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2017-03-18T14:00:22.883 | <p>I am considering a regenerative braking system implemented with a AC permeant magnet motor with an efficiently of ~96%. It is easy to determine via energy equations the kinetic energy of the car and therefore how much energy is possible at max efficiently but regenerative systems also rely on the brakes of the car to slow it down to a stop. Is there a particular rule or reference to how much conventional braking is required for a given scenario? I wonder if it is possible to determine how much the motor would actually decelerate the car and then work out energies from there. Could anyone offer some advice?</p>
| |mechanical-engineering|electrical-engineering|automotive-engineering| | <p>An important energy storage method that is being given very little consideration is gas compression. The kinetic energy of the vehicle can drive an air pump or compressor that would compress air into a tank specially designed for pressures and heat of compressed air. Later the energy stored as compressed air can be released through a regulated nozzle to convert to electrical or mechanical energy.</p>
<p>The air pump would be connected mechanically directly to the drive train or the wheels or axle to have the braking effect. The storage process is not efficient as much heat is generated with the compression. However, the energy can be stored in the compression tank for a very long time with little loss. The capacity for energy storage is only limited by the volume of the storage tank and the maximum design pressure.</p>
<p>Special consideration has to be made for the lubricating oil used for the air pump/compressor. It has to be non-combustible other wise it will ignite with the oxygen in the heated compressed air. This would rule out most hydrocarbon based oils.</p>
| 14304 | How to determine how much energy recovered from regenerative braking |
2017-03-18T18:53:00.597 | <p>I am not sure where else to post this, so here goes.</p>
<p>I have a velocity profile ($v_{\phi}$) for a radial distance ($r$). From this velocity profile I am trying calculate the shear rate ($y$) which can be calculated by ( Eq 3 in Image):
$$
y = \frac{\partial v_{\phi}}{ \partial r} - \frac{v_{\phi}}{r} = r \frac{\partial}{\partial r} \left( \frac{v_{\phi}}{ r}\right )
$$
Please find the image of the equation from the papers attached,</p>
<p><a href="https://i.stack.imgur.com/swMBh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/swMBh.png" alt="enter image description here"></a></p>
<p>Screenshot of the equation for the paper for your kind consideration.</p>
<p>I have implemented the equation as such,</p>
<pre><code>for i = 1: size(v_phi,1)
y (i) = ( r(i) ) * (( ( v_phi (i) / r(i) ) - ( v_phi (i+1) / r(i+1) ) ) / ( r(i+1) - r(i)) );
y1 (i) = ( ( ( v_phi(i) / r(i) ) - ( v_phi(i+1) / r(i+1) ) ) * ( v_phi (i) / r(i) ) );
end
</code></pre>
<p>However the y and y1 values calculated are very different. I am just wondering, whether I am doing something wrong in the implementation and looking for help if possible.</p>
<p>Many thanks.</p>
<pre><code>Data >>>>
v_phi = [36.99 37.00 35.78 31.43 26.91 23.32 19.46 16.97 14.49 12.12 10.42 8.52 7.25 6.06 5.11 3.92 3.30 2.54 2.11 1.42 1.46 1.46 0.97 0.63 0.63 0.51 0.40 0.91 0.24 0.36 0.63 0.71 0.45 0.26 0.74 ];
r = [3.59 3.76 3.94 4.12 4.29 4.47 4.65 4.82 5.00 5.18 5.35 5.53 5.71 5.88 6.06 6.24 6.41 6.59 6.76 6.94 7.12 7.29 7.47 7.65 7.82 8.00 8.18 8.35 8.53 8.71 8.88 9.06 9.24 9.41 9.50 ];
</code></pre>
| |matlab|feedback-loop| | <p>Biswajit Banerjee already indicated that your code has bugs, which is why your results are so far off. I will explain in more detail where these bugs are:</p>
<p><code>v_phi</code> is a row vector, so <code>size(v_phi, 1)</code> always returns <code>1</code> because there is only one row. You want <code>size(v_phi, 2)</code> or better yet, <code>numel(v_phi)</code>.</p>
<p>Because you're using forward differences to compute the derivatives, you can only iterate up to the <em>second-to-last</em> element in each of the input vectors, so the loop declaration should start like:</p>
<pre><code>for i = 1:size(v_phi, 2) - 1
</code></pre>
<p>After that, you mix up the order of subtraction at least once or twice. That is, where you should be doing <code>y(i+1) - y(i)</code>, you instead do <code>y(i) - y(i+1)</code>, for example.</p>
<p>Also, as a matter of style, loops can often be replaced by vector operations in Matlab. For example, the following one-liner computes $\frac{dv_{\phi}}{dr}$ over the whole data set:</p>
<pre><code>dvdr = diff(v_phi) ./ diff(r);
</code></pre>
<p>Your equation can actually be solved without using any loops.</p>
| 14306 | Calculating a derivative in a loop? |
2017-03-20T09:44:42.580 | <p>I want to mount two patio heaters on my terrace in front of a window wall. I cannot attach anything directly to the window wall so I am going to run an aluminium box beam between two brick columns. The beam will be 5000mm long. The beam dimensions are 80mm x 40mm with a 4mm thick wall, 13kg in weight. The heaters weigh 6.5kg each and are 1000mm long. They will be 1000mm from each end and therefore 1000 from each other. I need to drill 8 x 8mm holes for mounting brackets (2 brackets per heater, 2 holes per bracket).</p>
<p>I am sure the beam will deflect a little - any serious risk of collapse?
<a href="https://i.stack.imgur.com/xNptP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xNptP.jpg" alt="Patio Heater Attachment Points"></a>
<a href="https://i.stack.imgur.com/AHzvy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AHzvy.jpg" alt="Proposed Positioning"></a></p>
| |mechanical-engineering|building-design|construction-management| | <p><a href="https://i.stack.imgur.com/2rUzK.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2rUzK.gif" alt="Load equations"></a></p>
<p>Using the 5th equation from the bottom the displacement is: </p>
<p>disp = Wa^3b^3/3EIL^3</p>
<p>We can superimpose it for each of your loads. Essentially your beam is loaded with 3.25kg (~33N) at 4 points (1,2,3,4m along).</p>
<p>I.e. disp(total) = disp(1st load) + disp(2nd load) ... + disp(4rth load)</p>
<p>For all the loads W = 33N </p>
<p>for the first and fourth a = 1 and b = 4
for the second and third a = 2 and b = 3</p>
<p>E = 69 x 10^9 (Elastic Modulus of Aluminium)</p>
<p>I = 7.11x10^-7 (Second moment of Area)</p>
<p>Ends up being: </p>
<p>disp = 0.001m</p>
<p>So at the middle x = 2.5</p>
<p>So displacement = 0.0025m </p>
<p>Sounds like it should be fine. </p>
| 14325 | Aluminium Beam - is it fit for purpose? Can it easily support this load? |
2017-03-20T14:06:22.553 | <p>I accidentally inserted a mesh edit item and tried moving a node, but now can't remove it. The context menu doesn't have any helpful options and I don't see any relating tools in toolbars. How to remove it?</p>
| |ansys|ansys-workbench|meshing| | <p>I figured it out. First right click your mesh in the project tree and then click "Clear Generated Data". Then, right click, say your "Node Move", there should be a "Delete", click "Delete", then right click your "Mesh Edit" and there should also be a "Delete", click that.</p>
| 14329 | How to remove a 'mesh edit' item in ANSYS Workbench? |
2017-03-21T00:27:27.837 | <p>I'm doing some vibration tests on planetary gear sets and a lot of the information I´ve read mentions a lot about mode shapes and multiplicities of the natural frequencies. I understand how one can derive the natural frequencies from the equations of motion of the gears and even the eigenvalues, but I don't understand what the multiplicity of the natural frequencies mean. Does it refer to the harmonics of each natural frecuency? If for example there's a fault it will probably be a multiple of the 1st harmonic? Also, does the eigenvalues have to do with the modal modes?</p>
| |mechanical-engineering|gears|vibration|modal-analysis|frequency-response| | <p>"Multiplicity" usually refers to several different modes with the same natural frequency. For example, ignoring any fault conditions and tolerances, if there are several identical planet gears, each one will have the same mode shapes and frequencies.</p>
<p>If you are measuring the vibration response, you may not be able to distinguish these modes - you will measure one "mode" which is a combination of all of them. </p>
<p>If there is a fault involving just one of the nominally identical parts, then you may be able to identify different modes for that part.</p>
<p>This situation can occur in any structure which is symmetrical - it's not restricted to planetary gearboxes.</p>
| 14337 | What does multiplicity of a multiple natural frequency mean? |
2017-03-21T11:33:21.197 | <p>I am trying to recreate in Microsoft Visio a diagram of a photovoltaic power plant that would look like the one below.</p>
<p>But I have some issues for doing so, in particular : </p>
<ul>
<li>I don't know how to create a "grid" symbol like the one that is at the bottom right of this picture</li>
<li>I don't know how to make an "abbreviated line" with tildes like the ones you can see Inside the PV modules at the extreme left of the picture.</li>
</ul>
<p>Can someone give me some help with this?</p>
<p><a href="https://i.stack.imgur.com/fhtft.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fhtft.png" alt="Diagram of a photovoltaic power plant"></a></p>
| |photovoltaics|diagram|electrical| | <p>Eventually I finished my flowchart with Visio. Here's how it looks like:</p>
<p><a href="https://i.stack.imgur.com/ceIfv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ceIfv.png" alt="Plant scheme"></a></p>
<ul>
<li>The PV modules were a template already present in Visio.</li>
<li>For the inverters, the square box with a diagonal line was an existing template. I added the signs for "DC" and "AC" by finding them in vectorial on Google images.</li>
<li>The transformers were an existing template.</li>
<li>I used the option "pattern" to have a grid-pattern for the rectangle that represents the grid.</li>
<li>The "waves" that means "abbreviation of a line" were made by hand.</li>
<li>The battery sign was an existing template.</li>
<li>All the rest is only basic geometrical shapes, lines, arrows.</li>
</ul>
| 14341 | Electrical flowchart in Microsoft Visio : issues |
2017-03-21T13:04:48.787 | <p>I want to make a "gearbox" that will transform input rotational movement to an output rotational movement with the same speed but in the different direction (clockwise / counterclockwise) on the same axis. How can this be done?</p>
<p><a href="https://i.stack.imgur.com/F4FBr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/F4FBr.png" alt="mystery gearbox"></a></p>
| |mechanical-engineering|gears| | <p>Option 1 - just reverse direction without affecting the speed: Bevel gears in differential gear setup, with the carrier (the axle on which the intermediate gears are located) fixed to the chassis. </p>
<p><a href="https://i.stack.imgur.com/cdcnH.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/cdcnH.jpg" alt="enter image description here"></a></p>
<p>Option 2: reverse direction and change speed by a large amount. Planetary/epicyclic gear, fixed carrier (the green part in the image below) fixed to the chassis with the second shaft connecting to the ring gear (pink). Changing gear ratio between the ring gear and the sun gear (yellow) changes the speed and torque on top of reversing the direction. </p>
<p><a href="https://i.stack.imgur.com/2syd1.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/2syd1.jpg" alt="enter image description here"></a></p>
<p>Option 3: Use two universal joints to offset the shafts relative to each other, so that they are no longer in line, then use a standard gear to reverse the direction, possibly changing speed a little in the process.</p>
<p><a href="https://i.stack.imgur.com/1PJVo.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/1PJVo.gif" alt="enter image description here"></a></p>
| 14343 | How to reverse the direction of rotational movement on the same axis? |
2017-03-21T13:49:07.710 | <p>How do i calculate the load of a glass facade and is it placed uniformly on the edge of the slab or should i place a point load on each edge of the panel of the facade?
And Can anyone give me a link on an example on how to calculate the load and how to place it.
Thank you in advance,</p>
| |structures|glass| | <p>The value of the distributed load is trivial to determine: it is equal to the linear weight of the glass. The linear weight is itself equal to the product of the thickness, height and specific weight of the glass (depends on the type of glass, usually between 24-28 kN/m<sup>3</sup>).</p>
<p>How this load should be placed, however, depends on how the glass is mounted on the structure. Does it rest on a support structure directly on the slab? In this case, apply it as a uniformly distributed load. If, however, it is actually supported by vertical structures which themselves rest on1 the slab, then the load should be concentrated at the supports. In this case, the value of the concentrated force at each support is equal to the distributed load times half of the span between supports (the other half is handled by the other support).</p>
| 14346 | Glass Facade Load on edge of slab? |
2017-03-21T16:11:53.370 | <p>I have used the Equivalent Lateral Load Method from the IBC code to distribute forces and moments on each storey (level) but what I need to know is how to distribute the moment on each shear wall.</p>
<p>This maybe a stupid question, but I'm just a student. Do I distribute the force & moment on each storey to each shear wall individually or do I distribute shear & </p>
| |design|seismic| | <p>When you said moment in your question, I got confused. You need to say how to distribute shear. You are not distributing any moment. You are distributing shear forces to shear walls, which would cause "torsional moment" or "torsion" with respect to the geometric center of the floor. Even if you were talking about moment, you are talking about torsional moment here, which you aren't. You are talking about distributing shear forces.
Other than that yes the shear walls will resist the shear with respect to their stiffnesses. Remember, stiffness attracts force. </p>
| 14352 | How to distribute moment from Lateral Load Method (Static Analysis IBC for example) to Each shear wall individually? |
2017-03-21T20:29:30.993 | <p>I understand that coils have properties over a single wire in producing a magnetic field. But I'm a hobbyist not an expert. I know electricity was discovered from a wire carrying a charge when presented in a magnetic field. So I wonder if that is why we use wire and wire coils to pass through the magnetic field. (Because that's what was used at the time of discovery.) </p>
<p>Solid copper presents a massive magnetic field, as seen when dropping a strong magnet through the inside of a length of copper pipe. </p>
<p>Has anyone tried using solid copper plates to generate magnetic fields instead of coiled wire? Would that work or not? Why?</p>
<p><a href="https://i.stack.imgur.com/f70E3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f70E3.png" alt="Coiled Wire vs. Copper Plate electrical generation"></a></p>
| |generator|electrical| | <p>They are also used in transformators for changing an ac voltage to another value by inducing a current without any metal parts touching each other, just by the coils’ magnetic field. I<em>E=E</em>I</p>
<p>İn your question it would work but the coils would generate more magnetic field because they have turns.</p>
| 14355 | Why are coils used to generate magnetic fields |
2017-03-21T21:55:15.587 | <p>I understand the Transmission Error is the displacement error of the gear shafts, but if measured how would the signal over time show and why? if its a sinusoid would that mean that only a few times per cycle the displacement occurs or is it always present?</p>
| |mechanical-engineering|gears|dynamics| | <p>Transmission error would usually be at the tooth mesh frequency. i.e. if you have N teeth on the gear, then you would see N cycles of oscillation over one revolution. It is probably not a pure sinusoid, depending on your type of gear, but it is periodic. If you just google "dynamic transmission error", you can find many papers with plots of time histories. e.g. <a href="https://core.ac.uk/download/pdf/337497.pdf" rel="nofollow noreferrer">https://core.ac.uk/download/pdf/337497.pdf</a></p>
| 14357 | What is the Gear Transmission Error and How would a Signal Analysis look? |
2017-03-22T02:15:31.103 | <p>I notice from <a href="https://engineering.stackexchange.com/q/10789/3353">this question</a> that the moments at joint connection are somehow "conserved", or being "transferred" from one element to the others. In other words, there are no "missing moment".</p>
<p><a href="https://i.stack.imgur.com/np5DW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/np5DW.png" alt="enter image description here"></a></p>
<p>Note that at joint for left image:</p>
<p>$$2.767\text{ kNm}=(2.37+0.397)\text{ kNm}$$</p>
<p>And for bending moment diagram that I examined so far (using structural software), this seems to be <em>always</em> the case.</p>
<p>So I wonder whether this relationship is true for all kinds of joint connection? Or is there a deeper physics principle behind it (such as <a href="https://en.wikipedia.org/wiki/Conservation_of_energy" rel="nofollow noreferrer">the conservation of energy</a>)? How can we derive it from the first physics principle, if it there is a first principle?</p>
| |structural-engineering|structural-analysis|moments| | <p>Well, such an equilibrium (be it of moment, or shear or axial force) is necessary for any static system, and can be trivially demonstrated with Newton's second law. For forces, that is the classic $F=ma$ (force equals mass times acceleration), while for moments it is equal to $M = I\alpha$, where $M$ is moment (or torque), $I$ is the <a href="https://en.wikipedia.org/wiki/Moment_of_inertia" rel="noreferrer">rotational inertia</a> (also known as the moment of inertia with dimensions $ML^2$, as opposed to the moment of inertia we commonly discuss in structural engineering, with dimensions $L^4$, which is usually called the <a href="https://en.wikipedia.org/wiki/Second_moment_of_area" rel="noreferrer">second moment of area</a>), and $\alpha$ is the rotational acceleration.</p>
<p>If you could calculate the force at a given point as equal to anything other than zero, that would mean that the entire system would accelerate in the net force's direction according to $F=ma$ (if $F\neq0$, then $a\neq0$). Equivalently, if the total moment around a point is non-null, then the entire system will have a rigid-body rotational acceleration (if $M\neq0$, then $\alpha\neq0$). This happens all the time in the real world, of course, but we call those mechanisms, not structures.</p>
| 14359 | What are the principles (if there are any) behind the conservation of bending moment in frame analysis? |
2017-03-22T02:40:14.760 | <p>If I have a planetary gear set consisting of one sun and N planets, what does it mean if it says that typical vibration modes of 6 natural frequencies always have multiplicity m = 1 for different N, except for the zero natural frequency? How would this multiplicities look in a frequency domain plot?</p>
| |mechanical-engineering|gears|dynamics|vibration| | <p>I cannot find a freely available PDF of it, but if you have access to ASME journals, then this paper lays out the basics of planetary gear natural frequencies (at least in the most basic case, there are many later papers with different variations in assumptions).</p>
<p>Lin J, Parker RG. Analytical Characterization of the Unique Properties of Planetary Gear Free Vibration. ASME. J. Vib. Acoust. 1999;121(3):316-321. doi:10.1115/1.2893982.</p>
<p>As outlined in this paper, planetary gears have three types of modes: rotational modes, translational modes, and planet modes. Rotational modes always have multiplicity 1, translational modes always have multiplicity 2, and planet modes have multiplicity N-3. What the multiplicity means in that there is more than one vibration mode (eigenvector) with the same natural frequency (eigenvalues). </p>
<p>For translational modes, here is how to think about the multiplicity: there could be one mode where everything is translating vertical (e.g. just up down... in reality it might not be exactly up/down, but let's just assume it is for now). There is a completely separate mode where everything is translating laterally (just left right). Now, let's call the stiffness in the vertical direction $k_v$ and the stiffness in the horizontal direction $k_h$. The mass is the same for both cases, let's call it $m$. So the vertical natural frequency is $\sqrt{k_v/m}$ and the horizontal natural frequency is $\sqrt{k_h/m}$. But because planetary gears are axisymmetric, $k_v = k_h$. So the two natural frequencies are identical. They are two completely separate modes, that just happen to have the same frequency. That is what is meant by multiplicity 2. If you were to make the bearings or some other part of the structure slightly stiffer in one direction, such that $k_v>k_h$, then it's no longer symmetric and you would have two different frequencies each of multiplicity 1.</p>
| 14360 | What does natural frequencies with multiplicities m=1,2 or n mean? |
2017-03-22T19:31:06.240 | <p>If I have a transfer slab in the basement level how to find the tributary area for the columns because the columns in the upper floors have different location and tributary area than that of the basements.</p>
<p>Please if possible to show an example </p>
<p><a href="https://i.stack.imgur.com/NbDlS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NbDlS.png" alt="enter image description here"></a></p>
| |structural-engineering|columns| | <p>As I understand your question, you are asking how to determine how the loads from the upper floors are divided between the lower columns. For instance, column C14 is going to transmit its loads to C12, C17, C18 and the unnamed wall near the core (assuming its load-bearing). How much of the load will go to each of the columns?</p>
<p>Curiously enough, the answer to this question depends on the chosen foundation for your structure and whether they allow rotations (such as shallow foundations) or don't (such as multi-pile foundations).</p>
<hr>
<h2>"Pinned" Foundations</h2>
<p>In the case of foundations which allow rotations, then you can roughly approximate the answer in this case (where the load is relatively centered between the four surrounding columns) by dividing the load equally between the four surrounding columns.</p>
<p>This is a rough estimate, because the load actually isn't centered between the four columns. Also, the core wall's vertical stiffness is actually much larger than that of the other columns, so it will certainly pull in more of the load.</p>
<p>For a simply supported beam with a concentrated load $P$ anywhere on the span (at a distance $a$ from the left support and $b$ from the right), the reactions are equal to</p>
<p>$$\begin{align}
R_{left} &= P\dfrac{b}{a+b} \\
R_{right} &= P\dfrac{a}{a+b}
\end{align}$$</p>
<p>Perhaps one could use something similar to better estimate the distribution of the load given that it isn't centered in the slab, but I don't know what that equation would be. It also would do nothing to solve the problem (which I believe might be significant) of the core wall's significantly larger vertical stiffness.</p>
<p>To solve that problem, you'd need to use a FEM program to model the structure.</p>
<hr>
<h2>"Fixed" Foundations</h2>
<p>Tributary areas aren't really adequate in this case.</p>
<p>The orientations of the supporting columns aren't the same. This means that their rotational stiffnesses around the x and y axes will be different: column C12 will be far more rigid than C17 and C18 for rotations around the y-axis, while C17 and C18 will be more rigid around the x-axis. And this doesn't even mention the fact that the core structure will be insanely stiff.</p>
<p>For instance, let's look at how the load would be distributed between C17 and C18 if it were perfectly centered between them. Each of the columns can be represented by a rotational spring which represents their rotational stiffness around the relevant axis:</p>
<p><a href="https://i.stack.imgur.com/nmexH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nmexH.png" alt="enter image description here"></a></p>
<p>The values themselves don't matter, but it is clear that the equal stiffness of the columns around the relevant axis means that the load is equally distributed between them (assuming it is centered, which it actually isn't).</p>
<p>Looking at the distribution between C18 and C12, however, the former will be far stiffer (I'd guess some 10 times stiffer):</p>
<p><a href="https://i.stack.imgur.com/nVccE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nVccE.png" alt="enter image description here"></a></p>
<p>C18 therefore absorbs some 50% more of the load than C12, even though the load is centered between them. The same concept applies for C17 and the core, though in this case, the core would be approximately infinitely stiff:</p>
<p><a href="https://i.stack.imgur.com/GIREq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GIREq.png" alt="enter image description here"></a></p>
<p>The core'll therefore absorb more of the load than C17.</p>
<p>Considering just these local effects already shows that the distribution is non-trivial. However it isn't adequate to consider the isolated structure around each of the upper columns; one must consider global effects. The edge columns will always pull less of a reaction than the central ones due to continuity effects.</p>
<p>For instance, let's look at how the load from C14 is distributed between C18 and C12, but remembering that there are also C4 and C9 in the distance.</p>
<p><a href="https://i.stack.imgur.com/5i6Tp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5i6Tp.png" alt="enter image description here"></a></p>
<p>The continuity effects of C4 and C9 mean that C12 absorbs 30% more load than we thought by looking at only C18 and C12 in isolation, while C12's reaction is slightly reduced.</p>
<p>All of this is to show that this is very non-trivial; the models I've shown here are obscenely simplified, since they look at the load distribution along a line, without taking into consideration the influence of nearby structural elements. For instance, the continuity effect from C18-C12-C9-C4 is probably even stronger, since the nearby presence of the massively stiff core will probably make the span from C12-C9 even stiffer.</p>
<p>The only reasonable recommendation in my opinion is to use a FEM program to model the entire structure, including the slab, at least floor-by-floor. Any "manual" method will be highly simplified and inadequate by modern standards.</p>
<hr>
<h2>Why the difference?</h2>
<p>The easiest way to demonstrate this difference is with another model, now actually showing the support columns and the "foundations":</p>
<p><a href="https://i.stack.imgur.com/sHtPY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sHtPY.png" alt="enter image description here"></a></p>
<p>In the model to the left, the pinned foundations mean that the load is equally distributed, even though the columns have different stiffnesses (but equal area). To the right, however, the fixed foundations cause the load distribution to be uneven, with more load being absorbed by the stiffer column.</p>
<p>This can also be seen in the difference of the deflection of the models:</p>
<p><a href="https://i.stack.imgur.com/AryvE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AryvE.png" alt="enter image description here"></a></p>
<p>The pinned supports allow the columns to rotate freely, eliminating any potential rotational stiffness. With the fixed supports, however, the columns do resist the beam's attempt to rotate the nodes.</p>
| 14367 | How to find tributary area for a column when I have a transfer slab? |
2017-03-23T23:27:57.097 | <p>Here's the excercise: A tensile test is carried out on a mild steel specimen of gauge length 40mm and cross-sectional area 100mm<sup>2</sup>. The results obtained for the specimen up to it's yield point are given below:</p>
<pre><code>Load (kN) | 0 | 8 | 19 | 29 | 36
Extension(mm) | 0 | 0.015 | 0.038 | 0.060 | 0.072
</code></pre>
<p>I need to determine the gradient of the load / extension graph.
When the data is plotted (in the book) on a graph it gives a straight line therefore the gradient should be any load (of the given values) divided by the
corresponding extension gradient = load/(extension/1000) except e.g. the first one (8kN) gives 533*10<sup>6</sup> while the third one (19kN) gives 500*10<sup>6</sup>.</p>
<p>Why and which one do I use for the calculations?</p>
| |mechanical-engineering| | <p>Theoretically, the elastic range of a material is linear. But as Yogi Berra said, "In theory there is no difference between theory and practice. In practice there is." This data is obtained from a physical experiment, so errors are natural.</p>
<p>Perhaps the piece or the strain gauge was incorrectly placed. Or, even more likely (given the data), you were dealing with a specimen which simply wasn't perfect (none ever is). Slight microimperfections meant that even in the elastic regime you might have a result which isn't perfectly linear.</p>
<p>So you need to find a best estimate of what the "theoretical" Young's modulus is.</p>
<p>Here are the results for each of the points, take your pick:</p>
<pre><code>disp (mm) | F (kN) | F/d (kN/mm)
----------+---------+------------
0 | 0 | -
0.015 | 8 | 533.3
0.038 | 19 | 500
0.060 | 29 | 483.3
0.072 | 36 | 500
</code></pre>
<p>It is visually clear that the result is hovering around 500.</p>
<p>I'm unsure of how serious the statistical analysis needs to be.</p>
<ul>
<li>You can simply say 500 looks reasonable.</li>
<li>You can take an average, which gives you 504</li>
<li>You can do a least-squares analysis, which gives you 490.6, but with a line crossing the y-intercept at 0.25, as opposed to zero, as would be expected.</li>
<li>You can force the least-squares to cross through zero, which gives you 495.</li>
</ul>
<p>They all give you the following results:</p>
<p><a href="https://i.stack.imgur.com/rTzzS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rTzzS.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/kpix6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kpix6.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/pqEFS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pqEFS.png" alt="enter image description here"></a></p>
| 14386 | How to determine gradient of load/extension graph? |
2017-03-23T23:33:01.417 | <p>Could the rotor be shaped like a fan inside the stator of an AC induction motor. And could the stator act as a duct.</p>
<p>Would this lead to a high airflow:materials used.</p>
<p>Thanks </p>
| |electrical-engineering|motors|ac| | <p>It could, yes. But the tradeoff would be worse performance. The strength magnetic field is a function of distance from the coil, so increasing the volume by turning your rotor into a fan shape would reduce the efficiency.</p>
<p><a href="https://en.wikipedia.org/wiki/Internal_fan-cooled_electric_motor" rel="nofollow noreferrer">Typically, this is dealt with by close mounting a fan, sometimes inside the motor case</a>. The motor stays compact and efficient while the cooling is provided by a fan specially built for the purpose.</p>
| 14387 | AC induction motor |
2017-03-24T05:59:31.990 | <p>I was wondering why there isn't a 9/20" diameter bolt and then realized the bolt sizes start at what ever size, the next size is + 39/1250 of an inch. Is there any reason for this? Is there any place to find a 9/20" hardware?</p>
| |mechanical-engineering| | <p>0.03125 = 1/32 (exactly). The USA (and the UK before metrication) had a long tradition of subdividing basic measurements by factors of 2. For example the standard sizes of drills are usually state in 32ths or 64ths of an inch. This principle was used for other measures as well as lengths - e.g. gallons, quarts, pints - and sometimes for larger measures as well as fractions - e.g. pecks (4 gallons) and bushels (8 gallons), or weight units like hundredweights (8 stones).</p>
<p>The decimal conversions are more convenient to use with modern measuring devices than fractions, but anything less than 0.1 thousandth of an inch is not of much practical significance.</p>
<p>The list of standard sizes has small enough increments that there would be no particular use for an "odd size" like 9/20" = 0.45 in, when there is already a standard size 7/16" = 0.4375 in. Most likely, this would just lead to confusion and accidental use of the incorrect size bolts, which could cause problems if the wrong replacements parts, or tools like wrenches, were under or over size.</p>
<p>Of course in some very specialized applications, bolts and other fasteners might be designed to <em>any</em> size to meet the required specification, if the cost of the parts is not a significant issue. Such parts are likely to be made from special materials, as well as in special sizes.</p>
<p>Standard metric bolt sizes are more straightforward, because the bolt diameters and head sizes are all integer numbers of millimeters.</p>
| 14395 | why does sae increment by .0312 |
2017-03-24T15:56:29.570 | <p>I have a tube with a circular cross-section, this tube protrudes from the ground vertically for a section, then arcs over to one side. At the end of the tube, a flat plate is mounted to the end of it. The arc is of a constant radius but is not necessarily a 90 degree bend. (i.e. the angle of the flat plate to the ground plane is not necessarily 90 degrees). How would you go about calculating, by hand, how much load the flat plate could hold before the tube bends beyond it's yield point. </p>
<p>Assuming all geometries are known and the material properties are known and constant. </p>
<p>I've been working on a solution using a combination of formulas from Roarks book. But, I'd like to see how others would approach this problem. "Solve using FEA" is not an answer i'm looking for, but rather a reasonably accurate hand calculation approach.</p>
<p><strong>Edit:</strong>
-dimensions and loading direction added</p>
<p>Approximate tube dimensions:</p>
<p>Vertical straight section: 1m</p>
<p>'Bent' section arc radius: 125mm</p>
<p>Flat end, 10 degrees to floor</p>
<p>Tube OD - 25mm</p>
<p>Tube ID - 23mm</p>
<p>Loading direction will be in the negative z-axis i.e. gravitational loading only.</p>
<p><a href="https://i.stack.imgur.com/GIgrr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GIgrr.png" alt="Tube Sketch"></a></p>
| |structural-analysis|stresses| | <p>I'd check for a maximum overturning moment which will cause yield stress at the base of the member where it is attached to a base plate or enters the foundation as a flag pole, of course with any required safety. Then It is a simple arithmetic $M=\sigma\cdot s$</p>
<p>and</p>
<p>$$ s=\pi\frac{d_o^4-d_i^4}{32\cdot d_o^2},$$</p>
<p>in which $d_i$ is the inner diameter and $d_o$ is the outer diameter.</p>
<p>A more detailed diagram as to the attachment of the post to the base plate or the configuration of the plate at the top would allow more detailed observation and checking for local buckling, shear, normal loading.</p>
<p>I am adding this clarification as to mode of failure.</p>
<p>The pipe will start to rotate counter clockwise and undergo elastic defleection which will add to the moment arm, but as we add the vertical loading it will yield from the base which already has the greatest bending moment of approximately m = P.16 cm.</p>
| 14404 | Circular tube stress loading |
2017-03-24T19:43:45.777 | <p>I want to build a professional CNC router for my garage.</p>
<p>I want to use the residential split phase configuration for power to my VFD and industrial power supplies.</p>
<p>I cannot find any straight answer regarding connecting 240V 1ph (L/N/E input) apparatus to 120V/120V (L1/L2/E).</p>
<p>Can I feed the other line to neutral terminal?</p>
<p>Step-up transformer is out of question since it's high current draw.</p>
| |electrical-engineering|power-electronics|power|electrical| | <p>I have bugged people who claim to know about this, and am no expert but here are my 2 cents:</p>
<p>Residential 110 voltage is taken from the 240 supply. The transformer is middle tapped to make 2 110 legs. Look at it as a single long coil that is 240 volts from top to bottom but if you connect at the middle you can get 2 legs at 110.</p>
<p>For alternating current that changes polarity with a sine wave, you are getting half the wave in each leg. If you look at it on a graph, I think you get the positive half in one leg and the negative in the other.</p>
<p>The problem with 3 phase is you need the third phase to not be in sync with the others. So the only way to get the frequency right is to generate your third leg. You can run a motor on 2 phase that generates the third phase by coupling that motor with a generator. I think it might be possible to use a variable frequency device instead of the motor-generator couple but I now less about VFD.</p>
| 14406 | Split phase on industrial devices |
2017-03-25T12:52:38.153 | <p>I have to determinate the power (Watts) for an oven to reach the Temperature Reference. And I have been given only the function to apply power to the oven and read the temperature.</p>
<p>I've reading the Ziegler-Nichols method. And the first step is to obtain the tangent in the inflection point, and then check the values of:</p>
<ol>
<li>L : The point where the tangent intersect the x axis</li>
<li>T : The point where the tangent intersect the reference line</li>
</ol>
<p>But I do not understand one thing, the range of the power I can apply power from 0 to 10,000(W).</p>
<p>Do I have to calc this tangent for a static random Power? But which value, because it will be the same in "production mode", when the power is vary because of the controller?</p>
<p><a href="https://i.stack.imgur.com/xKaXH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xKaXH.png" alt="Calculation of L & T for Ziegler-Nichols PID method"></a></p>
| |electrical-engineering|embedded-systems|pid-control| | <p>There are two answers here. First answer is the theoretical one. It doesn't matter what static power you apply. You could apply any number. The reason is that the plot of y(t) that you end up drawing will be normalized to an input of 1. </p>
<p>In other words, compare slides 12 and 13 in this presentation</p>
<p><a href="http://www.hh.se/download/18.7502f6fd13ccb4fa8cb2f6/1360676928481/PID.pdf" rel="nofollow noreferrer">PID control - Simple tuning methods</a></p>
<p>If you apply an input power of 1, then the output plot you get gives you the value $K_p$ (you called it just K in your plot above). But if you input a power of $\Delta$u, then the output plot gives you the value $\Delta$u$K_p$. So to find the true value of $K_p$ you will divide by whatever power you input. </p>
<p>Second answer is the practical one. In reality you will never measure your parameters perfectly. There will always be some error or noise in the measurement. If you use a step of 1W on a 10,000W oven, you aren't going to get a very good measurement. So higher power will always be a better measurement. But, on the other hand, if you are unfamiliar with this system, you might not want to try full power right away. You might want to have a reasonable idea of the open loop system behavior before going to full power. If it were me, I would probably start by doing a 1000W (i.e. 10% full scale) step, making sure that I understood how the system was behaving and that I liked the answer. Then I would work up to full power in a few steps. I would take the final Z-N parameters from a run at full power.</p>
| 14413 | Tuning PID Controller |
2017-03-25T15:04:54.873 | <p>If this is the wrong place for this question, please migrate it to someplace where it's more suitable.</p>
<p>So one of my old teachers from my previous school sent me a contraption to try and figure out what it was and how it worked. Pictures <a href="https://i.stack.imgur.com/bPpP2.jpg" rel="noreferrer">here</a>.</p>
<p>I've played around with it for a while. I tested to see if it did something when spinning really, really quickly by attaching a motor, tried taking it apart numerous ways, and plenty of things in-between.</p>
<p>As far as I've gathered, here's what I think it's doing. The propeller drives a shaft that is attached to something that converts the circular motion into linear motion. At this point the glass is frosted or there's some kind of glue on the inside that makes it very hard to see clearly what's going on. The top is connected to the bottle via two pipes that end in a hole in the side of the bottle. The cap itself has a hole in the top to which the top thing is screwed to, possibly also leaving space for water to go through -- it's unclear. The purple part is sticky on one side, suggesting that the whole thing sticks to the wall or maybe to something that moves like a car.</p>
<p>Or maybe the propeller isn't what does the driving at all. Maybe it's filled with water and stuck to a wall or something facing downwards, and gravity would cause the water to cause the whole thing to leap to life! But, I can't find any way of filling it up -- there's no existing holes as far as I can tell. I may be able to take off the whole red cap but that's made difficult with the yellow pipes.</p>
<p>So I'm stuck. Does anyone have any idea about what this mysterious thing is?</p>
<p><strong>UPDATE</strong>:</p>
<p>The propellers have arrows on them indicating the direction they should turn. This suggests that the propeller is indeed the driver and not the drivee.</p>
| |mechanical-engineering|fluid-mechanics|motors|mechanisms|linear-motors| | <p>It looks like it might be a air-powered propeller. You pump up the bottle with air, and the high pressure is used to run a air motor that spins the propeller. This might have been part of a toy airplane, with the wings and tail now gone.</p>
| 14416 | What is this contraption? |
2017-03-25T23:53:05.430 | <p>If you've been watching Bones series, there is an episode in which Bones and Hodgins were trapped beneath the ground, and they managed to send an SMS by wiring the phone to the car honk. Is this possible? What is the exact mechanism used?</p>
<p>I'd appreciate it if someone points out if the question is in the wrong site. </p>
<p><a href="https://vimeo.com/210057258" rel="nofollow noreferrer">https://vimeo.com/210057258</a></p>
| |mechanical-engineering|electrical-engineering|telecommunication| | <p>From the clip you gave (which I glossed over the first time), I think they needed to power the phone so they wired into the circuit that draws power from the car battery for the car horn. That circuit doesn't require the car to be on in any way so that would be the reason to use that circuit.</p>
| 14422 | Sending SMS using car horn |
2017-03-26T18:03:03.327 | <p>There is a steel plate with length of 26.5cm (10.4 inch) and width of 4cm (1.57 inch). the steel plate is attached on one end, at the other end there is a load of 30kg, what thickness should be the plate so it will not bend ? </p>
<p><a href="https://i.stack.imgur.com/8Gh6z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8Gh6z.png" alt="enter image description here"></a></p>
| |steel|stresses| | <p>It will always bend as steel has finite stiffness. You need to decide how much deflection is acceptable. </p>
<p><a href="http://www.engineersedge.com/beam_calc_menu.shtml" rel="nofollow noreferrer">This page</a> lists the bending equations for a variety of loading situations. This case is a <a href="http://www.engineersedge.com/beam_bending/beam_bending9.htm" rel="nofollow noreferrer">cantilever with load at one end</a></p>
<p>These equations do rely on some assumptions but should be fine for this situation, be aware though that they apply to static loads only, if the load moves the stresses involved can be significantly higher. </p>
<p>These equations also assume that stresses are within the elastic limit of the material, if the yield stress is exceeded then the beam will be permanently deformed and may fail completely. </p>
<p>As a rough guide I would say that you are looking at about 8mm thick or more </p>
<p>It is also worth nothing that a flat plate is a very inefficient way to support a cantilevered load and unless there is a strong reason why the support needs to be very slender you are much better off either adding an arch type support web below the beam or using a hollow section. For example a 25x25mm square tube with wall thickness of 2mm would support 30 kg no trouble with very small deflection at 27cm cantilever length and weigh much less than an equivalent 4cm wide strip. </p>
<p>you also need to consider the fact that there will be a large torsion load where the plate meets the thing it is attached to which is a common point of failure. </p>
| 14436 | Steel plate thickness for load |
2017-03-27T14:44:03.420 | <p><a href="https://i.stack.imgur.com/vYP02.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vYP02.png" alt="enter image description here"></a></p>
<p>If i first assume the member(far left) to not be zero and use method of joint at both points of constraint, with both constraints having unknown vertical reaction, it cannot be solved.</p>
<p>Load and lengths are known, horizontal reaction can be found.</p>
| |statics| | <p>In this situations, it's worth remembering <a href="https://en.wikipedia.org/wiki/Hooke%27s_law#For_linear_springs" rel="nofollow noreferrer">Hooke's Law</a>, which tells us that internal forces and deformations are related.</p>
<p>A beam can deform either because a load is directly applied to it or because one (or both) of its extremities is being deformed by a load on another beam somewhere.</p>
<p>If, however, a beam has no load applied directly to it and both of its ends are constrained (can't move in any direction), then it will never deform in any way shape or form. And if it doesn't deform, then we know by Hooke's Law that it won't suffer any internal forces.</p>
<p>That is the case for your truss: there are no loads applied to the member between both supports, and since it is between two supports, it will never deform and therefore never suffer any axial forces. And since this is a truss, any rotations suffered by the surrounding members won't be transmitted to the member either, so neither will there be any bending moment (or shear).</p>
<hr>
<p>For the more visually inclined, imagine the deformed configuration of the structure: all the other members will be deformed as the load pulls the cantilever down. This member, however, won't be deformed at all. And therefore, neither will it suffer any internal forces.</p>
<p><a href="https://i.stack.imgur.com/MriYd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MriYd.png" alt="enter image description here"></a></p>
| 14454 | Why is the truss member between two fixed supports zero? |
2017-03-27T15:31:17.183 | <p><a href="https://i.stack.imgur.com/dnPjL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dnPjL.png" alt="enter image description here"></a></p>
<p>I have to answer the following question which is (image shown above)
"what is the force to keep the rectangular gate closed? in N/M" </p>
<p>This is all the information I have to solve the question. I understand there will be use of moments. I just require help understanding the question.</p>
<p><a href="https://i.stack.imgur.com/GZ4Sw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GZ4Sw.png" alt="enter image description here"></a></p>
| |fluid-mechanics|moments| | <p><img src="https://i.stack.imgur.com/f3Eap.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/8jCQT.jpg" alt="enter image description here"></p>
<p>Hope you understand!
If any questions, feel free to ask.</p>
| 14456 | Fluid mechanics- Force required to keep the gate closed |
2017-03-27T16:37:07.063 | <p>(I posted this in the Sound-Design section but got advised to post here instead).</p>
<p>I just purchased a 3d printer and I am working on a product with a box that has a hole in the bottom and the top. Inside this box I have a few electronics that has a pretty loud sound. What I now look to do is to kill that sound as much as possible so it is not possible to hear it as loud. </p>
<p>At home I have a few acoustic panels looking like this and as they are very lightweight it seems like a very good solution for this design: <a href="https://www.thomann.de/se/eq_acoustics_classic_wedge_foam_tiles_grey.htm?sid=dbe7318446923736e3fdee98e5ba1174" rel="nofollow noreferrer">EQ Acoustics Classic Wedge 30 Tile grey</a> </p>
<p>If I cover the interior of the box on all the walls would this make the sound not leave the box as much which would result in a lower sound? I am very novice when it comes to this area so any help/tips is very appreciated! </p>
<p><a href="https://i.stack.imgur.com/ZJD3x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZJD3x.png" alt="enter image description here"></a></p>
<p>Width and length is around 30cm.</p>
| |civil-engineering|sound-isolation| | <p>If you can remove or reduce the openings you will have the most success in reducing sound. Failing that, if you need airflow, baffles will help to reduce the sound (at a cost of reducing the airflow). Once you've addressed the design of the holes, adding mass will be a good way to reduce sound transmitted through the material. For a good example, look at the vinyl sound deadening liners used in the automotive industry. In fact, you could try applying these to the inside of this product. They are much thinner than acoustic panels, albeit heavier. One popular aftermarket brand name is Dynamat.</p>
| 14459 | How can I kill the sound as much as possible in a rounded box? Acoustic panels/other option? |
2017-03-28T06:55:07.067 | <p>I have a wall with the bottom end of it fixed to the ground. I want to derive from first principles the frequencies and mode shapes of the structure.</p>
<p>I know that I will need to start with the following formula:</p>
<p>$K-\omega^2M=0$ </p>
<p>And I know how to do that for a cantilever column.</p>
<p>But wall is a different thing than column because it is an area element instead of a line element. And I don't know how to generate the correct expression for $K$ and $M$ for a continuum area element. </p>
<p>How to proceed with the analytical solution for wall eigenvalue analysis? </p>
<p>Note: I am not looking for a FEM solution. </p>
| |structural-analysis|modal-analysis|eigenvalue-analysis| | <p>Your "wall" would be referred as a "plate" in the technical literature. The eigenvalue problem for a plate is significantly more complicated than a beam, but similar ideas. </p>
<p>I would suggest that you look for a book on "vibrations of continuous systems", for example: <a href="http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471771716.html" rel="nofollow noreferrer">http://www.wiley.com/WileyCDA/WileyTitle/productCd-0471771716.html</a> </p>
<p>There are also specific books dedicated entirely to plates, e.g. <a href="https://www.crcpress.com/Vibrations-of-Shells-and-Plates-Third-Edition/Soedel/p/book/9780824756291" rel="nofollow noreferrer">https://www.crcpress.com/Vibrations-of-Shells-and-Plates-Third-Edition/Soedel/p/book/9780824756291</a> however they may be hard to understand without sufficient background. </p>
| 14467 | Analytical solution for wall eigenvalue analysis |
2017-03-28T16:02:13.370 | <p>This may be a dumb question, but is a solid block 100% of the time stronger than a 'strategically hollow' part. Strategically hollow meaning, the internal structure is design in such a way to support the loading direction?</p>
<p>My thinking is originating from both 3D printing internal supports but also from bio-mimicry designs. Looking at the internal structure of bones, they aren't solid, would they definitely always be stronger if they were solid? Intuitively I would say 'obviously' however could an internal structure is designed in such a way that the surface energy, and load paths create a stronger structure than a single solid block?</p>
<p>If there is a way to design the internal structure, how would you go about designing/analyzing this, if not is there a way to prove that it's not?</p>
| |mechanical-engineering|structural-engineering| | <p>Seems like you might not be considering the importance of the <strong>strength to weight ratio</strong>.</p>
<p>A solid bar will be stronger, but it will also weigh more. This additional weight will put extra stress on other components, and also have it's own internal stresses associated with supporting itself (imagine a long solid cantilever beam vs. a hollow one).</p>
<p>Often times engineers will use hollow members or designs with webbing. The general purpose of the designs is to put material where it needs to be for strength, but try not to overdo it, or else then you have weight that makes it harder to move (or causes more internal stress between the members).</p>
<p>Often this involves making things hollow, especially since things like bending and torsion get the most resistance from the furthest out points. You maximize your material where you would see the most stresses (think of I beams for bending and hollow poles for twisting).</p>
<p>Nature has this down to a pretty good science. Bones and such have strength where they need them, but there are enough cavities and such to fit other vital parts of the anatomy instead of being full of bone without as much structural benefit.</p>
| 14475 | Solid versus strategically hollow |
2017-03-29T12:51:28.420 | <p>I am currently making software to interact with a conductive rubber cord. I am very interested in figuring out the percentage of a stretch of the cord, to determine how the resistance changes.</p>
<p><a href="https://i.stack.imgur.com/QIK3M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QIK3M.jpg" alt="enter image description here"></a></p>
<p>Example:</p>
<p>Stretch the cord 1 cm, see the difference in the resistance:</p>
<p>Stretch the cord 3 cm, see the difference in resistance:</p>
<p>Programmatically, I can tell this change. When I use my hands, I can see the resistance flow change as my pressure increases on stretching the cord. However, as my hands are probably not the best instrument for accurate stretching, I am in need of an alternate machine to do this for me.</p>
<p>Does a such device exist, which can hold one end of this material for me, and pull the other end 1 cm? 3 cm? etc.
Since I am a software engineer, and certainly lack in the field of mechanical engineering, is a good approach to take here to use motors and a raspberry pi to possibly create my own stretch machine?</p>
| |measurements|distance-measurement| | <p>Yup, </p>
<p>You could use something as simple as a bar clamp. Most hardware stores will carry some version. The type you'd need is the type that can be reversed to a spreader (95% of them can do this just double check). </p>
<p>With one of these you'd need something to connect your cord to it. But still fairly straight forward (more clamps for example.</p>
<p>This wont be super accurate but it'll work. </p>
<p>For something more accurate you're looking at something more like a load frame. Most engineering universities will have a few of these (mechanical or civil departments). Very easy to use, but they take some set up time (a half hour ish). Every university I've worked with is usually pretty open to doing external testing, though you'll be limited to their schedule. There are other material testing places that will have them but they charge oodles usually.</p>
| 14484 | Stretching material accurately with a machine |
2017-03-29T23:19:36.893 | <p>Can we bring engineering to the paper in paper mache? It's a mix of a binder and a reinforcing matrix. PVA is common for the binder and is strong. That's fine as I think it's strong enough and readily available. Can we do better for the traditional news paper? I think that glass /carbon fibres would be incompatible as PVA doesn't readily adhere. Is there another filler? I read somewhere that they used to make air plane drop tanks and boats from some sort of paper mache, demonstrating it's engineering potential.</p>
<p>I realise that this is more materials science than pure mechanical engineering, but there are commonalities and it's all we have on SE. I'm deliberately avoiding Arts & Crafts as I'm not looking for a child friendly, mix a long with mom type of material. I'm looking for the dirtiest, leanest meanest (probably cancer causing) reinforcement that's as strong as possible drawing on modern technology. The only stipulation is that it should be available to the domestic purchaser (so bags of nano tubes are out). Is there something stronger than newspaper?</p>
| |materials|composite| | <p>Absolutely. Paper mache at it's core is an early predecessor of carbon fiber. The paper is the fiber structure and the paste (not sure of the technical term) is the equivalent to the epoxy. </p>
<p>If you increase the 'strength' of the fibers you'll eventually reach a point that the paste is the weak point, beyond which increasing the strength of the paper won't do much. If you increase the strength of the paste to a better adhesive you'll then be able to increase the strength of the final part.</p>
| 14490 | Can we make engineering grade paper mache by improving on the paper? |
2017-03-30T09:07:10.117 | <p>I am asked to find the member force of AB and DE
<a href="https://i.stack.imgur.com/4qeDo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4qeDo.png" alt="fgfg"></a>
<a href="https://i.stack.imgur.com/Orhvk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Orhvk.png" alt="32323"></a>
I make an imaginary cut of the question and consider the upper part of the cut . Bur , i didnt get the same ans provided . However , when i consider the lower section of the cut , i get the same ans with the book . Anything wrong with my working ?
Taking moment about A = 20(4) +30cos45 (3) + DE(4) = 0 , DE = -35.9 upwards </p>
<p>btw , the vertical reaction at a is 51.3 kN upwards , at E , vertical reaction = 35.9 kN upwards , and 21.2kN to the left ...</p>
| |structural-engineering|structural-analysis| | <p>As noted in the comments to your question, you forgot to include the moment due to the internal force in BE.</p>
| 14493 | cut section method |
2017-03-30T16:04:47.427 | <p>Which dimensions of columns supporting flat slabs do not require a punching shear check (according to ACI code)?</p>
| |structural-engineering|civil-engineering|reinforced-concrete| | <p>As has been mentioned in the comments, there is no specific combination of dimensions that mean that punching shear is not a problem. An Engineer may make a judgement that a check isn't required based on experience, but a check would always be the safe option.</p>
<p>ACI (American Concrete Institute) even has a similar question in its Frequently Asked Questions: <a href="https://www.concrete.org/tools/frequentlyaskedquestions.aspx?faqid=876" rel="nofollow noreferrer">Punching shear check for shear wall foundations</a></p>
| 14508 | Punching Shear in Slab? |
2017-03-30T16:16:01.880 | <p>Here is the question, I am stuck on part B:</p>
<p><a href="https://i.stack.imgur.com/RAxBr.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RAxBr.png" alt="Question"></a></p>
<p>Basically, this is my thought process so far, as $x_a \rightarrow 0$, the Henry's law line becomes tangential to the actual partial pressure curve.
<a href="https://i.stack.imgur.com/tVoz0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/tVoz0.png" alt="Graphs"></a> Henry's law states that:<br>
$$
p_{\alpha} = Hx_{\alpha}
$$</p>
<p>Similarly, using the activity coefficient model:
$$
p_{\alpha} = \gamma_{\alpha}x_{\alpha}p^*_{\alpha}
$$</p>
<p>Where H is Henry's constant and $p^*_{\alpha}$ is the vapour pressure of the pure component. So at small $x_{\alpha}$ values one can assume:
$$
H = \gamma_{\alpha}p^*_{\alpha}
$$</p>
<p>From this I can get values for the $\gamma$ (I've been given values for pure component pressure). Then I can get values for $\ln \gamma_1, \ln \gamma_2$. My next step would be to substitute in values of $x_2$ and $x_1$ into the given correlations. So, for $\gamma_1$ to be correct, $x_1$ must be very small, assume it is zero therefore $x_2 \approx 1$, therefore:</p>
<p>$$
\ln \gamma_1 = \frac{A \times 1^2}{\bigg( (1-1)^2\frac{A}{B} + 1 \bigg)^2} \implies \ln \gamma_1 = A
$$</p>
<p>Clearly, this is wrong because if I run through the same process for $\ln \gamma_2$, I get a different A value. Can someone please advise, thanks.</p>
| |thermodynamics|chemical-engineering| | <p>There was an error in the question, the model in the question is the van Laar activity model, the coefficient for $\ln \gamma_2$ on the numerator should be B, in this case, it is easily solvable.</p>
| 14509 | Calculation of Activity Coefficient Model parameters |
2017-03-30T18:37:33.663 | <p>As I know, a thinner mat may be possible by providing a nominal amount of vertical reinforcement as compared to a mat without vertical reinforcement, the vertical reinforcement is used to resist punching shear at d/2 from columns or shear walls at foundation level.</p>
<p>Now my question is to what distance (or offset) from the column to stop placing these vertical reinforcement is it at a distance d/2 ?</p>
<p>Please if possible provide reference in ACI Code.</p>
<p><a href="https://i.stack.imgur.com/lz4bi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lz4bi.png" alt="critical section outside slab shear reinforcement"></a></p>
<p>What i need is the critical section outside slab shear reinforcement from which distance from the column ?</p>
| |structural-engineering|civil-engineering|geotechnical-engineering| | <p>For a more complete discussion see: <a href="https://www.concrete.org/store/productdetail.aspx?ItemID=421108" rel="nofollow noreferrer">ACI 421.1R-08 - Guide to Shear Reinforcement for Slabs</a>.</p>
<p>But in principle you need to continue to provide punching shear reinforcement until the shear capacity of the slab is sufficient to prevent punching shear at that perimeter.</p>
<p>From ACI 421.1R-08 this means:</p>
<p>You need to check at $d/2$ from the column face to determine if shear reinforcement is required. If it is then you need to continue to check perimeters at a distance of $\alpha d$ from the column face until you find that (in SI units)
$$\frac{\nu_u}{\phi} \leq \frac{0.17\lambda\sqrt{{f'}_c}}{2}$$</p>
<p>Then the outermost perimeter of shear reinforcement must be at least $\alpha d - \frac{d}{2}$ from the column face.</p>
<p>$\nu_u$ is maximum shear stress due to factored forces. $\phi$ is the strength reduction factor = 0.75. ${f'}_c$ is the specified concrete compressive strength. $\lambda$ is the modification factor for the reduced mechanical properties of lightweight concrete (=1 for normal weight concrete). $d$ is the effective depth of slab.</p>
| 14514 | To Which distance Punching Shear Reinforcement is placed in Mat Foundation? |
2017-03-31T07:34:07.857 | <p>In electrical domain, we have resistors, capacitors, and inductors as fundamental constitutive elements. Similarly, in mechanical domain, we have masses, springs and dampers.</p>
<p>In most of the literature, we can immediately see that they have the one-to-one relation, perse i.e., we can go from one form of representation to other, or even mix them up (mechatronics). However, the problem I see is, in the electrical domain, we have the inductor which is an energy storage element, representing an integrator, collectively speaking. In contrast, when we look it up in the mechanical domain, all we have are mass-spring-damper equivalent, and there is no explicit representation for an inductor.</p>
<p>So, my question is this, does there exists a fundamental equivalent for an inductor in mechanical domain i.e., a mechanical equivalent of an integrator with a physical meaning in itself. If so, how it is called?</p>
<p>I have tried a lot of literature to find one but failed in the quest. So, all your help and suggestions are welcome.</p>
| |mechanical-engineering|electrical-engineering| | <p>Consider an analogy where voltage and force are "efforts" and current and velocity are "flows". We can choose either the effort or flow as the input, and the other as the output.</p>
<p>A spring is a mechanical integrator in the sense that it integrates its velocity to generate a proportional force. (It integrates a flow input to generate an effort output). Alternatively, a mass can be considered a mechanical integrator in the sense that it integrates the applied force to generate a change in velocity. (It integrates an effort input to generate a flow output).</p>
<p>The analogy to the electrical domain should be clear. </p>
<p>If we consider "flow" as the input and "effort" as the output, then a capacitor integrates electrical flow (current) to generate an electrical effort (voltage), i.e., $e = \frac{1}{C}\int i\,dt$. Similarly, a spring integrates mechanical flow (velocity) to generate mechanical effort (force), i.e., $F = k\int v\,dt$.</p>
<p>Alternatively, if we consider the "effort" as the input and the "flow" as the output, then an inductor integrates electrical effort (voltage) to generate a change in electrical flow (current), i.e., $i = \frac{1}{L} \int e\,dt$. Similarly, a mass integrates mechanical effort (force) to generate a change in mechanical flow (velocity), i.e., $v = \frac{1}{m} \int F\,dt$. </p>
| 14517 | Does a Mechanical equivalent of an Integrator exist |
2017-03-31T17:43:05.230 | <p>I am trying to use an Arduino Uno R3 as a tachometer (It is going to measure the RPM of my motorcycle). Basically it only has to count how many times an input pin goes HIGH in one second.
The problem is pulseIn is too slow for it...
The pin will be on for up to 230 times/ second (14.000 RPM).
So that would be 1000/230 (approx. 4 ms per rotation, i believe).
What is the easiest way to do the job? I can't figure any solution yet...</p>
| |mechanical-engineering|electrical-engineering|embedded-systems| | <p>Similar to <a href="https://engineering.stackexchange.com/a/14535/110">this</a> response, I would suggest using an interrupt, but also start a timer, and then capture the time elapsed on the next interrupt. </p>
<p>Alternatively you could start a timer on a interrupt, and count the number of rising edges in for the duration of the timer. So of the timer was set to 1 sec, and there was x rising edges, the RPM can be now calculated. </p>
<p>In the automotive's are magnet and hall effect sensor is used to calculate RPM.</p>
<p>Good luck.</p>
| 14525 | How to measure signal period with Arduino? |
2017-04-02T17:41:49.210 | <p>"By mounting a gyro on a set of gimbal rings, the gyro is able to rotate freely in any direction. Even though the gimbal rings are tiled, twisted, or otherwise moved, the gyro remains in the plane in which it was originally spinning" </p>
<p>Now I've seen that statement in several different textbooks, but I'm having trouble understanding the terminology used. It's stating that the gyro is able to rotate freely in any direction, but then it also says that the gyro remains in the plane it's rotating. If it can rotate freely in any direction, then it isn't in it's original plane of rotation. So if anyone could simplify that, it'd be helpful. </p>
<p>Also the gimbals allow things to rotate around it while the gimbal doesn't rotate, but then wouldn't you need the gyro to move for precession to take effect? </p>
<p>For an example if I said that I applied a force with regards to precession taking place, would the gyro maintain it's plane of rotation but that force would take affect on the gimbal? </p>
<p>Thank you! </p>
| |instrumentation| | <p>The confusion arises in terminology. What instrumentation engineers call a <em>gyroscope</em> is actually an <em>integrating gyroscope</em> or <em>rate gyro</em>, and what they call a <em>gimbal</em> in <em>gimbal lock</em> isn't the gimbal used to mount the gyroscope itself - the gimbals are used to keep a <a href="http://inertialnavigations.blogspot.com/" rel="nofollow noreferrer">platform housing three gyroscopes level</a>. </p>
<p>A <a href="http://www.tpub.com/neets/book15/63a.htm" rel="nofollow noreferrer">standard gyroscope</a> will point in the same direction at all times, while a <a href="http://www.tpub.com/neets/book15/63e.htm" rel="nofollow noreferrer">rate gyroscope</a> is used for measuring angular velocity:</p>
<p><a href="https://i.stack.imgur.com/wzFNw.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wzFNw.gif" alt="small and large tilts on a rate gyro"></a></p>
<p>The rate gyro will deflect the springs in proportion to how fast the frame is spinning - the faster the spin, the larger the springs deflect, which via strain gauge or similar measurements engineering, can be turned into an electrical signal. Measure the change in rate to derive the angular acceleration. Have a computer integrate the rate over time to derive current angular position. </p>
<p>Note the pivoting gimbals for this device though - they are in the exact configuration of a gimbal lock. A rate gyro's housing can only measure changes in a single axis. If the frame housing the gimbal moves away from the axis, then the device wouldn't measure properly. So, three <em>rate gyroscopes</em> are required to measure angular movement in all three angles. These are mounted on a platform that maintains a fixed orientation in space, so the rate gyroscopes can function properly. <strong>In other words, motors specifically move the platform that the rate gyroscopes are mounted onto so it stays on the same plane.</strong> </p>
<p><a href="https://i.stack.imgur.com/SzjeV.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SzjeV.gif" alt="Gyroscopes on a platform."></a></p>
<p>When the rate gyroscope feeds a porportional signal to the instrumentation panel, it also feeds this signal to three motors. They turn the platform in response to the signal, keeping the platform level.</p>
| 14561 | How do gimbals affect gyros? |
2017-04-02T20:06:20.020 | <p>What is the simplest way if you have a shaft rotating clockwise for it to have an extension that rotates anti clockwise. Least loss of energy is best so i think a mechanical mechanism would be best.</p>
<p>I am thinking is there a way you could use the main shaft in a turbo fan engine to drive another fan in the opposing direction in front of the original fan.</p>
| |gears|mechanisms| | <p>Least loss of energy would be using one gear with adequate lubrication.</p>
<p>I prefer planetary gears because they last much longer and distribute the load better. Plus you can do it with a hollow shaft. The center shaft turns clockwise, the planet gears turn counterclockwise as does the outer shaft.</p>
<p>Here is a link that should show you a gif of the gear in action.</p>
<p><a href="https://giphy.com/gifs/yXQrz0IscQeWs" rel="nofollow noreferrer">https://giphy.com/gifs/yXQrz0IscQeWs</a></p>
| 14563 | Shaft to swap direction of rotation |
2017-04-03T02:31:04.823 | <p><a href="https://i.stack.imgur.com/OGull.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OGull.png" alt="Image showing Ackermann steering where all of the projected wheel axles meet at the center of rotation"></a></p>
<p>Source: <a href="https://en.wikipedia.org/wiki/Ackermann_steering_geometry" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ackermann_steering_geometry</a></p>
<p>The image above shows a vehicle with <a href="https://en.wikipedia.org/wiki/Ackermann_steering_geometry" rel="nofollow noreferrer">Ackermann steering</a>. The front wheels are connected by a mechanism so that they turn at different angles such that the projected axles of the wheels meet at the same point, the center of rotation.</p>
<p><a href="https://i.stack.imgur.com/KYvue.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KYvue.png" alt="Image showing non-Ackerman steering where the projected wheel axles do not all meet at the same location"></a></p>
<p>Source: <a href="https://www.google.com/patents/US6634109" rel="nofollow noreferrer">https://www.google.com/patents/US6634109</a></p>
<p>This image shows a case where are four wheels do not have the same center of rotation.</p>
<p>If the Ackerman condition is not satisfied as shown in the figure for a car taking a turn, then what would be the equivalent centre of rotation for the vehicle? </p>
| |automotive-engineering|dynamics| | <p>This is quite easy:</p>
<ul>
<li>Choose one of the wheels</li>
<li>build a second wheel so it would satisfy Ackerman condition</li>
<li>Get the First rotation point ( O<SUB>1</SUB> )</li>
<li>Do the same with the second wheel, and get a second rotation point ( O<SUB>1</SUB>' )</li>
</ul>
<p>The actual rotation point can be anything between these two points.</p>
<p>If you choose, for instance, the first point as rotation point, it means that the first wheel has no issue but the second wheel therefore has an additional damping to its movement.</p>
<p>To find the real rotation center you could compute including the damping on each wheel, but In a driving situation equal damping is quite unlikely, more over it can change in function of the directions. On top of this, this Damping changes in function of the object dynamics. So you may most likely end up with an irrelevant model.</p>
| 14572 | Ackerman Conundrum |
2017-04-03T18:24:11.433 | <p>The heat of a lightning bolt causes the surrounding air to rapidly expand causing thunder in the process.</p>
<p>A reciprocating combustion engine ignites an air fuel mix to use the rapid expansion created in the process to build up pressure and use that to move pistons in order to create rotating motion.</p>
<p>Would a reciprocating engine that uses only air heated by a very high powered electric arc work? </p>
<p>It surely would be very ineffective, hard to supply with enough power without cables and would probably melt the engine block, but could it function in principle?</p>
| |mechanical-engineering|automotive-engineering|electric-vehicles| | <p>Yes, but it would probably be a very poor engine.</p>
<p>As you noted in your question, there are huge practical issues with using the arc alone as your power source, compared to using it to ignite the fuel mixture in a conventional spark-ignition engine. But you are correct to observe that both the electric arc and the combustion process cause a rapid expansion inside the chamber, and that this rapid expansion could drive a piston.</p>
<p>What you propose is to convert electrical energy to thermal energy to kinetic energy and useful work (you want to move the car). Implicit in this is some form of energy storage, which we can assume to be chemical for the sake of comparison with conventional technologies. So, chemical -> electrical -> thermal -> mechanical. You have to go through at least three conversions of energy from one form to another in order to make this technology work.</p>
<p>Contrast that with electric motor technology, which stores energy in batteries (or fuel cells) and produces work relatively efficiently. Chemical -> electrical -> mechanical.</p>
<p>Contrast it also with the conventional internal combustion engine, which stores energy relatively densely in hydrocarbon fuels and produces relatively more work per refueling cycle. Chemical -> thermal -> mechanical.</p>
<p>Your proposed solution adds another conversion step to the electric motor solution. It seems to combine the worst characteristics of both established solutions, since you have the same storage challenges to work with as the electric car does. So I wouldn't call it a <em>practical</em> technology; but you might be able to produce a working prototype of this sort of engine, for use as a novelty or teaching aid.</p>
| 14587 | Could a reciprocating engine be powered by electric arc instead of combustion? |
2017-04-04T09:08:01.143 | <p>So today when I was taking a taxi and looked at the vehicles, I observed that the majority of the light vehicles (i.e. 16-person minibus/ 4-6 person cars) use 5 (or less frequently, 6) wheels studs for each wheel. </p>
<p>Intuitively, 6 evenly distributed components are more stable as it has 3 axes of symmetry, as compared to 5.</p>
<p>Is there any mechanical reason for that? </p>
| |automotive-engineering|wheels| | <p>To some extent, the number of bolt studs depends on the anticipated torque on the wheel. That's why small cars may have 4, larger cars 6, and semitrailer trucks have 11 or 13. </p>
<p>There's certainly a cosmetic contribution when considering sporty or luxury cars, but "form follows function" here as well. The designer might add a bolt or two for looks but would never under-design the fitting. </p>
| 14598 | Reason of Using exactly 5 Wheel Studs per Wheel |
2017-04-04T11:31:59.130 | <p>I'm having trouble drawing an isometric view of this shape:</p>
<p><a href="https://i.stack.imgur.com/UPdLj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UPdLj.png" alt="enter image description here"></a></p>
<p>I've only just started an engineering course and so far it's involved a lot of drawing which I usually can do but I can't seem to figure out how to 'place' this object to make an isometric view. Am I able to make the base of the object in my drawing whichever face I want?, as I feel perhaps if I rotate it so that the face with the hole is facing the ground might make it easier to draw, or do I have to use the 20x20 square as the base like it is in the current view.</p>
<p>This is also kind of confusing me as well in regards to drawing it in parallel oblique view but that's another issue I guess which I'll probably be able to figure out once I wrap my head around how to attempt the isometric view.</p>
<p>The other shapes I've drawn (see <a href="https://imgur.com/a/xsm2r" rel="nofollow noreferrer">example</a>) felt a lot easier to simply start and don't mess with my brain like this particular shape does. It isn't being marked but I still want to understand how to do it since I'll probably have to deal with these types of shapes later on anyway and would rather have the basics down now.</p>
<p>I'm sorry if my explanation doesn't make sense but I'm not really sure how to word my issue at the moment.</p>
<p>Any help would be greatly appreciated, thanks.</p>
| |mechanical-engineering|drawings| | <p>Don't think, look at the numbers! Drawing isometric, or any projective images for that matter is very straightforward. Just as long as you do not try to find a trick to make it easier. Just work out the distance numbers. </p>
<p>Start by sketching your primary axes. These are the directions you are going to measure against in all parallel projections.</p>
<p><a href="https://i.stack.imgur.com/ehqfv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ehqfv.png" alt="enter image description here"></a></p>
<p>Then start measuring lines along those directions according to your measurements. Do not try to eye them just measure one axis at a time. So for example from the bottom back corner to the top corner there is 40 units, to the beginning if the miter theres another 40 in the -30 degree direction (axis toward right)</p>
<p><a href="https://i.stack.imgur.com/2RTWX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2RTWX.png" alt="enter image description here"></a></p>
<p>Just continue mechanically form there to measure all known points. First one plane then from that plane and so on. Don't worry about the circle at first, just find its center and radius as if it were a box. Just faithfully measure units on your ruler (or whatever your drawing software if any is using)</p>
<p><a href="https://i.stack.imgur.com/lWVqZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lWVqZ.png" alt="enter image description here"></a></p>
<p>Don't worry about hiding lines, you can do that later.</p>
<p><a href="https://i.stack.imgur.com/0wokO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0wokO.png" alt="enter image description here"></a></p>
<p>Then once you have one or a few solids done. Hide some lines to make things clearer to see. If you missed any lines it will become apparent at this point.</p>
<p><a href="https://i.stack.imgur.com/Ob1Ip.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ob1Ip.png" alt="enter image description here"></a></p>
<p>Then block the next solid. Now you need to use some draftsman's assumptions in the circle, a draftsman would assume the circle is in middle of the shape at least vertically form that image.</p>
<p><a href="https://i.stack.imgur.com/LOvmI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LOvmI.png" alt="enter image description here"></a></p>
<p>And finally:</p>
<p><a href="https://i.stack.imgur.com/OuImW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OuImW.png" alt="enter image description here"></a></p>
| 14602 | How to draw isometric view of this shape? |
2017-04-04T12:41:40.323 | <p>I am currently building a box filled with electronics and I now look to isolate the sound a bit with the lightest and best material out there.</p>
<p>What I am looking for is a thin, lightweight and very soundproof material. I came across a material called Sortbothane and it seems like a very good fit.</p>
<p><a href="http://www.ebay.com/itm/Isolate-It-Sorbothane-Acoustic-Vibration-Damping-Film-40-Duro-0-04-x-4-x-/172548443510?hash=item282cb04d76:g:EqcAAOSwZVlXnC9J" rel="nofollow noreferrer">Sorbothane Acoustic & Vibration Damping Film 40 Duro</a> </p>
<p><code>0.04" of Sorbothane Will Absorb up to 4 dB</code></p>
<p>That number is pretty impressive I must say and the material seems to be reasonable light and thin as well.</p>
<p>So to my question, do you think Sortobothane is the way to go or is there any other materials out there with the same/better soundabsorbing?</p>
<p>I am also thinking to combine a few materials to make it very solid. For instance I am planning to use a small layer of silicone as well.</p>
| |sound-isolation| | <p>Isolate and absorb are two different things:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Vibration_isolation" rel="nofollow noreferrer">Vibration isolation</a> means let one side of the material vibrate but keep the vibrations from transmitting through.</li>
<li><a href="https://en.wikipedia.org/wiki/Absorption_(acoustics)" rel="nofollow noreferrer">Acoustic absorption</a> means to dissipate vibration energy into heat at the audible frequencies.</li>
</ul>
<p>When used in the proper geometric configuration, a thin layer of <a href="https://en.wikipedia.org/wiki/Sorbothane" rel="nofollow noreferrer">Sorbothane</a> could work well for either of these applications. Its just important to approach your problem with an intent. It will be best used at the interface between two masses, like the enclosure and the mounting location or the door of the enclosure and the enclosure, etc. Selecting this interface, and changing the amount of mass can direct toward isolating or absorbing the concerned frequency. This will take some math and/or some experimentation. Other means of attachment across this interface like bolts, screws, adhesives, etc will influence how well the absorbs/isolates behaves.</p>
<p>Lower-energy and higher frequency sounds that are already airborne (like a buzzing electronic component) may be better absorbed with <a href="https://en.wikipedia.org/wiki/Acoustic_foam" rel="nofollow noreferrer">open celled acoustic foam</a> inside the enclosure.</p>
| 14607 | What is the best thin/light material to kill/absorb sound? Sorbothane? |
2017-04-04T15:46:29.610 | <p>I want to make a shaft rotate in 21 discrete steps, from a motor that is turning smoothly. </p>
<p>I am thinking of some kind of simplified clock escapement type of mechanism. Or non circular gears. </p>
<p>I know it could be done by programming a stepper motor. But it will be easier and more fun for me to do it mechanically with 3D printed nylon parts. </p>
<p>This is for a low torque low speed application. About 1 rpm and less than 0.1 Nm torque. </p>
| |mechanical-engineering|gears| | <p><a href="https://en.wikipedia.org/wiki/Geneva_drive" rel="noreferrer">Geneva drive</a><br>
<a href="http://www.thingiverse.com/thing:192448" rel="noreferrer">3D Printed Geneva Drive</a></p>
<p><a href="https://i.stack.imgur.com/BXJqM.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/BXJqM.gif" alt="Geneva drive"></a>
<a href="https://i.stack.imgur.com/5RB9m.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/5RB9m.gif" alt="enter image description here"></a></p>
<p><br><br>
<a href="https://en.wikipedia.org/wiki/Anchor_escapement" rel="noreferrer">Anchor Escapement (Clock Mechanism)</a><br>
<a href="http://www.thingiverse.com/thing:1128317" rel="noreferrer">3D Printed Anchor Escapement</a></p>
<p><a href="https://i.stack.imgur.com/E8SQZ.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/E8SQZ.gif" alt="enter image description here"></a></p>
| 14614 | What mechanism can make a shaft rotate in steps |
2017-04-04T17:54:56.897 | <p>I just modeled a High-rise building of 43 stories with an area of 700 square meter, Using ETABS as a software to model the building I got the Static and Dynamic Base Shear as the image below, my Question is the following:</p>
<p>Is it Reasonable that the dynamic base shear (Specx and Specy) is bigger than the static base shear (Ex and Ey) ? How can i know if this result is logical or not ?
If the result is Correct i don't need to scale the dynamic base shear ?</p>
<p><a href="https://i.stack.imgur.com/Kw63A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kw63A.png" alt="ETABS Dynamic and Static Base Shear"></a></p>
| |structural-engineering|civil-engineering|structural-analysis|dynamics| | <p>It is desirable but as far as Software concern, It is not possible. Something went to wrong with "Response Spectrum Function Defination" I.e. Define period 0 - 0.036, 0.1 - 0.09, and so on.
Also, check Load type is acceleration type or not.
<img src="https://i.stack.imgur.com/0uCt1.jpg" alt="enter image description here"></p>
| 14615 | Can Dynamic Base shear be greater than Static Base shear? |
2017-04-04T23:14:48.497 | <p>Is there any way to know whether or not an unknown system has positive or negative zeros just from looking at the step response? This is the step response:
<a href="https://i.stack.imgur.com/DcQVy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DcQVy.jpg" alt="enter image description here"></a></p>
| |control-engineering| | <p>Here is my approach to identifying this system.</p>
<p>Because the steady-state response to a step input is a nonzero constant, we can infer that the system must not have a zero at the origin (because the steady-state response to a step would be zero), and it must not have a pole at the origin (because the steady-state response to a step would be a ramp). </p>
<p>Typically, when systems start off in the "wrong" direction, we can infer the presence of a right-half-plane (or "nonminimum phase") zero. The location of the zero determines the degree of "undershoot". I started off by guessing the zero was at +1. </p>
<p>Next, I estimated the location of the poles. The dominant time constant appears to be about a second. Playing around with the transfer function in MATLAB suggested that a better estimate would be s = -1.5. </p>
<p>The nonzero initial condition suggested to me that there might be <em>another</em> zero. To keep the transfer function proper, I increased the order of the pole to 2. </p>
<p>I found that I got a reasonable fit with the transfer function $G_p(s) = \frac{(s-1)^2}{(s+1.5)^2}$, though you could obviously do better by tweaking parameters. </p>
<p><a href="https://i.stack.imgur.com/YX2rP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YX2rP.png" alt="Step response."></a></p>
| 14619 | Extract information about zeros of system from step response |
2017-04-05T20:01:00.530 | <p>I'm finding myself doing quite a lot of thread cutting in stainless steel lately, particularly 304 and 316, often these are moderately deep threads ~ 25 mm or so and at M12 it is becoming a bit of a chore. I'm using a decent tap wench and cutting oil with carbon steel taps (taper, second cut and finishing/plug) but I was wondering if anybody has any experience or knowledge of any ways to make this a bit easier eg by a different tap material or coating. </p>
<p>I'm not doing enough volume to justify any major investment in tooling as this is all prototype work but it's that difficult area of a dozen or so at a time. </p>
<p>Also it is specifically stainless which is the problem and these are stressed welded components so free machining stainless isn't really an option (as far as I know) and I do appreciate that this is a bit of a 'how do I have my cake and eat it' type question. </p>
<hr>
<p>The main issue is the taps getting progressively tighter, I haven't broken any yet but they often feel like they are getting to the point where they might. I generally start the taps off in a lathe (not under power) with a sliding chuck in the tail-stock so I'm fairly confident that they are going in pretty square. </p>
| |machining|threads| | <p>Thread percentage my friend. Drill dia.=thread dia. -(1.229XPITCHX.075%) 0.75 = Percentage of thread. Can substitute with 0.60 or 0.65% for harder tough to tap materials such as C276 or Duplex.</p>
| 14636 | Hand thread taps for stainless steel |
2017-04-06T08:37:29.457 | <p>In Windows is it possible to write a simple program to read the binary value of a USB port without going thru a USB driver? </p>
| |electrical-engineering|embedded-systems| | <p>It assumed that this question is a continuation from <a href="https://engineering.stackexchange.com/questions/14628/how-to-transition-from-battery-to-usb-in-a-microcontroller">How to transition from battery to USB in a microcontroller?</a>. It is also assumed that your goal is to collect data using a PIC micro controller and transfer the data to a Windows base computer. </p>
<p>An alternate is to use configure a UART port on the PIC micro controller to transfer data serially. Then use a FTDI based USB serial cable such as the one in the image to transfer data.</p>
<p><a href="https://i.stack.imgur.com/V3wWo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V3wWo.jpg" alt="enter image description here"></a></p>
<p>The data can be accessed using a serial COM port on the PC. The USB port will be enumerated as a RS232 standard COM port. This will also help you create simple C program too.</p>
<p>This alternate solution will solve your 3V and 5V predicament from the other question. </p>
<p>References:</p>
<ul>
<li><a href="https://www.sparkfun.com/products/9717" rel="nofollow noreferrer">FTDI Cable 5V VCC-3.3V I/O</a></li>
</ul>
| 14642 | Is it possible to read a USB port directly? |
2017-04-06T09:16:10.287 | <p>If you spin the stator of an AC induction motor than would the rotation of the rotor increase? Or would it have a different affect such as moving with a higher force?</p>
| |electrical-engineering|motors|ac| | <p>A standard induction motor has to turn at the speed determined by the current frequency. If you attempt to apply external torque to drive it faster it will act as a brake resisting that torque.
If you apply sufficient torque to force the rotor out of phase with the driving current it would effectively stall ( the "braking force" dropping way off and or becoming erratic )</p>
| 14644 | Mechanically increasing RPM of ac induction motor |
2017-04-06T18:48:58.757 | <p>I have the following problem most of which I have calculated but have a difficulty with the beam under angle. The angle is 56,30 degress. I just cannot figure out how the forces there are acting - the equations I am making have three unknown variables in two equations, which cannot be calculated, so apparently there are forces I "cannot see" there as well and the equations are wrong.</p>
<p>The disributed loads have been replaced with concentrated loads: the point where they act is indicated.</p>
<p><a href="https://i.stack.imgur.com/jRKti.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jRKti.jpg" alt="enter image description here"></a></p>
<p>Thank you very much for advice.</p>
| |structural-analysis|applied-mechanics|statics| | <p>It is worth noting that this can be treated as two separate structures:</p>
<p><a href="https://i.stack.imgur.com/zPU8R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zPU8R.png" alt="enter image description here"></a></p>
<p>I assume that's how you solved the reactions of support A. Given that, you can ignore the entire left-hand structure and act like you only have the right-hand side.</p>
<p>For this, let's use the standard equations:</p>
<p>$$\begin{align}
\sum F_x &= B_x + C_x + 29.4 = 0 \\
\sum F_z &= B_z + C_z - 24.5 - 7 = 0 \\
\sum M_B &= -29.4 \cdot 6 + 6C_z = 0 \\
\therefore C_z &= 29.4\text{ kN} \\
\therefore B_z &= 24.5 + 7 - C_z = 2.1\text{ kN}
\end{align}$$</p>
<p>It might seem like you're now stuck without solving for $F_x$, but you can notice that the diagonal member is a truss, with only axial loads. That means that the vertical and horizontal forces applied to it must be proportional to its tangent. Therefore we have:</p>
<p>$$\begin{align}
C_x &= -C_z \cdot \dfrac{6}{9} = -19.6\text{ kN} \\
\therefore B_x &= -29.4 - C_x = -9.8\text{ kN}
\end{align}$$</p>
<p>To check our work:</p>
<p><a href="https://i.stack.imgur.com/1vkPV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1vkPV.png" alt="enter image description here"></a></p>
| 14664 | Reaction force on a beam |
2017-04-06T19:19:31.900 | <p>is there anyway of converting this machine into giving me digital readings?</p>
<p>I bought it on ebay the other day and I am planning on cleaning it up and getting it looking good as new.
<a href="https://i.stack.imgur.com/A4EUU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A4EUU.jpg" alt="enter image description here"></a></p>
| |electrical-engineering|design| | <p>I can think of a few different ways of doing something like this.</p>
<ol>
<li>I believe that type of dial uses a magnetic field to move the needle. A well placed Hall effect sensor could do the trick. This is not invasive and also relatively hidden.</li>
<li>A hall effect sensor may be a bit complex to place and calibrate. Another option would be to tap into the reading and pull the voltage generating the current for the needle. Link this to an ADC in a micro-controller (Arduino, MSP430, etc.) and you should get an decent result. <a href="https://www.youtube.com/watch?v=ruuxn2u3yao" rel="nofollow noreferrer">Electron microscope image capture via microcontroller (with drill bit animation)</a>, I highly recommend this video, very similar to your project.</li>
<li>Use a Raspberry-Pi and a Pi-camera. If you are more of a fan of programming as opposed to circuits you can have a pi-camera attached to device and read out where it is pointing with some relatively simple code.</li>
</ol>
| 14665 | Anyway of converting this to outputting digital readings? |
2017-04-07T05:53:54.157 | <p>I want to build an electric motorcycle ( more of an electric bike, actually), but i don't want to use a conversion kit.</p>
<p>I want to take out the pedals and put a motor there (either 12v DC or 220v AC).
My problem is how to transfer the power from the engine to the rear wheel:
Do I need some sort of gearbox?
Let's say I would use a 0.7kw, 3000rpm 220v AC motor:
How do I connect the spinning rod of the motor to a chain or belt (it doesn't necessarily have to be chain driven€ <a href="https://i.stack.imgur.com/Xd8TJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xd8TJ.jpg" alt="enter image description here"></a></p>
<p>Is this a bad idea? Can the motor drive the wheel directly? The whole thing (including the rider) would weigh less than 150kg. 0.7kw is enough for the speed I desire. (I am interested in building a dirt bike actually...)</p>
<p>Excuse my English, I am not a native speaker.</p>
| |motors| | <p>A transmissions main job is to manage torque. What you have to do is look at the motor spec-sheet and see how much torque it generates.
Ensure it is sufficient to get you moving, calculate the friction coefficient with your weight and make sure the torque isn't so high you get wheel spin.
Finally tune it to make it comfortable.</p>
<p>But really, this would be more for a "finely tuned machine". Sounds like you want something fun and easy!!! In reality you should just get a motor with a metal rotor(part that spins), weld/bolt it onto the sprocket, and pump some current through it. If your bike already has multiple gears I would say leave them on as you can adjust the torque ratio to accommodate a wider range of electric motors.</p>
| 14671 | Transmission for electric motor |
2017-04-07T21:47:13.767 | <p>I am an occasional hobbyist "DIYer" (not an engineer) who needs help choosing a steel plate for a small home improvement project:</p>
<p>I will use it to create a 20" circle, suspended from a bolt at the center. It should be able to hold 350lbs pretty much evenly distributed over its surface.</p>
<p>There is already a generous safety margin built into the 350lbs estimate. </p>
<p><strong>What is a reasonable choice of steel plate (steel type + thickness) for me to go with?</strong> </p>
<p>Just to be clear, this is not a critical application -- ie, no one will be hurt (fall/get squashed) if this fails.</p>
| |structural-engineering|materials|structures|steel| | <p>The weight on the bolt will be 350 lbs plus the weight of the plate. Depending on what size washer and bolt head you use the force at the hub/bolt overlap could be 700 psi. 1/4" plate is probably ok for the full disk, but I would double it with 1/4" thick by 3" diameter inner disk/washer.</p>
| 14684 | choice of steel plate for simple application |
2017-04-08T05:59:33.897 | <p>I saw the following extract from Hibbeler's 'Mechanical Engineering - Statics, 13th Ed';</p>
<p><a href="https://i.stack.imgur.com/GIxbu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GIxbu.png" alt="relation between distributed load and shear"></a></p>
<p>The <em>general</em> result $\frac{dV}{dx} = w(x)$ seems to rely on the shear forces at the interface acting in opposite directions. If my force distribution function along the top of the beam is linear with gradient 0, then my $\Delta F$ would act in the centre of $\Delta x$ and the shear force at the right and left interface of $\Delta x$ would be identical vectors, both acting downwards. My vertical force equilibrium would then be described as follows;
$$
\Delta F - V - (V + \Delta V)=0
$$</p>
<p>This then breaks the general relationship described by (7-1). What am I misunderstanding?</p>
| |structural-engineering|structural-analysis| | <p>I think the misunderstanding is because the book carelessly ignores the difference between the shear <em>stress</em> in the beam, and the shear <em>force</em> at a point where you cut the beam.</p>
<p>The shear <em>stress</em> at the two cutting planes in Fig 1-13(b) has a consistent sign convention. But since the outward normal to the two planes are in opposite directions (one in the positive X direction, the other in the negative X direction), that flips the sign of one of the <em>forces</em> on the cutting planes as shown in the figure.</p>
<p>Fig 1-13(b) show the shear <em>forces</em> on the two cuts, but "shear diagrams" show the shear <em>stress</em> - hence your confusion about the signs.</p>
| 14688 | Relation Between Distributed Load and Shear |
2017-04-08T12:15:53.317 | <p><a href="https://i.stack.imgur.com/LgMH3m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LgMH3m.jpg" alt="enter image description here"></a> <a href="https://i.stack.imgur.com/IVLO9m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IVLO9m.jpg" alt="enter image description here"></a></p>
<p>Do you think, one could split a mercury drop into two drops with the magnetic force? Mercury is a liquid metal, hence if I put a current through it, it will become a electromagnet (a wire with a current induces a magnetic field). </p>
<p><a href="https://i.stack.imgur.com/je8hhm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/je8hhm.jpg" alt="enter image description here"></a></p>
<p>If I then place two magnets on each side, but sufficiently separated, I thought that the fluid "electromagnet" will split and both sides will travel to the magnets.</p>
| |magnets|electromagnetism| | <p>To add to Eric's Answer, you can see a visual demonstration from <a href="https://www.youtube.com/watch?v=pB-qAwkgfFQ" rel="nofollow noreferrer">Physics Girl</a> - this really does work with another magnetic liquid - Oxygen</p>
| 14689 | How to split mercury with a magnetic force? |
2017-04-08T21:51:46.627 | <p><img src="https://i.stack.imgur.com/dHflo.jpg" alt="Vacuum Tube"></p>
<p>Does the current flow from the bottom left pin to the bottom right pin, or from the bottom pins to the cap on top?</p>
| |electrical-engineering|vacuum|audio-engineering| | <p>This may be short, but might answer the question:-
<a href="https://i.stack.imgur.com/HHEiJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HHEiJ.png" alt="pinout"></a></p>
<p>It's an <a href="http://www.radiotechnika.hu/images/811A.pdf" rel="nofollow noreferrer">811</a> power triode. There is no cathode per se as it's also the heater filament.</p>
| 14695 | How does electricity flow through this kind of vacuum tube? |
2017-04-10T17:16:33.633 | <p><a href="https://i.stack.imgur.com/XUnp1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XUnp1.png" alt="enter image description here"></a></p>
<p>The part I'm stuck on is the last part. Basically, the question is to obtain the following equation for the entropy of vaporisation using the Redlich-Kwong equation:
$$
\Delta S = R\Bigg[ \ln \frac{V_2 -b}{V_1 - b} \Bigg] + \frac{0.5a}{bT^{1.5}}\ln \Bigg[ \frac{V_2(V_1-b)}{V_1(V_2-b)} \Bigg]
$$</p>
<p>Solution Attempt:</p>
<p>I think I should start by using the known fact that at equilibrium, the free energy of both phases must be the same. Using Gibbs free energy:
$$
dG = -SdT + VdP \implies -S_1dT + V_1dP = -S_2dT + V_2dP
$$
Therefore, I can write an equation for change in entropy due to vaporisation:
$$
S_1 - S_2 = (V_1 - V_2)\frac{dP}{dT}
$$
However when I simply differentiate the given EoS and multiply by $V_1 - V_2$ I don't get the same result. Clearly, in the answer they have integrated wrt v at some point but I just don't get why or how. Any help will be appreciated, thank you.</p>
| |mechanical-engineering|thermodynamics|chemical-engineering|energy| | <p>I arrived to an answer very similar to what it says it should end up being except for a difference in the sign of a term. </p>
<p>Here is how I proceeded:</p>
<p>The total differential of a two-variable function can be written as
$$ dS(T,V) = \left( \frac{\partial S}{\partial T} \right) _V dT + \left( \frac{\partial S}{\partial V} \right) _T dV $$ </p>
<p>Since a phase change occurs at constant temperature, $ dT = 0 $. Hence, the equation becomes simply</p>
<p>$$
dS = \left( \frac{\partial S}{\partial V} \right) _T dV
$$</p>
<p>Replacing the right hand side of the above equation using other thermodynamic relations for the partial derivative in terms of $ P $ and $ T $, we get </p>
<p>$$
dS = \left( \frac{\partial P}{\partial T} \right) _V dV
$$</p>
<p>Now we differentiate $ P $ with respect to $T$ at constant $V$:
$$
\left( \frac{\partial P}{\partial T} \right) _V = \frac{R}{V-b} + \frac{a}{2T^{3/2}} \frac{1}{V(V+b)}
$$</p>
<p>Noting that $ \Delta S = \int dS $,</p>
<p>$$
\Delta S = S_2 - S_1
= \int_{V_1}^{V_2} \left( \frac{\partial P}{\partial T} \right) _V dV \\
= \left[ R \ln (V-b) + \frac{a}{2b T^{3/2}} \ln \left( \frac{V}{V+b} \right) \right]_{V_1}^{V_2} \\
= R \ln \frac{(V_2-b)}{(V_1 - b)} + \frac{a}{2b T^{3/2}} \ln
\left[
\frac{V_2}{V_1} \left( \frac{V_1+b}{V_2+b} \right) \right]
$$</p>
<p>Either the question is wrong in their signs or I made a mistake with the signs in my solution. Either way, this should hopefully guide you in the right direction.</p>
| 14720 | Derivation: Entropy of Vaporisation using Redlich-Kwong EoS |
2017-04-10T18:10:11.917 | <p>Why does the Ultimaker 3D Printer has a Heater + Heater transfer plate (aluminium) + Glass?</p>
<p>I wonder why a glass plate, and if is possible to remove the glass and print directly in the aluminium plate adjusting the heating.</p>
<p><a href="https://ultimaker.com/en/products/ultimaker-3" rel="nofollow noreferrer">Link to the ultimaker</a>.</p>
<p>Pictures:</p>
<p><a href="https://i.stack.imgur.com/hBsTd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hBsTd.jpg" alt="Ultimaker view"></a></p>
<p><a href="https://i.stack.imgur.com/uf2vc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uf2vc.jpg" alt="PCB Heater view"></a></p>
| |aluminum|3d-printing|heating-systems|glass| | <p>The heater PCB is not suitable for printing on for multiple reasons.</p>
<p>A PCB is not as durable as the supplied glass is, nor is it as flat. The surface of a PCB held up by the corners with no other supports would droop in the middle, and no current Ultimaker has a way to compensate for that i.e. automatic mesh bed leveling. A PCB is also not as versatile as a swappable build plate. It's a bit silly to swap out a heater every time you scratch the PCB, eh? (You also probably don't want to scratch and expose the heater leads in the PCB.) With Ultimaker's system, you simply unclip the glass plate and take it out.</p>
<p>Because most PCBs are not flat, as discussed earlier, the Ultimaker uses the aluminum transfer plate that you mentioned. If you simply place the glass on top of the PCB, the two will not make full contact at the center, resulting in uneven heating of the build plate. Assuming the glass and alu plates are both reasonably flat, if you set them on top of each other, there won't be an air gap between them. If you use adhesive or something to adhere the PCB to the alu plate, then you have an evenly heated build surface. And you can swap the glass out without having to rip off all that adhesive.</p>
<p>As for printing directly onto the aluminum plate, that's a question more appropriate for the <a href="https://3dprinting.stackexchange.com/questions/3838/why-does-the-ultimaker-3d-printer-has-a-heater-heater-transfer-plate-aluminiu">3D printing exchange</a>.</p>
| 14722 | Why does the Ultimaker 3D Printer has a Heater + Heater transfer plate (aluminium) + Glass? |
2017-04-11T19:29:16.190 | <p>I have the following robot arm:</p>
<p><a href="https://i.stack.imgur.com/zaa5O.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zaa5O.png" alt="enter image description here"></a></p>
<p>Now how do I obtain the equation of motion:</p>
<p>$$ M \ddot\theta + \vec{b} + \vec{g} = \vec{\tau} $$</p>
<p>I am familiar with forces, torque, velocity and so on. I know that this equation follows from force and torque equations. However, I don't know how to establish these force and torque equations.</p>
<p>The torque is applied by motors in the joints, so that the arm does not move. No payload is attached to the end-effector.</p>
<p>As far as I understood, $\vec{\tau}$ contains the two scalar torques for each of the two joints. $\vec{g}$ should be the force vector in negative y-direction, although I am not quite sure. I have no idea what $\vec{b}$ might be.</p>
| |mechanical-engineering|torque|robotics| | <p>Define $\vec{\tau}_1$ on the first joint and $\vec{\tau}_2$ on the second joint. The first arm is well defined. The traditional $\sum \vec{F} = m\vec{a}$ and $\sum \vec{M} = I\vec{\alpha}$, valid only when looking at the center of mass, can be applied. However, the moment equation is only valid relative to the center of mass. The general formula for motion when described by another point P on the body (travelling with the body) is $\sum \vec{M}_P = I_p\vec{\alpha} + \vec{p} \times m\vec{a}_P$, where $\vec{p}$ is the distance vector from point P to the center of mass, and $\vec{a}_P$ is the acceleration at P.</p>
<p>Since the pinned point doesn't have an acceleration, it can be established for pinned bodies that, with P being the point of the pin, $\sum \vec{M}_P = I_P\vec{\alpha}$. The second arm doesn't have this luxury, and we must establish an equation of motion for the body. A major note is that we need an equation to describe $\vec{p} \times m\vec{a}_J$. While $\vec{a}_J$ the acceleration of the joint, in terms of $\theta_1$, in Cartesian coordinates, is quite complex:</p>
<p>$$\vec{a}_J = L_1\left[\left(-\ddot{\theta_1}\sin(\theta_1)-\dot{\theta_1}^2\cos(\theta_1)\right)\hat{i} + \left(\ddot{\theta_1}\cos(\theta_1)-\dot{\theta_1}^2\sin(\theta_1)\right)\hat{j}\right]$$</p>
<p>Since $\theta_2$ is defined by the relative angle, this can be simplified as we evaluate across the cross product of the two vectors:</p>
<p>$$\vec{p} \times m\vec{a}_J = \frac{mL_1L_2}{2}\left(\ddot{\theta_1}\cos(\theta_2)-\dot{\theta_1}^2\sin(\theta_2)\right)$$</p>
<p>So, the equation of motion for the second body would be:</p>
<p>$$\vec{\tau_2} = I_J\ddot{\theta_2}+ \frac{mL_1L_2}{2}\left(\ddot{\theta_1}\cos(\theta_2)-\dot{\theta_1}^2\sin(\theta_2)\right)$$</p>
<p>Where $I_J$ is the inertia of the second body relative to that joint. Since $\vec{\tau_1(t)} = I_P\ddot{\theta_1}$, which implies that $\dot{\theta_1} = I_P^{-1}\int_0^t\vec{\tau_1(t)}dt$, we can simplify:</p>
<p>$$\vec{\tau_2(t)} = I_J\ddot{\theta_2}+ \frac{mL_1L_2}{2I_P}\left[\vec{\tau_1(t)}\cos(\theta_2)-\left(\int^t_0\vec{\tau_1(t)}dt\right)^2\frac{\sin(\theta_2)}{I_P}\right]$$</p>
<p>This simplification allows for solving in $\theta_2$ only, without solving for $\theta_1$ simultaneously. Assuming the input function $\tau_1(t)$ is known, the resulting equation can be simple or vastly complex.</p>
| 14746 | Equation of motion for 2-link arm |
2017-04-12T00:07:24.663 | <p>If you wanted highest efficientcy and speed from an AC motor what would you do?
Could the rate of alternation of current be optimizes?
More windings?
Materials?
More lamination?
Other factors I haven't thought of?</p>
| |electrical-engineering|motors|ac| | <p>Regarding efficiency:</p>
<ol>
<li><p>Fit ceramic bearings. There are many videos on YouTube which demonstrate how much less friction these bearings provide.</p></li>
<li><p>If you have a large budget; Make use of active magnetic bearing technology which provides zero friction as the shaft effectively hovers within the bearing. You will also need to factor in the power usage of the active magnetic bearing controller into this, as such, this would more likely pay off in a large motor setup. </p></li>
<li><p>Place the motor within a strong vacuum. This will eliminate drag. </p></li>
<li><p>Use a permanent magnet motor which is approximately 20% more efficient than an exciter based motor.</p></li>
</ol>
<p>Regarding speed:
You could increase the voltage within tolerance which should increase the RPM speed. For example, if you run 220V 50hz motor, you could supply 240V or 250V to get a slight boost in speed. Note, that the motor would draw more current under load and you could damage the motor.</p>
| 14748 | Increasing speed and efficiency of ac induction motor |
2017-04-12T14:58:11.200 | <p>I am trying to find the COP of a VCR system. I can reach the COP with using the EES(Engineering Equation Solver), without dealing with the U<em>A values, i.e. just with the energy balances for the cycle. How the COP can differ, if I incorporate the U</em>A values of the Heat Exchangers into the model? Is there any need for this?</p>
| |thermodynamics|energy-efficiency| | <p>My opinion is, no, there's not any need to incorporate UA into your calculations. Think about it this way:
The cooling provided by the evaporator really only comes from boiling refrigerant. The heat absorbed by vapor is almost negligible.
To absorb more heat, you need to boil <strong>more</strong> refrigerant rather than getting refrigerant to boil <strong>sooner</strong>.
To boil more refrigerant, you need to run your compressor at a higher speed, which means adding more work. Since: $$COP = \frac{Q_{evaporator}}{W_{compressor}}$$, you won't gain much clarity adding UA to the model.<br>
Caveat: if refrigerant is not exiting the condenser as pure liquid, efficiency improvements can be made by increasing UA of the condenser.</p>
| 14755 | How the COP differs for a Vapor Compression Chiller without the U*A values of Evaporator/Condenser? |
2017-04-13T00:41:10.803 | <p>speed measured by pitot tubes is that of the wind speed relative to the aircrafts. Now, this itself depends on weather conditions which can affect wind speeds. So, then why employ pitot tubes at all if they can give deceiving readings? Can not satellite imaging be used rather for measuring accurate speeds?</p>
| |aerodynamics| | <p>There are two speeds that the pilot is concerned about. First is airspeed, and the second is ground speed.</p>
<p>Ground speed (speed of the aircraft relative to the ground) can and is checked by GPS. This speed tells the pilot how long the flight will be and if they are staying on track with their fuel calculations for the flight, crucial, but less critical compared to airspeed</p>
<p>Airspeed, (Speed of aircraft relative to the wind that is passing over the wing) is checked by the pitot tube. This speed is a more critical measurement for actual flight. The aircraft dynamics require that the air passing over the wing to be within certain ranges for different types of flight maneuvers (take off and landing have their own critical speeds, cruising flight another, stall speed is another). These can only be measured directly through a pitot tube.</p>
<p>Where the challenge (conceptually) is, is in understanding the difference between the two speeds. If your aircraft is flying into the wind at 60 knots (airspeed) and the wind speed is also 60 knots, then your GROUND speed, will be zero. Your aircraft is moving forward relative to the wind at the same speed as the wind is pushing it back. If you've ever seen a bird on a windy day flapping really hard, but not moving anywhere relative to you standing on the ground, this is what is going on. </p>
<p>Now consider the pilot turns 180 degrees and is now flying with the wind. Once the plane is at a steady flight speed, his airspeed is 60 knots, but his ground speed will be 120 knots! </p>
<p>Note, do not fly in 60 knot winds if your cruising speed is 60, this is ridiculously extremely insanely dangerous to do. </p>
| 14763 | Deceptional Pitot Tube |
2017-04-13T02:36:00.443 | <p>I'm making a double pendulum for an art installation, and I would like to add energy to the system to compensate for friction and air drag losses so it can run more or less indefinitely.</p>
<p>The design so far has the top pendulum connecting to an axle at the top which is parallel to the ground and can rotate freely. So far I've sketched out a design which uses a rotational position sensor and a microcontroller to determine the pendulum shaft's current speed and position. A small, reversible electric motor will connect to the axle with a somewhat "limp" drive belt around the consistency of a rubber band (to lessen the impact of rapid chaotic changes mismatching with power input direction). As the axle rotates unpredictably, the control software will reverse or advance the motor to add a slight force in the current direction of rotation. Once tuned, the hope is that it will not perceptibly interfere with the natural motion of the pendulum while adding enough energy to keep the pendulum swinging with the same total energy as it started at.</p>
<p>Questions:
1) Are there mechanically simpler ways of adding energy to such a chaotic system?
2) Will energy added to the top pendulum distribute to the bottom pendulum in a way that would leave a casual observer unaware, or would the system develop obviously unnatural motion?
3) How can I roughly estimate (+-50%, to size the drive motor) the average energy input required to overcome friction and drag losses?</p>
| |power-engineering| | <p>To show a complete free falling double pendulum, a clutch separating your motor would work sufficiently. </p>
<p>I'd still recommend a position and angular velocity sensor - when the magnitude of rotation slows down below a specified set point, have the motor begin speeding up the clutch to a speed beyond the angular rotation of the motor. When the angular rotation is in the correct direction, engage the clutch. Then have a random number generator dictate how long the clutch stays engaged, to ensure every second of the demonstration is new. During this energization, this will be noticeable to the eye, but it will make for a perpetual double pendulum that acts truly and completely random.</p>
| 14764 | Power assist mechanism for double pendulum? |
2017-04-13T14:13:38.350 | <p>I am working on a building that would like to used compressed air for some pneumatic actuators. They would like to bring the air in from an adjacent building that has more than enough capacity, but the air quality coming over is not sufficient and I need to add a dryer and filters to bring it to the correct quality.</p>
<p>Depending on the usage in the other building, I could feasibly see air that has run through the dryer and filters in the new building being effectively siphoned back to the other building. Is there a reason that I could/should not install an inline check valve where the piping comes into the new building right before the filters and dryer?
Check valves are used extensively on the compressors and receivers, but a brief look online didn't turn up anywhere where people were using them as I propose.</p>
| |compressed-air|industrial-engineering| | <p>The problem comes down to pressure. If your supply building (building 1), which has capacity, is far enough away, the pressure at the end of that line may be significantly lower than the line in new building, (building 2), especially before the filters. If this is the case, the new line won't have enough force to open the check valve against the old line, and none of the new air will get into the new line.</p>
<p>As such, a careful consideration and monitoring of the pressures should be done before beginning. Monitor the use of building 2 during slow times, during peak usage, and then model if for the new higher usage. Ideally use 3-4 points, knowing the flow rates and pressures for each. Plot them on a curve in excel, using a <a href="https://www.youtube.com/watch?v=kFasmbrDG5I" rel="nofollow noreferrer">quadratic regression,</a> and figure out where the pressure will be at the new flow rate at the new use rate.</p>
<p>You should note that the compressor's pressure should drop on that curve. Then measure building 1's usage as well, ideally at similar flow rates and pressures if possible. Finally, use some basic fluid dynamics losses to find the pressure loss after routing from so far away.</p>
<p>If $Q_{Overcapacity}$, is the new flow rate, $P(Q_{Overcapacity})$ is the pressure at the new flow rate, and $P(0)$ is the pressure at no flow rate, the system will only provide new air if the following equation holds true:</p>
<p>$$P_1(0) > P_2(Q_{Overcapacity})$$</p>
<p>If this equation does hold true, then the actual flow rate of the new line can be determined by when the following equation is true:</p>
<p>$$P_1(Q_{Makeup}) = P_2(Q_{Remainder}) + P_{losses}(Q_{Makeup})$$</p>
<p>In this scenario, $Q_{Makeup}$ is the flow rate of the makeup line, $Q_{Remainder}$ is the remainder to meet the complete flow demand. If this pressure is to low for the equipment to operate, then the entire system needs to be scrapped and a higher pressure source is required anyways. </p>
| 14773 | Check valve in compressed air pipe network? |
2017-04-13T16:45:43.217 | <p>A gauge pressure sensor measures a pressure at a port against ambient pressure.<br>
A unidirectional differential pressure sensor measures a pressure at a "high" port against a "low" port.</p>
<p>Is a gauge pressure sensor the same thing as a unidirectional differential pressure sensor (simply using ambient pressure as the "low" side)? Is this just a matter of where your reference point is? </p>
<p>I'm guessing you could turn a differential sensor into a gauge sensor by using ambient as your "low" side. Can you turn a gauge sensor into a differential sensor by using a pressure port rather than ambient?</p>
| |pressure| | <p>Differential sensors are essentially gauge sensors with a low side port (gauge sensors usually do not have a port where you can connect a tube).</p>
<p>If you connect the low side port of a differential pressure sensor to ambient pressure, you will essentially have a gauge sensor.</p>
<p>A useful application of this case is when you want to connect your gauge sensor to ambient pressure but in your system there is no ambient pressure reference in the immediate environment.</p>
<p>Therefore, you have to use a tube and connect the low side port at an ambient pressure reference further away.</p>
| 14778 | Pressure sensors - gauge vs unidirectional differential |
2017-04-13T21:45:20.537 | <p>There are many new cars now that automatically turn off their engines when stopped at traffic lights. They then automatically turn them on when moving off. These systems have many names along the lines of <em>active</em> or <em>eco</em> . </p>
<p>The manufacturers claim that these systems are to save fuel, reduce running costs and save polar bears. At the same time there are no published figures detailing the economies involved. Is this just marketing hyperbole? It seems to me that switching your engine off when it's at a warmed up idle is absolutely the last place that would generate any fuel saving worth printing. Most cars idle at ~ 0.3 gallons per hour anyway which is nil in the context of a typical trip. These systems also require a larger alternator, battery, and starter motor. It's a con init?</p>
| |automotive-engineering|environmental-engineering| | <p>The main purpose for ECO modes in newer vehicles is actually driven by the EPA in an effort to reduce emissions. Manufacturers sell it as fuel economy/efficiency - which is not totally misleading but that sure is not the complete story. The biggest downside to auto off when idle is wear and tear on your engine with regards to lubrication. Engine oil settles, then gets distributed again so very small increments of friction and wear occur at every start. With older vehicles, this type of wear usually only occurs during a "cold" start, now (even though marginal) the wear occurs at every red light.</p>
<p>This link may help with some background info: <a href="https://www.epa.gov/regulations-emissions-vehicles-and-engines/regulations-greenhouse-gas-emissions-commercial-trucks" rel="nofollow noreferrer">https://www.epa.gov/regulations-emissions-vehicles-and-engines/regulations-greenhouse-gas-emissions-commercial-trucks</a></p>
| 14782 | Do cars that automatically turn off engines at idle save any significant fuel? |
2017-04-13T22:30:43.940 | <p>It is easy to find a very flat surface, "perfect" straight edge, or squares for ensuring things are perpendicular. The machines that make these need to be flat/orthogonal to do so, but how were the first squares and straight edges that were fairly accurate made?</p>
<p>"Fairly accurate" is subjective, and I don't know exactly how to define it, but in my mind I am imagining measurements made in a time before the age of heavy machinery and modern manufacturing.</p>
| |measurements| | <p>Perpendicularity can be achieved by using any item of constant length.</p>
<p><a href="https://en.wikipedia.org/wiki/Knotted_cord" rel="nofollow noreferrer">Knotted cord</a> was used by the Greeks, Romans, and Ancient Egyptians. Sticks work well too.</p>
<p>The simplest way to establish perpendicularity (and one I frequently use in fabrication) is to measure diagonals and confirm they are equal. Even though I own a large steel square, this method has much less error on projects that are larger than the square. As joojaa pointed out in the comments, this only works if opposing sides are also equal as shown by the congruency symbols on the diagram below.</p>
<p><a href="https://i.stack.imgur.com/Iub9n.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Iub9n.gif" alt="perpendicularity via equal diagonals"></a></p>
<p>A 3:4:5 triangle also has a 90 degree corner, and can be more convenient that constructing a whole rectangle if it doesn't already exist in the design.</p>
<p><a href="https://i.stack.imgur.com/3THP0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3THP0.png" alt="3 4 5 triangle"></a></p>
<p><br>
The <a href="https://en.wikipedia.org/wiki/Perpendicular#Construction_of_the_perpendicular" rel="nofollow noreferrer">wiki "construction of the perpendicular" section</a> and the links below have the compass method and a few others I have not used.</p>
<p><a href="https://i.stack.imgur.com/IaJtM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IaJtM.png" alt="perpendicularity via compass"></a></p>
<p><a href="http://www.instructables.com/id/handy-tricks-to-find-square/" rel="nofollow noreferrer">Instructables, tricks to find square</a><br>
<a href="http://www.bbc.co.uk/schools/gcsebitesize/maths/geometry/polygonsrev2.shtml" rel="nofollow noreferrer">Quadrilaterals</a><br>
<a href="https://matheducators.stackexchange.com/questions/9847/given-a-3-4-5-triangle-how-do-you-know-that-it-is-a-right-triangle">Math Educators SE 3:4:5 Triangle Question</a><br>
<a href="https://www.mathsisfun.com/geometry/triangle-3-4-5.html" rel="nofollow noreferrer">MathIsFun 3:4:5 Triangle</a></p>
| 14783 | How were 90 degree angles made? |
2017-04-14T17:53:39.287 | <p>I have a vessel that has a pressure measuring device. The value it shows as the pressure inside the vessel is 6 inches of water column. Is inches of water column meant to always be used as gauge pressure, or does one have to specify gauge vs atmospheric? </p>
| |pressure|unit| | <p>Inches of water is conventionally used as gauge pressure, not absolute. You could conceivably specify it as inches of water absolute, but that would be an unusual use - sub-atmospheric pressures are more commonly measured in inches or mm of mercury. </p>
| 14792 | I need clarification on how pressure is measured |
2017-04-14T17:59:55.733 | <p>I need someone to confirm or deny my understanding in the following situation.</p>
<p>I have an empty vessel that is in equilibrium with the atmosphere and I have a pressure gauge on the vessel that displays gauge pressure. The gauge should read 0 atm, correct? If I then close this vessel, pull vacuum on the vessel to completely evacuate it (assume this is possible), then fill it with it 2 atmospheres of nitrogen, the gauge will show 1 atm, correct?</p>
<p><strong>EDIT -</strong>
There is some specific volume of nitrogen that at ambient temperature will exert 2 atmospheres of pressure within my vessel. In my scenario, I mean to say I have a supply line that can deliver as much nitrogen as I need to the vessel at ambient temperature to meet this condition.</p>
| |pressure|measurements| | <p>There's no such quantity as 2 atmospheres of nitrogen. If you pulled a vacuum to -1 atm, then connected the tank to a nitrogen tank of the same size at 2 atm, you'd have 2 tanks of nitrogen at 1 atm pressure (assuming that nitrogen is an ideal gas)</p>
<p><strong>EDIT-</strong> from the ideal gas law: $$m \sim \frac{PV}{T} $$, assuming volume, $V > 0$, temperature, $T < \infty $. Once the vacuum is pulled, $P =0\; atm (abs), m = 0$. By adding some nitrogen such that $P = 2\; atm (abs)$, $ m = 0 + \frac{2V}{T}$, then $P = 2\;atm(abs) = 1\;atm(gauge) $. I think you're ok. </p>
| 14793 | Measuring vessel pressure scenario |
2017-04-14T19:46:24.843 | <p>For example,
How does one compare 100 g's for 6 milliseconds to 45 g's for 11 milliseconds. These are two different shock scenarios with different magnitudes and times, so how can I compare them? If I have something that is shock rated at 100 g's for 6 ms does this imply that it will withstand 45 g's for 11 ms?</p>
<p>I have tried finding a "shock equivalence" chart or a metric for comparing two shock scenarios but I haven't found anything.</p>
<p>I was thinking it may depend on the mass of the equipment that is being shock rated:</p>
<p>$$v=at$$
$$F=ma$$
$$Power=Fv$$</p>
<p>An energy metric like power may give me a reasonable estimate to compare, but I am just guessing here. Is there a standard way to compare two scenarios? A chart or equation would be nice..</p>
<p>Thank you for the help!</p>
| |mechanical-engineering|shock|specifications| | <p>For car exidents there is a meassure, which is called Head Injury Criterion. It is a measure to estimate the possibility of an head injury during a car accident.</p>
<p>The HIC for the time interval $t_1,t_2$ is given by</p>
<p>$$HIC_{t_1,t_2} = (t_2-t_1)^{-1.5}\int_{t_1}^{t_2}a(t)dt.$$</p>
<p>The final HIC is given by the maximum value of the previuos expression, which can be obtained by sliding the interval over the total crash time.</p>
<p>But depending on your application other similar measures can be used.</p>
| 14795 | Comparing 2 shock scenarios with different magnitudes and times |
2017-04-15T01:35:15.480 | <p>Under-steering is caused when, at a corner, the front wheels lose traction and as a result the car does not take the corner as quick as it should. So basically what we should do is try and get both the tyres to have a better grip on a corner.</p>
<p>Now what an anti-roll bar does is it allows both tyres to remain on ground during a turn so that we get maximised traction.</p>
<p>So why don't we increase the stiffness of the anti-roll bar when faced with the problem of under-steering?</p>
| |mechanical-engineering| | <p>"Now what an anti-roll bar does is it allows both tyres to remain on ground during a turn so that we get maximised traction."</p>
<p>Actually just the opposite. The roll bar shifts load from the inside wheel to outside, effectively lifting the inside wheel. With a sufficiently stiff roll bar the inside wheel can be lifted off the ground (Watch most any front wheel drive race event and you will see the inside rear wheels being routinely lifted in the air.)</p>
<p>In less extreme case you have increased the contact pressure on the outer wheel and correspondingly lowered it on the inside. You might expect the net available cornering force to be a wash but it turns out the gain on the outside is less than the loss on the inside, so increasing roll stiffness results in a net loss of available traction. The extreme case is useful to think about as those FWD racers are deliberately tending to oversteer cornering on three wheels.</p>
<p>Of course there are competing effects related to real would suspensions geometry and how the body roll effects keeping the tire flat to the ground, so its not quite so simple. This is why folks typically increase the rear stiffness to reduce understeer, rather than lowering the front stiffness.</p>
| 14797 | Why do we remove (or reduce stiffness of) anti-roll bars from the front side to prevent under-steering? |
2017-03-25T21:05:49.670 | <p>I'm currently building a sonar running off an arduino. My radar includes an ultrasonic sensor that will detect the distance an object is sitting at, and a servo motor that the sensor will sit on top of. I was wondering if it was at all practical to use a PIR sensor with this setup so that the radar will be able to detect people instead of just inanimate objects. I'm worried that while rotating, the sensor might just constantly stay high if it sees any moving person. </p>
<p>I have no prior PIR sensor experience and that is why I was hoping someone with some experience could let me know if this is worth pursuing or if will be too much of a problem.</p>
<p>The range of this sonar isn't too important. Although I would like it if it could scan several meters, under 1 meter would still be fine.</p>
| |electrical-engineering| | <p>Perhaps.</p>
<p>A hobbyist level PIR sensor is actually a digital output device. It <em>sees</em> parts of 2D space split into various patterns and detects thermal transitions /movement between them. If the thermal emitter moves from zone a to zone b within a certain time, a positive detection event is output. Just like you see with your PIR burglar alarm. If the person doesn't move, there is no transition with a static sensor. But your sensor is rotating. However, this might even happen with no one around as the thermal background radiation will be constantly changing wrt the sensor. Again these rules govern the siting of PIR burglar alarm detectors. One thing. You might have to screen the sensor as they have very wide fields of <em>view</em>. That might increase angular accuracy. They only cost a few quid. Try it.</p>
<p>I mentioned that the detector is 2D which seems counter intuitive, but it has no ability to judge distance as far as you're concerned. All it sees is a map of zones spread out over the x and y axes. There is no depth perception that you can use. It's not analogue like the ultrasonics.</p>
<p>Note. If you have access to a scientific thermal sensor or camera you could judge distance as well so get 3D scanning as with ultrasonics, but I'm assuming that this is out of scope here.</p>
| 14801 | Can I use a PIR sensor on a home made sonar? |
2017-04-15T19:36:29.217 | <p>How do you calculate the torque on the cam, with a force placed on the follower?</p>
<p>Two possibilities:</p>
<ol>
<li>Linear cam (schematic)</li>
</ol>
<p><a href="https://i.stack.imgur.com/NdLTU.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NdLTU.gif" alt="enter image description here" /></a></p>
<ol start="2">
<li>Radial cam (schematic)</li>
</ol>
<p><a href="https://i.stack.imgur.com/sxwX1.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sxwX1.gif" alt="enter image description here" /></a></p>
| |torque| | <p>Does this help?</p>
<p><a href="https://i.stack.imgur.com/k7syl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k7syl.png" alt="vectors"></a></p>
<p>Take the [Angle] that's most prejudicial to opposing F, and consider the lever arm to the camshaft centre. It looks like my screen grab is the worst case scenario, but with the changing lever arm to point of contact, you should really repeat this for sever angles of cam rotation. I'm sure there's software for this, but it can be done graphically by hand by plotting a surface development of the cam profile against shaft rotation, say 0 - 180 degrees. Repeat the calculation every 15 degrees and Fanny's your Aunt. </p>
<p>One thing. You can't possibly have a fully algebraic solution unless you have an equation for the profile of the cam as you won't be able to determine [Angle] other than by numerical /graphical means. I might be tempted to scan the cam profile and use a graphics package to segment and develop the cam profile. Or use a rule.</p>
| 14806 | Calculate torque for a cam |
2017-04-17T04:08:12.613 | <p>I have a 120mmx120mm brushless fan with an Aluminum frame, which as you can imagine, isn't very thick. I want to thread the base, so I can mount it with a fairly large screw. For that I need to add more Aluminum so the screw has more to grip on.</p>
<p>I'm thinking of creating a casting mold around the exterior side of the Aluminum frame (proof sealed) and then pouring molten Aluminum in it. I've seen videos of people pouring Aluminum and it seems to have a relatively low Heat Capacity compared to other metals (i.e. it will cool down and solidify before it melts the surface of Aluminum frame), as I've seen Anthill Art being made without so much as scorching the earth. Also, as Aluminum is a good Thermal Conductor, the heat would dissipate throughout the frame pretty quickly.</p>
<p>How would I alleviate the two problems of low Heat Capacity and being a good Thermal Conductor? I know that one always have to pour excess molten metal to the block being added to, but do you guys have better ideas??</p>
| |metals|aluminum|molding| | <p>Your ultimate goal is not clear - you've asked how to do a task rather than how to achieve a result.</p>
<p>For example: if your fan is not terribly massive, then what matters is the surface area of contact between the screw head and the frame. This is more easily achieved with a wide flat washer or two (under the screw head and perhaps under the matching nut on the far side). </p>
<p>Another possibility is to follow ChrisJ's answer and bond a piece on, using an epoxy rated for metal-to-metal. </p>
| 14816 | Pouring aluminum into another block of Aluminum |
2017-04-17T13:34:21.253 | <p>The problem is that when I try to solve the following equations of J2 perturbed gravity:</p>
<pre><code>function xdot = Gravity( ~ , x )
Re = 6378.137;
J2 = 1.08262668e-3;
mu = 398600.4418;
r = norm(x(1:3));
% xdot = [x(4:6)
% -mu/r^3*x(1:3)];
xdot = [ x(4);
x(5);
x(6);
-mu*x(1)/r^3*(1+1.5*J2*(Re/r)^2*(1-5*(x(3)/r)^2));
-mu*x(2)/r^3*(1+1.5*J2*(Re/r)^2*(1-5*(x(3)/r)^2));
-mu*x(3)/r^3*(1+1.5*J2*(Re/r)^2*(3-5*(x(3)/r)^2))];
end
</code></pre>
<p>using any of the solvers (ode45, ode113, etc.), the propagated position and velocity has a considerable error compared to a trusted commercial software (STK), and the error keeps growing. I've tried various settings of tolerances, initial steps, and max steps, but the problem persists!</p>
<p>It might be surprising, but when I wrote a fixed step 4th order Runge-Kutta code as below, it didn't get much better.</p>
<pre><code>function [t,x] = RK4( fn,tspan,x0 )
% An implementation of fixed step 4th order Runge-Kutta method
f = str2func(fn);
t = tspan;
h = tspan(2)-tspan(1);
x(:,1) = x0;
for i = 1:size(t,2)-1
k1 = f(t(i),x(:,i));
k2 = f(t(i)+h/2,x(:,i)+k1*h/2);
k3 = f(t(i)+h/2,x(:,i)+k2*h/2);
k4 = f(t(i)+h,x(:,i)+k3*h);
x(:,i+1) = x(:,i) + (k1 + 2*k2 + 2*k3 + k4)*h/6;
end
end
</code></pre>
<p>any idea of what the problem might be? I also have to mention that to a great surprise, I did get a result using RK4, but the same code with the same conditions misbehaved later!!</p>
<p>A sample result:
<a href="https://i.stack.imgur.com/Cnwsx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cnwsx.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|aerospace-engineering|dynamics|numerical-methods| | <p><a href="https://www.mathworks.com/matlabcentral/fileexchange/55167-high-precision-orbit-propagator?s_tid=srchtitle" rel="nofollow noreferrer">https://www.mathworks.com/matlabcentral/fileexchange/55167-high-precision-orbit-propagator?s_tid=srchtitle</a></p>
<p><a href="https://www.researchgate.net/publication/340793133_High_Precision_Orbit_Propagator_C_code" rel="nofollow noreferrer">https://www.researchgate.net/publication/340793133_High_Precision_Orbit_Propagator_C_code</a></p>
<p>The motion of a near-Earth satellite is affected by various forces. One of these forces is the Earth's central gravitation and the others are known as perturbations. These perturbations are classified into gravitational and non-gravitational forces. In this case, the equation of motion can be written as:
r ̈=-(GM/r^3)*r+γ_p</p>
<p>γ_p is the vector of additional accelerations induced by the disturbing forces.
γ_p=r ̈_E+r ̈_S+r ̈_M+r ̈_P+r ̈_e+r ̈_o+r ̈_D+r ̈_SP+r ̈_A+r ̈_emp
r ̈_E = Accelerations due to the non-spherically and inhomogeneous mass distribution within Earth (central body)
r ̈_S, r ̈_M, r ̈_P = Accelerations due to other celestial bodies (Sun, Moon, and planets)
r ̈_e, r ̈_o = Accelerations due to Earth and oceanic tides
r ̈_D = Accelerations due to atmospheric drag
r ̈_SP, r ̈_A = Accelerations due to direct and Earth-reflected solar radiation pressure
r ̈_emp = Accelerations due to unmodeled forces
Here I have used the following integrator and force model to simulate the satellite's perturbed motion:
Integrator: Variable-order Radau IIA integrator with step-size control
Force Model:</p>
<ul>
<li>gravity field of the Earth (GGM03S model)</li>
<li>gravity of the solar system planets (positions of the planets are computed by JPLDE436)</li>
<li>drag effect using Jacchia-Bowman 2008, NRLMSISE-00, MSIS-86, Jacchia 70 or modified Harris-Priester atmospheric density model (in Accel.m you can uncomment your favorite model)</li>
<li>solar radiation pressure using geometrical or cylindrical shadow model</li>
<li>solid Earth tides (IERS Conventions 2010)</li>
<li>ocean tides</li>
<li>general relativity</li>
<li>ECEF2ECI and ECI2ECEF transformations using IAU 2006 Resolution</li>
</ul>
<p>The Simulation starts by running the test_HPOP.m. In the InitialState.txt set initial values for your favorite satellite; lines 2-7 are the state vector of satellite/spacecraft in the International Terrestrial Reference Frame (ITRF). Lines 8 to 12 are the satellite parameters related to the forces from atmospheric drag and solar radiation pressure. Lines 8-10 are in units m^2 and kg. Line 11: Cr is the radiation pressure coefficient (Cr = 1 + reflectivity of satellite). Line 12: Cd is the atmospheric drag coefficient of the satellite. In the test_HPOP.m you can consider different perturbations by setting them 1 as follows:</p>
<p>AuxParam.n = 70; % maximum degree of central body's gravitation field</p>
<p>AuxParam.m = 70; % maximum order of central body's gravitation field</p>
<p>AuxParam.sun = 1; % perturbation of the Sun</p>
<p>AuxParam.moon = 1; % perturbation of Moon</p>
<p>AuxParam.planets = 1; % perturbations of planets</p>
<p>AuxParam.sRad = 1; % solar radiation pressure</p>
<p>AuxParam.drag = 1; % atmospheric drag</p>
<p>AuxParam.SolidEarthTides = 1; % solid Earth tides</p>
<p>AuxParam.OceanTides = 1; % ocean tides</p>
<p>AuxParam.Relativity = 1; % general relativity</p>
<p>References:</p>
<p>Montenbruck O., Gill E.; Satellite Orbits: Models, Methods and Applications; Springer Verlag, Heidelberg; Corrected 3rd Printing (2005).</p>
<p>Montenbruck O., Pfleger T.; Astronomy on the Personal Computer; Springer Verlag, Heidelberg; 4th edition (2000).</p>
<p>Seeber G.; Satellite Geodesy; Walter de Gruyter, Berlin, New York; 2nd completely revised and extended edition (2003).</p>
<p>Vallado D. A; Fundamentals of Astrodynamics and Applications; McGraw-Hill, New York; 3rd edition(2007).</p>
<p>NIMA. 2000. Department of Defense World Geodetic System 1984. NIMA-TR 8350.2, 3rd ed, Amendment 1. Washington, DC: Headquarters, National Imagery, and Mapping Agency.</p>
<p><a href="http://sol.spacenvironment.net/jb2008/" rel="nofollow noreferrer">http://sol.spacenvironment.net/jb2008/</a>
<a href="https://ssd.jpl.nasa.gov/planets/eph_export.html" rel="nofollow noreferrer">https://ssd.jpl.nasa.gov/planets/eph_export.html</a>
<a href="https://celestrak.org/SpaceData/" rel="nofollow noreferrer">https://celestrak.org/SpaceData/</a></p>
| 14822 | J2 orbit propagation problem in MATLAB (solver instability) |
2017-04-17T13:51:24.903 | <p>I am trying to understand how to make generation of electrical power more efficient by using Faraday's law to directly convert the kinetic energy of the ionized or easily ionizable exhaust gases of combustion (oxidation of hydrocarbons or to make things simpler just hydrogen) into electrical energy. In this system, the only real moving parts would be the motion of the ionized or plasma exhaust gas medium through the surrounding faraday coils. We may need a fan to coax the pre-combustion gases in a linear flow. The usable current would be induced in the Faraday coils from the accelerating or decelerating ionized gas from the exhaust.</p>
<p>My main question is how to completely ionize the exhaust gases at the point of combustion. My understanding is that these are already ionized at the instant of combustion but rapidly de-ionize when the oxidation reduction process is complete.</p>
| |electrical-engineering|fluid-mechanics|thermodynamics|combustion|experimental-physics| | <p>Hydrocarbon combustion does produce ions, due to chemiionization mechanisms. This does make flames more highly ionized than other gases at the same temperature. That having been said, the concentration of ions is still very small (rarely more than $10^{-6}$ by volume). You can augment this somewhat with metal salts but even that has limits. Approaching complete ionization is unlikely.</p>
<p>Not saying you can't generate some amount of electricity this way, it just may not be very efficient.</p>
| 14823 | Magnetohydrodynamics, ionizing combustion exhaust then converting its kinetic energy to electricity |
2017-04-17T15:32:15.047 | <p>I'm trying to find online the derivation for the 2 parameter margules equation but cant find it anywhere.</p>
<p>basically need to prove
$$
\frac{d}{dn1} (x_{1}x_{2}(A_{21}x_{1} + A_{12}x_{2})) = x_2^2(A_{12} + 2(A_{21}-A_{12})x_{1})
$$</p>
<p>but everytime I try just end up with a page full of rubbish. I've searched online for the derivation but cant find it anywhere.</p>
<p>EDIT: the A values are functions of temperature only and can be considered constants</p>
| |chemical-engineering|vapor-pressure| | <p><a href="https://i.stack.imgur.com/JSqS2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JSqS2.jpg" alt="Hope this helps!"></a></p>
<p>Hope it makes a lil more sense. Be careful to multiply your excess gibbs by n & xlx2 from the activity coefficient relationship with gibbye! Also, be super grateful you didn't have to do the same thing for the Margules 3 parameter that I literally just worked out like 15 minutes before I saw this!</p>
| 14826 | 2 Parameter Margules derivation |
2017-04-18T00:42:13.250 | <p>I know the reasons why old cars backfire, but I've noticed that new fuel injected supercars do this as well in the video game Driveclub. I know a video game doesn't seem like a reliable source, but I researched and found that the game creators simulate each car and the in-game driving experience very closely to their real life counterparts, even claiming they recorded the engine sounds of every car in real life to make sure that the in-game car audio reflects the sounds of the actual engine.</p>
<p>That given, I noticed that lots of the supercars/hypercars backfire a lot. Sometimes this happens during acceleration, but never during pure linear acceleration, mostly when coming out of a turn. This is usually indicated by somewhat subtle flames coming out of the exhaust and sometimes a soft popping noise. More prominently however, I've noticed it during deceleration, making two loud pops in quick succession accompanied by two quick bursts of intense flames shooting out the exhaust.</p>
<p>I've researched this online myself but I haven't found a complete explanation. I have also seen some explanations as why this happens in drag racing cars, but I wouldn't suspect the same explanation would hold for street legal cars since they are constructed and purposed differently. So, why do supercars backfire during acceleration, but not during linear acceleration? And why is the backfiring so much different during deceleration (specifically the two pops, louder noise, and bigger fire)? </p>
| |automotive-engineering| | <p>Ryan - "Backfires" are the result of non-combusted fuel vapors making their way into a hot muffler. The pops you describe in your post are DOWNSHIFTS, moments when the transmission is being shifted into a lower gear, producing a higher torque gear ratio, but also sending the engine RPMs abruptly upward. The onboard system then reacts by increasing the duty cycle of the fuel injectors to direct pressurized flow into the combustion chambers. </p>
<p>Each car is different in its precise system configuration but generally all modern supercars operate Drive-By-Wire (DBW), which means the driver's throttle input is mediated by a computer module, as opposed to earlier technology where a mechanical connection attached the footpedal to the throttle plate to force it open commensurate with pedal travel. You may notice that older muscle cars and hot rods are less likely to exhibit the characteristic backfire in question. </p>
<p>This is because modern supercars are highly tuned to maintain "stoichiometric" Air to Fuel ratio. This is of course very difficult on a car that produces high peak power at low RPM and is intended to be driven "spiritedly". </p>
<p>When the system is supplied more fuel than it needs, it is called RICH. When the system is supplied less fuel than it needs, it is called LEAN. A lean air/fuel mixture burns at a higher temperature than a rich one, so running LEAN can lead to accelerated thermal wear and component failure. Therefore, since the A/F ratio is fluctuating with driver input, manufacturers err on the side of caution and set their fuel maps with a tendency to run rich. That is why supercars all backfire wildly.</p>
| 14831 | Why do fuel injected supercars still backfire? |
2017-04-18T20:46:49.103 | <h1>3D Positioner</h1>
<p>Recently I used a <a href="http://www.pmk.de/en/products/3d_positionierer" rel="nofollow noreferrer">"3D positioner" made by PMK Mess</a>. It is pretty awesome -- basically you move it to any position, turn the locking knob, and it stays rock solid.</p>
<p><a href="https://i.stack.imgur.com/kQ5Gl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kQ5Gl.jpg" alt="3D positioner"></a></p>
<h1>How might it work internally?</h1>
<p>I want to know how it works. I would like to be able to answer the following questions.</p>
<ul>
<li>How can the joints be locked in place so well using one knob in the center? </li>
<li>Are there canonical mechanisms that solve this sort of issue? </li>
<li>What are some good resources or references for designing something like this?</li>
</ul>
<p>It consists of 2 ball joints and 1 DOF rotational joint. Pictures of each joint are shown below. I would take it apart, but it's not mine.</p>
<h1>Pictures of the joints</h1>
<h3>Bottom ball joint</h3>
<p><a href="https://i.stack.imgur.com/PybNu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PybNu.jpg" alt="Bottom ball joint"></a></p>
<h3>Center 1 DOF rotational joint</h3>
<p><a href="https://i.stack.imgur.com/1IGwh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1IGwh.jpg" alt="Center 1 DOF rotational joint"></a></p>
<h3>Top ball joint</h3>
<p><a href="https://i.stack.imgur.com/iA0pB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iA0pB.jpg" alt="Top ball joint"></a></p>
| |mechanical-engineering|design| | <p>They could be locked by friction with part of the socket (circled) pressed against the ball. That force could be applied via a push rod by a cam on the knob's shaft.</p>
<p><a href="https://i.stack.imgur.com/CFMPT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CFMPT.png" alt="friction surface"></a></p>
<p><a href="https://i.stack.imgur.com/6eMXp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6eMXp.png" alt="cam mechanism"></a></p>
<p>The force might also be applied by a toggle mechanism, which is two linkages that snap into a straight line, producing a very high axial force from a relatively small lateral force.</p>
<p>The same force that locks the ball joints could also lock (by friction) the each arm onto the knob's shaft simply because that's where the reaction force to it is.</p>
| 14846 | How can multiple rotational joints be locked with one mechanism? |
2017-04-19T03:46:31.500 | <p>If we apply a simple push to rotate a system of gears to generate electricity, what is the maximum practical and theoretical power efficiency of the system, assuming the push as the only source of energy? What if the gears are to be rotated continuously?</p>
| |mechanical-engineering|electrical-engineering|power|energy-efficiency|power-engineering| | <p>In theory, we can get arbitrarily close to 100%.</p>
<p>In practice, the higher the efficiency you want, the more the generator will cost.</p>
| 14849 | What is the maximum efficiency of a mechanical electric generator? |
2017-04-19T09:17:00.713 | <p>I need a way to move air which is somewhere in between a compressor and a fan: something along the lines of 20-40 cfm at 2-3 psi (500-1000 liters/minute at 10-20 kPa).</p>
<p>Typical small compressors (of the right power rating) seem to be approx 0.5 cfm at 30 psi. Typical centrifugal blowers are approx 60 cfm free flow but only 0.05 psi max static pressure. </p>
<p>I'm having trouble visualizing what I need, let alone sourcing it. Leaf blower? Shopvac? Supercharger? Do I want a piston pump, diaphragm pump, screw compressor, turbopump, some type of higher pressure centrifugal fan, or ???</p>
<p>For this application energy efficiency at the described operating point is most important; size/cost/noise is much less important. <strong>UPDATE</strong> It also needs to be oil-free.</p>
| |pumps|compressed-air|compressors| | <p>To perform the amount of mechanical work described here, it would take about $330W$ of power, if you consider the maximum pressure and airflow.</p>
<p>$$Power = \Delta P \cdot Q$$</p>
<p>This is assuming 100% efficiency, which will never be the case. Does your power budget allow this?</p>
| 14852 | What type of air blower/compressor to use? |
2017-04-19T14:26:06.287 | <p>My background is in electronics, not mechanical engineering and fluid dynamics, so please keep that in mind. Also, this is not a homework or academic assignment.</p>
<p>Q: A dewar is partially filled with liquid nitrogen (LN2) and is open to the atmosphere. If a container, open on the top, has a hole in the bottom and is lowered into the liquid, how quickly will the liquid fill the container?</p>
<p>Specifics:
The dewar has an inside diameter of 14" and has 8" of LN2 ( volume ~1200 cubic inch).
Second container is 4" outside diameter, 3.75" inside diameter, 7" height.
A small hole, ~40 mils, is drilled in the bottom of the container.
Container will be lowered into the LN2 to a depth of 6".
Atmospheric pressure: Sea level.</p>
<p>I need to determine the rate of flow into the secondary container so I know how quickly it will fill to a 6" depth (~66 cubic inch).</p>
| |fluid-mechanics|cryogenics| | <p>I am at work and cannot perform the calculations for you, but if you solve the following equation you will find the velocity $v$ being the square root of 2 times Earth gravity times depth $x$:</p>
<p>$$v = \sqrt{2 \cdot 9.81 \, \mathrm{m/s^2} \cdot x}$$</p>
<p>If I remember correctly, from the velocity you should be able to calculate the volume rate $\dot{V}$ by multiplying the area of the opening $A$ with the velocity $v$:</p>
<p>$$\dot{V} = A \cdot v$$
$$\mathrm{\frac{m^3}{s} = m^2 \cdot \frac{m}{s}}$$</p>
<p>then just convert to the volume units of your choice.</p>
<p><a href="http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html" rel="nofollow noreferrer">http://www.engineeringtoolbox.com/bernouilli-equation-d_183.html</a>
Read the example on Bernoulli's equation and the vented tank example.</p>
| 14857 | Rate of flow of liquid nitrogen from a static tank through an orifice in a submerged vessel |
2017-04-20T15:02:48.300 | <p>I am tasked with finding the radius of a hole at the bottom of a tank to obtain the desired outflow rate when the tank is full.</p>
<p>The tank is a cylinder with dimensions Height: 4.8m and Radius: 1.79m. It holds roughly 48316.6897 liters of water.</p>
<p>I have already determined $C_d$ (The coefficient of discharge) to be 0.6372.</p>
<p>The outflow rate I desire is 3500 liters per hour.</p>
<p>I know that $\frac{dh}{dt}= \frac{a_0C_d\sqrt{2gH}}{A}$. where $a_0$ is the radius i need to obtain, and $A$ being the surface area.</p>
<p>From chain rule I can get $\frac{dh}{dt}$ in terms of radius and $\frac{dv}{dt}$(The desired outflow rate)</p>
<p>$\frac{dh}{dt} = \frac{\Delta{V}}{\pi r^2} $</p>
<p>Now I let them equal each other</p>
<p>$\frac{\Delta{V}}{\pi r^2} = \frac{a_0 C_d\sqrt{2gH}}{A}$</p>
<p>This gives me the equation
$$a0=\frac{\Delta{V}}{C_d\sqrt{2gH}} $$</p>
<p>Now, substitute in what we know</p>
<p>$3500$ Litres per hour is $\frac{35}{36000} \frac{m^3}{s} $</p>
<p>$$ a0 = \frac{\frac{35}{36000}}{0.6372\sqrt{2\cdot 9.8 \cdot 4.8}} $$</p>
<p>According to this $a_0$ is $0.157304491$ millimeters in radius.</p>
<p>I am doubting my calculations because $3500$ liters going past a hole of radius $0.157304491$ millimeters every hour seems absurd. </p>
| |mechanical-engineering|fluid-mechanics|fluid| | <p>To expand on previous answers, you're solving Bernoulli's equation which describes the conservation of mechanical fluid energy.</p>
<p>$$P_1+\frac{\rho V_1^2}{2}+\rho g h_1 = P_2+\frac{\rho V_2^2}{2}+\rho g h_2$$</p>
<p>Assuming atmospheric pressure at the exit is 0, fluid in the tank has zero velocity, $h$ is the depth of the tank at the outlet and change in potential energy is neglected, the formula simplifies to:
$$P_1 = \frac{\rho V_2^2}{2}$$ where $P_1$ is equal to $\rho g h$ and simplifies to:
$$g h=\frac{V_2^2}{2}$$
You can solve for the exit velocity and you already know your flow rate:
$$V_2 = \sqrt{2gh}{}$$
$$Q_{actual}=C_d A V_2=C_d\frac{\pi D^2}{4}\sqrt{2gh}$$
Now you can solve for the diameter of the outlet:
$$D=\sqrt{\frac{4Q_{actual}}{C_D \pi\sqrt{2gh}}}$$</p>
| 14876 | Calculating the radius of a hole for a desired outflow rate |
2017-04-20T15:51:54.193 | <p>So I am making a tech demo demonstrating fairly realistic penetration physics. I Have run into some confusion due to my lack of knowledge. If I have the resistance of the armor in MPa, that means I have the amount of joules acting in a square meter, so i divide by the area of the cross section of the shell, in meters (assuming it's a cylinder) then multiply by the thickness of the armor, also in meters, to get the required joules to punch through the armor of the given thickness with a round of the given cross sectional area?</p>
| |materials| | <p>A value in MPa is likely to be a yield stress or ultimate tensile stress. This is fine for static forces and conditions where strain rates are low eg if you are punching a hole in a plate with a hydraulic press. </p>
<p>However in the context of armour we re often looking at very high strain rates and in this case material properties for static conditions may not apply. For example with kinetic energy penetrators you can see viscous flow as well as shockwave induced effects like spalling. </p>
<p>So realistic models tend to be complex, based on empirical data and highly contextual. </p>
<p>What might be better is too look at empirical standard for armour plating eg <a href="http://www.ssab.co.uk/products/brands/armox/recommended-plate-thickness-for-different-protection-levels" rel="nofollow noreferrer">this graphic for Armox</a> Obviously a lot of the fine detail of this sort of data is proprietary but you should at least be able to get order of magnitude figures. </p>
| 14881 | penetration of rolled homogeneous steel, issues with finding the joules required to punch through |
2017-04-21T08:04:44.737 | <p>Titanium is more durable and stronger than steel. I checked the price of titanium and it is not too high. Despite that, I could find only one car made of titanium costing, and it cost 2 million. </p>
<p>Does it have anything to do with aerodynamics?</p>
| |mechanical-engineering|materials|automotive-engineering|metals| | <p>Price average Titanium alloys <a href="https://www.ownyourownfuture.com/how-much-is-titanium-worth-per-pound/" rel="nofollow noreferrer">costs</a> 70-80 dollars a pound.
Steel cost 0.03 dollars a pound.
A unibody car chassis suddenly becomes 2000x more expensive for the raw material alone. <a href="https://i.stack.imgur.com/1J1iU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1J1iU.jpg" alt="enter image description here" /></a></p>
<p>Workability, titanium is hard to weld without inert gas or friction welding, both expensive process compared to the standard welding or robotic welding techniques common in most auto plants.</p>
| 14897 | Why are car bodies not made of titanium? |
2017-04-22T15:59:03.613 | <p><a href="https://i.stack.imgur.com/YGTY6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YGTY6.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/4gBX8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4gBX8.jpg" alt="enter image description here"></a></p>
<p>I made my code like this,</p>
<pre><code>function DK = pendulum_cartesian(t,K)
g = 9.8; l = sqrt(K(1)^2 + K(2)^2);
DK = zeros(4,1);
DK(1) = K(3);
DK(2) = K(4);
DK(3) = ( (-K(1)*K(3)^2) - (K(1)*K(4)^2) + K(1)*K(2)*g )/(l^2);
DK(4) = ( (-K(2)*K(3)^2) - (K(2)*K(4)^2) - (K(1)^2)*g )/(l^2);
theta = atan(K(2)/K(1));
plot(t,theta);
end
</code></pre>
<p>and i typed,</p>
<p>[t,K] = ode45(@pendulum_cartesian, [0,10], [0.1,0.00017,0.1,0])</p>
<p>but there is no value in theta, and plots nothing.</p>
<p>where is a problem?</p>
| |matlab| | <p>Assuming your equations and your code is correct.</p>
<p>This is what I used for the function:</p>
<pre><code>function DK = pendulum_cartesian(t,K)
DK = zeros(4,1);
DK(1) = K(3);
DK(2) = K(4);
DK(3) = ( (-K(1)*K(3)^2) - (K(1)*K(4)^2) + K(1)*K(2)*9.81 )/(K(1)^2 + K(2)^2);
DK(4) = ( (-K(2)*K(3)^2) - (K(2)*K(4)^2) - (K(1)^2)*9.81 )/(K(1)^2 + K(2)^2);
end
</code></pre>
<p>Then run with:</p>
<pre><code>[t,K] = ode45(@pendulum_cartesian, [0,10], [0.1;0.00017;0.1;0]);
theta = atan(K(:,2)/K(:,1));
plot(t,theta);
</code></pre>
| 14921 | i have problem with cartesian coordinates pendulum EOM in matlab |
2017-04-23T00:34:21.947 | <p><a href="https://i.stack.imgur.com/Rag1B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rag1B.png" alt="Question"></a></p>
<p>b) I dont understand how to take the integration like its integral of ft from 0 to 2 pi = integral of velocity, and the integral of the first is the area under the graph= 8j and then I dont know how to get the answer fron there cause dont get what to plug in ?
Any help is greatly appresiated. </p>
| |dynamics|acceleration| | <p>For the first question, you can split up the equations of motion into normal and tangential direction:</p>
<p>$$ma_n=F_n$$
$$ma_t=F_t.$$</p>
<p>The tangential force $F_t=12/\pi \text{ N}$. Hence, $a_t=\frac{12}{10\pi} \frac{\text{m}}{\text{s}^2}$. For the normal force (centripetal force) you have to remember that $F_n=m\frac{v^2}{r}$. Hence, $a_n=\frac{v^2}{r}=1\frac{\text{m}}{\text{s}^2}$. The total acceleration is given by $a=\sqrt{a_t^2+a_n^2}=1.07\frac{\text{m}}{\text{s}^2}$.</p>
<p>So, now it's up to you. Try the other questions and tell us what you have tried or where you got stuck.</p>
<p><strong>Edit:</strong>
For b) we note that $$ma_t=F_t\implies \frac{F_t}{m}=a_t=\dfrac{dv}{dt}.$$
This can be rewritten by the chain rule (Note, that $ds/dt=v$) as:
$$\frac{F_t}{m}=\dfrac{dv}{ds}\dfrac{ds}{dt}=v\dfrac{dv}{ds}\implies \int_{s=0}^{2\pi}F_t ds=m\int_{v_0}^{v_{2\pi}}vdv.$$</p>
<p>The integral of on the right-hand side is just the area in your $s-F_t$-diagram. It turns out to be $8 \text{ J}$. Can you continue from here?</p>
| 14928 | Theory on tangential acceleration |
2017-04-23T04:31:44.683 | <p>I'm trying to use the <a href="https://en.wikipedia.org/wiki/Impedance_analogy" rel="nofollow noreferrer">mechanical impedance analogy</a> to describe a spring mass damper system as an electrical circuit. </p>
<p>Mechanical Impedances for linear motion are</p>
<p>$$ Z_{lin}(s) = \frac{F(s)}{\dot X(s)} = Ms + B + \frac{K}{s} $$</p>
<p>where M has units of kg, B has units of N-s/m and K has units N/m. The units of mechanical impedance then are N-s/m.</p>
<p>When I translate this to the rotational domain, I start having problems.</p>
<p>Rotational impedance will again be the ratio of an effort (torque) to a flow (rotational velocity).</p>
<p>Similarly, we have</p>
<p>$$ Z_{rot}(s) = \frac{\tau(s)}{\omega(s)} = Js + B_{rot} + \frac{K_{rot}}{s} $$</p>
<p>Based on dividing torque by angular velocity, the units of rotational impedance are [N-m-s/rad], or $\frac{ML^2}{T}$. When I try to verify that the units of rotational impedance are consistent, I get </p>
<p>$$J\cdot s = \frac{kg\cdot m^2\cdot rad}{s} = N\cdot m\cdot s \cdot rad = \frac{ML^2}{T}$$</p>
<p>$$B = N\cdot m\cdot\frac{s}{rad} =\frac{ML^2}{T} $$
and
$$\frac{K}{s} = \frac{N\cdot m\cdot s}{rad^2}=\frac{ML^2}{T}$$</p>
<p>I can't sleep with all those radians being all over the place! It seems awfully inconsistent to have them in different place and I haven't been able to find any sources on the web or in my textbooks that adequately deal with this since mechanical impedance is such a niche topic. In doing this, I've assumed that the $s$ has units of [rad/s], which some people dislike and just use $s= \frac{1}{T}$ Is there a way for me to make this a little more consistent, possibly by using radians in the units for rotational inertia and the spring constant?</p>
<p>I've also looked at the discussions here:</p>
<ol>
<li><a href="https://physics.stackexchange.com/questions/36079/unit-of-torque-with-radians">https://physics.stackexchange.com/questions/36079/unit-of-torque-with-radians</a></li>
<li><a href="https://physics.stackexchange.com/questions/11500/simple-harmonic-motion-what-are-the-units-for-omega-0">https://physics.stackexchange.com/questions/11500/simple-harmonic-motion-what-are-the-units-for-omega-0</a></li>
<li><a href="https://physics.stackexchange.com/questions/33542/why-are-radians-more-natural-than-any-other-angle-unit">https://physics.stackexchange.com/questions/33542/why-are-radians-more-natural-than-any-other-angle-unit</a></li>
</ol>
| |mechanical-engineering|control-engineering|robotics|circuit-design| | <h2>Radians and unit analysis</h2>
<p>The comment by <a href="https://engineering.stackexchange.com/users/7627/mryoumath">MrYouMath</a> is correct. For the purposes of dimensional analysis, the radian is considered a dimensionless quantity and therefore does not need to be included. This is because the radian is defined by an arc length (L) divided by a radius (also L) to give (L/L). The units of the Laplace domain complex number, $s$, are technically rad/s which, as the OP states, are considered to be $1/T$ for dimensional analysis.</p>
<p>$$ J\cdot s = \left[\frac{N\cdot m \cdot s \cdot rad}{rad}\right] = \frac{ML^2}{T}\left(\frac{L}{L}\right)\left(\frac{L}{L}\right) = \frac{ML^2}{T}$$</p>
<p>So from a dimensional analysis perspective we are fine. You can double check the units for the other terms as well.</p>
<p>Mathematically, radians are often ignored/assumed because they are so convenient (you don't end up with a bunch of constant factors in your equations). In engineering, radians serve mostly as a reminder that we are using them instead of degrees! So you can ignore them until you are figuring out what the units of your final answer are.</p>
<h2>Mechanical impedance to blame?</h2>
<p>The reason that the units don't work out has nothing to due with the fact that this is a mechanical system. Check out the units for the transfer function of an RLC circuit:</p>
<p>$$ \frac{V}{I} = Ls + R + \left(\frac{1}{C}\right)\frac{1}{s} $$</p>
<p>$$\left[ \frac{kg\cdot m^2}{s^3 \cdot A^2}\right] = \left[\frac{kg \cdot m^2}{s^2 \cdot A^2}\right]\cdot\left[\frac{rad}{s}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right] + \left[\frac{kg\cdot m^2}{s^3\cdot A^2}\right]\cdot\left[\frac{s}{rad}\right]$$</p>
<p>Notice once again those pesky radians sticking their noses where they don't belong! In all seriousness though, you will see this "problem" regardless of what domain you work in. </p>
<p>The fact of the matter is simply this: when doing unit analysis in the Laplace domain (transfer functions, etc.), ignore the radian units for the complex number $s$. If you do this, the units <em>will</em> work out as long as you use consistent units for your physical constants $J$, $B$ and $K$ (you'll have to check it for the mechanical system yourself, I've wasted far too much time at work already).</p>
<p>Final note: Just remember that when you do your frequency analysis the units of frequency are still rad/s (because we use $\sin{\omega t}$ instead of $\sin{2\pi ft}$).</p>
| 14929 | Units for rotational impedance |
2017-04-23T05:30:47.860 | <p>I am looking at Hibbeler's section on a `Cable Subjected to a Distributed Load' in Engineering Mechanics - Statics, Ed13, as shown in the attached images.</p>
<p><a href="https://i.stack.imgur.com/GlUhl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GlUhl.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/EiveO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EiveO.png" alt="enter image description here"></a></p>
<p>I am getting stuck deriving equation (7-7). I derive the horizontal equilibrium equation just fine;</p>
<p>$$
|\vec{T} + \vec{\Delta T}| \cdot \cos(\theta + \Delta \theta) - |\vec{T}|\cdot \cos(\theta) = 0
$$</p>
<p>I rearrange to describe horizontal tension component as a function of $\Delta x$ (given $\Delta T$ and $\Delta \theta$ are ultimately functions of $\Delta x$);</p>
<p>$$
|\vec{T}|\cdot \cos(\theta) = |\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)
$$</p>
<p>Then I express the differential;</p>
<p>$$
\frac{\Delta(|\vec{T}|\cdot \cos(\theta))}{\Delta x}
=
\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}
$$</p>
<p>Then when I go to express the derivative, I come up with a limit which does not exist. </p>
<p>$$
\frac{d(|\vec{T}|\cdot \cos(\theta))}{dx}
=
\lim_{\Delta x \to 0}\frac{|\vec{T} + \vec{\Delta T}|\cdot \cos(\theta + \Delta \theta)}{\Delta x}
$$</p>
<p>Further to this, it is not obvious to me how the resultant force from the distributed load acting over $\Delta x$ can be defined as $w(x)\Delta x$ (in line 8) as the author suggests. I feel it should be defined as;</p>
<p>$$
\Delta F = \int_{x}^{x+\Delta x}w(x)dx
$$</p>
<p>becuase the loading intensity function is not constant. Any pointers as to where my errors are? Appreciate any advice offered. Thanks.</p>
| |structural-engineering|structural-analysis|structures| | <p>You have</p>
<p>$$-T\cos \theta + (T+\Delta T) \cos( \theta + \Delta \theta)=0.$$</p>
<p>It is important to note that $T$ and $\theta$ are both depending on $x$. Dividing the previous expression by $\Delta x$ and rewriting $T+\Delta T = T(x+\Delta x)$ and $\theta + \Delta \theta = \theta(x+\Delta x)$, will lead to:</p>
<p>$$\frac{T(x+\Delta x)\cos [\theta(x+\Delta x)]-T(x)\cos[\theta(x+\Delta x)]}{\Delta x}=0.$$</p>
<p>Now, take the limit $\Delta x \to 0$. As the above is the definition of the derivative of $T(x)\cos[\theta(x)]$ we obtain:</p>
<p>$$\dfrac{d}{d x}\left[T(x)\cos[\theta (x)]\right]=0 \implies T(x)\cos[\theta (x)] = \text{const.}$$</p>
| 14930 | Expressing Derivative of Cable Tension Component |
2017-04-23T19:32:15.217 | <p><strong>Problem</strong></p>
<p>Consider the following 2D rigid body in the $x$-$y$-plane, which is rigidly rotated around the $z$-axis with the angle $\beta$:</p>
<p><a href="https://i.stack.imgur.com/hzwQ2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hzwQ2.png" alt="Rigid Body Rotation - Kinematics"></a></p>
<p><em>(Image Source: <a href="http://www.continuummechanics.org/smallstrain.html" rel="nofollow noreferrer">Continuum Mechanics - Small Scale Strains - Limitations of Small Strain Equations - Effect of Rotations on Strains</a>)</em></p>
<p>A point $[X, Y]^T$ will then be rotated to a new point
$$[x, y]^T = \ldots$$
$$\ldots = [X \cdot \cos(\beta) - Y \cdot \sin(\beta), \ldots$$
$$\ldots X \cdot \sin(\beta) + Y \cdot \cos(\beta)]^T$$</p>
<p>The engineering strains for this rigid body rotations can thus be given as
$$\epsilon_{xx} = \cos(\beta)-1$$
$$\epsilon_{yy} = \cos(\beta)-1$$
$$\epsilon_{xy} = 0$$</p>
<p>This means that normal strains will be induced for (large) rigid body deformations. For $\beta=90°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -1$. For $\beta=180°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -2$.</p>
<p>In a text book, I found a corresponding picture, which shows a FEM simulation based on engineering strains. As a result of the rigid body rotation, the body grew in size:</p>
<p><a href="https://i.stack.imgur.com/3HVUx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3HVUx.png" alt="Rigid Body Rotation - FEM"></a></p>
<p>This seems to correspond to a <a href="https://de.wikipedia.org/wiki/Datei:Drehung.gif" rel="nofollow noreferrer">gif</a> I found on the German Wikipedia site (here a static version) showing the von Mises stresses:</p>
<p><a href="https://i.stack.imgur.com/DCe6o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DCe6o.png" alt="Rigid Body Rotation - von Mises stresses"></a></p>
<p><strong>Question</strong></p>
<p>Why does the body grow in size instead of getting smaller? The normal strains are always smaller than zero, which means that the stresses are also smaller than zero (for a linear stress-strain-relationship). Would that not entail compression?<br>
What do I miss?</p>
| |stresses| | <p>The question seems to be "why does the FEM seemingly produce results that are contradictory to theory?"</p>
<p>Consider a square with nodes </p>
<pre><code>id x y
1 -1.0 -1.0
2 1.0 -1.0
3 1.0 1.0
4 -1.0 1.0
</code></pre>
<p>The FE displacement field is
$$
u_x(x,y) = \sum_{j=1}^4 u_x^j N_j(x,y) ~,~~
u_y(x,y) = \sum_{j=1}^4 u_y^j N_j(x,y)
$$</p>
<p>The strain field is
$$
\varepsilon_{xx}(x,y) = \sum_{j=1}^4 u_x^j \frac{\partial N_j(x,y)}{\partial x} ~,~~
\varepsilon_{yy}(x,y) = \sum_{j=1}^4 u_y^j \frac{\partial N_j(x,y)}{\partial y}
$$
and
$$
\varepsilon_{xy}(x,y) = \tfrac{1}{2}\sum_{j=1}^4 \left[u_x^j \frac{\partial N_j(x,y)}{\partial y} + u_y^j \frac{\partial N_j(x,y)}{\partial x}\right]
$$
The nodal shape functions are
$$
N_1(x,y) = \frac{(1-x)(1-y)}{4} ~,~~
N_2(x,y) = \frac{(1+x)(1-y)}{4}
$$
$$
N_3(x,y) = \frac{(1+x)(1+y)}{4} ~,~~
N_4(x,y) = \frac{(1-x)(1+y)}{4}
$$
The gradients of the shape functions are
$$
\begin{align}
G_{1x} := \frac{\partial N_1(x,y)}{\partial x} & = -\frac{(1-y)}{4} ~,~~
G_{1y} := \frac{\partial N_1(x,y)}{\partial y} = -\frac{(1-x)}{4} \\
G_{2x} := \frac{\partial N_2(x,y)}{\partial x} & = \frac{(1-y)}{4} ~,~~
G_{2y} := \frac{\partial N_2(x,y)}{\partial y} = -\frac{(1+x)}{4} \\
G_{3x} := \frac{\partial N_3(x,y)}{\partial x} & = \frac{(1+y)}{4} ~,~~
G_{3y} := \frac{\partial N_3(x,y)}{\partial y} = \frac{(1+x)}{4} \\
G_{4x} := \frac{\partial N_4(x,y)}{\partial x} & = -\frac{(1+y)}{4} ~,~~
G_{4y} := \frac{\partial N_4(x,y)}{\partial y} = \frac{(1-x)}{4}
\end{align}
$$
Therefore the strain field is
$$
\begin{bmatrix}
\varepsilon_{xx}(x,y) \\ \varepsilon_{yy}(x,y) \\ 2\varepsilon_{xy}(x,y)
\end{bmatrix}
= \begin{bmatrix}
G_{1x} & G_{2x} & G_{3x} & G_{4x} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & G_{1y} & G_{2y} & G_{3y} & G_{4y} \\
G_{1y} & G_{2y} & G_{3y} & G_{4y} & G_{1x} & G_{2x} & G_{3x} & G_{4x}
\end{bmatrix} \mathbf{u}
$$
where
$$
\mathbf{u} = \begin{bmatrix}
u_x^1 & u_x^2 & u_x^3 & u_x^4 & u_y^1 & u_y^2 & u_y^3 & u_y^4
\end{bmatrix}^T
$$
Consider the case where the square is rotated by 90 degrees clockwise around node 1 so that the new positions of the nodes are</p>
<pre><code>id x y
1 -1.0 -1.0
2 -1.0 -3.0
3 1.0 -3.0
4 1.0 -1.0
</code></pre>
<p><em>Note that we are not assuming any deformation of the square.</em></p>
<p>Then the displacements are</p>
<pre><code>id u_x u_y
1 0.0 0.0
2 -2.0 -2.0
3 0.0 -4.0
4 2.0 -2.0
</code></pre>
<p>If we plug in the values, at node 1 we get</p>
<pre><code>B1 = [[G1x(1) G2x(1) G3x(1) G4x(1) 0 0 0 0];...
[0 0 0 0 G1y(1) G2y(1) G3y(1) G4y(1)];...
[G1y(1) G2y(1) G3y(1) G4y(1) G1x(1) G2x(1) G3x(1) G4x(1)]]
= -0.50000 0.50000 0.00000 -0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 -0.50000 -0.00000 0.00000 0.50000
-0.50000 -0.00000 0.00000 0.50000 -0.50000 0.50000 0.00000 -0.00000
</code></pre>
<p>and the nodal displacement vector is</p>
<pre><code>u = [0 -2 0 2 0 -2 -4 -2]
</code></pre>
<p>The strain at node 1 is then </p>
<pre><code>eps1 = B1*u' = [-1 -1 0]'
</code></pre>
<p>Therefore, <strong>the finite element solution is identical to your solution</strong> and just says that <strong>stresses will develop in the element due to pure rigid body rotation even if the element does not deform</strong>.</p>
<p>The finite element method is typically implemented in displacement form. So you specify the displacements and then find the stresses in the element. </p>
<p>An alternative would be to specify moments that would rotate the element. That's tricky to do. I'm more inclined to believe that the figure in the textbook is just meant to be an illustration.</p>
| 14944 | Rigid Body Rotations and Engineering Strain |
2017-04-24T13:12:31.937 | <p>While reading about airplane engines on the internet, it is common to find figures of their thrust, in kilo Newton. However, in the case of ships and naval vessels, when talking about the engine, the only parameter mentioned is the horsepower of the power plant driving the propeller.</p>
<p>So, my question is, what is the typical thrust generated by a large naval ship (such as the nuclear-powered USS Enterprise aircraft carrier, with a power of 210 MW)? How can this be calculated?</p>
| |ships|thrust| | <p>This gives a complete analysis of how to calculate : <a href="http://web.mit.edu/13.012/www/handouts/propellers_reading.pdf" rel="nofollow noreferrer">http://web.mit.edu/13.012/www/handouts/propellers_reading.pdf</a></p>
| 14951 | Thrust generated by a ship |
2017-04-26T10:57:48.177 | <p>I currently have a ditch that fills with water draining out into an area lower than the water (via a hose), but the hose for most of the travel extend above the hight of the water.</p>
<p>As long as the hose stays filled with water, which it should assuming the water doesn't evaporate enough, this system works. Unfortunately, the water doesn't drain as quickly as I'd like after a rain storm.</p>
<p><strong>Does the piping at all effect (e.g. the volume it holds) at what level the the basin will drain to?</strong></p>
<p><strong>If I increase the diameter of the piping will the volume of water flowing out per minute increase?</strong> </p>
<p>Bonus points for providing the physics principles being applied here. I'm not familiar with the names.</p>
<p><a href="https://i.stack.imgur.com/BOyBv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BOyBv.png" alt="basin drainage diagram"></a></p>
| |fluid-mechanics| | <p>While the other answers give a relationship, there are factors that need to be included such as friction factor, inlet & outlet coefficients and length, so this link gives a more practical approach : <a href="http://files.engineering.com/getfile.aspx?folder=23da7aca-8762-4e14-b029-09" rel="nofollow noreferrer">http://files.engineering.com/getfile.aspx?folder=23da7aca-8762-4e14-b029-09</a></p>
| 14975 | Will water level of remain level between two points |
2017-04-26T14:49:38.753 | <p>I have a 10 x 10cm square plate of metal I would like to insert into a hole in the wall of plastic water tank (cube shape), I'm planning on 3D printing the tank itself (only 1-2 litres). </p>
<p>What would be the best design to stop water from leaking between the plate and the plastic body? </p>
<p>So far all I can think of is to make the hole in the tank smaller (i.e. 8 x 8cm) than the metal sheet and perhaps using a sealant and affix it using screws possibly, but as there's live wiring around it I need to be sure it has the best possible seal. Any design ideas would be much appreciated </p>
| |materials|design|3d-printing|waterproofing| | <p>It sounds like you just need a small <a href="https://tankliners.com.au/" rel="nofollow noreferrer">liner</a> on the plate itself. Foam sealants are typically used when the space that needs sealing is greater than 1/4inch. For smaller spaces, waterproof caulking may be used instead. Just Do not allow the foam sealant to come into contact with skin; it adheres to the skin and can pull off the top layer of the epidermis when removal is attempted.</p>
<p>Other quick and easy temporary measures for stopping pipe leaks include wrapping waterproof tape over the bad spot or rubbing the hole with a stick of special compound. Applying epoxy paste or inserting a self-tapping plug into the hole are other alternatives. When using waterproof tape, be sure to dry the pipe thoroughly before you start wrapping.</p>
<p>Here's an interesting site I found about the topic too:
<a href="https://haynes.com/en-us/tips-tutorials/how-trace-interior-water-leaks-your-car" rel="nofollow noreferrer">https://haynes.com/en-us/tips-tutorials/how-trace-interior-water-leaks-your-car</a></p>
| 14982 | How to water seal a metal plate into the wall of a plastic water tank? |
2017-04-27T11:50:03.547 | <p>Could anyone tell me what the variables in this formula are? </p>
<p>$$\tau=\frac{S\;a\; \bar{v}}{I\; t}$$</p>
<p>I believe it is used for shear stress and that $I$ is moment of inertia but unsure on the other variables.</p>
| |mechanical-engineering|stresses| | <p><img src="https://i.stack.imgur.com/SzMtI.png" width="400"></p>
<p>The sign convention I currently use is as shown ($x$ "out of the display" towards you, $y$ to the left, $z$ down)
The way I was taught the formula is:
$$ \tau=\frac{V_z\cdot Q_y}{I_y \cdot t} $$</p>
<p>$V_z$ … shear force at position $x$ in $z$-direction <br>
$Q_y$ … first moment of area ($\int zdA$) of the blue area wrt neutral axis<br>
$I_z$ … second moment of area of ($\int z^2dA$) entire cross section wrt neutral axis<br>
$t$ … thickness of cross section at point where $\tau$ has to be determined</p>
<p>$Q_y$ can also be written as $A\bar{z}$, where $A$ is the blue area and $\bar{z}$ the $z$-coordinate of its centroid.</p>
| 15006 | Shear Stress Formula |
2017-04-28T22:38:27.050 | <p>I read that a Tesla Model S P100D achieved 730 horsepower on a dynamometer. Would increasing battery size increase the horsepower? If not, what would?</p>
| |motors|automotive-engineering|battery|electric-vehicles| | <p>A Tesla battery is made from many cells, there is a max current rating for each cell, a larger battery has more cells and therefore give out a larger peak current. (Ian ringrose)</p>
<p>It is also linked to the naximum output of the electric motor. Which produces the rotation.</p>
| 15039 | How do you increase the horsepower of an electric motor? |
2017-04-29T11:46:39.323 | <p>Where is L measured from in the formula for the Reynolds Number. I know it is a linear dimension of the system measured in metres but unsure of how to measure it.</p>
| |fluid-mechanics|thermodynamics| | <p>Reynolds number is a dimensionless constant used to describe the flow characteristics of a flow system. It is a tool for ccomparative analysis. So the absolute value of Reynolds number does not hold any physical meaning unless the flow conditions are specified.
The L is a parameter used to quantify the extent or size of flow.It can be choosen to be any length which gives an idea of the extent of the flow.But once we compute Reynolds number of a flow by choosing a particular length(say, the chord for a wing), it should be made sure that we have to use the same length in computing the Reynolds number of another flow if both are intended to be compared.
Hope you got it.</p>
| 15044 | Reynolds Number |
2017-04-29T18:42:40.083 | <p>I'm doing a gear design project, and I'm using a multivariate mixed integer non-linear optimization suite known as <code>Couenne</code> distributed by COIN-OR. Within Python, I use this solver to solve for all aspects of the gears in a 2 ratio reverted gear-train given 2 target gear train ratios (4.3 fwd, 9.1 rev). My classmates are telling me that the diametral pitch for each gear pair must be an integer value, or a value with a well known fraction in decimal form. Is this true? I've found that placing this constraint makes it much harder to approach the 2 target ratios. Perhaps it was a consideration about manufacturing, because It would be exceedingly hard to machine a diameter that is an irrational number. However, couldn't the same be said for any diameter? Manufacturing shouldn't depend on the dimension because for any given manufacturing precision we will never know what is after the decimal place. For both irrational and integer diametral pitches, the manufacturing error would be the same. So why would gear diametral pitches have to be integer values?</p>
<p>Without this constraint, I'm able to achieve a total gear train ratio error of 1.3*10^-8 percent from the target ratios.</p>
<p>EDIT: when I refer to "diametral pitch" I am referring to the gear module, number of teeth per inch</p>
| |gears|torque| | <p>That's why it's a STANDARD.</p>
<p>Using reasonably available tools you can measure part X, round the results to nearest standard-allowed values, and pick a matching part from catalogue or manufacture following a simple set of standard-defined parameters, and it will fit. With weirdo sizes you have a weirdo system where every element needs to be custom-calculated and custom-made because it fits nothing in the world except what it was made for.</p>
<p>It's an arbitrary restriction to curb anarchy of a billion custom-purpose standards that don't match each other, allow various manufacturers to provide standarized parts that match each other and fulfill all reasonable expectations.</p>
<p>So, the diametral pitch doesn't have to be integer. Engineers will hate you for it, and people will criticize you for vendor lock-in practices, plus you'll need to manufacture all the gears on your own, but there's no law (legal, or of physics) that forbids it.</p>
| 15049 | Does gear diametral pitch have to be an integer? (teeth/diameter) |
2017-04-30T05:28:36.507 | <p>I want to build a home automation system by hooking up various sensors (e.g., passive IR motion sensors, magnetic door contact switches, vibration detectors, etc.) around the house to detect things then send the sensor signals to a central command center / control station.</p>
PIR Motion Sensor
<p><a href="https://i.stack.imgur.com/gfZxW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gfZxW.jpg" alt="enter image description here"></a></p>
<p>I want to do this <em>wirelessly</em> and <em>solderlessly</em> with a budget of roughly $10 per sensor. So, for example, I could hook up a breadboard circuit that receives the sensor signal and sends it to a Wifi network.</p>
<p>Is there a simple circuit component or IC that can convert the analog sensor signal to a wireless signal then transmit that signal to the network via wifi without, say, hooking up an entire microprocessor (e.g., Raspberry Pi) to each sensor? A Raspberry Pi for each sensor would blow my budget.</p>
<p>I'm looking to install a simple component part in, say, a solderless breadboard circuit at the sensor level. Then read all the signals off the network at the control station using a Raspberry Pi or laptop.</p>
<p>What kind of components and configuration would anyone here recommend?</p>
<h2>Edit 1</h2>
<p>Apparently, a USB wireless adapter or a network adapter might be the component I'm looking for. But would it be possible to build up a separate breadboard circuit for each sensor and pipe all the signals to the wifi network? If so, how? I want to keep my budget to about $10 per sensor.</p>
USB Wireless Adaptor
<p><a href="https://i.stack.imgur.com/23etr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/23etr.jpg" alt="enter image description here"></a></p>
<h2>Edit 2</h2>
<p><a href="http://www.electronicszone.eu/index.php/en/12-english/199-infrared-motion-detector" rel="nofollow noreferrer">I found this article</a> which reflects what I'm trying to do with the sensor. But it says nothing about a wireless (analog-to-digital) conversion.</p>
PIR Motion Sensor Schematic
<p><a href="https://i.stack.imgur.com/E2xw7.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E2xw7.gif" alt="enter image description here"></a></p>
Breadboard Hookup
<p><a href="https://i.stack.imgur.com/NShw9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NShw9.jpg" alt="enter image description here"></a></p>
<h2>Edit 3</h2>
<p><a href="https://openhomeautomation.net/getting-started-esp8266/" rel="nofollow noreferrer">I like the suggested ESP8266 chip described here</a>. (Could this really be $ 5-10 per sensor?)</p>
ESP8266 chip
<p><a href="https://i.stack.imgur.com/YFYPl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YFYPl.jpg" alt="enter image description here"></a></p>
Breadboard configuration
<p><a href="https://i.stack.imgur.com/5c3yo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5c3yo.jpg" alt="enter image description here"></a></p>
Schematic
<p><a href="https://i.stack.imgur.com/FvrLI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FvrLI.png" alt="enter image description here"></a></p>
| |electrical-engineering|sensors|embedded-systems|wifi|iot| | <p>The cheapest option for you to consider would be to use Esp8266 modules. They technically cost 1 USD. You can import a bulk (lot) order from aliexpress to make it cheaper. For you command centre you can use a NodeMCU (Esp8266 V12E) costing about 10-12 USD (800 INR). <br/></p>
<p>Program the NodeMCU as a server and the smaller Esp8266s as clients. It is pretty simple and you can choose from many libraries on 'github'. Their GPIOs receive inputs from each sensor and transmit the data over Wi-Fi as there are all connected to a network hosted by your router/modem to your command centre. <br/></p>
<p>The controllers need a 3V input which is not common, so you need to use level controllers (costs ~2 dollars) step down standard voltages. Once you have received all the data on your server you can process it as desired and program it to give necessary feedback to your sensors.</p>
<p>Cost (excluding senors) for approx 5 sensors = 20 USD.<br/>
Refer the links below for more information.<br/>
<a href="http://www.nodemcu.com/index_en.html" rel="nofollow noreferrer">http://www.nodemcu.com/index_en.html</a><br/>
<a href="http://esp8266.net/" rel="nofollow noreferrer">http://esp8266.net/</a><br/>
<a href="https://iot-playground.com/blog/2-uncategorised/42-esp8266-wifi-pir-motion-sensor-arduino-ide" rel="nofollow noreferrer">https://iot-playground.com/blog/2-uncategorised/42-esp8266-wifi-pir-motion-sensor-arduino-ide</a></p>
| 15054 | Send sensor signals to WiFi network |
2017-04-30T10:56:15.203 | <p>In my text book , it is written
"Power generated by the turbine can be increased by using a gradual expansion at the turbine outlet"
The picture below explains the meaning of gradual expansion at the turbine outlet <a href="https://i.stack.imgur.com/09X1l.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/09X1l.jpg" alt="enter image description here"></a></p>
<p>i.e the cross sectional area of the pipe at the turbine outlet is increasing gradually m, causing gradual expansion.. </p>
<p>How does this thing cause the power generated by the turbine to increase ? </p>
| |mechanical-engineering|fluid-mechanics|turbines|turbomachinery| | <p>The turbine referred to, in your text book must be a specific type of turbine, namely reaction turbine. Reaction turbines utilise mainly the pressure energy of flowing water to produce Power. The flowing fluids consist of both pressure energy(a function of fluid pressure) and kinetic energy(a function of flow velocity). The reaction turbines are rotated by the pressure difference of water on both sides of turbine blades. The reaction turbines convert most of the pressure energy to mechanical power, but they are unable to convert much of the kinetic energy in such a way. So the kinetic energy brought in by the water is left to flow away through the outlet
But if a pipe whose diameter is small at first and then increasing slowly is used at outlet, then the pressure at the pipe beginning will be lowered significantly. This happens due continuity conditions and Bernoulli's principle.
According to continuity equation if the cross sectional area of a flow decreases the flow velocity increases. But according to Bernoulli's principle if flow velocity increases the pressure decreases, given there is no change in elevation, losses, external work etc.
Thus the pressure at outlet side of turbine decreases and the pressure difference between both sides of the turbine increases. Thus the driving force increases and hence the power. So such A pipe actually converts some of the kinetic energy into pressure energy for better utilisation in reaction turbines.</p>
<p>Now, the water cannot flow outside if its pressure remains below atmospheric pressure. So in order to increase the pressure the diameter is increased slowly. In this case also Bernoulli's principle is utilised. The increase in diameter of the outlet pipe is kept gradual to avoid losses (eg: Eddy formation due to sudden expansion). Such pipes with gradually increasing diameters used in reaction turbines are called Draft Tubes.</p>
| 15059 | Does the turbine outlet affect the power generated by the turbine ? |
2017-05-01T10:44:09.803 | <p>I am working on a project using springs. What is the maximum load per deflection of springs available in market?</p>
| |mechanical-engineering|springs| | <p>There is no sensible answer to this. Springs are designed to have a relatively high deflection relative to their to ultimate strength. Once you want very stiff springs you can just use solid bars of steel. </p>
<p>The trick with spring design is to end up with a product which has a <em>specific</em> set of properties and it is possible to make arbitrarily stiff or light springs. What may be more difficult is combining a desired stiffness and capacity with a particular set of dimensions and lifetime. </p>
<p>Equally any answer may also depend on what you define as the difference between a spring and a structure as all solid materials are essentially springs for a given load range. </p>
| 15074 | Load per deflection of springs |
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