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2020-07-13T21:10:36.937 | <p>An accepted method of leak testing a test piece is by using a differential measurement between it and a reference such as in the schematic <a href="https://www.tqc.co.uk/our-services/leak-testing/useful-information/leak-testing-basic-principles/" rel="nofollow noreferrer">here</a> - why might one bother doing this as opposed to just comparing your test piece alone to a predefined or recorded leak rate, or curve?</p>
| |product-testing| | <p>Pressures change in a vessel based on ambient conditions. The reference volume removes that source of error.</p>
| 36698 | Why use a reference volume for leak testing? |
2020-07-10T00:23:05.320 | <p>I am building an electric hoist.
Design requirements:</p>
<ul>
<li>Lightweight</li>
<li>Quiet</li>
<li>Relatively low lifting requirements (100 kg)</li>
<li>Relatively fast lifting speed (80 fpm -> .4 m/s)</li>
<li>Low cost</li>
<li>Simple</li>
</ul>
<p>Force required to overcome gravitational pull = 100kg * 9.8m/s2 = 980 N</p>
<p>Wire spindle radius = 4 cm</p>
<p>Torque required = 980 * .04 = 40 Nm?
Power required = 40 Nm * .4 m/s = 16 watts?</p>
<p>Output RPM of spindle -> .4 m/s * 60 = RPM * spindle_circumference = RPM * (.04 * 2 * pi) = 95 rpm = ~100 rpm.</p>
<p>So either I find a direct drive pancake motor that can output 40 Nm at 100 rpm or I use a gearbox.
My other option is getting say a 4000 rpm motor with a 1 Nm torque and a 40:1 reduction.</p>
<p>Pros and cons?
Other thoughts given design requirements?</p>
| |mechanical-engineering|electrical-engineering|motors| | <ul>
<li>Relatively low lifting requirements (100 kg)</li>
<li>Relatively fast lifting speed (80 fpm -> .4 m/s)</li>
</ul>
<p>The first part of your power calculation was wrong. There's no need to go into pulley diameters at this early stage of the calculation. Let's start at the load hook.</p>
<p>100 kg ~ 1 kN. Speed > 0.4 m/s. So power > 400 Watts.</p>
<p>That's mechanical power, after a gearbox, after motor losses. So the motor will need to be at least 1 kW electrical input power. That's based on pure lifting speed. We can add a little more for load acceleration, or ignore it with little error.</p>
<p>Motors tend to have a 'nice' speed for decent power output. Too slow and you have to work at delivering a lot of torque, which increases the weight of the motor. Too fast and they scream and need careful balancing. The 2000 to 3000 rpm range is reasonable for this sort of motor power. 2500 rpm is about 250 rad/s, so a speed of 400 mm/s would need a shaft of radius 400/250 = about 1.6 mm radius to wind directly.</p>
<p>3 mm diameter is clearly impractical for a 1 kN load, so you need a reduction gearing to a drum of practical size. The range 50 mm to 100 mm radius feels about right to me, so you'd need a speed of 400 mm / 80 mm = 5 radians per second.</p>
<p>With a motor doing 250 rad/s and the drum requiring 5 rad/s, you need a reduction gearing of around 50:1.</p>
<p>Obviously different motors and different drum sizes would result in different gear ratios, speeds etc, but at least we have a good ballpark to start from, which is ...</p>
<p>1 kW (electrical) 2500 rpm motor<br />
50:1 gearbox<br />
160 mm diameter winding drum</p>
<p>50:1 is quite a ratio to do in a single stage, even epicyclic. Worm drives are inefficient, so probably best avoided. An important side effect of their inefficiency is that they don't back drive, so could do away with the need for a brake, so maybe worth thinking about. Compactness might sit nicely with low weight, so suggest an inverted epicyclic stage within the winding drum, driven by a toothed rubber belt reduction stage from the motor. That's if you have the option to build it custom of course.</p>
| 36699 | Building an ultralight electric hoist, what gearbox reduction / motor combination for torque / power requirements? |
2020-07-15T18:22:15.820 | <p><a href="https://i.stack.imgur.com/uZo4n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uZo4n.png" alt="enter image description here" /></a></p>
<p>Above is a chart from a Eurocode handbook. It gives a correct buckling curve depending on cross section and its dimensions. I noticed that for hollow sections, the choice depends on whether the section is hot finished or cold formed. <strong>Why does it matter how the beam is manufactured?</strong> Shouldn't the buckling behavior depend just on cross section geometry and the material used? Why does it matter how the beam is made?</p>
| |structural-engineering| | <p>Manufacturing methods for structural steel shapes result in stresses that remain in the material after the section has been formed. Those residual stresses have an impact on the buckling curve and are a major reason the design curves are distinct from the theoretical Euler buckling curve.</p>
<p>In hot-rolled sections and welded sections, the residual stresses are principally due to uneven cooling. For example: in a hot-rolled wide flange beam, the flanges, being thicker than the web, cool more slowly. But the flange tips also cool more quickly than the flange where it joins to the web because the tips have more surface exposed to air. The result is residual compressive and tension stresses in the section, and this means that when the member is subjected to a compressive load, the fibers in the section are not all under the same stress. This gives rise to the inelastic buckling portion of the buckling curve. This <em>is</em> being accounted for in the rolled sections listed in the table you included, but all those sections will be hot-rolled so there's no need to distinguish between manufacturing methods.</p>
<p>The figure below (<a href="https://www.civilengineeringx.com/structural-analysis/structural-steel/residual-stresses/" rel="nofollow noreferrer">obtained here</a>) shows the typical arrangement of residual stresses in sections fabricated by various methods.</p>
<p><a href="https://i.stack.imgur.com/YOfi7l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YOfi7l.png" alt="residual stresses" /></a></p>
<p>The cold forming process by which HSS are produced also introduces residual stresses, particularly at the corners. The <a href="https://steeltubeinstitute.org/resources/heat-treating-hss/" rel="nofollow noreferrer">Steel Tube Institute</a> has a nice overview of this. If the piece is subsequently hot finished, this will relieve some of these residual stresses and the buckling curve will be changed. I don't possess a copy of Eurocode 3, but my expectation is that buckling curve "a" for the hot finished sections will lie closer to the theoretical elastic buckling curve than curve "c" for the cold formed sections.</p>
| 36715 | Why do cold formed sections buckle differently from hot finished? |
2020-07-15T23:57:51.237 | <p>How do oblique shockwaves differ from 2D to 3D problems? More specifically, how are the conditions after the shock calculated for a shockwave formed on a cone (e.g. the inlet cone on a ramjet)?</p>
<p>I am familiar with the process/calculations of the after-shock conditions of obliques shockwaves at a ramp, namely for the deflection angle:</p>
<p><a href="https://i.stack.imgur.com/lD2QW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lD2QW.png" alt="Deflection angle calculation" /></a></p>
<p>where <a href="https://i.stack.imgur.com/0rVLd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0rVLd.png" alt="theta" /></a> is the deflection angle, <a href="https://i.stack.imgur.com/9Ct74.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Ct74.png" alt="beta" /></a> is the angle between the shockwave and the normal, <a href="https://i.stack.imgur.com/nZfr6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nZfr6.png" alt="M" /></a> is Mach number and <a href="https://i.stack.imgur.com/qGFLV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qGFLV.png" alt="gamma" /></a> is the ratio of specific heats.</p>
<p>Firstly, is there a difference in the conditions when the ramp is revolved into a cone? Secondly, if they are different, what are these differences, specifically in the calculation of the deflection angle and velocity after the shock?</p>
<hr />
<p>EDIT: Looking at the link suggested and a few textbooks, I found Modern Compressible Flow: with Historical Perspective by John D. Anderson explained conical flow very clearly. To summarise, the ODE to solve is the following:</p>
<p><a href="https://i.stack.imgur.com/HkaYU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HkaYU.png" alt="Taylor-Maccoll" /></a></p>
<p>Finding the solution of the flow involves being given either the cone angle and surface Mach number, or the shock angle and the surface Mach number, and then numerically integrating the ODE above.</p>
<p>As for the other conditions after the shock, the shock wave relations for <em>immediately</em> after the shock still hold. Similarly, after the solution to the Taylor-Maccoll equation has been found, the usual isentropic relations for density, pressure, temperature etc. can be used.</p>
| |fluid-mechanics|compressible-flow| | <p>To reiterate my edit and close the question:</p>
<p>Looking at the link suggested and a few textbooks, I found Modern Compressible Flow: with Historical Perspective by John D. Anderson explained conical flow very clearly. To summarise, the ODE to solve is the following:</p>
<p><a href="https://i.stack.imgur.com/HkaYU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HkaYU.png" alt="Taylor-Maccoll" /></a></p>
<p>Finding the solution of the flow involves being given either the cone angle and surface Mach number, or the shock angle and the surface Mach number, and then numerically integrating the ODE above.</p>
<p>As for the other conditions after the shock, the shock wave relations for <em>immediately</em> after the shock still hold. Similarly, after the solution to the Taylor-Maccoll equation has been found, the usual isentropic relations for density, pressure, temperature etc. can be used.</p>
<hr />
<p>EDIT:
The comment I was referring to regarding a link was deleted so I'll link the website here: <a href="https://www.grc.nasa.gov/WWW/K-12/airplane/coneflow.html" rel="nofollow noreferrer">https://www.grc.nasa.gov/WWW/K-12/airplane/coneflow.html</a></p>
| 36720 | How are the conditions after the shock calculated for a shockwave formed on a cone? |
2020-07-16T10:16:20.280 | <p>I have a theoretical question about harmonics and DC.</p>
<p>Given I have a circuit which has a power supply providing an AC input(230Vrms @ 50Hz), this then goes into a load which, takes this supply rectifies it to DC and performs a DC to DC conversion for a DC load.</p>
<p>If the system (load)imposes some DC onto the AC supply it will be seen as a DC offset for instance 1V DC offset. If I were to measure the DC voltage over the Line and neutral going into my load.</p>
<p>Could this hypothetical DC offset be caused by harmonics in the load say the 2nd,3rd,4th harmonic but, appearing as a DC offset.</p>
| |electrical-engineering|power|power-electronics| | <p>In general, a load which is "symmetrical" with regard to the positive and negative half-cycles of the mains will only generate odd harmonics. If the load is not symmetrical and the current waveforms are different on the positive and negative cycles then you will see even harmonics and you may read a DC current component on the line.</p>
| 36726 | AC Harmonics on a system appearing as DC |
2020-07-16T13:41:27.610 | <p>I need to design a simply supported symmetric I-beam (for a school problem). However, I can't find much information about the selection of the section or if there exist any standards for that. the only thing I could find is AASHTO I-beams of which there appears to only 5 types (sizes) and all of them are too large for my needs (since they are used in bridges I guess). the beam I need to design has a span of 9m and carries DL 2.2 kN and LL 11 kN.</p>
| |beam|concrete|prestressed-concrete| | <p>Let's assume we have prestressed the beam enough so the concrete under the load has zero stress at the bottom and the and maximum at the top. and let's assume it is a 350kg/cm^2 concrete which is a lower range of prestress concrete strength.</p>
<p><span class="math-container">$$ M= 1.2\cdot M_{deadload}+1.6\cdot M_{live load}$$</span></p>
<p>I assume your loads are distributed loads and you forgot to mention that.</p>
<p><span class="math-container">$$ M= \frac{1.2 *2.2kN*9^2}{8}+\frac{1.6* 11kN*9^2}{ 8}=26.73 kNm+178.2 kNm=205 kNm$$</span>
<span class="math-container">$ assuming \ \phi=0.85, \ 205/0.85=241 kNm$</span></p>
<p>As a first test size, we pick <span class="math-container">$\text{ h = L/20 = 45cm and b=25 cm.}$</span></p>
<p>The stress triangle is a 45cm height by 25 cm base and with 350kg/cm^2 on top of the beam and zero at the bottom.</p>
<p><span class="math-container">$$C=T= 350 *45/2 *2/3*45*25= 5906250kgcm *9.8/100= 578.8kNm$$</span>
So we are over designing by a factor of roughly 2.5.</p>
<p>We can now iterate down to find the right size, albeit with an eye on controlling the deflection.</p>
<p>This is a very basic first rough step to get an idea for the size of the beam, we need to know a lot more as to the type of strand or concrete before we can start the formal design.</p>
| 36729 | How to choose the dimensions for a prestressed I-beam? |
2020-07-17T13:05:09.093 | <p>recently I had an exercise where I had to build and measure an analog PID controller/filter. The starting point was this circuit.</p>
<p><a href="https://i.stack.imgur.com/eBhfX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eBhfX.jpg" alt="enter image description here" /></a></p>
<p>This is a tamed PID controller implemented with a single opamp. Resistor and capacitor values are calculated regarding the crossover frequency and a 40 degree phase margin.</p>
<p>First question I have:
When this circuit is connected to a power supply and given an input voltage(sine wave while manually editing the frequency) the output voltage slowly but surely goes up to the supply voltage level where it stands which draws the conclusion that the analog PID controller is unstable right? Or am I interpreting something wrong here? If I am right, that means that it is BIBO unstable(Since the only thing bounding the output is the opamp not the controller)</p>
<p>However, after a new Resistor is added parallel to R3 and C2 which makes it a PID-Controller/filter(not sure about this formulation when it is mentioned as an analog filter) it becomes stable and these are the bode plot and step response of the circuit with a new resistor.</p>
<p><a href="https://i.stack.imgur.com/YqzZb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YqzZb.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/mEyO8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mEyO8.jpg" alt="enter image description here" /></a></p>
<p>My next question would be, does an analog controller has delay? Looking at the step response I would conclude that there is no delay. Nevertheless the delay from the step response should be calculated and compared to the delay measured from the bode plot.</p>
| |pid-control|stability| | <p>After consulting with a colleague following conclusions were made.
Controller without resistor in the feedback part has this transfer function</p>
<p><span class="math-container">$PID(s)=\frac{K(s-z_1)(s-z_2)}{s(s-p_1)}$</span></p>
<p>this makes it unstable(borderline stable since one pole is zero)</p>
<p>To move this pole along x axis in the negative direction a resistor is added which produces this transfer function</p>
<p><span class="math-container">$PID(s)=\frac{K(s-z_1)(s-z_2)}{(s-p_2)(s-p_1)}$</span></p>
<p>Now the controller is stable.</p>
<p>Regarding delay, analog controller has no delay. Response to an input is immediate, However its digital counterpart has a delay which can be measured from it step response or bode plot</p>
| 36744 | Analog PID Controller/Filter delay and stability |
2020-07-18T00:38:59.870 | <p><a href="https://i.stack.imgur.com/6LTCX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6LTCX.png" alt="enter image description here" /></a></p>
<p>I understand equation (1), but I really don't see the connection between equation (2) and the picture.</p>
<p>For starters, shouldn't the <span class="math-container">$b_0$</span> constant be multiplied by</p>
<p><span class="math-container">$$net(n)-(a_1 \cdot net(n - 1) + a_2 \cdot net(n-2))$$</span></p>
<p>Edit: Honestly even an "I'm not sure" would be helpful</p>
| |control-engineering|control-theory| | <p>Well as a fresh start: net(n) is the value before the node, this can be seen by equation (1). The value after the node is the following (lets call this <span class="math-container">$x$</span> for simplicity):
<span class="math-container">$$x(n) = net(n) - a_1x(n-1) - a_2x(n-2)$$</span>
Therefore:
<span class="math-container">$$\tilde{y}(n) = b_0x(n)+b_1x(n-1)+b_2x(n-2)$$</span>
Now substituting the first equation into the second one yields the following:
<span class="math-container">$$\tilde{y}(n) = b_0net(n) - b_0a_1x(n-1) - b_0a_2x(n-2)+b_1net(n-1) - b_1a_1x(n-2) - b_1a_2x(n-3)+b_2net(n-2) - b_2a_1x(n-3) - b_2a_2x(n-4)$$</span>
<span class="math-container">$$ = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1\left(b_0x(n-1) + b_1x(n-2) + b_2x(n-3)\right) - a_2\left(b_0x(n-2) + b_1x(n-3) + b_2x(n-4)\right)$$</span>
From here, it can be observed the parts multiplied with <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span> are actually equal to a shifted version of the original equation of <span class="math-container">$\tilde{y}(n)$</span>! by replacing them with the these shifted outputs, your equation in question will show up:
<span class="math-container">$$\tilde{y}(n) = b_0net(n)+b_1net(n-1)+b_2net(n-2) - a_1\tilde{y}(n-1) - a_2\tilde{y}(n-2)$$</span></p>
| 36748 | Can someone explain how the output of this control system is derived? |
2020-07-18T18:31:44.550 | <p>Let me just get this out of the way: this is a total newbie question. But I appreciate any advice I can get!</p>
<p>I'm rebuilding / retrofitting a wood-and-steel trailer, and trying to identify affordable fasteners that won't turn into solid lumps of rust. I live in a relatively dry area, but if there's any moisture at all, it's going to sprayed onto the underside of a trailer, along with road grime, salt, etc.</p>
<p>From everything I've found, the (readily available) materials that are most corrosion-resistant in their own right are stainless steel and hot dipped galvanized steel. However, I'm bolting onto/through a bunch of mild steel, and stainless is appreciably more anodic and than mild steel, while zinc (plating) is significantly more cathodic. Is this a problem? Is the zinc plating just meant to be sacrificial and shouldn't worry about it? Am I looking at the whole question wrong?</p>
<p>I'm guessing I'm far from the first person to run into this question, but an hour communing with Google hasn't gotten me very far. Any suggestions?</p>
| |materials|steel|corrosion| | <p>The major cause of corrosion for a wood/steel trailer will be moisture in the wood keeping the steel damp; any salt will increase the affect. The most practical action would be put in galvanized . Stainless is good but unnecessary. Stainless should not be active ( it will be inert) so will not accelerate or slow corrosion on the carbon steel. Especially bad in either case is an acidic salt containing chloride in damp wood.</p>
| 36756 | Galvanic compatibility with mild steel |
2020-07-20T06:13:41.300 | <p>Consider a rectangular stock section with a dowel hole and other features of size (FOS) on it. The dowel hole serves to locate the entire part.</p>
<p>Should my choice of datum features be what is most convenient for the machinist or for the Q&A engineer? Given that it is rectangular, it is easiest to set up the datums as the 3 orthogonal faces/edges of the part.</p>
<p>But the external faces/edges serve no engineering purpose and is pointless to dimension any FOS from the edges. They should be dimensioned from the dowel hole instead since this determines the part's final position/orientation during assembly</p>
<p>In this scenario, can I have more than 3 datum features, i.e. including the dowel hole as a datum FOS to give 4 datums?</p>
| |mechanical-engineering|design|tolerance| | <p>Typically in my experience, your 3 orthogonal edges of the rectangular blank are your first 3 datums. Your dowel hole would be the first machined feature and since it locates the rest of the features, this would be your 4th datum. Your subsequent features would reference that 4th datum, as well as any of the original 3 to tolerance the features.</p>
| 36771 | Choice of Datum Features and Number of Datum Features |
2020-07-20T12:07:20.283 | <p>I have a few values in Newton metres and I need to make sure each value is an SI unit for calculations. However when I try to search for the units I am given, I can only find similar, but not quite the same, units.</p>
<p>Can someone help clear up the confusion?</p>
<ul>
<li>I'm given the unit Nm<span class="math-container">$^{-1}$</span>, the unit of a spring constant. As far as I can tell this is already an SI unit but I'm confused as I can only find the unit Nm online. Are Nm and Nm<span class="math-container">$^{-1}$</span> the same? I would say that Nm is a newton metre whereas Nm<span class="math-container">$^{-1}$</span> is newtons <strong>per</strong> metre, which is why I'm confused about this</li>
<li>Similarly, I'm given Nsm<span class="math-container">$^{-1}$</span> but I can only find Nsm or Nsm<span class="math-container">$^{2}$</span> online, leading to more confusion over whether or not I have the correct units.</li>
</ul>
| |springs|unit| | <p>There are seven fundamental SI units (seconds (time), metres (length), kilograms (mass), ampere (current), kelvin (temperature), mole (amount of substance) and candela (luminous intensity) - <a href="https://en.wikipedia.org/wiki/International_System_of_Units" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/International_System_of_Units</a>.</p>
<p>There are a bunch of <em>derived units</em> and acceptable prefixes (for orders of magnitude) (see that Wikipedia page). When you see a measurement expressed purely in terms of one or more of those units (like Nm or N/m), then it is an acceptable SI unit. But, if you mix and match prefixes in a unit (gram per cubic centimeter, for example), then you've fallen off the SI truck. As the Wiki page puts it: <em>When prefixes are used to form multiples and submultiples of SI base and derived units, the resulting units are no longer coherent.</em></p>
<p>One thing that is important to note is the importance of dimensional analysis. As others have pointed out, Nm and N/m are very different, as are <span class="math-container">$Nsm^{−1}$</span>, <em><strong>Nsm</strong></em> and <span class="math-container">$Nsm^2$</span>. When you multiply or divide two quantities, carry the corresponding units around and do a similar multiplication or division of the units to figure out what the resuling unit should be. For example, the specific heat of water at standard conditions is about <span class="math-container">$4180\ Jkg^{−1}K^{−1}$</span> (joules per kilogram per kelvin). It I want to raise the temperature of 10 kg of water 2 kelvin, I will need <span class="math-container">$$10kg\ \cdot \ 2K\ \cdot \ 4180\ Jkg^{−1}K^{−1}$$</span>.</p>
<p>Multiplying the numbers is easy (10 x 2 x 4180 = 83,600). But, then you need to consider the units of the result. Multiplying the units out yields: <span class="math-container">$$kg \cdot K \cdot J \cdot kg^{−1}K^{−1}$$</span> Then go through and delete units that "cancel themselves out" (<span class="math-container">$kg \cdot kg^{-1}$</span> and <span class="math-container">$K \cdot K^{−1}$</span>). What's left is <em>Joules</em>. So the answer is 83,600 J. Since our expectation is that some amount of energy is needed to raise the temperature of water, we can be more confident in the answer.</p>
| 36774 | Questions about newton metre units |
2020-07-20T17:45:24.233 | <p><a href="https://i.stack.imgur.com/5CRft.png" rel="noreferrer"><img src="https://i.stack.imgur.com/5CRft.png" alt="enter image description here" /></a></p>
<p>Quite often this kind of arrangement can be seen when beams are connected together. A rectangular plate is welded into a beam, and the plate is bolted to the other beam. I suppose one purpose would be simply that bolting is easier to do in some cases at the site rather than welding the beams together directly at construction site, <strong>but is there any other purpose?</strong></p>
<p>I remember once hearing this kind of plate refered to as a <em>moment plate</em>, but I couldn't really find much information on such a thing. Is there some kind of moment-related purpose for this plate or am I just remembering wrong?</p>
| |structural-engineering| | <p>Welding a beam to a column can deform the column at its bending axis.</p>
<p>The axial load on a steel column is determined by multiplying the moment by a bending factor. The bending factor is the area of the cross section divided by the section modulus.</p>
<p>These factors are tabulated in the AISC Manual for the x and y axis.</p>
<p>When a portion of that cross section is modified by welding, the column is no longer as strong.</p>
| 36779 | What is the purpose of end plates when connecting beams together? |
2020-07-21T12:17:46.510 | <p>I have a noisy neighbor (unreasonable and hostile) generating low frequencies that are disturbing. I'm at the point where I realize I need to just build some kind of barrier, and also buy better windows and doors.</p>
<p>So I am thinking of building a short wall (maybe 7 ft tall) to absorb or redirect the low frequencies (40-80Hz). Building material will be double brick, high density. The impact on my house is worst in the section directly facing the brown wall, hence why the wall is short.</p>
<p><a href="https://i.stack.imgur.com/x7muX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x7muX.png" alt="illustration showing my house and the neighbor" /></a></p>
<p>The horizontal brown line shown above is meant to be the wall.</p>
<p>Before I start, I am concerned that the barrier will maybe allow the sound to pass around (or above). I suppose some sound is to be expected, but I want to eliminate most of it.</p>
<p>I just want to find out if my approach makes sense, or whether I should also be looking at some kind of DIY "bass trap" shape of this wall to absorb even more.</p>
| |acoustics|sound-isolation| | <p>As others have mentioned, a brick wall of any reasonable mass or dimension will be almost completely ineffective.</p>
<p>If the goal is to reduce the SPL at only a specific, relatively compact location in your home (e.g. at the head of your bed), active noise cancellation may be feasible at very low frequencies. (Note that this will not control vibration, though, if this is also an aspect of the problem.)</p>
| 36796 | Will a short brick wall help shield against bass or will it go around? |
2020-07-23T14:41:11.777 | <p>When specifying the depth of a cut, should the dimension start from the datum surface or from the surface where the cut will be made? In the example below, a cut of 5mm is made from the surface opposing datum surface A. Is the appropriate dimension 5 or 10?</p>
<p><a href="https://i.stack.imgur.com/0zxIp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0zxIp.png" alt="Example" /></a></p>
<p>I'm told to do it from the datum surface but wouldn't it be more convenient for the machinist if it were from the surface from which the cut will be made? The former requires additional maths whereas the latter is more obvious in terms of "what has to be done on the part"</p>
<p>Ok, the maths is not hard, but imagine having to do it 10 times and scribbling all over the drawing.</p>
| |mechanical-engineering|technical-drawing|tolerance|drawings| | <p>If the request is coming from whoever will be machining or inspecting the part, I would side with them. They know what they need to see to make sure the part ultimately meets the print.</p>
<p>Also, since your first datum is on the opposing face this implies it is getting machined first, hence why it is a reference datum. The geometry of the face you're making the cut in question (in terms of flatness, parallelism, etc) is not defined (at least not in the image you have shared), so measuring off that surface is less precise than measuring off Datum A.</p>
<p>For inspection purposes it is also much easier to have the distance from 'A' explicitly stated as this is the reference surface for the feature parallelism. I imagine surface 'A' would be on an inspection plate, a CMM or indicator would be set to measure the feature surface depth nominally 10 mm from the reference face and measure parallel. No extra math for them required, less room for error.</p>
<p>Generally if someone in manufacturing with more experience asks for something that doesn't immediately make sense to you and you don't have a strong argument against it, there's probably a good reason for it. Like wearing a mask in public.</p>
| 36826 | Do drawing dimensions have to start from datum? |
2020-07-24T14:09:16.417 | <p>I am having a small 2.5cc methanol 2-stroke engine that I am building an electric starter mechanism for. From my research I was under the impression that for a 2-stroke engine like the one that I have which is completely mechanical using a port for fuel intake it should not matter what direction I am starting the engine, it will just continue to move in that direction.</p>
<p>Image of the engine:</p>
<p><a href="https://i.stack.imgur.com/cmxof.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cmxof.jpg" alt="enter image description here" /></a></p>
<p>In order to make sure my DC motor used to start the engine is not suddenly "powered" by the much faster methanol engine I am using a one way bearing on the gear connected to the methanol engine. When I did a test run I could hear the methanol engine engage (due to the noise), but it cut out practically immediately afterwards and I could see kinda heavy abrasion on the gear of the DC motor (it is 3d printed, not some metal gear). I am getting the feeling that the methanol engine starts but wants to turn into the opposite direction of how it got started and then dies out pretty quickly because of the resistance.</p>
<p>This goes contrary to my believe that these 2-stroke engines should be able to run in any direction. What are the parameters that influence this for an engine and how would I go about determining that direction except ruining my starter gear and do it empirically? How does this work with a pull start, i.e. how do they ensure that the motor never engages in the other direction?</p>
| |motors| | <p>I have finally completed the starter mechanism for the RC engine (not the one in the picture, but a vertex .18 engine). I can add another anecdotal evidence, that the engine turns counter clock wise (seen from the front/output shaft). I could not get it started when using a clock wise starter.</p>
| 36848 | Starter direction for methanol engine |
2020-07-25T11:07:19.170 | <p>I have recently designed and had fabricated the following two sumps for my Aquarium (See image).</p>
<p><a href="https://i.stack.imgur.com/ODpnY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ODpnY.jpg" alt="Sump Design" /></a></p>
<p>I have a pump in the right hand container (section 4), and the return is coming back into the left hand container (section 1) - both flowing at equal rate (~1000lp/h).</p>
<p>Two 32mm flexi-pipes connect the two containers.</p>
<p>I was expecting the water level in both containers to equalise, but the left container is consistently running near to maximum whilst the right container water level is - as I would expect it to be - at the baffle height for 'section 1'.</p>
<p>Why am I not seeing equal water heights in both?</p>
<p>See below a picture of the situation occurring with RED lines indicating water height, BLUE lines indicating baffle heights, GREEN arrows indicating water flow direction. Please ignore the fact that I currently have one of my bypass valves turned on - it's all I can do to drop the water height in the left container a few centimetres.</p>
<p><a href="https://i.stack.imgur.com/Pyx71.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pyx71.jpg" alt="Sumps attached and with water flowing" /></a></p>
<p>Note - Reducing flow rate, and removing water from the system does not seem to have any positive effect on the left hand container. Filter sponges/socks etc were added after this effect was happening. The system has been running stable for 24 hours with this effect visible.</p>
| |fluid-mechanics|fluid| | <p>There must be pressure ( water level) difference to provide force to push water through the system. If you stop input and output for a few minutes the levels will even out , except where a baffle may prevent it .With no flow ,no pressure differential is need to move water . It is not much pressure , height of 4 " equals about 0.17 psi. Unrelated , have you considered a light to grow algae or plants like a salt water refugium ?</p>
| 36861 | Unequal water level in connected containers - Aquarium Sump |
2020-07-25T13:08:05.907 | <p>I have a question regarding how are the vertical forces on a spur gear shaft are supported. I am currently doing my undergraduate (in final year) and have only dealt with gearboxes where the shafts axis lies in a horizontal plane. I am designing a device that requires the shaft to be vertical.</p>
<p>I understand how the bearing would support the axial and radial forces but what about the vertical forces, how could I support these (from the weight of the shaft and gears).</p>
| |design|gears| | <p>A vertical force is either a axial, radial or combination of both depending on what part of it is aong the shaft and what is not. These describe all forces there can be on the simplified shaft system from bearings point of view.</p>
<p>If you turn the axis of shaft running vertically then the weight just becomes a axial load if the axis is on angle then part of the load us axial and part of the load radisl as per standard force composition rules. If you have hard time understanding this draw the image and all force arrows then turn it around so its in the exact same orientation you did analysis before. Same result.</p>
<p>The reason we do this is that the bearing is axisymnetric it does not matter as such which way the force is in the radial direction just its magnitude. Offcourse the bearing is very different in the axial direction.</p>
<p>So you would design it the same way but with possibly more axial load allowance.</p>
| 36862 | Vertical gear shaft support |
2020-07-25T19:26:34.727 | <p>I am new to designing rugged + waterproof products. I am curious about how to dissipate heat when the electronics are running and the ambient environment is even hotter. It seems like the mode of heat transfer is limited to conduction through the shell and then convection with the ambient air. My product will be small (like GoPro), need to be outdoor in Dallas all day (37°F to 96°F), and be able to protect against dust and water submersion (IP67). I am concerned about where the heat will go and the thermal cycling as the device experiences ~30°F change daily.</p>
<p>From a design stand point, what options (material, geometry...) do I have to solve this problem?</p>
| |mechanical-engineering|design|heat-transfer|consumer-electronics|convection| | <p>Normally, you would mount a the heat-generating components of the PCB to a heatspreader, and then mount the heat spreader to the inside of a thermally conductive (aka metal) case, possibly with a heatsink on the other side.</p>
<p>If the ambient is even hotter than your device, you either need to do what a fridge does or use components rated for higher temperatures. Or a peltier, I guess.</p>
| 36869 | As a rugged and waterproof device, how does gopro dissipate heat? |
2020-07-25T20:52:06.200 | <p>For those unaware of what a DC stepper motor is, it boils down to an electronic motor where the rotation can be precisely controlled, i.e. you can control the rotation of the motor in fractions of a degree. For example the motors I have currently allow you to move the motor just 1/2048 of a full rotation. This enables very precise movement.</p>
<p>I am trying to achieve the same thing, however with a motor, that is not powered by electricity but by gasoline. This motor is connected to a large and heavy flywheel that enables more or less constant RPM once the motor has settled. Using gear reduction I can get a very constant very low RPM which works fine for precisely moving a piece of machinery, however the problem is that I cannot really engage and disengage that movement reliably.</p>
<p>Since the gasoline engine also powers a 500W DC generator obviously the first choice would be to just use that electricity for a stepper motor, but that would take the fun out of the project (its a fun project, nothing serious). So far my ideas were the following:</p>
<ul>
<li>Use strong electromagnets to simulate a controlled clutch where the energized electromagnets would engage the clutch more or less immediately and have some powerful springs to retract it once the electromagnets are no longer energized. Combined with more or less constant RPM this would make it possible to create precise movement by engaging the magnets for a defined time. Problem is that while I am able to create the clutch and circuit I don't have electromagnets that are able to push that much weight and dont overheat after a minute or two (literally).</li>
<li>Use some hydraulic oil with an oil pump powered by the gasoline engine. On the surface this is perfect, however it has some issues: I would be using a geared oil pump which needs to be engaged on demand, essentially just transferring the problem downstream. Using a DC motor to control the geared oil pump seems like cheating...</li>
</ul>
<p>So long story short, does anyone have any ideas how to replicate a DC stepper motor with a gasoline engine (assuming constant low RPM) that does not require any DC motor on its own?</p>
| |mechanical-engineering|motors|hydraulics|engines| | <p>A hydraulic solution should work fine. You can control the start / stop motion with a 4-way valve like this:</p>
<p><a href="https://i.stack.imgur.com/MABRr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MABRr.png" alt="enter image description here" /></a></p>
<p>The default valve position sends oil back to tank. Shifting the valve right / left gives you forward / reverse motion. The image above shows a cylinder, but you can swap out the cylinder for a hydraulic motor with no problem. Typical valve shifting times are 10 - 50 ms depending on which type you get.</p>
<p>Couple of other details on the diagram. Section 2 shows a pressure relief valve for safety. It also shows a pressure gauge with a shutoff valve (opposed triangles symbol). No filter is shown on the diagram, but you would need one installed in the tank return line.</p>
<p>The last thing you would need is a simple timer control to actuate the valve solenoids.</p>
| 36870 | Replicate stepper motor with gasoline engine |
2020-07-26T04:58:13.943 | <p>A new home (completed in early 2020) in San Jose California does not have GFCI outlets in all the bathrooms. I thought this was a requirement (according to <a href="https://www.redwoodcity.org/home/showdocument?id=15416" rel="nofollow noreferrer">https://www.redwoodcity.org/home/showdocument?id=15416</a>). There are also some kitchen outlets that don't seem to follow this code. I'm not sure whether the builder has overlooked this or there is some loophole to the rule (such as when the builder got approval for the plans).</p>
<p>In general, I'd like a better understanding of how building code is defined and enforced. E.g. what level is it defined (municipal/state/federal) and how it is applied to buildings. Ultimately I'd like to know if the builder made a mistake or if I'm uninformed.</p>
| |civil-engineering|building-design|electrical| | <p>New homes are inspected by local authorities to comply with local building requlations. As Mike says, give them a call.</p>
<p>The National Electrical Code (NEC) are standardized guidelines for electrical construction created by the National Fire Protection Association (NFPA). The NEC are guidelines, not laws. The NEC is adopted into law by states and local jurisdictions.</p>
<p>Different states adopt different versions of NEC guidelines as shown by <a href="https://www.nfpa.org/NEC/NEC-adoption-and-use/NEC-adoption-maps" rel="nofollow noreferrer">NFPA NEC Enforcement</a>.
<a href="https://i.stack.imgur.com/vJGFF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vJGFF.png" alt="enter image description here" /></a></p>
<p>The 2017 NEC is the base model code for the 2019 California Electrical Code (CEC) of the California Building Standards Commission (CBSC). CBSC adopt statewide ammendments to deal with statewide issues with the NEC.</p>
<p><a href="https://www.dgs.ca.gov/-/media/Divisions/BSC/06-News/Information-Bulletins/BSC-Bulletin-20-01-FINAL.pdf" rel="nofollow noreferrer">California Building Standards Commission - Information Bulletin 20-01</a></p>
<p>Local authorities adopt statewise references with local ammendments to deal with local issues.</p>
<p>From <a href="https://library.municode.com/ca/san_jose/codes/code_of_ordinances?nodeId=TIT24TECO_CH24.01AD_PT2DE_24.01.233ELCO" rel="nofollow noreferrer">San Jose, California - Code of Ordinances Title 24 - Technical Codes</a></p>
<blockquote>
<p>24.01.233 - Electrical code.<br />
"Electrical code" means the California Electric Code or CEC, 2019 edition, based on 2017 National Electric Code promulgated by the National Fire Protection Association, as amended and set forth in the California Building Standards Code,</p>
</blockquote>
<blockquote>
<p>24.06.100 - Adoption of technical provisions of CEC.</p>
</blockquote>
<blockquote>
<p>24.06.110 - Portions of CEC which are not approved, adopted or incorporated by reference.</p>
</blockquote>
<p>This is a top-down, bottom-up approach. Standards come from the top, but issues with these standards are found/tweaked at the local level. Local can only change local. Local ammendments will drive state ammendments. Hopefully, incorporated
at some point at the national level. Hence, revisions of NEC.</p>
<hr />
<p>The use of <strong>all</strong> and <strong>some</strong> in the OP's question, implies there are some GFCI's in the home. The code does not require <strong>all</strong> GFCI protected outlets to be GFCI's. One GFCI can protect multiple downstream outlets.</p>
<p>Test these GFCI outlets and see if other outlets are powered. Plug a light into a non-GFCI outlet and press Test on GFCI. If lamp goes out, then it is powered by GFCI. Press Reset to reset GFCI.</p>
<p>From <a href="http://www.electrical101.com/gfci-line-load-wiring.html" rel="nofollow noreferrer">GFCI Load Wiring - Electrical 101</a>
<a href="https://i.stack.imgur.com/Ug7vz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ug7vz.png" alt="enter image description here" /></a></p>
| 36874 | How is electrical building code defined and applied to new constructions? |
2020-07-27T04:08:52.583 | <p>I'm writing a few elementary paragraphs about switches and switch technology, and I cannot for the life of me remember what a particular kind is called, to give it an honorable mention.</p>
<p>Here is a sketch of the switch in question:</p>
<p><a href="https://i.stack.imgur.com/CSviG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CSviG.jpg" alt="roller/strip switch" /></a></p>
<p>It is constructed of at least one roller, past which a strip of spring is pulled or pushed. The spring has cut-outs of various shapes, which can be regular or irregular, making the spring strip "prefer" some positions over others, requiring a certain force to overcome them. Various combinations of perforations and rollers could be used to create very complex "preference" patterns. You can see the usefulness: you can put the switch in any one of it's "stop points" and it just stays there until you move it away; and it gives wonderful oportunity (from a design standpoint) for what is now called "haptic feedback".</p>
<p>I learned about them in grade school, studying simple and complex machines, but now I can't remember any sort of name for them. I've combed the LEXICON encyclopedia, where I seem to remember having seen the thing in the first place, and all the switches I come up with are <em>electrical switches</em>. In Google, ditto, along with a great number of maddening diagrams of gamers mechanical keyboard switches and their various merits.</p>
<p>I'm not a well-formed engineer, so I beg you have patience if this isn't the right place to ask; or tell me what other information I should give and I will try.</p>
<p>I'm looking generally for example uses, but I'll be quite happy if someone kindly just gives the name, as I will probably be able to do the rest of footwork myself. Cheers!</p>
<p><strong>EDIT1: 27-JUL-2020</strong>:
Kevin gave the right name, and now I see that my recollection went astray: the rollers actually roll on <em>each other</em> with only the band between them, and they have (in many cases, at least) mobile axes, constrained by the transversal rigidity of the band.</p>
| |mechanical-engineering|machine-elements| | <p>Kevin Reid answered the question, and I've accepted it-thanks, Kevin! This answer is to post a copy of relevant parts of the original article that had become rather blurry in my memory. I'll probably post other related info as I dig it up in order to catch the best selection of search keywords possible in case anyone else ever looks for sliding mechanical switches on Eng.SE. It's from the Lexicon Universal Enciclopedia. The figure is especially pertinent because this is what I was trying to depict in the sketch that I posted in my question, only my memory omitted some important details. Just looking at the strip with a series of round perforations, I can almost feel the smooth "sinusoidally varying resistance"!:</p>
<p><img src="https://i.stack.imgur.com/C3RPf.jpg" alt="figure_rolamite" /></p>
<blockquote>
<p><strong>rolamite</strong>
... It was developed in 1966 Donald Wilkes as a result of his search for a reliable miniature mechanical switch. The switch that Wilkes designed was one-eighth the size of the previous switch and had about half the number of parts.<br />
In its simplest form the rolamite consists of a rectangular frame, two rollers, and a flexible band. The rollers are suspended within the 5-shaped loops of the metallic band When the band is tightened the rollers are in pure rolling contact with the band and there is no slippage. This configuration allows the rollers to roll within the guide rails with as low as one-tenth the friction of the best ball and roller bearings in a similar application. In this rudimentary form the rolamite functions as an almost frictionless suspension system for the rollers By using bands with varying widths or cutouts, the rollers can be made to seek preferred positions between the guide rails. This effect is caused by the tendency of each band loop to straighten itself out as long as the band width is constant, the straightening forces are equal and the rollers move freely If the bandwidth on the rollers are different, the roller with the wider band will dominate, the band will try to unwind and a driving force proportional to the difference in band widths will be created. By varying the shapes of the cutouts and the diameters of the rollers, rotates, with a wide variety of springlike characteristics can be obtained. ...<br />
--ALEXANDER COWIE</p>
</blockquote>
<p><strong>OTHER INFORMATION RE: ROLAMITE:</strong></p>
<ul>
<li>
<blockquote>
<p>Bad things happen when the ribbon breaks. It overcomes a problem seldom seen. Bad things happen when the ribbon stretches. Compound bearings are well understood and available. Did I mention bad things happen with that ribbon? Imagine what would happen if somebody put a penny or FOD of any kind on the ribbon, frinstance. Probably why D9 Cats use segmented tracks instead of ribbon.<br />
-- A thread at <a href="https://bbs.homeshopmachinist.net/forum/general/24321-what-happened-to-rolamite?t=23745" rel="nofollow noreferrer">homeshopmachinist.net</a></p>
</blockquote>
</li>
<li>It looks like a lot of hype--almost like perpetual motion, with real believers writing stuff like: "virtually zero friction". I've never seen an application where the thing was actually in use, and the only one that is claimed to exist is in a ground-penetrating nuclear bomb, rather conveniently out of my reach, unfortunately.</li>
</ul>
| 36885 | What is the name of this particular mechanical linear position switch made of perforated spring strip? |
2020-07-27T14:57:41.950 | <p><a href="https://i.stack.imgur.com/WtOsF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WtOsF.jpg" alt="3D Diagram" /></a></p>
<p>Let's say I have a wedge which can only be rotated about a fixed axis perpendicular to the ground. There is a cam follower mechanism constrained to vertical movement with a ball roller on the end which is applying a vertical force on this wedge at a fixed distance L from the center of rotation. As a function of F, theta, and L, how do I find the resultant torque? This is not schoolwork, it's based on a nutating mechanism I'm designing for my job.
<a href="https://i.stack.imgur.com/LDVTr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LDVTr.png" alt="Transparent 3D View" /></a></p>
| |statics|mechanisms| | <p>Lets callthe cam force as per your sketch P. P will have 2 components acting on the surface of the wedge, <span class="math-container">$$P* cos(\theta)_ \text{paralel to surface} , \ P sin(\theta)_ \text{perpendicular to surface} $$</span></p>
<p>Bur both of these components pass through the center of rotation, meaning they will not produce any torque, hence any rotation.</p>
<p>If P is too big, the shaft at the center can bend or break but wont rotate.</p>
<p><strong>EDIT</strong></p>
<p>After you modified your question. And if we disregard the original sketch.</p>
<p>there is a varying lateral force that varies as the cam moves around.</p>
<p>This force F multiplied by L is your torque.</p>
<p>, <span class="math-container">$F= Psin(theta)*sin(\alpha) \ , $</span> Sin of the local slope angle.</p>
<p>That angle Alph is a simple sine function, I let you figure it.</p>
| 36892 | Torque acting on a rotating wedge due to a linear force parallel to the axis of rotation |
2020-07-27T17:12:41.327 | <p>Given the space, weight, and power constraints of an RV it is useful to make as much use as possible out of every component. Is there an engineering reason why a refrigerator / freezer and an air conditioner could not share a single compressor and condenser?</p>
| |mechanical-engineering|hvac|refrigeration| | <p>To be able to run a freezer, the refrigerant must be colder than the freezer, so at most-10 degrees C or so. Such a temperature on an evaporator coil for regular HVAC would freeze the condensate and result in a non-functioning device (the coil would ice up). Also, a refrigeration cycle turns on and off according to the temperature. Placing two different temperature requirements on the system would result in one or the other not getting enough cooling.</p>
| 36896 | RV HVAC and Fridge Freezer on the same coolant loop? |
2020-07-30T10:50:56.793 | <p>I have a sliding gate installed on my compound wall. I want to control the gate opening/closing with a motor. I have the electronic circuitry to control motor start/stop. I m thinking about attaching rack on the gate and pinion on the motor to achieve this. But the gate should be able to open/close manually in case of a power failure and I don't think this is possible with a gear motor. I have been searching for some mechanism to achieve this and I found planetary gear system. Does this solve my issue or is there any other mechanism to achieve the same?</p>
| |motors|gears|mechanisms| | <p>As long as you're able to lock one of the planetary elements while the motor is powered that solves your problem but that doesn't seem like the most efficient solution. I would try an electromagnetic clutch between the motor and the pinion. When you have power, the clutch is engaged and the motor can turn the pinion. When you lose power, the clutch disengages and you can manually turn the pinion with a crank or equivalent.</p>
| 36922 | Gear mechanism for automatic gate |
2020-07-31T12:53:17.650 | <p><strong>Disclaimer:</strong> I am not an engineer - just a DIY guy looking for some advice or help please, so please forgive me if I use incorrect terminology :)</p>
<p>I am building a cabinet that has a hidden inner cabinet that is meant to rise up out of the former. The inner cabinet needs to rise <code>500mm</code> and needs to support a maximum weight of <code>40kg</code> (including the weight of the inner cabinet carcass itself).</p>
<p>I am on a budget with this build at home, so I'm trying to use either parts I already have or low-cost ones. I figure that I have 3 options available:</p>
<p><em><strong>1. Motorised Scissor Lift</strong></em>
<a href="https://i.stack.imgur.com/FQlr2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FQlr2.png" alt="enter image description here" /></a>
This will just simply be 2 pairs of arms, where the bottom pivot points are connected to a shaft that rides horizontally along a 16mm stainless steel threaded rod (powered by a small geared 12V DC motor). The top and bottom pivot points will ride along steel rails on bearing wheels. The arms will be made from 50mm wide mild/carbon steel with a thickness of 5mm.</p>
<p><em><strong>2. Poor-man's Linear Actuator</strong></em>
<a href="https://i.stack.imgur.com/xdFPH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/xdFPH.png" alt="enter image description here" /></a>
Off-the-shelf linear actuators are horrifically expense where I'm from, so I took to understanding the mechanics involved and sought to try building one myself. My small proof of concept using the same 16mm stainless steel threaded rod (riding inside a seam-less steel pipe for sturdiness which is not shown in the image above), connected to the same motor works. However, I'm uncertain if the motor is powerful enough to turn the rod.</p>
<p><em><strong>3. Scissor Lift alternative</strong></em>
<a href="https://i.stack.imgur.com/Kh0Nu.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Kh0Nu.png" alt="enter image description here" /></a></p>
<p>The above is an alternative although I'm not convinced this will be as sturdy as the first option. Plus the motor will have to ride up and down.</p>
<hr />
<p>With each option, I aim to ensure that the inner cabinet slides along some sort of linear rail to avoid any "wobbling" as it rises and drops.</p>
<hr />
<p>The motor I'm using rated as follows:
DC GEARED ROUND BRUSH MOTOR, 12VDC, 0.6A, 1:30 REDUCTION 270RPM, 2KG/CM</p>
<p><a href="https://i.stack.imgur.com/ai5Xb.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ai5Xb.jpg" alt="enter image description here" /></a></p>
<hr />
<p>I fear that my motor might be under-powered to achieve the necessary lift, so would truly appreciate the community's help on this and the which of the above options would be suitable. Also, please advise me if the standard stainless steel threaded rod is okay, or if a trapezoidal lead screw would be more suited.</p>
<p>My sincerest thanks in advance.
<em>PS - sorry for the long post!</em></p>
| |motors|actuator|linear-motion| | <p>Either one would probably work as long as you gear it properly; but like Eric Shain, I'd advise you to go with the scissor jack because it's simpler. However, because the force varies with the angle of the scissors, you'll need to gear it down somewhat more than my suggestion.</p>
<p>Lifting the setup is around 201 joules (<code>41 kg * 0.5 m * 9.8 m/s (g)</code>), and it would exert about 402 newtons. Your motor takes 7.2 watts, which means it should theoretically take around <strong>25 seconds.</strong> at minimum. For the motor to have enough force to lift the object, you need to jigger the ratios so that it only lifts the payload by about 20 millimeters per second.</p>
<p>Good luck!</p>
<p>Oh, and if anyone in the community sees any mistakes or miscalculations, please feel free to correct me in comments! (Thank you!)</p>
| 36938 | Lifting a cabinet using actuators or scissor lift |
2020-08-02T20:13:25.277 | <p>I have an unknown-history Lenovo Yoga 3 Fold-Back Keyboard Laptop that I am trying to charge/test/repair I and don't have a OEM or aftermarket power adapter. The power input site looks like a usual finger-thick usb power inlet but with an additional small "wing," however it will physically accept a regular USB male. The bottom of the laptop says "20V 2A."</p>
<p>I have searched the web looking for the power spec I should input in order to power up the laptop, but I found nothing. Nor did I find a name this connector may be called (USB-Wingy? USB-Flaggy? USB-Perky!?).</p>
<p>I tried wiring a common USB male pigtail (having presumably the Red +5V power, white -Data, green +data, and Black -Ground) to a 70W 20V DC power supply (from a generic Universal Laptop Power Supply kit), and the following attempts failed (meaning no light on the "Battery" light on the laptop):</p>
<ol>
<li>Red to 20V+, Black to Ground</li>
<li>Red and white to +20V, Green and Black to Ground.</li>
</ol>
<p>What are the correct specs and wiring, and what is the name of this connector's shape so that we can put a name to talk/query about it?</p>
<p>Pic attached of the input site and the common USB pigtail I used.
<a href="https://i.stack.imgur.com/h9e2d.jpg!%5Bbla%5D" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h9e2d.jpg!%5Bbla%5D" alt="LENOVO YOGA 3 Mystery USB 20V Inlet" /></a></p>
<p>Update Here's a better pic and the Model is 80JH Lenovo Yoga 3.14 and yea there are another two stor shorter pins deep in on the side of 1 and 4 for a total of 6.
[<img src="https://i.stack.imgur.com/mJDaa.jpg!%5Basdf%5D" alt="LENOVO YOGA 3 Mystery USB 20V Inlet 2" />]</p>
| |power-electronics|connections| | <p>The image is poorly focused and laptop name is not specific to identify model, but it appears to be a Lenovo YOGA 3 Pro, 6 pin USB/charging port.</p>
<p>The USB port is dual purpose. USB and charging (20V). By default, it is set up as a USB port.</p>
<p>This is a problem because the voltage changes dependent on mode. The 5.2V rail has to be disabled to become the 20V charging rail. Hard to say what applying 20V randomly has done to the USB port.</p>
<p>Connect a USB cable and it works as a standard USB 2.0/3.0 port. If you connect a USB 6-pin charging cable (cable is notched so you cannot connect it to wrong port), it disables the 5.2V rail and connects the pin to the 20V battery.</p>
<p>This flavor of question has been asked on the Lenovo forums with no responses. <a href="https://download.lenovo.com/consumer/mobiles_pub/lenovo_yoga_3_pro_1370_hmm.pdf" rel="nofollow noreferrer">Lenovo YOGA 3 Pro-1370 Hardware Maintenance Manual</a> has no info. Lenovo forums does have relevance but images are removed.</p>
<p>From <a href="https://www.kjctech.net/yoga-3-pros-2-in-1-charging-port/" rel="nofollow noreferrer">KC Blog's Yoga 3 Pro’s 2-in-1 Charging Port</a>:</p>
<p><a href="https://i.stack.imgur.com/jTmTR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jTmTR.png" alt="Lenovo 2-in-1 Charging USB port" /></a></p>
<blockquote>
<p>The power adapter that comes with the laptop has a special charging connector that has two extra pins on each side, matching two extra pins on the Yoga’s charging port.</p>
</blockquote>
<p>From <a href="https://forums.lenovo.com/t5/Lenovo-Yoga-Series-Notebooks/What-s-with-the-Yoga-3-Pro-power-adapter-it-s-a-USB-charger-too/m-p/1824182?page=9" rel="nofollow noreferrer">Lenovo forums Re: What's with the Yoga 3 Pro power adapter - it's a USB charger too?</a></p>
<blockquote>
<p>These sideband signals (SB5 and SB6) are used to communicate the voltage switches on the notebook and the power adapter.</p>
</blockquote>
<blockquote>
<p>The Lenovo USB [charging] cable only has three conductors with the ends using proprietary Lenovo USB type-A plugs with the additional sideband signals (See FIGURE 1). At the Plug the sideband signals are tied together so that they only use a single conductor in the cable.</p>
</blockquote>
<blockquote>
<p>With the cable unplugged, the notebook will be asserting 5V onto the SB5/SB6 signals (through a 103 KΩ resistor) and 5V onto the V-Bus line, while the EPS has the SB5, SB6 signals at ground and asserting 5V onto the V-Bus line.</p>
</blockquote>
<blockquote>
<p>When the cable is plugged into the notebook and the EPS, both systems are driving V-Bus to 5V initially. The sideband signal is connected to ground through a 200 KΩ resistor, forming a resistor ladder which drives the sideband signal to 3.3V. Upon detecting 3.3V the Notebook will stop driving 5V onto the V-Bus line, while the EPS will start driving 20V onto the V-Bus line.</p>
</blockquote>
<p>So the images are missing and the source is unknown, but looks official. Lenovo use what they call USB Sideband signaling to switch between USB and charging.</p>
<ul>
<li>Pin 1 - 20V</li>
<li>Pin 2 - NC</li>
<li>Pin 3 - NC</li>
<li>Pin 4 - GND</li>
<li>Pins SB5 or SB6 - 200kΩ connected to Pin 4.</li>
</ul>
<p>You will need a special connector. I'd connect the 200kΩ and see what happens to Pin 1 output.</p>
| 36966 | What are the LENOVO YOGA 3 USB-like inlet 20V 2A wiring/signal/voltage specs? |
2020-08-03T15:07:58.097 | <p>I’ve got a reasonably strong grasp of Newtonian mechanics, but as far as material science goes I don’t know which of the many material properties I need to use to solve my problem:</p>
<p>I’m trying to mount a heavy object at the end of a lever arm. I’ve got a 1 cm (0.4 in) diameter steel rod, and I’m planning to embed half of it in a concrete wall to hold it in place, and the other half will extend 2 m (6.6 ft) parallel to the ground with a rope holding 50 kg (120 lbs) sitting in a notch at the end. I’d like to buy a thinner rod if possible, but picturing this in my head it seems with about 980 Newton-meters (eh, 1330 ft-lbs?) of torque I’m likely to bend the rod.</p>
<p>Which material science properties do I use to figure out how to minimize elastic deformation and avoid plastic deformation entirely?</p>
<p>Note: This question was originally posted to Physics SE, and was moved here on request from @Gert :)</p>
| |stresses|strength|material-science| | <h2>Material Properties</h2>
<p>For the linear-elastic analysis you describe, the key pieces of material information are the elastic modulus (E) to calculate deflection and the yield stress (Fy) to check if the material remains elastic under the given loading. For steel, E is commonly around 200 GPa. The Fy can vary significantly depending on the grade of steel, so you'll want to check for your particular product, but between 250 MPa and 350 MPa is a reasonable assumption for mild steel.</p>
<p>The E and Fy values will often be directly available but the actual <a href="https://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve" rel="noreferrer">stress-strain curve</a> is helpful for developing understanding. (The image below is taken from the linked Wikipedia article.)</p>
<p><a href="https://i.stack.imgur.com/PerdZ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/PerdZ.png" alt="stress strain curve" /></a></p>
<p>This sort of plot can be generated by tension-testing to failure a small coupon of metal. The initial linear portion of the plot is where the material is still elastic. The slope of this line is the elastic modulus (aka Young's Modulus or Modulus of Elasticity). In this region, the strains are recoverable and if the load were removed the section would return to its original shape. The yield point is where you see the curve deviate from that initial linear portion. Beyond the yield point, the behavior is inelastic and permanent deformations occur. Some materials do not have a well defined yield point and Fy will be estimated by an <a href="http://www.engineeringarchives.com/les_mom_offsetyieldmethod.html" rel="noreferrer">offset method</a>. Many resources are available online or in introductory Material Science textbooks if you're interested in learning more about stress-strain curves.</p>
<p> </p>
<h2>Analysis Process</h2>
<p>Provided your cantilever beam is kept in the elastic region and the deflections are small, simple equations can be used to analyze beam response. It's important to recognize that for inelastic behavior or large deflections it would be necessary to employ a different methodology than the one presented below.</p>
<ol>
<li>Choose a material and cross section. This establishes E, Fy, and the moment of inertia, I, for the cross section.</li>
<li>Calculate the maximum applied bending stress in the section and compare to Fy. Adjust the section as needed to ensure the beam remains elastic.</li>
<li>Check deflection. Again, adjust section as needed in order to limit deflection.</li>
</ol>
<p> </p>
<h2>1. Material and Cross Section</h2>
<p>Assume E = 200 GPa, Fy = 250 MPa</p>
<p><a href="https://en.wikipedia.org/wiki/List_of_second_moments_of_area" rel="noreferrer">Moment of Inertia</a> for your 1 cm diameter cross section:</p>
<p><span class="math-container">$$ I = \frac{\pi r^4}{4} = 4.91 x 10^{-10} \,m^4$$</span></p>
<p>Note that for beams in uniaxial bending, a solid circular cross section will be one of the least efficient sections. You want to concentrate material where bending stresses will be highest, which in this case is the lower edge of the beam cross section. (This is why I-shapes are so popular for beams in buildings and bridges.) For a given area of steel, a circular tube will offer improved efficiency over a solid bar and a rectangular tube with the long leg placed vertically would be even better.</p>
<p> </p>
<h2>2. Calculate Maximum Bending Stress and Check Against Fy</h2>
<p>For a cantilever beam, the maximum moment will occur at the fixed end. Your problem statement indicates a 50kg mass and 2m cantilever length. Convert kg mass to N force by multiplying by acceleration due to gravity (<span class="math-container">$9.81 m/s^2$</span>)</p>
<p><span class="math-container">$$ M_{max} = PL = (0.49kN)(2m) = 0.98 \, kNm$$</span></p>
<p>For bending stress in a cross section via Euler-Bernoulli beam theory:</p>
<p><span class="math-container">$$\sigma = \frac{My}{I}$$</span></p>
<p>And take <span class="math-container">$y = (bar\, radius)$</span> since the maximum stress will be at the edge of the bar. Thus,</p>
<p><span class="math-container">$$\sigma_{max} = \frac{(0.98 kNm)(0.005m)}{4.91 x 10^{-10} \, m^4} = 9992 \,MPa$$</span></p>
<p>This far exceeds our assumed Fy of 250 MPa, so the 0.01 m diameter bar would definitely not remain elastic. You'll need to choose a section with a higher moment of inertia. I'd suggest setting up a spreadsheet to play around with various cross sectional geometries.</p>
<p> </p>
<h2>Check Deflection</h2>
<p>For an elastic beam experiencing small deflections, the following equation predicts deflection at the free end of an end-loaded cantilever.</p>
<p><span class="math-container">$$\delta_{max} = \frac{PL^3}{3EI}$$</span></p>
<p>Now, as discussed above, you'll need to change the section from the 1 cm diameter bar, but for purposes of illustration...</p>
<p><span class="math-container">$$\delta_{max} = \frac{(0.98 \,kNm)(2 \,m)^3}{3(200\,GPa)(4.91 x 10^{-10} \,m^4)} = 13.3 \,m $$</span></p>
<p>Granted, this isn't physically possible for a 2 m long beam, but in practice the key thing to note is simply that this beam doesn't meet the requirements to even utilize this equation. The beam isn't elastic and the deflection is definitely not small. The theoretical max deflection decreases fairly rapidly with increasing bar diameter - as user NMech noted, a 9 cm diameter bar experiences 2 mm end deflection. However, I'd expect you're not that keen on using a 9 cm diameter bar. If you really need a single 2 m long cantilever end-loaded beam, it'd be recommended to explore more efficient cross sections than a circular bar.</p>
<p> </p>
<h2>Bottom Line</h2>
<p>While a 1 cm diameter steel bar is not adequate for your load configuration, hopefully this information can give you a good start on selecting a suitable section.</p>
<p>Also note that the above neglected the self-weight of the beam. If you'd like to include self-weight in your analysis, it can be easily done by employing the principle of superposition (valid only for linear-elastic analyses). Essentially just add up the load effects from a cantilever beam with a uniformly distributed load (self-weight) and a cantilever beam with your point load at the free end. To that end, standard beam equations such <a href="https://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf" rel="noreferrer">these</a> come in very handy for simple analyses.</p>
| 36974 | Beginners material science for handling a load |
2020-08-03T18:51:09.677 | <p>I heard this term from a colleague, however I wasn’t sure what it really meant.</p>
| |terminology| | <p>Not sure if it truly applies to Engineering. But I have seen it as a ship moves through the commissioning process, especially military. Deficiencies during testing are documented so they can be fixed.</p>
<p>Commissioning process:</p>
<ul>
<li>Factory Acceptance Tests/Trials (FAT)</li>
<li>Shipboard Tests</li>
<li>Harbour Acceptance Tests (HAT)</li>
<li>Sea Acceptance Tests (SAT)</li>
</ul>
<p>From <a href="https://www.navsea.navy.mil/Portals/103/Documents/SUPSHIP/SOM/Ch10-TestTrialsAndDelivery-01Nov17.pdf?ver=2019-03-22-120308-147#page=24" rel="nofollow noreferrer">SUPSHIP Operations Manual (SOM) - Chapter 10 – Testing, Trials and Delivery</a></p>
<blockquote>
<p>Electronic Trial Cards will be utilized to document known noncompliant discrepancies, including items identified in outstanding Quality Deficiency
Reports/Corrective Action Reports. Deficiencies disclosed during all trials will be documented on Electronic Trial Cards and the status maintained in the management data base.</p>
</blockquote>
| 36978 | What does the term “trial card” mean in the field of engineering? |
2020-08-04T14:00:57.677 | <p>I have to deal with a situation where I have two parallel and completely independent rods whose motion is dependent only on external inputs, and I want to add up the lengths traveled by each rod.<br />
My first thought was to use two rack-and-pinion mechanisms by which I move two gears, one by each rod, and then couple them to a single rack. But this is not going to work since the rack when moved by one gear, will cause the other to move as well.</p>
<p>This seems to be an easy thing to do on a macro scale but I wish to 3D print this mechanism and they are going to be very small, so I expect to build a very simple mechanism.</p>
<p>Since the description is not so clear, here is a diagram:<br />
<a href="https://i.stack.imgur.com/k5KT6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k5KT6.jpg" alt="The Basic Idea" /></a></p>
<ul>
<li>When Rod A and Rod B are at A0 and B0 respectively, I want Rod Y to be at Y0</li>
<li>When either Rod A or Rod B is at state 1 and the other at state 0, I want Rod Y to be at Y1</li>
<li>When Rod A and Rod B are at A1 and B1 respectively, I want Rod Y to be at Y1</li>
</ul>
<p>Rods A and B should be absolutely independent of each other. And the gap between the lower face of Rod A and the upper face of Rod B should be around 35 mm, with all rods being about 3 mm wide. How can this be solved?</p>
| |mechanical-engineering|gears|machine-design| | <p>If I hadn't messed up understanding your idea then you can connect all rods to achieve the desired result.</p>
<p>I've considered two bars connecting to maintain all rods paralel to each other.</p>
<p>The starting position will be with Rod A and Rod B are inactive (position 0-0) and rod Y will remain in <span class="math-container">$Y_0$</span>.
<a href="https://i.stack.imgur.com/uJEjM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJEjM.png" alt="Initial position" /></a></p>
<p>The intermediary position Rod A <strong>or</strong> Rod B will be active, and the other is inactive. Rod Y will stay in position <span class="math-container">$Y_1$</span>.<br />
<a href="https://i.stack.imgur.com/fS7gh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fS7gh.png" alt="Intermediary position" /></a></p>
<p>The final position occur when both Rods A <strong>and</strong> B are active. Rod Y will stop in the position <span class="math-container">$Y_2$</span>.<br />
<a href="https://i.stack.imgur.com/6xHIR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6xHIR.png" alt="Final position" /></a></p>
| 36984 | Adding the effects of two rods to move a single rod |
2020-08-04T16:14:49.247 | <p>I am having trouble locating the calculations for finding the torque, in ft/lbs, required to turn a 1/2" screw with 13 threads per inch and pull 2000lbs 6 inches.<br />
.</p>
| |mechanical-engineering|mathematics| | <p>The circumstance of your screw is d<em>r = 0.5</em>3.141= 1.57 inch. So the slope of the thread is (1/13 )/1.57= 0.048 since this is a small slope we can assume sin of the slope angle = angle</p>
<p>If we ignore the friction which can be substantial, we need 2000* 0.048= 98lbs tangential force.</p>
<p>So your torque is</p>
<p>T= f*r</p>
<p>T = 98 *0.5/2= 24 lbs. inch</p>
| 36987 | How do I compute the torque in ft lbs to turn a screw and move a 2000 lb weight? |
2020-08-04T16:18:25.517 | <p>Here are screenshots of the data-sheets of two separate products for a reference. They both are nickel plated, used for coupling nuts. The <a href="https://www.turck.de/datasheet/_en/edb_6932999_gbr_en.pdf" rel="nofollow noreferrer">former</a> is shielded, the <a href="https://www.turck.de/datasheet/_en/edb_6626491_gbr_en.pdf" rel="nofollow noreferrer">latter</a> is not, I wonder if this is the main reason to chose one over the other.</p>
<p>GD-Zn coupling nut:</p>
<p><a href="https://i.stack.imgur.com/1S5Wn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1S5Wn.png" alt="GD-Zn product" /></a></p>
<p>CuZn coupling nut:</p>
<p><a href="https://i.stack.imgur.com/oJLFD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oJLFD.png" alt="CuZn product" /></a></p>
<p>I believe GD stands for Gadolinium (Gd), but I might be wrong. I could not find much on the internet, apart from the <a href="https://link.springer.com/chapter/10.1007/0-306-47090-X_75" rel="nofollow noreferrer">following</a>:</p>
<blockquote>
<p>Alloying Gd with Zn significantly reduces melting temperature of the alloys (the eutectic alloy melts at ∼860°C) compared to that of pure Gd (1313°C) and also improves the ductility over the GdZn intermetallide.</p>
</blockquote>
<p>This information however is about comparing Gadolinium Zinc alloy with pure Gadolinium, not pure Zinc. It hints us about the manufacturing process of such an alloy, and I suspect it has more to do with itch electro-magnetic properties, which are unclear to me.</p>
<p>What is the purpose of GD-Zn alloy? What advantages they provide over CuZn (Brass) alloy?</p>
| |materials|automotive-engineering|metallurgy|alloys| | <p>When I look on the net I found a surprising result : GD seems to stand for German Din (material specifications). GD Zinc is a zinc die cast alloy with 4 Cu and 1 Al. Equivalent to AG 41A alloy in the US. So one of the components is brass ,the other is an ( ordinary) zinc die cast.</p>
| 36989 | Purpose of a GD-Zn alloy |
2020-08-05T00:45:35.560 | <p>I have seen <a href="https://conferences.computer.org/vr-tvcg/2020/pdfs/VR2020-2f8MzUJjtCXG6Ue9RYFSN2/560800a382/560800a382.pdf" rel="nofollow noreferrer">this paper</a> about the <a href="https://en.wikipedia.org/wiki/Omnidirectional_treadmill" rel="nofollow noreferrer">Omnidirectional Treadmill</a> With CAD model ( with tiny rollers on it):</p>
<p><a href="https://i.stack.imgur.com/9oTPs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9oTPs.png" alt="enter image description here" /></a></p>
<p>Which compered its prototype with <a href="https://infinadeck.com/" rel="nofollow noreferrer">infinadeck</a> as shown below:</p>
<p><a href="https://i.stack.imgur.com/jXlGh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jXlGh.jpg" alt="enter image description here" /></a></p>
<p>but don't see those roller in <a href="https://infinadeck.com/" rel="nofollow noreferrer">infinadeck</a> as seen <a href="https://youtu.be/GoVAOfU8UJQ" rel="nofollow noreferrer">here</a>:</p>
<p><a href="https://i.stack.imgur.com/KmePS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KmePS.png" alt="enter image description here" /></a></p>
<p>So does the rollers in <a href="https://www.techspot.com/news/56821-is-this-the-best-omnidirectional-treadmill-for-gaming-yet.html" rel="nofollow noreferrer">infinadeck (more zoomed video)</a> are so small that can not be seen or that company use another mechanism for this device, for example like <a href="https://www.youtube.com/watch?v=NK41x5kenO4" rel="nofollow noreferrer">this</a>:
<a href="https://i.stack.imgur.com/FIYM6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FIYM6.jpg" alt="enter image description here" /></a></p>
<p><strong>Update:</strong></p>
<p>By seeing <a href="https://www.youtube.com/watch?v=fvu5FxKuqdQ&app=desktop" rel="nofollow noreferrer">this video</a> i guess Thanks, infinadeck don't have tiny rollers and only have the chain show in the CAD model of paper like this:</p>
<p><a href="https://i.stack.imgur.com/Q95yV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q95yV.png" alt="enter image description here" /></a></p>
<p>But i don't know i am right?!</p>
<p>Thanks.</p>
| |mechanisms| | <p>The infinadeck is literally a treadmill made of treadmills. The sideways motion works in the same way as a regular treadmill with belts stretched over rollers and driven by a motor. See <a href="https://m.youtube.com/watch?v=fvu5FxKuqdQ" rel="nofollow noreferrer">https://m.youtube.com/watch?v=fvu5FxKuqdQ</a> for more details</p>
| 36998 | How Infinadeck Omnidirectional Treadmill works |
2020-08-05T12:43:27.777 | <p>I have been melting aluminum in a DIY furnace for a few months in order to make lost wax casts. I was thinking that since bismuth has a much lower melting point than aluminum, that I could add a small amount of bismuth to lower the melting point (9 parts Al, 1 part Bi). Qualitatively this seems to be working. But I am confused.</p>
<p>Looking up a <a href="https://media.springernature.com/original/springer-static/image/art%3A10.1007%2Fs11669-014-0300-3/MediaObjects/11669_2014_300_Fig1_HTML.gif" rel="nofollow noreferrer">phase diagram for Al-Bi</a>, I find:</p>
<p><a href="https://i.stack.imgur.com/4kgJ7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4kgJ7.png" alt="phase diagram for Al-Bi" /></a></p>
<p>Which if I understand it correctly seems to mean that the melting point actually went up. Am I reading it correctly? If both metals separately have a lower melting point than the alloy at that ratio (and I am not reaching the new melting point), then what am I making?</p>
<p>Besides lowering the melting point, my hope was also to lower the viscosity so that the mix fills in my mold better. Will adding bismuth help accomplish this goal? Should I be adding tin instead? Is there a tin-bis-al ratio that would yield minimum viscosity at temperatures below 1500 °F (ideally mostly made out of Al)?</p>
| |metallurgy|alloys| | <p>You are not making anything ,usable. Bi does not raise the melt point until over 3.4 % Bi, that is lost in this diagram. If your aim is to make aluminum castings , look in a book and do it the way the rest of the world does, add Si. At about 12% Si the melting point is down to about 998 F, about as low as you will get . And,suprise, this is a common composition for aluminum alloy castings.</p>
| 37005 | What alloy am I making (Al-Bi) |
2020-08-05T14:46:30.807 | <p>While reading about the <em>Korvette K130 Braunschweig</em> class of corvettes for the German navy, I noticed that almost all of the weapon systems are held in what I'd call a <em>wire bearing</em>.
I'd like to know what its actual designation and its purpose are.</p>
<p>Examples:</p>
<ul>
<li><a href="https://de.wikipedia.org/wiki/MASS_(T%C3%A4uschk%C3%B6rperwerfer)#/media/Datei:Pori_MASS_decoy.JPG" rel="nofollow noreferrer">MASS</a></li>
<li><a href="https://de.wikipedia.org/wiki/Korvette_K130#/media/Datei:RBS_15_Mk3_-_Seezielflugk%C3%B6rper_-_Abschussrampe_(6059656292).jpg" rel="nofollow noreferrer">RBS 15</a></li>
<li><a href="https://de.wikipedia.org/wiki/Korvette_K130#/media/Datei:Korvette_Magdeburg_F261_(6059116043).jpg" rel="nofollow noreferrer">RAM</a></li>
</ul>
| |bearings|naval-engineering| | <p>These are called “wire rope dampers” or “wire rope isolaters”, and their purpose is to increase the time over which recoil energy is transferred into the mounting, thus reducing the peak force.</p>
| 37006 | Wire bearings on actuators on military ships |
2020-08-05T18:44:26.283 | <p>Im looking for alloys that have a melting point lower than 600°C and as high as possible hardness. Is there a search engine I could use to find those alloys with their physical properties?</p>
| |metallurgy|alloys| | <p>You can use MatWeb for a preliminary search of materials with desired physical properties.</p>
| 37012 | Where to find a list of hard low melting point alloys? |
2020-08-06T08:14:42.307 | <p>When we design a pipe or pressure vessel, we look at hoop stress and longitudinal stress and then look at the material properties. We assume that the material reacts to the load as a solid. However <a href="https://engineering.stackexchange.com/a/16308/61">This answer</a> and comments claim:</p>
<blockquote>
<p>Wall thickness doesn't matter if you try to contain a pressure that's greater than the yield strength. The material will permanently deform no matter how thick it is.
It's like i'm trying to contain a gas bubble in a soft mud. No matter how much mud I add around it - i can't contain it.</p>
</blockquote>
<p>Is this true? Do metals behave significantly different when subjected to pressures beyond their yield strength?</p>
<p>This could be tested experimentally: To arrive at a very thick walled pipe we could drill a hole into a block of gold (<a href="https://www.azom.com/properties.aspx?ArticleID=598" rel="nofollow noreferrer">ultimate tensile strength 220 Mpa</a>). Build the block large enough and the pipe wall thickness should withstand the immense pressure - So when we pressurize the hole to something > 220 MPa (a pressure common in water cutting, but at much other places.)</p>
<p>If the statement cited above is correct, we should see a different failure mode than a thin walled bursting vessel - for example creep. Or not, if material behaviour is dictated by the stress in the vessel material, not in the medium contained.</p>
<p>If we look at the pressures involved for even a soft metal like gold, we see that this a mostly academic question - for the vast majority of applications, the pressures encountered are well below the yield strneghts of the materials so we simply don't need to consider additional failure modes.</p>
<p>The above experiment is just an example, to show that such a situation could indeed be engineered. I'm sure other experiments can (and have) been done. My intution is that the material will start to flow in some way (towards the least resistance) but I'd like to know for certain.</p>
| |pressure-vessel|material-science| | <p>Maybe; Like most things it depends on specifics. Even at pressure below that which would cause general yield , there is local yield at stress concentrations and imperfections. This is shown when using acoustic emission to monitor a vessel during hydrotest; Typically 1.5 x design pressure. Sounds are produced by local plastic strain. Then methods such as ultrasound (UT) and radiography (RT) are used to evaluate the source. Any case I know of did not reveal a significant problem. There has probably been at least one case of failure of the vessel caused by overpressure. Over 60 years ago (long before I was an engineer) a pressure vessel disintegrated in a refinery in Whiting IN, (Standard Oil , IN) It was some kind of hydrocracker and had an internal explosion; I do not recall ever seeing an estimated peak pressure. Unfortunately it had low toughness even at the elevated operating temperature. Pieces were blown many hundreds of feet. I happen to have a fist size shard that was being disposed of many years later which reminded me of the accident.</p>
| 37024 | How does a pressure vessel fail when the interal pressure exceeds the yield stress of the material? |
2020-08-06T12:25:39.863 | <p>I'd like to clean my bicycle chain with as little effort as possible, so I thought an ultra-sound cleaner would be best. Professional oil-based ultrasound cleaners are way too expensive for a hobbyist, but water-based ultrasound cleaners are pretty cheap. So I could just run it with oil.</p>
<p>Could this work or is it doomed to fail because you need different kinds of transducers (or something like that)?</p>
<p>I understand that the risk is all mine, so I won't sue any of you for damages. I'd just like a more qualified opinion than my own. I'm assuming here that the worst that could happen is that it doesn't work and I end up with an oil-smeared device that I need to properly dispose of. Oil is less conductive than water, so I see no risk of shorting the device.</p>
| |ultrasound| | <p>Yes, it is possible.</p>
<p>I actually went ahead and bought a simple large (for an end-user device) ultra-sound cleaner and filled it with low-viscous oil. I operated it twice in the space of a few weeks. It still works and it does clean. The main challenge is to get the dirt out of the oil in the device's basin.</p>
| 37030 | Can I use a water-based ultrasound-cleaner with oil? |
2020-08-07T16:25:45.350 | <p>I am looking for a career advice as I am interested in chemistry and electronics at the same time but the universities somehow have the two as different major. Is there any branch of engineering that combines the two?</p>
| |electrical-engineering|chemistry| | <p>You could follow either Chemistry (Science), Chemical Engineering or Electrical Engineering and purse a career in Semiconductors. There are many variation in Semiconductor that fall either into chemistry or electronics. Below is small sample of options</p>
<ul>
<li><a href="https://www.nikon.com/about/technology/product/semiconductor/index.htm#:%7E:text=A%20semiconductor%20lithography%20system%20undertakes,substrate%20known%20as%20a%20wafer." rel="nofollow noreferrer">Semiconductor Lithography Systems</a></li>
<li><a href="https://www.mouser.com/applications/mems-technology/" rel="nofollow noreferrer">MEMS Technology</a></li>
<li><a href="https://eesemi.com/fa-techniques.htm" rel="nofollow noreferrer">Semiconductor Failure Analysis</a></li>
<li><a href="https://www.waferworld.com/silicon-wafer-processing-how/#:%7E:text=Reduction,takes%20six%20to%20eight%20hours" rel="nofollow noreferrer">Silicon Wafer Processing</a></li>
<li><a href="https://en.wikipedia.org/wiki/Microfluidics" rel="nofollow noreferrer">Microfluidics and microfluidic devices</a></li>
</ul>
| 37053 | What is the field on engineering that combines chemistry and electronics together? |
2020-08-07T18:22:45.640 | <p>So it is taught that arches are good to support structure and most bridges are build out of arches, so I am curious to know how we can determine the internal forces of such complex shapes. Is it only done by numerical methods such as FEA or there are analytical equations that help prove it.</p>
<p>Can you point to any famous analytical equations?</p>
| |mechanical-engineering|architecture| | <p>That is a good question.</p>
<p>The mechanical wisdom of arches manifested in the way the Loads flow, or directed, through the structure. If it is designed correctly, there are only compression loads all along the arch. In this case, when you design such a bridge, you most probably deal with the stability of the structure. I.e. studying its sensitivity to modifications in the estimated load distribution - which may come up with internal loads having moments or tensile components.</p>
<p>Please note that each load distribution has it own unique shape that "idealize" the internal loads (for a specific geometry like the span and the height of the bridge).</p>
<p>Designing arches and bridge is an ancient art, it has been established way before FEA was even a wild dream. One of the tricks of "seeing" it is done by a rope. a rope is quite stiff when you try to stretch it, But will deform easily in case you apply compression forces or try to bend it (apply a moment). Take a rope, pin it on both ends and hang weights all along. The rope will deform in a way it "feels" only tension.
Put only one weight in the middle of the rope and you will get a triangle. Arrange a uniform distributed weights all along the rope and you will get the classical hyperbolic shape.</p>
<p>If you arrange the weights in a way that represents the loads your upcoming bridge shall be subjected to - you can snapshot the rope shape, flip it, and here is your bridge. Now, since your bridge shape is inverted - It will "feel" only compression.</p>
| 37057 | What are the type of Equations that can determine if parabolas are better shapes to support structure or hyperbolas? |
2020-08-07T18:25:03.903 | <p>I have been testing basic circuits in <a href="https://www.simulide.com/p/home.html" rel="nofollow noreferrer">SimulIDE</a>. I created a basic exclusive or circuit that works most of the time.
There are two inputs.</p>
<p>When both inputs are off the bottom LED does not light up, that is correct.</p>
<p>When 1 of the 2 inputs are on, the bottom LED lights up, that is correct.</p>
<p>When both inputs are on, the bottom LED still lights up, even though an exclusive or circuit should only output ON when 1 of 2 inputs are on. In other words, if both inputs are on, the bottom LED should not light up.</p>
<p>I was wondering if timings could be the problem, but I'm not comfortable enough with circuits to trouble shoot this. Why does the bottom LED light up when both inputs are ON?</p>
<p><a href="https://i.stack.imgur.com/Lm79D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lm79D.png" alt="enter image description here" /></a>
(Bottom LED should be off)</p>
<p>Bottom LED should be offI hope this is the right place to ask, and that the tags are correct. If they are not, please let me know.</p>
<p>Edit 1:
So I began adding LEDs everywhere I can to test for an issue. Strangle adding an LED after the bottom OR fixes the issue. Adding an LED after the inverter also fixes the issue. The question now is, why is this happening?</p>
<p><a href="https://i.stack.imgur.com/2sZLV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2sZLV.png" alt="enter image description here" /></a>
(LED placed after bottom OR causes bottom right LED to turn off)</p>
<p><a href="https://i.stack.imgur.com/eYfHi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eYfHi.png" alt="enter image description here" /></a>
(LED placed after inverter also causes bottom right LED to turn off)</p>
<p>When I say "turn off" in the captions I'm referring to how that LED should behave in an exclusive or circuit.</p>
| |electrical-engineering|computer-engineering| | <p>After posting this question to the SimulIDE boards, I have found the answer. The LEDs internal property <code>resistance</code> had to be set to 100 for the circuit to function correctly. A user could also add resistors before the LED to achieve the same effect. <a href="https://www.simulide.com/p/forum.html" rel="nofollow noreferrer">This</a> is the link to the SimuleIDE forum. My question was posted under Help and was titled "Exclusive OR Circuit behaving strangely".</p>
<p><a href="https://i.stack.imgur.com/ahvdb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ahvdb.png" alt="enter image description here" /></a></p>
| 37058 | Exclusive Or Circuit in SimulIDE |
2020-08-08T07:03:57.957 | <p>This is related to <a href="https://engineering.stackexchange.com/questions/23612/deflection-of-a-cantilever-beam-composed-of-separate-not-bonded-planks">this question </a></p>
<p>It is the case of a simple cantilever beam having a uniform rectangular cross-section.</p>
<p>cantilever beam composed of separate planks - taken from <a href="https://i.stack.imgur.com/QHDLZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QHDLZ.jpg" alt="http://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/" /></a></p>
<p>The question in the original post was <strong>What happens if we split it along its longitudinal axis?</strong>
<a href="https://engineering.stackexchange.com/users/28221/sudheer-reddy-tenali">@sudhir</a> asked <strong>What would happen if the beam was loaded uniformly</strong></p>
| |mechanical-engineering|structural-engineering|beam|deflection|shear| | <p>In the original post the deflection of the coupled beam with the decoupled beam was</p>
<p><span class="math-container">$$ \delta_{decoupled, concentrated} = 9 \delta_{coupled,concentrated}$$</span></p>
<p>If you change to a uniform load then with respect to the relative deflection of the coupled/decoupled would remain the same. i.e. <span class="math-container">$\delta_{decoupled, uniform} = 9 \delta_{coupled, uniform}$</span></p>
<p>Of course, the total deflection would be less because the formula for <span class="math-container">$\delta$</span> will change(see <a href="https://www.chegg.com/homework-help/questions-and-answers/solve-deflection-cantilever-uniform-load--boundary-condition-x-0-y-0-x-0-dy-dx-0-q5606113" rel="nofollow noreferrer">link</a> ).</p>
<p>Assuming that the distributed load is <span class="math-container">$w=\frac{P}{l}$</span> (so that the total load is the same for comparison purposes:</p>
<p><span class="math-container">$$\delta_{coupled, uniform} = \frac{w\cdot l^4}{4\cdot I_{coupled}}=\frac{P\cdot l^3}{4\cdot I_{coupled}} = \frac{3}{4} \delta_{coupled, concentrated} $$</span></p>
| 37066 | Deflection of a cantilever beam composed of separate (not bonded) planks Uniform load |
2020-08-09T01:26:38.343 | <p>The oil spill in Mauritius would be caused by cracks in the hull. How can this happen? Is it possible to be caused solely by rough weather?</p>
| |marine-engineering| | <p>MV Wakashio ran aground on a coral reef off Mauritius on July 25. MV Wakashio is a 300m Bulk Ore Carrier. It was empty at the time (no cargo).</p>
<p>From BBC: <a href="https://www.bbc.com/news/world-africa-53702877" rel="nofollow noreferrer">Mauritius declares emergency as stranded ship leaks oil</a>:</p>
<p><a href="https://i.stack.imgur.com/VCuYC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VCuYC.png" alt="enter image description here" /></a></p>
<p>You can clearly see the reef and direction of waves hitting reef. No oil, when this picture was taken.</p>
<p>From IPR <a href="http://www.ipreunion.com/actualites-reunion/reportage/2020/08/09/-live-desastre-ecologique-a-maurice-le-mv-wakashio-continue-de-deverser-du-fuel-dans-le-lagon,122884.html" rel="nofollow noreferrer">Mauritius: oil spill from damaged tank is over</a>:</p>
<blockquote>
<p>Since Thursday August 6, 2020, the bulk carrier MV Wakashio has been emptying its fuel in the lagoon of Pointe d'Esny, in Mauritius. Stranded since July 25, the Mauritian government was slow to react. After nearly twelve days off the coast, a crack finally formed in one of the holds.</p>
</blockquote>
<p>From Press release covered by ThisNews: <a href="https://ythisnews.com/mauritius-on-ship-leaking-tonnes-of-oil-in-the-indian-ocean/#:%7E:text=Nagasaki%20Shipping%20Company%20of%20Japan,due%20to%20the%20weather%20effects.&text=He%20added%2C%20%E2%80%9CDue%20to%20the,has%20escaped%20into%20the%20sea." rel="nofollow noreferrer">Mauritius on ship leaking tonnes of oil in the Indian Ocean</a>.</p>
<blockquote>
<p>The Minister disclosed that the bulk carrier was travelling with 200 tonnes of diesel, 3,894 tonnes of bunker fuel in the form of low-sulphur fuel oil and 90 tonnes of lube oil.</p>
</blockquote>
<blockquote>
<p>Due to the bad weather and constant pounding over the past few days, the starboard side bunker tanker has been breached and an amount of fuel oil has escaped into the sea.</p>
</blockquote>
<p>From video on FleetMon <a href="https://www.fleetmon.com/maritime-news/2020/30533/major-oil-spill-mauritius-island/" rel="nofollow noreferrer">Major oil spill off Mauritius Island</a></p>
<p><a href="https://i.stack.imgur.com/2a40r.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2a40r.jpg" alt="enter image description here" /></a></p>
<hr />
<p>Grounding had a role to play, but hull breach was caused by storm wind and waves leveraging grounded 300m hull. All ships flex and twist due to wave action. But if hull is grounded, the laws of physics take over.</p>
<p>An investigation will have to take place, but empty ship and grounding about midship will be significant to findings.</p>
<p>YouTube: <a href="https://www.youtube.com/watch?v=Nrnnc4AsEec" rel="nofollow noreferrer">Radio One: La ‘fissure’ du MV Wakashio vue de prés</a></p>
| 37072 | How do cracks form on a ship's hull? |
2020-08-09T16:14:32.217 | <p>A number of years ago, I remember hearing of a technique to manufacture the leading edge surface of a hypersonic jet (or something aerospace related) where multiple sheets of metal were welded together in a specific pattern to produce a laminate. Before welding, a layer of some sort of was placed between the metal so that when heated it would evolve a gas, inflating the unwelded pockets between the two sheets to create a super-thin cooling channels between the two layers.</p>
<p>What is this called? Was it a concept or is it used in industry?</p>
| |aerospace-engineering|manufacturing-engineering| | <p>Aside from "evolving a gas by heating" (which sounds improbable IMO) the sounds like superplastic forming and diffusion bonding, which are standard manufacturing techniques.</p>
<p>In SPF the metal layers are bonded together in a pattern with "holes" in the bond, and inert gas is used to "inflate" the component into its final shape within a mold.</p>
<p>Diffusion bonding has been a "solution looking for a problem" for about 40 years, until the aerospace industry had the need to make metallurgically "perfect" bonds between large surfaces with complex shapes. The principle is very simple: make two surfaces with a high degree of accuracy, hold them together under moderate pressure, heat them up so that the atoms can diffuse across the boundary. The main disadvantage is that it is a relatively slow process - the diffusion takes minutes, compared with seconds for welding etc.</p>
| 37075 | Technology used to create chambers in metal laminates |
2020-08-10T06:10:32.037 | <p>I recently saw a video where a glassblower was making something, and one of the first steps was to push the wad of molten glass down into a shaper with a number of vertical spikes, such that the result vaguely resembled a gear.</p>
<p>I am aware that gears can be made of plastics and wood (e.g. <a href="https://woodgears.ca/gear_cutting/index.html" rel="nofollow noreferrer">https://woodgears.ca/gear_cutting/index.html</a>) as well as metals, but would glass (perhaps a stronger variant such as tempered or borosilicate glass) be suitable to make gears out of? Are there any historical examples of this?</p>
<p>Obviously with modern materials and techniques steel is probably the best choice in an industrial setting, but I am curious as to it's suitability as an intermediate step between cheap but quickly worn down wooden gears for rapid prototyping and metal gears for serious use. I would imagine it could be a lot easier and cheaper to melt glass in a home workshop and pour it into a mold or shape it appropriately as compared to doing the same thing with metals (keeping in mind equipment costs e.g. having to build a foundry capable of melting steel).</p>
| |gears|glass| | <p>Another point, that hasn't been mentioned: Over the time you will have some wear, which means scratches on the surface. These scratches can be starting points for cracks. That's why gorilla glass uses pressure stress in the surface.</p>
| 37086 | Feasability of glass gears? |
2020-08-10T19:04:16.277 | <p>In case a plane stalls, what should stall first, the tip or the root of the Wing?
Why?</p>
| |aerospace-engineering|aircraft-design| | <p>It depends on wing geometry. You need to look at the aerodynamic center of the wing, which is the center of effort for perturbation forces. The center of lift (unstalled) will usually be well forward of the aerodynamic center. As Cl increases, it is desirable to have the center of lift shift aft. This produces speed stability by lowering the nose, thus gaining speed and restoring the original Cl. This is a static stability condition - it says nothing about how the pertubation gets damped.</p>
<p>At local incipient stall, the local drag goes up and the local lift drops precipitously. That lost lift will have to be picked up by the rest of the wing.</p>
<p>In a swept back wing, if the root stalls first, the lift lost is forward of the lift that replaces it, shifting the center of lift aft.</p>
<p>In a swept forward wing, if the tip stalls first, the lift lost is forward of the lift that replaces it, shifting the center of lift aft.</p>
<p>Here's a doc on the stall characteristics of a flat swept-forward wing which naturally begins to stall at the root. - <a href="https://www.abbottaerospace.com/downloads/naca-tn-1797-a-study-of-stall-phenomena-on-a-45-swept-forward-wing/" rel="noreferrer">naca-tn-1797 A Study of Stall Phenomena on a 45° Sweot Forward Wing.pdf</a> {I can spell <em>swept</em>, but whoever scanned and converted it couldn't ;)}</p>
| 37096 | In case a plane stalls, what should stall first, the tip or the root of the Wing? Why? |
2020-08-11T04:27:49.073 | <p>I have a system determined by 3 Euler angles that describe the orientation with respect to a fixed coordinate system XYZ.</p>
<p>The angular velocities are: <span class="math-container">$\omega_2$</span> (precession), <span class="math-container">$\omega_1$</span> (nutation) and <span class="math-container">$\omega_z$</span> (spin).
The angular acceleration of <span class="math-container">$\omega_z$</span> is <span class="math-container">$\alpha_z$</span>. The other angular accelerations are 0.</p>
<p>How can I find the acceleration of a point (e.g. A on the spinning reference frame xyz)?
I am looking for the procedure to follow.</p>
<p><a href="https://i.stack.imgur.com/8A4lx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8A4lx.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|kinematics|acceleration| | <p>First of all you need to be about which reference frame you are estimating the acceleration of A. I assume you need the accelation of A wrt to the inertial frame of reference.</p>
<p>The most generic way is to link the acceleration of A to B. Then find the acceleration of B with respect to O.</p>
<p>See the following <a href="https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec25.pdf" rel="nofollow noreferrer">link for a starting point </a></p>
<p>The bottom line is that if you need the acceleration of P with respect to O, then you'll probably just need the following two equations:</p>
<p><span class="math-container">$$\vec{v}_P =\vec{v}_O + \vec{\omega}\times \vec{r}_P$$</span>
<span class="math-container">$$\vec{a}_P =\vec{a}_O + \dot{\vec{\omega}}\times \vec{r}_P + \vec{\omega}\times \left( \vec{\omega}\times \vec{r}_P\right)$$</span></p>
| 37104 | Euler rotations and acceleration |
2020-08-12T14:59:23.047 | <p>First off non-engineer apologies,</p>
<p>A plate of 5251 <a href="http://www.series2club.co.uk/forum/forum/pruned_attachments/5251.pdf" rel="nofollow noreferrer">Aluminium</a> is used to support the load of a centrally seated human I estimate a load of 1 kilonewton.</p>
<p>The beam measures a length of 1000mm, 100mm wide.</p>
<p>For a thickness 6mm I calculate the <a href="https://www.omnicalculator.com/math/moment-of-inertia" rel="nofollow noreferrer">Moment of Inertia</a> to be 1800mm4 and given a Elastic modulus of 70000MPa the <a href="https://www.omnicalculator.com/construction/beam-deflection" rel="nofollow noreferrer">maximum deflection</a> to be about 14cm.</p>
<p>Assuming the beam will fail under tension of 130-240 N/mm2 will it break or permanently deform ?</p>
<p>How thick would the plate have to be before the process would be elastic ?</p>
<p>(6mm is the thickest plate though I could stack them; I can get 5mm being much cheaper)</p>
| |stresses|beam|deflection| | <p>Using the properties you mentioned, the equation for bending stress is</p>
<p><span class="math-container">$$ \sigma = \frac{M}{I_{xx}}\cdot y_{max}$$</span></p>
<p>where :</p>
<ul>
<li><span class="math-container">$M=F\frac{l}{2}=0.5kNm$</span> (that's a lot)</li>
<li><span class="math-container">$I_{xx} $</span>= 1800mm4</li>
<li><span class="math-container">$y_{max} = \frac{thickness}{2}=3[mm]$</span></li>
</ul>
<p>The results is a whopping 833.[MPa], so it will fail.</p>
<p>Assuming you use a solid beam (same material), you can estimate the minimum thickness from the following equation:</p>
<p><span class="math-container">$$ h_{min}= \sqrt{\frac{6 M}{b\cdot \sigma_{max}}}$$</span></p>
<p>The only thing new thing here is that <span class="math-container">$\sigma$</span> is substituted for <span class="math-container">$\sigma_{max}$</span>. Al does not have a clear yield area like steel has. Therefore "elastic region" is a bit vague (normally you'd settle for proof stress). However, given the ultimate tensile properties you posted and the intended use, I'd settle for at least 100[MPa], which yields a minimum thickness of 17.5[mm].</p>
<p>OK, now for the tricky part. If you plan to stack them, then you'd need a lot more. A rough estimation is between 8 or 9 plates of 6[mm]. For why that happens you might want to read the following <a href="https://engineering.stackexchange.com/questions/23612/deflection-of-a-cantilever-beam-composed-of-separate-not-bonded-planks/37043#37043">question</a></p>
| 37123 | simply supported beam with a midspan load |
2020-08-12T17:24:49.403 | <p>I have spent the past few days thinking about a good way to get power/RPM from a fixed wheel delivered to a wheel that is able to move in a 2D plane relative to the fixed wheel. I am not too good at CAD, but I tried to make a schematic to make it easier to understand:</p>
<p><a href="https://i.stack.imgur.com/nslmh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nslmh.png" alt="enter image description here" /></a></p>
<p>The orange "wheel" is fixed and is driven by a shaft. I want to get that RPM to the purple wheel in the middle. The purple wheel will be moved within the red frame (but not up/down). Some other constraint or actually simplification: I dont care if the RPM fluctuates during movement, so it can get lower/higher if the purple wheel is moving. Also the position of the orange wheel can be changed as needed, but once it is attached, it can no longer move. The orange wheel can be outside of the red frame too. The purple wheel drives a tool that needs to work against some resistance (say drilling through wood).</p>
<p>I was thinking about using a belt or a chain with a tensioner. My idea was to install another wheel either diagonally opposite of the orange wheel or at the bottom of the frame that is attached to spring loaded tensioner, but I failed to figure out how long my belt would have to be and how to make sure there is always enough tension to make sure the purple wheel gets enough torque to be able to deliver enough force to the tool it drives.</p>
<p>Are there some well known patterns for my problem? Or what would be a good way to tackle that problem?</p>
| |gears|power-transmission| | <p>you can use a pantograph belt. They use them in large machines to transfer rotation to a moving drill to cut squares and what not.</p>
<p>Basically they are several pullies assembled on a pantograph with links designed to move as prescribed.</p>
| 37128 | Delivering RPM from a fixed wheel to a moving wheel |
2020-08-13T08:39:25.407 | <p>The following device is generally used in washing machines to dispense water and cleaning solution in the right amounts whenever needed. How does this device work?</p>
<p><a href="https://i.stack.imgur.com/0XKEN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0XKEN.jpg" alt="device" /></a></p>
| |mechanical-engineering|electrical-engineering| | <p>Building on the responses from <a href="https://engineering.stackeSxchange.com/a/37135/8055" rel="nofollow noreferrer">Mahendra Gunawardena</a> and <a href="https://engineering.stackexchange.com/a/37142/8055">Niels Nielsen</a> solenoid valve basically has a solenoid body and valve body as shown below.</p>
<p><a href="https://i.stack.imgur.com/bQTLc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bQTLc.png" alt="enter image description here" /></a></p>
<p>They are many different type of solenoid valves. direct acting, pilot-operated, two ways and many more. Below is an image showing the basics internal of a solenoid.</p>
<p><a href="https://i.stack.imgur.com/HAlsm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HAlsm.png" alt="enter image description here" /></a></p>
<p>At high level there are normal close and normally opened a solenoid. Below is the high level mechanisms of a normally closed solenoids</p>
<blockquote>
<p>Inside the valve we have the armature. The solenoid is placed over
this and completely surrounds the armature so that it’s at the center
of its magnetic field. Inside the cylinder of the armature is the
plunger and spring.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/14Gsi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/14Gsi.png" alt="enter image description here" /></a></p>
<blockquote>
<p>The spring pushes the plunger down in a normally closed type valve.
Because the plunger is pushed by the spring, it will sit in the down
position to close the valve indefinitely. But, if the coil receives an
electrical current then it will generate an electromagnetic field and
this magnetic field passes through the plunger and will cause it to
move upwards against the spring therefore opening the valve</p>
<p>Once the electrical current is stopped then the magnetic field
disappears and the spring will force the plunger down again to close
the valve.</p>
</blockquote>
<p>Below is the high level mechanisms of a normally opened solenoids.</p>
<p><a href="https://i.stack.imgur.com/2KBbo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2KBbo.png" alt="enter image description here" /></a></p>
<blockquote>
<p>With normally open valves we again have the coil sit around the
armature but this time the spring pushes the plunger in the upwards
position so that the valve is always open unless the solenoid coil is
powered.</p>
<p>If we then pass a current through the coil it again creates an
electromagnetic field but this time the field pushes the plunger
instead of pulling it. When the plunger is pushes it will close the
valve and stop the flow of fluid in the system.</p>
<p>When the electrical current is stopped the spring will then force the
plunger back into the upward position and open the valve again.</p>
</blockquote>
<p>To learn more about solenoid valves check out <a href="https://theengineeringmindset.com/how-solenoid-valves-work/" rel="nofollow noreferrer">How Solenoid Valves Work</a>. Additionally. <a href="https://youtu.be/-MLGr1_Fw0c" rel="nofollow noreferrer">How Solenoid Valves Work - Basics actuator control valve working principle</a> video might be useful too.</p>
| 37134 | This device is used in washing machines to dispense water and cleaning solution in the right amounts. What is the working principal of it? |
2020-08-13T13:31:47.513 | <p>In the <a href="https://en.wikipedia.org/wiki/State-space_representation#Linear_systems" rel="nofollow noreferrer">state space representation</a>, the state equation for a linear time-invariant system is:</p>
<p><span class="math-container">$$
\dot{\mathbf{x}}(t) = \mathbf{A}\mathbf{x}(t)+\mathbf{B}\mathbf{u}(t)
$$</span></p>
<p>This state equation can be derived by decomposing an <span class="math-container">$n^{th}$</span> order differential equation into <span class="math-container">$n$</span> first-order differential equations and then choosing the state variables <span class="math-container">$x_1(t),x_2(t),...,x_n(t)$</span> and their derivatives <span class="math-container">$\dot{x}_1(t),\dot{x}_2(t),...,\dot{x}_n(t)$</span>. The state equation essentially describes the relationship between the state variables and the inputs in <span class="math-container">$\mathbf{u}(t)$</span>.</p>
<p>Additionally, the output equation for a linear time-invariant system is:</p>
<p><span class="math-container">$$
\mathbf{y}(t) = \mathbf{C}\mathbf{x}(t)+\mathbf{D}\mathbf{u}(t)
$$</span></p>
<p>However, I am not sure how this output equation is derived. More precisely, what is an "output"? Is it the set of state variables and inputs that need to be observed by the engineer or another system downstream? If that is true, then if I have a mass-spring-damper system, where the displacement of the mass is represented by the state variable <span class="math-container">$x_1(t)$</span>, the velocity of the mass is represented by the state variable <span class="math-container">$x_2(t)$</span>, and an externally applied force on the mass is represented by the input variable <span class="math-container">$u_1(t)$</span>, and I was interested in observing/measuring the displacement of the mass, would my output equation then be:</p>
<p><span class="math-container">$$
y(t) = x_1(t)
$$</span></p>
<p>Alternatively, if I was interested in observing both the displacement of the mass and the externally applied force, then would my output equation be:</p>
<p><span class="math-container">$$
\mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix} = \begin{bmatrix} x_1(t) \\ u_1(t) \end{bmatrix}
$$</span></p>
<p>So far, neither the state variables nor the inputs have been scaled in my output equation. Because of this, I don't understand the purpose of the <span class="math-container">$\mathbf{C}$</span> and <span class="math-container">$\mathbf{D}$</span> matrices. Could they be used to linearly transform the state variables and inputs for another system downstream? From this image on a typical state space representation:</p>
<p><span class="math-container">$\hskip2in$</span> <img src="https://upload.wikimedia.org/wikipedia/commons/e/eb/Typical_State_Space_model.svg" alt="Text" /></p>
<p>It seems that what I am saying is correct, but I would prefer a better explanation.</p>
| |control-engineering| | <p>I am not entirely sure whether your question has to do with C and D matrices, or with how and why to select the output variables. I'll try to tackle both.</p>
<p><strong>Regarding the latter (how and why you decide on the output variables):</strong></p>
<p>You are right, in that for a simple system, there is not much point in developing the <span class="math-container">$\mathbf{y}$</span> vector and corresponding equation. In terms of logistics, you already have the relevant data.</p>
<p>However, in more complex system you might be interested in the response of only a few of the state variables, or in linear combinations of them.</p>
<p>So I tend to think the <span class="math-container">$\mathbf{y}$</span> as a way to perform a "boil down to essentials". However, there are other reasons, which I believe you suspect as linear combination of solutions to obtain a transformed solution</p>
<p><strong>Example:</strong> think of the following system with two mass springs you can select either the absolute <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> displacements of the mass.
<img src="https://www.researchgate.net/profile/Natalie_Baddour/publication/262957549/figure/fig1/AS:280381904441384@1443859663568/2DOF-spring-mass-system.png" alt="2 dof mass system" /></p>
<p>Another equivalent representation is <span class="math-container">$x_1$</span> and <span class="math-container">$x_2-x_1$</span> (essentially the deformation of the spring). If you are only interested in the deformation of the string you might create a <span class="math-container">$C=[1, -1]$</span> and you are done. However, it might be easier to see, that it is easier to construct the equations for <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span> because their odes are similar (while constructring the ode for the spring deformation will be different).</p>
<p><em>Bottom Line:</em> state representation makes much more sense in more complex systems.</p>
<p><strong>Regarding the use of C and D matrices</strong></p>
<p>The first good reason that comes to mind for the use of C and D matrices, is to perform the Observability and Controlability Tests. Its in the same link you provided <a href="https://en.wikipedia.org/wiki/State-space_representation#Controllability" rel="nofollow noreferrer">wikipedia link</a>.</p>
| 37136 | How do I determine the output equation in the state-space representation? |
2020-08-14T16:33:53.683 | <p>A spool has a mass of 500 kg and a giration radius of <span class="math-container">$k_G = 1.30m$</span>. The
spool lies on the surface of a conveyor belt, the static coefficient of friction is <span class="math-container">$\mu_S=0.5$</span>.
The spool is not moving at the beginning.</p>
<p>Determine:</p>
<ul>
<li>the largest acceleration <span class="math-container">$a_T$</span> of the conveyor belt, such that the spool does not slip.</li>
<li>the pulling force in the rope.</li>
</ul>
<p><a href="https://i.stack.imgur.com/E0hnh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E0hnh.png" alt="enter image description here" /></a></p>
<p>The equations I have are:</p>
<p><span class="math-container">$$I_G = m \cdot r^2 = 500 kg \cdot (1.3 m)^2 = 845 kgm^2 $$</span></p>
<p><span class="math-container">$$\sum_i \vec{F_i} = m_G \cdot \vec{a_G}$$</span></p>
<p><span class="math-container">$$\vec{N} + \vec{T} + \vec{F_z} + \vec{F_w} = m \cdot \vec{a_G} $$</span></p>
<p>After projection I get:</p>
<p><span class="math-container">\begin{equation}
\begin{split}
T - F_w & = m_G \cdot a_G \\
N - F_z & = 0
\end{split}
\end{equation}</span></p>
<p>For the moments I get:</p>
<p><span class="math-container">$$\sum \vec{M_G} = I_G \cdot \vec{\alpha}$$</span></p>
<p><span class="math-container">\begin{equation}
T \cdot 0.8 - F_w \cdot 1.6 = I _G \cdot \alpha \\
\end{equation}</span></p>
<p>Other relations are</p>
<p><span class="math-container">\begin{equation}
\begin{split}
-a_G + 1.6 \cdot \alpha & = a_T \\
a_G & = -0.8 \cdot \alpha \\
F_w & = N \cdot \mu
\end{split}
\end{equation}</span></p>
<p>The solution from the teacher is:</p>
<p><a href="https://i.stack.imgur.com/lSO2p.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lSO2p.png" alt="enter image description here" /></a></p>
<p>This solution does not satisfy</p>
<p><span class="math-container">$$a_G = -0.8 \cdot \alpha$$</span></p>
<p>My question is: "Is this last equation wrong?"</p>
| |mechanical-engineering|friction|acceleration| | <p>Point <span class="math-container">$B$</span> is the system's initial instantaneous center. So the acceleration of point <span class="math-container">$A$</span> is equal and opposite to the acceleration of <span class="math-container">$G$</span> at <span class="math-container">$t=0$</span>. This is the trick to solving this problem.</p>
<p>There is a force couple of <span class="math-container">$2452.5\,N$</span> at slip inception with an arm of <span class="math-container">$0.8\,m$</span> to the instantaneous center <span class="math-container">$B$</span>. There is some additional tension on the rope needed to provide the acceleration of mass to the right. Use parallel axis theorem to calculate <span class="math-container">$I$</span> about <span class="math-container">$B$</span>.</p>
<p><span class="math-container">$\alpha = F_s* 0.8\,m)/(I_g + mass*.8\,m^2)$</span>
<span class="math-container">$\alpha = (2452.5\,kgm/sec^2 * 0.8\,m)/(845\,kgm^2 + 500\,kg*.8\,m^2)$</span> = <span class="math-container">$1.684\, rad/sec^2$</span> clockwise.</p>
<p>acceleration of <span class="math-container">$G =\alpha* 0.8\,m$</span><br />
acceleration of <span class="math-container">$G =1.684\,rad/sec^2 * 0.8\,m = 1.347\,m/sec^2$</span> to the right.<br />
that takes a force of <span class="math-container">$1.347\,m/sec^2 * 500kg = 673.65\,N$</span></p>
<p><span class="math-container">$F_s = 2452.5/,N$</span> to the left.<br />
<span class="math-container">$T = 2452.5\,N + 673.65\,N = 3126.1\,N$</span> to the right.</p>
<p>acceleration of belt = <span class="math-container">$1.6\,m * 1.684\,rad/sec^2 - 1.347\,m/sec^2 = 1.347\,m/sec^2$</span> to the left.</p>
<p>The answer provided in the question seems to have made the mistake of assuming that <span class="math-container">$G$</span> goes to the right half as fast as A goes to the left. They didn't realize that <span class="math-container">$G$</span> is taking <span class="math-container">$A$</span> with it. That is really the entire point behind this classic problem.</p>
<p>This is how you were meant to work the problem -</p>
<ol>
<li>Gather the torque terms and write an acceleration equation in terms of the
unknowns at the no-slip limit.</li>
<li>Gather the force terms and write an acceleration equation in terms of the
unknowns.</li>
<li>Relate G acceleration <span class="math-container">$a_g$</span> to belt acceleration <span class="math-container">$a_b$</span> via the unknown rotational
acceleration <span class="math-container">$\alpha$</span>.</li>
<li>And lastly, note that <span class="math-container">$a_b = - a_g$</span>.</li>
</ol>
<p>Four equations and four unknowns <span class="math-container">$\{T,a_b,a_g,\alpha\}$</span> - solve them.</p>
<ol>
<li>Taking moments about <span class="math-container">$G$</span>: <span class="math-container">$\quad\alpha =
[(-2452.5\,N)*1.6\,m+T*0.8\,m]/I_g$</span></li>
<li>Summing forces on spool: <span class="math-container">$\quad a_g=[(-2452.5\,N) + T]/M_g$</span></li>
<li>Relating <span class="math-container">$a_b$</span> to <span class="math-container">$a_g$</span>:<span class="math-container">$\quad a_b=a_g + \alpha*1.6\,m$</span></li>
<li>Relating <span class="math-container">$a_b$</span> to <span class="math-container">$a_g$</span>:<span class="math-container">$\quad a_b= -a_g$</span></li>
</ol>
| 37157 | Acceleration of spool on conveyor belt |
2020-08-15T09:20:51.243 | <p>My professor told us that it is possible to see the friction in a mass-friction-spring as the contribution of a closed loop control system. He wrote the following formulas:</p>
<p>Transfer function:
<span class="math-container">$$G(S)=\frac s {s^2+\frac k M}$$</span>
gain:
<span class="math-container">$$\rho=h/M$$</span>
where k=spring constant, h=friction constant, M=mass</p>
<p>Can you recognize these formulas and give me some indication about this method?
I think it would be useful to understand what is the meaning of the transfer function is this case?</p>
| |control-theory|transfer-function|feedback-loop| | <p>Based on the information you've given, I believe your professor is suggesting that a friction term can be represented as shown in the following block diagram.</p>
<p><a href="https://i.stack.imgur.com/aFtqb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aFtqb.png" alt="A block diagram showing G(s) as the plant and rho in the feedback path." /></a></p>
<p>The transfer function <span class="math-container">$G(s)$</span> relates force (<span class="math-container">$F(s)$</span>, the input in the diagram) to velocity (<span class="math-container">$sX(s)$</span>, the output in the diagram) for a mass-spring system. The damping (<span class="math-container">$\rho$</span>) is represented in the feedback loop as a proportional gain acting on the velocity. We can use the feedback rule to create a single, closed loop transfer function:</p>
<p><span class="math-container">\begin{align}
G_{cl}(s) &= \frac{G(s)}{1+\rho G(s)},\\
&= \frac{\frac{s}{s^2+k/M}}{1 + \left(\frac{h}{M} \right)\frac{s}{s^2+k/M}},\\
&= \frac{s}{s^2 + \frac{h}{M}s + \frac{k}{M}}.
\end{align}</span></p>
<p>As you can see, the closed loop transfer function is the same as the general mass-spring-damper transfer function.</p>
| 37167 | mass-friction-spring system with closed loop |
2020-08-15T12:09:01.800 | <p>Although curing of concrete under plastic sheets is a recognized technique, I have some doubts about it.</p>
<p>I think that the plastic sheets may trap heat coming from the exothermic reaction, which will result in important temperature rise in fresh concrete.</p>
<p>I am concerned about heat coming from inside and not outside.</p>
<p>My question is: is this a real issue?</p>
| |civil-engineering|concrete| | <p>When casting large machine foundations at sites in tropical conditions we always covered the concrete with sacking that was constantly kept wet to absorb the heat of the curing concrete. We never used non-absorbant covering like plastic.</p>
| 37168 | Trapping heat by curing of concrete under plastic sheets |
2020-08-16T16:39:33.273 | <p><a href="https://i.stack.imgur.com/uW6pr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uW6pr.png" alt="enter image description here" /></a></p>
<p>According to Building codes, when calculating the maximum load a fillet weld can take, one checks only that the stress in pure shear is below the maximum shear strength. We know that shear yielding stress and tensile yield stress are related (using the Von Mises Criterion for the onset of yield):</p>
<p><span class="math-container">$$\sigma_s = \frac{\sigma_y}{\sqrt(3)}\approx0.6*\sigma_y$$</span></p>
<p>where <span class="math-container">$\sigma_s$</span> is the yield stress in yield and <span class="math-container">$\sigma_y$</span> is the yield stress in tension.</p>
<p>But why do we assume the weld is in a state of pure shear? <strong>Why is this a valid assumption?</strong></p>
| |stresses|welding| | <p>First of all, one small but important note:</p>
<p>The relationship between shear yield stress <span class="math-container">$S_{sy}$</span> and the (tensile) yield stress <span class="math-container">$S_y$</span> is dependent on the failure theory.</p>
<ul>
<li>Von Mises: <span class="math-container">$S_{sy} = 0.577 S_y\approx 0.6 S_y$</span></li>
<li>Tresca: <span class="math-container">$S_{sy} = 0.5 S_y$</span></li>
</ul>
<p>I.e. <em>the Tresca is a more conservative criterion.</em>. That is probably the reason that it is preferred for materials with brittle failure. And altough normally steel can be considered as ductile, the Heat Affected Zone (HAZ) around the weld usually exhibits a more brittle failure. Therefore, Tresca seems to be more appropriate.</p>
<p>Also I don't know if the Building code you are referring to explicitly states the Von Mises relation, or is just saying "shear stress"</p>
<p>Let's proceed to the calculation, the total force passing through each weld is <span class="math-container">$\frac F 2$</span>.</p>
<p>Also let's assume a length of weld equal to l.</p>
<p>The force needs to pass through every cross-section that passes from the lower left corner of the blownup image of the weld. We can examine the following 3 cases.</p>
<ol>
<li>horizontal crosssection (cross-section area <span class="math-container">$\sqrt 2 a l$</span>) normal stress</li>
<li>diagonal cross-section (cross-section area <span class="math-container">$a l$</span>) combination of normal and shear</li>
<li>vertical cross-section (cross-section area <span class="math-container">$\sqrt 2 a l$</span>) shear stress</li>
</ol>
<p>In the following analysis I will use the following equation for simplicity <span class="math-container">$$\sigma_0= \frac{F}{2\sqrt 2 a l}$$</span>
If you calculate the stress for:</p>
<p><strong>1. horizontal cross-section:</strong>
<span class="math-container">$$\sigma_1 = \frac{F/2}{\sqrt 2 a l}= \frac{F}{2\sqrt 2 a l}=\sigma_0\le S_y$$</span></p>
<p><strong>3. vertical cross-section:</strong>
<span class="math-container">$$\tau_3 = \frac{F/2}{\sqrt 2 a l}= \frac{F}{2\sqrt 2 a l}=\sigma_0 \le S_{sy}$$</span></p>
<p>Finally, <strong>case 2 for the combined normal and shear stress.</strong></p>
<p>From the geometry (<span class="math-container">$45^\circ$</span> plane) the total force of <span class="math-container">$\frac F 2$</span>, has a normal component with magnitute <span class="math-container">$\frac{F}{2}\frac{\sqrt 2}{2}= \frac{F}{2\sqrt{2}}$</span> and a shear component of equal magnitute. Therefore for case 2, you can calculate</p>
<p><span class="math-container">$$\sigma_2 =\frac{\frac{F}{2\sqrt{2}}}{a l}=\frac{F}{2\sqrt{2} a l}=\sigma_0, \quad \tau_2 =\frac{\frac{F}{2\sqrt{2}}}{a l}=\frac{F}{2\sqrt{2} a l}=\sigma_0$$</span></p>
<p>using the von Mises criterion for the equivalent general plane stress</p>
<p><span class="math-container">$$\sigma_{v,eq} = \sqrt{\sigma_2^2 + 3\tau_2^2}= \sqrt{\sigma_0^2 + 3*\sigma_0^2}= 2 \sigma_0<=S_y$$</span></p>
<p>If summarise the results the equations are:</p>
<p><span class="math-container">$$\begin{cases} (1.) \quad\sigma_0\le S_y\\
(2.) \quad2\sigma_0\le S_y\\
(3.) \quad\sigma_0\le S_{sy}\end{cases} \rightarrow
\begin{cases} (1.) \quad\sigma_0\le S_y\\
(2.) \quad2\sigma_0\le S_y\\
(3.) \quad\sigma_0\le 0.5 S_{y} (Tresca)\end{cases} $$</span></p>
<p>It is obvious that (2.) and (3.) are equivalent and they are also more conservative than case (1.). Also the calculations of (3.) are simpler.</p>
<p><strong>Bottom line</strong>: The pure shear stress is as stringent as any other state of stress encountered at any plane of the weld, and its easier to download. (thanks <a href="https://engineering.stackexchange.com/users/13311/jonathan-r-swift">@Jonathan R Swift</a> )</p>
| 37181 | Why is fillet weld assumed to be in a state of pure shear stress? |
2020-08-18T03:39:25.980 | <p>I am wondering if rotating flyweights would slow down the descent of a object that is free falling down to the Earth's surface.</p>
<p>Please reference the conceptual drawing below:</p>
<p><a href="https://i.stack.imgur.com/vAA2I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vAA2I.png" alt="enter image description here" /></a></p>
<p>This drawing is showing a cross-sectional view of a sealed hollow cylinder. Inside this hollow cylinder would be two DC electric motors with each having two rods attached to their rotor shaft and would have two thrust bearings sandwiching each rod which would allow each rod to freely pivot up or down as the motor shaft rotates. At the end of each rod will be a flyweight.</p>
<p>There would be a battery attached to the bottom of the cylinder to power the DC motors and to also act as a dead-weight so the cylinder remains in this upward position as it free falls down to Earth's surface. The drawing is also showing a metal plate with a hole in it located at the top of the cylinder to enable this to be dropped from a balloon or helicopter. The DC motors would be counter-rotating to each other in order to keep the cylinder from spinning around.</p>
<p>A balloon or helicopter would take this cylinder up to a certain height and then release it. Just before releasing it, the two DC motors will be turned on and the flyweights will be rotating at a 90 degree angle to the motors' rotor shafts.</p>
<p>Once the cylinder is released and starts to accelerate downward, the rotating flyweights should pivot upwards to a certain degree (as shown in the drawing) and should act as a drag on each of the motors' rotor shafts and thus a drag on the falling cylinder, thus slowing down its descent. I believe the faster the flyweights rotate, the more drag they will produce and the slower the cylinder will descend.</p>
<p>Would rotating flyweights slow down the descent of an object that is free falling down to the Earth's surface?</p>
<p><strong>EDIT</strong></p>
<p>I have created a revised design that I believe should slow down the descent of the falling object.</p>
<p><a href="https://i.stack.imgur.com/7JYq9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7JYq9.png" alt="enter image description here" /></a></p>
<p>The revised drawing above is showing that a round disc with an O-Ring around it which been inserted into the cylinder. This disc has two metal posts attached to the bottom of it. Each post has a thrust bearing and a shaft bearing on top of it. Attached to the shaft bearings are steel cables that are attached to the rods that are attached to the motors' rotor shafts. The inner lining of the cylinder would have some kind of lubricant on it to allow the disc to slide up or down.</p>
<p>The concept is that the air pressure inside the cylinder, below the O-Ring, could be set at ground level pressure (14.7 psia) while the air pressure above the O-Ring will be lower due to the lower ambient air pressure outside of the cylinder. There will be holes in the top of the cylinder to allow air to enter in or exit out of the top section of the cylinder.</p>
<p>In order to create a decelerating force, the DC motors would need to be turned on and off in a continuous loop. When the motors are turned off, the disc will move towards the top of the cylinder and when the motors are turned on, the disc will be pulled back down. A temporary decelerating force should be created during the time periods when the disc is being pulled down.</p>
| |mechanical-engineering|control-engineering|design|applied-mechanics| | <p>I don't quite understand what your rotating things are doing exactly, but the only thing a closed system like this can do is change the center of gravity of the box. If your rotating things moved the center of gravity down you could theoretically make the bottom of the box hit the ground a fraction of a second slower.</p>
<p>This is the same concept as a high-jumper whose center of gravity actually passes below the bar.</p>
| 37197 | Would a centrifugal flyweight governor system slow down the descent of an object that is free falling down to the Earth's surface? (Revised design) |
2020-08-18T16:49:35.667 | <p><strong>Current State</strong></p>
<p>Currently I have designed a coffee roaster that mostly uses the stainless steel tri clamps / tubes that many brewers use. The system can get up to 450 degrees F (232°C) even with moving air.</p>
<p><strong>Question</strong></p>
<p>I was wondering if someone has come across a cheaper material or system than the tri clamp one? The requirements are to withstand 450 degrees F (232°C) or more, and to be food safe. Current material is stainless steel which works for food safe but the pipes with all the clamps are up to $300 or so which is more than the rest of the project.</p>
<p><strong>What I have investigated so far</strong></p>
<p>I looked at many plastics available and while some can handle higher heats, those are generally more expensive. I also looked at using HVAC ducting but learned that galvanized steel can let off gasses when heated up. So are the stainless steel tri clamps the best solution for the money or is there something cheaper out there that I am not thinking of?</p>
| |materials|cost-engineering| | <p>I suggest aluminum: able to take temperatures long term and at stress up to 600 F, cheap, strong, very machinable, readily available, and food grade.</p>
| 37209 | Better / Cheaper Material Choice? |
2020-08-18T17:16:32.873 | <p>consider the following case:
A straight tube, with a constant mass flow rate of water <span class="math-container">$\dot m_{in}=\dot m_{out}$</span> , and with a linear power entering in it <span class="math-container">$\dot Q [\frac W m]$</span>. And the water is the liquid phase in all the tube.</p>
<p><a href="https://i.stack.imgur.com/hrycJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hrycJ.jpg" alt="enter image description here" /></a></p>
<p>My professor told us that in this case the incompressible fluid is a good approximation if the velocity of the water is much less than the velocity of the sound. Can you explain me why this should be a good criteria?
In particular what confuses me is that the density should change depending on the temperature region.</p>
| |fluid-mechanics| | <p>The question confuses two separate concepts. One is the idea of incompressible flow, and the other is constant density flow.</p>
<p>The professor is referring to a criterion that allows you to use the incompressible flow equations, without heat addition. When you derive the more general flow equations, using Newton's Second Law, Conservation of Mass, and Equation of State, you find that there's an important parameter called Mach Number, M, defined as fluid velocity divided by local speed of sound. Even more, M appears as M^2, and the latter often appears in terms such as (1 - M^2). When you study these equations, you find that if you neglect M^2 when compared to unity, you find that there can be no variation in density. Thus, if M^2 is << 1, you can use the incompressible flow equations, without heat addition. Practically this means for flows where approximately
M^2 < 0.1, or M < 0.3.</p>
<p>With heat addition, you need to invoke in addition to the principles mentioned above the Energy Equation. These are a much more complicated set, and it's often advantageous to look for less accurate, but very useful simplifications, unless it's obvious that changes in density - for whatever reason - are important features of the flow.</p>
| 37212 | incompressible fluid approximation and fluid vs sound velocity |
2020-08-18T20:31:54.767 | <p>I have a turbomolecular pump from Osaka vacuum that is quite dirty, and I'd like to clean the blades off to improve balance.</p>
<p>I'm not quite sure how to approach this, I've tried carbon cleaner for turbochargers for cars to no success, and my next thought is to walnut blast the blades, but I'm worried about bending them.</p>
<p>What method should I use to clean these unknown deposits off the blades of this pump?</p>
<p><a href="https://i.stack.imgur.com/BV1Fo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BV1Fo.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/DWCMc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DWCMc.jpg" alt="enter image description here" /></a></p>
| |turbines|turbomachinery| | <p>Acetone or other ketones are used to clean auto engine internals of "varnish"( which looks similar to your deposits). This is for research purposes ,not general engine cleaning. The material looks organic so solvent seems preferred to mechanical methods like nut shell blast. Corrosion would not be a factor for the metal if solvent is used.</p>
| 37218 | How should one clean the blades of a turbomolecular pump |
2020-08-19T09:07:58.043 | <p>I'm an software dev and my job is to convert a .dxf file into a .svg file</p>
<p>My code actually works, but there are some issues with the dimensions or something similar. <strong>Unfortunately, i have no background in technical drawing whatsoever.</strong></p>
<p>The finished .svg looks different from what expected.</p>
<p>When using an online .dxf to .svg converter it works. The .svg looks as expected from the customer.</p>
<p>My guess is that there seems to be something in the header of the .dxf file that I'm ignoring.
I've read through several documents but i can't figure out why my .svg looks different.</p>
<p>Expected:</p>
<p><a href="https://i.stack.imgur.com/wr9fB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wr9fB.png" alt="enter image description here" /></a></p>
<p>What i am getting:</p>
<p><a href="https://i.stack.imgur.com/akJka.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/akJka.png" alt="enter image description here" /></a></p>
<p>Its much smaller and its flipped. Ignore the color, i filled every hexagon on purpose!
Looking at the converted .svg in detail, all coordinates are different from what it says in the .dxf</p>
<p>This is the header of the .dxf file:</p>
<p>header:</p>
<pre><code><span class="math-container">$ACADVER: "AC1012"
$</span>AUNITS: 0
<span class="math-container">$CELTYPE: "BYLAYER"
$</span>CLAYER: "0"
<span class="math-container">$DIMALT: 0
$</span>DIMALTF: 0.0393700787
<span class="math-container">$DIMASZ: 3.5
$</span>DIMBLK: "_FILLED"
<span class="math-container">$DIMCEN: 5.346
$</span>DIMDLI: 22.572
<span class="math-container">$DIMEXE: 1
$</span>DIMEXO: 0
<span class="math-container">$DIMGAP: 0.625
$</span>DIMLFAC: 1
<span class="math-container">$DIMLIM: 0
$</span>DIMSCALE: 1
<span class="math-container">$DIMTFAC: 1
$</span>DIMTM: 0.01
<span class="math-container">$DIMTOL: 1
$</span>DIMTP: 0.01
<span class="math-container">$DIMTXT: 3.5
$</span>DWGCODEPAGE: "ANSI_1252"
<span class="math-container">$EXTMAX: {x: 594, y: 420, z: 0}
$</span>EXTMIN: {x: 0, y: 0, z: 0}
<span class="math-container">$HANDSEED: "CEF6"
$</span>LIMMAX: {x: 594, y: 420}
<span class="math-container">$LIMMIN: {x: 0, y: 0}
$</span>LTSCALE: 32.67
<span class="math-container">$PDMODE: 3
$</span>PDSIZE: 0.05
<span class="math-container">$TILEMODE: 1
$</span>VIEWCTR: {x: 297, y: 210}
$VIEWSIZE: 420
</code></pre>
| |technical-drawing|drawings| | <p>SVG and images in general define a coordinate system with</p>
<ul>
<li>positive x towards the right</li>
<li>positive y towards the bottom</li>
</ul>
<p>I.e. the pixels with coordinate y=1 are near the top.</p>
<p>On the other hand dwg, defines a coordinate system with</p>
<ul>
<li>positive x towards the right</li>
<li>positive y <strong>upwards</strong></li>
</ul>
<p>All you need to do is flip your svg along the x axis, either on the original or in the final svg.</p>
| 37229 | Convert .dxf file to .svg |
2020-08-19T10:21:41.677 | <p>I need help to check a calculation I did. I want to know if it is possible to use this method or if I am using an assumption which is wrong. Let me explain the problem, a beam with length <span class="math-container">$l$</span> is fastened in one end. A force <span class="math-container">$F$</span> an a moment <span class="math-container">$M_v$</span> is applied at the end of the beam, see figure below. The beam has a circular cross-section. Due to the force, the end of the beam will deform a length <span class="math-container">$\delta$</span>. Only the deflection is known and the geometric parameters, such as the length and diameter.</p>
<p>Using <em>Euler-Bernoulli beam theory</em> the deflection can be expressed as:</p>
<p><span class="math-container">$$\delta = \frac{Fl^3}{6EI} \tag{1}$$</span></p>
<p>Where <span class="math-container">$E$</span> is the Young's Modulus of the material and <span class="math-container">$I$</span> the inertia, which is <span class="math-container">$I=\frac{\pi d^4}{64}$</span> for a circular scross section. Here <span class="math-container">$d$</span> is the diameter of the beam.</p>
<p>Inserting the inertia in (1) and rearranging it as an expression of <span class="math-container">$F$</span> gives:</p>
<p><span class="math-container">$$F = \frac{3 \delta \pi d^4 E}{32l^3} \tag{2}$$</span></p>
<p>This can be inserted in the general formula for maximum bending stress in a cross-section</p>
<p><span class="math-container">$$\sigma_{max}= \frac{Fl}{\frac{\pi d^3}{32}} = \frac{32Fl}{\pi d^3} \tag{3}$$</span></p>
<p>Here the bending resistance for a circular-cross section has allready been inserted in the formula and the bending moment has been replaced for the maximum moment which is <span class="math-container">$Fl$</span>.</p>
<p>This is the part which I am not so sure on, I use the force from (2) and insert it in (3) to get the maximum stress. Please let me know if this is possible or if I am making an error.</p>
<p>Furthermore, the shear stress can be calculated from <span class="math-container">$\tau = \dfrac{M_v}{W_v}$</span> where <span class="math-container">$W_v = \dfrac{\pi d^3}{16}$</span>, which is the torsion resistance in the material. I then proceed to use the <em>von Mises yield criterion</em> to get an estimate of the maximum stress in the material.</p>
<p><span class="math-container">$$\sigma_{von\ Mises} = \sqrt{\sigma^2+3\tau^2}$$</span></p>
<p>As I asked before, I am mainly interested if this is a possible way to proceed with solving this problem or if I am using some methods/assumptions which are wrong.</p>
<p><a href="https://i.stack.imgur.com/ASB2G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ASB2G.png" alt="Beam loading case" /></a></p>
| |mechanical-engineering|beam|solid-mechanics| | <p>In general what you are doing is ok. Assuming you have small enough deflections (either through bending or torsional), you can independently solve the problems. I.e.:</p>
<ul>
<li>Calculate the Force required to obtain the bending exactly as you have done. <span class="math-container">$$F = \frac{3 \delta \pi d^4 E}{32l^3}$$</span></li>
<li>Calculate the magnitude of the shear stress.</li>
</ul>
<h1>Caveats</h1>
<p>However, from that point on there are a few caveats. <strong>Regarding:</strong></p>
<p>a) <strong>bending</strong>: the maximum magnitude of normal stress you are calculating is at the top and bottom of the beam. Any point on the neutral axis should have a magnitude of zero.</p>
<p>b) <strong>torsional shear</strong>: The magnitude at the distance <span class="math-container">$\frac d 2$</span> is constant but the direction changes. see the following image:</p>
<p><img src="https://i.stack.imgur.com/bpdgD.png" alt="Torsional stress at differet point of the crosssection" /><a href="https://i.stack.imgur.com/bpdgD.png" rel="nofollow noreferrer">1</a>.</p>
<p>the magnitute of the maximum torsional stress is correctly:</p>
<p><span class="math-container">$$\tau_t = \frac{M_u}{\frac{\pi d^3}{16}}$$</span></p>
<p>c) <strong>Shear</strong>: Although usually discarded there is also a shear stress associated with <span class="math-container">$$\tau_s = \frac{F}{\frac{\pi d^2}{4}}$$</span>. Normally, that is very small, but also it has a constant direction (downwards in this occasion).</p>
<p>The point you need to take is that you need to add as vectors <span class="math-container">$\tau_s$</span> and <span class="math-container">$\tau_t$</span>. Therefore, at different point in the material you'd have different values. Given image <a href="https://i.stack.imgur.com/bpdgD.png" rel="nofollow noreferrer">1</a> and taking points A,B,C,D anti-clockwise, the resultant shear stress will be:</p>
<ul>
<li>at the rightmost point ( Point A (+x, y=0) will be <span class="math-container">$$\tau_{A, res} = \tau_s - \tau_t$$</span>.</li>
<li>at the topmost point ( Point B (x=0, +y) will be <span class="math-container">$$\tau_{B, res} = \sqrt{\tau_s^2 + \tau_t^2}$$</span>.</li>
<li>at the leftmost point ( Point C (-x, y=0) will be <span class="math-container">$$\tau_{C, res} = \tau_s + \tau_t$$</span>.</li>
<li>at the bottommost point ( Point D (x=0, +y) will be <span class="math-container">$$\tau_{D, res} = \sqrt{\tau_s^2 + \tau_t^2}$$</span>.</li>
</ul>
<h1>Maximum stress</h1>
<p>So the main thing is regarding your equation of the Von Mises. Which values do you plug for <span class="math-container">$\sigma$</span> and <span class="math-container">$\tau$</span>.</p>
<p>You'd need to go through each point and apply the corresponding stress:</p>
<ul>
<li>Point A, use <span class="math-container">$\sigma_{A} = 0$</span> and <span class="math-container">$\tau_{A, res} = \tau_s - \tau_t$</span></li>
<li>Point B (and D), use <span class="math-container">$\sigma_{B} = \frac{32Fl}{\pi d^3}$</span> and <span class="math-container">$\tau_{, res} =\sqrt{\tau_s^2 + \tau_t^2}$</span></li>
<li>Point C, use <span class="math-container">$\sigma_{A} = 0$</span> and <span class="math-container">$\tau_{A, res} = \tau_s + \tau_t$</span></li>
</ul>
<p>Unfortunately, these are not the only points you need to check. For example, you should check at least at <span class="math-container">$\pm 135$</span> degrees (in that quadrature in the image <span class="math-container">$\tau_s $</span> and <span class="math-container">$\tau_t$</span> do not cancel each other). But that is the idea.</p>
| 37231 | Torsion and Bending stress calculated from deflection |
2020-08-19T14:18:41.747 | <p>The Darcy friction factor is defined as:
<span class="math-container">$$f_D =\frac{64}{Re}$$</span>
with <span class="math-container">$Re$</span> I mean Reynolds number (global).</p>
<p>I was wondering why Darcy in laminaire flow is independent of pipe roughness (wall roughness)?
That is not the case in turbulent and transient zone, the iterative methodes are much more complex and the friction factor is a function of wall roughness.</p>
| |fluid-mechanics|friction|pipelines| | <p>In contrary to what has been discussed, the friction factor <strong>depends</strong> on the roughness height also in <strong>laminar flow</strong>, which has been shown in literature (Gloss and Herwig 2010. "Wall roughness effects in laminar flows: An often ignored though significant issue", amongst others), but is often <strong>not mentioned</strong> in <strong>standard textbooks</strong> (like the Moody diagram shown above). The simple reason being that an obstacle in a laminar flow produces drag, so the size of the roughness matters. The commonly used equation f = 64/Re therefore only applies if D >> k.</p>
| 37238 | Why is the friction factor in laminar flow independent of pipe roughness? |
2020-08-20T16:35:13.730 | <p>I've read in more than one place (e.g., <a href="https://legacyac.com/how-to-use-a-swamp-cooler/" rel="nofollow noreferrer">here</a> and <a href="https://www.newair.com/blogs/learn/how-to-make-swamp-cooler-colder" rel="nofollow noreferrer">here</a>) that a swamp cooler (a.k.a., evaporative cooler) can be effectively paired with a dehumidifier to cool an enclosed space, but this makes no sense to me. It seems like the evaporative cooling would be cancelled out by the heat resulting from condensing the water vapor. No energy is leaving the system as a result of these processes, and waste heat is generated by each device, so it seems the only effect would be to make the room hotter.</p>
<p>The only exception I can imagine is if the water resulting from dehumidification holds the bulk of the heat from that process and is quickly drained, while the evaporative cooler has its supply replenished with cooler tap water. But the articles don't specify that.</p>
| |hvac|cooling|evaporation| | <p>This method would work in humid climates. If you live in a humid climate a swamp cooler will not work because there is too much humidity in the air to begin with. Swamp coolers take hot dry air and turn it into cool moist air. The idea is to dehumidify the air going into the swamp cooler so that it will produce cooler air coming out. In theory this should work if you have a good dehumidifier and aim it to the back of the swamp cooler. This should help a swamp cooler cool more effectively in a humid region. In essence you are taking hot moist air, turning it to hot dry air so you can get COOL moist air. If the humidity going into the swamp cooler is significantly dropped, the swamps cooler should cool a room as much as 30 degrees below the room temperature. It would lower the temperature more than the dehumidifier would raise the temperature.</p>
| 37252 | Why do a swamp cooler and dehumidifier effectively cool a room when paired together in an enclosed space? |
2020-08-21T17:05:11.493 | <p>I have a feeling that if my stovetop was made out of regular glass, it would have cracked from rapid cooling/heating a long time ago – especially if you consider accidental water spills. What kind of glass is able to withstand such abuse?</p>
<p>(I am just looking for terms to google for further reading.)</p>
| |glass| | <p>Stovetops are made of glass-ceramic which has extremely low thermal expansion, hence no cracking from temperature change. In fact, the coefficient is with <span class="math-container">$0.1 \cdot 10^{-6} \ 1/K$</span> even lower than that of Borosolicate glass at <span class="math-container">$3.3 \cdot 10^{-6} \ 1/K$</span>. Since glass-ceramic can reach a negative coefficient of thermal expansion, getting even closer to zero is just a matter of engineering (thanks to @Volker Siegel for this interesting fact).</p>
<p>The brand name for one kind of glass-ceramic, Ceran (by German company Schott) is in German often used to generally describe glass-ceramic stovetops. Borosilicate glass is, to my knowledge, used only for heat-resistant cookware, but not the stovetops themselves.</p>
| 37273 | Why doesn't stovetop glass crack from thermal shock? |
2020-08-22T04:48:45.783 | <p>Why are most jigsaw blades designed to cut on the up stroke with teeth that point towards the shank? Why are reverse teeth jigsaw blades less popular? <a href="https://i.stack.imgur.com/BbnIs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BbnIs.jpg" alt="jigsaw blade" /></a></p>
| |mechanical-engineering|manufacturing-engineering|material-science| | <p>To expand on Solar Mike's answer and to help explain part of niels nielsen's answer, the saw blade is normally designed to essentially squeeze the material between the teeth and the shoe. This generally makes for a more controlled cut. Circular saws, saws-alls, and most other mechanical saws work on this principle, too.</p>
<p>If you were to use the blade you have instead, you're looking at more "jumps", where the blade catches the material without cutting. This can happen with a normal jigsaw blade, but generally only if you try to force-feed the saw, turn too sharply, the material delaminates, the blade gets pinched, or a small handful of other reasons.</p>
<p>Because of the "squeeze", you don't have to put downward pressure on the saw to keep it cutting and avoid jumping. With your downward facing teeth, you'll have to go slower with cutting to avoid jumping or put force on the tool to keep it in contact with the material. Going slower is generally more acceptable, unless the material is just super hard and you have to force the blade into it to cut. But laminate, like Adam talks about, should be quite soft.</p>
<p>The other part of this where any tear-out is going to happen. With a normal blade, tear-out/splinters/ragged edges are going to happen on the top of the material, facing you. With downward facing teeth, the tear-out is going to be on the bottom of the material. Where you need/want that tear-out to happen depends on your project. If you are making a normal cut, you can simply flip the material over and cut from a drawing on the "bottom" of your material, leaving the good surface free from line marks, scratch marks from the saw shoe, tear-out, and most other marks that can come from building a project.</p>
<p>However, with laminates or veneers, like a kitchen counter or piece of furniture, you are going to be cutting from the top of the material, not the bottom, so you would want the downward facing blades to avoid chipping the top surface. Personally, I'm not sure I'd want to use a jigsaw for this purpose, but it might be the best tool out of a lot of bad options, too.</p>
| 37287 | Why are most jigsaw blades designed to cut on the up stroke and have teeth that point towards the shank? |
2020-08-24T10:27:26.673 | <p>I've just bought myself an electrical chainsaw. Specifications of the chain velocity of my saw read 55 km/hr.</p>
<p>I've read online that many wood-cutters have encountered chain breaks and want to protect myself against unfortunate accidents.</p>
<p>So, my idea is to always cut with the underside of the blade. This way if a break occurs at the point of contact with the wood and since the direction of motion of the chain is towards me from the underside, the chain will move away from me when it snaps.
I am basing this on the fact that the saw has a small sprocket underneath giving motion to the chain. When it breaks the sprocket cannot pull it towards me anymore and given the momentum of the other outward sprocket on the head of the blade, the chain will be pulled outwards, up and away from me. All this supposing that I am always cutting with the underside of the blade.</p>
<p>I do not know if I am correct though! Do you know of any experiments that have been conducted to demonstrate this? Since this is a matter of life and death, I would expect much information about this, but I can't find any!</p>
| |mechanical-engineering|chain| | <p>Another perspective : A friend retired as a manager with a large lumber company ( Louisiana Pacific ). He had spent much of his career overseeing cutting crews. They did not try to figure out when or how a chain might break , he just wore a heavy leather apron whenever he used a chainsaw . I understand his crews also used the aprons. We had a volunteer group that helped clear downed trees on the TX coast after storms. He pretty much did all the chainsaw work.</p>
| 37321 | Protection against accidents when chainsaw chain breaks |
2020-08-24T17:09:39.973 | <p>Let's say I design a simple piece of carbon fiber to be manufactured, in the shape of a simple cylinder with a diameter of 2cm and a height of 3cm (pretty small).</p>
<p>Is it possible and/or practical to try to mold this piece of carbon fiber in such a way that it could be hollow on the inside?</p>
<p>And if so, is it possible to mold the piece of carbon fiber inside of a vacuum chamber, so the resulting epoxy sealed carbon fiber part contains significantly less air in the hollow when completed?</p>
| |manufacturing-engineering|vacuum|carbon-fiber| | <p>As far as I know, you can't mold a sphere or cylinder unless the material is on the outside of the mold and you keep the mold in place. Think paper machee over balloons.
There is no way to get the balloon out. There is no way to get material to the inside of a hollow mold unless you can spin the material onto the walls, which won't work for carbon fiber.</p>
<p>The easiest solution is the best, which is to create an object via forming two halves together and then evacuating the contents.</p>
| 37330 | Is it possible/practical to mold a single hollow piece of carbon fiber with a sealed vacuum in the middle? |
2020-08-25T02:14:57.527 | <p>When a real world string is plucked hard, like a guitar string, it enters a very brief period of noise generation after the release of the plectrum. Analysis of the output appears as pure chaotic noise. Then the harmonic content follows and you hear a musical note evolve.</p>
<p>What relationship between the increased tension or kinetic energy from high amplitude vibrations and the transition to noise generation would exist? How would you predict at which point the noise generation begins and where it transitions from harmonic to noise and back?</p>
| |mechanical-engineering| | <p>When you really dig into a string with a pick and pull the string far off-center with it, the string is not in a sine-wave shape or anything like it: it is two lengths of straight string with the pick tip at the point where they meet. Those segments, along with the trace of the string in its unperturbed location, form a triangle.</p>
<p>Then, when the pick starts sliding off the string, a short burst of <em>pick noise</em> is generated, followed immediately by a lot of high-frequency vibrations that travel away from the pick release point towards opposite ends of the string as the sharp kink in the string caused by the pick "snaps back". Those high frequency waves are not even multiples of the string fundamental and hence sound harsh and strident ("anharmonic"), and they propagate back and forth along the entire length of the now-released string until the components of that noise which aren't harmonics of the fundamental get suppressed by cancellations and die out.</p>
<p>By this point the string has begun vibrating at its fundamental, the initial burst of high-frequency, random crackle has been quenched, and what remains superimposed on the fundamental are all the higher harmonics that were left over.</p>
<p>High-speed videos of bass players slamming a roundwound string with a pick show this effect clearly.</p>
| 37336 | Why do vibrations of a string create inharmonic noise past a certain amplitude of excitation? |
2020-08-25T18:47:42.390 | <p>What shape propeller with a large surface are, would be best propelled from the wind from behind and from the side as well as the front while facing the sun?
I am thinking a pinwheel, but maybe there is a better design?
<a href="https://i.stack.imgur.com/9aDLt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9aDLt.png" alt="enter image description here" /></a></p>
| |design|wind-power| | <p>You are looking for a “vertical axis wind turbine”.</p>
<p>See below for some design examples:</p>
<p><img src="https://i.stack.imgur.com/li0b6.jpg" alt="enter image description here" /></p>
| 37349 | Wind turbine that is efficiant from all angles |
2020-08-26T08:25:35.633 | <p>Looking to install OEM active valve mufflers on my 2019 Mustang EcoBoost.</p>
<p>Wondering if the valves opening and closing, thus changing exhaust back-pressure would cause the engine to run lean/rich and do long term harm to my bone stock car.</p>
<p>Long term reliability has priority over sound for me but if it has ZERO effect then I will go ahead and do this.</p>
<p>Thanks in advance</p>
| |automotive-engineering| | <p>An OEM setup will be fine. Ford will have different fueling maps set up for each setting. Doing this without the electronics built in would be a mess but it's quite safe as long as the car knows what is going on with the exhaust.</p>
| 37357 | Long term engine issues by using Active Exhaust? |
2020-08-26T21:02:17.140 | <p>Scenario: Object 1, a bridge freezes over do to the surface area exposed to the below freezing air. Object 2 has the same surface area exposed to the same air, but its base sits on the ground that is above freezing. Would object 2, with the same amount of surface area exposed to the same air freeze or would it pull heat from the ground through conduction to not freeze? If object one had more contact with the ground to draw heat, would that prevent a bridge from freezing?</p>
| |structural-engineering|materials|thermodynamics|building-physics|thermal-conduction| | <p>Heat is the <a href="https://www.quora.com/What-determines-the-direction-of-a-heat-flow" rel="nofollow noreferrer">transfer of energy from a hotter region to a cooler region</a>.</p>
<p>The surface in contact with the earth enjoys the benefit of conductive heating. Being cooler, the energy of the warmer earth will be transferred to the earth-bridge, ostensibly until the energy imbalance is removed.</p>
<p>The air-bridge is going to become colder due to the colder air and will also continue to get colder until the imbalance is removed. Huge masses of the planet earth can require extensive time to remove such an imbalance. Poor conductivity means a slower transfer but local weather conditions such as sunshine and seasons are considerations.</p>
<p>The temperature difference between the earth and the air is a factor. If the cold air is chilling the earth-bridge at a rate higher than the earth is warming it, the air wins.</p>
<p>It might require active or passive energy transmission devices (heat pipes) to provide enough energy to the air-bridge to prevent freezing. <a href="https://climaterealityproject.org/blog/how-exactly-does-geothermal-energy-work" rel="nofollow noreferrer">Geothermal energy</a> is based on this concept, loosely speaking.</p>
<p>There is much more complex physics involved in this type of question. Your question was phrased at a layman's level, hence my layman's answer. I didn't do well enough in physics class to provide formulae.</p>
| 37367 | Is it just surface area alone that causes bridges to freeze? |
2020-08-27T14:17:58.840 | <p>According to this phase diagram, the melting point of 30% copper aluminum bronze is around 548°C.</p>
<p>Lets say I create molten aluminum at 680°C, can I add small pellets of copper and swirl them around to create a molten alloy? or will the alloying process not start until Copper's melting temp has been reached at around 1080°C?</p>
<p><img src="https://www.imetllc.com/wp-content/uploads/2014/06/Al-CuPhaseDiagram.jpg" alt="Text" /></p>
| |aluminum| | <p>Maybe try small amounts silicone, phosphorus, or both (less than1%)
Lead even if you're not opposed to the idea and some things don't always want to homogenize easily but repeated Heating should remedy that especially if it doesn't take you a real long time to get to temp but it's hard to say for sure considering it I don't recall you saying what you were using it for</p>
| 37379 | What temperature required to alloy copper and aluminum? |
2020-08-27T23:57:54.813 | <p>I am new to mechanical engineering so I don't know the terms for different bearings.</p>
<p>I'm trying to make a honey extractor. I have a rotating basket made out of stainless steel that is going a maximum of about 100 RPM. Through the center of the basket, I have a 1/2 inch rod going vertical. I want to figure out a good stand for this rod to rotate on. The basket needs to be taken out since the whole unit needs to be cleaned after use. Everything needs to be stainless steel or at least food grade.</p>
<p>I don't know what kind of bearing to set it on. I initially thought of a transfer bearing but they look like they are very expensive for one that can hold the weight of the basket plus honey frames, about 200 lbs.</p>
<p>I also considered putting it on a large ball bearing in a cup but I don't know how I could build it to make the ball bearing stay in the right spot.</p>
<p>I also considered just having the rod rotate on a stainless steel plate but I am not sure if 200 lbs will wear on the surface over time.</p>
| |mechanical-engineering|design|bearings| | <p>For that mass, use a large diameter roller bearing supporting the drum directly - just like the lazy susan rotating serving plates used at table.</p>
<p>Of course, if you want a low friction bearing then a circular mercury bath is also a solution - used in some lighthouses to get a very smooth rotation for the lenses around the light. But purchasing that amount of mercury might be costly, let alone the permits to have it as the Health & Safety Police may be all over you like flies around dogs xxxx. And as you are dealing with food...</p>
| 37393 | Looking for specific bearing |
2020-08-28T05:28:02.887 | <p><a href="https://i.stack.imgur.com/CIKDQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CIKDQ.png" alt="enter image description here" /></a></p>
<p>So, I have a crank that has a lever attached to a fulcrum, which rotates about an axis. But I also have a handle attached perpendicular to this lever, and the effort force is applied to the handle 90° downward.
I am trying to calculate torque = R<em>F</em>sin(\theta)
So, my question is: what is the radius in this instance. Is it the distance from the axis of rotation to the edge of the lever, or is it the distance from the axis of rotation to the edge of the handle where the force is applied? See image for more clarity.
If it were the latter distance, then trigonometry would need to be used to find the diagonal.
Thanks in anticipation of your help.</p>
| |mechanical-engineering|torque| | <p>The torque is <span class="math-container">$\tau= RF \ ,$</span> no other term or sin needed if you apply force trying to rotate the lever.</p>
<p>The torque of a force about an axis is its perpendicular projection to a radial passing through the center or axis we want to measure the torque for. And no matter how long the round handle is as long as it is perpendicular to the handle it doesn't affect the torque.</p>
<p>Let's say your diagonal would be 45 or 60 degrees it does not make a difference. Of course, the handle where you grab and turn acts as a cantilever and has bending stress, but it is orthogonal to the force so it soes not matter.</p>
| 37396 | How do I calculate the torque of a crank? |
2020-08-29T03:31:14.930 | <p>I am wondering how a helicopter manages to stay balanced and stay in one spot while it is in the air.</p>
<p>I understand that there is a second rotor that is used to counteract the angular momentum and also the use of swashplate to direct the helicopter forwards and backwards.</p>
<p>Assuming the angular momentum of the main rotor is not an issue,
Is there a sensor on the helicopter that can detect if it is not upright? Is the swashplate then adjusted to tilt the copter into an upright position?</p>
| |mechanical-engineering|aerodynamics|aircraft-design| | <p>Most helicopters of modern manufacture are dynamically unstable while hovering- they require constant, "hands-on" control inputs from the pilot to keep things in balance. This task is very demanding and requires a lot of practice to master.</p>
<p>Hiller helicopters (which have been out of production for many years) were one of the rare examples of dynamically stable helos, which could actually hover "hands-off".</p>
<p>Hovering can also be performed by an automatic pilot system or something called a <em>stability augmentation system</em> (SAS) which assists the pilot by reducing the degree of dynamic instability, rendering the helicopter more easily controllable. Both types of systems require sensors to detect tilt and accelerations.</p>
<p>BTW many radio-controlled helicopter models are SAS-equipped to make them less difficult for hobbyists to fly.</p>
| 37408 | How does a helicopter hover |
2020-08-29T13:18:40.353 | <p>I would like to work out the elastic modulus of a specific grade of titanium.</p>
<p>Further to this I would like to learn how to calculate the maximum tensile strength of that metal in relation to its cross-sectional area.</p>
<p>Here is a disc of 3 inches diameter that will be spun up to high velocities. I need to calculate the elastic range and maximum tensile strength before it self destructs.</p>
<p>Amy help would be greatly appreciated.</p>
| |stresses| | <p>Centripetal force is <span class="math-container">$F= m \frac{v^2}{r}$</span></p>
<p>Therefor the acceleration <span class="math-container">$\alpha=\frac{v^2}{r}$</span></p>
<p>Let's say the titanium tensile stress is 128000psi and ignore its elongation.
it means for an infinitesimally outer ring to separate we need a force of 180000psi per linear inch of the arc of that ring. disregarding the tangential tensile stress just to get an idea.</p>
<p>Considering the very thin arc and a very small mass of it, this means unrealistic rotation speeds unachievable by our technology. Long before that supersonic speed of the disk will cause it to vibrate and fly off in a fiery scorching spiral.</p>
| 37415 | Calculate elastic modulus of titanium |
2020-08-29T23:20:21.143 | <p>If I have an I beam which I subject to some bending load, how can I calculate when the thinner middle part of the I beam will buckle? I though I could use Euler's critical load formula here, but it's unclear to me how exactly I calculate the compression load of the middle part of the beam
<a href="https://i.stack.imgur.com/dpT7S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dpT7S.png" alt="enter image description here" /></a></p>
| |beam|strength|buckling| | <p>Shear buckling of beam web happens when the shear at a section of the beam under consideration surpasses the controlling combination of factored shear.</p>
<p>The following three equations are the LFRD method for:</p>
<ol>
<li>no web instability</li>
<li>Inelastic web buckling</li>
<li>Elastic web buckling</li>
</ol>
<p>These depend on the ratio <span class="math-container">$h/t$</span> of the web. If this ratio is bigger than 260 web stiffeners are required.</p>
<p>Design for shear per AISC (LRFD SPEC F2)</p>
<ul>
<li><span class="math-container">$\varphi v$</span> resistance factor for shear (0.9)</li>
<li><span class="math-container">$h$</span> web height</li>
<li><span class="math-container">$t_w$</span> web thickness</li>
<li><span class="math-container">$V_u$</span> controlling combination of factored shear</li>
<li><span class="math-container">$V_n$</span> nominal shear strength <span class="math-container">$=0.60F_yA_w$</span></li>
<li><span class="math-container">$F_{yw}$</span> yield stress of the web (ksi)</li>
<li><span class="math-container">$A_w$</span> web area, the overall depth d times the web thickness wt</li>
</ul>
<p>Design equation for <span class="math-container">$\dfrac{h}{t_w} \leq 260$</span> :</p>
<p>The design shear strength of the unstiffened web is <span class="math-container">$\varphi V_n$</span>, where <span class="math-container">$V_u \leq \varphi V_n$</span>.</p>
<p><span class="math-container">$$V_n = \begin{cases}
0.6 F_{yw}A_w & \text{if } h/t_w \leq 2.45\sqrt{\dfrac{E}{F_{yw}}}=59\ (\text{for 50ksi steel})\\
0.6 F_{yw}A_w\left(2.45 \dfrac{\sqrt{E/F_{yw}}}{h/t_w}\right) & \text{if } h/t_w \in \left(2.45 \dfrac{\sqrt{E/F_{yw}}}{h/t_w},\ 3.07 \sqrt{\dfrac{E}{F_{yw}}}\right] \\
A_w \dfrac{4.52E}{(h/t_w)^2} & \text{if } h/t_w \in \left(3.07 \sqrt{\dfrac{E}{F_{yw}}},\ 260\right]
\end{cases}$$</span></p>
| 37420 | Beam buckling criterion? |
2020-08-31T13:52:15.400 | <p>Is design of "Budweiser's bow-tie shaped beer can " feasible & unique from engineering point of view?
Can there be a better design than it?</p>
<p>Like the shape of can is inclined at ten degree & appears easy to crush it? Is there any other advantage/disadvantage of this design? Please help me. <img src="https://i.stack.imgur.com/WJGka.jpg" alt="enter image description here" /></p>
| |mechanical-engineering|design|product-engineering| | <p>Budweiser made some design modification to regain its marketing and branding endeavours which had seen huge decline in sales threatening Budweiser's status as America's Best selling beer. The new design emphasized the iconic bowtie and updated its appearance giving it a eye-catching look, providing good holding experience which was easy to grip.</p>
<p>After can modification Due to Can's slimmer middle and sleek design it contained lesser amount of bear than its traditional design.
In packaging terms, it took twice as much aluminium as normal canes which increased its production cost.</p>
<p>So the new modification failed to impress the customer, instead of focussing on can appearance they would have focussed on taste of beer and should not make compromise on quantity of beer.</p>
| 37432 | Is design of "Budweiser's bow-tie shaped beer can " feasible & unique from engineering point of view? |
2020-09-01T01:19:44.080 | <p>With the start of the leaf raking/blowing season just a month away, I have been thinking of a way in which I could make my old leaf blower more effective so I can spend less time doing leaf blowing. I know that I can go out and buy a more powerful leaf blower, but before I do that there is an idea that I have that I believe will increase the effectiveness of my current leaf blower.</p>
<p>It would involve building an air amplifier attachment for my leaf blower which would increase the overall air flow by making use of the Coanda Effect. I was inspired with this idea after recently watching a YouTube video on how the Dyson air multiplier fans work.</p>
<p>Please reference the drawing below:</p>
<p><a href="https://i.stack.imgur.com/cn5KI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cn5KI.png" alt="enter image description here" /></a></p>
<p>I want to point out that this is a cylinder-shaped attachment which would be made out of a lightweight yet durable plastic and that it would be hollow. Also, it would be designed to attach to the standard leaf blower tube which is 3.5 inches in diameter.</p>
<p>Would this air amplifier attachment increase a leaf blower's effectiveness?</p>
<p><strong>EDIT</strong></p>
<p>This attachment would most likely need to be screwed on to the tube or held on with a couple of latches otherwise the back pressure inside of the attachment would likely cause it to pop off.</p>
| |mechanical-engineering|fluid-mechanics|applied-mechanics|aerodynamics| | <p>probably not, here is why.</p>
<p>the coanda effect describes how a moving mass of air entrains nearby air and sets it in motion. So if we start with an extremely fast-moving but small jet of air, with it we can urge a much larger mass of air into motion. However, energy conservation dictates that the kinetic energy of the large mass of air set into motion be less than or equal to that of the small high speed jet, and this means that the large air mass will be moving <em>slower</em> than the high-speed jet.</p>
<p>Key to making a leaf-blower effective is that the dynamic forces exerted on particles of stuff by the moving air be greater in magnitude than gravitational forces on those same pieces of stuff. If this condition is met, then particles of stuff are successfully propelled by and carried along with the air flow. The dependence of this condition on the flow velocity is nonlinear and extremely steep, with a relatively small flow rate increase having a very large effect on how heavy the particles can be while still getting carried in the flow.</p>
<p>This means the key to effective leaf-blowing is making the nozzle exit velocity of the air jet as high as possible, rather than distributing the jet flow over the largest possible area of the ground with the leaves on it.</p>
| 37436 | Would this air amplifier attachment increase a leaf blower's effectiveness? |
2020-09-01T09:55:59.830 | <p>I have data from a number of high frequency data capture devices connected to generators on an electricity grid. These meters collect data in ~1 second "bursts" at ~1.25ms frequency, ie. fast enough to actually see the waveform.</p>
<p>The meters are collecting voltage and current from each of the 3 phases. An example of the data (plot and tabular) is shown below, with one phase shown in each colour.</p>
<p>I want to roll this waveform data up to some summary statistics at a lower frequency (20ms). Specifically, I am looking to calculate:</p>
<ul>
<li>Active power, reactive power and power factor</li>
<li>The grid frequency as it changes over time</li>
</ul>
<p>Apologies but I'm a mechanical engineer and this is not my strong suit! All of the references I can find refer to idealised situations, where the phase angles etc are pre defined. I could fit idealised sin curves to each of the timeseries, but I feel there is a better solution. Are there any simple techniques to calculate the above directly from the timeseries?</p>
<p><a href="https://i.stack.imgur.com/yKLh1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yKLh1.png" alt="" /></a></p>
<p><a href="https://i.stack.imgur.com/38J3s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/38J3s.png" alt="" /></a></p>
<p>Here's a "toy" data set of the first few waves of one voltage phase as a pandas Series for those who are interested:</p>
<pre><code>import pandas as pd, datetime as dt
import pandas as pd, datetime as dt
ds_waveform = pd.Series(
index = pd.date_range('2020-08-23 12:35:37.017625', '2020-08-23 12:35:37.142212890', periods=100),
data = [ -9982., -110097., -113600., -91812., -48691., -17532.,
24452., 75533., 103644., 110967., 114652., 92864.,
49697., 18402., -23309., -74481., -103047., -110461.,
-113964., -92130., -49373., -18351., 24042., 75033.,
103644., 111286., 115061., 81628., 61614., 19039.,
-34408., -62428., -103002., -110734., -114237., -92858.,
-49919., -19124., 23542., 74987., 103644., 111877.,
115379., 82720., 62251., 19949., -33953., -62382.,
-102820., -111053., -114555., -81941., -62564., -19579.,
34459., 62706., 103325., 111877., 115698., 83084.,
62888., 20949., -33362., -61791., -102547., -111053.,
-114919., -82805., -62882., -20261., 33777., 62479.,
103189., 112195., 116380., 83630., 63843., 21586.,
-32543., -61427., -102410., -111553., -115374., -83442.,
-63565., -21217., 33276., 62024., 103007., 112468.,
116471., 84631., 64707., 22405., -31952., -61108.,
-101955., -111780., -115647., -84261.])
</code></pre>
| |electrical-engineering|mathematics| | <h2>regarding Total, active, reactive Power and <span class="math-container">$\cos \phi$</span></h2>
<ol>
<li>Estimate the energy in a timeperiod T for every timestep (<span class="math-container">$t \in [t_0,t_0+T]$</span>) by using the formula</li>
</ol>
<p><span class="math-container">$$P = \frac{1}{T}\sum_{t=i} I_{A,i} V_{A,i}dt + \frac{1}{T}\sum_{t=i} I_{B,i} V_{B,i}dt + \frac{1}{T}\sum_{t=i} I_{C,i} V_{C,i}dt$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$dt$</span>: is the timestep (e.g. 1.25ms)</li>
<li><span class="math-container">$T$</span> is the integration period</li>
<li><span class="math-container">$\frac{1}{T}\sum_{t=i} I_{k,i} V_{k,i}dt$</span>: the active power of each line k <span class="math-container">$\in \{A,B,C\}$</span></li>
</ul>
<ol start="2">
<li>Find the RMS values for Voltage and Current (<span class="math-container">$V_{k, RMS}, I_{k, RMS}$</span>) for each line (A,B,C) in the time period.</li>
</ol>
<p>The <strong>total power</strong> for each <em>line</em> is <span class="math-container">$$P_{k,T} = V_{RMS} * I_{RMS}$$</span></p>
<p>make sure you don't confuse line with phase (see star and delta configurations)</p>
<ol start="3">
<li><p>The <strong>reactive power</strong> will be the difference between the two (Total-Active).</p>
</li>
<li><p>if you want the <span class="math-container">$cos \phi$</span> for each line then just do</p>
</li>
</ol>
<p><span class="math-container">$$\phi_k = \arccos \left(\frac{P_{k,Active}}{P_{k,Total}}\right)$$</span></p>
<p><strong>Caveat</strong>:
This is a numerical estimation. Depending on the duration of integration, you might get some strange results (e.g. Total Power less than Active). That is why you should prefer complete periods if possible only one at a time (longer time periods will tend to average too much the data).</p>
<h2>Regarding the grid frequency:</h2>
<p>what you should do is perform an fft and find the peak frequencies. For a signal like that, you would need to add a window (usually Hanning, or Hamming) and also get a longer time period (e.g. 10 or more standard grid frequency periods).</p>
<h2>Python Code</h2>
<p>I am also adding some python code, just for verification of the above:</p>
<pre><code>#%%
import numpy as np
import matplotlib.pyplot as plt
# %%
Tmax = 1/50 #integratin period
f_g = 50 # grid frequency
dt = 0.00125
ts = np.arange(0, Tmax, step=dt)
# %% Plot 3 line voltages
Vmax = 10
V_As = Vmax*np.cos(2*np.pi*f_g*ts)
V_Bs = Vmax*np.cos(2*np.pi*(f_g*ts + 1/3))
V_Cs = Vmax*np.cos(2*np.pi*(f_g*ts + 2/3))
plt.plot(ts, V_As, label='A')
plt.plot(ts, V_Bs, label='B')
plt.plot(ts, V_Cs, label='C')
plt.legend()
# %% Estimation of Power in line A, for a given impedance R_A (complex)
R_A = 2+1/2*np.pi*1j
angle = np.angle(R_A) # actual cos phi
print(angle)
I_As = np.real(Vmax/np.abs(R_A) * np.exp(1j* (2*np.pi*f_g*ts + np.angle(R_A))))
# plots V and I for verification
plt.plot(ts, V_As, label='V_A')
plt.plot(ts, I_As, label='I_A')
plt.legend()
plt.grid()
# %%
def calc(Vs,Is):
P_activ = np.sum(Vs*Is)*dt/Tmax
Vrms = np.std(Vs)
Irms = np.std(Is)
P_total = Irms*Vrms
phi = np.arccos(P_activ/P_total)
print ('Active :{:.3f}'.format(P_activ))
print ('Vrms :{:.3f}'.format(Vrms))
print ('Irms :{:.3f}'.format(Irms))
print ('P_total :{:.3f}'.format(P_total))
print ('phi :{:.3f}[rad] = {:.3f}'.format(phi, phi*180/np.pi))
calc(V_As, I_As)
</code></pre>
| 37443 | Calculating features of a 3 phase power timeseries |
2020-09-01T21:15:20.370 | <p>In wikipedia there is the following</p>
<blockquote>
<p>With the speed of instruction fetch reduced by 50% in the 8088 as
compared to the 8086, a sequence of fast instructions can quickly
drain the four-byte prefetch queue. When the queue is empty,
instructions take as long to complete as they take to fetch. Both the
8086 and 8088 take four clock cycles to complete a bus cycle; whereas
for the 8086 this means four clocks to transfer two bytes, on the 8088
it is four clocks per byte.</p>
</blockquote>
<p>And the following example:</p>
<blockquote>
<p>Therefore, for example, a two-byte shift or rotate instruction, which
takes the EU only two clock cycles to execute, actually takes eight
clock cycles to complete if it is not in the prefetch queue.</p>
</blockquote>
<p>I don't understand how the example works</p>
| |architecture| | <p>Internally, 8086 and 8088 are the same. They have an 8 byte instruction queue. When the queue is empty, processor stalls while instruction is fetched.</p>
<p>Difference is BIU Bus Interface Unit, which fetches bytes for 8088 and words (2 bytes) for 8086. Each processor takes 4 clock cycles to access RAM.</p>
<p>If we have a fast word instruction, the 8088 will take 8 clock cycles to execute it. The 8086 will take 6 clock cycles if the instruction is word addressed or 8 clock cycles if the instruction is across two bytes. The wiki is assuming optimum storage.</p>
<p>4 clock cycles to fetch and 2 to execute. For a word aligned instruction, the first byte tells processor the type of instruction, and the second specifies the action.</p>
<p>The execution by the EU, knows the instruction is a register based instruction, so it hides execution time for 8088 and odd byte aligned 8086, which means 8 clock cycles.</p>
| 37449 | How to calculate the clock cycle in relation to bytes per clock bus? |
2020-09-04T14:45:29.267 | <p>Throttles, chokes, dampers, and valves all restrict fluid flow. Aside from specific terminology/usage in internal combustion engines, are these really interchangeable terms, or do they have some specific meaning? Perhaps choke and throttle only apply to internal combustion engines and have no standard meaning outside that context? Damper might then be a special case of valve that applies to gases?</p>
<p>Dictionary definitions are not tremendously helpful. From Wiktionary:</p>
<p><strong>Choke</strong>: 1. A control on a carburetor to adjust the air/fuel mixture when the engine is cold.</p>
<p><strong>Throttle</strong>: 1. A valve that regulates the supply of fuel-air mixture to an internal combustion engine and thus controls its speed; a similar valve that controls the air supply to an engine.</p>
<p><strong>Damper</strong>: 1. Something that damps or checks:
A valve or movable plate in the flue or other part of a stove, furnace, etc., used to check or regulate the draught of air.</p>
<p><strong>Valve</strong>: 1. A device that controls the flow of a gas or fluid through a pipe.
2. A device that admits fuel and air into the cylinder of an internal combustion engine, or one that allows combustion gases to exit.</p>
| |mechanical-engineering|fluid-mechanics|automotive-engineering|chemical-engineering|hvac| | <p>Choke is usually just applied to starting fuel/air mix regulation for ICE engines, as far as I'm aware.</p>
<p>Throttle is used to actively manage fluid flow while a device is operating.</p>
<p>Dampers typically only refer to means of controlling air/exhaust flow in non-ICE, like power plants or HVAC systems. As in your definition a damper is usually just a movable plate or plates (ie louver).</p>
<p>Valve is just a general term for any sort of flow control device that is usually comprised of an inlet, outlet and means of limiting or preventing flow. Myriad configurations available.</p>
| 37482 | Throttle vs Choke vs Damper vs Valve? |
2020-09-05T13:38:33.057 | <p>I was looking at <a href="https://engineering.stackexchange.com/questions/32834/differentiating-between-two-orthographic-projections">this previous question</a> and I got bored and started drawing it up just for something to do.</p>
<p>The image in the question shows a nice front view and an elevation/section as shown below.</p>
<p><a href="https://i.stack.imgur.com/QT3lb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QT3lb.jpg" alt="Original Image" /></a></p>
<p>Now my Background with drafting is:</p>
<ul>
<li>1 class in high school which was very basic hand drafting;</li>
<li>1 class in University which was more advanced drafting and intro to CAD at the tail end;</li>
<li>4 years of drafting mainly buildings for structural purposes;</li>
<li>6 months of Mech/Elec drafting for HVAC, plumbing, and basic circuits; and</li>
<li>18 years of bridge design and drafting.</li>
</ul>
<p>And in case some of the comments/answers vary by region, my background is all in Canada.</p>
<p>So after 2 decades what I have been taught is very rusty to say the least, and what I have been using and applying in the transportation structures field does not necessarily cross over to other fields 100%.</p>
<p>So I am looking at trying to clarify some terminology and figure out if I am not the only one finding some MINOR issues with the elevation that is drawn up in the example.</p>
<h1>Question 1 Section Versus Elevation</h1>
<p>The first thing that bugs me when I look at that elevation is that its hatched like its a section. To me an Elevation is taken from outside of the object looking at it. Because there is no cut through the material, there should be no hatching. A section on the other hand cuts through the material. Where the material is solid, you apply hatching, and where the material is void/space, you leave it as white?</p>
<h1>Question 2 Hidden lines</h1>
<p>I find this to be a bit of a judgement call. I find that in structural drafting and some architectural drafting if a bit inconsistent/situational. For the most part, elevations are basically the near face. Depending on how artistic you want to get you may or may not include skew affect for for for surfaces going into the page. You are not going to show every wall, door, column, stair case behind the outside wall/face/plane of the building. However I do find that we will show wall/foundation outline below ground as hidden. When it comes to a bridge, the backface of the abutment may or may not be shown as hidden behind the wingwalls. However in the back of my mind when I was doing all those basic block with hole, slanted face drawings in high school, I recall that all hidden line work was included, not just the near face.</p>
<p>When it come to sections I have heard two schools of thought this. School A is basically just like the computer and you only include what is in the cutting plane. This means you would never have a hidden line in a section. Kind of the way computers cut sections through solids, at least in the early days. School B says when you have a section you also include include nearby hidden items, but not things that are far away/have no bearing on what you are trying to convey. I buy into this and it why certain details get included in a bridge deck cross section and other like the abutment and girder bearing supports are not included. Too far away from the section cut, AND not pertinent to the information that needs to be relayed.</p>
<p>So when it comes to this drawing above, should all hidden lines be included because the piece is relatively small, and simple? or should no hidden lines be shown? I tried to illustrate this with Section A-A option 1 and 2 below.</p>
<h1>Question 3</h1>
<p>Based on what I see in the "Side View" above, to me there is a rectangular block below the middle circles. I based this on the filets we can see in "Side View" and the solid to the left of the 7 measuring the lower wall thickness. What bothers me, is that based on the rest of the elevation above, it appears to be taken through the centre line. That being the case, that vertical block should be hatched? Is it a bad idea to mix hatching and hidden lines (see option 1 below)</p>
<h1>Question 4</h1>
<p>Why in the elevation/section is the base bolt hole centre line the only reference given, and not the hidden outline for the holes? The holes are closer the bottom of the protruded curve shown above the 6 in the top right part of the elevation. So if you are showing that, wouldn't the bolt holes warrant inclusion since they are closer? Basically the difference between Option 1 and Option 2 below.</p>
<h1>Question 5</h1>
<p>Does it really matter which view dimensions show up on as long as there are enough dimension to properly locate and size everything? The centre cirle, it has its outside dimension referenced in the "Side View" above, and its interior dimension in the elevation view. Wouldn't it be better to hve both dimensions in the same view for quicker referencing?</p>
<h1>Question 6</h1>
<p>Is it common to show the bevel countersink in plan view (top view), or is it more common to show a single circle and give a not describing the bevel or countersink for that matter?</p>
<h1>Question 7</h1>
<p>I know when you have a change in angle in the face, ie a slope surface meets another plane surface as an angle, a line of intersection is drawn. Does the same hold true for curved surface. Should a line be drawn when you change from a flat surface to a curved surface? If so, should there be a line at a change in curvature as well, ie two curved surfaces meeting or transitioning from 1 radius to another? Back of my head says yes, but I am questioning myself more and more as I write this set of questions!</p>
<h1>Question 8</h1>
<p>Almost completely unrelated. Back in the day when I was being taught hand drafting, I recall being told that when a dash line touches another line, you always start a solid dash from that point of intersect. Easy enough to do when you hand drafting. Now in the age of CAD programs, and I have not seen them all, it appears to me that dashes and dots are just proportioned between end points of the lin with no regard if they intersect with a another dashed line. This mean you occasionally get white space where lines cross. Not necessarily a good look. Alternatively one can break the line so it has an end point at the point of intersection. This forces a solid line to appear at intersection point. Looks better, but now if you want to get the length of the line the line is broken into small segments potentially. I know in AutoCAD you could use a polyline with linetype generation turned on/off, but that also may not cover all situations.</p>
<p>Is this lines crossing at a solid dash actually a thing? Am I remembering wrong? Or is it being pushed aside in modern CAD as having a continuous line from end point to end point is more important?</p>
<p><a href="https://i.stack.imgur.com/l9KIE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l9KIE.png" alt="Sample Drawing" /></a></p>
<p>In the side view and section I drew up, which one would be preferred (has the fewest errors)?</p>
<p>What changes if any would be recommended?</p>
<p>Update: I just realize, my section cut it looking the opposite direction</p>
| |technical-drawing|drawings| | <p>Your question is much broad in my opinion, I'll stick with the most important aspects IMO.</p>
<p>Computer 2D drawing has limitations (no detection of crossing of invisible lines, dimension lines touching the solid lines, and lots of others), and cad programs simply don't care about these kind of things (and I hate this aspect, but we work with what we got).<br />
<strong>Technical drawing is made to garantee that both individuals</strong> (you and whoever read the drawing later) <strong>can understand what is the object (even the details)</strong>. Isn't uncommon to bend the rules a little bit maintainning this mindset.</p>
<p>One point about the handmade drawing rules is, normally there's no different colors (all drawing lines is made with good old graphite). So the rules are made to improve the compreension of who will see later.<br />
My usual method is to think about who will read later and probably with <strong>no colors</strong> (much common to print with tonner so only black and white)</p>
<p>By what I can see in your drawing (the one you made) let-me point my tougths:</p>
<ol>
<li>Top view without dimension lines, <em>if this one is made by me at least 5 dimensions would be here (probably more)</em>. This view would be good for dimensions balancing (not overdimensioning the others views, making it harder to understand).</li>
<li>Front view, dimensions using invisible lines, realy? This is one definite no-go by what I know. And dimensions inside of the piece is not a good idea too (dificult readability).</li>
<li>Right side view dimensions referencing invisible lines (not good). And dimensions in bad position (easy to misunderstand the red line with green if all are the same color).</li>
<li>Section A-A. If you use Right side view, and good dimensions balancing between views there's no need for section. But some rules that I follow are:
<ul>
<li>No invisible lines in section, and absolutelly no partial (only close) invisible lines this is asking for misunderstanding.</li>
<li>The option 2 looks better, but you need one of the two options. 1) Delete the yellow lines and consider this as only the cut area. 2) Include in the top-right the back of piece border (the same you made in the right view and option 1 view).</li>
<li>Hatching or not the reinforcing part is a good point. There's one rule about not hatching (similar to drawing bolts and rivets), but this one isn't too common to me. Usually I consider who will read later and decide about hatching or not.</li>
</ul>
</li>
</ol>
<p>Now answering the other questions:</p>
<blockquote>
<p>Question 1 Section Versus Elevation
The first thing that bugs me when I look at that elevation is that its
hatched like its a section. To me an Elevation is taken from outside
of the object looking at it. Because there is no cut through the
material, there should be no hatching. A section on the other hand
cuts through the material. Where the material is solid, you apply
hatching, and where the material is void/space, you leave it as white?</p>
</blockquote>
<p>Elevation is a new term to me (I'm not from civil) but for me the cut section (hatched section) without the cut line is uncommon too.</p>
<blockquote>
<p>Question 2 Hidden lines
I find this to be a bit of a judgement call. I find that in structural
drafting and some architectural drafting if a bit
inconsistent/situational. ... So when it comes to this drawing above,
should all hidden lines be included because the piece is relatively
small, and simple? or should no hidden lines be shown? I tried to
illustrate this with Section A-A option 1 and 2 below.</p>
</blockquote>
<p>No hidden lines shown. I'm from school A, but for metal hatching (like this piece) using hidden lines would be harder to locate, so not showing would be the best.<br />
In larger frames with concrete maybe the rules are a bit different (I don't know), but for metal parts with lots of details certainly no invisible lines inside hatching is the best.</p>
<blockquote>
<p>Question 3
Based on what I see in the "Side View" above, to me there is a
rectangular block below the middle circles... What bothers me, is that
based on the rest of the elevation above, it appears to be taken
through the centre line. That being the case, that vertical block
should be hatched?</p>
</blockquote>
<p>For reinforcing parts in hatching the rule is the same as bolts and rivets, and don't need to hatch. Don't ask how to understand when is good to hatch or not (I don't know), usually I try to not cut these reinforcing joints at all.</p>
<blockquote>
<p>Question 4
Why in the elevation/section is the base bolt hole centre line the
only reference given, and not the hidden outline for the holes?</p>
</blockquote>
<p>Like I said, no invisible lines in hatch section. You can put the center line or not, if necessary.</p>
<blockquote>
<p>Question 5
Does it really matter which view dimensions show up on as long as
there are enough dimension to properly locate and size everything? ...
Wouldn't it be better to have both dimensions in the same view for
quicker referencing?</p>
</blockquote>
<p>The secret here is dimension balancing, you distribute the dimensions between views to improve readability. Whoever is reading your drawing need to understand first the object and only after the dimension.<br />
A good rule of thumb is <strong>ever</strong> use 3 or more views (even for simple objects).</p>
<blockquote>
<p>Question 6
Is it common to show the bevel countersink in plan view (top view), or
is it more common to show a single circle and give a not describing
the bevel or countersink for that matter?</p>
</blockquote>
<p>If the countersink is necessary then you show in the drawing. Show in top view (like you've done) and show in front view with invisible lines (or in partial cut like in the example you've seen)</p>
<blockquote>
<p>Question 7
I know when you have a change in angle in the face, ie a
slope surface meets another plane surface as an angle, a line of
intersection is drawn. Does the same hold true for curved surface.</p>
</blockquote>
<p>Is not true, the normal procedure is to draw as you see the view (no line because there's no real edge), if you go way older then you would draw kind of the shadow on the curvature but for hand drawing isn't necessary.<br />
With at least 3 views (the minimum number of views necessary) the radius will be easily recognized even without a line.</p>
| 37502 | Drafting techniques for various views |
2020-09-05T14:47:15.980 | <p><a href="https://i.stack.imgur.com/NznM4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NznM4.png" alt="enter image description here" /></a></p>
<p>After solving I got
<span class="math-container">$$T(s) = \frac{s^{2}+3s+3}{(s+1)(s+2)}$$</span></p>
<p>and for the stability of the system I got:
<span class="math-container">$$ s_1 = -1 $$</span></p>
<p><span class="math-container">$$ s_2 = -2 $$</span></p>
<p>So, is the system stable?</p>
| |control-theory|transfer-function|stability| | <p>Unfortunately, your solution is not correct regarding the final transfer function. Here is a step by step solution:</p>
<p><a href="https://i.stack.imgur.com/0L5g7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0L5g7.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ W(s) = X(s) - Y(s) $$</span></p>
<p><span class="math-container">$$ Z(s) = W(s)\cdot \frac{s}{s+1} $$</span></p>
<p><span class="math-container">$$ R(s) = Z(s)\cdot \frac{1}{s} $$</span></p>
<p><span class="math-container">$$ Y(s) = R(s) - W(s)\cdot \frac{1}{s+2} $$</span></p>
<p>By working out (easily) the math you can reach to the final transfer function, which is:</p>
<p><span class="math-container">$$ T(s) = \frac{1}{s^2+3s+3} $$</span></p>
<p>Now, regarding the stability of the whole system. A system is stable if and pnly if the poles of the closed loop transfer function have negative real part. The poles of the above transfer function appear to be complex conjugate numbers:</p>
<p><span class="math-container">$$ p_{1,2} = -1.5 \pm 0.866i $$</span></p>
<p>So, <span class="math-container">$Re(p_{1,2}) < 0 \Rightarrow $</span> System is stable. Until you get used to deriving transfer function from block diagrams, I strongly encourage you to write is as I have by reducing it to simpler signals and finally combine them all together at once. You can also use Simulink to create the complicated block diagram and the simple one with only one transfer function block and compare the step responses. You will find out that they are identical.</p>
| 37504 | Did I get the correct result? Block reduction problem |
2020-09-05T18:04:38.423 | <p>I know that I can get gyro angles by integration, but that are not the angles I need.</p>
<p>I need the same angles that I calculate from the accelerometer (extrinsic euler xyz angles) to apply properly a complementary filter.</p>
<p>Background: I am building a flight controller for a quadcopter.</p>
| |acceleration| | <p>I found the answer with the help of the answer of this question: <a href="https://stackoverflow.com/questions/39441900/how-use-raw-gryoscope-data-s-for-calculating-3d-rotation">https://stackoverflow.com/questions/39441900/how-use-raw-gryoscope-data-s-for-calculating-3d-rotation</a></p>
<p><strong>I represent the rotation as a quaternion and then convert it to any angle I want.</strong></p>
<p>But I still don't understand why there is no problem in interpreting the gyro angles as Euler angles and converting them into a (delta) quaternion. The gyro angles don't have any order in which they were moved so the end position should always differ depending on the order the gyro was moved in the respective angle.</p>
| 37507 | How to convert gyro data to extrinsic/global/world euler xyz angles? |
2020-09-06T17:12:38.920 | <p>I am trying to understand why a bigger swept area on disc brakes provides higher braking force for the same force being exerted on the brake lever/pedal by the rider.</p>
<p><a href="https://i.stack.imgur.com/Gy15I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gy15I.png" alt="enter image description here" /></a></p>
<p>If we use the same exact brake pad material, same thickness, same disc material/surface treatment etc. and the only variable we change is the width of the brake pads and therefore increase the swept area, would this result in higher braking force for the same lever/pedal input force?
If so, please explain why.</p>
<p>Larger swept area means better cooling of the pads during braking but that shouldn't make a difference when the brakes are still cold.
Larger swept area also means the pad material can be softer since the brake pads are larger so won't wear as quickly as the smaller variety, is this a significant reason why larger swept are provides better braking performance?</p>
<p>Would increasing the length of the pads (thus not increasing swept area) also increase braking performance (braking force higher for same lever/pedal force)? If so, would it have the same effect as making the pads wider?</p>
<p>source: <a href="https://engineering.stackexchange.com/questions/33114/hydraulic-brakes-why-motorcycle-brakes-more-powerful-than-bicycle-brakes">Hydraulic Brakes - Why motorcycle brakes more powerful than bicycle brakes?</a> (check first answer)</p>
<p>Thank you for any input, it is greatly appreciated!</p>
| |heat-transfer|dynamics|friction| | <p>Brakes primarily convert kinetic energy to heat energy. So a large area can absorb more heat lowering the peak temperatures ;of course this is strongly affected by the thickness/mass of the discs and other factors. AND the larger area can get rid or more heat . High temperatures cause deterioration of pad materials ,so lower ( not as high) temperatures extend pad life. I anticipate brake usage and apply brakes "lightly" to increase stop time which reduces peak pad temperatures. As a result , when I last looked at my Murano pads at about 70,000 miles they showed very little wear. Years ago a friend did amateur road racing which required extreme braking and new pads and rotors every racing weekend.</p>
| 37522 | Disc Brakes - Swept Area , why is larger better? |
2020-09-07T15:56:54.080 | <p>I have no engineering background at all, but when I bumped my small electric heater and it started whining and creating a burning smell, (since it's labor day with no stores open to buy a new one) I unplugged it and took it apart. Once I got down to the electric motor housing, I could spin the fan with a finger and hear the same whine, and noticed the axle (or whatever you call a rod that rotates a fan) attached to the fan was slightly crooked where it was housed within this piece.</p>
<p>Further inspection revealed that the brass-colored bearing (if that's the correct term) isn't securely seated within the star-shaped housing, but rather just held down by the thin tin flaps, and beyond that pressure moves freely, allowing it to seemingly cause the friction / disturbance creating the whine.</p>
<p>Is there a reason for the bearing to be held in place by this flimsy star chamber rather than a solid seating?</p>
<p>I tried to straighten it out and it appeared straight, but once I put it back together, the whine had only been reduced marginally, and the burning smell still occurred.</p>
<p><a href="https://i.stack.imgur.com/8gCW0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8gCW0.jpg" alt="enter image description here" /></a></p>
| |electrical-engineering|bearings| | <p>Apparently the bronze bushing/plain bearing is self-centering in the cadmium plated steel holder. The bearing can move slightly to accommodate the position and angle of the motor shaft. Oil would likely help it.</p>
| 37534 | Is there a reason for the bearing to be held in place by this flimsy star chamber rather than a solid seating? |
2020-09-08T12:13:55.067 | <p>I have a heat exchanger with known flows on priamry and secondary sides - <span class="math-container">$\dot m_p$</span>, <span class="math-container">$\dot m_s$</span>, known temperature in <span class="math-container">$t_{si}$</span> and a desired temperature out on the secondary side <span class="math-container">$t_{so}$</span>. How to find the required <span class="math-container">$t_{pi}$</span> analytically? So far I've always done these calculations numerically with the excel solver.</p>
<p>This is where I get stuck in solving analytically - from the energy balance we find:</p>
<p><span class="math-container">$$ t_{pi}=t_{po} - \frac{\dot m_s c_s}{\dot m_p c_p} * (t_{so} - t_{si})$$</span></p>
<p>but we need <span class="math-container">$t_{po}$</span>, when I try this (from the heat transfer equation) I think I'm stuck:</p>
<p><span class="math-container">$$ t_{po} = \frac{kA \Delta T_m}{\dot m_p c_p} + t_{pi} $$</span></p>
<p>because to solve I'd need to extract my temperatures from the <span class="math-container">$ln$</span> in the denominator of <span class="math-container">$\Delta T_m$</span> and that's where I'd rather look in a textbook than do the math myself. Except my textbook (Perry) doesnt have that detail. <span class="math-container">$A$</span> is known and <span class="math-container">$k$</span> can be estiamted, so is known for the purpose of this questions.</p>
<p>So, <strong>for given mass flows, and desired outlet temperature on the secondary side, what is the required inlet temperature on the primary side?</strong></p>
<p>The <em>ultimate</em> problem I want to solve is that I want to ask a question about HX control and I feel it would help to have this equation.</p>
| |heat-transfer| | <h1>Background</h1>
<p>Normally the process would go line this. Let's say the primary is hot, and the secondary is the cooler fluid.</p>
<p>The heat transfer rate <span class="math-container">$\dot{Q} = -\dot{Q}_p = \dot{Q}_s $</span>. I.e.:</p>
<ul>
<li><p>The cooler fluid (s) gains
<span class="math-container">$$\dot{Q} = m_s\cdot C_{p,s}(T_{s,o}- T_{s,i}) $$</span></p>
</li>
<li><p>The hot fluid (p) loses:
<span class="math-container">$$\dot{Q} = - m_p\cdot C_{p,p}(T_{p,o}- T_{p,i}) $$</span></p>
</li>
</ul>
<p>Therefore the change in temperature is :</p>
<p><span class="math-container">$$ T_{p,o} = T_{p,i}- \frac{m_s\cdot C_{p,s}}{m_p\cdot C_{p,p}}(T_{s,o}- T_{s,i}) $$</span></p>
<p>Normally, if you get to that point then you need to calculate the length of the exchangerm and therefore the <span class="math-container">$A$</span> but since you know it I'll press on.</p>
<p>Here, I'm going to assume counterflow. At that point you need the <em>logarithmic mean temperature difference <span class="math-container">$\Delta T_{lm}$</span></em>. (please note that for different type of flows parallel, cross etc you need to change this).</p>
<p><span class="math-container">$$\Delta T_{lm} = \frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\Delta T_1 = T_{p,i}-T_{s,o}$</span> : temperature difference at one exit</li>
<li><span class="math-container">$\Delta T_2 = T_{p,o}-T_{s,i}$</span> : temperature difference at other Exit</li>
</ul>
<p>Then you can apply:</p>
<p><span class="math-container">$$ kA\cdot\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}= \dot{Q} = m_s\cdot C_{p,s}(T_{s,o}-T_{s,i}) $$</span></p>
<p><span class="math-container">$$ kA\cdot\frac{T_{p,i}-T_{s,o}-T_{p,o}+T_{s,i}}{\ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right)}= \dot{Q} $$</span></p>
<h2>Iterative solution</h2>
<p>Here you can, solve with respect to <span class="math-container">$T_{p,i}$</span>
<span class="math-container">$$ T_{p,i}-T_{s,o}-T_{p,o}+T_{s,i}= \frac{\dot{Q}}{kA\cdot} \ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right) $$</span></p>
<p><span class="math-container">$$ T_{p,i}= \frac{\dot{Q}}{kA} \ln \left(\frac{T_{p,i}-T_{s,o}}{T_{p,o}-T_{s,i}}\right) +T_{s,o}+T_{p,o}-T_{s,i}$$</span></p>
<p>From that point the easiest think of Excel is to iterate in order to find the solution (set a guess for <span class="math-container">$T_{p,i}$</span> apply it on the right hand, get a new <span class="math-container">$T_{p,i}'$</span>, which you plug into the equation until <span class="math-container">$T_{p,i}-T_{p,i}' \rightarrow 0 $</span> ).</p>
<h2>close form solution</h2>
<p>The other option if you need a close form solution (which is probably what you are asking), you can look at the <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert W function</a>. In that case, you start from:</p>
<p><span class="math-container">$$ kA\cdot\frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}= \dot{Q}$$</span></p>
<p>After replacing <span class="math-container">$\Delta T_1, \Delta T_2 $</span>,the solution for <span class="math-container">$T_{p,i}$</span> takes the following form:</p>
<p><span class="math-container">$$ T_{p,i} = T_{s,o} - \frac{
Q }{A k} ProductLog[-\frac{A k(T_{p,o}-T_{s,i}) e^{-\frac{A k }{Q}(T_{p,o}-T_{s,i}) } }{Q}]$$</span></p>
<p>Where:</p>
<ul>
<li><span class="math-container">$ProductLog[x]$</span> is the Lambert W function.</li>
</ul>
<p>However Excel does not have this function builtin (at least to my knowledge). So you need to either do it another language/system (e.g. octave, or python) or find a macro for lambert w.</p>
| 37545 | What's the setpoint of the temperature on the primary side of a heat exchanger? |
2020-09-08T22:34:51.210 | <p>I am on a student design team planning to build an electric motorcycle and we are planning to use an EMRAX 208 motor. We want to have liquid cooling for the motor but I cannot find any resources on where to buy the cooling system. Are there any companies that deal with this sort of stuff? If not, do you have any suggestions on how to make a cooling system? Thank you for your help in advance</p>
<p>Note: I am not asking for you to tell me which one to purchase, I am just trying to be directed in the right direction. Also this is my first time asking a question here so I apologize in advance if I break rules.</p>
| |mechanical-engineering|cooling| | <p>You should design a cooling system according to the <a href="https://emrax.com/wp-content/uploads/2017/10/user_manual_for_emrax_motors.pdf#%5B%7B%22num%22%3A36%2C%22gen%22%3A0%7D%2C%7B%22name%22%3A%22XYZ%22%7D%2C47%2C732%2C0%5D" rel="nofollow noreferrer">motor manual</a> (this is just one I found, make sure you use the correct manual).</p>
<p>Before you get into liquid cooling, make sure that the air cooling requirements cannot be satisfied.
The motor doesn't perform any worse with air cooling, as long as you can meet the requirements.</p>
<p>For the liquid cooling system, start by determining the required coolant flow and heat dissipation.
Use that to spec a pump and a heat exchanger (probably a radiator).
The radiator might be passively cooled or have fans.</p>
<p>The cooling section in the manual provides design requirements and guidelines for either cooling method.
Make absolutely sure that your cooling system meets requirements, or you might have warranty problems.</p>
<p>If you do design a liquid cooling system, you should consider using it to cool other systems, such as batteries.
You may not have any other systems that need active cooling, but it can simplify your design if you do.</p>
| 37554 | Liquid cooling system for Emrax 208 motor |
2020-09-09T11:24:25.310 | <p>Do off-the-shelf air compressors have minimum driving speeds, below which they cannot build the necessary pressure? Reciprocating compressors and scroll compressors are available—is there a difference in minimum working speed?</p>
<p>In my application, I would like to drive a compressor at very slow speeds (1–10 rpm) to store wind energy over a long period of time, but I'm concerned internal leakage inside the compressor may make that nonviable.</p>
| |mechanical-engineering|fluid-mechanics|compressed-air|compressors| | <p>Google dugout aeration windmill system.</p>
<p>There are various makers. Here's one: <a href="https://koenderswatersolutions.ca/" rel="nofollow noreferrer">https://koenderswatersolutions.ca/</a></p>
<p>The compressor on them does one cycle per revolution of the windmill. While not as slow as 1 rpm the one I have is only rarely over 50 rpm.</p>
<p>Internally they are a diaphragm pump. This means you only have the leaking of the valves to contend with.</p>
<p>The unit I have is fairly low pressure. The pond is 14 feet deep, so the pressure is only about 7 PSI plus friction losses in the 1000 feet of 5/8 tubing between the windmill and the pond.</p>
| 37561 | Slow Driving an Air Compressor |
2020-09-10T16:59:33.837 | <p>I'm a total novice and self taught, currently designing and building a workbench for woodworking with an adjustable height top. This is the last major part to finish and I'm having trouble determining which bearing configuration to use.</p>
<p><strong>EDIT:</strong> I essentially am trying to do this, except with the drive mounted below the floor instead of over it: <a href="https://www.youtube.com/watch?v=URrEdGeoDBs" rel="nofollow noreferrer">https://www.youtube.com/watch?v=URrEdGeoDBs</a>
<a href="https://i.stack.imgur.com/QDJ2G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QDJ2G.jpg" alt="enter image description here" /></a></p>
<p><strong>Overview:</strong>
There are four legs (4x4s) that travel in a cuff. I'm designing for the top to carry and lift about 400lbs, but more likely will be under 100lbs (with each leg being able to lift 100 lbs max / 25 lbs). Total amount of lift is going to be between 15-17" depending on the bearing configuration. After the bench is lifted to different heights, I'll have spacers to hold it in place and then back off the force of the lift (so if I'm hammering something, I won't be hammering against the threaded rods)</p>
<p>The rising mechanism is driven by four threaded rods (3/4"). Below the table, each rod will have a gear, and all four will be connected by one roller chain driven by one drive gear (which I'll use an electric drill to turn). On the threaded rod is a square nut, that fits inside a square tube held in place by two tension pins (so as the rod turns, it will push up against the tension rods/square tube which pushes the top of the workbench up and down).</p>
<p><strong>Some sketches:</strong></p>
<ul>
<li>Underneath view with gears and chain: <a href="http://prntscr.com/uf1bup" rel="nofollow noreferrer">http://prntscr.com/uf1bup</a></li>
<li>Front view, leg cuffs, threaded rod / square tubing. Bottom right, the bearing issue (see the "?"): <a href="http://prntscr.com/uf241j" rel="nofollow noreferrer">http://prntscr.com/uf241j</a>
<a href="https://i.stack.imgur.com/rYrVN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rYrVN.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/nBcZz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nBcZz.jpg" alt="enter image description here" /></a></li>
</ul>
<p><strong>Problem Description and Research:</strong></p>
<ul>
<li>Because the gears must be located under the bench (the inside of the bench is used for some tools and whatnot), I need a way to brace the forces of the rod/workbench top against the bench floor.</li>
<li>From what I understand, radial bearings are designed to handle perpendicular forces to the shaft, while thrust bearings are designed for mostly axial/thrust forces parallel to the shaft. Also, the more hardware I have above the workbench top, the less rise I can get, so being really economical with what I use above the workbench is important.</li>
</ul>
<p>I bought a bunch of different radial and thrust bearings to try out an assortment of options, they all have pros and drawbacks. (For underneath I'm using a flanged bearing along with the gear and collar clamp to hold it in place, I'm struggling with the above floor bearings unless you have other ideas on the underneath part)</p>
<p><strong>Option 1):</strong> Flanged / pillow radial bearing.
<a href="https://rads.stackoverflow.com/amzn/click/com/B01NCUJH8I" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/gp/product/B01NCUJH8I/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1</a>
These bearing are rated for Static (axial) Load Rating: 6.65 kN/1500 lbs and Dynamic (radial) Load Rating: 12.8 kN/2900 lbs.
On the pro side, the flange plus set screws will allow me to hold the threaded rod in place, along with a collar it should hold well and not slip.
Con, I don't think this is the right kind of bearing. I tapped one with a hammer and it started to come out of the race a little...but maybe for these types of lifting forces it'll be fine?</p>
<p>Prototype picture: <a href="http://prntscr.com/uf1sa9" rel="nofollow noreferrer">http://prntscr.com/uf1sa9</a>
(2x4 represents the workbench floor, flange bearing above with collar above.)
<a href="https://i.stack.imgur.com/LrU4u.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LrU4u.jpg" alt="enter image description here" /></a></p>
<p><strong>Option 2): Thrust Bearings.</strong>
<a href="https://rads.stackoverflow.com/amzn/click/com/B01EBKYSWA" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/gp/product/B01EBKYSWA/ref=ppx_yo_dt_b_search_asin_title?ie=UTF8&psc=1</a>
If I use thrust bearings, which seem to be the correct type of bearing to use, they bind and rub on the floor of the bench because the inner race is proud. I've tried drilling a wider hole, but the tolerances are too tight to solve the problem. I feel like there's a configuration that this could work but I've spent days playing with different types and haven't found a solution.</p>
<p>Quick prototype pics:</p>
<ul>
<li>Recessed bottom plate: <a href="http://prntscr.com/uf1u9y" rel="nofollow noreferrer">http://prntscr.com/uf1u9y</a></li>
<li>Thrust bearing: <a href="http://prntscr.com/uf1vsp" rel="nofollow noreferrer">http://prntscr.com/uf1vsp</a></li>
<li>Thrust bearing and collar (this is rubbing despite the recessed bottom): <a href="http://prntscr.com/uf1uy5" rel="nofollow noreferrer">http://prntscr.com/uf1uy5</a></li>
<li>Using the race from one as a bushing (this works, but I feel like there must be an easier way and uses up a ton of real estate): <a href="http://prntscr.com/uf20rx" rel="nofollow noreferrer">http://prntscr.com/uf20rx</a>
<a href="https://i.stack.imgur.com/K1fE4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K1fE4.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/mOi2D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOi2D.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/94bFx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/94bFx.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/cBYs8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cBYs8.jpg" alt="enter image description here" /></a></li>
</ul>
<p><strong>Option 3:</strong> I also have a bunch of these radial bearings...but pretty sure they're not the right type for this application? <a href="http://prntscr.com/uf22dd" rel="nofollow noreferrer">http://prntscr.com/uf22dd</a>
<a href="https://i.stack.imgur.com/U3EP9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U3EP9.png" alt="enter image description here" /></a></p>
<p>So, should I go with Option 1, 2, 3, or is there another / easier configuration that would make a lot more sense? If you know of a better bearing design a link would be really useful as a visual as well.</p>
<p>Thank you!</p>
| |bearings| | <p>Unless you're spinning the rods more than 100 rpm bushings should work, are cheaper and easier to work with than bearings.</p>
<p><a href="https://www.mcmaster.com/flanged-bushings/oil-embedded-flanged-sleeve-bearings-7/" rel="nofollow noreferrer">Examples</a></p>
<p>Comment added to answer: the video makes it much more clear what you're trying to accomplish, thanks. Option 1 should work for you. The bearing in that housing is designed for radial loads, but typically a deep groove radial ball bearing can handle an axial load equal to 50% of it's rated radial load. Or just steel washers sandwiching bushings, but it seems you already have all the pillow block bearings.</p>
| 37589 | Radial vs thrust bearings for assembling a lift mechanism |
2020-09-11T23:59:35.760 | <p>I am having a hard time trying to understand the effect of geometry on a specimen's yield and tensile strength or if there even is one.</p>
<p>If there is an effect or not, why is it so?</p>
<p>Thanks in advanced!</p>
| |mechanical-engineering|structural-engineering|materials|applied-mechanics|mechanical-failure| | <p>And the question you did not ask ;what about shape/crossection . That is, round and rectangular are both very common specimens .And rectangular with a curved crossection- full wall thickness pipe body. They also make very little if any difference. Note specifications often permit round tensile bars or full thickness tests with the same required mechanical properties.</p>
| 37600 | Does a change in specimen dimension affect the measured yield strength and tensile strength? |
2020-09-12T16:02:29.863 | <p>I just finished a problem, and got the correct answer, however I am confused on one of the mathematical operations that must be performed in order to get the correct answer.</p>
<p><a href="https://i.stack.imgur.com/FLrPv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FLrPv.png" alt="Piston Connecting two cylinders" /></a></p>
<p>The problem is as reads: We have a piston connecting two cylinders A and B of diameters 100mm and 25mm, respectively. The pressure acting on the outside of the entire piston is 100kPa, and the mass of the piston is 25kg, note that there is specific gravity present. A hydraulic lift pump at Cylinder A is pumped up to 500kPa. What is the gas pressure in cylinder B ?</p>
<p>So the formula for the answer is :</p>
<p>PressureB = [(PressureA × AreaA) - (MassPiston × 9.81) - (PressureOutside × (AreaA - AreaB ))] / AreaB</p>
<p>This all makes sense to me except for the difference in AreaA - AreaB multiplied by external Pressure. Why wouldn't we add the two areas instead to find the net force that the external pressure exerts ?
Any explanation will be appreciated.</p>
| |fluid-mechanics|thermodynamics|pressure|hydraulics| | <p>you take the difference because the force due to pressure acting on the top surface has direction upwards, while the force on the bottom surface has the opposite direction (same direction as gravity).</p>
<p>So the two forces cancel each other. Becauce the <span class="math-container">$F_{top}= P_{Ext}\cdot A_{top}$</span> and <span class="math-container">$F_{bot}= P_{Ext}\cdot A_{bot}$</span> the net force is:</p>
<p><span class="math-container">$$F_{net}=F_{bot}- F_{top} = P_{Ext}\cdot (A_{bot}-A_{top})$$</span></p>
<p>Hence the difference.</p>
| 37608 | Hydraulic lift connected to two pistons |
2020-09-12T17:16:21.193 | <p>Has there ever been any experimental verification of the benefits of designing a gear pair to have a "hunting tooth"?</p>
<p>The idea is to even out wear by having every tooth on the pinion meshing with every tooth on the gear.</p>
<p>Tooth hunting can be achieved by ensuring no common factors > 1 between the pinion and gear.</p>
<p>Thanks</p>
| |mechanical-engineering| | <p>Depends on the quality of your finished gear, both micro (surface finish) and macro (tooth runout, spacing, etc). Improper meshes will ultimately cause vibration in your system, if it's bad enough it will be pretty loud. Also, if you have an oversized tooth on your pinion and the mating tooth space on the gear is undersized you'll increase the stresses and likelihood of failure.</p>
<p>For lower quality gears hunting would be preferred so that the pinion and gear wear evenly.</p>
<p>For decent quality gears with respect to load it's less of an issue. And typically everyone else working with them will prefer an integer ratio.</p>
| 37610 | Gear Design - Hunting tooth |
2020-09-12T18:51:44.147 | <p>I'm not sure if this is a good place to ask this question, but I'm currently taking Control System Design and considering everything is this deriving of equations I'm having a hard time actually seeing what is actually happening. I'm just curious as to what Lagrange Equations actually do in a space-state system, and how it would apply to an actual real life situation.</p>
| |control-theory| | <p>IMHO, the application of Lagrange Equation and the state-space representation are two separate things.</p>
<p>The Lagrange equation is used in order to derive the mathematical models of a system (i.e. the equations of motion for a mechanical system).</p>
<p>Once you obtain the mathematical model, then you can convert between all different types of represenations:</p>
<ul>
<li>differential equations of motion</li>
<li>Transfer Function</li>
<li>State Space representaition</li>
</ul>
<p>So, to sum up, I consider the application of Lagrange equation to a system as a preprocessing step, in order to obtain the mathematical model. One of the facets of the mathematical model is State-space representation.</p>
| 37611 | What does the Lagrange Equation do for State-space systems |
2020-09-12T19:39:31.630 | <p>How to decide what is in momentum and what is out momentum in using shell momentum balance?
My main motive is to understand why at 25:26 in the balance equation at <a href="https://www.youtube.com/watch?v=sMsw8BPT578" rel="nofollow noreferrer">https://www.youtube.com/watch?v=sMsw8BPT578</a> the professor is doing <span class="math-container">$\tau_{xy}|_{x} - \tau_{xy}|_{x+\Delta x} $</span> and not the other way round?</p>
<p>Even on wikipedia, <a href="https://en.wikipedia.org/wiki/Shell_balance" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Shell_balance</a> it isn't written what is in and what is out. If you consider always that the ground/wall kind of releases momentum then consider this <a href="https://www.youtube.com/watch?v=2DHR96x21GY" rel="nofollow noreferrer">https://www.youtube.com/watch?v=2DHR96x21GY</a> at 17:26 where in is taken as being released from the center of the circular pipe and not the walls.</p>
| |fluid-mechanics|chemical-engineering| | <p>We made an assumption at the start that <span class="math-container">$ \tau_{xy} $</span> means momentum flows in x direction which means the "in" is in the x direction. So as long as "in" is taken in the positive x, y, z direction for appropriate coordinates the analysis w'd be right.</p>
<p><strong>Reference Book:</strong></p>
<p>Transport Phenomena</p>
<p>Textbook by Edwin N. Lightfoot, Robert Byron Bird, and Warren E. Stewart</p>
<p>Page 43
<a href="https://i.stack.imgur.com/sCfM3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sCfM3.png" alt="enter image description here" /></a></p>
| 37615 | Determine in and out momentum in shell momentum balance |
2020-09-14T06:22:27.623 | <p>I have a small machine shop and need to size a beam for a gantry crane. The design is a 4 post fixed design with 2 horizontal beams running parallel across the shop ends with 1 beam to be mounted perpendicular on trolleys on top of the two parallels making a crane that can traverse x and y travel with an electric hoist rasing and lowering loads.</p>
<p>The span is approximately 28 feet and the biggest hoist I have is a mere 1 ton but would like to size it so that I have 3000-4000lbs load capacity. Also consider that the x and y axis trolleys are not motorized so a small dynamic/torsional load will be imparted in the x and y when the payload is pulled around.</p>
<p>I'm thinking a wide flange 12*45 would be a decent size. I'm unsure though if in beam calculators they account for the weight of the beam or if it is just the payload. With a centered load and 4000lbs the deflection shows .311" over 28 feet which seems reasonable enough if I understand what I'm looking at.</p>
<p>I'm thinking I'm looking for a moment of inertia in the x axis if the beam is standing in the traditional direction. I'm also not too clear on deflection limitations on steel.</p>
<p>Thanks for the help.</p>
| |beam| | <p>Although this is probably a question best answered by @kamran (I'm not familiar with imperial units or with safety factors in US) I'll give it a shot (just to improve my game).</p>
<p>Assuming the wide flange properties I found are correct</p>
<ul>
<li><span class="math-container">$I_{xx}= 350 [in^4]$</span></li>
<li><span class="math-container">$h = 12.06 [in]$</span> depth of beam</li>
<li><span class="math-container">$m_{beam}= 45 \frac{lb}{ft} $</span></li>
</ul>
<p>Although the mass of the beam is distributed, I am assuming that it is located in the center to make a more conservative estimation. the total load should be:</p>
<p><span class="math-container">$$P_{tot}[lbf] = m_{beam}g L + P_{Load} $$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$ m_{beam}\cdot L \cdot g = 1260 [lbf]$</span></li>
<li><span class="math-container">$ P_{beam}\cdot L \cdot g = 4000 [lbf]$</span></li>
</ul>
<p>The maximum bending moment is at the center :</p>
<ul>
<li><span class="math-container">$ M_b = P_{tot}*\frac{L}{2}= 73640 [lbf\cdot ft]$</span></li>
</ul>
<p>Then the maximum stress is
<span class="math-container">$$\sigma_b = \frac{M_m}{I_xx}\cdot \frac{h}{2} = 15.2245 [ksi] = 105[MPa]$$</span></p>
<p><span class="math-container">$$\delta = \frac{P \cdot L^3}{48\cdot E \;I_xx}= 0.4 [in]$$</span></p>
<p>For the static calculation, this is marginally ok. The safety factor seems to be about 2-3. However since you mention dynamic loads, I would feel prefer it to be slightly more beefy.</p>
<p>In any case, I would wait for @kamran's verdict.</p>
| 37638 | Beam for shop gantry |
2020-09-14T16:38:44.900 | <p>I'm trying to solve a design problem involving the dimensioning of a gearmotor that drives a rack & pinion system to move a cart.</p>
<p>The cart weighs 600 kg and is supported by 4 V-shaped steel wheels such as those in the picture which are in contact with an L-shaped steel profile.</p>
<p><a href="https://i.stack.imgur.com/VE0an.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VE0an.png" alt="enter image description here" /></a></p>
<p>Now, the design speed for the cart is 0.5 m/s and the acceleration has a ramp of 0.5 m/s<sup>2</sup>.</p>
<p>I want to find the maximum tangential force on the pinion in order to choose the appropriate motor, and I know that it has two components:</p>
<ul>
<li>acceleration * mass</li>
<li>mass * friction coefficient (600 * 9.81 * u)</li>
</ul>
<p>Which is the appropriate value for the friction coefficient? I believe it should be the rolling friction between steel and steel (i.e. 0.005), but do I also need to take into account the radius of the wheel or something?</p>
<p>Thank you for any advice.</p>
| |mechanical-engineering| | <p>Unless your wheels are skidding, they're in rolling contact. Use the rolling coefficient.</p>
| 37649 | Rack and pinion drive for cart |
2020-09-15T19:12:18.523 | <p>I am trying to calculate the final temperatures as stated above. There is hot gas flowing through a tube that is insulated and inside of a room (no wind). I will try to walk you through my thought process in doing so while typing out the formulas I'm using.</p>
<p>I'm starting with a visual resistor method (shown below). The main issues I believe I'm having is with both convective forces (inside the tube and outside, assuming there is convection inside). The formula for <em>hrad</em> (shown below) is one I found online but I'm not sure if that is correct.</p>
<p>NOTE: I'm not using the Nusselt equations shown as I did find one for a cylinder in cross flow that I'm using.</p>
<p>I am using the Rayleigh number for the external convection while also trying to use radiation as a factor. The problem arises when I try to run the system without insulation the outer surface temperature is MUCH lower than it should be (900F inside, 100F outside , its showing 550F surface temp). I know that this should be near the 900F mark as metal is a terrible insulator. I'm losing most of my heat transfer from my convection internally so I'm assuming that I'm doing something wrong there but can't figure it out.</p>
<p>Working on this today and when I completely remove the convection from the inside flow the temperatures seem to be a lot better. Another question that now arises is do the k-values for steel change with temperature? I'm assuming this is a definite yes but now am looking for a correlation between the temp and k values. This may be digging too deep now and basically need a full experiment to do but maybe its something that's already done? Thank you!</p>
<p>I've been trying to create an excel document that will be able to calculate these variables for any inputs for the system.</p>
<p>I was hoping that someone who knows more than me about this can look over what I'm using and hopefully tell me whether I'm using the correct equations/values.</p>
<p>If anyone could explain my wrong assumptions/thinking that would help immensely.</p>
<p><a href="https://i.stack.imgur.com/1VUun.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1VUun.jpg" alt="enter image description here" /></a></p>
| |fluid-mechanics|thermodynamics|heat-transfer| | <blockquote>
<p>The formula for <span class="math-container">$h_{rad}$</span> (shown below) is one I found online but I'm not sure if that is correct.</p>
</blockquote>
<p>You can try turning off radiation entirely to see the effect. You can also try fixing the surface temperature and estimating the relative heat flows by convection compared to radiation, since the two equations are well-defined. Use that estimate to substitute a multiplying factor <span class="math-container">$M$</span> for the outer resistivity. Rather than</p>
<p><span class="math-container">$$\frac{1}{R_{outer}} = \frac{1}{R_{conv}} + \frac{1}{R_{rad}} $$</span></p>
<p>write as</p>
<p><span class="math-container">$$\frac{1}{R_{outer}} = M\frac{1}{R_{conv}} $$</span></p>
<blockquote>
<p>I'm losing most of my heat transfer from my convection internally so I'm assuming that I'm doing something wrong there but can't figure it out.</p>
</blockquote>
<p>You have one of two approaches for the case that you remove the insulation. Are you holding the internal temperature constant and allowing the total heat flow to the gas in the tube to increase? Or are you fixing the heat flow to the gas in the tube and thereby observing that the internal temperature decreases?</p>
<blockquote>
<p>Another question that now arises is do the k-values for steel change with temperature?</p>
</blockquote>
<p>Yes. See one of these two links and/or search also for thermal conductivity versus temperature for the steel that you use.</p>
<p><a href="https://physics.stackexchange.com/questions/330158/why-does-the-thermal-conductivity-of-pure-metals-decrease-with-increase-in-tem?rq=1">https://physics.stackexchange.com/questions/330158/why-does-the-thermal-conductivity-of-pure-metals-decrease-with-increase-in-tem?rq=1</a></p>
<p><a href="https://nvlpubs.nist.gov/nistpubs/jres/12/jresv12n4p441_A2b.pdf" rel="nofollow noreferrer">https://nvlpubs.nist.gov/nistpubs/jres/12/jresv12n4p441_A2b.pdf</a></p>
<blockquote>
<p>I was hoping that someone who knows more than me about this can look over what I'm using and hopefully tell me whether I'm using the correct equations/values.</p>
</blockquote>
<p>Your approach appears to be correct for a specific case. You have not accounted for the temperature change of the gas inside the tube. An example of the formulation for this is <a href="https://engineering.stackexchange.com/questions/25985/cooling-hot-nitrogen-with-an-aluminum-block/25996#25996">shown at this link</a>. An example of the equation for heating a fluid in a solar concentrator is <a href="https://engineering.stackexchange.com/questions/21266/solving-heat-loss-equation-associated-with-solar-trough-concentrators/24108#24108">shown at this link</a>.</p>
| 37675 | Determining the final surface temperature and final gas temperature flow through a tube in a room |
2020-09-16T11:33:58.333 | <p>I am currently designing a ground based digging drill, that hopefully will be accepted by a commercial entity. I am currently working on the preliminary design of the robot and am hoping to move to the modeling phase soon, however I am slightly stuck on my Torque calculations.
<strong>Can anyone please confirm if my calculations below are correct?</strong></p>
<p><em>Background:</em></p>
<p>The drill will be relatively automatic, small, and simple. This autonomous digging drill, will be an updated model of a larger similar model, that I am using for my design process, which requires 2 motors to operate the drill: One to rotate the drill head, and one to push the drill head into the material.</p>
<p><em>Info</em></p>
<p>The larger drill design has a diameter 2x greater than the new design (25cm and 12.5cm respectively) and they both have simple right cone profiles. <strong>To properly scale and select the appropriate motor for this updated drill, I need to utilize the original design to make new estimates.</strong></p>
<p><em>Data:</em></p>
<p>The large drill has a <strong>torque requirement 1000-1200Nm.</strong>
Since the diameter of my drill is ½ of the original drill, I believe <strong>my drill will have a torque requirement of 500 – 600Nm based on T = Fr</strong>.</p>
<p><strong>The large drill requires 750 – 1000N to push into the material.</strong>
For this force estimate I thought I should use P = F/A (where A is the surface are of the cone (without the base)) as both drills should apply the same pressure on the material.</p>
<p><strong>I get that the force requirement for my drill is 187.5 – 250N.</strong></p>
<p><strong>Can anyone please confirm if this sounds correct?</strong></p>
<p>I just need to know if I am on the right track here.</p>
<p>Thank you for reading</p>
| |mechanical-engineering|motors|design|torque|drilling| | <p>Very roughly, ignoring the nails or any twist in the angle of the blade, if your cutting cones are just two or three blades that cut the dirt basically acting as angled knives then the torque required is dispensed over all the length of the angle.</p>
<p>In reality, the attack angle of the blades is twisting and the blades are of a spiral shape which makes it easier to cut.</p>
<p>For any point along the length of the blade, the torque is <span class="math-container">$$\tau= R*dr*\sigma\tau$$</span></p>
<p>and for the entire drill</p>
<p>So <span class="math-container">$$\tau=\Sigma (\frac{RdR*t}{2})=R^2/2 *\tau$$</span></p>
<p>Therefore for half R, you need 1/4 torque.</p>
<p>For the vertical force your assumption in correct.</p>
| 37684 | Help with scaling torque and forces? |
2020-09-16T14:48:30.333 | <p>I am considering how a spaceframe would work as a rigid batten in a sail. Tubes are most often used, so I can't imagine this would be cost effective; it's mostly a thought experiment. I'm not a mathematician in anyway, or even an engineer, just a hobbyist, so if you want to go deep, just know I need a TL;DR at some point. :}</p>
| |structural-engineering| | <p>Sail battens have to be flexible adapt to the angle of the sail and the wind and the configuration the boat captain sets.</p>
<p>Unless one designs an entire boat altogether from scratch.</p>
| 37686 | Are tetrahedron spaceframes strong versus flex? |
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