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2019-11-14T09:33:59.257 | <p>On the image below there is a regular prism which is angled by <span class="math-container">$\alpha$</span>. <a href="https://i.stack.imgur.com/AyUOn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AyUOn.png" alt="enter image description here"></a></p>
<p>The (black) mass m1 applies a torque <span class="math-container">$M_k$</span>. The mass m2, which is the rest of the Body applies the torque <span class="math-container">$M_s$</span> In the opposite direction. The black dot on the bottom right corner is the pivot point.
So i wanted to calculate the normal force distribution on the bottom surface which is completely touching the ground.
Because the force F1 has to be smaller than F2 right?
My final Goal is to calculate the friction of the prism.
So my question was: Does it matter if i calculate the normal Force by <span class="math-container">$ F_N = m * g$</span>. Or do i have to integrate <span class="math-container">$M$</span> over the whole bottom surface and thus derive the mean of the normal force.</p>
<p>Another problem is: If the prism is accelerated by a another torque <span class="math-container">$M_a$</span> is applied. So you definitely cannot calculate the normal force by <span class="math-container">$m*g$</span> anymore.</p>
| |applied-mechanics| | <p>You just need to find the CG of the parallelogram if there is no external torque, just add the overhanging mass of black triangle. This is typically called off-center column or P-Delta loading or a column with both axial and moment loading.</p>
<p>if there is external torque other than the black triangle mass you add it treating a+b as a beam with moment <span class="math-container">$M_k \ $</span> applied at the right support of the beam. </p>
<p>So scaling your sketch we assume the height of the part as a, and then you have a base a+b with its CG at <span class="math-container">$$\bar X =\frac{1/2ab*b/2 +a^2*(b/2+a/2)+1/2ab*(a/2+2*b/2)}{1/2ab+a^2+1/2ab} $$</span></p>
<p>So the distribution of the forces on the base will be the sum of the normal weight divided by the base surface which is uniform stress, say q, and the triangular stress, say p, do to overturning moment which is <span class="math-container">$ \sigma_{right}=\frac{6M( \bar X- 1/2(a+b))}{ (a+b/2)^2} \quad \sigma_{left}=0 $</span></p>
<p>So we add these 2 stresses, q and the triangular p together and we get a trapezoid stress surface, with the right hand bigger than left hand stress.. </p>
<p>Note: I assumed the depth and mass density as 1.</p>
| 31927 | Normal Force distribution of Prism |
2019-11-14T17:04:49.823 | <p>In the pharmaceutical environment, the drop height from the hopper to the area where it collects must be minimized. Tubes can be used to minimise the height. I know that it affects powder segregation. But do not understand the mechanism of it. How does it affect powder segregation ? As long as the flow is optimal (eg. no bridging), then why will powder still segregate if drop height is not minimal ?</p>
| |design| | <p>Small powder particles react with the air flow differently, such as small water droplets which never fall and remain suspended in the air and create haze. </p>
<p>So a long shoot will give a chance to the powder to separate.</p>
| 31932 | what impact does drop height have on powder segregation? |
2019-11-15T18:47:47.797 | <p>When we twist a shaft/beam/rod, shearing stresses are produced. The angle by which the end of the shaft rotates can be found as:</p>
<p><span class="math-container">$${\displaystyle \theta ={\frac {TL}{GJ}}}$$</span></p>
<p>where <span class="math-container">$T$</span> is torque, <span class="math-container">$L$</span> is the length, <span class="math-container">$G$</span> the shear modulus, and <span class="math-container">$J$</span> a torsion constant depending on the geometry of the cross section. For circular cross section, <span class="math-container">$J$</span> equals the second moment of area. For non-circular shafts there is an effect called "warping" present where the cross sections don't remain planar during twisting; if we had a beam with a rectangular cross section and held it radially away from us and twisted it, the cross sections would bend towards us/away from us. Do I understand warping correctly?</p>
<p>But shafts with circular cross sections do not experience this. Planar cross sections remain planar, and deform only in 2D. <strong>Why is this?</strong></p>
| |torque|shear|deformation| | <p>Let's imagine a solid cylinder attached to a fix support at one end and free at other. We twist it at the other end by a torque.</p>
<p>It will rotate about it's axis without warping, ( small angles condition).</p>
<p>Now we cut the cylinder with laser longitudinally in 4 cuts into 8 triangle sections, radiating from the center of the circle, so if you look at the cross section it look like a circle with 4 diagonal lines at 45 degrees. </p>
<p>Now if we twist this cylinder with the same torque we see the triangles wind around each other like a cable and the end of them is not a plane any more. it looks like a concrete drill bit. The circular triangles have warped and if you measure the strain half of each triangle is shortened and half elongated.</p>
<p>The reason is the strain energy stored into this axial deformation is more than what would be if there was only shear strain.</p>
<p>Now if you weld the seams back together you deny the triangles the opportunity to wind about. So the plane of the rotated cylinder remains plane and the strain stored is just shear.</p>
| 31949 | Why torsional warping does not occur for shafts with circular symmetry? |
2019-11-16T13:01:02.730 | <p>Lord Evulz just stole the Earth's atmosphere. Our hero must race his trusty old Mini - which he had already prepared to be airtight, to ensure the tires do not explode, with pre-packaged oxygen to supply his fuel with, etcetera... - to the last escape rocket before his air supply runs out.</p>
<p>Assume regular Earth highway conditions - (with no traffic because everybody asphyxiated to death) - with the only difference that there is no air resistance. What are the limiting factors on the car's top speed? And what velocity might be attained?</p>
<p>If the specific model matters much, take the 1977 British Leyland Mini 1000 - because it is Mr Bean's car and I fully expect him to find a way to drive it into orbit.</p>
| |automotive-engineering|vacuum|car| | <p><strong>TLDR:</strong> Top speed increases from 81 to 89 mph. </p>
<p>The air drag force is 0.5 x air density x drag area * speed^2.
(<a href="https://en.wikipedia.org/wiki/Automobile_drag_coefficient" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Automobile_drag_coefficient</a>). Therefore drag power is 0.5 x air density x drag area * speed^3. The cube term on speed is very important, i.e. doubling speed requires 8 times more power to overcome air resistance. At highway speeds, air resistance becomes dominant. </p>
<p>According to <a href="https://www.automobile-catalog.com/auta_details1.php" rel="nofollow noreferrer">https://www.automobile-catalog.com/auta_details1.php</a>, the 1977 Mini has an <em>Estimated</em> drag area of 0.788 m^2 </p>
<p>Doing the math, this means that at the top speed of 81 mph, overcoming air drag requires about 31 HP. According to same website, the 1977 Mini has 39 net HP. Therefore, at top speed, 79% of the engine's power is being used to overcome air resistance.</p>
<p>So, in theory, removing air resistance would give a huge amount of extra available power. However, based on the same site again , max horsepower occurs at 4750 RPM, whereas the engine redline is 5200 rpm. Therefore, only a modest increase of 9% is possible. Top speed would be 89 mph.</p>
<p>To achieve a higher max speed, you would need to redesign the gearbox to take advantage of the reduced air drag. I.e. give a higher max velocity for the same engine speed by using a different gear ratio. </p>
| 31957 | How fast could a Mini Cooper go in a vacuum? |
2019-11-16T13:56:22.540 | <p>The picture (which I found on the web without any accompanying explanatory text) shows a home-made z-axis micro-adjust lead-screw add-on to the vertical milling attachment of a vintage hobbyist lathe.</p>
<p>The greenish motor-and-spindle-mount casting rides up and down on the black post, sandwiched between two plates, which ride up and down on two smaller posts. </p>
<p>I can't quite make out whether there is a threaded bushing of some kind where the arrows are pointing. How would that aspect of the design typically be engineered? </p>
<p>I'm thinking that if those plates were simply threaded, without some kind of insert, you could have a somewhat loose and sloppy sandwich.</p>
<p><a href="https://i.stack.imgur.com/Bhs6Z.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bhs6Z.jpg" alt="Z-axis micro-adjust lead-screw"></a></p>
| |mechanical-engineering|mechanisms|threads| | <p>It's difficult to determine from the photo which parts are moving and which parts are dependent on another part for specific movement. If you get proportionately smaller travel per revolution of the main knob, the other parts of the assembly could be a <a href="https://en.wikipedia.org/wiki/Differential_screw" rel="nofollow noreferrer">differential screw mechanism</a>.</p>
<p>Differential screw mechanisms are used in micrometers, in which the main knob rotates a substantial amount and the spindle moves a fraction of the distance. It is a construction in which an inner rod is threaded to one specification and the outer rod is threaded to a slightly larger or smaller specification.</p>
<p>I've constructed a 3D model in which the travel of the spindle is 0.25 millimeters for each rotation of the main knob, which is threaded to 1.0 mm pitch, while the inner thread is 1.25, a difference of 0.25.</p>
<p>The reason for such a mechanism is that it is easier to build a 1.0 mm and a 1.25 mm pitch thread compared to a 0.25 mm pitch, which is what would be required if one desired to bypass the differential threading option.</p>
<p><a href="https://i.stack.imgur.com/ZAjMl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZAjMl.png" alt="differential threading image"></a></p>
<p><a href="https://www.thingiverse.com/thing:2279953" rel="nofollow noreferrer">Thingiverse</a> provided me the model to learn more of this construction.</p>
<p><a href="https://i.stack.imgur.com/OAcsO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OAcsO.jpg" alt="differential threading model from Thingiverse"></a></p>
<p>The outer threads of this mechanism would be engaging the brackets you've noted in the photo. Note also the key and matching keyway between the inner threaded rod and outer sleeve. It's critical that the portion does not rotate in order to provide the fractional movement desired.</p>
<p>In your photo, it's logical to expect that only one of the brackets contains a differential mechanism, while the other provides an alignment bushing of sorts.</p>
| 31959 | how does this z-axis micro-adjust lead-screw design work |
2019-11-17T04:04:34.427 | <p>A friend and I were debating whether a biplane/propeller plane has a faster turning rate than jet aircraft. I argued that the propeller planes often do have better turning rates than jets, but I could not explain why. Does anyone know why this is?</p>
| |mechanical-engineering|aerospace-engineering|aerodynamics| | <p>Interesting Question:</p>
<p>(Information from my own knowledge as well as from <a href="https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80t-80.pdf" rel="nofollow noreferrer">Aerodynamics for Naval Aviators</a>. I do feel like this question is better suited to the aerodynamics area.)</p>
<p>Formally, turn rate is defined as a unit of angular measurement over a unit of time; degrees per second is a common unit of turn rate. The other measure used to define turning performance is turn radius, otherwise explained as the size of a turn at a predefined speed and altitude.</p>
<p>Turn rate can be defined in many different ways: instantaneous, sustained, maximum, and a few others I've probably forgotten. Instantaneous turn rate regards performance if the pilot pulls as hard as possible on the stick without destroying the airplane: generally, planes lose speed at their instantaneous turn rates, as they use more energy in the turn than their engines can produce. Sustained turn rate, on the other hand, is the maximum rate of turn that can be achieved while maintaining a steady altitude. Sustained turn rate assumes that there is a speed at which the drag and engine power will become equal. Maximum turn rate is a term added to both instantaneous and sustained turn rate to, commonly, note that at a certain speed, a plane has the best rate (degrees per second) of turning. This is most often seen with sustained turn rate, because instantaneous turn rate generally increases with speed (but not always).</p>
<p>Generally, (theoretical) comparisons between low wing loaded airplanes (generally lighter, slower propeller planes) and high wing loaded airplanes (high speed jet aircraft) result in the jet aircraft being much faster, but with less of a turn rate. The general explanation behind this is that planes which fly at a lower speed are limited less by human endurance: G-forces increase proportionally to the flying speed and change in velocity during a turn. Slower planes can increase the change in velocity (pulling the turn tighter) because their speed is lower, if the G limits are the same.</p>
<p>In the sustained turn department, modern fighter jets tend to be faster than biplanes; in practical combat, it can be argued that biplanes use their low speeds and higher instantaneous turn rates to have long lasting duels that start at altitude and spiral towards the ground.</p>
<p>Theoretically, if you strap the engine from an F16 onto a biplane, and replace the human with a computer, the aircraft could turn 180 degrees in a couple of seconds.</p>
<p>Interesting question, but the issue in comparison can be the less efficient engines on biplane designs, and more efficient engines on modern aircraft.</p>
| 31962 | Turn rate of a jet vs. biplane |
2019-11-17T20:00:32.107 | <p>I'm looking for a few basic devices such as converting translational to rotational motion, 20:1 gearbox, etc. Found nothing at Home Depot. Could you suggest where does one buys/orders such components?</p>
| |mechanisms| | <p>Here are my go to sources:</p>
<ul>
<li><p><a href="https://www.pic-design.com" rel="nofollow noreferrer">Pic Design</a> (Good for gears, screws and other components)</p></li>
<li><p><a href="https://www.sdp-si.com" rel="nofollow noreferrer">Stock Drive</a> (Similar to Pic Design)</p></li>
<li><p><a href="https://www.wmberg.com" rel="nofollow noreferrer">WM Berg</a> (Good collection of parts. Particularly good for chain and belt drives and alternatives)</p></li>
<li><p><a href="https://www.mcmaster.com" rel="nofollow noreferrer">McMaster-Carr</a> (Broad range of products)</p></li>
<li><p><a href="https://www.centuryspring.com/" rel="nofollow noreferrer">Century Spring</a> (Awesome for springs)</p></li>
</ul>
| 31971 | Where do I get basic mechanical components? |
2019-11-17T20:18:59.053 | <p>The lower viscosity of water reduces friction. Besides, the water can transfer heat way better than oils. So why not use water as a lubricant in wind or hydro turbines, especially on equipment which is operating in the ocean ...</p>
<p>It seems to be a perfect material. It's the most abundant liquid on Earth, and it causes no ecological damage. </p>
| |mechanical-engineering|water-resources| | <p>Great question! Water absolutely <em>is</em> used as a lubricant in some power generating systems. For example many Francis style turbines use their feed water as the primary lubricant and coolant of the bottom main bearing. (In the old Loeffel turbines, that bearing was typically made of lignum vitae wood, and could last up to 100 years in near constant use). Good engineering takes advantage of the cheap availability of water whenever applicable.</p>
<p>But water freezes and expands, blowing apart your engines, so it's not suitable for every application. This is enough of an everyday issue that automobiles are designed with "freeze plugs" and block heater ports and typically use toxic and environmentally unfriendly antifreeze chemicals in their liquid coolant systems. If your generator fails in cold weather, you don't want it to freeze solid before you can manage to repair it.</p>
<p>(A freeze plug, also called a Welch plug, is a core hole plug purposely made weak enough that it will fail before the block cracks due to frozen coolant. Core holes are left by the block casting process.)</p>
<p>And although absolutely pure water is an electrical insulator in the laboratory, in real life it's not, because even the slightest amount of contamination makes water an excellent electrical conductor. Oils, on the other hand, typically have consistently low conductivity until they are so loaded with metal contaminants that they have lost lubricity. If your generator has an oil leak, the oil won't provide a conductive path to repair technicians and other system parts.</p>
<p>And finally, as others have already noted, water is very corrosive, especially seawater. Since rust expands to five times the size of the steel it came from, you can't use any of the cheap iron alloys in water, which drives up the cost of materials pretty quickly.</p>
<p>Petroleum oils are a superb lubricant, able to be purified and recycled many times. At least as early as 1915 intelligent people were writing that burning fuels like petroleum is a criminal waste of a fantastically valuable and limited resource; it's robbing future generations in order to turn our shared treasure into atmospheric poisons. It's nice to see people like yourself thinking about these things!</p>
| 31972 | Why is oil used as the lubricant in power generators, while water is the most available, cheapest and accessible lubricant? |
2019-11-18T13:58:11.933 | <p>Is there any heat energy lost to the mechanical force of pushing a piston in a combustion engine?</p>
<p>In other words, would a natural gas burning as a flame produce more heat than the same amount of natural gas that was burned when driving a combustion engine?</p>
<p>The context of this question is around fuel efficiency of natural gas water boilers. </p>
<p>My current understanding is that a naked natural gas flame is burning inside the unit, and heats water that is contained within a series of pipes, which in turn is cooled in a heat exchanger, and that same heat exchanger heats the incoming cool water.</p>
<p>I'm wondering, would if it make any sense at all to burn that natural gas inside a combustion engine and use the heat of the combustion engine used to heat the water to the same extent as a naked flame.</p>
<p>More generally my goal is to solve the question, can we heat water the same extent by burning the same amount of natural gas <strong>and</strong> get free mechanical energy?</p>
<p>Thanks</p>
| |thermodynamics|combustion| | <p>Check out combined heat and power systems... generates electricity and makes use of the waste heat to heat water - but is only really efficient when there is sufficient hot water demand ie heating load.</p>
| 31982 | Recovering mechanical energy from natural gas water boiler |
2019-11-18T23:12:58.797 | <p>We usually use induction motors to turn electrical energy to mechanical, however synchronous motors are the machines on the other side of the grid (generation).</p>
<p>A week ago we had a student visit to the hydroelectric and ... centrals and afterward we paid a visit to a steel company. I noticed that they use the combination of the synchronous and induction motors to do the exact same job, so the output power of the motors were exactly the same so the task and the hours of operation. </p>
<p>Why just not using induction motors ?</p>
| |electrical-engineering|motors|industrial-engineering| | <p>As the name suggests, induction motors are inductive, it means they require inductive power input, or reactive power.</p>
<p>The problem with reactive power consumption is, if you max the monthly limitation out, the providers are going to charge you a lot. Why? then this is another question. Hint: Think about the phase relationship in AC circuits. </p>
<p>The synchronous motors can receive active power from the grid, and at the same time (under some circumstances) generate reactive power and inject it into the grid. </p>
<p>Usually, by this method engineers can compensate for almost all reactive power that those induction motors consume so, the units don’t have to pay any fine to the provider, and providers will be happy.</p>
| 31988 | Why large industrial units combine induction and synchronous motors? |
2019-11-19T00:58:03.597 | <p>The <a href="https://en.wikipedia.org/wiki/Kangaroo" rel="nofollow noreferrer">kangaroo</a> has incredible jumping abiltiies. Its <a href="https://en.wikipedia.org/wiki/Gastrocnemius_muscle" rel="nofollow noreferrer">Gastrocnemius muscle</a> is a major part of generating its jumping force. Has any engineer based a design, probably a vehicle, upon the kangaroo's locomotion system? </p>
| |transportation| | <p>If you mean wether we have similar structure and articulation of bones and muscles, yes we do, as your link indicates. And some of us by training can jump rather high.</p>
<p>It is more than a muscle or a joint and muscles. Each of our species after long period of time has evolved for certain tasks, obviously.</p>
<p>I think a kangaroo jumps by a nerve impulse which once triggered organizes an ordered sequence of muscle contraction and wave of impulse throughout it's entire body, like a set of struts and springs calibrated to achieve a mechanical feat. In fact all animals do this in sorts for locomotion and doing tasks.</p>
<p>It's in their genes, the path of nerve impulse, the call to instantaneous tightening of back and neck, prohibiting them to lag limply as a damper, but work as a wiplash and and amplifier for the progression of smooth arc of jump.</p>
<p>This is part of the subject of the field of neural networking. </p>
<p>I used to run at my college, and we had learned that you run with your whole body not your legs.</p>
| 31991 | Has the kangaroo's jumping ever been copied in human engineering? |
2019-11-19T13:45:59.377 | <p>with masses declared as grams and static stiffness as N/mm, how do I dimension them for Hz ?</p>
| |mechanical-engineering| | <p>Units of N, mm (for the stiffness k), and g (for the mass m) are not consistent. You need to convert these values to use in the basic equation for the frequency: <span class="math-container">$$f=\frac{1}{2*\pi} \sqrt \frac{k}{m}$$</span> Knowing that <span class="math-container">$1 \, N = 1 \, kg*m/s^2$</span>, <span class="math-container">$1 \, m = 1000 \, mm$</span>, and <span class="math-container">$1 \, kg = 1000 \, g$</span>, you get the following:
<span class="math-container">$$f=\frac{1}{2*\pi} \sqrt \frac{N/mm*((1 kg*m/s^2)/N)*(1000 mm/m)}{g*(1 kg/1000 g)}$$</span>
All of the units cancel except for <span class="math-container">$\sqrt {1/s^2}$</span>. Doing the math gives the following result:
<span class="math-container">$$f = \frac{1000}{2*\pi} \sqrt \frac{k \; in \; N/mm}{m \; in \; g}$$</span>
where f is in units of cycles/sec = Hz.</p>
| 31997 | How do I dimension the natural frequency formula? |
2019-11-19T16:49:39.670 | <p>I am trying to prove with Matlab that if I have an improper system and I place poles at higher and higher frequencies the performances of the system improves. In particular I am considering the following two degree of freedom scheme:</p>
<p><a href="https://i.stack.imgur.com/I5YwC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I5YwC.png" alt="enter image description here"></a></p>
<p>where <span class="math-container">$C_{f}= \frac{(1+s)(1+0.05s)^{2}]}{(1+\tau s)^{3}}$</span></p>
<p>My code is the following:</p>
<pre><code>s = tf('s');
P = 1/[(1+s)*(1+0.05*s)^2];
C = (s+1)/s;
tau_1 = 0.1;
CF_1 = [(1+s)*(1+0.05*s)^2]/((1+tau_1*s)^3);
tau_2 = 0.01;
CF_2 = [(1+s)*(1+0.05*s)^2]/((1+tau_2*s)^3);
tau_3 = 0.001;
CF_3 = [(1+s)*(1+0.05*s)^2]/((1+tau_3*s)^3);
T1 = (C+CF_1)*P/(1+P*C);
T2 = (C+CF_2)*P/(1+P*C);
T3 = (C+CF_3)*P/(1+P*C);
figure;
bodemag(T1,'r',T2,'b',T3,'g'),grid
legend('tau = 0.1','tau = 0.01','tau = 0.001')
</code></pre>
<p>so what I expected is that the performances with respect to the reference tracking increse as tau gets smaller, but if I do the Bode plot, what I get is:</p>
<p><a href="https://i.stack.imgur.com/kT8fZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kT8fZ.png" alt="enter image description here"></a></p>
<p>from which I don't really see much of an improvement. Moreover if I change some values:</p>
<p><a href="https://i.stack.imgur.com/TOWnj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TOWnj.png" alt="enter image description here"></a></p>
<p>which is somenthing that to me does not makes sense because I should have that with <span class="math-container">$\tau =1$</span> I should have better performances than with <span class="math-container">$\tau =10$</span>, this because with <span class="math-container">$\tau =1$</span> the pole is at higher frequencies that with <span class="math-container">$\tau =10$</span>.</p>
<p>Can somebody please help me solving this problem? </p>
<p>Thanks in advance.</p>
<p>[EDIT] If I plot the step responses I see the same problem:</p>
<p><a href="https://i.stack.imgur.com/NkXPt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NkXPt.png" alt="enter image description here"></a></p>
<p>[EDIT 2]For completeness, I post the image of the step response for the first choise of tau's:</p>
<p><a href="https://i.stack.imgur.com/pJvua.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pJvua.png" alt="enter image description here"></a></p>
<p>for these values of <span class="math-container">$\tau$</span> there is a clear improvement for overshoot and a faster response. </p>
<p>I also tried other values smaller than 1 for <span class="math-container">$\tau$</span> and all of these show what I expected in the step response. While for values bigger than 1, I obtain something similar at the other situation.</p>
<p>Does somenone know why this happens? Thanks.</p>
<p>[EDIT 3] With the last one, so values of <span class="math-container">$\tau$</span> smaller than 1, I have also noticed an increase in phase margin, so this should mean that the system is performing better.</p>
<p>While, if I consider the values <span class="math-container">$\tau=1$</span> <span class="math-container">$\tau=10$</span> i get that for the first one the phase margin is 125 deg and for the second is 170 deg. So this should be in according to the fact that <span class="math-container">$\tau=10$</span> performs better.</p>
| |electrical-engineering|control-engineering|control-theory|matlab| | <p>Your closedloop crossover frequency (when the magnitude of <span class="math-container">$P(s)\,C(s)$</span> is equal to one) lies at roughly 1 rad/s. This means that the feedback controller already causes the system to track reference signals that have a frequency content sufficiently below that. So adding feedforward that only approximates the inverse of the plant (<span class="math-container">$C_f(s)\,P(s)\approx1$</span>) below the crossover frequency should not aid in the tracking performance, so that is why for both <span class="math-container">$\tau=1$</span> and <span class="math-container">$\tau=10$</span> the magnitude of the overall transfer functions both start to drop off around 1 rad/s, similar to if you wouldn't use any feedforward (<span class="math-container">$C_f(s)=0$</span>).</p>
<p>However, for <span class="math-container">$\tau=1$</span> the magnitude of <span class="math-container">$C_f(s)\,P(s)$</span> near 1 rad/s is still close to 0 dB, but the phase at 1 rad/s has already dropped to -135°, so the actual feedforward is actually doing partially the opposite of what the ideal feedforward would do and thus steering away from better reference tracking. For <span class="math-container">$\tau=10$</span> the magnitude of <span class="math-container">$C_f(s)\,P(s)$</span> near 1 rad/s is already close to -60 dB. So, even though the phase is already way lower compared when using <span class="math-container">$\tau=1$</span>, the magnitude is so low that the feedback controller can counteract the little disturbance the feedforward term is causing near that frequency. So that is why <span class="math-container">$\tau=10$</span> performs better (less overshoot) than <span class="math-container">$\tau=1$</span>. Though, it can be noted that for <span class="math-container">$\tau=10$</span> there is a 0.3 dB peak near 0.06 rad/s and an overshoot of 0.03 after 20 seconds, while the system with only feedback and no feedforward has no overshoot.</p>
| 32002 | Problem with performances of a control scheme |
2019-11-20T22:19:51.513 | <p><a href="https://i.stack.imgur.com/BAoAz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BAoAz.jpg" alt=""></a></p>
<p>The 40 kg load on the rope accelerates from 0 to 50km / h in 6 seconds. </p>
<p>The rope is tightened with a force of 500kg while the whole assembly is on it.
The middle pulley is driven by a dc motor. </p>
<p>I want to calculate the optimum angle between the middle (drive) pulley and the end pulley (pulley 2 on the picture) for driving without slipping or without too much load on the motor.</p>
<p>An angle between two pulleys that is too small would cause slippage on the rope, while too big angle would cause a higher load on the motor, and a higher power consumption.</p>
<p>The pulleys are made of aluminum and the rope has a coefficient of friction of 0.1</p>
<p>The current angle between the two pulleys is 15 degrees, as seen in the picture.</p>
<p>Motor spec are: 24V, 3560 oz/in stall torque, 4.3 HP </p>
<p>If you need more information, please ask.</p>
| |motors|friction|pulleys| | <p>Sorry if my explanation is simple (anything wrong please point out).<br>
With my understanding of your situation the right way should be something on this line.</p>
<p>Your objective velocity is <em>50Km/h (13,88m/s)</em><br>
To obtain this velocity in 6s you'd need calculate aceleration<br>
<span class="math-container">$$a=v/t$$</span>
<span class="math-container">$$a = \frac{13,88[m/s]}{6[s]}$$</span>
<span class="math-container">$$a = 2,315[m/s²]$$</span><br>
The force needed to obtain our aceleration is
<span class="math-container">$$F=m*a$$</span>
<span class="math-container">$$F = 40[Kg] * 2,315[m/s²]$$</span>
<span class="math-container">$$F = 92,6N$$</span></p>
<p>To acelerate without slipping you need estatic friction force higher than your aceleration force so <span class="math-container">$F\mu_{est} > F$</span></p>
<p><span class="math-container">$F\mu_{est} = F_{motor} * \mu$</span> ( <span class="math-container">$F_{motor}$</span> for traction force on wheel)<br>
<span class="math-container">$$92,6N = F_{motor} * 0,1$$</span>
<span class="math-container">$$F_{motor} = 926N$$</span></p>
<p>Now, for the angle you want we use trigonometry.<br>
Considering the cable pulling force is 500Kg total, I'll consider 2.500N traction in each side.</p>
<p><span class="math-container">$$sin\theta = 930/2500$$</span>
<span class="math-container">$$\theta = 21,84º$$</span></p>
| 32021 | Calculating optimal angle on a rolling pulley wheel, which holds weight, on a horizontal cable |
2019-11-21T06:51:02.477 | <p>The amount of light that an object reflects is called its <a href="https://en.wikipedia.org/wiki/Albedo" rel="nofollow noreferrer">albedo</a>; a scale of 0-1, 0 being a black body ad 1 reflecting all light. This question is probably going to be closed [It is "subjective" ?? No idea how] so I may as well be brief. </p>
<p>In the 1990's some scientists were suggesting that if you changed the albedo of an asteroid then it could be hotter on one side. That would change the curve of orbit. I have heard nothing since about this idea. I imagine that various ways would exist such as spraying the asteroid with paint or putting a solar panel on it. Is this possible? [I would ask if anyone is trying to do it but the rules of stack overflow say I can only ask one question.] </p>
| |aerospace-engineering| | <p>Yes, in theory, it is possible to use albedo modification as a way to influence the orbital parameters of a body through <a href="https://en.wikipedia.org/wiki/Radiation_pressure" rel="nofollow noreferrer">solar pressure</a>. This is the same principle used in solar sails, a propulsion method that relies on radiation pressure from the Sun to propel space probes and even adjust their attitude. The theoretical basis has been known since the late 19th century, the principle was experimentally demonstrated in 1899 by Lebedev and successfully deployed on a spacecraft in 2010 by the <a href="https://en.wikipedia.org/wiki/IKAROS" rel="nofollow noreferrer">IKAROS</a> mission.</p>
<p>It can be applied to asteroids by simply making them more reflective, which will increase the effects of solar pressure. Over enough time, the minute but constant acceleration will add up to significant changes in their trajectory. If the aim is to avoid a planetary impact, it might not even be necessary to coat a specific side of the asteroid.</p>
<p><a href="https://oaktrust.library.tamu.edu/bitstream/handle/1969.1/ETD-TAMU-2011-08-10184/GE-THESIS.pdf?sequence=2&isAllowed=y" rel="nofollow noreferrer">Here is a Master's Thesis</a> on the related subject of designing a payload that can acomplish said albedo modification. From the abstract:</p>
<blockquote>
<p>The development of the Surface Albedo Treatment System (SATS) onboard a
spacecraft mission to the near earth asteroid (NEA) Apophis in 2012 is an innovative concept of deflecting NEAs from possible impact with the Earth through altering the <a href="https://en.wikipedia.org/wiki/Yarkovsky_effect" rel="nofollow noreferrer">Yarkovsky effect</a>, a non-secular force in the solar system due to uneven surface thermal emission most profoundly affecting small rotating bodies subjected to sunlight. Though this force is small, its magnitude can be dramatic if extended over a period of time and if it uses the close approach of an asteroid near Earth to magnify the perturbation.</p>
<p>The payload dispenses colored powder called albedo changing particles (ACPs)
onto the surface changing its albedo and indirectly the surface temperature which
changes the Yarkovsky effect. This study gives an in-depth description of both
computational and experimental parts of the design of this system with primary focus on
initial ground test setup. The initial experiments proposed to design the SATS is outlined in detail and justified by the mission criterion of interest as well as modeling the actual dispersal on the surface. </p>
</blockquote>
| 32025 | Can you change an asteroid's albedo to take it off a collision course? |
2019-11-21T16:02:30.133 | <p>I need to find the response of my FTIR spectrometer. I do have a calibrated black-body source (area-emitter) which I measured at various temperatures.</p>
<p>The system operates in inverse cm, cm-1. My approach so far was the following:</p>
<ul>
<li>1) Measure spectra at temperatures <span class="math-container">$T_i$</span> </li>
<li><p>2) Scale the spectra i.e. </p>
<ul>
<li>the x-axis <span class="math-container">$x_{in~um} = 1e4/x_{in~cm-1}$</span></li>
<li>the y-axis, <span class="math-container">$y_{in~um} = y_{in~cm-1}/x_{in~cm}^2$</span>
to account for the dispersion relation between measured energy spacing and desired wavelength spectrum. At this point an example for scaled measurement (blue) and theoretical blackbody (orange) at the same nominal temperature looks like this (CO2 and H2O lines are at the correct positions):
<a href="https://i.stack.imgur.com/aYwfl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aYwfl.png" alt="enter image description here"></a></li>
</ul></li>
<li><p>3) calculate the response function <span class="math-container">$R=I_{BB}(T_i)/y(T_i)$</span> (I think, here in a first approach I assume that the background radiation does not contribute significantly. I would have expected similar response functions for the different temperatures; however what I get looks like:</p></li>
</ul>
<p><a href="https://i.stack.imgur.com/RvlSl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RvlSl.png" alt="enter image description here"></a></p>
<p>The "response function" drifts with temperature. I am unsure now if I have a misconception in the defintion of the response function, or if I picked up an artefact. I hope someone with more experience can help me out?</p>
| |measurements|optics|radiation|thermal-radiation|spectrometry| | <p>For anyone coming later:</p>
<p>I ended up using a procedure described here:</p>
<p>Kevin C. Gross, Phenomenological model for IR emission from high explosive detonation fireballs</p>
<p>In one sentence: I ended up measuring the blackbody spectrum at increasing temperatures. Then fitting a linear relation between measured and theoretical value for emission intensity (per wavelength) to obtain system gain and offset. So far, this leads to consistent results when measuring validation samples. Thanks @ Jeffrey J Weimer for the tip.</p>
| 32031 | Calculate FTIR-system response functions from measurements of a calibrated black body |
2019-11-22T12:06:49.093 | <p>Exactly what are "ceramics"? I found a lot of similar "definitions" of that term online, e.g. on Wikipedia and <a href="https://www.sciencelearn.org.nz/resources/1769-what-are-ceramics" rel="nofollow noreferrer">here</a>. </p>
<p>These (and other) sources essentially state that ceramics are non-metallic, non-organic materials. And they state the most prominent examples.</p>
<p>But these only seem to be necessary properties. <strong>Are there sufficient criteria based on which one can definitely decide whether a certain material is ceramic? Is this list of criteria comprehensive?</strong></p>
<p>Is there even a widely accepted clear definition at all?</p>
| |materials|ceramics| | <p>The classic (materials science) definition of a ceramic is a compound that is held together by a network of <em>covalent bonds</em>. The fact that these bonds are directional means it is <em>non-metallic</em>- deformation by dislocation-mediated slip does not occur (so no ductility mechanisms are active at ambient temperature) and the substance is a poor conductor of both heat and electricity (since the electrons responsible for holding the material together are immobile). </p>
| 32047 | Exact definition of ceramics |
2019-11-24T15:25:13.577 | <p>Some laps have patterns of holes drilled into them - two examples are below. In both cases the user claims that the holes (among other techniques) help ensure that the laps stays close to round when they're adjusted. I've seen it in enough places that I'm willing to believe there's some truth to it. Beyond vaguely saying that it redistributed the stress, I just don't see how though. </p>
<p><em>How do these hole patterns help ensure roundness when the lap changes sizes?</em></p>
<hr>
<p><strong>Examples of the claims (time to start on the relevant sentence):</strong></p>
<p><a href="https://www.youtube.com/watch?v=H2SeLcbcYkE&feature=youtu.be&t=188" rel="nofollow noreferrer">https://www.youtube.com/watch?v=H2SeLcbcYkE&feature=youtu.be&t=188</a>
<a href="https://www.youtube.com/watch?v=AEsaSN4LLU4&feature=youtu.be&t=178" rel="nofollow noreferrer">https://www.youtube.com/watch?v=AEsaSN4LLU4&feature=youtu.be&t=178</a></p>
<p><strong>Examples of the laps:</strong></p>
<p><a href="https://i.stack.imgur.com/pKzI7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pKzI7.png" alt="OD Lap"></a>
<a href="https://i.stack.imgur.com/gjLiV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gjLiV.png" alt="ID Lap"></a></p>
| |machining| | <p>You can think of the top picture as two semicircular curved beams joined at one end and with equal and opposite forces applied by the clamp at the other end.</p>
<p>If you apply a force at the end of a uniform cantilever beam, it doesn't bent into a circular arc but part of a cubic curve with more curvature at the root than at the tip. This will give an more constant curvature along the length of the "beam".</p>
<p>To counteract that, the holes are smaller at the "root" of the beam than at the tip, which is equivalent to a change in the moment of inertia along the length of the beam - greater at the root (small holes) and smaller at the tip (large holes).</p>
<p>For the second picture, a solid ring with no slots would simply be far too stiff to deform it by expanding the center. The slots and holes mean that the only continuous material which can support hoop stress is a thin ring at the outside of the disk.</p>
| 32075 | Do the hole patterns drilled in laps really help ensure uniform expansion and contraction? |
2019-11-24T18:01:41.163 | <p>I am trying to figure out what the exact term is for the clutch system used in a roll-up window shade. </p>
<p>I thought they were called centrifical clutches, but that typically gives me the opposite results like what is used in an automatic vehicle. Where speed causes it to engage.</p>
<p>Basically I want to look up some existing designs because I'm trying to create a product that uses one.</p>
| |mechanical-engineering|power-transmission| | <p>It is a spring-loaded ratchet and pawl system. the pawl mechanism is unlocked by the centripetal force of the initial rolling of the shade by a user.</p>
<p>Here is a diagram. <a href="https://i.stack.imgur.com/qwcNd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qwcNd.jpg" alt="diagram."></a></p>
| 32076 | What is name of clutch system in a roller shade? |
2019-11-27T13:12:09.577 | <p>I was reading some documents that said
25 wt. %, 500 ml<em>L<sup>-1</sup>
as well as
20 g</em>L<sup>-1</sup></p>
<p>I think 25 wt. %, 500 mlL<sup>-1</sup> is referencing 25% weight of 500 ml of water so that would be 125 ml of water</p>
<p>20 g*L<sup>-1</sup> would be 20 g per liter but since the reference is 500 ml then I should reduce this formula to 10 g to fit the 500 ml.</p>
<p>Is that correct?</p>
<p><span class="math-container">$$\text{NiSO}_4\cdot6\text{H}_2\text{O}\quad\left(20\ \text{g}\cdot\text{L}^{-1}\right)$$</span>
<span class="math-container">$$\text{NH}_3\quad\left(25\ wt.\%,\ 500\ \text{ml}\cdot\text{L}^{-1}\right)$$</span></p>
| |measurements| | <p>No. The nickel sulfate hydrate needs to be dissolved in water so that you have 20g per litre.
The ammonia-water solution needs to be present as 500ml per litre. The 25% is just relating to the ammonia concentration to use (33% being the maximum at RTP).<br>
The ratio to use should be in your article as it looks like you're making an ammoniacal nickel precipitate.
However your question was using this example to obtain a generic understanding. If you see a % sign, if it's telling you how to make a concentration, it will usually have w/w, v/v or w/v after it(weight-weight, volume-volume and weight-volume respectively).
For this sort of question, I would post in Chemistry. </p>
| 32097 | Weight percentage and grams per liter |
2019-11-27T13:56:53.970 | <p>I have a discrete control system with known dead time and time constant, where a process variable is sampled at regular discrete time intervals and fed back to the controller. I want to find out an optimal, or at least good, sampling interval.</p>
<p>I came accross a few websites saying that "Sample time should be 10 times per process time constant or faster", e.g. <a href="https://controlguru.com/sample-time-is-a-fundamental-design-and-tuning-specification/" rel="nofollow noreferrer">sample-time-is-a-fundamental-design-and-tuning-specification/</a>
But I cannot find any theory supporting this statement, so I guess it may be a rule of thumb. </p>
<p>Is this statement correct? and can someone explain it?</p>
<p>====Edit====</p>
<p>This Edit is after Jeffrey's answer.</p>
<p>I want to provide more details about my control system. It is a discrete system with a reference input. The goal is to keep the output close to the reference input (like a room temperature control system). The difference between the output and the reference input is the error fed to the controller, so the controller has control output accordingly. The system has a dead time, as well as a time constant, both at the scale of <span class="math-container">$1$</span> to <span class="math-container">$10\mu s$</span>. The fastest sampling interval of the output is at the sacle of <span class="math-container">$0.1\mu s$</span>.</p>
<p>My question is: is the fastest sampling rate good enough to achieve a stable control? </p>
| |control-engineering| | <p>After I spend a very long time investigate on this, I found that everything traced back to the work flow of designing control system.</p>
<p>As we design the control system, there are two approach,</p>
<ol>
<li>Design the controller in continuous domain and discretize</li>
<li>Discretize the system and design the controller in discrete domain.</li>
</ol>
<p>As we have plenty of tools to design, analyse and optimize the control system in contunuous domain, then we prefered to first approach. However, we need the discretize step to be good enought to approximate or represent the dynamics of continuouse system. That where the rule of thumb of 10x sampling frequency came from.</p>
<p>In other hand, if you design or tuning the controller in discrete domain, even if the sampling period very slow down to 0.1 of deadtime and the system still stable and controllable with proper gain.</p>
<p>For optimallity, there are trade of between performance and computational power. The sampling frequency also limited by hardware like sampling time of ADC. If you sampling very fast compare to the deadtime, then the system will behave close to continuous system which allow you to use continuous design. At some point, increasing sampling frequency doesn't has a significant impact anymore. some paper use this as sweet spot, also 10X of ultimate frequency. But if you have a constraint by any other limitation you can just use highest sampling frequency as possible.</p>
<p>Note: if you have too slow sampling time, it might change your plant models
For IPDT it not changed much but for FOPDT the plant model might become static.</p>
| 32099 | Rule of thumb for choosing a best sampling rate |
2019-11-27T17:35:14.763 | <p>We're planning on shipping some items from the US to Australia as you cannot get them here, that being:</p>
<p><a href="https://www.chewy.com/sentry-stop-that-noise-pheromone-dog/dp/56486" rel="nofollow noreferrer">https://www.chewy.com/sentry-stop-that-noise-pheromone-dog/dp/56486</a></p>
<p>I have been in contact with a freight forwarder who can ship them to Australia, but obviously they will need to be shipped as a <strong>hazmat</strong> item and go inside a particular sized box and be shipped by FedEx or DHL.</p>
<p>Now, as I understand it, the reason these are classified as hazmat is because of the chance of bursting as well as being flammable.</p>
<p>As per the link above:</p>
<blockquote>
<p>Contents under pressure. Do not puncture or incinerate container. Do not set on stove or radiator, expose to heat or store at temperatures above 120°F (49°C), as container may burst.</p>
</blockquote>
<p>I have no idea, but I assume they wouldn't be exposed to high temperatures such as that during shipping, but my question is...</p>
<p>Assuming we got them ok and they didn't burst, do pressurised cans such as these still pose a risk to bursting in "normal" conditions AFTER they have been exposed to high heat? Such as if they <em>had</em> been exposed to high heat during shipping, but we got them ok, would they still pose a risk to us with the possibility of bursting?</p>
<p>The manufacturer provides a Material Safety Data Sheet on their website, which you have to email them or register to get it sent to you, but if it's of any interest you can download it <a href="https://www.mediafire.com/file/y76wedx4pi2diqt/Sentry_Stop_That%21_Behavior_Spray_for_Dogs_and_Cats.pdf/file" rel="nofollow noreferrer">here</a> from MediaFire.</p>
<p>Alternatively you can find instructions to download it via <a href="https://www.sentrypetcare.com/products/sentry-stop-that-noise-and-pheromone-spray/" rel="nofollow noreferrer">their site</a> - see under "PRODUCT SAFETY INFORMATION".</p>
<p>If you don't want to download anything then you can see screenshots of the document I uploaded to an Imgur gallery here: <a href="https://i.stack.imgur.com/44umr.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/44umr.jpg</a></p>
<p>Note though, the above data sheet was provided via <a href="https://www.sentrypetcare.com/products/sentry-stop-that-noise-and-pheromone-spray/" rel="nofollow noreferrer">their site</a> that shows their old packaging, the link from Chewy above shows their new packaging, as far as I know the packaging is the only thing that has changed.</p>
| |pressure|chemical-engineering|gas|safety|compressed-gases| | <p>Assume that high temperature causes no chemical reactions in the contents or between the contents and the can.</p>
<p>High temperature will increase the pressure in the can. Assume the can is fully unaffected in its mechanical integrity up to the point that it fails. </p>
<p>With all of the above, when you get the can back at room temperature, it will have no change in its potential to fail or not.</p>
<p>One weak link will be the potential for increased fatigue at the seams on the can as the contents go to high pressure and back to lower pressure. Another weak link will be in the potential for over-stress on the plastic cap seal on the top of the container.</p>
<p>When the contents react at high temperature and that temperature is exceeded, all bets are off.</p>
<p>In summary, this system affords no absolute guarantee that it will have no increased probability to fail when temperature cycles to a high value and returns to room temperature. The biggest uncertainty is what exactly is meant by high temperature (i.e. 100 °C or 500 °C) and for how long (i.e. immediately or for a few minutes or for a few days). The practical issue is whether and to what extent your shipment will reach the bounds.</p>
<p>Based on what is stated on the Website, the failure seems to be immediate at a point at or just above 50 °C. I would have to say that whatever company is ready to accept these cans knows that they have to keep them below that temperature during shipping. But, what you might want to know is whether they were at a high temperature just below that for an extended time,</p>
<p>I might suggest to mark the cans before shipment with a paint or dye or wax that changes color irreversibly when a certain temperature is exceeded. Or include a temperature probe with the shipment that tracks temperature with time over the entire shipment. Then, you can decide at the receiving end of the shipment whether the cans where shipped to your satisfaction.</p>
| 32102 | Do pressurised cans still pose a risk to bursting in normal conditions, but after being exposed to heat? |
2019-11-28T19:33:24.537 | <p>I am looking for borosilicate glass tubes which can be used to contain a vacuum, so the top is sealed and then there is a vacuum pump at the bottom where the edges are also sealed, effectively making a vacuum in a glass tube.</p>
<p>The only thing to mention is that the walls need to be somewhat thicker than normal to protect against implosion due to the fact that there will be a vacuum inside, so I need an idea of how thick the glass needs to be bearing in mind that the inner diameter is 20 cm.</p>
<p>The pressure outside the cylinder is atmospheric and the pressure inside is very low and can be taken to be 0. Is there a formula or computation which would give me a good approximation for how thick the tube needs to be? I also appreciate that there could be some shear stress due to the tube needing to be so thick.</p>
| |pressure|stresses|vacuum| | <p>Let's call the diameter of your glass pipe, D, and assume the top and bottom of the tube are domes in a gentle transition from cylinder so we can ignor the local stress concentration.</p>
<p>The pressure, P, being radial around the pipe causes only compression and no moment or shear in the glass.</p>
<p>If we cut the glass longitudinally we have two half cylinders such that the thickness, t, of the glass has to counter total pressure per unit length. Let's call the required stress of the glass, S.</p>
<p>t× S= P×D</p>
<p>t= P×D/S= 1atm× D/S</p>
<p>The compression strength of normal glass is in the range of 1000psi, but it can vary depending on many factors so you check it, then you plug it in the above formula and use the required safety factor in your industry to get the thickness, t.</p>
| 32114 | Calculating the Thickness of a Glass Tube which contains a Vacuum |
2019-11-29T09:52:00.197 | <p>In my thesis, I deal with heating circuits, where several heat generators are present, e.g. a combined heat and power unit and a supporting boiler.
Usually, the boiler is placed in front of the heat storage tank, but I have also repeatedly seen heating circuits where the boiler is placed behind the heat storage tank. </p>
<p>For example, the structure of the heating circuit of an <a href="https://ieeexplore.ieee.org/document/6872090" rel="nofollow noreferrer">IEEE paper</a>:</p>
<p><a href="https://i.stack.imgur.com/kviXO.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kviXO.gif" alt="enter image description here"></a></p>
<p>What are the advantages and disadvantages?</p>
<p>Do I understand correctly that if the boiler is in front of the heat storage tank, the water of the heat storage tank is heated and if it is behind it, it has its own reservoir of water which it heats?</p>
| |heat-transfer|heating-systems|energy-storage|boilers| | <p>In the diagram shown, then if the water temperature of the storage tank falls below the set value, the aux boiler comes in to augment the water temperature.</p>
<p>Other possibilities include the use of a diverter valve so when the tank water temperature is too low, the valve changes to demand water heated by the boiler, if that boiler has a rapid response then direct, if not then it may well have a storage tank of its own.</p>
| 32119 | Placement of supporting boiler in heating circuit |
2019-11-30T03:46:09.470 | <p>Looking for a material that:</p>
<ol>
<li>Is transparent/clear - the clearer the better</li>
<li>Can be made to about 80cm long x 8cm wide by somewhere between 2-5cm thick (likely 2-3cm, exact details will depend on the material's properties)</li>
<li>Can be shaped on the cross section, either during manufacturing, or after</li>
<li>Is at least a bit flexible - I'd estimate no more than 20 degree bend along the length, without too much force required - i.e. something that could be bent by hand, like a plastic ruler.</li>
<li>Has a refractive index for visible light of 1.5 or above (higher is better)</li>
<li>Is cheap, or at least not too expensive</li>
<li>Is resistant to chipping or snapping</li>
</ol>
<p>Can anyone help?</p>
<p>Partial matches to the above might be welcomed, but it would be useful to know why they are only partial matches. The key points are (1), (2), (3), (4) - other than those, the higher refractive index the better, the cheaper the better, the more resistant to chipping and snapping the better, but those are continuums rather than fixed points! The method of shaping also matters, ie something that can theoretically be shaped but in practice requires a really expensive or rare process, or hand shaping, is not practical.</p>
<p>For example, I have access to injection molded Polyester resin, but I understand that might not be flexible.</p>
| |materials| | <p>There are optically clear silicone products often used for encapsulation. These are usually two part mixtures that are then cast. They are quite flexible. I doubt they are machinable so you would have to cast to shape using a form or mold of some kind. <a href="https://www.quantumsilicones.com/2012/08/qsil-229lv/" rel="nofollow noreferrer">Here</a> is an example, but there are many others. Just do a <a href="https://www.google.com/search?q=optically%20clear%20silicone&oq=optically%20clear%20sili&aqs=chrome.0.0j69i57j0l4.5251j1j4&sourceid=chrome&ie=UTF-8" rel="nofollow noreferrer">Google search</a>.</p>
<p>I've also found optically clear urethanes. <a href="https://www.smooth-on.com/product-line/clear-flex/" rel="nofollow noreferrer">Here is an example</a>. Urethanes can be found with a range of stiffnesses.</p>
| 32125 | Seeking transparent, somewhat flexible material |
2019-12-01T16:13:46.457 | <p>Please explain the meaning of this statement with close reference to Precipitation hardening,"The relative rates of nucleation and growth are controlled by temperature. The size and dispersion of the secondary phase is controlled by time, with coarsening occurring as time proceeds."What I do not understand is the difference between the usage of "growth" and "size" as different parameters. They say they are dependent on temperature and time respectively.Arent they basically the two sides of the same coin.Please explain me with close reference the the precipitation hardening in Al-4.5wt%Cu alloy.</p>
| |mechanical-engineering|materials|metallurgy|material-science| | <p>More precipitation sites initiate the higher the temperature. So you want a higher temp to have many sites. The precipitates grow faster ( become coarse) at higher temperature. But the objective is many precipitates ( high temp.) with small size ( low temp). So the temperature selection is a compromise. I think some newer precipitation stainless steels have a two temperature age : start high , then a second lower temperature. Digressing ; one of my favorite metallurgical terms "anti-phase domain" is the undefined zone ( several atoms wide) between the matrix and the precipitate.</p>
| 32150 | Precipitation Hardening |
2019-12-02T21:59:52.177 | <p>I am studying the mixed sensitivity design for control systems, and I have seen that the control effort has limitations given by the bandwidth of the process and by the bandwidth of the sensitivity function. In particular I have seen that if we plot the control sensitivity function, we have that if I am above the bandwith of the process I have an increase of the control effort, while if I am below it decreases. And also I have a lower limitation given by the bandwidth of the sensitivity.</p>
<p>But I don't understand why. Also, does the controller bandwidth impose some limitations? Can please someone explain this to me? Thanks in advance.</p>
<p>[EDIT] I recognize that my question was to shallow, so I will try to do more reasonings and add more argumentations as I search for answers.</p>
<p>I will use an approch based on the mixed sensitivity for my discussion.</p>
<p>Suppose I have the following plant and controller:</p>
<pre><code>s = tf('s');
G = 10/((s+10)*(s+1));
K1 = 0.5/s;
</code></pre>
<p>Now I define the following mixed sensitivity controller, where I used a weight for the sensitivity in order to achieve my desired specifications, which are a 60 dB attenuation at high frequency, a cross over frequency of 0.66353 and a maximum resonance peak of 2 dB. I also inroduce a weight for the control effort, which I use to limit the amount of effort the control has to do. In particular, I use a constant weight, so by doing so I am hopefully trying to limit the control effort to increase too much:</p>
<pre><code>W_bs2 = 0.66353; %same cross over frequency as the previuos point
M = 2; %peak of the sensitivity
A = 0.001; %attenuation
Ws2 = (s/M + W_bs2)/(s+W_bs2*A);%sensitivity weight
%bodemag(1/Ws),grid;
Wu = tf(1);
[K2,CL,GAM2] = mixsyn(G,Ws2,Wu,[]); %define the controller with the mixed
sensitivity
display(GAM2);
K2 = minreal(K2); %define a minimal order controller
</code></pre>
<p>some of the comments may not make sense because they refer to some previous code.</p>
<p>Now, I define the control effort:</p>
<pre><code>Q1 = K2/(1+G*K2); %control effort
</code></pre>
<p>and I plot the control effort and the weight of the control effort, which should impose a limitation:</p>
<p><a href="https://i.stack.imgur.com/Jgu7m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jgu7m.png" alt="enter image description here"></a></p>
<p>in this case I used a control effort weight equal to 1, but probably an higher value would have been better.</p>
<p>Now, the bandwith of the controller is 0.9879, so if I go above it the control effor shoulb be increasing instead of decreasing. Let's try:</p>
<p>I define a second sensitivity weight which gives an higher cross over frequency to the control effort when defining the controller:</p>
<pre><code>W_bs3 = 2; %same cross over frequency as the previuos point
M = 2; %peak of the sensitivity
A = 0.001; %attenuation
Ws3 = (s/M + W_bs3)/(s+W_bs3*A);%sensitivity weight
%bodemag(1/Ws),grid;
Wu = tf(1);
[K3,CL,GAM3] = mixsyn(G,Ws3,Wu,[]); %define the controller with the mixed
sensitivity
display(GAM2);
K3 = minreal(K3); %define a minimal order controller
</code></pre>
<p>and if I plot the control effort of it and confron it with the previous result:</p>
<p><a href="https://i.stack.imgur.com/hT36v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hT36v.png" alt="enter image description here"></a></p>
<p>the second mixed sensitivity controller gives me the yellow line, while the first gives me the red line.</p>
<p>Now, I am not sure I am going to the right direction, but hopefully I am getting closer.</p>
<p>Can someone help me? Thanks again.</p>
<p>[EDIT 2] So,after a lot of searching and thinking I think the point is that I haven't really understood the concept of bandwisth and how it influences the system. I am in particular looking at sensitivity bandwidth and complementary sensitivity bandwidth.</p>
<p>At this point, my idea is that the sensitivity bandwidth an the complementary sensitivity bandwidth should be limited by the fact that <span class="math-container">$S+T=1$</span>. But still not really clear what happens. So it should be that if S is large, T cannot be too large. But is the bandwidth related to this? </p>
| |electrical-engineering|control-engineering|control-theory| | <p>The reason the closed-loop bandwidth is related to control effort is pretty clear: once you're outside the bandwidth of the plant, the higher the frequency of the command, the higher the controller command amplitude (i.e., controller effort) needs to be to get the same result. So it should just flow from intuition that if you're pushing the system loop closure above the plant's bandwidth that the controller effort will go up.</p>
<p>I'm not getting your comment about the sensitivity bandwidth being linked to control effort -- there's more to it than that, because sensitivity is at least in part a function of the plant itself.</p>
<blockquote>
<p>Also, does the controller bandwidth impose some limitations?</p>
</blockquote>
<p>Not in theory-land. There's an implicit assumption that you're going to define a controller of arbitrary bandwidth, then give it to some poor slob (possibly me) to implement; and that your job is then done. While real-world designs sometimes <strong>do</strong> feature drivers or sensors of limited bandwidth, it's more common that there are real-world limitations on how hard you can push the physical devices (or how much you can ask of power supplies); that limits the available control effort.</p>
| 32172 | Limitations on the control effort |
2019-12-03T12:35:53.290 | <p>I'm doing my first machine design for a project and I have a Honda GX160 combustion motor which has the following specs for power and speed:
<a href="https://i.stack.imgur.com/WUNKx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WUNKx.jpg" alt="enter image description here"></a></p>
<p>I'd like to use 0.24 HP at 2000 rpm. Is this possible? Or I can just get 2000 rpm at 2.1 HP as the NET POWER curve shows?</p>
<p>I'm using this motor because it's the only one I have available for this project.</p>
<p>(It would also work for me at 1300 rpm if it has at least 0.24 HP, but I don't think its recommended)</p>
<p>Thank you!</p>
| |motors|machine-design| | <p>The chart shows the maximum power with the throttle wide open.</p>
<p>You can run at lower powers simply by having the throttle partly closed.</p>
<p>If you want to use a low power output like 0.24HP, you will probably need some sort of automatic control system to keep the RPM down to 2000. A simple throttle lever with manual adjustment might be too sensitive to adjust successfully.</p>
<p>A small engine like this may not run smoothly below 2000RPM. Any sudden change in the load might cause it to stall, unless you add a bigger flywheel to your design.</p>
| 32178 | Reading Motor Datasheet |
2019-12-03T16:39:24.117 | <p>Consider a four-wheel steering vehicle in which:</p>
<ul>
<li>The front wheels follow Ackerman condition</li>
<li>The rear wheel are rotated with a fixed angle (<em>dir</em> == <em>dor</em> in the picture below, and such angles do not change if the driver moves the steering wheel)</li>
</ul>
<p>Actually, this is similar to a classic front wheel steering vehicle, but the rear wheels are not "straight", they are angled.</p>
<p>What kind of trajectory would this vehicle follow? Is there any closed-form equation to describe it?</p>
<p><a href="https://i.stack.imgur.com/VA8cv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VA8cv.png" alt="enter image description here"></a></p>
| |automotive-engineering|wheels| | <p>It means there is still some side slipping and dragging of the wheels if the rear tires and front tires are not all aligned to track circles around one common center.</p>
<p>In your case, the rear wheels axis are parallel and never meet.</p>
<p>If a vehicle needs for steering at both the rear axle as well as the front, then the geometry of the steering linkage both at the front and the rear axle should direct inner wheels to turn more so that all the wheels axis meet at the center of rotation of the vehicle. </p>
| 32183 | Car trajectory with angled rear wheels (non-Ackerman) |
2019-12-04T14:45:27.317 | <p><a href="https://i.stack.imgur.com/mEugv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mEugv.png" alt="enter image description here"></a></p>
<p>If you consider a motor attached to a spur gearbox with a reduction ratio of say <span class="math-container">$r=50:1$</span> which is then back-driven so that the motor acts as a generator, what would be the effective moment of inertia of the system? If the motor has a rotor inertia of say <span class="math-container">$1*10^{-5} kgm^2$</span> and the gearbox has a mass inertia of <span class="math-container">$1*10^{-6} kgm^2$</span> would I be correct in calculating the overall inertia through </p>
<p><span class="math-container">$I_{total}=r^2I_{rotor}+I_{gearbox}=0.025 kgm^2$</span></p>
| |mechanical-engineering|electrical-engineering|motors|applied-mechanics|mathematics| | <p>That should be correct. The rotational velocities and accelerations in the motor, caused by turning the output shaft, increase by r. Along with the acceleration, the reaction torque in the motor increases by the same amount.
The reduction ratio will also apply on this torque and and hence increase it again by r at the output shaft. </p>
| 32192 | Impact of Gearbox on Motor Inertia |
2019-12-04T15:17:37.480 | <p>I have been asked to provide some of our technical docs in both metric and US units. For the most part, that is easy enough:</p>
<ul>
<li><p>approx 1cm → approx 1/2"</p></li>
<li><p>500μm → 0.02" </p></li>
</ul>
<p>But then I start getting stuck:</p>
<ul>
<li>50μm → 0.002" or 2 thou or 2 mils? Is there a common symbol for thou/mills?</li>
<li>20nm → ???</li>
</ul>
<p>I imagine that people who care about nanometers have mostly gone metric now anyway, but it would still be good to provide US measurements for everything. What units are commonly used for lengths smaller than 0.001" in the US, and what symbols are used for them?</p>
<p>If it varies across different fields, the relevant ones would be optics, precision machining and mechanical engineering.</p>
| |distance-measurement|unit|international| | <blockquote>
<p>50μm → 0.002" or 2 thou or 2 mils?</p>
</blockquote>
<p>I've seen both 0.002" and 2 mils. On a drawing it would always be 0.002". In a specification document it could be either. I've never seen 2 thou written in a formal specification (but I have heard people say it in the shop). But that might just be my experience. There could be variation from industry to industry.</p>
<blockquote>
<p>Is there a common symbol for thou/mills?</p>
</blockquote>
<p>I'm not aware of any. We just always wrote "mils" (not "mills")</p>
<blockquote>
<p>What units are commonly used for lengths smaller than 0.001" in the US, and what symbols are used for them?</p>
</blockquote>
<p>Colloquially, I know that people in precision machine shops refer to 0.1 mils (0.0001") as "tenths" (as in one ten thousandth of an inch). I've never seen that written in a formal specification document though, just as talking between machinists.</p>
<p>For US customary units, there's nothing smaller than an inch, you just start adding more zeros, applying prefixes, or using scientific notation. For example, it would be common to specify surface roughness in microinches.</p>
<blockquote>
<p>20nm → ???</p>
</blockquote>
<p>I would write it as "7.87e-7 inches"</p>
| 32193 | US customary units below 1 mil/thou? |
2019-12-04T15:37:39.203 | <p>Belleville disc springs are usually spec'd with a free height, loaded height, and loaded force. From this information you can calculate the average spring rate of a belleville spring. However, what I would like to know if if the force is actually linear (or nearly linear).</p>
<p>I assume it is not, but don't know what curve the force might follow. Is there an equation that can predict the force-curve (plot of force versus height) of a belleville spring given ID, OD, free height, and thickness?</p>
| |mechanical-engineering|springs| | <p>The curve is not linear.</p>
<p><span class="math-container">$s = $</span> actual deflection (mm)<br>
<span class="math-container">$h_o = $</span> total possible deflection (mm)<br>
<span class="math-container">$F = $</span> actual load (N)<br>
<span class="math-container">$F_c = $</span> designed limit load (N) </p>
<p><a href="https://i.stack.imgur.com/YI9yT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YI9yT.jpg" alt="enter image description here"></a><br>
<a href="https://www.spirol.com/library/sub_catalogs/dsc-Deflection_and_Load_Characteristics_us.pdf" rel="nofollow noreferrer">Here</a> you can see the original archive with the ploted graph from one industry (I usualy use this material for reference about these type of spring).</p>
<p>I don't know the exact formula. But the reasoning should be (this is my guess): </p>
<ul>
<li>From 0% Force to ~70% force should be affected mostly by the
compression between the inner diameter and outer diameter.</li>
<li>From ~70% Force to 100% should be affected mostly by bending.</li>
</ul>
| 32195 | Is the Spring Rate for a Belleville Spring Linear? |
2019-12-05T04:56:01.123 | <p>I'm looking for a data that what is the typical application range for a common uni directional strain gauge like used in weight scales?
I've read about around 200 ppm in a research gate article, can I measure 40 ppm with a cheap setup (arduino nano plc, industrial grade op amp, soldered connections)?</p>
| |measurements|deformation| | <p>I think it will come down to the quality of your amplifier. Let's put some numbers together. </p>
<p>With 10 volt excitation (typical) and a gage factor of 2.2 (typical), and full bridge wiring as per line 9 in this reference (<a href="https://www.hbm.com/en/7163/wheatstone-bridge-circuit/" rel="nofollow noreferrer">https://www.hbm.com/en/7163/wheatstone-bridge-circuit/</a>, there are other ways to do a full bridge but this is common), 40e-6 in/in would give an output of 0.5 mV. </p>
<p>In other to make a good measurement, you'll want a minimum resolution on your analog to digital converter to at least 10 times less than this (i.e. if your minimum resolution is 4e-6, then you can distinguish 40e-6 from 36e-6 or 44e-6, but no finer. That's probably the minimum acceptable and might want something a lot better depending on your requirements). So let's stay you need to be able to read 0.05 mV (i.e. 50 microvolts) with your ADC. Averaging multiple values can help increase resolution, but only to a point. you need a decent enough resolution to start with. </p>
<p>Arduino nano has a 10 bit ADC with a 5V range (<a href="https://store.arduino.cc/usa/arduino-nano" rel="nofollow noreferrer">https://store.arduino.cc/usa/arduino-nano</a>). So 1 least significant bit is 5 / 1024 ~= 5 mV.</p>
<p>Thus you can read 5 mV and you want to be able to read 0.05 mV. So your amplifier needs to be able to provide an amplification of 100:1 (without introducing any non-linearity, distortion, noise, etc). You'll also need to keep the noise on the input side of the op amp to on the order of 0.05 mV (or else the op-amp will just amplify the noise along with the signal). </p>
<p>A professional quality strain gage amplifier which would meet those specs (e.g. Omega DMD-466) is going to run \$400-\$500 or more. I doubt you'll be able to meet the same performance with a "cheap" setup, although I won't say for sure that you can't do it. </p>
| 32210 | what is the range of the measurable relative elasticity with a commonly used strain gauge |
2019-12-06T08:57:49.137 | <p>I need to work on a project where multiple objects need to exchange trough wireless protocol and they are near to High frequency TIG welding machine.
All objects works by pair (one emitter and one receiver) and they are up to 20 pairs in the same area (something like 30 or 50 square meter)
Each pair do not have to interfer with other pair of objects and they have to not be interfered by the TIG welding machine.</p>
<p>Should I better use WIFI or Bluetooth protocol?
Range is not an issue as each object of a pair is not far from the other object of the same pair.</p>
<p>For what I found, High frequency TIG welding is given to 1Mghz </p>
| |welding|wifi| | <p>Don't worry about the frequency of the applied power. Worry about the fact that a spark gap is a fairly efficient way (and can be a <em>really</em> efficient way) of generating radio frequency radiation. It's the way Heinrich Hertz made <a href="https://en.wikipedia.org/wiki/Heinrich_Hertz#Electromagnetic_waves" rel="nofollow noreferrer">the first man-made radio waves</a>.</p>
<p>I would test. I would start by getting a spectrum analyzer in the same room as your arc welder and making estimates of the signal degradation from that, then I would actually do some end-to-end tests.</p>
<p>There are some areas of study -- and RF propagation is one of them -- where <strong>all</strong> computational answers end up being inexact because the interaction with the environment is too convoluted. Unless you're an absolute RF genius, you just want to take measurements and experiment.</p>
| 32235 | High frequency TIG welding and Bluetooth or WIFI frequency |
2019-12-07T02:23:22.123 | <p>no code involved and no technical equipment is going to involved in experiment and implementation for business/tech startup that is, in brief, a way for me to exercise during my business studies! Building branded CROSSFIT exercise equipment with readily available materials. With reasons, I chose each material following, bicycle tires (cheap rubber known for durability), milk cartons (hoping to combine with chemicals to create ultimate durable and form factor retention), filler</p>
<p>Creating a recipe will take some effort I realize, also i did previous research and there are two routes to go, 3dimensional structures make plastic being recycled more durable mechanically, but the better opportunity is to melt rubber into ingots, treat the milk cartons with low-budget chemicals, and then combine the three aforementioned parts into weight lifting plates.</p>
<p>Help if you can, Im looking for the low budget route that will maximize output with low effort. </p>
<p>What chemicals make plastic more durable with form factor rentention the most possible, and what lubricants and filler will make the combination as durable as possible. </p>
| |recycling| | <p>So if I understand what you want to do correctly, your plan is to melt down bicycle tires, melt down milk cartons, and then combine those two with some third thing, and create a new product. Your question is what should that third thing be. If that is incorrect please edit the question.</p>
<p>Your biggest problem here is that you can't melt down bicycle tires. The process of creating them is irreversible. Just google "can I melt down bicycle tires" and you'll get a bunch of pages explaining why you can't do it. To quote one of the top hits: "it's like un-baking bread".</p>
<p>now milk cartons, you could melt down. Those are usually just HDPE. You can modify HDPE by adding other components to tweak certain properties. However, I doubt you'll be able to do it cheaply, and you certainly won't be able to do it without a lot of "technical equipment" and specialized knowledge.</p>
<p>Sorry to rain on your parade but I don't think what you have in mind is feasible. </p>
| 32256 | Will specific fillers or chemicals make simple plastic milk cartons more durable, enough to be used as weightlifting plates |
2019-12-07T20:31:29.437 | <p>In Rocket Lab's presentation video, <a href="https://www.youtube.com/watch?v=eZE043cKcD4&t=21s" rel="nofollow noreferrer">at around 00:21</a>, you can see some engineer work with CAD software. Here's a screenshot:</p>
<p><a href="https://i.stack.imgur.com/YQoj9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YQoj9.png" alt="CAD software used by Rocket Lab"></a></p>
<p>Can anyone recognize which software they're using? Asking for a friend.</p>
| |cad|software| | <p>I think the hint comes from the tab that says SOLIDWORKS in the left column, near the top of the block of actions. It's not visible in this screen shot, but is clearly visible in the video linked in the post.</p>
| 32264 | What CAD software is this? |
2019-12-08T11:13:41.343 | <p>Original Question was <a href="https://earthscience.stackexchange.com/questions/18625/how-much-does-the-weight-of-urban-structures-buildings-affect-the-compaction">here</a>.</p>
<p>I live in Van, Turkey. Van City is situated on an alluvial plain beside Lake Van. General geological structure of the area can be seen on pages 43-44 in <a href="http://www.dsi.gov.tr/docs/sempozyumlar/van-g%C3%B6l%C3%BC-hidrolojisi-ve-kirlili%C4%9Fi-konferans%C4%B1-van.pdf?sfvrsn=2" rel="nofollow noreferrer">this conference book</a>. One of the images is below.</p>
<p><a href="https://i.stack.imgur.com/0gFEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0gFEm.png" alt="enter image description here"></a></p>
<p>According to the article in the link, sediment layers has a slope of 15-30 degrees toward the lake.</p>
<p>With the construction of every building, the pressure over the sediments increases. Obviously, this increase is greater in downtown area which has a lot of apartment blocks (mainly 5 to 7 story high reinforced concrete buildings). Downtown area is roughly in the middle of the plain.</p>
<p>I want to know that how much does the weight of urban structures (buildings) affect the compaction (permeability/porosity/density?) of alluvial sediments below a city and how does this affect the flow of underground water?</p>
<p>How much the stress caused by the weight of the building decreases with depth? (of course if at least a broad value can be given to help to explain the size of the change). I wonder to understand that either the effect of the weight of urban structures <strong>is generally negligible, or not.</strong></p>
| |geotechnical-engineering|soil|soil-mechanics| | <p>In many cases, the weight of the building on the soil is roughly equal to the weight of the layer of the soil excavated for the foundation, subterranean parking, utility rooms, storage, etc.</p>
<p>Let's assume the buildings are made of 20cm thick concrete slabs, </p>
<p><span class="math-container">$0.2*1m/2400kg.m^3= 480kg/m2$</span></p>
<p>and a live load of <span class="math-container">$100kg/m^2 \quad total=580kg/m^2 \quad 580*6floors= 3480kg/m2$</span></p>
<p>We estimate the soil weight is around <span class="math-container">$1600kg/m^2$</span> which is not
unreasonable.</p>
<p>Let's assume the depth of foundation and parking 3m, excavation. <span class="math-container">$ 3*1600= 4800kg/m^2$</span> which is even a bit larger than the weight of the building added to the soil.</p>
<p>And also many sites need grading to create a flat pad for building and its surrounding landscaping and access roads. And because usually grading is done favoring cut rather fill, that takes some more weight from the top of the soil.</p>
<p>Usually, these kinds of jobs require a soil and geology report and one of the reasons for the report is checking into your concern. </p>
| 32270 | How much does the weight of urban structures (buildings) affect the compaction (permeability, porosity, density) of alluvial sediments below a city? |
2019-12-09T04:19:36.737 | <p>I was thinking of building a free standing structure out of plumbing pipe that would go flush against the wall to hang jackets and other items.</p>
<p>Attached is a sketch of the general thought of what the structure would look like.</p>
<p><a href="https://i.stack.imgur.com/lr8KI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lr8KI.png" alt="Sketch"></a></p>
<p>Picturing it in my head, it seems like it should be all good. But then I just got curious about that actual math to get a rough idea of what force it would take to tip it over.</p>
<p>This is assuming that the pipe is strong enough to hold.</p>
<p>I started looking into the calculations for levers, but it seems to be like there are a couple levers chained together. At that point it became beyond me.</p>
<p><em>Note: I plan on taking other safety precautions like attaching to the wall. But I am more curious theoretically now.</em></p>
| |mechanical-engineering|structural-engineering|structures| | <p>This structure as it is is not going to tip over no matter how much force you apply on the top where you have the arrow at 12 inches.
Unless the force is so much that the pipe is going to break or crush under the compression stress.</p>
<p>Because the arrow falls within the footprint of your hanger. Let's do a quick calculation:</p>
<p>Let's say you hang a weight of P lbs from the arrow and let us call the point on the floor at the left corner A and the one on the right corner at the end of 18inch B.</p>
<p>Taking the moments about A and knowing that for equilibrium the sum of moments should be zero, we have:</p>
<p><span class="math-container">$$ \Sigma M_a=0 ,\quad P*12+ B_v*18=0 \\ B_v=- \frac{P*12}{18}=- \frac{2}{3}P$$</span></p>
<p>We note that a reaction vertically pointing up ( by the negative sign) by the amount of 2/3 of any weight we hang on top is sufficient to keep the hanger balanced. And <span class="math-container">$\ A_v = -\frac{1}{3}P $</span></p>
<p>In real life though if you hang a really heavy object from the arrow, your vertical pipe can bend forward under the bending moment, or the corner on the bottom may give a bit so the hanging point moves horizontally forward and then the hanger's balance can become shaky.</p>
<p>As a general rule whenever you have the weight's vertical projection fall inside the footprint of the support the structure is balanced. It could be a huge weight of a high-rise building, still, it is vertically balanced and will not tip over. </p>
<p>In the case of the buildings, there are some safety factors that have to be applied depending on the use and type of the building.</p>
| 32281 | What calculation to find how much force before tipping over? |
2019-12-09T11:55:25.197 | <p>I am trying to calculate the internal forces and their nature in this truss, however, I cannot seem to do so as both the method of joint and the method of section do not work. I think this is because there are not enough conditions of equilibrium that can be used. Is there any way to actually calculate the forces in the members even if this is the case?</p>
<p>For example, at joint A, there are 3 unknown members, and all the members acting through it are only concurrent at that specific point, making it impossible for me to use the equilibrium condition of the sum of moments being 0.
<a href="https://i.stack.imgur.com/Y5fcu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y5fcu.png" alt="The diagram of the truss"></a></p>
<p>It would be extremely appreciated if someone could inform me about how to calculate the forces in the members of this truss. Thank you.</p>
| |structural-engineering|structural-analysis|statics| | <p>The structure is not statically determinate, unless you remove bar v. The internal forces will depend on the relative stiffness of the members.</p>
<p>The practical way to do the analysis is make a finite element model.</p>
| 32284 | Truss analysis of internal forces in members |
2019-12-09T17:28:49.253 | <p>Does a house heating system affect indoor air humidity?
Or to be more precise, does it affect your perception of how dry the air in the house feels (for instance dry is your mouth waking up)?</p>
<p>(I searched for this but could only find explanations from furnace company websites)</p>
<p>Every (lay)person I ask this seems to think house furnaces make the air drier (at least the perceived dryness). One friend claims that the air gets even dryer than the outside air.</p>
<p>My intuition is that heating can only increase the evaporation inside the house (from the little surface liquid water sources (shower, taps) there may be), therefore making humidity go up. </p>
<p>I suspect people think this because furnaces are only on in the winter when the outside air is dry leading to dry air inside as air exchanges. Thus it could be due to this spurious correlation.</p>
<p>Another possible explanation is that the higher air temperature increases the air's water carrying capacity, which in the absence of an abundant water source, extracts water from your body faster. Therefore increasing perceived dryness. Even if these two countervailing effects are true, can we use first principles to bund which effect dominates?</p>
<p>EDIT (after seeing the answers and discussions below):
Perhaps perceived humidity is too loose to define. What if we define the water loss of a human over a period as our measure? But my ideal answer would have two numbers to compare: </p>
<ul>
<li>Baseline_loss = how much water does a person looses sleeping in a cold sealed room (say at 1C temperature) with 100% relative humidity and no air movement. </li>
<li>Total_loss = how much water does a person loses if the same sealed
room was heated up to 22C (a nice in-home temperature).</li>
</ul>
<p>Then Heating_induced_loss = Total_loss- Baseline_loss</p>
| |heat-transfer|heating-systems|dehumidification| | <p>To summarize the comments given here: During the winter, the cold air outside can hold only a small amount of water in solution, which means it takes very little moisture in cold air to get it to 100% relative humidity. </p>
<p>But then, when your furnace draws in some of that cold outside "makeup" air, it heats it up and its relative humidity then drops significantly- and you begin feeling "dry" because the warm air in your house begins drawing moisture out of your skin, clothing, furniture, plants, goldfish bowl, etc. </p>
<p>Because of this, it is common for houses to feel uncomfortably dry in the winter unless you put a kettle of water on the stove to simmer whenever the heat is running. </p>
| 32291 | Does a house heating system affect perceived and true indoor air humidity? |
2019-12-09T18:53:32.307 | <p>I am trying to mate a gear with a shaft via a key.key coincident with gear and cocentericity of gear with shaft cannot both be made. </p>
<p><a href="https://i.stack.imgur.com/bDn97.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bDn97.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|gears|solidworks| | <p>Your issue is that you're assuming the key and key way will have exactly the same dimensions. In reality the key will have to be slightly smaller than the keyway.</p>
<p>What yo want to do is start with the <strong>concentricity</strong>, then use the <strong>width</strong>, or <strong>parallel</strong> mate to line up the key and keyway. This way you're not over-constraining the assembly.</p>
<p>Solidworks sometimes lets you get away with that, but even when it does it's not a great idea and it can break unexpectedly later.</p>
| 32293 | solidworks mating gear with shaft problem |
2019-12-10T03:15:56.590 | <p>My task is to create a 3D Sketch of a trombone like the one below</p>
<p><a href="https://i.stack.imgur.com/1UcTV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/1UcTV.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/Pp4Je.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Pp4Je.png" alt="the end of the trombone"></a></p>
<p>My professor recommended defining the points 1-4 and putting a spline through them. I did that, but my resulting design looks weird. It curves inward between points 1 and 2 before curving out towards the rest of the points. I'm not sure what I'm doing wrong, and I don't know if I have enough given constraints to curve it outward manually myself by editing the spline</p>
<p><a href="https://i.stack.imgur.com/VhGYC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/VhGYC.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/jKVqL.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jKVqL.png" alt="enter image description here"></a></p>
| |mechanical-engineering|design|cad|solidworks| | <p>In addition to my original answer, I thought it worth adding an example of how you might model the full trombone. I noticed that in your screenshot you had "Plane 3" visible, indicating to me that you've been adding reference geometry outside of the three base planes?</p>
<p>This model can be made using just two sketches and four features, with all of the dimensions within those sketches being exactly copied from the reference drawing in their positioning. This removes a lot of potential for error.</p>
<p>Step 1) Sweep path: I have taken the flared end to be the datum, and built a sketch following the path of the tube. The drawing is dimensioned to the centreline so this is easy to match. </p>
<p><a href="https://i.stack.imgur.com/YsGGfm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YsGGfm.png" alt="sweep path"></a></p>
<p>Step 2) Sweep body: Select the path, and Sweep a body. Select "Circular Profile" and enter in the outer diameter of the tube [I'm aware the drawing in in inches and my model is in millimetres - I didn't think it worth changing my settings just for this demo!)</p>
<p><a href="https://i.stack.imgur.com/vcAx2m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vcAx2m.png" alt="Sweep Body"></a></p>
<p>Step 3) Bell Profile: We covered this in the original answer. The only thing to add, is that I've set the left-most anchor point to be coincident with the tube surface. This means that if you were to edit the tube diameter that we set in Step 2, then this sketch would automatically update. <em>This</em> is where defining the curve of the bell as a parabola instead of with coordinate points would come into its own, as the whole curve would be able to adjust to accommodate the change. Note again how all dimensions are linked to the same datum that we picked before.</p>
<p><a href="https://i.stack.imgur.com/6V4Nam.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6V4Nam.png" alt="Bell Profile"></a></p>
<p>Step 4) Revolve Bell: Pretty self explanatory, revolve. It's best practice to put a centreline into your sketch, and SolidWorks should recognise this as the Axis of Revolution automatically.</p>
<p><a href="https://i.stack.imgur.com/tWCgrm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tWCgrm.png" alt="Revolve Bell"></a></p>
<p>Step 5) Shell Body - shell the body to the specified wall thickness.</p>
<p><a href="https://i.stack.imgur.com/8jQCWm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8jQCWm.png" alt="Shell Body"></a></p>
<p>Step 6) Clean Rim: Because the drawing specifies that the rim has a horizontal external diameter, rather than being perpendicular to the spline, this meas that the shell command leaves a small 'step' on the inside of the rim. This is not an error - try reducing the wall thickness down to a smaller value to see where this comes from. Using the 'Delete and Patch' tool will remove these two faces, and extend the inner surface of the bell to meet the external diameter.</p>
<p><a href="https://i.stack.imgur.com/YNKLcm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YNKLcm.png" alt="Clean Rim"></a></p>
<p>Step 7) Admire your work.</p>
<p><a href="https://i.stack.imgur.com/0jyJW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0jyJW.png" alt="Full Demo"></a></p>
| 32301 | SOLIDWORKS - 3D Sketch of Trombone |
2019-12-10T07:02:00.563 | <p>I am having trouble understanding how indoor temperature plays into what a healthy indoor relative humidity level should be. Most sources I have come across searching the topic online provide a range based on the outdoor temperature. For example, according to <a href="https://www.thermastor.com/relative-humidity-and-your-home/" rel="nofollow noreferrer">Therma-Stor</a>, "The ideal relative humidity for health and comfort is about 40–50%. In the winter months, it may have to be lower than 40% RH to avoid condensation on the windows." I understand this is important for preventing condensation that can lead to mold and other problems, but I have not seen condensation forming on any windows and I think the air may be too dry in my case.</p>
<p>I live in a home that was built in 1938. It is not well-insulated and there have been several additions in its history. It is currently winter in my location, so the furnace is running. I do not feel any drafts coming from outside, so I believe most of the heat escapes due to the poor insulation -- the closer I am to exterior walls, the colder it feels. The thermostat in my home is set to 64 degrees Fahrenheit (17.8 <span class="math-container">$^oC$</span>). Increasing the temperature beyond that has a minimal effect on the downstairs temperature, while the upstairs temperature increases drastically. I have looked at the air ducts, but I cannot find any place where I am able to adjust the airflow.</p>
<p>The outdoor temperature ranged from 19-49 degrees Fahrenheit (-7 to 9.4 <span class="math-container">$^oC$</span>) today, with a current outdoor temperature of 21 degrees Fahrenheit (-6 <span class="math-container">$^oC$</span>) and humidity of 79%. I just measured the temperature and humidity in my bedroom, which is downstairs, and it is 57 degrees Fahrenheit (13.9 <span class="math-container">$^oC$</span>) with a humidity of 37%. In the office, which is upstairs, the temperature is 64 degrees Fahrenheit (17.8 <span class="math-container">$^OC$</span>) with a humidity of 35%. I placed the thermometer/hygrometer in the center of each room.</p>
<p>Considering the temperature of my bedroom is quite low, I am assuming that the 37% humidity level is pretty dry -- the thermometer/hygrometer also provides a reading of "Low" next to the humidity level. It seems that most sources I have found assume a more reasonable indoor temperature in the range of 68-72 degrees Fahrenheit (20 to 22 <span class="math-container">$^oC$</span>). I do have a portable humidifier I could run in my room, and I am wondering what would be an appropriate humidity level under these outdoor and indoor conditions? Is there some sort of an equation I could use, maybe something similar to how the thermometer/hygrometer determines when to display a reading of "Low," "OK," or "High?"</p>
| |hvac|temperature|air| | <p>In the world of heating/ventilating/air conditioning engineering (a.k.a. "HVAC") there is a number called the <em>comfort index</em> (which you can search on) which takes both humidity and temperature into account to determine where most humans would feel comfortable. This will tell you the range of optimum humidity percentages to shoot for at a given temperature setting.</p>
<p>Regarding the use of the humidifier: generally speaking, at lower temperatures, the more humidity, the better your comfort. I recommend adding humidity until you just start getting condensation on your windows, then back off by 10%. Note that mold problems commonly arise in humid houses during the winter with the main culprit being chests of drawers and tall wardrobes backed up against exterior walls. The narrow gap between the furniture and the (poorly-insulated) wall and the poor circulation there will cause the temperature of the air in it to drop below the condensation point and moisturize the wallboard, leading to the rampant growth of mildew on the wall in a pattern which exactly duplicates the outline of the chest or wardrobe. It's a good idea to regularly check for mold behind the furniture, and also to pull those things a few inches away from the walls to improve circulation and prevent condensation. </p>
| 32302 | How does indoor temperature factor in when determining ideal relative humidity in a home? |
2019-12-11T16:29:00.010 | <p>could anybody help me by providing an internal involute spline data sheet similar to the attached picture for an 1-3/8" splined shaft with 21 teeth, 16DP and 30 degree pressure angle.
Thanks for your help.</p>
<p><a href="https://i.stack.imgur.com/twAhf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/twAhf.jpg" alt="enter image description here"></a></p>
| |mechanical| | <p>From Machinery's Handbook for a flat root side fit spline. There are a few other different styles, this is what I use. This is a class 5 tolerance (typical industry level) per ANSI B92.1-1970</p>
<p><a href="https://i.stack.imgur.com/tJBjZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tJBjZ.png" alt="21T16P30A"></a></p>
| 32327 | Internal involute spline 1.375 - 21T |
2019-12-13T01:51:27.927 | <p>I'm trying to make an equation driven conical spiral, you can download the file <a href="https://www.dropbox.com/s/14u0kdhxeui9px8/Coil%20spring%20no%20disk.SLDPRT?dl=0" rel="nofollow noreferrer">here</a>. I know that I can use the spiral tool to achieve this but there are reasons why I want to do it like this. I have five global variables that control the characteristics of the spring: the initial radius ri, the final radius rf, height H, number of coils nc, and diameter of wire d (it's a square wire so the diameter is the width/height of the square). The equations for generating the coil are:</p>
<pre><code>x=("ri"-t*(("rf"-"ri")/"H"))*cos(t*((2*pi*"nc")/"H"))
y=("ri"-t*(("rf"-"ri")/"H"))*sin(t*((2*pi*"nc")/"H"))
z=t
</code></pre>
<p>with limits from t=0 to t=H. The problem is that when I change the variables, the spring does not rebuild, even when I use Ctrl+Q several times. I have no idea why this is. I have another small coil above the spring which I'm going to use to make an end coil and it seems to update ok but the main coil doesn't. Any help would be appreciated.</p>
| |solidworks| | <p>I can't answer your original question re: why using CTRL+Q did not successfully rebuild for you when using an equation driven curve - because when I downloaded your file, I could not recreate the error.</p>
<p>What I can do is demonstrate how you could apply your global variables to the built-in helix/spiral tool, such that the you can edit your variables in exactly the same way, and it automatically rebuilds whenever you edit a global variable value. (SW sends a CTRL+B when you press enter in the value/equation box).</p>
<p><a href="https://i.imgur.com/MgukWXh.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/MgukWXh.gif" alt="Editing Global Variables on Coil Spring"></a></p>
<p>I'm not just suggesting this 'just because I can' - it's a more robust way to model this object. Your supplied model using the equation curve takes almost <strong>four times</strong> as long to rebuild. (I have deleted features in your model after the initial sweep to keep it a level playing field)</p>
<p><a href="https://i.stack.imgur.com/QoLnD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QoLnD.png" alt="Rebuild times for helix/spiral vs equation curve"></a></p>
<p>EDIT: Here is my file to download if you would like: <a href="http://www.filedropper.com/conicalspringglobalvariablesdemo" rel="nofollow noreferrer">http://www.filedropper.com/conicalspringglobalvariablesdemo</a> </p>
| 32340 | Equation driven conical spring not rebuilding properly in Solidworks 2019 |
2019-12-13T14:35:38.070 | <p>I want to attach a 12 toothed spur gear with a bore diameter of 6mm onto the shaft of this <a href="https://www.maxongroup.com/maxon/view/product/488607" rel="nofollow noreferrer">motor</a> (6mm diameter). How would I go about doing this? From what I see, the shaft is smooth so is my only option to find a gear with a set screw? Or is it sufficient to attach a gear with no set screw and simply rely on the friction of the bore? The torques will be low. </p>
<p>Are there any websites that one would recommend for finding gears like this? I'm finding it impossible to find gears with a module of 1.0, bore diam. of 6mm, and a thickness of less than 20mm to fit onto the shaft.</p>
<p>Any help would be greatly appreciated. </p>
| |mechanical-engineering|motors|gears|power|power-transmission| | <p>Boston Gear or McMaster-Carr might have suitable off the shelf gears.
You could also press the gear onto a hub that has a set screw. I would avoid pressing anything directly onto a motor shaft, you could bend the rotor and/or damage the internal bearings.</p>
| 32345 | Attaching Gear onto Motor |
2019-12-13T17:10:12.890 | <p>Hello I am building a door and I have taken a large rectangle piece of wood for the base.</p>
<p>At the top-right point of the rectangle I am planning to connect it with the top right point of a large rectangle piece of wood vertically which would be the door.</p>
<p>This will create a rotation axis correct or am I wrong?</p>
<p>I usually trust myself before building something like this but this time I am undecided , i dont know so I came to this site for help whatever.</p>
| |applied-mechanics| | <p>If I understand you correctly, your idea would work. I would just try to focus on what you're using to connect the two components with. Because of the structure of your design, I'd recommend a Hinge Bolt as there will be a lot of stress on whatever is used. I'd also suggest making sure the lower part is firmly connected to whatever is beneath it, perhaps screwed into place, as you don't want the whole thing to flip over. I hope this helps.</p>
| 32354 | Door functional or not? |
2019-12-13T19:55:28.937 | <p>I'm doing a project that involves spinning up a (roughly) 2.28 kg object to 5000 rpm (via a DC motor) and then letting it continue rotating under its own influence (without any more input from the motor).</p>
<p>See my image:<img src="https://i.stack.imgur.com/JwgAB.png" alt="![enter image description here">]<a href="https://i.stack.imgur.com/JwgAB.png" rel="nofollow noreferrer">1</a></p>
<p>For my project to work, friction and other forms of resistance must be reduced to a minimum. My initial thought was simply to to use the DC motor itself, but in my experience, most such motors generate excess internal friction. And clearly, a motor that goes on spinning without electrical input will become a generator, thus creating more resistance.</p>
<p>So, does anyone have any suggestions? Preferably something that I could simply buy.</p>
| |design|motors|prototyping| | <p>Option 1: A plain old ordinary motor with a clutch of some sort that is completely non-contact when it is disengaged. If losses when you're spinning the thing up don't matter too much, then a rubber drive wheel on a motor that engages the outside circumference of your object (or a wheel on the axle on top) would work.</p>
<p>Option 2: DC motor have "excess" losses because they have permanent magnets in them, that continue to cause losses in the armature iron as they spin. So use a motor that doesn't have permanent magnets, such as a wound-field DC motor (hard to get, unless you get a universal motor), or an induction motor driven from a VFD to get you above synchronous speed.</p>
<p>There will be a trade-off here: most inexpensive motors of this sort are still not aerodynamically perfect, and may not be happy at 5000 RPM. Moreover, finding suitable motors that are both off the shelf and as small as your application demands will be a challenge. You'll have to shop around for a suitable motor (and possibly VFD), and you may not be able to avoid aerodynamic losses unless you're going to operate this in a vacuum.</p>
<p>I think that if I were designing this for production and could specify the motor, I'd want a small induction machine or a variable-reluctance motor -- probably the induction machine.</p>
| 32355 | Designing A Freely Spinning Object |
2019-12-14T03:25:28.810 | <p>Looking at a part like this
LM2596</p>
<p>It is rated for 3Amps with a voltage range from 3.3-12 volts.</p>
<p>Does that mean that this voltage regulator will only output a maximum of 3amps? </p>
| |power|power-electronics|power-engineering|regulations|current| | <p>That is an oversimplification.</p>
<p>From the data sheet <a href="http://www.ti.com/lit/ds/symlink/lm2596.pdf" rel="nofollow noreferrer">http://www.ti.com/lit/ds/symlink/lm2596.pdf</a> there are two relevant features, internal over-temperature protection and a maximum current limiter (Typical value 4.5A, minimum 3.6A).</p>
<p>The bottom line is that trying to get more than 3A <em>continuous</em> current from the regulator will probably not end well. </p>
| 32361 | Maximum amps from a switching voltage regulator |
2019-12-15T11:30:34.493 | <p>How many general methods are there for transferring electricity from the railway to a train? I could see that some trains are connected by a pantograph and some have a third rail. </p>
<p>Are there any other methods? What is the general engineering principle behind each? What are the basic differences (pros and cons) of each method? I was looking for these questions but could not find a good review. If anyone has a reference or could answer them kindly do. </p>
<p>Also, is it possible to build a small model of such a system and scale it up? I am not talking about a commercial model train. I am aiming at self-designing the same engineering principals and demonstrating them on a scalable small model. If so, I would be happy to have a nice tutorial reference. </p>
<p>I am focusing on <a href="https://en.wikipedia.org/wiki/Electric_multiple_unit" rel="noreferrer">Electric multiple units (EMU)</a>.</p>
| |rail| | <p>I didn't see anyone mentioning a keel system. In this system which was used in the past by DC Transit (DC = District of Columbia, nothing to do with voltage), a keel extends from beneath the car through a slot midway between the tracks. A shoe on the bottom of the keel contacts the power rail. This was designed for public safety and the absence of an overhead wire contact. In some respects it must be similar to no. 13 listed above by H.M. Muller.</p>
| 32376 | What are the engineering principles for a train to get electricity from the railway |
2019-12-16T02:51:18.247 | <p>What are the nominal voltage ratings and operating tolerances for 60-Hz electric power systems above 100 volts up to a maximum system voltage of 1 kV in Brazil?
I am looking for a document like ANSI C84, that describes typical distribution service and equipment nameplate voltage ranges and tolerances.</p>
<p>I know about 60% of the population is served by a 127 Volt system and a minority is served by a 220V system and both systems use the same type N plug.</p>
<p>For example is the 127 Volt system split phase 127/256 like North American 120/240?
Is the 220 Volt system split phase 220/440 ?
What are the standard three phase voltages, 220Y127, 380Y220 ?
Thank you</p>
| |electrical-engineering| | <p>In Brazil, the most common (domestic distribution) system is a three phase 4 wire one, with the most common voltages being 127/220 and 220/380 (with many exceptions).
On 127/220, some houses will have a connection to two phases and the neutral, giving the choice to use either voltages.
On 220/380, most houses will probably work with only one phase, as pretty much all domestic products works at either 127 or 220, while 380 is used mostly on industry.</p>
<p>Not completely sure about the next info, but it seems that 110/220 used to be a thing in the past, probably with a split phase like in the US, but sometime at around 1999 there was a resolution that made it change to the actual system. The funny point is that the usual voltages are called "110 and 220" up to this day, even when the "110V" is actually 127V and most Brazilians don't even know it.</p>
<p>Information on that is decently hard to find even in the native language (PT-BR), but some decent overall info can be found <a href="https://www.mundodaeletrica.com.br/diferentes-niveis-de-tensao-quais-sao/#:%7E:text=Onde%20tem%20tens%C3%A3o%20el%C3%A9trica%20de,voc%C3%AA%20tem%20os%20380%20Volts." rel="nofollow noreferrer">here</a>, and a list with the three phase voltages by State can be found <a href="https://www.portalsolar.com.br/tabela-de-trifasico-no-brasil-220v-e-380v" rel="nofollow noreferrer">here</a>.</p>
| 32384 | Nominal voltage ratings in Brazil? |
2019-12-16T14:36:15.800 | <p>Both circulation and vorticity have to do with rotation of a fluid element.How are vorticity and circulation are related.</p>
| |fluid-mechanics|fluid| | <p>No. If no vorticity then, no circulation. </p>
<p>Mathematically,
The flux of vorticity is circulation.</p>
<p><span class="math-container">$ \Gamma = \int\int \omega~ ds$</span> </p>
<p>So, <span class="math-container">$\Gamma \rightarrow $</span> 0 when <span class="math-container">$\omega \rightarrow $</span> 0.</p>
<p>Physically, vorticity is just not a rotation of a fluid element. But it is the rotation of that element about its own axis (spin). (Please know about free/forced vortex for a kind of imagination).</p>
<p>So, circulation is non-zero only when there is a flux of vorticity. </p>
| 32392 | Is it possible to have circulation without vorticity? |
2019-12-16T20:16:23.693 | <p>I want to make a sprung latch mechanism which keeps a door in place until a certain force is applied to the door. In the diagram attached, the force the door must release at is Fd, and the latch spring force necessary to hold it until Fd is reached is Fs. </p>
<p>My questions are these:
1. How do I work out what force spring I need?
2. Assuming the contact is frictionless, and that distance d stays the same, does it make a difference what the angle j is?</p>
<p>Many many thanks for any help!</p>
<p><a href="https://i.stack.imgur.com/zMsrJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zMsrJ.png" alt="latch free body diagram"></a></p>
| |mechanical-engineering|applied-mechanics|mechanisms|mechanical|solid-mechanics| | <p>The force of resistance of the latch varies. at the time the door touches the latch it is zero and when it is pushed fully back it is k*d.</p>
<p><strong>Edit</strong></p>
<p>why F*tan j is the horizontal component? after OP's comment.</p>
<p>If we draw the perpendicular, <span class="math-container">$F_d*cos j,$</span> and <span class="math-container">$F_d* Sin j \ , $</span> components of Fd on the surface of latch, the hypotenuse of the triangle is the tan of that angle and is the horizontal component factor of the Fd.</p>
<p>We have ignored the round corner of the door on the diagram and assumed it's a sharp point.</p>
<p>End of edit.
<span class="math-container">$$ F_d*tanj=k*d \\F_d=\frac{k*d}{tanj }$$</span></p>
<p>The steeper the angle J the less force is needed to push the door open.</p>
| 32400 | Latch Mechanics |
2019-12-17T10:20:50.810 | <p>I've been asked to modify an existing zimmer/walking frame for a tall person. Bashing the bits of metal is the easy bit, just use tube to make longer legs to replace the original short removable adjustable ones, using tube that's actually thicker wall than the original. </p>
<p>The tricky bit is now they've asked me for a new weight rating, given that the frame is larger. The original frame was rated for use by people up to 180kg. It would be quite easy to say that as the new height is <span class="math-container">$x$</span>, we should derate the weight by <span class="math-container">$x$</span>, or by <span class="math-container">$x^2$</span>, or by <span class="math-container">$x^3$</span>, but justifying a PIDOOMA for the exponent or the weight is more difficult.</p>
<p>What are the expected failure modes of the frame? Here's a picture of it after fitting with the extended legs, the lowest 500mm.</p>
<p><a href="https://i.stack.imgur.com/brQq4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brQq4.png" alt="enter image description here"></a></p>
<p>I'm assuming two failure types are relevant, prompt failure resulting in the user ending up on the ground, and increased wear meaning a weakening after several years' use.</p>
<p>For prompt failure, if I assume that all joints are hinges, then the weakest appear to be on the cross member holding the two side frames together, which is why I guess there are corner braces there. </p>
<p>The moment on those hinges, should the user slump sideways on the frame, will be increased slightly by the longer legs, but will still be dominated by the width of the frame, so not even by a factor of <span class="math-container">$x$</span>. By my reckoning, it's the smallest height of the triangle formed by the width and height of the walker, and a diagonal from one handle to the floor contact on the other side.</p>
<p>As the new legs are stronger than the original, and use the original spring pins into the same size holes, there should be no reason at all to expect any reduction in simple vertical strength.</p>
<p>The fretting moment at the area of tube overlap where the adjustable legs couple to the frame, held in place by spring pins, will increase directly in proportion to the adjustable leg length. However, the overlap region has been increased in length, more than proportionately, to reduce the contact force caused by this moment. So it may be no worse.</p>
<p>A colleague has suggested a simple reduction to a nice round 100kg, which would certainly be conservative. My concern is that I wouldn't be able to justify it, and that it may exclude other potential users for no good reason.</p>
<p>Are there industry standard tests for this sort of thing? How would the original manufacturer have rated it for 180kg users?</p>
| |structural-analysis| | <p>Your last paragraph is the key one: Yes, there industry standard tests, and they were able to rate it for 180kg by passing those tests. You can do theoretical calcs until the cows come home to make sure you pass those tests first time, but <strong>you must undertake and pass these tests prior to adding a rating sticker</strong> or you open yourself right up to liability.</p>
<p>I looked at BS EN ISO 1119-1:1999 "Walking aids manipulated by both arms - Requirements and Test Methods" Part 1: Walking frames</p>
<p>I have picked out some key points, but you should read the whole document (or local equivalent) yourself.</p>
<p>One walking frame shall be tested, through Stability, Static Load, Fatigue, and Static Leg-strength in that order.</p>
<p>Stability: Forward, Backward and Sideways tests have different tilt angls, then a vertical load of 250N is applied to the centre of the hand grips.</p>
<p>Static Load: A load of 1,500N is applied, with an additional 15N per kilogram of maximum user weight above the standard maxium of 100kg. You would need to test to 2,700N (+/-2%) to demonstrate equivalence</p>
<p>Fatigue: A cyclic load of 800N plus 8N per kg over 100kg (1,440 for equivalence) applied downwards on the handles, 200,000 times, slower than 1Hz.</p>
<p>Static Leg-strength: Applied to each of the legs in turn, applied a minimum of 10mm further out than the lower end of the leg, and directed towards the centre of the frame. done by removing the rubber foot, and putting in a tightly fitting plug up to 100mm inside the leg. Height adjustment of frame must be at maximum. The standard <em>requires</em> a loading force for this of 300N, applied gradually over 2s, and held for minimum of 5s. <strong>but</strong> it also recommends testing at 500N, to provide a safety factor "for reasonable foreseeable misuse".</p>
| 32406 | Failure modes of a zimmer frame |
2019-12-18T10:09:47.217 | <p>I want to get a rough estimate for the transmission efficiency of two different gear trains. One is a one-stage gear train with 11.67:1 ratio (140/12 teeth) and the other is a compound two-stage gear train with overall gear ratio of 18:1 (72/12 + 36/12 teeth). </p>
<p>I read in a forum that one can expect a 10% loss per stage of gear train in this case but I wanted a second opinion. Any help would be greatly appreciated.</p>
| |mechanical-engineering|gears|energy-efficiency| | <p>Assuming you mean gears and not chain and sprocket systems, it really depends on the size and cost and duty of the gear reduction.</p>
<p>The simplest case is a colinear drop box that has a propshaft input and a propshaft output with no side loads at all. For modest ratios, the reduction should be around 3 percent of the <em>maximum rated torque</em>. For example, if the gearbox is rated for 1000 ft lbs, because your 600 ft lb diesel requires a 30% derating, and you are operating at around 400 ft lbs most of the time, you still loose 3% of 1000, or 30 ft lbs. That's 7.5%.</p>
<p>Other load factors and physical arrangements can further reduce this. Overhung loads (side load associated with a spocket on the input or output shaft) may require additional bearings or a lower rating and tend to be a bit less efficient.</p>
| 32415 | How Much Efficiency is Lost in a Gear Train |
2019-12-19T03:12:30.150 | <p>In this tutorial <a href="https://www.youtube.com/watch?v=EVL7TzCde8I" rel="nofollow noreferrer">here</a></p>
<p>There's a current limiting application where two op amps are being used to limit the max output current.</p>
<p>The first differential opamp with 4 equally sized resistor values; connected to the output through a shunt resistor to measure the current draw.</p>
<p>The output from this opamp is then fed into the non inverting input of the second opamp while the inverting input of the opamp has a variable resistor to limit the output current.</p>
<p>Can anyone explain the ratio of this second opamp inverting input resistor?</p>
<p>Lets say I wanted to limit the current from 1-5 amps or 0-1 amps or even 2-20amps.</p>
<p>What's the ratio?</p>
| |current| | <p>Below is the schematics that is discussed in your question. </p>
<p><a href="https://i.stack.imgur.com/YbEP0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YbEP0.jpg" alt="enter image description here"></a></p>
<p>I believe the resistor for the opamp in the bottom right hand corner can be any type of potentiometer, example it could be a 1K multi turn potentiometer like the one below</p>
<p><a href="https://i.stack.imgur.com/MrtVH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MrtVH.jpg" alt="enter image description here"></a></p>
<p>Below is better representation of the above circuit. </p>
<p><a href="https://i.stack.imgur.com/zCSqe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zCSqe.jpg" alt="enter image description here"></a></p>
<p>For the circuit work for ampere rangers outside of 0-1A you might have resize the resistors on the TLC272 opamp. This would require some work and I will leave that for you do the analysis and present to the community for feedback. </p>
<p>Alternatively I would try the following circuit with the LM358 for additional ampere rangers. </p>
<p><a href="https://i.stack.imgur.com/v4Pyy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v4Pyy.jpg" alt="enter image description here"></a></p>
<p>I suggest taking a looking at <a href="https://www.youtube.com/watch?v=ywFpNdvSC7A" rel="nofollow noreferrer">Adjustable Switching Power Supply Using LM2576 [Buck Converter, CC-CV]</a> for details on how this circuit works. </p>
<p><strong>References</strong></p>
<ul>
<li><a href="https://www.bourns.com/pdfs/3296.pdf" rel="nofollow noreferrer">3296 - 3/8 ˝ Square Trimpot® Trimming Potentiometer</a></li>
<li><a href="https://www.smbaker.com/lm2576-constant-voltage-constant-current-switching-power-supply" rel="nofollow noreferrer">LM2576 constant voltage / constant current switching power supply</a></li>
<li><a href="https://www.pcbway.com/blog/technology/How_to_build_an_adjustable_switching_power_supply_using_LM2576_[Buck_Converter__CC_CV].html" rel="nofollow noreferrer">How to build an adjustable switching power supply using LM2576 [Buck Converter, CC-CV]</a></li>
<li><a href="http://www.ti.com/product/TLC272/technicaldocuments" rel="nofollow noreferrer">TLC272 Dual Single Supply Operational Amplifier</a></li>
<li><a href="http://www.ti.com/lit/ds/symlink/tlc272.pdf" rel="nofollow noreferrer">TLC272, TLC272A, TLC272B, TLC272Y, TLC277 LinCMOS PRECISION DUAL OPERATIONAL AMPLIFIERS</a></li>
<li><a href="http://www.ti.com/lit/ug/sbou192/sbou192.pdf" rel="nofollow noreferrer">DIYAMP-SC70-EVM</a></li>
<li><a href="http://www.ti.com/lit/ds/symlink/lm2576.pdf" rel="nofollow noreferrer">LM2576xx Series SIMPLE SWITCHER® 3-A Step-Down Voltage Regulator</a></li>
<li><a href="https://www.pcbway.com/project/shareproject/How_to_build_an_adjustable_switching_power_supply_using_LM2576.html" rel="nofollow noreferrer">How to build an adjustable switching power supply using LM2576?</a></li>
</ul>
| 32431 | current limiting through opamp |
2019-12-19T18:37:57.440 | <p>Assume there is a rectangular beam of length L resting on a rigid horizontal surface. The beam is of a real homogeneous material, say aluminum. Assume the surface is infinitely rigid. I apply a vertical force F on each end of the beam, clamping the beam to the surface.
1) How do the clamp forces on the surface distribute along the beam?
2) What would be the shape of the curve for the bottom of the beam that would result in equal distribution of forces?
This has real applications in gluing of large flat objects together, where standard clamps cannot reach the interior areas.
Thanks
Kevin</p>
| |statics| | <p>Let's say you want to apply a uniform load of q over the length of the beam pressuring equally along the length. clamps are holding the ends with the force of F= qL/2 at each end.</p>
<p>Imagine there was no floor supporting the beam and these were the reactions of a distributed loading, <span class="math-container">$q= 2F/l.$</span> This q is the pressure you want to apply uniformly upward from the ground.</p>
<p>That load would deflect the beam into a parabola with maximum deflection of <span class="math-container">$5qL^4/384EI, \\ \text{and the equation of the beam's centerline deflection is} \ y= -\frac{qx}{24EI}(L^3-2Lx^2+x^3)$</span></p>
<p>This curve is the deflection of a simply supported beam under the uniformly distributed load of q, which looks like a rope hanging at 2 ends.</p>
<p>You have to camber your beam so that the centerline of it is deflected down by the above eq. and form it by the above deflection curve to have the floor distribute a uniform load, q, to it when it is clamped and forced down to perfect straight line touching the surface.</p>
| 32445 | load distribution |
2019-12-20T07:43:13.647 | <p>I would like to control a 1-phase 3HP motor using a SSR ( the reason being, I need to switch it on/off using a signal (0-5VDC).</p>
<p>Are normal SSRs (ex:Fotek 40A model) capable of handling the starting current of a 3HP motor? </p>
<p>What are the protection devices that I need to use in the circuit.</p>
<p>I guess a Motor MCB is required. Is anything else required?</p>
<p>All opinions/suggestions are welcome.
Thanks in advance</p>
| |electrical-engineering|control-engineering|motors|pumps|electrical| | <p>You can also just use a VFD similar to this <a href="https://www.mroelectric.com/product/ATV312HU15N4" rel="nofollow noreferrer">Schneider Electric</a> one. Looks to be only 2HP but there should be 3HP available.</p>
| 32453 | AC Motor On/OFF Control using SSR instead of contactor |
2019-12-20T08:11:51.523 | <p>So for any structure, simply supported beam, three hinge frame, cantilever etc. where there is a uniformly distributed load. For example, for a simply supported beam with uniform load of w along the length of x. Why is the bending moment equation w(x)(x/2)? I understand that the magnitude of the force is given by w(x), but is the reason for the (x/2) is because the magnitude acts halfway through the load? </p>
<p>Now I know that shear force and bending moment are related, with the shear force being the derivative of the bending moment, so I can just integrate the shear force equation to get w(x)(x/2), but I want to confirm through fundamentals as to why the equation is as it is. </p>
<p>Thanks.</p>
| |moments| | <p>consider an infinitesimally small section of a beam from coordinate <span class="math-container">$x_1$</span> to <span class="math-container">$x_1-dx$</span> loaded with a distributed load w. The moment of this load about point <span class="math-container">$x=0 \ is \\ M= w*x*dx.$</span></p>
<p>Now if we have the load area extended from <span class="math-container">$x_0 \ to\ x,$</span> we need to integrate the moment over the length of beam loaded with w: <span class="math-container">$$M= \frac{1}{2}wx^2$$</span></p>
<p>Or as you noted because the CG of load rectangle w*x is at its middle the moment is the area times the distance of CG to <span class="math-container">$x_o$</span>, <span class="math-container">$\ M= wx*x/2=wx^2/2$</span> </p>
| 32454 | Why is the bending moment equation for a uniformly distributed load (w) across a length x on a structure given by w(x)(x/2)? |
2019-12-20T08:22:02.220 | <p>I need to hold small pieces of wood in place to be drilled by a mini computer-numerical control (CNC) mill. I have a <a href="https://rads.stackoverflow.com/amzn/click/com/B07459P7BF" rel="nofollow noreferrer" rel="nofollow noreferrer">book stand</a> (no affiliation) that has two "springed pegs" fastened with screws:</p>
<blockquote>
<p>Springed pegs allow very thick book</p>
</blockquote>
<p>I think these would do very well and I could not find such springed pegs separately. Are they available under a different name?</p>
<p><strong>Update</strong>: My current solution prevents the pieces from shifting horizontally by slotting them in a pocket with their exact size with 5 mm of depth. I have been holding the pieces in place with double-sided tape and it works. A milling job takes about a minute and I have hundreds of pieces, so I am looking for a quicker solution. The machine uses a downcutter that presses the wood into the bed, so I think light tension would be enough.</p>
| |manufacturing-engineering|cnc| | <p>Not what you're asking for, but maybe even more suitable? These are also available in a low-profile version. Google "toggle clamp".</p>
<p><a href="https://i.stack.imgur.com/gPJef.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gPJef.png" alt="enter image description here"></a></p>
<p>Or these? Google "spring plunger".</p>
<p><a href="https://i.stack.imgur.com/9mgVo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9mgVo.png" alt="enter image description here"></a></p>
| 32455 | Where to find springed pegs with screws for a CNC mill? |
2019-12-20T13:37:23.197 | <p><a href="https://i.stack.imgur.com/MXicV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MXicV.png" alt="Cross-section of cylinder"></a>
The situation I'm imagining is the one shown above, which is the cross section of a cylinder made of alternating layers of material. The material axes are shown where 1 and 2 are in the same plane and 3 is parallel to the stacking direction. If a cylinder composed of this material is subjected to a torque around its central axis as shown, would it be subject to axial stresses and therefore result in torsional warping? I read <a href="https://engineering.stackexchange.com/questions/31949/why-torsional-warping-does-not-occur-for-shafts-with-circular-symmetry">here</a> and in textbooks that circular members don't have torsional warping but they are usually assumed to be made of isotropic materials. Would this be the case for this type of cylinder too?</p>
| |materials|applied-mechanics| | <p>Yes, it will warp.</p>
<p>Imagine the blue layer material stiffer, with higher Young modulus E and shear modulus G and the white material as less stiff. </p>
<p>The blue material will be the dominant material in reacting to the torsion and will warp in developing both St Venanat shear stresses and warping stress which are axial tensile and compressive stresses and additional shear stresses.</p>
<p><a href="https://i.stack.imgur.com/ZrExl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZrExl.png" alt="warped torsion"></a> </p>
<ul>
<li>Torsion causing warping in the blue laminates.</li>
</ul>
<p><a href="https://i.stack.imgur.com/kxlaM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kxlaM.png" alt="axil stresses and shear due to torsion."></a></p>
<ul>
<li>Additional axial stresses due to the warping.</li>
</ul>
<p>The white layer will undergo similar deformation forced by the blue material deformation and will be stressed less, with a wavy step like undulations at the intersection line with blue material.</p>
<p>In these diagrams, I have not shown the white material for clarity.</p>
| 32460 | Would an orthotropic cylinder subjected to torsion have warping? |
2019-12-20T17:31:56.117 | <p>I am looking at a control system which has an unstable pole in the process. The transfer functions of process and controller are the following:</p>
<pre><code>G = 10/((s+10)*(s-1));
K1 = 4*(s+1)/s;
</code></pre>
<p>where G is the process and K1 is the controller. I have found with the Routh Criterion that the closed loop is stable for <span class="math-container">$K>1.125$</span> and so I have chose the value 4 since it gives good performances. Now, when I plot the root locus of the closed loop:</p>
<pre><code>T1 = K1*G/(1+K1*G)
figure;
rlocus(T1)
</code></pre>
<p>I get the following:</p>
<p><a href="https://i.stack.imgur.com/0GS1h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0GS1h.png" alt="enter image description here"></a></p>
<p>What I don't undestand is from where the zero that cancels the unstable pole comes from, since the closed loop transfer function is:</p>
<p><span class="math-container">$\frac{(40(s + 1))}{(40 - 10s + 40s + 9s^2 + s^3)}$</span></p>
<p>can somebody help me?</p>
| |control-engineering|control-theory|systems-engineering| | <p>rlocus() takes the <em>open loop</em> transfer function as an argument, not the closed loop. i.e. G not T. Look at the documentation for rlocus here <a href="https://www.mathworks.com/help/control/ref/rlocus.html" rel="nofollow noreferrer">https://www.mathworks.com/help/control/ref/rlocus.html</a></p>
| 32462 | Doubt on zero pole cancellation |
2019-12-20T20:25:36.127 | <p>I am currently studying a robotics paper, where a normalized frobenius norm of a matrix is calculated as part of determining the condition number of the corresponding linear system of equation:</p>
<p><a href="https://i.stack.imgur.com/Am5HJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Am5HJ.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/22Z6C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/22Z6C.png" alt="enter image description here"></a></p>
<p><span class="math-container">$\mathbf{W}$</span> is defined as
<span class="math-container">$$\mathbf{W} = \frac{1}{n} \mathbf{I}$$</span> and inserting this definition into <span class="math-container">$\left|\left|\mathbf{J}\right|\right|$</span> results in
<span class="math-container">$$\left|\left|\mathbf{J}\right|\right| = \sqrt{\text{tr}\left(\mathbf{J}\mathbf{W}\mathbf{J}^{T}\right)} =\sqrt{\text{tr}\left(\mathbf{J}\frac{1}{n}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{I}\mathbf{J}^{T}\right)} = \sqrt{\frac{1}{n}\text{tr}\left(\mathbf{J}\mathbf{J}^{T}\right)} .$$</span></p>
<p>The normalization factor is therefore independet of <span class="math-container">$\mathbf{J}$</span>. Therefore, what is the purpose of scaling <span class="math-container">$\left|\left|\mathbf{J}\right|\right|$</span> down by a factor proportional to its dimension? </p>
<p>In terms using <span class="math-container">$\kappa(\mathbf{J})$</span> to optimize a mechanism, this would mean that the respective value of each iteration would simply be scaled down by constant factor, and therefore not influence the results. Am I missing something here?</p>
<p>Thanks in advance!</p>
| |design|robotics|kinematics| | <p>It is because you are using the trace of the matrix, which is equal to the <em>sum</em> of the singular values. The ideal <span class="math-container">$\mathbf{JJ^T}$</span> has singular values all equal to 1, but for an <span class="math-container">$n \times n$</span> matrix this sum would be equal to <span class="math-container">$n$</span>, not <span class="math-container">$1$</span>, and the text is defining the ideal matrix norm as 1.</p>
| 32467 | Purpose of normalization of Jacobian in a parallel manipulator |
2019-12-21T13:38:10.390 | <p><a href="https://i.stack.imgur.com/X9jqz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X9jqz.png" alt="enter image description here"></a></p>
<p>What do they do? </p>
<p>The aux contacts are normally open and the device is MR PQM 500 reactive power compensation system.</p>
| |electrical-engineering|power-electronics|electrical|power-engineering|electronics| | <blockquote>
<p>The aux contacts are normally closed ...</p>
</blockquote>
<p>This is unlikely. I suggest that they are normally open and close before the main contacts do.</p>
<p>You haven't provided a link for the MR PQM 500 system but I suspect that it is a capacitor bank with a controller which switches the contactors to switch in capacitance as required to bring the power factor closer to unity.</p>
<p>There is problem when switching in capacitors. If the mains voltage is close to peak at the instant the contacts close a very high current will flow with resultant arcing, etc. because the discharge capacitor appears like a short-circuit. To prevent this a special arrangement is employed:</p>
<ul>
<li>The auxiliary contact block at the front is driven by the main contactor movement but the auxiliary contacts close slightly before the main ones.</li>
<li>The three coils on top are inductors which have enough inductance to limit the current into the discharged capacitor to a safe value. This is enough to bring the voltage up so that the main contacts don't see the big surge they otherwise would.</li>
</ul>
| 32471 | What's the function of those coils there? |
2019-12-21T17:58:48.313 | <p>How much steam is emitted from an electric arc furnace in a unit of time? I have not been able to find any data on this which is strange considering the amount of water that is used to cool an electric arc furnace.</p>
| |electrical-engineering|structural-engineering|steel|welding| | <p>No steam at all from the furnace ; even the charge is preheated so that it is dry. Water may be used to cool the power transformers.</p>
| 32476 | Electric Arc Furnace Steam Production |
2019-12-21T19:32:58.843 | <p><a href="https://i.stack.imgur.com/X2IJv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X2IJv.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/wfl0C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wfl0C.png" alt="enter image description here"></a></p>
<p>If I understand correctly, one of the star's teeth is bent into one of 4 nut's slots to secure the nut. What happens if none of the teeth align with the slots? Do you unscrew the nut a bit? Or it is somehow mathematically made that there will always be one slot that aligns properly?</p>
| |mechanical-engineering| | <p>See <a href="https://www.skf.com/us/products/bearings-units-housings/bearing-accessories/lock-nuts/requiring-keyway/installation-removal/index.html" rel="nofollow noreferrer">https://www.skf.com/us/products/bearings-units-housings/bearing-accessories/lock-nuts/requiring-keyway/installation-removal/index.html</a> for installation instructions.</p>
<p>As you noted, the spacing can be set so that things are always pretty close. There are 19 tabs on the washer and four slots on the nut. The slot in the tread shown in figure 3 is wide enough to allow the washer to rotate slightly to relative to the bearing race, and this means the washer can always have one tab bent over.</p>
<p>A Starlock washer is a different item entirely - the teeth are on the inside and serve to limit axial movement of the washer in one direction, when placed over a circular shaft.</p>
| 32479 | What happens if a star lock washer doesn't align properly with the nut? |
2019-12-22T06:07:53.660 | <p>These parts were salvaged from a special mechanical calibration rig. I don't see anything that looks like that in McMaster-Carr, but perhaps I don't know where to look. </p>
<p>The aluminum part is probably custom but the rod (300-series stainless steel) and the plastic (acetal, I think) threaded part with the brass handle look like standard parts. The aluminum part has an internal thread so the depth of the plastic part can be adjusted by screwing it in or out. The plastic part has a conical lead-in and a bore that fits the stainless steel rod with some play. I believe it acted as a pivot with springs to take out the play. </p>
<p><a href="https://i.stack.imgur.com/2RSfz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2RSfz.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>Okay, I found an engineer who was familiar with the way they were used and it's not particularly exciting.</p>
<p>The three parts were screwed in to immobilize the motion platform while the cryostat etc. was being loaded and unloaded via a crane, then screwed out to free the motion. There was no deliberate damping other than that created by driving the electromagnetic shakers through a low-impedance source. </p>
<p>Frankly, I like the other ideas better. </p>
| 32484 | What are these mechanical components called? (assuming they are standard components) |
2019-12-22T18:13:12.947 | <p>They say that in the near future high cosmic radiation is possible. In addition, the Earth’s magnetic field is likely to weaken.
I want to create something like a camera that would protect against electromagnetic radiation. What will be effective for this? Useful articles, real examples? I want to protect equipment, for example, my phones, watches, if at least somehow.
Thanks in advance</p>
| |electromagnetism|radiation| | <p>There are 2 effects to look out for:</p>
<ol>
<li><p>Magnetic induction into wiring from e.g. <a href="https://en.wikipedia.org/wiki/Coronal_mass_ejection" rel="nofollow noreferrer">coronal mass ejections</a>. The electrical grid is already <a href="https://www.nbcnews.com/mach/space/how-we-ll-safeguard-earth-solar-storm-catastrophe-n760021" rel="nofollow noreferrer">largely protected</a> against these events. To protect equipment in your home, unplug it.</p>
</li>
<li><p>Radiation damage, in the event that Earth's magnetic field collapses. In computers, this manifests first as <a href="https://en.wikipedia.org/wiki/Single-event_upset" rel="nofollow noreferrer">Single Event Upsets</a>: components in a chip that change their state due to being hit by a charged particle. This can corrupt stored data and disrupt operations. Most of these are temporary (the component still works), but occasionally a component can be destroyed if the charged particle has enough energy and the feature size of the chip is small enough.</p>
<p>There are radiation-tolerant designs (built for the military and for spaceflight applications), but this is not something you can retrofit to an existing design. Your best bet is to disconnect the equipment and bury it - a few meters of water or soil <a href="https://what-if.xkcd.com/29/" rel="nofollow noreferrer">should be enough</a> to protect your equipment.</p>
</li>
</ol>
<p>There is <a href="https://en.wikipedia.org/wiki/Earth%27s_magnetic_field#Future" rel="nofollow noreferrer">no indication</a> Earth's magnetic field will collapse. Field strength is decreasing at the moment, but that variation is within the normal range.</p>
| 32493 | Protect equipment against electromagnetic radiation |
2019-12-23T16:41:38.067 | <p>Being a non-engineering type, I'm trying to calculate what power and torque I would need from a DC stepper motor to execute a particular task. I'm certain that my question will open me up to potential observations regarding my lack of knowledge in this field. That said, if your critical observations use humor to make the point, I will appreciate them for that. However, I do expect to gain some good knowledge from this group. Here goes nothing.</p>
<p>I am building a device that will be used to lift, and hold 200 pounds approximately 2 feet in elevation. The mechanical elements of this device are 100% the same as a standard worm-shaft driven, scissor jack for a motor vehicle. </p>
<p>Envision a letter "A" as the two arms on the top of the scissor, and the left arm is anchored to the substrate, and the right arm is attached to the threaded block that travels along the worm-shaft. As the worm-shaft is rotated, the threaded block will travel along its axis which will in-turn cause the "A" to get taller and skinnier. This achieves the lift needed for the device.</p>
<p><strong>Question 1:</strong><br>
In order to achieve the required power and torque to lift and hold 200 lbs without having to manually turn the worm-shaft (as one would when jacking up a car), should I couple the motor to the worm shaft concentrically, or use one gear on the motor shaft, and one on the worm-shaft?</p>
<p><strong>Question 2:</strong><br>
Depending on the direct-couple method, or the gear ratio method, how do I determine the power and torque of the motor?</p>
<p>I really appreciate any help I can get with this challenge.</p>
<p><a href="https://i.stack.imgur.com/7xyVs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7xyVs.jpg" alt="Scissor Jack"></a></p>
| |torque|stepper-motor| | <p>The length of each arm of your jack should be comfortably bigger <span class="math-container">$12"(inch)> 2ft/2$</span> because if they are close to 12" they lock at horizontal position.</p>
<p>We chose 18" and set the pegs at the lowest level so the arm is 20 degrees when not deployed meaning the jack can expand from 6" to 36" with a travel course of 30" which is more than you need but is more practical.</p>
<p>The horizontal force at the worm-screw is <span class="math-container">$$F=200/2*cot(20)= 100lbs*2.74= 274.7 lbs$$</span></p>
<p>From here on F becomes smaller if we lift the load.</p>
<p>Let's say the diameter of your worm-screw is 1/2" and its pitch is 1/32", meaning the slope of the thread is approximately <span class="math-container">$$ 1/2*\pi*32= 1/50$$</span> and the force on it is <span class="math-container">$$f_{screw}=274.7/50= 5.49 lbs \\ \text{and the torque needed is}\\ T=5.49*1/4"= 1.373 lbs.ft $$</span> </p>
<p>This is the case with no friction and 100% efficiency. I would use a load factor of 1.5 and check the manufacturer's recommendation.</p>
| 32504 | Determining the required specs of Stepper Motor |
2019-12-24T01:06:21.373 | <p>When analysing and designing a structure for the serviceability limit state the deflections of each beam must be found. When considering a multistory rigid jointed structure (as below) and applying first-order elastic analysis, what method must be applied in order to find the magnitude of the maximum lateral displacements on each beam (e.g. <span class="math-container">$\delta$</span> on the deflected shape)?</p>
<p>Essentially my question is, how do I find the value of <span class="math-container">$\delta$</span> on beam BE from the deflected structure. </p>
<p>Am I correct in idealising the beam as the simply supported beam with the correct concentrated moments on either side (acting as fixed support).
<a href="https://i.stack.imgur.com/8vpIy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8vpIy.png" alt="Example Of Structure To Find Deflection"></a></p>
| |structural-engineering|structural-analysis|deflection| | <p>From a straightforward, academic perspective (by which I mean disconsidering seismic and other code-enforced conditions), your approach is either exactly or roughly correct (depending on your definition, as described below).</p>
<p>Obviously, you must first calculate the entire structure and determine all the internal forces. As you've stated, that will include both the loads directly applied to BE but also any nodal loads transmitted to BE from the columns. This will give allow you to make a perfectly accurate representation of the internal forces on the beam and therefore of the deflections it faces "in isolation".</p>
<p>However, what your analysis is currently missing is that the beam actually <strong><em>isn't</em></strong> in isolation. It belongs to a larger structure which is also deforming, and this deformation also applies what can be effectively considered an imposed displacement to the nodes.</p>
<p>In this particular case, both nodes of BE will be moved to the right, while the left-most node will also move up and the right-most node, down. This means that BE's "straight line representation" in the deformed configuration is actually slanted.</p>
<p>And so it becomes a matter of definitions and possibly codes. What is the deflection to be considered when checking the servicibility limit state?</p>
<p>I'm Brazilian and therefore follow code NBR 6118 for concrete structures and NBR 8800 for steel structures. Both of these<sup>1</sup> are unclear as to which deflection to use, stating only that elements should be "analysed as isolated, simply-supported beams under loading"<sup>2</sup>. I interpret this as allowing your method of disconsidering imposed deflections, but it's not air-tight (imposed deflection could be considered a load, but they do use the word "isolated"...).</p>
<p>Another interpretation is to consider the relative displacement of the nodes: both nodes drop by some amount, but that cancels out, so the only thing that might increase users' unease (the basis for SLS) is the additional deflection of one of the nodes.</p>
<p>Or you might want to consider the total deflection of both nodes: after all, if your columns are very "compressible", your users might become uncomfortable if high variable loading causes the ceiling to drop by half a meter.</p>
<p>That depends on your relevant code (and possibly how you interpret it). If you need to consider these imposed deflections, then you're basically forced to use the stiffness matrix method: you can certainly do it by hand, but it'll just give you a headache.</p>
<p>See below the model I used to look at this problem, and the three ways you can interpret the deflection of BE:</p>
<ul>
<li>The top isolated beam is your method, ignoring the imposed displacements</li>
<li>The middle beam adopts the relative displacement of the nodes, concentrated on the right node</li>
<li>The bottom beam adopts the full displacement of both nodes and is equivalent to the actual result obtained from the frame (the slight discrepancy is a rounding error due to my inputs).</li>
</ul>
<p><a href="https://i.stack.imgur.com/hAW00.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hAW00.png" alt="enter image description here"></a></p>
<hr>
<p><sup>1</sup> and therefore the Eurocode as well, since they're the inspiration for the Brazilian codes</p>
<p><sup>2</sup> That's my own literal translation from the Portuguese.</p>
<p><sub>Diagrams obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis program.</sub></p>
| 32513 | Deflection In Multistory Structure |
2019-12-25T02:09:47.683 | <p>My friend has proposed that the buildings in the images below have been sinking into the ground over time and that they're not planned with basements.</p>
<p>He thinks that the bottom of the building was once been at ground level and that some of them have windows and doors left from the past. I'm skeptical to this, I can't possibly think that a building can sink two floors down in the ground.</p>
<p>He's arguing that so many old buildings with basements have "misplaced" windows too close to ground level. I'm thinking that it's not possible to have windows or doors underground, as that would pose a risk for mold, water damage, and shattered windows, not to mention the fact that a building would sink 3 or more meters into the ground, how would the engineers not be able to design a foundation that is not able to support the weight of the building.</p>
<p>We really need an answer to this, we're not engineers, and can't find any information online. Please leave sources as well if possible, thanks!</p>
<p>Here is an image </p>
<p><a href="https://i.stack.imgur.com/TDCql.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TDCql.jpg" alt="enter image description here"></a></p>
<p>More images can be found at <a href="https://i.stack.imgur.com/ex9l3.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/ex9l3.jpg</a></p>
<p>EDIT: When doing a reverse image search for one of the images, it leads me to some conspiracy theory called mud flood, it's the most stark image of them all where it looks like the building has sunken 8 meters into the ground or more it can't have possibly been built like this. <a href="https://www.google.com/search?q=mud%20flood&hl=en-SE&sxsrf=ACYBGNRoTlujntkxR-pdhKhcoca-r2hFqg:1577290787017&source=lnms&sa=X&ved=0ahUKEwjT6MSImtHmAhXuAhAIHXkFAa8Q_AUIDSgA&biw=1858&bih=926&dpr=1" rel="nofollow noreferrer">Reverse image search on google</a></p>
<p><a href="https://i.stack.imgur.com/dzv8L.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dzv8L.jpg" alt="enter image description here"></a></p>
| |structures|building-design| | <p>The building in the first image is Vrubel Museum of Fine Arts in Omsk, Russia. In 2017 a restoration of the 0th floor has revealed an underground <a href="https://en.wikipedia.org/wiki/Long_gallery" rel="nofollow noreferrer">gallery</a> (not to be confused with an art gallery) that was used as storage space for goods in its initial years as a commercial center.</p>
<p><a href="https://i.stack.imgur.com/X9MBZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X9MBZ.jpg" alt="enter image description here" /></a></p>
<p>The windows are in service of a pretty clever <a href="https://en.wikipedia.org/wiki/Pavement_light" rel="nofollow noreferrer">system of prisms</a> adopted to light up the basement and said corridor gallery</p>
<p><a href="https://i.stack.imgur.com/JuNXT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JuNXT.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/6nLeK.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6nLeK.gif" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/qQqhg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qQqhg.jpg" alt="enter image description here" /></a></p>
| 32525 | Buildings sinking into the ground |
2019-12-27T23:47:05.573 | <p>I have the coordinates of 3 points thrgouh which, a circle should pass . Having the coordinates of the points in 3D, how could I have the coordinates of the center of circumscribed circle ?
also : if one of the points has some deviations and causes a circumscribed circle couldn't pass through the 3 points, is there a way to determine the required coordinates of the third point in a way that the circle could be constructed to find the deviation in space ?</p>
| |applied-mechanics|mechanisms|geometry| | <p>Solve the following four equations; here, A, B and C are the 3D-vectors of your three points. M is the center of the circle (what you are looking for), r is its unknown radius. x means "vector product".</p>
<pre><code>[1] (A-M)² = r²
[2] (B-M)² = r²
[3] (C-M)² = r²
[4] ((C-A)x(C-B))(C-M) = 0
</code></pre>
<p>The first three equations say that the distance from the each point to the center must be r. These three conditions will, in 3D, have as solution a straight line of the centers of all spheres that go through points A,B,C.</p>
<p>To find the center of a <em>circle</em>, one must force the solution to the plane where A,B,C lie. There are many ways to accomplish this; the following one seems the easiest one: First, one creates the plane's normal vector: Two vectors in the plane are C-A and C-B, and their cross product is a normal vector. To force C-M (and hence M) also into the plane, the scalar product with normal vector must be 0 (so that they are at right angles, i.e., the normal vector of the plane is orthogonal to C-M).</p>
<p>Example:</p>
<p>A=(1,1,1), B=(2,3,1), C=(3,3,2)</p>
<p>M=(x,y,z)</p>
<p>Here are the four equations:</p>
<pre><code>[1] (1-x)² + (1-y)² + (1-z)² = r²
[2] (2-x)² + (3-y)² + (1-z)² = r²
[3] (3-x)² + (3-y)² + (2-z)² = r²
(2,2,1) x (1,0,1) = (2,-1,-2), so
[4] 2(3-x) - (3-y) - 2(2-z) = 0
</code></pre>
<p>With all parentheses expanded:</p>
<pre><code>[1] 1-2x+x² + 1-2y+y² + 1-2z+z² = r²
[2] 4-4x+x² + 9-6y+y² + 1-2z+z² = r²
[3] 9-6x+x² + 9-6y+y² + 4-4z+z² = r²
[4] 6-2x - 3+y -4+2z = 0
</code></pre>
<p>To get rid of all the squares in [1]...[3], subtract subsequent equations:</p>
<p>[1]-[2] is a linear equation in x,y,z</p>
<pre><code>-3+2x -8+4y = 0
</code></pre>
<p>[1]-[3] is also a linear equation in x,y,z</p>
<pre><code>-8+4x -8+4y -3+2z = 0
</code></pre>
<p>[4] is anyway a linear equation in x,y,z</p>
<pre><code>6-2x - 3+y -4+2z = 0
</code></pre>
<p>This is now a simple system of three linear equations with three variables, which is easy to solve. The solution yields the coordinates of M:</p>
<pre><code>x = 13/6, y = 5/3, z = 11/6
</code></pre>
<p>r can be computed from any of [1]...[3]:</p>
<pre><code>r = square root of ((1-13/6)² + (1-5/3)² + (1-11/6)²) = square root of 5/2
</code></pre>
<p>And as joojaa pointed out above, you can always find a circle through 3 points (unless they are on a straight line, which will make the equations above unsolvable).</p>
| 32563 | How could I get the coordinates of the circumscribed circle giving the 3 points coords? |
2019-12-30T06:01:00.477 | <p>Around what percent of energy provided by gasoline is lost as heat due to rolling friction in vehicles? Is this number higher for larger vehicles likes trucks than smaller vehicles like bikes? </p>
<p>Ultimately, I want to know if energy loss due to ground friction is a problem and whether an attempt to decrease rolling friction can help to resolve the inefficiency? </p>
| |automotive-engineering|heat-transfer|friction|traction| | <p>There are 2 standards. One of them is called towing resistance and means obviously the slow towing force on flat solid pavement required to keep a loaded wheel rolling. </p>
<p>It is a fraction of wheel load and usually for cars and trucks it is in the range of 0.0045 to 0.008 and for bicycles with 120psi tire pressure 0.002. if we multiply this by 100 we get the percent factor.</p>
<p>Some tire manufacturers produce performance tires with less rolling resistance. </p>
<p>See the table in this Wikipedia article.</p>
<p><a href="https://en.m.wikipedia.org/wiki/Rolling_resistance" rel="nofollow noreferrer">Wikipedia rolling resistance.</a></p>
| 32579 | What is the average energy loss due to rolling friction in contemporary motor vehicles? |
2019-12-30T21:47:00.253 | <p>I've been reading the <a href="https://www.comsol.com/blogs/what-is-geometric-nonlinearity/" rel="nofollow noreferrer">Comsol post on geometric nonlinearity</a>. The article introduces situations where the small angle approximation and linear analysis of structures gives inaccurate results. For example, the case of a beam attached from both ends is discussed, specifically when the deformation is so large that the horizontal supporting force must be taken into account and the stiffness of the beam is no longer constant. Buckling also comes up. Here is a quote:</p>
<blockquote>
<p>Buckling, or the loss of stability when the load reaches a certain critical value, is caused by geometrically nonlinear effects.</p>
</blockquote>
<p><strong>Why is buckling caused by geometrically nonlinear effects?</strong> What is meant by this statement?</p>
| |structural-engineering|buckling| | <p>The real world doesn't know anything about approximations you make in mathematical models of it. If always behaves in a fully 3-dimensional way.</p>
<p>For Euler buckling of a column, you are really studying <em>perturbations</em> about the steady stress condition, which is just an axial direct stress in the column. You want to find the end load at which a small perturbation becomes unstable. </p>
<p>If you try to make a mathematical model that only includes first-order approximations to the behaviour of the perturbed structure, the results say that the column is stable for any compressive load. Since that doesn't correspond to what happens in real life, the over-simplified model is just wrong, and useless.</p>
<p>As an example of why first-order approximations don't always work, consider a bar of length <span class="math-container">$L$</span> along the <span class="math-container">$x$</span> axis, from <span class="math-container">$x = 0$</span> to <span class="math-container">$x = L$</span>, with no loads applied to it. The strain and stress in the bar is obviously zero.</p>
<p>Now, rotate the bar through 180 degrees, so the end at <span class="math-container">$x = L$</span> is now at <span class="math-container">$x = -L$</span>. Obviously the strain and stress in the bar is still zero. </p>
<p>But if you analyse this as a simple column, you have changed the original length of <span class="math-container">$+L$</span> to a length of <span class="math-container">$-L$</span>, creating an axial strain of <span class="math-container">$-2$</span>, and therefore there is a huge compressive stress in the bar. Clearly something went wrong with the math!</p>
<p>To make a model with the correct behaviour (or at least, <em>closer</em> to the correct behaviour than "completely wrong") you have to include second-order effects. For example, if the column is perturbed into a curved shape, <em>the point of application of the end load position moves corresponding to the deformed shape of the column.</em> If the end of the column moves "sideways" from the axis, that means the end load is now applying a bending moment to the base of the column as well as an axial load. </p>
<p>Also, since the perturbed shape of the column is a curve, and the length of the center line of the column does not change <em>measured along the curve</em>, that implies that the end of the column moves <em>axially</em> as well as sideways. That means that the applied load does work (= the appled force <span class="math-container">$\times$</span> the axial component of the displacement).</p>
<p>The problem with understanding most "simple" derivations of Euler buckling is that you meet them <em>before</em> you have learned the general concept of how to model arbitrary (large) displacements of a body. Therefore, the derivation often seems to be "pulling rabbits out of a hat" in order to get to the accepted answer, but doesn't provide a convincing explanation of why it was done that way.</p>
<p>An article like your blog link doesn't really help. It doesn't say anything that is <em>wrong</em>, but it can't replace a university-level course on continuum mechanics that is a follow up to the "small displacements and small strains" approximations that are made in most first courses.</p>
| 32583 | Why is buckling caused by nonlinear geometry? |
2019-12-31T03:59:09.550 | <p>I have tried to compare two courses, and there seems to be some overlap, but not as much. Which method is better when controlling a nonlinear system? Also, what is the main difference between the two?</p>
| |control-engineering|control-theory|optimal-control|nonlinear-control| | <p>In nonlinear control theory, you will recognize most concepts such as controllability and observability where the linear case is often a special case of the nonlinear case. I would highly recommend digging into linear control theory first if you have not done so. Depending on the course you take, concepts such as Lyapunov stability are discussed here, which is a very important concept.</p>
<p>In contrast, optimal control is rather independent of the underlying system, be it linear or nonlinear and has therefore overlap with both, depending on what type of optimal control problem you look at. I would say that it has a broader scope since you will find optimization problems everywhere in different fields of research. Especially important is model predictive control which solves optimal control problems on a receding horizon and is one of the most-used control schemes in the industry (apart from PID control) because it can handle state and input constraints explicitly (which most other types of control fail to achieve).</p>
<p>All in all, I would recommend going this path:</p>
<p>Linear control theory => Nonlinear control theory => Optimal control theory</p>
| 32586 | Which course to take? Optimal control? Nonlinear control? |
2020-01-02T06:23:34.000 | <p><a href="https://i.stack.imgur.com/xIzOe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xIzOe.jpg" alt=""></a> could there ever be a structure that would require something like this? </p>
| |mechanical-engineering|design|structural-analysis|steel| | <p>It looks like a mock-up models for advertising purposes.
Imagine placing one of these at the entrance of the bolt factory. hahaha</p>
| 32603 | Could a bolt this size actually be used on a project? |
2020-01-02T14:52:03.403 | <p>what is the physical interpretation of poles and zeros of a mechanical system?
I'm asking things like the meaning if I have only 0 or only complex numbers. And about the number of the poles and zeros, it should be equal? What can I infer about the stability?</p>
| |control-engineering|control-theory|applied-mechanics| | <p>The poles and zeros of a LTI plant or system in relation to their eigenvalues determine how quickly a system will stabilize or destabilize and if it oscillates. </p>
<p>When looking at these in the complex plane, the Re(a+ib) determine how quickly a system will exponentially stabilize or destabilize (the left vs right half plane) while the Im(a+ib) determine how the system oscillates. (High/low freq and amplitude)</p>
<p>The poles and zeros do not have to be equal, however imaginary systems always appear in pairs namely a+ib and a-ib. </p>
<p>A system is absolutely stable if all poles are at the left half plane =<0. </p>
<p>Other stability descriptions can be named depending on specific criteria such as conditional, marginal, or unstable or chaotic depending on where poles lie. </p>
<p>You can find a lot of infos on the web about this, such as t<a href="https://web.mit.edu/2.14/www/Handouts/PoleZero.pdf" rel="nofollow noreferrer">his pdf from MIT.</a> (warning pdf)</p>
| 32608 | what is the physical interpretation of poles and zeros of a mechanical system? |
2020-01-02T14:53:38.167 | <p>This is an HVAC mechanical engineering question on a document from a manufacture where they convert units of electrical energy to therms of natural gas.</p>
<p>Carrier has writeups for building energy simulation modeling, and <a href="https://www.shareddocs.com/hvac/docs/1004/public/0a/hap_ehelp_031.pdf" rel="nofollow noreferrer">this link</a> is a PDF for a writeup using the HAP software.</p>
<p>Question:
The example on page 2 as a high level feasibility they convert units of NG to kWh:</p>
<blockquote>
<p>In a certain locale the price for electricity is 0.0960 dollar per
killowatt hour and the cost of natural gas is 0.937 dollar per therm.
If we convert the gas price to use units of kWh, we get a gas price of
0.0320 $/kWh. Therefore a kWh of electricity is three times as expensive as a kWh of natural gas. If the average annual COP of the
DHRC is more than three times that of the gas boiler, then heating
cost will be reduced by using the DHRC. If the COP of the DHRC is less
than three times that of the boiler, heating cost will increase. Note:
This is a single example for illustration. Energy prices vary widely
by location and over time. You should use your own local utility
prices to assess heat recovery viability for your projects.</p>
</blockquote>
<p>I know 1 therm = 29.3001 kWh or 1 kWh = 0.03413 therm.</p>
<p>But what’s the math behind coming up with “convert the gas price to use units of kWh, we get a gas price of 0.0320 $/kWh”?</p>
| |mechanical-engineering|hvac| | <p>I think the problem is just that you're expecting the solution to be more elegant; it looks like you already have it pretty much figured out.</p>
<p><a href="https://i.stack.imgur.com/rOzlo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rOzlo.png" alt="equation with terms dropped out"></a></p>
| 32609 | energy unit conversions |
2020-01-02T17:14:21.543 | <p>I have a workflow of two functions. Both of them aggregate the items that are receiving and only the first is adjustable. I mean that I can decide how many items to aggregate on the first function <code>F1</code> in order to not exceed the capacity (bandwidth) of the channel that connects to the second function <code>F2</code>. In other words, I control the throughput that the second function is receiving items. If I aggregate more at <code>F1</code> I decrease the throughput. If I aggregate less at <code>F1</code> I increase the throughput. My goal is to tune the system for its maximum throughput while the input workload varies and I have a limited channel between <code>F1</code> and `F2. This it the picture of my system:</p>
<p><a href="https://i.stack.imgur.com/qSNe6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qSNe6.png" alt="enter image description here"></a></p>
<p>The data stream of items (workload) that <code>F1</code> receives varies along the time. I want to control the throughput output of <code>F1</code> in order to not surpass the channel capacity.</p>
<p>I am planning to use PID control to decide how many items to aggregate on the first function before to emit the items to the second function. I am a newbie in control systems. As far as I understood I have to start with the Proportional controller (P) which is basically the current error. Then I have to decide if I will use the Integral controller (I) which is the historic cumulative value of the errors. And then I have to decide if I will use the Derivative controller (D) which is based on its current rate of change.</p>
<p>For me, it is not clear yet if I really need the Integral and Derivative controllers. I mean, if I find the error between the actual throughput and the desired throughput I just set the first aggregate function to aggregate items in order to have the desired throughput. In my mind this is enough, isn't it?</p>
<p>Something could influence is the time that I take to aggregate items on the first function. Suppose that I have a slow processor. But, in my use case what is the insight that I have to have in order to use the Integral or the Derivative controllers?</p>
<p>Thanks,
Felipe</p>
| |control-theory|pid-control| | <p>Even if you can tune a function as per your needs, using proportional control(P) only you can never fully eliminate error from the system. Error is inherent in P control.</p>
<p>Using I control along with P, you can fully eliminate the error in the system, but now your system has a specific transient response. D control may be used if you want to control the transient response of the system, you want it faster or slower, how much overshoot you can manage to achieve a faster response, that depends on your system and need.</p>
<p>The amount of error in the system using only the P control depends on system parameters like the OpAmps you might be using. So if the error is tolerable, you need not add I and D. But if you need zero error at least I is needed, while D will let you vary the speed of the response</p>
| 32615 | Is it necessary to use PID control in my use case? |
2020-01-03T00:30:42.207 | <p>I am interested in putting a small (3"x3"x2") assembly under X-ray, following a shock test, in order to evaluate failures of an internal ceramic piece. Unfortunately, this assembly utilizes quite a bit of potting material (epoxy) and as such, cannot easily be disassembled. </p>
<p>I'm considering reaching out to some local dentist practices to see if I can purchase X-ray time from them, however, I'm concerned that the material makeup of my assembly will not allow adequate penetration and will result in a poor image. </p>
<p>The general makeup of the assembly from the outside and moving inwards is...</p>
<ul>
<li>0.125" thick aluminum outer casing</li>
<li>0.25" to 0.5" foam-like epoxy </li>
<li>some small (but strong) magnets</li>
<li>a ceramic "skeleton" structure (this is what I'm most interested in evaluating for failures)</li>
</ul>
<p>Should note that the magnets are not necessarily in the line-of-sight of the ceramic, so 'penetration-ability' of those is a non-issue.</p>
<p>However, regarding the aluminum, with a HVL of 0.06" at 120 keV (<a href="http://www.sprawls.org/ppmi2/RADPEN/" rel="nofollow noreferrer">source</a>), I'm worried that I may not be able to sufficiently penetrate the outer shell and produce a good image.</p>
<p><strong>MY QUESTIONS ARE:</strong> </p>
<ul>
<li><strong>Penetration aside: Is it reasonable to suspect that I will be able to gather useful information and identify fractures in the ceramic?</strong></li>
<li><strong>Google tells me that the typical energy capability of a dentistry x-ray is <200 keV. For someone with experience in the radiation field, do you expect this would be sufficient to penetrate my test article and produce a good image?</strong></li>
</ul>
| |structural-analysis|failure-analysis| | <p>X-rays are a poor way to test for cracking in a ceramic. A crack in a ceramic part which has not shattered to pieces is a very small feature, of order ~0.001" wide or less, which commonly-used x-ray equipment cannot resolve. Second, a tiny crack in a ceramic part does not present a significant difference in x-ray absorptivity, inelastic scattering, or diffraction between the cracked and uncracked regions to the incoming x-rays which means that a tiny crack will not produce <em>x-ray contrast</em> in the imaging apparatus. </p>
<p>In addition, the materials which enclose your ceramic part will have the effect of attenuating and scattering the incoming x-rays which degrade their ability to resolve useful contrast in that portion of the "sandwich" you need to find cracks in. </p>
<p>In light of this, you would be better off trying another sort of nondestructive testing technique than x-radiography.</p>
| 32621 | Determining If A Test Article Is Suitable For X RAY Following Shock Testing? |
2020-01-03T12:44:13.660 | <p>I am using Nastran for buckling analysis, and I'm new to all of this so please bear with me. </p>
<p>For a linear buckling analysis, I understand that you must first apply any compressive load in the static subcase. Then solve the eigenvalue buckling problem in the second subcase using the first static subcase. The buckling load is then the amount of compressive load applied multiplied by the resulting eigenvalue.</p>
<p>However, I now want to apply a second fixed force of exactly 50N laterally and see what the buckling load would be with this additional 50N force. Now if I apply both the compressive load and the lateral load to the static subcase, and then solve the eigenvalue buckling using that subcase, the resulting buckling load varies a lot depending on the compressive load applied in the static subcase. </p>
<p>Example 1:</p>
<ul>
<li>Apply 1N compressive force, eigenvalue is 100. So buckling load is 1N*100 = 100N</li>
<li>Apply 10N compressive force, eigenvalue is 10. So buckling load is 10N*10 = 100N
This makes sense, however:</li>
</ul>
<p>Example 2:</p>
<ul>
<li>Apply 1N compressive force and 1N lateral force, eigenvalue is 90. So buckling load is 1N*90= 90N.</li>
<li>Apply 10N compressive force and 1N lateral force, eigenvalue is 9.5. So buckling load is 10N*9.5=95N, which is higher than 90N. </li>
</ul>
<p>This just proves that I am doing something wrong, and I am hoping someone can clarify what it is.
Any help would be appreciated, thank you!</p>
| |mechanical-engineering|structural-engineering|finite-element-method|buckling|nastran| | <p>In your example 2, Nastran is scaling <em>both</em> loads as part of the buckling load. </p>
<p>In your first case, you have buckling with <span class="math-container">$90\times 1 = 90$</span>N axially plus <span class="math-container">$90\times 1=90$</span>N laterally.</p>
<p>In the second, you have buckling with <span class="math-container">$9.5\times 10 = 95$</span>N axially plus <span class="math-container">$9.5\times 1 = 9.5$</span>N laterally.</p>
<p>From example 1 you have a buckling load of 100N axially plus zero laterally.</p>
<p>You could plot a graph with those three points, and estimate the axial load when the lateral load is 50N. Then set up the Nastran input for those values and repeat if necessary.</p>
<p>Alternatively, since you have now bracketed the answer between 90N and 95N, you could run analyses with 91+50, 92+50, 93+50 and 94+50, and estimate when the buckling factor is exactly 1.0.</p>
<p>In a general situation, this sort of loading is likely to be incompatible with the assumptions of linear buckling. You could model it as a nonlinear analysis. First ramp up the side load from 0 to 50N, then increase the axial load and see what happens. You may find the structure doesn't "buckle" in the sense of a catastrophic instability, but just gets more flexible as you increase the axial load and lateral deflection increases.</p>
| 32626 | How would I do a buckling analysis with an additional fixed known lateral force? |
2020-01-03T20:17:34.063 | <p>Consider the following problem from Hibbeler's Engineering Mechanics, Statics. (13th edition)</p>
<p><a href="https://i.stack.imgur.com/rwIjV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rwIjV.png" alt="enter image description here"></a></p>
<p>and the associated free body diagram and equilibrium equations</p>
<p><a href="https://i.stack.imgur.com/Lv0Wh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lv0Wh.jpg" alt="enter image description here"></a></p>
<p>The problem is solved by equating <span class="math-container">$N_B$</span> to zero. Let's say I was asked to determine all support reactions and the force in cable CD (as in the picture, not requiring the ramp to just start being lifted) It seems that 3 equilibrium equations are not sufficient, as there are 4 unknowns <span class="math-container">$F\, ,N_B\,,A_x\,and\, A_y $</span> </p>
<p>What renders this statics problem unsolvable, is it statically indeterminate?</p>
| |statics| | <p>Imagine there is no rope - determinate.
Imagine there is no floor - determinate.</p>
<p>As you gradually increase tension in the rope the reaction force at B will gradually fall as the tension rises, until it reaches zero. If the tension rises further it will start to move. So, you see the reaction force is directly driven by the tension, provided the tension is with the range between zero and when the bridge starts to lift.</p>
<p>So yes, it's indeterminate in this interim state, but there is not any reason to find the static equilibrium forces in this state since the worst case forces are in the two determinate states that I described above.</p>
| 32634 | Statical equilibrium in 2D |
2020-01-04T12:02:26.363 | <p>I recently visited a railway museum with a lot of Soviet-era locomotives. The picture below shows the wheels on a locomotive from the 1930s or 1940s. Some of the wheels are thicker on one side (as it it were a partially filled cup). What is the purpose of that?</p>
<p><a href="https://i.stack.imgur.com/JbStW.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/JbStW.jpg" alt="Picture of the wheels of a Soviet-era locomotive"></a></p>
<p>(Now that I am looking at the picture again I wonder if they are balancing the weight of the coupling rods in some way.)</p>
| |rail|wheels| | <p>Slightly different answer - aside from the hammer effect of having out of balance connecting rods, there is also a different problem set. </p>
<p>Since the left and right hand rods are at connected at 90 degrees wrt each other (deliberately, as this is how you avoid the dead center starting problems referred to above), what happens is that as they rotate, the whole engine starts to sway from side to side due to the unbalanced rotating load. If you happen to hit a resonant frequency with the rails, the engine (and train) can derail entirely. The faster you go, and the more driving wheels that the engine has, the worse the problem. 2-10-4 ("Texas"class) locomotives were notorious for having this problem, for example.</p>
<p>Finally, having one side 'pushing' whilst the other is 'pulling' can cause the axle boxes to work loose and this will severely increase wear on the axle boxes, wheels and brakes etc and in extreme cases, it can cause the loco chassis to fail (bend or even break) due to repeated flexing. Since both Berkshire (2-8-4) and Texas locomotives were designed to pull heavy freight loads, this can also cause the loco to 'fishtail' at any speed, which causes (different) problems again.</p>
<p>Rick</p>
| 32639 | What is the purpose of these “partially filled in” locomotive wheels? |
2020-01-05T03:59:52.357 | <p>I want to do a non-linear analysis (possibly SOL 400) on Nastran. I used PyNastran (python module) to set-up the BDF file for a Linear buckling analysis (sol 105). However I feel like it might be too difficult to set everything up correctly with non-linear analysis due to the little information I found. Tutorials on the subject all recommend a pre-processor like Patran, which I don't have. </p>
<p>Does anyone know a free pre-processor that can actually produce the nastran input file (BDF)?</p>
<p>I know of salome-platform.org, but my understanding is that it cannot produce the BDF file. Or is there another solution to my problem?</p>
<p>Any help to solve my problem would be appreciated. Thanks!</p>
| |structural-engineering|structural-analysis|applied-mechanics|finite-element-method|nastran| | <p>As far as I know, there is no free preprocessor for Nastran.</p>
<p>According to the docs (<a href="https://docs.salome-platform.org/5/smesh/user/importing_exporting_meshes_page.html" rel="nofollow noreferrer">https://docs.salome-platform.org/5/smesh/user/importing_exporting_meshes_page.html</a>) Salome will export to Nastran format, but it's just the mesh (nodes and elements). It won't do the case controls and all the nonlinear solution control parameters and all the other things in the deck</p>
<p>If you want the full deck, I think your options are patran, hypermesh, NX (or write the deck yourself in a text editor... That's free but big learning curve) </p>
| 32649 | Free pre-processor for Nastran? |
2020-01-06T12:21:20.040 | <p>Suppose I have a cogeneration plant that runs on natural gas and produces heat and electricity. The heat effeicieny is <span class="math-container">$\eta_{th} = 0.4$</span> and the electricity efficiency is <span class="math-container">$\eta_{el} = 0.4$</span>.</p>
<p>The burning of natural gas produces 215 g/kWh of CO2.</p>
<p>For the calculation of the CO2 emission of the generated electricity from the cogeneration plant, one could now calculate <span class="math-container">$\frac{215}{0.4} = 537.5$</span> g/kWh.</p>
<p>However, I would like to calculate the CO2 emissions for the electricity generation of the cogeneration plant, taking into account that CO2 emissions are saved because another heat generator has to generate less heat.</p>
<p>A natural gas boiler with an efficiency of <span class="math-container">$\eta_{th} = 0.9$</span> is taken as a reference for heat generation. So the 1 kWh of heat is generated with <span class="math-container">$\frac{215}{0.9} = 238.88$</span> g of CO2.</p>
<p>How do I calculate the CO2 emissions for the electricity generation in this case?</p>
<hr>
<p>My calculation attempt:</p>
<p>First I would calculate how much CO2 emssions the boiler has for 0.4 kWh:</p>
<p><span class="math-container">$238.88 \cdot 0.4 = 95.55$</span> g of CO2</p>
<p>Then I would subtract this value from the previously calculated co2 emission of the electricity:</p>
<p><span class="math-container">$537.5 - 95.55 = 441.95$</span></p>
<p>The electricity generated by the cogeneration plant would then have a CO2 emission of 441.95 g/kWh.</p>
| |electrical-engineering|energy|heating-systems|energy-efficiency|power-engineering| | <p>Your basic logic is sound: You compare the case of the cogeneration plant + boiler vs. boiler alone. Your math looks correct to me.</p>
<p>Where did you get the value 215 g/kWh from? All (german language!) sources I found give 202g/kWh for natural gas, which may well be because actual of differences in grid gas composition. When I do the math myself I arrive at 197 g/kWh if using the lower heating value, only 178 g/kWh for the HHV. </p>
<p>The <a href="https://en.wikipedia.org/wiki/Heat_of_combustion" rel="nofollow noreferrer">difference between HHV and LHV</a> could be crucial since many boilers today are condensing boilers (that utilize the heat of evaporation of the exhaust water) but few to none cogeneration plants are, so you should possibly use different carbon intsities for both.</p>
<p>Lastly, the effiencies for the cogeneration plant don't look like real life numbers but are not too far of (I'm just now looking at a datasheet for a small CHP with 38% and 43% respectivly, in my experience this is a rather high <span class="math-container">$\eta_{el}$</span>)</p>
| 32664 | Calculation of the specific CO2 emission of a cogeneration plant for the produced electricity |
2020-01-06T15:33:44.703 | <p>Recently I have seen a question as follows:</p>
<p>Dimension of the hole is <span class="math-container">$50 +.02/-.00$</span> mm and shaft is <span class="math-container">$50 +.02/-.00$</span> mm.The minimum clearance is?</p>
<p>My answer is .00 but the correct answer as per the website is -.02</p>
<p>Where am I going wrong?</p>
<p>Please help me.</p>
| |tolerance| | <p>The minimum clearance occurs when the largest possible shaft meets the smallest possible hole. That would be in this case a shaft of <span class="math-container">$50.02 mm$</span> in a hole of <span class="math-container">$50.00 mm$</span>. The difference between the two is the <span class="math-container">$-0.02 mm$</span> from the solution you found.</p>
| 32667 | Difference between min/max clearance and min/max tolerance and relation between them |
2020-01-07T01:59:22.180 | <p>I am trying to calculate the maximum braking force/deceleration for a bicycle.
Note: the question assumes <strong>only the front brake is used</strong></p>
<p><strong>Free body diagram:</strong></p>
<p><a href="https://i.stack.imgur.com/qlNjm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qlNjm.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/pYLXE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pYLXE.png" alt="enter image description here" /></a></p>
<p><strong>Typo:</strong> in the image above, "<span class="math-container">$F_{a,\ max(dry)} = g X_3 / h$</span>" is wrong, since it is an acceleration and it should be written as "<span class="math-container">$a_{max(dry)} = g X_3 / h$</span>"</p>
<p>For dry tarmac where I know there is enough grip to get the rear wheel to lift I managed to calculate the max deceleration (result=6.8m/s^2). (this is the deceleration where all weight is transferred onto the front wheel (rear wheel will start lifting up at higher deceleration)).</p>
<p><strong>From the moment equilibrium equation (using the front tire contact patch as the origin) I got the following equation for maximum deceleration before the rear tire starts lifting off (assuming there is enough grip):</strong>
<a href="https://i.stack.imgur.com/1cgVz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1cgVz.png" alt="enter image description here" /></a><br />
I used this to calculate max deceleration in dry conditions</p>
<p>However I am now trying to calculate the max deceleration in wet weather (lower coefficient of friction) where the front wheel slipping should be considered as an option.</p>
<p><strong>I tried expressing Nf (normal force on front wheel contact patch) in terms of a (deceleration) and got the following result:</strong></p>
<p><a href="https://i.stack.imgur.com/5y8ht.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5y8ht.png" alt="enter image description here" /></a><br />
Which when simplified results in:</p>
<p><a href="https://i.stack.imgur.com/fnwFD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fnwFD.png" alt="enter image description here" /></a><br />
With this formula I can calculate Nf</p>
<p>Assuming that there won't be enough grip (in the wet) to make the rear wheel lift up.. I can then use Nf to determine Fbr (braking force).
To determine maximum deceleration (assuming wheel slip occurs before rear lift up) I can use the equilibrium of forces in x-direction:</p>
<blockquote>
<p>ΣFx=0=Fa-Fbr =m<em>a-μ</em>Nf --> a=μ*Nf/m</p>
</blockquote>
<p>My questions:</p>
<ul>
<li>Are my assumptions/calculations correct? If so: is there a better/faster way of doing this?
To check if the rear wheel lifts up or if the tire slips first I will compare the max deceleration for slipping tire scenario and max deceleration for lifting rear wheel scenario and see which deceleration has a lower deceleration value to see which scenario will happen first</li>
</ul>
<p>When braking some weight is transferred from the rear to the front wheel, but if the deceleration is too high the front wheel will slip.
Gently increasing brake pressure will lessen the chance of front wheel slipping...</p>
<p>How should I go about determining max deceleration without the front wheel slipping, given a certain friction coefficient μ ?
I can calculate the max braking force without slipping for a given Normal force on the front wheel but <em><strong>as soon as you start braking this normal force changes due to weight redistribution</strong></em> and I don't know how to account for this.</p>
<p>Thank you very much for any input! It is greatly appreciated</p>
| |statics|dynamics|friction|bicycles| | <p>At a quick readthrough, your calculations seem to be correct and they also seem to answer the hard parts of your question. I also doubt the calculations can be made significantly simpler than your version.</p>
<p>I did a recalculation of the "wet condition" as follows:</p>
<ul>
<li>With no lift of the rear wheel, there will moment equilibrium around the center of mass, i.e. <span class="math-container">$N_R x_2 + F_{BR}h = N_F x_3$</span></li>
<li>With no vertical acceleration of the rear wheel, we also have vertical equilibrium, i.e. <span class="math-container">$F_G = N_R + N_F$</span></li>
<li>Assume we are breaking hard enough to reach the friction coefficient of wheel and ground, <span class="math-container">$F_{BR} = \mu N_F$</span></li>
<li>Solving for the front wheel normal force gives <span class="math-container">$N_F = \frac{F_G x_2}{x_1-\mu h}$</span> as you already got. This is normal force adjusted for weight redistribution caused by braking.</li>
<li>So to calculate the corresponding deceleration, simply check the assumption that the rear wheel stays on the ground, <span class="math-container">$N_F < F_G$</span> and calculate <span class="math-container">$a=\frac{\mu N_F}{m}$</span>.</li>
</ul>
| 32674 | Calculate max deceleration without front wheel slipping (bicycle) |
2020-01-07T08:33:41.830 | <p>Is it possible to model a heat pump cycle without using tools like EES, but using Excel? I am considering to use Refprop application as an add-in in Excel, and model a heat pump cycle. Another alternative would be to create a P-h database for various fluids, and go from very complicated series of equations to find pressure, temperature, and density values.</p>
| |thermodynamics|modeling|refrigeration| | <p>You can do this.</p>
<p>I set up a sheet using vlookup to interpolate through fluid characteristics and because the source data I used (from steam tables) had a 5 degree increment I had to build an interpolation function using several vlookups / I will let you do that as it is good experience.</p>
| 32679 | Modelling a refrigeration cycle in Excel |
2020-01-07T18:34:09.180 | <p>Is there any meaning of using Kalman Filter for cases when you do not have good statistical model of the system?</p>
<p>For example, if you have a drone and it has IMU sensor and GPS sensor. You do not have any statistical information, like variance, covariance, etc. If you want to estimate your current position, is there any advantage of using Kalman filter over just averaging the two sensors? </p>
<p>For the above problem, if in practice the Kalman filter has some advantage, what are the reasons behind it that I did not consider in my above example?</p>
<p>Thanks in advance</p>
| |sensors|robotics|kalman-filters| | <p><strong>You can't just "average the two sensors"</strong> </p>
<p>The acceleration readings are essentially meaningless unless you know their direction, you can't know their direction unless you have the platform orientation, and you can't know the platform orientation from the IMU (unless it has a <em>reliable</em> magnetometer).</p>
<p>For the IMU-alone case, your system basically has the information that you'd have if you were stuck inside of Einstein's elevator. The system knows -- with errors -- how rapidly its velocity (not position) is changing, and how rapidly -- with errors -- it is turning. That's <em>all</em>. </p>
<p>The best that you can do is to:</p>
<ul>
<li>Maintain a 3D orientation estimate by integrating the gyro readings -- basically this gives you a rotation from your vehicle frame to North East and Down, or whatever your preferred "global" reference frame is.</li>
<li>Maintain a 3D velocity estimate in your global reference frame by rotating the accelerometer readings into the global frame with the orientation estimate, and then integrating.</li>
<li>To complicate things, you need to know your local acceleration due to gravity and you need to correct for it in the vertical direction -- so any errors in your estimate of "down" translate immediately into an offset in the effective acceleration reading in the North or East directions.</li>
<li>Maintain a 3D position estimate in your global reference frame by integrating the velocity readings.</li>
</ul>
<p>You can say at this point "Yes, but I'm going to keep everything in the vehicle frame" -- fine, but life will be just as complicated and inaccurate. You're not going to gain any information by doing so.</p>
<p>Now you have a 9-state estimate (rotation, velocity, position, each in 3D), with a rather nasty nonlinear relationship between rotation and the rest. </p>
<p>This means that any error in starting state, plus noise, drift or offset in the IMU readings, gets integrated multiple times. It's pretty much twice for accelerometer readings, but rather complicated for gyro readings because the orientation estimate is used to rotate the accelerometer readings. It is very difficult to do. In general, with typical consumer-grade IMUs, the state estimate will be utterly useless within seconds, <strong>if</strong> it starts out perfect.</p>
<p>There actually <em>are</em> (or at least were) purely inertial navigation systems in the 1960's and 1970's that used this technique -- but they needed hugely expensive IMU hardware, they needed very careful pre-launch calibration, they needed maps of the local graviometric variations for their route, and even the ones in strategic bombers and submarines were backed up by people (or machines, in the case of nukes) taking celestial readings (star or sun shots with a sextant).</p>
<p>For typical consumer-grade IMU hardware, with perfect starting calibration, you can free-run for maybe a second or so before the error gets into the meters.</p>
<p><strong>Need for a statistical model</strong></p>
<p>In the specific case of a drone (or anything else) with an IMU and GPS, you can side-step the need for a statistical model of the vehicle, or at least you can make a model without prior knowledge of the vehicle's motion. I've done this, and it works.</p>
<p>Basically, you treat the IMU output as the system input (just treat the system as having mass subject to accelerations and rotations), then you compare the system's position to the GPS readings.</p>
<p>So your model is <span class="math-container">$\dot x = f \left ( x, \hat u \right)$</span>, <span class="math-container">$y = x_{position}$</span>, where <span class="math-container">$\hat u(t)$</span> is the output of the IMU. (Note that this is the state-space representation for a system that is nonlinear in the state evolution, but linear in the output). </p>
<p>The only statistical variation you need to account for is the IMU and GPS error. For the IMU you let <span class="math-container">$\hat u(t) = u(t) + w(t)$</span>, where <span class="math-container">$w(t)$</span> is your expected IMU noise, and <span class="math-container">$u(t)$</span> is the actual body motion. Similarly, for the GPS you let <span class="math-container">$\hat y_{position} = y_{position} + r(t)$</span>, where <span class="math-container">$y_{position}$</span> is your actual position, <span class="math-container">$\hat y_{position}$</span> is your actual GPS reading, and <span class="math-container">$r(t)$</span> is your expected GPS error.</p>
<p>So there is a "statistical model", but it's one that you can gain from a knowledge of the properties of the IMU and of GPS.</p>
<p><strong>Orientations from the fusion</strong></p>
<p>Because you're keeping track of everything with that covariant matrix, the Kalman filter will sort out varying accelerations into the platform orientation, position, and velocity. It doesn't actually have to know anything about the platform dynamics, because those are being directly read by the IMU.</p>
<p>You can reason this out intuitively. When you add GPS into the mix, the system is now in the same situation as you would be if you were stuck inside of Einstein's elevator <em>but with a position readout</em>. If you're in Einstein's elevator <strong>and you know that you are not moving relative to the earth</strong> then you know that the "down" that your accelerometer is telling you is also "true down".</p>
<p>Now you know which way is up. Let's say that you're watching your position readout, your gyro, and your accelerometer. You notice some tugs on the elevator -- your gyro says you haven't moved, the accelerometer vector has tilted toward the front wall and then the back, and the GPS is now reporting that you are some distance east of where you started. <strong>Now</strong> because you know that you moved straight east <strong>and</strong> that the sideways acceleration was in the direction of the front of the elevator. So you can face the front of the elevator, say "I am pointing to the East", and you have successfully oriented yourself.</p>
<p>This is what the Kalman filter is doing with it's seemingly inadequate statistical model -- if you get <span class="math-container">$f(x, u)$</span> right, because of the way the covariant matrix ends up evolving, just sitting still will rapidly let the Kalman know which way is down (because it's not moving, yet the accelerometer is reporting acceleration "upward"). Then, if the Kalman sees a changing lateral acceleration coupled with motion in the GPS reporting, it'll deduce lateral direction. Moreover, this will continue to happen -- the combination of GPS and acceleration will <em>always</em> give information about orientation along the line of acceleration, so if the acceleration direction continually changes, the Kalman filter will know orientation.</p>
<p>There's probably refinements that could be had by actually modeling the platform dynamics and using the control inputs to the platform to increase the information flowing into the filter -- but I've done it the way I describe above, and it worked great (after considerable work).</p>
| 32685 | Kalman filter for sensor fusion — what is the advantage? |
2020-01-07T18:44:24.737 | <p>I have received a CO2 cylinder from an online seller fitted with a protective cap over the valve:</p>
<p><a href="https://i.stack.imgur.com/COqRM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/COqRM.jpg" alt=""></a></p>
<p>I understand these should unscrew, but this one appears to be stuck very firmly. Is there a standard or recommended way to remove these safely? The biggest challenge seems to be securing the cylinder itself to stop it slipping while attempting to turn the cap with a strap wrench or similar. Any tips?</p>
| |mechanical-engineering|safety|pressure-vessel| | <p>It’s all about leverage in this case. I used the crowbar (the hook side with the split hammer claw) Insert the split into the cap hole and create a leverage by pulling on the end of the crowbar. The longer the crowbar the easier to unscrew. Dude who took an angle grinder is clearly on a death wish.</p>
| 32686 | How to remove protective valve cap from CO2 cylinder? |
2020-01-08T09:46:37.417 | <p>Two reservoirs are connected by two pipes with different diameters in series. The elevation of Reservoir A is 40m and Reservoir B is 35m to respective water surfaces. The first pipe is 500 m long with a diameter of 0.075m and the second pipe is 2000m long with a diameter of 0.1m. What is the flow between the two reservoirs? The absolute roughness is 0.01 for both pipes and teh pipe elevation is constant.</p>
<p><a href="https://i.stack.imgur.com/XSxpG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XSxpG.png" alt="Schematic of the Problem Outlined"></a></p>
<p>What I have tried so far is by trial and error in the Moody Diagram find a Friction Factor to use with the Darcy-Weisbach Equation. However, I cannot get a number matching my first guess of the Reynolds Number. The pipes have different diameters so that means I have to do a trial for each pipe. Also when calculating I cannot use the Head loss of 5m but of 1m since that is the correspondent Head Loss for that length of pipe? How do I go about solving for the problem above?</p>
<p>Thank you for your help.</p>
| |civil-engineering|hydraulics| | <p>Here's how I'd attack the question:</p>
<ul>
<li><p>write down the Darcy Weisbach equation, with two terms on the right sides for the two different pipes. Head losses add up.</p></li>
<li><p>exchange flow velocity for a term dependent on volume flow and pipe diamter in both terms on the right side.</p></li>
<li><p>The pipe is really long and really thin and the head is really small. Start by using the laminar flow friction factor (64/Re)</p></li>
<li><p>Because of this you could solve this algebraically but I'd use any numerical tool (excel target value search is fine). </p></li>
<li><p>Check the actual resultant Re in both pipe segments, if the assumption (laminar flow) holds, if not take the friction factor from the moody chart and reiterate.</p></li>
</ul>
| 32696 | Determining steady flow between two reservoirs with two different pipes in between |
2020-01-09T23:27:11.607 | <p>I see these frequently on tall power lines like the one in the image below, sometimes a single one at each end, sometimes pairs as are shown here. Sometimes on one side of the tower but not the other.</p>
<p>I vaguely remember reading once that they are used to reduce swinging in high winds, and where I live typhoons come regularly.</p>
<ol>
<li>What are these called?</li>
<li>Are they used to reduce swinging in high winds?</li>
<li>How do they do this? </li>
<li>When is a pair used instead of a single; why not just use a bigger one?</li>
</ol>
<p>From <a href="https://www.wired.com/story/iran-apt33-us-electric-grid/" rel="nofollow noreferrer">Wired</a> original: <a href="https://i.stack.imgur.com/pLy9u.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/pLy9u.jpg</a></p>
<p><a href="https://i.stack.imgur.com/ejd3a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ejd3a.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/QJWgN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QJWgN.png" alt="enter image description here"></a></p>
| |mechanical-engineering|fluid-mechanics|vibration|aerodynamics| | <p>In order to supplement <a href="https://engineering.stackexchange.com/a/32714/6264">@fred_dot_u's answer</a> and add a few details on the nature of the problem that these devices address I'll add a bit of additional material.</p>
<p>For a description of a numerical model of how Stockbridge dampers work and ANSYS simulation results, see for example <a href="https://journals.sagepub.com/doi/pdf/10.1177/0954406212452064" rel="nofollow noreferrer">Aeolian vibration of a single conductor with a Stockbridge damper</a>, Oumar Barry, Donatus CD Oguamanam and Der Chyan Lin, Proc IMechE Part C: J Mechanical Engineering Science, 227(5) 935–945, DOI: 10.1177/0954406212452064</p>
<hr />
<p>Also, from J-L Lilien's <a href="http://www.tdee.ulg.ac.be/userfiles/file/Vibrations_eoliennes_intro.pdf" rel="nofollow noreferrer">Power Line Aeolian Vibrations</a>.</p>
<blockquote>
<p>Aeolian vibration is a low amplitude (conductor diameter) high frequency (5 to 150 Hz) phenomenon. Aeolian vibration is one of the most important problems in transmission lines because it represents the major cause of fatigue failure of conductor strands or of items associated with the support, use, and protection of the conductor. In this phenomenon, conductor strand fatigue failures occur at the suspension clamps or at the clamps of the other devices installed on the conductor such as spacers, spacer dampers, dampers and other devices.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/KiBEL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KiBEL.png" alt="Fig 2.1 typical broken strand due to aeolian vibration after removal of suspension clamp" /></a></p>
<blockquote>
<p>Fig 2.1 typical broken strand due to aeolian vibration after removal of suspension clamp.</p>
<p>Forces induced by vortex shedding are the cause of this type of vibration (Blevins1990, Buckner 1968, Claren et al 1969 & 1974).</p>
</blockquote>
<hr />
<p><a href="https://i.stack.imgur.com/kWrP7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kWrP7.png" alt="Fig 2.2 Flow visualization of vortex shedding" /></a></p>
<blockquote>
<p>Fig 2.2 Flow visualization; identification according to (row, column): (1,1) R=1.1 (Taneda), (1,2) R=9.6 (Taneda), (2,1) R=26 (Taneda), (2,2) R=140 (Taneda), (3,1) R=2000 (Werlé & Gallon), (3,2) R=10000 (Corke & Nagib)</p>
</blockquote>
<hr />
<blockquote>
<p>The fatigue mechanism</p>
<p>The fatigue in power line conductors are caused by fretting between different layers
of strands at their contact area (fig 5.1). Some cracks appear and propagate. The
stick/slip elliptical contact area is depending on the radius of curvature taken by the
conductor. During alternating bending stress (due to oscillations) that radius of
curvature is changing continuously. It is particularly important near clamp ends at the
last point of contact.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/rZXd9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rZXd9.png" alt="Fig 5.1 the inter-strand contact area where first crack may appear (Epri-1979)" /></a></p>
<blockquote>
<p>Fig 5.1 the inter-strand contact area where first crack may appear (Epri-1979)</p>
</blockquote>
<hr />
<p>From <a href="http://preformed.com/images/pdfs/Energy/Transmission/Motion_Control/VORTX_Vibration_Damper/EN-ML-1007-4AeolianViBook.pdf" rel="nofollow noreferrer">Aeolian Vibration Basics</a></p>
<p><a href="https://i.stack.imgur.com/G1dFn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G1dFn.png" alt="Figure 14 – VORTX™ Damper" /></a></p>
<blockquote>
<p>Figure 14 – VORTX™ Damper</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/gpGwv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gpGwv.png" alt="Figure 16 – Damper Response Test Set-Up" /></a></p>
<blockquote>
<p>Figure 16 – Damper Response Test Set-Up</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/RmyrT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RmyrT.png" alt="Figure 17 – Damper Response Curve" /></a></p>
<blockquote>
<p>Figure 17 – Damper Response Curvee</p>
<p>Figure 17 shows the power (in watts) delivered to the damper by the shaker (vertical axis) over a range of frequencies. The results are shown for three of the same damper design, along with the average. This plot shows four distinct peaks in the power curve. These represent the resonant frequencies of the damper weights. Each weight has two resonant frequencies: one where the furthest end of the weight has the maximum movement, and the other where the end nearest the clamp has the maximum movement (higher frequency). The VORTX Damper shown in the curve has two different weights, each with different resonant frequencies, which accounts for the four peaks in the curve</p>
</blockquote>
| 32712 | How do those "dipole-shaped" weights attached to sections of electrical power lines work? |
2020-01-11T02:02:46.467 | <p>I am trying to control a mechanical system with some transfer function T (output is position & input is voltage). I have implemented a state feedback controller in order for the output of the system to reach a certain position. The actuator of the system is an RC Servo Motor which is actuated by PWM signals. I want to convert the output of the controller into proper PWM in order to control the system. Are there any guidelines which I should follow or is there a specific way in order to manage to convert the signal ?</p>
| |mechanical-engineering|control-engineering|control-theory| | <p>First, go take a look at some documentation on the servo PWM signal. The gist of it, is that a PWM signal is a pulse between 1000us and 2000us, repeated every 20ms. The signal is interpereted as 0 when the pulse it 1500us. More than that is positive, and less negative (or the reverse, can't remember).</p>
<p>How you convert your output signal to this format will depend on the device. On an arduino there are numerous examples on the internet.</p>
<p>You are trying to control the servos position, and the servo takes a <em>position</em> signal in the form of PWM. So all you have to do is convert your position value into servo position and send that to the servo.</p>
<p>There are many ways to do this, so I will suggest a simple one. Use a table.</p>
<p>Simply move the servo through it's range, and at several points record the servo position using your system as well as the pulse width you're sending. This creates a table converting your position value directly to a PWM value. Between these points you can interpolate.</p>
| 32727 | How to convert output of controller (control signal) into PWM signal? |
2020-01-11T05:50:59.090 | <p>As in the first image,
Moment at any point between C and B should be 0.</p>
<p>But when I try to solve it by substituting equal and opposite forces at end C so that anti-clockwise moment of <span class="math-container">$F(l-a)$</span> and force <span class="math-container">$F$</span> act at C, moment diagram comes out different.
Why is there discrepancy?
<img src="https://i.stack.imgur.com/riis7.jpg" alt="Actual Bending Moment and shear force diagram"><img src="https://i.stack.imgur.com/NfmYh.jpg" alt="enter image description here"></p>
| |structural-engineering|beam|moments| | <p>This is cheating a bit, but let's start backwards by looking at the correct diagrams you've already shown us and comparing them to your solution. This will hopefully let you notice something which hints at the problem with your solution.</p>
<p>Your solution is the bending moment equation <span class="math-container">$M = F(\ell - a) - Fx$</span>.</p>
<p>Now, what is the bending moment equation for the correct moment diagram? Think about it.</p>
<p>Well, that's a trick question. What I should have asked is: what are the bending moment <strong><em>equations</em></strong> (plural!) for the correct moment diagram?</p>
<p>There is an obvious discontinuity in the slope of the bending moment diagram: it is linear from A to B, and then constant at 0 from B to C. That's my point: you could never solve this with a single equation, you'll always need two: one from A to B, another from B to C.</p>
<p>But what's the source of that discontinuity? Well, for that we need to remember that bending moment is the integral of shear force. Since you haven't asked about the correct shear diagram, I'm going to assume you understand why it looks as shown in the correct answer (simply put, it's constant and equal to <span class="math-container">$F$</span> between the support and the force, and zero elsewhere).</p>
<p>So, if the shear equation is discontinuous, that means the derivative (slope) of the bending moment equation will also be*. And since the shear equation gives us the slope of the bening moment equation at any point, we can trivially see that the slope of the line will be equal to <span class="math-container">$F$</span> from A to B, and zero from B to C.</p>
<p>Given this, we can derive the two bending moment equations (here using <span class="math-container">$x=0$</span> at A):</p>
<p><span class="math-container">$$M = \begin{cases}
Fx - Fa &\text{if }x\in[0, a]\\
0 & \text{otherwise}
\end{cases}$$</span></p>
<hr>
<p>* Note that a discontinuous slope means the slope can change instantaneously, but the value of the bending moment equation on both sides of the discontinuity must be equal; the slope is discontinuous, not the value.</p>
| 32729 | Problem while solving for the moment of a cantilever beam |
2020-01-11T21:19:06.960 | <p>Does the stacking of peltiers device shows promising results in achieving the cryogenic temperature? I have tried doing the experiment, the overall temperature of the system increased around 80ºC and then the dissipation of that heat was the prime issue? Heatsinks usually fail to dissipate this high amount of heat.
Is there any alternative technology to achieve cryogenic temperature electrothermally without producing much amount of heat?
Can any modifications in stacking help?</p>
| |mechanical-engineering|thermodynamics|heat-transfer|refrigeration|cryogenics| | <p>The problem is the temperature. It’s a stable semi conductor that will have constant results. I’ve tested many & all have a 60 - 75 degree temp split, you can’t get around it then it breaks down losing electrons or something because it slowly diminishes & the life span of a Peltier under full load is roughly 24 hours</p>
| 32733 | Can a Peltier's device (i.e.Thermoelectric cooler) achieve cryogenic temperature efficiently? |
2020-01-13T18:19:27.910 | <p>I am looking for a linear pneumatic cylinder to raise and lower a fixed object. Everything will be submerged in oil (for electrical purposes). I am new to engineering and pneumatics in particular, and I don't know if I can completely submerge the cylinder without damaging it.</p>
| |mechanical-engineering|pneumatic| | <p>If it is a double-acting cylinder then possibly fine, BUT you will need to check the seals are not going to be damaged by the oil and make sure that the pressure inside the cylinder, on both sides of the piston, is greater than the oil pressure so oil is less likely to make its way past the seals.</p>
<p>A single acting cylinder needs to be able to fill or expel air on the low pressure side of the piston which will be an issue when submerged.</p>
| 32761 | Can I submerge a pneumatic cylinder in oil? |
2020-01-14T14:59:42.033 | <p>Power output of nuclear reactors is controlled by control rods that sit in between the fuel rods: </p>
<p>These are pulled out to increase fission rates - and lowered to decrease them.</p>
<p>The control rods are made from neutron absorbing materials - such as cadmium.</p>
<p>When lowered completely, so many neutrons are absorbed, that no chain reaction is maintained. Pulling them out increases the neutrons that are available for fission.</p>
<p>My question:</p>
<p>The lower tips of the fuel rods will almost always be exposed to neutrons, so these should burn down much faster (have their uranium fissioned) than the upper parts that rarely have the control rods removed from them. Is there some engineering going on to have the fuel rods burn down at the same rate along their whole length?</p>
| |nuclear-engineering| | <p>Your understanding is correct. In a PWR reactor, the core usually operates with the control blades fully withdrawn, but some reactors operate with the rods slightly inserted into the top of the core. In both cases, the tips of the control rods will burn out faster than the rest of the control rod. Even when the rods are fully withdrawn, the tip of the control blade is going to see enough flux to affect the lifetime.</p>
<p>BWR reactors operate with the blades inserted, some blades are "deeply inserted", especially at beginning of life where there is a lot of excess reactivity in the core. The tips of the BWR blades will usually burn out faster than the rest of the blade and be most limiting.</p>
<p>Once the control rods or control blades have experience enough exposure (neutron fluence), they must be replaced.</p>
| 32771 | How do nuclear power plants prevent fuel rods from depleting nonuniformaly? |
2020-01-15T03:16:36.240 | <p>I have a plant <span class="math-container">$G(s)$</span> which gives position and velocity as output
<span class="math-container">$$ G = [G_{ru}(s)\quad G_{vu}(s)]^T $$</span>
where <span class="math-container">$G_{ru}$</span> is the integral of <span class="math-container">$G_{vu}$</span>, and <span class="math-container">$u$</span> is the input to the system. Now if I want to control the system with a proportional-derivative law I can write</p>
<p><span class="math-container">$$ u = C e_r $$</span>, with
<span class="math-container">$$C = (k_p +s k_d)$$</span> where <span class="math-container">$e_r$</span> is the error with respect to the reference position (e.g., a step command),</p>
<p>I can build the sensitivity and complementary transfer functions as</p>
<p><span class="math-container">$$ S = (1+G_{ru}C)^{-1} $$</span></p>
<p><span class="math-container">$$ T = G_{ru}C(1+G_{ru}C)^{-1} $$</span></p>
<p>When I instead consider the system as having 2 outputs, </p>
<p>and I define a new controller as </p>
<p><span class="math-container">$$C_{2} = [k_p\quad k_d]$$</span>, </p>
<p>and the corresponding sensitivities as
<span class="math-container">$$S_2 = (I_2 + GC_2)^{-1}$$</span>
<span class="math-container">$$T_2 = GC_2(I_2 + GC_2)^{-1}$$</span></p>
<p>we can see that <span class="math-container">$S_2$</span> and <span class="math-container">$T_2$</span> are now <span class="math-container">$2 \times 2$</span> transfer matrices. </p>
<p>The question is: why <span class="math-container">$T_2(1,1)$</span> differs from <span class="math-container">$T$</span> since they have the same controller and represent the same thing (in this case the how the position behaves given a step command)?</p>
<p>The corresponding bode plots are depicted here below. Only the dc gain is the same, but the transient is quite different. Why does this happen?</p>
<p><a href="https://i.stack.imgur.com/qN2dR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qN2dR.png" alt="frequency domain"></a></p>
<p><a href="https://i.stack.imgur.com/PRWNu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PRWNu.png" alt="step response"></a></p>
| |control-engineering|control-theory|pid-control| | <p>Both the (negated) output and reference are needed to calculate the error. So in the scalar case the reference also gets multiplied by <span class="math-container">$s\,k_d$</span>, while <span class="math-container">$T_2(1,1)$</span> only considers the contribution of <span class="math-container">$r$</span> to <span class="math-container">$y_1$</span>. In order to get the correct results you should also add the contribution of <span class="math-container">$s\,r$</span>, which could be expressed as <span class="math-container">$s\,T_2(1,2)$</span> or <span class="math-container">$T_2(2,2)$</span>.</p>
| 32774 | analysis of PD controller vs static-gain |
2020-01-15T03:50:53.757 | <p>I am trying to find the forces in hinches <span class="math-container">$B$</span>,
<span class="math-container">$C$</span> and <span class="math-container">$D$</span>. The exact solution is not important for me. I introduced the following internal forces to solve the problem: <span class="math-container">$A_y$</span> in <span class="math-container">$A$</span>, <span class="math-container">$B_x$</span> and <span class="math-container">$B_y$</span> in B, <span class="math-container">$C_x$</span> and <span class="math-container">$C_y$</span> in <span class="math-container">$C$</span>, a force <span class="math-container">$F_{DE}$</span> in the direction <span class="math-container">$\vec{DE}$</span>. But I think I am missing something. I have currently 6 unknowns. But I need equilibrium for the subsets ABE (grey), ABF (yellow) and CDG (yellow). Are there any other internal forces / torques I should introduce?</p>
<p><a href="https://i.stack.imgur.com/9G6ey.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9G6ey.png" alt="">Pruning shears</a>
<a href="https://i.stack.imgur.com/jsXKy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jsXKy.jpg" alt="enter image description here"></a></p>
<p>First subsystem:</p>
<p><span class="math-container">$C_x + F_{DE} \cos(45) = 0$</span></p>
<p><span class="math-container">$C_y + F_{DE} \sin(45) + 20= 0$</span></p>
<p><span class="math-container">$F_{DE} \sin(45) \times 25 + 20 \times 150= 0$</span></p>
<p>Second subsystem:</p>
<p><span class="math-container">$B_x - C_x = 0$</span></p>
<p><span class="math-container">$B_y - C_y - A_y - 20= 0$</span></p>
<p><span class="math-container">$-F \times 150 + A_y \times 60 - C_x \times 30 = 0$</span></p>
<p>Third subsystem:</p>
<p><span class="math-container">$-B_x - F_{DE} \cos(45) = 0$</span></p>
<p><span class="math-container">$A_y -B_y - F_{DE} \sin(45)= 0$</span></p>
<p><span class="math-container">$-A_y \times 60+ F_{DE} \sin(45) \times 55 = 0$</span></p>
<p>I have 9 equations but only 6 unknowns. Are there any linear interdependencies between the equations?</p>
| |statics| | <p>Writing down a system of equations for all the unknowns is generally not the best strategy. Instead, look for strategic opportunities to apply one equation of equilibrium, to one component, where it will involve one unknown.</p>
<p>A good starting point is often to sum moments about some location that is in line with one or more of your unknowns. That way, those unknowns are not part of the equation.</p>
<p>For example, if you look at part CDE, and you take the moment about C, you only have to deal with force DE as an unknown. With one equation, you solve for one unknown, force DE.</p>
<p>Then, stay with part CDE. Sum horizontal forces, and that will give you the horizontal component of the force at C. Sum vertical forces and that will give you the vertical component of the force at C.</p>
<p>Now you know all the forces that touch part CDE, so move on to different part.</p>
<p>Again, use the strategy of summing moments at a location that is in line with one or more unknowns.</p>
<p>Look at part ABF, and sum moments about B. You know the forces at point C, so the only unknown will be the force at point A.</p>
<p>Keep going, and you will get all the unknowns.</p>
| 32775 | Forces on pruning shears |
2020-01-15T04:51:37.297 | <p>In solidworks , is it possible to trim a part against another part in assembly level</p>
| |solidworks| | <p>Yes, it is possible. The cut must be added to the feature tree of the part to be trimmed, but the location of the body remains defined within the assembly.</p>
<p>I wouldn't usually recommend this as best practice, however, since there are a lot of interlinking references. (Suggesting an alternate referencing structure is outside the scope of this question, however).</p>
<ol>
<li>In the assembly, click Edit Component for the part to be cut</li>
<li>Insert - Feature - Cavity</li>
<li>Select the part that you wish to do the cutting with.</li>
<li>Hide the cutting tool if necessary. You can't suppress or delete it, as the cavity feature is referencing it.</li>
</ol>
<p><a href="https://i.stack.imgur.com/Xrioa.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xrioa.gif" alt="Assembly Boolean Demo"></a></p>
| 32776 | Trim parts against parts in assembly level solidworks |
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