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2018-08-09T22:03:35.710 | <p>I have a peltier (TEC) device capable of 400W Qmax, a 160W thermal load that I'd like to keep at ambient, and <a href="https://customthermoelectric.com/water-block-3-0-x-3-0-x-0-85.html" rel="nofollow noreferrer">this water block</a> running at about 1.5 gpm w/ room temp water. I think this system is right on the border of what is possible, but I'm struggling to determine the exact efficiency of my peltier unit. </p>
<p>The datasheet for <a href="https://customthermoelectric.com/19911-5m31-28cz-thermoelectric-peltier-module.html" rel="nofollow noreferrer">my peltier</a> indicates a Qmax of 400W and claims that maximum applied power is about 690W -- does this mean that I have a 57% efficiency? To me that seems extraordinarily high for a TEC. </p>
<p>Because I don't believe 57% is right, let's assume a 30% efficiency. In that case, my peltier will need to be operated at around 533W in order to dissipate the 160W load + the load from the peltier. And now, I have 693W total being dumped into my water block. </p>
<p>Am I doing this calculation correctly? </p>
<p>Other notable factors:
- This system will be operated in a vacuum environment so convective cooling is a no-go.
- There is a thin (~1/8") aluminum plate between the heat load and the peltier.</p>
| |thermodynamics| | <p>Your water is at 20 C and the block is at -15 C. Temperature differential is 35 C. </p>
<p>That's a pretty big gradient for the TEC. The plot of Q vs I says that you can shift no more than about 125 W but you'll need 28 A and that will cost you about 675 W. (this is from the plots on the data sheet)</p>
<p>If the source of heat is 25 W or so it will require 10 A and cost you about 100 W (10 V * 10 A). This is where I would use the TEC you specified.</p>
<p>This seems to indicate that you'll need either:</p>
<ol>
<li>To settle for a lower amount of heat shifted <em>or</em></li>
<li>More TECs in series to reduce the temperature gradient of each one <em>or</em></li>
<li>More TECs in parallel to increase the heat pumped <em>or</em></li>
<li>Settle for a smaller temperature differential</li>
</ol>
<p>or a combination of the above.</p>
<p>PS: You'll need to do option 3 when you apply option 2 because now you need to get rid of the energy used to pump the heat from the colder stage.</p>
| 23164 | Peltier Calculation Conundrum |
2018-08-10T22:23:22.430 | <p>I have recently been working on a gearbox for a robot. I have this pattern of 4 gears with each gear connected to its 2 neighbors. It looks like this:</p>
<p><a href="https://i.stack.imgur.com/ry2HY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ry2HY.png" alt="Early Drawings"></a>
You will probably notice something though. Large gear on the right and the small one in the center do not properly mesh. I made sure that the center to center distance of the all the gears was the sum of their respective pitch radii so that is not the issue:
<a href="https://i.stack.imgur.com/1iHns.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1iHns.png" alt="The issue"></a>
There only seems to be a few positions where the gears engage one another properly. By moving things around and eyeballing it I got a set of constraints where everything appears to mesh properly but it is unlikely that it is perfect:
<a href="https://i.stack.imgur.com/aeQy3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aeQy3.png" alt="eyeballed fix"></a>
<a href="https://i.stack.imgur.com/lUMAz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lUMAz.png" alt="sketch"></a>
So my question is how do I calculate distance between the small gear in the center and the large one on the bottom so the gears will all mesh properly. I don't like eyeballing it because I don't think the gears will mesh perfectly.
Also, it makes it harder to improve my design if I can't make these calculations.</p>
<p>Thanks,<br>
Ben</p>
<p>Some specs on the gears:</p>
<p>Large Gears: 44 tooth, 2.2" pitch diameter, 2.3" outer diameter, 14.5 deg pressure angle, 20 Diametrical Pitch.<br>
Small Gears: 12 tooth, 0.6" pitch diameter, 0.7" outer diameter, 14.5 deg pressure angle, 20 Diametrical Pitch</p>
<p>A similar question that got no answer: <a href="https://engineering.stackexchange.com/questions/2637/how-to-find-angle-of-timing-gears?rq=1">How to find angle of timing gears</a></p>
| |mechanical-engineering|gears|robotics|mechanical|kinematics| | <p>Let’s consider gears as being represented by a pitch circle on which lie several equally spaced dots with a number equal to the number of teeth. These dots can either represent the teeth or the gaps between teeth, and they follow the rule that, as a pair of meshing gears rotate, the dots of one gear will coincide with the dots of the other gear as the dots pass the point of tangency of the pitch circles:</p>
<p><a href="https://i.stack.imgur.com/Klw0A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Klw0A.jpg" alt="enter image description here"></a></p>
<p>The distance between neighbouring points measured along the pitch circle is the circular pitch <span class="math-container">$p$</span> for both gears.</p>
<p>If we know the location of a dot on one gear, we can infer the location of a corresponding point on a meshing gear:</p>
<p><a href="https://i.stack.imgur.com/tnGhu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tnGhu.jpg" alt="enter image description here"></a></p>
<p>That is, if a dot on gear 1 is an arclength <span class="math-container">$s$</span> away from the pitch circle tangency point, there will exist a corresponding dot on gear 2 that is also an arclength <span class="math-container">$s$</span> away. We know this to be true since rotating the gears result in moving the dots by the same distance along the pitch circles; corresponding dots will coincide at the tangency point, and a rotation of the gears will move both dots by <span class="math-container">$s$</span>. Care is to be taken in the directions of rotation.</p>
<p>We can use this information to consider tooth spacing for a closed loop of four arbitrary gears:</p>
<p><a href="https://i.stack.imgur.com/QeXWO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QeXWO.jpg" alt="enter image description here"></a></p>
<p>Here, for <span class="math-container">$i=1, 2, 3, 4$</span>, each gear has <span class="math-container">$N_i$</span> teeth, and the centres of the pitch circles connect to form a closed polygon with internal angles <span class="math-container">$\theta_i$</span>.</p>
<p>Let’s consider a dot on gear 1 at the point of tangency of pitch circles 1 and 2. A corresponding dot will occur for gear 2 coincidently. From the above information, we can use the dot on gear 2 to infer a corresponding dot on gear 3, and then use the dot on gear 3 to infer a corresponding dot on gear 4, and finally use gear 4’s dot to obtain another dot on gear 1. This should appear as follows, with arclengths of interest indicated:</p>
<p><a href="https://i.stack.imgur.com/mm1t9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mm1t9.jpg" alt="enter image description here"></a></p>
<p>Note how the consideration of a single dot on gear 1 led to the discovery of a second dot on the same gear. If these are to be two dots on the same gear, they must be separated by an integer number of tooth spacings <span class="math-container">$p$</span>, i.e. they must be separated by <span class="math-container">$zp$</span>, where <span class="math-container">$z$</span> is an integer.</p>
<p>Note how the arclengths are related to the radii of the pitch circles <span class="math-container">$r_i$</span> and the internal angles <span class="math-container">$\theta_i$</span>. This yields the equations:</p>
<p><span class="math-container">$$s_A = r_1 \theta_1$$</span></p>
<p><span class="math-container">$$s_A + s_B = r_2 \theta_2$$</span></p>
<p><span class="math-container">$$s_B + s_C = r_3 \theta_3$$</span></p>
<p><span class="math-container">$$s_C + zp = r_4 \theta_4$$</span></p>
<p>Elimination of the unknown arclengths leads to the following condition:</p>
<p><span class="math-container">$$r_4 \theta_4 - r_3 \theta_3 + r_2 \theta_2 - r_1 \theta_1 = zp$$</span></p>
<p>By noting that the radius of each gear <span class="math-container">$r_i = N_i p/2\pi$</span>, this condition simplifies to</p>
<p><span class="math-container">$$N_4 \theta_4 - N_3 \theta_3 + N_2 \theta_2 - N_1 \theta_1 = 2\pi z$$</span></p>
<p>Therefore, for gears of given numbers of teeth, you need to alter the interior angles of the closed polygon such that the above condition holds for some integer <span class="math-container">$z$</span>, which may be positive, negative or zero.</p>
<p>For your particular case where <span class="math-container">$N_1 = 12$</span>, <span class="math-container">$N_2 = N_3 = N_4 = 44$</span>, this becomes</p>
<p><span class="math-container">$$44( \theta_4 - \theta_3 + \theta_2) - 12 \theta_1 = 2\pi z$$</span></p>
<p>Find a geometrically valid set of internal angles that obey this rule, and the teeth should all mesh. Once you know the internal angles, you can determine the distance between the top and bottom gear.</p>
<p>This rule can also be generalised to any closed loop of an even number of gears, where the rule can be expressed as</p>
<p><span class="math-container">$$\sum_{i=\textrm{even}} N_i \theta_i - \sum_{i=\textrm{odd}} N_i \theta_i = 2\pi z$$</span></p>
| 23174 | 4 Gears in a Circular Pattern - Calculating the Proper Distances for Meshing |
2018-08-11T00:41:09.713 | <p>From wikipedia:</p>
<blockquote>
<p>Before opening either door, the air pressure of the airlock—the space between the doors—is equalized with that of the environment beyond the next door to open</p>
</blockquote>
<p>Anyone knows how this work? What the mechanism make possible "equalize" the air pressure between the doors? For example, in the ISS (International Space Station).</p>
| |pressure|air| | <p>each "end" of the airlock has a valve in it which, when opened, allows the pressure between the inside of the airlock chamber and the pressure inside the adjacent chamber to equalize. Once this is done, the equalizing valve is shut and then the door to the adjacent chamber can safely be opened. </p>
| 23182 | How an airlock works? |
2018-08-11T15:37:40.647 | <p>Below the handrail of a staircase is some kind of gridwork to prevent people or objects from falling off the staircase into the space below. What is the term for this gridwork? - perhaps "handrail grating"?</p>
| |terminology| | <p>The first and last are called newel posts and the intermediate ones are called spindles.</p>
<p>But whether these names apply to materials other than wood is a different question.</p>
| 23186 | What is the protective part of a staircase handrail called? |
2018-08-11T15:40:39.023 | <p>I currently have <a href="https://www.omega.com/pressure/pdf/LC703.pdf" rel="nofollow noreferrer">this</a> load cell, which has a 10V excitation voltage. However, there is a very nice/convenient load cell amplifier that works with an excitation voltage of 5 VDC.</p>
<p>What are the consequences of operating a cell at a voltage lower than the normal excitation? Just increased noise in the signal?</p>
| |signal-processing|instrumentation| | <p>The load cell you linked to has a line item on the specs that says "Output: 2 mV/V nominal". Let's just say that you picked the version with 100 lbf full scale output. What that means is this: if you use a 1 V excitation, you'll get a response something like this</p>
<pre><code>Load Output
0 lbf 0 mV
50 lbf 1 mV
100 lbf 2 mV
</code></pre>
<p>Now if you use a 10 V excitation, you'll get this</p>
<pre><code>Load Output
0 lbf 0 mV
50 lbf 10 mV
100 lbf 20 mV
</code></pre>
<p>So if you want to use a 5V excitation, it will work perfectly fine. Just do the math on the expected output. </p>
<p>Regarding your comment on noise. The signal to noise ratio will be likely be better with a higher excitation voltage. So yes, if you use 5 V instead of 10V, you probably have a slightly worse signal to noise ratio. Depends on your application and the precision required as to whether that will be acceptable. </p>
| 23187 | Operating a load cell below excitation voltage |
2018-08-12T04:06:59.500 | <p>As far as I know a treatment plant's job is to treat black water to become effluent. Part of the process is removing dangerous microbes, grease and other environment damaging substances. However, is it possible that the odor remains in the effluent? Or that safe effluent can still contain odorous compounds unless scrubbed?</p>
<p>NOTE: I live few kms away from one that discharges into a river that has moderate odor but at that distance away from residences, I guess it doesn't matter.</p>
| |waste-water-treatment| | <p>Where I live (Germany), WWTP are required to clean the water to a degree so it does not harm the receiving water. This is far from drinking water quality. In your area, restrictions may be more lax.</p>
<p>In my experience, WWTP stink, I know one you can smell kilometers away! Especially the larger ones, even when most tanks are covered and there's an odor control. The stinkiest parts are usually near the inflow, rakes etc. Sewage sludge pre-fouling is also very stinky. Secondary clarifiers should not be smelly.</p>
<p>So it may be hard to tell apart if the whole WWTP is stinking or only the effluent, if the <em>effluent</em> stinks this could prompt further questions:</p>
<ul>
<li>What are the (legal) requirements for WWTPs in your locale?</li>
<li>How is the ecological health of the receiving water body?</li>
<li>Does the effluent smell on all days, or only after strong rain events? Or only ater longer dry spells?</li>
</ul>
| 23195 | Is the sewer outfall discharge supposed to have foul odor? |
2018-08-12T14:24:16.143 | <p>I am using Bernoulli's equation in order to calculate a relative area between two points in a pipe. Problem is, based on the solution I have obtained, I end up with imaginary numbers and not sure what to make of this practically. Here is my math:</p>
<p>$$\begin{gather}
P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2 \\
v_1 A_1 = v_2 A_2 \\
\dfrac{A_1}{A_2} = \sqrt{\dfrac{v_2^2}{v_2^2+\dfrac{2}{\rho}\left(P_2-P_1\right)}}
\end{gather}$$</p>
<p>I know my velocity at location 2, density, and pressures. I have no change in height so I got rid of third element in Bernoulli's equation. I am trying to obtain an area ratio between both points. </p>
<p>It is very possible in my model that I will have a low P2 and high P1 leading to a negative denominator and an imaginary solution. I feel like I am missing something practical or mathematical here that is simple but my mind has drawn a blank. </p>
<p>As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow. </p>
| |pressure|airflow| | <blockquote>
<p>As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow.</p>
</blockquote>
<p>The assumptions that $\dot{V}_1 = v_1 A_1 = v_2 A_2 = \dot{V}_2$ and that $\rho =$ constant are only valid for incompressible flow. Air is compressible, not incompressible. Use the ideal gas law with $\rho = M_{air}(p/RT)$ and $v = \dot{V}/A = \dot{n}RT/pA$. This gives</p>
<p>$$\dot{n}_1 = \dot{n}_2 = \dot{n}$$
$$\frac{RT_1}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_1}{p_1 A_1}\right)^2 = \frac{RT_2}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_2}{p_2 A_2}\right)^2$$</p>
<p>The proper analysis for a compressible (ideal) gas must consider pressure and <em>temperature</em> not pressure and <em>velocity</em>. Corrections for non-ideality may be applied depending on how drastically the gas is compressed and/or how accurate the answer really must be.</p>
<p>The ratio of areas as a function of velocities for ideal gas flow is expressed as</p>
<p>$$ \frac{A_1}{A_2} = \frac{\dot{V}_1 v_2}{\dot{V}_2 v_1} = \frac{T_1 v_2 p_2}{T_2 v_1 p_1}$$</p>
| 23204 | Area ratio using Bernoulli's equation |
2018-08-13T10:26:32.957 | <p>I have a beam, simply supported with 2 points loads and 2 UDLs. The loads are symmetric about the centre of the beam.</p>
<p>I've tried to derive an expression for the moment along the beam and then via 2 integrations obtained expressions for slope and deflection respectively.</p>
<p>Could someone please let me know whether I'm close with what I have or help me to create an overall moment equation, which i can plot in MathCAD preferably, and also help me with finding the maximum deflection of the beam? Attached is the FBD for the problem.</p>
<p><a href="https://i.stack.imgur.com/xBXVm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xBXVm.jpg" alt="enter image description here"></a></p>
<p>Below is my attempt at a solution for deflection (In the case where x = L).</p>
<p><a href="https://i.stack.imgur.com/zVRH3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zVRH3.jpg" alt="enter image description here"></a></p>
<p>Any help is much appreciated.</p>
| |structural-engineering|statics|beam|deflection| | <p>Note: I'll be using the following notation:</p>
<ul>
<li>$q_1, q_2$ are the distributed loads (where you use $w$, which I reserve for deflections)</li>
<li>$L = c + 2(a+b)$, the total length of the span</li>
<li>$\ell = a + b$</li>
</ul>
<p>I'll also be making use of the structure and loading's symmetry, and adopting the following model:</p>
<p><a href="https://i.stack.imgur.com/0XLPm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0XLPm.png" alt="half of beam modelled by placing Rz support at midspan"></a></p>
<p>First let's get the reactions. Since the structure and the loading is symmetrical, this is trivial, with each support getting half of each load.</p>
<p>$$R = -\dfrac{q_1(L-2a)}{2} - \dfrac{q_2L}{2} - P$$</p>
<p>Let's then use singularity functions to describe the loading on the beam and integrate our way up to the deflection:</p>
<p>$$\begin{align}
\newcommand{\a}[1]{\langle #1 \rangle}
q &= R\a{x}^{-1} + q_1\a{x-a}^0 + q_2\a{x}^0 + P\a{x-\ell}^{-1} \\
Q &= R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\
M &= R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\
EI\theta &= \dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\
EIw &= \dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x + C_2 \\
\end{align}$$</p>
<p>Now let's use the boundary conditions to find those constants.</p>
<p>$$\begin{align}
\theta\left(\dfrac{L}{2}\right) = EI\theta ={}& \dfrac{R}{2}\left(\dfrac{L}{2}\right)^2 + \dfrac{q_1}{6}\left(\dfrac{L}{2}-a\right)^3 + \dfrac{q_2}{6}\left(\dfrac{L}{2}\right)^3 + \dfrac{P}{2}\left(\dfrac{L}{2}-\ell\right)^2 + C_1 = 0 \\
\therefore C_1 ={}& q_1\left(\dfrac{(L-2a)}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}-a\right)^3\right) \\
&+ q_2\left(\dfrac{L}{4}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{6}\left(\dfrac{L}{2}\right)^3\right) \\
&+ P\left(\dfrac{1}{2}\left(\dfrac{L}{2}\right)^2 - \dfrac{1}{2}\left(\dfrac{L}{2}-\ell\right)^2\right) \\
={}& q_1\left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\
&+ \dfrac{q_2L^3}{24} \\
&+ \dfrac{P\ell(L-\ell)}{2} \\
w(0) = C_2 ={}& 0
\end{align}$$</p>
<p>We therefore have the full description of the entire beam. Personally, I prefer splitting the functions up for each beam segment. Therefore, we have:</p>
<p>For $x\in[0,a]$:</p>
<p>$$\begin{align}
Q ={}& R\a{x}^0 + q_2\a{x}^1 \\
={}& \dfrac{q_1(2a - L)}{2} \\
&+ q_2\left(x - \dfrac{L}{2}\right) \\
&- P \\
M ={}& R\a{x}^1 + \dfrac{q_2\a{x}^2}{2} \\
={}&\dfrac{q_1(2a - L)}{2}x \\
&+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\
&- Px \\
EI\theta ={}& \dfrac{R\a{x}^2}{2} + \dfrac{q_2\a{x}^3}{6} + C_1 \\
={}&q_1\left(\dfrac{2a - L}{4}x^2 + \dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\
&+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\
&+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\
EIw ={}& \dfrac{R\a{x}^3}{6} + \dfrac{q_2\a{x}^4}{24} + C_1x \\
={}&q_1\left(\dfrac{2a - L}{12}x^3 + \left(\dfrac{a^3}{6} - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right)x\right) \\
&+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\
&+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right)
\end{align}$$</p>
<p>For $x\in(a,\ell]$:</p>
<p>$$\begin{align}
Q ={}& R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 \\
={}&q_1\left(x - \dfrac{L}{2}\right) \\
&+ q_2\left(x - \dfrac{L}{2}\right) \\
&- P \\
M ={}&R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} \\
={}&q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\
&+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\
&- Px \\
EI\theta ={}&\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + C_1 \\
={}&q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\
&+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{L^3}{24}\right) \\
&+ P\left(-\dfrac{1}{2}x^2 + \dfrac{\ell(L-\ell)}{2}\right) \\
EIw ={}&\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + C_1x \\
={}&q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\
&+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\
&+ P\left(-\dfrac{1}{6}x^3 + \dfrac{\ell}{2}(L-\ell)x\right)
\end{align}$$</p>
<p>And finally, for $x\in(\ell,\frac{L}{2}]$:</p>
<p>$$\begin{align}
Q ={}& R\a{x}^0 + q_1\a{x-a}^1 + q_2\a{x}^1 + P\a{x-\ell}^0 \\
={}&q_1\left(x - \dfrac{L}{2}\right) \\
&+ q_2\left(x - \dfrac{L}{2}\right) \\
&+ 0 \\
M ={}&R\a{x}^1 + \dfrac{q_1\a{x-a}^2}{2} + \dfrac{q_2\a{x}^2}{2} + P\a{x-\ell}^1 \\
={}&q_1\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x + \dfrac{a^2}{2}\right) \\
&+ q_2\left(\dfrac{1}{2}x^2 - \dfrac{L}{2}x\right) \\
&-P\ell \\
EI\theta ={}&\dfrac{R\a{x}^2}{2} + \dfrac{q_1\a{x-a}^3}{6} + \dfrac{q_2\a{x}^3}{6} + \dfrac{P\a{x-\ell}^2}{2} + C_1 \\
={}&q_1\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{a^2}{2}x - \dfrac{a^2L}{4} + \dfrac{L^3}{24}\right) \\
&+ q_2\left(\dfrac{1}{6}x^3 - \dfrac{L}{4}x^2 + \dfrac{q_2L^3}{24}\right) \\
&+ P\left(-\ell x + \dfrac{L\ell}{2}\right) \\
EIw ={}&\dfrac{R\a{x}^3}{6} + \dfrac{q_1\a{x-a}^4}{24} + \dfrac{q_2\a{x}^4}{24} + \dfrac{P\a{x-\ell}^3}{6} + C_1x \\
={}&q_1\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{a^2}{4}x^2 + \left(\dfrac{L^3}{24} - \dfrac{a^2L}{4}\right)x + \dfrac{a^4}{24}\right) \\
&+ q_2\left(\dfrac{1}{24}x^4 - \dfrac{L}{12}x^3 + \dfrac{L^3}{24}x\right) \\
&+ P\left(-\dfrac{\ell}{2}x^2 + \dfrac{L\ell}{2}x - \dfrac{\ell^3}{6}\right)
\end{align}$$</p>
| 23220 | Bending equation and max deflection of UDL and multiple point loads |
2018-08-13T13:14:53.690 | <p>It's just a matter of curiosity, but I find hard to justify why every single watch keep the hour hand below the minute one; from a user perspective, it seems a poorly design decision for those times when they overlap the hands.</p>
<p>Some manufacturers – notably Rolex with its Mercedes-like symbol – even added larger stuffs to the hour hand to be seen when such occurrences happen.</p>
<p>I suppose it may be done to simplify the internal gears, but I also suppose that by placing the hour-minute-seconds hands in a reverse order shouldn't be a problem…</p>
<p>IMO, a more reasonable user experience should show from top-to-bottom: seconds, hour, [GMT], minute hands. Isn't this achievable ?</p>
<p>Can someone satisfy my curiosity ? :)</p>
<p>Thanks in advance!</p>
| |mechanical-engineering|design|gears|mechanisms| | <p>At current time - probably only tradition. </p>
<p>The origin would be historical, when original single-hand clocks gained the minute hand. There would be existing, prior clockwork near the (back side of the) face of the clock, just where it was originally, and then new clockwork would be built, to propel the new hand, placed further back from the face.</p>
<p>And if you propel an axis-inside-axis transmission, the inner axis must extend further from both ends of the outer (tubular) axis and can be propelled on the extended part. So you have the tube reaching to the hour work, and the inner axis reaching into the new gear on top. And so, the hands on the opposite side follow, hour hand attached to the shorter tube axis, minute hand on top, on the longer inner axis.</p>
<p>Since there was never really any significant reason to challenge this, and the clockmaking industry is quite conservative and traditional, it wasn't changed. A minor argument might be the amount of friction (and so, frequency of need to wind the clock/watch) - the faster hand only moves against one surface, the slower is working between two - but that's a really minor issue.</p>
<p>Sure it could be done. It would most likely involve mirroring the clockwork against the plane of the clock face (or just moving the face to the opposite side) and switching the roles of the two axis (plus obvious minor diameter adjustments. At current time though... an LCD screen displays both hands on the same plane.</p>
| 23222 | Why all watches hands are stacked in that specific order ? Can it be changed? |
2018-08-13T20:40:22.183 | <p>Is Solidworks Motion capable of finding the torque of the planet carrier (output) in a planetary gear set when in steady state? The sun and ring gears are both inputs and all gears are rotating at a constant angular velocity.</p>
| |automotive-engineering|simulation|solidworks| | <p>Yes, it is possible, if you define all physical properties (mass, moment of inertia ...) of the rotating object. You need to run the simulation first, notice put the motor at the component which you want to find the torque of.</p>
<p>1- Click on results.</p>
<p>2- Under property manger (under results) select 'forces'.</p>
<p>3- Select 'motor torque' under subcategory.</p>
<p>4- Select 'magnitude' of whatever component you are interested in.</p>
<p>5 click on the green check mark.</p>
<p>You'll see the plot in function of time.
This is also true for finding power, forces, acceleration ... . </p>
| 23228 | Solidworks Motion capabilities |
2018-08-13T22:18:21.543 | <p>I'm driving a small, benchtop vibe table (<a href="https://controlledvibration.com/product-item/12-inch-platform-shaker/#tab-id-3" rel="nofollow noreferrer">shaker</a>) with two Crown DSI-1000 PA's, and also cleaning my signal with a DBX 231s EQ. Currently running the system at 6.8 GRMS, but I'm looking to increase my overall driving level. </p>
<p>With the test article mass/ test levels I'm shooting for, I'm conflicting with the specified peak capability of my shaker.</p>
<p>To be more precise, I would like to drive a ~5.5 lb object to ~10 grms (random vibe). According to my back-of-the-envelope calcs (Eqn 1), I don't <em>think</em> there is much I can do besides purchasing a larger shaker. This shaker tops out at 200 lbf. However I figured it would be worthwhile to probe the community for ideas in case there is some sort of <em>engineering voodoo</em> I can employ. Will also post in EE stack exchange.</p>
<p><strong><em>TL:DR, Is there any way for me to increase the peak driving force/ capabiltiy of my shaker table without the risk of damaging it?</em></strong> </p>
<p>My PA's and EQ are also getting close to their maximums, at just 6.8 grms. </p>
<p>Desired profile: NASA GEVs STD-7000a: <a href="https://i.stack.imgur.com/ss5E2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ss5E2.jpg" alt="enter image description here"></a></p>
<p>Eqn 1: <em>GRMS = [PEAK_SHAKER_FORCE / (ARTICLE_MASS + ARMATURE_MASS)] / 2</em> . --> GRMS = (200 lbs/ 13.5 lbs) / 2 ...................................................................... --> GRMS_MAX = 7.4</p>
| |aerospace-engineering|vibration|power| | <p>No, there is not really much you can do. If you want to get 10 Grms, you'll need a bigger shaker. The only trick you could play would be to decrease the bandwidth of your test spectrum. i.e. instead of testing 20 - 2000 Hz, just test 20 - 1000 Hz. The g^2/Hz stays the same, but the Hz range is smaller, so now you don't have to drive 10 Grms anymore, but a smaller number. You could perhaps split your testing into multiple segments. Segment 1 test 20 - 1000 Hz, and then segment 2 test 1000 - 2000. Depending on the intent of the test, this may or may not be acceptable. </p>
| 23231 | Vibe Table: Increase Peak Force Capability? |
2018-08-15T21:21:18.190 | <p>For <a href="https://en.wikipedia.org/wiki/Pressure_vessel#Stress_in_thin-walled_pressure_vessels" rel="nofollow noreferrer">pressure vessels in the shape of circular cylinders</a>, we can use $\sigma_{hoop}=\frac{pr}{t}$ to find the minimum skin thickness by setting the hoop stress the maximum allowed value, and then solving for t. But, what about (the more complicated) <a href="https://barbaramatthews.files.wordpress.com/2012/08/ellipsetank.jpg" rel="nofollow noreferrer">elliptical cylinders</a>?</p>
<p>Since ellipses have a <a href="https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes" rel="nofollow noreferrer">semi-major axis and a semi-minor axis</a> (as opposed to circles with just a single radius), it stands to reason that an elliptically cylindrical pressure vessel would have two minimum thickness (instead of just one). Is this correct? If so, how does one go about finding them?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|pressure-vessel| | <p>You can use a finite element code to simulate the stress distribution in the wall.
Simulating elastic problems under pressure is straight forward to model and you get the advantage of getting the stress distributions in 2 or 3 dimensions in contrast with the other methods mentioned.</p>
| 23257 | Minimum thicknesses of an elliptical cylinder pressure vessel |
2018-08-16T15:47:35.127 | <p>$\newcommand{\a}[1]{\langle#1\rangle}$
As you can see, the left end of the beam is fixed, and a distributed load is applied to the beam. I calculated the shear forces by conventional means, but when I calculate it with singularity function I get weird results. Here are my calculations:</p>
<p><a href="https://i.stack.imgur.com/SiTTe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SiTTe.png" alt="enter image description here"></a></p>
<p>$$q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - \frac{2000}{3}\a{x-0}^1 + \frac{2000}{3}\a{x-0}^0$$</p>
<p>The first term comes from reaction torque and the second term comes from reaction normal force. The shear force is just the integral of the $q(x)$.</p>
<p>$$v(x) = 3000\a{x-0}^{-1} - 3000\a{x-0}^{0} + \frac{2000}{6}\a{x-0}^2 - \frac{2000}{3}\a{x-0}^1$$</p>
<p>As you can see the results are false. I'll appreciate any help. </p>
| |structural-engineering|beam| | <p><span class="math-container">$\newcommand{\a}[1]{\langle#1\rangle}$</span>
<strong>Loading</strong></p>
<p>Take a look at the last two parts of your loading equation.</p>
<p><span class="math-container">$$\frac{2000}{3}\a{x-0}^0 - \frac{2000}{3}\a{x-0}^1$$</span></p>
<p>Would look something like this:</p>
<p><a href="https://i.stack.imgur.com/XzppF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XzppF.png" alt="enter image description here"></a></p>
<p>If we were to superimpose the two graphs they would not look like the triangular loading you have in your picture. If we wanted that loading, we would use:</p>
<p><span class="math-container">$$-2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1$$</span></p>
<p>This would make sure that we had -2000kN/m at x=0m, and 0kN/m at x=3m.</p>
<p><strong>Integration</strong></p>
<p>From</p>
<p><span class="math-container">$$
\int \langle x-a\rangle ^{n}dx={\begin{cases}\langle x-a\rangle ^{{n+1}},&n\leq 0\\{\frac {\langle x-a\rangle ^{{n+1}}}{n+1}},&n\geq 0\end{cases}}
$$</span></p>
<p><span class="math-container">$$
q(x)=-3000\a{x-0}^{-2} + 3000\a{x-0}^{-1} - 2000\a{x-0}^0 + \frac{2000}{3}\a{x-0}^1
$$</span></p>
<p><span class="math-container">$$
v(x) = - 3000\a{x-0}^{-1} + 3000\a{x-0}^{0} - 2000\a{x-0}^1 + \frac{1000}{3}\a{x-0}^2
$$</span></p>
<p>Since this is continuous over the entire length, we can simplify to</p>
<p><span class="math-container">$$
v(x) = 3000 - 2000x + \frac{1000}{3}x^2
$$</span></p>
<p>Which is exactly what we would expect, since there is 3000kN at x=0m, and 0kN at x=3m.</p>
<p>And we can also plot in <a href="http://www.wolframalpha.com/input/?i=y%20%3D%203000-2000x%2B(1000%2F3)x%5E2%20from%20x%20%3D%200%20to%20x%20%3D%203" rel="nofollow noreferrer">wolfram alpha</a>.</p>
| 23271 | Shear forces along a beam using singularity functions |
2018-08-16T18:17:32.757 | <p>I am looking to classify ECG signal using with neural network. I have never done signal analysis before.</p>
<p>From the samples of data I have, I have noramlised, transformed using the FFT , get the pics, amplitudes and the frequencies.
Now, How do I feed those values, to my ANN (Artificial Neural Network)? This looks like a tuple(pair of value) for each row.</p>
<p>Sorry. I have no much experience in engineering but I am recently challenge myself on area that is not particularly mine.</p>
| |electrical-engineering|matlab|signal-processing| | <p>Since you haven't mentioned exactly what you want to classify or how you're building your network I'll give a more general explanation of how you prepare data for a classification task and the general shape of the network.</p>
<p>The important part of any classification problem using a neural network is to figure out what your classes are going to be. Sometimes you know the classes in advance (i.e. you can label your data), sometimes you need to identify the classes after collecting your data using a clustering algorithm or some other prcess.</p>
<p>Since you're dealing with ECG FFT data (I assume you're referring to an electrocardiogram?) some examples of classes you might have would be specific people (i.e. heartbeat info for person A, person B, and person C) and you want your ANN to classify which person it is based on heartbeat. Another case could be a binary classification between a "healthy" heartbeat and an "unhealthy" heartbeat. For these sorts of examples you would probably be able to label the data in advance (ex. you know that dataset 17 was measured from participant #27, who was healthy).</p>
<p>Once you have labelled data you can think about how to use it in your network. Let's use the example of having three people who you want to identify based on heartbeat (person A, person B, and person C). You have $n$ frequency data per test, and $N$ total tests (with multiple tests per person I hope, otherwise you won't be able to do anything meaningful). We need to express the classification as some vector, so with three classes you should have a three-element vector ([1 0 0] means person A, [0 1 0] means person B, and [0 0 1] means person C). </p>
<p>Now you know your input (a vector containing $n$ magnitudes corresponding to a set of frequencies) and your output (a three-element vector indicating the class). Therefore you need a network with $n$ input nodes (one for each data point in a given test set) and three output nodes (one for each class). When you input a given frequency set to your network you'll get a 3-element output, for example [1 0 1]. You compare that to your label (let's say this set was taken from person A, so your label should be [1 0 0]) and take the difference between the two as your error, which is fed back to the network to help it learn. This process is repeated many times until the network gets good at classifying the data.</p>
<p>Most packages handle the minutia of the learning process for you, so all you need to do is specify the size of your network, give it an input set and the corresponding output set (labels).</p>
| 23274 | classify ECG signal using neural network |
2018-08-17T05:42:37.287 | <p>Near where I live there's a construction site for a new supermarket. The last week or so they did things I don't quite understand.</p>
<p>Some background: </p>
<ul>
<li>The ground in our area (Ruhrgebiet) is rather clayey</li>
<li>The site is near a small hilltop, the area generally drains into the Ruhr</li>
<li>The site was excavated maybe 2m deep (below ground level) and filled in with 1.5m of sand gravel mix.</li>
</ul>
<p>Now the part that I don't understand:</p>
<ul>
<li>An excavator with a drill drilled through the sand-gravel and the soil beneath, maybe 3m total depth (below ground level), I guess 0.5 m diameter</li>
<li>The holes are set in rows, maybe 2m apart</li>
<li>A huge machine that on first glance looks a bit like the ones they set sheet pile walls with drives up to each hole, inserts a pipe and (appears to) fill the hole with gravel - not the sand gravel mix, but fairly large, roundish individual stones with no small infill material</li>
<li>I see no way that a binder or water could be added to the gravel - there's just a funnel for the gravel and a loooong pipe that is entered into the hole. I don't think it's for a pile foundation.</li>
</ul>
<p>Picture of the machine (not from "my" site, but <a href="https://www.bauforum24.biz/forums/topic/15880-keller-raupe-110/" rel="nofollow noreferrer">from here</a> - sadly no explanation what the ultimate purpose is):</p>
<p><a href="https://i.stack.imgur.com/CmJrQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CmJrQ.jpg" alt="enter image description here"></a><br>
The funnel can be hoisted up near the end of the pipe, or lowered to be filled by front loader.</p>
<p>My theory is that this is for surface water drainage - water will percolate through the sand-gravel mix and then into lower ground layers though the highly porous gravel. There are a few other sites close by dedicated to infiltrate storm water into the ground, so maybe this is a thing in my locale. The image source has someone describing the machine purpose built for "Kieseleinbausäulen" (Gravel infill piles) but I don't find that term anywhere else. </p>
<p>Ultimately I don't know (and the only way to really find out is to ask one of the guys on site, but so far they've been to far away from the fence when I came by), my question is: To what end does one drill holes and fill them with gravel? Is my hunch correct, are there other explanations?</p>
| |civil-engineering| | <h1>You might be right</h1>
<p>Your first thought about using the holes for drainage could be right. This would be similar to a <a href="https://en.wikipedia.org/wiki/French_drain" rel="noreferrer">French Drain</a> or a <a href="https://en.wikipedia.org/wiki/Dry_well" rel="noreferrer">Dry Well</a>.</p>
<p>French drains are usually horizontal and dry wells aren't usually installed on hills.</p>
<h1>The likely option</h1>
<p>What is more likely is that the contractor is installing a form of compacted aggregate pier. See this for a description from a company that does them <a href="https://www.haywardbaker.com/solutions/techniques/vibro-aggregate-piers" rel="noreferrer">(vibro aggregate pier)</a>.</p>
<p>The main idea is that a hole is drilled and then filled with compacted aggregate. This provides a firmer foundation without treating the entire area or installing steel or concrete piles. The aggregate piers are sometimes used instead of micropiles, but they aren't as good of replacement for large concrete piers or driven steel piles.</p>
| 23286 | Drilling holes in ground and filling with gravel - to what ends? |
2018-08-17T09:24:38.197 | <p>If a rod is turned around it's own axis, force/torque is needed. </p>
<p>Let's assume the rod it self has a diameter $d$, a length $l$, the mass $m$ and should be turned by a motor on one side. The other side is laying on a support, that prevents it from moving sideways which is not frictionless $\mu_{rod-supp}$. </p>
<p><img src="https://i.stack.imgur.com/I74m7.jpg" alt="Sketch"></p>
<p>How strong must this motor be?!</p>
<p>So far I tried around a bit with Moment of Inertia. But I don't know how to go from there.</p>
<p>It is a real life problem I need to solve. The rod needs to turn once in 3-8 minutes and I have to find a fitting stepper motor for that, that's why I need the torque.</p>
| |mechanical-engineering|applied-mechanics| | <p>If you assume that 100% of the weight is carried by the support, then the normal force is just $mg$, so the friction force is $mg\mu$, and the torque required to overcome friction is $mg\mu d/2$. If you want to be able to acceleration the rod quickly, then the polar moment of inertia will come into play. However, since you indicated a rate of 1 rev every few minutes, the accelerations involved here will be trivial. The friction force will dominate. Take the calculation for friction and add a little margin and call it a day. </p>
| 23292 | Calculate the torque needed to turn a rod it's own axis |
2018-08-18T15:53:11.623 | <p>I read that it "is a positive pressure
against the environment". But I am not too sure what that means.</p>
| |thermodynamics|refrigeration| | <p>This isn't always true. Industrial low pressure ammonia vapor refrigeration was usually operated below external pressure so you didn't poison people or contaminate stuff in the event of a leak. It was relatively easy to remove atmospheric leakage into the systems, and it didn't pose any great threat to the system. With modern equipment, you don't want anything getting into the system. The systems are designed to be hermetically sealed and they don't suffer contamination well. Nobody wants to purge leaked atmosphere out of there refer system once a day like was common practice in the past. These old systems basically used to require an operator to run them, like old boilers and everything else back then. Old icemakers used to be a problem because many operated near atmospheric pressure on the low side, and if there was a high side leak, the low side would suck air and you lost your cool. This is why they all had little bubble windows in them.</p>
<p>See table 2.9 in the following FAO report on seawater ice making for fisheries.
<a href="http://www.fao.org/docrep/006/Y5013E/y5013e05.htm" rel="nofollow noreferrer">http://www.fao.org/docrep/006/Y5013E/y5013e05.htm</a> </p>
<p>And as a practical matter, detecting a leak is easier if the low side is below atmospheric, but locating the leak is easier with positive pressure, and except for residential comfort air, you won't often be shutting a system down to go on a leak hunt.</p>
| 23315 | Why is the saturation pressure of the refrigerant desired to be slightly greater than atmospheric pressure in an evaporator? |
2018-08-18T22:43:54.013 | <p>I found a technical drawing of a popular guitar shape online and I'm having trouble to understand a certain annotation on some arcs in the drawing. It basically says stuff like "0.43R 2PL" or "3PL" (see attached).</p>
<p>any idea what this could mean? My current guess is that it means that there are X PL other curves in the loop that have the same radius.</p>
<p><a href="https://i.stack.imgur.com/3uquT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3uquT.jpg" alt="enter image description here"></a></p>
<p><a href="http://www.tdpri.com/threads/d-size-tele-body-blueprint-files-here.74504/page-30#post-6860876" rel="nofollow noreferrer">Drawing Source</a></p>
| |technical-drawing| | <p>Your guess is correct. "0.430R 2PL" means there are two places where there is a curve with the given radius. Usually it is obvious from looking at the complete drawing where all the curves are - for example in the OP's complete drawing, <a href="http://www.tdpri.com/attachments/tele_body_drawing_reve-1-pdf.347484/" rel="nofollow noreferrer">http://www.tdpri.com/attachments/tele_body_drawing_reve-1-pdf.347484/</a>, the part being referred to is symmetrical (though it is oriented at an angle to the symmetry axes of the complete object).</p>
<p>This is more commonly used for things like small holes, where you might see something like "0.1R 36PL" rather than each one of the 36 holes being dimensioned individually. There are some examples of this on the OP's complete drawing.</p>
| 23320 | 2PL / 3PL annotation on arc in technical drawing |
2018-08-19T14:22:30.980 | <p>I am trying to interpret the hole specifications in the attached diagram. What does each component mean? I understand what 'Thru All' means, does 4x mean 4 holes? What is the diameter of the hole? is it 5mm or M6 = 6mm? What is 6H?</p>
<p>Thanks in advance
<a href="https://i.stack.imgur.com/kRGPm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kRGPm.png" alt="enter image description here"></a></p>
| |mechanical-engineering|technical-drawing|drawings| | <p>4 x means it applies to the four instances that look similar on the drawing.</p>
<p>Ø 5 means the hole diameter (drill size) is 5mm.</p>
<p>“Thru all” is added here to remove the requirement for a section view to show hole depth. You can’t tell by the outlines whether it’s through, or you can see the bottom of the hole</p>
<p>M6x1.0 - 6H is the thread and tolerance to be tapped into the drilled hole - again, through all. With blind holes it’s often the case that a minimum thread depth is specified that’s less than the hole depth to avoid multiple tapping operations using a bottoming tap.</p>
<p>This drawing would likely be better using a centreline and symmetry to dimension the hole positions. Even if distances to the bottom right corner are critical, it still needs some sort of ‘construction lines’ to e.g. show that the top left hole is at the same vertical height as the top right. This is a fairly simple part, but in more complex geometry you can’t assume things like that unless they are explicitly shown. Adding centrelines would do this.</p>
| 23328 | Understanding hole specifications |
2018-08-19T17:05:52.503 | <p>I know that on the moon, while the daytime surface temperature can be above 270° F, in the shade but a few feet above the surface temperature can get quite low. Wondering whether a similar phenomenon can be observed if a vaccum is created around an object which is suspended in he middle of it.</p>
<p>Say for example, a steel ball is suspended in the middle of a plexiglass cube, by 3 plexiglass threads, and the interior volume of the cube is 1 cubic meter, and the ball has a diameter of 10cm, the box walls are 5cm thick. The cube has been evacuated to the practical limit of modern vacuum pumps. Let's say the steel ball is chilled to a temperature of -20 c (measured at the time the evacuation is complete). Let's say it's 70° F outside, and the box is sitting on a plexiglass tripod in the shade, shielded from direct sunlight. </p>
<p>What would be the temperature reading of the steel ball, approximately, after say 2 weeks? (i.e. at equilibrium).</p>
<p>(I don't mean to be overly prescriptive, but trying to be sufficiently specific to avoid answers like "it depends on how thick the walls are" and similar)</p>
| |thermodynamics|thermal-conduction|thermal-insulation|thermal-radiation| | <p>Imagine the system below at equilibrium.</p>
<p><a href="https://i.stack.imgur.com/ERaVc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ERaVc.png" alt="ball in box"></a></p>
<p>The energy balances are as follows:</p>
<p>At the ball ...</p>
<p>$$\epsilon_b T_b^4 =\epsilon_w T_w^4$$</p>
<p>At the wall ...</p>
<p>$$\epsilon_w A_w \sigma T_w^4 = h_a A_w (T_a - T_w) + \epsilon_a A_w \sigma T_a^4$$</p>
<p>In these balances, $\epsilon_j$ are emissivities, $A_j$ are areas, $\sigma$ is the Stefan-Boltzmann constant, $T_j$ are temperatures, and $h_a$ is the convection coefficient of air.</p>
<p>Choose all materials and sizes to specify all $A_j$ and $\epsilon_j$ values. Choose a still or windy day to specify $h_a$. Define the air temperature. This leaves two unknowns ($T_b$ and $T_w$) with two equations. The problem can be solved.</p>
<h2>Example</h2>
<p>Take the case where the ball and walls are the same material so their emissivities are the same. We find that $T_b = T_w$. The radiation heat flows inside the box are balanced. Now assume the air has no radiation and the walls are perfect black bodies ($\epsilon_w = 1$) to obtain $\sigma T_w^4 = h_a (T_a - T_w)$. Take stagnate air with $h_a = 5$ W/m$^2$ K and $T_a = 275$ K. The wall and ball are at 238 K. In this case, the air is pumping heat to the walls by convection, and the (black body) walls are radiating back to the air. As the walls (and ball) go toward gray bodies (as $\epsilon_w \rightarrow 0$), the wall temperature increases. In the limit where the walls emit no radiation, we end with the simple case that $T_b = T_w = T_a$. No heat flows from anything at equilibrium.</p>
| 23331 | How cold for lack of heat? |
2018-08-20T06:32:54.580 | <p>In Oldham coupling,is it possible that the same angular velocity is transferred from one shaft to the other when the centre disc is not sliding?If so,why do we want the centre disc to slide?I know that the shafts are misaligned but it is occurring to me that the same motion is possible when the centre disc is held back from sliding?</p>
| |mechanical-engineering|mechanisms| | <p>If you plot the center of the third disc in one revolution you'll see it moves in a circle with diameter $δk$ with $δk$
the radial deviation of the two shafts. So if you fix the disc, it can't compensate the radial deviation at all. If i remember the frequency of the disc is about twice as it frequency of the shafts.<a href="https://i.stack.imgur.com/M2J9A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M2J9A.png" alt="enter image description here"></a> </p>
<p><a href="https://i.stack.imgur.com/fAAOP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fAAOP.png" alt="enter image description here"></a></p>
<p>The Oldham coupling is a kind of cross disc coupling (German). The second figure shows the general principal behind this soort of couplings.</p>
| 23336 | Transfer of angular velocity in Oldham's coupling with the centre disc not sliding |
2018-08-20T13:26:07.347 | <p>I know how to translate torques and vertical forces into singularity functions, but may i also express the horizontal forces with singular function? if yes then what would be that expression? <a href="https://i.stack.imgur.com/K6Z4o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K6Z4o.png" alt="enter image description here"></a> </p>
<p>Here the green beam is fixed at point A, and a distributed load acts on the beam, and a horizontal force acts at the upper edge of the beam.</p>
| |structural-engineering|beam| | <p>Yes, it is possible to describe horizontal loading with singularity functions.</p>
<p>However, in order to do so you must change the loading equation from scalar to vector-form:</p>
<p>$$\{q\} = \left\{\matrix{q_x \\ q_y}\right\}$$</p>
<p>So you'll handle the horizontal and vertical load components separately.</p>
<p>Obviously, how you then interpret the singularity functions changes, since beams behave differently under transversal and axial loads (the fourth integral of transversal loading is the deflection, which isn't true for axial loading).</p>
<p>Given this, I don't know how useful singularity functions are when dealing with horizontal loading. However, your question is whether it is possible to express the horizontal loading with singularity equations and the answer to that question is "yes".</p>
| 23341 | Is is possible to use singularity function to expresse the horizontal forces ? |
2018-08-20T14:11:47.627 | <p>Is the length of a steel dowel pin standardized to a certain tolerance? McMaster-Carr provides tolerances on the diameter of dowel pins but not the length. In the event where I want a dowel pin to fit into two blind locating holes, I'd like to know if I need to leave a little extra "wiggle room" in the hole depth.</p>
<p>For example, I have two parts that need to meet as close together as possible. One part has a 5 mm deep locating hole for a dowel pin. If I have a 10 mm dowel, can I specify a 5 mm deep locating hole on the second part (assuming the dowel is actually <10 mm long)? Or should I choose something deeper (5.5 or 6 mm perhaps)?</p>
| |mechanical-engineering|design|machining| | <p>Usually the nominal length of the dowel is the total overall including the actual ground, pin section + the rounded end and + beveled end. Usually lengths are not toleranced so you should have a deeper hole then the nominal length as they run slightly larger. </p>
<p>ALSO....If you are using dowel pin to locate two blind holes, make sure to use a pull dowel and grind a "vent" on the side on the side of the pin so that it can be removed. Better if one of those holes doesn't have to be blind so the pin can be pushed out. A regular dowel pin, pressed into a blind hole is a real chore to try and remove. </p>
| 23345 | Tolerance on dowel pin length |
2018-08-20T14:20:58.463 | <p>I am wondering why turbines are not implemented with controllable pitch blades (on the rotor system)? I assume the reason is that the coupling between the stator and rotor systems posed by typical controllable pitch mechanisms poses an unacceptable frictional loss for such a high speed/ low torque application.</p>
| |fluid-mechanics|turbines| | <p><a href="https://www.asme.org/engineering-topics/articles/turbines/turbine-fuel-efficiency-fitting-a-pitch" rel="nofollow noreferrer">Turbine Fuel Efficiency: Fitting a Pitch</a></p>
<p>Here is an excellent answer to my question.</p>
| 23346 | Why aren't turbos/turbines controllable pitch? |
2018-08-21T05:12:31.460 | <p>In books of dynamics, I normally see topics in kinematics and then in kinetics, and finally a last chapter on mechanical vibrations dealing with SDOF and MDOF systems. But I do not see any book mentioning how the topic of vibration fits in dyynamics? It feels natural to say to mechanical vibrations would be an extension of kinetics.</p>
<p>What is the correct classification of mechanical vibrations?</p>
| |dynamics|vibration| | <p>"Kinetics" is just a synonym for "dynamics," and most English books use the term "dynamics" rather than "kinetics." There are plenty of mechanical engineering textbooks with "dynamics" in the title, but few with "kinetics" (I'm ignoring books on topics like "kinetics of chemical reactions", etc, which are not about <em>mechanical</em> engineering)</p>
<p>The difference between "kinetics" and "kinematics" is that kinematics deals with the motion of object <em>without considering what causes that motion</em> (for example the mathematical relationships between position, velocity, and acceleration of a particle moving along a given curved path in space), but kinetics/dynamics <em>does</em> consider the forces causing the motion (i.e. Newton's laws of motion).</p>
<p>Since a fundamental topic in studying mechanical vibrations is the fact that for linear elastic materials which obey Hooke's law, Newton's laws mean that the motion for small displacements about an equilibrium position <em>is</em> simple harmonic motion, I would classify most of the subject of "mechanical vibrations" to be part of dynamics/kinematics, rather than kinetics.</p>
| 23364 | Mechanical vibrations: kinetics of kinematics? |
2018-08-21T09:47:38.257 | <p>I am an economist, so please keep replies simple ;)
I am searching for the relationship between heat rate and capacity for gas-fired power plants. Here is what I refer to as heat rate, i.e. the inverse of efficiency: <a href="http://en.wikipedia.org/wiki/Heat_rate_(efficiency)" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Heat_rate_(efficiency)</a></p>
<p>From an IEA publication I got the following approximate relationship (for a CCGT plant):</p>
<pre><code>maximum heat rate = 1.1 * minimum heat rate
</code></pre>
<p>Maximum heat rate holds whenever the power plant is operated at minimum capacity, and</p>
<pre><code>minimum capacity = 0.4*maximum capacity
</code></pre>
<p>I'd like to know what the heat rate is when I run the power plant at a medium capacity of let's say:</p>
<pre><code>medium capacity = minimum capacity + (maximum capacity-minimum capacity)/2
</code></pre>
<p>So</p>
<pre><code>medium heat rate = ?
</code></pre>
<p>Is the relationship linear (i.e. = minimum hr+ (maximum hr-minimum hr)/2) or how would you estimate the heat rate here?</p>
| |power|gas|power-engineering| | <p>I will take a stab at an answer in what I hope might be simple and general terms.</p>
<p>What is defined as heat rate is akin to a <em>relative inefficiency</em> or <em>relative loss</em> rather than <em>heat rate</em>. The value itself has no units. Multiple by 100 and you get a percentage.</p>
<p>You first statement says, the greatest relative loss we expect in a plant is 10% over its minimum. The minimum relative loss occurs when the plant is operating at it greatest efficiency. So, a plant that is 85% efficient in the best case has a heat rate (relative loss) of $1/0.85 = 1.2$ or 120% in the best case. We expect this plant to have a heat rate (relative loss) of $1.1 * 1.2 = 1.3$ or 130% at its worst case operation. </p>
<p>Your next statements say, the worst case (130% heat rate) happens when the plant operates at its lowest capacity. Capacity is akin to the actual amount of electricity that is produced. So, a plant will be least efficient (have the highest heat rate) when it produces the lowest amount of actual electrical energy.</p>
<p>Your next statement says, the lowest capacity of a plant is 40% of its maximum capacity. Basically, this says, we do not operate power plants to produce less than 40% of their maximum rated output of electricity. Let's suppose the maximum capacity of the example power plant from above is $C_{max} = 100$ (with whatever units). The lowest capacity is $C_{min} = 0.4C_{max} = 40$. At 40, the plant operates with a heat rate of 130%. Now for the tricky part ... <em>We apparently have no information about the relationship between increasing capacity and decreasing heat rate</em>. We do not know whether the minimum heat rate of 120% is at maximum capacity ($C = 100$) or perhaps even before. All we know is, as we increase capacity, we can expect that heat rate decreases.</p>
<p>The finding leaves us with this conclusion. Absent any further information, we are free to assume any model we want for a relationship between increasing capacity and decreasing heat rate.</p>
<p>In real life, power plants are not operated continuously at their physical maximum capacity. They are operated below this physical maximum limit, by example continuing from above perhaps at $C_{operation} = 80 - 90$. In real life, inefficiencies also kick in above this operational design "set point".</p>
<p>My instinct to tackle this absent any other information would be to do the following:</p>
<ul>
<li>Set the maximum heat rate at the lowest capacity. This is defined by reference.</li>
<li>Set the minimum heat rate somewhere at 80% - 90% of maximum capacity. This <em>can</em> be defined by further research (i.e. look up typical operating capacities of power plants relative to their rated maximum capacity).</li>
<li>Define the relationship between increasing capacity and decreasing heat rate as linear over the above range. This is <em>your</em> model. Publish it.</li>
<li>Do not model the behavior of systems below their minimum capacity.</li>
<li>Do not model the behavior of systems above the 80% - 90% their rated maximum capacity.</li>
</ul>
| 23369 | Relationship heat rate with capacity |
2018-08-21T17:36:42.193 | <p>I'm attempting to understand the difference between a pressure relief valve and a pressure regulating valve. From my research far, I've come to the following conclusion:</p>
<ul>
<li><p>Pressure relief valves are primarily used for limiting pressure build ups and are often safety related. Pressure relief valves vent the fluid when the pressure exceed some setpoint.</p></li>
<li><p>Pressure regulator valves on the other hand are used to regulate a pressure via step-down. These regulator are typically non-venting.</p></li>
</ul>
<p>So I'm somewhat confused as to whether or not I should be using a pressure relief valve or pressure regulator in my application. It seems like a pressure relief valve would do the same exact thing as a regulator except it would vent the gas. Do pressure regulator offer better regulation than a relief valve? </p>
| |pressure|pneumatic|compressed-gases| | <p>Sort of. </p>
<p>Certainly things like pressure cookers and turbochargers use blow off vales as a routine way to regulate pressure. </p>
<p>Generally this is to make sure that internal pressure doesn't exceed some design maximum rather than for fine control and it is a bit contextual as it tends to imply that whatever working fluid you are dealing with goes to 'waste'. Obviously in both of the above examples this doesn't really matter much. </p>
<p>In the case of a turbo charger both the fluid and the energy to compresses it are more or less 'free' once you have the physical plant in place so a blow off valve is a simple and reliable way to regulate pressure. </p>
<p>In contrast if you have something like industrial argon as a welding shielding gas you don't want to vent that to atmosphere if you don't need to so you would use a more sophisticated regulator to control the flow-rate indeed y ou mmight even use two, one for course pressure regulation and one for fine volume flow-rate control. </p>
<p>Equally in a more general case if you are using bottled gas the unregulated pressure is directly related to the mass of gas in the cylinder so are often forced to use a regulator unless you can accept a drop-off in available pressure with usage. </p>
<p>As an aside several Robot Wars Robots have used direct bottle pressure (CO2) ie just a solenoid valve and no regulator to fire weapons, flippers etc. This tends to give fairly fierce performance while the bottle is full and is simple to implement, especially by remote control but every time you use it the available pressure drops and so can be a bit inefficient. </p>
| 23378 | Is it standard practice to use a pressure relief valve as a pressure regulator? |
2018-08-22T08:36:53.953 | <p>I'm reading the derivation of the equation of the hanging catenary from Wikipedia:</p>
<p><a href="https://en.wikipedia.org/wiki/Catenary" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Catenary</a></p>
<p>In the same article the parabolic equation governing the shape of the main cable of a suspension bridge is discussed, and its equation is obtained as following:</p>
<p>$$y={\frac {w}{2T_{0}}}x^{2}+\beta$$</p>
<p>Here $w$ is the density of the deck and $T_{0}$ is the horizontal tension on the cable at lowest point.</p>
<p>Let's say I'm designing a suspension bridge with the length and density of the deck and its other features are known. I would be interested in knowing the equation of the main cable to calculate the length of the secondary cables (the vertical ones). <strong>But how do I get $T_{0}$?</strong></p>
| |structural-engineering|civil-engineering| | <p>Since $x=0$ marks the midpoint of the parabola, let $y(0)=0$ at this point to eliminate $\beta$. Then, we only need to know the sag $s$ of the cable (which is typically set during the design process). The sag is equivalent to the value of $y$ at one of the symmetric towers (where $x=L/2$, where $L$ is the span or horizontal distance between the towers). </p>
<p>Thus, if we know the sag $s$ and span $L$ (and the distributed weight $w$), then we can immediately obtain $T_0$ from $y=\frac{w}{2T_0}x^2$. </p>
<p>If we know the length $l$ of the cable but not the sag, then we can use the formula for the <a href="http://mathworld.wolfram.com/ParabolicSegment.html" rel="nofollow noreferrer">length of a parabola</a> $$l=\sqrt{\left(\frac{L}{2}\right)^2+4s^2}+\frac{L^2}{8s}\sinh^{-1}\left(\frac{4s}{L}\right)$$ to obtain the sag $s$. (We'll probably have to solve this equation numerically.)</p>
| 23395 | Equation of the parabolic suspension bridge cable when the deck mass is known |
2018-08-22T20:30:38.350 | <p>When widening a road, why does construction seem to inevitably include a "smoothing" out of the curves on the road, essentially making it more straight? </p>
<p>I've seen this personally on roads where I travel, and you can even catch a glimpse on Google satellite imagery of roads that are in the widening process: Google overlays a "ghost" lane to show where the new lanes of the road will be. It's faint, but you can see an example here on a road near Flowery Branch, GA:</p>
<p><a href="https://i.stack.imgur.com/xO3WZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xO3WZ.png" alt="enter image description here"></a></p>
<p>Is this the engineers' way of correcting previous mistakes, or a response to devleopment that's sprung up around the road since its initial construction, or safety reasons? Or something else entirely?</p>
| |highway-engineering| | <p>I am familiar with this effect; with careful observation you can count at least three rounds of road-straightening on highway 17 between the cities of San Jose and Santa Cruz in California.</p>
<p>The first road was built over 100 years ago and carried horse-drawn wagons. At the time, moving large volumes of earth and rock was more expensive than surfacing the road and so it followed sharp curves in the mountainsides instead of being extensively built on fill from nearby road cuts. </p>
<p>Then came automobiles and more and faster traffic, and when widening the original stage road the sharpest turns were rounded off to carry faster vehicles and cut-and-fill was used so the mountain contours didn't have to be closely followed. </p>
<p>At each step in this decades-long process, the cost of earthmoving fell and the cost of pavement rose, and so it became economical to use more and more cut-and-fill in conjunction with straighter and straighter road paths.</p>
<p>The stage road is now a 4-lane freeway which still has a few somewhat sharp turns in it but much of it is built by cut-and-fill and can be traversed at 50 to 60 MPH. </p>
| 23404 | Why are roads straightened when widened? |
2018-08-23T00:26:59.840 | <p>Reading an article about multistage rockets (not educated at all on the topic), and from the get go it seems to assume they are used/need to be used.</p>
| |aerospace-engineering|rocketry| | <p>Another reason is that each motor stage can be designed for different purposes to match the flights characteristics.</p>
<p>So, a relatively short high thrust burn to escape gravity</p>
<p>And a longer or continuous burn to accelerate over a course / trajectory</p>
<p>Can save weight and complexity so adjustable nozzles are not always needed.</p>
| 23407 | What is the purpose of building multistage rockets, rather then packing more fuel into a single stage? |
2018-08-23T01:11:52.797 | <p>If I have a square plate of height h supported along one side (an L-bracket, modeled as fixed support), and there is an evenly distributed force acting over the face of the plate. From load analysis I am able to determine the resulting force (acting in the centre of the plate) to be P, thus the distributed load over the plate is P/A = P/(L^2)</p>
<p>My question is if we are considering the bending of the plate, can we model this as a cantilever beam with UDL of P/L or will it be something else? I am leaning to the former, and my calculations with such an assumption seems to yeild a reasonable result, but I wanted to double check.</p>
<p>Cheers</p>
| |structural-analysis| | <p>A plate supported along one side, subject to a uniformly distributed load, will behave exactly the same as a cantilever beam. You are correct that the UDL will be P/L, as it will be P/A*w where w is the width.</p>
<p>With a non-uniform load, treating it as a beam would give you the <em>average</em> stress, but miss the fact that there will be regions of higher stress and regions of lower stress.</p>
| 23409 | How to model a UDL over a plate for bending? |
2018-08-23T02:09:27.597 | <p>Looking at the teeth of a chain saw, I believe the chain moves from <em>right to left</em> while cutting.</p>
<p>It has two wedges <strong>A</strong> and <strong>B</strong>. </p>
<p>I've never owned a chain saw, so I've been trying to visualize how these two wedges work. </p>
<p>And I feel stuck because the wedge <strong>A</strong> seems unnecessary. Also it doesn't look that sharp and it seems to block the sharper wedge <strong>B</strong>. I'm dying to know the purpose of wedge <strong>A</strong> here. Appreciate any help. Thanks!<br>
<a href="https://i.stack.imgur.com/cf4J2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cf4J2.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>The first issue: the most important cutting edge is actually on top of the link, a segment curved towards the middle:</p>
<p><a href="https://i.stack.imgur.com/pmk4t.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/pmk4t.jpg" alt="enter image description here"></a></p>
<p>The blade on the side is secondary, just for clearing any remains of whatever might have sprung out due to material flexibility.</p>
<p>These links alternate the sides, so that the overhang isn't excessively long but the cut provides a gap wide enough for the entire chain to fit. Otherwise you'd be stuck with a very shallow cut and unable to get anywhere deeper than what the blade sticks above the chain!</p>
<p>Next: Cutting wood with a chainsaw is subject to the same set of rules as any machining; lathe, CNC mill, drill on a boring machine - all the same rules:</p>
<p><a href="https://i.stack.imgur.com/5uIKj.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/5uIKj.gif" alt="enter image description here"></a></p>
<p>There's the blade angle, speeds&feeds, material properties - and there's the cutting depth, which is an important parameter that accounts for power, material durability, chip properties and so on.</p>
<p>And the blade can't be right above the chain, because there would be no room left for the chip! It must be offset from the chain (by the distancer "A") to fit all the material it removes before it exits the cut and can be ejected!</p>
| 23411 | Shape of chain saw teeth |
2018-08-23T15:38:59.700 | <p>When browsing through pneumatic parts, I notice that some vendors specify different components for pressure regulator and vacuum regulators. <a href="https://www.smcusa.com/products" rel="nofollow noreferrer">SMC for example does this</a>.</p>
<p>I'm curious why vendors would actually need to differentiate vacuum regulators from pressure regulators. It seems like a vacuum regulator is simply a pressure regulator whose inlet port is connected to something wished to be vacuumed, and the outlet port is connected to a vacuum supply.</p>
<p>Is there more going on here?</p>
| |compressed-air|vacuum|pneumatic| | <p>The difference is the side on which pressure is regulated, and the reaction curve.</p>
<p>Typical pressure regulator takes arbitrary input pressure (from compressor) and provides specified (lower) pressure on output, locking the valve as output pressure rises; not allowing it to drop below preset level. The vacuum regulator assures the input pressure is as specified, while accepting arbitrary output (vacuum pump), by closing as the input pressure drops to preset. In the regulator feedback loop, the "readout" is performed on different sides of the valve. Since the regulators are usually unidirectional devices, not allowing for backflow, you can't just flip one around, and expect it to work like the other - if e.g. you put a vacuum pump in place of the compressor connected to a tank through a standard pressure regulator - it won't work.</p>
<p>You can imagine a tank that needs to maintain a pressure in specific range; the fill input will have a pressure regulator that allows filling the valve to preset pressure. But if for other reasons (e.g. the tank being heated by sunlight) the pressure rises, the relief valve will disengage, releasing the excess build-up; a vacuum regulator works similar to the relief valve, except its set point is somewhere below 1 bar, while a typical relief (safety) valve will open when internal pressure rises considerably above atmospheric.</p>
| 23420 | What's the difference between a vacuum regulator and pressure regulator? |
2018-08-24T16:40:11.570 | <p>Gears are called positive gears and belt drives are called negative drives. Explain in detail.</p>
| |mechanical-engineering|automotive-engineering|gears| | <p>The negative drive is an informal expression to describe 'friction drive'. The belt drive is based on friction between the disc and the belt, however positive drive is based on movements without slipping. </p>
<p>Not all belt drive system are 'negative drive'. We also have a positive drive belt system, in this special case, the belt doesn't slip over the driving disc but it rolls over teeth. </p>
| 23437 | What are positive drive gears? |
2018-08-25T20:10:18.967 | <p>I have Sketch 1. It's on Plane 1. There's a different Plane 2, and I want to make Pketch 2 which is simply the entities in Sketch 1 copied to Sketch 2. I don't want to copy-paste, because I want to have their constrains preserved.</p>
<p>Doing "Convert Entities" doesn't help, because the shapes become different, since they are projected to the plane. </p>
| |solidworks| | <p>To copy a sketch you already must have a sketch so I created one.Now I am gonna close this sketch.
1. Now I am going to select the sketch on the feature tree at the left and I will press Right Ctrl+C, you need to hold Ctrl and then press C.
Now I have made a plane as a reference and moved it a little bit to make it more clear, you do not need to create one, you can copy sketch to whatever plane you want.
I am going to start a sketch on this plane, and then I will just press ight Ctrl+V, hen the sketch should appear.
As you can see the same sketch was copied to a different plane, it is an easy way to save your time, if you want the same sketch on parallel planes you can use Convert Entities.</p>
| 23455 | Solidworks: Make a copy of a sketch on a different plane |
2018-08-25T20:14:20.660 | <p>I am a freshman student and I am very interested in using solar energy to break water into hydrogen and oxygen. So, I want to know which areas are important in this regard..</p>
| |chemical-engineering|photovoltaics| | <p>At your level, start with the two semesters of general chemistry. You will need a solid basis in chemical dynamics, redox chemistry, and chemical thermodynamics. Next take physical chemistry, especially when it includes electrochemistry. This will expand your understanding of the three core topics. Finally, take a course that covers aspects of heterogeneous catalysis, for example as offered in chemical engineering departments. This will explain how to translate the core topics from bulk systems to surfaces. As you move to graduate school, you should take courses in catalysis, solid state chemistry/physics, and electrochemistry.</p>
<p>Undergrad curricula in chemistry, chemical engineering, or materials science (not materials engineering) should provide what you need. Graduate curricula would be in the same fields.</p>
<p>Do <strong>not</strong> neglect to take the associated laboratory courses as an undergraduate. You <em>must</em> learn how to do experiments in general chemistry and physical chemistry. You <em>must</em> learn how to keep certifiable notebooks, review literature, write well-structured technical reports, and prepare publishable-level documents on your research. These skills are built from the groundings in the undergraduate laboratories, they do not just magically appear because you show up one day fully enthused and excited to do your research work in a laboratory.</p>
| 23456 | what classes should I take so that I can perform research into finding catalysts to split water into oxygen and hydrogen? |
2018-08-25T21:13:53.227 | <p>I have the following differential equation:</p>
<p>$$\ddot{\psi_t} + \omega^2 \psi_t = \omega R_s i_f - \omega u_f + \dot{u}_t - R_s \dot{i}_t$$</p>
<p>where $w$ is the electrical speed of the motor.</p>
<p>I have considered R.H.S. as $\Delta\omega$ and would be feed-forwarded, so I obtain the following transfer function:</p>
<p>$$\frac{\psi_t}{\Delta\omega} = \frac{1}{s^2 + \omega^2}$$</p>
<p>Any tips for controlling the above undamped second order system? Can I use PI controller for this?</p>
| |electrical-engineering|control-engineering|control-theory| | <p>So let's just start with a pure proportional controller. What does success (i.e. error=0) look like for the controller? Basically its when $\psi=0$ (or equal to a setpoint, but let's just say 0 for simplicity). And a proportional controller can drive the system to $\psi=0$. When $\psi=0$, the control effort will be zero, but $\dot\psi$ will not, in general, equal zero. Let's just call it $\dot\psi=v$. So your system will not stop at $\psi=0$. It's going to overshoot. The controller will notice this, and correct, and drive the system back to $\psi=0$. But at this point, now $\dot\psi=-v$, so the system is going to undershoot. It will keep oscillating like this forever. Now, if you had a damped system, the damping would take out a little energy each time, so that you don't come back to $\dot\psi=v$, but maybe $0.9v$, and eventually the oscillations will die out and you reach steady state. But if the damping is zero, the oscillations never die, they will continue at the same amplitude forever. </p>
<p>Now if you have a PI controller, it's probably even worse. When the system hits $\psi=0$, you probably have some integrated error built up, so the control effort is not zero, but non-zero. So the overshoot is even higher than with a P controller.</p>
<p>One way to overcome this is to use a PD or PID controller. The D term essentially acts on $\dot\psi$. Now when $\psi=0$, the error term and thus control effort will be zero only if $\dot\psi=0$ too. </p>
<p>Probably a better way, however, is to use two nested controllers. One controller acts on $\psi$ and the other controller acts on $\dot\psi$. The inner controller compares $\psi$ to the setpoint and generates a desired $\dot\psi$ that will achieve that (e.g. if $\psi>0$ then the $\dot\psi$ setpoint is negative). The outer controller takes the setpoint for $\dot\psi$ as an input, and applies control effort (in your case $\Delta\omega$) in order to drive $\dot\psi$ to the desired value. Both controllers could be P or PI.</p>
| 23457 | Control of second order undamped system |
2018-08-27T10:38:57.380 | <p>Question is on the meaning of input variable TID within material model 163 of LS-Dyna 971.</p>
<p>Definition of TID variable.</p>
<p>TID: Table ID defining yield stress versus volumetric strain at different strain rates.</p>
<p>Remark of TID: Rate effects are accounted for by defining a table of curves using *DEFINE_TABLE. Each curve defines the yield stress versus volumetric strain for a different strain rate. The yield stress is obtained by interpolating between the two curves that bound the strain rate.</p>
<p>What I understand of the yield stress is a single point on a stress vs volumetric strain curve for each individual strain rate. So wouldn't that give me a yield stress vs volumetric strain point for each individual strain rate.</p>
<p>Is the yield stress referring to just the stress. Having trouble understanding the inputs, any clarification would be helpful.</p>
| |mechanical-engineering|materials|applied-mechanics|stresses|simulation| | <p>While the yield stress is a well defined point in materials such as steel or other metals, that is not the case when we talk about foams.
Here they are calling "yield stress" to the whole volumetric strain-stress curve.
So they are basically asking for a number of strain-stress curves at different strain rates.</p>
<p>Or you can use material 63 if strain rate effects are not important.</p>
| 23473 | LS DYNA Material Model 163, Modified Crushable Foam |
2018-08-28T14:36:55.343 | <p>The common way to model or estimate gas (e.g. CO2) desorption rate from a liquid solution is by using an equation similar to this one</p>
<p>$J_{CO_2} = -k_L a(H·C^G_{CO_2} - C^L_{CO_2})$</p>
<p>where $J_{CO_2}$ is the molar (or mass) flowrate, $k_L$ is the mass transfer coefficient, $a$ is the specific contact surface, $H$ is one of the many expressions used for Henry's constant, and the G and L scripts stand for gas and liquid phase respectively.</p>
<p>The problem with this approach is that all the correlations I know used to estimate $k_L$, $a$, or most commonly $k_La$ imply that there is a gas stream with a given gas velocity (e.g. aeration or sparging).
How could I estimate $k_La$ for a still solution, for instance a glass containing beer where the gas bubbles are stripped slowly? Furthermore, if I wanted to assess the effect of stirring the beer with a spoon, how could I estimate the effect? </p>
| |chemical-engineering|modeling|process-engineering|multiphase-flow|mass-transfer| | <p>In perfectly stagnate gas systems, transport of a component through a gas collapses to diffusion of the component through the gas. In this case, $k_L a \propto (D / \delta)^n$ where $D$ is diffusion coefficient, $\delta$ is a film thickness, and $n$ can be correlated or derived from first principles (typically $n \rightarrow 1$ in film theories). References can be taken from correlations of the Sherwood number to the Grashof and Schmidt numbers.</p>
<p>A perfectly stirred liquid says $C^L$ is uniform throughout. A perfectly stagnate liquid requires that transport through the liquid also be modeled by diffusion. The approaches for diffusion in gas and diffusion in liquid are similar in concept.</p>
<p>The transport models to review are film theory, penetration theory, and two-film theory.</p>
<p>When you really are considering flow of a gas dissolved in a liquid, solubility can be an additional concern. The process of gas bubbles forming from dissolved gas is nucleation + growth. The bubbles become buoyant. This motion is <em>not</em> modeled by diffusion, rather by a process that might be equivalent to free convection. The combined transport of gas bubbles from a dissolved gas in a stagnate liquid must be modeled to include nucleation + growth of bubbles + free convection of the bubbles. I would have to wonder whether the net result (nucleate + grow + rise out of liquid) tends to a homogenized bubble concentration in the container at some point. Considerations made in the absence of gravity will, as will stirring.</p>
<h2>References</h2>
<p>Here are some links that I found in a search "stagnate air, mass transfer rate" with or without "stirred liquid".</p>
<p><a href="https://doi.org/10.4319/lo.1995.40.2.0290" rel="nofollow noreferrer">Methane transfer across the water-air interface in stagnate wooded swamps ...</a></p>
<p><a href="https://pubs.acs.org/doi/pdf/10.1021/es60111a012" rel="nofollow noreferrer">Rates of evaporation of low-solubility contaminants from water bodies ...</a></p>
<p><a href="https://pubs.acs.org/doi/pdf/10.1021/es00057a018" rel="nofollow noreferrer">Seasonal variations in air-water exchange of PCBs in Lake Superior</a></p>
<p><a href="https://pubs.acs.org/doi/pdf/10.1021/ac960814r" rel="nofollow noreferrer">Mass transfer characteristics of solvent extraction ...</a></p>
<p><a href="https://pubs.acs.org/doi/pdf/10.1021/ie50498a055" rel="nofollow noreferrer">Significance of liquid-film coefficients in gas adsorption</a></p>
<p><a href="https://www.sciencedirect.com/science/article/abs/pii/S0255270103001004" rel="nofollow noreferrer">Gas-liquid mass transfer coefficient in stirred tanks ...</a></p>
<p><a href="https://www.sciencedirect.com/science/article/abs/pii/S0255270104000388" rel="nofollow noreferrer">Gas-liquid mass transfer coefficient in stirred tanks ... eddy structure ..</a></p>
| 23493 | Modelling gas desorption from a still solution (e.g. a beer) |
2018-08-29T01:19:30.683 | <p>I have a question about the use of a servo motor.</p>
<p>I need to lift a load of $130\ kg$ with two servo motors with $13\ Nm$.</p>
<p>The diameter of the shaft is $Ø40\ mm$, ie, $r=\ 20mm$.</p>
<p>$1\ N = 1\ Kg\ *\ 1\ m/s^2$</p>
<p>$1\ Nm = 1\ Kg\ *\ 1\ m/s^2*\ 1\ m$</p>
<p>Therefore:</p>
<p>$25.506\ Nm = 130\ Kg\ *\ 9.81\ m/s^2*\ 0.02\ m$</p>
<p>By my reasoning, two servo motors could raise this load if the point of application of the load is exactly the diameter of the shaft.</p>
<p>If I use a cable that every time its diameter increases inside a pulley, at some point I will not be able to lift that load anymore?</p>
| |dynamics| | <p>In addition to what Jonathan has said, there is several other practical things to consider:</p>
<ol>
<li>not only torque is limited, but power also; since power is torque times angular velocity (P=T*omega), your motor specification will limit the maximum speed you will be able to lift the load at</li>
<li>depending on your application, you will not necessarily have to apply all the weight force by your motors; in typical elevators there is a counterweight (usually the size of the average load) which balances the <strong>weight force</strong>, so the motor will only have to apply the <strong>inertial force</strong> (of load + counterweight, plus any deviations from the average weight)</li>
<li>so, if you are building a kind of elevator, you don't have to wind the cable around a pulley either; because the counterweight pulls at the cable, you only need to take care that there is enough traction at the cable sheave</li>
<li>if you really want to lift all the weight by the motors, you could also use a gear or a block and tackle mechanism, which can translate torque into a more manageable range (without resorting to ridiculously small shaft diameters)</li>
</ol>
<p>So your pulley diameter of 20mm is not set in stone after all. But remember: if you use a mechanical advantage of some sort (like in bullet 4), you will only change force/torque, whereas power stays the same (or almost, considering friction effects). So the power limitation of your motors is usually the more severe one, and this depends on what ultimate speed you want to move the load at.</p>
| 23499 | Power decreases as the diameter increases |
2018-08-29T15:02:50.963 | <p>I need to thermally insulate two metal (say aluminum) parts, the first of which is at 200-400 °C (473-673 K), and the second one is at room temperature. It should be possible to screw the parts together or something, to form one solid assembly. The typical dimensions are like 10-40 mm. The insulator should be able to be created in a DIY setting.</p>
<p>The insulation material has to be reasonably stiff/strong to enable a rigid compound construction, so all plastic foams etc. are out and, due to the high temperatures, almost all (easily available) solid plastics are also out.</p>
<p>Comparing thermal conductivities I have found that all mineral-based materials are reasonably insulating at ~1 W/(mK) when compared to metals (~10^2 W/(mK)).</p>
<p>What has come to my mind (in order of ideation):</p>
<ol>
<li>porcelain</li>
<li>glass</li>
<li>gypsum</li>
<li>fine-grained concrete</li>
</ol>
<p>Anything obvious I have missed?</p>
<p>Gypsum seems very practical in that it can be easily cast with very fine structures. However I am not so sure whether it resists the heat (crystal water? gets more brittle at higher temp?). Anyway it is certainly a bad material for threading a metal screw in.</p>
<p>Glass is stiff and temperature resistant, but it is also sensitive to high temperature gradients and breaks easily under mechanic stress.</p>
<p>Fine-grained concrete (don't know if this exists at all as a product) - probably similar to gypsum, but a little stronger. But the question remains, if it can resist repeated heating/cooling/dehydration.</p>
<p>Porcelain (or other sintered minerals): sounds perfect, but can I reasonably expect to create a shaped part with threads/holes etc. at home?</p>
<p>Are there any best industry practices for parts like that?</p>
| |thermal-insulation| | <p>The aluminum can barely support itself at 400 C."Lightweight castable" refractory ( can be handled like concrete) should do the job; but you would need to find a supplier .It is what a petrochemical or power-plant might use . Portland cement/concrete type material will not tolerate temperatures of 400 C. Johns-Manville and Harbison-Walker made product like this in the past.</p>
| 23504 | DIY thermally insulating construction material |
2018-08-30T16:39:58.293 | <p>In this solution <a href="https://www.drdo.gov.in/drdo/English/index.jsp?pg=helicoptor-fire-control-system.jsp" rel="nofollow noreferrer">Helicoptor Fire Control System</a></p>
<p>CMA is used. For motion analysis we use computer vision and image processing. But what is special about CMA.</p>
| |sensors|aerospace-engineering| | <p>From a quick skim of a few monographs and book excerpts (which, by the way, are easily found via web search -- something worth doing before posting a wide-open question), it appears that CMA is an active system - sonar or radar, for example, which provides a wealth of information on object range, size, and motion. Further, any active system can operate regardless of local illumination, as opposed to a vision system which shuts down after sunset (roughly speaking).</p>
| 23528 | Contact Motion Analysis (CMA) |
2018-08-30T22:42:52.153 | <p>I have a planar object whose center of mass is not at the geometric centroid, and I want to roughly estimate the magnitude of four hypothetical point masses that at the four corners that would result in the center of mass. In summary, I want to backsolve the center of mass equation with reasonable assumptions. I currently have 2 equations for the center of mass in the x and y direction, but I need two more equations to be able to solve for all four. </p>
| |statics| | <p>let's say the off-center of center of mass from geometrical centroid is dx and dy, and total mass is M.</p>
<p>We assume the masses from top left counting clockwise m1,m2,m3,m4,located with their CGs equally located on four corner of a vertical rectangular. First let's assume the object is symmetrical, with b as the base of the rectangle and a as its vertical long side.</p>
<p>so we have $$\bar{x} =\frac {m1\times a/2 + m2\times a/2 +m3\times a/2 +m4\times a/2}{M} $$ and </p>
<p>$$\bar {y}= \frac {m1\times b/2 + m2\times b/2 +m3\times b/2 +m4\times b/2}{M} $$</p>
<p>now we back calculate dx and dy into the above EQs to figure the difference in mass of individual parts.</p>
<p>We observe that if we add a very small differential amount dm1 to m1 the center of gravity will move up and left away from center of geometry by a factor of</p>
<p>$$ dx=\space \thicksim \frac{dm1\times a/2} {M+dm1}\space and\space \space dy =\space\thicksim \frac{ dm1\times b/2}{M+dm1} $$
This rough approximation should give us intuition to figure out the discrepancy in mass or if we have an indication about the geometry we can set up the differential discrepancy as Dx for that.</p>
| 23532 | Estimating planar mass distribution |
2018-08-31T19:38:05.407 | <p>I need to save some engineering files that need the thicknesses in the filename. Windows hates "/" or "\", so I can't use a simple fraction. Symbols like this...</p>
<p>⅓, ⅔, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅛, ⅜, ⅝, ⅞</p>
<p>...are nice, but don't cover the full gamitt of dimensions (ie. 1/16").</p>
<p>In addition, Windows hates <strong>"</strong> in the filename, so I assume <strong>in</strong> is used instead?</p>
| |measurements|standards|nomenclature| | <p>This generator can seemingly make any fraction you want in unicode:</p>
<p><a href="https://qaz.wtf/u/fraction.cgi" rel="nofollow noreferrer">https://qaz.wtf/u/fraction.cgi</a></p>
<p>I put ¹⁄₁₆ in my Windows filename just fine. [Ironically enough, before finding the converter, I came up with using Shayne Turner's left bracket (<code>1[16</code>) idea.]</p>
<p>As for indicating the inches measurement, I'm just using 2 single quotes instead of a double.</p>
| 23547 | What is the engineering standard for filenames with fractions and inches? |
2018-09-03T14:18:08.680 | <p>I'm designing a diagnostic system that simulates an environment and user input for a dummy vehicle. All possible critical situations are simulated in order to find out whether any errors are triggered, or if the software contains any faults or bugs.</p>
<p>It's a vehicle for state defense, it doesn't have OBD like normal cars, but a display with buttons with which you can seek and troubleshoot errors. I have to design a procedure for testing each new software release. I was wondering if any standard exists for testing/validating software. Just like driving cycles(NEDC, WLTP) exist for emission tests.</p>
<p>Do any standards(SAE, DIN, etc.) exist for this kind of thing?</p>
| |automotive-engineering|simulation|software|standards|car| | <p>Since your project is defense software related I suggest taking a look at <a href="https://en.wikipedia.org/wiki/MIL-STD-498" rel="nofollow noreferrer">MIL-STD-498</a>. The one that is close related to your question is <a href="https://web.archive.org/web/20110616201638/http://www.fisica.uniud.it/~cabras/swe/MIL-STD/498/STP-DID.PDF" rel="nofollow noreferrer">SOFTWARE TEST PLAN (STP)</a></p>
| 23582 | What standards exist regarding diagnostic systems in automotive engineering? |
2018-09-04T03:49:32.517 | <p>While I'm reading the paper, it said<br>
"The I control can makes the phase lag, so it can makes stability margin reduced."<br>
Words are not same as that, but the meaning is the same.</p>
<ol>
<li>How the I control makes the phase lag in the PID controller?<br>
If the plant's transfer function is $G_p (s)$, the open-loop transfer function $G(s)$ is<br>
$G(s) = \left(\frac{k_i}{s} + k_p + k_d s\right)G_p (s)$<br>
when we use the PID for the controller. After substitute $s=j\omega$,<br>
$G(j\omega) = \frac{\left(k_i-k_d\omega^2 \right)j+jk_p\omega}{\omega} G_P (s)$<br>
So, how can I find that the I controller makes phase lag? </li>
<li>How the phase lag makes stability margin reduced?<br>
In the paper, it wrote "stability margin". I think that's phase margin. Anyway, how can I get the relation between 'stability margin' and 'phase lag'? </li>
</ol>
<p>Thank you for advising!</p>
| |control-engineering|pid-control| | <p>In this answer stability is covered first, thereafter the effect of the PID-controller on the stability.</p>
<p>Consider the following simple control system
<img src="https://i.stack.imgur.com/OvINx.png" width="400" /></p>
<p>which has a closed loop transfer function of $$\Gamma(s)= \frac{C(s)G(s)}{1+C(s)G(s)}=\frac{L(s)}{1+L(s)}.$$
Here $L(s) = C(s)G(s)$ is called the loop.</p>
<p>The system is <strong>unstable</strong> if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|>1(=0$dB$)$.</p>
<p>Similarly, the system is <strong>stable</strong> if at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|<1(=0$dB$)$.</p>
<p>Furthermore, neutral (or marginal) stability is at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$, the magnitude $|L(j\omega)|=1(=0$dB$)$.</p>
<p>The stability margin is ''the space'' you have until the system becomes unstable (i.e. robustness).
Since stability is determined by using two measures: the gain and the phase, consequently there are also two stability margins (the gain margin and the phase margin).</p>
<p>The gain margin is ``the space'' you have at frequency $\omega$ where $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until $|L(j\omega)| = 0$.</p>
<p>The phase margin is ``the space'' you have at frequency $\omega$ where $|L(j\omega)| = 0$ until $\angle L(j\omega) = \arg\left(L(j\omega)\right)=-180^\circ$ until.</p>
<p>The gain, phase and corresponding margins can be computed by hand, however I would recommend using software (e.g. MATLAB), since they can be very extensive.
To find effect your controller $C(s)$, you can draw Bodediagram and see the effect of the various parameters (MATLAB command <code>bode()</code>):
<a href="https://i.stack.imgur.com/Fvt94.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fvt94.png" alt="Bodediagram PID-controlers"></a>
The margins of the complete system can also be visualized using MATLAB (command <code>margin()</code>), which should give more insight in the effect of the chosen controller and your stability margins.
<a href="https://i.stack.imgur.com/lxeX9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lxeX9.png" alt="Stability margins"></a></p>
| 23591 | The relationship between the I control and the stability margin |
2018-09-05T05:47:25.223 | <p>In the case of a simple cantilever beam having a uniform shape and being loaded at its end - the calculation of the deflection at any point along the beam is well known.
What happens if we split it along its longitudinal axis?</p>
<p><a href="https://i.stack.imgur.com/6Xiu2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Xiu2.jpg" alt="cantilever beam composed of separate planks - taken from http://www.bu.edu/moss/mechanics-of-materials-bending-shear-stress/ "></a></p>
<p>I realize we are eliminating the shear forces along that axis between the beam segments, and this allows each segment to slide relatively to its neighbors. However, I'm not sure how this phenomena affects the beam deflection. How can I calculate it?</p>
| |mechanical-engineering|structural-engineering|beam|deflection|shear| | <p>Although I agree with @kamran, I have another way of thinking about it</p>
<p>The deflection of the structure is</p>
<p><span class="math-container">$$\delta=\frac{P\cdot L}{3\cdot E\cdot I}$$</span></p>
<p>The only difference in this problem between then bonded and the decoupled is the I. Assuming b: breadth, and h: height</p>
<p>For the coupled: <span class="math-container">$I_{coupled}= \frac{b\cdot h^3}{12}$</span></p>
<p>For the decoupled, the things slightly more complex. The resulting <span class="math-container">$I_{decoupled}=3 I_{(single\ board)}$</span>. Each board has a <span class="math-container">$h_{single}=\frac{h}{3}$</span>. Therefore the total moment of inertia is (Exactly as @kamran):</p>
<p><span class="math-container">$$I_{decoupled}=3 \frac{b\cdot (\frac{h}{3})^3}{12} = \frac{1}{9} \frac{b\cdot h^3}{12} = \frac{1}{9} I_{coupled}$$</span>.</p>
<p>Therefore, the deflection of the coupled (bonded) compared to the decoupled (for the same load) is:</p>
<p><span class="math-container">$$\delta_{decoupled} = 9 \delta_{coupled}$$</span></p>
<p>An interesting <strong>historical</strong> fact about your problem is that the Viking shipbuilders new about that and used it in dragon ships for the bow.
I.e. it was difficult to find a properly shaped tree to shape the bow of the ship. So what the "decouple" the different layers and it was possible (and easier) to bend without breaking.</p>
<p>Makes me wonder what other craftsmen have intuitively found out, without the need of maths.</p>
| 23612 | Deflection of a cantilever beam composed of separate (not bonded) planks |
2018-09-05T07:03:06.910 | <p>Im talking about 10k 5k 2k 1k 800 400 grit papers, wax polishing bars and polish paste.</p>
<p>Do they have a universal color code which corresponds to grit? </p>
<p>Many times the letters are either not on the paper or hard to read.</p>
| |polishing| | <p>No, they do not. I've got red, yellow, green, and grey 800 grit in a drawer in my garage!</p>
<p>I also have various different grit values that are all grey.</p>
| 23613 | Do fine grit sand papers and wax polish bars and pastes have a universal color code? |
2018-09-05T07:48:49.757 | <p>I ordered this model:</p>
<p><a href="https://www.ebay.com/itm/coil-winding-mechanical-tensioner-MT-01-wire-diameter-0-04-0-12-tension-device/181965432380?hash=item2a5dfc263c:g:KFIAAOSw~4NZt4lY" rel="nofollow noreferrer">https://www.ebay.com/itm/coil-winding-mechanical-tensioner-MT-01-wire-diameter-0-04-0-12-tension-device/181965432380?hash=item2a5dfc263c:g:KFIAAOSw~4NZt4lY</a></p>
<p>I downloaded this diagram to get a better idea of the parts(This from a different model but they share a majority of components):
<a href="https://i.stack.imgur.com/Pxs2y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pxs2y.png" alt="enter image description here"></a></p>
<p>I'm just curious about the 1-10 knob on the back that puts a metal pad either closer or farther away from the device. I assume this varies the magnetic resistance</p>
<p>I'm also curious about the back-tension button. I see that when you tighten the screw the tension arm takes more pressure to pull down.</p>
<p>All-in-all, I just want to know how the knob on the back works in conjunction with the back-tension</p>
<p>Also, If you've used a Coil winding magnetic tensioner before, any info on its functions I would love to learn about.</p>
<p>I'm reaching out because there is basically nothing online about how these things work and all the labels on my tensioner are in Chinese so I'm clueless</p>
| |mechanical-engineering| | <p>I just bought the same tensioner. It took 20 mins (well, maybe longer!) to sus out but I reached the same conclusion. The back-tensioner is just an adjustable overtension alarm where the microswitch wires in to the coilwinder to stop it if the tension gets too high.
The wire feeds in from below, normally off the wire drum standing on its end and through an upturned funnel guiding the wire. The wire enters between a couple of felt pads to stabilise it from its rotation off the wire drum, then around the magnetic tension spool to a free-wheeling pulley and thence over the alarm arm pulley to the coilwinder.
The Magnetic tensioner works a bit like a car speedo, where a rotating magnetic drags a cup around against a spring the faster it goes. I guess by using a magnetic feild to add drag to the tensioning pulley, set by the big knob on the other side, it is a more consistant tension compared to using a slipping clutch based on friction materials. </p>
| 23614 | Coil winding magnetic tensioner: How does it Work? |
2018-09-05T19:04:26.607 | <p>Gapless railroads have their joints welded together to reduce noise. The lack of a gap between rail sections also reduces wear on both the track and the wheels. However, because the weld essentially produces one very long track, the individual sections have no room to expand as they get warm on hot days. What prevents the track from warping beyond usability? (Yes, I have heard of tracks warping beyond safety margins on extremely hot days, but I'm wondering about just normal expansion between normal weather extremes.)</p>
| |rail| | <p>To take it back to the basic engineering principle: a change in temperature of something changes its stress-free length. That is, the length it would be if there was no stress on it.</p>
<p>For gapless rails (I'd call them Continually Welded Rails or CWR in the UK), the position of the rail is restrained by the connection of the rail to the sleeper or the slab below it. So the length is unable to change.</p>
<p>If you change the stress-free length without changing the <em>actual</em> length, you create stress. If your rail has got colder, it wants to get shorter, and this causes tensile stress; if your rail gets hotter it wants to get longer, and this causes compressive stress.</p>
<p>Failure occurs when the rail supports can't provide enough restraint against this tensile or compressive stress. On curved track this will manifest itself as the curve radius decreasing under tensile stress / cold weather, or radius increasing under compressive stress / hot weather. For a straight section of track this usually only happens under compressive stress / hot weather, where the track will buckle, snaking left and right, which makes the path of the rail longer, hence relieving the compressive stress.</p>
<p>So, what prevents the track warping is friction between the track and its connection to the sleeper, and friction between the sleeper and the ballast. When too excessive a temperature change occurs, the failure point is normally the friction between the sleeper and the ballast.</p>
<hr>
<p>Note that as per <a href="https://engineering.stackexchange.com/a/23624/255">user190081</a> and <a href="https://engineering.stackexchange.com/a/23628/255">blacksmith37</a>'s answers, the stress-free length at installation may not be the distance between sleepers, as the rail may either be mechanically stretched or heated. This is as failure in hot weather is far more of a problem than failure in cold weather, so a distance between the sleepers equal to the stress-free length at a relatively high temperature makes failure less likely. There is obviously a limit to this though, as failure at cold temperature extremes is still physically possible.</p>
| 23623 | How do gapless rails deal with thermal expansion? |
2018-09-06T14:24:37.237 | <p>If someone were to invent a device that uses the hydrodynamic energy contained within the city water supply to generate enough electricity to reduce their monthly electric bill (however unlikely this would be), would it legal to do so?</p>
<p>In other words, are there any existing federal/state laws that specifically state that a home owner cannot use any device that generates electricity from the hydrodynamic energy contained within the city water supply?</p>
<p>I have an idea for such a device but I don't want to invest the time and money to construct it if say sometime in the future my local water company finds out about it and then sends one of their lawyers to my home informing me that I have to stop using it or face heavy fines.</p>
| |mechanical-engineering|electrical-engineering|hydraulics|electrical| | <p>Interesting problem.</p>
<p>Let’s see, if we have a cylinder 12 inches in diameter and 8 feet high that would be </p>
<pre><code>(pi*(0.5f)^2*8f)*7.48 g/ft3=47 gallons per operation.
</code></pre>
<p>At 3 GPM, that would be about 15 minutes to fill the cylinder all the way.</p>
<p>If it were used to lift a 6780 pound weight, 8 feet up, that would be 8*6780 foot pounds of work, in 15 minutes. That would be </p>
<pre><code>(8 feet*6780 pounds/(15 minutes*60 seconds/minute) = 60 foot pounds per second.
</code></pre>
<p>Converting that to watts, we get </p>
<pre><code>60 fps*1.36w/fps=81.7 watts in 15 minutes.
</code></pre>
<p>Converting that to watt hours, which is how the electric company bills, we get </p>
<pre><code>81.7W*0.25 hours=20.4 WH or 0.0204 KWH.
</code></pre>
<p>Where I live, we pay about $0.12 per KWH so you are generating </p>
<pre><code>(0.0204KWH*$0.12/KWH)=$0.00254 of electricity.
</code></pre>
<p>To do that, you need 47 gallons of water. In my area, we pay about $0.008 per gallon of water. That comes out to </p>
<pre><code>47g*$0.008/g=$0.38.
</code></pre>
<p>If you use this 100 times, you will use 38 dollars of water to generate 25 cents of electricity.</p>
<p>But what if your water is free? In that case, if you run this continuously, it will cycle 96 times in 24 hours (15 minutes per cycle*24 hours) and use 96*47, about 4500 gallons of water to produce about enough to power to run a few LED light bulbs. If you can get the equipment for free and nobody is going to notice if your water consumption goes up to like 135,000 gallons per month, go for it. In the meantime, I suggest that if you start using that much tap water, somebody is going to notice and if it was not illegal when you started, it would be after a short time.</p>
| 23640 | Is it illegal to use the hydrodynamic energy contained within the city water supply to generate electricity? |
2018-09-06T18:34:09.333 | <p>Can two adiabatic process paths intersect on a plot? This question is for 2 cases-
1.) Can two irreversible adiabatic curves intersect.
2.) Can a reversible & an irreversible adiabatic curve intersect?</p>
<p>There are some answers for this question on Quora, but they mostly address the intersection of two reversible adiabatic processes.</p>
<p>Also if the answer to either of the above 2 questions is 'yes', then I would like to know that if we do the proof given in the first answer of Quora link below with a reversible & an irreversible adiabatic process (or even two irreversible one) instead of two reversible adiabatic processes, it appears to be in violation of Kelvin Planck's statement. That is if we take an initial state and expand the system reversible adiabatically to a temperature 'T', now we take same system with same initial state but this time expand it irreversible adiabatically to same temperature 'T' (but with different P & V ofcourse) & join these two final states by an isothermal curve, we get a cycle with 3 processes (diagram in Quora link below, first answer). Now considering this cycle, we get a net positive work since it's area on a P-V diagram is positive but the system interact with surrounding only in 1 of the process, i.e the system recieves/rejects heat in only one of the processes, i.e the isothermal one. So it appears that we have a device which produces work continuously in a cycle just by interacting with a single thermal reservoir thus violating Kelvin Planck's statement of the second law of thermodynamics. how is that possible?</p>
<p>Also I want a physical interpretation of the processes that if they don't intersect then why don't and what it would be physically for a system.</p>
<p>Here are answers on Quora-
<a href="https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other" rel="nofollow noreferrer">https://www.quora.com/Why-do-two-adiabatic-curves-never-cut-each-other</a></p>
| |thermodynamics|heat-transfer| | <p>A simple approach considers only the first law for a system.</p>
<p><span class="math-container">$$
dU = \delta q + \delta w
$$</span></p>
<p>Internal energy change <span class="math-container">$dU$</span> is caused by heat flow <span class="math-container">$\delta q$</span> and work flow <span class="math-container">$\delta w$</span> across a boundary between the system and surroundings. Heat leaving the system (an exothermic process) decreases internal energy. Work done by the system leaves the system (and is negative), thereby also decreasing the internal energy. Recall that <span class="math-container">$U$</span> is a state function. For a closed system, it depends only on <span class="math-container">$T, V$</span>.</p>
<p>A reversible process allows that any step that increases <span class="math-container">$dU$</span> can be identically and exactly reversed to decrease <span class="math-container">$dU$</span> just by changing the signs of <span class="math-container">$\delta q$</span> and <span class="math-container">$\delta w$</span>. So, when heat leaves and work enters to increase <span class="math-container">$dU$</span>, the reversible step is to have <strong>exactly</strong> the same heat enter as left and <strong>exactly</strong> the same work leave as entered. These heat and work flows are <strong>always</strong> across a boundary between the system and surroundings.</p>
<p>An irreversible process has an additional heat flow <span class="math-container">$\delta q_{irr}$</span> that always leaves the system. That heat flow can be due to friction or internal interactions in the material in the system. That heat flow is <strong>never</strong> returned during the reverse path. It also does NOT flow (leave or enter) through the boundary between the system and surroundings. It is either directly absorbed in the system or surroundings (as a consequence of having internal interactions of the fundamental particles -- i.e. as a consequence of using a "real" substance rather than an ideal gas) or it is absorbed at the boundary (i.e. as friction). Because we loose f <span class="math-container">$q_{irr}$</span>, to return to the same internal energy <span class="math-container">$U$</span> as the starting point after taking an irreversible step, we must supply more work during the irreversible path than during the reversible path.</p>
<p>An adiabatic process has no <em>reversible</em> heat flow. A reversible adiabatic process has no heat flow. An irreversible has irreversible heat flow <span class="math-container">$\delta q_{irr}$</span>.</p>
<p>The work done <em>by the system</em> during a compression or expansion process is <span class="math-container">$\delta w = -p_{ext} dV$</span>. We can allow that <span class="math-container">$p_{int} = p_{ext}$</span> during the process regardless that it is reversible or irreversible (reversible processes must have <span class="math-container">$p_{int} = p_{ext}$</span> by definition).</p>
<p>On a graph of <span class="math-container">$p$</span> versus <span class="math-container">$V$</span>, work is the area under the curve. During a reversible, adiabatic process, that area will be <span class="math-container">$dU$</span> <strong>exactly</strong>. During an irreversible, adiabatic process, that area will be <span class="math-container">$dU \pm \delta q_{irr}$</span>. When the two curves are drawn on a graph, they can intersect either at one common starting point or at one common ending point.</p>
<p>In summary, the total work done during a reversible, adiabatic process is never the same as the total work done during an irreversible, adiabatic process that either starts or ends at the same set of <span class="math-container">$T, V$</span> conditions (i.e. starts or ends at the same <span class="math-container">$U$</span>). Therefore, adiabatic curves for the two different processes, reversible and irreversible, will diverge from a common starting point or intersect at a common end point.</p>
<p>Any concerns about whether this violates the Kelvin-Planck form of the second law cannot be addressed here. This discussion is about the intersection of individual paths. The Kelvin-Planck statement is for cycles.</p>
| 23645 | Adiabatic paths |
2018-09-06T19:22:37.270 | <p>I’ve been trying to figure out what this connector is and everyone has no clue. There isn’t enough identifying information to find anything on google. All I see is a label “lwai”. It’s located in a panel with power outlets, rj45 jacks, and some audio jacks. I think I found that it might be related to some hvac system but I can’t find any description or picture that confirms that. <a href="https://i.stack.imgur.com/5kgSk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5kgSk.jpg" alt="mystery connector"></a></p>
| |outlet| | <p>This is a 3 phase power connector with neutral and earth according to the IEC 60309. It's made for 120V/208-144V/250V 50-60Hz Systems. The contact in the top should be ground. The contact in the center could be a pilot contact which might indicate a 63A connector</p>
| 23647 | What is this connector |
2018-09-06T23:41:33.870 | <p>The crux of a PID controller is:-</p>
<p><em>output = Kp * error + Ki * integral + Kd * derivative</em></p>
<p>So typically there is proportionality and a continuously varying (analogue) output. In the case of a simple process like a domestic gas boiler, the boiler is either fully on or fully off. Or, say a cooling fan that can only be switched on or off to keep something cold. </p>
<p>In these cases, can a PID controller be used at all? There is still an error term. Or would an alternative form of control be necessary?</p>
| |pid-control|feedback-loop| | <p>I was looking for a similar question and found this entry. So Transistor answer is right and give me all the hints to implement a version of this in Structural Text.</p>
<p>I am using a cross by zero SSR for my heating element and PT100 for temperature measurement. Control for my application is having a delta of temperature of +/-0.3C. I use my own PID equation from C for MCU. A minimum 2 seconds for switching was placed to avoid fast switching of SSR. The system usual set temperature is 45C but the PLC reads 450 as the temperature to include a decimal point.
Find the code below. Hopefully this helps others:</p>
<pre><code>MeasureDelay(IN:=TRUE,PT:=T#1000MS);
error:=SetTemperature-Temperature;
IF MeasureDelay.Q THEN
MeasureDelay(IN:=FALSE);
cumError:= cumError+error * 1;//T#1000MS
IF cumError<-100 THEN
cumError:=-100;
END_IF
IF cumError>100 THEN
cumError:=100;
END_IF
rateError := (error - lastError)/1;//T#1000MS
output := LREAL_TO_INT(1.5 * error + 0.7 * cumError + 5 * rateError);
lastError:= error;
END_IF
IF output>300 THEN
ControlPort:=TRUE;
ELSIF output<0 THEN
ControlPort:=FALSE;
ELSE
time1:=DINT_TO_TIME(output*100);
IF time1<T#2000MS THEN
time1:=T#2000MS;
END_IF
time2:=DINT_TO_TIME(30000-output*100);
T1(IN:=TRUE,PT:=time1);
IF T1.Q THEN
T1(IN:=FALSE);
T2(IN:=TRUE,PT:=time2);
IF T2.Q THEN
T2(IN:=FALSE);
ControlPort:=TRUE;
ELSE
ControlPort:=FALSE;
END_IF
END_IF
END_IF
</code></pre>
| 23651 | Can a PID controller have a simple on/off output? |
2018-09-07T08:51:52.007 | <p>now I'm working on a project which involves a dc motor and high voltage. I'm developing a machine consists of a motor to rotate a platform disc and high voltage will be applied during the rotation. The problem is when I supply a positive high voltage at motor shaft during rotation, the motor stop working. Do you guys have any idea, how I can supply the high voltage properly the high voltage supply?<a href="https://i.stack.imgur.com/eq3fh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eq3fh.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering|motors| | <p>Don't apply the high voltage to the motor shaft, isolate the disc and only apply the high voltage to that.</p>
<p>Edit based on comments:</p>
<p>isolate ie insulate, electrically separate the disc from the motor shaft while keeping the mechanical drive to make it rotate. Slip rings can be used to apply the voltage to the, now insulated, disc.</p>
<p>See here for an explanation of slip rings: <a href="https://en.wikipedia.org/wiki/Slip_ring" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Slip_ring</a></p>
| 23658 | how to avoid a clash of magnetic charge and electric charge |
2018-09-07T14:42:06.660 | <p>3d printing is a slow process and for my solution I need fast printing. So I print with no infill and I fill inside with a two part, 10 min drying polyurethane resin. I mix them in a disposable cup and use a syringe to put it inside. It gives super results as it is both sturdy and lightweight so I recommend this method.</p>
<p>My problem is that if the print has a high volume, the mixture begins solidifying inside syringe. So I need a device which can mix parts and apply it on the spot. Below is a crude drawing of what I need. Is there a name for such a device readily avaible or will I have to build it myself? Or any other advise is appreciated.</p>
<p><a href="https://i.stack.imgur.com/OBkvg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OBkvg.jpg" alt="Mixer"></a></p>
<p>What I have tried so far:</p>
<ul>
<li>If I use a slower drying formula then most of the solution leaks from small holes in printed object. In faster drying formulas it is managable.</li>
</ul>
| |fluid-mechanics|applied-mechanics| | <p>in the construction industry, there are many companies which sell two component epoxy and its applicator. which happens to be a double cylinder pump with a mixing nozzle.</p>
<p>These pumps do exactly what you need, they mix the two components at the final stage in a mixing nozzle. Some are hand operated, some have work with air pressure and some are electrical.
I add the battery operated epoxy gun by "Read Head" Co. <a href="https://i.stack.imgur.com/L2mH8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L2mH8.jpg" alt="here"></a></p>
<p>The nozzle is a plastic cone with spiraling inside blades designed to mix the two components and deliver to the fasteners.</p>
| 23661 | Polyurethane mix-on-the-spot system |
2018-09-07T22:23:47.623 | <p>I have a medical device I am designing which will lie under the patient bed.</p>
<p>I need to add a "connector/joint" on this device to permit the attachment of some consumable.</p>
<p>Here is a quick overview of the process:</p>
<p><a href="https://i.stack.imgur.com/cAC3s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cAC3s.png" alt="basic assembly schema"></a></p>
<ul>
<li>Blue part is low cost injected plastic disposable part that will be assembled and disassembled only 2-3 times before being disposed.</li>
<li>Orange part is the device, can contain more complex parts (springs...).</li>
<li>Orange part must last thousands of assembly/disassembly.</li>
<li>Orange part is not movable, only blue part can be moved when assembling.</li>
<li>When assembled, it must resist to at least 100N force in all direction, including rotation around the "snap axis".</li>
<li>Orange part must be around 3cm which is a good side to apply pressure with two fingers.</li>
<li>Disassembly pressure force should be low as only two fingers will be use, other hand will retain blue part.</li>
<li>It should be easy to assemble/disassemble with "closed eyes" as the device will be under a bed and there might be no visibility on the device itself.</li>
</ul>
<p>As my drawing might be a bit confusing, here is a common device that is similar (aka a buckle).</p>
<p><a href="https://i.stack.imgur.com/FJ1rE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FJ1rE.jpg" alt="buckle"></a></p>
<p>The main problem with buckle design is the require pressure force for disassembly which is way too high, it's also too hard to assemble and not robust enough. But it is a good example for the "general principle".</p>
<p>What I am looking for is not for someone to design the assembly for me, but I am looking for exiting systems that are an alternative to a snap in joint and any pointer that would help me design it.</p>
| |mechanical-engineering|joining| | <p>I would use a collar lock with a slide action locking gate. Solves the durability issues and can be used thousands of times. The collar can be square and slide into a dual keyway to support the blue item while a cylindrical gate (spring loaded or not) shuts behind it. Collar thickness determines load bearing capacity.</p>
| 23663 | Overcoming limitations of a snap in joint |
2018-09-08T02:25:57.647 | <p>A couple of books on elasticity, e.g. Gould and Soutas-Little, in the section on 2d elasticity in cylindrical coordinates, mention the case of <em>'quasi-axisymmetric'</em> problems. These are problems where the stress and strain retain and axial symmetry (i.e., no dependence on the angular coordinate $\theta$), whereas the displacements $u_r$ and $u_\theta$ may carry $\theta$ dependence. </p>
<p>In particular, on writing the expression for $u_\theta$:
$$
u_\theta = A*r\theta + B*\cos\theta + C*\sin\theta + D*r
$$
with A,B,C,D constants, both mention that the last three terms rigid body motions, and hence do not contribute to any strain. </p>
<p>This statement per se I understand. Rigid body motions cannot give rise to strains or stresses. However, while I can imagine that the $\cos\theta$ and $\sin\theta$ represent rotations, I do not see how the $r$ term cannot give rise to strains! If $u_\theta$ is (say) zero at the inner radius and finite at the outer radius, it must surely give rise to strains! What is going on? What am I misunderstanding?</p>
<p><strong>Addition to answer below</strong> --</p>
<p>Please see my comment to the answer for an explanation of the $r\theta$ term as well.</p>
| |mechanical-engineering| | <p>Consider a (small) rigid rotation $\alpha$ about the axis of the cylindrical coordinate system.</p>
<p>The displacement of a point at radius $r$ from the axis is
$$\begin{align} u_r &= 0 \\ u_\theta &= \alpha r. \end{align}$$</p>
<p>That's where your $D \star r$ term comes from.</p>
<p>The $\cos\theta$ and $\sin\theta$ terms come from rigid body <em>translations</em> in the $\theta = 0$ and $\theta = \pi/2$ directions, not from rotations. For example, consider a (small) rigid translation $a$ in the $\theta =0$ direction. Describing this is cylindrical coordinates gives
$$\begin{align} u_r &= a\cos\theta \\ u_\theta &= a\sin\theta. \end{align}$$</p>
<p>A rigid translation in the $\theta = \pi/2$ direction is similar, but with the $\cos$ and $\sin$ terms interchanged.</p>
<p>But the $A * r\theta$ term looks wrong. It doesn't make any sense, because it doesn't even represent a continuous displacement field. The values of $u_\theta$ when $\theta = 0$ and $\theta = 2\pi$ must be the same if the deformed shape of the structure "fits together" properly, but they are not!</p>
| 23667 | Meaning of quasi-axisymmetric case in 2D elasticity |
2018-09-08T23:42:48.040 | <p>I am currently building myself a "hobby" vacuum chamber from 1 inch tickness acrylic sheets (I am very aware that acrylic is not the proper material). Inner dimensions of the the chamber is 40x40x40mm. After a few simulations in Solidworks, safety factor and displacements seemed okay.</p>
<p>What I need is to feed 8 cables through the sheets. After a research, I couldn't find any airtight cable gland or any low cost vacuum feedthrough. What is your suggestion for a low cost feedthough?</p>
| |vacuum| | <p>A sparkplug is air tight and provides an electrical feedthru for very high voltages. </p>
<p>They're also threaded and taps for the thread are common. </p>
<p>Also they are cheap!</p>
<p>(The body of the plug is resistive ( between about 40 $\Omega$ and 1 k$\Omega$) and sometimes inductive - check the specs)</p>
| 23678 | Low cost feedthrough for an acrylic vacuum chamber |
2018-09-09T16:16:07.773 | <p>I'm trying to upgrade a multiple gas flow controller, which must be capable of controlling the mass flowrates of three gases independently through one outlet line to our system, and am in need of a bit of advice on programming it. I'm using an arduino and a few DACs to tell the <a href="https://en.wikipedia.org/wiki/Proportioning_valve" rel="nofollow noreferrer">proportioning solenoid valves</a> how open they should be. The current setup schematic is below, though I have more pressure sensors and mass flow sensors at my disposal if necessary. </p>
<p><a href="https://i.stack.imgur.com/MScS8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MScS8.png" alt="enter image description here"></a></p>
<p>My colleagues' advice is to use PID loops for each of the gases. This would be easy, particularly as there's an <a href="https://playground.arduino.cc/Code/PIDLibaryBasicExample" rel="nofollow noreferrer">arduino code already available</a> - however I'm not convinced that's either necessary or the best option, because:</p>
<ol>
<li>There is hardly any inertia in the system, unlike the thermal inertia in an oven for example. If it were just one gas, with one inlet pressure and one outlet, just the P part of a PID would be enough I think. </li>
<li><p>The flow rate of each gas is dependent on the pressure differential before and after the valve, which is then dependent on the mass flow of the other gases and the outlet pressure, hence a PID could reach the optimal flow for one gas but in doing so change the flow rates of the other gases - leaving oscillation in the system. (the pressure in the system at the outlet may be changing slightly also) </p></li>
<li><p>I don't know if its possible, or if so then how, to write a PID loop for simultaneous multiple input/outputs.</p></li>
</ol>
<h2>Hence I would appreciate any ideas, or solutions that exist which I was unaware of, for how my program should go to control the proportioning solenoid valves?</h2>
<p>P.S. I'm not asking anyone to actually write a code for me, but just a basic idea of how I should do it would be great please? </p>
<p>P.P.S. (There is a formula for gas flow rate <a href="https://www.pipeflowcalculations.com/gasleak/" rel="nofollow noreferrer">here</a>, so I suppose you could try to work out theoretically what the optimal signal to the valves would be based on some kind of simultaneous solution of this formula for the three gases, taking into account friction factors for all the little fittings and parts in the system and expansion factors for the particular gas mixtures concerned (I don't think any of the gas cylinders are absolutely pure mixtures) etc. but I would think it's possible not to go that complex here!)</p>
| |control-engineering|pressure|gas|pid-control|flow-control| | <p>My best stab at this would be to attach parallel inputs of:<br><pre>
Gas1(100psi) -> Flow Meter1 -> Valve1 -> Manifold Input1<br>
Gas2(100psi) -> Flow Meter2 -> Valve2 -> Manifold Input2<br>
Gas3(100psi) -> Flow Meter3 -> Valve3 -> Manifold Input3<br></pre>
...where:<br></p>
<ul>
<li>I picked "100psi" to stress that the input pressure should be >> the output pressure. Could be 50psi, or whatever...</li>
<li>Valve[n] is controlled by your spiffy Arduino code taking Flow Meter[n] as its input.</li>
<li>All Valve[n] outputs go to a manifold that effectively connects together all inputs.</li>
<li>The Manifold output (not shown) connects to your artificial lung.</li>
</ul>
<p>In addition:<br></p>
<pre>
Manifold Output -> Lung -> Px Sensor<br>
' | Purge Valve<br></pre>
<h2>Strategy:</h2>
<ul>
<li>Open purge valve.</li>
<li>Control Proportional Valve[n] so each gas has its proportional mass flow input. (This is where the spiffy software comes in, more on that later). </li>
<li>Allow purge to stay open while all 3 gasses are flowing, until steady state has been achieved. </li>
<li>Close the purge valve. </li>
<li>Monitor the Px sensor while the pressure builds up. While in the background, maintain proper flow proportions. </li>
<li>Once the target Px is reached, close all Valve[n] valves simultaneously.</li>
</ul>
<h2>Control Algorithm</h2>
<p>I would suspect that you could get away with a straight P controller. I assume that you are not too concerned with small errors (you didn't say). I do not think that you will need to work the pressure sensor data into any of your calculations. However, if delta-p starts getting too small, then maybe you would need to build in a higher order control system, or perhaps the Px data to scale the valve outputs as delta-p goes down. TBD.</p>
<p>You will spend 80% of your time prototyping 1 control system to work. Then another 80% of your time integrating the hardware.</p>
<p>Getting the proportional valve to work as you expect, well that's the golden goose. Stability will be dependent on a few things, but macroscopically I would be concerned about:</p>
<ul>
<li>Valve Cv value (Ahh, Cv numbers. Mean different things to different people, and nothing to most.) Are the valves ported properly for the desire flow rates? Are the valve orifices a suitable size for the flow rates and control range?</li>
<li>Arduino limitations. We love Arduino's. They're fun. They also suck. Especially when you want to do something real, and real fast. I'd bet dollars-to-donuts that you could get this to work with some flavor of Arduino. I'd also bet that you will end up working around some wonky limitation that you could have avoided by spinning your own electronics, and programming in native c. But heck, it's fun. The parts, and more importantly your time, are really cheap - so why not.</li>
<li>Is it all really stable under all conditions? Everyone will scream "Model it!". Instead, just try it and find out. A few hundred times. It will be a good experience. Even if you do take the modeling route, you will still need to try it out a few hundred times, so leave the expensive modeling software to the interns, where it'll do less damage.</li>
</ul>
<p>Purchase a $30 adjustable pressure relief valve, in case something goes awry. No need to blow the poor fella's lungs apart.</p>
<p>It sounds like a fun project.</p>
<p>Good luck!</p>
| 23682 | Programming a gas mass flow controller rig |
2018-09-10T04:55:30.933 | <p>The project is to rotate a load using a motor with a varying speed between like <em>2000-10000 rpm</em>.</p>
<p>I'm using a <strong>flange coupling</strong> to attach the cylindrical tube to the motor. Since I will rotate it to high speed, I think I need to tightly and safely attach the load to the motor. </p>
<p>Do you have any suggestion on what coupling or what is the best way to attach the cylindrical tube to the motor <strong>safely</strong>?</p>
<p>NOTE: the shaft and load are the same lines. this machine is about centrifugal electrospinning which is a machine to fabricate a nanofiber. </p>
<p><a href="https://i.stack.imgur.com/JsGkQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JsGkQ.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|motors| | <p>Normally, you would use two grub screws at 90 degrees to each other, to secure the load onto the motor shaft. I have done this to attach a stainless steel flywheel on a motor spinning at 6,000 rpm and it's holding just fine.</p>
<p>To give it extra strength, you can also use something like Loctite 638 to glue the flange coupling, or whatever coupling you should choose, to the shaft. This probably means the coupling needs to be metal, not plastic (although I'm sure there'll be other glues to bong plastic to metal). Note that Loctite 638 is pretty much permanent: once you've glued your coupling onto the shaft, it won't be coming off again (easily).</p>
| 23685 | How to attach a cylindrical load to a motor safetly |
2018-09-11T04:48:09.740 | <p>The only one I know is the three point bending test for flexural strength.</p>
| |concrete|strength| | <p>There are two methods to test flexural strength, three point and four point method. </p>
<p>Three point method: <a href="https://i.stack.imgur.com/RFnzR.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RFnzR.png" alt="enter image description here"></a></p>
<p>and the corresponding moment line: <a href="https://i.stack.imgur.com/dA9WX.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dA9WX.png" alt="enter image description here"></a> </p>
<p>four point load: <a href="https://i.stack.imgur.com/g5Gwm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/g5Gwm.png" alt="enter image description here"></a></p>
<p>and its moment line: <a href="https://i.stack.imgur.com/c8j7N.png" rel="noreferrer"><img src="https://i.stack.imgur.com/c8j7N.png" alt="enter image description here"></a></p>
<p>The maximum flexural stress happens to concentrate only in a point in three point test, if you are testing a non-homogeouse material as it's always so in reality, it's not handy at all, brittle materials in three point test are prone to premature fracture, however the stresses in four point test, extended across the beam (construction ...), it means in case of not-homogenous material béton armé (Eng. Reinforcement concrete), we can have more accurate map of material behaviour. </p>
<p>Usually these two methodes are considered as destructive test. </p>
<p>Another destructive test is <strong>splitting tensile strength test</strong>, <a href="https://www.youtube.com/watch?v=gC7XB0A3F4M" rel="noreferrer">here</a>, you can find more about this test.</p>
<p>Pure axial (tensile, compression) tests (destructive tests) are quite popular. Universal testing machines or hydraulic presses are two familiar examples of instrument testing machines.</p>
<p>Besides hardness tests (semi-destructive) such as Brinell, Rockwel and Vickers are among the most popular tests these days in Europe. </p>
| 23700 | What methods are there to test the flexural, tensile and compressive strength of concrete? |
2018-09-12T04:45:44.323 | <p>The curing process for regular cement concrete is sprinkling water fog or using a moisture retaining fabric. I was wondering what the process is for polymer concretes because they don't use any water.</p>
| |civil-engineering|concrete| | <p>Note that the curing process for ordinary concrete is a chemical reaction between the ingredients of the cement mix and water which is stirred into that mix. Water is sprinkled onto the curing concrete to make sure that enough water is available to make the curing reaction go to completion and also to cool the concrete, since the reaction releases heat. </p>
<p>In the case of polymer cement, "curing" is a chemical reaction between the resin and the curing agent which are mixed together before pouring. Sometimes the curing agent is air, which fuels the chemical reaction that causes the resin to turn into a solid, and in some cases the air-curing chemical reaction is assisted by moisture in the air. </p>
| 23707 | What is the curing process for polymer concrete if it doesn't use any water? |
2018-09-12T05:12:22.380 | <p>When I was younger, I had interned in a civil/structural engineering firm specializing in bridges that had many PE civil engineers on staff and a couple licensed mechanical engineers.</p>
<p>Now as a soon-to-be licensed mechanical engineer, interested in starting his own business, I'm wondering what is limit of a mechanical PE. Can a mechanical PE subcontract from a civil firm? Can the mechanical PE only work on certain tasks/items of a civil project? Of the little I remember, the mechanical engineers were the experts on moving bridges (bascule, swing). My specialty is finite element analysis.</p>
<p>I'd be curious if anybody knows where exactly this line is, and even better, if they can point to laws/regulations/policy documents that describe this. For specifics, I am located in California, but would be interested in all United States-based answers.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|licensure| | <p>I'm a licensed civil engineer in California, and I don't think the answer is a black and white yes or no. My civil license allows me to supervise and/or stamp work that I have taken personal responsibility for. It allows me to stamp only work that I have direct work experience in. I have no specific seismic experience, for example, so I'm not legally allowed to stamp and work with seismic calculations, even though that work rightly falls under civil engineering. Likewise, I'm not aware of any regulation in the Business and Professions Code that might prevent me from stamping work that might be considered mechanical engineering, if I can show relevant and recent experience in that field. In fact, I've done some pump station work in the past, which is just as much a mechanical discipline as a civil one. (Ironically, my engineering degree is mechanical, not civil). Most, if not all public agencies do require a registered civil engineer's stamp for public works type projects, but that is up to the agency, not part of the state regulations. That might also be true for agency contracts for mechanical/structural type work as well. But that would be a separate agency requirement, not a state regulation. FEA is most certainly as much a mechanical discipline as anything else, so I suspect you'd be fine doing that work with a mechanical license. </p>
<p>My advice would be to read the Business and Professions Code for Professional Engineers available at the state's website. This is a good place to start: <a href="https://www.bpelsg.ca.gov/laws/pe_act.pdf" rel="nofollow noreferrer">https://www.bpelsg.ca.gov/laws/pe_act.pdf</a> If you need to ask a specific question to someone at the state level, you could start here: <a href="https://www.bpelsg.ca.gov/" rel="nofollow noreferrer">https://www.bpelsg.ca.gov/</a> There are also professional engineering associations around the state that might also give you good guidance. All of the above is from my memory so take it with a grain of salt, and good luck!</p>
| 23708 | What can a licensed Mechanical Engineer do in the civil/structural field? |
2018-09-12T15:31:04.703 | <p>I have seen many times that people gets the frequency response function using DTFT. Can I use the Fourier series, and when I do that?</p>
| |frequency-response| | <p>Examining a physical phenomena (expressed as a mathematical function) in the frequency domain can be very fruitful sometimes.
In case you are capable to perfectly express your function mathematically, i.e. by writing down an analytical function, you can manually use a Fourier series to find the amplitudes of any specific frequency.
However, this is a mathematical approach which is quite irrelevant to the most of engineering tasks. Since you probably do not know exactly what you are looking for, trying to analyze what is going on will be very tedious this way.
More often, you find yourself with a lot of recorded digital data you would like to analyze. You will be interested to find the dominant frequencies (by which the amplitude is the largest) and will be glad to obtain a plot showing you all the data arrange along the frequency axis. This is exactly what DFT (or FFT and etc,,) does.</p>
| 23714 | Fourier series with frequency response |
2018-09-12T18:27:13.360 | <p>I want to make a <strong>geodesic dome greenhouse</strong> (16ft wide, type: 3V 5/8). </p>
<p><a href="https://i.stack.imgur.com/DXMt6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DXMt6.jpg" alt="enter image description here"></a></p>
<p><sup>Source: <a href="http://geodesicdomes.ca/wp-content/uploads/2009/12/domes-fullsize.jpg" rel="nofollow noreferrer">http://geodesicdomes.ca/wp-content/uploads/2009/12/domes-fullsize.jpg</a></sup></p>
<p>I would cover a geodesic dome frame with clear poly plastic sheeting. </p>
<p>I can purchase clear poly plastic in almost any rectangular dimension (either as <a href="https://www.homedepot.ca/en/home/p.240-in-x-1200-in-cgsb-approved-vapour-barrier.1000166576.html" rel="nofollow noreferrer">narrow house wrap</a> or as wider greenhouse poly). </p>
<hr>
<p><strong>Problem:</strong></p>
<p>When I've researched this sort of setup online, I've found lots of photos of attempted wrapping jobs. Unfortunately, when the poly is wrapped around the dome, it ends up looking as <strong>ugly as sin</strong>. There ends up being lots of overlap and/or loose areas, which have the added disadvantage of being noisy in wind. </p>
<p>Example:</p>
<p><a href="https://i.stack.imgur.com/ssDXN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ssDXN.jpg" alt="enter image description here"></a></p>
<p><sup>Source: <a href="https://www.instructables.com/id/20-Geodesic-Dome-Greenhouse/" rel="nofollow noreferrer">https://www.instructables.com/id/20-Geodesic-Dome-Greenhouse/</a></sup></p>
<p><strong>Question:</strong></p>
<p>What method of wrapping a geodesic dome would result in the least amount of overlapping and loose areas?</p>
| |geometry| | <p>Also seeing a net of nylon straps that wraps over the plastic. Lighter and easier to pack than rope.</p>
| 23718 | Wrap a geodesic dome with a poly sheet: Minimize overlap & loose areas |
2018-09-12T20:57:20.827 | <p>I am studying gauges for measurements and I can't find the answer to what device I would use to set angles in 2 plates.</p>
| |measurements| | <p>Although a photo or diagram or more detailed description would be helpful, you may find what you seek in a machinist's protractor. There are a few different designs, including a few electronic digital models.</p>
<p><a href="https://i.stack.imgur.com/EDBlg.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/EDBlg.jpg" alt="digital protractor"></a></p>
<p>I have the above model, but have yet to use it for anything requiring extreme precision. It has a tendency to flex a bit even after tightening the angle lock.</p>
<p><a href="https://i.stack.imgur.com/rPtKF.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/rPtKF.jpg" alt="protractor 320"></a></p>
<p>This model appears often in searches and is much more accurate, according to my previous reading via a machinists' email list. Much if the discussion is that the typical home hobby machinist does not require this much precision, as these devices are rather expensive, upwards of US$300.00</p>
<p><a href="https://i.stack.imgur.com/6RPML.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/6RPML.jpg" alt="simple protractor"></a></p>
<p>Less precise and less expensive, the above type provides perhaps one-half degree accuracy, but does qualify as a possible solution to your question.</p>
| 23722 | What device would you use to set angles in two plates? |
2018-09-13T04:45:44.023 | <p>I'm designing a chair from scratch and I'd like it to have a swivel base. For this my initial idea was to press fit a bearing (thrust bearing) in the shaft of the base and connect the seat by press fitting a small shaft through the middle. </p>
<p>But I'm having my doubts if this can handle the load of a person sitting on the chair.</p>
<p>Looking for a purpose built or better suited bearing for this. </p>
<p>Also opened to suggestions for a better way to build this. </p>
<p>Thanks</p>
| |mechanical-engineering|bearings| | <p>Most 'Swivel Chairs' use simple thrust bearings, which consist of a low-friction layer (either ballbearings, or a flat disc of plastic), sandwiched between a lower disc attached to the swivel legs and an upper disc attached to the chair base. the shaft coming down out of the chair base goes through the sandwich and extends down to where the legs branch off, and somewhere in there is a bushing that holds the shaft centered in the tube which supports the bottom disc as described above. </p>
<p><a href="https://i.stack.imgur.com/HG3gFm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HG3gFm.png" alt="Swivel Chair Thrust Bearing"></a></p>
| 23726 | What is the type of bearing used on swivel base chairs? |
2018-09-13T11:24:02.270 | <p>I have a 4-wheeled vehicle that is 10 x 10 cm. Each wheel has its own motor. The wheel diameter is 3.1 cm. The total weight of the vehicle is 500 grams. The friction coefficient of the rubber wheels to concrete is 0.6-0.85 (found this online, from the research I did I think it is needed for calculating torque). How much torque is needed on the wheel before the motor becomes too powerful and just cause itself to slip. </p>
<p>If someone share the steps and formulas needed to solve the problem it would be of great help. </p>
<p>So far I only calculated the weight on each wheel (I'm not even sure if this is correct)</p>
<p>$$\begin{align}
W &= \text{wheel}_1 + \text{wheel}_2 + \text{wheel}_3 + \text{wheel}_4 \\
500\text{ g} &= 4\text{wheel} \\
125\text{ g} &= \text{wheel} \\
\end{align}$$</p>
<p>so 125 g per wheel (assuming weight is distributed equally).</p>
| |torque| | <p>Before start to solve the problem, i like to use your assumption about load distribution. And i assume you found the static coefficient of friction. Notice there is a difference between static and dynamic coefficient of friction. The dynamic coefficient is slightly smaller in amplitude than static coefficient. </p>
<p>Look at the free body diagram of one of the wheels: <a href="https://i.stack.imgur.com/Vgtno.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vgtno.png" alt="enter image description here"></a></p>
<p>To make the wheel move, you need to apply enough torque to cancel out the force $F_t$:
$$F_t = \mu F_N$$
$$F_N = mg$$
Here $\mu$ is the static coefficient of friction, $F_N$ normal or reaction force on the wheel, $g$ is constant, about $9.81 \frac {m}{s^2}$ and $m$ is the mass of the vehicle in this case $m$ is just a fraction of it nl. one fourth. </p>
<p>Now we can calculate the torque: </p>
<p>$$\tau = F_t \frac{d}{2} =\frac{d}{2} mg \mu $$</p>
<p>Here $d$ represent the diameter. Notice this is enough to bring the vehicle in movement, but if you want to accelerate the vehicle then this amount of torque is not enough.</p>
| 23731 | How to calculate the torque of 4 wheeled vehicle before slipping |
2018-09-13T11:35:55.807 | <p>There are thermo-shrinking plastics, commonly used in electronics, which stay shrunk after cooling down. There are thermally expanding foams. The closest to 'expanding when cooled' I know is water, with ice expanding, but only by a small margin, a scarce couple percent.</p>
<p>Is there a material that would start expanding significantly when cooled? May shrink back when heated, or may stay expanded. It doesn't need to be as forceful as ice (which can explode rocks) but it should have at least 30% expansion ratio, the more the better. </p>
| |materials|thermal-expansion| | <p>While there's a list of (rare) materials with negative thermal expansion, that 30% expansion ratio excludes most of them and narrows down your search to polymers or structures.</p>
<p>Within the realm of polymers, <strong>two-way reversible shape memory polymers</strong> might be your option. They can have relatively high reversible strains and can be thermally activated:</p>
<p><a href="https://i.stack.imgur.com/wxslM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wxslM.png" alt="enter image description here"></a></p>
<p><em>WM Huang, Thermo-Moisture Responsive Polyurethane Shape Memory Polymer for
Biomedical Devices, The Open Medical Devices Journal, 2010, 2, 11-19</em></p>
<p><a href="https://i.stack.imgur.com/ovTII.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ovTII.png" alt="enter image description here"></a></p>
<p><em>C. Liu, H. Qinb, P.T. Mather, Review of progress in shape-memory polymers, Journal of Materials Chemistry</em></p>
<p>You may also want to look for <strong>structures</strong> which are made of positive thermal expansion materials, but when assembled they have a the opposite effect: </p>
<p><a href="https://i.stack.imgur.com/K9DA3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K9DA3.jpg" alt="enter image description here"></a></p>
<p><em>Christopher Spadaccini, Mechanical Metamaterials: Design, Fabrication, and Performance, Frontiers of Engineering: Reports on Leading-Edge Engineering from the 2015 Symposium (2016)</em></p>
| 23732 | Are there any cold-expandable materials of significant expansion factor? |
2018-09-15T11:02:13.700 | <p>I would like to calculate the deflection of the following geometry with a 3D load in Y direction:</p>
<p><a href="https://i.stack.imgur.com/38eTt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/38eTt.jpg" alt="enter image description here"></a></p>
<p>Do you know of any textbook, website etc. which discusses how to do that analytically ? I cannot find any suitable reference.</p>
| |mechanical-engineering| | <p>let's call the cross section width B, base and its height H.
And assume the length of the section supported by the wall L1 and the short elbow L2 and call the load P.</p>
<p>the deflection at the end of L2 will be the sum of 3 components: the deflection of L2 as a simple cantilever loaded with P, the deflection of L2 due to torsional rotation of L1 under the torque P(L2), and the deflection of L1 loaded as a cantilever beam with load P.</p>
<p>The torsion strain is <span class="math-container">$$\tau_{max} = \frac {P\times L2} { \alpha\times B\times H^2\times G }$$</span>
<span class="math-container">$$\phi = \frac { TL1} {\beta BH^3G } $$</span>
where G is the section shear modulus, alpha and beta are constants depending on the proportion of the base to height of the section. In this case they can be approximated for 0.246 and 0.249 assuming b is twice the height.</p>
<p>And we know the deflection of a cantilever beam with a single load at the end </p>
<p><span class="math-container">$$\delta = PL^3/3 EI $$</span></p>
<p>So we calculate the 2 components of the deflection and add <span class="math-container">$\phi \times L2 $</span> to it.</p>
| 23765 | Deflection of L shaped plate |
2018-09-15T21:33:32.117 | <p>My limited knowledge in engineering tells me:</p>
<ul>
<li><p>Titanium is expensive, stronger and heavier than aluminium, lighter and weaker than high yield steel.</p></li>
<li><p>The most significant advantage of titanium is the good strength to weight ratio.</p></li>
<li><p>In case of submarines, weight is not really a big concern, as long as the weight is manageable (less than the displacement and bigger than displacement - maximum ballast mass).</p></li>
</ul>
<p>The mentioned points exclude the use of titanium. Since the designers decided otherwise, and surely had some good reasons, I wonder where my misunderstanding is?</p>
| |materials|metals|marine-engineering|naval-engineering|submarines| | <p>All Soviet industries were state owned, including titanium production.
So Soviet made titanium submarines to showcase their technological might.
THough titanium is strong, corrosion resistant and lighter than steel as well as non-magnetic. This would seem advantageous to any navy....still.</p>
<p>It has numerous drawbacks.....</p>
<ul>
<li>Titanium is not easy to shape.</li>
<li>Welding requires inert gasses to displace oxygen, meaning welders would need oxygen masks through the whole work shift.</li>
<li>Titanium was horribly expensive, so much so, the designs for ALFA class consumed nearly 1% of Soviet GDP and they only built 6, one of which was scuttled because reactor issues.</li>
<li>Titanium is tough, but it's not flexible. It has a terrible Young Modulus ie. It's "Give" or ability to withstand deforming stresses with ability to return to basic level of normalcy is limited, Steel is superior in this regard.
<a href="https://i.stack.imgur.com/EJLWX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EJLWX.png" alt="enter image description here" /></a></li>
<li>Titanium is corrosion resistant, but the titanium hulls still required extensive inspection due compression stress after each mission dive. With Many of the subs experiencing hull cracks.</li>
<li>The US developed "Austensic Steels" Which were non-magnetic too. Also developed deeper diving torpedoes that relegated any advantage the Alfa had in deep water</li>
</ul>
| 23777 | Why did the Soviet Alfa-class submarines have a titanium hull? |
2018-09-17T11:30:26.127 | <p>I am new to the Underfloor heating system, and a bit confused on the terminology for Operative Temperature.</p>
<p>Is this the temperature that is measured by the sensor or is this the set-point temperature on the thermostat?</p>
| |heat-transfer|heating-systems| | <p>That may well depend on who did the writing.</p>
<p>The set point is the desired temperature and the sensor should measure the ambient temperature so that it heats, or not, according to where the ambient temperature is wrt to the set point.</p>
| 23791 | Operative Temperature in Underfloor Heating? |
2018-09-17T16:28:04.107 | <p>As an energy engineer focused on energy efficiency in the built environment, I am familiar with the standards and codes covering energy use in buildings; specifically <a href="https://www.ashrae.org/technical-resources/bookstore/standard-90-1" rel="nofollow noreferrer">ASHRAE 90.1</a>, the "Energy Standard for Buildings," and <a href="https://basc.pnnl.gov/resources/2018-iecc-international-energy-conservation-code" rel="nofollow noreferrer">IECC 2018</a>, the "International Energy Conservation Code."</p>
<p>Recently several areas of increasing concern are intersecting with this work, such as indoor air quality, but also <strong>stormwater management</strong>:</p>
<blockquote>
<p>Stormwater management is the effort to reduce runoff of rainwater or melted snow into streets, lawns and other sites and the improvement of water quality, according to the United States Environmental Protection Agency (EPA). <a href="https://www.eecenvironmental.com/what-is-stormwater-management/" rel="nofollow noreferrer">(Source)</a></p>
</blockquote>
<p>My question is <strong>what engineering codes and/or standards govern stormwater management in and around buildings?</strong></p>
<p>Some teams we work with are concerned about how to manage the water which falls on the roof of the buildings they are designing. Are there guidelines related to collecting and/or treating this water before diverting it?</p>
<p>Note: my work is within the U.S., but I'm also interested to know of standards or codes that are in use in other countries. </p>
| |standards| | <p>I answer this with a perspective from the United Kingdom. With an annual problem of flooding of low-lying areas especially in the Worcester area West of Birmingham the national government has written new laws that specified new housing estates and construction sites must hold 50% of water runoff and prove it before building planning can be approved. As far as I am aware there are no building codes in the eurocodes that specifically deal with stormwater runoff. More often than not stormwater runoff is dealt with by means of concrete structures that are inserted underground such as pipeworks and drainage ditches and even urban rivers lined with concrete. from this perspective, such structures are covered by Eurocode 1 and 2.</p>
<p>This document maybe of use to you in this regard. [Stormwater management][1]. An important consideration is that maybe stormwater should be dealt with in a more effective ways such as pipes that have holes to allow them to act as soakaways on route to a river and thereby allowing the ground that they are buried in to absorb a fair amount of water before actually reaches the river system</p>
<p>[1]: <a href="https://ec.europa.eu/research/participants/documents/downloadPublic?documentIds=080166e5a5630bbf&appId=PPGMS" rel="nofollow noreferrer">https://ec.europa.eu/research/participants/documents/downloadPublic?documentIds=080166e5a5630bbf&appId=PPGMS</a> ""</p>
| 23795 | What standard and/or code covers building stormwater management? |
2018-09-17T20:36:12.367 | <p>I would be grateful if someone could explain me what does the value of zeros of a transfer function tell about the dynamics of the system.</p>
| |control-engineering|control-theory|transfer-function|systems-design|regulations| | <p>The zeros of a transfer function influence the dynamics of the system in multiple ways. </p>
<ol>
<li>Consider the output of a given system for $u(t) = 0$ as a weighted linear combination $$y_a(t) = \sum_{i=1}^n R_i e^{P_it}$$ where $P_i$ are the poles of the system. The weights $R_i$ are influenced by the zeros.</li>
<li>For a zero at $s = N_i$, a signal with the complex frequency $N_i$ is not transmitted anymore, i.e. input signals $u(t) = e^{N_i t}$ are "blocked" by the system.</li>
<li>Zeros are invariant towards state feedback, i.e., while the poles of a system can easily be moved by a corresponding state feedback, that is not possible for the zeros. </li>
<li>It can be shown that a system with a zero in the right half-plane will always be unstable for high feedback gains and that the possible control performance for such a system is limited.</li>
<li>The zeros of the transfer function of the open loop describe an internal dynamic of the system, which is not observable. This internal dynamic was initially described by Isidori as <strong>zero dynamics</strong>. </li>
</ol>
<p>The zero dynamics describe the internal dynamic of a system for the special case of an input $u(t)$ and an initial condition $x(0)$ such that the output $y(t) \equiv 0$ for all times $t \geq t_0$. </p>
<p>For further reading on this pretty interesting topic, I'd start right at the source with Isidori's book "Nonlinear Control Systems: An introduction"</p>
| 23799 | Zeros of a Trransfer Function |
2018-09-19T13:54:24.097 | <p>How would one describe the difference between measuring and gauging? As I understand it at my company, we use a fixed tool to <code>gauge</code> something, but we will use a set of calipers to <code>measure</code> something. Is this a correct understanding? </p>
| |measurements| | <p>I think you have it correct.</p>
<p>We used "go / no go" gauges to check some things and made measurements for others.</p>
| 23819 | Measuring vs gauging |
2018-09-19T13:54:42.010 | <p>Apologies if I've got the terminology wrong, I'm in now way involved in engineering, but I work in a high rise in a city with a lot of construction going on. I see a lot of different types of cranes, and it's got me wondering.</p>
<p>The two most common types I see are (I think) both tower cranes, however Some have the classic T shape, where the arm stays level, whilst the others have an are that raises up and down. Both types seem to be working on the same types of sites, so I've been wondering what the difference between the two types are, and why you'd pick one over the other.</p>
<p>Here are examples of what I'm talking about:</p>
<p><a href="https://i.stack.imgur.com/QgGqB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QgGqB.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/zc8pN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zc8pN.jpg" alt="enter image description here"></a></p>
| |building-design| | <p><strong>The key benefit of Luffing Jibs, is their ability to park at night in a smaller radius</strong>. When a crane is 'out of service', it's a requirement that it is able to 'weathervane'. This means that in high winds, the jib must always rotate such that the jib is facing downwind, and the ballast serves to counteract the rotation force of the wind. Consequently, it must be impossible for the jibs to hit one another, whatever direction they are facing. For multiple 'saddle jibs' (classic T shape), they get around this by setting the jibs at different heights, so that they overlap. They still can't have the towers close together than the length of the lower jib, however.</p>
<p><a href="https://i.stack.imgur.com/4OUa8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4OUa8m.png" alt="Overlapping Saddle Jibs"></a></p>
<p>With the increase in pre-fab, it's often desirable to fit more towers onto a single site. Furthermore, strict overhanging regulations in cities may prevent you from siting a saddle jib close to the boundary of your site, in case it swings over (or hits) neighbouring buildings when free-slewing at night. A luffing jib can be parked at a steep angle, and thus with this smaller out-of-service radius, you can fit more cranes onto the site, and stay within the overhanging regulations.</p>
<p><a href="https://i.stack.imgur.com/l2lDim.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l2lDim.png" alt="Luffing jibs on site"></a></p>
<p>There is a trade-off, however. If you try to park your jib at <em>too</em> steep an angle (over ~60degrees), the jib no longer has enough leverage to rotate the slew ring around under wind load. This is why you now sometimes see a solid 'rectangle' in the tip of a luffing jib. This is a 'windsail', and is literally a sheet of steel (sometimes fabric) designed to increase the surface area that the wind can push against, allowing an even smaller park radius compared to a traditional luffing jib.</p>
<p><a href="https://i.stack.imgur.com/AzCqim.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AzCqim.png" alt="Fixed windsail on top of Shard in London"></a></p>
<p>If you want to go tighter still (over ~75degrees), you need more area again, but, if you have a large fixed sail, that would make driving the crane in service very difficult under a light wind, and would limit the maximum wind speed under which it would be safe to operate the crane. The solution is to have a retractable windsail, which is deployed at night, and retracted when the crane is in use. </p>
<p><a href="https://i.stack.imgur.com/MxRqym.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MxRqym.png" alt="Retractable Windsail"></a></p>
<p>his innovation was a key part in Select Plant Hire being offered the contract to supply cranes for the Shell Centre on London's south bank, which had 21 cranes on a single site, 17 of which required retractable windsails. This allowed them to offer a faster production schedule than other companies that couldn't get so many cranes into the space.</p>
<p><a href="https://i.stack.imgur.com/q8kIAm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q8kIAm.jpg" alt="Shell Centre Site with multiple retractable windsails"></a></p>
<p>As far as I'm aware, saddle jibs are just more simple, and consequently a bit cheaper to use, and are used where possible by default. When you see a mixture of cranes on site, it's likely that the luffing jibs are sited near the edges, where overhanging regulations are an issue, with saddles in the middle to save money.</p>
<p><a href="https://i.stack.imgur.com/RTMB8l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RTMB8l.png" alt="Saddle jib in middle, luffing at edges"></a></p>
| 23820 | What's the difference/advantages/disadvantages to a tower with a moving jib, vs a flat jib |
2018-09-20T16:34:09.007 | <p>I have been given a spherical drug particle that dissolves by surface erosion into the bloodstream. The rate of dissolution is proportional to the exposed surface area of the particle. I have to model the volume of the dissolving particle as a function of time, which I did, and I got</p>
<p>$$V(t) = \left(\dfrac{3V_0^{1/3} - kt}{3}\right)^3$$</p>
<p>where $V_0 = V(t=0)$, the initial volume of particle. When I plot using Excel with $V_0=100$, I get a reasonable downward sloping curve which reaches 0 in about 3 sec.</p>
<p>Now I have to derive the rate of drug release as a function of time (g/s or mol/s) from the volume function. There is no other information given. How do I do this? Please help!</p>
| |chemical-engineering| | <p>This is actually more of a calculus question than anything.</p>
<p>That function gives you the volume over time. You are looking for the rate of change in volume over time.</p>
<p>That is the definition of the derivative of that function. So all we need to do is get <span class="math-container">$\dfrac{\partial V}{\partial t}$</span>.</p>
<p>For that, we define <span class="math-container">$V(x) = f(g(x))$</span>, where <span class="math-container">$g(x) = 3V_0^{1/3}−kt$</span> and <span class="math-container">$f(x) = \left(\dfrac{g(x)}{3}\right)^3 = \dfrac{1}{27}g(x)^3$</span>. Via the <a href="https://en.wikipedia.org/wiki/Chain_rule" rel="nofollow noreferrer">chain rule</a>, we know that <span class="math-container">$\dfrac{\partial V}{\partial t} = f'(g(x))\cdot g'(x)$</span>. Therefore, we have that</p>
<p><span class="math-container">$$\begin{alignat}{4}
\dfrac{\partial V}{\partial t} &= &&\dfrac{1}{9}(3V_0^{1/3}−kt)^2\cdot(-k) \\
&= -&&\dfrac{k}{9}(3V_0^{1/3}−kt)^2
\end{alignat}$$</span></p>
<p>The result here is negative (assuming a positive <span class="math-container">$k$</span> and that the result between parentheses is also positive), which indicates that the volume is dropping over time. Obviously, the volume lost is precisely what's released into the bloodstream. So if that latter interpretation (rate of release) is most relevant, just remove the negative sign.</p>
| 23842 | How can I convert volume as a function of time to mass or mole as a function of time for a dissolving drug particle? |
2018-09-20T16:53:29.830 | <p>I am a sophomore in mechanical engineering and am unable to understand the purpose and functioning of Oldham's Coupling.</p>
| |mechanisms| | <p>Have a watch of this video: <a href="https://www.youtube.com/watch?v=utEKKox2WHA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=utEKKox2WHA</a> It's a marketing video for a specific make/model of oldham coupling, but shows good footage of the coupling running at a variety of RPM's, including failure.</p>
<p>Oldham couplings are used to allow transmission of torque between two parallel shafts, which may be misaligned (non-coaxial) . The allowable misalignment will be shown on the data sheet for each specific coupling.</p>
<p>This may be a static misalignment, or dynamic, such as a driven roller that moves up/down according to the thickness of stock that it is running on top of.</p>
<p>In failure an Oldham Coupling will not transmit any torque, to the output shaft. This contrasts with flexible Jaw Couplings (which can also accommodate some misalignment, although not as much), which, if the insert fails, will continue to transmit torque, albeit with some backlash. Which of these is appropriate depends on which failure mechanism is safer for the system at hand.</p>
| 23843 | Which is the best tutorial on Oldham's Coupling? |
2018-09-21T23:32:00.943 | <p>I was recently walking down a road when a car came my way (I'm an inveterate jaywalker). I moved over to the sidewalk, only to notice it clearly wasn't a sidewalk at all. The pavement looked like this:</p>
<p><a href="https://i.stack.imgur.com/RFDyd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RFDyd.jpg" alt="enter image description here"></a></p>
<p>The paving stones are embedded into and perpendicular to the ground, making it rather uncomfortable to walk on!</p>
<p>Now, what is the purpose of such pavement? Is this layout meant to improve drainage in case of rain (this is in London, so it's always raining)? But then why pave the ground at all instead of simply using grass or greenery to anchor the soil and prevent erosion?</p>
| |civil-engineering|highway-engineering|pavement| | <p>Antivehicular ones. But also anti pedestrian, The goal is to keep pedestrians and people, bikers, etc off this section. Namely because if it should be in proximity to moving vehicles.</p>
| 23866 | What's sort of "sidewalk" pavement is this? |
2018-09-22T15:16:45.920 | <p>In order to conduct study, I need to produce controlled vibrations with amplitude (maximum) approx 200 um (micro) at frequency between (100-150) Hz. Vibration amplitude must be controlled with input voltage (Battery operable) while frequency remain constant.</p>
<p>I tried many things :</p>
<p>1) Eccentric Motor - frequency and amplitude, both varied with voltage.</p>
<p>2) Solenoid - difficult to control with voltage. (If any way exists, please suggest me).</p>
<p>3) Voice Coil Motor - Its very costly, I need solution around 10-20 $ max. as project constraint.</p>
<p>4) Linear Resonant Actuator - It is the best solution, but I am not able to find any LRA which can produce vibrations in um (micro metre).</p>
<p>Please suggest me any product exists or ways to produce controlled vibrations.</p>
| |motors|vibration|linear-motors|linear-motion| | <p>You can make an AC electromagnet pretty cheaply. Wrap some enamel coated wire around a steel bolt. Then get a steel cantilever beam with a resonant frequency equal to the frequency of interest. Place the magnet just under the tip of the beam and drive it with an ac voltage at the resonant frequency. 200 um should be achievable with a few volts if you have a sufficiently big magnet. </p>
| 23872 | Need to produce vibration amplitude of 200 um (approx) at natural (resonant) frequency, can be any frequency between 100 to 150 Hz |
2018-09-22T19:56:47.653 | <p>The same geometry and the same free body diagram. However, if we replace the material from steel to glass, for example. How does that influence the analysis?</p>
<p>Two things I think are
1. steel is ductile and glass is brittle. This means the cross section when it fails will be different.
2. glass has lower yield stress, which means it is much easier to fail.</p>
<p>Is there anything else that I omit? Thank you very much.</p>
| |structural-analysis|statics|stresses| | <p>If you are dealing with a simple Back-of-the-envelope calculation, then it should not matter, from stress perspective, what type of material you chose. All you have to do is to compare the results to the material allowable strength.</p>
<p>However, if you are talking about a FEA problem, you should take care closely of the material you choose. Tools like ANSYS calculate the object deformation (which is totally affected by the material Young's modulus and Poisson's ratio) and then derive the stresses out of it.</p>
| 23876 | Stress analysis for the same configuration but different material |
2018-09-23T02:07:19.017 | <p>What causes the pressure change in a horizontal pipe with no energy input. Is it the head loss? If so, how?</p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>We take the case of a horizontal pipe open on one end and driven with a pressure source on the other, flow is subsonic, fluid is newtonian, and incompressible. </p>
<p>Pressure diminishes with length away from the source due to frictional dissipation from viscosity. This is commonly referred to as "head loss". In the absence of energy input- that is, we stop the pump- then there is no flow, the viscosity terms are zero, no energy is dissipated, and there is no head loss.</p>
| 23880 | Pressure rise in horizontal pipe |
2018-09-25T12:11:31.413 | <p>I am currently looking ways to control or siwtch a max 220v 35A load. Simple magnetic relay rated for such load would work, but i am exploring the other alternatives too.</p>
<p>here i stumbled into triacs, in which solid state relays are made of. I stumbled into this ic</p>
<p><a href="https://www.st.com/resource/en/datasheet/t16t.pdf" rel="nofollow noreferrer">https://www.st.com/resource/en/datasheet/t16t.pdf</a></p>
<p>and i could not get the grasp on how a small triac such as this one.</p>
<p>is able to switch a high load of 220v 16amps</p>
<p>Thank you!</p>
| |electrical-engineering| | <p>35A needs a heavy duty contactor. It's a relay but designed for switching the high current.</p>
<p>A triac latches on until the current drops to 0. This means that the actual switching off happens at the zero crossing point. Switching on is still done whenever the gate voltage exceeds the turn on threshold. This can result in a inrush current spike but as long as that isn't sustained the triac will be able to deal with it. The one you linked can handle 120A for one cycle.</p>
<p>While the triac is off the isolation is enough to stop current flow.</p>
| 23904 | How are small triacs able to "switch" or control high power AC loads |
2018-09-26T14:31:10.217 | <p>I am attaching the model part as an Image below for your reference:-
<a href="https://i.stack.imgur.com/7SS81.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7SS81.png" alt="bent plate model so far"></a></p>
<p>I need the pattern like this on the above model.
<a href="https://i.stack.imgur.com/ikoc1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ikoc1.jpg" alt="desired embossed pattern"></a></p>
<p>I have the model shown above, and want to put embossed lines 0.2 mm thick with 0.2 mm spacing in between on the top, sides, and the bends. The feature should be parallel to the horizontal section. I tried using emboss but I don't know how to emboss all three sides.</p>
| |cad|software|autodesk-inventor| | <p>The solution to this, is simply to add the "embossed lines" <em>before</em> the bend. </p>
<p>Have a look at the modelling process below:</p>
<ol>
<li>Extrude a flat plate. Because this model will be symmetrical this is only half the size of the final object.</li>
<li>Extrude a 'master rib'. This is 0.1mm away from the YZ plane, so that there's an 0.2mm gap once it's mirrored</li>
<li>Pattern the rib to cover the full plate (using the 'rectangular pattern' tool, with one direction defined</li>
<li>Bend the plate, using the 'Bend Part' tool. You will need to make a sketch on the surface of the plate, and draw a line at the point where you want the bend to start. The line defines the "end of the flat top", so if you want the overall width to be 20mm, with an internal bend radius of 2mm, and a plate thickness of 5mm, you would need the line 13mm from the YZ plane.
N.B. You may have to click the word "Modify" on the ribbon to reveal this command if you have it hidden, or just use the "Search Help & Commands" window at the top right</li>
<li>Use the Extrude Tool with Cut Boolean to make the 'angled cutoff', as required</li>
<li>Finally mirror the part around the YZ plane to complete your component.</li>
</ol>
<p><a href="https://i.stack.imgur.com/vBmEl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vBmEl.png" alt="Modelling with embossed lines"></a></p>
| 23926 | How do you implement a pattern over a bent plate in Autodesk Inventor? |
2018-09-26T16:13:20.240 | <p>Do you have any idea on why the <code>Calculation 1</code> below doesn't match?</p>
<p>Reference:</p>
<p><a href="https://developer.mozilla.org/en-US/docs/Web/API/Web_Audio_API/Basic_concepts_behind_Web_Audio_API#Audio_buffers_frames_samples_and_channels" rel="nofollow noreferrer">Audio buffers: frames, samples and channels</a></p>
<pre><code>VLC - Media Information:
num channels: 2 (stereo)
sample rate: 44100 Hz
bits per sample: 32 => bytes per sample: 4
bitrate: 192 kb/s
duration: 7 mins, 23 secs => 443 secs
Windows - File Properties:
size: 10,642,422 bytes => 10393 KB
</code></pre>
<p><strong>Calculation 1 (without using bitrate) (using rounded values for simplicity)</strong></p>
<pre><code>1 second: 44100 samples => 44100 * 4 = 176400 bytes => 172 KB (looks a lot for just 1 sec)
This way: 443 secs => 443 * 172 KB = 76196 = 74 MB, but the file just 10 MB (!!! Contradiction !!!)
</code></pre>
<p><strong>Calculation 2 (using bitrate)</strong></p>
<pre><code> bitrate: 192 kb/s => 24 KB/s
1 second: 24KB => 443 seconds: 443 * 24 KB = 10632 KB = 10393 KB (like above) + [some metadata] (!!! Match !!!)
</code></pre>
<p>Thanks!</p>
| |audio-engineering|sound-isolation| | <p>Because the file is compressed. The sample rate and bits per sample are parameters of playback after decompression. The file size is for the compressed format, the amount of data that needs to be downloaded, not the amount of data that reaches the digital-to-analog converter.</p>
| 23929 | Using the bitrate value on a calculation is causing contradiction, why? |
2018-09-28T14:22:22.627 | <p>I'm planning to build an air-dropper for a small drone - a device to release small payloads hanging under the drone.</p>
<p>This is a commercial example of such a system, used with a much bigger drone - for heavier payloads, pretty heavy itself, and using the drone's built-in systems for control:</p>
<p><a href="https://www.youtube.com/watch?v=e81eQ0pYLCs" rel="nofollow noreferrer">https://www.youtube.com/watch?v=e81eQ0pYLCs</a></p>
<p>(7:18 - system in action. 5:44 - the part, for which counterpart I'm asking.)</p>
<p>I'm planning to build something similar, but much smaller. Payloads of up to 100-150 gram (but often smaller too), a completely independent control system. I already have working remote control electronics capable of outputting 3.3V at 40mA (driving a tiny coreless motor like <a href="https://www.alibaba.com/product-detail/DC-3-7V-dia-4mm-micro_60459075751.html" rel="nofollow noreferrer">this</a>; I also have such motors), under 10 gram together with the battery.</p>
<p>I still have no idea what sort of actuator I could use to release the payload. I'd prefer it to remain unpowered when 'latched', needing only power to open (drop the payload). It should be as lightweight as possible, because any of its weight takes away from payload capacity of the drone. And not being very expensive would be a big plus.</p>
<p>For the ideas I had so far:</p>
<ul>
<li>Most servos I've seen are rather on the heavy side, or provide so little torque I'm afraid any latch I come up with won't suffice. </li>
<li>I considered pairing a permanent magnet and an electromagnet that would cancel out the magnet's field, but first I'm not sure if I can output a field strong enough, and then the cancellation won't be perfect and if my payload is very light, it won't drop.</li>
<li>I don't think I have the current to drive a linear electromagnetic actuator to unlatch a hook (not sure about that though). </li>
<li>I don't have means to build more complex gearboxes to convert the high RPM of the tiny motor to torque - at least from scratch; if something of that kind is commercially available and small enough, it might be good. </li>
<li>...also, it would be great if it was fully reusable and safe. I seriously considered just a small relay to put full current from a battery through a resistive spiral to burn through the line with the payload. I'm not enthusiastic about this idea.</li>
</ul>
<p>I'm not specifying the housing, 'hook', line, attachment methods etc here - these are to be made such as required for it to work.</p>
<p>(also worth considering: a fully mechanical connector with no electronics; disengaging on a maneuver not encountered in normal flight, e.g. a rapid drop. I can't generate negative-G, but I can make the drone -almost- drop out of the sky. If this could disengage the hook, that would also be a workable solution. This is not as preferable as it requires considerable vertical space and is rather risky.)</p>
| |mechanical-engineering|actuator| | <p>I would recommend using servo (or other actuator) to pull a pin that is holding the load in shear. That way, as long as the friction of your system is low, the actuator will only need a fraction of the force (like a mousetrap). You would attach the pin to the servo arm such that it is as close to the rotation point as possible to give you the most torque. Also, make sure the attachment point can rotate for proper actuation and release. On the other end you will want to use something thin (or with a shallow hole) to support the pin. This way the stroke does not have to be very long.</p>
<ul>
<li>Here is a <a href="https://hobbyking.com/en_us/turnigytm-tgy-1440a-analog-servo-v2-0-8kg-0-10sec-4-4g.html" rel="nofollow noreferrer">4.4 gram servo</a> that should work well.</li>
<li><a href="https://hobbyking.com/en_us/radio-servos/servos.html?price=0-12&wrh=2%2C3&unit_weight_filterable=0-8" rel="nofollow noreferrer">Hobbyking</a> has a pretty good specifications filter now.</li>
<li>The pin can be made out of <a href="https://hobbyking.com/en_us/push-rod-and-sleeve-set-5pcs-bag.html" rel="nofollow noreferrer">rc push rod material</a> so you don't have to reinvent the wheel.</li>
</ul>
<p>Here is a cartoon image of the proposed idea. Things are a bit exaggerated; you will want to reduce the distance between your servo and the pin support so that the pin as is short as possible. A shorter pin means it can be a thinner rod without bending or buckling under the load.</p>
<p><a href="https://i.stack.imgur.com/nFiUE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nFiUE.png" alt="enter image description here"></a></p>
<p>Also note that the servo will need pwm control. You can get controllers for this, or if you want discrete input you can cut open the servo and just power the dc motor directly. Make sure you have a delay or a limit switch (possibly in its circuitry already) so you don't fry the motor.</p>
| 23958 | air-dropping actuator for a small drone |
2018-09-28T20:43:13.650 | <p>I am aiming to program a quadcopter flight controller as a personal project.
To be able to do this I understand one of the fundamentals is a PID control system . So my question is does anyone have any recommendations for books to read or will google be good enough?</p>
<p>I have found the following books , has anyone read these?:
"Advanced PID control" - <a href="https://www.amazon.co.uk/Advanced-PID-Control-Karl-Astrom/dp/1556179421/ref=sr_1_4?s=books&ie=UTF8&qid=1538160392&sr=1-4&keywords=pid+control" rel="nofollow noreferrer">here</a>
or ,
"PID Control: New Identification and Design Methods" - <a href="https://www.amazon.co.uk/PID-Control-Identification-Design-Methods/dp/1849968977/ref=sr_1_14?s=books&ie=UTF8&qid=1538160392&sr=1-14&keywords=pid+control" rel="nofollow noreferrer">here</a></p>
| |control-engineering|control-theory|pid-control| | <p>As @niels nielsen mentions the topic is very broad and successful implementation and design require additional techniques not necessarily directly associated with control design. Depending on your specific application this might include dynamic modeling, system identification, model validation and updating, etc. </p>
<p>There are numerous books that deal with PID control and, given the above, it should be clear that it is impossible to provide an exhaustive list of references. I have not used the books you mention and although their table of contents looks decent their applicability will depend on your particular application. Some alternative references are mentioned below.</p>
<p>To get a quick overview of what a PID controller actually does the <a href="https://en.m.wikipedia.org/wiki/PID_controller?wprov=sfti1" rel="nofollow noreferrer">Wikipedia page on this subject</a> is actually quite decent. However, I would not consider this sufficient information for any serious implementation.</p>
<p>For SISO control systems I’ve used both “Feedback Control of Dynamic Systems” by Franklin and “Feedback Systems: An Introduction for Scientists and Engineers” by Astrom and Murray. The former is somewhat outdated but covers the basics well and also includes some material on dynamic modeling. The latter is more up to date, covers some more advanced materials and is also available <a href="http://www.cds.caltech.edu/~murray/amwiki/index.php/Second_Edition" rel="nofollow noreferrer">online</a>.</p>
<p>For control of MIMO systems I personally like “Multivariable Feedback Control” by Skogestadt. However, if you decide to go into this direction, be prepared for the mathematical approach that is inherent to understanding the design and analysis of MIMO control systems.</p>
<p>Another good reference, which approaches PID control from an implementation point of view, is “Applied Control Theory for Embedded Systems” by Tim Wescott. I think this is SISO only but it does discuss PID tuning based on measured transfer functions and how to measure these transfer functions. If you are looking for rigorous mathematical detail this is not the book for you.</p>
| 23967 | Books for learning PID control? |
2018-10-01T10:10:30.490 | <p>Nowadays deep vacuum pumps are fairly expensive, so I'm wondering: how could people reach the deep vacuum required for <a href="https://en.wikipedia.org/wiki/Crookes_tube" rel="nofollow noreferrer">crookes tube</a> or <a href="https://en.wikipedia.org/wiki/Geissler_tube" rel="nofollow noreferrer">Geissler tubes</a> in the 19th century ? (I'd be interested in building one myself)</p>
| |vacuum|vacuum-pumps| | <p>Geissler invented the so-called "liquid piston" pump using mercury for the piston. The basic idea of the design is to entrap some of the air between two volumes of mercury in a pipe, and then force the mercury along the pipe to release the air into the atmosphere.</p>
<p>Geissler achieved 0.1 Torr with that design of pump in the 1850s. Sprengel invented an improved version of the same idea. The first models were capable of <span class="math-container">$10^{-2}$</span> Torr which was subsequently improved <span class="math-container">$10^{-5}$</span> Torr.</p>
<p><a href="https://cds.cern.ch/record/455984/files/p281.pdf" rel="nofollow noreferrer">https://cds.cern.ch/record/455984/files/p281.pdf</a> has more history, and diagrams (but unfortunately, not of very good quality).</p>
| 23990 | deep vacuum in the 19th century |
2018-10-01T22:30:10.377 | <p>I have this diagram:</p>
<p><a href="https://i.stack.imgur.com/ChVQm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ChVQm.png" alt="enter image description here"></a></p>
<p>And I have done these calculations:
<span class="math-container">$$\begin{align}
R_x&=-20\sin(30)+30\cos(35)+80\cos(45)=71.14\text{ lb} \\
R_y&=20\cos(30)+30\sin(35)-80\sin(45)=-22.04\text{ lb}
\end{align}$$</span></p>
<p>The magnitude of this resultant force is then:
<span class="math-container">$$|R|=\sqrt{71.14^2+(-22.04)^2}=74.48\text{ lb}$$</span></p>
<p>Here's the part I need help with. I can get a theta angle for where the resultant force's angle <em>is,</em> but I want to figure out how one would get this angle measured counterclockwise from the positive x-axis:
<span class="math-container">$$\theta=\tan^{-1}\left(\frac{-22.04}{71.14}\right)=-17.21°$$</span></p>
<p>How can I get this angle measured counterclockwise from the + x-axis? I don't get this and would really appreciate some help. This comes up a lot and I do not understand it.</p>
| |statics|mathematics| | <p>I haven't bothered checking your math since that doesn't seem relevant to your question.</p>
<p>Now, assuming that your answer of -17.21° is correct, that is in fact the counter-clockwise angle. It means that you need to go negative 17.21° counterclockwise, which is equivalent to rotating 17.21° clockwise.</p>
<p>Now, if what you actually mean is how to get the positive angle equivalent to -17.21°, well, that's just basic geometry: a circle has 360°. So to go from a negative angle to the equivalent positive angle, just add 360°.</p>
<p><span class="math-container">$$-17.21 + 360 = 342.79°$$</span></p>
| 23996 | Statics problem: Finding resultant force angle measured counterclockwise from positive x-axis |
2018-10-02T08:23:20.397 | <p>I am trying to perform static structural analysis of a turbine rotor which is rotating at a given angular speed, say 1000 rad/s. I also know the pressure distribution from CFD analysis. What are the boundary conditions for static structural analysis? I can use angular velocity as inertial load, pressure distribution from CFD as applied load. Do I have to restrain the translation/rotation of surface 1, shown in figure? If yes, why?</p>
<p><a href="https://i.stack.imgur.com/mZVYV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mZVYV.jpg" alt="enter image description here"></a></p>
| |finite-element-method|nastran| | <p>If your CFD analysis for the pressure loads was correct, the resultant of all the pressure loads should contain axial load and a torque on the complete assembly. If there is no torque, the turbine is useless! The axial load is not useful for the complete machine, but you can't avoid it. </p>
<p>Whether it is rotating or not, the turbine wheel can move in space as a rigid body. So you need to restrain it to prevent that, and create reaction forces that balance the applied axial load and torque.</p>
<p>The best way to do this is to restrain it at the location where it would be attached to the turbine shaft. You will then get the "correct" stress distribution in the disk. If you restrain it at some other arbitrary point or points, you will probably get an unrealistic high stress at those points.</p>
<p>(The turbine disk is probably <em>not</em> attached to the shaft at your "surface 1", unless it is a shrink fit on the shaft - for example there is no practical way that you could bolt it to the shaft through that surface.)</p>
<p>Actually, a much more efficient way to model this would be to model a sector of the turbine wheel containing just one blade, and derive the correct boundary conditions from the fact that the deflections of all the other sectors are identical (when measured in cylindrical polar coordinates) - but going into the details of how to do that is outside the scope of your actual question.</p>
| 23999 | Boundary Conditions fro Static Structural Analysis of a Turbine Rotor Using FEM |
2018-10-02T16:22:51.447 | <p>I wonder, whether an equation</p>
<p><span class="math-container">$$R = 2P + S$$</span></p>
<p>should give a geometrically sound design, where <span class="math-container">$R$</span> is the tooth count for the ring; <span class="math-container">$P$</span>, for the planets; and <span class="math-container">$S$</span>, for the sun.</p>
<p>I have tried to model a planetary gears with <span class="math-container">$R = 47$</span>, <span class="math-container">$P = 16$</span> and <span class="math-container">$S = 15$</span>. They do look well as for diameters, but the teeth do not match - it cannot be assembled.</p>
<p><a href="https://i.stack.imgur.com/hIUZf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hIUZf.png" alt="enter image description here"></a></p>
<p>Does the equation above guarantees, that the gears will fit each other? I now tend to think, that it does not take into account assemble-ability. Or perhaps I made a mistake somewhere.</p>
<p>Module is 0.9. Centre circle of planets was calculated as:</p>
<p><span class="math-container">$$D = \frac{m(Z_1 + Z_2)}{2} = \frac{0.9(15 + 16)}{2} = 13.95$$</span></p>
<p>I double checked everything and didn't find any mistake yet.</p>
<p><strong>EDIT</strong></p>
<p>The gears should be precise, I used "Involute Gears" macro in FreeCAD, which should provide a proven correct gears. The carriage is based on precise fully constrained sketch and the parts are assembled in fully constrained assembly. Thus, there should be no imprecisions here.</p>
<p>Where I would look for possible mistake (but didn't findany) is the position of planets against outer ring and sun (calculated as shown above), imperfections of outer ring, which is produced as subtract shape of involute gear with 47 teeth from a disk. And starting rotations of the gears for assembly.</p>
<p><a href="https://i.stack.imgur.com/2R2Sr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2R2Sr.png" alt="Assembly"></a>
<a href="https://i.stack.imgur.com/ZGsP0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZGsP0.png" alt="Carrier"></a></p>
<p>But what I'm thinking is, if it is even that easy. The outer ring has 47 teeth, which is a <strong>prime number</strong> (and selected on purpose, because correct gears should employ prime numbers and have to care about a common divider), meaning that one third (we have 3 planets) has <strong>15,6 teeth</strong>. And that means, that if you set the carrier sou that it points into the middle of the tooth with one of it's points, the other two will be offset from a center of a tooth by 6/10 of a tooth. While in the same time, the Sun has 15 teeth which are perfectly dividable by 3 and we can have it set with all three i.e. gaps pointing in the direction of all three carrier lines. This is confirmed by creating three lines constrained at 120° and rotating with one (other two follow), I cannot find any position at which would each line point at the same place on a tooth.</p>
<p>Now, I'm thinking, that this prohibits same condition for every planet. Meaning this is not possible to assemble (as is). But perhaps I'm missing something; the pointed simulator does depict same configuration assembled, but then it's only infographics which could cheat... </p>
<p><a href="https://i.stack.imgur.com/sTPDv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sTPDv.png" alt="Gear Simulator"></a></p>
<p>So issue is still open.</p>
<p><strong>EDIT 2</strong></p>
<p>It is also good to point out what should be possible to assemble then. Well, one configuration which should work is <span class="math-container">$S = 15$</span>, <span class="math-container">$P = 15$</span> and <span class="math-container">$R = 45$</span>, but then it is a <em>very bad</em> design due to <span class="math-container">$S = $</span>P (having the worst common divisor possible). Adding 3 to 5 I get 48, which won't fit the initial equation, but adding another 3 I get <span class="math-container">$S = 15$</span>, <span class="math-container">$P = 18$</span> and <span class="math-container">$R = 51$</span> which doesn't seem to be that bad. The greatest common divisor of a Sun and Planet is 3, meaning sun will turn 5 times before the same tooth will meet same tooth on the other gear.</p>
| |gears| | <p>This formula provides a close to equal spacing of planet gears.</p>
<p>where...</p>
<p>n = number of planet gears</p>
<p>s = number of teeth in the sun gear</p>
<p>r = number of teeth in the ring gear</p>
<p>d = the number of degrees between planet gears</p>
<p>d = 360 * floor((s + r) / n) / (s + r)</p>
<p>In the case where r = 47 and s = 15 and n = 3, d ~= 116.129...
There will be a slightly larger spacing between the first and last planet gear.</p>
<p>In the case where (s + r) / n is an integer, the spacing will be equal.</p>
| 24004 | Functionally Matching Gears (teeth count) in Planetary Gearbox |
2018-10-02T18:16:53.503 | <p>I'm not sure if the upper question is clear enough to understand but here's what I've got:</p>
<p>I've made a model of excavator in Adams View and got trajectory of it's motion (boom, stick, bucket) using trace marker.
This image shows pretty much what I've done {not mine, from Wikipedia}:
<a href="https://i.stack.imgur.com/ncdlm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ncdlm.png" alt="what I mean"></a></p>
<p>Now, my question is how I possibly can get the surface area of it's uhh.. trajectory in Adams View? Or the area that is circled in purple (dashed line) - I'm not sure what to call that.</p>
| |mechanical-engineering|modeling| | <p>I've used a feature in <a href="https://inkscape.org/en/" rel="nofollow noreferrer">Inkscape</a> that allows one to select a polygon or series of joined paths which will return the area enclosed.</p>
<p>I'd had to use this feature in the recent past and had found the answer in the <a href="https://graphicdesign.stackexchange.com/questions/16312/quantify-the-area-of-a-shape-in-inkscape">Graphics StackExchange.</a></p>
<p><a href="https://i.stack.imgur.com/7OGQ0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7OGQ0.png" alt="area of polygon image 1"></a></p>
<p>In the first image, I've roughly traced the path with the Bezier tool, then selected Extensions, Visualize, Measure Path. In the drop down box for Measurement Type, I selected Area. Be sure to select the item you wish to measure. Click Apply.</p>
<p>The program works slowly, presents (Not Responding) on a regular basis, but works!</p>
<p><a href="https://i.stack.imgur.com/0yDsi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0yDsi.png" alt="area of polygon image 2"></a></p>
<p>In this case, my units are set to millimeters. It would be necessary to have a reference scale object in order to have an accurate representation.</p>
| 24006 | How to get the surface area of trajectory [MSC Adams]? |
2018-10-03T05:24:05.110 | <p>Is understanding tensors a prerequisite for understanding plastic deformation (metal forming) equations, plane stress, plane strain conditions, etc..? If so which book should I use to understand tensors?. Because on the internet people suggest that a good resource to learn tensors are the physics books that deal with relativity, etc. So I am not sure whether that perception will be suitable for studying the stress tensors.</p>
| |structural-analysis|deformation|metal-folding| | <p>In a mechanical engineering curriculum at university, you would probably take a series of 3 classes. First would be called "Strengths of Materials" (or maybe "mechanics of materials"). This would be a sophomore level class. No tensors, just linear algebra (matrices). The second class would be "theory of elasticity". This is going to involve tensors. Third class will be "Continuum mechanics". This will be very heavy on tensors. After that, there would probably be a specialized class on plasticity. Depending on your comfort level with the math and your background, pick a text book on one of these subjects.</p>
| 24013 | Understanding plastic deformation |
2018-10-04T11:16:18.493 | <p>Sometimes we need a strong threaded joint in some part that is produced using molding. Say, this part itself is plastic. Then we design the part with a hole and insert there a.. "threaded bushing"? I don't know the proper name for this element in English. In Russian it is "закладная втулка" (insertable bushing/cylinder/etc - втулка/vtulka is a catch-all term for all things cylindrical that are used in an "axial manner" - something is inserted into them, or they are inserted somewhere, or both). I'm looking for a proper term in English.</p>
<p>I googled for <strong>закладная втулка</strong> and found this example relevant to my text: </p>
<blockquote>
<p><a href="https://i.stack.imgur.com/mHNOzm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mHNOzm.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>In the text that I'm translating into English, these brass insertable threaded elements in a plastic part produced using molding look like this: </p>
<blockquote>
<p><a href="https://i.stack.imgur.com/dX4Zj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dX4Zj.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>I found a DIN on <a href="http://www.fastbolt.ru/din16903" rel="nofollow noreferrer">one Russian page</a> describing "закладная втулка" and this DIN led me to English terms "<a href="https://www.linn-praezision.de/en/products/din-16903.html" rel="nofollow noreferrer">threaded bushing</a>", "threaded insert". Are these okay? </p>
| |terminology|molding| | <p>This is called "<em>threaded insert</em>" in English.</p>
<p>For more context, here are product page for threaded inserts for plastic: <a href="https://www.pemnet.com/fastening-products/si-inserts-for-plastic/" rel="nofollow noreferrer">this</a> and <a href="https://www.stanleyengineeredfastening.com/fasteners/inserts/heat-ultrasonic-inserts/dodge-ultrasert-iv-flanged" rel="nofollow noreferrer">this</a>. Both of are major manufacturers based in US.</p>
<p>Additional examples of plastic parts with threaded inserts: <a href="https://www.flickr.com/photos/46038819@N07/15722721032" rel="nofollow noreferrer">this</a> and <a href="https://www.flickr.com/photos/46038819@N07/22177155596" rel="nofollow noreferrer">this</a>.</p>
| 24030 | Looking for translation of "закладная втулка" (threaded bushing?) |
2018-10-04T15:50:24.237 | <p>Attempting this kinematics question but not really understanding if what I'm doing is right or completely wrong. Should I be starting by finding Vp? Any help appreciated. Not an assignment, just practicing.</p>
<p><a href="https://i.stack.imgur.com/YYfPY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YYfPY.png" alt="Question"></a>
<a href="https://i.stack.imgur.com/LhGeB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LhGeB.jpg" alt="Attempt"></a></p>
| |mechanical-engineering|dynamics| | <p>Let us work out the geometric relations first and then do the kinematics. I will follow a purely algorithmic approach to show the details. You can arrive at the same result from trigonometric relations.</p>
<p>Let us choose a fixed coordinate frame that has its origin at A, with the three unit vectors <span class="math-container">$\mathbf{E}_1 = (1, 0, 0)$</span>, <span class="math-container">$\mathbf{E}_2 = (0, 1, 0)$</span>, and <span class="math-container">$\mathbf{E}_3 = (0, 0, 1)$</span> where <span class="math-container">$\mathbf{E}_1$</span> is along AB, <span class="math-container">$\mathbf{E}_2$</span> is perpendicular to AB (in the plane ABP), and <span class="math-container">$\mathbf{E}_3$</span> is perpendicular to the plane ABP.</p>
<p>The position vectors of the three points are
<span class="math-container">$
\mathbf{x}_A, \mathbf{x}_B, \mathbf{x}_P \,.
$</span>
Then the direction vectors of interest are
<span class="math-container">$$
\mathbf{r}_{BA} = \mathbf{x}_B - \mathbf{x}_A ~,~~
\mathbf{r}_{PA} = \mathbf{x}_P - \mathbf{x}_A ~,~~
\mathbf{r}_{PB} = \mathbf{x}_P - \mathbf{x}_B \,.
$$</span>
From the figure, if <span class="math-container">$\theta_A, \theta_B$</span> are the angles made by PA and PB with AB, and <span class="math-container">$r_A, r_B$</span> are the lengths of PA and PB, we have
<span class="math-container">$$
\begin{align}
\mathbf{r}_{BA} &= b \mathbf{E}_1 \\
\mathbf{r}_{PA} &= -r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2 \\
\mathbf{r}_{PB} &= -r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2
\end{align}
$$</span>
Since <span class="math-container">$\mathbf{r}_{PB} = -\mathbf{r}_{BA} + \mathbf{r}_{PA}$</span>, we have
<span class="math-container">$$
-r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2 = -b \mathbf{E}_1 - r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2
$$</span>
or
<span class="math-container">$$
\begin{align}
-r_B \cos\theta_B & = -b - r_A \cos\theta_A \\
r_B \sin\theta_B & = r_A \sin\theta_A
\end{align}
$$</span>
In matrix form
<span class="math-container">$$
\begin{bmatrix}
\cos\theta_A & -\cos\theta_B \\
\sin\theta_A & -\sin\theta_B
\end{bmatrix}
\begin{bmatrix} r_A \\ r_B \end{bmatrix} =
\begin{bmatrix} - b \\ 0 \end{bmatrix}
$$</span>
The determinant of the <span class="math-container">$2 \times 2$</span> matrix is
<span class="math-container">$$
D := -\cos\theta_A \sin\theta_B + \sin\theta_A \cos\theta_B = \sin(\theta_A - \theta_B)
$$</span>
Therefore, inverting the matrix, we have
<span class="math-container">$$
\begin{bmatrix} r_A \\ r_B \end{bmatrix} = \frac{1}{\sin(\theta_A -\theta_B)}
\begin{bmatrix}
-\sin\theta_B & \cos\theta_B \\
-\sin\theta_A & \cos\theta_A
\end{bmatrix}\begin{bmatrix} - b \\ 0 \end{bmatrix}
$$</span>
or
<span class="math-container">$$
\boxed{
\begin{align}
r_A &= \frac{b\sin\theta_B}{\sin(\theta_A -\theta_B)} \\
r_B &= \frac{b\sin\theta_A}{\sin(\theta_A -\theta_B)}
\end{align}
}
$$</span>
Plugging in <span class="math-container">$b = 300$</span> mm, <span class="math-container">$\theta_A = 60^\circ$</span>, <span class="math-container">$\theta_B = 20^\circ$</span>, we have <span class="math-container">$r_A = 159.6$</span> mm and <span class="math-container">$r_B = 404.2$</span> mm.</p>
<p>Now that we know the geometry, we can do the kinematics. Choose a rotating frame that has origin
at point A, and the coordinate axes
<span class="math-container">$$
\begin{align}
\mathbf{e}_1 &= \sin\theta_A \mathbf{E}_1 + \cos\theta_A \mathbf{E}_2 \\
\mathbf{e}_2 &= -\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2 = \mathbf{r}_{PA}/||\mathbf{r}_{PA}||\\
\mathbf{e_3} &= \mathbf{E}_3.
\end{align}
$$</span>
or, inverting the relation,
<span class="math-container">$$
\begin{align}
\mathbf{E}_1 &= \sin\theta_A \mathbf{e}_1 - \cos\theta_A \mathbf{e}_2 \\
\mathbf{E}_2 &= \cos\theta_A \mathbf{e}_1 + \sin\theta_A \mathbf{e}_2 \\
\mathbf{E_3} &= \mathbf{e}_3.
\end{align}
$$</span></p>
<p>Then the position vector of P in that coordinate system is
<span class="math-container">$$
\mathbf{r}_{PA} = r_A \mathbf{e}_2
$$</span>
and the velocity of P relative to the rotating frame is
<span class="math-container">$$
\mathbf{v}_{P/R} = \dot{r}_A \mathbf{e}_2
= \dot{r}_A \left[-\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2\right]
$$</span>
The time derivative of <span class="math-container">$r_A$</span> is
<span class="math-container">$$
\dot{r}_A = \frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} \dot{\theta}_B +
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}
(\dot{\theta}_A - \dot{\theta}_B)
$$</span>
or, using <span class="math-container">$\omega_A = \dot{\theta}_A$</span> and <span class="math-container">$\omega_B = \dot{\theta}_B$</span>,
<span class="math-container">$$
\boxed{
\dot{r}_A = \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}
\omega_A +
\left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]
\omega_B
}
$$</span>
where <span class="math-container">$\boldsymbol{\Omega}_A = \omega_A \mathbf{E}_3 = \omega_A \mathbf{e}_3$</span> and <span class="math-container">$\boldsymbol{\Omega}_B = \omega_B \mathbf{E}_3 = \omega_B \mathbf{e}_3$</span> are the angular velocities at A and B of AP and BP, respectively. </p>
<p>Plugging in the values that we know, we have
<span class="math-container">$$
\dot{r}_A = -190.24 \omega_A + 628.81 \omega_B \,.
$$</span></p>
<p>At P, the velocity is
<span class="math-container">$$
\mathbf{v}_P = \boldsymbol{\Omega}_A \times \mathbf{r}_{PA} + \mathbf{v}_{P/R}
= \boldsymbol{\Omega}_B \times \mathbf{r}_{PB}
$$</span>
Therefore, we have
<span class="math-container">$$
\begin{align}
& (\omega_A \mathbf{e}_3) \times (r_A \mathbf{e}_2) + \\
& \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}
\omega_A +
\left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]
\omega_B\right]\mathbf{e}_2 \\
&= (\omega_B \mathbf{e}_3) \times (-r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2)
\end{align}
$$</span>
Now
<span class="math-container">$$
\begin{align}
\mathbf{e}_3 \times \mathbf{e}_1 &= \mathbf{e}_2 \\
\mathbf{e}_3 \times \mathbf{e}_2 &= -\mathbf{e}_1 \\
\mathbf{e}_3 \times \mathbf{E}_1 &= \sin\theta_A \mathbf{e}_3 \times \mathbf{e}_1 -
\cos\theta_A \mathbf{e}_3 \times \mathbf{e}_2
= \sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1\\
\mathbf{e}_3 \times \mathbf{E}_2 &= \cos\theta_A \mathbf{e}_3 \times \mathbf{e}_1 +
\sin\theta_A \mathbf{e}_3 \times \mathbf{e}_2
= \cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1
\end{align}
$$</span>
Therefore, we have
<span class="math-container">$$
\begin{align}
& -\omega_A r_A \mathbf{e}_1 + \\
& \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}
\omega_A +
\left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]
\omega_B\right]\mathbf{e}_2 \\
& = \omega_B r_B \left[-\cos\theta_B(\sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1)
+\sin\theta_B(\cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1)\right]\\
& = -\omega_B r_B \left[\cos(\theta_A-\theta_B) \mathbf{e}_1 + \sin(\theta_A-\theta_B) \mathbf{e}_2
\right]
\end{align}
$$</span>
Comparing the components along <span class="math-container">$\mathbf{e}_1$</span> and <span class="math-container">$\mathbf{e}_2$</span>, respectively, we have
<span class="math-container">$$
-\omega_A r_A = -\omega_B r_B \cos(\theta_A-\theta_B)
$$</span>
and
<span class="math-container">$$
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}
\omega_A +
\left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} -
\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right]
\omega_B = -\omega_B r_B \sin(\theta_A-\theta_B)
$$</span></p>
<p>Plugging in <span class="math-container">$\omega_A = 10$</span> and the previously computed values of <span class="math-container">$r_A$</span> and <span class="math-container">$r_B$</span> into the first equation above, we have <span class="math-container">$\omega_B =$</span> 5.1555 rad/s.</p>
<p>Therefore, the relative velocity of the slider block is
<span class="math-container">$$
\dot{r}_A = -190.24 \omega_A + 628.81 \omega_B = 1339.4~\text{mm/s} \,.
$$</span></p>
| 24033 | Two Dimensional Kinematics |
2018-10-05T13:58:30.133 | <p>I have a task to create a software that calculates how much evaporative coolers are needed to maintain a certain temperature in a given room. I need to create a simple form, where variables are next :</p>
<p>1 - Location (i can fetch dry and wet bulb temperatures) 2 - Desired Temperature 3 - Air changes per hour 4 - Area of the room (m2) 5 - Number of people in the room</p>
<p>I am working with AOLAN evaporative coolers, and i have all the specification from them for the devices used. I just can't wrap my head around the calculations. So my question(s) are rather simple : How do i calculate the amount of coolers needed to cool down an area with all the given parameters up there.</p>
<p>I would love if someone could give me guidelines into how to achieve this.</p>
| |cooling|ac|evaporation| | <h1>Foundations (Energy Balance)</h1>
<h2>For the Room</h2>
<p>Consider the system shown in the picture below. Air flows through a room volume (area and height) with a residence time. Heat is provided by people inside the room.</p>
<p><a href="https://i.stack.imgur.com/zHum0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zHum0.png" alt="picture of room"></a></p>
<p>The steady state energy balance equation becomes </p>
<p><span class="math-container">$$
\dot{m}_a\left( \tilde{H}_{a,out} - \tilde{H}_{a,in} \right) = \dot{q}_p \\
\left(\frac{A h}{t_{TO}}\right) \left(\frac{M_a p}{RT_{in}}\right) \tilde{C}_{p,a}\left(T_{out} - T_{in} \right) = N_p\hat{\dot{q}}_p
$$</span></p>
<p>This says, the heat generated causes an enthalpy change in the air flow (out - in). The enthalpy change of an ideal gas causes it to undergo a temperature change.</p>
<p>You have the room area. You also need its height. This gives volume. Divided by residence time gives volumetric flow. With ideal gas density you get mass flow. The specific heat and molar mass of air are known. The pressure of the inlet air is known. The output temperature of your AC unit is the inlet temperature to the room. You might also adjust the flow rate of air in to the room. This is the residence time. You know the number of people and the heat per person. The remaining factor is the outlet temperature. This is the desired room temperature.</p>
<p>Your goal is to have a set room temperature. As more people enter the room, you will need either to increase the flow rate (decrease residence time) at the same inlet air temperature or you will need to decrease the inlet air temperature to the room (the outlet temperature of your AC unit) at the same air flow rate.</p>
<h2>Sizing the AC Unit(s)</h2>
<p>The minimum cooling load on the AC units is <span class="math-container">$\dot{q}_p$</span>. You must remove at least the heat generated by the people in the room. The CoP of an AC is <span class="math-container">$CoP = \dot{q}/\dot{w}$</span>, where <span class="math-container">$\dot{w}$</span> is the work required. This is to first order the electrical power input multiplied by an efficiency factor for the heat pump. Each AC takes <span class="math-container">$W$</span> watts power at an efficiency of <span class="math-container">$\epsilon$</span>. The net result to establish the minimum number of units is</p>
<p><span class="math-container">$$
\dot{q}_p = N_u\ CoP\ \epsilon\ W = N_p \hat{\dot{q}}_p
$$</span></p>
<h1>Summary</h1>
<p>The approach might be as follows:</p>
<ul>
<li><p>Determine the minimum number of AC units to meet the cooling load based on the efficiency, CoP, and power load of a unit as well as the number of people in the room and their heat output.</p></li>
<li><p>Determine whether the AC units will meet the demand to maintain a desired room temperature based on the rated output temperature and air flow rate of the units.</p></li>
</ul>
| 24041 | Calculate amount of evaporative coolers needed to cool down an area |
2018-10-06T09:45:11.187 | <p>I have an Aluminium square cross sectional rod of <span class="math-container">$20mm$</span> X <span class="math-container">$20mm$</span>. Length 700mm.
At the end of this rod there is a mass of <span class="math-container">$660$</span> <span class="math-container">$g$</span> attached.
My question is what torque do i need to lift this mass up assuming my pivot point is at the other extreme of the rod? And how do i consider the mass of the rod in the calculations!</p>
| |mechanical-engineering|motors|torque| | <p>The angular momentum of a rod or beam rotating about one end is its angular momentum about the center of mass of the rod plus mass times (l/2)^2, in this case, parallel axis momentum theory.
<span class="math-container">$$L = m\frac{l^2}{12} +m\times (l/2)^2 = ml^2/3 $$</span>
Where m is the mass and l the length of the aluminum rod, 700mm.</p>
<p>So the torque needed is </p>
<p><span class="math-container">$T= 1/3(ml^2) + 660l$</span></p>
| 24054 | Understanding Motor Torque calculations |
2018-10-07T02:59:01.597 | <p>Simple bending equation is derived assuming pure bending(when a couple is applied onto a beam with no shear stress).We tend to use the above derived equation when moment is caused due to shear stress?How far is this correct?</p>
| |mechanical-engineering|applied-mechanics| | <p>We only can ignore the shear when the effects it has on shape and displacement are small enough to not be a concern. In general, we <em>can't</em> ignore them. If we ignore shear, then we are saying that the beam ends (and all cross sections) are constrained to remain perpendicular to the beam's deflected surfaces. Real shear stresses don't let that happen. But for a long beam relative to the thickness, the shear strain can't develop very quickly along the length and the perpendicular assumption is mostly true. But for short, thick beams, shear strain has a large impact on the deformed shape.</p>
<p>One of the first comprehensive approaches to handle this was <a href="https://en.wikipedia.org/wiki/Timoshenko_beam_theory" rel="nofollow noreferrer">Timoshenko beam theory</a>.</p>
| 24064 | Why do we use simple bending equation when bending moment is caused due to shear force? |
2018-10-07T11:09:59.537 | <p>How do we get the moment when there's a distributed load?</p>
<p>For example, the picture below has 9[(60)(18)]. I get that 60 is the distributed load and 18 is the total length of the load, my question is how did it get the 9?</p>
<p><a href="https://i.stack.imgur.com/pTXhS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pTXhS.jpg" alt="enter image description here"></a></p>
| |moments| | <p>The resultant of distributed loads always acts on the centroid of the distributed load geometry, here the distributed load is uniform so its centroid lies half the way. If the distributed load varies linearly from zero at one end to a maximum value at the other end, then its centroid would lie at <span class="math-container">$\frac{1}{3} L$</span> from the "max load" end and <span class="math-container">$\frac{2}{3}L$</span> from the "zero load" end, with <span class="math-container">$L$</span> the side length.</p>
| 24069 | How to solve for moment with uniform distributed load |
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