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2021-01-09T14:31:38.133 | <p>I am investigating fuzzy control and I am a little confused, I have come across 3 models and I can't quite fathom out what is going on.</p>
<p><a href="https://i.stack.imgur.com/HPi4k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HPi4k.png" alt="enter image description here" /></a></p>
<p>In this PD+I controller, I can see the error signal being scaled by Kp and the derivative of the error being scaled by Kd and being fed into the fuzzy controller. I can the integral of the error being scaled by Ki and being fed into the summing junction along with the controller output scaled by Kh.</p>
<p><a href="https://i.stack.imgur.com/oh6Jh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oh6Jh.png" alt="enter image description here" /></a></p>
<p>In this PID controller, I can see the error is fed into GE and GE4 which are the proportional scaling factors, Discrete Derivative Filter 1 and 2 which are the derivatives of the error which are being scaled by GCE and GCE1 respectively and are then fed into their respective fuzzy controller. This is where it becomes confusing, the output of Fuzzy logic controller 1 is scaled by GCU and then integrated which would seem to be the exact same as the PD+I controller in figure 1, it seems that controller 1 is a PID controller 2 is a PD controller?</p>
<p><a href="https://i.stack.imgur.com/aPzrH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aPzrH.png" alt="enter image description here" /></a></p>
<p>Finally, this controller is the 1 that really bamboozled me, I can see the error e(t) being scaled by Ki (shouldn't this be Kp?) and the derivative being scaled by Kp (shouldn't this be Kd?) and the output from the controller is integrated with no scaling factor, apart from the absence of the scaling factor before the integral, this looks the same as the PID.</p>
<p>I suspect that my understanding of Fuzzy controllers are not correct so I would be very appreciative if someone could shed some light on this. Thanks in advance.</p>
| |control-engineering|control-theory|pid-control| | <p>So I have managed to shed some light on this little conundrum so I thought I would share my findings, hopefully, it will be useful for others.</p>
<p>As @joojaa pointed out in the comments, the first one is, in fact, a Fuzzy PD-type controller with an additional non-Fuzzy integral, the second example is actually a Fuzzy PI+PD controller and the last example is a Fuzzy PI-type controller.</p>
<p>Now to clear up the confusion,</p>
<p>In the first example, if we remove the non-Fuzzy Integral, we have a Fuzzy PD-type controller, the error, the change of error and output have membership functions which rule base is determined with consideration to the type of controller, Kp and Kd are the proportional and derivative tuning factor respectively.</p>
<p>In the second example, Fuzzy logic controller 2 is clearly the same as the first example when we remove the non-Fuzzy integral, with GE4 and GCE4 being the proportional and derivative tuning factors respectively. Fuzzy logic controller 1 is the Fuzyy PI-type controller, the subtle differences are the membership function rule base will reflect the fact that this is a PI-type controller, GE and GCE are the integral and proportional tuning factors respectively which is shown in the third example, also pay attention to the output U, the PI and PD type controller is CU and U respectively, the G is just a gain scaling factor.</p>
<p>In the third example, this is a Fuzzy PI-type controller, the only difference in this controller to the ones above is that the output CU doesn't have a scaling factor.</p>
<p>To summarise, the main difference between a Fuzzy PI and PD-type controller is an integrator for PI-type controllers for the CU output, the location of the tuning factors and the way the membership function rule base is defined.</p>
<p>This is to the best of my knowledge.</p>
| 39620 | What's the difference between Fuzzy PI, PD and PID controllers? |
2021-01-10T11:08:31.433 | <p>A thin walled double pipe counter flow heat exchanger is to be used to cool oil (Cp = 2198 J/kg°K) from 150°C to 40°C at a rate of 2.27kg/s by water (Cp = 4187 J/kg°K) that enters at 21°C at a rate of 1.36 kg/s. The diameter of the tube is 0.127m and its length is 60m. Determine the overall heat transfer coefficient of this heat exchanger using (i) the LMTD method and (ii) the ε-NTU method</p>
<p>My attempt:</p>
<p>Given:</p>
<p>Mass flow rate of oil: <span class="math-container">$\dot{m}_{o} = 02.27 kg/s$</span></p>
<p>Mass flow rate of water: <span class="math-container">$\dot{w}_{o} 1.36 kg/s$</span></p>
<p>Specific heat of oil: <span class="math-container">$ C_{p,o} = 2798 J/Kg/.k $</span></p>
<p>Specific heat of water: <span class="math-container">$C_{p,w} =4187 J/kg.k $</span></p>
<p><span class="math-container">$$C_o = m_oC_{p,o} = 2.27 \cdot 2198 = 4989.46J/s K$$</span>
<span class="math-container">$$C_w=m_wC_{p,w} = 1.36 \cdot 4187 = 5694.32 J/s K$$</span>
<span class="math-container">$$C_o = C{min} = 4989.46 J/Ks$$</span>
<span class="math-container">$$C_i = C_{max} = 5694.32 J/Ks$$</span></p>
<p>From heat balance:
Heat released by oil = heat gain by water</p>
<p><span class="math-container">$$Co (Ti - To) = Cw (Tco - Tci)$$</span></p>
<p><span class="math-container">$$Ti = 150 Degrees$$</span>
<span class="math-container">$$To = 40 Degrees$$</span>
<span class="math-container">$$Tci = 21 Degrees$$</span></p>
<p><span class="math-container">$$ 4989.46 (150-40) = 5694.32 (Tco - 21)$$</span>
<span class="math-container">$$Tco - 21 = ?$$</span></p>
<p>I am stuck on calculating <span class="math-container">$$Tco$$</span> How do i calculate <span class="math-container">$$Tco$$</span></p>
<p>Can anyone help please??</p>
| |thermodynamics| | <h1>calculation of exit temperature</h1>
<p>The heat transfer rate between <span class="math-container">$\dot{Q}$</span> oil and water should be equal to the change in heat capacity in the medium.</p>
<p><span class="math-container">$$\dot{Q}= m_oC_{p,o}\delta T_{o} = - m_wC_{p,w}\delta T_{w}$$</span></p>
<p>you can solve this algebraically and obtain:</p>
<p><span class="math-container">$$ 2.27 \cdot 2198 \cdot(150−40)= 1.36\cdot 4187 (T_{wo}−21)$$</span></p>
<p><span class="math-container">$$ T_{wo} =\frac{2.27 \cdot 2198 \cdot(150−40)}{ 1.36\cdot 4187} +21 [^oC]$$</span></p>
<p>if my numerical calculations are correct this should be:</p>
<p><span class="math-container">$$T_{wo} = 117[^oC]$$</span></p>
<h2>LMTD method</h2>
<p>For a counter flow heat exchanger the heat transfer rate <span class="math-container">$\dot{Q}$</span> is equal to:</p>
<p><span class="math-container">$$\dot{Q} = kA\cdot\Delta T_{lm}$$</span></p>
<p>where:</p>
<ul>
<li>The <span class="math-container">$\Delta T_{lm}$</span> is the temperature difference for counterflow, which is given from the following equation.</li>
</ul>
<p><span class="math-container">$$\Delta T_{lm} = \frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}$$</span></p>
<p>For this particular example</p>
<ul>
<li><span class="math-container">$\Delta T_1 = T_{o,i}-T_{w,o} = 150-T_{w,o} = 33$</span> : temperature difference at one exit (at the center of the drawing)</li>
<li><span class="math-container">$\Delta T_2 = T_{o,o}-T_{w,i} = 40 -21=19[C] $</span> : temperature difference at other Exit (bottom of the drawing).</li>
</ul>
<p>Because <span class="math-container">$\dot{Q}= m_oC_{p,o}\delta T_{o}$</span>, it is possible to calculate k for LMDT method as:</p>
<p><span class="math-container">$$ k_{ry}= \frac{\dot{Q}\ln (\Delta T_1/\Delta T_2)}{A(\Delta T_1-\Delta T_2)} $$</span></p>
<p><span class="math-container">$$ k_{ry}= \frac{\dot{Q}\ln (33/19)}{A(33-19)} =21642 \frac{1}{A} [\frac{W}{m^2*K}] = \approx 904[\frac{W}{m^2 K}]$$</span></p>
<p>A is the exchange area and its <span class="math-container">$A = 60\cdot 2\cdot \pi\cdot r_{tube}$</span>.</p>
<h1>NTU method</h1>
<p>Maximum possible heat rate <span class="math-container">$q_{max}$</span></p>
<p><span class="math-container">$$q_{max} = C_{min} (T_{o,i}- T_{w_i}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$ C_{min} = m_oC_{p,o} = 4989.46$</span></li>
<li><span class="math-container">$T_{o,i}= 150 [^oC]$</span></li>
<li><span class="math-container">$ T_{w_i}= 21 [^oC]$</span></li>
</ul>
<p>The effectiveness is defined as:
<span class="math-container">$$\epsilon = \frac{\text{Actual heat transfer}}{\text{max heat transferred}}$$</span></p>
<p><span class="math-container">$$\epsilon = \frac{ m_oC_{p,o}\delta T_{o}}{4989.46(150-21)}$$</span></p>
<p>However <span class="math-container">$\epsilon$</span> for a parallel flow can be given from equation:</p>
<p><span class="math-container">$$\epsilon = \frac{1- e^{- NTU \cdot (1-c)}}{1- c\cdot e^{-NTU \cdot (1-c)}}$$</span></p>
<p>where <span class="math-container">$c = \frac{C_{min}}{C_{max}} = 4989.46/5694.32=0.8762$</span></p>
<p>Substituting should get you
<span class="math-container">$$\epsilon \approx 0.8527$$</span></p>
<p>solving for <span class="math-container">$NTU$</span> should get you:</p>
<p><span class="math-container">$$NTU = \frac{\ln(\frac{\epsilon - 1}{c*\epsilon - 1})}{c - 1} = 4.3654$$</span></p>
<p>From there, because :</p>
<p><span class="math-container">$$ NTU=\frac{UA}{C_{min}}$$</span>
you can solve for U:
<span class="math-container">$$U = =\frac{NTU \cdot C_{min}}{A}\approx 909[\frac{W}{m^2 K}]$$</span></p>
<p><a href="https://i.stack.imgur.com/EjfGa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EjfGa.png" alt="enter image description here" /></a></p>
| 39634 | Thermodynamics - heat exchanger |
2021-01-10T22:38:49.013 | <p>For example one layer (one-atom thick) of a superconductor, and the next layer (also one-atom thick, of an insulator, next layer again an atom-thick layer superconductor etc.. until the overall thickness of 1 mm is reached.</p>
<p>If so, how?</p>
| |materials| | <p>There are crystalline compounds that do exactly this - bi-elemental cubics such as face centered cubic, body centered cubic are like this, plus many others.</p>
<p>Common table salt has this structure.</p>
<p>And you really do want to let nature do this for you at it's own pace. If you try to do this forcibly, that force is going to come back out sooner or later. And it adds up fast when you are counting atom by atom.</p>
| 39646 | Is it possible to create a material with alternate atom-size layers of 2 different elements? |
2021-01-11T10:35:38.823 | <p>How can I calculate some properties of a stack of 2 bonded cantilever beams like in the image? Especially I am interested in the spring constant and deflection with force on random locations. However, what confuses me is their lengths are not same. Otherwise I would recalculate the moment of inertia according to cross section of beam stack and then results can be acquired with well known formulas.</p>
<p><a href="https://i.stack.imgur.com/oRjYJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oRjYJ.jpg" alt="example beam" /></a></p>
| |mechanical-engineering|structural-engineering|beam| | <p>I would first calculate the deflection <span class="math-container">$\delta$</span> and rotation<span class="math-container">$\theta$</span> for The point at the end of the double-layer assuming the rest of the cantilever is rigid, using the equivalent area method, like in concrete beam design, Wasabi explained it.</p>
<p>Then calculate the deflection and rotation for the rest of the beam and add it to the first part.</p>
<p>This method of freezing and releasing sections can be used for more complex arches and tapered beams as well.</p>
| 39651 | How can deflection and spring constant of cantilever beam stack be calculated? |
2021-01-11T11:11:22.923 | <p>I am currently designed a structure that require hexagonal nuts to be placed side by side in honeycomb pattern, welded upon sheet. Other than doing it in old-fashioned way, is there any way/technique/machine/tools i might have missed to achieve the goal quickly?</p>
<p>Thank you in advance for your input, really appreciate any</p>
<p><a href="https://i.stack.imgur.com/rsNym.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rsNym.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|welding| | <p>The fastest method to attach hex nuts to sheet metal would be to modify a stud welder. A commercial stud welder is grounded to the metal plate and the threaded stud is held in the opposite electrode. Applying pressure to the stud against the plate closes a switch, dumping a rather large current through the stud, effectively spot welding it to the plate. A quick search returned a custom-made nut spot welder <a href="https://www.youtube.com/watch?v=je9j20b8bcA" rel="nofollow noreferrer">video</a> of about a minute length. The same search returned a number of "spot welding nuts" videos and related links.</p>
<p>Note in the linked video, the square nuts have convex bottoms, causing the corners to contact the surface first, concentrating the current in those locations, ensuring that a weld is formed.</p>
<p>A flat hex nut would require substantially more energy/current than one with, say, a triple point profile.</p>
<p>Another answer is brazing, as one can place the nuts are required, on a surface pre-coated with flux and apply sufficient heat to secure the nuts. The brazing would not require the entire sheet to be heated at once, but the heat can be applied in a moving wave. Big torches, plenty of gas, I think it could be done.</p>
| 39652 | Quicker way to weld hexagonal nuts in honeycomb pattern |
2021-01-11T13:03:52.337 | <p>I am looking for an equation to provide me the angle of view based on a distance from a window. I'm only interested in the horizontal not the vertical.</p>
| |geometry| | <p>This is basic trignometry. Use the <em>tan</em> function, <span class="math-container">$ tan\ \theta = \frac o a $</span> where <em>o</em> is the opposite (half the window width) and <em>a</em> is the adjacent (the distance to the window).</p>
<p><span class="math-container">$$ \theta = 2 \ tan^{-1}(\frac w 2 \times \frac 1 d) $$</span></p>
<p>where <span class="math-container">$w$</span> = window width and <span class="math-container">$d$</span> = distance to window.</p>
| 39654 | horizontal angle of view based on distance from window |
2021-01-11T16:15:04.280 | <p><a href="https://i.stack.imgur.com/qN9vB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qN9vB.png" alt="A rigid body with a translating (not rotating) reference frame, x'y'z', attached at point A, relative to the fixed frame, xyz, at O." /></a></p>
<p>My book, written by Beer, states that the rate of change of a vector is the same with respect to a fixed frame and with respect to a frame in translation.
But while it says the derivative <strong>˙</strong>rB/A represents the rate of change of rB/A with respect to the frame Ax′y′z′, <em>as well as</em> with respect to the fixed frame, it says the derivative <strong>˙</strong>vB/A defines the acceleration aB/A of B relative to the frame Ax′y′z′.(aB/A of B relative to the frame Oxy isn't mentioned on the same page of the book.)</p>
| |dynamics| | <p>Please note that in all likelyhood, I am repeating the book text.</p>
<p>If the coordinate system A (<span class="math-container">$Ax'y'z'$</span>) translates parallel with a constant velocity and acceleration from the fixed system O(<span class="math-container">$Oxyz$</span>), then the following equations hold:</p>
<p><span class="math-container">$$r_{A|O} + r_{B|A} = r_{B|O}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$r_{A|O}$</span> is the position of A with respect to O (usually this is abbreviated to <span class="math-container">$r_{A}$</span> because O -the fixed system- is implied)</li>
<li><span class="math-container">$r_{B|O}$</span> is the position of B with respect to O</li>
<li><span class="math-container">$r_{B|A}$</span> is the position of B with respect to moving frame A</li>
</ul>
<p>When you differentiate that over time</p>
<p><span class="math-container">$$\dot{r}_{A|O} + \dot{r}_{B|A} = \dot{r}_{B|O}\rightarrow v_{A|O} + v_{B|A} = v_{B|O}$$</span></p>
<p>i.e. <strong>the velocity of point B with respect to the fixed system O is equal to the vector addition of the the velocity of point A with respect to the fixed system O AND the velocity of point B with respect to the fixed system A</strong></p>
<p>Respectively (by differentiating again over time):</p>
<p><span class="math-container">$$\ddot{r}_{A|O} + \ddot{r}_{B|A} = \ddot{r}_{B|O}\rightarrow $$</span>
<span class="math-container">$$a_{A|O} + a_{B|A} = a_{B|O}$$</span></p>
<p>i.e. <strong>the acceleration of point B with respect to the fixed system O is equal to the vector addition of the the acceleration of point A with respect to the fixed system O AND the acceleration of point B with respect to the fixed system A</strong></p>
<h2>example</h2>
<p><a href="https://i.stack.imgur.com/JL281.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JL281.png" alt="enter image description here" /></a></p>
<p>Assume this man walking in the wagon of a train which is moving with 6[m/s]. His velocity relative to the train is 1[m/s].</p>
<p>Then:</p>
<ul>
<li><span class="math-container">$v_{A|O}= 6[m/s]$</span> the relative velocity of the train with the fixed observer</li>
<li><span class="math-container">$v_{B|A}= 1[m/s]$</span> is the relative speed of the man wrt to the train</li>
</ul>
<p>then the velocity of the man wrt to the fixed observer is</p>
<ul>
<li><span class="math-container">$v_{B|O}= v_{A|O} + v_{B|A}= 6+1=7 [m/s]$</span> is the velocity of the man wrt to the ground.</li>
</ul>
| 39661 | Is the relative acceleration with respect to a translating frame equal to that with respect to a fixed one? |
2021-01-12T13:21:42.350 | <p>Most fans (small and large) have inner hubs (see image left).
Would there be a gain in efficiency by removing that center hub, thus creating more blade space?</p>
<p>You could fix the blades to an outer ring (see image right - yes awesome paint.net skills) and put the components necessary for rotation in the outer frame.</p>
<p>Are there fans that work this way?
Would they work at least comparably?</p>
<p>Are there reasons beyond airflow efficiency that this wouldn't work like the engine design being not worth it, using more electricity, etc.?</p>
<p><a href="https://i.stack.imgur.com/oSgia.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oSgia.png" alt="Fan rework" /></a></p>
| |mechanical-engineering|electrical-engineering|airflow| | <p>There is another factor here that is not being discussed.</p>
<p>Primarily in turbo fan gas turbine engines the central hub is also a mount for the 'spinner' or 'cone'. This allows the incoming fan drawn air to be inducted in such a way as to created a convergent air flow increasing air velocity in to the fan.</p>
<p>This process more than makes up for any induction losses by lost blade space at the central hub.</p>
| 39670 | Benefit of removing the center "engine hub" inside a fan |
2021-01-12T23:51:52.070 | <p>Can I create a mate in solidworks that allows for a part to move up and down but stop at a certain distance from another part?</p>
<p>I have an L-bracket mated to an aluminum rod (black). If it is possible I would like the whole black rod and L-bracket combo to move up but not exceed a certain distance from that horizontal rod below the L-bracket. This mate should be defined from one of edges of the horizontal rod and the bottom side of the L-bracket.</p>
<p>Is this possible?</p>
<p><a href="https://i.stack.imgur.com/ho4hB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ho4hB.png" alt="Pic1" /></a></p>
| |solidworks| | <p>Have a look at some of the advanced mates. For example limit mates : <a href="https://help.solidworks.com/2018/english/SolidWorks/sldworks/t_Limit_Mates_SWassy.htm" rel="nofollow noreferrer">https://help.solidworks.com/2018/english/SolidWorks/sldworks/t_Limit_Mates_SWassy.htm</a></p>
<p>I reckon that'll do what you need. Or you can get even fancier by defining a mate along a path. Good luck :
<a href="https://help.solidworks.com/2018/english/SolidWorks/sldworks/t_Path_Mate_SWassy.htm" rel="nofollow noreferrer">https://help.solidworks.com/2018/english/SolidWorks/sldworks/t_Path_Mate_SWassy.htm</a></p>
<p><a href="https://help.solidworks.com/2018" rel="nofollow noreferrer">https://help.solidworks.com/2018</a> is your friend!</p>
| 39680 | Create a mate that allows a part to move within a specified distance from another part |
2021-01-13T11:09:18.783 | <p>I'm hoping for some clarification regarding converting torque to force at point of contact.
I'm tightening some threaded components, the end faces of each are machined and will form a metal-metal seal under compression. I would like an idea of the force being applied at the interface due to a prescribed torque.</p>
<p>Using T = F.r (neglecting friction) the following values apply:
T = 5N.m (setting of torque wrench)
r = 0,0275 m (lever length of torque wrench)</p>
<p>I calculate F = 181,82 N but I think this force is referring to how hard my arm is pushing the wrench to achieve the desired torque.</p>
<p>Can anyone clarify this and suggest how I might go about finding the force at the interface?</p>
<p>Thanks in advance.</p>
| |torque|seals| | <p>It sounds like you want the clamp force, or axial force applied by the screw threads based on your input torque (radial force x lever arm). <a href="https://www.engineersedge.com/calculators/torque_calc.htm" rel="nofollow noreferrer">Here's a useful calculator</a></p>
| 39688 | Determine applied force due to torque |
2021-01-13T14:59:41.300 | <p><a href="https://i.stack.imgur.com/BMcR1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BMcR1.png" alt="enter image description here" /></a></p>
<p>Above is a picture I found online, showing a prestressed concrete beam. It mentions a "hyperstatic moment".</p>
<p>When a beam is prestressed, it deflects upwards. According to the picture, this deflection needs to be counteracted by an additional reaction at the center, which in turn creates the hyperstatic reactions at the ends. But isn't this upward deflection the whole idea of using prestressed concrete? Isn't this deflection counteracted by the working load on the beam? Isn't the idea to design the prestress so that when the intended load on the beam occurs, it balances the deflection from prestress and the beam is straight again? Why do we need to discuss this hyperstatic reaction as something separate?</p>
| |civil-engineering| | <p>To understand the hyperstatic moment, we need to first remember that prestress is, by definition, the application of an internal stress state on the beam. Therefore, it cannot generate external loads on the beam. That is, if you had a beam floating in space and then you applied some prestress to that beam, it would deform but not present any rigid body motion.</p>
<p>Without loss of generality, this can be easily seen with a simplification of your case, where instead of parabolic prestress we have polygonal prestress which only applies concentrated force at the points of curvature (just get the maxima and minima of your parabolas and then connect them with straight lines).</p>
<p>If we then apply this prestress to a simply-supported beam, we see that no reaction forces are generated, even though we have a non-zero bending moment diagram:</p>
<p><a href="https://i.stack.imgur.com/y1Mg7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y1Mg7.png" alt="enter image description here" /></a></p>
<p>However, if we then look at the deflection of the beam, we can see that even though the midspan has a downwards force applied to it, the beam as a whole still displays a positive camber.</p>
<p><a href="https://i.stack.imgur.com/bUFUq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bUFUq.png" alt="enter image description here" /></a></p>
<p>But then we remember that you aren't dealing with a simply-supported beam, but a hyperstatic (or statically indeterminate) one, with a support at the middle. So what'll happen in the real case? Well, the support won't allow the camber at midspan to occur, which it does by generating a downwards force (reaction) on the beam at it's "midspan".</p>
<p><a href="https://i.stack.imgur.com/P45Jh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P45Jh.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/52035.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/52035.png" alt="enter image description here" /></a></p>
<p>If we simply isolate the effect of that midspan reaction (by replacing it with a concentrated force), we get the hyperstatic moment diagram*; note that if we add it to the bending moment of the isostatic case, we get exactly the same as in the hyperstatic case (the values are actually a bit off — 4.0 at midspan here and 3.8 above — because of rounding, I should've used more decimal places when defining the force):</p>
<p><a href="https://i.stack.imgur.com/0lYfg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0lYfg.png" alt="enter image description here" /></a></p>
<p>If we were to look at the deflections, we'd note the deflection at midspan is precisely equal and opposite to that found in the isostatic case, ensuring the support's boundary condition is satified.</p>
<p>So here we can see that the hyperstatic moment is the moment generated by a beam's boundary conditions to counteract the deflections caused by the prestress. If you came up with a prestressing configuration that satisfied all boundary conditions (for example, generating zero deflection at the midspan support), then there'd be no reaction forces to that prestress and therefore no hyperstatic moment. But in practice that never happens.</p>
<p>An isostatic (statically indeterminate) beam, however, will never present a hyperstatic moment because it can simply accommodate these deflections instead of resisting them.</p>
<hr />
<p><sup>* Those who are paying attention will notice that my choice of the central support as the one to be replaced by a concentrated force was entirely arbitrary. I could just as easily have chosen the left or right supports instead, creating simply-supported beams with cantilevers. However, the choice is irrelevant; whichever one is chosen, the diagram will be identical.</sup></p>
<hr />
<p><sup>Diagrams created with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free educational 2D frame analysis tool.</sup></p>
| 39691 | What is "hyperstatic moment"? |
2021-01-14T17:02:06.643 | <p>A uniform bar weighing 343N leans on a heightened plane (see figure). What force P is needed to start the bar moving?</p>
<p>Coefficient of friction (on all surfaces) = 0.30</p>
<p><a href="https://i.stack.imgur.com/3X4pA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3X4pA.jpg" alt="enter image description here" /></a></p>
<p>The blurred length is most probably 5m.</p>
<p>I've done several revisions on answering this problem: adding a vertical friction force on the edge, adding another one to the horizontal of the edge, considering the weight on the other end of the bar, putting the weight at the midpoint of the bar, etc.</p>
<p>It's kind of tricky, that my equations did't add up — different answers on different equations of the same variables.</p>
<p>The given answer is: P = 246N</p>
<p>~
Problem solved. Turns out I just had to put the bar on the horizontal, and get the component of every force thereafter. Problems with triangles can get confusing.</p>
| |friction| | <p>This is a diagram of the reaction forces at points A and B. The reaction forces are perpendicular to the surface of contact for A. In the case of B, again the reaction force is perpendicular to a surface (think of it the reaction).</p>
<p><a href="https://i.stack.imgur.com/PKz6a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PKz6a.png" alt="enter image description here" /></a></p>
<p>So you have the following forces:</p>
<ul>
<li><span class="math-container">$N_A$</span>: reaction force at point A</li>
<li><span class="math-container">$T_A$</span>: Friction force at point A</li>
<li><span class="math-container">$N_B$</span>: reaction force at point B <span class="math-container">$\left(\frac{N_{By}}{N_B{x}}=\frac{3}{4}\right)$</span></li>
<li><span class="math-container">$T_B$</span>: Friction force at point B <span class="math-container">$\left(\frac{T_{By}}{T_B{x}}=\frac{-4}{3}\right)$</span></li>
</ul>
<p>Regarding the friction forces they are drawn this way, because that will be direction of force when the force P is large enough for them to start moving. Additionally, the magnitude of the friction forces will be (just before movement is initiated):</p>
<p><span class="math-container">$$T_A = \mu N_A, \quad T_B=\mu N_B$$</span></p>
<p>In order to calculate <span class="math-container">$N_B$</span> we can estimate by the equilibrium of moments about point A:</p>
<p><span class="math-container">$$\sum M_A = 0 \rightarrow N_B\cdot 5m - W\cdot \left(4\cdot\frac{3}{5}\right) = 0$$</span></p>
<p>Where:</p>
<ul>
<li>W is the weight of the bar and its applier right at the middle of the beam (4[m] from the edge). The horizontal distance from point A is <span class="math-container">$\left(4\cdot\frac{3}{5}\right)$</span></li>
</ul>
<p><span class="math-container">$$N_B = \frac{12}{25}\cdot W$$</span></p>
<p>The equilibrium of forces on Y axis is:</p>
<p><span class="math-container">$$\sum F_y = 0 \rightarrow N_A + N_{By} -T_{By}-W = 0 $$</span>
<span class="math-container">$$N_A + N_{B}\frac{3}{5} - \mu\cdot N_{B}\cdot \frac{4}{5} -W = 0 $$</span>
<span class="math-container">$$N_A + \frac{12}{25}\cdot W(\frac{3}{5}- \mu\cdot \frac{4}{5}) -W = 0 $$</span>
<span class="math-container">$$N_A =\left(1 + \frac{12}{25}(\mu\cdot \frac{4}{5}-\frac{3}{5} \right)\cdot W $$</span>
<span class="math-container">$$N_A =\frac{517}{625}\cdot W $$</span></p>
<p>In order to calculate <span class="math-container">$P$</span> we can estimate by the equilibrium of moments about point B:</p>
<p><span class="math-container">$$\sum M_B = 0 \rightarrow -N_A\cdot 3m + (P- T_A)\cdot 4m + W\cdot \left(1\cdot\frac{3}{5}\right) = 0$$</span></p>
<p><span class="math-container">$$ P=\frac{1}{4}\left(N_A\cdot 3m - W\cdot \left(1\cdot\frac{3}{5}\right) \right) +T_A$$</span></p>
<p><span class="math-container">$$ P=\frac{1}{4}\left(\frac{517}{625}\cdot W\cdot 3m - W\cdot \left(1\cdot\frac{3}{5}\right) \right) +\mu \cdot \frac{517}{625}\cdot W$$</span></p>
<p><span class="math-container">$$ P=\left(\frac{1}{4}\left(\frac{517}{625}\cdot 3 - \frac{3}{5}\right) +\mu \cdot \frac{517}{625}\right)\cdot W$$</span></p>
<p><span class="math-container">$$ P=\frac{4491}{6250} \cdot W$$</span></p>
| 39716 | Required force to start the motion of a leaning bar |
2021-01-14T21:25:28.230 | <p>I need to choose where to drill an agricultural well. I am trying to get as much water as possible (at least 100 gpm).</p>
<p>I am considering hiring a guy who claims to be able to find water using a device called an ABEM WADI. My understanding is that this devices uses VLF signals from far away transmitters to identify fractures in the ground. He says he walks in a straight line in a candidate area for well drilling, and looks for fractures noted on the screen of the WADI. He claims that areas with a certain type of fracture (I think he said 45° fractures were the best) are more likely to yield a lot of water.</p>
<p>To what extent is this method scientifically valid? It sounds legit, but methods of "dousing" or "witching" for water are often little more than divination, so I'm skeptical. I googled around and found some scientific papers about VLF-EM scanning, but I lack the science/engineering chops to interpret them.</p>
| |geotechnical-engineering|electromagnetism|water-resources|hydrology| | <p>The answer above is wrong. VLF-EM equipment isn't using the VLF signal as 'radar' and looking for scattering. THE VLF signal induces an electromagnetic response in conductive rocks, and the measuring equipment looks for variations in the strength and phase of this response. As such it is useful to hydrogeologists for detecting fractures in terrain where these are water bearing.</p>
| 39723 | Is VLF-EM scanning for fractures a valid way to find groundwater for a well? |
2021-01-14T23:37:09.537 | <p>So I'm looking into how stuff is made from scratch, starting with a surface plate which has been lapped flat using the three plate method. But almost everything I read on the subject on "the history of precision" and such stops there and glosses over the rest by simply stating that it's possible. How does one go about making perfect 90 degree corners from simply a flat surface?</p>
<p>To clarify, when I say "from scratch" I mean going from no tools at all to a perfectly square block. I'm asking about the history of machining, how the first precision machines were made.</p>
<p>If anyone has any good resources on the subject which explains it in further detail I would appreciate it very much :)</p>
| |measurements|machining|metrology| | <p>The "three plate" method has been used as the basis for precision manufacturing/inspection. <a href="https://ericweinhoffer.com/blog/2017/7/30/the-whitworth-three-plates-method" rel="nofollow noreferrer">This</a>
is a decent summary, it's pretty ingenious and only requires dye and abrasives.
<a href="https://pearl-hifi.com/06_Lit_Archive/15_Mfrs_Publications/Moore_Tools/Foundations_of_Mechanical_Accuracy.pdf" rel="nofollow noreferrer">This</a> will teach you everything you need to know and then some, "Foundations of Mechanical Accuracy".</p>
| 39732 | How are perfectly square blocks made from scratch? |
2021-01-16T07:44:53.000 | <p><strong>Question:</strong>
Why are there no flat power generators like in the picture below, that work on the surface of shallow, but steadily flowing rivers ? (As a floating micropower plant.)</p>
<p><a href="https://i.stack.imgur.com/kEWb4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kEWb4.png" alt="enter image description here" /></a></p>
<p>The picture shows a conveyer belt with vanes/blades(?) attached to it. The water flow moves the conveyer belt. A generator could be attached to the front and back "wheel" of the belt.</p>
<p><a href="http://%5B!%5Benter%20image%20description%20here%5D%5B1%5D%5D%5B1%5D" rel="nofollow noreferrer">Here's</a> a video of something similar. I would just build it on a larger river.</p>
<p><strong>Why would I ask this?</strong></p>
<p>There are much more flat rivers than waterfall-like structures on this planet. Using them looks like a much more non-nature-inversive, cheap solution. Having a longer surface should supply better drag by flowing water.</p>
| |power-engineering| | <p>When you want to solve a problem, the best start is to look at previous attempts. To provide some perspective, I'm doing that for you now. You are not looking at a typical hydro power plant where a dam provides a high head, and the flow is ducted onto a a francis or pelton turbine. You are describing a microhydropower installation with a floating turbine.</p>
<p>Floating hydrpower allows capturing some power without building a dam. The turbine could be placed in or near the middle of the river, where the current is fastest. An installation with a damn will always harvest vastly more power from the same river.</p>
<p>Before electrical power transmission became widespread, there used to be boat mills - workshops with machinery driven by water wheels, placed on boats.</p>
<p><a href="https://i.stack.imgur.com/rapwX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rapwX.png" alt="enter image description here" /></a></p>
<p>(Boat mill in Servia, 1900, Image from <a href="https://www.lowtechmagazine.com/2010/11/boat-mills-bridge-mills-and-hanging-mills.html" rel="nofollow noreferrer">lowtechmagazine page on boat mills</a>)</p>
<p>Improvised versions have also been used for electricity generation.</p>
<p>Floating hydro power is, AFAICT, an ongoing area of developement. The two most common turbine shapes appear to be a propeller hanging from buoys:</p>
<p><a href="https://i.stack.imgur.com/MYAsy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MYAsy.jpg" alt="enter image description here" /></a></p>
<p>(<a href="https://siamagazin.com/smart-hydro-powers-floating-river-canal-turbines-to-provide-electricity-to-remote-locations/" rel="nofollow noreferrer">Image source</a>)</p>
<p>... Or some sort of flat paddle wheel:</p>
<p><a href="https://i.stack.imgur.com/lgW6T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lgW6T.png" alt="enter image description here" /></a></p>
<p>Vertical axis turbines also exist.</p>
<p>I think <a href="https://engineering.stackexchange.com/a/39766/61">Kamran</a> explains quite well why a propeller or a paddle wheel is used, rather than a conveyor belt. I will just add this: Look at the water wheel in the direction of flow: You want to maximise area here.</p>
| 39763 | Why are hydropower plants always wheel-shaped and not flat? |
2021-01-16T19:13:32.347 | <p>CO2 is used as fertilizer in aquariums where it's brought into the water with diffusors from compressed form in cylinders. <a href="https://en.wikipedia.org/wiki/Carbon_dioxide_removal#Direct_air_capture" rel="nofollow noreferrer">Direct Air Capture</a> machines for CO2 have been presented for experimental geo-engineering purposes.</p>
<p>I was wondering whether it's possible to use the technology on a much smaller scale which allows one to supply the fertilizer as a constant CO2 stream (a few g per hours) or somehow compress the CO2 into a cylinder to produce it at home. This should avoid the the need for at least transport and energy for compression and allows to near endless supply for free.</p>
<p>Could I emit CO2 certificates? Just kidding...</p>
| |chemical-engineering|chemistry| | <p>It is possible and this is currently being done at the OpenAir Collective. There are two direct air carbon capture (DACC) machines under active development, Violet and Cyan. Both are small-scale, open-sourced, non-commercial systems that could be used to take CO2 out of the air and to supply it as fertilizer to plants. These units do not offer CO2 compression into a cylinder however.</p>
<p>Violet is OpenAir's version of a prototype system by Prof. Tao Wang; his prototype captures about 1 kg CO2 per day and releases it into a greenhouse to improve plant growth. Here is a photo of his system. It operates in cycles so some storage will be required to supply CO2 as a constant stream.</p>
<p><a href="https://i.stack.imgur.com/c8P6s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c8P6s.png" alt="Photograph of the moisture-swing prototype" /></a></p>
<p>Violet is still being developed at present but the prototype system above is currently in service.</p>
<p>Cyan is an even smaller system buildable for under $100 that presently captures a few grams per day and locks the CO2 into calcium carbonate. The CO2 can be released through combination with vinegar or other acids. Cyan is a very portable unit that can be used at home.</p>
<p><a href="https://i.stack.imgur.com/NmtFO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NmtFO.png" alt="Photograph of the outside of the Cyan unit" /></a></p>
| 39767 | Is it possible to use direct air capture technology for CO2 on a non-commercial level? |
2021-01-17T17:15:25.543 | <p>can someone explain me what 1/8" or 1/16" means in the image below?
<a href="https://i.stack.imgur.com/Ugo6S.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ugo6S.jpg" alt="enter image description here" /></a></p>
| |technical-drawing| | <p>1/16" refers to the thickness of the circle containing the letters, which appears to be 4 3/8" diameter. 1/8" refers to the radius of the cylinder top under the lettered circle.</p>
<p>If your question refers to the " notation, Pete W's comment addresses that aspect. " is inches Imperial measurement.</p>
| 39782 | Notation of a technical drawing |
2021-01-17T17:59:46.183 | <p>I was given the following formula to relate the change of area against the change of length of an electric wire with a Poisson ratio:</p>
<p><span class="math-container">$ {\Delta A \over A} = -2 \nu {\Delta L \over L} $</span></p>
<p>where <span class="math-container">$ \Delta A \over A $</span> represents the change in cross-sectional area of the wire due to the transverse strain as the wire gets pulled longitudinally stretching length <span class="math-container">$ L $</span> to <span class="math-container">$ L+\Delta L$</span>.</p>
<p>I don't get how this equation is derived. The Poisson ratio is defined by <span class="math-container">$ \nu = -{\epsilon_{lateral} \over {\epsilon_{longitudinal}} } = - {{\Delta d / d} \over {\Delta L / L}} $</span> where <span class="math-container">$ d $</span> is the diameter of the cross section. Then the ratio of the area:</p>
<p><span class="math-container">$ {\Delta A \over A} = {{0.25\pi(d+\Delta d)^2 - 0.25\pi d^2} \over {0.25\pi d^2}} = {{2d \Delta d} \over d^2} + {{\Delta d^2} \over {d^2}} = -2 \nu { \Delta L \over L} + \big( \nu {\Delta L \over L} \big)^2 \neq -2 \nu {\Delta L \over L} $</span></p>
| |mechanical-engineering|stresses|strength| | <p>Since you are essentially using infinitesimal changes, then higher order differences can be neglected.</p>
<p>I.e. following from your equation
<span class="math-container">$${\Delta A \over A} = {{0.25\pi(d+\Delta d)^2 - 0.25\pi d^2} \over {0.25\pi d^2}} = {{2d \Delta d} \over d^2} + {{\Delta d^2} \over {d^2}} = -2 \nu { \Delta L \over L} + \big( \nu {\Delta L \over L} \big)^2 $$</span></p>
<p>because <span class="math-container">$\left(\frac{\Delta L}{L}\right)^2$</span> is a second order difference, you can assume that <span class="math-container">$\left( \nu {\Delta L \over L} \right)^2\approx 0$</span>.</p>
<p>Therefore:
<span class="math-container">$$ -2 \nu { \Delta L \over L} + \underbrace{\left( \nu {\Delta L \over L} \right)^2}_{\approx 0}\approx -2 \nu { \Delta L \over L} $$</span></p>
| 39785 | Poisson relationship between area and length of electric wire |
2021-01-18T12:14:04.803 | <p>My book, Vector Mechanics for Engineers: Statics and Dynamics 12th Edition, states in the plain motion of a rigid body, the motion of the rigid body is completely defined by the resultant force and resultant moment about G acting on the body. But it also states</p>
<blockquote>
<p>We also note, as we did earlier, that the system of the external forces
does not, in general, reduce to a <strong>single</strong> vector m¯
a attached at G. Therefore,
in the general case of the plane motion of a rigid body, the resultant of the
external forces acting on the body does not pass through the mass center
of the body.</p>
</blockquote>
<p>Is it because the resultant moment is not zero at G in general, and if the resultant force doesn't pass through G, we can replace it with the same vector attached at G and the couple ? Is that right?</p>
| |mechanical-engineering| | <p>Yes, in general, the sum of external forces applied to a rigid body, do not necessarily pass through the center of the mass, but a distance away from it, causing</p>
<p>A linear acceleration
<span class="math-container">$$ΣF = ma_G$$</span></p>
<p>And a moment</p>
<p><span class="math-container">$$ΣM_G = I_Gα$$</span></p>
<p>It is only in particles that this sum of forces always passes through the CG, of the particle, obviously.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/aE9BU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aE9BU.png" alt="sum of external forces." /></a></p>
| 39799 | In the general case of the plane motion of a rigid body, the resultant does not pass through the G of the body. Why is that? |
2021-01-18T16:20:44.380 | <p><a href="https://i.stack.imgur.com/QgvgP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QgvgP.png" alt="enter image description here" /></a></p>
<p>In prestressed concrete, we add a tensioned steel cable at the bottom half of the beam to balance out the tensile forces resulting from the applied load. But before the load is added, doesn't the beam deflect upwards and cause tensile cracks now at the top of the beam (situation 5 in the above picture)? Let's say we tension the cable and cast the concrete and then release the cables into compressive stress. I don't see how in practice we can immediately apply the designed load to balance the situation before the beam bows upwards and cracks at the top.</p>
| |structural-engineering|prestressed-concrete| | <p>For starters, what makes you think the concrete doesn't crack at the top?</p>
<p>Let's assume for a second that it does crack. Well, if we were dealing with a pure concrete beam, that'd be a big problem since the crack would propagate down the height of the beam until it falls apart.</p>
<p>Thankfully, prestressed concrete beams aren't just concrete and a prestressing cable. They are also given some light longitudinal rebar (usually just the minimum required by the relevant code on each face). So if the beam starts to crack, it'll activate the rebar which will impede the beam's collapse.</p>
<p>And is this a big deal? Well, no, not really. After all, the beam will soon be under the expected load, at which point the downwards forces will counteract the camber, the top face will go under compression and those cracks will all seal up. No harm, no foul.</p>
<p>So, even if the concrete were to crack a bit on the top, it's honestly not the end of the world (assuming all safety requirements are met, of course).</p>
<p>But does the concrete crack?</p>
<p>Well, not necessarily. And I'd say it usually doesn't.</p>
<p>The reason has already been mentioned in <a href="https://engineering.stackexchange.com/a/39814/1832">@Andrew's answer</a>: prestress doesn't just generate a bending moment, but also a compressive force. So the prestress doesn't simply generate the exact reverse of the expected bending moment so that they cancel out: it generates the reverse of a fraction of the bending moment and then leaves the rest to be canceled out by simple compression.</p>
| 39808 | Why doesn't prestressed concrete crack at the top before the load is applied? |
2021-01-18T16:56:37.520 | <p>In an introduction to field-programmable gate arrays (FPGAs) Sparkfun describes that FPGAs aren't really 'programmed' (see below for excerpt) even though this is often used in every day language.</p>
<p>If this is 'not entirely correct', <strong>what would be the correct terminology for putting instructions onto an FPGA, used by the experts in the field?</strong> I would presume FPGA's are 'flashed', but I'm not sure if 'configured' would be the right word since putting instructions onto an FPGA is literally changing the hardware.</p>
<p>(sorry if this is not the right platform to ask this, it's more of a meta-question for engineering)</p>
<p>Citation from the <a href="https://learn.sparkfun.com/tutorials/programming-an-fpga/all" rel="nofollow noreferrer">Sparkfun introduction Jan 2021</a>:</p>
<blockquote>
<p>But you aren’t writing a program. You are creating a circuit. You
don’t use programming languages to create circuits, you use hardware
description languages (HDLs).</p>
</blockquote>
| |electrical-engineering| | <p>To configure FPGA a HDL language such as Verilog or VHDL is used. For example VHDL stands for very high-level design language. Designers use HDL to describe an electronic circuit.</p>
<p>Below is a schematic of a half adder:</p>
<p><a href="https://i.stack.imgur.com/Zgs0Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zgs0Z.png" alt="enter image description here" /></a></p>
<p>Below is an implementation of a half adder in VHDL</p>
<p><a href="https://i.stack.imgur.com/DdkKI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DdkKI.png" alt="enter image description here" /></a></p>
<p>Combinations of such digital logic circuits are used to design product such as a calculator. These designs are programmed into to the FPGA. In most cases the program is stored in ROM and at powerup the digital logic is loaded to the FPGA to create a calculator as in this example.</p>
<p>Generally most experts refer to this as <strong>loading logic into FPGA</strong>.</p>
<p>Flashing can be used, but this is mostly using in embedded systems, using a micro-controller.</p>
<p>Hope this helps</p>
<hr />
<p><strong>References:</strong></p>
<ul>
<li><a href="https://www.nandland.com/vhdl/modules/module-half-adder.html" rel="nofollow noreferrer">Half Adder Module in VHDL and Verilog</a></li>
</ul>
| 39809 | Correct terminology for 'putting instructions onto an FPGA' |
2021-01-19T19:53:34.413 | <p>I am trying to design something for a friend, but am struggling to come up with the mechanics. Can you help?</p>
<p>So imagine a long extendable pole, like a <strong>bow staff</strong>. Imagine after pulling out a "peg" the bow staff extends using springs. Now say I wanted to <strong>automatically</strong> close this extended pole. What would be a good way of doing this?</p>
<p>My idea is a wire/string inside the staff that runs from the top to the bottom. Pulling this wire can compress the staff. In my head, some kind of motor can wrap up the string to compress the staff. My worry is that the motor wouldn't be able to pull against the strength of the springs. Can this be solved through a gear mechanism allowing the motor to be safe from the resistance of the springs? If so, how?</p>
<p>My second idea is to eliminate the motor. Instead, the wire is simply pulled from the bottom to compress the staff. But I wouldn't want to pull the string the entirety of the bow staff. Potentially through some pully or gear system, a semi-short pull of the string can compress the entire staff. Like the string would only have to be pulled out half or a quarter of the distance of the bow staff, but still, compress the entire thing.</p>
<p>Which method is more doable? Is there a better way to automate the compression of the staff?</p>
<p>This is only in the planning stages so springs are not chosen, though their strength will be similar to that of the springs used inside an umbrella. Here is a picture that should provide some basic dimensions. Fully extended, each section is about <strong>13cm</strong>. The inner diameters are <strong>9.5mm</strong>, <strong>6.5mm</strong>, and <strong>3.5mm</strong>.
<a href="https://i.stack.imgur.com/acKTS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/acKTS.jpg" alt="basic pic" /></a>
I look forward to hearing your answers. I don't have a lot of experience with gears which is why I am asking. Thanks!</p>
| |motors|gears|mechanisms| | <p>Both of these ideas are doable, however if you go with the second idea that is by simply pulling the string with some kind of pulley or lever mechanism, you still would need some kind of actuator to pull the string. It maybe a piston or a solenoid but still an actuator. plus you would need a mechanism to pull the whole staff with just one forth pull of string and that would be a bit difficult to manage it in such a small space.</p>
<p>with the first method that is by using a motor and a brake wire, according to the type of the spring you mentioned that is one which is used in a umbrella, I don't think that compressing that spring would be such a problem. A small dc motor with brake wire attached to its shaft would do it. This won't need much space it will be having very low weight and can be done in very low cost.</p>
<p>To calculate the torque required to compress the spring, multiply the stiffness of string by distance by which it needs to be compressed. That will be your force which is needed to be applied by the motor.
Force = stiffness * compression distance.
Now you can calculate the torque required for the compression by using simple formula
Torque = force * perpendicular distance.
Now that you have required torque you can select a motor with given rated torque and other specifications like rpm, power etc.</p>
<p>if you have concern that motor will take the load or it will damage the motor then you can go for simple gear arrangement to pull the string. This way the load will not be directly applied on the motor shaft and you can also select a appropriate gear ratio which will be able to overcome the force given by springs. And you can also provide your motor with a simple bearing housing so that the load coming on to the shaft will be taken by the bearing and motor will be safe.</p>
<p>If you are deciding to automate this process you will need a feedback of the rotation of the motor so that you can check how much the staff has compressed. You can either use a rotary feedback that is a encoder or a Potentiometer they will give the angle by which motor has rotated which can be converted to the distance traveled or compressed. or you can use a linear feedback like limit switch or a proximity sensor. according to me a limit switch or a proximity would be easier to install on to mechanism, would be easy to program on an Arduino and also be cost effective.</p>
<p>so you can go for both the ideas, both will work but i am not sure about how second idea is going to work as i think putting a pulley system into the mechanism is bit difficult. where as first idea is quite sure that it will work, you will find the motor of desired specs and it would be easier to automate the mechanism if it has a motor and a some kind of sensor. So i would suggest that you go for the first idea..</p>
| 39825 | Automatic compression of springs |
2021-01-20T06:52:38.477 | <p>Longitudinal axis (x) is the axis orthogonal to its lateral directions (y and z), but what is transverse direction? Is transverse just another term for lateral?</p>
| |mechanical-engineering|materials| | <p>According to Webster's dictionary, "lateral" means generally "sideways", while "transverse" means "at right angles to the side".</p>
| 39834 | Is transverse synonym of lateral? |
2021-01-20T16:26:43.890 | <p>For example, characteristic combination found <a href="https://eurocodes.jrc.ec.europa.eu/doc/WS2008/EN1994_3_Hanswille.pdf" rel="nofollow noreferrer">here</a>:</p>
<p><a href="https://i.stack.imgur.com/AzeBr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AzeBr.png" alt="enter image description here" /></a></p>
<p>It can be seen that the dead loads <span class="math-container">$G_k$</span> are taken as the characteristic values instead of multiplying them by some factors of safety as in the case of ultimate limit states. Why are loads serviceability limit states taken as characteristic values instead of applying partial safety factors into their values?</p>
| |structural-engineering| | <h2>There's a reason why they're called "factors of safety"</h2>
<p>The answer to your question is in its title.</p>
<p>When designing a structure, we need to make sure of two things:</p>
<ul>
<li>that the structure will remain a structure and not end up as a pile of rubble (i.e. that the structure is safe)</li>
<li>that the structure can actually be used as intended.</li>
</ul>
<p>The first is easy enough to understand: you either have a structure or a pile of rubble; do the math to ensure it's the former even under extraordinary circumstances (as described by factors of safety)</p>
<p>The second is more subjective... what does it even mean, really? Well, basically, that human beings will be able to perform whatever tasks they wish while using the structure.</p>
<p>Would you be able to sleep at night if you look up and the concrete beams are all cracked? If your bed bounces with every step your neighbor takes during their late-night walk to the kitchen for a cup of milk? Would you feel safe if you get to your office and the beam at the reception is drooping like Atlas' shoulders?</p>
<p>Of course not. So that's the purpose of serviceability limit state checks; they aren't there to make sure the structure is safe, that's the job of ultimate limit state checks.</p>
<p>Serviceability limit state is merely about whether the beam will suffer large deflections, and/or vibrations (and other checks, depending on the situation, such as cracking in concrete). And you only care about these under "normal" circumstances, which are described by the characteristic load values (if the structure undergoes an extreme, once-a-generation loading, do you really care if it deflects a bit too much for comfort?).</p>
| 39847 | Why loads in serviceability limit states usually don't include factors of safety? |
2021-01-20T21:26:02.553 | <p>Having used some big, old, manual Bridgeports in school, I was looking into getting a small mill (manual or CNC) for hobbyist home use on metal and wood. I was surprised to see many CNC milling kits on Amazon for quite cheap ($200-600 USD) whereas the the cheapest manual mini mills seem to be roughly 400-1000+ USD. My question is why the cheapest <strong>CNC mills</strong> cost less than the cheapest <strong>manual mini mills</strong>? (I am presuming the entry level for manual and CNC mills are the same except obviously the key automated part --- ie that RPM, vibration, stability are similar, but please comment on that in your answer if that's a confounding variable).</p>
<p>EDIT: My question originally had two parts with the second asking if there was also a performance difference between entry level CNC and entry level manual mills, but since none of the answers addressed that in much detail and it's sort of a separate (but only possibly related) question I have removed it.</p>
| |machining|cnc| | <p>Making a machine CNC ready has a neglible cost. Electronics cost nearly nothing, and are easy to do when your not aiming to super quality. All of the moving elements can be simplified because you dont need to present user interface through mechanical couplings and dont need to worry about user safety.</p>
<p>So a bit counter intuitively cnc machine can be simpler than a equivalent manual machine. Now that is to say the CNC does not even need to be equivalent to a manual machine. It can also be made for much lighter loads than a manual machine would, since there is a limit on how slow a human can be expected to go. So a cnc that just takes really slow passes on a PCB does not need to be much suturdier than a 3D printer.</p>
<p>Then theres the economics. A super cheap and light CNC is actually semi useful as one operator can batch many machines. Whereas a manual is still limited by the cost of a operator so it does not make sense to invest less into that use. But again this depends on market, CNC is obviously good for production runs. But not necceserily for very small scale or toolroom, modelmaking (die) or prototype work where a manual tool is still invaluable. Especially when paired with a CNC or two.</p>
| 39856 | Why are entry-level CNC mills cheaper than entry-level manual mini mills? |
2021-01-21T12:22:24.750 | <p>Why a technology like freewheeling not used in cars to save fuel? We all know that when we move to neutral gear (coasting), the car travels farther than when it is in gear because there is no engine braking effect. I understand there is a safety issue there - but is that the only reason for not using it? What I am trying to say is - when the foot is off the gas pedal, stop engine braking effect using some technology to achieve better mileage.</p>
| |mechanical-engineering|thermodynamics|automotive-engineering| | <p>It overloads brakes. The original Buick dynaflo auto trans had essentially free wheeling ( many decades ago) . They needed new drum brakes twice per year, depending on miles driven. Buick re-engineered the Dynaflo in a year or two to give engine braking.</p>
| 39858 | Why coasting is not used to save fuel? |
2021-01-21T22:13:17.067 | <p>Calculating limit stress on a solid pin loaded in shear is straightforward, and there are several handy calculators which will do <a href="https://www.engineersedge.com/material_science/bolt_single_shear_calcs.htm" rel="nofollow noreferrer">single shear</a> and even <a href="https://www.engineersedge.com/material_science/bolt_double_shear_calcs.htm" rel="nofollow noreferrer">double shear</a>. However, everything I have found assumes that the pin is solid.</p>
<p><a href="https://i.stack.imgur.com/R7PF9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R7PF9.jpg" alt="enter image description here" /></a></p>
<p>For weight savings reasons, one might want to replace a solid pin with a hollow pin. However, stress flow in a hollow object is different from in a solid object, so it seems dangerous to assume that the yield stress is purely inversely proportional to area without consideration of the nature of the cross-section.</p>
<p>How to take into account the cross-section when calculating failure stress in single-shear and double-shear loading for a pin with arbitrary ID? Is it sane/safe to use the existing shear eqn. above with the hollow cross-section area and then bump the required safety factor by, say, 2x?</p>
<p>UPDATE: The best answer includes design tables and/or guidelines. Pure theoretical approaches are interesting but ultimately unsatisfactory, as 1) they can be easily simulated with FEA as well and 2) they might not include real-world experience.</p>
| |solid-mechanics|shear| | <p>You are right in thinking that a thin walled pin will fail primarily by compression/wall buckling (I tend to think about more as crushing), however that does not change the calculation of the shear stress.</p>
<p>Shear stress is defined in a very specific way: by dividing Forces acting on a direction <em>parallel</em> to a surface, divided by the surface area.</p>
<p>Of course that increases when considering shear stress which are induced due to bending depending on the cross-sections, or because of stress concentrations (see here <a href="https://www.fracturemechanics.org/hole.html" rel="nofollow noreferrer">https://www.fracturemechanics.org/hole.html</a>). Despite these cases (and maybe some other I am forgetting) where the nominal shear stress is factored, the baseline calculation of shear stress remains the same.</p>
<p>If you want to understand failure, you'd consider all types of stress (normal and shear stresses) and the use some failure theory (Tresca, Von Mises etc) to determine if failure occurs. If you are worried about buckling you'd need to carry out a different type of analysis.</p>
<hr />
<p><strong>UPDATE</strong>:</p>
<p>From an additional look I've taken, my belief that hollow pins:</p>
<ul>
<li>made out of common engineering materials</li>
<li>and <span class="math-container">$R_{in} <\frac{2}{3}R_{out}$</span>,
are not much in danger of a transverse collapse has been reinforced.</li>
</ul>
<p>The closest I've found where references with respect to bending loading and thin walled structures. To be honest, this subject feels more of a hot topic in solid mechanics <em>science</em>, rather than the typical <em>rule-of-thumb engineering</em>. Nevertheless, just a couple of references I've found that seem relevant:</p>
<ul>
<li><a href="https://publications.waset.org/1191/an-analysis-of-collapse-mechanism-of-thin-walled-circular-tubes-subjected-to-bending" rel="nofollow noreferrer">Analysis of collapse mechanism of thin walled circular tubes subjected to bending</a></li>
<li><a href="https://www.tandfonline.com/doi/abs/10.1080/14786442408634359?journalCode=tphm17" rel="nofollow noreferrer">Calculation of bending stresses on thin-walled tubular beam</a></li>
</ul>
<p>You can also have a look at this book:</p>
<ul>
<li><a href="https://link.springer.com/chapter/10.1007/978-90-481-2516-6_8" rel="nofollow noreferrer">Thin walled beams</a></li>
</ul>
| 39867 | How to calculate failure stress on a hollow pin loaded in shear |
2021-01-22T21:34:36.617 | <p>I'm having <strong>trouble understanding the derivative part</strong> of a PID controller, because it sometimes seems to <strong>react the opposite way</strong> that I would like.</p>
<p>Let's use a simple example where :</p>
<ul>
<li>the controlled variable is a <strong>vehicle position</strong>, on a 1 dimension axis (in m)</li>
<li>the actuator signal is the <strong>vehicle speed</strong> (in m/s)</li>
<li>the setpoint is 100 m</li>
<li>the sampling time is 1 s</li>
</ul>
<p>Now let's analyze two cases:</p>
<ol>
<li>The current position (<em>at t = Ns</em>) is 80 m, and the previous position (<em>at t = (N-1) s</em>) was 60 m.<br />
Hence, we've made 20 m of progress toward the setpoint of 100 m (we're going the right way).<br />
The current error is (100 - 80) = 20 m, while the previous error was (100 - 60) = 40 m: this gives a derivative error of (20 - 40)/1s = <strong>-20 m/s</strong>.</li>
<li>The current position (<em>at t = Ms</em>) is 140 m, and the previous position (<em>at t = (M-1) s</em>) was 120 m.<br />
Hence, we've made 20m of regress from the setpoint (we're going the wrong way!).<br />
The current error is (100 - 140) = -40 m, while the previous error was (100 - 120) = -20 m: this gives a derivative error of (-40 - (-20))/1s = (-40 + 20)/1s = <strong>-20 m/s</strong>.</li>
</ol>
<p>In both cases, the <strong>derivative has the same value</strong>, so the controller's derivative behavior will be the same. But the <strong>situations are very different</strong>: in the first case we're getting closer to the goal, while in the second we're getting away from it.<br />
Why is the derivative part making the same adjustment for two situations that are so different?</p>
| |control-engineering|control-theory|pid-control| | <p>The "D" in PID is often used when PI response produces overshoot in step response. Sometimes, definitely not always, there is misunderstanding of the problem.</p>
<p>I don't really know how to explain this in time domain. Here is the frequency domain explanation: For the most common case of plant, "single dominant pole" (also includes a plant which is an integrator), the PI controller itself introduces an unwanted "closed loop zero" in the passband, which means overshoot in time domain. The D term in PID is a somewhat clumsy way to correct for this.</p>
<p>There are other options to eliminate the zero from PI. The set point signal can simply be low-pass filtered, to directly cancel the unwanted zero. Rate limits on set point do this to a degree as well. A more involved option is the much maligned PDF control, or a mathematically equivalent arrangement, such as putting the zero in feedback path, rather than forward path as PI would do. (an additional higher frequency feedback pole may be necessary too, in practice, because can't have infinite high freq gain). You don't see this often, but it's not hard in software. The benefit is more intuitive tuning, compared to PID.</p>
<p>The implementation, compared to PID, is as follows: instead d/dt of error, compute d/dt of <em>both</em> the set-point input and plant output. Give them both their own Td coefficients, Td_i and Td_o, and throw them into the integrator along with the error term. (yes, we are integrating something we just differentiated). Often, both the P term and the d/dt(input) term are unnecessary (Kp=0 and Td_i=0), leaving just Ki * integral(error - Td_o * deriv(output))</p>
| 39894 | PID controller - Counterproductive derivative |
2021-01-23T03:53:19.617 | <p>I want to make a strong connected weight-scale Arduino project, for which I already own 3×350Kg sensors (red). I plan to make the plate (green) around 30cm×30cm in 99.5 Brut aluminum, with my three sensors placed in T (two on rear of left and right sides and the third one in the middle of the front side), and the local pressure on the plate should not exceed 5 Mega Pascal (Yellowish). However, I know nothing about strength of materials computation.
<a href="https://i.stack.imgur.com/L5VBO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L5VBO.png" alt="enter image description here" /></a>
What minimal thickness would I need in order for my plate to bear such a load ?</p>
<p>I have read somewhere that I should avoid the deformation to be more that 0.1%×dimensions (i.e : 0.3mm), and, according with Wikipedia,</p>
<ul>
<li>Young's modulus = 70 GPa</li>
<li>Shear modulus = 26 GPa</li>
<li>Bulk modulus = 76 GPa</li>
<li>Poisson ratio = 0.35</li>
</ul>
<p>However, I have no idea what to do with those numbers. All explanations I found on the net were much too complicated in regard with the problem and to know what to do with it.</p>
| |structural-engineering|aluminum| | <p>My reputation is too low to comment so I'll give some pointers in an answer :)</p>
<p>Looking at your design your structures weak points are the areas where the plate and the sensors interface, especially the ones in the corners.
This is because you have a very long lever from your load to the feet of your sensors and a very small contact area between the sensor and plate.
If you could move the sensors feet closer together this would dramatically reduce the stress on the middle plate, as close as you can without making the scale too unstable.</p>
<p>Regarding stress and deformation they actually don't matter for your application as long as the sensors remain level and nothing but the sensors feet touch the ground/table. This assuming you are not planing on loading this scale millions of times and failure due to fatigue start to matter.</p>
<p>As it is right now the question can not be answered because the dimensions of the sensors, and the details about how the plate will/can interface with the sensors, are unknown.</p>
<p>If you could provide some more details about your sensors, preferably a model name and datasheet, perhaps you could get a more elaborate answer.</p>
<p><strong>EDIT:</strong>
Based on your feedback I did some simple simulations to illustrate how your plate will behave and give you some numbers to work with.
I'm not sure what material you are planing on using because 99.5 Brut seems to be related to purity and does not specify any specific alloy.</p>
<p>In my examples i use Alloys 1060 and 6061-T6.
The deflections in the simulation are highly exaggerated for illustrative purposes, in reality you can barely see that it bends.</p>
<p>Deflection 8mm, 6061-T6, 3500N evenly distributed.
<a href="https://i.stack.imgur.com/20cMA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/20cMA.png" alt="Deflection z-axis" /></a></p>
<p>Stress (Von Mises). 8mm, 6061-T6, 3500N evenly distributed.
<a href="https://i.stack.imgur.com/r51gP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r51gP.png" alt="Von Mises Stress" /></a></p>
<p>As you can see, the plate is barely loaded and the maximum deflection is 1.3 mm which is fine for your case unless your feet are a fraction of a millimeter.</p>
<p>Since both materials have the same Modulus of Elasticity, about 70 GPa, the stress and deflections are valid for both.
However, 1060 has a yield stress of 27,6 MPa meaning it's no longer elastic deformation, it's plastic (permanent) deformation.</p>
<p><a href="https://i.stack.imgur.com/xpPnZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xpPnZ.png" alt="1060 yield" /></a></p>
<p>Simulating plastic deformation is a lot more complicated and nothing I will go elaborate further on here, it means however that the result is not valid for 1060 and a 8mm 1060 plate would not satisfy your requirement.</p>
<p>A 10mm 1060 plate would however work but be on the limit.</p>
<p>For 6061-T6 you can go a lot thinner since it can handle much higher stress.
Here are som plate thicknesses and their corresponding deflections.
6mm => ~3mm
5mm => ~5mm
4mm => ~10mm
3mm => ~22mm</p>
<p>At 3mm thickness we are really at the limit of what 6061-T6 can handle.
Even though 3mm 6061 can handle the load, at 22mm deflection your sensors are no longer upright and their output will no longer be valid for your application.</p>
<p>I hope this gives you some more insight into how the plate thickness (and choice of aluminum) would affect your scale :)</p>
| 39901 | What thickness do I need for a 30x30 (cm) aluminium plate to hold 350Kg on three points? |
2021-01-23T14:05:21.430 | <p>Why are the two acceleration considered as a1 and a2 here? What is the reason for this even though the tension of both the strings are same.<a href="https://i.stack.imgur.com/FZ86z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FZ86z.jpg" alt="enter image description here" /></a></p>
<p>T is equal to 5g<em>Tcos 53 for left side and 5g</em>Tcos37 for the right side</p>
| |applied-mechanics|acceleration| | <p>As posed, you have three unknowns - <span class="math-container">$a1$</span>, <span class="math-container">$a2$</span>, and <span class="math-container">$a3$</span>. The tension on the cable is everywhere uniform, but the accelerations are different because the static force components affecting the tension are different.</p>
<p><span class="math-container">$F=ma$</span> at each mass. So for mass <span class="math-container">$m1$</span>, the tension is<br />
<span class="math-container">$m1*(g*cos53 +a1)$</span>. Likewise<br />
<span class="math-container">$m2*(g*cos37 + a2)$</span> and<br />
<span class="math-container">$1/2*m3*(g*cos90 +a3)$</span>.</p>
<p>These are all equal. And the length of the line doesn't change.</p>
<p>You can assume that you start from a stationary condition (all velocities are initially zero) - that doesn't change the accelerations. You can choose any initial length that like - that doesn't change the accelerations either.</p>
<p>So <span class="math-container">$m1(g*cos53 + a1) = 1/2 m3(g*cos90 + a3) = m2(g*cos37 +a2) $</span> from equal tension.</p>
<p>And <span class="math-container">$a1 = 2*a3 - a2$</span> from constant length.</p>
<p><span class="math-container">$ 5 kg (0.600g + a1) = 3 kg (1g + a3) = 5 kg ( 0.799g + a2)$</span><br />
<span class="math-container">$0.01 kg + 5 kg * a1/g = 3 kg * a3/g = 0.993 kg + 5 kg * a2/g$</span><br />
<span class="math-container">$0.0033 + 1.666 a1/g = a3/g = 0.331 + 1.666 a2/g$</span></p>
<p><span class="math-container">$0.0033 + 1.666 (2a3 - a2)/g = a3/g$</span><br />
<span class="math-container">$0.0033 - 1.666 a2/g = -2.332 a3/g$</span><br />
<span class="math-container">$-0.0014 + 0.7144 a2/g = a3/g $</span></p>
<p><span class="math-container">$-0.0014 + .7144 a2/g = 0.331 + 1.666 a2/g$</span><br />
<span class="math-container">$-0.3324 = .9516 a2/g$</span></p>
<p><span class="math-container">$a2 = -0.349 g$</span><br />
<span class="math-container">$a3 = -0.2509 g$</span><br />
<span class="math-container">$a1 = -0.1529 g$</span></p>
| 39913 | Why are the two acceleration considered as a1 and a2 here and find acceleration of 6kg mass? |
2021-01-23T16:12:52.453 | <p>Why is the atmospheric pressure given as the value of length of mercury column in barometer. Since the density for the both the cases are different.</p>
<p>Is it that by experiment we found the value to be same or it has some scientific relation with mercury and atmosphere</p>
| |fluid-mechanics| | <p>Because of the fear of being accused of witchcraft, of all things.</p>
<blockquote>
<p>The mercury barometer is the oldest type of barometer, invented by the Italian physicist Evangelista Torricelli in 1643. Torricelli conducted his first barometric experiments using a tube of water. Water is relatively light in weight, so a very tall tube with a large amount of water had to be used in order to compensate for the heavier weight of atmospheric pressure.</p>
<p>Torricelli’s water barometer was more than 10 meters (35 feet) in
height, which rose above the roof of his home! This odd device caused
suspicion among Torricelli’s neighbors, who thought he was involved in
witchcraft. In order to keep his experiments more secretive,
Torricelli deduced that he could create a much smaller barometer using
mercury, a silvery liquid that weighs 14 times as much as water.</p>
</blockquote>
<p><a href="https://www.nationalgeographic.org/encyclopedia/barometer/#:%7E:text=The%20mercury%20barometer%20is%20the,using%20a%20tube%20of%20water" rel="nofollow noreferrer">https://www.nationalgeographic.org/encyclopedia/barometer/#:~:text=The%20mercury%20barometer%20is%20the,using%20a%20tube%20of%20water</a>.</p>
| 39915 | Why is the atmospheric pressure given as the value of length of mercury column in barometer |
2021-01-23T16:15:19.750 | <p>Water is filled in a flask up to a height of 20 cm. The bottom of the flask is circular with radius 10 cm. If the atmospheric pressure is <span class="math-container">$1.013 \times 10^5$</span> Pa,find total force exerted on the bottom of flask. Take g = 10 ms<sup>-2</sup> and density of water = 1000 kgm<sup>-3</sup>. P0 is atmospheric pressure here.</p>
<p>Now, here if I draw the FBD of the base of flask. Then, there are three pressures acting on the base. Atmospheric pressure acting from top and below because fluids exerts pressure in all directions. Then is the hpg pressure.</p>
<p>Total pressure on the bottom of flask is <span class="math-container">$P_0 -P_0 + hpg$</span>. Is this correct if it asked total pressure acting on the base? </p>
| |fluid-mechanics|fluid| | <p>From the inside of the flak the pressure is <span class="math-container">$P_0+ hg\rho$</span>.</p>
<p>If you are looking for the force that secrets force on the base then that force is <span class="math-container">$(P_0+ hg\rho- P_0 )A$</span>.</p>
| 39916 | Finding all the pressure acting on base of cylidner |
2021-01-23T17:36:42.860 | <p>Why is the direction of pressure always perpendicular to surface area of a body for fluids ? We also assume that it is an ideal fluid here.</p>
<p>So , pressure acts in all directions because fluid has a tendency to flow.</p>
<p>Now , in book it says that for a block in water. Pressure must only act perpendicular to the surface area of block because otherwise , there is no friction between adjacent layers of the fluid.
Also ,pressure must act in perpendicular direction on the sides on the tub if a block is put inside a tub which has water contained in it.</p>
<p>How does pressure being perpendicular does not cause that. How can we prove that pressure always act perpendicular in this way.</p>
<p>Also , in real life. This law must not be valid right since it only for making calculations easy? Is that true so.</p>
| |fluid-mechanics|fluid| | <p>This is actually something not easily answered because it is part of the definition of pressure. I will instead point you to other answers which hopefully make sense.</p>
<p>The following is an excerpt from Lumen Physics.</p>
<p>*The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. (See the tire in Figure , for example.)</p>
<p><a href="https://i.stack.imgur.com/b9lO6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b9lO6.png" alt="enter image description here" /></a></p>
<p>Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides. (See Figure 3.)</p>
<p><a href="https://i.stack.imgur.com/479Hj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/479Hj.png" alt="enter image description here" /></a>*</p>
<p>Basically, a key part of the above is saying that pressure is always perpendicular it is because if there were a component of force parallel to the surface, then the object will also exert force on the fluid parallel to it as a consequence of Newton's third law.</p>
<p>Additionally, there is a similar question to <a href="https://physics.stackexchange.com/questions/500341/why-the-force-exerted-by-a-fluid-on-an-object-submerged-in-it-is-always-perpendi">physics stackexchange</a>.</p>
| 39921 | Why is the direction of pressure always perpendicular to surface area for fluids? |
2021-01-23T22:54:45.513 | <p>The question given is</p>
<blockquote>
<p>A uniform rod, of mass <span class="math-container">$m$</span> and length <span class="math-container">$2a$</span> smoothly hinged to a vertical wall is connected to a point on the wall above the hinge by a light inelastic string. Find the magnitude and direction of the force on the rod from the hinge.</p>
</blockquote>
<p>The diagram given is</p>
<p><a href="https://i.stack.imgur.com/4w1ME.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4w1ME.png" alt="Diagram 1" /></a></p>
<p>I have constructed the forces on to the freebody diagram as such</p>
<p><a href="https://i.stack.imgur.com/Pqlxh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pqlxh.png" alt="Diagram 2" /></a></p>
<p>Using the vertical component of net force:
<span class="math-container">$$F_x=T \sin(20)-mg+R_V=0$$</span>
<span class="math-container">$$\implies T\sin(20)=mg-R_V$$</span></p>
<p>Using the horizontal component of net force:
<span class="math-container">$$F_y=T \cos(20)-R_H=0$$</span>
<span class="math-container">$$\implies T\cos(20)=R_H$$</span>
<span class="math-container">$$\implies T=\frac{R_H}{\cos(20)}$$</span></p>
<p>Finding the moment around the hinge A:
<span class="math-container">$$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$</span>
<span class="math-container">$$\implies T\sin(20)=\frac{mg}{2}$$</span>
<span class="math-container">$$\implies T=\frac{mg}{2\sin(20)}$$</span></p>
<p>Then, equating the two equations for <span class="math-container">$T\sin(20)$</span> I get:
<span class="math-container">$$\frac{mg}{2}=mg-R_V$$</span>
<span class="math-container">$$\implies 2R_V=mg$$</span></p>
<p>Finally equating both equations for <span class="math-container">$T$</span>:
<span class="math-container">$$\frac{mg}{2\sin(20)}=\frac{R_H}{\cos(20)}$$</span>
<span class="math-container">$$\implies \frac{2R_V}{2\sin(20)}=\frac{R_H}{\cos(20)}$$</span>
<span class="math-container">$$\implies \frac{R_V}{R_H}=\frac{\sin(20)}{\cos(20)}$$</span></p>
<p>And since <span class="math-container">$R_V,R_H$</span> are the vertical and horizontal components of the same force respectively, the <span class="math-container">$tan^{-1}$</span> of this ratio is equal to the angle between the resultant and the horizontal hence:</p>
<p><span class="math-container">$$\tan^{-1}(\frac{\sin(20)}{\cos(20)})=20°$$</span></p>
<p>Which is wrong as the answer given in the book is that the angle is <span class="math-container">$82.7°$</span> and the reaction force is <span class="math-container">$0.474mg$</span>, I am aware that this entire method is a hackjob but I have no clue on how to tackle this question properly, are the reaction forces I drew the problem or is it my working onwards?</p>
<p>How would I tackle this problem properly?</p>
| |mechanical-engineering|statics| | <p>Something is not right. Please check the calculation below.</p>
<p><a href="https://i.stack.imgur.com/DEn52.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DEn52.png" alt="enter image description here" /></a></p>
| 39930 | Finding the reaction force at the hinge |
2021-01-24T03:43:40.930 | <p><a href="https://i.stack.imgur.com/1kgSA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1kgSA.jpg" alt="enter image description here" /></a>This is an unsoldered circuit component that had two metal pieces that weren't making contact inside a thin glass tube. What is this? I haven't seen anything like it before.</p>
| |circuits|electronics| | <p>It looks to me like a <a href="https://en.wikipedia.org/wiki/Reed_switch" rel="noreferrer">reed switch</a> (I can't be really sure though because the image is not clear enough).</p>
<p><a href="https://i.stack.imgur.com/O1SJP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/O1SJP.png" alt="enter image description here" /></a></p>
<p>A reed switch is an electrical switch operated (usually) by an applied magnetic field. A common example of a reed switch application is to detect the opening of a door or windows, when used as a proximity switch for a security alarm.</p>
| 39934 | Unknown circuit component with glass encapsulated contacts |
2021-01-24T08:43:15.777 | <p>I would like to automate a door lock. For reasons I won't get into, I'm not allowed to modify the lock itself, I can only attach things to the "safe side" of the door. On this safe side, instead of a keyhole, there's a knob. I've already attached a continuous servo to this knob (the lock needs multiple turns between its locked and unlocked position). The problem now is that when opening the door manually with a key from the other side instead of via the servo, the microcontroller loses track of what position the servo is in.</p>
<p>I would like to solve this with limit switches in the locked and unlocked position: that way, the microcontroller can instruct the servo to keep turning until the limit switch is depressed. If the lock needed less than a single turn between locking and unlocking, this would be easy to do, but because it need multiple turns, I can't really think of a way to do this.</p>
<p>I am looking for a way that allows two limit switches. I considered using a multiturn potentiometer, but decided against it because I would like to limit additional torque.</p>
<p>An ideal solution would be low-torque, cheap and ideally 3D-printable.</p>
| |design|mechanisms| | <p>You could try this mechanism (please pardon the mspaint drawing):</p>
<p>Notice how the inner shaft has to rotate more than 360 degrees to hit both limit switches. You can use multiple stages to get more rotations.</p>
<p><a href="https://i.stack.imgur.com/nWpNO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nWpNO.png" alt="mechanism drawing" /></a></p>
| 39936 | Multiturn rotational limit switches |
2021-01-24T20:41:39.977 | <p>I wanted to be able to transmit data from a wireless module to an Arduino. It is possible that a human body could come between the wireless module and the Arduino or someone might cover up the module with their hands. So I was looking for some kind of wireless module around 100 MHz because the human body transmits these signals without interference or absorption. I have found many 433 MHz radio wave modules online but nothing around 100 MHz range.</p>
<p>I was wondering if there is any off the shelf component or is there a simple strategy for creating a wireless module around this frequency range on my own?</p>
| |manufacturing-engineering|wireless-communication| | <p>The reason you can't find many modules around 100 MHz is that it's not an open frequency and you generally need a permit to transmit at those frequencies.
Another reason is that you will face some serious interference from FM radio broadcasts that are all in this frequency region.</p>
<p>I'm not sure what size you imagine your module being but at 100 MHz the antennas start to become unpractical, lower frequencies means longer wavelengths which require longer antennas.</p>
<p>I would recommend starting with a 433 MHz radio and see if you encounter problems before you go to the lower frequencies.
Note also that two 433 MHz modules can have vastly different performance depending on their antenna design, amplification circuit, pcb-design, etc. so don't dismiss 433 MHz just because you have a poor experience with one module. A popular choice is the RFM65/69 (69 is 915 MHz) modules.
Also have a look at LoRa, they offer excellent range but I have no idea about how they are affected by being blocked by a body.</p>
| 39947 | Wireless communication modules transmitting through body |
2021-01-25T02:16:29.243 | <p><a href="https://i.stack.imgur.com/Kqimq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kqimq.jpg" alt="enter image description here" /></a></p>
<p>Relying on my past knowledge on how to attack the problem, I should use the equation
Moment of B about AC = <span class="math-container">$r_{AB} \times B • n^ AC$</span>
To find the least force on B.</p>
<p>I just don't know what position vector I should use for B, I am certain that I should not use points A OR C as a reference to get its position vector. That's what I believe. Or should I use it? Or the way I know is not the right way of solving? Is there any other way?</p>
<p>Sharing what I've done so far, I tried solving the midpoint of line AC, thinking force B could be from point B to the midpoint of line AC. Say, I've got my position vector for force B, I continued solving for <span class="math-container">$r×B$</span>, then the dot product of the answer and unit vector AC, and I've got force B as 28 lb., and since my process is unclear surely my answer was wrong, too.</p>
| |statics| | <p>You need to compute the length of the moment arm from <span class="math-container">$B$</span> to <span class="math-container">$\overline{AC}$</span></p>
<p>That is the height of the triangle, and can be computed directly from tuples as <span class="math-container">$\frac{\|\overline{AC} \times \overline{AB} \|}{\|\overline{AC}\|}$</span></p>
<p>So <span class="math-container">$\overline{AC}\times\overline{AB}= \{72,108,54\}\qquad\|\overline{AC}\times\overline{AB}\|= 140.58\,sqin$</span></p>
<p>and <span class="math-container">$\overline{AC}=\{-9,6,0\}\qquad\|\overline{AC}\|=10.82\,in$</span></p>
<p><span class="math-container">$140.58\, sqin/10.82\,in=13\,in$</span></p>
<p><span class="math-container">$260\, in\, lbf/13\, in = 20\, lbf$</span></p>
| 39950 | Least force required |
2021-01-26T09:35:14.473 | <p>While grounding power sockets in modular prefabricated steel buildings (see Light Gauge Steel Frame Technology), can we just connect the ground wire to the steel frame of the building itself? as the entire frame is grounded. Would that eliminate the need of other systems required for grounding?</p>
<p>No one can actually “touch” the steel frame, if you’ve concerns about the entire structure being on live power. There are layers of insulation: rock wool between the panels of the steel frame, then OSB boards or thermopine, then entire structure is covered with water proof membrane, then cladded with steel, and on the inside there’s thermopine and wood/paint on cement sheet.</p>
<p>Can someone tell me how can I get to know if such practices are allowed in the US? Which local authority to contact?</p>
| |electrical-engineering|materials|building-design| | <p>In the US, electrical installations are primarily governed by the National Electrical Code, also known as NEC or NFPA 70 (It's official designation.) You can read the code for free (with an account) on NFPA's website. Section 250.104(C) of the 2017 code defines the requirements for bonding of "Structural Metal." It requires bonding of exposed structural metal that is likely to become energized. One can argue the definitions of <em>exposed</em> and <em>likely</em> with the inspector, but if there is insulation-on-stud contact, abrasion is possible, and if there are screws penetrating the layers of "insulation" you reference, then it's possible for a person to make electrical contact with the wall framing. Do note that even without screws penetrating, rock wool, OSB, and waterproofing membranes are not usually tested and rated for an insulating value. While they have some insulation value, if you can't prove that the breakdown voltage is per code, they will not be considered as insulators. True insulators are usually rated and listed by an NRTL such as UL.</p>
<p>For residential applications, the International Residential Code also applies, and it has some specific electrical requirements (see section E3609) but they are not specific on this matter. If your prefab buildings aren't one or two family residences, then this code won't apply anyway.</p>
<p>In all cases, the safest thing is to discuss it with the electrical plan checker at the department of building ad safety for the municipality or county that the project will be installed in. As long as the plans clearly show what will and won't be bonded, and the plan checker agrees that you're complying with the code, you can usually explain the logic to an inspector.</p>
| 39972 | Is this alternate grounding practice practically implementable? |
2021-01-26T19:29:26.170 | <p>Reading <a href="https://www.steelconstruction.info/Member_design" rel="nofollow noreferrer">here</a> on Eurocode.</p>
<p>It says that when a cross section is subject to both normal force and moment their effect should be combined like this:</p>
<p><span class="math-container">$$\frac{N_{Ed}}{N_{Rd}} + \frac{M_{y, Ed}}{M_{y,Rd}} + \frac{M_{z, Ed}}{M_{z, Rd}} \leq 1$$</span></p>
<p>Why this? Shouldn't we individually check that the ratio of neither design force to design resistance for force nor ratio of design moment to design resistance for moment are less than one?</p>
<p>Why are they summed together like this? I just don't get the mathematical reason for this.</p>
| |structural-engineering| | <p>If we look at each of those fractions individually, their meaning is obvious. Obviously, for a structure under axial load (or bending moment), the applied force must be lower than what the structure can hold.</p>
<p>But at a fundamental level, what matters isn't the force or moment that's applied, but the stress the structure is under. If you have a beam with a bending moment that's 99% of what the beam can support and then you apply tension equal to 50% of the resistance, you'll obviously have some areas where the combined tension from the bending moment and force is greater than what the beam can take and it'll collapse.</p>
<p>So the question is how to handle these combined cases.</p>
<p>For a simplistic calculation, we just need to think about how all of these loads are cumulative in at least one point of the beam: axial loads are uniform across the section, bending around the horizontal axis generates (for example) tension at the bottom fibers and moment around the vertical axis generates tension on the right-side fibers. So the bottom-right fibers get put into tension from all of these applied loads.</p>
<p>So we need to make sure that the combined stress from all of these loads is lower than the resistance (which I'll simplify to the yield strength <span class="math-container">$f_y$</span>).</p>
<p>And to do so we just need to make the simple observation that each of those fractions describes what fraction of <span class="math-container">$f_y$</span> is taken by each of those solicitations. For example, for normal forces:</p>
<p><span class="math-container">$$\begin{gather}
\frac{\sigma_N}{f_y} = \frac{N_E/A}{N_R/A} = \frac{N_E}{N_R} \\
\therefore \sigma_N = \frac{N_E}{N_R}f_y
\end{gather}$$</span></p>
<p>With bending moments, the function is different but the conclusion is the same.</p>
<p>So all we're saying with that sum of fractions is that the fraction of the resistance taken up by the normal force plus the fraction taken up by the horizontal and vertical bending moments must be less than one. If you want to make that explicit:</p>
<p><span class="math-container">$$\frac{N_{Ed}}{N_{Rd}}f_y + \frac{M_{y, Ed}}{M_{y,Rd}}f_y + \frac{M_{z, Ed}}{M_{z, Rd}}f_y \leq f_y$$</span></p>
<p>There are some interactions we can take into consideration that allow us to go over this threshold to a certain extent, but this is a straightforward conservative solution.</p>
| 39977 | Why are actions combined in this way Eurocode? |
2021-01-27T10:46:27.513 | <p>Hello everyone this is my first post to the forum. I want to control 2 stepper motors using RS-232 as a way of directly communicating with step drivers from my desktop. I am new to communication standards such as RS-232, 422, 485. I have two questions. One, is RS-232 like an I2C network where Devices such as sensors, servo drivers(PCA9685) are given an address and then you can program what happens to that device via its address? If not how does RS-232 work in a use case such as robotics where you have to control step motors? Thank you for any and or all help. My goal is to be able to control 2 nema17 or 23 motors and write python code to control them. I want to use a solution other than usb/arduino for the connection that gets made between motor controllers and the computer.</p>
| |electrical-engineering|motors|robotics|computer-engineering|stepper-motor| | <p>Try researching USB to UART solutions as an alternate. RS232 ports on desktop computers are thing of the past, thus you are limiting your design. USB port are pretty common with computers thus a USB -> UART might be simpler modern solution with added features.</p>
<p><a href="https://i.stack.imgur.com/RzRnj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RzRnj.jpg" alt="USB -> UART" /></a></p>
<p>Note this is still a 1 - 1 communication, but with additional USB port on the desktop, the design can be scaled easily.</p>
<p>Note: One of the differences between UART and RS232 is the voltage levels. UART is 3.3V or 5.0V, RS232 is +/- 15V</p>
<hr />
<p><strong>References:</strong></p>
<ul>
<li><a href="https://electronics.stackexchange.com/questions/110478/difference-between-uart-and-rs-232">Difference between UART and RS-232?</a></li>
<li><a href="https://www.sparkfun.com/products/12977" rel="nofollow noreferrer">USB to TTL Serial Cable</a></li>
</ul>
| 39984 | how can I control 2 stepper motors with RS-232? |
2021-01-27T14:50:03.573 | <p>In ASCE 7-16 there is a load combination as:
1.2D + 1.0E + L + 0.2S
but then there is also another combination as:
0.9D + 1.0E
In both of them earthquake load is the same so why would we use the second one, while the first one would create the worse combination with 1.2 instead of 0.9 for dead load? What am I missing exactly?</p>
| |structural-engineering|civil-engineering| | <p>Dead loads can, in some cases, be beneficial to the structure's safety.</p>
<p>For instance, say you have a simply-supported beam with a cantilever and you want to find its bending moment envelope. Depending on the length of the cantilever, it's possible the simply supported segment will mostly be under negative bending moment. So when calculating the positive bending moment envelope (from dead and live loads), the dead loads will actually "cancel out" some of the live load's results. In such cases, you'd then want to use 0.9 instead of 1.2 as the dead load factor.</p>
| 39988 | The difference between two load combinations in LRFD |
2021-01-27T20:58:29.207 | <p>Consider the problem stated as follows:</p>
<p>A signal y passes through a high pass filter <span class="math-container">$\frac{s}{s + ω }$</span>. A high pass filter with cutoff frequency ω isolates the variations of this optimized variable from its average value. The state that represents the high pass filter is denoted by η.
In the literature, the equation is represented as: <span class="math-container">$\frac{d η}{d t } = - ω η + ω y$</span>.</p>
<p>My question is, which principle did they use to arrive at that equation.
I know that the inverse Laplace transform of <span class="math-container">$\frac{s}{s + ω }$</span>. will result in a differential equation, as it is being multiplied by s. However, I don’t know how to compute <span class="math-container">$\frac{s}{s + ω } * y$</span>.</p>
<p>I have attached a screenshot of the control diag<a href="https://i.stack.imgur.com/1nMAX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1nMAX.png" alt="enter image description here" /></a>ram. I will really appreciate it if someone can please explain this concept to me.</p>
| |control-engineering|control-theory|robotics|nonlinear-control| | <p>It can be noted that the high-pass filter can also be written as</p>
<p><span class="math-container">$$
\frac{s}{s + \omega} = 1 - \frac{\omega}{s + \omega},
$$</span></p>
<p>such that</p>
<p><span class="math-container">$$
\frac{s}{s + \omega} Y(s) = Y(s) - \underbrace{\frac{\omega}{s + \omega} Y(s)}_{\mathcal{L}\{\eta(t)\}(s)},
$$</span></p>
<p>with <span class="math-container">$Y(s) = \mathcal{L}\{y(t)\}(s)$</span> denoting the Laplace transform of <span class="math-container">$y(t)$</span>.</p>
<p>In order to show that the second term indeed matches <span class="math-container">$\mathcal{L}\{\eta(t)\}(s)$</span> one can use that multiplying by <span class="math-container">$s$</span> in the <span class="math-container">$s$</span>-domain is equivalent to taking the derivative with respect to time in the time-domain. Therefore, the following equations are all equivalent</p>
<p><span class="math-container">\begin{align}
\mathcal{L}\{\eta(t)\}(s) &= \frac{\omega}{s + \omega} \mathcal{L}\{y(t)\}(s), \\
(s + \omega)\,\mathcal{L}\{\eta(t)\}(s) &= \omega\,\mathcal{L}\{y(t)\}(s), \\
\frac{d\,\eta(t)}{dt} + \omega\,\eta(t) &= \omega\,y(t), \\
\frac{d\,\eta(t)}{dt} &= \omega\,(y(t) - \eta(t)).
\end{align}</span></p>
<hr />
<p>One could also try to calculate <span class="math-container">$\frac{s}{s + \omega} Y(s)$</span> directly by first differentiating <span class="math-container">$y(t)$</span> with respect to time and then passing it through a low-pass filter. However, in general one can not calculate the derivative of every input <span class="math-container">$y(t)$</span>. Therefore, splitting the high-pass filter into a direct feedthrough minus a low-pass filter makes for a more robust of a calculating the high-pass filter output.</p>
| 40003 | High pass filter and differential equation relationship |
2021-01-28T09:12:51.173 | <p><a href="https://i.stack.imgur.com/bXJjh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bXJjh.png" alt="a tapered bar" /></a></p>
<p><a href="https://classes.mst.edu/civeng2210/concepts/05/elongation/index.html" rel="nofollow noreferrer">https://classes.mst.edu/civeng2210/concepts/05/elongation/index.html</a></p>
<p>If the bar is separated in to segment AC and segment CB, to maintain its equilibrium, Shouldn't the N(x)s be directed in opposite?</p>
<p>In Static, I learned that if we apply two equal, opposite forces at the ends of a member and cut it at the middle, there should be internal forces equal and opposite to each forces at the ends of the segments like this.</p>
<p><a href="https://i.stack.imgur.com/a5iMh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a5iMh.png" alt="a member cut into two pieces" /></a></p>
<p>PS:I should've been clearer. What I'm asking is why all the forces(P and N(x))of the portion AC in the first pic are pointing right, whereas F and F' of the portion AC in the second pic are in different directions. Why is that? I'm totally confused.</p>
| |mechanical-engineering| | <p>From both images you have the internal forces (<span class="math-container">$N(x)$</span>) are opposite for facing cross-sections.
<a href="https://i.stack.imgur.com/BvuRA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BvuRA.png" alt="enter image description here" /></a></p>
<p>The portion denoted as C, is a very thin cylindrical disk with two opposite facing surfaces.</p>
<p><strong>PS:</strong> This is not a real answer. I need to show the image. If the question updates, I will probably update this post to something more meaningful.</p>
<hr />
<p><strong>UPDATE</strong></p>
<p><code>PS:I should've been clearer. What I'm asking is why all the forces(P and N(x))of the portion AC in the first pic are pointing right, whereas F and F' of the portion AC in the second pic are in different directions. Why is that? I'm totally confused.</code></p>
<p>I assume you are referring to the green encircled arrows in the image below.</p>
<p><a href="https://i.stack.imgur.com/IpKeJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IpKeJ.png" alt="enter image description here" /></a></p>
<p>The reason is that the author of that image wants to point out that when P is applied at the end, then at any point between P and the support there is a tensile force.</p>
<p>Essentially that the beam "<strong>carries</strong>" along the force between the point of application (B), and the support (A). In my native language, a more generic term to describe the beam (or any system that "carries" the force) is literally translated as "carrier" (I've not seen it in English though).</p>
<p>The point is that when you take a sliver (a thin disk of material) along AB , at any point you would get C the same behavior as the sliver C. Actually, it doesn't have to be a thin disk of length dx. A portion of any thickness between A and B would exhibit the same behavior (such as the bottom figure with FF' at the original post).</p>
| 40012 | The direction of the internal axial force of a tapered bar |
2021-01-28T11:04:21.147 | <p>In LRFD strength reduction factors depend on the material and the applied load (shear, axial moment) too? Or just material? Are they also called resistance factor?</p>
| |structural-engineering| | <p>In general speaking, the strength reduction factor reflect the confidence on a structural element's ability to withstand a certain type of loading. It is derived statistically from both the research experiments and the past experiences; and it is based on the structural response/behavior of a certain type of material subjected to a certain type of load. In this sense, you are correct in calling it a "resistance factor", which is always less than 1.</p>
| 40015 | What affects LRFD strength reduction factors? |
2021-01-28T11:37:07.530 | <p>I'm trying to make an evaluation of the available 3D printing options of polypropylene and I'm having difficulties comprehending material characteristics that make PP PP-like.</p>
<p>I realize this sounds a bit non-exact, but please bear with me. Polypropylene is very beneficial for this specific product because it handles repeated large plastic deformations very well and won't break. When you look at the material properties and compare them to other plastics for injection moulding, one characteristic stands out - elongation at break, which can be up to 150% or more for PP for injection moulding (this value is from a generic database, actual materials have similar values).</p>
<p>When I checked the HP's PP mateiral and Stratasys's Durus PP-like mateiral, their elongation at break was very low, around 20%, which is less than PA11 for SLS 3D printing.</p>
<p>Maybe elongation at break is not a good parameter to evaluate the response to large plastic deformation I mentioned. Maybe someone has more insight into this.</p>
<p>EDIT: I tested the PP printed part for bending fatigue. The part was printed on a HP fusion jet printer and the results are very similar to the injection molded part.</p>
| |materials|3d-printing| | <p>I tested the PP printed part for bending fatigue. The part was printed on a HP fusion jet printer and the results are very similar to the injection molded part.</p>
| 40016 | Polypropylene material characterization |
2021-01-28T12:18:08.893 | <p>My background is in Environmental Science, although I have recently been taught in the basics for energy engineering especially for solar panels.</p>
<p>I'm currently using PVsyst to build solar panels on a given building at a given location. However, I'm finding it difficult to explain and understand this diagram with my limited circuit knowledge.</p>
<p>Can someone offer a detailed explanation of the components along with what they represent in the network and what the entire picture means? I know that 1. represents a diode, and it looks to me that 2. represents a European-resistor, I partially know this because of basic practice in LtSpice.</p>
<p><a href="https://i.stack.imgur.com/NlbZd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NlbZd.png" alt="enter image description here" /></a></p>
| |electrical-engineering|circuits|solar-energy|circuit-design|photovoltaics| | <ol>
<li><strong>Diode</strong> Like you assumed</li>
<li><strong>Panels</strong>: These are the PV panenls that produce current. They produce DC current I and DC voltage V</li>
<li><strong>Inverter</strong>: the inverter converts the DC current and Voltage from the panels to AC voltage and current, which is in sync with the Grid. This part is responsible for converting the energy from the panels into the energy that the grid can transfer.</li>
<li><strong>User</strong>: the user is a load that is not in the grid. This makes sense in the case of net metering, where the producer of energy can use some of the energy for their own needs.</li>
<li><strong>Grid</strong>: this is the electrical distribution grid. This transfers the energy to other parts where it maybe consumed.</li>
</ol>
| 40017 | Explanation for photovoltaic circuit |
2021-01-28T12:58:03.400 | <p>There is few material on net related to stroboscope applications. since it is an optical device, it is best used to control the vibrations of rotary devices like turbine blades electromotors etc.</p>
<p>but is it also useful if you want to control the vibrations of a lump mass ( big and heavy structure) doing very small vibration in its place in very high frequencies ? I don't give a number of how much or how big ( up to you to consider the proportions)</p>
| |mechanical-engineering|automotive-engineering|vibration| | <blockquote>
<p>Is stroboscope a useful device to control the translational vibrations?</p>
</blockquote>
<p>No.</p>
<blockquote>
<p>A stroboscope, also known as a strobe, is an instrument used to make a cyclically moving object appear to be slow-moving, or stationary. It consists of either a rotating disk with slots or holes or a lamp such as a flashtube which produces brief repetitive flashes of light. <a href="https://en.wikipedia.org/wiki/Stroboscope" rel="nofollow noreferrer">Wikipedia</a>.</p>
</blockquote>
<p>A classic example is the stroboscope used to set the timing on a petrol engine. The strobe lamp is triggered by the ignition system and the light pointed at the timing belt pulley which has a mark on it. The effect is to "freeze" the timing mark against the timing scale mounted on the engine block so that you can read the timing angle.</p>
<blockquote>
<p>There is few material on net related to stroboscope applications.</p>
</blockquote>
<p>I'm sure there are thousands of articles and videos.</p>
<blockquote>
<p>Since it is an optical device, it is best used to control the vibrations of rotary devices like turbine blades electromotors etc.</p>
</blockquote>
<p>No. It is used for inspection, not control, although it is feasible that a vision system could use the data to correct a process.</p>
<blockquote>
<p>But is it also useful if you want to control the vibrations of a lump mass (big and heavy structure) doing very small vibration in its place in very high frequencies?</p>
</blockquote>
<p>Probably not.</p>
<p>Have a look at <a href="https://www.youtube.com/watch?v=rEoc0YoALt0&t=720s&ab_channel=SteveMould" rel="nofollow noreferrer">Amplified video motion</a> which will give you some ideas for search terms.</p>
| 40018 | Is stroboscope a useful device to control the translational vibrations? |
2021-01-28T13:54:59.493 | <p>I'm trying to understand an efficiency curve from PVsyst, using this <a href="https://files.pvsyst.com/help/index.html?inverter_euroeff.htm" rel="nofollow noreferrer">documentation</a></p>
<p>However, I'm not sure what this sentence means:</p>
<ul>
<li>This production is penalized by an <strong>ohmic loss</strong><span class="math-container">$^1$</span> of the internal components (transformer and transistors), which increases <strong>quadratically</strong><span class="math-container">$^2$</span> with power <strong>(like R · I²)</strong><span class="math-container">$^3$</span>. *</li>
</ul>
<ol>
<li>What is meant by ohmic loss in this instance?</li>
<li>When it mentions that it increases quadratically, does it mean in quadratic form (equation) i.e. <span class="math-container">$ax^2 + bx + c = 0$</span> and how does this relate to the efficiency curve of an inverter, relative to power loss?</li>
<li>What does the equation represent, and what does it suggest when either parameter <span class="math-container">$R$</span> or <span class="math-container">$I^2$</span> is greater than the other?</li>
</ol>
| |electrical-engineering|energy-efficiency|solar-energy|photovoltaics| | <p>I <em>believe</em> what this sentence is trying to convey is that all circuit components have a small resistance associated with them if you look at them in enough detail. For example a transformer circuit can be viewed as (this is from <a href="https://electricalacademia.com/transformer/equivalent-circuit-transformer-referred-primary-secondary-side/" rel="nofollow noreferrer">electrical academia website</a>:</p>
<p><a href="https://i.stack.imgur.com/wi3VZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wi3VZ.png" alt="enter image description here" /></a></p>
<p>The reason that the resistance is there is because there are wires in the system (unless of course you happen to have superconductive wires). When current pass through that resistance energy is converted to heat, and cannot be reclaimed. The higher the current passing through the system the higher the energy lost.</p>
<p>Regarding the <strong>quadratic</strong>. If you double the current through the system, then the heat losses will quadruple.</p>
<p>Regarding the third part of your question I am not certain what equation you refer to, and what is the exact meaning of your question.</p>
| 40020 | explaining ohmic loss and understanding inverter efficiency curve |
2021-01-28T15:35:51.900 | <p><a href="https://i.stack.imgur.com/VwN0C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VwN0C.png" alt="a statically determinate truss" /></a><a href="https://books.google.co.kr/books?id=1dUKAAAAQBAJ&pg=PA151&lpg=PA151&dq=because%20the%20truss%20is%20statically%20determinate%20both%20bars%20are%20free%20to%20lengthen&source=bl&ots=5Sg6fZ7Msy&sig=ACfU3U2Bmwdr0AJjQU6zawiccFi7h5R3aQ&hl=ko&sa=X&ved=2ahUKEwjoq7bK-L7uAhWW-mEKHX5eC2YQ6AEwCXoECAcQAg#v=onepage&q=because%20the%20truss%20is%20statically%20determinate%20both%20bars%20are%20free%20to%20lengthen&f=false" rel="nofollow noreferrer">source</a></p>
<blockquote>
<p>Because the truss is statically determinate, both bars are free to
lengthen or shorten, resulting in a displacement of joint B.</p>
</blockquote>
<p>Why is being statically determinates related to being free to lengthen or shorten?</p>
| |mechanical-engineering| | <p>In an indeterminate truss, the strain and geometry of the members play a significant role. So if we make one shorter or longer the balance of geometry and the distribution of forces will drastically change.</p>
<p>imagine a cantilever beam supported by a prop at the free end. Depending on the length of that prop the cantilever beam can be:</p>
<ul>
<li><p>Partially supported</p>
</li>
<li><p>Not supported at all, even being pulled down,</p>
</li>
<li><p>Or the prop is so tall that it not only cancels the loads on the beam but pus the beam under prestress.</p>
</li>
</ul>
<p>But in the truss in your figure, if you increase the length of one of the members, that node moves accordingly, up, down, or to the side. there is no resistance from other members to the movement of the node, they can guide its movement though. of course, if you change the geometry you have to calculate new forces.</p>
| 40023 | Why if the truss is statically determinate, both bars are free to lengthen or shorten? |
2021-01-29T05:34:02.477 | <p>It says <span class="math-container">$V\sin \alpha$</span> (component of velocity perpendicular to OP) is the cause of angular displacement. How is it?</p>
<p>if only <span class="math-container">$V\cos \alpha$</span> existed, we need not turn over head to always to look at a particle. What does this mean?</p>
<p>Also, Can we say that <span class="math-container">$V$</span> is the linear speed? Because that is the one tangential in direction.</p>
<p>How is <span class="math-container">$PQ = OP \cdot \Delta \theta$</span> ?</p>
<p><a href="https://i.stack.imgur.com/pHAC1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pHAC1.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|applied-mechanics|kinematics|linear-motion| | <p>P is a particle subjected to the linear velocity vector V.</p>
<p>this vector has two orthagonal components, <span class="math-container">$V_{sin} \alpha \ and \ V_{cos}\alpha , $</span></p>
<p><span class="math-container">$ V_{cos} \alpha, $</span> will move the particle up radially but has no effect on its angular movement because it is orthogonal to the angular rotation.</p>
<p><span class="math-container">$ V_{sin }, $</span> on the other hand, can and is the only component that causes angular rotation of particle P because it is perpendicular to OP, Your figure does not show this clearly and that may be the cause of your confusion.</p>
<p>I guess your text means if there was only <span class="math-container">$ V_{cos}\alpha$</span> it meant your particle will only move our and up like a rocket and you'd need to look back to see the particle.</p>
<p>And the answer to your other part is yes, V is the linear speed.</p>
<p>And <span class="math-container">$PQ=OP*\Delta \theta $</span>, not <span class="math-container">$ \ PQ= OP* V_{sin} \alpha$</span> because the linear speed is not necessarily always slanted at an angle of <span class="math-container">$ \alpha $</span>, it can change and cause a change in rotation speed. So how much turn we have after sometime depends on the curve of the slant with the tangent to rotation.</p>
| 40037 | Did not understand this regarding angular velocity |
2021-01-29T07:07:06.860 | <p>I have a glass here.<a href="https://i.stack.imgur.com/q3vPJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q3vPJ.jpg" alt="enter image description here" /></a></p>
<p>This is the bottom face of the glass.
<a href="https://i.stack.imgur.com/kTUeF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kTUeF.jpg" alt="enter image description here" /></a></p>
<p>Now this is on what I had kept my glass. There is a little much of water spilled on it. Let us call this thing as C</p>
<p><a href="https://i.stack.imgur.com/MSR2A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MSR2A.jpg" alt="enter image description here" /></a></p>
<p>Now , when I lifted the cup. The C on which I had kept my glass also lifted up. The bottom portion of glass and the top portion of the thing were in contact. They did not leave each other.</p>
<p>Now , the bottom portion of the cup is flat. So, I kept my glass on that C. The cup and the C both lifted up. Why is it like that ? They should not do this.</p>
| |fluid-mechanics|fluid| | <p>The surface of C where the glass sat is smooth, then the glass may have a slightly concave bottom and the water, with capillary action, managed to form a seal between the two surfaces. Even if the glass bottom and surface C are both flat, capillary action can still form a seal around the edge leaving a void in the centre, causing the same effect of lower v. higher pressure.</p>
<p>If the air trapped in there then cools down a few degrees - the outside air pressure holds the two items together. Until of course mass or vibration takes over and the item C falls off.</p>
<p>Happens often with coasters and glasses.</p>
<p>A hot cup of tea can show bubbles when there is some spilt tea in the saucer - the trapped air is being heated, until the tea in the cup cools and we have the same situation as above.</p>
| 40038 | Why does this glass tend to stick on surfaces |
2021-01-29T10:30:14.270 | <p><a href="https://i.stack.imgur.com/LZU2s.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LZU2s.jpg" alt="enter image description here" /></a></p>
<p>Is units of angular displacement never a degree and is always a radian ?</p>
| |applied-mechanics| | <p>I suspect this book is related to motion dynamics. The reason why it states that angular measurements are in radians only, is probably because it tries to avoid confusion and ambiguity.</p>
<p>The problem is that in most technical subjects degrees are used for angular measurements. However, when you start delving into physics and calculus- which are a prerequisite in your case since you are probably reading about dynamics- using radians make more sense. There are two main reasons (+1 which some times is as important) for that.</p>
<ol>
<li>There is very straight forward relationship between the <strong>arc of a circle <span class="math-container">$L$</span></strong> and the <strong>angle</strong> <span class="math-container">$\theta$</span> when it is represented in radians.
<span class="math-container">$$L = r\cdot \theta[rad]$$</span></li>
</ol>
<p><a href="https://i.stack.imgur.com/0CNfI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0CNfI.png" alt="enter image description here" /></a></p>
<p>The same equation in degrees would be: <span class="math-container">$L = r\cdot \theta[deg] \cdot\frac{\pi}{180}$</span>. Notice the extra factor <span class="math-container">$\frac{\pi}{180}$</span></p>
<ol start="2">
<li>Similarly (and more importantly) angles in radians produce much simpler forms for angular displacement derivatives.</li>
</ol>
<p>So for example:</p>
<p><span class="math-container">$$\frac{d }{dx} \sin\theta[rad] = \cos\theta[rad]$$</span></p>
<p>The equivalent for degrees would be:</p>
<p><span class="math-container">$$\frac{d }{dx} \sin\theta[deg] = \frac{\pi}{180} \cos\theta[deg]$$</span></p>
<p>It gets even worse for the double derivative with respect to time because then you have:</p>
<p><span class="math-container">$$\frac{d^2 }{dx^2} \sin\theta[deg] = -\left(\frac{\pi}{180}\right)^2 \sin\theta[deg]$$</span></p>
<p>while when you are using radians its simply:</p>
<p><span class="math-container">$$\frac{d^2 }{dx^2} \sin\theta[rad] = -\sin\theta[rad]$$</span></p>
<ol start="3">
<li>Small angle approximation <span class="math-container">$\sin\theta\approx \theta$</span> works better with radians (because <span class="math-container">$1 [deg]\approx \frac{1}{56}rad$</span>).</li>
</ol>
<p><strong>Bottom Line</strong>: You can use both degrees and radians for angular displacement, however when you need to take derivatives wrt to time, radians make it easier.</p>
| 40045 | Is dimension of angular displacement never degree? |
2021-01-29T12:34:01.373 | <p>I saw this in my textbook that</p>
<p><span class="math-container">$\dfrac{\text{d}K}{\text{d}t} = \dfrac{\text{d}(1/2mv^2)}{\text{d}t}= F_t v$</span> where <span class="math-container">$F_t$</span> is the resultant tangential force on the body.</p>
<p>Now, let <span class="math-container">$F_t= F\cos\theta$</span></p>
<p><span class="math-container">$$\dfrac{\text{d}K}{\text{d}t} = F\cos\theta v = Fv = \dfrac{F\text{d}r}{\text{d}t} = F\text{d}r$$</span></p>
<p>So , by writing resultant tangential force . Does it mean that resultant force on the body only or this <span class="math-container">$F_t$</span> is a component of resultant force .</p>
<p>If it is resultant force and not a component of it. Then why to write tangential ?</p>
<p>Do we also have radial force then. A resultant must have two components where <span class="math-container">$F\sin\theta$</span> is the radial force or 2nd component. So , we can’t use dot product here but have to use magnitude of cross product right.. So , will that be right way to solve it ? .</p>
<p><span class="math-container">$F \times \text{d}r$</span> (vector cross product and radial force)?</p>
<p><a href="https://i.stack.imgur.com/kzcKg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kzcKg.jpg" alt="enter image description here" /></a></p>
| |power|energy| | <p>To answer your initial question "Can we say kinetic energy is tangential K.E?": <strong>No</strong></p>
<p>You have to remember that kinetic energy is a scalar quantity while velocity is a vector quantity. So you can have spatial components for velocity but not for kinetic energy. For a particle, kinetic energy is depended only from velocity's <strong>magnitude</strong> (which happens to be the Euclidean norm of the velocity <span class="math-container">$\|v\|=\sqrt{v_x^2+v_y^2+v_z^2}$</span>).</p>
<p>Of course Kinetic energy can have "components" but they have a totally different meaning. E.g. for a rigid body kinetic energy can be either translational <span class="math-container">$\frac{1}{2}mv^2$</span> or rotational <span class="math-container">$\frac{1}{2}I \omega^2$</span>.</p>
<p>Now regarding, the equation you are showing, this is most likely from a chapter with respect to Langrange's method for obtaining the equations of motion. In the case you are showing, you are probably describing a planar curvilinear motion. In that case, the resultant force on the body is non zero. The force is also a vector (like velocity). The force can be analysed to components parallel to the velocity and perpendicular.</p>
<ul>
<li>the parallel component will change the magnitude of the velocity (see Kinetic Energy)</li>
<li>the perpendicular component will change the direction (does not affect kinetic energy).</li>
</ul>
| 40049 | Can we say kinetic energy is tangential K.E? |
2021-01-29T14:43:09.850 | <p>given that handy stroboscope can also check translational vibration of parts, is it a substitute for a test that can be done by accelerometer for stationary vibrating equipment?</p>
| |mechanical-engineering|vibration| | <p>No. Strobes are useful for what they can do, but they are not substitutes for accelerometers.</p>
<p>A strobe can give you a "picture", but it doesn't <em>measure</em> anything except maybe frequency. It doesn't measure amplitude, harmonic content of the vibration, etc.</p>
<p>Also a strobe is only useful for amplitudes which are big enough to see. Accelerometers can measure high frequency vibrations with amplitudes of micrometers or smaller.</p>
| 40054 | Is stroboscope a substitution for accelerometer testing? |
2021-01-30T09:29:28.263 | <p><a href="https://i.stack.imgur.com/20XAo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/20XAo.png" alt="statically determinate truss" /></a></p>
<p><a href="https://i.stack.imgur.com/BfUuY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BfUuY.png" alt="statically indeterminate truss" /></a></p>
<p><a href="https://books.google.co.kr/books?id=1dUKAAAAQBAJ&pg=PA151&lpg=PA151&dq=because%20the%20truss%20is%20statically%20determinate%20both%20bars%20are%20free%20to%20lengthen&source=bl&ots=5Sg6fZ7Msy&sig=ACfU3U2Bmwdr0AJjQU6zawiccFi7h5R3aQ&hl=ko&sa=X&ved=2ahUKEwjoq7bK-L7uAhWW-mEKHX5eC2YQ6AEwCXoECAcQAg#v=onepage&q=because%20the%20truss%20is%20statically%20determinate%20both%20bars%20are%20free%20to%20lengthen&f=false" rel="nofollow noreferrer">source</a></p>
<p>I assume those supports in the first figure are pins so there are 4 reactions in total.
And as shown in the second figure, there are 3 reactions in total.</p>
<p>I thought for a structure in plane to be statically determinate, it must have only 3 reactions, but these are quite contradictory to me.</p>
| |mechanical-engineering| | <p>We usually describe structures as either statically determinate or indeterminate.</p>
<p>However, this determinacy is actually two-fold: there's internal and external determinacy.</p>
<p><em><strong>External static determinacy</strong></em> describes whether the supports can be determined via static equilibrium equations. This is what you were thinking of when you are looking at the number of supports. However, both examples are actually externally statically determinate: the simple cantilever truss has four supports, but it also has a hinge, which allows us to perform bending moment calculations to one side of the hinge, granting us a fourth equilibrium equation; and the truss frame only has three supports and is therefore trivially easy to solve for its reactions.</p>
<p><em><strong>Internal static determinacy</strong></em> describes whether the internal forces of the members can also be determined via static equilibrium equations. This is what the textbook is actually talking about in this case. As described by <a href="https://engineering.stackexchange.com/a/40065/1832">@kamran's answer</a>, this is a function of the number of members, reactions, and joints (a statically determinate truss has <span class="math-container">$m + R \leq 2j$</span>).</p>
<p>The first truss satisfies this equation (<span class="math-container">$2 + 4 \leq 2\cdot3$</span>), while the second doesn't (<span class="math-container">$6 + 3 \leq 2\cdot4$</span>). Therefore the first truss is internally statically determinate while the second is internally statically indeterminate.</p>
<p>An externally statically indeterminate structure will always be internally statically indeterminate. However, an externally statically determinate structure can be either internally statically determinate or indeterminate.</p>
| 40064 | Is static determinacy regardless of the number of reactions? |
2021-01-30T14:53:26.347 | <p>A man threw a ball with force 10 N to two displacement of 5 m and 10 m.</p>
<ul>
<li><em>F = ma</em>, so let mass be 2 kg.</li>
<li>a = 5 m/s<sup>2</sup>.</li>
</ul>
<p>Work done = 10 × 5 and 10 × 10 = 50 Nm and 100 Nm.</p>
<p>Now, since a = 5, I assumed that after 5 m velocity = 25 [m/s].</p>
<p>Now, when I do K.E f - K.E initial for final work done i.e 50 Nm for now.</p>
<ul>
<li>Initial K.E. = 0.</li>
</ul>
<p>So , then <em>K.E final</em> is not equal to <em>work done final</em>.
25 × 25 is not equal to 50 Nm.</p>
<p>Where am I wrong?</p>
| |power|energy| | <p>The mistake is in your assumption.</p>
<p><strong>Now, since a = 5, I assumed that after 5 m velocity = 25 m/s.</strong></p>
<p>the correct would have been</p>
<p><em>Now, since a = 5, I assumed that after <strong>5 [s]</strong> velocity = 25 m/s.</em></p>
<p>The way you can calculate it is from the following form of the equation for acceleration:</p>
<p><span class="math-container">$$a = u \frac{du}{ds}$$</span></p>
<p>you can obtain that <span class="math-container">$a\cdot 5m = \frac{u_{out}^2}{2} \rightarrow u_{out} = \sqrt{2 \cdot a\cdot 5[m]}$</span>.</p>
<p>The rest I leave to you.</p>
| 40071 | Why am I getting wrong value here? |
2021-01-30T18:24:12.967 | <p>I'm a layman engineer who's decided to take another step forward in my research and work with McKibbens Pneumatic Artificial Muscles. I’m new to creating custom rubber components and laser cutting, so please bear with me. My intent is to be able to order sheets of rubber, which can then be cut in a laser cutter into strips. I would then like to melt the long edges of the strip and fold the rubber strip onto itself, connecting the edges to form a rubber tube. The melting part is for fusing the two sides together. I was wondering what types of rubber would be best suited for this? I was thinking natural or silicone rubber, but again I’m new to making my own rubber components.</p>
| |materials|pneumatic|lasers|material-science| | <p>There are silicone specific adhesives, as silicone does not adhere well to any other material. Appropriately, there are rubber adhesives for genuine rubber as well. The method you describe is going to severely test the adhesion of the material, as it places the most stress along the weakest points in the system.</p>
<p>Have you considered using surgical rubber tubing as the foundation of the system build? I've used this tubing for elastic properties as in a slingshot, but also in an inflation mode (without the sleeve). It works well in the latter instance, as long as the system is not exposed at length to sunshine or chemical fumes of any sort to degrade the silicone. The inflation mode was with water and the resulting rupture was more amusing than dangerous.</p>
<p>A "finger-trap" sleeve as reference in the documentation I've found would help with the longevity of the system.</p>
<p><a href="https://www.mcmaster.com/surgical-tubing/" rel="nofollow noreferrer">McMaster-Carr</a> carries such tubing from inside diameters of 1/16" to 1" and 2 mm to 8 mm.</p>
<p>Response to comments: If joining surfaces is preferred to adhesives, your method of heating the surfaces will determine the strength and is likely to be stronger if the process is done properly.</p>
<p>The information I've seen in this respect use a heated blade. It's typically mounted in a jig which directs the "tubular" form of material in such a way as to contact the heated blade simultaneously, while the jig compresses the shape at the contact point.</p>
<p>It's critical to have sufficient energy imparted to the blade to continually compensate for the heat removed by the melting process and to have a well-regulated speed over the blade.</p>
<p>I've laser cut silicone sheet and the mechanical characteristics of the sheet were not changed, although the edges were "cleaner" as a result. I've not cut latex rubber but a quick search indicates that it is not a problem. Cutting any product containing polyvinyl chloride compounds is dangerous, but latex rubber should not have such a compound.</p>
<p>Consider to edit "patters" to "patterns" as I didn't understand at first the reference, not having read all the papers found for this item you desire to construct.</p>
<p>Another option you may wish to consider is to create your specific shapes using 3D printed molds and casting the latex rubber within. This would provide for your custom modifications as well as reduce/eliminate adhesives and bonding complications. The realm of possibilities with a 3D printed mold is exhaustive.</p>
<p>I use OpenSCAD for creating 3D models, which is useful for repetitive mathematical designs, although many of the CAD programs such as Fusion 360 permit parametric modeling as well.</p>
<p>Due to the flexibility of the rubber, you'd be able to create molds that might otherwise not release the part. Alternative methods would include using a water soluble support filament such as PVA which washes away with water, for more complex designs. Vacuum processing during the mold pour may be necessary for more intricate models, but I'm getting ahead of myself.</p>
<p>Even though marked as accepted, additional thought of the matter also suggests that one may want to consider 3D printed versions using TPU filament. It's very "stretchy" and would allow for infinitely complex structures. I've not tested porosity but the substance melts during the printing process and does not delaminate when stretched.</p>
| 40073 | What Rubber is Best for Laser Cut McKibbens PAMs? |
2021-01-31T14:12:47.357 | <p>In the figure attached for reference, it is given that if <span class="math-container">$R_1$</span> is the resistance element of valve 1 and τ ∝ <span class="math-container">$R_1$</span>, but I am not able to understand it intuitively.</p>
<p>If <span class="math-container">$R_1$</span> is high, output flow through that valve must be less, this means it is easier to store the liquid in the tank. Hence, τ ∝ <span class="math-container">$\frac{1}{R_1}$</span>.</p>
<p>Please tell where I am wrong in my logic. TIA.</p>
<p><a href="https://i.stack.imgur.com/bQKJp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bQKJp.png" alt="enter image description here" /></a></p>
| |electrical-engineering|control-engineering|instrumentation|flow-control|liquid| | <p>If flow is q, and pressure is p, The linear flow characteristic "R" is defined such that p = (q)(R)</p>
<p>This is directly analogous to V = IR in ohms's law, in the electrical analogy.</p>
<p>If we differentiate p=qR, we get dp = (dq)(R)</p>
<p>In this case, pressure is being reckoned in units of height, for which it should be mentioned that p = (ρ)(g)(h) where the greek letter rho (ρ) is the liquid density, g is the gravity (9.8m/s^2), and h is height.</p>
<p>Thus we write dh = (dq)(R)</p>
<p>We substitute this into equation 2.68:</p>
<p>(A)(dh/dt) = qi - q1</p>
<p>which becomes</p>
<p>(A)(R)(dq/dt) = qi - q1</p>
<p>which is a first order ODE, so the solution is exponential form, and the time constant is AR, which is of course proportional to R.</p>
<p>On an qualitative/intuitive level, bigger R means more constricted flow, which means less flow for any given pressure aka height, which means it takes longer for the tank to drain, which means larger time constant. Similarly, bigger A means there is more liquid to be drained for a given height, which means it takes longer, which means larger time constant.</p>
| 40082 | Liquid Level Control System Resistance |
2021-01-31T19:52:52.677 | <p>Looking at both the <a href="https://www.ifixit.com/Teardown/Nest+Learning+Thermostat+2nd+Generation+Teardown/13818" rel="nofollow noreferrer">Nest iFixit teardown</a> and this <a href="https://www.youtube.com/watch?v=iMbIrkTV0fE" rel="nofollow noreferrer">Nest bezel replacement tutorial</a>, it looks like the Nest stainless steel bezel mounts directly to the PCBs and displays inside.</p>
<p>The Nest bezel is able to rotate freely though (and <em>very</em> smoothly, I should add) while the components inside remain fixed - how is this possible? I was under the impression that this smooth, frictionless connection between a rotating body and a stationary body was only really achievable through ball/roller bearings.</p>
| |mechanical-engineering|bearings|product-engineering|prototyping| | <p><a href="https://i.stack.imgur.com/8LVyz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8LVyz.jpg" alt="enter image description here" /></a></p>
<p>Just prior to this point in the video you can see that the display has been inserted from the front and this portion will be added from the back and screwed in. All that is required is to ensure that there is enough space between the two parts for the bezel flange not to be nipped. Many plastics have a low coefficient of friction with steel or nickel plating. Try sliding a stainless steel kitchen knife across various hard plastics and see what you find. (I've never touched a Nest so I don't know what I'm comparing with.)</p>
<p>Links:</p>
<ul>
<li><a href="https://www.tstar.com/blog/coefficient-of-friction-plastics-vs-steel" rel="nofollow noreferrer">Coefficient-of-friction-plastics-vs-steel</a>.</li>
<li><a href="https://www.igus.ie/iglidur/plain-bearing?tab=1" rel="nofollow noreferrer">Igus plastic bearings</a>.</li>
</ul>
| 40086 | How does the Nest Learning Thermostat bezel rotate without bearings? |
2021-02-01T06:38:44.880 | <p>Everyone.</p>
<p>I want to make the transformation form Straight motion to Rotational motion ,like [this].<a href="http://507movements.com/mm_094.html" rel="nofollow noreferrer">1</a>
<a href="https://i.stack.imgur.com/3KgFu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3KgFu.png" alt="enter image description here" /></a></p>
<p>So I made .</p>
<p><a href="https://i.stack.imgur.com/48i1D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/48i1D.png" alt="enter image description here" /></a></p>
<p>But it did not work.
Probably, this was mistake of Groove size on spiral groove.
But I thought I must know the formula between red arrow and blue arrow of this mechanism.</p>
<p>So Would you tell me the formula of this?
And Any clue for this mechanism welcome.</p>
<p>Thank you & Sorry for poor English in advance .</p>
| |solid-mechanics| | <p>This is a quick sketch of a potential mechanism. It may not be 1:1, but is just an idea. Recommend you use this (and others you think of) for brainstorming potential solutions.</p>
<p>As long as the forces being driven can be overcome.</p>
<p><a href="https://i.stack.imgur.com/HDIh3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HDIh3.jpg" alt="Quick sketch, perhaps not linear, but you get the idea. Brainstorm from here." /></a></p>
| 40105 | Transformation of 8mm straight motion to 210 degree motion |
2021-02-01T15:56:19.597 | <h1>What am I trying to control?</h1>
<ul>
<li>I have a <strong>container</strong> with approximately <code>3*2*6</code> meters (width, height, length).</li>
<li>I am controlling through PID the <code>Temperature 2</code> (temperature inside a tubing system) to get <code>Temperature 1 and 3</code> (temperatures near floor and celling of the container) to the desired values (see graphs below).</li>
<li><code>Temperature 2</code> is coming from a sensor inside a tube system through which the air is circulating and <code>Temperature E.</code> is coming from a wooden box inside the container.</li>
<li>Box and tubing is a closed system with some minimal temperature leaks.</li>
<li>The PID’s output is connected to <code>PWM</code> of triacs which heat the air in the tubing system.</li>
<li>EO% is describing how much % of 1 second (the PWM period) the triacs will be set to <code>True</code> = therefor making heat. So if EO is 50, then the triacs will be making heat exactly 0.5 seconds</li>
</ul>
<p><strong>Sketch of my system:</strong></p>
<p><a href="https://i.stack.imgur.com/xGgid.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xGgid.png" alt="enter image description here" /></a></p>
<p><strong>Step response of the system (X axis format is <code>HH:MM:SS</code>):</strong></p>
<p><a href="https://i.stack.imgur.com/3n1SC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3n1SC.png" alt="enter image description here" /></a></p>
<h2>What have I tried so far?</h2>
<ul>
<li>I am running the software on <code>PLC</code> from local distributor (not an Arduino board) where I have implemented this <a href="https://github.com/br3ttb/Arduino-PID-Library/blob/master/PID_v1.cpp" rel="nofollow noreferrer">PID model from Arduino library.</a>. But in the future I will be using Beckhoff PLC's.</li>
<li>I have tried implementing the basic PID controller with little accuracy.</li>
<li>Then I have tried implementing PID controller with <a href="http://brettbeauregard.com/blog/category/pid/coding/" rel="nofollow noreferrer">proportional on measurement and derivative on measurement</a> and my accuracy went higher (link is to a blog talking about the two methods).</li>
<li>Now I can control temperatures with precision to almost <code>0.5 °C</code> and little oscillation.</li>
<li>PWM frequency is 1 HZ and PID sample time is 2 seconds.</li>
</ul>
<h2>What is the problem?</h2>
<ul>
<li>You can see on the pictures below that I slowly increment my PID <code>setpoint</code> through time (PID’s <code>setpoint</code> is the <strong>black line</strong>).</li>
<li>When my system has a temperature in a range from 20 to 55 °C my accuracy is quite good.</li>
<li>But once I get above 55 °C, PID’s is not able to control the temperature accurately anymore and the PID output always result in a spike and therefor even larger inaccuracy.</li>
</ul>
<h2>What can I do to eliminate this problem? And what is causing this behavior? Are there any better methods then the simple PID controller?</h2>
<hr />
<h2>Other observations:</h2>
<ul>
<li>I think the problem lies in the <code>I</code> component of the PID controller.</li>
<li>When the temperature is low the <code>P</code> component manages to control the temperature just right. But when the error is getting bigger and bigger the <em>I</em> component is cumulating and then it overshoots.</li>
<li>When I am running the PID on all temperatures (25, 40 even 58 °C) the accuracy is as I said 0.5 °C and oscillation is minimum. The problem only occurs when I slowly increment the PID’s <code>setpoint</code> and therefore increase the temperature.</li>
</ul>
<hr />
<h1>Pictures (X axis format is <code>HH:MM:SS</code>):</h1>
<p><strong>Picture 1 and 2 (temperatures and the PID's output):</strong></p>
<p><a href="https://i.stack.imgur.com/O6VVM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O6VVM.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/eGm4u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eGm4u.png" alt="enter image description here" /></a></p>
<p><strong>Picture 3 and 4 (temperatures and the PID's output):</strong></p>
<p><a href="https://i.stack.imgur.com/Nn8EV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nn8EV.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/9H4uw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9H4uw.png" alt="enter image description here" /></a></p>
| |control-engineering|heat-transfer|control-theory|airflow|pid-control| | <p>I suggest that your problem is between the PID (I assume you're using only P & I like 99% of industrial uses) and the physical limitations of your system. Of course you're going to overshoot, because you don't have instant feedback. Air in the box has to mix and then return to the inlet of your measuring point. I seriously doubt it has anything to do with high temperature - it's just the system reaching it's setpoint.</p>
<p>You've also made no mention of how you have attempted to tune the loop. You can measure how your system reacts to setpoint changes and tune accordingly.</p>
| 40108 | Why does accuracy of temperature PID controller drop over time and how to eliminate it? |
2021-02-01T17:18:53.183 | <p>I'm not having luck with the Trim Entities feature, which is just for sketches. I just want to shave the top part off completely at the angle that I've drawn.
<a href="https://i.stack.imgur.com/QNnE5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QNnE5.png" alt="enter image description here" /></a></p>
| |solidworks| | <p>There's a couple of ways to do this - I've illustrated both @StainlessSteelRat's Plane Cut method, and @NMech's Cut/Extrude method.</p>
<p>Both are valid, and as always with these questions - there's not enough context to recommend which of these two, or the numerous other potential methods is best for your situation.</p>
<p><a href="https://i.imgur.com/RLjqM5f.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/RLjqM5f.gif" alt="enter image description here" /></a></p>
| 40112 | Quick Solidworks question: How to cut a 3D part along a line in a sketch? |
2021-02-01T20:31:11.053 | <p>I am trying to build a door to a hidden room and want to use pneumatics to open the door. I have purchased some equipment to test with, namely a <a href="https://rads.stackoverflow.com/amzn/click/com/B00BUA1NUA" rel="nofollow noreferrer" rel="nofollow noreferrer">12 inch Parker air cylinder</a> and a <a href="https://rads.stackoverflow.com/amzn/click/com/B00VSCCEIU" rel="nofollow noreferrer" rel="nofollow noreferrer">U.S. Solid 5 way 2 position valve</a>. After hooking everything up I have the following configuration:
<a href="https://i.stack.imgur.com/tArNW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tArNW.jpg" alt="Arduino connected to relay, relay connected to valve and barrel plug." /></a>
When I connect the barrel plug to a <a href="https://rads.stackoverflow.com/amzn/click/com/B01GC6VS8I" rel="nofollow noreferrer" rel="nofollow noreferrer">24v 5amp DC power brick</a>, attach to pressure (currently have the compressor set to somewhere like 5-10 psi but have tested with 20-30), and run a simple "turn the pin on and off" Arduino code the relay clicks as expected and the valve sounds like it is engaging, however the airflow doesn't change. It remains coming out of the A output. I contacted U.S. Solid and have already got a replacement valve but the same issue occurred with their replacement so I'm beginning to think I'm doing something wrong.</p>
| |pneumatic| | <p>The <a href="https://ussolid.com/u-s-solid-1-8-5-way-2-position-pneumatic-electric-solenoid-valve-dc-24-v.html" rel="nofollow noreferrer">specification</a> for that valve states:</p>
<blockquote>
<p>Working Pressure: 0.15 ~ 0.8 Mpa</p>
</blockquote>
<p>For those still using colonial units, that's a minimum pressure of 28 psi.</p>
<p>The reason it doesn't move is that it's a pilot operated valve. A small airline is tapped off the pressure inlet and goes across the valve to the pilot. The solenoid opens the pilot allowing air to push the valve spool to the left in your photo against the force of the return spring. You haven't got enough pressure to beat the spring.</p>
<p><a href="https://i.stack.imgur.com/JWkjj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JWkjj.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. A pilot-operated 5/2 valve in the off position. Image source: <a href="https://www.hafner.pl/en/general-informations-about-pneumatics/spool-valves" rel="nofollow noreferrer">Hafner</a>.</em></p>
<p>Note in Figure 1:</p>
<ul>
<li>The spring pushing the spool into the home position.</li>
<li>The cross-valve pilot air path ending in the solenoid stopper. When the solenoid is operated air is allowed through the white hole and into the spool chamber to push it to the right.</li>
</ul>
<p>The advantage with this type of configuration is that a low current is sufficient to operate the valve and so they are much more efficient than direct solenoid operated valves, smaller and run cooler.</p>
<p>The downside is that they can't switch low pressure or vacuum without a separate pilot supply - if the feature is available on the valve.</p>
<hr />
<p>Tip: Amazon is not a good source for component data. "No datasheet? No sale!"</p>
| 40119 | Pneumatic valve clicks but doesn't actuate |
2021-02-01T20:50:32.097 | <p><a href="https://ardushop.ro/303-thickbox_default/attiny85-usb-development-board.jpg" rel="nofollow noreferrer">This</a> dev board. 78M05G regulator, attiny85 MCU.</p>
<p>Similar (ON MC78Mx05) regulators' droupout voltage is listed at 2V0. USB voltage is specced at 5V0 (+0V25)(-0V55). The ATTiny85 can take 2V7 to 5V5, though for some variants the 4V45 low end of the USB spec straddles a speed grade.</p>
<p>By my understanding you're operating in undefined territory if you are below the sum of output voltage and droupout voltage, which would be 7V here. So why the linear regulator? Alternatively, what am I missing?</p>
| |electrical-engineering| | <p>Here is the corrected schematic for the uxcell Digispark Kickstarter Attiny85 Micro USB Development Board. <a href="https://diyodemag.com/_images/5bfe6676c672e0c74e3374b7" rel="nofollow noreferrer">Original schematic</a>. I say corrected because parts of the internet has gotten the direction of D3 wrong. Correct orientation of D3 (in Red) is shown.</p>
<p><a href="https://i.stack.imgur.com/JXFKq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JXFKq.png" alt="enter image description here" /></a></p>
<p>Attiny85 requires 1.8V to 5.5V to operate.</p>
<p>The Attiny85 Development Board can be powered in three ways:</p>
<ul>
<li>+4.3V: via USB (no JP2). +5V - D3 = +4.3V</li>
<li>+5V: via JP2 & 78M05, pins 1 & 2. 7V to 20V dropped down to 5V via 78M05.</li>
<li>+5V: via JP2, pins 3 & 2. External 5V source on JP2.</li>
</ul>
<p>D3 allows USB and JP2 to be safely plugged in at the same time. If JP2 and USB are plugged in, D3 will have +5V'ish on both sides. D3 is not forward biased, which prevents damage to USB +5V.</p>
<p>If D3 was reversed, then the USB connection could not power the board.</p>
<p>The MC78M05 is only required if the user wishes to power it from a 7V to 20V external source.</p>
| 40120 | Why is there a 5V linear regulator on this USB ATTiny85 dev board? |
2021-02-03T02:01:50.727 | <p>Assuming i have a circular disc of a significant mass and diameter. Neglecting friction, if a motor is put at its center how do i calculate the torque required for the wheel to just start to rotate?</p>
<p>Since the center of gravity is at its middle point, the only vectors i can work with dont provide any moment</p>
<p>i've seen in posts such as <a href="https://engineering.stackexchange.com/questions/23151/how-can-we-calculate-torque-required-to-rotate-the-wheel-mass-unknown">this</a> saying a very low torque can rotate the wheel but how low?</p>
| |motors|applied-mechanics|torque|moments| | <p>let me take a stab at this.</p>
<p>Similar to linear motion of a mass m on a frictionless surface or in space, that a force <span class="math-container">$F$</span> no matter how small will move a mass no matter how big, albeit with smaller acceleration if the mass is bigger or F smaller.</p>
<p><span class="math-container">$$F=m\alpha$$</span></p>
<p>A torque no matter how small will turn a disk or a randomly shaped piece of rock or an ice skater or even a massive cloud of plasma as those that turn around usually a black-hole center in space and later become galaxies. you toss a hammer in the air and the slightest pitch of your wrist will force it to spin around an unintuitive center of rotation.</p>
<p>The equivalent of mass for rotation is second area moment or moment of inertia of the object which is related to how far its mass is distibuted from its CG, and equivalent of linear acceleration is angular acceleration, and the force changes to torque.</p>
<p><span class="math-container">$$\tau= I \alpha$$</span></p>
<p>For example, in order to measure G, the famous newton's gravity factor they suspended a wide, heavy dumbbell from a long string and after it was carefully set to stand still, it started to turn when they placed a massive object near one end of it and thus they measured G. And we know G is really a very small number.</p>
| 40135 | Torque required to start rotating a wheel about its center |
2021-02-03T02:03:35.727 | <p>I have been talking to some structural engineers trying to find one to design foundations for my contemplated bespoke building based on hi-cube 40-ft shipping containers. The initial concept is just two containers on concrete slab with garage space between them:</p>
<p><a href="https://i.stack.imgur.com/K0kJF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K0kJF.png" alt="enter image description here" /></a></p>
<p>Ultimately, I would like to build extra levels on top of the containers. But at this point I do not want to design that. For now I just want what is on the picture, albeit with foundations strong enough to carry the extra load of say 2 additional levels.</p>
<p>I have been trying to explain the above to structural engineers, specifically mentioning that any extra levels will not be directly connected to the concrete foundation but to the containers only (which would be reinforced if needed). Essentially, the containers will be the foundations for the extra levels.</p>
<p>But so far I have been told that they need the design of the extra levels.</p>
<p>Can someone please explain: why is it not sufficient to <em>just assume extra load</em> on the containers to calculate the concrete foundations? Why would the exact design of the extra levels make any difference (as far as the containers' concrete foundation is concerned) beyond its weight?</p>
| |structural-engineering|concrete|building-design|foundations| | <p>Foundation's task is not just to support gravity loads, it should resist lateral loads, wind and earthquake, as well.</p>
<p>Wind loads are dependent on the geometry of the building and seismic loads are dependent on the mass distribution throughout the buliding.</p>
<p>For example of you will show access to the roof of your third floor container for its use as a deck, you have to add 2.5 times live load more than floor loads for that roof (100lbs/square foot versus 40 for a floor).</p>
<p>If this deck is over just half of the roof of the container, it will introduce torque forces due to unsymmetrical loading at the time of the earthquake.</p>
<p>Therefore detailed plans of upper floors are needed for proper structural designs in general and foundation design in particular.</p>
| 40136 | How do extra building levels affect the foundation design apart for the extra load? |
2021-02-03T08:51:17.020 | <p>How do I show this is clearly not edge triggered</p>
<p><a href="https://i.stack.imgur.com/d08Rm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d08Rm.png" alt="enter image description here" /></a></p>
<p>The truth table for this is as follows</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>E</th>
<th>R</th>
<th>S</th>
<th>Q</th>
<th>notQ</th>
</tr>
</thead>
<tbody>
<tr>
<td>0</td>
<td>0</td>
<td>0</td>
<td>prevQ</td>
<td>prevnotQ</td>
</tr>
<tr>
<td>0</td>
<td>0</td>
<td>1</td>
<td>prevQ</td>
<td>prevnotQ</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>0</td>
<td>prevQ</td>
<td>prevnotQ</td>
</tr>
<tr>
<td>0</td>
<td>1</td>
<td>1</td>
<td>prevQ</td>
<td>prevnotQ</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>0</td>
<td>prevQ</td>
<td>prevnotQ</td>
</tr>
<tr>
<td>1</td>
<td>0</td>
<td>1</td>
<td>1</td>
<td>0</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>0</td>
<td>0</td>
<td>1</td>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
<td>0</td>
<td>0</td>
</tr>
</tbody>
</table>
</div>
<p>Edge triggering definition: In edge triggering the circuit becomes active at negative or positive edge of the clock signal. For example if the circuit is positive edge triggered, it will take input at exactly the time in which the clock signal goes from low to high</p>
<hr />
<p>I'm confused how to show this</p>
| |electrical-engineering| | <p>It is clear in the last three lines of your truth table that the R and S inputs directly influence the output when E is high (no edges). Therefore, not edge-triggered.</p>
| 40139 | How to show gated sr latch isn't edge triggered? |
2021-02-04T14:52:42.383 | <p>While watching How It's Made episode on CCDs, this just springed to my mind... I always though of CCDs as of some kind of alien technology. Only now I realized that it's just an array of silica crystals that, same that we use for generating power but all connected to CPU for processing. Is this correct?</p>
| |sensors|solar-energy|photovoltaics| | <p>Yes, the actual light sensor in any CCD or CMOS imaging device is an ordinary silicon photodiode. However, there's just one silicon crystal (wafer), not an "array of silica crystals". Silica is silicon dioxide, which is an insulator, not a semiconductor.</p>
<p>The difference is that in a solar cell, you make one big diode that covers the entire chip/wafer so that you can get lots of current, but in an imaging device, you make millions of tiny diodes, along with the circuitry to sequentially read out the charges that they produce.</p>
| 40155 | Are cameras' CCDs just elaborate tiny solar cells? |
2021-02-05T06:15:00.270 | <p>When I read about stress analysis and specifically in the context of strain gauge, I often see the phrase <code>resistance per unit strain</code> or <code>voltage per unit strain</code>. I understand the electronics and how some of the equations are derived or come about. I have a basic understanding of what stress and strain are, but I don't get what <code>per unit strain</code> is and why we're using it.</p>
<p>Take the simplest example - a strain gauge connected in series with a constant current source, where <span class="math-container">$ E $</span> is the voltage drop across a strain gauge, <span class="math-container">$ R_g $</span> is the resistance of the strain gauge, <span class="math-container">$ G_F $</span> the gage factor, and <span class="math-container">$ \epsilon $</span> the strain:</p>
<p><span class="math-container">$$ E + \Delta E = I(R_g + \Delta R_g) $$</span>
<span class="math-container">$$ \Delta E = I \Delta R_g = I R_g {\Delta R_g \over R_g} = I R_g G_F \epsilon$$</span></p>
<p>Everything makes sense until the equation is rearranged and call it the <code>potential drop per unit of strain</code>:</p>
<p><span class="math-container">$$ {\Delta E \over \epsilon} = I R_g G_F $$</span></p>
<p>One unit is a measure of unity. One unit of strain therefore is <span class="math-container">$ \epsilon = \Delta L / L_0 = unity = 1 $</span>. That implies <span class="math-container">$ \Delta L = L_0 $</span>, or the change in length equates its initial length. If the object is under tensile stress, <span class="math-container">$ \Delta L = L $</span> is a 100% elongation. So the term <code>potential drop per unit of strain</code> would logically mean the voltage drop per each 100% elongation! Most strain being measured in the elastic region is well below a few percent. 100% elongation (or my take on <code>per unit strain</code>) feels .... out of context here.</p>
| |mechanical-engineering|electrical-engineering| | <blockquote>
<p>I don't get what <code>per unit strain</code> is and why we're using it.</p>
</blockquote>
<p>When you have a quantity expressed as "per unit (of) X", you get two things.</p>
<p>First, you can recover the value of the quantity for <em>any</em> X by just multiplying (assuming that the relationship is linear). That's what multiplication does - it just <em>scales</em> some base value up or down. So you can express some relationship that's relative to X in a more universal way by fixing the quantity to the value it would have if X were 1, and then using that as a multiplication factor (the slope). So you're designating a particular value to be the reference value.</p>
<p><a href="https://i.stack.imgur.com/i6rPH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i6rPH.png" alt="enter image description here" /></a></p>
<p>This is what the term "scalar" should be evocative of.</p>
<blockquote>
<p>So the term <code>potential drop per unit of strain</code> would logically mean the voltage drop per each 100% elongation! Most strain being measured in the elastic region is well below a few percent.</p>
</blockquote>
<p>It doesn't matter if X actually obtains the value of 1 in practice. What matters is that the mathematical relationship is correct.</p>
<p>Take the spring force equation (Hooke's law): <span class="math-container">$F = -kx$</span>. The spring constant describes force magnitude per unit displacement - letting you recover the value for any displacement. It doesn't matter whether or not displacements of +/- 1m make sense for that particular spring; there's a range of values where the model is applicable.</p>
<p><a href="https://i.stack.imgur.com/od3St.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/od3St.png" alt="enter image description here" /></a></p>
<p>You can also invert it (take the reciprocal) and use it as "displacement per unit Force", to infer the actual displacement if the force is what you can measure. So "potential drop per unit of strain" and it's inverse let you recover one or the other depending on what you can measure, by just plugging in the measured number - as long as you can assume the linear relationship.</p>
<p>The second thing you get is the ability to do comparisons in a normalized way. E.g., you're not interested in absolute values, but instead you want to compare two things relative to each other, or to some "standard" - "per unit X" can fulfill that role. E.g., you can compare the strength of the gravitational force (of two different stars) per unit mass (a 1kg "test mass"). Or the stiffness of two springs. Or the sensitivity of two strain gauges.</p>
| 40174 | What is "per unit strain"? |
2021-02-05T22:49:08.703 | <p>So I've been wondering this for a while; is PID just a simple but slightly outdated control method / isn't state-space (SS) a far superior method in pretty much every way?</p>
<p>SS allows for MIMO, PID just SISO. SS allows for non-linear models, PID does not do non-linear models. SS allows you to see and control the states of your system, this is kind of lost in the laplace transform of PID... So given this; why do I encounter PID everywhere? Is it simpler and cheaper and good enough if it works?</p>
<p>I've asked this question to my mechatronic systems design professor, but I did not really find his answer all that satisfying: "State space is not going to do magic if loopshaping way of designing PID reaches its edge. State-space is time domain to design linear control while loopshaping is frequency domain. With both you can design PID."</p>
<p>His research group is focused on the high-tech side of things, and has made a 'fractional order' PID controller toolbox (<a href="https://www.sciencedirect.com/science/article/pii/S240589631830449X" rel="nofollow noreferrer">FLOreS - Fractional order loop shaping MATLAB toolbox</a>, essentially you place a bunch of zeros and poles close to each other to mimic something which is between e.g. s^1 and s^2). In any event, I think he might be biased so I wish to get some perspective.</p>
| |control-engineering|pid-control| | <p>I do not think so. Instead of looking directly at PID, lets look at Frequency domain loop shaping in general compared to State-space control (eg, Lead-Lag filters, Notch filters and PID). On top of the fact that in the current industry over 90% of all controllers are PID's or controllers I just mentioned isnt just because they are much easier to learn, but also for a few more reasons:</p>
<ol>
<li>The bodeplot gives an amazing insight in how the controlled system is going to respond. The bandwidth, gain margin and phase margin show instant bounds in robustness and operating range. On top of that, the sensitivity bode-plot gives direct margins on how well the controlled system functions when influenced by noise. While there are some methods to assess controller robustness in State-space, these are incredibly complex to use and to analyze.</li>
<li>Controller elements directly "shape" the bode plot (hence the name loop-shaping). eg: do you want a larger phase-margin? add a lead filter, do you want to reduce or remove the resonance peak? add a notch filter and so on. Tuning these filters use frequency values you can read directly from the current bode plot. In state-space, you have no clue whether your designed LQR-filter did remove the resonance peak other than doing a frequency analysis. If it didnt, you have no clue what to change to ensure it does.</li>
<li>In a practical environment, the model of the system is often obtained using system-identification. Using frequency response identification, you obtain in essence the bode plot of the system, so even without an actual Laplace model of the system, you can already design a controller for it. On top of that, the nyquist plot can be used to assess stability, so really no model is required. State-space domain identification requires first of all the estimation of the number of states. On top of that, the only thing known from the returning model is that the output responds the same to the input as the actual system, what happens to the states on the other hand, is not known.</li>
<li>While I agree that controlling MIMO systems using State Space is indeed much easier, it is in fact possible to create frequency domain controllers of a MIMO system by either decoupling or sequential loop shaping. These methods are however a tad more complex to understand and properly apply.</li>
</ol>
<p>Of course, there are many application to which a state-space controller would outperform a frequency domain controller, but then again, there are many applications to which implementing a simple frequency domain controller that just does the job is much cheaper and faster.</p>
| 40192 | Is the PID just an outdated control method and state-space is superior? |
2021-02-06T05:48:59.897 | <p>Why does certain parts of metal show more corrosion at the place where there is high dislocation density in comparison to the place where the dislocation density is less? I do sincerely feel that this is particularly because of the fact that the activity is less in that particular localized region where the corrosion is taking place. I think the situation is similar where grain boundaries act as stressed region but what has this to do with the activity. A detailed answer to the above situation will most certainly be welcome.</p>
| |materials|metallurgy|metals|material-science|corrosion| | <p>In the real world ,I don't believe I have seen CORROSION at STRESSED areas. Corrosion may be accelerated at STRAINED areas; the strains introduce energy to the microstructure which may promote corrosion. Cracking is another story - which you did not ask about . Hydrogen cracking of high strength steels is likely the most common case . It can often occur with no apparent corrosion ; In particular cathodic protection which prevents corrosion ,can introduce hydrogen which causes cracking.</p>
| 40200 | Why does corrosion take place at certain stressed regions? |
2021-02-06T18:15:01.443 | <p>I am using Engineering Equation Solver, and it has been telling me the units of my distribution are inconsistent.</p>
<p>According to a the book I am reading, I think it should be unitless (as it is a probability).
However EES, says <a href="https://i.stack.imgur.com/trIB3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/trIB3.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ h = \frac{k}{c}\cdot \left(\frac{v}{c}\right)^{k-1}\cdot exp\left(-\left(\frac{v}{c}\right)^{k}\right)$$</span>
In theory, v and c have the same units [m/s] and k is unitless. I do not understand why the units are not working out.</p>
<p>Hope you can help me out.</p>
| |power|energy|turbines|wind-power|unit| | <p>If you have defined the same units for v and c then the equation should work just fine.</p>
<p>In all likelihood, h is the probability density so you its units should be <span class="math-container">$\frac{s}{m}$</span></p>
<p>In order to see where the problem lies, I would suggest breaking up the terms and checking their units independently within the software. I.e. I would find the units in EES for:</p>
<ul>
<li><span class="math-container">$\frac{k}{c}$</span> : the units should be <span class="math-container">$\frac{s}{m}$</span></li>
<li><span class="math-container">$\frac{v}{c}$</span> : It should be unitless</li>
<li><span class="math-container">$\left(\frac{v}{c}\right)^{k-1}$</span> : It should be unitless</li>
<li><span class="math-container">$\exp\left(\left(\frac{v}{c}\right)^k\right)$</span> : It should be unitless</li>
</ul>
<p>Then I would try to build up the equation to see, if the problem continues.</p>
| 40206 | Weibull distribution Units - ENGIEQSOL |
2021-02-07T08:14:04.240 | <p>I have a nonlinear system:</p>
<p><span class="math-container">\begin{cases} x'=f(x)+u \\ y=f(x) \end{cases}</span></p>
<p>where <span class="math-container">$f(x)$</span> - gradient of some one-extremal function (for example <span class="math-container">$f=e^{-(x)^2}$</span>), i.e. <span class="math-container">$\frac{df}{dx}$</span>.</p>
<p><strong>Task:</strong>
I want construct a continuous control <span class="math-container">$u$</span> that ensures the following condition:</p>
<p><span class="math-container">$y(t)=y(0) e^{-\beta t}, \beta>0$</span></p>
<p><em>Which method to use for the solution: MRAC, asymptotic output tracking, feedback linearization, or something else?</em></p>
<p>I am not an expert, please do not pass by and give advice.</p>
| |control-engineering|control-theory|optimal-control|nonlinear-control|adaptive-control| | <p>Taking the time derivative of <span class="math-container">$y$</span> yields:
<span class="math-container">$$
\dot{y}=\frac{\partial f}{\partial x}(f(x)+u)
$$</span>
We need <span class="math-container">$y(t)=y(0) e^{-\beta t}$</span> but this is possible if and only if <span class="math-container">$\dot{y}=-\beta y $</span> , if the hessian is invertible then this is possible if and only if <span class="math-container">$u=-f(x)-\left(\frac{\partial f}{\partial x}\right)^{-1}\beta y$</span>. To get an intuition on why the hessian must be invertible assume that <span class="math-container">$y,x$</span> are scalars then if hessian is not invertible this means that
<span class="math-container">$$\frac{\partial f}{\partial x}=0 \implies \dot{y}=0$$</span>
This means that your system is not output controllable even though it is state controllable.</p>
| 40211 | Which way of solving from nonlinear control to choose? |
2021-02-07T08:31:14.170 | <p>Is there a compact device which can trade flow rate for lift height without too many moving parts?</p>
<p>My particular application is a 10 meter deep and 10 centimeter wide well with shallow water level, therefore the usual high-powered well pumps run dry too quickly. Submersible pumps with low flow rate, high working lifetime and sufficient pressure do not seem to exist. But even if such pumps did exist, I would still be interested in an answer from an academic point of view. Maybe there is a solution without any moving parts at all? That would be neat!</p>
<p>Ideas I discarded so far:</p>
<ul>
<li><p>A <a href="https://en.wikipedia.org/wiki/Hydraulic_ram" rel="nofollow noreferrer">hydraulic ram pump</a> seems to requires a high flow rate and long feed lines.</p>
</li>
<li><p>Daisy-chaining multiple small pumps with low pressure multiplies failure rate. (I tried five pumps of type AD20P-1230D with a head of 2 meter each, but the rated 50,000 hours lifetime is highly optimistic.)</p>
</li>
<li><p>A water wheel driving a chain pump sounds like a maintenance nightmare due to the mechanical nature of the chain pump and it is also not very compact (disregard the cow in the following illustration).</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/fyh2P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fyh2P.png" alt="water wheel and chain pump" /></a></p>
| |fluid-mechanics|pumps|hydraulics| | <p>Here's a crackpot idea: Positive displacement pump with compressed air as the piston. I've seen these documented <em>somwhere</em> on the internet, and they are a common trade fair gimmick to show off check valves.</p>
<p>The basic idea is:</p>
<ul>
<li>water flows freely into container through check valve (as large as fits)</li>
<li>pressurized air is pumped into container, water is pushed out through a pipe near the low end (and another checkvalve, probably DN15 or DN25 or so)</li>
<li>when water is pumped out, the air is allowed to escape from the container and water re-enters</li>
</ul>
<p>The container can be a piece of pipe, DN80 or so. The tricky part is to actutate the compressed air - I've seen a device with two swimmers (again, don't know where anymore) that could fit down a narrow hole. With off the shelf components, probably two electic level switches and solenoid valve like you would use for a single acting pneumatic cylinder. when using flexible tubing for air and water, it should be easy to lift the whole device from the well.</p>
<p>Big downsides:</p>
<ul>
<li><p>efficiency! compressed air is a rather expensive form of energy. You'd want a air supply with 1 bar + enough to overcome flow resistance, anything more is wasted energy. Even then total efficiency will be lower than a normal pump</p>
</li>
<li><p>it's a somewhat experimental project, not a tried and tested solution</p>
</li>
</ul>
| 40212 | Low maintenance device to trade flow rate for lift height? |
2021-02-07T21:16:47.597 | <p>In framing the question, I can't be more specific as I have no idea (though a suspicion) about what this came from.</p>
<p><a href="https://i.stack.imgur.com/HQG0R.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HQG0R.jpg" alt="enter image description here" /></a></p>
<p>I have a number of them that are several different diameters from 125mm to about 300mm.</p>
<p>They are very thin steel (< 1mm) and each size came in a tight stack, as though laminated. They don't seem to have been glued together as they come apart easily. We have had them for about 20 years. Note the notch at 7 o'clock possibly to align or secure a stack.</p>
<p>I suspect they are part of some sort of electrical winding, but I would have thought they would have "teeth" on the outside.</p>
<p>Any ideas what they came from?</p>
<p>Nowadays they are part of an art project.</p>
<p><a href="https://i.stack.imgur.com/kvd6d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kvd6d.jpg" alt="enter image description here" /></a></p>
| |electrical-engineering|mechanical| | <p><a href="https://i.stack.imgur.com/Wm1v5.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Wm1v5.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. Silicone steel stamped lamination, stator and rotor. Image source:<a href="https://www.hsmagnets.com/product/silicon-steel-stamped-lamination-stator-and-rotor/" rel="noreferrer">HS Magnets</a></em></p>
<p><a href="https://i.stack.imgur.com/e1TIK.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/e1TIK.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 2. On their way to becoming motors.</em></p>
| 40226 | What is this object? |
2021-02-08T00:02:07.383 | <p>So in my bachelors and throughout my engineering degree, we did everything in MATLAB. But I've heard that many companies use python over matlab. Trying Python out for myself, it seems harder to use in every way; you have to load in modules while in MATLAB they're included by default, plotting is not included by default, indexing starts at 0 (okay cool, but why?), there is no transfer functions by default, and there is no nice way of carrying everything over between MATLAB and something like SIMULINK... So MATLAB pretty much seems superior in every way, why do so many companies apparently use Python instead?</p>
| |matlab|software| | <p>While I agree that MATLAB is far easier to use than Python with having a bunch of 'extensions' (if you will) already included, variable sharing / incorporation with simscape and SIMULINK (for control of multibody systems you're going to have a far harder time in python).. I think you are overlooking one major factor, and that is price:</p>
<p><a href="https://nl.mathworks.com/pricing-licensing.html" rel="nofollow noreferrer">https://nl.mathworks.com/pricing-licensing.html</a><br />
On what I can find directly on the mathworks site, 800-2000 euros for 1 license key; for company wide licensing you need to request a quote. If your company is small, and multiple people will have to share files/data, the cost will start piling up. Python... is completely free, and you won't have to mess around with any licenses. So if you're not using specific toolboxes that MATLAB only has / don't have the capital, and you can get by with Python, it is the more practical alternative.</p>
| 40228 | Why use python over MATLAB? |
2021-02-08T16:11:57.307 | <p>Imagine a body which is allowed to rotated around its centre of gravity. I place a rotating moment, <span class="math-container">$M_d$</span> on a point which is a distance <em>d</em> from the body's centre. <span class="math-container">$M_d$</span> is only required to put the body in motion, i.e. overcome the friction and intertia required for the body to start rotating around its centre of gravity.
Compare this moment to a moment <span class="math-container">$M_0$</span> located at the centre of gravity. Same conditions apply for <span class="math-container">$M_0$</span>, it only has initiate rotation. Is <span class="math-container">$M_0 = M_d$</span>? Or is one bigger than the other?</p>
<p>Theoreticaly they should be the same but intuitively when compared to screwing a screw, it is easier to screw the screw if you apply the moment directly on the screw axis of rotation.
Could someone please explain if there is a theoretical difference or if it is only my intuition which is wrong.</p>
| |dynamics|moments| | <p>A pure torque applied to a rigid body doesn't have a point of application. But if the torque is the result of some forces to that body more likely there will be a force F passing through the Center of the mass of the body and a torque applied to the body.</p>
<p>note how the resultant of the forces have simplified to torque and a force F applied at the CG of the rigid body in the figure.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/Olh63.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Olh63.png" alt="rigid body forces" /></a></p>
| 40239 | Moment placed off-centre on a rotating body |
2021-02-08T20:49:18.520 | <p>I have a beam that has a given limit of 1690 lbs uniform load for a 24" span. I will be subjecting this beam to multiple point loads at different locations on the beam. In order to determine if the loads will exceed the beam allowable limit (1690 lbs) I was wondering if I could add up all these point loads into a single uniform load to compare to the load allowable. Is this possible?</p>
| |mechanical-engineering|structural-engineering|structural-analysis| | <p>Your design is governed by the moment, the equivalent uniformly distributed load, <span class="math-container">$$ W = \frac{2*M}{L^2} = 17.6 lb/in$$</span></p>
<p>If you want to place multiple point loads on the beam, you should sum all moments due to the point loads, then compare it with the limit given by the manufacturer.</p>
| 40244 | Can I convert multiple point loads into a single uniform distributed load? |
2021-02-09T16:26:52.487 | <p>I'm not entirely sure if this is the right site for this question, but I recently went on a Greek Myth Binge and came across Daedalus again, and his famous first hanglider, made from feathers. Now, obviously this wouldn't work, even if we ignore using wax as a glue and the heavy structure and short wingspan, the feathers of normal birds are likely to small for human sized wings, and too fragile. And today, we of course have hangliders, but they're made from fabric or similar things, and I was just wondering why we don't use modern technologies to imitate feathers, of course optimized for human use. Would these hangliders be less effective than our ones? Or too expensive/complicated to build and maintain? Is it even possible to build a feather based hanglider? Why or why not?</p>
| |aerospace-engineering|aerodynamics| | <p>Nature created a light structure (hollow bones in birds), materials (feathers) and a precision control system (brain, nerves, muscles and tendons etc) to achieve flight.</p>
<p>Matching that with the tech we have is challenging since hollow tubes do for the bones and hang gliders have that.</p>
<p>The materials used in terms of silks etc are light but any matching of feathers will need precision control and trying to build a detailed control system will need too much mass for the job in hand.</p>
| 40258 | Why don't we use feather like structures for hang gliding? |
2021-02-09T22:10:16.213 | <p>They look like snap buttons and used for stacking pagers for charging.</p>
<p>Closest things I can find are socket screws and cap nuts.</p>
<p><a href="https://i.stack.imgur.com/pgtek.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pgtek.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/Za1b7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Za1b7.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|electrical-engineering|industrial-engineering|circuit-design|connections| | <p>The following pins appear to be pogo pins or spring loaded pins</p>
<p><a href="https://i.stack.imgur.com/D49Vb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D49Vb.png" alt="enter image description here" /></a></p>
<p>Below are example of pogo pins</p>
<p><a href="https://i.stack.imgur.com/RiOaQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RiOaQ.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/MN0zm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MN0zm.png" alt="enter image description here" /></a></p>
<p>The following appear to be <a href="https://www.google.com/search?bih=680&biw=1424&rlz=1C1CHBF_enUS892US892&hl=en&ei=IR8jYPz4Jaa7ggfXvKG4Dg&q=swage%20mount%20pcb%20pins&oq=Swage%20Mount%20&gs_lcp=CgZwc3ktYWIQARgAMgIIADIGCAAQFhAeMgYIABAWEB4yBggAEBYQHjIGCAAQFhAeMgYIABAWEB46BwgAEEcQsANQ2OcBWNjnAWCB_QFoAXACeACAAVuIAZgBkgEBMpgBAKABAaoBB2d3cy13aXrIAQjAAQE&sclient=psy-ab" rel="nofollow noreferrer">swage mount</a> connectors</p>
<p><a href="https://i.stack.imgur.com/NAxsX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NAxsX.png" alt="enter image description here" /></a></p>
<p>Here is an example</p>
<p><a href="https://i.stack.imgur.com/2t4Zh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2t4Zh.jpg" alt="enter image description here" /></a></p>
<hr />
<p><strong>Reference:</strong></p>
<ul>
<li><a href="https://www.mill-max.com/about" rel="nofollow noreferrer">Mill-max</a></li>
<li><a href="https://youtu.be/jDT_uRF3Klw" rel="nofollow noreferrer">How to pair the Retekess SU-668 Restaurant Paging System</a></li>
</ul>
| 40270 | What are the charge/power connector of restaurant order pagers / buzzers called? |
2021-02-10T17:31:36.583 | <p>This question relates to a previous question (<a href="https://engineering.stackexchange.com/questions/38942/how-to-calculate-the-required-torque-in-a-static-equilibrium-system/38946?noredirect=1#comment70008_38946">How to calculate the required torque in a static equilibrium system</a>) where I have enquired about the start-up torque required to rotate a system. However, the parameters have changed slightly since then, and as a result I am not entirely sure if the system is still considered to be in "static equilibrium" - somehow I believe it still is, and I am speculating that the only factor that has changed is gravity, considering the additional weight added.</p>
<p>However, I do not know what other forces are now at play in the new system.</p>
<p><strong>Question:</strong> What is the required start-up torque to rotate the system with the parameters provided below ?</p>
<p>This is what I have:</p>
<p>2x bars perpendicular to each other, intersecting each other at their centres. Something like this:</p>
<pre><code> 10kg (F3)
|
|
|
|
10kg (F1) ----------+---------- 10kg (F2)
|
|
|
|
10kg (F4)
Bar weight : 0.5kg (500g)
Axle weight : 0.5kg (500g)
Max RPM : 3 RPM
F1, F2, F3, F4 : 10kg each
Time to Max RPM : 10sec
*** Ignore Friction and Drag.
The motor torque specification is given in kg.cm.
</code></pre>
<p>A motor (whose start-up torque requirements I need) is attached to an axle. The motor is required to turn the axle at a constant speed of about 3RPM. What is the required start-up torque to reach 3RPM after 10 sec ?</p>
<p>(I am hoping to hear from Kamran again)</p>
| |motors|torque| | <p>This is a quick answer till further detail by the OP.</p>
<p>The 10 kg forces directed downward at the ends of the bars cancel out as for the rotation of the system, they just impart 40kg force at the axel.
Let's say the length of the bars is 100cm. and we ignore the axel for now because it is too close to the rotation center.</p>
<p>Then the moment of inertia of each bear if it's a uniform thickness rod is:</p>
<p><span class="math-container">$$I= \frac{1}{12}mL^2 \ for \ two \ I =\frac{1}{12}2mL^2$$</span></p>
<p><span class="math-container">$$\tau = I\alpha= \frac{1}{12}1kg10000cm *3.33*2\pi $$</span></p>
<p>I let you do the rest. Please clarify If I misunderstood your question.</p>
| 40285 | How to calculate the Torque required to rotate two bars perpendicular to each other |
2021-02-10T20:23:02.627 | <p>I need a digital output from a raspberry to power a stepper driver that takes 5v as input.
After that, the output from the stepper driver must be converted back to 3.3V because a Pi can't handle 5v on the GND-pins.</p>
<p>I bought the following level shifter: <a href="https://www.spikenzielabs.com/Catalog/adafruit/breakouts/4-channel-i2c-safe-bi-directional-logic-level-converter-bss138" rel="nofollow noreferrer">https://www.spikenzielabs.com/Catalog/adafruit/breakouts/4-channel-i2c-safe-bi-directional-logic-level-converter-bss138</a></p>
<p>But since it is I2C-safe, im not sure if it will allow being turned into 5V because the "master" only handles 3.3V?</p>
| |electrical-engineering|power-electronics|power-engineering| | <p>The pi does not "handle" voltages on the ground pins, ground is just ground.
The pi, the driver and the level shifter must all have common ground for communication to work.</p>
<p>For you application:</p>
<ul>
<li>Connect the shifter GND to the pi GND and to the GND pin on your driver.</li>
</ul>
<p>If the driver has separate grounds for signals and power, use the signal GND.</p>
<ul>
<li>Connect 3.3v to the LV-pin, the pi has a 3.3v output.</li>
<li>Connect 5v to the HV-pin, this can come from either you pi or driver, it's just a reference.</li>
<li>Connect the signal pins from the pi to the A-pins on the shifter.</li>
<li>Connect the signal pins from the driver to the B-pins on the shifter.</li>
</ul>
<p>Power everything up and have fun with your project :)</p>
<p><strong>EDIT:</strong></p>
<p>To answer the question in the title.
The level shifter will not convert power in the sense that you can input 3.3v and output 5v to power something with the 5v.
It will "translate"/change the voltages of the signals you connect to A and B.</p>
<p>It has a low voltage side (A) and a high voltage side (B).
If you send a signal to A1 at the A side voltage the same signal will be output at B1 but at the B side voltage.
This shifter is bidirectional so inputting B level signals at the B pins will output the same signals at A level on the A pins.</p>
<p>The A level is set by what you input on the LV-pin and the B level on the HV-pin. The LV and HV voltages can come from anywhere, but you need to supply both. Since the pi has both it's probably easiest to take them both from the pi.</p>
| 40291 | Can a level shifter convert power from 3.3v to 5v and then back to 3.3v? |
2021-02-10T20:38:36.130 | <p>I want to build myself a pull-up bar. I am confident enough that I can guesstimate how much material I need for it to be strong enough, but this time I would like to challenge myself to properly calculate all the strengths needed.</p>
<p>However I am stuck at the first point -- how do I properly estimate the loading on such a thing? By this I mean only the maximal instant force that my body would apply to the bar, safety factors will be added later.</p>
<p>Is there some standard that tells you what dynamic forces to expect from a human body?</p>
<p>My best guess so far goes something like this: I can hang on single hand, but not with another person pulling me down, so the max total force that I can apply to the bar will be somewhere between 2x and 4x my body weight, anything more and my hands will certainly slip. Therefore I design for 4x my body weight and add safety factors on top of that. Does that make sense?</p>
| |structural-analysis| | <p>Since you want to challenge yourself with the design process,
it would be good to use the force information not just for strength
of the bar, but also the deflection. For example when you design a
bow for archery you want deflection, and when you design a
bridge, you don't. Olympic weightlifting bars are somewhere
in between So it is worth considering how
flexibility will influence your use of the bar.</p>
| 40292 | Estimating dynamic load from human body |
2021-02-11T00:10:30.897 | <p>Disclaimer: amateur mechanic, not an engineer of any kind.</p>
<p>I just bought a spring compressor for my car. It's the factory recommended type so I'm assuming it's strong enough, but I'm wondering exactly how strong - since it looks (to my untrained eye) like there's a clear weakest link. The spring gets sandwiched between cast iron plates. The bottom plate rests on a threaded handle that looks really thick and sturdy. The top plate however is held in place by a hemispherical piece ('Upper Ball' in Figure 1), which is in turn held onto the main shaft by a small dowel pin ('Pin' in Figure 1). The upper ball fills a tapered hole in the upper plate, which presses down onto the coil. Everything seems like a well-machined fit - the pin slides in easily with no play.</p>
<img src="https://i.imgur.com/3iqA6pT.jpeg" width="378" height="504">
<p>Here's an image of the actual pieces:</p>
<img src="https://i.imgur.com/JvKmoYx.jpeg" width="378" height="504">
<p>The pin is 7.75mm in diameter and 56.0mm in length. The upper ball has an ID of 18.6mm and an OD of 44.3mm, so during use the pin sticks out both sides as per the image. I can't tell what it's made of, but I'm assuming either grade 5 or grade 8 steel or the equivalent metric grade.</p>
<p>Two questions:</p>
<p>(a) am I correct in assuming that this is the weakest part of the assembly?</p>
<p>(b) is it possible to calculate an approximate yielding/breaking strength of this part of the assembly, assuming either grade 5 of grade 8 steel?</p>
<p>For reference, the total force I would routinely put on this compressor is 1000 lbs (~4500 N).</p>
| |structural-analysis|automotive-engineering| | <p>I would have thought the nut and screw screw thread was more likely to be the failure point, but you are looking at the part and I'm only looking at a sketch of it.</p>
<p>The yield strength of grade 5 steel is around 600 MPa. The area of the pin is about 47 mm^2. So to permanently bend the pin by shearing it would take about 28000 N or about 6300 lbf. (Or arguably twice that, since the load goes to both sides of the pin).</p>
<p>Nothing much to worry IMO.</p>
| 40296 | How strong is this spring compressor? |
2021-02-11T06:27:26.767 | <p>Hi Guys wanted some help for breaking down a block diagram which can be seen below. I know that typically with feedback systems be it negative or positive you would employ using the following:-</p>
<p>Close Loop = <span class="math-container">$C.L.T = \frac{open loop} {closed loop}$</span></p>
<p>However the signal <span class="math-container">$q_{i-1}$</span> is confusing me typically without it, the closed loop would be</p>
<p><span class="math-container">$\frac{q_i}{d_r,i} = \frac{PDG_i(s)}{1-PDG_{i}(s)}$</span> as a result of positive feedback I'm not sure how to treat that <span class="math-container">$q_{i-1}$</span> signal in the system can anyone help help derive the transfer function.</p>
<p><a href="https://i.stack.imgur.com/uJckA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uJckA.png" alt="enter image description here" /></a></p>
| |control-engineering|control-theory|transfer-function| | <p>Suppose we acknowledge <span class="math-container">$q_{i-1}$</span> as an independent signal, then using the block diagram, the following equations can be derived:
<span class="math-container">$$e_i = d_i - d_{r,i} = - q_i+q_{i-1} - d_{r,i}$$</span>
<span class="math-container">$$q_i = G_i(s)u_i = PDG_i(s)e_i$$</span>
substitute the first in the latter:
<span class="math-container">$$q_i = PDG_i(s)(- q_i+q_{i-1} - d_{r,i})$$</span>
<span class="math-container">$$(1+PDG_i(s))q_i = PDG_i(s)(q_{i-1} - d_{r,i})$$</span>
<span class="math-container">$$q_i = \frac{PDG_i(s)}{1+PDG_i(s)}(q_{i-1} - d_{r,i})$$</span>
Which represents a transferfunction with <span class="math-container">$(q_{i-1} - d_{r,i})$</span> as input and <span class="math-container">$q_i$</span> as output. The troublesome part here is this <span class="math-container">$q_{i-1}$</span>, of which I do not know its meaning (is <span class="math-container">$i$</span> a discrete time index, or is <span class="math-container">$q$</span> a set of systems where <span class="math-container">$q_i$</span> is an part of it).
Even thought one could substitute
<span class="math-container">$$q_{i-1} = \frac{PDG_{i-1}(s)}{1+PDG_{i-1}(s)}(q_{i-2} - d_{r,i-1})$$</span>
into the equation, the problem is only shifted to <span class="math-container">$q_{i-2}$</span> and so on. As such, the complete transfer matrix (where the input is every instance of <span class="math-container">$d_{r,i-n}$</span> and the output is every instance of <span class="math-container">$q_{i-n}$</span>) will be the full definition of your system. However, if <span class="math-container">$i$</span> is a discrete time index, you could try to convert the model and controller to the Z domain and describe <span class="math-container">$q_{i-1} = z^{-1}q_i $</span></p>
| 40299 | Closed Loop transfer Function |
2021-02-11T06:56:30.830 | <p>I'm working on a project that requires very precise, but low pressure controls. I'm using a pump to charge a chamber to <15kPa, then a manifold of proportional valves to control output pressures from a few channels at around 3kPa with low flow. I've been using the tubing from the pump to the manifold as a reservoir, but I'd like more volume, in the range of 0.2-0.5L. I'd imagine such a reservoir exists for <$50.</p>
<p>A <a href="https://www.mcmaster.com/3681T74/" rel="nofollow noreferrer">squeeze bottle</a> with some barb fittings is in the ballpark, but I'm curious if there are other ideas. I'm hoping someone can help me find the right terminology here, as all I can find are large tanks when I search 'reservoir' or 'accumulator'.</p>
| |pneumatic|product-engineering|machine-design| | <p>I use this kind of thing frequently. The $50 price range is about right for 100mL ish, might be more if you are looking at 0.5L. The max design pressures are generally much higher, but that doesn't matter.</p>
<p>You can buy such things from companies that make pneumatic accessories, and also sometimes companies that make miniature pneumatic cylinders (the "tanks" are sometimes just empty cylinder bodies with an endcap). SMC, Festo, Clippard in the US etc. The terminology varies and internet searches will result in too big sizes. Call up a sales person for a company that deals with miniature pneumatics or industrial automation components. (pneumatics for factory-automation is the most common customer).</p>
<p>It is also fairly trivial to make your own from e.g. from cylindrical tube and custom endcaps that you clamp and seal with o-rings, or from pipe and endcaps and standard fittings. For nicer integration, you can combine with the manifold and have the volume machined into a larger solid block, and simply cap it with a flat plate and a face-seal o-ring. With a split manifold, a number of extra options open up too, you can get a very nice presentation if are trying to sell it. The overall part count will still go down and this can make up for the raw material cost and - the extra machining is quite simple. I have done all these things, it's a matter of what is right for your project.</p>
| 40300 | What is the name for a low pressure, low volume air reservoir? |
2021-02-11T08:22:43.650 | <p>When talking about control systems I sometimes hear the term "typical", however I have no clear understanding what it is. It's also mentioned in some codes (so far without definition), here's an example image from the DIN EN 62424 (IEC 62424:2016):</p>
<p><a href="https://i.stack.imgur.com/lYtUR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lYtUR.png" alt="enter image description here" /></a></p>
<p>From context, I think that it refers to software building blocks or common functions of the device (sensor or actor) itself but I'm not sure. We don't have a control engineer in house whom I could ask.</p>
<p>I tried googling the term but typical is such a common word.</p>
<p>Edit to add: I wondered if "typical" is a weird anglicism, but found this in the DIN:
<a href="https://i.stack.imgur.com/Gi2pd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gi2pd.png" alt="enter image description here" /></a></p>
<p>"Acronym and identification of a grafical diagram in a database, or group of signals or grouped PCE-tasks"</p>
<p>It doesnt really answer my question what a tpyical does or is for.</p>
| |control-engineering|terminology| | <p>Optical resonators answer is correct. However here's an illustrative example I found:</p>
<p><a href="https://i.stack.imgur.com/mU4GY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mU4GY.png" alt="enter image description here" /></a></p>
<p>The part in the cloud is "typical 11", the number "11" to the left of the motor's bubble in the lower right tells us that all the stuff in the cloud is implemented with this motor:</p>
<ul>
<li>local start/stop</li>
<li>remote stop</li>
<li>remote displays for current, Fault, runtime</li>
</ul>
| 40301 | What is a "typical" in the context of control engineering? |
2021-02-12T05:57:05.147 | <p>I hesitate to ask this on the EE site because it's not really an EE question. I'm more interested in the mechanical forces that keep an USB plug from slipping out of the socket.</p>
<p>I am asking because of the common issue of USB-c plugs becoming loose over time, compared to USB-a plugs which do not seem to have this issue. I wonder what is reason behind this. Is the mechanical structure of a USB-c plug considerably different than a USB-a?</p>
<p>Also, last time I went to have mt phone checked up, I was told that the "sensor" of my phone is misaligned, hence why my plug keeps falling out. I don't understand what the "sensor" being referred to was, nor can I find it on the typical USB-c pin layout. Some explanations on this would be much appreciated.</p>
| |mechanical-engineering| | <p>CORRECTION per @Jonathan R Swift -- The "USB C" is held by the spring tabs inside the plu, on the narrow sides. <a href="https://www.myfixguide.com/wp-content/uploads/2018/03/Lenovo-ThinkPad-PD-adaptor-ADLX65YCC2A-4.jpg" rel="nofollow noreferrer">photo</a></p>
<p>I dealt with a similar issue in a low-volume-manufactured device with pcb-mount "mini-B" generation USB sockets and plugs. There was not enough interference fit between the panel-mount socket and the plug of the cheap panel-mount extension cable that went into it, leading to intermittently flaky connection. The solution was to very gently squeeze the sockets with a pair of pliers, to increase the mechanical interference.</p>
<p>On a phone, it might be possible to gently pry apart the faces of the plug instead, but before doing this, check for is that there isn't dirt or debris in there that prevents the plug from going all the way in and mechanically latching. -- UPDATE not for USB-C</p>
<hr />
<p>This question inspired me to fix my own phone plug (microUSB), and the problems (failure to mechanically latch and the intermittent connection) seemed both caused by debris.</p>
<p><strong>!! Remove battery before poking around !!</strong></p>
<p>My micro-B socket had an astonishing amount of pocket lint, seemingly multiple times the volume of the space into which it was packed. Cleaned it out with a combination of a small safety pin, a piece of index card material cut to ~1.5mm width (to sweep out the clearance below the "tab", which is too narrow for the safety pin diameter), and ultra fine point tweezers.</p>
| 40324 | How does a USB plug fit into the receptacle |
2021-02-12T09:09:47.157 | <p>Just curious and working on some hypothetical problems on my own...I'm currently studying for my PE license. Also, trying my best here with the Latex stuff.</p>
<p>Let's say there's a crude oil pipeline and it somehow gets a hole in it and begins to leak. If you wanted to calculate the flow rate through the hole, my understanding is that you would approach the problem in the following manner:</p>
<p><span class="math-container">$$ Q = CA_0\sqrt{2g_c\frac{P_1-P_2}{\rho}}$$</span></p>
<p>Where:</p>
<p><span class="math-container">$A_0$</span> = cross-sectional area of orifice (leak hole) = <span class="math-container">$\frac{\pi*d^2}{4}$</span> = <span class="math-container">$\frac{\pi(2)^2}{4}$</span> = <span class="math-container">$3.14 in^2$</span></p>
<p><span class="math-container">$P_1$</span> = 1,000 psi (pressure in pipeline)</p>
<p><span class="math-container">$P_2$</span> = 14.7 psi (atmospheric pressure on outside of pipe)</p>
<p><span class="math-container">$\rho$</span> = density of crude oil = 870<span class="math-container">$\left(\frac{kg}{m^3}\right)$</span> <span class="math-container">$\left(\frac{2.2 lb}{kg}\right)$</span> <span class="math-container">$\left(\frac{1m}{3.28ft^3}\right)^3 $</span> <span class="math-container">$\left(\frac{1ft}{12in}\right)^3$</span> = <span class="math-container">$\frac{0.031lb_m}{in^3}$</span></p>
<p><span class="math-container">$g_c$</span> = 32.2<span class="math-container">$\frac{lb_m*ft}{lb_f*s^2}$</span> = 386.4<span class="math-container">$\frac{lb_m*in}{lb_f*s^2}$</span></p>
<p>So I get the following:</p>
<p><span class="math-container">$$Q = C3.14in^2\sqrt{\left(772.8\frac{lb_m*in}{lb_f*s^2}\right)\left(\frac{\frac{985.3lb_f}{in^2}}{\frac{0.031lb_m}{in^3}}\right)}$$</span></p>
<p>Is this the correct approach?</p>
| |bernoulli| | <p>This is a correct method, given you have a good estimate of the discharge coefficient (C). There are a lot of resources out there on estimates for C for an orifice in a cross flow configuration such as this. "Handbook of Hydraulic Resistance" is a good resource, although, I am not sure what you would have available to you during the PE exam. A value such as 0.7 may be a good initial guess for C.</p>
| 40327 | Calculating Pipe Leak Flow Rate |
2021-02-13T04:13:12.720 | <p>Hi Guys I am trying to evaluate the following transfer function below can anyone verify if this is correct</p>
<p><a href="https://i.stack.imgur.com/kWyXS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kWyXS.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$e_i = -d_{r,i} + d_i $$</span>
<span class="math-container">$$d_i = q_{i-1}-q_i$$</span>
<span class="math-container">$$d_{r,i} =H_iq_i$$</span>
<span class="math-container">$$Therefore, e_i = -Hiq_i + q_{i-1} - q_i$$</span>
<span class="math-container">$$q_i = e_iPDG_i$$</span>
<span class="math-container">$$q_i = PDG_i(-H_iq_i+q_{i-1}-q_i)$$</span>
<span class="math-container">$$q_i = -PDG_iH_iq_i+PDG_iq_{i-1}-PDG_iq_i$$</span>
<span class="math-container">$$\frac{q_i}{q_{i-1}} = \frac{PDG}{1+HPDG+PDG}$$</span></p>
<p>Im thinking this is incorrect but I'm hoping someone can verify this for me?</p>
| |control-engineering|control-theory|transfer-function| | <p>Shouldn't the error equation have <span class="math-container">$-H_iq_i$</span> instead of <span class="math-container">$H_iq_i$</span>?
<span class="math-container">$$e_i=−d_{r,i}+d_i$$</span>
<span class="math-container">$$d_i=q_{i−1}−q_i$$</span>
<span class="math-container">$$−d_{r,i}=−H_iq_i$$</span>
Therefore (modified equation):
<span class="math-container">$$e_i=-H_iq_i+q_{i−1}−q_i$$</span>
<span class="math-container">$$q_i=e_iPDG_i$$</span>
<span class="math-container">$$q_i=PDG_i(-H_iq_i+q_{i−1}−q_i)$$</span>
<span class="math-container">$$q_i=-PDG_iH_iq_i+PDG_iq_{i−1}−PDG_iq_i$$</span>
<span class="math-container">$$\frac{q_i}{q_{i−1}}=\frac{PDG_i}{1+H_iPDG_i+PDG_i}$$</span></p>
| 40341 | Transfer Function |
2021-02-13T21:50:59.367 | <p>I'd like to create a 3d printed system to damp the vibrations on the phone when we are on the road. The idea is to be able to hold the phone for my little kid when we travel</p>
<p>First I tried a spring system that connects the fixed part to the phone holding part, with no success, the vibrations were not only eliminated but increased.</p>
<p>Also tried inserting sponge/rubber material in between the phone and the phone holder, but seemed useless.</p>
<p>At this point I don't know what else to try, and I didn't find any commercial product.</p>
<p>How would you do it?</p>
| |dynamics|vibration| | <p>I agree totally with Jonathan R Swift's answer but I will try to give a insight as to why it is not appropriate to use a shock absorber.</p>
<p>The main issue is that the excitation frequency of the vibration is not constant. The vibrations frequency will change depending on how fast the car will travel and on the roughness of the road. All things being equal if you increase the speed of the car the excitation frequency increases, and vice versa.</p>
<p>As the excitation frequency changes the amplitude of the vibration will change. To get a rough idea of the change, below is a graph of the <strong>transmissibility ratio</strong> for different coefficients of the damping ratio.</p>
<p><a href="https://i.stack.imgur.com/XLWI9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XLWI9.png" alt="enter image description here" /></a></p>
<p>The transmissibility ratio shows how much forces and displacements are amplified to a mass if the foundation is excited. The value on the y-axis shows the amplification factor, so for example if Transmissability ratio (T for short) is 2, then if the excitation at the foundation has an amplitude of 1.2 [cm] then the excitation on the examined mass would be 1.2 *2 =2.4[cm]. (Small note - for those who are more knowledgeable-, T is applicable to the steady state response). So to have proper damping we need lower values of T.</p>
<p>One thing to note is the dependence on <span class="math-container">$\zeta$</span>. <span class="math-container">$\zeta$</span> is the damping ratio and indicates how fast the vibrational energy is converted to other forms of energy (usually heat). The higher the value of <span class="math-container">$\zeta$</span>, the higher the damping. However you will see that for high values of <span class="math-container">$\zeta$</span>, you will see that the transmissability ratio remains high (while the lower values have pronounced peak and then reduce significantly. So the bottom line is that by increasing <span class="math-container">$\zeta$</span> you will not always have less vibration.</p>
<h3>Additional issues</h3>
<p>The problem gets worse, because, what I showed you is for a simple system with one dof. Real structures have many more degrees of freedom. The end result is that you will get a graph more like:</p>
<p><a href="https://i.stack.imgur.com/axACW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/axACW.png" alt="enter image description here" /></a></p>
<p>In that graph you will notice that the amplification drops and increases, and doesn't really drop of below unity.</p>
<p><strong>transient response</strong></p>
<p>I've mentioned above that T applies to the steady state response. However, there is another response, <strong>transient response</strong>, which in engineering problems is usually neglected (because machines in many cases run at specific rpms for very long periods of time). In this case the transient response is a significant portion of the vibrational response and, cannot be safely removed.</p>
<h3>bottom line.</h3>
<p>As others mentioned, the best way would be to rigidly connect (as much as possible) the pad to the frame/seat.</p>
| 40357 | Damp vibration on phone attached to car seat |
2021-02-14T07:03:54.890 | <p>Hello all of you engineering geniuses. Awhile ago a very talented individual on here helped me work out the conceptual mechanics of this design. As you can see from the model, there is a lead screw assembly that pushes a rod forward which elevates the platform. The total travel is 0° to 60° (theta). The equation shown is the linear travel of "Xblock" as a function of theta. If you were to plot this function, you will see it is not quite linear. The terms, Wp, Ht, Hp, and Lrod are all constants. I am trying to determine an equation for angular velocity of the platform, which is actually decelerating as Xblock moves forward at a constant velocity. Of course the linear travel is easy enough, simply the rps of the motor x lead x time. The lead is 8mm but the lead and rps can be treated as constants. I could simply take the derivative of function, but it's a function of position as a function of another position, not time, so I'm not entirely sure where to go from here. <a href="https://i.stack.imgur.com/tl9lT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tl9lT.jpg" alt="concept" /></a></p>
| |mechanical-engineering|dynamics|positions-measurement| | <p>As PeteW mentioned you can differentiate both sides of the equation with respect to t, and then solve for <span class="math-container">$\dot\theta(t)$</span>.</p>
<p>If I done correct the calculations, you'll end up with</p>
<p><span class="math-container">$$\dot\theta(t) = -\frac{X'(t)}{a(\theta(t)) + b(\theta(t))}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$a(\theta(t))= -H_{p} \cos(\theta(t))+
\frac{1}{2} W_{P} \sin(\theta(t))-L_{rod} \arcsin\left(\frac{H_T-H_{p} \cos(\theta(t))+\frac{1}{2} W_{P} \sin(\theta(t))}{L_{rod}}\right) \sin(\theta(t)) $</span></li>
<li><span class="math-container">$b(\theta(t))= \frac{\cos(\theta(t))}{\sqrt{1-\frac{\left(H_T-H_{p} \cos(\theta(t))+
\frac{1}{2} W_{P} \sin(\theta(t))\right)^2}{L_{rod}^2}}} \left(\frac{W_{P}}{2} \cos(\theta(t))+H_{p} \sin(\theta(t))\right)$</span></li>
<li><span class="math-container">$X'(t)=\frac{2\pi n}{60} l_{thread}$</span>: the velocity of the cart:</li>
<li><span class="math-container">$n$</span> is the revolutions per minute [rpm]</li>
<li><span class="math-container">$l_{thread}=8[mm]$</span> is the thtread size in [meters]</li>
</ul>
<p>After, all the simplifications the angular velocity can be calculated as a function of <span class="math-container">$n [rpm]$</span>, and <span class="math-container">$\theta [rad]$</span></p>
<p><span class="math-container">$$\dot\theta(n, \theta) = -\frac{2\pi l_{thread}}{60(a(\theta) + b(\theta))}n$$</span></p>
<p>where:</p>
<ul>
<li><p><span class="math-container">$a(\theta)= -H_{p} \cos(\theta)+
\frac{1}{2} W_{P} \sin(\theta)-L_{rod} \arcsin\left(\frac{H_T-H_{p} \cos(\theta)+\frac{1}{2} W_{P} \sin(\theta)}{L_{rod}}\right) \sin(\theta) $</span></p>
</li>
<li><p><span class="math-container">$b(\theta)= \frac{\cos(\theta)}{\sqrt{1-\frac{\left(H_T-H_{p} \cos(\theta)+
\frac{1}{2} W_{P} \sin(\theta(t))\right)^2}{L_{rod}^2}}} \left(\frac{W_{P}}{2}\cos(\theta)+H_{p} \sin(\theta)\right)$</span></p>
</li>
</ul>
| 40363 | How to determine the angular velocity of the platform as a function of motor rotations per second? |
2021-02-15T14:23:25.523 | <p>It is stated on <a href="http://www.concrete.org.uk/fingertips-document.asp?id=474" rel="noreferrer">here</a> that:</p>
<blockquote>
<p>Air-entraining admixtures cause small stable bubbles of air to form uniformly through a concrete mix. The bubbles are mostly below 1 mm diameter with a high proportion below 0.3 mm. The benefits of entraining air in the concrete include <strong>increased resistance to</strong> <strong>freeze-thaw degradation</strong>, increased cohesion (resulting in less bleed and segregation) and improved compaction in low-workability mixes.</p>
</blockquote>
<p>This does not make sense to me. By introducing more voids into the concrete there are more paths of ingress for water which can then expand and crack the concrete.</p>
<p>Why do air entrainment admixtures improve the freeze-thaw resistance of concrete?</p>
| |civil-engineering|concrete| | <p>The air entrained in the concrete causes discontinuous voids, which will not cause water to penetrate the concrete. Water will permeate through the bleeding pores which are continuous cavities. The entrained air bubbles are small and will only allow space for the freezing water to expand.</p>
| 40384 | Why do air entrainment admixtures improve the freeze-thaw resistance of concrete? |
2021-02-15T15:22:29.857 | <p>Currently here it is 9 degrees Farenheit. My home is heated by electric resistance heating (not a heat pump, just plain heating coils). Am I correct in believing that it's more energy efficient to keep our wooden blinds and curtains closed and turn on LED lighting than to open our blinds and curtains for light?</p>
<p>Our windows are double-paned. My local utility says it's better to open the windows than turn on an LED light, but I can't believe that's true under current conditions.</p>
<p>My belief is:
Loss of heat due to opening blinds and curtains * electricity needed to replace that heat via resistance > electricity needed to power LED lights (say 8 watts per window).</p>
<p>NOTE: This is when the sun is <em>not</em> shining directly in thru the window. Direct solar radiation can be very heating, as detailed in an answer below.</p>
| |energy-efficiency|thermal-insulation| | <p>The reason your utility company recommends to let the sunshine in is not to save on lighting energy. it is to take advantage of solar radition energy.</p>
<p>a window can let the solar radiation in two ways, direct radiation and scatter radiation.</p>
<p>The calculation of radiation energy has to do with many factors such as the geographical altitude and azimuth of the sun directon. usually 60% of a hypothetical normal ray vector radiates and heats up the interior furniture and walls and floors.</p>
<p>Interestingly the solar radiation energy is independent of the ambient outside temperatures.</p>
<p>The electrical company has calculated the windows heat loss and heat gain and based on combined average seasonal weather offer their advices.</p>
<p>Usually their recommendations is sound.</p>
<p>This is an in-depth article on the solar radiatin energy calculations, <a href="https://energy-models.com/heat-gains-and-losses-windows-and-skylights-glass" rel="nofollow noreferrer">heat gains and losses</a></p>
| 40386 | More efficient to use LED lighting than to open blinds for light and thereby lose heat when very cold outside? |
2021-02-16T13:38:03.670 | <p>When reading about turbulent flows and turbulence spectrum I frequently come across the terms 'wavelength' 'wavenumber'. I understand that the energy is transferred from larger eddies to smaller ones and so there is a spectrum of different sized eddies spinning at different velocities. But what does the wavelength and frequency of turbulence mean?
Is wavelength the diameter of the various sized eddies? Does the frequency of turbulence mean the angular velocity of vortices or the frequency of variation of different state variables?</p>
<p>I have a feeling that the wavelength is related to some characteristic length of the eddies but why term this 'length' the 'wavelength'? How is it related to waves?</p>
| |fluid-mechanics|turbulence| | <p>Wavelength is with reference to Kolmogorov's theory of turbulence. Kolmogorov postulated a theory for how turbulence is conveyed from large to small scale eddies, as you described. The wavelength and frequency origins are buried deep in the mathematics of the theory. Basically, the theory postulates an energy spectrum for turbulence, which divides the energy spectrum by eddy wavelengths. This originates from a Fourier series that is used to characterize the distribution of perturbations over different length and time scales, followed by more complicated mathematics to obtain the energy spectrum (<a href="https://atmos.washington.edu/%7Ebreth/classes/AS547/lect/lect2.pdf" rel="nofollow noreferrer">https://atmos.washington.edu/~breth/classes/AS547/lect/lect2.pdf</a> page 4). The result is an energy spectrum, which can be integrated to determine the turbulent kinetic energy composition of a particular wavelength. The author in the attachment describes the energy spectrum of a particular wavelength as the result of all eddies whose characteristic length l is 1/2 of the corresponding wavelength. I am not the best with turbulence, but I hope this provides some insight.</p>
<p>Other sources:
<a href="https://clouds.eos.ubc.ca/%7Ephil/courses/atsc500/pdf_files/09-kolm.pdf" rel="nofollow noreferrer">https://clouds.eos.ubc.ca/~phil/courses/atsc500/pdf_files/09-kolm.pdf</a>
<a href="https://www.dartmouth.edu/%7Ecushman/courses/engs250/Kolmogorov.pdf" rel="nofollow noreferrer">https://www.dartmouth.edu/~cushman/courses/engs250/Kolmogorov.pdf</a></p>
| 40400 | What is 'wavelength' with reference to turbuence? |
2021-02-16T14:01:44.163 | <p>I have a question about work measured in foot-pound.</p>
<p>Let's say we have a 10 pound-mass (<span class="math-container">$10$</span> <span class="math-container">$\text{lb}_m$</span>) object. Then
since that 10 pound-mass (<span class="math-container">$10$</span> <span class="math-container">$\text{lb}_m$</span>) object gives 10 pound-force (<span class="math-container">$10$</span> <span class="math-container">$\text{lb}_f$</span>) due to gravity, if we raise that object
2 feet, the work is 20 foot-pound (<span class="math-container">$20$</span> <span class="math-container">$\text{ft}\cdot\text{lb}_f$</span>).</p>
<p>I was wondering if I am right.</p>
| |mechanical-engineering| | <p>The conversion 1 lb<span class="math-container">$_m$</span> = 1 lb<span class="math-container">$_f$</span> only works on earth. When the gravitation constant is different (e.g. on the moon), the direct conversion is not valid. Here are some points of reference.</p>
<ul>
<li><p>by definition: <span class="math-container">$1$</span> lb<span class="math-container">$_m = 0.453592387$</span> kg</p>
</li>
<li><p>standard gravity on earth <span class="math-container">$g = 32.174$</span> ft/s<span class="math-container">$^2 = 9.80665$</span> m/s<span class="math-container">$^2$</span></p>
</li>
<li><p>by definition: <span class="math-container">$1$</span> lb<span class="math-container">$_f$</span> <span class="math-container">$= 32.174$</span> ft/s<span class="math-container">$^2$</span></p>
</li>
</ul>
<p>By example, an object with a mass of 5 lb<span class="math-container">$_m$</span> has a mass of 2.2680 kg regardless of where it sits (e.g. what planet). But this mass</p>
<ul>
<li>exerts 5 lb<span class="math-container">$_f$</span> as its weight on earth</li>
<li>exerts 0.827 lb<span class="math-container">$_f$</span> as its weight on the moon with gravitation 5.32 ft/s<span class="math-container">$^2$</span></li>
</ul>
<p>By inversion, an object that exerts 5 lb<span class="math-container">$_f$</span> as its weight on the moon has a mass of 30.2 lb<span class="math-container">$_m$</span>.</p>
| 40401 | pound force and pound mass |
2021-02-16T17:44:12.613 | <p><strong>Question:</strong> <em>If aboard a train being driven along a track by centripetal force, would it feel the same as if you were on a regular straight track?</em></p>
<p>I hope this isn't subjective, I think there is a right or wrong answer here. Either it does feel similiar or it doesn't so hopefully this question isn't closed. I don't see it as an opinion based answer.</p>
<p>I recently did a little khan academy course because I wanted to understand the physics of centripetal force. I was a little unsure of some things so I devised this thought experiment to help me clear things up in my own head. I was hoping you fine folks might be able to help clarify a few things for me.</p>
<p><strong>The Scenario:</strong></p>
<p>You board a train cart, sit down and make yourself comfortable inside your carriage. This train is special because it doesn't have an engine, the carriage is driven along a rail sitting on a rotating plane, your carriage starts close to the axis of rotation and is pushed towards the edge of the plane and away from the rotating center axis by centripetal forces. A lot like being inside a centrifuge!</p>
<p><a href="https://i.stack.imgur.com/xDwwa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xDwwa.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/1xUJj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1xUJj.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/qa4jr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qa4jr.png" alt="enter image description here" /></a></p>
<p><strong>Details:</strong></p>
<p>Now I know as your radius changes the centripetal force changes, so the speed of our train changes as given by the formula: Centripetal force = mass x velocity^2 / radius. Constant acceleration/deceleration isn't good for passengers so for this scenario were going to presume the rpm of the plane is being altered to induce a generally constant speed like a normal train.</p>
<p>The track in the picture is a little short so please use your imagination to extend it, it's just to get the general idea across.</p>
<p><strong>What I don't understand:</strong></p>
<p>What I'd like to know, is once the train is at it's correct speed and my body has caught up to that speed, could I hypothethically get up and walk around on the train like a person would in a normal train or passenger plane, drink a cup of tea or go to the bathroom. Because I'm moving forward I'm not dealing with the forces imposed by centripetal forces, that's an asuumption I'm making which I presume is correct but what I'm not sure about is because the angle of my velocity vector is changing constantly (The direction I would fly off in if were to jump of the train.) does my body have to be for lack of a better word consistently updated by the carriage, pinning me against the train window at a suffiently high rpm? I don't know if that force is there or not. So the real question can I have a cup of tea on board or not?</p>
| |mechanical-engineering|dynamics|experimental-physics|linear-motion| | <p>Your question is related to rotating frames of reference and the Coriolis Force.</p>
<p>As a passenger you are going to be experiencing a lateral force which will be greater the the faster you are travelling. The magnitude of the acceleration you'd feel would be</p>
<p><span class="math-container">$$a_{coriolis} = -2\omega\times v_{train} $$</span></p>
<p>and the force would be :</p>
<p><span class="math-container">$$F_{coriolis}= -2\cdot m\cdot \omega\times v_{train} $$</span></p>
<p>where:</p>
<ul>
<li>m : is the mass of the object</li>
<li><span class="math-container">$\omega$</span>: is the angular velocity</li>
<li><span class="math-container">$v$</span>: is the velocity of the train in this example</li>
</ul>
<p>To understand what happens you have to start from Newton's law. Basically, whenever there is no force applied on a mass (or if the net force is zero), the mass tends not to change its kinetic state. Therefore, (in the absence of gravity), it would tend to travel in a straight line. In order for the mass to change its course, you need to apply a force on it.</p>
<p>Once you are on the train and you are travelling with the train, if the train is not rotating then you wouldn't feel anything. However, if the carousel starts to rotate, a force would need to be applied on you to keep you up with the train. In particular example that force would be friction.</p>
<p>This is exactly the same example, with someone who is at the center on a merry-go-round and wants to travel to the edge. Try that when the merry-go-round is not rotating and everything is fine. Try it when its spinning, and there is a very good chance that you loose a couple of teeth.</p>
<p>There is a <a href="https://www.youtube.com/watch?v=dt_XJp77-mk" rel="nofollow noreferrer">very nice video</a> which gives a very nice explanation about this.</p>
<h2>is the force there?</h2>
<p>Well... that depends on your frame of reference. :-)</p>
<p>Coriolis is one of the <em>pseudo-forces</em>. If you are on an inertial frame of reference (Newtonian), then you don't need the Coriolis force. To make that clearer (this will make more sense if you watch the video), if you throw a ball from a rotating frame of reference, then you don't need Coriolis to describe its movement.</p>
<p>However, when you are observing from the rotating frame of reference, then Coriolis is necessary. Therefore the ball that was thrown in the video, will need the existence of a force in order to describe its trajectory.</p>
<p>Hope that makes sense!</p>
| 40405 | If aboard a train being driven along a track by centripetal force, would it feel like you were on a regular straight track? |
2021-02-16T19:17:41.180 | <p>I'm looking for right angle gearbox, but everything is horrible expensive or in parts like two bevel gears from aliexpress.
Where I can find or how to name transmission like this in the picture but in box case (if possible with mounting holes)?</p>
<p>I need right angle transmission 1:1, no matter what type of gear or something else. Looking for compact size, outer size of one arm, max 40mmx40mm. Maybe I don't know proper name to find it.</p>
<p>Also if I don't find good out of the box solution, how to proper design that type of transmission? I know I need bearings on shaft and screw, but how to avoid skipping on teeth?</p>
<p><a href="https://i.stack.imgur.com/dAo3t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dAo3t.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|gears|mechanisms|power-transmission| | <p>these things are sold cheaply (less than $20) for bending the output of an electric drill through 90 degrees, so you can drill holes in cramped places. they can handle from 1/4 to 1/3 horsepower typically. hardware stores sell them. they are called <em>right angle drill drive attachments</em>.</p>
| 40408 | Right angle gearbox, proper name or design |
2021-02-17T19:56:51.697 | <p>How is the load torque on a turbine pump handled. Generally, with motors, there is a load attached to the shaft which resists the output torque of the motor however, with a pump increasing the flow rate of water, I assume the water should act as a smaller load.</p>
<p>Currently, I am calculating this load torque using the approach of a point load at the end of a cantilever beam. I know in a more realistic sense, the water would be a distributed load but I am going for a simplistic answer and I'm wondering if this is a valid approach to calculating this load.</p>
| |pumps| | <p>Pump torque can be calculated via the power consumed and the rotational speed in rpm:</p>
<pre><code>Power = Torque * Rotational Speed
</code></pre>
<p>Rearranging this equation, the torque is</p>
<pre><code>Torque = Power/Rotational Speed
</code></pre>
<p>Without details about the type of pump, this may be the most straightforward calculation to determine the torque. If you know more about the type of pump (i.e. centrifugal, positive displacement, etc), there are specific pump type equations you can use to estimate the power and then the torque (if you have pump rpm). Even easier than that, if you have the pump characteristic curves you may be able to find the power as a function of another parameter, like flow rate or rpm.</p>
<p>I would advise against using the method you suggested if you're simply looking for pump torque (unless you know more about the shaft and applied force). The methods I listed above are more conventional for pump analysis/selection.</p>
| 40428 | Calculating the load torque on a turbine pump |
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