found a relatively common part for tractors called a <strong>stabilizer stub pin.</strong> This is used in the three-point hitch that attaches towed accessories to the tractor:</p>
<p><a href="https://i.stack.imgur.com/Ybhre.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Ybhre.jpg" alt="Stabilizer Stub Pin"></a></p>
<p>These pins have a smooth bore, so the adjusting screw could slide into it, <strong>BUT</strong> there isn't any way to get it to work with the neck/shoulder on the adjusting bolt. I thought that I might be able to use just a standard bolt if it had a machined groove in it; my idea was that I could drop the bolt into the stabilizer stub pin and then use a circlip/snap ring/retaining ring to act as the shoulder on the Misumi adjusting bolt. </p>
<p>Googling around led me back to Misumi's site, actually, for their "bearing shaft screws":</p>
<p><a href="https://i.stack.imgur.com/5sEd6.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/5sEd6.jpg" alt="Bearing shaft screws"></a></p>
<p>Ultimately I'm backing out and am going to take a different approach entirely to the problem. I need relatively low-cost assemblies, but they need to be ALL stainless steel because of the installation environment. The only way to get the pivot on the bolt is to use the stabilizer stub pin and the bearing shaft screws, but the stub pin is not stainless and the bearing shaft screw is prohibitively expensive because it's machined to actually provide a bearing surface. </p>
<p>For the record, too, you can find these assemblies pretty easily if you look for "chainsaw tensioners," but again I <em>need</em> the stainless steel material so they were out of the question for me. </p>
<p><a href="https://i.stack.imgur.com/ed3sU.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ed3sU.jpg" alt="Chainsaw tensioner"></a></p>
| 25826 | What is this captive screw and barrel nut adjustment assembly called? Picture attached |
2019-02-05T17:25:15.373 | <p>I'm thinking of making a RC tank that should be simple as possible. But I'm not sure how to make proper steering for the tank. I googled about tank steering and found (for me) complex solutions as dual differential steering.
Now, my question is - Is it possible to steere the tank with momentum?
Idea for project is - Left motor will power left Wheele(tracks are too complicated to design), right motor will power right Wheele. Powered Wheele are at the back. So if I want to steere the tank to the left side, right motor will be on, left will be off. Does this method Works? Should right motor go forward but left go backward?</p>
<p>Here is the sketch. This is sketch for rotating/steering from place.
I guess it will work fine, but what if tank moves forward/backward and wants to steere?</p>
<p><a href="https://i.stack.imgur.com/FKrPo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FKrPo.png" alt="The sketch"></a></p>
| |mechanical-engineering|mechanisms| | <p>You want to either use swivel pivot on the front wheels like shopping cart front wheels, or place the CG of your toy near the rear end. </p>
<p>Or else the steering will not be very effective.</p>
<p>If you do that by applying different power to left or right rear wheels you get turned to right or left.</p>
<p>If you watch delivery fork lifts of say Home Depot, you see they can maneuver precisely into tight corner. Their front wheels are free to pivot. And back wheels are attached to a joy stick.</p>
| 25850 | Steering with momentum |
2019-02-07T09:42:21.687 | <p>Please suggest some effective repair solution for spalling overhead slab due to high corrosion in rebars.</p>
<p><img src="https://i.stack.imgur.com/mEp6V.jpg" alt="Spalling Slab with corroded rebars"></p>
| |structural-engineering|civil-engineering|reinforced-concrete| | <p>It's eminently dangerous. I would immediately inform the authorities of unsafe condition of the structure and request immediate evacuation of the building, had I observed something like this.</p>
<p>Next step would be shoring up by a qualified contractor and adding beams or demolition and reconstruction of the slab per instructions of an engineer.</p>
<p>I would try to find out the reason for this defect to correct it for reconstruction.</p>
| 25865 | Finding good and effective solution for overhead RCC slab spalling due to corrosion in Rebars |
2019-02-09T16:19:03.940 | <p>requirements:</p>
<ul>
<li>a panel of wood 4 ft long by 2 ft wide laying flat</li>
<li>this panel should be able to tilt side to side (2 way)</li>
<li>the weight on the panel will be about 250 lbs up to 400 lbs.</li>
<li>controlled lift of each side should be max 3 inches</li>
</ul>
<p>Solutions investigated:</p>
<ul>
<li>pneumatic cylinders (price unknown)</li>
<li>air suspension (for cars) (roughly $250)</li>
</ul>
<p>Any ideas how this can be done without breaking the bank?</p>
| |mechanical-engineering|control-engineering|pneumatic| | <p>Fix a lever to the panel and lift or lower one side by hand. Hook it into stops at all the positions you want to hold it at.</p>
| 25900 | Ideas needed: how to tilt a panel side to side |
2019-02-10T13:35:36.860 | <p>For a two node beam element there are four shape functions for four degree of freedom:
<a href="https://i.stack.imgur.com/7TqiA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7TqiA.png" alt="Hermitian Shape function for two node beam"></a></p>
<p>For a straight three node beam element how shape functions are?</p>
<p>Please note that beam is straight and not curved.</p>
| |structural-engineering|beam|finite-element-method| | <p>If the nodes are at <span class="math-container">$\xi = -1, 0, +1$</span> you can find the shape functions using <a href="https://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow noreferrer">Lagrangian polynomial interpolation</a>.</p>
<p>In fact you don't need to work through the general procedure, since you can write down the general form the shape functions must take with only a few unknown parameters, and then solve for the unknown values.</p>
<p>Consider the shape functions that are non-zero, and have non-zero derivative, at <span class="math-container">$\xi = 1$</span>. They must be zero, and have zero derivatives, at <span class="math-container">$\xi = -1, 0$</span>. Therefore they must have the general form
<span class="math-container">$$N(\xi) = (\xi+1)^2 \xi^2 (a\xi + b)$$</span>
for some values of <span class="math-container">$a$</span> and <span class="math-container">$b$</span>.
Using the product rule, the derivative is
<span class="math-container">$$N'(\xi) = 2(\xi+1)\xi(2\xi + 1)(a\xi + b) + (\xi+1)^2 \xi^2 a.$$</span>
At <span class="math-container">$\xi = 1$</span> we therefore have
<span class="math-container">$$\begin{align}N(1) &= 4a + 4b \\
N'(1) &= 16a + 12b\end{align}$$</span></p>
<p>For one shape function we want <span class="math-container">$N(1) = 1, N'(1) = 0$</span> and for the other, <span class="math-container">$N(1) = 0, N'(1) = 1$</span>. </p>
<p>Solving the simultaneous equations for <span class="math-container">$a$</span> and <span class="math-container">$b$</span> gives the two shape functions as
<span class="math-container">$$\begin{gather}(\xi+1)^2 \xi^2 (-3\xi+4)/4 \\
(\xi+1)^2 \xi^2 (\xi-1)/4 \end{gather}$$</span></p>
<p><a href="https://i.stack.imgur.com/nv75Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nv75Q.png" alt="enter image description here"></a></p>
<p>You can get the shape functions that are nonzero at <span class="math-container">$\xi = -1$</span> simply by changing <span class="math-container">$\xi$</span> to <span class="math-container">$-\xi$</span> (be careful with the sign of the non-zero slope shape function!)</p>
<p>The shape functions for the middle node can be found in the same way, but if you see what is going on you can just write them down by inspection:
<span class="math-container">$$\begin{gather}(\xi-1)^2(\xi+1)^2 \\
(\xi-1)^2 \xi (\xi+1)^2 \end{gather}$$</span></p>
| 25909 | Shape Functions of beam element with 3 nodes (quadratic element) |
2019-02-10T17:46:32.620 | <p>The Washington DC water treatment plant appears to use sedimentation tanks for both treating water before consumption and for treating sewage that is then discharged.</p>
<p>My question: What happens to the sedimentation after it is collects? None of the articles that I have read discuss this.</p>
<p>It seems to me that the sediment in the water source won't necessarily be toxic, but I suspect that the sediment in the sewage contains a brew of toxic metals and organics. So does that need to be treated as toxic waste? Is the source sediment used for fertilizer, or is it just sent to a landfill? Is there an issue with dust?</p>
| |water-resources|waste-water-treatment| | <p>I was part of a team of contractors to remove and spread the dried sediment from a water treatment facility - the sediment was a red cake/ powder and was spread on the fields as fertiliser. </p>
| 25913 | What happens to the sedimentation collected in water treatment plants? |
2019-02-11T11:03:40.087 | <p>I have a project which involving several mating parts and also shaft guided through bearings, my issue here is that I don't have much experience with tolerances and manufacturing processes and I cannot determine the perfect balance between cost and accuracy. I am using Aluminium 6061 for raw material. For example I have some guidance "shafts" which are "made" by removing a large material from the part, I made some research and I saw that a +- 0.025 mm is a typical tolerance for a decent CNC machine. My question is, could I go down with that value, in terms of cost or is already a tight tolerance ?</p>
| |mechanical-engineering|tolerance|cnc|prototyping| | <p>It's going to be an uphill battle from that point on. While you can get every part of the machine and the process to tighter tolerances, their errors may compound to the sort of tolerance of 0.025; you may manufacture relatively simple parts with lower error but whether you succeed or not is more up to luck than your efforts - whether the tolerances of the tool, the table drive, the head drives, the frame rigidity, the sliders rigidity, and the material itself add up and ruin the accuracy or cancel each other resulting in near-perfect workpiece.</p>
<p>One nasty property to consider is the internal stresses of the material, resulting from the manufacturing process. If you remove large parts of it, the stresses that were canceled out by say, two outer surfaces pulling with equal strength against the central bulk pushing, will result in the workpiece ending as soon as you unlatch it. I had worked with 7mm brass sheets, and upon removing most of material from one side, I't observe good 2cm of deflection on a 30cm bar. </p>
<p>Regardless, if you aim at sub-0.01mm accuracy in all domains you can control (tool shape, material alignment, table flatness and machine being level), and the machine is good quality, you can expect better than 0.03mm accuracy in the result. If you want better, you'll need to get into the "guts" of the process - drives of the head and table, bearings, vibration dampening etc.</p>
| 25920 | Mechanical tolerances on a CNC machine |
2019-02-12T18:22:14.380 | <p>for this question: "What is the difference between strength, hardness and toughness in materials?" i have searched and have found these following definitions</p>
<blockquote>
<p>Strength refers to resistance to deformation, and also to a large
elastic range. In the Elastic region of the stress-strain
relationship, the relationship is described by a linear function, such
that σ = E ϵ, where σ is the stress, E is the Elastic modulus, and ϵ
is the strain.</p>
<p>Toughness is the resistance to failure or crack propagation. It is
somewhat related to strength. Very strong materials will have low
toughness, i.e. low tolerance for flaws or defects, i.e. incipient
cracks.</p>
<p>Reference
<a href="https://www.physicsforums.com/threads/difference-between-toughness-and-strength.67220/" rel="nofollow noreferrer">https://www.physicsforums.com/threads/difference-between-toughness-and-strength.67220/</a></p>
</blockquote>
<p>i don't understand those definitions. Aren't deformation and failure one? Are toughness, strength and hardness both the ability to resist external forces?</p>
| |materials| | <ul>
<li>Strength = Stress required to produce failure. It is often used to determine how much force an object can take without breaking, but there are many measures of strength.</li>
<li>Hardness = Force / Area. The larger the area produced by a given force, the lower the hardness. It can be correlated to, but is not the same as strength.</li>
<li>Toughness = Energy required to form a crack. Ceramics have low toughness because cracks form easily.</li>
</ul>
| 25943 | Strength vs. Hardness vs. Toughness |
2019-02-12T21:24:57.607 | <p>I have a question that I can't solve.
Coefficient of drag= x,
Coefficient of Lift= y,
it is given that x= 0.015 + 0.05*(y^2), FInd Max Lift to drag ratio</p>
<p>and I know that Lift to drag ratio is L/D= Coefficient of lift/Coefficient of drag. Can anyone tell me how to solve this</p>
<p>Thanks in advance</p>
| |aerodynamics| | <p>As has been mentioned above by Eric Shain the correct answer is to take the partial with respect to y= CL and equate it to zero. Let's Call the max L/D, M.</p>
<p><span class="math-container">$$M = \frac{C_L}{C_D} = \frac{C_L}{C_{D_0} +K*C_L^2}, \quad C_{D_0}=0.015\ and\ K=0.05 $$</span></p>
<p><span class="math-container">$$ \frac{\partial M}{\partial C_L} =\ \frac{\partial}{\partial C_L}
[ \frac{C_L}{0.015 +0.05*C_L^2}]\ =\quad \frac{[0.015 + 0.05*C_L^2
]-C_L [0 + 2(0.05)C_L^2]}{[0.015 + 0.05*C_L^2
]^2} =0$$</span></p>
<p><span class="math-container">$$ \Rightarrow 0.015- 0.05C_L^2=0\\ C_L^2=0.015/0.05=0.3\\C_{L_M}=\sqrt{0.3} \quad and\ C_{D_M}= 2*C_{D_0}=0.03$$</span></p>
<p>And <span class="math-container">$$M=\frac{C_{L_M}}{C_{D_M}}= \frac{\sqrt{0.3}}{0.03}=18.257$$</span></p>
<p>Maximum lift to drag ratio is 18.257. please double check my arithmetic, but it looks correct.</p>
<p><strong>Edit</strong>
The following is a quote from Wikipedia on lift to drag ratio. </p>
<blockquote>
<p>A House sparrow has a 4:1 L/D ratio, a Herring gull a 10:1 one, a Common tern 12:1 and an Albatross 20:1, to be compared to 8.3:1 for the Wright Flyer to 17.7:1 for a Boeing 747 in cruise.[4] A cruising Airbus A380 reaches 20:1.[5] The Concorde at takeoff and landing had a 4:1 L/D ratio, increasing to 12:1 at Mach 0.95 and 7.5:1 at Mach 2.[6] A Helicopter at 100 kn (190 km/h) has a 4.5:1 L/D ratio.[7] A Cessna 172 glides at a 10.9:1 ratio.[8] A cruising Lockheed U-2 has a 25.6 L/D ratio.[9] The Rutan Voyager had a 27:1 ratio and the Virgin Atlantic GlobalFlyer 37:1</p>
</blockquote>
| 25947 | Max Lift to drag ratio |
2019-02-14T00:57:04.287 | <p>I work in a pharmaceutical plant where saline solution is produced and bottled. The environment has to be clean (by pharma standards), which rules out many greases and other protective coatings.
As expected from a saline environment (from my experience it can be as agressive as a marine one), the corrosion over all parts in the room is heavy.</p>
<p>I have mostly controlled the corrosion of structures (all of them have to be made of stainless steel: 316L for those parts in contact with the product, 316 for some others and some structural parts of the equipment are made of 304 SS) by carefully selecting materials, specially material pairs (when I have to), introducing electrical discontinuities in structural joints (if I have a bronze piece, for example, in contact with a SS one, I introduce teflon washers/separators to reduce galvanic corrosion), and at least in one case by using a sacrifice anode.</p>
<p>But the corrosion of electrical components and wiring is something I haven't been able to control, to the point that I've had to replace all the wiring in one of the machines in as little as 5 months. Wires (not only power lines, but also control ones, such as termocouple wiring) get rusted not only in the terminals (where it's evident), but also under their plastic sleeves. In the case of control wiring, this causes noisy signals that result in machine stoppages and downtime (my most recent case is a termocouple that sent a correct signal of 180°C but showed (unreal) peaks of up to 270°C that, of course, triggered an alarm in the PLC).</p>
<p>I can only blame it on corrosion, as the problem goes away by replacing all the wiring and terminals. The wires I took off show corroded under the sleeves, and some of them were brittle. (If you think it's not corrosion, I'd like to hear your opinion).</p>
<p>I can't hide all the wiring from the product (in particular some of the termocouples my get some product spray) and my electrical boxes are as separated from product and tigth as possible (even they don't hold to any IP standard).</p>
<p>So my questions are:</p>
<ol>
<li>Is there anything I can do to prevent this sort of corrossion in electrical components and wiring?</li>
<li>Is there anything akin to a sacrifice anode that I can use in electrical parts / wiring?</li>
</ol>
| |electrical|corrosion|maintenance| | <p><strong>Conformal Coatings are your Friend!</strong><br>
I suggest you look into "conformal coatings" used to protect electrical and/or electronic components from corrosion. These coatings are commonly used on Naval electronics -- especially for gear used aboard submarines. Use as directed (well ventillated, away from potential spark/ignition sources until fully dry, etc.) on pretty much everything except electrical contacts and moving parts. </p>
<p>So long as you have different metals in contact, you've gotta keep moisture away or else you'll get electro-chemical corrosion (AKA ad-hoc battery.) FYI, such coatings do a pretty good job, but they make probing circuitry harder. I've had to grind sharp points on a set of DVM leads to pierce these coatings; they dry pretty tough, by design.</p>
| 25964 | corrosion in electrical parts and wiring |
2019-02-14T11:15:57.127 | <p>Below are two cropped views of "<a href="https://en.wikipedia.org/wiki/Aerial_telescope" rel="nofollow noreferrer">Johannes Hevelius's 8 inch telescope with an open work wood and wire "tube" that had a focal length of 150 feet to limit chromatic aberration.</a>" from <a href="https://iiif.lib.harvard.edu/manifests/view/ext:aHR0cHM6Ly9paWlmLmxpYi5oYXJ2YXJkLmVkdS9tYW5pZmVzdHMvaWRzOjE2NDcwMzE1$1i" rel="nofollow noreferrer">Harvard University, Houghton Library, pga_typ_620_73_451_fig_aa</a> (found <a href="https://hollis.harvard.edu/primo-explore/fulldisplay?context=L&vid=HVD2&search_scope=everything&tab=everything&lang=en_US&docid=01HVD_ALMA211764767520003941" rel="nofollow noreferrer">here</a>) the first of which I've zoomed and enhanced the contrast.</p>
<p>I've used these images in two other questions in Astronomy SE:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/29317/7982">How did Johannes Hevelius' long telescope work? Why all the round holes?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/29597/7982">How does making a refracting telescope very long reduce the chromatic aberration of an uncorrected lens?</a></li>
</ul>
<p><a href="https://astronomy.stackexchange.com/a/29324/7982">This answer</a> provides a <a href="https://archive.org/stream/Machinacoelesti1Heve#page/410/mode/2up" rel="nofollow noreferrer">link to the original text and image</a> but the discussion was published in 1673 and may be difficult to understand or paste into google translate.</p>
<p>However, the structure itself may be familliar to engineers. I'm getting the feeling that the series of square blocks with round holes and string or lines connecting them (presumably under tension?) have no specific optical function, and are purley part of a system to keep the long "telescope tube" stiff.</p>
<p>Is this combination of blocks and lines a recognizable implementation of some named structure or technique? </p>
<hr>
<p><a href="https://i.stack.imgur.com/iFTbX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iFTbX.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/1I0eh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1I0eh.jpg" alt="enter image description here"></a></p>
| |structural-engineering|terminology|engineering-history| | <p>Lateral stiffness is from the diamond stays and notched spreaders. The weight is distributed along the beam using a running "crow's foot" arrangement. The square blocks and holes are presumably there to aid in telling the rigger what they need to do to keep it all straight and true.</p>
| 25969 | Is the combination of blocks and lines in this old, very long telescope an implementation of some named structure or technique? |
2019-02-14T17:10:12.113 | <p>I have data from some ships, basically the base speed for a list of RPM. But it is know that this speed is not constant and it will change mainly based on whether and load of the ship. Is it any rule of thumb to stime the effect of this varialbes into a ship speed without going to heavy on the math.</p>
| |mechanical-engineering| | <p>Speed of a ship and power demand and fuel consumption have to do with many factors. </p>
<p>They are roughly enumerated as follows.</p>
<p>Ship's engine power and draft versus speed curve.</p>
<p>Seas' conditions: draft, roughness, undercurrents.</p>
<p>Wind and meteoritical conditions.</p>
<p>These information are sett into many navigation equipment on board of even small yachts.</p>
<p>They are taught as part of marine engineering courses.</p>
<p>Here is one article covering some of the general concepts and charts, <a href="https://ac.els-cdn.com/S2468013315300127/1-s2.0-S2468013315300127-main.pdf?_tid=a6cb3bd2-99a9-48b1-a97b-b00096995428&acdnat=1550170555_633ee92829157e4dd3a6b28a73d9d197" rel="nofollow noreferrer">On the estimation of ship’s fuel consumption and speed curve: A statistical approach</a></p>
| 25980 | How to estimate a ship speed based on weather conditions? |
2019-02-14T22:13:59.673 | <p>I'm designing a cooling system to cool nitrogen gas stream at 300°C with a mass flow rate of 15.6 g/s moving through an undulating channel in a cylindrical aluminum ring block with external fins. How much mass of aluminum or what special feature is needed to cool 15.6 g of nitrogen moving through and exiting to 25°C EVERY SECOND. It's to be noted that the block is immersed in a large pool of water, and that should be factored in as a cooling advantage. I'll love a method of calculation that allows flexibility with variables like mass flow rate, temperature, and channel diameter.</p>
| |thermodynamics| | <h1>Cooling a Gas in a Channel (Tube)</h1>
<p>Assume we have a cylindrical channel that has gas flowing in it. The channel is in a material that itself is cylindrical. We have a single pass tube heat exchanger or alternatively the equivalent of a 1-D extended system but with gas flow. The outside of the tube is exposed to air or water at a temperature <span class="math-container">$T_\infty$</span>. The energy balance at a given point <span class="math-container">$z$</span> for the flow of the gas through this heat exchanger is basically as below.</p>
<p><span class="math-container">$$ \dot{m}\tilde{C}_p \frac{dT}{dz} = - U P (T - T_\infty) $$</span></p>
<p>In this, <span class="math-container">$P$</span> is the perimeter area of the inside of the tube, <span class="math-container">$P = 2\pi r_{i}$</span>. The factor <span class="math-container">$U$</span> is the <a href="https://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html" rel="nofollow noreferrer">overall heat transfer coefficient</a> (based from the inside of the tube). It is a combination of the convection of the gas in the tube, the conduction through the tube, and the convection outside the tube.</p>
<p>Recast this in a dimensionless form with <span class="math-container">$\Theta = (T - T_\infty)/(T_h - T_\infty)$</span> and <span class="math-container">$Z = z/L$</span> where <span class="math-container">$T_h$</span> is the entering hot temperature and <span class="math-container">$L$</span> is the tube length. Allow <span class="math-container">$\beta = UPL/\dot{m}\tilde{C}_p$</span>.</p>
<p><span class="math-container">$$ d\ln(\Theta) = - \beta\ dZ $$</span></p>
<p>With the boundary condition <span class="math-container">$\Theta(Z=0) = 1$</span>, the answer is <span class="math-container">$\ln(\Theta) = -\beta Z$</span>. This gives the temperature profile of the gas in the tube. See <a href="https://engineering.stackexchange.com/questions/25937/how-can-i-calculate-the-amount-of-heat-my-copper-water-trap-can-dissipate/25977#25977">this post</a> for a comparable question and answer.</p>
<p>For the system proposed, <span class="math-container">$T_h = 300$</span> <span class="math-container">$^o$</span>C. The value of <span class="math-container">$T_\infty$</span> should be BELOW 25 <span class="math-container">$^o$</span>C. Otherwise, cooling could eventually require an infinite area heat exchanger (in essence an infinite length tube). Assume <span class="math-container">$T_\infty = 0$</span> <span class="math-container">$^o$</span>C, and the goal is <span class="math-container">$\Theta = (25 - 0)/(300 - 0) = 0.083$</span>. Use a 1 mm diameter tube with values of <span class="math-container">$\dot{m} =$</span> 16 g/s, <span class="math-container">$\tilde{C}_p = 1$</span> J/g <a href="https://www.engineeringtoolbox.com/nitrogen-d_977.html" rel="nofollow noreferrer">for nitrogen</a>, and <span class="math-container">$U = h_a = 50$</span> W/m<span class="math-container">$^2$</span> <span class="math-container">$^o$</span>C. This latter assumption says that gas flow in the channel is the limiting value. The result is</p>
<p><span class="math-container">$\beta = UPL/\dot{m}\tilde{C}_p = (50)\pi(0.001)L/((16)(1)) = 0.010 L$</span></p>
<p><span class="math-container">$\ln(0.083) = -0.010 x \Rightarrow x = 250$</span></p>
<p>You need a 1 mm diameter tube that is 250 m long with walls that are cooled at 0 <span class="math-container">$^o$</span>C.</p>
<p>Every 10x increase in heat transfer coefficient cuts the required length by 10x. Every 10x decrease in flow rate cuts the required length by 10x. Decreasing the external temperature <span class="math-container">$T_\infty$</span> gives a modest decrease in required length.</p>
<h1>Flowing Gas Thru a Tube</h1>
<p>The velocity of the gas through the tube is <span class="math-container">$v = \dot{V}/A$</span>. Use <span class="math-container">$A = \pi D$</span> where <span class="math-container">$D$</span> is the diameter. The volumetric flow of an ideal gas is related to its mass flow as below with <span class="math-container">$T$</span> as temperature, <span class="math-container">$M$</span> as molar mass, and <span class="math-container">$p$</span> as pressure.</p>
<p><span class="math-container">$$ \dot{V} = \dot{m} \frac{RT}{Mp} $$</span></p>
<p>Consider 1 bar at 300 <span class="math-container">$^o$</span>C with nitrogen to obtain</p>
<p><span class="math-container">$$ \dot{V} = \dot{m} \frac{(8.314)(573)}{(28)(101325)} = 0.00170 \dot{m}$$</span></p>
<p>Use the value 16 g/s to obtain <span class="math-container">$\dot{V} = 0.027$</span> m<span class="math-container">$^3$</span>/s. In a 1 mm diameter tube, the required gas velocity becomes 8.6 m/s.</p>
<p><span class="math-container">$$ v = 0.027 / ((\pi)(0.001)) = 8.6$$</span></p>
<p>At this point, we are left to find the required pressure drop to sustain 8.6 m/s through a 250 m long tube with a 1 mm internal diameter. This involves a <a href="https://en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation" rel="nofollow noreferrer">review of the friction factor</a> for fluids in pipe flow. The gas contracts as it cools. This demands further investigation too, because the basic calculations with incompressible fluids will fail with such cases.</p>
<h1>Other Options</h1>
<p>The analysis used a 1 mm diameter tube. Increasing the tube diameter will decrease the required length and decrease the required velocity. The issue at some point will be that a gas does not sustain a uniform temperature profile. It will cool along the wall of the tube but less at the center.</p>
<p>The analysis used only one tube. An alternative is to flow the same mass (volume) of gas through a multi-channel system. With <span class="math-container">$N$</span> tubes, the contact area increases by <span class="math-container">$N$</span>. This means, the required tube length decreases by <span class="math-container">$N$</span>. This alternative says, rather having the gas flow through one longer channel at higher speed, you have the gas flow through N channels at 1/N speed because area increases by N. The contact time calculated as area per volumetric flow is the same. The appreciation of why we flow gases through mesh grids to heat or cool them rapidly strikes home. As for the pressure drop, this issue should be addressed in a separate discussion.</p>
<p>A gas can be cooled by <a href="https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect" rel="nofollow noreferrer">adiabatic expansion</a>. An effective design may just be to allow the hot gas at high pressure to expand rapidly to lower pressure.</p>
<p>Finally, analysis <a href="https://engineering.stackexchange.com/questions/26086/calculating-heath-exchange-for-custom-boiler/26096#26096">at this problem</a> shows an example of how a heat exchanger design equation is used for fluid flow in a pipe surrounded by a constant temperature fluid. This may be an easier approach to take from a practical standpoint.</p>
<h1>Cooling a Gas Flowing in a Channel through a Block</h1>
<p>The first part of the analysis above gives you the LENGTH OF A TUBE that is required to cool a gas to a given end temperature when the tube walls are held at a constant (lower) temperature.</p>
<p>The second part of the analysis above gives you the GAS VELOCITY that is required to pass the gas through the tube at a specific mass flow rate (g/s).</p>
<p>Assume that you will make your channel with a diameter <span class="math-container">$D$</span>. Assume the block is cooler than your desired end temperature. Determine the length of channel that you need using the first part (the equation for <span class="math-container">$\Theta$</span> versus <span class="math-container">$Z$</span>). Determine the gas velocity you need to support the desired mass flow rate using the second step. Finally, follow up with pressure drop calculations for gas flow in pipes. I suspect you will find out that you likely will not be able to get the velocity that you need in a single channel because the pressure drop that you will need across such a long channel is so high as to be physically unreachable. Also, any curves in the channel only act to increase the required pressure drop. So, a "curvy" channel will be far longer but require far higher pressure to force the gas through.</p>
<p>In the end, you may have to realize that a better option to cool the amount of gas you want as rapidly as you want is to run the gas through a multi-channel (porous) solid that is held at a lower temperature than you need.</p>
<h1>Heating a Mass of Solid in a Given Time (Optional)</h1>
<p>Admittedly, the first part of the question is likely not addressed in the above. Specifically, what time is needed for a mass of a solid to absorb a specific amount of heat (from whatever source).</p>
<p>The approach to answer this first question depends on making one or another of many different assumptions. When the gas at constant temperature is in full contact with the solid, the problem is an unsteady state convection + conduction problem. It may require analysis comparable to the use of <a href="https://en.wikipedia.org/wiki/Heisler_chart" rel="nofollow noreferrer">Heisler charts</a>. When the gas is in full contact with the solid but also changing its temperature, the Heisler chart analysis is replaced by an incremental or numerical integration approach (solve the Heisler chart at t, increment by dt and solve again). Finally, when the gas is flowing through the solid and the solid is not maintained at a constant temperature, one goes to the need for a Heisler chart or its foundations plus time integration plus position analysis.</p>
| 25985 | Cooling Hot Nitrogen with an aluminum block |
2019-02-15T17:39:55.580 | <p>I am working on a 1.5-meter satellite dish with a custom receiver cantenna-like feed. The dish didn't come with feed arms, so I need a way to hold the feed into place. What I tried was using a 1-meter threaded rod. My threaded rod was 1m (length) x D=14mm (thickness). I used nuts to hold the feed in place and held the threaded rod from the little hole in the center of the dish (again, using nuts). This has the advantage of allowing me to adjust the position of the feed (focal length) so I can play around and experiment to see where my signal strength is at its peak.</p>
<p>Another advantage is that I don't need to use metallic arms (like many large dishes do) to keep it in place, as they can block a (small) portion of the radio waves from hitting the reflector and reduce its gain.</p>
<p>But there's a problem which leads me to come here and ask for help: even slightly touching the feed (or the edge of the rod) will have it wobble around for quite a while. This means that if there's a small wind, the feed will be wobbling around, which is not good.</p>
<p>My question is: what can I do about this? Can I use another type of rod, or make any adjustments to my current threaded rod in order to stop it from wobbling back and forth by the slightest touch? Could using a thicker threaded rod (the hole is 1 inch = 2.54 cm, but I could always drill a larger hole if needed), or would that not fix the problem?</p>
<p>A few pictures of the dish:</p>
<p><a href="https://i.stack.imgur.com/jqfkJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jqfkJ.jpg" alt="https://cdn.discordapp.com/attachments/417338610191302656/530112919938269194/image0.jpg"></a></p>
<p><a href="https://i.stack.imgur.com/CwSft.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CwSft.jpg" alt=""></a></p>
<p><a href="https://i.stack.imgur.com/kwWqG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kwWqG.jpg" alt="https://cdn.discordapp.com/attachments/417338610191302656/530112852120567829/thumbnail_20181202_164746_resized.jpg"></a></p>
<p>Any ideas are appreciated!</p>
| |mechanical-engineering|applied-mechanics| | <ol>
<li><p>Increasing the rod thickness would be very effective. If I've visualizing the configuration correctly, the rod is acting as a cantilevered beam (i.e., a beam rigidly attached at one end):
<br><img src="https://i.stack.imgur.com/oxlSk.png" width="300"><br> The lateral stiffness (which governs the vibration amplitude of such a cantilevered beam) scales up with the area moment of inertia, which remarkably has a <em>fourth-power</em> dependence on the diameter. This means that you could double the stiffness with only a 19% increase in rod diameter, and doubling the diameter would increase the stiffness (and suppress the vibration amplitude) by 16×.</p></li>
<li><p>What about stabilizing the end of the rod by attaching nonconductive tensioned cord or string stretching back to a support?<br><img src="https://s3.amazonaws.com/answer-board-image/0d9258c3-11a4-4b3a-a062-77f7f3792f7f.png"><br> This is how tall and narrow towers (which would otherwise be unacceptably laterally compliant) are stabilized. The limit of the tensioning load would be the buckling threshold of the rod (if the tension points back to the base; therefore, try to orient the tension loads so that they tend to elongate the rod).</p></li>
<li><p>(Whimsical) If you have substantially excess time and money, you might add a set of strain gauges and actuators to the length of the rod to detect vibrations and suppress them, respectively. One example of such a "smart structure" is <a href="https://patents.google.com/patent/US5631437" rel="nofollow noreferrer">a cannon barrel that can stabilize itself against vibration</a> as the shell moves through it, thus increasing accuracy, for example. Overkill?</p></li>
</ol>
| 25999 | Stopping/reducing wobbliness in threaded rod |
2019-02-16T08:28:12.957 | <p>I have one tile that came off. I got a replacement tile, and some grout. But the online guides say you need mortar to glue the tile to the ground. Can I use the grout as mortar? I want to save money so I only want to buy one thing.</p>
| |civil-engineering| | <p>Yes you can do it, is simple answer, if what you have is sanded grout.</p>
<p>If you can add a little acrylic glue available in 4oz tubes at building material centers for a couple of dollars it will be even better.</p>
<p>Alternatively you may be able to ask to take a torn bag of ThinSet for free, such as I have done. They actually encourage this to keep the shelves clean.</p>
<p>The mortar or what is known by many in trade as Readymix, is for applications where you may need to level more than one inch thickness. or your tiles or marble is bigger than 12x12 inches for it offers structural volum integrity and easier trawling.</p>
| 26007 | Can I use grout as mortar? |
2019-02-16T11:29:18.870 | <p>I am currently working on desalination project for my final year. The design is based on flashing of hot liquid in low pressure vessel and then condensing the extracted the steam to produce pure water.</p>
<p>The problem I am facing is that there doesn't seem to be any thermodynamic data (under saturation dome) available for saline water. The need for saturation data is because I need to calculate the required mass flow rate based on daily desalination goal.</p>
<p>For example, if the goal is to produce 100 L/day with average sunlight of 8 hrs, then desalination rate must be 0.0035 kg/s. If I know the quality (percentage of steam produced) of the liquid gas mixture, I can easily calculate the mass flow rate of saline water corresponding to desalination goal.</p>
<p>The only data I could find related to this problem was from a MIT resource <a href="http://web.mit.edu/seawater/" rel="nofollow noreferrer">Thermophysical properties of seawater</a>, but unfortunately it only provides a single value for enthalpy at any given saturation temperature or pressure.</p>
<p>Right now, I can see only three options:</p>
<ol>
<li>Use steam table, and consider the values an approximation and incorporate the error produced in my design by slightly overdesigning</li>
<li>Consider the enthalpy of seawater (from MIT resource) as enthalpy of saturated liquid (i.e. <span class="math-container">$h_f$</span>), and enthalpy of vapor (from steam table) as <span class="math-container">$h_g$</span> and then perform my calculations</li>
<li>Thirdly, I could take the enthalpy of seawater (from MIT resource) as the value that is halfway between <span class="math-container">$h_f$</span> and <span class="math-container">$h_g$</span> (from steam table), and then calculate the approximated value of <span class="math-container">$h_f$</span>.</li>
</ol>
<p><span class="math-container">$h_f=$</span> Enthalpy at saturated liquid line;
<span class="math-container">$h_g=$</span> Enthalpy at saturated vapor line</p>
<p>Any help/guidance in this regard will be highly appreciated</p>
| |thermodynamics|water-resources| | <p>The EES software library files provided on MIT website is not recognized (there seems to be some error, atleast on my PC). The MATLAB files works just fine.</p>
<p><span class="math-container">$h_f$</span> can be approximated from enthalpy of sub-cooled liquid near saturation temperature using <code>SW_Enthalpy</code> function and the value of <span class="math-container">$h_{fg}$</span> can be easily obtained by using the function <code>SW_LatentHeat</code>.</p>
<p>P.S. I was searching for the required data under the keyword of 'enthalpy' and that's probably the reason I was unable to find the required data.</p>
| 26011 | Is thermodynamic data of fresh water a good approximation for saline water? |
2019-02-16T22:00:53.707 | <p>Say a stepper motor has fixed value of "<code>units</code>", given as a holding torque related to distance of the center of its shaft. Like <strong>1kg/cm</strong> or <strong>14oz/inch</strong>.</p>
<p>I have a GT2 16-teeth pulley on the motor shaft. (I'm using a 5mm GT2 belt as a load transfer if that matters) Knowing that the motor can hold 1kg weight, 1cm away of the center of its shaft, and the 16-teeth pulley "outher-teeth" diameter is 9.7mm, I can easily assume that it can hold about 1kg of load safely (at least trusting manufacturer's datasheet). </p>
<p>But, what if I add a GT2 48-teeth pulley on the load part (I don't know the outer diameter)?</p>
<p>Is there any algorithm I could use (to atleast <em>roughly</em>) calculate load capabilites of the motor, based only on the pulley teeth count?</p>
| |stepper-motor| | <p>Circumference is a fixed ratio (π) times the diameter. Therefore the number of teeth is in proportion to the diameter.</p>
<p>If everything is 100% efficient then the torque will step up by the pulley ratio. 1 kg·cm with 16 teeth on the motor and 48 on the load will give 3 kg·cm available on the load.</p>
<p>Note that torque is force <strong>times</strong> distance so it's kg·cm, not kg/cm.</p>
| 26017 | Stepper motor torque calculation based on pulleys ratio |
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