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2021-10-27T23:54:51.370 | <p>I recently went on a tour to a local water company plant where we were shown a sludge tank which is where the sludge in the water is removed and stored here.</p>
<p>The sludge can be acidic and so can cause damage to the tanks such as cracks and therefore leakages.</p>
<p>I was told that the company only knew about the damages when they actually occurred and not before.</p>
<p>What could be some ways to detect early signs of these tank failures in order to spot them before the leakages actually occur ?</p>
| |mechanical-engineering|materials|chemical-engineering| | <p>Nondestructive testing methods such as Ultrasonic, Magnetic Flux Leakage, and Eddy Current are routinely used to inspect tank interiors that are in operation.</p>
<p>My advice to address the cause would be to coat the tank in a refractory material that is not vulnerable to the damage</p>
| 47924 | Preventing leakage in sludge tank |
2021-10-28T06:03:52.857 | <p>If I have a piece of rubber between two steel plates, and four bolts (at equal preload) compressing the plates, will the total compressive force on the rubber be Preload x 4?<a href="https://i.stack.imgur.com/dw3kV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dw3kV.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|statics|finite-element-method|mechanical| | <p>Let's say we have the two steel plates and rubber in place with the 4 bolts and nuts. Now we tighten the nuts while holding the steel plates by some spacer blocks between them not letting the sandwich assembly compress.</p>
<p>Now we apply torque turning the bolts until they are preloaded by a force Plbs.</p>
<p>If the bolts were free to move they would have shortened by;</p>
<p><span class="math-container">$$\delta=\frac{Pl}{AE}$$</span></p>
<p>L is the length of the bolts and A its area. E is the Young modulus.</p>
<p>But as soon as we remove the spacer blocks the rubber plate will shrink allowing the bolts to relax significantly.</p>
<p>Let's say the stiffness of the rubber plate is 2/3 the sum of the 4 bolts.</p>
<p>Then we the sandwich will shrink by releasing some of the preloaded tension on the bolts until there is equilibrium, meaning the leftover tension in the bolts is balanced by compression force by the rubber.</p>
<p>In our case for each incremental shrinkage in the bolts, it will compress the rubber by <span class="math-container">$1/(2/3)=1.5\delta_s$</span></p>
<p>At equilibrium, the shrinkage of rubber is <span class="math-container">$2/3\delta_s\ (1/2+1/3)$</span>
And the bolts have lost 0.66% of their preload.</p>
<p>This is assuming zero deflection on steel plates.</p>
| 47929 | How to find clamping force from bolt preload? |
2021-10-28T14:47:57.477 | <p>I'm planning a basement 3.6m depth, ~0.5m of which extended above ground surface level, in a deep clay/alluvial terrain (bedrock inaccessible). Torrential rainfall during rainy season saturates the region, causing considerable flooding in low-lying areas. The basement, while not where flooding would occur much above ground level (no more than about 20 cm./8 in.), would have no option for drainage for anything below grade. Power outages are not uncommon, especially during a strong storm, making a sump pump an unreliable option. I expect, therefore, to essentially be building an inverted pool.</p>
<p>To prevent water entrance, 2 or 3 mm. thick steel plates could be welded together to create a waterproof liner for the concrete walls. I had planned to use either 8" blocks, poured full, or to pour the entire wall outside the liner. But as concrete is porous, and could admit some water, I then realized the need to account for hydrostatic pressure potential, and, quite likely, would need to have the steel as exterior cladding to the wall. Sooner or later, ground penetration of the water would be inevitable, and rising water levels would create a lateral load against the basement wall. Assuming the maximum pressures (worst case scenario), what is the minimum wall strength/thickness required to maintain the wall's integrity?</p>
<p><strong>Basic Considerations</strong></p>
<ul>
<li>Basement cladded by steel plating for waterproofing</li>
<li>Alluvial/clay soil, no bedrock</li>
<li>Basement dimensions about 8m x 12m (about 26ft x 40ft) and about 3m/10ft below grade</li>
<li>Heavy monsoon rains--Limited pooling of water may be expected above grade, no more than about 20 cm./8 in.</li>
<li>No viable option for drainage below grade **</li>
</ul>
<p><strong>Further details/considerations</strong></p>
<ul>
<li>Two floors are planned above the basement, with exterior walls of 8" block (very uncommon for this region, and hard to obtain--the common block size being 3").</li>
<li>Would like to consider the potential for seismic activity and accompanying liquefaction.</li>
<li>Steel is expensive, but concrete blocks and cement are not.</li>
<li>Termites are ubiquitous and will ravage most woods in a short time; i.e. wood is not a good option.</li>
</ul>
<p><strong>Question</strong></p>
<ul>
<li>What are the wall requirements to withstand the worst-case hydrostatic pressures?</li>
</ul>
<p>** I took measurements of the grade today and asked a <a href="https://engineering.stackexchange.com/questions/48271/is-it-practical-to-drain-surface-water-to-a-canal-at-about-800-meters-distance">new question HERE</a> regarding the possibility of drainage. Perhaps there's a small chance of it.</p>
| |steel|reinforced-concrete|masonry|hydrostatics| | <p>A retaining wall to support wet soil (saturated, maybe) has to withstand this loading plus any additional earthquake loads which are usually a factor of 1.15 to 1.25 in our city of Los Angeles.</p>
<p><span class="math-container">$P=\frac{H^2}{2}(\gamma'_s + \gamma_w)1.25$</span></p>
<ul>
<li><span class="math-container">$\gamma'_s= \text{design bouyant lateral load of the soil}$</span></li>
<li><span class="math-container">$\gamma_w= water density= 62.4lbs/feet^3$</span></li>
</ul>
<p>The design lateral pressure of soil can be found in the ASCE 7-05 3.2 table 3.2-1 added below.</p>
<p>They act at H/3.</p>
<p>In the absence of a soil report, it is conservative to consider the full density of the wet soil and add the water hydrostatic pressure.</p>
<p>You have to consider the uplifting force of the water. I have seen the basements that were pushed out of the earth in the rainy season,</p>
<p>I strongly recommend you seek the advice of a local engineer.</p>
<p>Here is the ASCE 7-05 3.2 table 3.2-1</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/RfPry.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RfPry.png" alt="table" /></a></p>
| 47935 | How thick must a concrete or block wall be to withstand hydrostatic pressures up to three meters' height? |
2021-10-28T16:13:01.857 | <p>I think the component name is a rheostat- is that what allows you to control the voltage output please? I want to control it in steps close enough to 0.1 ish volts? So I would like to attach a knob switch (sorry not sure what it’s called, the one where you can rotate it to turn thing up and down) to precisely vary that voltage?</p>
<p>Rheostat or something simpler? I haven’t thought about resistance and stuff, if I had motors involved can I get some advice please on how it would work? So what we thought was using motors on display, and have it so others can play and see how the volts speeds them up or something interactive like that?</p>
<p>4v source is 2 batteries</p>
<p>Sorry I don’t think I explained this well before-
I just want to have a motor or two, (or maybe an Light) might be simpler where kids can turn the ‘knob’ clockwise and anti-clockwise and see the light get brighter and dimmer?</p>
<p>(It’s for my sons science fair)</p>
| |electrical-engineering|electrical|circuit-design| | <p>A potentiometer can be used as a voltage divider to control voltage. It however is a device for precise control of resistance rather than voltage. Voltage is equal to resistance multiplied by current and that means when the current drawn by the motor varies, so too will the voltage.</p>
<p>For a cheap project, consider getting a cheap multimeter. it can be used to measure the voltage. That way even if the motor draws a different amount of current, you just turn the potentiometer until you get to the desired voltage. This is a cheap method of using a feedback loop where a human is in the loop.</p>
<p>You still need to figure out how to measure speed. If the motor is slow enough, you could count how many turns it spins in a known period of time. You could get a sensor. A more expensive multimeter may even be able to give you a frequency measurement for a sensor. You could have the motor wind up a spool of thread and measure how much thread it winds in a known amount of time. (This also gives the interesting option of hanging different weights such as a basket of coins from the thread to change load on the motor)</p>
| 47938 | Precisely control voltage output |
2021-10-28T19:57:37.223 | <p>Consider a spring mass dashpot system subjected to a harmonic excitation force of</p>
<p><a href="https://i.stack.imgur.com/XwidN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XwidN.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$F(t)=F_0 sin(\omega t)$$</span></p>
<p>The response of a system assuming sufficient time has been passed and the complementary part of the solution has become zero will be</p>
<p><span class="math-container">$$x(t)=Xsin(\omega t-\phi)$$</span></p>
<p>where <span class="math-container">$$tan\phi=\frac{2\zeta r}{1-r^2}$$</span> <span class="math-container">$$r=\frac{\omega}{\omega_n}$$</span></p>
<p>when <span class="math-container">$r=1, \phi=90^0$</span>
which means at <span class="math-container">$t=0, x(0)=-X$</span></p>
<p>I'm not able to make sense of these results physically. The results say that block at t=0 will be at -X when the excitation force is 0. But I started (t=0) the movement of block at x=0, at which time the force was zero. Results are contradicting my actual conditions. Results say at t=0 x=-X but I started moving the block when x=0</p>
| |mechanical-engineering|vibration| | <p>I think the confusion is due to the fact that you have not understood the significance of "The response of a system assuming sufficient time has been passed and the complementary part of the solution has become zero".</p>
<p>The response of a simple harmonic oscillator (like the one in the image), can be distinguished (one of the many different categorisations), to <em>transient</em> and <em>steady state</em>.</p>
<p>The transient state is the part which is associated with a <span class="math-container">$e^{-a\, t}$</span> term (where a is a positive number), and eventually "becomes zero after a long time has passed".</p>
<p>The steady state is the portion or the response that follows the excitation.</p>
<p>The following graph, shows a force excitation of 20N @ 24.9 rad/s with red, and the partial (steady state), and homogeneous (transient) response. As you can see the homogeneous dies out (although initially is of equal but <em>opposite</em> significance), and eventually dies out (after about 0.7 of a s). From that point on the total response coincides with the partial solution.</p>
<p><a href="https://i.stack.imgur.com/WRxYb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WRxYb.png" alt="enter image description here" /></a></p>
| 47945 | Interpretation of results in a spring mass dashpot system subjected to harmonic excitation force |
2021-10-28T22:25:49.663 | <p>Suppose that I have a thermally insulated tank with hot air at a temperature of 400 C, and the tank is stored in a room with an ambient temperature of 25 C. I also have another identical insulated tank filled with hot air at 50 C that I store in another room with the same ambient temperature. Which tank would have a faster rate of heat transfer? Also, what if a third tank was filled with cold air, perhaps cooled down to 0 C? Would the heat transfer rate for the cold air be same as that for the 50 C since the absolute temperature difference in relation to the ambient temperature is the same?</p>
| |thermodynamics|heat-transfer| | <p>A concept that can be useful in clearing up relevant concepts to the question is <a href="https://en.wikipedia.org/wiki/Thermal_resistance" rel="nofollow noreferrer">thermal resistance</a>. The good thing is that thermal resistance does not distinguish between conduction and convection and thus, treats a wall in a much simpler way.</p>
<p>The general equation is
<span class="math-container">$$\Delta T = \dot{Q}\times R$$</span>
where:</p>
<ul>
<li><span class="math-container">$\Delta T$</span>: is the total temperature drop (or difference)</li>
<li><span class="math-container">$\dot{Q}$</span> : the heat flow rate</li>
<li><span class="math-container">$R$</span>: the absolute thermal resistance.</li>
</ul>
<p>If you are familiar with Ohm's law (<span class="math-container">$V=IR$</span>) there is a very straightforward analogy between:</p>
<ul>
<li><span class="math-container">$\Delta T$</span> and <span class="math-container">$V$</span>,</li>
<li><span class="math-container">$\dot{Q}$</span> and <span class="math-container">$I$</span> and finally</li>
<li>the resistances.</li>
</ul>
<p>From the above its obvious that <strong>the temperature difference and heat transfer rate are proportional</strong> for the cases involving conduction and convection.</p>
<hr />
<p>Regarding the last part of the question, if the transfer rate is the same in a room with ambient temperature of 25C between two tanks at 50 and 0 C degrees, I'd like to note the following:</p>
<ol>
<li>the absolute heat transfer rate is the same but obviously the direction the heat "travels" is different.</li>
<li>For small temperature differences, the heat transfer will be approximately the same, however, as temperatures increase, both heat conductivity and convection can have a temperature dependence, which can have an effect on the values. E.g. the following is an image for steel</li>
</ol>
<p><a href="https://i.stack.imgur.com/WPuoz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WPuoz.png" alt="enter image description here" /></a>
<strong>figure: <a href="https://www.researchgate.net/publication/299592587_Numerical_Predictions_for_the_Thermal_History_Microstructure_and_Hardness_Distributions_at_the_HAZ_during_Welding_of_Low_Alloy_Steels/figures?lo=1" rel="nofollow noreferrer">Temperature-dependent thermal conductivity Jose Adilson De Castro</a>)</strong></p>
| 47946 | Does a higher temperature difference result in a faster rate of heat transfer? |
2021-10-28T22:46:17.070 | <p>Considering the applied forces <span class="math-container">$F_y=F_y=0$</span> and <span class="math-container">$F_x=400$</span> lbf, what type of stresses does <span class="math-container">$F_x$</span> produce? I understand that <span class="math-container">$F_x$</span> causes tensile and bending stress but not sure about the transverse shear stress (even though it may be very small).</p>
<p>Based on the definition, the transverse shear stress is always accompanied by bending stress but following the formula <span class="math-container">$\tau = \frac{VQ}{Ib}$</span>, we realize that <span class="math-container">$F_x$</span> doesn't have any shear force (<span class="math-container">$V=0$</span>). So is the transverse shear stress zero?</p>
<p>Thanks!</p>
<p><a href="https://i.stack.imgur.com/C7Iui.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C7Iui.png" alt="enter image description here" /></a></p>
| |stresses|machine-design|shear| | <p>You are right the handle doesn't have any shear.</p>
<p>The only member that has shear is member BC. It has a constant shear in the plane XZ of</p>
<p><span class="math-container">$V=400lbs$</span></p>
| 47948 | Does F_x generates "Transverse Shear Stress" shear stress at the fixed end A? |
2021-10-30T10:19:38.907 | <p>At <a href="https://sdp-si.com/products/motors-gearheads-and-motion-control/frameless-brushless-dc-motors/Index.php" rel="nofollow noreferrer">this random motor shop</a> and elsewhere (in the results of
STMicroelectronics' motor profiler), I find a K<sub>e</sub> constant expressed in Vms/krpm.</p>
<p>If this were Vrms/krpm, I would guess that "rms" would stand for root mean square, and be happy with it, but I am at loss with "ms". Since V/krpm is an acceptable unit for a motor's electrical constant, I don't see how taking a mean square can lead to a valid electrical constant. I don't think interpreting "ms" as meter.second would make sense either.</p>
<p>So what am I looking at here?</p>
| |electrical-engineering|motors| | <p>The consensus seems to be that Vms/krpm is just a spelling mistake, and that the unit meant was Vrms/krpm, i.e. V/krpm for the root mean square voltage.</p>
<p>And for the sake of honesty, I have to mention that when starting ST's motor profiler one more time, I can now see Vrms/kRPM. It is just a little hard to distinguish because of the font and size they have picked:</p>
<p><a href="https://i.stack.imgur.com/3Tug6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Tug6.png" alt="enter image description here" /></a></p>
| 47972 | BLDC motor electrical constant in Vms/krpm? |
2021-10-30T13:26:39.203 | <p>The question is: <strong>Which force causes the reaction moment at a fixed joint ?</strong></p>
<p>Lets consider a very simple example, a beam with one end connected by a fixed support to a wall.
<a href="https://i.stack.imgur.com/q46Fi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q46Fi.png" alt="enter image description here" /></a></p>
<p>I understand that at A, there will be a vertical reaction force, a horizontal reaction force and a reaction moment. My understanding is that <strong>a moment about an axis is always caused by a force</strong>. Which force <strong>F</strong> causes the reaction moment?</p>
<p>Focusing on this example, suppose <strong>F</strong> is the horizontal reaction force. From the force equilibrium condition along the x direction, we get that this force has zero magnitude and thus cannot generate the reaction moment.</p>
<p>Suppose <strong>F</strong> is the vertical reaction force. Apparently, when we cut the beam at some C and apply the moment equilibrium condition at C, we are supposed to <strong>include both the reaction moment and the moment generated by the vertical reaction force</strong>, which indicates that the moment generated by the vertical reaction force and the reaction moment at A are two different moments caused by different forces.</p>
<p>So, to conclude, if the moment at A is caused neither by the vertical nor the horizontal reaction force, by which force is it caused?</p>
| |moments|bending| | <p>IMHO you are confusing the "moment of a force about a point A", with the the "reaction moment" that develops at the support of a structure, and more specifically how they are calculated. Although they are related there are not the same thing.</p>
<p>The first one is a very fundamental concept in physics. Moment of a force is the product of force times the distance of the point from the force carrier (colinear line to the the force).</p>
<p>The reaction moment that develops is developed by the structure. Although in statics we assume rigid structures (i.e. small or nonexistent displacements), in real life all structures deform. The deformation of the structures <strong>may</strong> have a significant effect on the structure.</p>
<p>What I am trying to arrive at is that "the carrier structure as well as the load determines the support reaction".</p>
<hr />
<p>Additionally, regarding the part of your question</p>
<blockquote>
<p>Apparently, when we cut the beam at some C and apply the moment equilibrium condition at C, we are supposed to include both the reaction moment and the moment generated by the vertical reaction force, which indicates that the moment generated by the vertical reaction force and the reaction moment at A are two different moments caused by different forces.</p>
</blockquote>
<p>This sentence is wrong on many levels, so it would be best if you actually did the math so that we could tell you where you got it wrong. I'll try to explain it with a small example: Assume C is right at the middle of the beam in your post.</p>
<p>In that case the CB section has a length of 1m, and the total transverse force on it is 10N. The reaction that develops on C is based on those values, so on C there is a Transverse force of 10 N and a 5 Nm bending.</p>
<p>On the AC section, again the length is 1m, and the total transverse force on AC section is 10N, however on C, there are also present the reactions from the section CB. Those forces are a Transverse force of 10 N and a 5 Nm bending (with opposite direction that above). So although there is a cut at C, the reaction forces that develop at C are such that the entire structure is in static equilibrium.</p>
| 47973 | Which force generates the reaction moment in a fixed support |
2021-11-01T00:00:28.643 | <p>If larger ships are less affected by ocean waves, and tend to ride smoothly through them, would that same principle influence the survivability of a larger building when weathering the seismic waves travelling through the ground?</p>
<p>Note that for the purposes of this question, the building would be assumed to have been strengthened in similar fashion to the lengthy hull of a large ship. The question is really about whether the building could ride the seismicity more smoothly given a larger contact area with the ground waves.</p>
| |building-design|building-physics|seismic| | <p>It all depends on the period of the ground shaking vs building. Long period earthquake waves shake tall buildings more as their periods are longer. The period of the ground shaking vs period of the building should not match.</p>
| 47997 | Would a larger building more safely or smoothly ride the seismic waves of an earthquake? |
2021-11-01T10:06:15.513 | <p>I am trying to make a pulley reduction system for my project. I need it to be as quiet as possible, so I am considering pulleys.</p>
<p>My project also has limited space, so I am trying to have as small a pulley on one end as possible. For example, <a href="https://www.technobotsonline.com/8mm-miniature-plastic-model-pulley.html" rel="nofollow noreferrer">this</a> is a pulley with 4mm radius. I think I've even found with 3mm radius.</p>
<p>For such small radii, the difference between the innermost and the outermost part of a pulley is significant. Let us assume the following pulley:</p>
<p><a href="https://i.stack.imgur.com/xaqYz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xaqYz.png" alt="enter image description here" /></a></p>
<p>In that case, what radius should I assume to calculate transmission speed? How does the belt actually work?</p>
| |pulleys|transmission| | <p>I would estimate The radius to be a bit larger than 4 mm. How much, depends on the friction and stiffness of your belt. Similar to continuously variable transmission, CVT.</p>
<p>If the belt is stiff enough and is not going to fill the grove fast, it will rotate around a circle a bit bigger than 4mm. If it is flexible and has less friction it will hug the pully and 4mm is the radius.</p>
<p>If you tighten the belt by a 3rd free turning pulley the radius will get very close to 4mm.</p>
<p><a href="https://i.stack.imgur.com/DA6jj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DA6jj.png" alt="CVT" /></a></p>
| 48008 | Small belt transmission: Radius difference? |
2021-11-01T11:32:09.550 | <p>Consider a prismatic bar loaded in tension as shown. I consider a point Q in the bar away from the ends. This point Q can be considered as an infinitesimally small element taken within the body.</p>
<p><a href="https://i.stack.imgur.com/8gTOb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8gTOb.jpg" alt="enter image description here" /></a></p>
<p>We can consider many different elements, representing point Q, which correspond to different rotations about the axis perpendicular to the plane of the screen. On each of these elements the stresses will be different and since we can obtain infinite number of elements by rotating at different angles we can have infinite number of stresses. Since all the elements correspond to point Q does that mean there are infinite number of stresses acting at point Q?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|stresses| | <p>Yes there is an infinite number of normal and shear stress combinations, one for each different rotation angle.</p>
<p>However, there is a relationship between the magnitudes of the normal and shear stress at all different orientations.</p>
| 48009 | Do we have infinite number of stresses at a point in a prismatic bar loaded in tenison? |
2021-11-02T13:20:01.940 | <p>Let’s say I want two motors in parallel in a circuit (so they get the same Voltage), and wanted one motor to spin anti-clockwise and the other clockwise- is it possible?</p>
<p>I thought by flipping one of the motors polarity, it would?</p>
<p>I want both motors to project a ball the same way, like a tennis ball launcher</p>
| |motors|circuit-design| | <p>I think you could reverse both polarity of Motor OR battery on that circuit Herbert- but reversing polarity of battery though it will reverse both motors- not just one!</p>
<p>That’s saying it’s two motors in parallel with power source</p>
| 48023 | How to make motor spin the other way |
2021-11-02T16:44:03.150 | <p>I'm interested in personal cooling devices and would like to know if there's a way to make a rough estimate of rate at which you would need to extract heat from the body of an average male walking in weather of say 34 deg C and relative humidity of 90% in order to reduce body temperature to what it would be if walking in the same clothing (say slacks and a T-shirt) at say 24 deg C and RH of 75%. I don't know if there's any kind of model out there or if you have to work it all out from scratch.</p>
| |heat-transfer| | <p>The implicit goal appears to be to keep the temperature of the object (the human body) constant independent of changes in the external conditions (clothing, air temperature, wind speed, humidity). The simplest model is a lumped approach.</p>
<p>Let's assume the body is always warmer than the surroundings. The heat extracted by the surroundings is</p>
<p><span class="math-container">$$\dot{q}_{x,s} = U A \Delta T$$</span></p>
<p>where <span class="math-container">$\dot{q}_{x,s}$</span> is the heat removed by the surroundings (W), <span class="math-container">$U$</span> is the effective heat transfer coefficient of the external surroundings (W/m<span class="math-container">$^2$</span> K), <span class="math-container">$A$</span> is the area of the object (m<span class="math-container">$^2$</span>), and <span class="math-container">$\Delta T$</span> is the temperature difference between the object and surroundings (K). The person produces <span class="math-container">$\dot{q}_p$</span> heat depending on the effort. Finally, your cooling device extracts <span class="math-container">$\dot{q}_D$</span> heat. The vector sum is zero.</p>
<p><span class="math-container">$$0 = - U A\ (T_o - T_s) + \dot{q}_p - \dot{q}_D $$</span></p>
<p>Your problem reduces to one of determining <span class="math-container">$\dot{q}_p$</span> and <span class="math-container">$U$</span> for different conditions. You could start with a constant "resting" heat production value <span class="math-container">$\dot{q}_{p,o}$</span>. Start also with estimated values for the overall convection heat transfer coefficient for a "naked" object sitting in different air conditions (humidity, wind speed). This will give you a ballpark on what you need for <span class="math-container">$\dot{q}_D$</span> for a naked person resting in different air conditions. Add clothes by increasing <span class="math-container">$U$</span>. Add walking to running effort by increasing <span class="math-container">$\dot{q}_p$</span>.</p>
<p>This ignores radiation. Radiation can be added back as an additional term in the energy balance as desired.</p>
| 48029 | Calculating the difference in heat loss from human body in different environments |
2021-11-03T13:02:07.683 | <p>I was watching a video <a href="https://www.youtube.com/watch?v=uvT3XvDSSQw" rel="nofollow noreferrer">https://www.youtube.com/watch?v=uvT3XvDSSQw</a> about piezoelectric materials.</p>
<p>They mentioned that "a current can flow when a pressure is applied to the material".</p>
<p>I have a problem with this expression. I'm not sure if my confusion is in its place.</p>
<p>I know that when a stress is applied on a piezoelectric material, it will generate an electric field due to the dislocation of the electric charges/ creation of electric dipoles.</p>
<p>But that doesn't mean that a current should flow from the material, because if a current flows, that means the material has lost electrons, therefore has lost some mass.</p>
<p>And in case it was true that the electrons were leaving the material instead of just emitting their electric field while still being connected with the materials molecules, then I think what they mentioned about connecting the material to a circuit and measuring the current flow will make sense.</p>
| |materials|measurements|electrical|current| | <p>the output signal of a stressed piezoelectric crystal is primarily a measurable <em>voltage</em> accompanied by an almost too-small-to-measure <em>current flow</em>. This is a general characteristic of <em>high-impedance</em> devices, of which the crystal is a good example.</p>
| 48039 | Do piezoelectric materials produce a current flow or just an electric field? |
2021-11-03T17:36:23.180 | <p>I'm an advanced industrial engineering student and I'm currently looking for some "final project" material. I have this idea of a product where you can plug in a type K thermocouple at the end of a meter long wire to your smartphone via the minijack port, as to mesure mainly kitchen-like temperatures (you know, how hot is the oven, is the meat ready, that kind of stuff) but it may also be used for other things. The concept would work initially for cooking and baking. It would need the output part of the minijack to get a steady and known electrical potential, and then to read the potential at the input part, where the difference found is proportional to the temperature measured. Can this be done? Can one get into writing and reading certain values of potential at the minijack port?</p>
<p>PS: I've got some basic knowledge of programming and null experience whatsoever with smartphone programming. If you can help me, please add links, references and material where I should look into and introduce myself further to the matter.</p>
| |electrical-engineering|thermocouple| | <p>Microphone plugs expect AC and the circuitry is AC-coupled, that prevents any possibility of it ever working. Thermocouples output a DC voltage, and very small, voltage at that which needs to be carefully amplified. Even if you had just a DC-coupled ADC, just connecting a thermocouple to it is unlikely to work due to the very low level signal. There needs to be signal conditioning.</p>
<p>Thermocouples also only measured the temperature difference between its ends. So to measure absolute temperature you need a temperature sensor on the "receiving" end or a way to set the temperature on that end to something known (like an ice-bath).</p>
| 48043 | Could the analogical input and output of the minijack on smartphones be used to mesure temperature via a type K thermocouple? |
2021-11-03T21:57:54.093 | <p>This is part of my on going saga of drafting question with respect to the <a href="http://www.digipeer.de/index.php?static=52" rel="nofollow noreferrer">V2 drawings</a> I am using for practice.</p>
<p>The following <a href="http://www.digipeer.de/index.php?media=DMA_FA_014_00072&size=2" rel="nofollow noreferrer">drawing</a> is titled "Container Overview". It has left me with 3 questions</p>
<p><a href="https://i.stack.imgur.com/wFhTh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wFhTh.jpg" alt="Container Overview" /></a></p>
<ol>
<li>Each part has dimensions pointing at the line which to me seem to indicate wall/material thickness. Either 3 or 2 sets of numbers are given. I was originally thinking tolerance ie 5/8/11 is 8 +/- 3. However that does not explain the 2 number system. ie 25/18 in second part. I did not take this as limits as I am guessing they would not change tolerancing style in the same part. <strong>Any ideas what these multiple numbers for the dimension are?</strong></li>
<li>Continuing with the assumption that the multiple numbers represent wall thickness. When given the various radii for curves, <strong>would those dimensions be measured to the inside face, outside face, or mid line usually?</strong> I was taking it to the face that the arrow is on. Same with the interior diameters in parts 2 and 4.</li>
<li>In Part 2 and 4, <strong>are they missing a dimension somewhere to position the top of the straight angled lines at the top of the parts?</strong></li>
</ol>
| |technical-drawing|drafting|drawings| | <p>This is still a bit of a guess, but the combustion chamber was made of two layers of steel with a gap between for the alcohol-water fuel to cool the chamber walls. And in the event of a burn through (a daily occurrence during the peak testing periods), I suppose it would prevent catastrophic explosions. I found that the inner wall dimension was 5 mm. The space and outer wall wasn't specified. I think the 5/8/11 is the thickness of the inner layer, the gap, and the outer layer - so 5 mm inner wall, 3 mm coolant passage, 3 mm outer skin.</p>
<p>see page 409, final paragraph. <a href="http://www.raketenspezialisten.de/pdf/jbisdruckvorlage.pdf" rel="nofollow noreferrer">http://www.raketenspezialisten.de/pdf/jbisdruckvorlage.pdf</a></p>
<p>Internet is crap at the moment and I can't get an excerpt image.</p>
| 48048 | V2 drawing interpretation |
2021-11-04T04:12:19.130 | <p>L = 10 in,
<span class="math-container">$T_{max}$</span> = 153 in-lb,
<span class="math-container">$T_{min}$</span> = 13 in-lb,
Carbon Steel 1020 CR,
<span class="math-container">$f_y$</span> = 57 ksi,
<span class="math-container">$f_{ut}$</span> = 68 ksi</p>
<p>I am not sure how to solve if I do not have the weight of the shaft. I really appreciate any help.</p>
| |mechanical-engineering|machine-design| | <p>You can either go</p>
<ol>
<li><strong>full analytical by setting a symbol d for the diameter.</strong></li>
</ol>
<p>Then you'd calculate weight as <span class="math-container">$\rho \cdot \frac{\pi\cdot d^2L}{4} \cdot g$</span> and the forces, then calculate transverse forces, bending moment and finally the stresses and the safety factor(again d will be in the formulas).
At the end you should arrive at an equation (or several representing the different points of check on the shaft) of the safety factor that is a function of the diameter. You can then solve for d.</p>
<p>This would be the most complicated in term of length of equations, and the most convoluted because of the nonlinear terms (square roots etc).</p>
<ol start="2">
<li><strong>Do an iterative approach:</strong></li>
</ol>
<p>Select a diameter e.g. 10[mm], then calculate the weight <span class="math-container">$\rho \cdot \frac{\pi\cdot (10\; \text{mm})^2L}{4} \cdot g$</span>, then calculate transverse forces, bending moment and finally the stresses and the safety factor. If the safety factor is not betwen 3 or 4, then guess another diameter and try again (e.g. if the safety factor is 2, then select larger diameter e.g. 16 mm).</p>
<p>if you automated this in an excel sheet, its very easy to do the calculation and arrive at a solution.</p>
<ol start="3">
<li><strong>Do the iterative in a smart way</strong></li>
</ol>
<p>What you can do is do the first iteration, by assuming the weight of the shaft is negligible. The equations are much simpler that way. Solve for a high safety factor (say 4), and then you arrive at a diameter. Use that diameter as a starting point. Chances are that unless there is no external load on the shaft then you'll be within the 3 to 4 margin for the safety factor.</p>
| 48052 | Machine Design, finding the diameter of a shaft |
2021-11-04T19:58:40.637 | <p>I have a Nema23 that has a 1.8 degree stepping resolution. The motor has a GT2 pulley attached on the rotating element. This GT2 pulley is coupled with another pulley, using a belt, which is directly around the shaft of a lead screw.</p>
<p>I'd like to increase my stepping resolution, I already use micro-stepping, so I was thinking of using a 60 tooth pulley on my lead screw and a 20 tooth pulley on the stepper. Such that a step from my motor would result in less rotation of the lead screw. But what I am curious about is how will this affect the torque experienced by the motor? Will I be able to lift the same load, or maybe even more? Are there any drawbacks to such a modification?</p>
| |torque|stepper-motor| | <p>Yes. you're right. Your stepper motor will produce 3 times more torque, 3 times more resolution.</p>
<p>And 3 times slower angular speed on the lead screw.</p>
| 48061 | Will this pulley modification result in more torque on my stepper motor |
2021-11-05T01:33:21.133 | <p>I created the following planetary gear in FreeCAD. <a href="https://i.stack.imgur.com/dfBsR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dfBsR.png" alt="planetary gear" /></a></p>
<p>I calculated number of teeth using formula R = 2*P+S. In my case
R = 75
P = 5 (there are 3 of them)
S = 45.</p>
<p>Also all gear have the same pitch of 2.5 mm.
However picture has huge gap. I can not figure out what could be wrong. Here is link to FreeCAD file <a href="https://drive.google.com/file/d/1WjNELprq0dIOSDJ_27AkpRTnRfdNPXfM/view?usp=sharing" rel="nofollow noreferrer">link</a></p>
| |gears| | <p>The following image explains where the formula <span class="math-container">$R=2P+S$</span> comes from.</p>
<p><a href="https://i.stack.imgur.com/BrOU9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BrOU9.png" alt="enter image description here" /></a></p>
<p><strong>Figure: Planetary gear with 4 planets (<a href="https://woodgears.ca/gear/planetary.html" rel="nofollow noreferrer">https://woodgears.ca/gear/planetary.html</a>)</strong></p>
<p>Although it has 4 planets the spacing will be the same with 2 or more planets. (Three is used for optimal balance and redundancy).</p>
<p>Essentially, the diameter of the Ring <span class="math-container">$d_R$</span> should be equal to the Diameter of the sun <span class="math-container">$d_S$</span> plus 2 diameters of the planets <span class="math-container">$d_p$</span>. I.e. :</p>
<p><span class="math-container">$$d_R = 2d_P+ d_S \tag{eq:1}$$</span></p>
<p>However for gears, the diameter of a gear (<span class="math-container">$d$</span>) and the number of teeth (<span class="math-container">$z$</span>) are proportional.</p>
<ul>
<li>in the metric system usually the module m is used, <span class="math-container">$d= m\cdot z$</span></li>
<li>in the US system usually the Diametral Pitch (P) is used, <span class="math-container">$d= \frac{z}{P}$</span></li>
</ul>
<p>In both cases, the equation 1 becomes:</p>
<p><span class="math-container">$$z_R = 2z_P + z_S$$</span></p>
<p>So for <span class="math-container">$z_R=75$</span> and <span class="math-container">$z_S=5$</span>, the correct number for the teeth of sun is <span class="math-container">$z_S = 75-2*5=65$</span></p>
| 48065 | planetary gear - number of teeth |
2021-11-05T09:18:36.843 | <p>Assume I have a slender cantilever beam, fixed at one end and force applied on the other. Now, during the deformation, we know that the beam's stiffness is going to change and the force-displacement response is not going to be linear. In this case, Large Strain theory is works.</p>
<p>On the other hand, if I have a very short and thick cantilever beam fixed at one end and force applied on the other, so during the deformation, it can be assumed that the beam's stiffness is somewhat constant and so is displacement reponse to force (Ofcourse, if I don't increase the force to astronomical levels). Here, Small Strain Theory works.</p>
<p>My question is, can I use the generalized Hooke's Law (which linearly relates the stresses to strains, in order to obtain stresses from strains) for both of these cases or not? Is Hooke's Law valid for both of these situations, and if not, then for which situation it is valid for?</p>
<p><em><strong>P.S: Assume the body doesn't encounter any plasticity at all</strong></em></p>
| |mechanical-engineering|structures|solid-mechanics| | <p>If there is a non-linear material or even a linear material but under nonlinear stress-strain behavior we can not apply hooks law.</p>
<p>A simple example is your cantilever beam stressed beyond hooks law into elastic-plastic range.</p>
<p>where may be part of the section near the neutral axis is still in the elastic range but the top and bottom are in the plastic range.</p>
<p>Of course, we can not use the hooks law here.</p>
| 48071 | Should we use Hooke's Law (that linearly relates stresses to strains) if the stiffness of body is changing during deformation? |
2021-11-07T00:13:32.183 | <p>While I am continuing my practice with the German drawings of the V2 rocket, I have come across the PG designation a couple of times on different drawings. I originally thought it was for diameter, but its not lining up with some of the other dimensions in the drawing.</p>
<p>In the following drawing PG 21 appears to be dimensioning the the outside of the threads. however the hole through the bushing is is dimensioned at 21 as well. There is no physical way they can both be 21.</p>
<p><a href="https://i.stack.imgur.com/WN6Nk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WN6Nk.png" alt="enter image description here" /></a></p>
<p>In the example below the PG 21 notation is used for both male and female threads and they do appear to line up. However there seems to be more than a 1 mm difference between the interior 20 mm hole and the threads.</p>
<p><a href="https://i.stack.imgur.com/QOIZE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QOIZE.jpg" alt="enter image description here" /></a></p>
<p><strong>What does the PG notation mean?</strong></p>
| |technical-drawing|terminology|drafting|drawings| | <p>Apparently my google fu failed me last night. This mornings searches however were more fruitful.</p>
<p>For future seekers what I found is that PG is an <a href="https://www.gewinde-normen.de/en/pg-thread.html" rel="nofollow noreferrer">obsolete standard and you can read more about it here.</a></p>
<p>Further searching lead me to a website where I found an update for Inventors Thread.xls spreadsheet where a user has added the information to the built in data base. <a href="https://forums.autodesk.com/t5/inventor-forum/pg-threads/td-p/565938" rel="nofollow noreferrer">Scroll to message 13</a>.</p>
<p>The default Thread.xls file is located here:</p>
<pre><code>C:\Users\Public\Documents\Autodesk\Inventor LT 2021\Design Data\XLS\en-US
</code></pre>
<p>you will need to change Inventor LT 2021 to the version you are using. I renamed the original file before adding the new one in case I ever wanted to revert back.</p>
| 48112 | What does the Pg annotation mean |
2021-11-07T18:14:42.207 | <p>Suppose a cantilever beam (of rectangular cross-section) subjected to vertical transverse shear load at free end. If I consider any arbitrary cross-section then both bending and shear stress will be there, but now if replace this beam with another beam of same area but whose width is tending to zero and depth is very large then maximum bending stress will now be tending to zero because the section modulus will be very high and also the maximum shear stress will be same as in the previous case because <strong>Maximum shear stress = 1.5 * average shear stress</strong> and since area is not changing so maximum shear stress will not change.
So why we do not use such beam practically ?</p>
| |mechanical-engineering|structural-engineering|stresses|beam|solid-mechanics| | <p>Not all the beams are prismatic.</p>
<p>In places where changing the section geometry is meaningful and makes sense we certainly do that.</p>
<p>All the appropriate steel (or any material) structures take advantage of using the most effective profile as soon as they aren't typical beam and column grid work, like churches, auditoriums.</p>
<ul>
<li>Bridges use tapered beams, trapezoid outriggers, and even cables if it makes sense.</li>
<li>Machinery and tools use the shapes that work best even if it seems a complex geometry, and unwieldy.</li>
</ul>
| 48123 | Beam with width tending to zero |
2021-11-08T04:49:39.147 | <p>I understand the compressor exit pressure is the highest pressure in all of the turbomachinery, and the combustor is a constant pressure process, and since the combustor P is lower than compressor exit P, the working fluid is directed down towards the turbine and nozzle, but What I dont fully understand is this:</p>
<p>how/why does pressure drop from compressor exit to combustor ?? Especially since the combustor is releasing so much extra energy something has to compensate for that and then some to achieve the lower pressure.</p>
<p>It seems to me that it must be the extra volume of the combustor causing this pressure drop. This is where I get fuzzy because this seems to be contradictory to the basic function of a diffuser, where volume expansion causes increase in pressure not decrease.</p>
<p>Any clarification would be appreciated.</p>
| |thermodynamics|turbines|compressors|turbomachinery| | <p>The combustion process is taken as ideally constant pressure. Real life intrudes and losses occur. Mattingly has a simple analysis (reference below) that indicates the inlet Mach number is very important. He models the flow as having a drag on the walls. The compressor design team probably does not care how fast the flow is, but it needs to be compatible with the combustion section.</p>
<p>Mattingly, Jack W. <em>Element of Gas Turbine Propulsion</em> 1996 McGraw-Hill, p. 822ff. (Probably in 2nd edition, other books he coauthored)</p>
| 48131 | Confused about how combustor pressure is lower than compressor exit pressure in turbomachinery? |
2021-11-08T09:46:31.943 | <p>I am working in Inventor LT 2021 which means I do not have access to the sheet metal tool. Which would have been great as the example I am working on is sheet metal. <a href="http://www.digipeer.de/index.php?media=DMA_FA_014_19666&size=2" rel="nofollow noreferrer">The example drawing I am working with is located here</a>. I am assuming the part on the left is the cut from the sheet metal with those interesting -/- markings being the limits of fold line? According to google translate "Zuschnitt" is "Cutting"</p>
<p><a href="https://i.stack.imgur.com/w0pGr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w0pGr.jpg" alt="enter image description here" /></a></p>
<p>I started out by creating the end view and then extruding it the full length of the object.</p>
<p><a href="https://i.stack.imgur.com/F9OWP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F9OWP.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Ca94k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ca94k.png" alt="enter image description here" /></a></p>
<p>I then sketched on the top surface the angle cuts with the thought of extruding and deleting the small triangles. I then sketched the continuation of the angled cut on the face of the short legs.</p>
<p><a href="https://i.stack.imgur.com/neCBL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/neCBL.png" alt="enter image description here" /></a></p>
<p>This is when I realized that the second extrusion would lead to an "ugly" transition to the cut from the first extrusion. The face/plane of the cut should twist as the sheet metal was folded.</p>
<p>I attempted to work around this with loft. I drew 4 sections the area of the material to be removed at the base of the leg, then just before the curve for the fold line, then just after the fold line, then again mid way across the piece.</p>
<p><a href="https://i.stack.imgur.com/N9u3M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N9u3M.jpg" alt="enter image description here" /></a></p>
<p>in my first attempt I simply picked the 4 areas in order and things did not go well. The loft buldged in ways really not desired. I attempted to add tangency to the start and end pieces, but that failed. I then tried doing 3 separate subtractions lofts. I had a 2 out of 3 success rate.<br />
I basically lofted the 2 sections before and after the bend successfully, but the loft through the bend did not work. I even added rails which were the end edges of the original piece as it went around the corner. I do not know how to get the right corner rail lines at the cut corner.</p>
<p><a href="https://i.stack.imgur.com/Z1g3I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z1g3I.png" alt="enter image description here" /></a></p>
<p>The above is an end view of the loft with the red lines being generated at the cut end and the blue lines being the rails at the other end. The difference between the two inside curves results in left over material as seen below.</p>
<p><a href="https://i.stack.imgur.com/EZDlC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EZDlC.png" alt="enter image description here" /></a></p>
<p>The setting in loft were as follows:</p>
<p><a href="https://i.stack.imgur.com/OtPrB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OtPrB.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/2Eegx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Eegx.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/H8g7u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H8g7u.png" alt="enter image description here" /></a></p>
<p><strong>How can I make this piece have a smoother transition within the limitations of inventor light 2021 (ie. no sheet metal tools)</strong></p>
| |cad|technical-drawing|drafting|drawings|autodesk-inventor| | <p>It is possible to bend solid bodies within Inventor LT using the "Bend Part" command.</p>
<p><a href="https://knowledge.autodesk.com/support/inventor/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/Inventor-Help/files/GUID-98FDE447-C0A7-4694-ABD3-BA19A329DDF6-htm.html" rel="nofollow noreferrer">https://knowledge.autodesk.com/support/inventor/learn-explore/caas/CloudHelp/cloudhelp/2019/ENU/Inventor-Help/files/GUID-98FDE447-C0A7-4694-ABD3-BA19A329DDF6-htm.html</a></p>
<p>Using this will allow you to create the flat pattern per the right hand view, and to bend it using sketched definitions for the start of the bend per the -/- lines, which you rightly assume define the transition between flat and curved sections on the final part.</p>
<p>This is not recommended for use in sheet metal applications since it does will not automatically generate a flat-pattern configuration, or bend lines etc. to be automatically imported in a 2D drawing. The warning on the help page is there primarily to make it clear that "Bend Part" is not the same as "Bend".</p>
<p>"Bend Part" is in some ways more powerful than "Bend", in that it is able to stretch/modify more complex bodies. For example, any attempt to "Bend" the part in your drawing in a direction perpendicular to the long folds would fail to calculate using "Bend". With power comes responsibility, though - it's also not error checking for if something is possible in sheet metal. In real life with a sheet metal part, this would be difficult to achieve due to the stiffness from the existing folds having to 'stretch' - you would need to cut a relief notch in them first, which if done in CAD, would allow the "Bend" tool to work again. So, for sheet metal, using "Bend" can help dissuade you from creating parts that cannot be manufactured.</p>
| 48135 | Loft not quite making the cut |
2021-11-08T11:04:22.857 | <p>It is already known that the I-beam is very bending resistant (causes less deflection for the same transverse load and also results in lesser stresses relatively). But for an eccentric transverse load (i.e. the line of action of load does not pass through the axis of the beam), the I-beam can also be subjected to torsion. And it is believed that I-beam are not very strong in torsion.</p>
<p>Can anyone point out the reason that why I-beams are not very resistant and strong in torsion, but when subjected to a transverse load, they can resist the transverse shear stress? What is the difference between the shear stresses coming from torsion and coming from tranverse loading?</p>
| |beam|shear|bending| | <p>IMHO the most important issue, is that the IPE cross-section is considered an open section with respect to torsion. This reduces the resistance to torsion significantly.</p>
<p><a href="https://i.stack.imgur.com/adiiI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/adiiI.png" alt="enter image description here" /></a></p>
<p><strong>Figure : Shear stress due to pure torsion of an I beam (source <a href="https://www.aisc.org/globalassets/continuing-education/quiz-handouts/designing-members-for-torsion.pdf" rel="nofollow noreferrer">AISC seminars</a>)</strong></p>
<p>In essence the effect can be observed easier if you take a straw and cut it open along its length.</p>
<p>Then take another straw and compare the behaviour between the original straw (closed section) and the "gutted" straw (open section)</p>
<p><a href="https://i.stack.imgur.com/JRvNi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JRvNi.png" alt="enter image description here" /></a></p>
<p><strong>Figure: Shear stress distribution of Close vs open thin wall section under shear stress (source <a href="https://holooly.com/solutions/comparison-of-open-and-closed-thin-walled-sections/" rel="nofollow noreferrer">Holooly</a>)</strong></p>
<hr />
<p>Another reason, is that the torsional response is governed by the polar moment of area (<span class="math-container">$J_p$</span>) instead of the second moment of area (<span class="math-container">$I_{xx} ,I_{yy}$</span>).</p>
<p>And while these are related , i.e. <span class="math-container">$J_p =I_{xx} + I_{yy}$</span>, the torsional response is affected for the I beam.</p>
<p>E.g. In the following table:</p>
<p><a href="https://i.stack.imgur.com/m6QSO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6QSO.jpg" alt="enter image description here" /></a></p>
<p>for IPE80 the</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;"></th>
<th style="text-align: center;"><span class="math-container">$I_{xx}$</span></th>
<th style="text-align: center;"><span class="math-container">$I_{yy}$</span></th>
<th style="text-align: center;"><span class="math-container">$J_{p}|$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">IPE 80</td>
<td style="text-align: center;">80</td>
<td style="text-align: center;">8.49</td>
<td style="text-align: center;">~84.5</td>
</tr>
<tr>
<td style="text-align: center;">IPE 160</td>
<td style="text-align: center;">869</td>
<td style="text-align: center;">63</td>
<td style="text-align: center;">~935</td>
</tr>
<tr>
<td style="text-align: center;">IPE 240</td>
<td style="text-align: center;">3892</td>
<td style="text-align: center;">283</td>
<td style="text-align: center;">~4175</td>
</tr>
</tbody>
</table>
</div>
<p>You might notice a trend that the <span class="math-container">$I_{xx}$</span> is over 10 times larger than <span class="math-container">$I_{yy}$</span> and that trend intensifies for larger dimension.</p>
| 48136 | Why can't I-beam resist torsion but can resist shear stress due to transverse loading? |
2021-11-08T22:08:27.923 | <p>In <a href="http://www.digipeer.de/index.php?media=DMA_FA_014_20728&size=2" rel="nofollow noreferrer">the following drawing</a> I am looking at the lined that come in on either side of the teeth/tabs that are spaced 9 mm apart and extend 6 mm into the material.</p>
<p><a href="https://i.stack.imgur.com/aObmI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aObmI.png" alt="enter image description here" /></a></p>
<p>In my head these are cut or shear lines between alternating teeth/tabs. Short ones ben up slightly and the long ones bed down slightly. However there is no side view to confirm this. The material is 0.8 mm thick. The title according to google translate is "Locking Plate" in english.</p>
<p>When I leave the lines open like in the drawing I windup with just a solid piece. The extended lines disappear.</p>
<p><a href="https://i.stack.imgur.com/GJU7A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GJU7A.png" alt="enter image description here" /></a></p>
<p><strong>Would it be a reasonable assumption to make a 0.1 slice where those lines go in and then partially bend the teeth/tab alternating up and down?</strong></p>
| |cad|technical-drawing|drafting|drawings| | <p>As commented, that could be the part as cut in this process step only. In a subsequent operation it could have tabs bent or they might be bent when the part is pressed into an assembly - a one-way operation that would make it difficult to remove the part.</p>
| 48153 | Are these cut lines? |
2021-11-09T09:05:54.653 | <p><a href="https://i.stack.imgur.com/4sIkM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4sIkM.png" alt="enter image description here" /></a></p>
<p>I was working earlier on an example problem of a trussed beam. Above you can see a picture illustrating the situation. On top there is a simply supported beam, with a stiffening truss below. The truss consists of 3 members, all connected through pins to each other, the supports, and the middle member pin connected to the beam. The beam has a uniform load on top.</p>
<p>I'm having a hard time calculating the degree of indeterminacy of this system. I'm told that this system is <strong>first degree</strong> indeterminate.</p>
<p>If we take the left support as point <span class="math-container">$A$</span> and right support as point <span class="math-container">$B$</span>, we should have supporting reactions <span class="math-container">$A_x$</span>, <span class="math-container">$A_y$</span> and <span class="math-container">$B_y$</span>, as well as 3 normal loads for the truss members (since they only carry normal loads, no moments or shear). Using statics, we can form 3 equations: <span class="math-container">$\Sigma F_x$</span>, <span class="math-container">$\Sigma F_y$</span> and <span class="math-container">$\Sigma M_z$</span>. Because we have 6 unknowns, I would say this is 3rd degree indeterminate.</p>
<p>My reasoning is quite likely wrong, I'm having hard time for some reason understanding static determinacy beyond simple examples. Could somebody walk through the process of calculating the degree of indeterminacy for this structure in detail? Thank you very much!</p>
| |structural-engineering|structural-analysis| | <p>No of equation:</p>
<p>You have 3 equilibrium equations for the whole structure, plus
2 equations (Fx=0, Fy=0) if you consider equilibrium of the truss joint where the members meet.</p>
<p>i.e. Equations = 5</p>
<p>Number of Unknowns:</p>
<p>You have 3 unknown reactions, plus
3 unknowns, that is, one unknown for each truss member.</p>
<p>i.e. Unknowns = 6</p>
<p>Degree of Indeterminacy:</p>
<p>DI = Unknowns - Equations = 1</p>
| 48158 | Very confused as how to calculate the degree of static indeterminacy of trussed beam |
2021-11-09T16:12:54.683 | <p>In the design of beams, the principal stresses and absolute maximum shear stresses are not calculated. Instead, we calculate maximum bending stress using flexure formula and equate it to allowable stress to get dimensions of cross-section and then using these dimensions we calculate maximum shear stress on cross-section.
If maximum shear stress on cross-section comes out to be less than allowable stress, then design is safe. But there is a possibility that maximum shear stress in <strong>the beam</strong> (whose location is at the top and bottom of cross-section; Absolute max shear stress= (max. bending stress)/2) )exceeds allowable shear stress, then the beam will fail.
So, why do we use this design procedure if this gives unsafe design?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam|strength| | <p><strong>Because the allowable stress is always lower than the maximum shear stress possible.</strong></p>
<p>For a rectangular beam, the maximum shear can be calculated as <span class="math-container">$f_{v_{max}} = {2P_{act}}/{A}$</span>. Now let's set the maximum shear stress to its limit, <span class="math-container">$f_{v_{max}} = {2P_{act}}/{A} \leq f_y$</span>, then <span class="math-container">$P_{act} = 0.5f_yA$</span>. However, the "allowable shear stress" is limited to be <span class="math-container">$f_a \leq 0.4f_y$</span>, then for the same cross-section <span class="math-container">$P_a = f_aA = 0.4f_yA$</span>. So, compare <span class="math-container">$P_{act}$</span> and <span class="math-container">$P_a$</span>, we conclude that the design is always safe as <strong><span class="math-container">$P_a$</span> (allowable shear force) <span class="math-container">$\lt P_{act}$</span> (maximum shear force)</strong> with a 25% safety margin, as <span class="math-container">$P_{act} = 1.25P_a$</span>.</p>
<p>The above proof can be held for the bending stress of rectangle beams using the generalized bending stress formula (<span class="math-container">$f_b = P/A \pm M_xy/I_x \pm M_yy/I_y$</span>), however, for other shapes, it pays to check the principal stresses especially when the cross-section is subjected to complex loadings. In the US, AISC has made the check simple by providing necessary parameters about the principal axes.</p>
<p>The proof was based on the ASD (allowable strength design method), but the same holds for the USD (ultimate strength design method) and LSD (limit state design method).</p>
<p>Final note - the maximum shear occurs at the centroid of a cross-section, at which the static moment of area <span class="math-container">$Q$</span> is the maximum (<span class="math-container">$\tau = VQ/Ib$</span>).</p>
<p><strong>ADD - Figure below shows stresses in a beam of rectangular cross-section at various depths:</strong></p>
<p><a href="https://i.stack.imgur.com/c2H8t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c2H8t.png" alt="enter image description here" /></a></p>
<p>(a) show points in the cross-section.</p>
<p>(b) normal and shear stresses acting on horizontal and vertical planes at each point.</p>
<p>(c) principal stresses at each point.</p>
<p>(d) the maximum shear stresses at each point.</p>
<p>Ref: "Mechanics of Materials", Gere & Timoshenko, 2nd.</p>
| 48165 | Design procedure used in beam design |
2021-11-09T17:58:46.157 | <p>How can I find the Torque needed to rotate a wooden slab with no weight applied on either end of the slab if I have a servo motor placed at the middle of the slab?
<a href="https://i.stack.imgur.com/lluyR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lluyR.jpg" alt="diagram that shows a wooden slab connected to a motor" /></a></p>
| |electrical-engineering|motors|electrical| | <p>Torque is equal to angular acceleration times moment of inertia of your board, idealized to<span class="math-container">$ \ 1/12mL^2$</span>.</p>
<p>There are interesting points to check for though.</p>
<p>Whit a small defect in symmetry your board can turn into a propeller and create a lift.</p>
<p>If the face thickness is not insignificant the whirling of the air current can become very complex.</p>
| 48170 | How to find the torque required to rotate wooden slab |
2021-11-10T19:47:47.663 | <p>I went back to to the holes in the plate from <a href="https://engineering.stackexchange.com/questions/48154/what-is-the-best-way-to-handle-the-hole-pattern-in-this-drawing">this sketch</a>. Based on comments I am assuming the 2.5 dia. holes are equally spaced 4.25 mm apart covert the surface of the plate. I also made an assumption that no hole edge could be closer to the edge of the plate than 4.25/2.</p>
<p>I started out with the rectangular array command. Since the second col is half a spacing off and the array command does not allow for this I was faced with two choices at this stage. Either make two arrays (first column and second column) or create the second column hole and include it with the other hole and just make one array. I chose the later.</p>
<p>I then proceeded to create a square array that was large enough to cover the circle. And then decided on a way to center the array on the circle. This was all pretty straight forward in my mind.</p>
<p>Then I had to look up how to break the array so I could delete the excess holes, and all holes that were too close to an edge. That lead me to <a href="https://grabcad.com/questions/break-sketch-pattern-without-deleting-elements" rel="nofollow noreferrer">this Q&A</a>.</p>
<p>So after several mouse clicks later I wound up with a bunch of independent holes in a nice pattern.</p>
<p><a href="https://i.stack.imgur.com/ryGaw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ryGaw.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/0Z1Vy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Z1Vy.png" alt="enter image description here" /></a></p>
<p>Then I realized I should have done the holes last because when I started adding center lines and constraining things, the two large circles wound up shifting around and no longer centered in within the small hole pattern...let alone still having to add some leg notches.</p>
<p>I also did a bit of googling to see if I could array a pattern within a region. Similar to how one does a hatch pattern in AutoCad. My search did not provide me with any positive answers</p>
<p><strong>Is there a better way to approach this pattern? Is there away to achieve similar results while keeping it an array?</strong></p>
| |cad|autodesk-inventor| | <p>I wound up trying two separate arrays. I defined the start point for the second array using a 30, 60, 90 degree triangle, and made the second circle equal to the first which had a dimensionally controlled diameter.</p>
<p><a href="https://i.stack.imgur.com/v4a7B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v4a7B.png" alt="enter image description here" /></a></p>
<p>Of note here because it confused me the first time, I was able to place a dimensional constraint on the horizontal without it being a driven dimension. It turns out that a HORIZONTAL dimension is not over constraining the triangle, but an aligned dimension is. My second run through I think I constrained the horizontal part of the triangle to be horizontal and it immediately came up as a driven dimension.</p>
<p>I then set up the first array, with a count of 30 items and made the vertical spacing a function of the 4.25 diagonal dimension. I then assigned the horizontal part of the array to 17 or 18, and assigned the spacing to 2 * the driven dimension.</p>
<p>I repeated this process for the second array based on the second circle but reduced the vertical and horizontal item counts by 1.</p>
<p>I then drew an X through the resulting array field to find the center of it.</p>
<p>I then zoomed in on the center circle and selected it using the move tool. After selecting it's center as the base point, when I went to zoom with the mouse wheel, I proceeded to lock up my computer for a significant amount of time. After returning from running about 3 ours of errands, my screen was waiting for me to pick a destination point which I did. And I wound up getting a pattern that look like the following over my completely sketched part.</p>
<p><a href="https://i.stack.imgur.com/7d2eZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7d2eZ.png" alt="enter image description here" /></a></p>
<p>It was around about this time I found the edit pattern choice when <kbd>RMB</kbd> :</p>
<p><a href="https://i.stack.imgur.com/SqeTM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SqeTM.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/xASuZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xASuZ.png" alt="enter image description here" /></a></p>
<p>and I then proceeded to use the <kbd>>></kbd> button to expand the options and then I used the suppress button to go around and select a <em>whole whack</em> (I believe that is the appropriate technical term) of circles I did not want included. Since the circles were in two arrays I had to repeat the operation four times. 2 times to catch all the circles in each array, and then 2 more times to catch the ones I missed the first go round. When selecting the circles to be suppressed, they remain visible in the 2D sketch but change to a thin dashed line, as seen above. Unfortunately they have to be individually selected as the crossing windows/fence line method did not seem to work for me. The nice part about suppressing the circles is that they are not visible in when viewing the 3D model and making the sketch visible, as you can see below.</p>
<p><a href="https://i.stack.imgur.com/NbRGX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NbRGX.png" alt="enter image description here" /></a></p>
<p>When you turn the visibility of the sketch off you wind up with a final product that looks like this:</p>
<p><a href="https://i.stack.imgur.com/SPoB9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SPoB9.png" alt="enter image description here" /></a></p>
<p>So while this is the method I chose, I do not know if there is a better way of achieving this result. I look forward to seeing other potential answers.</p>
| 48190 | Best way to array/pattern something in a sketch so that it remains in a defined region |
2021-11-11T22:07:06.887 | <p>A horizontal cantilever with an inverted T cross-section is used as a hoist. The figure below shows
an overview of the beam loading conditions (left) and the beam cross-section (right). One end of the
beam is built in, and the vertical load is applied 1.0 m along the cantilever. If the maximum allowable
stress in the material is 330 MN/m2
, determine the maximum load that can be lifted. Neglect the
weight of the cantilever itself.<a href="https://i.stack.imgur.com/BTebk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BTebk.png" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>Let's calculate the section properties:</p>
<p><a href="https://i.stack.imgur.com/UpNO7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UpNO7.png" alt="enter image description here" /></a></p>
<p>Note: <span class="math-container">$d_c$</span> is the offset distance of the center of the segment/block in consideration with respect to the centroid of the cross-section.</p>
<p>From the equation <span class="math-container">$\sigma = My/I$</span>, we know the larger of <span class="math-container">$y_t$</span> and <span class="math-container">$y_b$</span> will yield the controlling (maximum) stress. <strong>For this case, the topmost face controls,</strong></p>
<p><span class="math-container">$M = \sigma_aI/y_t = 330*10^6*1.8*10^{-6}/0.0713 = 8330 N-m$</span></p>
<p><span class="math-container">$Load = M/L = 8330/1 = 8330 N-m = 8.33 kN-m$</span></p>
<p><strong>Please check my calculation and verify the results.</strong></p>
| 48205 | Hard cantiliver beam question (got I = 3.622x10^-6 and maximum load 23.9 KN) is this right I cant find the answere anywhere? Do you agree with me? |
2021-11-12T15:04:53.250 | <p>Disclaimer: I am a data scientist, not an engineer.</p>
<p>I have been tasked with forecasting failures of a given part and seeing how it affects inventory. Suppose I have historical yearly failure data for part A. That is, I have a count of how many times part A failed by year (100 failures in 2015, 200 failures in 2016, etc...). Paired with the historical failure numbers are the amount of part A currently in service, that is, how much of part A is currently being used in the "system". The number of failures are dependent on how many parts are in service. If we observe a change in in service numbers, then we expect to see a change in failure numbers as well. When a part fails, we assume a one to one depletion of inventory. That is, one failure will decrease inventory by one.</p>
<p>The problem now is to use this historical failure data to construct a model and to forecast future demand. Where the trouble lies is that when we replace a part from a failure, it may no longer follow the failure trend of historical demand. Therefore, we create a dichotomous model that captures the total number of new parts in service and the number of old parts in service. However, I am having trouble attempting to incorporate the different ratios of old to new parts in service.</p>
<p>Currently, I have fitted a simple linear regression line to the historical data, and then use the line to forecast future demand. This is not ideal as the historical demand is not necessarily linear, but it is a starting point. Now, the challenge is to somehow modify the predictions to include new parts that have been introduced to replace failed parts. My current attempt is to weight the slope and intercept by the proportion of old to new parts in service, as well as "jumping back in time" on the slope line for new parts. In other words, end year inventory may be calculated as:
<span class="math-container">$$y_i = [(m\cdot r_i)x_{\text{old}} + b\cdot r_i] + [(m\cdot r'_i)x_{\text{new}}+b\cdot r'_i]$$</span>
where <span class="math-container">$y_i$</span> is the year end inventory, <span class="math-container">$m$</span> is the slope of the regression line, <span class="math-container">$b$</span> is the intercept of the regression line, <span class="math-container">$r_i$</span> is the proportion of old parts in service, and <span class="math-container">$r'_i$</span> is the proportion of new parts in service. However, I constructed this rather heuristically and I am not sure if it is even useful to construct such a model. I am seeking advice on how to handle the dynamic in service numbers to create an accurate model to forecast the failures.</p>
| |systems-engineering|data|failure-analysis| | <p>Modeling future behavior using past behavior is fraught with pitfalls. This is why we test large numbers of parts before turning the factory on and stocking the product in the warehouse. That said...</p>
<p>In failure analysis, that dichotomy is called <em>bimodality</em> which reflects the fact that there are not one but <em>two</em> different possible failure modes that the parts can experience, and those modes represent two different <em>populations</em> within the main population.</p>
<p>Statisticians have to deal with this all the time, and a standard text on <em>descriptive statistics</em> will outline the correct methodology for you. The key point #1 is to prepare a graph called a <em>weibull plot</em> on which the cumulative failure rate for a given failure mode shows up as the slope of a line. Then, you note that different failure modes have differing <em>slopes</em> in that plot, and their effects on reliability can be assessed.</p>
<p>Key point #2 is to always be aware that several part populations can be randomly mixed into your inventory, which makes accurate <em>failure analysis</em> essential.</p>
| 48211 | Incorporating a dichotomous failure model for sustainment analysis |
2021-11-12T23:11:30.650 | <p>If I were to make a sculpture out of Gallium, Ga melts at about 85 °F I think, would I be able to spray it with some sort of solvent based or not solvent based coating to help prevent it from breaking or heating up when holding it?</p>
| |heat-transfer|heat| | <p>What you're asking for is a perfect insulator. There is no perfect insulator, let alone a thin film liquid kind. If ambient temperature is even on iota higher than the melting temperature, eventually it will reach that and melt.</p>
<p>If the magic spray you're looking for existed you could make ice cubes that would never melt at room temperature, or super conductors that would never heat up at room temperature!</p>
| 48216 | What can you coat/seal Gallium with? |
2021-11-13T07:14:22.253 | <p>A nautical navigation textbook on my table says that we can magnetise a non-magnetised needle by rubbing it with a silk cloth. I decided to cross check on the internet, and found some experts calling it off as bullshit. They said that it will make the needle charged with static electricity, as the friction will cause the electrons to transfer between surfaces, and the force of a magnetic field does not affect a static object. So if any of you can conform/reject this notion, then please do.</p>
<p>Apart from that, I tried to make sense of the fact that a capacitor creates an instantaneous current when connected to ground. I also know that a magnet can be made by placing a ferrous demagnetised object in the center of a copper coil that has an electric current running through it.</p>
<p>So if we're stranded at sea with our compass demagnetised, can't we make use of two dissimilar materials by rubbing them, then connecting the resultant static charged object to a copper coil that is ground, placing a demagnetised needle in the center of the coil and finally, closing the switch to let the charged object discharge?</p>
<p>As a needle has such low volume and mass, I believe a few such discharging cycles could effectively find us north again!</p>
<p>Will this work?</p>
<p>[EDIT #1] User @abel has noticed a very sweet phenomenon, and I thought it deserves some visual demonstration. I'll try my best to explain it, but any corrections are welcome.</p>
<p>When two objects have opposite electrostatic charge, they arrange their opposite charges as near to each other's opposite charges as physically possible. Thus, regardless of the two objects' orientations, the charges always stay close to each other. If we were to turn an object (say, a needle) near the surface of another oppositely charged object (say, silk cloth), we effectively move the charges relative to the object (needle) because the charges don't move and the object (needle) does! This imitates the electromagnet coil, when the charges (electrons) in the coil move around the core magnet in the center. I drew these diagrams, dedicated to Sir @abel and for future viewers of this post.</p>
<p>Rolling the (-) charged needle on a (+) charged (flat) silk cloth...
<img src="https://i.stack.imgur.com/Zfhqx.png" alt="enter image description here" /></p>
<p>This is a cross section view. The needle's negative charged excess electrons attract the positive particles in the silk cloth. The like charges are repelled away from the points of contact...
<img src="https://i.stack.imgur.com/RMe4o.png" alt="enter image description here" /></p>
<p>Despite turning the needle to the right, it's electrons stay in the same place, which can be thought of as the electrons turning to the left of the yellow point on a stationary needle! The electrons are moving around the needle, just like the electrons in an electromagnetic coil move around the core magnet!</p>
<p>Turned needle...
<img src="https://i.stack.imgur.com/OvR50.png" alt="enter image description here" /></p>
<p>Needle's charges (electrons) represented by cyan circle...
<img src="https://i.stack.imgur.com/R2ymR.png" alt="enter image description here" /></p>
<p>Needle's charges (electrons) can be thought of as having turned left of needle's yellow point...
<img src="https://i.stack.imgur.com/L1n50.png" alt="enter image description here" /></p>
| |electrical-engineering|materials|electrical|electromagnetism|circuits| | <p>Your idea is remotely possible but not practical. "Possible" because, surprisingly, I once saw an iron rod several feet long get magnetized by getting struck on an end while it was held in an NS direction. It was able to pick up iron filings. I can't imagine how you'd find a capacitor, or a length of suitable wire to do what you suggested.</p>
<p>An accurate compass isn't essential for lifeboat navigation. NSEW directions in daytime can be determined by the altitude/azimuth of the sun. A dab hand (with a watch) can do it. At night, stars suffice. You don't need to be very accurate.</p>
<p>The textbook section that started you thinking is wrong. I doubt you found anything online, from a reputable source, that says otherwise. "Reputable source" might be hard to find these days. There are some that say a person could be magnetized by a vaccine injection.</p>
| 48219 | Magnetising a needle with static electricity using silk cloth |
2021-11-14T10:19:14.200 | <p>I'm super new to this topic I'm posting here so please forgive the nursery question and content. I'm hoping to use any answers to help me begin my research and investigation on what I need to learn.</p>
<p>I've been looking around the interwebs on how to calculate what size electric engine I need to move this garden cart on the beach (<a href="https://www.gorillacarts.com.au/product-page/gorilla-carts-450kg-steel-mesh" rel="nofollow noreferrer">https://www.gorillacarts.com.au/product-page/gorilla-carts-450kg-steel-mesh</a>).</p>
<p>I'm not at all interested in speed. I just want max torque to achieve my goal.</p>
<p>As I understand the basics of it...</p>
<p>Torque(wheel) = Force x Radius</p>
<p>As mentioned. I'm new to all this so probably slow on the uptake. I'm wanting to understand what big factors do I need to consider if I want to move this cart through soft beach sand. What measurements should I be taking and how do I convert that data to then select an appropriate engine?</p>
<p>I've seen similar questions posted on here on how to calculate forces for most appropriate engine but I'm not sure how wanting to operate on soft beach sand changes the math and approach.</p>
<p>I bought a crane scale and pulled the cart across a variety of surfaces while at max intended load.</p>
<p>Max weight + unit = 158kg</p>
<p>Flat cement = 3.5kg peak inital force to pull</p>
<p>Flat cement 10%'ish uphill grade = 14kg</p>
<p>Soft Flat sand = 41 kg</p>
<p>I realise there are a bunch of factors which will be difficult to measure i.e engine/bat inefficiencies but what are the big things I should be figuring out and how to translate that to the hardware requirements?</p>
<p>What the best tyre type for soft sand will play a big role I'm sure in engine choice. My values were on the stock wheels. I need to understand what would be the best tyres for soft sand.</p>
<p>Thanks very much for any help anyone can share with their knowledge and/or experience in the topic.</p>
<p>Thanks,
Nick</p>
| |mechanical-engineering|torque|mathematics|electric-vehicles| | <p>You have already measured some facts. You need continuous pulling force = the weight of 41 kg on the sand. That's the case when the vehicle has been already lifted up from the notches which will be formed under the tyres soon after the vehicle is stopped. You must know also the needed initial pull. It depends on how deep notches there already are below the tyres and what acceleration you expect. And you need these with full load.</p>
<p>The acceleration needs force (in Newtons) = total mass (kilograms) x the speed increasing rate (meters/second in one second). That's an extra which must be added to what's needed to pull up the tyres from the notches below them and to win the continuous sand resistance. To be exact some more is needed to accelerate the wheel rotation, but I guess the rotational inertia of the wheels is small when compared to the translational inertia of the whole system.</p>
<p>It's easiest to measure the total start force when pulling the van to move.</p>
<p>Then you must know how much forward pushing force your tyres can deliver. Lock the rear wheels (I guess they will be powered) and check what's the needed force to pull the vehicle - do it both with light an full load.</p>
<p>I'm afraid pulling as 2 wheels locked shows that you need rear tyres with deep paddles. You simply do not get the needed traction with the originals which probably are designed for low rolling resistance on hard surfaces.</p>
<p>In theory any motor could produce as much forward pushing force as wanted if there's a gearbox with high enough ratio. To have also some driving speed you need power. You can calculate it in 2 ways:</p>
<ol>
<li><p>The power in watts = Driving speed (meters/s) x Pulling force (Newtons). Sorry for using SI units, but they make everything much less error prone.</p>
</li>
<li><p>The power in watts = Rotation speed of an axis (radians per second) x Torque (Newtonmeters).</p>
</li>
</ol>
<p>The torque calculations are essential, as you probably knew. You can calculate the axis torque in Newtonmeters by multiplying the pulling force (=Newtons) with the radius of your powered wheel (=Meters).</p>
<p>You can change the driving speed to axis rotation speed by calculating how many revolutions of a wheel is needed per second for the wanted driving speed. Multiply it with 2Pi (=6.28) to get the result in radians per second or with 60 to get revolutions per minute. Gears multiply the rotation speed and divide the the axis torque with the same number, so gears do not affect the power (except by having friction, you need some reserve).</p>
<p>It's up to you to find a motor which can produce the needed torque for starting to move when stalled. Old fashioned DC motors (with brushes) are best in that sense. Modern electronically controlled brushless motors can be close. In addition the motor must output the needed torque for nominal driving speed at motor's nominal rotation speed continuously.</p>
<p>It's a challenge to keep water and sand out of the motor and to keep the motor cool enough. Designing the easy to use and reliable electricity system with all needed protections is another challenge. I guess you should search for existing solutions when the power, speed and torque requirements are calculated. Have at least 25% reserve.</p>
<p><strong>Not asked</strong>, but do not even think that someone would sit in the van. It's already an unstable vehicle (high, narrow, no stabilization mechanisms) and a person + the safety cage would make it intolerable. It can work as a carriage where electricity only helps on the sand. The max speed you need is about 1,5 meters/second. To reach that speed in 0.5 seconds with full 158 kg (brt.) load you'll need acceleration force 474N (or about 48 kg as you may say). In addition you need what's needed to push the vehicle up from the notches and to win the sand resistance.</p>
| 48232 | How to calculate forces for most appropriate electric engine for soft sand? |
2021-11-17T17:30:14.433 | <p>When wooden walls are created and sheathed in place before erecting (sheathing does go all the way to sill and top plates) some studs don’t have a great connection to top or sill plates because of imperfections in the end cut of the studs or less than perfect construction. They’re locked in place by sheathing before a vertical load (the roof) can settle those gaps. Even after the roof has been added, I’ve still seen gaps between studs and plates. Will strong sheathing actually carry the vertical load for years, or even the lifetime of the structure? Or will settling effectively work on the sheathing/stud connections, until those weaken and move until the studs create full contact with the plates?</p>
| |structural-engineering|construction-management| | <p>A good carpenter would cut all the studs by a table saw or miter saw correctly.</p>
<p>Type 5 (light wood frame) construction, has built-in measures to deal with these potential imperfections. eg, 4 by post under beams, staggered overlap on the double plate, double stud under window headers, etc.</p>
<p>Non the less, due to some careless drilling by the plumber or electrician after the fact some studs may get weakened or totally destroyed at some locations.
A half-inch plywood can take the load acting as a web of a T-beam or the compression member of a C channel between the two good studs.</p>
<p>I have seen studs totally honeycombed by termite damage passing the load to the sheathing or stucco. But it remains to see what happens during an earthquake.</p>
<p>A quick rough estimation of the strength of a short stud joint is to assume plywood will crack in a triangle pattern connecting the tip of 2 adjacent studs to the top of the short stud, like the ultimate strength concrete slab crack method.</p>
<p>Also, there are many fasteners designed to reinforce crippled studs.</p>
| 48291 | Sheathing carrying vertical load more than studs |
2021-11-17T23:50:54.963 | <p>I am a chemical engineering student and I am having trouble understanding a passage in a separation process textbook I am reading. The passage is about the preparing solids for leaching, such as grinding and crushing ores. At the end of the paragraph, the textbook states "Grinding of the particles is not necessary if the soluble material (the desired product) is dissolved in solution adhering to the solid. Then, simple washing can be used, as in washing of chemical precipitates."</p>
<p>I have tried researching this myself, and the only relevant information I found was if the desired product is on the surface of the solid, no grinding is needed. Is this what the textbook is trying to say? I am having a hard time understanding the wording of the textbook, especially about "solution adhering to the solid".</p>
<p>Reference I found: <a href="https://nitsri.ac.in/Department/Chemical%20Engineering/Leaching.pdf" rel="nofollow noreferrer">https://nitsri.ac.in/Department/Chemical%20Engineering/Leaching.pdf</a></p>
| |metallurgy|process-engineering|mass-transfer| | <p>It might help to consider this with the mind set of a primary metallurgist, not a chemical engineer. Primary metallurgist try to recover minerals, not chemicals, but they usually use chemicals in trying to do so.</p>
<p>Most metalliferous minerals that are processed by primary metallurgists are sulfides. Sulfides of such minerals usually do not dissolve in water. Some metalliferous oxides, chlorides and fluorides can dissolve in water.</p>
<p>Water insoluble sulfides are usually crushed and ground so they can be initially processed in a froth flotation plant to obtain an initial bulk concentrate. The size of grinding depends on the size of the mineral being recovered. Large minerals only require a coarse grind, whereas small sized minerals require a finer grind.</p>
<p>The slurry of crushed rock and water is placed into an agitator tank, a frothing chemical is added and a collecting chemical, usually a xanthate, is also added to the tank. The xanthate attaches itself to the sulfide mineral making it hydrophobic. Agitation keeps everything in suspension. Forcing air through the bottom of the tank, with the frothing agent produces stable bubbles in the slurry. The sulfides being hydrophobic, because of the attached xanthate, attach themselves to the air bubbles and rise to the surface to be scooped off.</p>
<p>Water soluble minerals don't need to be reduced to a specific size for them to dissolve, but the smaller they are the easier they dissolve. Such minerals can be recovered using counter current processing plants where dissolving water based liquor is pumped from one tank to another. In moving from one tank to the next the concentration of required elements increases.</p>
| 48296 | Under what conditions is grinding unnecessary for liquid-solid leaching in mineral processing? |
2021-11-18T16:35:03.833 | <p><a href="https://i.stack.imgur.com/XZEo5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XZEo5.jpg" alt="enter image description here" /></a></p>
<p>Is it possible that this tiny engine work without timing belt? I have seen the schematics of it and it pretty much rely on holes in the middle of piston to get the hot gas out whereas the ignition is at top of piston.</p>
| |mechanical-engineering|automotive-engineering|combustion| | <p>For one thing, it's a diesel engine so there is no need to set a spark relative to piston position.<br />
For another, even when you are dealing with a four-stroke ICE, it's technically possible to have both the ignition timing and the valve timing done electronically.</p>
| 48307 | Why this engine can work without a timing belt? |
2021-11-19T14:35:56.557 | <p>I would like a suitable opto-coupler to use as isolation from a 12V and around 30A current and another for 5V and around 30A. Are there any good opto-couplers for those ratings?</p>
| |electrical-engineering|power-electronics|consumer-electronics|microelectronics| | <p>Optocouplers are made for milliamperes. You expect a switching circuit where a high current transistor or mosfet is controlled through an optocoupler or other isolation providing part.</p>
<p>Such circuits are sold as ready to use components. One for your purposes could have name DC solid state relay with low output voltage drop.</p>
<p>The unfortunate fact is that your low DC voltage 5 or 12 volts probably cannot stand much voltage drop. The common equivalent series resistance of low cost mosfets is say 60 milliohms which means 1,8 volt drop at 30A. Fortunately there also exists for low voltage and high current applications special low drop types which have say 2 milliohms equivalent series resistance. An example: Infineon IPD100N04S402ATMA1</p>
<p>That will drop at 30A only 60 millivolts, which can be acceptable (a guess only).</p>
<p>Let's assume you succeed to buy one. You need still the control electronics which generates to the mosfet gate-source voltage = +10 volts for on state and 0V for off state and makes the transition between the states as quickly as your application needs.</p>
<p>For sparse switching on and off the control circuit can be simple - only check that during the transition (=in the half conductive period) the mosfet will not get too hot. I guess you need an electronics designer to make the calculations. If your application is controlling something with PWM and the load is something more complex than a constant resistance the job needs an experienced designer, a casual hobbyist or a beginner simply cannot make it right.</p>
<p>Getting +10V to control the mosfet needs a separate DC source or a DC/DC converter - There's no other way to control the mosfet than generate somehow the needed gate-source voltage; in this case it's +10V for ON-state. Special care is needed to prevent any switching attempts before the control circuit is surely stabilized when the system is taken into use by connecting ON the operating voltages.</p>
<p>For these reasons I suggest to get a ready to use solid state relay. Unfortunately I do not know the market nor your application well enough to suggest any certain type.</p>
| 48319 | How to know which opto-coupler is suitable for high current application? |
2021-11-21T06:14:03.813 | <p><a href="https://en.wikipedia.org/wiki/Fission_fragment_reactor" rel="nofollow noreferrer">From Wikipedia</a></p>
<blockquote>
<p>a fission fragment reactor is a nuclear reactor that generates electricity by decelerating an ion beam of fission byproducts instead of using nuclear reactions to generate heat. By doing so, it bypasses the Carnot cycle and can achieve efficiencies of up to 90% instead of 40-45% attainable by efficient turbine-driven thermal reactors. The fission fragment ion beam would be passed through a magnetohydrodynamic generator to produce electricity.</p>
</blockquote>
<p>From the above we can see a couple of major advantages, namely simplicity and very high efficiency. So why has no such reactor been developed?</p>
| |nuclear-engineering|nuclear-technology| | <p>Several Reasons, but the biggest is that this technology isn't available yet.</p>
<p>Neither the fission beam tech is ready for commercialization, nor is the magnetohydrodynamic generator. The simplest answer to most questions about "why is this technology not used" is that it isn't commercially feasible.</p>
| 48338 | Why are fission fragment reactors not used? |
2021-11-21T11:19:09.543 | <p>I remember when we first got pushbutton phones -- I would say user experience is literally ten times better -- if you used a rotary phone extensively as part of your job, you would rapidly develop a callous on your dialing finger. Of course, you also push the same number (we still use the term dial, at least some people do) maybe ten times faster than a dial.</p>
<p>My question is, why was a dial ever used? Could not something like a pushbutton phone have been developed much sooner? What is the idea behind the dial "UI"?</p>
<p>Note that I know something about computer history, how incredibly hard it used to be, for example, to store a byte of memory using mercury delay lines -- so if I suggest that a musical note be sent (as maybe push button phones did it) instead of N pulses for the integer N I can guess that this is much easier said than done.</p>
| |mechanical-engineering|electrical-engineering|telecommunication| | <p>Rotary phones were controlled by sending pulses down the line. These pulses were the equivalent of hanging up the phone quickly - hence what you see in old movies and TV shows when they quickly tap the hang-up mechanism to get the operator. In the final versions of rotary phones, you could "dial" by quickly pressing down the hang-up button the right number of times for each number. All the rotary mechanism did was send the equivalent of a hang-up press quickly in succession the right number of times.</p>
<p>So why didn't tones work? Because mechanical relays won't send tones, the phones system listeners wouldn't accept tones, and because in general technology has to be invented before it works. You might as well ask why Mr. Bell didn't invent cell phones straight off.</p>
| 48340 | Rotary Phones vs Pushbutton -- why did we ever have rotary? |
2021-11-21T18:44:49.370 | <p>Is there a linkage mechanism allowing the summation of the throws of two crank and slider mechanisms oriented next to one another in order to control an output slider with a motion equal to the summed displacement of the two input crank?</p>
<p>Eg: if slider 1 is positions 5mm above it's equilibrium position at top dead centre and slider 2 is 3mm above it's equilibrium position at just below top dead center I want slider 3 (the output) to be 8mm above it's equilibrium position</p>
<p><a href="https://i.stack.imgur.com/EIMoJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EIMoJ.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|mechanisms|kinematics|linkage| | <p>Have three equal diameter hydraulic cylinders connected to a common fluid tank. It, of course is useless, if you expect no interaction between the inputs.</p>
| 48345 | Summing Displacements of two Cranks and Sliders |
2021-11-23T07:33:19.630 | <p>There is a given ring A, with a given cross section B. Ring A is placed on Columns C. The number of the columns is a variable, but minimum three.
The linear distance X between two columns can be calculated, as well as the distance along the arc length.</p>
<p>For a large ring with a high number of columns, the arc length may be approximated to a linear length.</p>
<p><a href="https://i.stack.imgur.com/uO2Hn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uO2Hn.jpg" alt="enter image description here" /></a></p>
<p>On that ring, a known point load D is placed right between two columns.</p>
<p>In the given Image, E ist the bending curve of the deflection displacement.</p>
<p>How can the deflection displacement f be calculated?</p>
<p>Any help is appreciated.A approximation method would also be helpful.</p>
<p>Thank you.</p>
| |mechanical-engineering|structural-engineering|structural-analysis|rail|bending| | <p>Since the ring is welded to the support posts. Assume the assembly is stable, the posts are non-yielding nor buckling, and the welded connections can be assumed fixed. Then you can use the solution provided by the "Roark's Formulas" - Table 9.4 Case 1 and 1e as shown below. You shall use Case 1e to find the support reactions, then use Case 1 to find the deflection. The terms and parameters are listed at the bottom of this answer. Please note, you may also need to know how to use the unit step function.</p>
<p><a href="https://i.stack.imgur.com/vboMr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vboMr.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/SlTN7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SlTN7.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/oQDdZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oQDdZ.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/TKttt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TKttt.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/D87PN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D87PN.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/fRtlc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fRtlc.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/QiJqu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QiJqu.png" alt="enter image description here" /></a></p>
| 48360 | How to calculate the deflection displacement of a circular looped, closed bending beam? |
2021-11-23T08:52:44.407 | <p>This is in regards to steady heat conduction taking place in a pipe.</p>
<p><a href="https://i.stack.imgur.com/UmXW4.png" rel="noreferrer"><img src="https://i.stack.imgur.com/UmXW4.png" alt="enter image description here" /></a></p>
<p>Within the pipe a hot fluid flows and heat transfer first takes place via convection to the pipe and then via conduction within the pipe and then finally via convection from the pipe to the environment. As can be seen in the diagram, the author has taken the temperature at the inner surface of the pipe as constant <span class="math-container">$T_1$</span> as well as the temperature of the fluid is taken constant at <span class="math-container">$T_\infty1$</span>.</p>
<p>However, I think when the fluid flows inside the pipe the surface temperature would be varying, it wont be constant at <span class="math-container">$T_1$</span>. Similarly the average temperature of the fluid at every section will also not remain constant at <span class="math-container">$T_\infty1$</span>(it will vary along the axial direction).</p>
<p>Is this only a sort of approximation? Is <span class="math-container">$T_1$</span> the average of the surface temperatures along axial direction and <span class="math-container">$T_\infty1$</span> the average of average fluid temperatures at cross sections along the pipe?</p>
<p>Excerpt from the book - <a href="https://drive.google.com/file/d/1WaBp4I-Nbx902G4stCCdRb01ZrDb_lLo/view?usp=sharing" rel="noreferrer">https://drive.google.com/file/d/1WaBp4I-Nbx902G4stCCdRb01ZrDb_lLo/view?usp=sharing</a></p>
| |mechanical-engineering|heat-transfer|thermal-conduction| | <p>Your assumption is correct in all real world situations. the author has idealized the case for learning purposes.</p>
<p>in real world the heat of the pipe is transmitted to the air by a mix of conduction and conventioin itially. creating an expanding onion like stratifyed contour of warm air around the pipe.</p>
<p>This hot air will surround the pipe and will rise like a flame and then move the heat by convection.</p>
<p>This 'flame' is hotter rhan the ambient air and is more intense at the beginning of the pipe and will gradually become smaller as the liquid or water in the pipe gets cooler, even when the steady regime is stablished.</p>
<p>Cooling systems try to perform in this phase so as to use smaller pumps and less liquid mass to move around.</p>
| 48361 | Why does the author assume the temperature to remain constant, in this case, along the axial direction? |
2021-11-23T15:46:05.730 | <p>okay so I’m trying to build a tennis ball launcher for my school so we can showcase it- my teacher helped me get motors and stuff setup similar to this image.</p>
<p>Question: How would I include a spring so that (after we manually compress it) anyone can use a button or something to release the spring and push the ball towards the wheels please? I can’t find anything simple? ![enter image description here]![enter image description here]<img src="https://i.stack.imgur.com/33FNR.jpg" alt="enter image description here" /></p>
<p>I was thinking of having a bung or something at the end of the spring??</p>
| |mechanisms|project-management| | <p>Another idea just for the sake of fun.</p>
<p><a href="https://i.stack.imgur.com/FfbRX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FfbRX.png" alt="enter image description here" /></a></p>
<p>Suggested locking mechanism.</p>
<p><a href="https://i.stack.imgur.com/npuJ3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/npuJ3.png" alt="enter image description here" /></a></p>
<p>Crossectional View</p>
<p><a href="https://i.stack.imgur.com/xr7aQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xr7aQ.png" alt="enter image description here" /></a></p>
| 48365 | Mechanism to control spring release |
2021-11-24T05:53:54.387 | <p><strong>BACKGROUND:</strong> I had previously dealt with 2 pipe coils with oddly bend ends. for the most part I was able to use the coil tool and was able to line up a bunch of 2D sketches to complete the odd end bits. When I was done I noted that while what I created matched the views I was working with perfectly, they also had kinks in the pipe that should not be there. I figured this was a mainly a result of the pipes circles remaining vertical instead of perpendicular to the end of the coil path. I figured when I learned more I would go back and try and fix things.</p>
<p>I then stumbled on <a href="http://www.digipeer.de/index.php?media=DMA_FA_014_02627&size=2" rel="nofollow noreferrer">this drawing</a> and 3 more just like coming up so I figured I should bite the bullet and find out the right way of doing things.</p>
<p><a href="https://i.stack.imgur.com/s4bNK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s4bNK.png" alt="enter image description here" /></a></p>
<p>So in order to avoid the kinks in the pipe where various sections meet, I thought I would plot the continuous center line for the whole thing, place the pipe cross section at the start then sweep the cross section along the path. Sounded simple enough.</p>
<p><strong>WHAT WAS TRIED</strong></p>
<p>I started out by plotting the centerline of the top view, adding some key points, and transferring some information from the front view to complete.</p>
<p><a href="https://i.stack.imgur.com/yLkCB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yLkCB.png" alt="enter image description here" /></a></p>
<p>I then proceeded to draw the bottom end of the front view. I assumed that the slope of the helix continued to the tangent point at the quarter mark and then be tangent with the 35 radius circle. This turned out to NOT be true as the line never intersected with the circle.</p>
<p>My next step was to essentially uncoil the 1/4 turn of the helix. I simply used PI/2*Radius to get the arch length. The thought being I would get the start point of the line that would be tangent to the circle. Though as I just typed that I realize my simple uncoil may not have been correct as it did not account for the helix's slope, just the circle. Though probably close enough for the time being.</p>
<p><a href="https://i.stack.imgur.com/xHxUP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xHxUP.png" alt="enter image description here" /></a></p>
<p>My next step I jumped into 3D sketch for the first time and discovered the Helical Curve tool which I thought would make life easier. The first helix I made was the main coil, and it worked great...it never had to be attached to anything. The next helix I made was the last bottom 1/4 turn at a slightly different pitch (4*Fx:5.597). I then attempted to make a 3rd helix which just plain failed in a major way (others may have failed, but not that obvious if they did). The third helix is almost a half turn going from the edge of the main spiral towards its center (35 radius).</p>
<p><a href="https://i.stack.imgur.com/ZUush.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZUush.png" alt="enter image description here" /></a></p>
<p>So it looks pretty good in front view. Everything lines up where it needs to be. Even the right view looks pretty good.</p>
<p><a href="https://i.stack.imgur.com/TNdIt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TNdIt.png" alt="enter image description here" /></a></p>
<p>Though I did notice a bit of a kink where the tangent met the start of the third helix.</p>
<p><a href="https://i.stack.imgur.com/D75dE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D75dE.png" alt="enter image description here" /></a></p>
<p>I was expecting the bottom helix to come off more like the top. However this was not where I really noticed the problem. I really noticed the problem in top view. The sideways parabolic like shape in the top right quadrant starts and ends at an angle to their tangent lines.</p>
<p><a href="https://i.stack.imgur.com/6OZld.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6OZld.png" alt="enter image description here" /></a></p>
<p>I believe they should be "more" like my initial sketch superimposed below. I say more because my sketch is not perfect. The sketch assumed it was a 1/2 turn when in reality its slightly less. I used a single ellipse when creating the sketch.</p>
<p><a href="https://i.stack.imgur.com/IBe2R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IBe2R.png" alt="enter image description here" /></a></p>
<p>I used the following parameters for creating the third helix:</p>
<p><a href="https://i.stack.imgur.com/qpsTp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qpsTp.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Vw6T0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vw6T0.png" alt="enter image description here" /></a></p>
<p><strong>QUESTION</strong></p>
<p>What is the proper way to model coiled pipes like this?</p>
<p>How do I fix the kink at the start and end of helix 3?</p>
| |solidworks|cad|technical-drawing|drafting|autodesk-inventor| | <p>I stumbled on to another stack question then I stumbled onto a youtube video and it gave me some ideas on how to progress. and I wound up with an acceptable to me solution, but I am still not sure if there is a more practical solution/method.</p>
<p><strong>STEP 1</strong></p>
<p>I drew the front view in order to get that nice 35 radius arc, the tangent line, and the unrolled point of the helix. The methodology for unrolling the helix is valid. Just needed to determine 1/4 of the circumference of the helix's diameter. The first sketch looks as follows:</p>
<p><a href="https://i.stack.imgur.com/aJ2U5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aJ2U5.png" alt="enter image description here" /></a></p>
<p><strong>STEP 2</strong></p>
<p>I then extruded the arc, and the tangent apparently (can't figure out how to limit the selection of joined lines), by the radius of the helix (and maybe a tad more to be on the safe side) to create a curved surface. The centerline has to be somewhere on this surface.</p>
<p><a href="https://i.stack.imgur.com/XhP8k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XhP8k.png" alt="enter image description here" /></a></p>
<p><strong>STEP 3</strong></p>
<p>I then proceeded to make two helixes. The first in green going up from the 77 dimension, matching the parameters set in the drawing for diameter, turns and pitch. I then created a second helix (purple) going down from the 77 dimension and the start of the first helix. I based the pitch of the driving Y dimension where the tangent line from sketch 1 crossed the Y axis. Multiplied it by the inverse of the turn amount (4 in this case for the 1/4 turn).</p>
<p><a href="https://i.stack.imgur.com/cJbcr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cJbcr.png" alt="enter image description here" /></a></p>
<p><strong>STEP 4</strong></p>
<p>I projected the end points of the tangent line to the work plane at the helix's perimeter. I was originally just going to project the line, but I was not sure if the projected line would work for being part of the path. I then just connected the end points with a straight line. This line shared the end point of the second helix in the previous step.</p>
<p><a href="https://i.stack.imgur.com/bSY13.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bSY13.png" alt="enter image description here" /></a></p>
<p><strong>STEP 5</strong></p>
<p>I knew I needed to connect from the end of the last tangent piece at the helix's edge to the bottom of the arc on the helix's centerline. I used a 3D sketch and drew a straight line (black) connecting these points. In the youtube video I watched they used an arc, but I was not sure how they came up with the radius for the arc. I then projected the straight line onto the arc's curved surface that was previous created and that resulted in a nice line (yellow) in my opinion. <strong>Not sure if its the right line though.</strong></p>
<p><a href="https://i.stack.imgur.com/zU0bv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zU0bv.png" alt="enter image description here" /></a></p>
<p>This is the right side view in my model</p>
<p><a href="https://i.stack.imgur.com/korq3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/korq3.png" alt="enter image description here" /></a></p>
<p>This also connected to the bottom arc tangent piece at centerline which was already drawn in sketch 1.</p>
<p><strong>STEP 6</strong></p>
<p>I then moved to the top of the helix. Here I made sketch 3 of the top view of the pipe runout (black lines). I did this at the dimension of the top pipe end which would simply my next step. Or at least I thought I had. Apparently I just projected the last 40 mm straight section to get the same end result.</p>
<p><a href="https://i.stack.imgur.com/0hKKL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0hKKL.png" alt="enter image description here" /></a></p>
<p><strong>STEP 7</strong></p>
<p>Looking at the front view of the drawing I see its a straight slope from the end of the helix to the start of the last 40 mm tangent line of the top pipe. My Eureka moment was realizing 3 points define a plane and crossing my fingers that I could project on to this plane. I created the plane by selecting the two ends of the 40 mm line and the last point on the helix. Attempt 1 was a failure as I decided to sketch on the plane and project geometry...while it did that, but it does it in the direction normal to the plane. NOT what I wanted. So I started a 3D sketch and tried to project sketch 3 on to a surface hoping the newly created plane would count as a surface. I got lucky and it worked!</p>
<p><a href="https://i.stack.imgur.com/cwk9i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cwk9i.png" alt="enter image description here" /></a></p>
<p><strong>STEP 8</strong></p>
<p>I created a plane normal to an axis at a point. I chose the 40 mm line and the end point of it. I then created sketch 4 on this plane and drew two circles. 10 mm diameter and then offset it 0.5 towards the middle. This gave me my pip cross section ready for extrusion. Sorry AutoCAD reference there where I would extrude along a path. For inventor it is SWEEP!</p>
<p><a href="https://i.stack.imgur.com/stdqG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/stdqG.png" alt="enter image description here" /></a></p>
<p><strong>STEP 9</strong></p>
<p>SWEEP!!!!! Sit back admire the final product and wonder if there <strong>was a better way to do it?</strong></p>
<p><a href="https://i.stack.imgur.com/nnPHo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nnPHo.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/o9FM9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o9FM9.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/TSRxU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TSRxU.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/fySU5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fySU5.png" alt="enter image description here" /></a></p>
| 48373 | Coiling and bending pipe in 3D model |
2021-11-24T06:11:17.397 | <p>After searching details regarding direction control valves, I can understand that the diagonal line indicates solenoid valve actuation. The zig Zag line represents the spring action so in this case the DCV will move towards right when solenoid is powered.</p>
<p>Only one question arises, what is the meaning of the triangle they have mentioned in the SOV diagram?</p>
<p><a href="https://i.stack.imgur.com/gKMoe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gKMoe.png" alt="enter image description here" /></a></p>
| |fluid|pneumatic|diagram| | <p>It is a two stage actuator where the triangle represents a pilot. That means that there is a pressure assist. The direction of the arrow indicates the direction of flow.</p>
| 48375 | Triangle symbol in pneumatic diagrams |
2021-11-25T02:24:11.297 | <p>What purpose do dead coils in the middle of a spring serve? For example, this spring from a ballpoint pen has two dead coils at each end and two in the center of the spring. Why add dead coils in the middle? What advantage does this create, or what disadvantage does this overcome? Is it a stability/buckling thing? Is it related to fatigue life? I've never seen this before and I couldn't find any other examples with a pretty extensive Google search, so I'm wondering if anyone has any insights. It's driving me bonkers as I just need to know. Thanks!</p>
<p><a href="https://i.stack.imgur.com/v64W0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v64W0.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/JY2YB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JY2YB.jpg" alt="enter image description here" /></a></p>
| |springs| | <p>The dead coils on the ends of the spring furnish an (almost) flat surface, perpendicular to the spring axis, for the spring to engage the mating parts with. The dead coils in the middle effectively break the spring into two smaller springs, which will both compress in length upon loading without buckling, as NMech points out.</p>
| 48396 | What is the purpose of dead coils in the middle of a compression spring? |
2021-11-25T10:04:41.770 | <p><a href="https://i.stack.imgur.com/yR75C.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yR75C.jpg" alt="ray-line diagram showing light bending in the center of a lens" /></a></p>
<p>As light enters a denser medium from a rarer medium, it bends towards the normal. Why does the light bend after passing through point F & not E?</p>
| |optics| | <p>You are right, rays refract at both the entry and the exit points. That drawing is just a simplified diagram, frequently used in textbooks about geometrical optics. It helps to illustrate concepts like parallel rays, focal point, object distance, image distance etc. where we do not really care what happens inside the lens.</p>
| 48404 | Why does light bend after travelling half of the lens? |
2021-11-26T19:47:05.353 | <p>I'm studying strength of materials at undergrad level and was taking a test in the same. One of the questions from the test goes like this</p>
<p><a href="https://i.stack.imgur.com/FyZKQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FyZKQ.png" alt="enter image description here" /></a></p>
<p>In my understanding since the coefficient of thermal expansion <span class="math-container">$\alpha$</span> of <em>rod</em> material is greater than the <span class="math-container">$\alpha$</span> of <em>tube</em> material, the situation would be like this:</p>
<p><a href="https://i.stack.imgur.com/Keb30.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Keb30.jpg" alt="enter image description here" /></a></p>
<p>The washer will leave the contact with tube and there would be no stresses in either of the <em>tube</em> and <em>rod</em> at the end of heating. However, since I ask this question this is clearly not the case as per the test solutions.</p>
<p>I argue that in these kind of problems the <em>tube</em> material needs to have a higher <span class="math-container">$\alpha$</span> than the <em>rod</em> material only then the washer will not lose contact with the tube.</p>
<p>Please guide me.</p>
| |mechanical-engineering|structural-engineering|stresses| | <p>Your assessment is correct, as the problem didn't indicate the washer is rigidly attached to the steel tube.</p>
<p>The remaining possibility is the stress between the washer and the gun metal, if the washer is a metal with a smaller thermal expansion coefficient than the gun metal.</p>
<p>The picture below is a plate with a plug in the middle, when both are the same metal, they expand together as shown; when different metals are put together, stress at the interface could result, if the plug (the gun) expands more than the plate that has a lower thermal expansion coefficient.</p>
<p><a href="https://i.stack.imgur.com/FSDiL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FSDiL.png" alt="enter image description here" /></a></p>
<p>You can find the difference in expansion by comparing the expanded areas of the plug made by steel and made by the gun metal using the equation below:</p>
<ul>
<li><strong>THERMAL EXPANSION IN TWO DIMENSIONS</strong>
For small temperature changes, the change in area ΔA is given by ΔA = 2αAΔT, where ΔA is the change in area A, ΔT is the change in temperature, and α is the coefficient of linear expansion.</li>
</ul>
| 48434 | A question on thermal stresses |
2021-11-27T11:59:05.717 | <p>A sufficiently slender column will fail under compression below yield stress by bowing, releasing it's axial elastic strain. This is called buckling. Depending on the slenderness of the column, the buckling may be fully elastic, or may involve some plastic deformation powered by the released elastic strain energy. If the column is too stubby, it will not buckle at all before reaching the yield stress.</p>
<p>One such column, which is ductile, yields in compression. What is the expected ultimate failure mode? Is it a fully plastic version of buckling, or something else?</p>
| |structural-engineering|mechanical-failure|buckling| | <p>American Institute of Steel Construction has this graph, (AISC Spec E2), for columns that are not crooked and do not have residual manufacturing stress, and are doubly symmetrical. There are safety factors that take into consideration a variety of imperfections and apply according to the specific load combinations.</p>
<p>The vertical axis is <span class="math-container">$F_{cr}/F_y \
$</span> and the horizontal axis is <span class="math-container">$\lambda_c=\frac{KL}{r\pi}\sqrt{\frac{F_y}{E}} \ $</span></p>
<p>.</p>
<p>The chart uses two equations noted there that meet tangentialy at coordinates, <span class="math-container">$\lambda=1.5. \ $</span> , <span class="math-container">$\frac{F_{cr}}{F_y}=0.39$</span></p>
<p>It assumes yielded <strong>short</strong> columns <strong>Crush</strong> under the load, and gradually as the column slenderness or its length increases the may buckle under plastic, elastic-plastic, and finally elastic buckling.</p>
<p><a href="https://i.stack.imgur.com/U3lET.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U3lET.png" alt="cloumn chart" /></a></p>
| 48443 | Do yielded columns buckle? |
2021-11-28T09:05:57.707 | <p>I am currently searching for an easy-to-use tool to understand/simulate multibody dynamics of rigid bodys.</p>
<p>I have no engineering or mechanics background (I am from IT) and I feel that I cannot imagine in my head how mechanics are behaving if they are moved in this or that direction.
To get a better understanding it would help me a lot to have some software where I could click the bodies and move them with the cursor in this or that direction and watch how they are behaving.</p>
<p><a href="https://i.stack.imgur.com/sJq8o.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sJq8o.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/ZPBp3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZPBp3.jpg" alt="enter image description here" /></a></p>
<p>Do you have any suggestion of such a software? I googled already for some hours, but all I found are very complex 3D simulation software-solutions.
I don't care about exact dimensions or weights or gravity or anything. 2D is also totally fine for me. All I need is something where I can draw lines (bodies) and connect them with other lines and then afterwards move them.</p>
<p>Any help is very appreciated!</p>
<p>BR, mezorian</p>
| |dynamics|software| | <p>You may find that <a href="https://solvespace.com/index.pl" rel="nofollow noreferrer">SolveSpace</a> (multi-platform, free) will do as you require. For your purposes, it's not particularly difficult to learn.</p>
<p>The site also has a <a href="https://solvespace.com/linkage.pl" rel="nofollow noreferrer">link to a tutorial</a> which directly addresses multi-bar linkages. The image below is directly from the tutorial page and is certainly more complex than you present, but it representative of the capabilities of the program.</p>
<p><a href="https://i.stack.imgur.com/Y884F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y884F.png" alt="multi-bar linkage" /></a></p>
<p>Additionally, the program allows movement of the linkages and tracing of a specific point if required.</p>
<p>Eric Buijs has a collection of tutorial videos on peertube, one specifically oriented to <a href="https://peertube.linuxrocks.online/w/3nq8n9wMLDsvFHSvCUEz83" rel="nofollow noreferrer">multi-bar linkages</a>. The linked video is more than 21 minutes long and clearly narrated. Movement is shown at about the ten minute point in the video.</p>
| 48459 | Easy software for : Multibody dynamics of rigid bodies? |
2021-11-28T21:24:23.817 | <p>I am studying a bunch of stuff related to mechanical engineering and am considering an eventual switch to the field after taking the necessary classes. I'm older, and I took drafting classes in high school and part of college. I'm familiar with the standard drafting procedure with paper, as well as <a href="https://en.wikipedia.org/wiki/AutoCAD" rel="noreferrer">AutoCAD</a>, and these days <a href="https://en.wikipedia.org/wiki/SolidWorks" rel="noreferrer">SolidWorks</a>. It's a shame they don't teach paper drafting much anymore!</p>
<p>But I am left with a question I don't have the answer to, nor the resources to answer it. These days it seems most mechanical engineers (at least the ones I know) are really performing the role of both the drafter and the engineer.</p>
<p>Prior to 1995 when SolidWorks came about, what was the duty of the mechanical engineer? If we go even further back before CAD was invented, what duties did a mechanical engineer perform?</p>
| |cad|drafting|software|engineering-history| | <p>CAD tools can define parts, check they fit together, and check whether the design fails in some way (e.g. stress exceeds acceptable limits). They can't answer the less black-and-white questions a design engineer (in any discipline) should be asking:</p>
<ul>
<li>Is this actually going to work:
<ul>
<li>Can the part be made (as others have said)?</li>
<li>Does it meet all the project requirements, including the non-engineering ones (like aesthetics) which may not have been formally captured?</li>
</ul>
</li>
<li>Is there a better way to do this (where better involves weighing up lots of considerations, technical and commercial).</li>
</ul>
<p>The other open-ended question a development engineer asks is "what should this do?" - and CAD is rarely involved in that question, even if it is very valuable in evaluating the answers.</p>
| 48467 | How did mechanical engineers work before SolidWorks? |
2021-11-29T11:54:47.883 | <p>Whilst watching a F1 race recently I noticed the commentators talk about how all the cars on that race day started observing more/faster wearing down of the tyres on the front left of the car relative to the other tyres on the car. It got me thinking about how the track lay out affects tyre wear.</p>
<p><strong>Question</strong>:
Can we predict on which side of the car (left or right & front or back) will the tyre wear down more based on the number of 'right/left turns' & total distance of 'straights'?</p>
<p><strong>My attempt to rationalize the question is as follows</strong>:
If there are predominantly right hand turns, the tyres on the left hand side should wear down more/faster as there will be a load transfer to the left hand side of the car during a right hand turn due to centrifugal effects.
Similarly if there are longer straights, I think the back tyres should wear down more/faster as during acceleration in the straights, the load transfer will occur towards the back of the car due to inertia.</p>
| |materials|automotive-engineering|friction|traction| | <p>Usually a right turn puts more weight of the left side of the car, however that is not the only parameter.</p>
<p>Equally important is the speed as you approach the curb and start to apply the brakes, and also the speed that you engage the turn. So if you had the same amount of left and right turns but the right turns are at higher speed then you'd expect the left set of tires to be more worn out.</p>
<p>Also another factor is how much does the driver go over the sausage curb.</p>
<p><a href="https://i.stack.imgur.com/Re3K2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Re3K2.jpg" alt="enter image description here" /></a></p>
| 48474 | On Which Side Will The Tyre Wear Out More? |
2021-11-29T17:30:13.497 | <p>Is it possible to detect human motion by using a single pyroelectric sensor?
According to me we can't do so using a single sensor.</p>
| |sensors| | <p>To detect the motion of a human or other heat source, the sensing element has to distinguish between general background heat radiation and that given by a moving heat source. A single pyroelectric sensor would not be capable of this and so a dual element is used.</p>
<p><a href="https://i.stack.imgur.com/kWa3y.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kWa3y.jpg" alt="enter image description here" /></a></p>
<p>One form has the sensing element with a single front electrode but two, separated, back electrodes. The result is two sensors which can be connected so that when both receive the same heat signal their outputs cancel. When a heat source moves so that the heat radiation moves from one of the sensing elements to the other, then the resulting current through the resistor alternates from being first in one direction and then reversed to the other direction. Typically a moving human gives an alternating current of the order of 10<sup>-12</sup> A. The resistance R has thus to be very high to give a significant voltage. For example, 50 GV with such a current gives 50 mV. For this reason a transistor is included in the circuit as a voltage follower to bring the output impedance down to a few kilo-ohms.
A focusing device is needed to direct the infrared radiation onto the sensor. While parabolic mirrors can be used, a more commonly used method is a Fresnel plastic lens. Such a lens also protects the front surface of the sensor and is the form commonly used for sensors to trigger intruder alarms or switch on a light when someone approaches.</p>
| 48478 | Question related to pyroelectric sensor |
2021-11-30T14:25:16.297 | <p>On most turbochargers that I've seen, the compressor wheels have 6 full-length blades and 6 splitter blades, 7+7, 8+8, or 9+9. Even on large marine engine turbochargers this generally holds, like the MAN TCR which has 8+8 blades on the compressor.</p>
<p>Then you have newer billet compressor wheels with 11 full-length blades and no splitters. What benefit does this bring over the "conventional" arrangements that I described above?</p>
<p>MAN's axial-flow turbine turbos have 11+11 blade compressor wheels. Generally, on V engines, MAN will use an axial turbo with a dual outlet compressor housing as the low-pressure turbo, then each outlet will feed a conventional (radial turbine) turbo which has 8+8 compressor blades before going to the intercoolers and the cylinder banks.</p>
<p>Now something really struck me: the turbos on EMD 710 engines (which are two-stroke Diesel engines) have 17+17 blades on the compressor. Why <em>do</em> EMD use so many blades? Since two-stroke Diesels need forced induction for scavenging, my initial guess was that they did this to make the turbo more efficient at the higher pressure ratios needed to simultaneously boost and scavenge. At low speeds the turbo becomes a centrifugal supercharger as there is a clutch that allows the crankshaft to drive the compressor, but at high speeds the clutch disengages and the turbo takes over. Might the large number of blades have to do with this?</p>
<p>My question is, why does the low-pressure turbo have more blades than the high-pressure turbo? Is it generally true that, in a compound turbo system, you want more blades on the low-pressure turbo than the high-pressure turbo?</p>
<p>From what I've found, people seem to say that fewer blades flow better at high pressure ratios but are not as efficient at low pressure ratios, and that more blades leads to better low-end response but not as good of a top end. Another source that I've read claims that more blades flows better at high pressure ratios but not at low pressure ratios. The responses I've found were all over the place, so I'd like to gain more insight on what more vs. less blades does on turbochargers (and centrifugal compressors in general).</p>
| |turbomachinery| | <p>Think of blades as something pushing on the fluid. There is a pressure side and a suction side to each blade. Pressure varies from the low on suction side to the high on the pressure side.</p>
<p>Any work done to move the fluid where you need it to go (axially "forward" or radially outward at the compressor) is useful. Anything else (work moving it tangentially, against a wall, or backflow around your blades) is waste. Having a lot of space between blades or blades and shroud allows more of the dreaded but inevitable backflow around your compressor (from its "output" side to its "intake"). On the other hand, more blades can mean more to balance, accelerate, and spend energy deforming. Spinning a disk (100% blade and 0% fluid) does you no good. Extremely thin blades will themselves deform too much from stresses between pressure and suction sides.</p>
<p>In summary, optimization can be thought of as a matter of making the blades thick enough for the loads, sparse enough to have space for the flow you need, and plentiful enough to hold back the pressure you generate.</p>
<p>The contradictory information you are finding is likely because of the choice of control variables. There's a lot to choose from between the geometry (angle of attack, diameter, etc), rpm, fluid properties, flow rates and pressures. To just look for a trend between pressure ratio and number of blades is futile because it depends on so much more. You can come up with some custom metrics to be your constant control variables to make it go either way.</p>
| 48492 | Why do EMD turbochargers have so many blades? |
2021-11-30T21:06:02.460 | <p>I have the book "Power system analysis and design" by Sarma.
In his intro to balanced LN voltages, in chapter 2 p62 he states:</p>
<p><span class="math-container">$$V_{LL} = \sqrt{3} V_{LN} \angle30º$$</span></p>
<p>but later in the chapter, p70, he states for balanced LN voltages:</p>
<p><span class="math-container">$$ V_{LL} = \sqrt{3} V_{LN} $$</span></p>
<p>This is a very different statement, why is he saying that after literally saying a few pages before there is a 30º phase shift.</p>
| |electrical-engineering|power|circuits|power-engineering|multiphase-flow| | <p><a href="https://i.stack.imgur.com/dao2om.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dao2om.png" alt="enter image description here" /></a></p>
<p><em>drawing by SSR</em></p>
<p>Because <span class="math-container">$V_{AB} = V_A - V_B$</span> (Vector Subtraction). Dashed line is <span class="math-container">$-V_{B} = -(120 \angle -120° V)$</span> <span class="math-container">$= 120 \angle 60° V$</span>. Vector addition of <span class="math-container">$V_{A}$</span> and <span class="math-container">$-V_{B}$</span> gives line voltage <span class="math-container">$V_{AB} = 208 \angle 30° V$</span></p>
<p>In a three-phase, wye connected system the line-to-line voltage (line voltage) <span class="math-container">$V_{LL}$</span> (<span class="math-container">$V_{AB}$</span>, <span class="math-container">$V_{BC}$</span>, <span class="math-container">$V_{CA}$</span>) is <span class="math-container">$\sqrt {3}$</span> larger than phase to neutral voltages (phase voltage) <span class="math-container">$V_{LN}$</span> (<span class="math-container">$V_{AN}$</span>, <span class="math-container">$V_{BN}$</span>, <span class="math-container">$V_{CN}$</span>) and leads the phase voltage by 30°.</p>
<p>In the first case the author is making emphasis that the line voltage is a vector (magnitude <span class="math-container">$\angle$</span> angle or <span class="math-container">$V_{LL} = \sqrt{3} V_{LN} \angle30º$</span>) and the second case talks about magnitude (<span class="math-container">$ V_{LL} = \sqrt{3} V_{LN} $</span>).</p>
<p>The phase angles are only relevant when looking at phasor diagrams, so they are usually left off for clarity.</p>
| 48498 | Line to neutral voltage in three-phase power systems |
2021-11-30T21:16:09.737 | <p>I was looking at this problem:</p>
<p>"A balanced delta connected impedance load with (12+j9)Ω per phase is supplied by a balanced three-phase 60 Hz, 208-V source "</p>
<p>I do not wish to ask for help at solving this, rather, I want to know how to interpret the "208 V source" statement. If someone gives me that statement I will immediately think it means<br />
<span class="math-container">$$ V_{an} = 208 V \angle 0º $$</span>
and we can just get <span class="math-container">$V_{bn}$</span> and <span class="math-container">$V_{cn}$</span> by adding <span class="math-container">$\angle +240º$</span> and <span class="math-container">$\angle +120º$</span> respectively.</p>
<p>However I have seen people interpreting this as <span class="math-container">$ V_{LL} = 208 V$</span>, what reason do they have to conclude it refers to LL and not LN, the problem does not sate it.</p>
| |electrical-engineering|power|electrical|power-engineering|power-transmission| | <p>Note the problem specifies the load is "delta connected". There is no neutral connection to measure phase voltage (<span class="math-container">$V_{LN}$</span>) against.</p>
<p>Also note the supply frequency is 60 Hz. 208 V is a common <a href="https://www.allaboutcircuits.com/textbook/alternating-current/chpt-10/three-phase-y-delta-configurations/" rel="nofollow noreferrer">line</a> voltage (<span class="math-container">$V_{LL}$</span>) in the US (which uses a 60 Hz supply), with corresponding phase voltage of 120 V. You are using a US text (Sarma <em>et al</em>, from another post) so you can expect to see examples of US practice.</p>
<p>So the problem is asking about <span class="math-container">$V_{LL} = 208 V\;$</span>.</p>
| 48500 | Power systems unclear voltages |
2021-12-01T04:41:26.333 | <p>I have an application where I need to rotate a solar panel on a moving cart around 2 axes. The solar panel is medium size (22x25in) and is only 50 watts. I plan to use an actuator to rotate the panel on axis1 to account for when the sun is high in the sky versus rising or setting. For axis2, I need to be able to rotate the base that the panel rests on by 360 degrees.</p>
<p>For axis2 is there a type of motor that holds its rotational state when powered down even when opposed by some load, similar to an actuator? An example of the application would be similar to the image below, the red portion. The blue portion is where I plan to use the actuator.</p>
<p><a href="https://i.stack.imgur.com/s9f5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s9f5m.png" alt="model" /></a></p>
| |motors|solar| | <p>Although the size of panel is relatively small, before deciding on the motor (or on the dimensioning of the gears) my advice is to perform a wind load analysis. Those loads tend to become quite high in certain areas. Plus there is the additional problem of dynamic loading which can really take its toll on the structure and on the gears. Mind you my experience is from 80 to 180 m^2 --850 to 2000 sq.ft-- solar tracker.</p>
<p>Once you have the win loads, then you can determine the Motor (so its very much depended on the location that this will be installed).</p>
<p>Now regarding some details you need to rethink about your project:</p>
<ul>
<li><p>keep in mind that if you use this arrangement in the picture with the spur gears you will need to keep the motor always engaged. This can lead to overheating. So you probably need to add some sort of failsafe locking mechanism. I.e. something that you activate when you want to move (and otherwise its braking).</p>
</li>
<li><p>Slewing ring or <a href="https://en.wikipedia.org/wiki/Slewing_drive" rel="nofollow noreferrer">slewing drives</a> (which use Worm gears as DKNguyen suggested) are probably the way to go ---if you can afford them -- for the azimuth direction (bottom gear). However, you can probably get away with something cheaper here.</p>
</li>
<li><p>The elevation direction is probably more demanding. If you plan to build this to sell it, I would strongly advice you not to use the spur gear you have in the image. In a matter of months in moderate weather you will begin to notice wear and tear on the gears. This will increase the wind induced oscillations and produce failure quicker.</p>
</li>
</ul>
| 48506 | Motor to rotate gear and hold place when powered down |
2021-12-03T14:44:58.433 | <p>This question might be better suited for this stack. In the physics stack, I posted about a question about siphoning. <a href="https://physics.stackexchange.com/q/677102/259268">https://physics.stackexchange.com/q/677102/259268</a></p>
<p>An 10 meter tall cylinder is sitting in a infinite pool of water. It has supports on the bottom attached to the base of the pool to maintain its vertical position. The top of the cylinder is closed and the bottom is open to the water. The bottom of the cylinder is not exposed to air and is submerged. There is a hose located 1 meter below the top of the cylinder which leads through the wall of the cylinder. The hose entrance is at a higher elevation (~9 m) than the hose exit (~8 m). Currently the hose exit is closed.</p>
<p>Water fills the cylinder and hose combination (the air is released, there is no air in the entire setup, only water). The atmospheric pressure outside the cylinder(and hose) and the properties of water allows for the 10 meter height of the water column. Suddenly the hose exit opens, starting the siphon process. Water drops 8 meters into a water turbine sitting level with the surface of the pool. Beneath the turbine is just an extension of the pool that the cylinder also occupies.</p>
<p>As the water falls from the hose exit, it starts pulling water from the top of the cylinder. The open bottom of the cylinder starts pulling water from the pool. The water goes through the turbine and re-enters the pool. The flow of water through the cylinder is continuous as it can not drain the entire pool of water.</p>
<p><a href="https://i.stack.imgur.com/vBNZ1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vBNZ1.jpg" alt="Siphon" /></a></p>
| |fluid-mechanics|pressure|hydraulics|siphon| | <p>The sketch is to help you to clear up your question. You should identify where the hose is terminated - (a) or (b)? Please feel free to copy and edit the sketch to match your thought.</p>
<p><a href="https://i.stack.imgur.com/Mgz1Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mgz1Q.png" alt="enter image description here" /></a></p>
<p><strong>Two Example Cases:</strong></p>
<p><a href="https://i.stack.imgur.com/hvySJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hvySJ.png" alt="enter image description here" /></a><a href="https://i.stack.imgur.com/i4NJ8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i4NJ8.png" alt="enter image description here" /></a></p>
<p><a href="https://en.wikipedia.org/wiki/Siphon" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Siphon</a></p>
| 48557 | Q: Could you siphon an 10 meter tall water column? |
2021-12-04T08:56:27.540 | <p>I am a completely new not-even-started starter in welding.</p>
<p>I would like to know is Lift-tig welding the same as Lift-arc welding?
If not, what are the differences?</p>
| |welding| | <p>Yes, Lift arc is a technique used for TIG Welding. Other technique used are: <a href="https://www.materialwelding.com/what-is-the-scratch-start-lift-arc-and-hf-ignition-technique-in-tig-welding/" rel="nofollow noreferrer">Scratch-Start, lift arc and HF Ignition technique</a>.
When we say arc welding, it a broader term to describe welding processes relies in welding arc to produce heat. example are: Stick welding, MIG, TIG welding.</p>
| 48574 | Is Lift-tig welding the same as lift-arc welding? |
2021-12-04T09:43:55.463 | <p>I've been reading up about reinforced concrete, as it is one if not the cheapest way how to build (with some solid foundation). I found out that it actually has quite poor durability (50-100 years or something). The reason for this, if I understand correctly, are the steel rebars, which work inside the concrete because of corrosion and deteriorate the construction by cracking the structure. If I also understand correctly, the rebar is used to make the structures more flexible and easier to build and bigger structures simply cannot be build just by using concrete (since then it would probably be easy to get longer durability as Pantheon seems to do just fine).</p>
<p><strong>It might be completely dumb question but why they don't use stainless steel instead?</strong> From what I've read the cost of stainless steel is only about 4-5x more expensive. I did some very rough calculations and considering the price of land, price of work, price of concrete, price of everything what you put in the building afterwards like other engineering materials, the cost of building demolition in its life cycle and so on it seems like financially very shortsighted decision.</p>
<p><strong>Or is it the case that the stainless steel would not prolong the lifetime so rapidly? If so, why?</strong></p>
<p><em>(stainless steel should not corrode and the thermal attributes of steel and concrete seem to be almost the same)</em></p>
| |steel|concrete|building-design|building-physics| | <p>There are multiple avenues through which this can be approached.</p>
<p>One is the economics of the construction. As you've mentioned, concrete structures are far cheaper to build than steel, though the difference is highly dependent on location and context. See <a href="https://engineering.stackexchange.com/a/7002/1832">my answer to another question</a> for more details on this.</p>
<p>Another which I also mention in that answer is that concrete requires effectively no maintenance during its service life, so long as it's done in a minimally competent manner. Steel, on the other hand, requires frequent maintenance vs. corrosion.</p>
<p>And while I've never seen a life-cycle cost analysis such as you've done, let's take your results at face-value, that concrete and steel end up costing roughly as much. But then, as mentioned in <a href="https://engineering.stackexchange.com/a/48624/1832">@tigerguy's answer</a>, you need to remember that costs 50 years in the future must be discounted into relative irrelevance.</p>
<p>But still, even if you disagree with <a href="https://en.wikipedia.org/wiki/Discounted_cash_flow" rel="nofollow noreferrer">DCF</a>, you need to question your primary assumption. Your entire point is that it's a problem that we need to demolish that structure 50 years from now. But is it a problem? Or is it an incentive for society to demolish that building and make one more adequate for its needs 50 years from now, using technology and installations more reasonable for that future? If its a bridge, it's an opportunity to build one that better matches demand at that time: if we're all in flying cars, demolish it and build a park instead. If we're still earth-bound, make it wider to accommodate ever more traffic (a bad idea!) or incorporate more modern safety measures, etc.</p>
<p>After all, when's the last time you've seen bridges or other infrastructure proactively updated to incorporate new safety measures or what have you? In my experience, having the bridge look like it's about to collapse is quite useful when trying to get politicians to modernize infrastructure.</p>
| 48576 | Reinforced concrete by steel, poor durability, why? |
2021-12-04T10:25:37.083 | <p><a href="https://youtu.be/w6aUzGaXShQ?t=135" rel="nofollow noreferrer">https://youtu.be/w6aUzGaXShQ?t=135</a></p>
<p>At time 2:15, what's the type of the welding tool he uses?</p>
<p>I have read a couple of types for welding tools but I cannot see it matches any type.</p>
| |welding|tools| | <p>It's a TIG welding hand torch. This article explains it well.</p>
<p><a href="https://www.alphaweld.com.au/blog/48-an-introduction-to-tungsten-inert-gas-tig-welding" rel="nofollow noreferrer">https://www.alphaweld.com.au/blog/48-an-introduction-to-tungsten-inert-gas-tig-welding</a></p>
| 48577 | What is this welding tool in this video? |
2021-12-04T10:51:02.127 | <p>Okay,</p>
<p>I have this value in mmH<sub>2</sub>O (2.650) and am not sure where to go next to convert it to aerodynamic parameter pressure coefficient please?</p>
<ol>
<li>mmH<sub>2</sub>O * 9.8066 = N/m<sup>2</sup></li>
<li>???? What do do next please?</li>
</ol>
<p>I have Patm and P from pitot tube</p>
| |fluid-mechanics|aerospace-engineering|aerodynamics| | <p>The pressure coefficient is a parameter for studying both incompressible/compressible fluids such as water and air. The relationship between the dimensionless coefficient and the dimensional numbers is:</p>
<p><a href="https://i.stack.imgur.com/UfU1o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UfU1o.png" alt="enter image description here" /></a></p>
<p><a href="https://en.wikipedia.org/wiki/Pressure_coefficient" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Pressure_coefficient</a></p>
<p>Does this help? Obviously, you'll need to know/have more than just a single parameter.</p>
| 48579 | Convert from mm of water to Coefficient of pressure (Cp) |
2021-12-04T21:18:31.840 | <p>We know that in <strong>Oblique Parallel Projection</strong> Point <span class="math-container">$(x,y,z)$</span> is projected to position <span class="math-container">$(x_p,y_p)$</span> on the view plane.Projector (oblique) from <span class="math-container">$(x,y,z)$</span> to <span class="math-container">$(x_p,y_p)$</span> makes an angle <span class="math-container">$\alpha$</span>
with the line (L) on the projection plane that joins <span class="math-container">$(x_p,y_p)$</span> and <span class="math-container">$(x,y).$</span>
Line <span class="math-container">$L$</span> is at an angle <span class="math-container">$\phi$</span> with the horizontal direction in the
projection plane.See this <strong>image1</strong>:
<img src="https://i.stack.imgur.com/ngxGa.jpg" alt="enter image description here" /></p>
<p>And in Oblique Parallel Projection Angles, distances, and parallel
lines in the plane are projected
accurately.For example see below <strong>image2</strong>:<a href="https://i.stack.imgur.com/waFgn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/waFgn.jpg" alt="enter image description here" /></a></p>
<p>My question is where is the angle <span class="math-container">$\alpha$</span> in <strong>image2</strong>, I mean I see the angle <span class="math-container">$\phi$</span> on the image , so where is <span class="math-container">$\alpha$</span> in that image to understand better?</p>
<p>N. B:1 -- I am following <a href="https://drive.google.com/file/d/1_nRZfRTqdkkKsQh2VrRvA-QAiQzeAIHx/view?usp=drivesdk" rel="nofollow noreferrer">Hearn and Baker book which screenshot like this. </a></p>
<p>N. B. -- I want to understand just intuition in easy way rather than details.</p>
| |technical-drawing|computer-engineering|drafting|computer-aided-design|computer| | <p><a href="https://i.stack.imgur.com/q60kR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q60kR.png" alt="enter image description here" /></a></p>
<ul>
<li>The green arrow is the out-of-plane vector</li>
<li>the light blue arrow is the plane vector</li>
</ul>
<h2>Calculation of <span class="math-container">$a$</span> angle</h2>
<p>If you got the coordinates <span class="math-container">$x,y,z$</span> and <span class="math-container">$x_p,y_p,0$</span> then the most generic way to use is through the dot product and the inverse cosine.</p>
<p>I.e.</p>
<ul>
<li>the vector in the projection plane will have direction:</li>
</ul>
<p><span class="math-container">$$v_{plane} = \begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$</span></p>
<p>and the unit direction vector would be:</p>
<p><span class="math-container">$$e_{plane} = \frac{1}{||v_{plane}||}\begin{bmatrix} x-x_p\\ y-y_p\\0\end{bmatrix}=\frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$</span></p>
<p>Similarly, the out of plane vector is:</p>
<p><span class="math-container">$$v_{out} = \begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}$$</span></p>
<p>and the corresponding unit direction vector would be:</p>
<p><span class="math-container">$$e_{out} = \frac{1}{||v_{out}||}\begin{bmatrix} x-x_p\\ y-y_p\\z\end{bmatrix}=\frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}$$</span></p>
<p>The dot product of the unit vectors would be:</p>
<p><span class="math-container">$$e_{out} \cdot e_{plane} = \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ z\end{bmatrix}\cdot \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}$$</span>
<span class="math-container">$$e_{out} \cdot e_{plane} = \frac{( (x-x_p)^2 + (y-y_p)^2 )}{\sqrt{((x-x_p)^2+ (y-y_p)^2 + z^2) ((x-x_p)^2+ (y-y_p)^2)}}
$$</span></p>
<p><span class="math-container">$$e_{out} \cdot e_{plane} = \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }}
$$</span></p>
<p>At the same time:</p>
<p><span class="math-container">$$e_{out} \cdot e_{plane} = ||e_{out}|| \cdot ||e_{plane}|| \cdot \cos(a)= 1\cdot 1\cdot \cos(a) = \cos(a) $$</span></p>
<p>therefore:</p>
<p><span class="math-container">$$\cos(a) = \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }}
$$</span></p>
<p><span class="math-container">$$a = \arccos\left( \frac{\sqrt{ (x-x_p)^2 + (y-y_p)^2 }}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + z^2 }}\right)
$$</span></p>
<p>(PS: please check the derivation for any mistakes/errors, because the final equation seems a bit too elegant)</p>
<hr />
<h2>Calculation of <span class="math-container">$\phi$</span></h2>
<p>Calculation for <span class="math-container">$\phi$</span> angle can be expressed by the same equation but you can select a point on the x axis e.g. <span class="math-container">$(x, y, z) = (1,0, 0).$</span></p>
<p>In that case the equation will have the following form:</p>
<p><span class="math-container">$$e_{out} \cdot e_{plane} = \frac{1}{\sqrt{(x-x_p)^2+ (y-y_p)^2 + 0^2}}\begin{bmatrix} x-x_p\\ y-y_p\\ 0\end{bmatrix}\cdot \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}$$</span></p>
<p><span class="math-container">$$\phi = \arccos\left( \frac{x-x_p}{\sqrt{(x-x_p)^2+ (y-y_p)^2 }}\right)
$$</span></p>
| 48585 | How to visualize angle of projection in Oblique Parallel Projection? |
2021-12-05T10:13:17.970 | <p>Assumed we have a <strong>fuel flask</strong> that is made of titanium. The bottle will be filled with white gasoline when used. At the bottleneck there is a standardized thread manufactured into the titanium.</p>
<p>Are there any dangers of (contact-)corrosion between the materials if the screwed-in bottle cap is made of aluminium instead of titanium?</p>
| |metals|aluminum|threads|corrosion|fuel| | <p><strong>Galvanic Compatibility</strong></p>
<p>Often when design requires that dissimilar metals come in contact, the galvanic compatibility is managed by finishes and plating. The finishing and plating selected facilitate the dissimilar materials being in contact and protect the base materials from corrosion.</p>
<p>For harsh environments, such as outdoors, high humidity, and salt environments fall into this category. Typically there should be not more than 0.15 V difference in the "Anodic Index". For example; gold - silver would have a difference of 0.15V being acceptable.</p>
<p>For normal environments, such as storage in warehouses or non-temperature and humidity controlled environments. Typically there should not be more than 0.25 V difference in the "Anodic Index".
For controlled environments, such that are temperature and humidity controlled, 0.50 V can be tolerated. Caution should be maintained when deciding for this application as humidity and temperature do vary from regions.</p>
<p><strong>Anodic Index</strong></p>
<p>Metallurgy Index (Volt)</p>
<p>Nickel, solid or plated, <strong>titanium and alloys & Monel (0.30)</strong></p>
<p>Aluminum, wrought alloys of the 2000 Series (0.75)</p>
<p>Aluminum, wrought alloys other than 2000 Series aluminum, cast alloys of the silicon type (0.90)</p>
<p>Aluminum, cast alloys other than silicon type, cadmium, plated and chromate (0.95)</p>
<p><a href="https://www.corrosion-doctors.org/Definitions/galvanic-series.htm" rel="nofollow noreferrer">https://www.corrosion-doctors.org/Definitions/galvanic-series.htm</a></p>
| 48595 | Aluminium Bolt in Titanium Casing: Any Dangers of (Contact-)Corrosion? |
2021-12-05T16:04:43.047 | <p>According its <a href="https://en.wikipedia.org/wiki/Tesla_Model_S" rel="nofollow noreferrer">Wiki</a>, a Tesla Model S battery pack can accumulate 100 kWh energy.</p>
<p>The wiki page of the <a href="https://en.wikipedia.org/wiki/Lithium-ion_battery" rel="nofollow noreferrer">LiON batteries</a> shows a 100-265 kWh/kg specific energy, and <a href="https://stealthev.com/product/tesla-module/" rel="nofollow noreferrer">this source</a> shows a Tesla battery pack with 210 kWh/kg (thanks @Solar Mike).</p>
<p>Calculating these, we get the required mass of a Tesla Model S battery to <span class="math-container">$\approx$</span> 600 kg.</p>
<p>This is obviousy much more than the reality. How is it possible? What is the real mass of a battery pack in a Tesla Model S?</p>
| |battery|electric-vehicles| | <p>Summarizing the comments: yes, the calculation is roughly correct. The weightiest part of the Teslas (and other electric vehicles) is their battery pack, and yes, they weight typically more hundreds kg.</p>
<p>Note, only a little part of it is actually lithium (for example, only <span class="math-container">$\approx 4.4\%$</span> of <span class="math-container">$\rm{LiFePO_4}$</span>) is lithium element.</p>
| 48600 | How can Tesla battery packs be so small? |
2021-12-05T18:38:44.473 | <p><a href="https://i.stack.imgur.com/K65gW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K65gW.jpg" alt="enter image description here" /></a></p>
<p>I'm using I-Joists to span this 27' vaulted living room and create a shed style roof. I've got two problems. Because of the angle, the joist won't make full contact with the walls (short of 20 years worth of settling), and of course it's got a tendency to want to slide. My ideas were to cut angles in the top plates of the walls so that the full surface area bears the load, and to fasten brackets from wall to joist on the outside of both walls to prevent the slide. Do these seem like reasonable solutions? Any better solutions?</p>
| |structural-engineering|fasteners|wood| | <p>Many fastener companies have a fastener made for this specific application. They test these fasteners under all kinds of situations, winds, earthquakes, extra loads.</p>
<p>Here is one made by Simpson Fasteners at a cost of about $7.59.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/Do1Uv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Do1Uv.png" alt="enter image description here" /></a></p>
| 48602 | Solutions for a slanted roof |
2021-12-06T00:57:51.137 | <p>I'm trying to create a parametric model of a container, using similar joints as this foldable bathtub. <a href="https://i.stack.imgur.com/rLgXL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rLgXL.jpg" alt="foldable bathtub" /></a></p>
<p>I do not need stress simulation, but need to see if my model can actually fold the full range of the intended motion with given parameters, or the material thickness or folding angle introduces constraints.</p>
<p>So my question is: are there good CAD tools for this kind of work?</p>
<p>Are FreeCAD or OpenSCAD good candidates?</p>
| |mechanical-engineering| | <p>If you have programming skills, OpenSCAD may accomplish your task. OpenSCAD allows animation, if one can perform the necessary code. The key feature of your question is "parametric" and your code should reflect that aspect.</p>
<p>I'm unfamiliar with FreeCAD, but there is a hobbyist version of Fusion 360, which also support parametric construction and constraints. Not so much coding, as it's primarily GUI based.</p>
<p>Another GUI option with parametric features is SolveSpace, multi-platform, free software, 2D-3D capable. Constraints can be created to allow movement to cover that part of your question.</p>
| 48608 | Best software for parametric design of foldable piece with rigid and elastic parts |
2021-12-06T16:48:24.643 | <p>I would like to build a small robot and was wondering if a particular configuration of steering would even be possible and stable.</p>
<p><a href="https://i.stack.imgur.com/Jy8hw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jy8hw.jpg" alt="enter image description here" /></a></p>
<p>What I would like is to have the two rigid axles mounted on a pivoting point like the image. The pivoting points would be free to move on their axes without restrain. Steering should be achieved via the different speed of the 4 wheels (one motor on each wheel).</p>
<p>Problem is, if I start accelerating the outer wheels, to me it looks like it would start going in a diagonal direction, with both axles keeping parallel to each other rather than achieving something like the picture, even though the outer wheels on a corner should go faster than the inner wheels. Is this configuration even possible?</p>
| |mechanical-engineering|dynamics|robotics| | <p>You have too many degrees of freedom to use it effectively without feedback. If you add a couple of potentiometers to detect the angles of the pivots, you should be able to compensate appropriately for it. Essentially the robot needs to know which way each wheel is pointing to determine which way and how much to try to turn it. The rest of "stable" steering (relative to the robot) will depend on whether or not you can process the feedback loop fast enough and your feedback algorithm.</p>
<p>In theory one person could try to learn how to process all 4 wheels at once while observing the robot for feedback, but to make it easy enough to call stable for the average person, it would involve creative controls. Humans tend to have 2 eyes and 2 hands. Steering wheel, gas pedal for car. Left, Right for tank drive. Are the twos a coincidence? I dare say I'm not cut out to be a puppeteer.</p>
| 48620 | Is this kind of steering possible? |
2021-12-06T22:19:07.800 | <p>I have a jig I am working on that accepts a circuit board. When the lever is pressed I would like the board to be ejected to the left - as shown in the image.</p>
<p>What I am finding is that the board is ejected directly upwards.</p>
<p>I have tried different shapes and length of arm to try and control this with no success. Does anyone know of modifications I can make to control the direction?</p>
<p>Requirements:</p>
<p>When the circuit board is installed it must be parallel to the table. I.e. I can't rotate the entire jig.</p>
<p>When the circuit board is installed there can't be any obstructions directly above the board.</p>
<p>Thanks!</p>
<p><a href="https://i.stack.imgur.com/nOlex.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nOlex.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|3d-printing| | <p>Solved by adding a small tab next to the lever with a 45 degree angled face towards the board. Found by trial and error this appears enough to consistently guide the board away from the lever instead of doing backflips directly upwards.</p>
<p><a href="https://i.stack.imgur.com/zK2xm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zK2xm.jpg" alt="enter image description here" /></a></p>
| 48629 | Eject Circuit Board From Jig |
2021-12-07T07:05:09.000 | <p>Suppose I have one plane <span class="math-container">$Ax+By+Cz+D=0.$</span> I want to test <span class="math-container">$(x, y, z)$</span> is inside of this plane or outside.</p>
<p><strong>My attempt:1</strong> Then draw normal N on that plane which point away from the observer <a href="https://i.stack.imgur.com/Y0nVH.jpg" rel="nofollow noreferrer">Image1.</a></p>
<p>If <span class="math-container">$Ax+By+Cz+D>0$</span>
or <span class="math-container">$(-A)x+(-B)y+(-C)z+(-D)<0$</span> then <span class="math-container">$(x, y, z)$</span> is outside of the plane.</p>
<p>And if <span class="math-container">$Ax+By+Cz+D<0$</span>
or <span class="math-container">$(-A)x+(-B)y+(-C)z+(-D)>0$</span> then <span class="math-container">$(x, y, z)$</span> is inside of the plane.</p>
<p><strong>My attempt:2</strong> Now draw the normal N on that plane which point towards the observer <a href="https://i.stack.imgur.com/XfZ7I.jpg" rel="nofollow noreferrer">image2.</a></p>
<p>If <span class="math-container">$Ax+By+Cz+D<0$</span>
or <span class="math-container">$(-A)x+(-B)y+(-C)z+(-D)>0$</span> then <span class="math-container">$(x, y, z)$</span> is outside of the plane.</p>
<p>And if <span class="math-container">$Ax+By+Cz+D>0$</span>
or <span class="math-container">$(-A)x+(-B)y+(-C)z+(-D)<0$</span> then <span class="math-container">$(x, y, z)$</span> is inside of the plane.</p>
<p>My question is my both attempts are right? Can anybody help me to understand.</p>
| |mathematics|computer-engineering|computer-aided-design|geometry| | <p>if a point is in a plane it means if you plug in x.y,z in the plains EQ. you get zero. And if you get nonzero the point is not in the plane.</p>
<p>Maybe I am not understanding your question. If you can rephrase your question I may get it.</p>
| 48636 | Inside/outside test of any point on 2D plane? |
2021-12-07T08:49:42.380 | <p>The book I'm referring to says that a shaft with a circular cross section in pure torsion will have it's cross sections remain flat during the loading. That is, the cross sections won't deform, they will just rotate.</p>
<p>It then says that this is not the case for non-circular cross sections. In a non circular c/s shaft, the c/s distort and are not flat during the loading.</p>
<p>I wanted to know why is that so?</p>
| |mechanical-engineering|structural-engineering|structural-analysis| | <p>We say and write what we observe. In the history, when an axisymmetric cross sectional beam (i.e. the beam's cross section still remains the same when rotated about its longitudinal centeroidal axis) was subjected to torsion, the cross sections of the beam remained plane/flat. This is true for both, solid beam and hollow beam. However, when a non-axisymmteric cross sectional beam (i.e. the beam's cross section DOESN'T remain the same when rotated about its logitudinal centeroidal axis) was subjected to torsion, it was observed that the cross sections didn't remain plane/flat. Infact, they wrapped when subjected to twisting. The general equation for calculating stresses due to pure torsion for a solid circular beam is shown below:</p>
<p><a href="https://i.stack.imgur.com/Xr9wN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xr9wN.png" alt="enter image description here" /></a></p>
<p>where T: Torque applied, r: distance from centroid, J: Polar moment of inertia. Now, the derivation of this equation involves an assumption that the plane section must remain plane, i.e. the cross section cannot warp. Therefore, this equation cannot be applied to any other non-axisymmetric cross sectional beams.</p>
<p>Warping basically refers to one half of cross section (above/below the neutral axis) being subjected to compression, and other half to tension. This behavior is what we usually see in a beam subjected to bending, however, this behavior is also observed in non-axisymmetric cross sections subjected to torsion only. Calculating shear stresses for a non-circular cross section is somewhat more complex and complicated than the circular ones.</p>
| 48638 | Torsion - Circular vs Non-circular shafts |
2021-12-07T17:19:09.497 | <p>Observe the figure below.</p>
<p><a href="https://i.stack.imgur.com/BneWz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BneWz.png" alt="enter image description here" /></a></p>
<p>Left end is pinned while on the right there is a roller support. The system is statically determinate, i.e. the reaction forces can be found using simple 3 equlibrium equations of statics. Now, while using these equations, I would use the distances as it is before the system comes into equilibrium. However, after the system comes into equilibrium, the roller support has translated along the horizontal axis, because as it names implies, it can without any resistance. So the contact point between the beam and the roller support has changed and is different than what it was at the beginning. The distance of the roller support from the pinned support (or from the centroid of distributed force) is now different. So it means that the force reaction calculated at the roller support is not the same as we had already calculated using the initial distances.</p>
<p>In case I don't have access to FEA, how would I use analytical methods to properly predict the force reactions along vertical axis at pinned and roller support, after the system comes into equilibrium?</p>
<p><em>EDIT:</em> So according to some comments and answers here, it was discovered that the rollers are fixed to the ground, so they couldn't move at all. What if I don't fix them to the ground (and make them kinda like a tyre), what would be the process of finding the reaction forces at each support now?</p>
| |statics| | <p>Again, you are overthinking and making a simple matter confusing.</p>
<p>As the beam is merely in contact with the roller, in order to move the beam horizontally and cause the roller to displace, two conditions must coexist - a horizontal load and the system has no horizontal restraint, the system shown does not meet both of the necessary conditions. A reminder - we are talking about rigid body statics and general beam theory. The former does not deal with internal stresses, and the latter was developed for beams with small deformation only.</p>
| 48647 | Can the general statics theory equations be used to calculate force reactions for pinned-roller support system? |
2021-12-08T06:19:27.497 | <p>I am using a hand spindle press to inject liquid into a pressurized autoclave through a stainless steel pipe. I was wondering is there any possible way to retrieve all of the remaining liquid out of the pipe without depressurizing the autoclave?</p>
<p>Also, I cannot use the hand spindle press to withdraw the remaining liquid. It would be nice if you can suggest to me some of literature about this problem.</p>
| |mechanical-engineering|pressure| | <p>You need an isolation valve at the autoclave and a vent valve there as well on the supply side.</p>
<p>Shut the isolation valve and open the vent and the liquid could gravity drain. You could then use the vent during fill to avoid injecting air into the autoclave. Leve the vent open until liquid comes out of it, shut the vent, open the isolation and begin injection.</p>
| 48657 | How to withdraw liquid out of a injector in a pressurized autoclave? |
2021-12-08T16:56:04.047 | <p>As part of a course I am taking I am required to draw assembly process diagrams for common devices, such as electric motors, lamps, and mechanical presses, and I am trying to get the intuition for this. Does anyone know where you can find videos that show how common devices like these are assembled?</p>
| |mechanical-engineering|manufacturing-engineering| | <p>Solar Mike's idea is best. Many (mechanical) engineers got started by taking complicated things apart when they were kids, and insights into how things get put together is best learned by taking things apart.</p>
| 48661 | Assembly process for common devices |
2021-12-08T17:22:34.473 | <p>I'm trying to learn more about the type of sheets I can get for a project, I would of course go for polycarbonate if not for its price, currently the project with PC will cost \$600-\$700
but I'm hoping for something cheaper.</p>
<ul>
<li>It does <strong>not</strong> need to be transparent, will go for solid black (doors will be PC)</li>
<li>It does <strong>not</strong> need to be highly scratch resistant / impact resistant</li>
<li>It needs to <strong>Withstand</strong> high temp (130 °C~)</li>
<li>Fire retardant, or at most won't contribute to a fire</li>
</ul>
<p>It' going be a used as a few things (don't judge :) )
Mainly homemade vacuumed curing station with medium heat (30 °C - 70 °C)</p>
<p>Materials I already thought about but found them to be either expensive or not fire safe enough (correct me if i'm wrong)</p>
<ul>
<li>PC</li>
<li>Metal sheets - expensive</li>
<li>ACM - expensive (mineral version for fire rating)</li>
<li>ABS - Won't withstand fire</li>
<li>PVC - Won't withstand fire</li>
<li>Glass (This one actually tick all the boxes but it's so hard to handle)</li>
</ul>
<p>It checked for 4mm sheets, 2050 mm X 1250 mm</p>
<p>Recommendation for other materials to look into?</p>
| |heat-transfer|metals|plastic|material-science|metal-folding| | <p>In Reaction to fire tests Calcium Silicate boards or gypsum boards are often used as substrate; these boards are classified as A1 or A2 according to the european EN 13501-1 standard.
These boards are</p>
<ul>
<li>not transparent, most often white,</li>
<li>are not really scratch resistant</li>
<li>can withstand a really high temperature (200°C, 300°C, some CaSi boards can withstand a temperatures as high as 1100°C which is obv overkill here)</li>
<li>as they are classified A or A2 according to the EN 13501-1 they don't contribute to a fire.</li>
</ul>
<p>You can find them in all size and prices.
You can also look into cement boards.</p>
<p>In the long term the CaSi boards can be sensitive to high changes in temperature (think of a fire against the plate, then cold water against the plate)</p>
| 48663 | Fire rated EN 13501-1 B-s1/s2, d0 sheets |
2021-12-09T22:42:10.913 | <p>Let's take for example hydroelectricity produced at a dam and not consumed by any household or industry... Will the electricity need to be stored? What will happen if it is not stored? Does it flow like water and get wasted back to the earth? What happens to it?</p>
| |energy|energy-storage| | <p>The main problem is that all the terminology associated with electricity was formalized before anyone knew what electricity was. Early on, there were three competing theories about what electricity was. Each contributed terminology that only made sense in terms of it's own theory, and all three of them were horribly wrong!</p>
<p>So we have an entire branch of engineering where the words don't mean what they mean in the other branches of engineering. Actually, we have two branches where this is true. We were building airplanes for 30 years before anyone figured out why they flew, and aero engineering has much the same problem.</p>
<p>When you <em>generate</em> electricity, you aren't creating or transforming anything at all. All you are doing is pushing on electrons. Whether they move or not depends on whether they have a path to travel along that doesn't push back as much.</p>
<p>So you have to understand that you don't actually generate electricity in a generator, you don't store electricity in a battery, and you don't consume electricity in a motor, that's just the unfortunate terminology that English adopted. It fails completely when you attempt to make analogies using these terms. You have to swap out the vocabulary when you make physical analogies to other physical systems.</p>
| 48676 | What happens to generated electricity if it is not used? |
2021-12-10T05:56:13.443 | <p>I am working on a belt-driven linear motion mechanism using drawer slides (I've attached a small video that shows the <a href="https://drive.google.com/file/d/1PCWtR78gSWcHuININ4XaOmcvChY2_g_z/view?usp=sharing" rel="nofollow noreferrer">Mechanism</a> for my First Tech Challenge Team.</p>
<p><a href="https://i.stack.imgur.com/XHubh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XHubh.png" alt="enter image description here" /></a></p>
<p>I was now wondering how much torque the motor would need given the constraints I defined (in the PDF linked below). I've already attempted to calculate it using my basic 1<sup>st</sup> semester AP Physics knowledge and Google, however, I believe that the calculations I did would be for a lever/arm mechanism and not for this. I'm not taking into account friction or air resistance in this calculation to try to limit the complexity.</p>
<p><a href="https://drive.google.com/file/d/1CbsT1_-bWOVDfrv7i_U5sMmEcfmetyIk/view?usp=sharing" rel="nofollow noreferrer">Constraints and Calculations PDF</a></p>
| |mechanical-engineering|motors|torque| | <p>If I am not mistaken, you have one fixed base part, to which the torque creating motor is attached, and 3 moving parts of the mechanism (the three drawer slides). This mechanism has one degree of freedom. You seem to be ignoring the masses of the three slides, so I would ignore them too. It is not clear to me whether you are lifting the load vertically with this mechanism or horizontally (i.e. presence of gravity or no presence of gravity).</p>
<p>I tried to apply Lagrangian formalism to this model, and in particular D'Alembert's principle.</p>
<p>Observe that by construction of the mechanism, the belt on the fixed base part is the one driven by the motor, so when in motion, we can introduce displacement <span class="math-container">$x$</span> of the said belt. That displacement displaces the first slide at distance <span class="math-container">$x$</span> from its initial position, the second slide is displaced at distance <span class="math-container">$2x$</span> from its initial position, and the third slide is displaced at distance <span class="math-container">$3x$</span> from its initial position. Since the load is on the third slide, the displacement of the load is <span class="math-container">$x_{load} = 3x$</span>.</p>
<p>It seems to me that, since the slides are massless, the Lagrangian of this system should be
<span class="math-container">$$L \, =\, \frac{m}{2} \left(\frac{dx_{load}}{dt}\right)^2 \, -\, m\, g\, x_{load}$$</span> (simply the kinetic minus the potential energy of the load, as the load is the only object with mass). Now, using the displacement <span class="math-container">$x$</span> as a generalized coordinate, the Lagrangian becomes
<span class="math-container">$$L \, =\, \frac{m}{2} \left(3\, \frac{dx}{dt}\right)^2 \, -\, m\, g\, \big(3\,x\big) \, =\, \frac{9\,m}{2} \left( \frac{dx}{dt}\right)^2 \, -\,3\, m\, g\, x$$</span></p>
<p>Since the torque <span class="math-container">$\tau$</span> in question generates a linear force <span class="math-container">$F$</span> along the belt on the fixed base part, the said force causes virtual displacement <span class="math-container">$\delta x$</span> of the belt. D'Alembert's principle then implies
<span class="math-container">$$\left(\frac{\delta}{\delta x} L\right) \delta x \,= \, F\delta x$$</span> which holds if and only if
<span class="math-container">$$\frac{\delta}{\delta x} L \,= \, F$$</span>
From Lagrangian mechanics,
<span class="math-container">$$\frac{\delta}{\delta x} L \, =\, \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \, -\, \frac{\partial L}{\partial {x}} \, = \, F$$</span>
(<span class="math-container">$\dot{x} = \frac{dx}{dt}$</span>) For this model, the equation of motion is therefore
<span class="math-container">$$ {9\,m} \frac{d^2x}{dt^2}\, + \,3\, m\, g \, =\, F$$</span>
Now, the linear force <span class="math-container">$F$</span> is related to the torque by the relation <span class="math-container">$$\tau \, =\, r \, F$$</span> Thus, finally we get
<span class="math-container">$$ {9\,m} \frac{d^2x}{dt^2}\, + \,3\, m\, g \, =\, \frac{\tau}{r}$$</span> or alternatively
<span class="math-container">$$ \frac{d^2x}{dt^2}\, \, =\, \frac{\tau}{9\,m\,r} \, -\, \,\frac{g}{3}$$</span>
Assuming that the torque <span class="math-container">$\tau$</span> is constant for half the time of the full extension of the mechanism, we can integrate the latter equation twice, obtaining the law of motion
<span class="math-container">$$x \, =\, \frac{t^2}{2}\left(\frac{\tau}{9\,m\,r} \, -\, \,\frac{g}{3}\right)$$</span>
Hence, the actual law of motion of the load is
<span class="math-container">$$x_{load} \, =\, \frac{t^2}{2}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$</span>
If the full extension of the mechanism is <span class="math-container">$x_{load} \,=\, D$</span> and the time to achieve it is <span class="math-container">$T$</span>, then one simple scenario is uniformly accelerating, applying constant toque <span class="math-container">$\tau$</span> in one direction , for half of the total time <span class="math-container">$T$</span>, then uniformly decelerating, applying constant toque <span class="math-container">$\tau$</span> in the other direction, for the remaining half of the total time <span class="math-container">$T$</span>. This means that we want when <span class="math-container">$t = \frac{T}{2}$</span> the displacement to be <span class="math-container">$x_{load} = \frac{D}{2}$</span>. So plug it
<span class="math-container">$$\frac{D}{2} \, =\, \frac{(T/2)^2}{2}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$</span>
<span class="math-container">$${D} \, =\, \frac{T^2}{4}\left(\frac{\tau}{3\,m\,r} \, -\, \,{g}\right)$$</span>
and if you solve for <span class="math-container">$\tau$</span> you get
<span class="math-container">$$\tau \,=\, \frac{12\, m\, r\,D}{T^2} \, + \, 3\, m\, r\,g$$</span>
So maybe you would want a torque at leas greater than this, i.e.
<span class="math-container">$$\tau \,\geq\, \frac{12\, m\, r\,D}{T^2} \, + \, 3\, m\, r\,g$$</span>
If gravity doesn't play a role, just turn it of by setting <span class="math-container">$g=0$</span>
<span class="math-container">$$\tau \,\geq\, \frac{12\, m\, r\,D}{T^2} $$</span></p>
| 48678 | Calculating Motor Torque needed for a belt driven linear motion mechanism? |
2021-12-10T10:03:51.893 | <p>I performed some tests with vibration isolators and have computed the PSD. The natural frequency of the isolators is 110 Hz in the axial direction. So my question is, should I expect to see noise damped below or above 110 Hz? Also should the natural frequency of the system be lower than or higher than the natural frequency of the isolator?</p>
| |vibration| | <p>This is not straightforward to answer IMHO. Depending on the mounting and the source of noise you might get different results.</p>
<p>In the most simple case, i.e. if the system can be modelled a sdof system resting on a single isolator, then you'd expect that excitation frequencies higher/greater than the natural frequency would be isolated (in this example 110 Hz). The higher the frequency the higher the isolation.</p>
<p>In that scenario you can see the following graph of transmissibility which shows the transmissibility for different damping ratios.</p>
<p><a href="https://i.stack.imgur.com/TUlnG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TUlnG.png" alt="enter image description here" /></a></p>
| 48680 | How to analyse results of vibration isolation test |
2021-12-10T11:05:33.347 | <p>I was carrying out a problem in which I have a fully actuated two link manipulator, so with an independent servo motor for each link that allows rotation in one direction and another (positive for counterclockwise rotation and negative for clockwise rotation).</p>
<p>Let's assume we start from a certain initial condition and arrive at a final condition through a feedback control of the torques applied to each motor. To do this, a counter-clockwise torque is first applied to each servo, then a clockwise torque to "brake" (no dissipation) the system.</p>
<p><a href="https://i.stack.imgur.com/LhQeD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LhQeD.png" alt="enter image description here" /></a></p>
<p>Assuming a time interval of 10 seconds, at each sampling instant (suppose it is 0.1 seconds) I have a different torque and / or angular velocity value (suppose we always have a positive angular velocity). Consequently, by projecting the <span class="math-container">$P = \tau \omega$</span> products, I obtain instant by instant the value of power generated by each motor.</p>
<p>Now, I have two doubts. The first one concerns the power value. Is it correct to obtain a negative power value, obtained by possibly multiplying a negative torque value for the angular velocity? Or should i consider the torque absolute value?
And then, is it correct to calculate the total energy value as the integral between 0 and 10 seconds of the power curve?</p>
<p>Of course I know that the model is very approximate and I have not even presented the dynamics and kinematics of the manipulator in question, but I would simply like to have a clear idea of the concept of power and energy.</p>
| |mechanical-engineering|applied-mechanics|energy| | <p>Just some general notes.
In general when one deals with dynamics problems vectors and their products can be quite a burden and a hassle.</p>
<p>My advice is clear up the notions of cross <span class="math-container">$\times$</span> and dot product <span class="math-container">$\cdot$</span> for vectors, and in early problem try to use summation of the moments which are calculated by vector form (this is especially true for 3d problems, although starting from simpler 2d like this one is the same).</p>
<hr />
<p>In your particular example, <span class="math-container">$P = \vec{T}\cdot \vec\omega$</span> is a simple dot product so the only thing that matters about the sign is that they have the same direction.</p>
<p>When the Torque and the angular velocity are in the same direction then the member will rotationally accelerate, while if one is opposite to the other then the <em>magnitude</em> of the angular velocity will decrease.</p>
<p>So depending of how your convention is the sign of power will indicate if energy is added to the system or not.</p>
<hr />
<h2>Alternative approach (work and energy)</h2>
<p>I can't help noticing that in this particular problem, instead of integrating torque and angular velocity, it might be easier to use work and energy principle. I.e. you can easily write the Langrangian of this system and calculate the total energy of the system based on the position, and from that derive the energy at different configurations. Then based on the time interval required to move from one to the other configuration you can estimate the total energy required.</p>
| 48682 | Energy consumed by a two-link planar manipulator with feedback control |
2021-12-11T15:44:02.097 | <p>Given the height and / or weight of a cut tree, what is the best way to calculate the width of stand required to keep it upright?</p>
<p>I have no engineering background, but have been trying to think of a way to work it out. I think the key elements are center of gravity, and I also need to decide what amount of force I am expecting (it will be indoor, so only accidental knocking) to counter.</p>
<p>The tree in question is a 15ft spruce, however if there is a ratio or method I can use to calculate stand width I can do the rest myself.</p>
| |centre-of-gravity| | <h3>This needs to be solved empirically.</h3>
<p>There isn't going to be a specification for "how out of balance is okay." The issue isn't the tree's center of gravity being off, it's really how stable the tree needs to be to support people knocking into it. Things like this are best solved by doing, which, as mentioned in the question comments, have already been solved by current manufacturers.</p>
| 48698 | How can I calculate the width of stand required for a Christmas tree based on its height? |
2021-12-12T13:29:18.477 | <p>I would like to know if a battery-powered electric vehicle such as a Tesla Model 3 automobile can be driven through a dust storm.</p>
<p>Back during the Dust Bowl days of the 1930s, driving through a dust storm could damage electrical equipment inside a car.</p>
<p><em>"...So much static electricity built up between the ground and airborne dust that blue flames leapt from barbed wire fences and well-wishers shaking hands could generate a spark so powerful it could knock them to the ground. Since static electricity could short out engines and car radios, motorists driving through dust storms dragged chains from the back of their automobiles to ground their cars..."</em> <a href="https://www.history.com/news/10-things-you-may-not-know-about-the-dust-bowl" rel="nofollow noreferrer">https://www.history.com/news/10-things-you-may-not-know-about-the-dust-bowl</a></p>
<p>Can a battery-powered electric vehicle be driven through a dust storm?</p>
| |mechanical-engineering|electrical-engineering|automotive-engineering|electrical|battery| | <p>I know of cars with the early electronic ignitions would misfire and cut out close to an army communications station.</p>
<p>After some of the updates to the early electronic ignition systems they stopped failing like that in that location.</p>
<p>So, given the electronics has moved on a <strong>lot</strong> since the late '70s then I believe that the electronic systems would be more stable. But knowing what the exact test regimes are now completed by the manufacturers would clarify the situation.</p>
| 48706 | Can a battery-powered electric vehicle be driven through a dust storm? |
2021-12-12T17:03:53.893 | <p>I've got a Bosch PSB 50 drill, which I'm using to drill through some apparently really tough wood. I don't know why, The drill seems to be struggling, and at times comes to a complete stop. I'm not sure what exactly is happening, maybe the drill isn't powerful enough, maybe there's a piece of metal inside, or perhaps the drill bit is full of junk. Whatever the case, the drill is struggling, but it is getting through, bit by bit. The motors getting awfully hot though, and I think this is a brushed motor, so now my question: Can it be damaged by this, ie. should I get a stronger drill, or can I attempt to brute force it until I get through?</p>
| |motors|torque|drilling|heat| | <p>The very first thing to check is the sharpness of the bit tip. a dull bit will not drill worth a d*mn and just waste your battery power. Note that it only takes one brief encounter with something hard to wreck the tip of a drill bit. Bits are cheap compared to your time- go out now and buy a sharp one, or get your bit resharpened!</p>
<p>Cordless electric drills from quality manufacturers like Bosch and Makita have a temperature sensor inside them which is supposed to automatically shut down the drill when the motor gets too hot.</p>
| 48711 | Damage to drill through torque? |
2021-12-13T04:44:52.127 | <p>I am looking for some guidance on a redneck cooling system.</p>
<p>I need to store seeding trees at a temperature just above freezing, and at a humidity of close to saturation.</p>
<p>I only need the system for about 2 months of the year.</p>
<p>At present I use snow.</p>
<ul>
<li>Let the room chill all winter so frost is deep in the ground.</li>
<li>Add 2 feet of snow at the end of winter.</li>
<li>Use a cheap box fan to circulate air, blowing it against the snow face.</li>
</ul>
<p>My limited understanding of heat exchangers is that they operate at a colder temperature that the set point of what they are cooling.</p>
<p>I think that a conventional air cooler would have constant problems with ice forming on the heat exchanger, since even a few degrees colder than the room will be below freezing. I inquired with a company called 'Cool Bot' that has a controller that uses a conventional air conditioner to cool a well insulated room. They agreed that my application would not be suitable due to ice formation.</p>
<p>The second thought is to use a freezer. Chill brine or antifreeze to some cold temperature, and run it through flexible pipe laid directly on the floor. If the floor had a liner so that in effect it was a 3" deep pond, then the seedling boxes can rest on pallets above the pond with the same fan circulating air against the pond.</p>
<p>I don't think a freezer moves enough BTUs for this. Substitute insulated tank + chilling unit. (Old electric water heater for the tank?). Propane tank. (means no salt.)</p>
<p>Now the problem becomes one of control.</p>
<p>My thought is to use a chest freezer outside, verify that it is water proof, and fill it with a solution that won't freeze at a given setting of the freezer. I think I would aim at about -10 to -15 °C. Fill the freezer with the solution. (Brine ok? Freezers inside are made of plastic.)</p>
<p>Run a coolant loop that has some large number of feet of tubing in the freezer with holes drilled in the lid for pass through. (0.700 ID x 1/16" wall thickness black polyethylene plastic is commonly used for irrigation, and is rated at 30 psi. 0.625 water line is rated at 75 psi, but wall thickness is twice the size, which would increase the amount of pipe needed for the same thermal transfer.</p>
<p>Addressing comments:</p>
<h3>Precision of temp control</h3>
<p>0.5C is a goal. But I have an electronic relay that registers to tenth of a degree, and calibrating it by putting it in a jar in an ice water bath is straight forward.</p>
<p>Trees are normally stored over winter an between -1 and -2 C. This is the temp that maximizes their survival, and leaves the most amount of sugars intact for spring growth. When I get them they have been out of the freezer long enough to start to thaw. Refreezing them is a bad idea. So you can rephase the goal as "As close to but not below freezing as possible."</p>
<h3>Similarity to Walk in Cooler</h3>
<p>There are two key differences: Most walk in coolers do not make an attempt to keep humidity at close to 100%. And most coolers do not attempt to keep the temperatures very close to freezing.</p>
<p>The other issue is one of cost. A 500 square foot walk in cooler not only costs north of a couple hundred grand, but it requires more power to operate than my entire farm electrical service (7 KVA) This for a system that is only used a few weeks a year. This is why this is "redneck engineering".</p>
<p>This is why the extreme thickness walls for insulation. This could also be done with a foot of styrofoam, but straw bales are about $25 each. A 24x24 internal space building only requires 42 bales, and they are the support walls too. There is a whole building literature about strawbale construciton.</p>
<h3>Tree storage temp.</h3>
<p>A tree in the wild routinely gets roots far colder than 0 C. I am not an expert on the physiology, but the trade practice is just below freezing. Dormancy is not a switch that is turned on and off. I suspect (but don't know) that dormancy requires a set of cues of light/day length/temperature to trigger. Since trees are harvested usually in late fall, storing them just below freezing balances them not getting the usual sequence of cues. Most growers do not grow entire freezer fulls of a single species harvested at a particular time, so this may be a general compromise.</p>
<p>I do know that respiration rate (metabolism) of trees in cold storage increases very rapidly with temperature. A day at 5 C can use more stored energy than a week at 1 C.</p>
<h3>Moding a conventional cooler.</h3>
<p>See comment above about costs.</p>
<h3>Strawbale heating</h3>
<p>Do NOT get straw bale mixed with hay bale. Much different critters. Hay is mowed green. allowed to dry some, then baled. If baled too wet they can spontaneously combust. If baled at moderate temps, they can heat up some.</p>
<p>The process isn't really ripening. It's bacteria taking advantage of a good culture medium, and generating heat as a byproduct.</p>
<p>Straw has much lower nitrogen content (protein) and there is no rush to bale it quickly. So it tends to be baled very dry. Respiration levels are low. I've yet to hear of a spontaneous straw bale fire.</p>
<p>Questions:</p>
<ul>
<li>Is this a workable solution?</li>
<li>What is the unit I need called?</li>
<li>How do I size it?</li>
<li>Pointers to calculating the amount of heat exchange pipe I need</li>
<li>What is the best way to control it so that I don't freeze my trees by mistake, nor allow them to get well above freezing?</li>
</ul>
<p>In passing: Building details:</p>
<p>24 x 24 interior dimensions. 32x32 exterior. Roof dimension 40x40 Proposed new cold room will be built out of 3x4x8 foot straw bales, laid brick style, with the 4' dimension being the width of the wall, and using several inch layer of lichtlehm (light clay) between bales to adjust for bale irregularities. Bales run about R1.5 to R2 per inch, so the wall insulation value is R72 to R96 depending on whose lies you believe.</p>
<p>4" vertical styrofoam to reduce underground thermal transfer. (Cut a trench as deep as reasonable with a ditch witch). Usual strawbale construction footings.</p>
<p>Roof sized for 2 feet of blown cellulose insulation. Roof has large overhangs (4 feet) to protect the walls from rain.</p>
| |hvac|heat-exchanger|refrigeration| | <p>At near-freezing temperatures, the amount of water vapor in the air to get 95% relative humidity is tiny, this is no problem.</p>
<p>I recommend filling 55 gallon poly drums on pallets with water, park them outdoors to get them near freezing, then move them into your storage unit with a pallet lift. Put the plants in the center and surround them with the drums. Use a small electric fan to keep the air moving very slightly inside the structure.</p>
<p>The cold water forms a tremendous "thermal flywheel" that releases heat as the temperature outside rises and soaks it up as it falls. with the huge amount of insulation you have, this should work passively except for the fan which does not need to be big, and could run off a solar cell array (solar fans are are used in motorhomes and trailers).</p>
<p>Experiment will tell you how many drums of water will be needed. Start with 20 and leave room for 10 or 20 more.</p>
| 48721 | Cooling system to maintain 0.5 °C at 95% humidity |
2021-12-13T16:00:58.957 | <p>I am trying to find the velocity of an impact test.</p>
<p>I am using a drop hammer impact machine to hit some composite specimens.</p>
<p>I know the equation for the velocity is V = V<sub>0</sub> + at ——————- (1)</p>
<p>And that V<sub>0</sub> = V<sub>f</sub> - (at) ——————(2)</p>
<p>And V<sub>f</sub> = <span class="math-container">$\sqrt{2gh}$</span> ————- (3)</p>
<p>Where <strong>V<sub>0</sub></strong> is the initial velocity, <strong>a</strong> is acceleration, <strong>t</strong> is time and <strong>V<sub>f</sub></strong> is the final velocity.</p>
<p>I do not have the height <strong>h</strong> between the impactor and the impact specimen so I would need to rearrange equation (3) to get this, however I do not know the final velocity so how would I find the height <strong>h</strong> ?</p>
<p>I assumed that the initial velocity is 0 but then I seen equation (2) and this confused me</p>
| |mechanical-engineering| | <p>How was the "test" setup? What measurement have you taken"?</p>
<p>Let's connect the dots:</p>
<p><span class="math-container">$E = m(V_f^2 - V_0^2)/2 = mg\Delta h$</span>; and</p>
<p><span class="math-container">$V_f = V_0 + at$</span></p>
<p>For free-fall (implied by <span class="math-container">$mg\Delta h$</span>), <span class="math-container">$V_0 = 0$</span>; <span class="math-container">$\Delta h$</span> = distance travelled = <span class="math-container">$h$</span>, so</p>
<p><span class="math-container">$V_f = \sqrt{2gh} = at$</span>, again, for free fall, <span class="math-container">$a = g$</span></p>
<p>Solving for <span class="math-container">$h$</span> as a function of <span class="math-container">$t$</span>, we got:</p>
<p><span class="math-container">$h = gt^2/2$</span></p>
<p>So, if you have timed the fall, you can get both the final velocity and the distance the object has traveled.</p>
| 48731 | Finding the velocity of an impact test |
2021-12-15T01:58:10.120 | <p><a href="https://i.stack.imgur.com/EaTbb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EaTbb.jpg" alt="enter image description here" /></a></p>
<p>I have the above situation with a Warren truss and a UDL acting along the bottom of the truss. The UDL is <span class="math-container">$60 kN$</span> per <span class="math-container">$3.75$</span> metres. The distance between the joints along the bottom is <span class="math-container">$3.75$</span> m. I am looking to determine the forces in each member of the truss. In my engineering class I have been told that an easier way of dealing with these UDLs is to find the total force acting on the truss and then make that force a point load acting directly in the middle of the truss. But does this method have repercussions or make assumptions? I would imagine that having 4 forces at each joint along the bottom would give a different result? Or do both methods give the same results for the forces all the members?</p>
| |structural-analysis| | <p>Yes, you are right.</p>
<p>We can not replace the four forces with only one equal to the sum of four, at the center.</p>
<p>The more joints a truss has with the same overall span and same UDL the less max moment or max stresses will be.</p>
<p>Let's compare it to a simply supported beam with a length L and UDL of <span class="math-container">$P/L$</span>.</p>
<p>If we handle this beam by applying UDL along the length, L, we get:</p>
<p><span class="math-container">$P_{total}= \frac{P}{L}L=P$</span></p>
<p><span class="math-container">$M= \frac{\frac{P}{L}L^2}{8} =\frac{PL}{8}$</span></p>
<p>But if we apply the sum of the UDL at the middle of the beam we get:</p>
<p><span class="math-container">$M=\frac{PL}{4}$</span></p>
<p>So by applying a concentrated equal load we get double the correct M.</p>
| 48754 | UDL vs Point Load on Truss |
2021-12-15T06:54:21.010 | <p>In the sensor catalogs, I cannot find this field. They only depict the frequency response range. For example, Bently Nevada's sensors. How we can find out their maximum sampling rate?</p>
| |sensors|frequency-response| | <p>For an analog accelerometer, you can try to sample as often as you'd like. It's up to the data acquisition card, if it can catch up. Of course you probably won't get useful data.</p>
<p>For digital accelerometers the documentation should provide the maximum sampling frequency.</p>
<hr />
<p>The maximum <strong>useful</strong> frequency is another matter. The following diagram shows the response of an accelerometer and the useful ranges (in log scale).</p>
<p><a href="https://i.stack.imgur.com/Ijjgy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ijjgy.png" alt="enter image description here" /></a></p>
<p><strong>figure: Frequency range of accelerometer (source: <a href="https://www.researchgate.net/publication/304094143_Brief_Review_of_Vibration_Based_Machine_Condition_Monitoring/figures?lo=1" rel="nofollow noreferrer">Dubravko Miljković</a>)</strong></p>
<p>Essentially you are trying to be on the flat portion of the sensitivity.</p>
| 48757 | How to find out the sampling rate of an acceleration sensor? |
2021-12-15T10:37:59.737 | <p>Consider a shaft which is fixed at one of the ends and is acted upon by a torque at the free end as shown.</p>
<p><a href="https://i.stack.imgur.com/5vUKY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5vUKY.jpg" alt="enter image description here" /></a>
We take a plane inside the shaft as shown in the figure before the torque is applied. <strong>I'm interested in knowing how the plane will look after the torque has been applied.</strong></p>
<p><a href="https://i.stack.imgur.com/BKfqU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BKfqU.jpg" alt="enter image description here" /></a></p>
<p>My intuition tells me that the plane will be bent (for the lack of a better word) after the loading as shown below. (</p>
<p><a href="https://i.stack.imgur.com/9xfTh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9xfTh.jpg" alt="enter image description here" /></a>
However if that is the case, the radial lines will curve at the end of loading and one of the major assumptions that the book I'm referring to for studying torsion in circular shafts, is that the radial lines must remain straight at all times.</p>
<p>Conversely, if the radial lines are to remain straight at all times, that would mean the plane after the loading doesn't bend and remains flat. However, is this possible physically, provided that the two edges (shown in red below) are fixed.</p>
<p><a href="https://i.stack.imgur.com/VUQD5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VUQD5.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|structural-analysis|stresses| | <p>One easy way of imagining the deformation of the plane you show in your sketch is to cut the shaft into infinitesimally thin disks with thickness, dL, and imagine these disks look like a clock with only an hour hand at 12oc.
Then rotate each disk by the amount of.</p>
<p><span class="math-container">$$ \theta = \frac{32dL T}{ (\pi G D^4)}$$</span></p>
<ul>
<li>T= torque</li>
<li><span class="math-container">$\theta$</span> = angle of rotation</li>
<li>D = dimeter of shaft</li>
<li>G = shear modulus of regidity</li>
<li>π D^4 / 32 = J, polar moment of inertia of shaft</li>
</ul>
<p>The hour hand now traces a curve and the plane in your sketch is now the plane generated by this curve and the central axis of the shaft.</p>
<p>The Important thing is <span class="math-container">$the\ thin\ disks $</span> remain <span class="math-container">$plane $</span> after deformation according to <span class="math-container">$St \ Venant.$</span></p>
<p>This is true only for small-angle deformations though.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/1Gld2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Gld2.png" alt="torque" /></a></p>
| 48759 | Deformation of a plane in Torsion of a Circular Shaft |
2021-12-15T11:42:24.020 | <p>I know that the equation for displacement is:</p>
<p>S = V<sub>0</sub> t + <span class="math-container">$\frac{1}{2}$</span>at<span class="math-container">$^{2}$</span></p>
<p>With S being displacement in metres, t is the time is seconds, V<sub>0</sub> is the initial velocity in ms<span class="math-container">$^{-1}$</span> and a is the acceleration in<br />
ms<span class="math-container">$^{-2}$</span>.</p>
<p>However, if the time used in the equation is milliseconds then what would be the unit of the displacement S?</p>
| |mechanical-engineering| | <p>The units of S (as was asked in the query) is controlled by what distance units are used in the velocity and acceleration (feet, meter, miles, etc.); not the time. Simply use one millisecond as .001 seconds and all will come out right. For example, for 5 millisecond time plug .005 for time into the equation.</p>
| 48761 | Converting displacement to correct units |
2021-12-15T16:05:17.563 | <p>How can I merge two input into one? (gear mechanism)
There are a tube, a shaft, and a gear.
The shaft is in the tube and both of them rotates in same direction.
And I have to deliver the power to the gear at the bottom.
How can I merge two input and deliver it to one gear without interrupting one's rotation?
There would be no reverse rotation of gear.</p>
<p><a href="https://i.stack.imgur.com/7vf7d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7vf7d.png" alt=" " /></a></p>
| |gears| | <p>you can use a mechanism like a car differential, connecting the tube and shaft to the sides but in a reverse sense. And the output is where it is in a car is the engine shaft.<a href="https://i.stack.imgur.com/83lWH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/83lWH.jpg" alt="diff" /></a></p>
| 48764 | How can I merge two input into one? (gear mechanism) |
2021-12-16T11:18:17.527 | <p><a href="https://i.stack.imgur.com/jJCVZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jJCVZ.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Zsg7A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zsg7A.png" alt="enter image description here" /></a></p>
<p>I saw this analysis online but I have noticed contradicting values from this method of analysis. The drawing of the simply supported beam comes from the crane structure above. All we desire is the reaction forces as well as the external forces being experienced at each of the three points which represent the joints of part of the truss. The person has placed <span class="math-container">$150N$</span> acting in the middle of the beam creating a moment of <span class="math-container">$22500 Nmm$</span> and creating reactions of <span class="math-container">$75N$</span> at each support.</p>
<p>But I was told on one of my previous questions that this is innorect due to it being part of a truss and not just a beam. Is the correct way to split up the <span class="math-container">$150N$</span> into <span class="math-container">$50N$</span> acting upon points <span class="math-container">$P_1$</span> <span class="math-container">$P_2$</span> and <span class="math-container">$P_3$</span> which would result in a total moment of <span class="math-container">$30000 Nmm$</span> and reactions of <span class="math-container">$100N$</span> on the right end and <span class="math-container">$50N$</span> on the left end? Which method is correct in this situation?</p>
| |structural-analysis| | <p>I agree with the answers given with the exception that the moment of the concentrated load of 150N should not be applied at a distance of 150mm. it should be broken into two loads of 75N and each applied respectively at 100mm and 200mm.</p>
<p>As far as the stress in members we also should work the horizontal component of the cable tension as an axial force to all three horizontal members.</p>
<p>So the members are under combined moment and normal stresses.</p>
| 48771 | Contradicting Results of a Truss Structure |
2021-12-16T20:07:19.660 | <p>Consider a close coiled helical spring which is acted upon by a torque about its longitudinal axis as shown. <strong>I'm interested in knowing the effect of this torque on the spring wire.</strong></p>
<p><a href="https://i.stack.imgur.com/suKzMm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/suKzMm.jpg" alt="Source" /></a></p>
<p>Here's my take: I made an imaginary cut via a plane (shown in red) to expose the cross section of the wire as below.</p>
<p><a href="https://i.stack.imgur.com/VhLIB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VhLIB.jpg" alt="enter image description here" /></a></p>
<p>The upper part of the spring (unshaded) has to remain in equilibrium, hence the torque from the support (ceiling) about the longitudinal axis has to be balanced by another torque developed about the longitudinal axis. This would happen if on the cross section internal resistive forces are developed in such a way that they cancel out producing no net force but producing a net moment about the point A. This moment of the internal resistive forces developed on the c/s about A, will balance the torque T from the upper end and bring the upper part of the spring in equilibrium.</p>
<p>In effect of the arguments above, the spring wire should experience bending. I do know that the correct answer to this question is bending, but is my reasoning correct?</p>
| |mechanical-engineering|structural-engineering|stresses|structures| | <p>The longitudinal torque causes a pure moment in the spring the same way as you show, except the neutral axis is not at the center of the coil it is near the vertical diagonal of the section. it is closer to the inside. See the figure.</p>
<p>The moment will be the same along the length of the coil all the way to the top, then it will affect a torque on the top vertical stub, which has to be canceled by the support at the ceiling.</p>
<p>This moment on the coil will tend to unwind it or wind it depending on the direction of the torque. If the torque is in the same direction as the coil it will wind the coil tighter and make it shorter. Otherwise, it unwinds the coil and makes it wider and longer.</p>
<p><a href="https://nptel.ac.in/content/storage2/courses/105106049/lecnotes/mainch10.html" rel="nofollow noreferrer">source</a></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/osDyU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/osDyU.jpg" alt="coil under pure moment" /></a></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/3P1K8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3P1K8.png" alt="stress in round section" /></a></p>
| 48779 | A close coiled helical spring with torque acting about longitudinal axis |
2021-12-17T01:04:17.317 | <p>I am designing a roll to roll web drive system where having smooth consistent very slow speed is important to quality of the product. I will be using a clearpath servo motor with a gearbox of 20:1 ratio.</p>
<p>It has been recommended to me that I might consider a Hypoid gear reduction for the smoothness characteristics but I cannot find any papers or specifications relating to smoothness of different servo grade gearbox types, nor do I know what this spec might be called.</p>
<p>If someone with more experience could tell me the words I might use to look for information on this that would be really appreciated.</p>
| |mechanical-engineering|gears|servo|nomenclature| | <p>The conclusion I've come to is to look at a spec called angular transmission accuracy, in conjunction with the backlash, as long as one is lower than my required minimum unexpected positional jump I believe I should be good.</p>
| 48785 | Looking for data or nomenclature that partains to "Smoothness/accuracy" of servo grade gearboxs for roll to roll application |
2021-12-17T16:22:40.000 | <p>According to a Meta Post I found this is the right place to ask this kind of question. I apparently can't find the post anymore but it pointed towards this subsite.</p>
<p>Yesterday I had an argument (for almost an hour) over whether in the end of Season 1 Chapter 2 of the Walking Dead game, this character called Andy was holding a rifle or a shotgun.</p>
<p>To show it there's two screenshots on imgur and a walkthrough video which is set to the specific time frame where the gun gets shown.</p>
<p><a href="https://i.stack.imgur.com/JFmXu.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/JFmXu.jpg</a></p>
<p><a href="https://i.stack.imgur.com/kM3Sb.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/kM3Sb.jpg</a></p>
<p><a href="https://youtu.be/_zjJqegIG8c?t=7456" rel="nofollow noreferrer">https://youtu.be/_zjJqegIG8c?t=7456</a></p>
<p>Here are my arguments that show that it's clearly a rifle:</p>
<ul>
<li>There's only one barrel</li>
<li>There's not enough place for slugs and loading extra ones</li>
<li>The "carving-in" is for holding the rifle</li>
<li>The understock looks like a rifle's</li>
<li>The muzzle and sound look and sound like a rifle's</li>
<li>There's a loading chamber for a rifle with the pulling thingy</li>
</ul>
<p>The arguments for a shotgun:</p>
<ul>
<li>Supposedly it has a shotgun loading mechanism</li>
<li>There is supposedly a compartment for the slugs, even though it's so thin</li>
<li>Sound doesn't matter, because the game designers made it so</li>
</ul>
<p>We agreed upon the fact it's a rather old gun and definitely not a modern one.</p>
<p>My questions would be:</p>
<ul>
<li>So which one is it?</li>
<li>What kind of model is it (if it's not fictional?)</li>
</ul>
| |specifications|firearms| | <h3>It's neither or both or whatever you want</h3>
<p>Evidence for rifle: It's drawn like a bolt action rifle with a small magazine</p>
<p>Evidence for Shotgun: The barrel is large like a shotgun</p>
<p>Evidence for neither: It's handled like a semi-automatic rifle (action wasn't cycled after the shot); there isn't room to eject a shell from the action; and the bolt cycle wouldn't be enough to put an actual shell into the action either manually or through the magazine.</p>
<p>This is a made-up weapon made up by people for whom accuracy wasn't a concern.</p>
| 48794 | Is this a rifle or a shotgun? |
2021-12-17T23:04:20.523 | <p>What is the optimum distance between a rack and centerline of a pinion? In the image below I have two racks opposite each other and the teeth surfaces are tangent to the faces. The center pinion is a 20 tooth 2 module gear. Is there some kind of rule of thumb or standard for this or do I actually mate the parts together like so?</p>
<p><a href="https://i.stack.imgur.com/zRH3T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zRH3T.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/cLplW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cLplW.png" alt="enter image description here" /></a></p>
| |gears| | <p>Part of the work involved in designing gearing includes a number of reference values.</p>
<p><a href="https://i.stack.imgur.com/KskMb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KskMb.jpg" alt="Working Gear Nomenclature" /></a></p>
<p>The image above is from <a href="https://khkgears.net/new/gear_knowledge/abcs_of_gears-b/basic_gear_terminology_calculation.html" rel="nofollow noreferrer">khkgears.net</a> and shows a number of useful references. The site also covers a number of aspects of creating gears.</p>
<p>Note in the image that the reference diameter for both intermeshing gears is also the point of contact of the teeth of the gear.</p>
<p>Formulae are provided on the linked site to enable the calculations of these values. This also applies to rack and pinion gearing as shown in your example. The site also provides additional links with useful guidelines and information, including <a href="https://khkgears.net/new/gear_knowledge/gear_technical_reference/calculation_gear_dimensions.html" rel="nofollow noreferrer">spur gear/rack combination</a>.</p>
| 48802 | How to determine optimal distance of rack and pinion? |
2021-12-19T17:00:38.700 | <p>Can a car ac compressor be used as a Hydraulic pump ?
Does it require some modifications to do so or is it impossible to begin with and the compressor can't handle the oil?</p>
| |hydraulics|hvac| | <ul>
<li>Air compressors work by <strong>compressing</strong> air, a <em><strong>gas</strong></em>.</li>
<li>Hydraulic pumps work by <strong>pumping</strong> hydraulic fluid, a <em><strong>liquid</strong></em>, which is not compressible.</li>
</ul>
<p>Attempting to run hydraulic fluid through an air compressor will most likely burst the compressor.</p>
| 48828 | Can an AC compressor used as a Hydraulic pump? |
2021-12-19T17:50:58.340 | <p>Standard rod used for CNC guides, outer diameter: 12mm.
Material: Ck55 steel, hardened to 60 HRC (according to store)</p>
<p>What would be the maximum length I can have before the rod starts to bend?</p>
<p>The guide will only need to hold about 2kg (a cordless router).</p>
| |steel|cnc|bending| | <p>Assuming a cantilever beam with a concentrated load at the end like the following:</p>
<p><a href="https://i.stack.imgur.com/wOsDx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wOsDx.png" alt="enter image description here" /></a></p>
<p>In that case, the beam will start to bend for any length L ; for small Ls it will be imperceptible.</p>
<p>In any case, the maximum deflection will be:</p>
<p><span class="math-container">$$\delta_{max} = \frac{P\cdot L^3}{3 \cdot E\cdot I}$$</span></p>
<p>where:</p>
<ul>
<li>E = Youngs Modulus (200 GPa)</li>
<li>I = Second Moment of Area <span class="math-container">$I = \pi \frac{D^4}{64}$</span></li>
<li>P = Point Load (~ 20 N)</li>
<li>L = Length of Cantilever</li>
</ul>
<p>So, if you know what is the maximum acceptable displacement you can solve for it:</p>
<p><span class="math-container">$$L= \sqrt[3]{\frac{3 \cdot E\cdot I\cdot \delta_{max} }{P}}$$</span></p>
<hr />
<p>for the <strong>simply supported case</strong>
<a href="https://i.stack.imgur.com/IHjP5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IHjP5.png" alt="enter image description here" /></a></p>
<p>(which is more likely what you will use):</p>
<p><span class="math-container">$$\delta_{max} = \frac{P\cdot L^3}{48 \cdot E\cdot I}$$</span></p>
<p>and</p>
<p><span class="math-container">$$L= \sqrt[3]{\frac{48 \cdot E\cdot I\cdot \delta_{max} }{P}}$$</span></p>
<hr />
<p><strong>One final thing</strong> is that the load is not only the weight of the router, but also forces from the cutting process.</p>
| 48831 | Max length of 12mm OD steel rod for CNC axis |
2021-12-22T18:00:38.487 | <p>How do ball bearings last? Most bearings in a standard hub (bottom of the image) are not actually taking a load. Only the single ball that happens to be at a particular location while spinning takes 100% of the load. If the bearings were cylindrical (top of the image) supporting a rolling plate, it would be slightly better, but not by much.</p>
<p>Regardless of the load (in a bike hub), and even if it's just cyclist+bicycle, that the area is infinitesimally small would suggest that the stress skyrockets towards infinity. And that's before even talking of the higher loads in a car's bearings.</p>
<p>How do ball bearings (in a bike hub) not crack from a load concentrated on an infinitesimally small point?</p>
<p><a href="https://i.stack.imgur.com/fuoSn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fuoSn.png" alt="ball bearings stress load" /></a></p>
<p>This is a sequel question to a question I just asked on <a href="https://bicycles.stackexchange.com/q/82183/48599">bicycles.SE</a> about how much grease I should use after I overhaul bike hubs—whether I should put so little to keep the wheel spinning freely with minimal drag, or put so much that water ingress would be unlikely (the red areas in the image).</p>
| |stresses|bearings| | <p>Ductility-- of the ball, the race, and the cone-- is indeed a key to the ball not cracking. My experience with bicycle bearings is of having the ball, the race, or the cone show some pitting, and/or for the race to show spalling or cracking. Usually during a repack I found that the balls were in flawless condition. From all this I came to the belief that (a) there was an advantage to the spherical shape (b) the balls were the hardest component, and (c) the relative advantage of the balls improved after some break-in on the race and the cone, as they developed "tracks" in them.</p>
| 48864 | How do ball bearings not crack from a load concentrated on an infinitesimally small point? |
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