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2020-09-16T22:29:13.867 | <p>In my limited engineering knowledge, and limited research skills, I couldn't find a complete answer for the reasoning behind the chosen tones for DTMF. I did find a hint that said something about resonance/interference, but that's not enough details to satisfy my curiosity.</p>
<p>On a possibly related note, could there be "Tri Tone Multi Frequency" signals? Or even Quad? Is there a limit?</p>
| |telecommunication|signal-processing| | <p>From: <a href="https://www.engineersgarage.com/tutorials/dtmf-dual-tone-multiple-frequency/" rel="nofollow noreferrer">Engineers Garage</a></p>
<p><a href="https://i.stack.imgur.com/KzktC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KzktC.png" alt="enter image description here" /></a></p>
<p>Dual Tone Multi-Frequency:</p>
<ul>
<li><span class="math-container">$f_1$</span> = 697Hz, 770Hz, 852Hz, 941Hz.</li>
<li><span class="math-container">$f_2$</span> = 1209Hz, 1336Hz, 1477Hz, 1633Hz.</li>
</ul>
<p>A key press on a DTMF keypad means a combination of <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> (Dual Tone) frequencies are heard at the same time. <span class="math-container">$f_1$</span> (rows) and <span class="math-container">$f_2$</span> (columns) have four frequencies (Multi-Frequency) associated with them, although 1633Hz was only used in military phones. <span class="math-container">$f_2$</span> frequencies are slightly louder than <span class="math-container">$f_1$</span> frequencies to compensate for high-frequency roll off of voice audio systems.</p>
<p>Frequencies were selected so harmonics would not be interpreted as a fundamental frequency. 2nd harmonic of 697Hz is 1394Hz, which is midway between 1336Hz and 1477Hz. Harmonics of <span class="math-container">$f_1$</span> can not be intrepreted as a fundamental <span class="math-container">$f_2$</span> frequency.</p>
<p>The combination of 941Hz and 1209Hz means that the sum (2150Hz) and difference (268Hz) are heard at the same time. 697Hz and 1477Hz means 780Hz and 2174Hz. None of these can be intrepreted as <span class="math-container">$f_1$</span> or <span class="math-container">$f_2$</span> frequencies.</p>
<p>Adjacent <span class="math-container">$f_1$</span> and <span class="math-container">$f_2$</span> frequencies were selected to have a 21/19 ratio (1.10), which is slightly less than a whole tone (musical reference <a href="https://physics.info/music/" rel="nofollow noreferrer">Music and Noise</a>) and can vary no more than ±1.5% (or ±1.8% depending on sources) from their nominal frequency.</p>
<p>The range of human hearing is 20Hz to 20kHz, most sensitive at 2 to 4kHz and the normal voice range is about 200Hz to 3.5kHz. Need 8 frequencies for 16 keys within 1950's phone's 4kHz bandwidth.</p>
<p>Human speech is unable to produce the combined tones, so you could implement modes, where users could talk and use touch tones to cause the system to react (operator).</p>
<p>As for the actual frequencies. From: <a href="http://bitsavers.informatik.uni-stuttgart.de/communications/westernElectric/books/Engineering_and_Operations_in_the_Bell_System_2ed_1984.pdf#page=293" rel="nofollow noreferrer">Engineering and Operations in the Bell System</a>, which I stole from <a href="https://dsp.stackexchange.com/questions/12958/how-were-dtmf-frequencies-determined-exactly-and-how-can-i-extend-them">how were DTMF frequencies determined exactly and how can I extend them?</a></p>
<blockquote>
<p>The tones have been carefully selected to minimize harmonic interference and the probability that a pair of high and low tones will be simulated by the human voice, thus protecting network control signaling.</p>
</blockquote>
<blockquote>
<p>The nominal voiceband channel is defined as 4 kHz although the speech signal is
essentially bandlimited to between 200 Hz and 3.5 kHz. The additional bandwidth allows for a guard band on either side of the speech signal to lessen interference between channels.</p>
</blockquote>
<p><a href="https://www.rfcafe.com/references/electrical/dtmf.htm" rel="nofollow noreferrer">DTMF Frequencies</a></p>
<hr />
<p>An abandoned patent from 2005 describes a triple tone modulation frequency (TTMF) tone system for <a href="https://patents.google.com/patent/US20060245364A1/en" rel="nofollow noreferrer">Bi-directional continuous voice and video quality testing system with TTMF tones</a>.</p>
<p>They used: 650Hz[f1], 750, 850, 950, 1050, 1150[f6], 1250[f7], 1350, 1450, 1550 and 1650.</p>
<blockquote>
<p>In order to avoid harmonics, the three frequencies comprising a TTMF tone may be chosen according to the following rules:
(a) no frequency is a multiplier of another frequency;
(b) the difference between any two frequencies is not equal to any of the frequencies; and
(c) the sum of any three frequencies is not equal to any of the frequencies.
Thus a permitted TTMF tone is a tone signal comprising, e.g., three frequencies such as ƒ1, ƒ6, and ƒ7 (as shown in the second column of Table 1).</p>
</blockquote>
<p>Digits: 0: f1, f6, f7; 1: f2, f6, f8; 2: f3, f6, f9; 3: f4, f6, f10; 4: f4, f6, f11; 5: f1, f6, f8; 6: f2, f6, f9; 7: f3, f6, f9; 8: f4, f6, f11; and 9: f5, f6, f7.</p>
<p>With other combinations for E-TTMF flag signals representing decimal digits from 10 - 99.</p>
<p>Digits 10-99 had two functions:</p>
<blockquote>
<p>(a) indicating the end of the played voice/video sample testing file; and (b) representing the two digit decimal portion of the voice/video quality measurement result.</p>
</blockquote>
| 37693 | How were the tones for DTMF chosen? |
2020-09-17T02:59:00.263 | <p>Pardon my paint skills, I did my best
<a href="https://i.stack.imgur.com/xkqTq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xkqTq.jpg" alt="enter image description here" /></a></p>
<p>My attempt is short and seems to fail, I have no idea why:</p>
<p><span class="math-container">$$\begin{align}
\alpha &= in - a_{1}\beta - a_{2}\gamma \\
\beta &= \alpha z^{-1} \\
\gamma &= \beta z^{-1} = \alpha z^{-2}
\end{align}$$</span></p>
<p>inputting 2nd and 3rd equation into the first one I get:</p>
<p><span class="math-container">$$\begin{align}
\alpha &= in - a_{1}\alpha z^{-1} - a_{2}\alpha z^{-2} \\
in &= \alpha + a_{1}\alpha z^{-1} + a_{2}\alpha z^{-2}
\end{align}$$</span></p>
<p>I can write the output as:</p>
<p><span class="math-container">$$\begin{align}
out &= b_{2}\gamma + b_{1}\beta + b_{0}\alpha \\
out &= b_{2}\alpha z^{-2} + b_{1}\alpha z^{-1} + b_{0}\alpha
\end{align}$$</span></p>
<p>I have output and input in terms of alpha, but I can't figure what to do from here.</p>
| |transfer-function|diagram| | <p>Just wanted to add that you could use <a href="https://www.diagrams.net/" rel="nofollow noreferrer">this</a> to draw diagrams next time. Much easier to use than paint. You can even use <a href="https://www.diagrams.net/blog/maths-in-diagrams" rel="nofollow noreferrer">Latex</a>.</p>
| 37698 | How to get a transfer function from this block diagram? |
2020-09-18T15:51:52.743 | <p>If <span class="math-container">$h=h(T, P)$</span>.</p>
<p>Does <span class="math-container">$ dh = c_pdT + \left[v - T\left(\frac{\partial v}{\partial T}\right)_P \right]dP \Rightarrow h_2 - h_1 = \int_{T_1}^{T_2} c_pdT + \int_{P_1}^{P_2}\left[v - T\left(\frac{\partial v}{\partial T} \right)_P\right]dP $</span> ?</p>
<p>If so, how?</p>
<p>I apologize for this, but I just haven't been able to find an appropiate justification for this operative behavior in any of the Calculus, Differential Equations and Thermodynamics books in my possession. I'm particular bugged by the "integration of differentials" and how it, before me, seems to damage the symmetry of the first equation in the statement.</p>
| |thermodynamics|chemical-engineering|mathematics| | <p>We can always write</p>
<p><span class="math-container">$$dh=\left(\frac{\partial h}{\partial T}\right)_P dT+ \left(\frac{\partial h}{\partial P}\right)_T dP;$$</span></p>
<p>this is just expansion in <span class="math-container">$T$</span> and <span class="math-container">$P$</span>. By definition,</p>
<p><span class="math-container">$$c_P\equiv\left(\frac{\partial h}{\partial T}\right)_P.$$</span></p>
<p>Then, we can write <span class="math-container">$h=g+Ts$</span> (by definition of <span class="math-container">$g$</span>) and <span class="math-container">$dg=-s\,dT+v\,dP$</span> (the fundamental relation) and thus obtain</p>
<p><span class="math-container">$$\left(\frac{\partial h}{\partial P}\right)_T=\left(\frac{\partial g+Ts}{\partial P}\right)_T=v+T\left(\frac{\partial s}{\partial P}\right)_T=v-T\left(\frac{\partial v}{\partial T}\right)_P=v(1-T\alpha),$$</span></p>
<p>where we've used a Maxwell relation to go from <span class="math-container">$(\partial s/\partial P)_T$</span> to <span class="math-container">$-(\partial v/\partial T)_P$</span>. This gives</p>
<p><span class="math-container">$$dh=c_P\, dT+ v(1-\alpha T) dP. $$</span></p>
<p>Make sense?</p>
| 37720 | Gaps in derivation of thermodynamic property equations |
2020-09-20T01:11:06.640 | <p>I'm trying to lift a platform with several guides using a linear actuator, but off the centre as follows:</p>
<p><a href="https://i.stack.imgur.com/mdDjam.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mdDjam.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/cQIuCm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQIuCm.png" alt="enter image description here" /></a></p>
<p>The load on the platform will be around 100kg, which is just within the linear actuator's 120kg load capacity. The platform will be around 50cm*50cm. My plan is to use 2-4 rods as guides with linear (or tapered?) bearings at each corner (or opposite sides if 2 rods) of the platform. Does anyone know if this would be a good idea? Would the linear actuator wear out more rapidly if the actuation was off-centre?</p>
<p>Any advice appreciated.</p>
| |mechanical-engineering|bearings|linear-motion|linear-motors| | <blockquote>
<p>Does anyone know if this would be a good idea?</p>
</blockquote>
<p>The proposed design has poor rigidity towards torsional loads. It doesn't matter whether the actuator is off-center or if the load is off-center: if load's center of gravity and actuator are not at the same position, the torsional load will likely damage the bearings.</p>
<p>In your example with two guide rails, torsional loads try to rotate the two linear bearings. The bearings are not usually designed to take much load in this direction, and some are in fact designed to be compliant of small rotations (alignment errors).</p>
<p><a href="https://i.stack.imgur.com/SKVr1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/SKVr1.png" alt="Torsional load on bearings." /></a></p>
<p>You can think of it like this: the actuator force is pulling on a long moment arm, while the bearing forces have a much shorter arm. Therefore the bearing has to handle many times larger force than is actually needed to keep the platform up. In your case of 100kg load, the bearing force would likely be hundreds of kilograms.</p>
<p><a href="https://i.stack.imgur.com/hdieX.png" rel="noreferrer"><img src="https://i.stack.imgur.com/hdieX.png" alt="Force diagram" /></a></p>
<p>The example with four guide rails is slightly better in that it has half the load per bearing because there are four of them. Because they are all in the same plane, the lever advantage is equal as with two guide rails.</p>
<blockquote>
<p>Would the linear actuator wear out more rapidly if the actuation was off-centre?</p>
</blockquote>
<p>If the linear actuator is rigidly connected, it might take some of the torsional force on itself and wear out its internal sliding surfaces more rapidly. But the best practice is to connect linear actuators in a way that does not transmit torque. In fact, most linear actuators (<a href="https://fi.rsdelivers.com/product/rs-pro/ld20-24-40-g4-300-c11-p09/rs-pro-electric-linear-actuator-24v-dc-300mm/1774513" rel="noreferrer">random example</a>) have only a single mounting hole at each end. Such a mounting prevents premature wear on the actuator - it would be the guide rail bearings that would break or wear out.</p>
<hr />
<p>kamran's suggestion of a scissor lift mechanism is a good choice, especially considering the quite large 100 kg load. The diagonal arms of a scissor mechanism transfer the torsional loads to the support of the scissor arm ends.</p>
<p>There are however other alternatives also. For example, one can mount four bearings on two guide rails like in the picture below. Now the moment arm for bearing forces is approximately the same as the moment arm for load forces, and horizontal load on bearings is thus smaller. The actual location of the actuator is not very important, as it only requires correctly sizing the guide rails and bearings to handle the load.</p>
<p><a href="https://i.stack.imgur.com/AmDvx.png" rel="noreferrer"><img src="https://i.stack.imgur.com/AmDvx.png" alt="Example of linear platform with better support." /></a></p>
| 37739 | Off-centre actuation |
2020-09-20T11:59:50.980 | <p>Given the transfer function: <span class="math-container">$$\frac{\omega_0}{s^2 + 2\omega_0\zeta s + \omega_0^2}$$</span></p>
<p>How would you derive the rise time of the system from 0% SS to 100% SS.</p>
<p><span class="math-container">$$t_r = \frac{\pi - \arccos{\frac{1}{\zeta}}}{\omega_0 \sqrt{1 - \zeta^2}}$$</span></p>
<p>I attempted to obtain this by using the solution to the 2nd order ODE: <span class="math-container">$$y(t) = A \exp{(-\omega_0\zeta t)} \cos{(\omega_0 \sqrt{1-\zeta^2}t)}$$</span> from <span class="math-container">$\dot{y}(0) = 0$</span>, <span class="math-container">$y(0) = -A$</span> and, <span class="math-container">$y(t_r) = 0$</span>, where <span class="math-container">$A$</span> is the steady state value, however, the solution is obviously incorrect.</p>
| |control-engineering| | <p>I realise now that the solution was wrongly assumed, using the BCs you would get <span class="math-container">$$y(t) = -A \exp{(-\omega_0\zeta t)}( \cos{(\omega_0 \sqrt{1-\zeta^2}t)} + \frac{\zeta}{\sqrt{ 1-\zeta^2 } }\sin{(\omega_0 \sqrt{1-\zeta^2}t)}) $$</span></p>
<p>Leading to</p>
<p><span class="math-container">$$\frac{\sqrt{1-\zeta^2}}{\zeta} + \tan{\omega_0 \sqrt{1-\zeta^2} t_r} = 0$$</span></p>
<p>where <span class="math-container">$ \arctan{\frac{\sqrt{1-\zeta^2}}{\zeta}} \equiv \arccos{\frac{1}{\zeta}}$</span></p>
<p>Then <span class="math-container">$$\omega_0 \sqrt{1 - \zeta^2} t_r = \pi -\ arctan{ \frac{\sqrt{1 - \zeta^2}}{\zeta} }$$</span> and the rest is trivial.</p>
| 37748 | Derivation of 2nd order damped rise time |
2020-09-21T05:49:10.070 | <p><a href="https://i.stack.imgur.com/RC21S.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RC21S.jpg" alt="wheel base" /></a></p>
<p>I have tried doing search by image on few popular Internet search engines but could not get it. (sorry, it picked up lot of dust and a bit ugly to see).</p>
<p>I want to know the name of such wheel and where can I probably purchase it by mentioning parameters. Are there any alternate designs?</p>
| |control-engineering| | <p>Technically its called a bump wheel, these are rarely seen anymore but so far can be found on ebay and aliexpress. I'll paste a link for you.
<a href="https://www.ebay.com/itm/224088720767?hash=item342cba857f:g:KrMAAOSwfWpfBaoI&amdata=enc%3AAQAHAAABABbKgMhnpIu%2Fmmi%2BDmgsxLJgY%2FLVD69CEw%2Bef6155ZvLLXw826FpEDIPLpVw5bZnCB9kE9aGjzJvgZJYv9A%2BrUFLNa2iclW3%2Ba6pjwhbbdAXiSHMllxsSZMs65dJf5FPg60Ml%2FskWeSGNKrMxbrDfqGm3H7EjNvKCr3BnK8m2%2BtuNqaQhH2f4n2I88%2Fdn7%2B6W18TEtVhbcW%2B4cC7vjN%2FApmqTkCiDLYN7JbkRLkv9VPYQxVwKq9ti%2FxaGdUgkH%2F%2B7GJeluzTCB5l8KJF%2Bz2gZdkhVhCxKZ0yM4kZWuWxc7BQKSA6FKezcRWXoosWZ%2FhwZZe3qquwdRNDw7gJofLS4ZA%3D%7Ctkp%3ABk9SR-b74tWbYQ" rel="nofollow noreferrer">https://www.ebay.com/itm/224088720767?hash=item342cba857f:g:KrMAAOSwfWpfBaoI&amdata=enc%3AAQAHAAABABbKgMhnpIu%2Fmmi%2BDmgsxLJgY%2FLVD69CEw%2Bef6155ZvLLXw826FpEDIPLpVw5bZnCB9kE9aGjzJvgZJYv9A%2BrUFLNa2iclW3%2Ba6pjwhbbdAXiSHMllxsSZMs65dJf5FPg60Ml%2FskWeSGNKrMxbrDfqGm3H7EjNvKCr3BnK8m2%2BtuNqaQhH2f4n2I88%2Fdn7%2B6W18TEtVhbcW%2B4cC7vjN%2FApmqTkCiDLYN7JbkRLkv9VPYQxVwKq9ti%2FxaGdUgkH%2F%2B7GJeluzTCB5l8KJF%2Bz2gZdkhVhCxKZ0yM4kZWuWxc7BQKSA6FKezcRWXoosWZ%2FhwZZe3qquwdRNDw7gJofLS4ZA%3D%7Ctkp%3ABk9SR-b74tWbYQ</a></p>
| 37758 | what is the name of this wheel found in toys which rotates freely? |
2020-09-21T11:21:31.170 | <p>In structural engineering we have to accept the fact that you can never be exactly sure how much load a structural member can resist, that the load carrying capability of an element is not one exact value but distributed with some probability distribution function (quality of laid concrete, for example, is quite dependent on many conditions on the worksite, such as the workers skill, etc.). The same goes for structural loads. Therefore we cannot say with absolute certainty whether some structural is stable, we can only calculate certain <em>probabilities</em>.</p>
<p>In the image below, the left curve represents the probability density of the possible loads, and the one of the right represents the probability density of the load carrying capacity of some structural element. The y-axis is the probability density, and the x-axis is the structural load. The red area where the curves overlap is a range of loads where structural failure is possible; before that range there is "zero" possibility of capacity being that low, therefore the capacity must be higher than any possible load (failure is defined as a scenario where the load is greater than the capacity). Beyond this range the load has "zero" possibility of being higher than the capacity therefore no chance of failure.</p>
<p><a href="https://i.stack.imgur.com/yVbYa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yVbYa.png" alt="enter image description here" /></a></p>
<p>The actual probability of failure can be calculated as follows, according to <a href="https://en.wikipedia.org/wiki/Structural_reliability" rel="nofollow noreferrer">Wikipedia:</a></p>
<p><span class="math-container">$${\displaystyle P_{f}=\int _{0}^{\infty }F_{R}(s)f_{s}(s)\,ds\qquad \mathrm {} }$$</span></p>
<p>where <span class="math-container">$F_R(s)$</span> is the probability the cumulative distribution function of resistance/capacity (R) and <span class="math-container">$f_s(s)$</span> is the probability density of load (S).</p>
<p>I have hard time understanding why the above formula gives the probability of failure. This is supposed to give the probability of the load being higher than the capacity/resistance, but I just can't wrap my head around how that works. <strong>Could somebody explain this to me in more detail?</strong> I am familiar with probability distribution functions and cumulative distribution functions, but I don't understand what's happening here. Why do you take the <em>cumulative</em> function of the load resistance but the probability density function of the load?</p>
| |structural-engineering| | <p>Like mentioned by @mart above, I believe there is something wrong with the curve of <span class="math-container">$F_R(s)$</span>. The text states cumulative function so it should have been something like.</p>
<p><a href="https://i.stack.imgur.com/eK3BG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eK3BG.png" alt="enter image description here" /></a></p>
<p>The way I would explain the Probability of failure is the following. There is failure if the load is greater (S) than the structural resistance capacity (R). So you want to iterate over all possible loads (thus the probability density of a load s).</p>
<p>Assuming you get a structural load s of S e.g. 1 [kN]. If the structure has R greater than S there is no failure. So you are interested for all the values, <strong>below</strong> S. That is the probability of
<span class="math-container">$$P(R\le S) = \int_{0}^Sf_R(s)ds = CDF_R(s)=F_R(s)$$</span>.</p>
<p>Since the two probabilities (A: load at S), (B: resistance R is less or equal to load S) are statistically independent, their probability of occuring simultaneously is given by:</p>
<p><span class="math-container">$$P(A\cap B) =P(A)\cdot P(B)= f_s(s)\cdot F_R(s)$$</span></p>
<p>Then all you need to do is integrate over the entire range of <span class="math-container">$s$</span>. That gets you:</p>
<p><span class="math-container">$$P_{f}=\int _{0}^{\infty }F_{R}(s)f_{s}(s)\,ds$$</span></p>
<p><strong>TL;DR</strong> I think the Wikipedia entry is incorrect and confuses.</p>
| 37765 | Probability of structural failure |
2020-09-21T12:54:50.297 | <p>Holes are made in brickwork walls for all kinds of reasons including gas flues, to route cabling and to fit windows. At some point that hole gets large enough that a lintel is required to support the brickwork above.</p>
<p>My questions are:</p>
<ol>
<li>My theory is that the height of the hole is not relevant as to
whether additional support is required. The only important factor is
the width of the hole. Is this correct?</li>
<li>Is a circular or rectangular hole preferred?</li>
<li>Does the type of construction i.e. cavity wall, double leaf solid wall, single leaf wall make a significant difference?</li>
<li>What is the maximum width of the hole before additional support is
required?</li>
</ol>
| |structural-engineering|civil-engineering| | <p>Direct answer to your questions:</p>
<ol>
<li><p>Yes, the opening depth is immaterial in support requirement determination.</p>
</li>
<li><p>Round opening is less likely to require support, due to arch action.</p>
</li>
<li><p>Wall type is critical when the brick layout pattern affects the available bond between bricks over the opening.</p>
</li>
<li><p>If depth of the wall above the opening permits arch action to occur, the bricks below the arch action zone will fall if the bond is insufficient to hold the weight of the bricks</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/kEr94.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kEr94.png" alt="enter image description here" /></a></p>
| 37767 | Maximum hole size in brickwork before a lintel is required |
2020-09-21T17:33:53.260 | <p>I have a pay-out winch. It is a spool of rope that has a disk brake attached to it, so I can control the torque required to rotate the spool. When the torque is very large, however, the rope begins to bury itself into the rope that is still on the spool, and this causes a lot of abrasion and wear on the rope. I tried a different winding pattern, but the rope still cuts into it.</p>
<p>I'm using 1 mile of 400 lb test kevlar. The spool diameter is 4 inches.</p>
<p>Does anyone have a suggestion? I was wondering if I can thread the rope through a mechanism as it leaves the spool to add friction (pink box in the picture). Are there any off-the-shelf solutions? For example the rope could go back and forth between two rows of pulleys, each of which has a little friction. The point is to avoid wearing out the rope with a lot of heat or abrasion.</p>
<p>Can you suggest a way to prevent the rope from burying itself? I guess I could make the spool larger and wider (decrease the depth of the rope still on the spool).</p>
<p><a href="https://i.stack.imgur.com/Sl692.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sl692.png" alt="enter image description here" /></a></p>
| |friction| | <p>I've loaded a winch with Spectra line, not quite as abrasive as Kevlar, but similar in strength and size. The key factor in such problems is to load the spool while the line is under tension. I used a heavy wheeled tractor as the load. As the winch turned the spool, the tractor was pulled across an even surface with brakes applied as required. The amount of line was similar to your circumstances, about a mile, but the easement of a state highway provided enough free distance to accomplish the task.</p>
<p>Once loaded, the volume of line was greatly reduced and was effectively a solid un-moving mass. The intended use of the winch ensured that it would always load under tension in the future, preventing the need for duplication of effort later.</p>
| 37771 | How to add friction to a rope without using abrasion |
2020-09-23T17:20:04.650 | <p>I've seen the term "kinematic" used qualitatively to describe machines several times in books and on YouTube. I think I understand what kinematics is as a study, but I'm not sure what people mean when they say a structure or machine "is kinematic".</p>
<p>Here are two examples:</p>
<ol>
<li><p>Multiple times I've seen people describe a tool with three feet as being kinematic. As in "This height gauge has three points of contact with the surface plate which makes it kinematic."</p>
</li>
<li><p>In "Precision Machine Design" there's a paragraph on machine structure where the author says "If the design is not kinematic, bearings can become overloaded by the forced geometric compliance between the structure to which the bearing rails or races are mounted and the smaller structure of the platten or spindle."</p>
</li>
</ol>
| |kinematics|machine-design| | <p>See <a href="https://en.m.wikipedia.org/wiki/Kinematic_coupling" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Kinematic_coupling</a></p>
<p>A kinematic coupling is constrained in (only) 6 degrees of freedom.</p>
| 37802 | What does it mean for a machine design to be "kinematic"? |
2020-09-26T17:34:55.673 | <p>I was reading about Automotive Brakes (Chapter 52, Automotive Mechanics by William H Crouse, Donald L Anglin, 10th edition). And in a section about drum-brake self-adjusters, it is said that the <em>incremental self-adjusters</em> are attached to the <em>secondary shoe</em> of a duo-servo brake; The adjustment is made when the vehicle is moving <em>backwards</em> and the brake is applied. And then the mechanism of how the adjusting lever and adjuster spring follows the text.</p>
<p>What I don't understand is why the <em>secondary</em> shoe? It only seems reasonable since the adjuster is doing its work while the vehicle is in backward motion. But then, why the adjustment only happens in <em>backward</em> motion?</p>
| |mechanical-engineering|automotive-engineering| | <p>The primary shoe is normally operated directly by the piston and the secondary one by the cylinder sliding in the back plate.</p>
<p>But this varies widely as there are versions using one cylinder with two pistons and others with two separate cylinders and pistons where both shoes are leading instead of having one leading and one trailing, twin leading shoe brakes provide more braking for less pedal force due to the self-servo action.</p>
<p>Without you providing the diagrams showing exactly how these are designed it is difficult to provide more information, but you have enough to give you material for further research.</p>
| 37846 | Drum-brake self-adjusters |
2020-09-27T07:32:02.067 | <p>It's the 21st century, anything is possible. Except for PID autotuning for a simple problem, heating/cooling system - make robust and ready to use solution. I've found only one library for PID autotune: <a href="https://github.com/br3ttb/Arduino-PID-AutoTune-Library" rel="nofollow noreferrer">https://github.com/br3ttb/Arduino-PID-AutoTune-Library</a>, but even its author says that it is not a robust approach. My question is:</p>
<ol>
<li>What types of algorithms for PID autotune for a simple heater/cooler application exists? What are their pros and cons?</li>
<li>Is there some open-source libraries for PID autotune?</li>
</ol>
| |pid-control| | <p>Perhaps you're interested in a PID alternative such as Model Predictive Temperature Control support (MPC). Marlin (3D printer firmware) has the ability to use MPC instead of PID for the extruder temperature control.</p>
<blockquote>
<p>Model Predictive Control (MPC) takes a different approach. Instead of trying to control against the measured temperature, it maintains a simulation of the system and uses the simulated hotend temperature to plan an optimal power output. The simulation has no noise and no latency, allowing for nearly perfect control. Thus it can compensate directly for extrusion speed and part-cooling. To prevent the simulated system state diverging from the actual hotend state, the simulated temperature is continually adjusted towards the sensor measurement. Although this does introduce a little noise and latency into the simulated system, the effect is far smaller than for PID.</p>
</blockquote>
<p>That quote is from this very comprehensible article about MPC here: <a href="https://marlinfw.org/docs/features/model_predictive_control.html" rel="nofollow noreferrer">https://marlinfw.org/docs/features/model_predictive_control.html</a>.</p>
| 37852 | Best PID autotuning algorithm/library for simple heater problem |
2020-09-27T10:44:25.220 | <p>I was watching a <a href="https://www.youtube.com/watch?reload=9&v=Ja03J1RQ3Hw" rel="nofollow noreferrer">YouTube video</a> to understand the concepts of <strong>hoop/circumferential stress</strong> and <strong>axial/longitudinal stress</strong> in <em>thin-walled cylinders</em> and their formula derivation.</p>
<p>So the formula of the <strong>Hoop Stress</strong>: <span class="math-container">$$\sigma_{hoop}=\frac{(p * r )}{t}$$</span></p>
<p>where <span class="math-container">$p$</span> is the internal pressure of the cylinder, <span class="math-container">$r$</span> is the internal radius of the cylinder, and <span class="math-container">$t$</span> is the thickness of the cylinder,</p>
<p>The formula of <strong>Axial Stress</strong>: <span class="math-container">$$\sigma_{axial}=\frac{(p*r)}{2t}$$</span></p>
<p><strong>My question is:</strong> They have mentioned in the video that since the axial stress is less than the hoop stress, then the thin walled cylinder is more likely to experience failure along its axis than along its circumference.</p>
<p>I still don't agree that just because the stress along its axis is smaller than that along its hoop, then it is more likely to undergo failure along its axis.</p>
<p>The stress is meant to resist deformation, and to resist the applied force. The axial stress is smaller just because the internal pressure applied along the axis is smaller than the pressure applied along the hoop.</p>
<p>So yes the axial stress is smaller than the hoop stress, but on the other hand the applied internal pressure is less along the axis than it is along the hoop of the cylinder.</p>
| |stresses|pressure-vessel|mechanical-failure|solid-mechanics| | <p>In the old days we used the formula- Stress = PD/2t , and the hoop stress was twice the axial /longitudinal stress . Pressure vessel heads were usually half the thickness of the walls because of the lower stress relative to the walls.</p>
| 37854 | Why is the thin-walled cylinder more likely to experience failure along its axis than along its hoop/circumference? |
2020-09-27T13:58:26.557 | <p>I have a switched-mode PSU for a gaming laptop.
It has a bad habit of burning out the switches on mains extension panels when I turn it on (two in the last four days).
The turn-on surge is always audible and spectacularly visible (sparks) if I plug in with the power on.
I don't think the PSU is faulty (I have two to compare), probably just not that well-designed.
I need to turn it on and off fairly regularly - it's not possible to leave permanently powered on.</p>
<p>I qualified in electronics (before SM-PSUs!) and write software these days, but I understand roughly what is going on.</p>
<p>Is there anything I can put in series on the input side to reduce the spike?
A ready made device would be great, but I'm happy to build something.
I'm guessing some sort of inductor will do the job, but I'm not sure of all the factors/trade-offs involved and I'd rather design well than just over-specify, plus I don't have any serious test equipment left.</p>
<p>Or is there another approach?</p>
<p>The PSU is specified as 100-240V / 1.7A AC in, and 19.5V / 6.15A / 120W DC out.
I am in the UK where we use the full 240V.</p>
| |electrical-engineering|power-electronics|circuits|consumer-electronics|electronics| | <p>OK, as it's been so quiet, I've looked at this in more detail.</p>
<p>The accepted solution seems to be an <a href="https://en.wikipedia.org/wiki/Inrush_current_limiter" rel="nofollow noreferrer">Inrush Current Limiter</a> based on a negative temperature coefficient (NTC) <a href="https://en.wikipedia.org/wiki/Thermistor" rel="nofollow noreferrer">Thermistor</a>.
This sits on an input line and can be before rectification (ac) or after (dc), but obviously must be before the capacitance that presents almost zero impedance at startup.
At room temperature, the thermistor has an impedance that significantly restricts the inrush current. As current flows through, it heats up and the NTC implies that its impedance drops with increasing temperature.
At some temperature, it reaches a balance where heat is being radiated at the same rate as it is being generated and a steady state is reached.</p>
<p>Turning to more quantitative issues:</p>
<ol>
<li>Typical impedances vary from less than 1Ω to 100+Ω at room temperature, typically given as 25°C in manuals.</li>
<li>The impedances can drop to fractions of an Ω at working temperatures which can easily be in excess of 200°C. This high temperature must be allowed for in terms of mounting and ventilation.</li>
<li>The dissipation constant (power lost per unit of temperature above surroundings) or "thermal inertia" of the device means that it can take quite a few seconds, even a few minutes, after power-off for the temperature to drop to a level where the impedance has significantly elevated again. If the PSU is turned on again within this time frame the original inrush current problem will at least partially manifest as the high-temperature/low-impedance combination will not restrict it enough.</li>
</ol>
| 37857 | How do I reduce the mains spike in a switched mode PSU? |
2020-09-27T20:11:41.757 | <p>I drew part and added some dimension on overall model and got value with uncertainty</p>
<p><a href="https://i.stack.imgur.com/RC7kO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RC7kO.png" alt="enter image description here" /></a></p>
<p>Why and how to fix?</p>
| |solidworks| | <p><strong>"Why"</strong> - This is not 'uncertainty' - it is displaying a tolerance. As far as the software is concerned, your part is exactly 14.40mm. The feature that you are using, however, is not used to communicate to the software how large the part should be (that will be the sketch that defined your body), but rather it's used to communicate to a <em>human</em> how large it's OK for the part to be when they make a physical copy. A full description of the reasons for and usefulness of tolerances is outside the scope of your question, however...</p>
<p><strong>"And how to fix"</strong> - There's a few ways to 'fix' this.</p>
<p>If you want to change just this once, then you can adjust the displayed tolerances within the 'Tolerance/Precision' dropdown when the dimension is selected. You may need to press the arrow (circled) to reveal the options.</p>
<p><a href="https://i.stack.imgur.com/PH8Ai.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PH8Ai.png" alt="Tolerance/Precision" /></a></p>
<p>If you want to change is on all new dimensions that you create, then you need to adjust it in the Document Properties</p>
<p><a href="https://i.stack.imgur.com/Htox7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Htox7.png" alt="DimXpert Tolerances" /></a></p>
<p>And, if you want to change it on all new dimensions in <em>all new documents</em> that you create, then you'll need to update your default template to include the changes made above to the Document Properties. This, too, lies outside the scope of your question - google how to save your current document as a tempate.</p>
| 37865 | Why do I have uncertainty in my dimension in Solidworks? |
2020-09-28T18:56:03.793 | <p>I'm new to control systems, and I'm attempting to analyze the stability of an open-loop transfer function. I know that by checking the locations of the poles (and ensuring all poles are in the LHP) I can ensure that my transfer function is stable.</p>
<p>However, I'd love to know <em>how</em> stable this transfer function is. Reading textbooks like "Modern Control Engineering" by Ogata and watching Brian Douglas YouTube videos, I've learned about the concept of gain and phase margins, and their application to determining how resistant a system is to changes in the gain or phase. This would be helpful in my design, as there are many elements of my system that are approximations, and I want to ensure that the TF will be robust even with these errors. From my understanding though, the math behind using Bode/Nyquist plots to determine stability margins relies on the fact that you're analyzing the stability of an open-loop system in the case that it was closed with unity feedback.</p>
<p>Is there any similar concept for strictly open loop transfer functions, if I truly don't care about ever closing the loop? Or do I have some fundamental misunderstanding of the concept of stability?</p>
| |control-engineering|optimal-control| | <p>The transient response (or homogeneous solution) of a linear ODE is
<span class="math-container">$$y_h(t) = \sum_{i=1}^N C_i e^{p_i\cdot t} $$</span>
where <span class="math-container">$p_i$</span> is the i-th pole of your system.</p>
<p>Assume you have a pole at location <span class="math-container">$p_i = \lambda + i\omega$</span>, with <span class="math-container">$i = \sqrt{-1}$</span>.
The real part (<span class="math-container">$\lambda$</span>) of this pole will determine the convergence rate towards zero, while the imaginary part <span class="math-container">$\omega$</span> represents the oscillation frequency.
This is visualized below</p>
<p><a href="https://i.stack.imgur.com/ptFKy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ptFKy.png" alt="Pole zero map" /></a></p>
<p>The pole with the largest real-part (<span class="math-container">$\lambda$</span>) has the slowest convergence towards 0 if <span class="math-container">$\lambda<0$</span> or will blow up <span class="math-container">$\lambda>0$</span>.
Hence, the pole with the largest real part is dominant and will tell you something about 'how stable the system is'.</p>
| 37883 | Is there a concept of gain and phase margin for a strictly open-loop transfer function? |
2020-09-28T19:16:48.723 | <p>I have already set millimeters in document propertis, nevertheless the display units are meters</p>
<p><a href="https://i.stack.imgur.com/IKyQv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IKyQv.png" alt="enter image description here" /></a></p>
<p>Why and how to fix?</p>
| |solidworks|unit| | <p>There are many possible culprits. One of them is the following.</p>
<p>Check in the lower right corner of your Solidworks window.
you should see</p>
<p><a href="https://i.stack.imgur.com/jC1Bi.png" rel="noreferrer"><img src="https://i.stack.imgur.com/jC1Bi.png" alt="enter image description here" /></a></p>
<p>if you click on it, you should see the following menu.</p>
<p><a href="https://i.stack.imgur.com/I1GUZ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/I1GUZ.png" alt="enter image description here" /></a></p>
<p>You might have pressed it accidentally.</p>
| 37884 | How to set measure display units in Solidworks drawing? |
2020-09-29T04:57:47.913 | <p>When you're trying to design a relatively long joint with only two fasteners, what's the best way to arrange them? Since fasteners effectively radiate a circle of clamping force around them, what's the best way to evenly distribute this force?</p>
<p>I originally thought of this question while thinking of the best positions for the binder clips holding my 3D printer's build plate on, but the core concept is surely applicable in larger applications too.</p>
<p>Some ideas off the top of my mind:</p>
<ul>
<li>1 fastener at 25% and the other at 75%. This is flawed in my mind because the 50% mark will receive far less clamping force than the rest of the joint.</li>
<li>1 fastener at 33% and the other at 67%. This is also not great because the middle section of the joint is receiving a double dose of clamping force, which is obviously not very evenly distributed.</li>
</ul>
<p>From these two examples, I can see that the ideal position is somewhere between 1/4 and 1/3 of the joint length for each fastener. What would this position exactly be?</p>
<p>What about a joint with more than two fasteners?</p>
| |mechanical-engineering|design|stresses|fasteners|joining| | <p>I think you might have a misconception regarding to how far the pressure from the fasteners extends. One subject you might want to have a look into is "bolt joint stiffness".</p>
<p>The most popular is the "Rotscher’s pressure-cone method". Essentially there is
a pressure cone which radiates outwards with an <em>pressure cone angle</em> a.</p>
<p><a href="https://i.stack.imgur.com/8hyIn.png" rel="noreferrer"><img src="https://i.stack.imgur.com/8hyIn.png" alt="enter image description here" /></a></p>
<p>According to Shigley using an <span class="math-container">$a= 30^\circ$</span> is usually suitable. The pressure cone angle is determined by the material properties, design of the fastener etc. However, the angle is less than <span class="math-container">$45^\circ$</span>. This means that it will be impossible for a long (>5 bolt diameters) and thin plate (less that 3 diameters) to be clamped uniformly with only two fasteners.</p>
<p>Having said the above, the clamping force and the friction should (more or less) remain the same irrespective of the position of the clamp. IMHO, what you should be more worried is <strong>where should I place the fasteners so that the whole system exhibits better strength</strong>.</p>
<p>For the strength, you can go the <em>analytical way</em> (see Shigley, Norton or other mechanical design textbook), or follow some rules of thumb (see <a href="https://mechanicalelements.com/understanding-bolt-choice/" rel="noreferrer">1</a>, <a href="https://fieldfastener.com/2016/09/22/rules-of-thumb-for-fastening-and-joint-design/" rel="noreferrer">2</a>, ... etc).</p>
| 37889 | What's the best way to design a joint with two fasteners? |
2020-09-29T13:41:01.757 | <p>I am familiar with using <span class="math-container">$\dot{Q}=c_p\cdot\dot{m}\cdot\Delta T$</span> to calculate the heat transfer rate of a fluid given a singular value for specific heat capacity (such as with water), but how do I go about calculating heat transfer rate for an aequeous solution such as <span class="math-container">$MgCl_2 (aq)$</span>? Do I somehow use the heat capacities of both water and salt together?</p>
| |thermodynamics|heat-transfer|cooling|refrigeration| | <p>The formula you are quoting is for estimating the heat exchange rate of a fluid that enters a control volume with rate <span class="math-container">$\dot m$</span> and has a change in temperature <span class="math-container">$\Delta T$</span>.</p>
<p>If you know</p>
<ul>
<li>the mass rate of the solution</li>
<li>the precise per weight ratio of your solution</li>
<li>the heat capacity of the elements</li>
<li>the temperature difference</li>
</ul>
<p>And provided there are not endothermal or exothermal reactions, then its basically a pretty straight forward sum of the parts.</p>
<p><span class="math-container">$$\dot{Q}_{total} = \dot{Q}_{water} + \sum _{i=1}^n \dot{Q}_{sub.1} $$</span>
<span class="math-container">$$\dot{Q}_{total} = \dot{m}_{water}c_{p,water}\Delta T + \sum _{i=1}^n \dot{m}_{sub.i}c_{p,sub.i}\Delta T $$</span></p>
<p>However, you will find that in most cases, because <span class="math-container">$c_p$</span> of water is so much greater that most other substances and the weight percentage in most solutions is much greater, you probably don't need to bother.</p>
| 37898 | How is heat transfer calculated for an aqueous salt solution? |
2020-09-30T06:47:07.177 | <p>Talking about a wire potentiometer, I was wondering about the way it could be used after installation. as you know, wire sensors have a roll of wire inside them that come out of a hole, and it sounds like that the wire should always remain direct and in the same orientation as it comes out.</p>
<p>for example in <a href="https://www.gefran.com/it/it/download/4625/attachment/en" rel="nofollow noreferrer">this link</a></p>
<p>you can see that the wire is always held along one axis , let's say X axis to verify the length of the wire:</p>
<p><a href="https://i.stack.imgur.com/4cO8O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4cO8O.png" alt="enter image description here" /></a></p>
<p>if my interest is the length change ( not the absolute current value) of the wire, and the wire does move not only in X direction, but let's say also in Y (and or) Z direction , while my interest is in <span class="math-container">$\delta L$</span> of the wire, not only in one certain axis, how much error would we introduce to the measurement ? is there a way to approximate that ?</p>
| |measurements|sensors|instrumentation| | <p>As far as I can see- I think your error in measuring <span class="math-container">$\delta L$</span> should be zero.</p>
<p>I mean apart from the increased friction, that might hinder the movement a bit, I don't think that you should get any error in the measurement of length.</p>
| 37908 | Orientation of wire in linear position wire sensors |
2020-09-30T09:51:44.037 | <p>I'm trying to make a adaptor to fit a nozzle about 5mm wide which only applies pressure <em>radially</em> during insertion. Pushing in a standard O Ring would apply some force <em>axially</em> up or down the inlet due to friction.</p>
<p>The adaptor should seal in the inlet, before injecting gas to the object, see sketch. I can't change the object in any way, only the adaptor. Considering the size, can anyone suggest something off the shelf I can use for this? Or how I might make it at all?
<a href="https://i.stack.imgur.com/u516R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u516R.png" alt="radial seal sketch" /></a></p>
| |pressure|pressure-vessel|seals| | <p>The idea that I was describing in the comments earlier is pretty similar to that described by Mart - I imagined the tube (red) flaring out underneath the O ring, and the opposing taper being either pushed down by hand or driven by a standard nut (black) depending on the sealing forces required.</p>
<p>The key difference is that instead of relying on the squashing of the seal to make it oval so it reaches the walls, you are stretching the seal onto a larger diameter.</p>
<p><a href="https://i.stack.imgur.com/Clti7.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Clti7.png" alt="seal idea" /></a></p>
| 37910 | Making a radial (only) seal inside a small tube |
2020-09-30T11:19:48.303 | <p>Once we study vector calculus, we are all set to describe physical fields such as velocity fields , electromagnetic fields etc. The study makes sense to me mathematically but I can not imagine how we would measure such quantities. How exactly do you continuously assign a number to each point in space? The amount of measurements would be way too much.</p>
<p>As a note, though I started with mathematics I expect an answer which describes the experimental procedures which are done.</p>
| |experimental-physics| | <p>I'm sure you are familiar with both ways, so I'll just point out the obvious (hoping not to insult your intelligence). There are many approaches, but I'll limit it to the few that I use more often. Namely:</p>
<ol>
<li>least squares</li>
<li>Kernel density estimation</li>
</ol>
<ul>
<li>1 bonus for completely deterministic system</li>
</ul>
<p>I'll give it a try with a 1D distribution function, but you can easily extend to more dimensions.</p>
<h1>Least Squares</h1>
<p>So assume you have the following set of data.</p>
<p><a href="https://i.stack.imgur.com/RrN7J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RrN7J.png" alt="enter image description here" /></a></p>
<p>Obviously there is trend there (and I will assume it follows a second order from my theoretical model). Then if I fit my data with a polynomial of second order I would get the following distribution.</p>
<p><a href="https://i.stack.imgur.com/ToWt9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ToWt9.png" alt="enter image description here" /></a></p>
<h1>Kernel density estimation/histogram</h1>
<p>The next case is when you have a lot of data which you gather and you want to get a distribution. The following is from a dataset with Earthquake Magnitudes above 4 Richter (quakes) around Fiji since 1964.</p>
<p><a href="https://i.stack.imgur.com/8PgYv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8PgYv.png" alt="enter image description here" /></a></p>
<p>If you are interested in the density distribution, then you would create either a <strong>histogram</strong> or a <strong>kernel density estimator (kde)</strong>, and you would plot the Earthquake magnitudes on the horizontal axis and probability/probability density on the vertical.</p>
<p><a href="https://i.stack.imgur.com/QrzP6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QrzP6.png" alt="enter image description here" /></a></p>
<h1>Bonus: Interpolation</h1>
<p>This is probably not likely the one that you would expect to use often when doing an experiment. Still, it has its uses. To use it you should get <em>everytime</em> the same results for X. The calculation is also trivial and has many variation (linear interpolation, log interpolation etc).</p>
| 37912 | How do you exactly measure distribution functions? |
2020-10-01T07:12:23.313 | <p>I was amazed when I got to know that there is such a simple mechanism know as Trompe to compress air.
It has no moving parts which make it cheap to construct and do not require any servicing.
Even the air being compressed has negligible moisture and its operation makes less noise unlike the compressors used in industries.
If this system has so many positive points so why it is not used widely? Is it due to its big size?</p>
| |compressed-air| | <p>Simply because it's so wasteful regarding energy. If you employ the flow and height difference of water for a hydroelectric generator to power a standard compressor (be it piston, centrifugal, or just a big fan), you can achieve vastly better pressures and air flow than with a trompe.</p>
<p>Additionally you won't need to have the water drop and air intake in the same place, so instead of air ducts and aqueducts/trenches/pipelines you can use much more convenient electric cables to bring the energy from the source to where the air flow is needed. Which is rather important, especially for larger installations - in case of the trompe you're limited to vicinity of rivers and streams with large height difference over short distance, which are on themselves rather rare, and you can't transport the air thus produced very far.</p>
<p>The definite upside of the trompe is that it's very fail-proof. Where the location allows, and there is no need for a more efficient solution, it can do its work very well for long time with minimal maintenance - other than removal of foreign objects brought by the water flow, it hardly needs any servicing. But the limitations make such situations rare, plus if it's a significant installation, you're still better off producing hydroelectric energy and using funds obtained from surplus sold to maintain a good compressor.</p>
| 37928 | Why air compressing mechanism "Trompe" is not popular in usage? |
2020-10-01T11:11:25.997 | <p>In Eurocode, two equations for loads are given for STR limit state:</p>
<p><span class="math-container">$$1.15K_{FI}G_{kj,sup}+0.9G_{kj,inf}+1.5K_{FI}Q_{k,1}+1.5K_{FI}\sum_{i>1}\psi_{0,i}Q_{k,i}$$</span></p>
<p><span class="math-container">$$1.35K_{FI}G_{kj,sup}+0.9G_{kj,inf}$$</span></p>
<p>Designer is required to calculate using both, and the design value of the load comes from the equation that gives the <em>least</em> favourable result. <strong>But what is the actual difference of these equations, why do we need two?</strong> I can see that the bottom equation (6.10a) only takes into account permanent (dead) loads (letter <span class="math-container">$G$</span>) and the one above considers temporary (live) loads as well (letter <span class="math-container">$Q$</span>).</p>
<p><strong>Why don't we need to consider the live loads in one of the equations?</strong> My lecture material says that the lower equation is rarely the deciding one, but it must always be checked. I've read my lecture materials and the Eurocode section on this, but I can't really get any good explanation of this. If somebody could explain the difference between the two equations and why we need them both I would be very grateful!</p>
| |structural-engineering|eurocodes| | <p>I am not sure which version of Eurocode you have this (I have access to an older version circa 2002), and 6.10a and b have a slightly different form. In any case I will answer and I will later update my answer.</p>
<h1>Equation A: Dominant secondary actions and combination of transients loads</h1>
<p>First of all: equation a is not only one equation. It represents all possible combination of temporary actions on a structure (could be wind, snow, fire, earthquake etc).</p>
<p>Since Qs are temporary loads, it is highly unlikely that all of them will occur together. So, what the Eurocode suggests, is to assume that any point in time only one of the actions will be dominant, while all the other are multiplied by a reduction factor (see <span class="math-container">$\psi_{0,i}$</span>). So if you got 4 temporary actions, you need to perform 4 times that calculation.</p>
<p>The Dead weights remain constant. You need to notice that, for <span class="math-container">$G_{kj, sup}$</span> the coefficient is slightly less in eq. a (set to 1.15) compared to equation B.</p>
<h1>Equation B: Dominant dead loads</h1>
<p>Equation B represents only dead loads, but it has a slightly higher coefficient for <span class="math-container">$G_{kj, sup}$</span>. In this scenario, the dominant load is the dead load, and no temporary loads are considered.</p>
<p>**Update:**I found a reference online regarding the finnish SFS-EN. What happens is that <span class="math-container">$G_{sup}$</span> is the upper design value, and <span class="math-container">$G_{inf}$</span> is the lower design value, which can be used if for example you have a building with maximum occupant capacity or completely empty. The two values create a type of average (leaning towards the upper limit, in order to make the calculation more direct.</p>
| 37932 | What is the difference between to Eurocode design equations (6.10a and 6.10b)? |
2020-10-01T14:33:09.683 | <p>I have system I want to move with a DC motor that requires about 1 kW of power when taking into account rolling resistance, drag resistance and when moving on an inclined plane with a specific velocity. I had trouble finding small enough motors to provide that power so I asked a friend for help and he told me to use a gear system with low power motors.</p>
<p>So I ended having a question in my mind: Does the motor power, say a 1 kW motor, need to match the 1 kW requirement of the system or is it the torque provided by this motor that needs to match the torque of the system I want to move, in this case ((9.55 * 1000 kW)/ 600 rpm) = 15.91 Nm? If it's the torque then that means I can use a 200 W with known torque and a gear system to multiply the torque and match that 15.91 Nm requirement? If it's the power then is my only option using enough motors to match that 1 kW value, say 5 200 W motors? I'm sorry if the question is not totally clear, I'll do my best answer any questions.</p>
| |motors| | <p>What you'll find is that most electric motor (not all) exhibit a curve like the following</p>
<p><a href="https://i.stack.imgur.com/f0ELi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f0ELi.png" alt="enter image description here" /></a></p>
<p>So at very low rpms they tend to have a very stable torque and low power. As rpm increase at some point Power tends to flat out and torque start to drop, because just like @jko said :
<span class="math-container">$$Power = M_t \cdot \omega=M_t \cdot 2\cdot\pi \frac{n}{60}$$</span></p>
<p>So <strong>there is always a tradeoff</strong>, that's the important thing you need to keep in mind.</p>
<p>I don't know the application (although I suspect it is some sort of vehicle) and/or how you arrived to the number of 1kW, however if you are certain about that value, there is no getting around the power of the motor. So you'd need at least a 1 kW motor.</p>
<p>Having said that, <strong>if you are not concerned about velocity</strong> (or how fast you get things done), I think you don't need to worry about the power, and you can focus on the Torque. In that case the tradeoff is time. You will be able to do something big with a small motor given enough time.</p>
| 37936 | On using gears and a low powered motor |
2020-10-01T19:56:40.827 | <p>So I have just started reading the book "<em>Introduction to Continuum Mechanics_third edition</em>" written by W. Michael Lai, David Rubin, and Erhard Krempl.</p>
<p>At the end of the book's introduction, the authors stated that <strong>"One important requirement which must be satisfied by all quantities used in the formulation of a physical law is that they be coordinate-invariant."</strong></p>
<p>I have read about <a href="https://en.wikipedia.org/wiki/Principle_of_covariance" rel="nofollow noreferrer">Principle of Covariance</a> in Wikipedia. And so I understood, in the example given of newton's 2nd law, how mass, velocity, and force are independent of transformations done between reference frames/ coordinate systems. I have imagined if mass were coordinate-variant, then a body's mass will change if we viewed it from another frame of reference, which is something odd to think about.</p>
<p>But then I thought and asked myself, are there any quantities that are variant when changing a coordinate system, if yes, like what? so this is my first question, and the second, why do all quantities used in the formulation of a physical law must be coordinate-invariant in "Continuum Mechanics" ?</p>
| |fluid-mechanics|materials|solid-mechanics| | <p>Here is why coordinate invariance is mandatory:</p>
<p>Imagine a lab containing a physical system upon which we make measurements in the interests of deriving a physical law. There is a special class of transforms we can apply to the system which leave its dynamics invariant. For each such special transform, called a <em>symmetry transformation</em>, there is associated with it a conserved quantity.</p>
<p>In the case of a transform that moves our laboratory across the street, we find that it is a symmetry transform and the conserved quantity is <em>linear momentum</em>.</p>
<p>In the case of a transform which rotates the lab around some arbitrary axis, we find that this is also a symmetry transformation and in this case the conserved quantity is <em>angular momentum</em>.</p>
<p>In the case of a transform which adds an arbitrary number of seconds to all the stopwatches in the lab that we use to time-stamp our measurements, this too is a symmetry transformation and the conserved quantity is <em>energy</em>.</p>
<p>So, for linear momentum to be conserved, our physical law must be coordinate-invariant: it must make no difference where we position our lab in 3-dimensional space, in order to have momentum conservation be a feature of the physical law we are working on.</p>
| 37939 | Why do all quantities used in the formulation of a physical law must be coordinate-invariant? |
2020-10-01T23:11:44.303 | <p>I know that current brushless motors, have a non rotating core made of copper coils, with permanent magnets on the outside rotating around it, and eventually connected to a shaft. But i was wondering, if it was possible to make the exact opposite ?</p>
<p>I mean having a rotating core made of magnets around a shaft, with non rotating coils around it. Would it work? Hallback arrays of magnets would still work ?</p>
<p>I mean, phases are the same, and Magnetic field are not unidirectional. I don't see why it couldn't.</p>
<p>Thank you in advance.</p>
| |motors|magnets| | <p>One reason you might not be able to envision why it doesn't work is because it does. There are compromises with respect to torque and motor size, but what you're describing is called an <a href="https://www.radiocontrolinfo.com/brushless-inrunner-vs-outrunner-motor/" rel="nofollow noreferrer">inrunner motor</a>.</p>
<p>From the linked site:</p>
<blockquote>
<p>The permanent magnets on the outrunner are placed on the rotor and the
rotor spins on the outside case. On the inside of the motor are the
stator windings which do not rotate, they are fixed in position.</p>
</blockquote>
<p>Additionally:</p>
<blockquote>
<p>On the inrunner motor, you essentially have the complete opposite true
for how it is built. On the outer side of the motor is the case. The
case in this situation does not rotate and is fixed. The stator
windings are placed on the inside face of the case. When you spin the
motor shaft of an inrunner, you are spinning the rotor which also
contains the permanent magnets much like the outrunner. The difference
of course being that they are now at the center of the motor.</p>
</blockquote>
<p>It's the first time I've attempted to validate what I believe and it's good to see this confirmation.</p>
| 37942 | Can you make the magnets rotate inside the coils of Brushless motor? |
2020-10-02T00:34:05.380 | <p>I've been looking for a way to size an electric motor for a vehicle that could be assumed as a box-shaped load with a mass of 60 kg, up a 10° incline, at 1.6 m/s, using continuous tracks with 1 motor per side, but for some reason while searching for the answer I found different formulas for this and all of them gave me different answers.</p>
<p>What is the actual formula I should use to find the power required to move this load? As detailed as possible since it's for a practical application not only a theoretical question.</p>
<p>The last one I found is:</p>
<p><span class="math-container">$$P = \frac{F_t\ v}{\eta}$$</span>
<br />
Where <span class="math-container">$F_t = Rolling\;Resistance\; +\; Gradient\; Resistance\; + Air\; Resistance$</span> <br />
and <span class="math-container">$\eta = Transmission\; Efficiency$</span>.</p>
<p>Is this the correct formula I should be using? Am I missing something? Should I consider Air Resistance since the vehicle is not moving too fast? I noticed both rolling resistance and gradient resistance have a friction coefficient, do I use the same one on both equations?</p>
| |motors|propulsion| | <p>At 1.6[m/s] the Air resistance is negligible so I would dismiss that term.</p>
<p>However, a word of caution: <strong>Rolling resistance Force</strong> is not the same as <strong>friction force</strong> if you are using wheels.</p>
<p>I would point you to the following page <a href="https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html" rel="nofollow noreferrer">for a rough estimation of Rolling resistance</a>.</p>
<p>The other factors that might come into play are <em>negligible</em> compared to what you have outlined (namely <strong>gradient</strong> and <strong>Rolling resistance</strong>).</p>
<p><strong>If you use tires</strong> (which I expect with a load of 60 kg), the equation should have a form like:</p>
<p><span class="math-container">$$ P = (c\cdot W\cos(\alpha)+ W\cdot sin \alpha)v$$</span></p>
<p>which can be approximated by the following (as you mentioned in your comment):</p>
<p><span class="math-container">$$ P = (c\cdot W + W\cdot sin \alpha)v$$</span></p>
| 37944 | Sizing electric motor |
2020-10-02T11:15:57.847 | <p>Recently dug myself into the theory of shock response spectrum analysis, but one thing is not clear for me.</p>
<p>The theory says that the peak response of the structure can be calculated as the product of the participation factor and the pertaining point of the shock response spectrum.</p>
<p>The participation factor is calculated as:
<span class="math-container">$$
\Gamma_{kj}=\phi_{j}^{T}\textbf{M}\textbf{1}_k
$$</span>
where <span class="math-container">$\phi$</span> is the eigenvector, <span class="math-container">$\textbf{M}$</span> is the mass matrix and <span class="math-container">$\textbf{1}$</span> is the corresponding rigid body motion vector.</p>
<p>Since the partcipation factor calculation includes the eigenvactors, which are not exact as they are scaleable, it would mean, that the response of the strcture would depend on the eigenvector normalization.</p>
<p>So my question would be that does it mean that it is assumed that for this type of analysis the eigenvectors are mass normalized and if so are they bounded to a specific value?</p>
<p>Reference:
<a href="https://www.comsol.co.in/multiphysics/response-spectrum-analysis" rel="nofollow noreferrer">https://www.comsol.co.in/multiphysics/response-spectrum-analysis</a></p>
| |vibration|modal-analysis|shock|eigenvalue-analysis| | <p>In the section "The Multiple DOF System" the Comsol document says</p>
<blockquote>
<p>It has been assumed that the mass matrix normalization of the
eigenmodes is used and that the damping matrix
can be diagonalized by the eigenmodes.</p>
</blockquote>
<p>Eigenvectors are assumed to be mass normalized in any mathematical derivation using them, unless somebody wants to be deliberately perverse.</p>
<p>The sensible way to define modal participation factors makes them dimensionless quantities. For mass normalized vectors, <span class="math-container">$\phi_j^{-1}\mathbf{M}\phi_j = 1$</span> so <span class="math-container">$\phi_j^{-1}\mathbf{M}$</span> has the dimension 1/length, and that is multiplied by a length (the rigid body motion vector) in your formula.</p>
<p>Note the participation factors can be negative, because the <em>sign</em> of a mass normalized eigenvector is arbitary even though its magnitude is fixed by the normalization.</p>
| 37953 | Are the modal participation factors bounded for shock response spectrum analysis? |
2020-10-02T15:24:01.493 | <p>One may want to ride an electric bike with the electronics turned off. There are many reasons to do this, i.e. flat battery, fitness (who would have thought)</p>
<p>When riding an ebike conventionally, is there a resistive friction caused by the DC motor?</p>
<p>If so, do e-bikes have a way to 'disengage' the motor from the wheel? Potentially, a clutch</p>
| |electrical-engineering|motors|electric-vehicles|bicycles| | <p>Different electric assist bikes will present different circumstances. For instance, a standard hub motor (front or rear wheel) will have permanent magnets which react with the iron in the stator. You can feel what is called cogging when rotating the wheel by hand, without power. This is energy expended by hand, but also applies when pedaling without applying throttle. It's not zero, but it's also not a substantial amount.</p>
<p>Geared hub motors will have an internal freewheel, which prevents this cogging. One can rotate the wheel (in the forward direction) and not feel interaction with the magnets. This particular design has less energy loss to consider when pedaling without applying throttle.</p>
<p>Mid-drive motors can be considered as geared hub motors, as they also have freewheels that permit pedaling without motor rotation and the description above applies there as well.</p>
<p>My now-sold velomobile used a <a href="https://www.clevercycles.com/cleverchimp-stokemonkey-human-electric-hybrid-driv.html" rel="nofollow noreferrer">Stokemonkey</a> electric assist which required me to pedal whenever power was applied by battery/throttle, but would not spin when I pedaled without application of throttle. There is a freewheel on the motor which had minimum resistance, but not zero.</p>
<p><a href="https://i.stack.imgur.com/KhugB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KhugB.png" alt="stokemonkey drive system with freewheel" /></a></p>
<p>Typically, there will be very little additional friction or load to consider, although the non-geared hub motor design is going to be the least efficient when operating without battery power applied.</p>
| 37961 | With ebikes, is there friction from the motor when riding with the electronics turned off? |
2020-10-02T19:23:06.083 | <p>how can I calculate how many newtons a 600[mm] long rod with a known yield strength (300 [MPa]), young modulus (190 [GPa]) and hollow (outer diameter = 16[mm], inner diameter = 10[mm]) will carry without buckling?</p>
<p>Thanks.</p>
<p>Edit: My main question is "How do I use the yield strength in the Euler buckling formula? What does it do and how do I determine "Factor Counting for End Conditions" for suspension pushrods?</p>
| |mechanical-engineering|buckling| | <p>For Euler Buckling the critical load is:</p>
<p><span class="math-container">$$ P_{cr} = \left(\frac{\pi}{K L}\right)^2 E I $$</span>
where:</p>
<ul>
<li><span class="math-container">$P_{cr}$</span> is the critical load</li>
<li><span class="math-container">$E$</span> is Young's modulus (for steel assume 200 [GPa])</li>
<li><span class="math-container">$I$</span> is the second moment of area. (<span class="math-container">$\frac{\pi}{64}(d_o^4- d_i^4)$</span></li>
<li><span class="math-container">$d_o, d_i$</span> is the outer and inner diameter respectively</li>
<li><span class="math-container">$L$</span> is the length of the rod</li>
<li><span class="math-container">$K$</span>: is a parameter that depends of how each end of the beam is constrained.</li>
</ul>
<p><a href="https://i.stack.imgur.com/KnVLO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KnVLO.png" alt="enter image description here" /></a></p>
<p>K is the most difficult to determine. For your case, would be 1.</p>
<p>Additionally, regarding the question of <strong>the yield stress <span class="math-container">$\sigma_y$</span></strong>. The way you use it, is that you need to check whether, the <span class="math-container">$P_{cr}$</span> is greater than the <span class="math-container">$\sigma_y\cdot A=\sigma_y\cdot \frac{\pi}{4}\cdot(d_o^2 - d_i^2) $</span>. If it is greater then yield occurs earlier than buckling.</p>
| 37969 | How do I calculate how many newtons a hollow rod can carry without buckling? |
2020-10-03T00:21:14.327 | <p>I am working on upgrading a research project on wind energy. They are trying to measure the performance of wind blade designs in a small wind tunnel (i.e. their models are relatively small).</p>
<p>What they were doing was measure the torque generated by the wind by mounting on the axis of the wind blade a bicycle discbrake and engaging it. (They have gone through many mechanical iterations)</p>
<p><a href="https://i.stack.imgur.com/G4IUy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G4IUy.png" alt="enter image description here" /></a></p>
<p>Currently they are using a <strong>servo motor</strong> to pull a <strong>bike cable</strong> that activates the clamps. What they do is: blade is accelerated, and then the brakes are applied and bring the blade to a halt. While the blase is slowing down, they are measuring the <em>torque</em> and <em>rpm</em>. In order to maintain a constant force on the clamp they are using a <strong>spring</strong> to maintain a constant tension of the bike cable.</p>
<p>I am upgrading the measurement software side, and two of the <em>wishlist</em> feature requests were :</p>
<ul>
<li>to control the system to apply enough clamping force that will maintain the number or rpms. (apparently they had tried that in the past with a PID but it did not work out).</li>
<li>to maintain a constant torque on the design (i.e. clamping force on the discbrake).</li>
</ul>
<p>When I saw the system I could see the following issues:</p>
<ul>
<li>given the discbrake design, the available travel between engaging and free running is very small (a few hundred microns). That makes the system the system very non linear and very difficult to control (in a few microns you go from 0 to max braking power - even with a lever).</li>
<li>there is a lot of <em>play</em> on the discbrake (its wobbly) - due to the mounting of the shaft- , however because the shaft is spinning at 3000 to 10000 rpm it tends to self align.</li>
</ul>
<p>Overall, I wasn't hopeful that I could achieve the level of control with their current system.</p>
<p>So my question (inspired while I was reading <a href="https://engineering.stackexchange.com/questions/37970/what-is-a-constricting-actuator-called">this question</a>) is <strong>what mechanisms can I look into to apply a controllable clamping force (or equivalenly clamping torque) on a 3[mm] spinning shaft which is rotating at 3000-10000 rpm?</strong></p>
| |control-engineering|design|torque|measurements|mechanisms| | <p>The position of the calipers isn't what you are adjusting, it's the <em>pressure</em> you exert on them.</p>
<p>Assuming the system is relatively slow (break force ramps up and down ~1 second), then a spring which is tensioned by a servo should work fine for this.</p>
<p><a href="https://i.stack.imgur.com/i5vMh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i5vMh.png" alt="enter image description here" /></a></p>
<p>If the spring is stretched further, the breaking force will be increased. Notice that even though the cable is barely moving, the force applied to it is changing.</p>
<p>You stated that PID control of this servo for constant torque or rpms "didn't work", but why didn't it work? Was the servo too slow to react? Was the control loop unstable? I think we need more specifics.</p>
| 37978 | Mechanism for controlling the clamping force |
2020-10-04T13:31:27.290 | <p>I want to verify the pump curve of a regenerative turbine pump. The manufacturer doesn't supply any pump curve of it.</p>
<p>This is the test setup.</p>
<p><a href="https://i.stack.imgur.com/0B20K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0B20K.jpg" alt="enter image description here" /></a></p>
<p>With the discharge valve shut off (plugged) and water line opened. I measured the following discharge pressure of 28 psi.</p>
<p><a href="https://i.stack.imgur.com/LEoKu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LEoKu.jpg" alt="enter image description here" /></a></p>
<p>This is the inlet or suction pressure of 14.3 psi.</p>
<p><a href="https://i.stack.imgur.com/nHsuw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nHsuw.jpg" alt="enter image description here" /></a></p>
<p>Is it correct to deduct the discharge pressure from the suction pressure to arrive at the head dead pressure? It is 28 psi - 14.3 psi = 13.7 psi.</p>
<p>So 13.7 psi is the pump dead pressure right?</p>
<p>I made a lot of measurements with pressure gauges and water flow rates to get some idea of the pump curve which the pump doesn't have</p>
<p>This is the pump specs. 8 is the meter in heads or about 11.7 psi.</p>
<p><a href="https://i.stack.imgur.com/s968U.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s968U.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/rpFxt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rpFxt.jpg" alt="enter image description here" /></a></p>
<p>This is my bathroom minipump connected to a heater and shower with 2 gauges label by A and B. The green arrow is the water pipe entry to the bathroom with 1/2 diameter hole. It can produce 10 liter per minute (2.64 gallon per minute) when directly measured with a container without all the hoses which are 1/4" in diameter (note all tankless heaters use the same 1/4" flexible hoses).</p>
<p>Without the pump turn on. I'm getting 4 liters per minute (about 1 gallon per minute) of water in the shower. With pump turn on. I'm getting 6 liters per minute (about 1.6 gallon per minute of water in the shower).</p>
<p>With and without the pump turned on. The gauge in the bathroom pipe entry labeled in A read 13 psi.</p>
<p>Without the pump turned on. The pressure gauge at B (after pump) reads 10 psi. With the pump turned on. The pressure gauge (after pump) reads 20 psi.</p>
<p>Do the actual test data above support the pump specs shown above?</p>
<p>From my existing water of 6 liter/minute (1.5 gallon per minute). I want it to be at least 2.5 gallon per minute (or 9.4 liter per minute). So i need to understand pump curve to be sure i got the right one.</p>
| |pumps| | <p>To build a pump cure (especially when you don't have many data points), first look at an existing pump curve for the pump technology you are inspecting. Here is one for a <a href="https://www.rothpump.com/regenerative-pumps-hydrocarbon-processing.html" rel="nofollow noreferrer">Regenerative Turbine Pump</a>.</p>
<p>Your 13.7 psi dead head pressure is correct based on your data. This is one point on the graph.</p>
<p>You have also evaluated one running point: 1.6gpm. In this condition the pressure was increased from 14.3 psi to 20psi. 20psi-14.3psi = 5.7psi differential. This is also a point on the graph.</p>
<p>Generally a pump curve is evaluated with lots more points, and going from atmospheric pressure suction to a gate valve that can be adjusted. This lets you easily move across the curve and doesn't require any math or second pressure measurement.
The last point (at minimum) you need is the unrestricted flow. With an atmospheric pressure intake and an atmospheric pressure outlet, measure the flow.</p>
<p>I have sketched a table and rough plot. Once you have your data you should plot it in a spreadsheet with a trend line. What I have drawn is just a guess and not to scale.</p>
<p><a href="https://i.stack.imgur.com/TTQd5m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TTQd5m.jpg" alt="enter image description here" /></a></p>
<p>Another thing to consider is that your shower head nozzle also has a pressure flow curve. Don't mix the two. It can be evaluated exactly the same way though by measuring pressure going into the nozzle (the other side is atmosphere). Then record the flow and pressure into a table. Repeat this for various pressures/flows and graph.</p>
| 38006 | Determining Pump Dead Head Pressure and Pump Curve |
2020-10-04T18:17:28.077 | <p>In particular drone propeller blades. I read that the noise we hear is the propeller compressing the air at the edge of it to produce a sonic boom. Is that true and is there a way to negate that?</p>
| |mechanical-engineering|design| | <p>in a subsonic air propeller blade, there are two sources of noise: first, the "hiss" of air moving past the blades and then <em>blade passage noise</em> which is a sound wave that sweeps around with the blade as it revolves, producing a sound pulse with each blade passage. It has a characteristic frequency of (prop revolutions per second) x (number of blades) and yields a buzzing hum that rises and falls in pitch as the prop speeds up or slows down.</p>
<p>The hiss is minimized by running the prop at slower speeds, which requires adding diameter or more blades to maintain the flow rate. the blade passage noise is minimized by widening the blades to the point where the leading and trailing edges of the adjacent blades begin to overlap.</p>
<p>Another trick to minimize blade passage noise is to cut a sawtooth pattern into the trailing edge of each blade right at the blade tip, which breaks up the sound wave pattern and muffles the hum- while increasing the hiss, however.</p>
| 38009 | How can propeller blades be made silent? |
2020-10-05T03:00:16.770 | <p>I hope I'm not too much off topic here. I'm hoping it can make sense as a mechanical engineering question.</p>
<hr />
<p>There is a crack in the front of the tub of my front loading washing machine.</p>
<p>As you can see in the drawing <code>#1</code>, there is a weight attached right where the crack formed. Seeing the machine do the spin cycle, I assume quite a bit of force is applied, noted as <code>F1weight</code> and <code>F2weight</code>.</p>
<p>I'm thinking of putting epoxy glue but I think <code>F1weight</code> will destroy the joint.</p>
<p>What I think I should do, is put some kind of fiber with a high tensil strength into the epoxy glue. When it dries, the glue will fix the leak and hold the fiber in place. The fiber will be responsible of ensuring the integrity of the glue and resist <code>F1weight</code>.</p>
<p>I'm thinking something like glass fiber fabric that would sink in the epoxy?</p>
<p>Any idea what could be used here?</p>
<p><a href="https://i.stack.imgur.com/wF3IS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wF3IS.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|materials|plastic|epoxy| | <p>I wouldn't worry about reinforcing the epoxy as it will be as strong or stronger than the plastic. What you do need to focus on is the bond between the plastic and the epoxy.</p>
<p>First make sure that the crack cannot flex. I have not taken the time to review this design but if you feel it is a high stress area I would mechanically reinforce it with screws or wire. It appears to be pretty small in your photos so it may be fine as is. One way to attach it if it does flex is drill either side and twist steel wire to secure it.</p>
<p>Then make sure all bonding surfaces are roughed up with 60 grit sandpaper.</p>
<p>Clean the surface with a dry towel or compressed air then use a plastic specific epoxy. I have had pretty good luck with <a href="https://rads.stackoverflow.com/amzn/click/com/B0044FBB8C" rel="nofollow noreferrer" rel="nofollow noreferrer">loctite plastic bonder epoxy</a>. Make sure to cover a large area since the bond strength will not be as good as we would like. Also make sure to push it down into the crack and around the wire if used. If this area of the device holds water make sure that the application is also watertight.</p>
| 38018 | Fix plastic tub (leak) and resist applied forces |
2020-10-06T12:42:49.660 | <p>We are a student team trying to design a low cost CNC plasma cutting machine.
The CNC will be a table with two rails and a bridge moving on them in which the head of the plasma torch will be.</p>
<p>Now we have to decide on the configuration of the steeper motors to move the bridge. It has to be pushed from each end of it, we have tought about:</p>
<ul>
<li>Two motors, one on each end of the bridge.</li>
<li>One bigger motor in one end (or the middle) and a mechanical transmission (a shaft and a couple of gears) to push from the other end.</li>
</ul>
<p>Given that the small motors cost about $15 each and assuming the big one is double the price, mass, torque, etc, what option would give the most precision with an acceptable cost?<br />
Shaft would be a metal rod, but I dont know how much money could be precise enough gears.</p>
<p>EDIT 1:
We finally went for double motors.
As the project is aimed at managing the project itself and setting up a production chain, and not at designing a good product, we didn't think much about the pros and cons, and thought that just doing two of the same was less work than designing the mechanical transmission and its production and assembly process.</p>
| |control-engineering|gears|stepper-motor|cnc|power-transmission| | <p>If you are looking to move the whole bridge (gantry might be a better term), and size of the length of the gantry is more that a couple tenths of a meter (>0.3[m]), then using a large motor with gear is not going to be economic at all.</p>
<p>With a single motor your only viable solution might be a synchro belt, or a chain.</p>
<p><a href="https://i.stack.imgur.com/sYqFD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sYqFD.png" alt="synchro belt" /></a></p>
<p>Having two motors will probably be more easy to go.</p>
<p>For other issues that might crop up you can have a look at this <a href="https://engineering.stackexchange.com/questions/37994/cnc-side-linear-rails/37995#37995">question also</a></p>
| 38042 | Precision of a steeper motor and gears vs multiple motors |
2020-10-07T04:31:56.287 | <p>I'm working with an <a href="https://store.arduino.cc/usa/due" rel="nofollow noreferrer">Arduino Due</a>'s two DAC pins.</p>
<p>I want to output to a single 4W 5.7Ohm speaker to test my code, with the intent of eventually outputting to a set of speakers (woofer, mid-range, and tweeter) for better quality for frequencies between 87.31Hz to 3520Hz+.</p>
<p>I'm using the DAC pins in <a href="https://www.arduino.cc/reference/en/language/functions/zero-due-mkr-family/analogwriteresolution/" rel="nofollow noreferrer">12 bit</a> mode, with output between <a href="https://forum.arduino.cc/index.php?topic=184100.0" rel="nofollow noreferrer">.55 - 2.75V</a> @ <a href="https://store.arduino.cc/usa/due" rel="nofollow noreferrer">3mA</a> according to the linked sources (I know forums aren't definitive sources, but I've cross checked and many forum posts and sites are around this range. 3mA comes from total for all I/O pins of 130mA divided by 42 I/O pins total, not counting the 12 PWMs)</p>
<p>I understand that I need a circuit like the following:</p>
<ol>
<li>DAC output to "<a href="https://en.wikipedia.org/wiki/Reconstruction_filter" rel="nofollow noreferrer">reconstruction filter</a>" to smooth the quantized output.</li>
<li>anti-aliasing filter, possibly the same as above, to remove noise (<a href="https://en.wikipedia.org/wiki/Nyquist_frequency" rel="nofollow noreferrer">Nyquist Frequency stuff</a>)</li>
<li>Use an audio-tapered potentiometer for volume control by reducing amplitude of signal.</li>
<li>Removal of both the <a href="https://en.wikipedia.org/wiki/DC_bias" rel="nofollow noreferrer">DC offset</a> and <a href="https://electronics.stackexchange.com/questions/233230/why-are-dc-signals-bad-for-loud-speakers">centering the signal on 0v</a> using a capacitor/filter.</li>
<li>Use an OpAmp to as a voltage follower to buffer the Arduino Due from the load of the speakers.</li>
<li>Use an OpAmp (possibly the same one as above) to increase the power of the signal (via an external power source probably, because the Due cant drive much) to a range suitable for the speaker.</li>
<li>A high pass filter to remove unwanted noise from the output.</li>
</ol>
<p>I will be playing a sample at 16kHz, which I assume means I will have a variable duty cycle and therefore cannot use anything sensitive to this to function properly.</p>
<p>My questions are the following:</p>
<ol>
<li>Is this overkill for this kind of circuit, or way under the bar for a proper setup?</li>
<li>How do I know where my high pass filter cutoff frequency should be, as my sample is of an acoustic instrument with natural overtones, meaning knowing just the frequency of the root note is not enough.</li>
<li>Without a power source providing positive and negative voltage, is it enough to offset with a capacitor to get a balance on 0v, and is this lost over the OpAmp if it does not have a negative voltage going in? Can I just balance it at the end with a capacitor then?</li>
<li>Is the OpAmp buffer/voltage follower enough to stop my Arduino from dying?</li>
</ol>
<p>Thanks for the help. I'm new to audio circuits, my experience is in DC and programming the Arduino.</p>
| |circuit-design|signal-processing|audio-engineering|ac|microelectronics| | <p>Your understanding of the issues is good.</p>
<p>Q1: It's not overkill at all. All of the elements you've mentioned are needed.</p>
<p>Q2: The <em>low-pass</em> filter frequency needs to significantly attenuate all frequencies above 1/2 the sampling frequency. It makes sense for a 'low fi' arrangement to reduce the filter frequency a bit so as to simplify filter design which otherwise tends to be complicated.</p>
<p>Consider a Chebyshev filter design.</p>
<p>Q3: You refer to a design style known as a 'false ground' of 1/2 supply voltage. You need to bias an amplifier so that the output is at 0.5*Vsupply. You can then capacitively couple the load.</p>
<p>Q4: Yes, the buffer is the solution.</p>
<p>In passing, please note than an op-amp will not have sufficient current output to drive a speaker. There are many suitable chip amplifiers for that purpose and some are configured for gain, filtering etc just the same way as an op-amp.</p>
<p>Hopefully you won't need a <em>low pass</em> filter at the output. The anti-aliasing filter ought to suffice.</p>
| 38057 | Arduino DAC to Speaker |
2020-10-06T23:10:47.430 | <p>My friend and I have an idea to make prototype of vending machine.
We need suggestions for making strong dropping mechanism.
The idea is:</p>
<ol>
<li>The user orders X item. Let's take a can of cola for example.</li>
<li>Then the cola can drops in container.</li>
<li>Then you place your hand below the container and the item drops in your hand.</li>
</ol>
<p>The problems we have to solve are point 2 and 3.</p>
<p><strong>How to make the dropping mechanism, strong and gentle?</strong>
We have 3d printer.</p>
<p>Sample drawing: <a href="https://i.stack.imgur.com/4gf1I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4gf1I.png" alt="enter image description here" /></a></p>
| |design|servo|machine-design| | <p>Instead of a Servo, this could be very easily be done by the combination of a spring and a linear magnetic actuator. The downside is, that the setup needs a minimum weight of the item to overcome the closing spring.</p>
<p>Upon detection of the hand, the actuator retracts the locking bar and the weight of the item pushes the door open against the spring. As the item falls off the door, the spring pushes the door back to the start position. Once the door is back in position, it triggers the actuator to lock the door again.</p>
<p>Instead of springs, a second actuator could be used: a linear actuator could pull the link back via a tilting link, then return to its extended position afterward. The actuator's connection to the link rides in a slot, so the gate can fall freely. If the slot is replaced by just a hole, the actuator could also act as the lock.</p>
<p><a href="https://i.stack.imgur.com/AxvJt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AxvJt.png" alt="enter image description here" /></a></p>
<p>A turning actuator could push the lid closed by simply doing about a quarter turn and having a peg that presses down the lever to return the gate to the position. And, if the resting position is down, it might even act as the Lock in the bottom position. With the tab chosen as the right length and bend at the end, a full rotation of the motor would release, then close the hatch with just one servo.
<a href="https://i.stack.imgur.com/dbMws.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dbMws.png" alt="Rotary Actuator to return" /></a></p>
| 38058 | Looking for dropping servo mechanism idea |
2020-10-08T00:04:53.273 | <p>Hi Engineering Exchange,</p>
<p>I have a beam with pinned-pinned boundary conditions (free to rotate but not free to translate). A legacy computer analysis provided me with the maximum mid-span acceleration the beam experienced. I know the mechanical properties of the beam (E, I, weight, length, area). Ideally I would have re-run the computer model to calculate the displacement, but unfortunately its lost to time.</p>
<p>Is there a method that can approximate the maximum mid-span displacement given that I know the:</p>
<ol>
<li>Maximum acceleration</li>
<li>Mechanical properties</li>
</ol>
<p>Thank you!</p>
| |beam|dynamics|vibration| | <p>If we assume a single degree of freedom vibration of the beam as a spring and mass system and substitute beam with half of its mass concentrated at its center, which is not grossly wrong and we also assume no damping, the beam will vibrate under its natural frequency and a rough estimate of its vibration amplitude can be substituted for its deflection.</p>
<p>We know the spring stiffness is</p>
<p><span class="math-container">$$k = \frac{48EI}{L^3} $$</span></p>
<p>Then the beam's natural frequency will be</p>
<p><span class="math-container">$$\omega_n = \sqrt{\frac{k}{1/2M}}=\sqrt{\frac{2*48EI}{m*l^3}} $$</span> and acceleration is</p>
<p><span class="math-container">$$ \alpha = -\omega^2A*sin\omega *t$$</span></p>
<p>Maximum acceleration happens at <span class="math-container">$ \omega=n\pi$</span></p>
<p>By pluging <span class="math-container">$n=1$</span> And <span class="math-container">$ t= (1/*2f) $</span> we can calculate <span class="math-container">$A$</span> the amplitude or deflection.</p>
| 38078 | Estimation of Beam Displacement from Acceleration |
2020-10-08T14:13:57.303 | <p>Does anyone have insights into configuring UART port on Renesas RX130 microcontroller for a bit rate of 115200? I have successfully configured for a 9600, 19200, and 38400. I can read from the MCU and write to the MCU via terminal programs such as tera term. I am using the code generator available through e2Studio IDE.. I am using Serial Communication Interface SCI1, from which I have configured RXD1 / P30 (Pin 18) and TXD1 / P26 (Pin 20). Note P30 is port 3 bit 0 and P26 is port 2 bit 6.</p>
<p>Below is the MCU package configuration view enter image description here
<a href="https://i.stack.imgur.com/aIhTb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aIhTb.png" alt="enter image description here" /></a></p>
<p>Below is Code Generator software component configuration view SCI1 enter image description here
<a href="https://i.stack.imgur.com/Jra9V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jra9V.png" alt="enter image description here" /></a></p>
<p>Has anyone run into similar issues? Thanks for your help!</p>
<p>Note: I am open to exploring Firmware Integration Tool (FIT) to configure UART ports</p>
| |electrical-engineering|digital-communication| | <p>Try typing 115200 for bit rate. I think the drop down accepts edited values. The internal software will configure the registers appropriately.</p>
<p><a href="https://i.stack.imgur.com/zuPMU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zuPMU.jpg" alt="enter image description here" /></a></p>
| 38087 | Configuring UART Port to 115200 on Renesas RX130 using Code Generator |
2020-10-09T04:44:30.713 | <p>Many composite beams combine material that has a good tensile strength with material that has a good compression strength (like concrete and rebar).</p>
<p>I was wondering if there has ever been a beam that used tension cables on the side in tension. It seems to me that this might be a cheaper solution than using I beams in certain cases, though it would be more involved construction since cabling pieces would be needed (like turnbuckles).</p>
<p>Could this ever be practical? It seems to me that it could be in very specific circumstances, but I'd like to hear from more knowledgable people because I feel like I'm overlooking something.</p>
| |structural-engineering|materials|structures| | <p>Pretty standard old technology; A moot point whether prestressed cable or rebar is used. However ; cold drawn cable can be stronger. The problem is civil engineers don't understand the possible problem of hydrogen stress cracking of high strength steels. There have been several TV mockumentaries ( eg. Engineering Disasters) of serious cable failures. Usually on a sea coast with more corrosion potential . The explanations are often poor as the liberal arts majors who write the dialogue do not understand the principles. A recent TV program comes to mind regarding the "cheese grater" building in London; However in that case it is high strength bolts , not cables , failed because of hydrogen stress cracking. Most highway bridges are built with prestressed concrete. When these beams are being hauled on trucks ,you can often see a bow because of the prestress.</p>
| 38098 | Is it practical to use tensile cables in a composite beam? |
2020-10-09T09:50:09.207 | <p>When reading about a harmonic oscillator, the most common form of damping is proportional to the velocity.</p>
<p><span class="math-container">$$F_D = c \;\dot{x}$$</span></p>
<p>Without giving much thought to viscous damping, one example I always thought of is the following. <strong>Taking my hand outside a car with velocity v. As the velocity increases the drag also increases.</strong></p>
<p>However, I also know that the aerodynamic drag of a vehicle is proportional to the square of the velocity. I.e. :</p>
<p><span class="math-container">$$F_A = \frac{1}{2} C_D \rho A (\dot{x})^2$$</span></p>
<p>So, either my example does not fall under viscous damping, or there is something fundamental that I don't understand.</p>
<p>I would very much appreciate your insight on this.</p>
| |fluid-mechanics|fluid|drag| | <p>The equation <span class="math-container">$F_{\textrm{A}} = C_{\textrm{D}}A\rho\dot{x}^2/2$</span> doesn't tell you that drag is proportional to the square of velocity. In itself, that equation doesn't tell you anything at all about how drag depends on velocity, because that equation is just a definition of the drag coefficient <span class="math-container">$C_{\textrm{D}}$</span>. What tells you how drag depends on velocity is the combination of:</p>
<ul>
<li>the insight that (for all mutually geometrically similar objects in incompressible Newtonian fluids) the value of <span class="math-container">$C_{\textrm{D}}$</span> depends on other variables (including velocity) <em>only</em> through the combination of variables known as the Reynolds number; and</li>
<li>the graph showing <em>how</em> the value of <span class="math-container">$C_{\textrm{D}}$</span> depends on Reynolds number.</li>
</ul>
<p>At small values of the Reynolds number, <span class="math-container">$C_{\textrm{D}}$</span> is inversely proportional to the square root of Reynolds number (Landau, Lifshitz and Pitaevskii, 1987, <em>Fluid mechanics</em>, Butterworth Heinemann). Since the definition of Reynolds number has a single factor of velocity in its numerator, this means that, in this regime, drag is proportional to the <span class="math-container">$3/2$</span> power of velocity. While this still isn't the form known in dynamical systems theory as "viscous damping", it's closer to it than would be the case if drag were proportional to the square of velocity. On the other hand, as the value of the Reynolds number tends to infinity, <span class="math-container">$C_{\textrm{D}}$</span> tends to a non-zero constant. In this regime, drag really is proportional to the square of velocity, and drag is definitely not of a viscous-damping form.</p>
<p>If you're interested in the micromechanics of <em>why</em> <span class="math-container">$C_{\textrm{D}}$</span> is inversely proportional to Reynolds number for small Reynolds numbers, but tends to a non-zero constant as the Reynolds number tends to infinity, there's a nifty explanation in Faber (1995, <em>Fluid dynamics for physicists</em>, Cambridge University Press), under the headings "skin friction" and "boundary layer separation".</p>
| 38108 | Does the aerodynamic force qualify as a viscous damping force in the Mechanical Harmonic Oscillator |
2020-10-10T09:56:38.773 | <p>For a scalar <span class="math-container">$c$</span>, the function:
<span class="math-container">$$f(x) = cx$$</span>
Is linear in that it satisfies the superposition property. If:
<span class="math-container">$$f(x_1)=cx_1$$</span>
<span class="math-container">$$f(x_2)=cx_2$$</span>
Then:
<span class="math-container">$$\alpha f(x_1) + \beta f(x_2) = \alpha(cx_1)+\beta(cx_2)=c(\alpha x_1 + \beta x_2) = f(\alpha x_1 + \beta x_2)$$</span>
However, for a scalar <span class="math-container">$b$</span>, the function:
<span class="math-container">$$g(x)=cx+b$$</span>
Is not linear but is affine, since:
<span class="math-container">$$\alpha g(x_1) + \beta g(x_2) = \alpha(cx_1+b)+\beta(cx_2+b)=c(\alpha x_1 + \beta x_2)+b(\alpha + \beta) \neq g(\alpha x_1 + \beta x_2)$$</span>
How, then, can I apply the same reasoning to show that a differential equation is linear?</p>
<p>For example, the linear (or affine?) differential equation for a <a href="https://en.wikipedia.org/wiki/Harmonic_oscillator#Driven_harmonic_oscillators" rel="nofollow noreferrer">driven harmonic oscillator</a> is:</p>
<p><span class="math-container">$$\frac{d^2x}{dt^2}+2 \zeta \omega_0 \frac{dx}{dt} + \omega_0^2 x - \frac{F(t)}{m} = 0$$</span></p>
<p>This equation can be re-written as:
<span class="math-container">$$\mathbf{w}^T\mathbf{x} - \frac{F(t)}{m} = 0$$</span>
Where:
<span class="math-container">$$\mathbf{w}=\begin{bmatrix} 1\\ 2 \zeta \omega_0\\ \omega_0^2 \end{bmatrix}$$</span>
<span class="math-container">$$\mathbf{x}=\begin{bmatrix} \frac{d^2x}{dt^2}\\ \frac{dx}{dt}\\ x \end{bmatrix}$$</span></p>
<p>Which suggests that this is indeed affine. I am not sure if this reasoning is correct so would appreciate some feedback. Also, is there a special name for <span class="math-container">$\mathbf{x}$</span>?</p>
| |control-engineering|signal-processing| | <p>A system is some kind of function that maps an input as a function of t to an output.
<span class="math-container">$$y(t) = H(u(t))$$</span>
This system is linear if the following holds:
<span class="math-container">$$y_1 = H(u_1), \quad y_2 = H(u_2)$$</span>
<span class="math-container">$$\alpha y_1 + \beta y_2 = H(\alpha u_1 + \beta u_2)$$</span>
for any scalar value <span class="math-container">$\alpha$</span>, <span class="math-container">$\beta$</span>.
Your driven harmonic oscillator is currently described in such a way the combination of the force input <span class="math-container">$F(t)$</span> and the position <span class="math-container">$x(t)$</span> equals 0. As such, you could describe the system as the following:
<span class="math-container">$$ 0 = H(F(t), x(t))$$</span>
Apply the theorem of linearity on this function:
<span class="math-container">$$0 = H(\alpha F_1 + \beta F_2, \alpha x_1 + \beta x_2)$$</span>
<span class="math-container">$$0 = \frac{d^2}{dt^2}(\alpha x_1 + \beta x_2) + 2\zeta\omega_0\frac{d}{dt}(\alpha x_1 + \beta x_2) + \omega_0^2(\alpha x_1 + \beta x_2) - \frac{\alpha F_1 + \beta F_2}{m}$$</span>
<span class="math-container">$$0 = \alpha\left(\frac{d^2x_1}{dt^2} + 2\zeta\omega_0\frac{dx_1}{dt} + \omega_0^2x_1 - \frac{F_1}{m}\right) + \beta\left(\frac{d^2x_2}{dt^2} + 2\zeta\omega_0\frac{dx_2}{dt} + \omega_0^2x_2 - \frac{F_2}{m}\right)$$</span>
<span class="math-container">$$ 0 = \alpha H(F_1, x_1) + \beta H(F_2, x_2)$$</span>
Because it is not intentional to have both the position and the force as inputs, the system could be described as the following:
<span class="math-container">$$m\frac{d^2x}{dt^2} + 2\zeta\omega_0m\frac{dx}{dt} + \omega_0^2mx = F(t)$$</span>
In this case, the position is the input of the system and the force the output. If you apply the properties of linearity on this system you will notice it is linear as well.</p>
<p>If a constant offset in a system that is logically not really an input (for instance its not controllable), the system is indeed not linear. For example:
<span class="math-container">$$y(t) = F(x(t)) = cx(t) + b$$</span>
This is often solved by linearizing the system such that the input takes this offset into account (supposing its known or can be approximated):
<span class="math-container">$$y(t) = F_l(u(t)) = cu(t), \quad u(t) = x(t) + \frac{b}{c}$$</span>
With this, the function is still not linear in <span class="math-container">$x(t)$</span>, but is linear in <span class="math-container">$u(t)$</span>.</p>
| 38133 | Shouldn't linear time-invariant systems be called affine time-invariant systems instead? |
2020-10-11T05:05:37.080 | <p>I'm reading a book on Control System Design and running into the same issue I had in Kinematic Design. When things are presented in a format without any numerical examples it just appears as gibberish to me and always has (I do have a learning disability which maybe my issue.) So I'm working the problems and trying to take the problem statement of <span class="math-container">$$ H(s) = \frac{(s+2)(s+4)}{(s+1)(s+3)(s+5)} $$</span> and place it into a block diagram as well as a matrix. The issue I'm running into is that the only examples I'm seeing involved <span class="math-container">$$ H(s)= \frac{b_0s^k+b_1s^{k-1}+b_k}{s^k+a_1s^{k-1}+a_k} $$</span> which is makes absolutely no sense to me. Any help would be appreciated.</p>
| |control-engineering|control-theory| | <p>Since <span class="math-container">$H(s)$</span> describes a transfer function, then:</p>
<p><span class="math-container">$$\tag{1}\label{eq1}
H(s) = \frac{Y(s)}{X(s)}$$</span></p>
<p>For some arbitrary input <span class="math-container">$X(s)$</span> and arbitrary output <span class="math-container">$Y(s)$</span>. It is usually the case that you are given the input <span class="math-container">$X(s)$</span> and the transfer function <span class="math-container">$H(s)$</span>, and so equation \eqref{eq1} is re-arranged as:</p>
<p><span class="math-container">$$Y(s) = X(s) \cdot H(s) \tag{2}\label{eq2}$$</span></p>
<p>So, the purpose of the transfer function is to describe the behavior (output) of a system to an arbitrary input <span class="math-container">$X(s)$</span>, as described by equation \eqref{eq2}. For example, suppose that <span class="math-container">$H(s)=(s+2)$</span> and <span class="math-container">$X(s)=s$</span>. Then:</p>
<p><span class="math-container">$$Y(s) = X(s) \cdot H(s) = s(s+2) = s^2 + 2s$$</span></p>
<p>Now, note that:</p>
<p><span class="math-container">$$H(s)= \frac{b_0s^k+b_1s^{k-1}+b_k}{s^k+a_1s^{k-1}+a_k}$$</span></p>
<p>Can be obtained by expanding:</p>
<p><span class="math-container">$$H(s) = \frac{(s+2)(s+4)}{(s+1)(s+3)(s+5)}$$</span></p>
<p>For example, for the numerator:</p>
<p><span class="math-container">$$(s+2)(s+4) = s^2 + 6s + 8$$</span></p>
<p>So:</p>
<p><span class="math-container">$$H(s) = \frac{s^2 + 6s + 8}{(s+1)(s+3)(s+5)}$$</span></p>
<p>For the denominator:</p>
<p><span class="math-container">$$(s+1)(s+3)(s+5)=(s+1)(s^2+8s+15)=s^3+9s^2+23s+15$$</span></p>
<p>So:</p>
<p><span class="math-container">$$H(s) = \frac{s^2 + 6s + 8}{s^3+9s^2+23s+15}$$</span></p>
<blockquote>
<p>and place it into a block diagram as well as a matrix.</p>
</blockquote>
<p>Referring back to equation \eqref{eq1}, the block diagram of this system is just:</p>
<p><a href="https://i.stack.imgur.com/VOwGo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VOwGo.png" alt="enter image description here" /></a></p>
<p>Where:</p>
<p><span class="math-container">$$H(s) = \frac{s^2 + 6s + 8}{s^3+9s^2+23s+15}$$</span></p>
<p>However, I am not sure what you mean by placing <span class="math-container">$H(s)$</span> into a matrix. Perhaps you mean convert <span class="math-container">$H(s)$</span> into its <a href="https://en.wikipedia.org/wiki/State-space_representation" rel="nofollow noreferrer">state space</a> representation?</p>
| 38147 | Transfer function into block diagram and matrix form |
2020-10-11T15:24:55.427 | <p>In a given simulation in MSC Adams, if I use a 1 or 0.1 step increment, I get the (favorable) result I intuitively assumed would occur, but if I use step increments of 0.01 or 0.001 a different (and unfavorable) result occurs. One would assume the finer-detailed result would generate fewer simulation false-result-inducing artifacts. Should I always assume the more granular simulation will result the more realistic and reliable result?</p>
| |simulation| | <p>This is a common issue on all numerical methods (not just multibody problems), i.e. the tradeoff between the following three types of error:</p>
<ol>
<li><a href="https://en.wikipedia.org/wiki/Round-off_error" rel="nofollow noreferrer">Round off error</a></li>
<li><a href="https://en.wikipedia.org/wiki/Discretization_error" rel="nofollow noreferrer">Discretization</a></li>
<li><a href="https://en.wikipedia.org/wiki/Truncation_error" rel="nofollow noreferrer">Truncation</a> (sometimes the 2 and 3 are included in the same term).</li>
</ol>
<p>All of them are forms of quantization error.</p>
<p>These three types have a dependence on mesh size (either spatial or temporal). You can see their dependence, in the following image.</p>
<p><a href="https://i.stack.imgur.com/zCmH9.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zCmH9.gif" alt="enter image description here" /></a></p>
<p>As you reduce the time step (or mesh size) the discretization error decreases, however then what happens is that you get the round off error to increase. Usually, it's the balance between those two that dominates the total error.</p>
| 38151 | Is a MSC Adams simulation with smaller steps always more reliable? |
2020-10-11T23:43:03.457 | <p>Given two points <span class="math-container">$(t_0,x(t_0)=x^{0})$</span> and <span class="math-container">$(t_1,x(t_1)=x^{1})$</span> in the <span class="math-container">$(t,x)$</span> plane, the objective is to find an optimal trajectory <span class="math-container">$x^{*}(t)$</span> such that the cost function</p>
<p><span class="math-container">\begin{equation}\label{eq:1}
J(x) = \int_{t_0} ^ {t_1} g(x,\dot{x},t) dt \tag{1}
\end{equation}</span></p>
<p>has a relative extremum with <span class="math-container">$g$</span> being a function with continuous first and second partial derivatives with respect to all its arguments. In this case where the end points are fixed, the necessary condition for minimizing the functional <span class="math-container">$J(x)$</span> is obtained by means of Euler-Lagrange equation</p>
<p><span class="math-container">\begin{equation}\label{eq:2}
\dfrac{\partial{g(x^{*},\dot{x}^{*},t)}}{\partial{x}} - \dfrac{d}{dt} \dfrac{\partial{g}(x^{*},\dot{x}^{*},t)}{\partial{\dot{x}}} = 0.\tag{2}
\end{equation}</span></p>
<p>Next, as I was studying the same problem (from Modern Control System Theory by M. Gopal) with the modified constraint - terminal time <span class="math-container">$t_1$</span> fixed but <span class="math-container">$x(t_1)$</span> free, I encountered a bit of difficulty (highlighted in bold texts) in understanding the flow of arguments. It goes as follows.</p>
<p>Given <span class="math-container">$x$</span> be any curve in the admissible class <span class="math-container">$\Omega$</span> and <span class="math-container">$\delta{x}$</span> represents the variation in <span class="math-container">$x$</span> which is defined as an infinitesimal arbitrary change in <span class="math-container">$x$</span> for a fixed value of variable <span class="math-container">$t$</span>.</p>
<p><em>The first variation in <span class="math-container">$J$</span> is given as</em></p>
<p><span class="math-container">\begin{equation}\label{eq:3}
\delta{J}(x,\delta{x}) = \dfrac{\partial{g}(x,\dot{x},t)}{\partial{\dot{x}}}\delta{x(t)}|_{t_0} ^ {t_1}+\int_{t_0}^{t_1}\Bigg\{\dfrac{\partial{g(x,\dot{x},t)}}{\partial{x}} - \dfrac{d}{dt} \dfrac{\partial{g}(x,\dot{x},t)}{\partial{\dot{x}}}\Bigg\} \delta{x}\ dt. \tag{3}
\end{equation}</span></p>
<p><em>For an extremal (minimal or maximal) <span class="math-container">$x^{*}(t)$</span>, we know that <span class="math-container">$\delta{J}(x,\delta{x})$</span> must be zero. In the following we show that the integral in \eqref{eq:2} must be zero on an extremal.</em></p>
<p><em>Suppose that a curve <span class="math-container">$x^{*}(t)$</span> is an extremal for the problem under consideration; <span class="math-container">$t_1$</span> specified and <span class="math-container">$x(t_1)$</span> free. The value of <span class="math-container">$x^{*}(t)$</span> at <span class="math-container">$t_1$</span> is say <span class="math-container">$x^{*}(t_1)=x^{1}$</span>. <strong>Now consider the fixed end-points problem with the functional <span class="math-container">$J$</span> in \eqref{eq:1} and the end-points <span class="math-container">$(t_0,x^{0})$</span> and <span class="math-container">$(t_1,x^{*}(t_1))$</span>. The curve <span class="math-container">$x^{*}(t)$</span> must be an extremal for this fixed end-points problem and therefore must be a solution of the Euler-Lagrange eqn. \eqref{eq:2}.</strong> The integral term in \eqref{eq:3} must be zero on an extremal and</em></p>
<p><span class="math-container">\begin{equation}\label{eq:4}
\dfrac{\partial{g}(x^{*},\dot{x}^{*},t)}{\partial{\dot{x}}}|_{t_1}\delta{x(t_1)}=0.\tag{4}
\end{equation}</span></p>
<p><em>Since <span class="math-container">$x(t_1)$</span> is free, <span class="math-container">$\delta{x(t_1)}$</span> is arbitrary; therefore it is necessary that</em></p>
<p><em><span class="math-container">\begin{equation}\label{eq:5}
\dfrac{\partial{g}(x^{*},\dot{x}^{*},t)}{\partial{\dot{x}}}|_{t_1} = 0. \tag{5}
\end{equation}</span></em></p>
<p><em>Equation \eqref{eq:5} provides the second required boundary condition for the solution of the second-order Euler-Lagrange equation.</em></p>
<hr />
<p>Now my question is- why should one consider a fixed-end points problem for this case? I am stuck at this problem for quite some time now and therefore sincerely appreciate any thoughts, or suggestions.</p>
| |control-engineering|control-theory|optimal-control|nonlinear-control| | <p>I think I have found an answer and please correct me if there are any other reasons apart from the following justification.</p>
<p>Since <span class="math-container">$x^{*}(t)=\{x(t)\in\Omega|J(x)<\min J(y), \forall { y \in \Omega}\}$</span> is optimal between all admissible trajectories <span class="math-container">$x(t)\in\Omega$</span> for <span class="math-container">$t\in[t_0,t_1]$</span> with <span class="math-container">$x^{*}(t_0)=x(t_0)=x_0$</span> and <span class="math-container">$x^{*}(t_1)\neq{x(t_1)}$</span>, it is still optimal for the fixed end point problem case <span class="math-container">$\Big((t_0,x_0)$</span> and <span class="math-container">$(t_1,x^{*}(t_1))\Big)$</span> as well and therfore the necessary conditions that hold for <span class="math-container">$x^{*}(t)$</span> in fixed end point case will hold in this case. Since, Euler-Lagrange condition was a necessary condition for <span class="math-container">$x^{*}(t)$</span> being optimal in fixed end point case, Eq. (2) holds true for this case and substituting this result in Eq. (3), the other condition in Eq. (5) can be found easily.</p>
| 38156 | Fixed end points optimal control problem |
2020-10-13T18:48:47.687 | <p>I want to design a system of linkages that will allow me to rotate the green element around the blue point/axis without entering the red circle.</p>
<p>The green triangle to go fully around the blue point. I only need about 120 degrees.</p>
<p>I have seen examples of it being done (some provided below).</p>
<p>I don't want to build a robot arm :), so it has to be purely mechanical, preferably with only one rotating axis as an input.</p>
<p>What is this idea/mechanism/process called? I have trouble googling for it.
Is there some kind of solver that would help me find relative lengths of all the linkages, etc.?</p>
<p><a href="https://i.stack.imgur.com/v97WW.png" rel="noreferrer"><img src="https://i.stack.imgur.com/v97WW.png" alt="what I want" /></a></p>
<p>Some examples</p>
<p>Elephant Compliant Mechanism
<a href="https://youtu.be/iAVV_7k79ZI?t=15" rel="noreferrer">https://youtu.be/iAVV_7k79ZI?t=15</a></p>
<p>Six-bar linkage for stretch and turn
<a href="https://www.youtube.com/watch?v=q8tVdjMo22E" rel="noreferrer">https://www.youtube.com/watch?v=q8tVdjMo22E</a></p>
| |design|mechanisms|linkage| | <p>The design that you illustrated is pretty much correct. See below:</p>
<p>In order for the end to move along a circular path, you want a four-bar-linkage with equal length arms.</p>
<p>Note how the length of the medium grey linkages (60mm) is equal to the radius of the red circle (50mm), plus some clearance (10mm).</p>
<p>Note how the distance from the centre of the red circle to the upper fixed pivot is equal to the distance from tip ('green triangle') to the upper moving pivot. This means that when the arms are horizontal, the tip will be horizontal vs the red circle centre.</p>
<p>The vertical spacing between arms is not critical, but must be equal at each end - it will define the available range of motion if you have the arms on the same plane, as illustrated. This provides a convenient 'end stop' at the lower position.</p>
<p><a href="https://i.stack.imgur.com/J3AO8.png" rel="noreferrer"><img src="https://i.stack.imgur.com/J3AO8.png" alt="enter image description here" /></a>
<a href="https://i.imgur.com/TDgAOtv.gif" rel="noreferrer"><img src="https://i.imgur.com/TDgAOtv.gif" alt="enter image description here" /></a></p>
| 38177 | Rotation around an axis outside of the mechanism |
2020-10-14T04:35:55.730 | <p><a href="https://i.stack.imgur.com/0wqAE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0wqAE.png" alt="Tilt Table Diagram" /></a></p>
<p>Hello everyone. This project is for a very specific application and I am trying to determine the force necessary to get the block from point A to B. Specifically the maximum force necessary and at what point it happens. Ideally this would be solved parametrically so that I could determine the force needed at any point along the travel (x axis), but I have included all of the measurements as well. A few things to note: the rolling block is securely fixed to a track with a quality bearing system. As the block is pushed forward, the platform increases in angle. The system is static at point A, then as the block moves forward most of the force is on the x axis, and then of course at some point in travel the majority force is transferred to the y axis. I do not have a coefficient of friction for the bearing system, and of course this will increase as more force is transferred to the y axis, but hopefully someone who can answer this will be able to address how to handle that in a general manner. It would be fantastic to have a differential equation derived for force as a function of the travel distance, but if anyone could show me how to simply determine the maximum force needed that would suffice. This is not a homework or exam question and I am not a mechanical engineer, obviously lol. I made this with Paint 3D. If you get really precise with the measurements I've given you will see that the trig doesn't exactly work out, but it's close. I simply need to know the method so I can dial in the measurements later. Thanks in advance!</p>
| |mechanical-engineering|torque|dynamics|systems-design| | <p>Since you will be using a guide, then my thoughts are the following. Assume at some point the rod forms an angle <span class="math-container">$\phi$</span>.</p>
<p><a href="https://i.stack.imgur.com/uXVGO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uXVGO.png" alt="enter image description here" /></a></p>
<h1>General Idea</h1>
<ul>
<li><p>Since you are pushing up the platform the downward component of the force is equal to the reaction on the pivot below the platform <span class="math-container">$R_{1y}$</span>.</p>
</li>
<li><p>this will create a horizontal component on the rod which will need to satisfy the following equation
<span class="math-container">$$\tan\phi = \frac{R_{1y}}{H}$$</span></p>
</li>
</ul>
<p>Where <span class="math-container">$H$</span> is the horizontal component of the force. (<span class="math-container">$H= \frac{R_{1y}}{\tan\phi}$</span>). Notice that the force reduces as <span class="math-container">$\phi$</span> approaches <span class="math-container">$90\deg$</span></p>
<p>So, the force H is the forces needed to overcome the weight. The only tricky part is that you need to estimate the angle <span class="math-container">$\phi$</span></p>
<h1>calculate <span class="math-container">$\phi$</span></h1>
<p>In order to calculate <span class="math-container">$\phi$</span> for an angle <span class="math-container">$\theta$</span> that the platform pivots from the horizontal you'd need (if you need a sketch drop me a comment and I'll sketch it for you tomorrow):</p>
<ul>
<li><span class="math-container">$H_{Total}$</span>: The total height between the platform top pivot and the pivot of the vehicle (I assume it is 11+24=35 in).</li>
<li><span class="math-container">$H_{Platform}$</span>: The platform height (11 in)</li>
<li><span class="math-container">$W_{Platform}$</span>: The platform width (12 in)</li>
<li><span class="math-container">$L_{rod}$</span>: The length of the rod. (35 in)</li>
</ul>
<p>Given the above and theta you can calculate <span class="math-container">$H_{Pl,\theta}$</span>, which is the <em>vertical</em> distance between the top pivot of the platform and the pivot below the platform. For me the simplest way to calculate this is using a rotation matrix (quantity <span class="math-container">$\color{red}{y_\theta}$</span> is <span class="math-container">$H_{Pl,\theta}$</span>).</p>
<p><span class="math-container">$$\begin{bmatrix}x_\theta\\ \color{red}{y_\theta} \\0 \end{bmatrix} =
\begin{bmatrix}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}\frac{W_{Platform}}{2}\\ -H_{Platform}\\0 \end{bmatrix}
$$</span></p>
<p>This reduces to :</p>
<p><span class="math-container">$$H_{Pl, \theta} = \frac{W_{Platform}}{2} \sin\theta - H_{Platform}\cos\theta$$</span></p>
<p><em>Note</em>: <span class="math-container">$H_{Pl, \theta}$</span> should be negative for angles less that 60 deg.</p>
<p>Then the angle <span class="math-container">$\phi$</span> (as a function of \theta) is given by:</p>
<p><span class="math-container">$$\phi = asin\left(\frac{H_{total}+H_{Pl, \theta}}{L_{rod}}\right)$$</span></p>
<p><span class="math-container">$$\phi(\theta) = asin\left(\frac{H_{total}+\frac{W_{Platform}}{2} \sin\theta - H_{Platform}\cos\theta }{L_{rod}}\right)$$</span></p>
<p><strong>Therefore</strong> you can now plot the force for all <span class="math-container">$\theta$</span> angles between 0 and 60 <span class="math-container">$\deg$</span> that is the limit of your movement.</p>
<h1>Calculate <span class="math-container">$R_{1y}$</span></h1>
<p>The following is the "free body diagram" of the top platform.</p>
<p><a href="https://i.stack.imgur.com/01XDS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/01XDS.png" alt="enter image description here" /></a></p>
<ul>
<li>Point 1: is the pivot below the platform</li>
<li>Point 2: is the pivot at the top of the platform</li>
</ul>
<p>The following equations describe the "balance" of the system:</p>
<ul>
<li><p>Balance on x axis
<span class="math-container">$$\sum F_x =0 \rightarrow R_{2x} - R_{1x}=0$$</span></p>
</li>
<li><p>Balance on y axis
<span class="math-container">$$\sum F_y =0 \rightarrow R_{2y} + R_{1y} - W=0$$</span></p>
</li>
<li><p>Moment around point 2
<span class="math-container">$$\sum M_2 =0 \rightarrow y_\theta R_{1x} + x_\theta R_{1y} - x_w W=0$$</span></p>
</li>
<li><p>relation between <span class="math-container">$R_{1x}, R_{2x}$</span></p>
</li>
</ul>
<p><span class="math-container">$$\tan\phi = \frac{R_{1y}}{R_{1x}}$$</span></p>
<ul>
<li><span class="math-container">$x_w$</span></li>
</ul>
<p><span class="math-container">$$\sin\theta = \frac{x_w}{H_{platform}}$$</span></p>
<p>5 equations, with 5unkwowns (<span class="math-container">$R_{1x},R_{1y},R_{2x},R_{2y}, x_w$</span>). They can be reduced to the following three:</p>
<p><span class="math-container">$$\begin{cases}
R_{2x} - \frac{R_{1y}}{\tan\phi}=0\\
R_{2y} + R_{1y} - W=0\\
- y_\theta \frac{R_{1y}}{\tan\phi} + x_\theta R_{1y} - \sin\theta H_{platform} W=0
\end{cases}
$$</span></p>
<p><em>NOTE:</em> <span class="math-container">$y_\theta$</span> will have negative values.</p>
<h1>Calculation of cart position <span class="math-container">$x_{cart}(\theta)$</span></h1>
<p>Since you now have a way of calculating the position of the pivot underneath the platform <span class="math-container">$(x_\theta, y_\theta)$</span>, and the angle <span class="math-container">$\phi$</span>, you can easily estimate the position of the bottom pivot by vector calculus. In order to calculate</p>
<p><span class="math-container">$$x_{cart}(\theta) = x_\theta - L_{Rod} \cdot \cos\phi$$</span></p>
<p>Then you can plot the force with respect to <span class="math-container">$x_{cart}(\theta)$</span>.</p>
<p>What I got is the following:
<a href="https://i.stack.imgur.com/XhAFU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XhAFU.png" alt="enter image description here" /></a></p>
<h1>Additional points</h1>
<p>There is also an additional force component (dynamic) which has to do with the centrifugal force that you need to overcome. The magnitude of the centrifugal force would be approximately 1.5[lbf], which would make its effect negligible (compared to the 385[lbf] of the mass).</p>
<p>Although, I am not entirely convinced that you don't need to consider acceleration in general (you have a constant angular velocity, and therefore you need to accelerate and decelerate the cart and the mass of 385lb is quite high)</p>
<h1>Python code</h1>
<pre><code># %%
import numpy as np
import matplotlib.pyplot as plt
# %%
H_tot = 35
H_plat= 11
W_plat= 12
L_rod = 35
# %%
theta = np.radians(90)
r_th = lambda theta: np.array(( (np.cos(theta), -np.sin(theta), 0),
(np.sin(theta), np.cos(theta),0) ,
(0,0,1)
))
# %%
v = np.array((W_plat/2, -H_plat,0)).T
# %%
def get_pivot1_coords(theta):
coords = r_th(theta).dot(v)
return coords[0:2]
# %% [markdown]
# # plot x,y theta
# xy contains two column with the coordinates of the pivot at the bottom of the platform
# %%
thetas= np.linspace(0,np.pi/3,60)
xy_raw = r_th(thetas).dot(v)[:2]
xy = np.vstack((xy_raw[0],xy_raw[1])).T
# %%
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(xy[:,0], xy[:,1])
ax.set_aspect('equal')
ax.set_title('Trajectory of the pivot 1 ( bottom of the platform)')
# %% [markdown]
# # calculate angle <span class="math-container">$\phi$</span> wrt <span class="math-container">$\theta$</span>
# %%
def calc_phi(theta):
xy1 = get_pivot1_coords(theta)
phi = np.arcsin((H_tot+ xy1[1])/L_rod)
return phi
# %%
phis = []
for theta in thetas:
# print(theta)
phis.append(calc_phi(theta))
phis = np.array(phis)
# %%
plt.figure()
plt.plot(thetas, phis)
plt.xlabel('<span class="math-container">$\\theta$</span> [rad]')
plt.ylabel('<span class="math-container">$\\phi$</span> [rad]')
plt.title('angle <span class="math-container">$\\phi$</span> as a function of <span class="math-container">$\\theta$</span>')
# %% [markdown]
# # calculate R1y
# %%
W = 385
def calc_R1(theta):
''' returns R1 (x, y) for a given theta
'''
xy1 = get_pivot1_coords(theta)
phi = calc_phi(theta)
R1y=(np.sin(theta)*H_plat*W)/(-xy1[1]/np.tan(phi) + xy1[0])
R1x= R1y/np.tan(phi)
return [R1x, R1y]
# %% Calculate R1s for all theta angles
R1 = []
for theta in thetas:
# print(theta)
R1.append(calc_R1(theta))
R1 = np.array(R1)
x_cart = xy[:,0]-np.cos(phis)*L_rod
# %%
plt.figure()
plt.plot(thetas*180/np.pi, R1[:,0], '.')
plt.xlabel('<span class="math-container">$\\theta$</span> [rad]')
plt.ylabel('<span class="math-container">$R_{1x}$</span> [rad]')
plt.title('<span class="math-container">$R_{1x}$</span> w.r.t. <span class="math-container">$\\theta$</span>')
# %%
plt.figure()
plt.plot(xy[:,0]-np.cos(phis)*L_rod,phis*180/np.pi)
plt.xlabel('<span class="math-container">$x_{cart}$</span> [in]')
plt.ylabel('<span class="math-container">$\\phi$</span> [deg]')
plt.grid()
plt.title('<span class="math-container">$\\phi$</span> w.r.t. <span class="math-container">$x_{cart}$</span>')
plt.figure()
plt.plot(x_cart, R1[:,0],label='<span class="math-container">$R_{1x}$</span>')
plt.plot(x_cart, R1[:,1], label='<span class="math-container">$R_{1y}$</span>')
plt.xlabel('<span class="math-container">$x_{cart} [in]$</span>')
plt.ylabel('<span class="math-container">$Force$</span> [lbf]')
plt.legend()
plt.title('x and y components for R w.r.t. <span class="math-container">$\\theta$</span>')
# %%
plt.figure(figsize=(10,8))
plt.plot(x_cart, R1[:,0], label='<span class="math-container">$R_{1x}$</span>')
plt.xlabel('<span class="math-container">$x_{cart}$</span> [in]')
plt.ylabel('<span class="math-container">$R_{1x}$</span> [lbf]')
plt.title('x components for R w.r.t. <span class="math-container">$x_{cart}$</span>')
plt.legend()
plt.show()
# %%
</code></pre>
| 38184 | What is the maximum force needed to move the roller block from point A to point B and where does that maximum force occur? |
2020-10-14T14:11:37.063 | <p>Does the position of a Gyroscope effect the sensor's reading? for example if I place a gyroscope at the centre of gravity of body and another at one corner, will the angle measurements be different for both Gyroscopes or position does not effect the sensor measurements?</p>
| |automotive-engineering| | <p>To my understanding Gyroscopes output angle changes relative to their frame. As such if you place them on a solid and undeformable body, it shouldn't matter where they are placed.</p>
<p>There are some cases, that Gyroscopes are coupled with magnetometers, in order to give absolute measurements - based on the earths magnetic field- however, that is not the norm.</p>
| 38189 | Position of Gyroscope on rigid body |
2020-10-15T06:51:26.350 | <p>For example, for a squirrel cage induction motor, the speed of operation is given by
Ns = 120*f/P
depending on the number of pole pairs and supply frequency. Now the supply frequency is 50/60Hz while the minimum number of pole pairs is 2. This limits a squirrel cage induction motor to a maximum speed of 3000/3600 rpm. However, Tesla's AC induction motors can run above 10000rpm. How is this possible?</p>
| |electrical-engineering|motors|electrical|electric-vehicles| | <p>Although I am by no means an expert on the subject, I think your promise is wrong in believing that the motor gets a 50/60 [Hz] voltage supply.</p>
<p>To my understanding most motors in electric vehicle, employ some type of <a href="https://en.wikipedia.org/wiki/Vector_control_(motor)" rel="nofollow noreferrer">Field Oriented Control</a>. A very crude description is they do PWM on different phases of the motor. What they are doing is they are trying to supply current in the coils is such a way so that the torque is maximum on the motor.</p>
<p><a href="https://i.stack.imgur.com/Ebnp2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ebnp2.png" alt="Image from Roboteq" /></a></p>
<p>So what they are doing, is based on the rpm they want to achieve, they maximizes the torque at different angles, and the rotor tends follow (either by accelerating or decelerating).</p>
<p>The algorithm cycles is in at least 10 [kHz] (i.e. they make adjustments at least every 0.1 ms), thus enabling them to adjust their speeds at much higher velocities than the 50/60 Hz.</p>
| 38203 | How do electric vehicles' motors run at high speeds above 10000rpm? |
2020-10-16T11:53:47.670 | <p>Hello there, hope you are all doing well.</p>
<p>Right now I am trying to design a Feedforward + PID Controller for KUKA youBot on simulation. Since I do not have the equation of the system, I am trying to tune my gains manually and it is a bit struggling for me. Is using MATLAB's System Identification Toolbox is a good approach to solve the situation. If so, where can I find good tutorials for that? I thought about setting the gains to 0 initially and evaluate the system just with Feed-forward control and then examine the new gains from the Feed-forward data. Does that make sense to you? I am kind of new to Control Theory, so I cannot think of many possible ways. Can you help me with that?</p>
<p>Hope you all have a great day.</p>
| |control-engineering|pid-control|robotics|feedback-loop| | <p>The System Identification Toolbox app is indeed the solution, but I can understand the amount of choices and options make it rather confusing at first. Especially if you have no prior knowledge to model identification. If you are going to use a PID controller, I simply suggest you perform a frequency identification on the system (as this estimates a transfer function of your model). In the essential, this is done using the following steps:</p>
<ol>
<li>Create a suitable input. a suitable input is an input signal that persistently excites the systems dynamics. A good example: Pure white noise. This excites every frequency equally and thus excites every dynamic in the system. Poor example: a sinusoid, this only excites one specific frequency, and therefore just one specific dynamic behaviour. In general, if you take the FFT of the input signal, the estimation's accuracy at a certain frequency is highly influenced by the magnitude of the input signal at this frequency.
One can create a suitable input signal in MATLAB using <code>idinput.mat</code> function, with preferably the PRBS or RBS options.</li>
<li>Use this input signal to measure the system in open loop (assuming the system is open loop stable) closed-loop identification is also possible, but requires a bit more insight, and really only required if it is unsure the open loop system will run into physical limits or is unstable.</li>
<li>Use the measured data and the input signal (remember to store both) to create an estimate of the system. From these vectors, create an <code>iddata</code> object using the function baring the same name. With this object, you can identify the model using the <code>etfe.m</code> function. I'd suggest looking up their references to find out what kind of optional tuning parameters are possible.</li>
</ol>
<p>And... well thats it! But I must note there are plenty of optional parameters that can be tuned to improve the identification, but I leave those for now to prevent unwanted confusion. I do have one note about identification: The highest frequency you can identify is equal to the nyquist frequency of the controller. This is half the sampling frequency of your system. The lowest frequency you can identify is roughly 1/Measurement length. Additionally, longer measurement result in more frequency points, thus higher estimation accuracy (this is specifically important for resonance peaks). Most algorithms work such that if the measurement length is increased to infinity, the true model can be extracted. Lastly some pages worth looking into:</p>
<ul>
<li><a href="https://nl.mathworks.com/help/ident/ug/frequency-domain-identification-estimating-models-using-frequency-domain-data.html" rel="nofollow noreferrer">https://nl.mathworks.com/help/ident/ug/frequency-domain-identification-estimating-models-using-frequency-domain-data.html</a></li>
<li><a href="https://nl.mathworks.com/help/signal/ref/tfestimate.html" rel="nofollow noreferrer">https://nl.mathworks.com/help/signal/ref/tfestimate.html</a> (this shows more information about window types and effects)</li>
<li><a href="http://www.publications.pvandenhof.nl/5SMB0/5SMB0-2019-3-FRF-identification_handout.pdf" rel="nofollow noreferrer">http://www.publications.pvandenhof.nl/5SMB0/5SMB0-2019-3-FRF-identification_handout.pdf</a> (this one is from my professor for the system identification course, an excellent in depth explanation). To my surprise, you can in fact also peek at the remaining lectures at his website: <a href="http://www.pvandenhof.nl/5smb0-system-identification-data-driven-modeling/" rel="nofollow noreferrer">http://www.pvandenhof.nl/5smb0-system-identification-data-driven-modeling/</a></li>
</ul>
<blockquote>
<p>I thought about setting the gains to 0 initially and evaluate the system just with Feed-forward control and then examine the new gains from the Feed-forward data</p>
</blockquote>
<p>Uhm, usually one tweaks the Feedforward parameters using the residual of the error for a certain reference. if you set your PID gains to 0, the loop is not closed and therefore the error you create is practically worthless. Additionally, the goal of feedforward is to polish the error to optimize reference tracking, but most of the work (such as approaching the reference initially) must be done by a feedback controller. As such, I advice to first tune a nice feedback controller (PID for instance) and use that to tune Feedforward.</p>
<p>If you have any further questions, or struggle with obtaining the results, feel free to ask further and if possible share some data (like bode plots) to improve my answers.</p>
| 38224 | Tuning Feedforward + PID Controller just with Feedforward Data |
2020-10-18T08:16:32.330 | <p>To calculate work in a cyclic process (say Carnot cycle) we find area under p-v curve,or use net heat transfer= net work done.
Why won't we simply add work calculated in each process of a cycle,like in Carnot: W1 +W2 +w3 + w4</p>
| |mechanical-engineering|thermodynamics|automotive-engineering|fluid| | <blockquote>
<p>Why won't we simply add work calculated in each process of a
cycle,like in Carnot: W1 + W2 + W3 + W4</p>
</blockquote>
<p>We can (But you need to be careful in calling energy of each process as work, <span class="math-container">$W_1$</span> & <span class="math-container">$W_3$</span> as you call them are heats added and taken from the system not mechanical work).</p>
<p>You already answered your question, the net work of the Carnot cycle is the area enclosed under the <span class="math-container">$p-v$</span> curve.</p>
<p><span class="math-container">$$ W_{net} = \oint_{\text{cycle}} P\ dV = \oint_{\text{cycle}} T\ dS$$</span></p>
<p>You can calculate the area under each process in <span class="math-container">$p-v$</span> diagram (some will be negative) and the sum will be the net work <span class="math-container">$W_{net} = Q_{in} - Q_{out}$</span> (enclosed in white in the following figure, but the axis are temperature and entropy, but you get the idea).</p>
<p><a href="https://i.stack.imgur.com/7was8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7was8.png" alt="enter image description here" /></a></p>
| 38247 | Work calculation in a thermodynamics cycle process |
2020-10-18T16:54:58.347 | <p>Given the fact that small container ships are easier to manufacture than big container ships.. then why is it considered better to build bigger ships instead of many smaller ships?</p>
<p>And to see why manufacturing is easier at smaller scale - consider the smallest and largest scale - I can build a 1 meter toy ship myself, but i can't build an OOCL Hong Kong like ship by myself. By induction this should be true for all levels. The biggest ship looks like this:</p>
<p><a href="https://i.stack.imgur.com/eZu5c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eZu5c.jpg" alt="OOCL Hong Kong biggest ship in the world right now" /></a></p>
<p>In this interview : <a href="https://youtu.be/Opnk-cPOM50?list=WL&t=776" rel="nofollow noreferrer">https://youtu.be/Opnk-cPOM50?list=WL&t=776</a> - Elon Musk explains that is more efficient to have a big rocket rather than a small one. And gives this example of big trucks and big ships.</p>
<p>But to me doesn't realy makes sense. With smaller size, manufacturing is easier, mass producing is possible, faster and cheaper, and materials can have greater strength for the weight. So much that at smaller size even plastics can be used.. at bigger sizes steels can be used.. and at extra big sizes - like a 10km long ship - nothing can be used since no such materials exist. So how come that bigger is better?</p>
<p>Indeed when we go to extremes then other things play a factor - factors like super small ships are smaller than the waves, and 1km long ships might not fit in port etc.. but outside of such non fundamental limitations, in principle, when we consider the cargo ship case, say this <strong>1 ship in the picture above with 2500 containers vs 10 ships with 250 containers each</strong>. Why is it better to go with the 1 big ship in this case? Seems to me is easier to build 10 smaller ships than one which is 10x bigger. Or even better - <strong>have 100 ships, each carrying 25 containers</strong>. So why am I wrong?</p>
<p>Does the same reasoning applies to trucks.. and why it doesn't apply to rocket engines? Why having 28 small Raptors for Starship is better than 5 big F1 engines like Saturn 5? How to properly think about this?</p>
| |design|machine-design|ships| | <p>This isn't hard at all.</p>
<ol>
<li><p>Longer boats/ships go through the water more efficiently. It's called hull speed.</p>
</li>
<li><p>1 Crew costs less than 10 crews</p>
</li>
</ol>
<p>It's mostly the same as why a bus ride costs less than a cab.</p>
| 38252 | Why one big cargo ship is more efficient than 10 or 100 smaller ships? |
2020-10-19T21:50:27.743 | <p>See the image below:</p>
<p>When I lay this out in Fusion, I can get the angles but I can't figure out how to calculate it mathematically.</p>
<p>The goal is to have the angled stretcher fully contact the upper 45"-long support and the floor(even though there will be excess to remove.)</p>
<p>I'm looking for a formula for calculating the stretcher angle(35.8 in this case) and the length that the 2x4 would have to be.</p>
<p><a href="https://i.stack.imgur.com/SC48h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SC48h.png" alt="Table Layout" /></a></p>
| |geometry| | <p>Here's my method:</p>
<p>Consider a rectangle that is <span class="math-container">$a \times b$</span>. This is the interior of your structure where the thickness of the supporting members is neglected.</p>
<p><a href="https://i.stack.imgur.com/00uUb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/00uUb.png" alt="enter image description here" /></a></p>
<p>Then use trig to write the sides of the outlined triangle in terms of <span class="math-container">$a$</span> and <span class="math-container">$\theta$</span>. In the small triangle at bottom-right, decompose it into two similar right triangles and write the relevant sides lengths in terms of <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, <span class="math-container">$w$</span> and <span class="math-container">$\theta$</span>.</p>
<p><a href="https://i.stack.imgur.com/IruSP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IruSP.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/IC1Ok.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IC1Ok.png" alt="enter image description here" /></a></p>
<p>Now you can represent the length of your beam as the sum of three lengths:</p>
<p><a href="https://i.stack.imgur.com/FrFyo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FrFyo.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$\begin{align} \text{Length} &= w\cdot \cot(\theta) + a\cdot \sec(\theta) + \frac{b-a\cdot \tan(\theta)}{\sin(\theta)} \\
&= w\cdot \cot(\theta) + a\cdot \sec(\theta) + b\cdot \csc(\theta) - a\cdot \sec(\theta)\\
&= w\cdot \cot(\theta) + b\cdot \csc(\theta)\end{align}$$</span></p>
<p>Now we have the total length of the beam in terms of just <span class="math-container">$b$</span>, <span class="math-container">$w$</span> and <span class="math-container">$\theta$</span>, so we need another equation in order to find <span class="math-container">$\theta$</span>.</p>
<p>Notice that the pink and yellow triangles are similar, so write an equation involving the ratios of their corresponding sides:</p>
<p><a href="https://i.stack.imgur.com/7dK2x.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7dK2x.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$$\frac{b}{a\cdot\tan(\theta)}=\frac{a+w\cdot\csc(\theta)}{a}$$</span></p>
<p>I <a href="https://www.wolframalpha.com/input/?i=solve+a*b+%3D+a%5E2*tan%28t%29+%2B+a*w%2Fcos%28t%29+for+t" rel="nofollow noreferrer">asked Wolfram|Alpha</a> to solve this for <span class="math-container">$\theta$</span> and it gave:</p>
<p><span class="math-container">$$\theta = 2 \tan^{-1}\left(\frac{\sqrt{a^2+b^2-w^2}-a}{b+w}\right)$$</span></p>
<p>Now we can find the angle from <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$w$</span>, and we can use the angle to find the length of the beam.</p>
<p><strong>I checked this with your numbers</strong>: <span class="math-container">$a=38$</span>, <span class="math-container">$b=31.75$</span> and <span class="math-container">$w=3.5$</span></p>
<p>This gave <span class="math-container">$\theta \approx 35.8265^{\circ}$</span> (which slightly concerns me, since your program rounded this <em>down</em> to 35.82) and <span class="math-container">$\text{Length} \approx 59.09 \text{in}$</span>.</p>
<p><strong>Last check</strong>: I found a long scrap of plywood in my garage that is 7.75 inches wide. I took it into my kitchen and laid it so it spanned a <span class="math-container">$2\times 3$</span> rectangle of floor tile. The formula gave 78.43 inches and it measured just over 78.5. Given the spacing of the grout between the tiles, I'm satisfied with this result.</p>
| 38271 | How to calculate miter angles for an angled stretcher |
2020-10-20T06:33:42.690 | <p>I want to modelling three buildings next to each other in opensees software . Duo to modelling leaning columns , buildings overlap with each other . Now my question is that for 3 adjacent buildings how can i consider leaning columns ?</p>
| |civil-engineering|software|columns|earthquake-engineering| | <p>The interference is graphical only, and since there are no common nodes, it is not a problem for the beams or columns to overlap with each other or intersect because the interactions are transmitted through nodes and elements, and it can be modeled normally</p>
| 38279 | How to model leaning columns in opensees? |
2020-10-20T20:59:04.597 | <p>Can someone explain what is laser saturation and laser saturation intensity? I don't find a proper definition and it seems to me similar to steady state intensity.</p>
| |lasers| | <p>The <strong>saturation effect</strong> of the laser-active medium occurs especially in the case of optically pumped lasers. To understand what is happening here, we need to look a bit deeper into the mechanisms behind the laser process.</p>
<p>For simplification, consider a rod laser with cross sectional area <span class="math-container">$A$</span> and length <span class="math-container">$L_L$</span>, the equations also work for disk and fiber lasers, as they are just extreme cases of rod lasers. The maximum power that we can extract from this laser rod is given by
<span class="math-container">$$P_{max \ extract.}=g_{kl}(v)I_sAL_L$$</span>
This equation can be derived from the equations describing the population inversion of the laser levels. For your question, the two remaining variables in the equation are important, <span class="math-container">$g_{kl}(v)$</span> is the <strong>small-signal gain coefficient</strong> and <span class="math-container">$I_s$</span> is the <strong>saturation intensity</strong>.
The saturation intensity, again, is described by
<span class="math-container">$$I_s = \frac{hv}{\sigma_{ou}\tau_o},$$</span>
so it is a function of the energy of the emitted photon <span class="math-container">$hv$</span> (-> output wavelength of our laser), the cross section <span class="math-container">$\sigma_{ou}$</span> (-> probability that a photon is emitted) and the mean life time <span class="math-container">$\tau_o$</span> (-> average time until a particle in the upper laser level spontanously emits a photon) and thus depends on values that are <strong>inherent to the laser-active medium</strong>.</p>
<p>To return now to your question, there is a saturation mechanism in our laser-active medium that occurs when the intensity of the laser beam that gets amplified in the laser-active medium <strong>reaches the saturation intensity</strong>.
In this case, the <strong>gain coefficient</strong> <span class="math-container">$g$</span> falls down to only half of the small-signal gain coefficient <span class="math-container">$g_{kl}$</span>. What a small gain coefficient means, is, that the population inversion, an absolutely necessary condition for the laser process, gets less and we can not amplify our laser beam as much as we would like to. If the intensity increases even more, the gain coefficient gets even lower. At this point, we see an effect of lowered overall efficiency.</p>
<p>A graph to illustrate this behaviour is shown below:</p>
<p><a href="https://i.stack.imgur.com/8JXbc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8JXbc.jpg" alt="enter image description here" /></a>
Source: Graf, Thomas; Laser - Grundlagen der Laserstrahlerzeugung (<em>Fundamentals of laser beam generation, title translation by me</em>); Springer 2015</p>
<p>Essentially, the saturation effect occurs because at the described point, there is so much stimulated emission that <strong>despite pumping</strong>, the population inversion reduces. To say it very casually, the upper pumping level and upper laser level get sucked dry because we amplify the laser beam in an unsustainable fashion.</p>
| 38291 | Laser saturation and laser saturation intensity |
2020-10-21T14:12:04.543 | <p>I am trying to calculate critical force <span class="math-container">$P$</span>, if two rods (same diameter and length - rigid fiberglass sandwiched on two steel plates) are parallel to each other.</p>
<p><span class="math-container">$$
P = \frac{\pi^2 EI}{(KL)^2}
$$</span>
<a href="https://i.stack.imgur.com/d2O8E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d2O8E.png" alt="Two Rods parallel to each other" /></a></p>
<p>I believe the configuration of both rods pinned would be <span class="math-container">$K=1$</span> (as shown in figure below - B). I am a bit confused if this configuration would behave as springs in this scenario? How would you calculate the the critical force besides using FEA?</p>
<p><a href="https://i.stack.imgur.com/ncx90.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ncx90.png" alt="Buckling of rod" /></a></p>
| |mechanical-engineering|structural-engineering|structural-analysis|stresses|buckling| | <p>The behavior of such a system isn't necessarily well-described by simple buckling.</p>
<p>As other answers have mentioned, the most important thing is how the rods are connected to the horizontal plates:</p>
<p>If the connections allow for small rotations, then buckling theory is valid and <span class="math-container">$K=1$</span>, as per your image.</p>
<p>If not, then it depends on the materials used. This is because this case will mean the rods won't be simply resisting an axial force of <span class="math-container">$P/2$</span>. They'll also have to resist a bending moment due to the force being eccentric to their axes. The magnitude of this bending moment will be a function of the stiffness of the vertical rods vs. the horizontal plates: if the plates are orders of magnitude stiffer, then the bending moment will be insignificant; if not, it'll need to be taken into consideration and you're out of the realm of simple buckling.</p>
<p>In this particular case, you state the rods are fiberglass and the plates are steel. This will likely fall into the category of "trivial bending moment", in which case you can still use simple buckling, but adopting <span class="math-container">$K=0.5$</span> (as per your image) since we're talking about the scenario where the connections are highly fixed.</p>
| 38294 | Buckling of Parallel rods |
2020-10-22T00:53:23.663 | <p>First off, this is not a homework question or anything like that. I'm trying to build a catapult to launch a payload using a flywheel as an energy device!</p>
<p>It goes like this:</p>
<ol>
<li>The flywheel spins up to maximum speed. All other components are at rest.</li>
<li>The catapult lever (in resting position, with a payload at the end of it) engages a tooth.</li>
<li>The flywheel tooth (always extended) connects with the catapult lever tooth and rotates it over a certain angle. During this time the flywheel is slowed down a certain amount and the lever gains a great amount of speed.</li>
<li>At the end of the interaction between the flywheel and the lever tooth (after some degree of rotation) the lever gets to the end of its stroke and abruptly stops, the payload continues flying upward and the remaining energy of the flywheel makes it continue to spin (it can freely "slip" past the lever tooth at the end of the movement).</li>
</ol>
<p>I've attached a couple pictures which describe the two states, T0 and T1 (basically step 3 and 4 respectively).</p>
<p><a href="https://i.stack.imgur.com/gflLJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gflLJ.jpg" alt="state T0" /></a></p>
<p><a href="https://i.stack.imgur.com/5Fcav.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Fcav.jpg" alt="state T1" /></a></p>
<p>My question is, how do I determine the final energy of the flywheel after this interaction given the following parameters: Moment of Inertia and initial Kinetic Energy of the flywheel, mass of payload, and the basic geometry between them. Assume the weight and inertia of the catapult lever are negligible and there is no friction between the interaction of the two moving parts.</p>
<p>I originally thought this problem would be as easy as assuming all the kinetic energy of the flywheel just went into the upward motion of the mass (the flywheel would come to a complete stop). However, after thinking about it for a while I realized it is probably not that simple at all...I smell some differential equations which scare me and has been a while since I've done any of that which is why I am asking for some help. Maybe it's not that complicated after all, but I am at a dead end. Anything would be appreciated.</p>
<p>Thanks in advance.</p>
| |mechanical-engineering|applied-mechanics|mechanisms|mathematics|energy| | <p>This is a very interesting problem.</p>
<h1>Energy approach</h1>
<p>At first I was inclined to solve it though energy i.e.:</p>
<p><span class="math-container">$$\frac{1}{2}I_{fly}\omega_0^2 = \frac{1}{2}I_{fly}\omega_1^2 + \frac{1}{2}I_{lev}\omega_1^2 + \frac{1}{2} m \cdot (\omega_1\cdot L)^2$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$I_{fly}$</span> is the moment of inertia of the flywheel <span class="math-container">$I_{fly} = \frac{1}{2} m_{fly}
r_{fly}^2$</span></li>
<li><span class="math-container">$I_{lev}$</span> is the moment of inertia of the lever <span class="math-container">$I_{lev} = \frac{1}{12} m_{lev} \left(L+ \frac{D_{fly}}{2}\right)^2 + m_{lev}\left(\frac{L-D}{2}\right)^2$</span></li>
<li><span class="math-container">$\omega_0$</span>, <span class="math-container">$\omega_1$</span>: are the angular velocities at the beginning and end of engagement respectively.</li>
<li><span class="math-container">$m_{lev}$</span>: mass of the lever</li>
<li><span class="math-container">$m$</span>: mass of the "missile"</li>
</ul>
<p>From this equation, it is pretty straight forward to obtain the <span class="math-container">$\omega_1$</span>:</p>
<p><span class="math-container">$$\omega_1 = \sqrt{\frac{I_{fly}}{I_{fly}+ I_{lev}+ m \cdot L^2}}\omega_0$$</span></p>
<p>Then the only thing you need to do is work out the launch angle (<span class="math-container">$\theta$</span>), break up into components and you can estimate maximum height and horizontal distance traveled.</p>
<p>Although, I believe the above mentioned method will give you a ball-park figure, I doubt it will be accurate. There are <strong>two assumptions 'with issues'</strong> here:</p>
<ul>
<li>there is full engagement during the duration of the slow down of the flywheel (i.e. there is no impact).</li>
<li>disengagement between the mass and the level occurs the moment the engagement of the tooth with the lever finishes. However, the time of disengagement of the mass will be depended on factors such as how its mounted on the level, coefficient of friction etc.</li>
</ul>
<h1>Impact</h1>
<p>The main problem with the above approach would be the speed of impact during first engagement.</p>
<p>If the lever and mass are small then the angular velocity of the level + mass will be greater than the angular velocity of the flywheel, i.e. there will be only brief contact at the beginning of the impact and maybe secondary impacts (which <em>might</em> not provide any extra energy to the mass.)</p>
<p>If the lever and mass are large then probably what will happen is that the flywheel will either recoil or slow completely down. In any case, this means that the travel angle will not be <span class="math-container">$\theta$</span> but something else entirely.</p>
| 38305 | How high will a flywheel catapult launch a payload? |
2020-10-23T20:17:31.410 | <p>I recently helped upgrade an old (±20years) 4ton boiler (heavy fuel oil) with some monitoring instrumentation. I decided to have a look at the data and I'm not sure if these kind of oscillations are normal for an older boiler? I haven't really worked with boilers before.</p>
<p>Ignore the sudden drop (a topic for a different question), I'm referring to the oscillating behavior in the steady region. It seems to be rather...excessive? but follows a relatively consistent profile which leads me to believe its just a really barebones control system.</p>
<p><a href="https://i.stack.imgur.com/EWv8G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EWv8G.jpg" alt="enter image description here" /></a></p>
| |heating-systems|steam|boilers| | <p>It is very difficult to cancel completely the oscillations in boiler pressure. I imagine that a boiler with an operating pressure between 8 ÷ 10 barg is small and most probably is heated with an on/off burner, in this case is impossible to eliminate the oscillations. But even when you can control the power of the burner and the feedwater flowrate you have to consider the shrink / swell effect that impacts the pressure oscillation.</p>
<p>Check <a href="https://controlguru.com/dynamic-shrinkswell-and-boiler-level-control/" rel="nofollow noreferrer">https://controlguru.com/dynamic-shrinkswell-and-boiler-level-control/</a> for reference.</p>
| 38353 | Are these oscillations normal in a boiler? |
2020-10-24T10:49:21.437 | <p>I was learning about the charpy impact test and came across the graph on <a href="https://ar5aes.wordpress.com/week-9-2/" rel="nofollow noreferrer">this webiste</a> that shows the impact energy for bcc and fcc structures as a function of temperature. I tried searching for an explanation as to why the graph differs for the two crystal systems but couldn't find any. Some places say it's because of the slip planes but I still didn't understand.</p>
| |energy|impact| | <p>It essentially depends on the fracture mechanisms available to the material at the temperature in question. In general, dislocations are more mobile at higher temperatures, enabling plastic deformation and ductile fracture.</p>
<p>FCC materials have more slip systems, or ways for dislocations to move, than BCC materials. A metal needs five independent slip systems to plastically deform. As you lower the temperature, certain slip systems will be “frozen out”, meaning that it is essentially impossible for dislocations to move according to that slip system. Because BCC metals have fewer slip systems to start with than FCC metals, they become brittle at higher temperatures than FCC metals.</p>
| 38357 | Why do some metals with bcc structure show ductile-brittle transition in charpy impact test? |
2020-10-24T23:13:05.863 | <p>I've recently come across pictures of segway/hoverboard like self balancing motorcycle type vehicles that were supposedly able to become as fast as a street legal scooter. I kept wondering how they would do an emergency brake from top speed.
If the wheel just blocked you would slam face down into the ground. So of course the vehicle had to try to lean the driver as far back as possible to use their weight for braking. But even then with enough speed and a subsequent brake my intuition tells me the driver would still just be lifted and slammed into the ground.
I really just wondered if my intuition is right and the top speed these things can go is reached quite early or if I am mistaken because I'm missing the insights in how these things work physically.</p>
| |electric-vehicles| | <p>There is no hard limit to the top speed of a self balancing vehicle.</p>
<p>Imagine the vehicle's drive system has unlimited torque and speed, and is operating in a vacuum (no drag). Starting from a balance, the vehicle can tilt slightly forward and will begin accelerating. With no drag, or speed limit on the drive system it will accelerate continuously and can reach arbitrarily high speeds. To stop, it simply has to tilt slightly backwards and repeat the process in reverse.</p>
<p>With drag, there is a limit, although it would be very fast. When the vehicle is tiled forward the vehicle mass exerts a torque, which counteracts the propulsion torque of the wheels. When the vehicle is tilted forward as far as possible (basically touching the ground), it's exerting the maximum torque. At some very high speed, the torque required to overcome drag will equal the maximum torque and the vehicle can accelerate no further.</p>
<p>Practically speaking though, as Tiger Guy said, the drive system will be the limit, and not one of these physics class type hypotheticals.</p>
| 38364 | Is there an upper limit to how fast a self balancing one wheeled vehicle can become while also being able to safely stop? |
2020-10-25T12:17:50.810 | <p>There was time (a decade ago) when I got <code>8 MB</code> (yes, <code>8 MB</code>, not <code>8 GB</code>) SD card for my Nokia cellphone ( At that moment, the maximum storage on the market was 32 MB, marketed as the "breakthrough in high-capacity" cards).</p>
<p><a href="https://i.stack.imgur.com/TuA5w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TuA5w.jpg" alt="enter image description here" /></a></p>
<p>However, as everything, those achievements left in history and then achieved 1GB...100GB... now 1000GB for the same sized <code>MicroSD</code>. I am interested:</p>
<ol>
<li>On what does this increase depends? transistor development or? why they didnt produce 100 GB cards 10 years ago?</li>
<li>Is there any maximum capacity limit of what technology can achieve within a single MicroSD card?</li>
</ol>
| |microelectronics| | <p>The main answer cover the question 2), but here is also something regarding question 1)</p>
<p>The increase of memory depends on the number of memory elements we are able to produce per area, there is a very general rule which is called the <a href="https://en.wikipedia.org/wiki/Moore%27s_law" rel="nofollow noreferrer">Moore's law</a> . This is not a physical law, just an accurate estimation to help guide investments, and set up milestones in semiconductor research.</p>
<p>This was not possible 10 years ago because we did not had the technology. to be accurate, We actually did had the theory mostly ready, but we were missing knowledge to able to implement it into a product.
Nowadays the Moore's law does not work that well anymore, because as we are scaling down we are facing more and more challenges, and today we are still working on the theory that could help us the next technology.</p>
| 38372 | What is theoretical (physical) limit a MicroSD card's capacity? |
2020-10-27T16:13:17.020 | <p>I have inherited a DIY build and I have some doubts about its design. Have a look at the following diagram:
<a href="https://i.stack.imgur.com/DVPxS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DVPxS.png" alt="enter image description here" /></a></p>
<p>1 is the linear actuator. It's quite beefy 2000 lbs 12" stroke. It is installed in an aluminium channel.
4- Is a carriage, essentially a box section with little rollers on all sides, so it has no play within the channel and moves quite smoothly. This is connected to the actuator's rod.
3. Is a flat steel bar that links to an identical channel (2) and carriage assembly, though this doesn't have an actuator in it.</p>
<p>The system is supposed to lift a max 220 lbs load that is evenly distributed along the bar.</p>
<p>My questions are:</p>
<ol>
<li>While there don't seem to be any bending forces to the actuator itself, is it a safe design, as far as longevity of the components goes?</li>
<li>is there anything I can do in channel 2 to support the system, e.g. a spring at the bottom, etc. to address any design problems?</li>
</ol>
| |mechanical-engineering|linear-motion|linear-motors| | <p>As a quick basic estimation approach we annotate the following:</p>
<ul>
<li><p>The length of bar 3 = L</p>
</li>
<li><p>Top and bottom bearing force on each box <span class="math-container">$F_b$</span></p>
</li>
<li><p>The height and width of boxes 4 and its counterpart on channel 2 H and W.
<span class="math-container">$$ \Sigma M=0 ,\quad 220*L/4 = H/2*4F_b \rightarrow \quad F_b=220*L/8H $$</span></p>
</li>
</ul>
<p>Now you have to figure if the bearings and their connections are okay to take this load multiplied by a safety factor of say 3.</p>
| 38406 | Single linear actuator lifting mechanism |
2020-10-27T20:03:54.873 | <p>I'm studying the laser input-output power relation. In particular I'm studying the laser slope efficiency <span class="math-container">$\eta_s$</span> (<span class="math-container">$\eta_s=\frac {dp_{out}} {dp_{in}})$</span>.
However, my textbook tells that <span class="math-container">$\eta_s$</span> can be divided in 4 factors: <span class="math-container">$\eta_s=\eta_p \eta_c \eta_q \eta_t$</span>.
I'm having difficulties in understanding <span class="math-container">$\eta_c$</span> that it is called "output coupling efficiency".
On my textbook, principle of lasers by O. Svelto, it is written that it represents the fraction of generated photons coupled out of the cavity, which one can call the output coupling efficiency.</p>
<p>Can you explain me <span class="math-container">$\eta_c$</span> in other words? What does it mean "coupled out" in this context?</p>
| |lasers| | <p>In a way, from the laser resonators point of view, we can consider two kinds of losses. The actual wasted losses (for example heat dissipation) and "useful losses", that is, the actual laser beam that leaves the resonator. While this laser beam is of course the reason to do the whole ordeal, it still means that we take a part of the laser beam that bounces through the resonator.</p>
<p>The output coupling efficiency relates these two kinds of losses, the wasted loss (<span class="math-container">$loss$</span>) and the "useful" loss, to each other.
<span class="math-container">$$\eta_{c} = \frac{T_{oc}}{T_{oc}+{loss}}$$</span></p>
<p>In context of a laser resonator, "coupling out" describes the concept of taking a part of the laser beam that gets amplified in the resonator and letting it leave the resonator through a mirror (the outcoupling or OC mirror) with a transmission of up to 5 % or even more (there are probably higher ones, but I have only ever worked with up to 5 %). While you usually want a mirror to reflect as much as possible, the outcoupling mirror is designed to let a certain exact amount of light pass. This transmission degree, <span class="math-container">$T_{oc}$</span> in the equation, is therefore also called outcoupling degree. When developing a laser, picking the outcoupling degree is an interesting parameter to play with, putting it too low makes your actually usable laser power too low, too high and your losses and with it the laser threshold increase too much.</p>
| 38410 | Laser efficiency, what is this term? |
2020-10-29T17:51:05.183 | <p>For example if the first gear has 32 teeth and the second 64, if the third also has 32 teeth would the output of the third retain the mechanical advantage created by the first 32 tooth gear powering the second?</p>
<p>32:64:32 or 1:2:1</p>
<p>Thanks for your time and help!</p>
| |gears|torque| | <h3>No. We can show this with math.</h3>
<p>The first gear pair has a ratio of 1:2, and the second pair has a ratio of 2:1. So, discounting losses, for every turn of the input shaft, the second shaft will make 1/2 of a turn. For the second pair, for every turn of the second shaft, the third shaft will turn twice.</p>
<p>We can then multiply these together to see how many turns the third shaft will make compared to the first shaft: 1/2 * 2 = 1. Therefore, for every turn of the first shaft, the third shaft will turn once, and there is no mechanical advantage for the overall system.</p>
| 38431 | Does a gear retain its mechanical advantage in a 1 2 1 ratio? |
2020-10-30T06:55:25.483 | <p>I've witnessed this phenomenon:
<a href="https://i.stack.imgur.com/f3Vmc.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/f3Vmc.jpg" alt="dry tips" /></a>
Why are only the tie tips dry? Why are only the stones around the tips dry? This is a railroad in Ukraine in the morning after a rain at night. Around 80% of observed tips looked like that, slightly elevated tips were not dry.</p>
| |concrete|rail|dehumidification|drying| | <p><a href="https://i.stack.imgur.com/KeV4n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KeV4n.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. A concrete sleeper mould with reinforcing steel. Concrete sleepers are reinforced with pre-stressed steel. Image source: <a href="https://www.bft-international.com/en/artikel/bft_Prestressed_concrete_sleepers_for_America_2032968.html" rel="nofollow noreferrer">BFT-International</a>.</em></p>
<p>The sleepers will act as heat sinks during the day soaking up heat from the ground. At night this heat is let off. The steel will conduct heat to the ends of the sleepers warming the end-plates and supplying more heat through the lower thermal resistance than at other points on the surface of the sleeper. The steel in the phots seems to have end plates which will improve the end-heating significantly.</p>
<p><a href="https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html" rel="nofollow noreferrer">Engineering Toolbox</a> gives the following conductivity values:</p>
<pre><code>Concrete 1.0 to 1.8 W/mK
Steel 36 to 54 W/mK
</code></pre>
<p>The steel is a far superior heat conductor than the concrete.</p>
| 38443 | Why railroad concrete tie tips become dry first after a rain? |
2020-10-30T10:51:16.660 | <p>A fan that consumes 20 W of electric power when operating is claimed to discharge air from a
ventilated room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s. is this claim reasonable?</p>
<p>I am getting maximum possible velocity of air around 12.64 so it's reasonable to me but all solution available on internet says maximum possible velocity of air is 6.3,so it's not reasonable.</p>
<p>Please help</p>
<p>The internet solution for the question:
<img src="https://i.stack.imgur.com/F6WyJ.jpg" alt="enter image description here" /></p>
| |mechanical-engineering|thermodynamics| | <p>If we solve for energy flow rate (E' = m' × e) of the air which is given by fan then we will get 0.25(8<sup>2</sup> / 2) = 0.25 × 64 / 2 = 8 J/s which is less than the electric input power 20 J given to fan that's why this claim is fair enough as the devices can have the low efficiencies but not 1 or greater than 1.</p>
<p>Here e = v<sup>2</sup> / 2 = ke<br />
from equation of energy balance<br />
W'<sub>in</sub> = m'(ke)<sub>out</sub> = m'v<sup>2</sup> / 2<br />
we get 12.6 m/s as the maximum possible velocity of the air.</p>
| 38445 | Thermodynamic energy balance |
2020-10-30T14:21:28.620 | <p>I'm designing a movable cover on rails under a spring and with added friction to dampen the movement. The model that describes it's movement the best is Coulomb damping model. The cover is moved from it's equillibrium position to one side and realeased. I want to calculate the time it's needs to get back to equillibrium position.</p>
<p>The paper on the link bellow defines the equations.
<a href="https://www.studocu.com/en-us/document/university-of-toledo/mechanical-vibration/lecture-notes/ln-6-coulomb-friction/3247643/view" rel="nofollow noreferrer">https://www.studocu.com/en-us/document/university-of-toledo/mechanical-vibration/lecture-notes/ln-6-coulomb-friction/3247643/view</a></p>
<p>The initial preload distance <span class="math-container">$x_0$</span> needs to be large enough so that the spring force is larger than the static friction.</p>
<p>My problem is that when I put the values in the equation and try to calculate the time mentioned above I get the error from the <code>acos()</code> part of the function, as the value in the <code>acos()</code> brackets is smaller than -1 even though that the force in the spring is larger than the friction. If I make the spring stiffness larger, then the equation works, when the part in the <code>acos()</code> brackets is larger than -1. The equation I'm using is on the picture in the link bellow.</p>
<p><a href="https://i.stack.imgur.com/8ara0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8ara0.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$x(t) = \left(x_0- \frac{\mu m g }{k}\right) \cos\left(\omega_n t \right) + \frac{\mu m g}{k}$$</span></p>
<p>I really do not understand what this limitation means?</p>
| |dynamics|vibration| | <p>The answer is given near the top of the page:</p>
<blockquote>
<p>The body will only move if the spring force is greater than the static friction force: <span class="math-container">$kx_0 > u_smg$</span>.</p>
</blockquote>
<p>If the spring constant is too small and the body doesn't move at all, you will get impossible values for the <span class="math-container">$\text{acos}()$</span> function when you assume it <em>did</em> move.</p>
| 38447 | Coulomb damping time calculation |
2020-11-01T13:33:02.040 | <p>This has been a burning question that I've been thinking about recently.</p>
<p>In statics problems like the one I've attached here where we are asked to find the angle that the resulting vector has with the positive y-axis, do we take the angle going clockwise or counter-clockwise from the y-axis to the resulting vector?</p>
<p>I remember from class that we calculate the angle between the positive x-axis with a vector going counter-clockwise 'always' however I'm unfamiliar with the rule when it comes to finding the angle with the positive y-axis.</p>
<p>I would appreciate any insight on this topic.</p>
<p><a href="https://i.stack.imgur.com/8mAUh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8mAUh.png" alt="enter image description here" /></a></p>
<p>Reference: Engineering Mechanics STATICS, J.L. Meriam, L.G. Kraige, 7th Edition</p>
| |mechanical-engineering|statics| | <p>By definition of dot product:
<span class="math-container">$$ \vec{a} \ . \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos (\theta) $$</span></p>
<p>Where <span class="math-container">$\theta$</span> is the angle between the two vectors <span class="math-container">$\vec{a}$</span> and <span class="math-container">$\vec{b}$</span>.</p>
<p>To find angle between your vector and +ve x-axis, set <span class="math-container">$\vec{a} = (40, -30)$</span>, <span class="math-container">$\vec{b} = \vec{i} = (1, 0)$</span> such that <span class="math-container">$\|\vec{i}\| = 1$</span></p>
<p><span class="math-container">$$ \theta = \cos^{-1}( \frac{(40, -30) \ . (1, 0)}{50 * 1 } ) = -36.86$$</span></p>
<p>You can calculate the angle between the vector and the +ve y-axis by setting <span class="math-container">$\vec{b} = \vec{j} = (0, 1)$</span>, which you can easily calculate as shown above.</p>
| 38467 | Calculating angle of a 2D vector with positive y-axis |
2020-11-04T01:49:46.487 | <p>I am trying to make a drawing in solidworks, with a view normal to plane offset from the top surface and at a 45 degree angle from the surface.</p>
<p>Here is an image showing my model with the plane. Sorry if the orientation of the view is weird from my screenshot.</p>
<p>I am using Solidworks 2019-2020.</p>
<p><a href="https://i.stack.imgur.com/dR7I4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dR7I4.png" alt="View 1" /></a>
<a href="https://i.stack.imgur.com/V6ZVC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V6ZVC.png" alt="View 2" /></a></p>
| |solidworks| | <p>I'm too late for the OP, but perhaps this will help someone who finds this in search. You are looking for Insert > Drawing View > <strong>Relative To Model</strong>. This option will allow you to manipulate the 3D model in another window and select a plane to make the view normal to.</p>
<p><a href="https://i.stack.imgur.com/yrps3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yrps3.png" alt="enter image description here" /></a></p>
<p>I use this a lot with injection molded parts where the overall product is done with a top-down approach so that the origin of each individual part is at an arbitration location and orientation. Then for the drawing I make the views relative to the pull direction of the part.</p>
<p>If you search the Solidworks Help for a "Relative View", you'll find any necessary details.</p>
<p>Another option is to make a "saved view" in the model that has your desired orientation, and then do the drawing w.r.t. this saved view.</p>
<p>There are pros and cons to both. The "relative to model" option updates parametrically, so if something in your model changes, the view should re-orient automatically. However it can't be "editted". You're stuck with the original definition and you can't even view what is was. It also can't be reused (other than by copy and paste).</p>
<p>The "saved view" option does not update parametrically, so you have to remember to manually update the "saved view" orientation if something about model changes such that the view orientation should be correspondingly updated. But unlike the Relative-to-Model option, you can effectively edit/update this sort of view by overwriting the saved view in the part/assy model.</p>
| 38503 | Can I make a drawing view normal to a plane in my model |
2020-11-04T07:22:17.040 | <p>I'm trying to remove the screw in the images below from a metal cover.</p>
<p><a href="https://i.stack.imgur.com/MXU2I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MXU2I.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/iqjb3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iqjb3.jpg" alt="enter image description here" /></a></p>
<p>The screw is locked in place by what I believe is called a locked washer. The screw can rotate, but I can't move it in and out. The height to the "lip" is about 12mm (see first image), which doesn't give me much room to get tools in there.</p>
<p>Any thoughts on the best way to remove this screw without damaging the cover? Ideally, I'd also prefer not to damage the screw too much. At least, I'd like to be able to perform some measurements on it after to find a replacement.</p>
<p>I've tried prying the washer out with needle nose pliers, but the washer is pretty solid. I've also tried cutting it off with a dremel cutting tool, but I've mostly just managed to cut some groove marks into the inside of the cover and I haven't done much damage to the washer. I've also considered using one of the thin, cylindrical cutters on the dremel to saw off the head of the screw, but I don't think I can do this without damaging the outside coating of the cover.</p>
| |fasteners| | <p>What you have is a circlip in a recess of the screw. You can see the ends of the "C" in the image, as well as a slight gap around the perimeter. A thin bladed screwdriver placed in the gap and rotated will cause the "C" to open, assisting in the release of the clip. If you have two thin screwdrivers, both gaps can be pried simultaneously to cause the release. There are circlip pliers designed for removing and installing these clips, but it's usually not practical for a one-off problem such as this.</p>
<p><a href="https://i.stack.imgur.com/EeuCL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EeuCL.png" alt="circlip" /></a></p>
| 38506 | How to remove/replace a screw with a lock washer in a tight place |
2020-11-04T15:54:30.270 | <p>I have a quick question guys. I am gonna select a brake but I am not sure.</p>
<p>A Crane Hoist motor has a rating of 300 HP and maximum speed of 1800 rpm. This motor is attached to a gearbox with a perfect (100% efficient) reduction ratio of 53.739:1. The API 7K specifications call for a safety factor of 2:1 (200% of motor full load torque) static holding torque from the brakes. The lining coefficient of friction is 0.40 (static) and the brake shoe face is 6.0 inches wide. Two (2) brakes need to be installed diametrically opposite each other on a large disc mounted directly to the cable drum (on the output side of the gear box).</p>
<p>Which disc diameter, and resultant brake torque is fine?</p>
| |hydraulics|regenerative-braking| | <p>The speed of the disk rotation is <span class="math-container">$1800/53.739=33.495$</span></p>
<p>The torque it creates is <span class="math-container">$$300*5252/1800=875.33 ft.lbf \quad 875.33*53.739= \\47040ft.lbf\rightarrow times 2=94079ft.lbf \ \text{at the disk}$$</span></p>
<p>R = disk Radius</p>
<p>If we assume the brake pad pressure <span class="math-container">$P/inch^2$</span> then its friction force <span class="math-container">$F_f=0.4*P$</span></p>
<p>and the disk diameter is related to this torque,</p>
<p><span class="math-container">$ \ 94079=2*F_f*6*R^2/2 \quad R^2=2*94079/2*F_f*6$</span></p>
<p>This is just the static and assuming the pads start from the center to the end of the disk.
We need to calculate the torque for kinetic friction during deceleration as well.</p>
<p>Check my arithmetic, please.</p>
| 38510 | Brake selection problem |
2020-11-05T11:34:11.920 | <p>I'm reproducing in Fusion 360 an object based on a technical drawing, but I have trouble understanding what this means:</p>
<p><a href="https://i.stack.imgur.com/DPLxe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DPLxe.png" alt="enter image description here" /></a></p>
<p>When I read the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Engineering_drawing_abbreviations_and_symbols#R" rel="nofollow noreferrer">Engineering drawing, abbreviations and symbols</a>, it states:</p>
<blockquote>
<p>Radius of an arc or circle. Flats and reversals (falling within the dimensional tolerance zone) are tolerated unless "CR" (controlled radius) is explicitly specified.</p>
</blockquote>
<p>But I still don't understand how this radius is being measured and therefore how to reproduce it.</p>
| |cad|technical-drawing| | <p>Measure 12.7 from both edges and the intersection gives the radius centre.</p>
| 38519 | How do I interpret a fillet indication on a technical drawing |
2020-11-05T23:11:59.123 | <p>After placing a dry solid block of nonporous plastic material in the freezer for a while, I was surprised to find out when I retrieved it that it had expanded globally by a few inches. Which type of plastic must this be and do plastics expand at colder temperatures?</p>
| |plastic| | <p>Materials with a negative thermal expansion (NTE) coefficient will increase in size as they are cooled down. Not all plastics have a NTE, but it is not a property seen only in one specific type of plastic. I don't think the plastic can be identified only knowing that it has a NTE.</p>
| 38527 | Does plastic expand at colder temperatures? |
2020-11-06T06:53:57.893 | <p>Some background: I am working for an ed-tech startup and we are hoping to release our first product soon but have been hitting a lot of hurdles with our <strong>current manufacturing strategy.</strong> We want to produce <strong>100-200 products to test our market</strong> before taking on more risk and going into injection molding, but we are running challenges with <strong>finding an economical</strong> method for producing the first 200 parts.</p>
<p>Every machining shop that we have visited have given us a quote of 25+ dollars for each set of pieces. It's hard to justify making that investment since there is a possibility that we would want to adapt our design to better fit market needs as we bring it out.</p>
<p>The solution that we have thought of has been to <strong>invest in a DIY machining kit</strong> to try and produce our product to meet demand and possible iterate on our product. <strong>Images on the parts that we want to create are listed below:</strong></p>
<p><a href="https://i.stack.imgur.com/7XX25.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7XX25.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/UjKoM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UjKoM.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/U3Vwb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U3Vwb.jpg" alt="enter image description here" /></a></p>
<p>If anyone has shared a similar experience and has a <strong>background in manufacturing/machining</strong> and can answer the following questions it would be very appreciated:</p>
<p><strong>What is the best material that we should use for these pieces?</strong></p>
<p>Criteria: We are mainly looking for a material that won't crack under a lot of pressure and feels and looks nice. Options from manufacturers we have spoken with: Delrin, Aluminum, High Density Polyetheline, plywood.</p>
<p>Some challenges that we have been facing have been to find an affordable way of finding the actual material. Most machining places say that the material would cost 1$ a set. However, online, it is super expensive for a pretty small sheet: <a href="https://www.interstateplastics.com/Delrin-Plastic-Black-Sheet-ACEBE5%7ESH.php?thickness=0.250&dim2=12&dim3=12&src=adwordspla&thisisforcallrail=1&campaignid=225228743&adgroupid=37851262274&creative=153445283219&matchtype=&network=g&device=c&keyword=sheets-ACEBE5SH-0.25-12.00-12.00&gclid=CjwKCAiA4o79BRBvEiwAjteoYFOOMRTpCPiFcAvXJTgjU4Hdt35RKLWdvqlCamLzXVEZEi2BTNjJzhoCWBAQAvD_BwE&" rel="nofollow noreferrer">https://www.interstateplastics.com/Delrin-Plastic-Black-Sheet-ACEBE5~SH.php?thickness=0.250&dim2=12&dim3=12&src=adwordspla&thisisforcallrail=1&campaignid=225228743&adgroupid=37851262274&creative=153445283219&matchtype=&network=g&device=c&keyword=sheets-ACEBE5SH-0.25-12.00-12.00&gclid=CjwKCAiA4o79BRBvEiwAjteoYFOOMRTpCPiFcAvXJTgjU4Hdt35RKLWdvqlCamLzXVEZEi2BTNjJzhoCWBAQAvD_BwE&</a></p>
<p>Followup: Where would be a good source to find a cheap material to work with?</p>
<p><strong>Are there any recommendations for DIY machine kits and what is the price range that we should spend for a machine?</strong></p>
<p>Criteria: Our quoted price for 100 product is 4,000 dollars and budget for the machining kit is Ideally less than 1,000 dollars. Would it be possible to work with plastics or aluminum with a kit in this price range and still maintain good quality products?</p>
<p><strong>How hard is it to learn how to machine for these products?</strong></p>
<p>Context: I know that the process involves actually coding the trajectory and path that the mill is going to take to produce the product. How difficult is it to learn the necessary coding for path planning and how hard is it to learn how to choose the right drill bit for each task? Would we, as beginners, be able to create each piece with either aluminum or plastics and make it good quality?</p>
<p><strong>Is there another manufacturing service that would be ideal for what we trying to accomplish?</strong> We have also investigated laser cutting and water jetting. However, It would be impossible to buy any of these machines.</p>
<p>Thank you so much,</p>
<p>Ryan</p>
| |mechanical-engineering|manufacturing-engineering|machining|robotics|3d-printing| | <p>Like other have suggested the cost is quite cheap for 25$ per set.
The cost of buying, setting up, learning how to properly use a small cnc machine will be at least 2 or 3 times greater if you factor in work hours, and delays.</p>
<p>Additionally, what I wanted to caution you about is the choice of materials. The options you mention, have very different properties and if there are any significant loads or dynamic loads, the design should be modified with respect to the material. For example using plywood on the <a href="https://i.stack.imgur.com/U3Vwb.jpg" rel="nofollow noreferrer">3rd component</a> with those sharp corners would be a bad idea (however, as I said I don't know if this is supposed to take any significant load or if its just a stand/holder).</p>
<p>A final thought, is that an investment on DIY equipment would make sense, if you are planning to scale production to probably something in the order of a few thousand pieces. However, even then <strong>it would probably make more sense to adjust design in order to accommodate for more cost effective manufacturing while maintaining function</strong>.</p>
<p>You'd be surprised how much cheaper it would be to manufacture the <a href="https://i.stack.imgur.com/UjKoM.jpg" rel="nofollow noreferrer">2nd component</a> from sheets of material rather than machining it with a cnc from a block.</p>
| 38535 | Small scale manufacturing |
2020-11-07T07:34:45.117 | <p>Is it possible to add nut using smart fastener feature in SolidWorks as shown below. Presently I have added nut using toolbox feature. I want to know if we can add nut and a washer to male thread directly using smart fastener feature rather than separately adding nut and washer from toolbox</p>
<p><a href="https://i.stack.imgur.com/nI7ad.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nI7ad.png" alt="enter image description here" /></a></p>
| |solidworks| | <p>Yes, it's possible.</p>
<p>This is known as a hardware stack, and the official SW help pages provides a clear explanation regarding setting these up: <a href="https://help.solidworks.com/2019/english/SolidWorks/sldworks/c_Smart_Fasteners_Hardware_Stacks.htm#voc1450446632618" rel="nofollow noreferrer">https://help.solidworks.com/2019/english/SolidWorks/sldworks/c_Smart_Fasteners_Hardware_Stacks.htm#voc1450446632618</a></p>
| 38547 | Smart fastener for male thread in solidworks |
2020-11-07T08:20:42.447 | <p>If I am drilling through bars of different material(ex: iron, steel etc.) of the same thickness and under the same conditions of drill operations(feed rate and rpm), how will the torque and thrust force measurement change for different materials? For example, let us assume that the materials in question are iron, steel and aluminium.</p>
| |mechanical-engineering|torque|drilling|thrust| | <p>What you are interested boils down to the "specific cutting force" (<span class="math-container">$N/mm^2$</span>) parameter - which varies by material. Not just steel or aluminum - but what alloy (e.g. addition of lead acts as a lubricant and makes it easier to machine), and hardness treatment it has. In approximate terms: Aluminum 800 MPa, Iron 1500 MPa, Steel 2500-3000 MPa. So very roughly 3x as much electricity or torque is required to cut steel vs aluminum (if same speed, feed, and geometry).</p>
<p>You may also see it expressed as "Specific Energy" (<span class="math-container">$J/mm^3$</span>) = how much energy is required to remove a cubic millimeter. According to Kalpajkian, the difference in specific energy can be more than 20x for some steel to aluminum alloys.</p>
<p>Good references for more detail:</p>
<ul>
<li><p><a href="https://www.sandvik.coromant.com/en-gb/knowledge/machining-formulas-definitions/pages/milling.aspx" rel="nofollow noreferrer">https://www.sandvik.coromant.com/en-gb/knowledge/machining-formulas-definitions/pages/milling.aspx</a></p>
</li>
<li><p><a href="http://www.mitsubishicarbide.net/contents/mhg/enuk/html/product/technical_information/information/formula4.html" rel="nofollow noreferrer">http://www.mitsubishicarbide.net/contents/mhg/enuk/html/product/technical_information/information/formula4.html</a></p>
</li>
</ul>
| 38548 | What is the relationship between the type of material(ex: steel, iron etc.), thrust and torque? |
2020-11-08T12:25:24.170 | <p>What is the <em>typical</em> quantity (in kg) of uranium-235 present at a given moment in a working nuclear plant?</p>
<p>How does this compare to the typical amount of uranium-235 in a typical nuclear bomb (typically one kg of uranium-235)</p>
| |nuclear-engineering|nuclear-technology| | <h1>Atomic Weapons</h1>
<p>The critical mass of metallic U-235 is about 50 kg[1] and the only U-235-based fission weapon deployed in combat--the Little Boy weapon--had about 64 kg of U-235[2]; there's more U-235 than that in nuclear power plants.</p>
<h2>Why this is not a good comparison</h2>
<p>Weapons-grade U-235 is highly enriched (think 80%+), meaning it contains mostly U-235 and correspondingly less U-238. A typical commercial nuclear reactor is running fuel with enrichments of up to 5%, which is far too low for weapons use. Of course, fresh fuel (fuel that hasn't been in the reactor) could be converted back into UF6 and further enriched if one had the capability.</p>
<p>However, once fuel has been irradiated in the reactor, fissions and neutron capture change the composition of the fuel. Neutron-absorbing U-236 is formed in the fuel and it becomes impractical to extract and further enrich just the U-235 from this fuel, which would be required constructing a fission weapon (there are, on the other hand, proliferation concerns surrounding reprocessing spent fuel related to the extraction of Pu-239). So even if there was enough mass of U-235 in a nuclear power plant to meet the required U-235 mass for a fission weapon, the composition of depleted fuel makes it unusable for this purpose.</p>
<h1>Nuclear Power Plants</h1>
<p>Nonetheless, to actually answer the question, let's estimate how much U-235 a typical PWR unit has.</p>
<h2>Mass of U-235 in the Plant's Core</h2>
<p>Let's assume we have a generic PWR plant with 157 fuel assemblies at an average enrichment of 3%, and with solid 0.3225" diameter pellets in a 17x17 assembly with 12' active fuel height. You can confirm that these are good round numbers in [3] which is also an excellent overall reference.</p>
<p>Each rod thus has a volume of <span class="math-container">$\pi(\frac{1}{2}0.3225 in.)^2 \times 144 in. = 11.7 in^3 = 192 cm^3$</span>. The density of UO2 is about 11.0 g/cm<span class="math-container">$^3$</span> and to use round numbers, <span class="math-container">$\frac{238}{(238+16+16)} = 88\%$</span> of that is going to be U. Multiply that all out and you get about 55 g of U-235 in one fuel rod.</p>
<p>A Westinghouse-style 17x17 assembly has 25 non-fuel locations (24 control rod guide tubes, 1 instrument tube again see [3]), so that gives us <span class="math-container">$157\times(17 \times 17 - 25) = 41448$</span> fuel rods, which gives us ~2300 kg of U-235 in the core.</p>
<h2>Spent fuel pool, casks, etc.</h2>
<p>After 36-54 months of operation (3x18 month cycles being pretty typical), fuel assemblies are discharged from the core into the plant's spent fuel pool. Although <span class="math-container">$\frac{2}{3}-\frac{3}{4}$</span> of the U-235 has been consumed by fission[4], a significant amount remains. Let's assume an average assembly is therefore left with about <span class="math-container">$\frac{1}{3} \times (17 \times 17 - 25) \times 55 gg = 5 kg$</span> of U-235.</p>
<p>After several years in the spent fuel pool, the fuel may be moved into casks holding several assemblies for further storage or transport. Depending on the country, the plant, etc. these may be held on-site pending the availability of a geological repository.</p>
<p>For a plant after 60 years of operation, that suggests up to ~2000 assemblies (assuming 18-month cycles with 1/3 of the core replaced at each reload) between the SFP and casks may be on site, which would be a further 10,000 kg of U-235.</p>
<h2>Summary</h2>
<p>So, our typical PWR unit might have on the order of 10-15,000 kg of U-235. A nuclear site may have 2 or more units (1 and 2 unit sites being predominate), so you'd have to multiply the above numbers depending on your definition of "one place" but to answer the question, our PWR has >>65 kg U-235.</p>
<h1>References</h1>
<ol>
<li><a href="http://blog.nuclearsecrecy.com/2015/04/10/critical-mass/" rel="nofollow noreferrer">http://blog.nuclearsecrecy.com/2015/04/10/critical-mass/</a></li>
<li><a href="https://www.livescience.com/45509-hiroshima-nagasaki-atomic-bomb.html" rel="nofollow noreferrer">https://www.livescience.com/45509-hiroshima-nagasaki-atomic-bomb.html</a></li>
<li><a href="http://www.xylenepower.com/the_westinghouse_pressurized_water_react.pdf" rel="nofollow noreferrer">http://www.xylenepower.com/the_westinghouse_pressurized_water_react.pdf</a></li>
<li><a href="https://www.radioactivity.eu.com/site/pages/Spent_Fuel_Composition.htm" rel="nofollow noreferrer">https://www.radioactivity.eu.com/site/pages/Spent_Fuel_Composition.htm</a></li>
</ol>
| 38568 | What is the *typical* quantity (in kg) of uranium-235 present at a given moment in a nuclear plant? |
2020-11-09T12:50:46.790 | <p>Usually, an MPC consists of discretizing the optimal control problem in time using some numerical quadrature scheme. So the infinite-dimensional OCP reads
<span class="math-container">$$\begin{aligned}J(\vec{u}) &= \varphi\left(\vec{x}(t_i+t_{hor})\right) + \int_{t_i}^{t_i+t_ {hor}}l(\vec{x},\vec{u})\text{d}t \\
\text{s.t.}\\
\dot{\vec{x}}&=\vec{f}(\vec{x},\vec{u}), \quad \vec{x}(t_i)=\hat{\vec{x}}_i\end{aligned}$$</span>
which can be transcribed to a static nonlinear program using, e.g., the trapezoidal rule for the integral and some other (or does it have to be the same?) integration scheme, e.g., the explicit Euler for the ODE, i.e.
<span class="math-container">$$\begin{aligned}
c(\vec w) &= \varphi\left(\vec{x}_N\right)+\frac{t_{hor}}{N}\sum_{k=0}^{N-1}\frac{l(\vec{x}_k,\vec{u}_k)+l(\vec{x}_{k+1},\vec{u}_{k+1})}{2}\\
\text{s.t.}\\
\vec{x}_{k+1}&=\vec{x}_{k}+\frac{t_{hor}}{N}\vec{f}(\vec{x}_k,\vec{u}_k), \qquad k=0,...,N-1\\
\vec{x}_0&=\vec{\hat{x}}_i
\end{aligned}$$</span>
where <span class="math-container">$\vec{w}=[\vec{x}_0^T\ ... \ \vec{x}_N^T \ \vec{u}_0^T\ ... \ \vec{u}_N^T ]^T$</span> are the decision variables, <span class="math-container">$t_{hor}$</span> is the MPC horizon, and <span class="math-container">$N$</span> is the number of discretization steps.
But often it can be observed that instead of the above sum, the cost function simply reads
<span class="math-container">$$c(\vec w) = \varphi\left(\vec{x}_N\right)+ \sum_{k=0}^{N}l(\vec{x}_k,\vec{u}_k).$$</span>
My question is, <strong>under what conditions is this possible</strong>? Can I always do this? I guess that it depends on whether the actual value of the integral matters or not but this implies that the integration does not change the optimal decision variables <span class="math-container">$\vec w^\ast$</span> since it involves just a scaling of the cost function? I could also imagine that it has to do with the function <span class="math-container">$l(\cdot)$</span> itself. Maybe this only works if <span class="math-container">$l(\cdot)$</span> is at most quadratic in the decision variables? Although I have seen this notation in nonlinear MPC.</p>
<p>EDIT 1: added ODE equality constraint and final costs</p>
<p>EDIT 2: Adding the final cost, I can see that one might need to have both discretizations to be the same or non at all for the continuous integral? Since in the discretized version, <span class="math-container">$\varphi\left(\vec{x}_N\right)$</span> depends on the final state which is the result of the ODE integration which, in this example, was done using the explicit Euler while the continuous integral has been approximated using the trapezoidal rule leading to different accuracies in the solution when they should probably be of the same order of accuracy? Therefore, just using the second sum leads to just adding up the values of the running cost term evaluated at the discretized steps which, in turn, depend on the ODE discretization. So this seems to make more sense to me than having two different integration schemes.</p>
| |control-engineering|control-theory|optimal-control|nonlinear-control| | <p>MPC finds the optimal input <span class="math-container">$u^*$</span> which is the input that minimizes the cost function <span class="math-container">$J$</span> or <span class="math-container">$c$</span>. This means that regardless of what this actual value is, its proven to be its minimum. As such, multiplying the cost function with any constant value does not change this minimum, it just scales the value. Therefore, <span class="math-container">$\frac{t_{hor}}{N}$</span> does not affect the optimal input and can be eliminated. If there is a terminal cost present, the terminal state should also be computed using this trapezoidal rule and is also multiplied with this constant.</p>
<p>Secondly, taking the cost over actual sampled values of <span class="math-container">$x_k$</span> or taking the average between two samples seems nearly a matter of perspective. In fact, if you write out both sums, you discover that the difference is something like this:
<span class="math-container">$$\frac{1}{2}\sum_{k=0}^{N-1}l(x_k,u_k) + l(x_{k+1}, u_{k+1}) = \frac{1}{2}(l(x_0,u_0)+ l(x_{N}, u_{N})) + \sum_{k=1}^{N-1}l(x_k,u_k)$$</span>
Additionally, if this cost function happens to have a initial cost term and terminal cost term you can imagine it begin fully equal to the other one. (albeit having a different initial and terminal cost value). However, if this <span class="math-container">$l(.)$</span> in your cost function is a continuous time one (I assume its not really linear in the states and inputs), I should not that the standard cost function probably uses <span class="math-container">$l(.)$</span> in a discretized way.</p>
<p>Hope this got you somewhere, I have not encountered your trapezoidal approach ever, so that makes it kinda interesting.</p>
| 38570 | Numeric quadrature vs summation of running costs in model predictive control |
2020-11-10T09:05:30.223 | <p>I am trying to work out how a recently acquired Arburg 220-90-350 from 1982 works so that I can build a new control system and get it working. It's a direct clamping machine with no toggle.</p>
<p>The injection side is quite simple, just 3 motors or cylinders each controlled by a separate valve.</p>
<p>The clamp and ejector side has me confused though, there is a directional valve controlling each cylinder (simple enough) but each cylinder has a 3rd line the same size as the others connected to another single valve labelled 'high pressure'. So to summarise, 2 hydraulic cylinders each with 3 hoses, 3 directional control valves controlling the 2 cylinders.</p>
<p>I can't begin to draw a schematic without further disassembling the machine so I am hoping someone can give me some reasons why a hydraulic cylinder would have 3 hoses of equal size, with 1 on each cylinder connected to the same valve.</p>
<p><strong>Edit:
On further investigation, I can confirm that non of the 3 hoses are connected together, i.e. there is not two in parallel.</strong></p>
<p><a href="https://i.stack.imgur.com/T890E.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T890E.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/i0jbX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i0jbX.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/BjP0u.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BjP0u.jpg" alt="enter image description here" /></a></p>
| |hydraulics|molding|injection| | <p>So it turns out that the ejector cylinder is just a standard double-acting cylinder with 2 hoses, the remaining 4 hoses are all used for the clamping cylinder.</p>
<p>Two of the large hoses do retract and return at low pressure, the two remaining hoses one large and one small do the high pressure clamping.</p>
<p>This is standard for a 2 platten direct clamping machine. The platen closes under low pressure and when it is closed the high pressure gives it the final clamping force.</p>
| 38577 | Why would these hydraulic cylinders have 3 large connections? |
2020-11-10T11:14:32.777 | <p>I am trying to understand the meaning of 'Trip force' of keypad dome on <a href="https://www.keyelco.com/product.cfm/product_id/14287" rel="nofollow noreferrer">these pages</a>, <a href="https://www.keyelco.com/product.cfm/product_id/14313" rel="nofollow noreferrer">and this</a>, <a href="https://www.keyelco.com/product.cfm/product_id/14318" rel="nofollow noreferrer">and this</a>.</p>
<p>Example: Trip Force: 14.1 oz. [400g]</p>
<p>Does it mean that when I put a mass of 400g on this dome then it will be depressed?</p>
<p>On another <a href="https://www.digikey.com/en/products/filter/tactile-switches/197?s=N4IgTCBcDaINIFMCeBnALgewHYIAQqwEMAHfAdwEs0BjAC1wA4A6AFgAYBbDkAXQF8gA" rel="nofollow noreferrer">website</a> this parameter has a different name 'operating force' and is given in units of gram-force (gf).</p>
<p>Does these 2 websites show same parameter with different names and units or these are different?</p>
| |mechanical-engineering| | <p>See the plot below: (Source <a href="https://www.snaptron.com/quality/test-procedures/" rel="nofollow noreferrer">https://www.snaptron.com/quality/test-procedures/</a>)</p>
<p>As you gradually push down on the dome, increasing the force that you apply, the switch will begin to displace - at a certain point, <span class="math-container">$F_{MAX}$</span>, the dome will 'pop' inside out - it is bi-stable, and this is what causes the 'click' and tactile feedback that you feel. After this snap, it takes a lower force to <em>hold</em> the switch down that it took to push it down in the first place.</p>
<p>The 'reverse snap', when the switch pops back to its original shape happens when you remove force from the switch, and reach as low as <span class="math-container">$F_{RMIN}$</span>.</p>
<p>To answer your question directly, the "Trip Force" will correspond to <span class="math-container">$F_{MAX}$</span> on this plot, and any sites that only list the force with units of grammes (a unit of mass, not force), will mean <span class="math-container">$\text{gf}$</span>. This is simply a typo/error/simplification on their part.</p>
<p><a href="https://i.stack.imgur.com/vr6U6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vr6U6.png" alt="enter image description here" /></a></p>
| 38578 | Trip force of a Keypad Dome |
2020-11-10T13:44:15.327 | <p>Here's the nameplate from a 5 HP single phase electric motor - used in an air compressor:</p>
<p><a href="https://i.stack.imgur.com/EgTc3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EgTc3.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/jq4HR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jq4HR.png" alt="enter image description here" /></a></p>
<p>Note that it reads:</p>
<blockquote>
<p>HIGH TORQUE GREASE BEACON 325</p>
</blockquote>
<p>which I interpret to mean that you should use high temperature grease of which "Beacon 325" is an example or recommended brand / product.</p>
<p>I'm familiar with electric motors which either have used permanently-lubricated sealed or shielded bearings, or alternatively bearings which required <em>oil</em>. But I haven't come across any which require periodic grease maintenance.</p>
<p>Although a permanently sealed bearing would of course have grease I'd be surprised if a motor was expected to be disassembled for regular grease service. And I don't see any grease fittings.</p>
<p>So what does this mean? What service is implied by the grease statement on the label? Could this refer to the motor application (air compressor) and not the motor itself? But it seems like a general purpose motor.</p>
| |motors| | <p>Grease gets used in roller and ball bearings which are well-sealed to prevent dirt intrusion and which are only rarely disassembled for maintenance (i.e., relubrication). Even when the bearing is hot, the grease does not seep out of the bearing and let it run dry the way that oil would. So (for example) automotive wheel bearings are greased but not oiled.</p>
<p>Oil gets used in bearings where they cannot be well-sealed against dirt intrusion and which use cheap bronze bushings instead of more expensive balls or rollers to carry the bearing loads. In this case, the porous bronze bushing is surrounded by an oil-soaked wick which keeps the gap between the bushing and the shaft running through it wet with oil, which slowly seeps out of that space over time.</p>
<p>So the specification for grease in your case means the bearings contain balls or rollers for long life and high loads, which are intended to be maintenance-free for thousands of hours of operation even in dusty conditions.</p>
| 38582 | Where/why would grease be required on this electric motor? |
2020-11-10T23:20:49.390 | <p>In many engineering descriptions I come across this term:</p>
<p>"Hardware/Software integration" or "Hardware/Software integration testing"</p>
<p>I came across couple of articles but they even made me more confused.</p>
<p>Can you give one or two examples which depicts a scenario for "Hardware/Software integration"?</p>
<p>If someone would ask me I would say "writing a piece of software or GUI which controls electronics". But I am not sure whether that would a correct example. Is there any typical example?</p>
| |electrical-engineering|terminology| | <p><strong>Example 1:</strong> <em>Hardware Integration</em></p>
<p>Some organization have Electrical Hardware building blocks that have been pre-designed, tested and validated. These building blocks could be for step-down regulator circuits, microcontroller controller circuits, audio processing circuits, video processing circuits, AM/FM RF circuits, BLE circuits, etc. In order to design a product electrical hardware engineers would select the appropriate electrical hardware building blocks and integrated them to create an electrical hardware product.</p>
<p><strong>Example 2:</strong> <em>Software Integration</em></p>
<p>Like the above some organization have software building blocks that have been pre-designed, tested and validated. These software building blocks are pre-designed, tested and validated. These software building blocks are designed to control and process data from electrical circuits similar to the above hardware building block list. A software integration engineer(s) would select the appropriate software building blocks to create a software design that would control the electrical hardware.</p>
<p>Most modern day electrical products require both Hardware and Software. Therefore, combining example 1 Hardware and example 2 software is considered hardware and software integration.</p>
<p>The purpose is to encourage re-use of proven building blocks to reduce product design time thus reduces time to market.</p>
| 38590 | What is really meant by HW/SW integration? |
2020-11-12T14:58:29.540 | <p><strong>I have a motor controller that operates at 10-50V and draws 5A controlling a 3V motor that draws less than 1A.</strong></p>
<p><strong>I had a an idea where I want to try to use a microcontroller to control this set up.</strong></p>
<p><strong>The motor controller has an auxiliary 5V output that can support loads that draw less than 10mA.</strong></p>
<p><em><strong>If I find a suitable microcontroller, can I use the motor controller to power it? Should I use the motor controller to power it?</strong></em></p>
<p><em>More info:</em> I am working on a design for a robot build. My system currently has two motors one that runs at 48V (and around 3A) and the aforementioned one (3V) both using the same type of motor controller. I'd prefer not to have two power supply lines running, so can I run the smaller motor system on the motor controller powered at 50V? Is that safe? (the motor controller can take a 56V max)</p>
<p>Thank you for reading, any input would be highly appreciated.</p>
| |electrical-engineering|motors|design|power|robotics| | <blockquote>
<p>If I find a suitable microcontroller, can I use the motor controller
to power it? Should I use the motor controller to power it?</p>
</blockquote>
<p>Yes, if your controller provide 5V 10mA (supposing sufficient quality in the DC output), you can run several MCU (e.g. I use the STM32F103, known for the arduinos, which consume around 10mA if configured at 8MHz).</p>
<p>Note, however that most low-power MCU run under 5V, so you can gain some mA by a proper conversion.</p>
| 38617 | Power a microcontroler with a motorcontroller? |
2020-11-13T09:52:29.910 | <p>There are 3 horizontal lengths of timber stretching across the centre-level of my attic/loft space and are attached at either end to the rafters with a single nail. I can wobble them by hand and they don't seem to be under any tension/compression.
The House is a bungalow in a T shape and in the picture attached, I am facing towards the top of the T from near the bottom of the T. The roof is a purlin roof and the supports under the purlins are on top of 2 central masonry walls (the hallway) and the house is a solid masonry structure.</p>
<p>From what I can tell, these horizontal members are either leftover temporary supports from the initial construction or are mid-level collar ties?</p>
<p>Could I remove these 3 timbers to open up the attic space for easier access & light storage? I would move them up higher, just under the ridge board, either way.</p>
<p>Any insight is appreciated as I'm not sure which way to go exactly and just looking for some guidance before I talk to an engineer.</p>
<p>Thanks,
Derek</p>
<p><a href="https://i.stack.imgur.com/kLWNU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kLWNU.jpg" alt="attic pic" /></a></p>
| |structures|building-design| | <p>Kamran is much more experienced, so I would go for his opinion (I also upvoted his answer). Especially given the circumstances (i.e. level 1 hurricane).</p>
<p>However if you are that desperate to use the space, one compromise which might be useful to you and maybe structurally sound at the same time (kamran might have something to say about that).</p>
<p><a href="https://i.stack.imgur.com/1wKfS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1wKfS.png" alt="enter image description here" /></a></p>
<p>You could try replacing the horizontal beams, with <strong>all</strong> of the following:</p>
<ul>
<li>a longer beam running along the bottom of the floor</li>
<li>two diagonal beam connected to the new beam</li>
</ul>
<p>Additionally, (but highly recommended) would be to use a steel wire across (where the old beams used to be. The point would be to use an adjustable cable (with a turnbuckle)</p>
<p><a href="https://i.stack.imgur.com/Yzn9Pm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yzn9Pm.png" alt="turnbuckle" /></a></p>
<p>alongside an easy release</p>
<p><a href="https://i.stack.imgur.com/dMOBv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dMOBv.png" alt="example of easy release" /></a></p>
<p>This way if you wanted to put something big in the space, you could untie the buckles and then tighten them back again. (NOT foolproof but if you are desperate for some space then it might be helpful).</p>
<p>However, as I said, I would have liked to hear from kamran, if this has a chance...</p>
| 38626 | Are these horizontal timbers removable in an attic/loft? |
2020-11-13T20:20:35.597 | <p>I'm trying to research additive (or 'bottom-up') fabrication techniques that rely on lasers or other advanced photonic systems. So far, the most prevalent technique I've come across is two-photon polymerisation, but I'm struggling to find any others. Any further examples would be much appreciated.</p>
| |manufacturing-engineering|lasers|photonics| | <p>Here is an incomplete list of laser-based additive manufacturing, based on my comment. I might expand on it later:</p>
<ul>
<li><p><strong>Selective laser melting/sintering (SLM/SLS):</strong> a laser beam melts/sinters a layer of powder at the desired spots, after solidification, a new layer is added and the process starts over.</p>
</li>
<li><p><strong>Stereolitography:</strong> basically the same as SLM, but instead of powder, you use a resin that solidifies when exposed to light. One of the oldest additive manufacturing processes.</p>
</li>
<li><p><strong>Laser deposition welding:</strong> a laser melts a spot on a base plate, additional material is fed into this melt pool in form of powder or wire (processes with both at once also exist, need to search a bit for a citation). From there on you build up a part in layers. This process is very close to the classic "3D printer" with plastic FDM.</p>
</li>
</ul>
| 38634 | What are some examples of laser-based additive manufacturing techniques? |
2020-11-14T23:12:02.213 | <p>I have an shaft/crank that will be turned by the user.
I want to precisely and dynamically control the torque required to do so.</p>
<p>For simplicity we can think of it as a exercise machine (so the torque will be pretty high, depending on the length of the crank arm).
User rotates a crank, computer controls how hard it is to do so.
Like a hand powered crane with a simulated load.</p>
<p>I'm looking for advice on how to do that.</p>
<p>Simplest solution, where the rope with a weight at the end is attached to the shaft lacks gradual control.</p>
<p>Solutions I was thinking about:</p>
<ol>
<li>Disc brake, even from a bicycle and a servo controlling the clamping force. - I think this will be very hard to precisely control (in terms of the torque needed to turn).</li>
<li>Backdrivable electric motor with gearing - But for high torque that would require a lot of gearing / a powerful motor,
plus that would mean the motor is constantly being backdriven, which would probably increase the risk of damaging it (or the controller).</li>
<li>Some combination of weights and motors where most of the torque is provided by the weight, but motor is there to adjust it.</li>
</ol>
<p>I need your opinions, suggestions and ideas.<br />
Thanks!</p>
| |motors|torque|mechanisms| | <p>Like you said there are many ways to do it. Computer controlled and usually expensive.</p>
<h2>Dynamometers.</h2>
<p>These are the tools built exactly for this type of task. Obviously they are amognst the most expensive for the job. The most difficult part would be to determine the specs (what is the maximum torque, what is the control resolution, is it computer controlled etc ). You can see what a dynamometer is at various links (e.g. <a href="https://www.google.com/search?q=Dynamometers" rel="nofollow noreferrer">1</a>.</p>
<h2>Electrical system with inverter and motor</h2>
<p>What you can do is connect your shaft to a generator/motor, and control though an inverter the torque. Basically, the motor is free to turn, but as you increase the electric current passing through the coils it is harder and harder to move the rotor.</p>
<p>Similar idea but reversed is the following. you could connect the shaft to a generator and connect it to a inverter which is connected to the grid. You could control how much power is absorbed (and therefore the torque), and you could put that energy to the grid (so you would have a human generator).</p>
<p>In both cases you'd need a encoder for the shaft rpm.</p>
<p>Pros:</p>
<ul>
<li>torque <em>should</em> be controlled quite nicely (barring friction and cogging torque)</li>
</ul>
<p>Cons:</p>
<ul>
<li>expensive</li>
<li>requires programming knowledge</li>
</ul>
<h1>Mechanical brakes</h1>
<p>There are many variations here. The main problem is controlling the torque accurately. Basically you will be relying on the coefficient of friction, which can change depending on <em>wear</em>, and other conditions e.g. humidity.</p>
<p>Wear might not seem relevant however, because you will be applying it constantly, it will become an issue sooner rather than latter.</p>
<ul>
<li>Drum brake</li>
</ul>
<p>you could use the computer to actuate the brake and control whether the applied torque.</p>
<p><a href="https://i.stack.imgur.com/hX5hs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hX5hs.png" alt="enter image description here" /></a></p>
<ul>
<li>Disc Brake:</li>
</ul>
<p>this is a similar to the drum brake.</p>
<ul>
<li>Belt Friction</li>
</ul>
<p>you can use a belt and apply tension on the belt in order to increase the friction force between belt and trum. If the drum is connected to the shaft it connect the torque on the shaft.
<a href="https://i.stack.imgur.com/bDrtY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bDrtY.png" alt="enter image description here" /></a></p>
<ul>
<li>Many other variations...</li>
</ul>
<h3>Constant torque and constant force in mechanical systems</h3>
<p>One main problem when using friction to mechanically control torque is that the torque might exhibit large variations. In order to mitigate that it is important to apply a constant force on the friction surface. One (good) way to do that is insert a spring between the actuator and the controlled surface. Then the displacement of the actuator will control the displacement of the spring and indirectly the force. If you have a sufficiently long travel for the spring then you can have a very good control on the force. A nice (but rather old) description is at this <a href="https://apps.dtic.mil/sti/pdfs/ADA434149.pdf" rel="nofollow noreferrer">link</a>.</p>
| 38647 | Controlling torque required to turn a shaft |
2020-11-14T23:55:41.543 | <p>I've started using SolidWorks migrating from other software, Creo Parametric (formerly Pro/E) and the Base/Boss terminology confused me. After significant research, I've found that a "base feature" represents the starting feature and a "boss feature" represents any features built off of that - <a href="http://help.solidworks.com/2020/english/solidworks/acadhelp/c_feature_based_models.htm" rel="nofollow noreferrer">http://help.solidworks.com/2020/english/solidworks/acadhelp/c_feature_based_models.htm</a>.</p>
<p>However, through using SolidWorks 2020, and from many different tutorials, I have never been able to make the first feature a base feature, e.g. "<strong>Base</strong>-Extrude1", but rather it alway shows as a boss feature, e.g. "<strong>Boss</strong>-Extrude1". Has the "Extrude Base" been deprecated from SolidWorks and Extrudes are always boss extrudes now or am I just doing it wrong? Are there any differences in the Property Manager for base features vs boss features?</p>
<p><a href="https://i.stack.imgur.com/M9Da4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M9Da4.jpg" alt="Example of Simple Extrude that shows as Boss-Extrude1" /></a></p>
| |cad|solidworks| | <p>There's no difference at all - an extrude is an extrude is an extrude (unless it's a cut)</p>
<p>SolidWorks differs from some other CAD in that Boss and Cut are separate tools - others have just one extrude tool with a Boolean toggle in the feature manager</p>
<p>You can (and should for your own sanity) rename features in the tree to better represent what they are to you. Call it Base if that helps you!</p>
| 38648 | Solidworks Extrude Base/Boss - Base Extrude Doesn't Seem to Exist Anymore? |
2020-11-15T13:18:12.780 | <p>I have tried to learn about "hysteresis" from different engineering knowledge sources and what i am able to understand that in simple words or nutshell, it is delay between output and input</p>
<p>Is there anything wrong with my understanding?</p>
| |mechanical-engineering|electrical-engineering|experimental-physics| | <p>I like to think of hysteresis as describing deadzones in the coupling between 2 variables.</p>
<p>A classic example is as SF said, a heater that turns on at 80 degrees and turns off at 90 degrees. Then on again once the temp falls back down to 80. There are 10 degrees of hysteresis here.</p>
<p>Another example would be pulling a cart using a rope. The carts motion is coupled to your motion, but if you reverse direction, the cart will stop briefly while the rope is slacked. If the rope is 1 meter long the hysteresis will be 2 meters.</p>
| 38654 | meaning of "hysterisis"in simple words? |
2020-11-15T13:46:40.360 | <p>I want to know if stud bolt dimensions are covered under ASME B18.2.1. I could not specifically find them in the standard or they may be included under some other technical name within the same standard or covered in a different standard. I am a newbie. Hence the doubt.</p>
| |mechanical-engineering|piping| | <p>Stud bolt dimensions are covered under several standards.</p>
<p>The stud bolt lengths are defined in the ASME B16.5 standard.</p>
<p>Bolts' threadings are defined in ASME B1.1 Unified Inch Screw Threads, (UN and UNR Thread Form).</p>
<p>There are several standards covering the material and other properties.</p>
<p><a href="https://i.stack.imgur.com/Xv9fY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xv9fY.png" alt="stud bollts and hex bolts" /></a></p>
<p>This article is helpful. <a href="http://www.wermac.org/bolts/bolts_stud-bolts_for-flanged-connections.html" rel="nofollow noreferrer">Bolts and nuts.</a></p>
| 38655 | Is stud bolts covered under ASME B18.2.1 |
2020-11-16T01:41:13.227 | <p><a href="https://i.stack.imgur.com/oxHbV.png" rel="nofollow noreferrer">differential drive vehicle in world frame</a></p>
<p>what would change in the kinematic model of the differential drive vehicle if i wanted to simulate it
using pygame module in python ( pygame or opencv or any other one )</p>
<p><strong><code>x_dot = (vl+vr)/2*cos(theta)</code></strong></p>
<p><strong><code>y_dot = (vl+vr)/2*sin(theta)</code></strong></p>
<p><strong><code>theta_dot = (vr-vl)/w</code></strong></p>
<p>vr is the velocity of the right wheel</p>
<p>vl is the velocity of the left wheel</p>
<p>theta is the heading angle</p>
<p>it uses a screen coords system which start at the top left corner of the screen insted of the lower left
my initial guess is since only the y axis is inverted then i should invert the sign of the y_dot</p>
<p>i hope i explained my question clearly the picture is in the link above</p>
| |modeling|simulation|kinematics| | <p>The kinematics equation should remain the same. The only problem I see is with the third equation which should be</p>
<p><code>theta_dot = (vr-vl)/W</code></p>
<p>What you need to consider on top of those equations since you are working with screen pixels is:</p>
<ul>
<li>relationship of x and y with screen pixels (units should be <span class="math-container">$\frac{\text{x units}}{pixel}$</span>, <span class="math-container">$\frac{\text{y units}}{pixel}$</span> or the inverse).</li>
<li>you'd need to add variables for monitoring the position (preferably floats). e.g:</li>
</ul>
<pre><code>x = 0.0
y = 0.0
</code></pre>
<ul>
<li>you'd need to integrate in time with respect to the position</li>
</ul>
<pre><code>x = x+ x_dot*dt
y = y+ y_dot*dt
theta = theta+ theta_dot*dt
</code></pre>
<p>where <code>dt</code> is the timestep you want to use.</p>
<ul>
<li>then you'd just need to redraw the vehicle at the orientation of theta (after you cast x, y to integers).</li>
</ul>
| 38663 | differential drive vehicle in screen coords |
2020-11-16T13:10:15.177 | <p>How to establish metric specifications for ergonomic suitability of a product? I meant like for a product say electric iron,it has a metric that it provides different temperatures range to iron, which can be measure in terms of temperature.</p>
<p>Similarly I have a question how to measure ergonomic suitability of a product i.e establishing Need matrices and Units.</p>
| |mechanical-engineering|product-engineering| | <p>To put it as an answer I might be able to expand it a little bit.</p>
<p>Measuring ergonomic suitability is really subjective. From my experiences there are obvious guidelines but even those are not clear matrices of numbers. They are more like: Make sure the handle is long enough to support various sizes of hands and is textured to provide grip. Avoid positioning buttons or handles in uncomfortable locations. place the power cord somewhere it doesnt get in the way etc. Do note that ergonomics not only include the ergonomics of using the device, but also about assembling the device. For instance, one should place a screw in a position that is easily reachable and with enough space to place a screwdriver.</p>
<p>However, depending on the classification of the device you are designing, certain "rules" have to be omitted to distribute the product (this also depends on the region you are living in). These rules are mostly about safety guidelines. For instance, in Europe you have the Conformité Européenne (CE) marking. Additionally, there is something like a Declaration of Conformity (again in Europe). This is a set of guidelines a product is expected to follow. For instance: CE dictates that a emergency button has to be colored in an distinctive color (compared to the colors of the device). The Declaration on the other had advices to use the color red with a yellow base (including exact color codes). I certainly hope I explained this correctly because its not that I know these things by heart.</p>
<p>So to come back to ergonomics, the best way to gain insight on how to improve the design is to build a physical model (think of like 3D printing the shape). Test it elaborately in a circumstance you would expect someone to use the device. So if its used in cold weather, expect the user to wear gloves. With a bit of trial and error you can easily define some of your own units to indicate ergonomic suitability.</p>
| 38667 | How to establish metric specifications for ergonomic suitability of a product? |
2020-11-16T14:16:43.067 | <p>As an EE, I am very comfortable with the mapping of thermal quantities to their electrical equivalents. However, understanding how thermal capacitances should be connected is driving me crazy.</p>
<p><a href="https://lpsa.swarthmore.edu/Systems/Thermal/SysThermalModel.html##section7" rel="nofollow noreferrer">this</a> example:</p>
<p><a href="https://i.stack.imgur.com/QXru1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QXru1.png" alt="enter image description here" /></a></p>
<p>Shows a "room" with a thermal resistance and capacitance. There is an external temperature excitation upon which the ambient temperature "rides". Why is it that when the system is modeled electrically, the bottom nodes for the R and C aren't common to one another? Why is it that the C is to the ambient reference, and the R is to the excitation? Shouldn't the C also connect to this excitation?</p>
| |modeling|circuits| | <p>In this example the point is that you have:</p>
<ul>
<li>an insulated (R) room</li>
<li>with some thermal capacitance (C).</li>
</ul>
<p>and the idea is that you have an external excitation, ie. you change the temperature of the boundary surface. <strong>The thermal resistance is related to the total heat flux (J/s)</strong> that enters on exit the system from all boundary walls.</p>
<p>For thermal resistance, the flow of energy is depended on the temperature difference. <strong>The thermal capacitance is related to the heat energy (J) stored in the system</strong>.</p>
<p>On the other hand, regarding any surplus of energy that enters the enclosed system gets stored and raises the temperature (i.e. raises the internal temperature). However, the accumulated heat energy (capacitance) does not change on the temperature of the boundary; rather it is an integration over time of the excess/deficit of heat flux.</p>
<p>So in terms of the capacitance, you <strong>need a fixed temperature point</strong>. On the other hand, for the resistance you need the temperature difference which in this example can be adjusted. So in order to compromise what happens in this example, is that the ambient temperature is used as a baseline, while the thermal excitation <span class="math-container">$\theta_e$</span> is defined based on the difference between ambient and the outer boundary of the system.</p>
| 38668 | Thermal Capacitance - Convention for Modeling of Electrical Equivalent |
2020-11-16T21:41:28.480 | <p>Today I came across this question and I can't sleep thinking about how to solve it. How do I find the principal values and sketch a shear force and bending please see my attempt as I'm not sure if this is correct. diagram on the below picture? <a href="https://i.stack.imgur.com/TmP9D.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmP9D.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/SE3Yb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SE3Yb.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering| | <p>So, in such problems, the first thing to do is calculate the support reactions. Thankfully you've already done that correctly, so we can move on from there.</p>
<p>Now, something to keep in mind when trying to figure this sort of thing out is the fundamental equation of Euler-Bernouilli Beam Theory:</p>
<p><span class="math-container">$$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI \dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$</span></p>
<p>or, in English, we must remember that shear forces are the integral of the applied loading and bending moment is the integral of shear (and more, but this is what's relevant to this problem).</p>
<p>So, let's take a look at your expected solution, adding some names to the points of interest:</p>
<p><a href="https://i.stack.imgur.com/OJoYs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OJoYs.png" alt="enter image description here" /></a></p>
<p>There are a few problems which are immediately apparent:</p>
<ul>
<li>your moment diagram is zero at the support, but we know that it must be 70 kNm from the reaction</li>
<li>you show zero shear in CD, but a bending moment going from 0 to 40 kNm in the same segment. However, if we remember that bending moment is the integral of shear, that makes no sense: if shear is constant at zero, that means the slope of the bending moment must also be zero. So at least one of those diagrams is wrong</li>
<li>You also have a change in the slope of the bending moment, from positive (as in, getting further away from zero) in CD to negative (moving towards zero) in AB. This would imply in a change of sign in the shear stress. Since the change in the slope of the bending moment is instantaneous, this would be equivalent to a discontinuity in the shear diagram. And since shear is the integral of the applied loading, this would imply in a concentrated force at that point, which we don't have.</li>
</ul>
<p>So, let's try again.</p>
<p>Let's start with the shear diagram. We know that it's the integral of the applied loading <span class="math-container">$q$</span>. In this case, however, <span class="math-container">$q(x) = 0$</span> everywhere. So we know the shear force will be constant everywhere. But constant and equal to what?</p>
<p>Well, while <span class="math-container">$q(x) = 0$</span> everywhere, it does have concentrated forces at A and D, which cause discontinuities in our shear diagram. So we know that the shear at A and D will be equal to 10 kN at those points (let's not think about whether that's positive or negative for now). And we've already concluded that the shear is constant throughout since <span class="math-container">$q(x) = 0$</span>, so I guess that means the shear diagram is simply 10 kN everywhere, right?</p>
<p>Not so fast.</p>
<p>In this case, we have the complication caused by the vertical segment BC. To handle this, we need to remember what the applied force this structure is supporting is. It's a vertical force. This vertical force is resisted by the horizontal member CD, but the force being resisted is vertical nonetheless. Wherever you are in the structure, all if it is supporting a vertical force.</p>
<p>And does a vertical force cause shear in a vertical element? Obviously not. It causes an axial load.</p>
<p>So what we end up having in this case is a shear diagram which is equal to 10 kN in AB and CD, but zero at BC, and an axial diagram which is zero at AB and CD but 10 kN at BC.</p>
<p>This doesn't break with the rule that shear must be constant when <span class="math-container">$q=0$</span> because the instantaneous change in direction that happens from AB to BC and from BC to CD behaves in exactly the same way as a concentrated force: it causes a sudden change to how the beam behaves under the external load: the horizontal segments resist it through shear, the vertical segment through axial. The direction of the internal force itself is always vertical, and that's what really matters. If BC weren't vertical but inclined, then we'd instead see only a drop in the shear diagram as some of the shear is transformed into axial. If BC were curved, then the shear diagram would change according to the tangent at any given point.</p>
<p>And so we move on to the bending moment diagram. This can be done by simply integrating the shear diagram. CD is 3 m long and suffers a constant shear of 10 kN, so the bending moment at C is of course equal to <span class="math-container">$3 \times 10 = 30\text{ kNm}$</span>. Notice this is also what you get from simply calculating the bending moment caused by D on C (just as you did to calculate the moment reaction at A). That's not a coincidence, it always holds true.</p>
<p>Indeed, think about what the reaction moment would be if AB were only 3 m long. In that case, the horizontal lever arm from A to D would be six meters, so the support's bending reaction would be 60 kNm. The bending moment is always 10 times the horizontal lever arm. And that also applied for the beam itself! The bending moment is always 10 times the horizontal distance from the desired point to D.</p>
<p>So, the bending moment at D is zero, at C, 30 kNm, at B, also 30 kNm, and at A it's the same as the reaction, 70 kNm. Basically, as far as the bending moment is concerned, that vertical step at BC is entirely irrelevant: the bending moment is basically identical to what it'd've been without BC (i.e. a simple cantilever).</p>
<p>Now, lastly, we need to talk about signs. Below we see the standard sign convention:</p>
<p><img src="https://i.stack.imgur.com/TfxCv.png" alt="" /></p>
<p>This shows the direction of positive internal forces depending on whether you're looking at the left or right side of the section.</p>
<p>We can see that a downward force to the right or an upward force to the left are both considered positive in shear. Since that's the case throughout the structure (being pushed down by a force to the right and pushed up by the support to the left), our shear will be entirely positive.</p>
<p>The downward force will obviously put BC in tension, which is positive in our convention.</p>
<p>However, the force creates a clockwise rotation to the right of the structure, which is resisted by the counter-clockwise moment at the support to the left. Both of these are negative in our convention, so we get negative bending moment throughout the structure as well.</p>
<p>And now, to check our work:</p>
<p><a href="https://i.stack.imgur.com/QMIU4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QMIU4.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/3vo4X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3vo4X.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jmSZ8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jmSZ8.png" alt="enter image description here" /></a></p>
<p>Note that though the bending moment plot is above the beam, it has a negative value. That's also standard convention: always draw the bending moment on the side which is under tension.</p>
<hr />
<p>Results obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis tool.</p>
| 38676 | Shear force and bending moment |
2020-11-17T17:05:17.800 | <p>I want to 3d print a custom computer case, I have the skillset to do so, however, I do not know where to find the technical drawings required. I need drawings for a 280mm Radiator, an ITX Motherboard, and various other parts.</p>
| |mechanical-engineering|electrical-engineering|3d-printing|drafting|computer| | <p>I would research component manufactures web sites. It is highly unlikely that you will find all the information in on location. Here our example web site links to get started.</p>
<ul>
<li><a href="https://sta3-nzxtcorporation.netdna-ssl.com/uploads/download/attachment/758/H210i_H210-manual-ver8-070720-web.pdf" rel="nofollow noreferrer">NZXT - Mini ITX Computer cases</a></li>
<li><a href="https://www.fractal-design.com/wp-content/uploads/2020/12/Era_Product-Sheet_EN.pdf" rel="nofollow noreferrer">Fractal Era ITX Computer cases</a></li>
<li><a href="https://web.archive.org/web/20110613223644/http://www.via.com.tw/en/downloads/whitepapers/initiatives/spearhead/ini_mini-itx.pdf" rel="nofollow noreferrer">Mini-ITX Mainboard Specification</a></li>
<li><a href="https://www.corsair.com/us/en/Categories/Products/Liquid-Cooling/Dual-Radiator-Liquid-Coolers/Hydro-Series%E2%84%A2-RGB-Platinum/p/CW-9060039-WW#tab-tech-specs" rel="nofollow noreferrer">Hydro Series™ H100i RGB PLATINUM 240mm Liquid CPU Cooler</a></li>
</ul>
<p>Another good place to look for is computer parts trade shows in your local area. With the current ongoing COVID-19 situation not sure if any of these trade fair active.</p>
| 38692 | Where can I find technical drawings for computer parts? |
2020-11-17T21:41:18.257 | <p>The book <a href="https://books.google.com/books?id=nEc2OxxT_uMC&pg=PA119&dq=momentum+entering+the+rotor+euler+equation&hl=en&newbks=1&newbks_redir=0&sa=X&ved=2ahUKEwjkoe3jwYrtAhWrzVkKHTFxDJoQ6AEwAXoECAcQAg#v=onepage&q&f=false" rel="nofollow noreferrer">"Gas Turbine Engineering Handbook"</a> by Boyce, on p. 117 states, the continuity equation,</p>
<p><span class="math-container">$ \dot m = \rho A V $</span>,</p>
<p>has as its differential form,</p>
<p><span class="math-container">$ 0 = \frac{d\rho}{\rho} + \frac{dA}{A} + \frac{dV}{V}$</span>.</p>
<p>How did the second equation get derived from the first?</p>
| |fluid-mechanics|mathematics| | <p>for <span class="math-container">$y = f(x_1, ..., x_n)$</span>, the sum of the partial differentials with respect to all of the independent variables is the total differential:
<span class="math-container">$$ dy = \frac{\partial y}{\partial x_1}dx_1+...+\frac{\partial y}{\partial x_n}dx_n $$</span></p>
<p>For our case:
<span class="math-container">$$ \dot{m} = \rho Av$$</span>
<span class="math-container">$$ d\dot{m} = \frac{\partial \dot{m}}{\partial \rho}d\rho + \frac{\partial \dot{m}}{\partial A}dA+ \frac{\partial \dot{m}}{\partial v}dv$$</span>
<span class="math-container">$$\frac{\partial \dot{m}}{\partial \rho} = vA$$</span>
<span class="math-container">$$\frac{\partial \dot{m}}{\partial A} = \rho v$$</span>
<span class="math-container">$$\frac{\partial \dot{m}}{\partial v} = \rho A$$</span></p>
<p>Substituting:
<span class="math-container">$$ d\dot{m} = Av \ d\rho + \rho v \ dA+ \rho A \ dv $$</span></p>
<p>for steady-state case, <span class="math-container">$\dot{m} = \text{const}$</span>, so <span class="math-container">$d\dot{m} = 0$</span>, then:
<span class="math-container">$$ Av \ d\rho + \rho v \ dA+ \rho A \ dv = 0 $$</span></p>
<p>Divide by <span class="math-container">$\rho A v$</span>:
<span class="math-container">$$ \boxed{\frac{d\rho}{\rho} + \frac{dA}{A}+ \frac{dv}{v} = 0}$$</span></p>
| 38698 | How to Derive the Differential Form of the Contintuity Equation |
2020-11-18T07:16:03.027 | <p>I've been spending an hour finding references for my research, but unfortunately I can't see useful information. So, I would really appreciate it if you could give me some references about the disadvantages of cylindrical residential buildings. Thank you!</p>
| |structural-engineering|architecture| | <p>I am currently working on a project, its a cylindrical and its more like a barrel shape.</p>
<p>It is a landmark project, future use for the city council for the Miri in Sarawak State, Malaysia.</p>
<p>As the main contractor of this project, I would share about several disadvantages of cylindrical shaped building.</p>
<ol>
<li><p>The Reinforced Concrete (RC) structure is time consuming to build, one thing is that the inner and outer beams are all curved: resulting in more time being consumed to cut and bend the reinforcement bar to the desired curvature. And this is the most crucial factor to the project schedule, because practical experience is required to estimate the time required for the tedious preparation work for curved beams, and may cause delay if the work schedule is not handled right.</p>
</li>
<li><p>Even worse, the radius of each floors are different, there is no viable option to utilize the formwork for these curve beams!</p>
</li>
<li><p>As the control points for setting out for each floor must include the center of the building, the drywall setting out needed to be done as soon as possible before any other work that may disturb the central point, some drywall having the layout out sitting along the radius is difficult to determine as some may affected by the setting out window-wall intersection the outer circle. And this may cause a lot of variation in the architectural finishes.</p>
</li>
<li><p>While the floor to floor radius varies and the building is round, the installation of temporary structure (scaffolds) for external architectural finishes is a challenge that needed to be assessed by a trained scaffolder, considering the safety for the workers who will be going to access the temporary structure and the integrity of the structure.</p>
</li>
</ol>
<p>One must account the practical disadvantages, even during design stage, that may arises for barrel/cylindrical shape building for better management of the construction and deliver a better quality of work to the client.</p>
| 38702 | What are the disadvantages of cylindrical residential buildings? |
2020-11-18T14:14:42.713 | <p>Firstly, pardon my ignorance about this topic because I am an electronics engineer.</p>
<p>I designed a 3 mm thick mounting bracket for one of my products. For this, I need to cut a galvanized metal sheet and bend it 90 degrees.</p>
<p>Maybe I am being over cautious, but (per my reasoning) since the galvanized coating is only in the outer section of sheet metal, the 3 mm wide edges will be uncoated, therefore susceptible to corrosion.</p>
<p>Also the 90 degree bent part will be stretched, so coating thickness will be reduced.</p>
<p>Are these effects real? Do they make any practical difference?</p>
| |metal-folding| | <p>Zinc protects by sealing the steel, but also protects it by acting as a sacrificial anode if that seal is breached.</p>
<p>So yes, exposed edges does make the steel more susceptible to corrosion, but not altogether unprotected as long as zinc remains to act as an anode. However, once the zinc is consumed then that protection no longer exists.</p>
| 38708 | Will laser cutting render my galvanized sheet metal susceptible to corrosion? |
2020-11-19T15:09:31.353 | <p><strong>Background:</strong>
The CS bolts are available with Black Oxide coating. I have read that the coating is applied using hot dipping and cold dipping methods. Also that hot dipped coating is more durable than cold dipped one.</p>
<p><strong>Problem:</strong>
How to check if a given bolt is hot dipped or cold dipped. I am looking for a crude method that does not involve any lab test, but for on spot check.</p>
| |materials|product-testing| | <p>I imagine you could just cut it in half and look at the boundary under a microscope or loupe. Compare against a known sample that is hot-dipped and one that was cold-dipped. I imagine it should be fairly obvious.</p>
<p>Another idea is to get some hardness testers and test them on some known samples.</p>
<p>If you can't find known samples (i.e. you can find bolts labelled as cold-dipped and hot-dipped black oxide), then then get some bluing compound and apply yourself to some clean steel and compare it against a a good black-oxide drill (which should be hot-dipped). I would hope the black-oxide coating on a drill would be even better than one on a hot-dipped bolt so keep that in mind.</p>
<p>That said, do any manufacturers actually use cold black-oxide on bolts? Because for a mass-produced item where you already have high tooling and setup costs, going with cold-dipped black oxide seems like a weird cost cutting measure that doesn't seem like it would actually cut costs by much. I would be really disappointed to find anyone using cold-dipping on something small and mass-produced like bolts.</p>
| 38724 | Identification of bolt material |
2020-11-20T16:51:28.643 | <p>When looking at glazings within various public databases I see the fields "tir" shown often. E.g as seen below:</p>
<pre><code>"glazing_data": {
...
"tir_front": 0.0,
"tir_back": null,
...
}
</code></pre>
<p>What does this field stand for?</p>
| |glass|solar| | <p>TIR: Thermal infrared (longwave) transmittance of the glazing layer.</p>
<p>Lawrence Berkeley National Laboratory has a windows program called THERM 6.3 / WINDOW 6.3
NFRC Simulation.</p>
<p>In pp 55 of this manual, there is a table referencing some properties of the glazing, including this.</p>
<p><a href="https://i.stack.imgur.com/9xN1L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9xN1L.png" alt="pp 55" /></a></p>
<p>here is the <a href="https://cdn.ymaws.com/www.nfrccommunity.org/resource/resmgr/technical_documents/nfrcsim6.3-2011-12-manual.pdf" rel="nofollow noreferrer">Therm 6.3 prog dowmload.</a></p>
| 38731 | What does TIR stand for in Window Pane Fenestration? |
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