CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2017-01-31T13:15:53.430 | <p>I am currently integrating a C-Code (written by a third person) in a Simulink model. In the C-Code various sensor signals are filtered, using a PT1 filter with the following equation:</p>
<p>$$y(t) = \dfrac{u(t) + Cy(t-1)}{C + 1}$$</p>
<p>Where $y(t)$, $u(t)$ and $C$ are the output (the filtered signal), input and a factor applied for each signal individually, respectively.</p>
<p>Looking at the step function response, I can see a typical PT1-behaviour with a gain factor of 1.</p>
<p>I cannot figure out the derivation of this function from the general formulation (the differential equation) of a PT1, which is:</p>
<p>$$Ty'(t) + y(t) = Ku(t)$$</p>
<p>Can any of you guys help me out?</p>
| |control-engineering|electronic-filters| | <p>The equation being used in code is a discrete-time implementation of the filter. The second equation you gave is a continuous-time expression of the filter dynamics. In order to create a discrete-time filter from the continuous-time dynamics the person who wrote the code used a first order approximation (backwards difference, specifically) of the derivative of $y$, like so:</p>
<p>$$\frac{dy}{dt} \approx \frac{y(t)-y(t-\Delta t)}{\Delta t}$$</p>
<p>Now, if we use the notation $k$ to denote a sample taken at time $t$ and $k-1$ to denote a sample taken at time $t-\Delta t$, then we can derive the equation used in code as follows:</p>
<p>$$T\frac{dy}{dt} + y(t) = Ku(t)$$
$$T \frac{y(k)-y(k-1)}{\Delta t} + y(k) = Ku(k)$$
$$\left(\frac{T}{\Delta t} + 1\right) y(k) = K u(k) + \frac{T}{\Delta t} y(k-1)$$</p>
<p>Now let $K=1$ and $C = \frac{T}{\Delta t}$, and you get:</p>
<p>$$\left(C + 1\right) y(k) = u(k) + C y(k-1)$$
$$ y(k) = \frac{u(k) + C y(k-1)}{C + 1}$$</p>
| 13516 | PT1 Filter without Derivative |
2017-02-01T08:50:01.280 | <p>I want to generate NACA 63824 airfoil coordinates to use them in my bachelor thesis.
I tried airfoil tools and java foil. But I can't generate 638xx series. </p>
<p>Can you recommend me a program or a code that generates these airfoil coordinates? </p>
| |airfoils| | <p>If you are a MATLAB or Oculus user, there is the <a href="https://www.mathworks.com/matlabcentral/fileexchange/23241-naca-5-digit-airfoil-generator/content/naca5gen.m" rel="nofollow noreferrer">NACA 5 digit Airfoil Generator</a>
program you can use. It works very well, considering it calculates the plot, as well as other main features of the airfoil, given the airfoil type. </p>
<p>Now, I am unaware of a program that gives plots based on the formula, but if you ever DO find out, please do not hesitate to post the link here.</p>
| 13526 | How to generate NACA 5 digit airfoil coordinates? |
2017-02-01T12:20:11.330 | <p>Let's say I have a large isotropic body in uniaxial tension, onto which a small rectangular piece of film is applied. Think of a strain gage on a big body, but the 'strain gage' is one single continuous isotropic layer.</p>
<p>Let x- and y-direction be on the surface of the big body and z-direction the normal (piercing trough the strain gage).</p>
<p>In the <em>big body</em> I get $\sigma_x = \sigma_0; \sigma_y=\sigma_z=0$.
Therefore</p>
<p>$$\begin{align}
\epsilon_x &= \frac{\sigma_x}{E} \\
\epsilon_y = \epsilon_z &= \frac{\sigma_x\nu}{E}
\end{align}$$</p>
<p>What is the strain in the small body?</p>
<p>My thinking was: Assuming I have an ideal boundary layer and negligible thickness of the film, the strain should be the same as in the big body. This is consistent with literature for $\epsilon_x$ and $\epsilon_y$.
But I keep reading an expression for $\epsilon_z$ in the <em>small body</em> involving $\nu_B$ of the big body and $\nu_s$ of the small body:</p>
<p>$$\epsilon_{z,smallbody}=-\frac{\nu_s(1-\nu_B)}{1-\nu_s}\sigma_x/E$$</p>
<p>Can someone explain how to get this result?</p>
| |mechanical-engineering|structural-engineering|applied-mechanics| | <p>I would like to start by stating that my answer is not exactly the same as the one you've found in the literature. It is almost the same, but one sign is switched. I don't know if you wrote it wrong or if I messed up somewhere (<strong>very possible</strong>, please correct me if you find my mistake). Regardless, I am so close that I'm sure the mistake is somewhere on this page (and not in the literature, of course). And it is close enough to show you where that equation comes from, which is your actual question.</p>
<hr>
<p>It is worth noting a key difference between the small and big bodies: the big body is in uniaxial tension, but the small one is under biaxial stress. It is under tension in the x axis but under compression in the y axis. Now, the deformation equations for biaxial stress are:</p>
<p>$$\begin{align}
\epsilon_x &= \dfrac{1}{E}(\sigma_x - \nu\sigma_y) \\
\epsilon_y &= \dfrac{1}{E}(-\nu\sigma_x + \sigma_y) \\
\epsilon_z &= \dfrac{-\nu}{E}(\sigma_x + \sigma_y) \\
\end{align}$$</p>
<p>Your small body is under known $\epsilon_x$ and $\epsilon_y$ (equal to those of the big body). From that we can calculate $\sigma_{x,s}$ and $\sigma_{y,s}$. (All variables with index $s$ are for the small body, $b$ for the big body)</p>
<p>$$\begin{align}
\epsilon_{x,s} = \epsilon_{x,b} &= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\
\dfrac{\sigma_{x,b}}{E_b} &= \dfrac{1}{E_s}(\sigma_{x,s} - \nu_s\sigma_{y,s}) \\
\therefore \sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\sigma_{y,s}\\
\epsilon_{y,s} = \epsilon_{y,b} &= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\
\dfrac{\nu_b\sigma_{x,b}}{E_b} &= \dfrac{1}{E_s}(-\nu_s\sigma_{x,s} + \sigma_{y,s}) \\
\therefore \sigma_{y,s} &= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s} \\
\therefore \sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\sigma_{x,s}\right) \\
\sigma_{x,s} &= \dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} \\
\therefore \sigma_{y,s} &= \dfrac{E_s\nu_b\sigma_{x,b}}{E_b} + \nu_s\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\
\sigma_{y,s} &= \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)
\end{align}$$</p>
<p>Now we just plug these in the $\epsilon_z$ equation. There's a lot to simplify here, so I'll show each of the steps.</p>
<p>$$\begin{align}
\epsilon_z &= -\dfrac{\nu_s}{E_s}\left(\dfrac{E_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{E_s\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\
\epsilon_z &= -\nu_s\left(\dfrac{\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \dfrac{\sigma_{x,b}}{E_b}\left(\nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right)\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2} + \nu_b + \nu_s\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s^2}(1+\nu_s) + \nu_b\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \nu_b\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b}{1-\nu_s} + \dfrac{(1-\nu_s)\nu_b}{1-\nu_s}\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\left(\dfrac{1 + \nu_s\nu_b + \nu_b-\nu_s\nu_b}{1-\nu_s}\right) \\
\epsilon_z &= -\dfrac{\nu_s\sigma_{x,b}}{E_b}\cdot\dfrac{1 + \nu_b}{1-\nu_s} \\
\end{align}$$</p>
<p>This is almost identical to the equation you've given from the literature, with the exception that my equation has $1 + \nu_b$, while the literature has $1 - \nu_b$. I can't figure out where I went wrong (or if you wrote it wrong), but this is so close that I believe it should demonstrate where the literature got this equation from.</p>
| 13528 | Poisson ratio boundary condition |
2017-02-01T14:04:17.380 | <p>Security valves with two connectors open if the pressure exceed a value x +/- tolerance on one side. I'd like to know if there's a mechanical valve which opens if the pressure on one side sinks below a certain value.</p>
<p>The use case is a system with a desired pressure d (e.g. 2 bar), but irregular pressure during operation (because it's based on bio mass fermentation) and a gas storage which stores extranous gas if the pressure is above 2 bar. This storage needs a value which can release pressure from the storage into the system if the pressure in the system sinks below the desired pressure.</p>
| |fluid-mechanics|valves| | <p>You're looking for a <a href="https://en.wikipedia.org/wiki/Pressure_regulator" rel="nofollow noreferrer">pressure regulator</a> (and this is indeed what you need in your use case, but that is another question.) The regulator works so that over a range of flows, the pressure on the outlet side is kept constant.</p>
| 13529 | Is there a (sort of inverse security) valve which opens if one connected side has a pressure below a certain value? |
2017-02-01T15:59:09.647 | <p>Here's the setup:</p>
<p><a href="https://i.stack.imgur.com/QBIKv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QBIKv.jpg" alt="Cantilever set-up"></a></p>
<p>The problem asks to find the relationship between the force $P$ and the displacement of B ($u_B$). I have done this correctly:</p>
<p>$$P = \left(\dfrac{E_1A_1}{L_1}+\dfrac{E_2A_2}{L_2}\right)u_B$$</p>
<p>Now imagine that members AB and BC both have yield stresses ($\sigma_{Y_1}$ and $\sigma_{Y_2}$) and are perfectly plastic after yield. </p>
<p>I need to find the relationship between $u_B$ and $P$ in the following cases:</p>
<ol>
<li>No member yields - ($0 < u_B < \sigma_{Y_1}$) & ($0 < P < P_{y_1}$) </li>
<li>Member AB yields, but not member BC - ($u_{Y_1} < u_B < u_{Y_2}$) & ($P_{y,1} < P < P_{y,2}$)</li>
<li>Both members yield - ($u_B > u_{Y_2}$) & ($P > P_{y,2}$)</li>
</ol>
<p>I'm having trouble finding explicit expressions for $P_{y_{1/2}}$ and $u_{Y_{1/2}}$. I'm not quite sure how to relate yield stress to elongation and displacement of the members. </p>
<p>To be honest, this problem has tied my brain in a knot and I believe my conceptual understanding (or misunderstanding) of the problem is limiting my ability to solve it mathematically. Any help at all would be greatly appreciated!</p>
<p>Edit: Here is the exact problem
<a href="https://i.stack.imgur.com/6cUI4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6cUI4.png" alt="Exact Problem"></a></p>
| |structural-engineering|statics|stresses| | <p>From Hooke's Law, we know that</p>
<p>$$\epsilon = \dfrac{\sigma}{E}$$</p>
<p>This is valid up until the yield point. Considering a perfectly plastic material, the deformation $\epsilon$ is constant after the yield point, which gives us the following graph:</p>
<p><a href="https://i.stack.imgur.com/pU5DH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pU5DH.png" alt="enter image description here"></a></p>
<p>So, to answer your question, what is the value of $u_{Y_1}$? Well, it is equal to the displacement under the yield stress:</p>
<p>$$u_{Y_1} = \dfrac{\epsilon}{L} = \dfrac{\sigma_{Y_1}}{E_1L_1}$$</p>
<p>Likewise for $P_{Y_1}$:</p>
<p>$$P_{Y_1} = \sigma_{Y_1}A_1$$</p>
<p>Now, for the rest of the question, I think it is worth expanding your equation:</p>
<p>$$P = \dfrac{E_1A_1u_B}{L_1} + \dfrac{E_2A_2u_B}{L_2}$$</p>
<p>This is equivalent to</p>
<p>$$P = \sigma_1A_1 + \sigma_2A_2 = P_1 + P_2$$</p>
<p>(where $\sigma_1$ and $\sigma_2$ are variables and proportional to $u_B$) which means that the force applied is equal to the sum of the forces applied by each member, which each are equal to the product of the stress and the area of the member. Keep that in mind.</p>
<p>Now, the question states that we can assume that AB will become plastic first. So we know that the situation will change when $u_B = u_1$. In this case, the equation above will change to</p>
<p>$$P = \sigma_{Y_1}A_1 + \sigma_2A_2 = P_{Y_1} + \dfrac{E_2A_2u_B}{L_2}$$</p>
<p>So, as of that point, member AB will no longer increase how much it resists deformation; it is no longer proportional to $u_B$, but is instead a constant equal to $P_{Y_1} = \sigma_{Y_1}$. Member BC, however, has yet to yield, and therefore remains proportional to $u_B$.</p>
<p>As the force increases, $u_B$ will eventually reach $u_2$, at which point BC's collaboration will also become a constant and equal to $P_{Y_2}$, so we have</p>
<p>$$P = \sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 = P_{Y_1} + P_{Y_2}$$</p>
<p>This tells us that once both members cross into the plastic region, the deformation will tend to infinity.</p>
<p>So, there's your answer:</p>
<p>$$P = \begin{cases}
\begin{aligned}
&\left(\dfrac{E_1A_1}{L_1} + \dfrac{E_2A_2}{L_2}\right)u_B &&\text{ for }u_B \in [0, u_1] \\
&\sigma_{Y_1}A_1 + \dfrac{E_2A_2}{L_2}u_B &&\text{ for }u_B \in [u_1, u_2] \\
&\sigma_{Y_1}A_1 + \sigma_{Y_2}A_2 &&\text{ for }u_B \geq u_2
\end{aligned}
\end{cases}$$</p>
<p>Which gives you the following behavior (the red line shows AB's contribution and the green, BC's; the black is the entire structure's resistance and is equal to the sum of both lines):</p>
<p><a href="https://i.stack.imgur.com/QlPdO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QlPdO.png" alt="enter image description here"></a></p>
| 13535 | Deformation of a fixed-fixed beam with perfectly plastic yield stresses |
2017-02-01T20:31:33.050 | <p>I can't find the name for this kind of hole. It allows the release of the plate by just loosening the bolt a bit, without having to remove it completely.</p>
<p>Searches using "teardrop" or "hole with oblong" do not return the desired results.</p>
<p><a href="https://i.stack.imgur.com/ZfK5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZfK5m.png" alt=""></a></p>
| |bolting| | <p>I would call that a 'key hole'.</p>
| 13543 | What is the name for a bolt hole with a slot that allows the bolt to be inserted or removed when the nut is loose? |
2017-02-01T22:18:53.947 | <p>Say I make a fiberglass/epoxy part on a plastic mold, and it cures at room temperature. The epoxy system calls for a postcure at 250F, but the mold cannot withstand 250F.</p>
<p>If I take off the part and postcure it separately, am I likely to see dimensional changes in the cured part? Is there any way to postcure the part off-the-mold without losing accuracy?</p>
| |mechanical-engineering|materials|aerospace-engineering|composite|epoxy| | <p>There is a chance 250F might be above the Tg of the part if it is cured at room temperature. If this is the case then you will definitely get sagging if you postcure without a ramp at 250F. You might want to try ramping slowly up to 250F so that the temperature of the oven never shoots past the Tg of the material( which is steadily increasing with degree of cure). </p>
| 13545 | Can you postcure epoxy composites outside the mold? |
2017-02-02T11:26:11.323 | <p>For this problem , why we dont have to consider the max shear occur under the concentrated load just like the case in <a href="https://engineering.stackexchange.com/questions/13531/max-shear-act-on-point/13537?noredirect=1#comment24328_13537">max shear act on point</a> ?
In the previous question , the author 'move the load' so that the the <strong><em>center of the beam is between the FR (resultant force ) and the load</em></strong> ...
Why in this case , we just need to consider 2 supports only ? Or the question is wrong ? <a href="https://i.stack.imgur.com/Z05fd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z05fd.jpg" alt="fds"></a></p>
<p><a href="https://i.stack.imgur.com/mdCBh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mdCBh.jpg" alt="fdf"></a></p>
| |structural-engineering|structural-analysis| | <p>This question is looking for the <strong>absolute</strong> largest shear, the previous question was looking for the largest shear <strong>at midspan</strong>.</p>
<p>If you draw the shear force diagram for a simply supported beam, subjected to loads in one direction only, you will notice that the largest shear is <em>always</em> at one of the supports. Hence there was no need to consider anything other than the supports in this question.</p>
<p>As a note: in my opinion the author of this question missed one possibility they should have considered: what happens if you move the load far enough to the left that the 5kN load drops off the beam, and move the first 20kN load to be directly over A? Answer: you get a shear at A of 60.5kN. Still not quite as big as found by the author at B, but bigger than what they found at A! <em>Of course, this only applies if its possible to move the loads off the beam. As I normally design bridges, I expect loads to be able to move off the end.</em></p>
| 13555 | absolute shear occur at support? |
2017-02-02T16:40:51.350 | <p>Given an equation of motion of an undamped system $\mathbf{M}\mathbf{\ddot{q}} + \mathbf{K}\mathbf{q} = \mathbf{f}$, $\mathbf{M}$ indicates the mass-matrix, $\mathbf{K}$ the stiffness matrix, $\mathbf{q}$ the time-dependent displacement, and $\mathbf{f}$ the applied force. The roots, $\omega^2$, of the equation $det(\mathbf{K} - \omega^2\mathbf{M}) = 0$ indicate the vibration frequencies.</p>
<p>My question is: what is the physical interpretation of the <strong>eigenvalues</strong> of the stiffness matrix $\mathbf{K}$?</p>
| |structural-engineering|finite-element-method| | <p>The matrix $\mathbf{K}$ simply represents the force response to a unit displacement on each of the degrees of freedom of the system.</p>
<p>Consider a 2D cantilever beam of length $\ell$ with two degrees of freedom. The end displacement $\delta$ and the end slope $\theta$. You can assemble a stiffness matrix of the form $\mathbf{f} = \mathbf{K} \mathbf{x}$</p>
<p>$$ \begin{vmatrix} F \\ M \end{vmatrix} = \begin{bmatrix} \frac{48 E I}{\ell^3} & -\frac{18 E I}{\ell^2} \\ -\frac{12 E I}{\ell^2} & \frac{6 E I}{\ell} \end{bmatrix} \begin{vmatrix} \delta \\ \theta \end{vmatrix} $$</p>
<p>The interpretation is the force/moment needed to achieve unit deflections $[1\;0]$ or $[0\;1]$ between the two columns of the matrix.</p>
| 13558 | What is the physical interpretation of the eigenvalues of the stiffness matrix in the finite element method? |
2017-02-02T23:16:31.107 | <p>This is a noob question, please don`t be too harsh on me. I would like to use a motor for turning a mixer. The shaft of the mixer should be removable and I though of 3d printing a case which holds the motor and a gearbox in place. I made a very simple drawing of how it should look like (or how i can imagine it). I was wondering if such a thing already exist but I do not find it because I do not know the name. It should be easy to remove the shaft. Some sort if quick release/fix. The motor will run at 300 to 1000 rpm at low torque.</p>
<p><a href="https://i.stack.imgur.com/lKmq6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lKmq6.jpg" alt="enter image description here"></a></p>
| |motors| | <p>Most general purpose gearboxes will have a fairly short stub shaft which is connected to the main shaft by some sort of coupling. Exactly how this is implemented will depend on the application. </p>
<p>Light duty couplings might just be a collar held on with grub screws or cotter pins heavier duty applications may use keys or splines. Splines are often preferred if the shaft needs to accommodate some axial movement and for very heavy duty applications such as vehicle transmission. </p>
<p>The version you have drawn has the added complication that a completely removable shaft needs to accommodate both the final drive gear and a set of bearings with a pretty good fit. </p>
<p>One solution would be to have the final gear running on a hollow, internally splines shaft but this would be rather complex and expensive to manufacture. There is also the consideration of keeping the gearbox internals reasonably well sealed to keep the grease in and any dirt and grit out. </p>
<p>In practice many mixers with removable blades are similar to electric drills, using a <a href="https://en.wikipedia.org/wiki/Chuck_(engineering)" rel="nofollow noreferrer">chuck</a> or similar quick release mechanism to allow easy fitting and removal. </p>
<p>A basic 3 jaw drill chuck will transmit moderate torque to a round shaft and somewhat higher torque to a hexagonal shaft and they have the advantage of being cheap and easily available. </p>
| 13563 | some kind of gearbox for driving a removable shaft; is there a name for it? |
2017-02-03T09:16:46.857 | <p>Doing a fun project to reconnect with my engineering training: I would like to design a control scheme for hovering in-place at a set altitude with a vertically-oriented propeller, rocket, etc. Right now I'm not even worried about lateral motion, just the vertical component. My plan is to use PID control and tune with the Ziegler-Nichols method for <a href="https://en.wikipedia.org/wiki/Ziegler%E2%80%93Nichols_method" rel="nofollow noreferrer">no-overshoot</a>. The system is so simple I can write down a model of the P-only controlled system rather than tune experimentally. Just using Newton's 2nd law and ignoring aerodynamic effects:</p>
<p>$$
\begin{align}
m x^{\prime\prime} & = g m + T \\
x^{\prime\prime} & = g + \frac{K(x - x_{target})}{m}
\end{align}
$$</p>
<p>Where $g$ is gravity, $m$ is mass, $x$ is height, and $T = K(x-x_{target})$ because my thrust is proportional to the error times the gain $K$. Using $sin(\omega t)$ as a trial solution I get:</p>
<p>$$
\begin{align}
\omega & = \textstyle\sqrt{\frac{K}{m}} \\
T_u & = 2\pi\textstyle\sqrt{\frac{m}{K}}
\end{align}
$$</p>
<p>However it seems like I get free choice of the ultimate gain $K$, since any choice of $K$ causes the system to oscillate, as far as I can tell. What am I missing?</p>
| |pid-control| | <p>Ziegler-Nichols tuning will not work for this system because there is no energy loss.</p>
<p>The reason that your system oscillates regardless of the value of $K$ that you choose is because there is no energy loss in your model. All of the energy is stored as potential gravitational energy (which can be ignored for now), kinetic energy in the mass, and "potential" energy in the proportional control (it acts like a spring in your system connecting the mass to the desired position). So if your system starts where $x \neq x_{target}$ you can think of it as pulling and releasing a spring (without any friction).</p>
<p>Naturally a real system will have some amount of air resistance, which I believe is proportional to the square of velocity. That will give you a non-linear system, which will require non-linear control techniques, so for a fun exercise I would suggest instead using the following equation to approximate the damping in your system:</p>
<p>$mx'' + bx' = gm + T$</p>
<p>As for what value to pick for $b$, who knows? It will depend largely on the aerodynamics of your aircraft. Ziegler-Nichols will work for this system.</p>
<p>However, I would recommend using more mathematical methods for tuning the system (like having a target closed-loop overshoot and rise time, for example) rather than Ziegler-Nichols, which is primarily an experimental tuning method for when we do not have an accurate model of the system plant.</p>
| 13572 | Model-based Ziegler-Nichols tuning of a hover PID controller |
2017-02-03T15:15:27.450 | <p>As is well known, a safety factor is often applied for engineering structures as a measure of precaution.</p>
<p>I would like to ask for which loads (if not all) does the safety factor apply for.</p>
<p>For instance, take an aircraft wing. The safety factor will need to be applied for the aerodynamic loads but does it need to be applied for any other loads such as the self-weight of the wing or the weight of the powerplant? </p>
| |structures| | <p>Yes, you have to multiply all the loads with safety factor, I think It is already based on the regulation worldwide in engineering field in general.</p>
<p>The purpose is to avoid the underestimation in calculation process and to human-error, and any other technical mistake in the field.</p>
| 13574 | What loads does the safety factor apply for? |
2017-02-03T15:16:45.803 | <p>I am not an engineering student, but can understand if explained, so excuse me if this is too basic.</p>
<p>For a personal DIY project I envisioned, I need to make an object, attached to a set of gears, fall at a certain speed - 6ft / 24 hours. This object will rotate the first driving gear and the resultant driven gear at 3000-5000 RPM. I estimate the ratio to be 1:100000.</p>
<p>So basically my question is whether the speed at which the object falls (from gravity) be regulated? The weight of the object doesn't matter, since I can set it to whatever is required for my rate of descent.</p>
<p>The setup will be similar to: <a href="http://s.hswstatic.com/gif/clock-weight-train.gif" rel="nofollow noreferrer">http://s.hswstatic.com/gif/clock-weight-train.gif</a></p>
| |mechanical-engineering|gears| | <p>A gearbox on its own won't really regulate speed, apart from the fact that it will have some internal friction but this is difficult to regulate. For example just having a 1000:1 gear ration won't make the object fall 1000 times slower unless there is some external torque applied. </p>
<p>A simple solution is to use viscous drag for example having the gearbox drive a fan or paddle wheel. The works because viscous drag increases with the square of speed so you will get to a point where the drag force equals the weight of the object at some reasonably constant speed. </p>
<p>There are also various types of mechanical governor which can be used to adjust the force on a mechanical brake in response to speed. A step further is to use an electronic system with speed sensors and actuators on a brake, this potentially allows a lot of flexibility and the ability to self adjust on the fly. </p>
<p>Another option is to use an escapement of they type used in clocks. This doesn't give true constant speed as the motion becomes a series of discrete steps but for the very low speeds you are talking about this may not be a problem. </p>
<p>Having said that the gear ratios you are talking about are huge, to put it into perspective if you did this with a single pair of gears and the driving gear was 5cm in diameter then the driven gear would need to be 5km in diameter </p>
<hr>
<p>Edit in response to comment: </p>
<p>If you want to use this arrangement to generate power then you need to bear in mind that gravitation potential energy has pretty poor energy density compared to other small to medium scale energy storage. </p>
<p>Specifically the total gravitational potential energy is 9.8J per kg per metre of drop. 24 hours is 86400 seconds so for a 2m drop this translates as 1.1e-4 Watts ie you would need 5000kg to provide 1W over 24 hours before considering losses. </p>
| 13575 | Can the speed of a falling object be regulated using gears? |
2017-02-03T15:38:20.683 | <p>I always wondered how it is possible to have electromagnets so strong that they keep a heavy maglev-train levitating without forcing anything metallic to be flinged away or to be pressed against the ceiling of the vehicle-cabin?
And how come a person inside that train which has a pacemaker implanted is not affected? With a pacemaker someone cannot have a MRI scan but still ride a maglev train?</p>
| |electromagnetism| | <p>First off there are many different types of maglev prototypes out there each with their pros and cons. In fact some designs have the drawback of needing shielded compartments for people with pacemakers! However, instead of going into much detail about all the different kinds of magnets that can be used on maglevs, I will try to explain how magnetic fields can be manipulated.</p>
<p>For example, some maglevs use an array of magnets called a Halbach array.
<a href="https://i.stack.imgur.com/lMz87.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lMz87.png" alt="Halbach Aray"></a> </p>
<p>On the left you can see that the magnetic field lines are somewhat uniform and on the right the maganetic field is not only on the top but also <em>much more condensed</em> than the image on the left. This is just one of the ways that enginerers can mitigate the problems caused by strong magnetic fields.</p>
<p>Also it is important to note that magnets will pull or repel each other with much more strength that when they are close together than when they are far apart. For example dipole magnetic fields decay at a rate of $1/d^3$ which means that as the distance $d$ increases magnetic field gets weak very fast.</p>
<p>Maglev trains exploit these properties of magnetic fields by fitting powerful magnets on the ground and on the bottom of the train. As the magnets approach each other the fields repel with a force strong enough to keep the train afloat while keeping the passengers from experiencing a strong magnetic field. </p>
<p>Hopefully this has shown you that magnetic fields can be shaped and that magnetic fields tend to decay quickly as you move away from them. As a final note consider how little magnetic material humans tend to carry around! Most metals in common use are not strongly attracted to magnetic fields like aluminum. So unless you are moving you weight set or you cast iron skillet you can ride the maglev in safety without prying your stuff from the ceiling!</p>
| 13577 | How is it possible electromagnets keep a heavy maglev train levitating without forcing anything metallic to be flinged away? |
2017-02-03T15:56:57.847 | <p>What type of machines are used to paint a bulk of small objects all over... things like buttons, keys, ping pong balls, paperclips etc.</p>
| |machining| | <p>In terms of general coatings there are a few options for bulk treatment of small objects. </p>
<ul>
<li>Powder coating : the part is electrostatically coated with a polymer powder which is then baked on. This produces quite a thick and tough coating which can cover complex surfaces well and doesn't require volatile solvents. </li>
<li>Dipping : the part is dipped is a rubber based medium which may also be foamed, coating the surface. This leaves a very thick, soft coating which is good for things like tool handles and lever arms. </li>
<li>Spray painting : can be mechanised by spraying from different directions on a conveyor type system, often going on to drying ovens. </li>
</ul>
<p>There are also various chemical techniques eg plating, passivating, anodising, phosphating, bluing which induce the formation of a relatively inert layer on a metal surface. For example most steel fasteners are bright zinc passivated. </p>
<p>Nickel, chrome, brass or bright zinc plating are fairly common on small metal parts. </p>
<p>Sheet metal parts may also be made from material which is pre-coated with a plastic film often over galvanising. </p>
<p>Aluminium parts are often anodised, an electrolytic process which encourages the formation of an even and stable layer of aluminium oxide which can also be dyed. tending to produce a semi-transparent colour. </p>
| 13579 | Machine that can paint a bulk of small objects all over |
2017-02-03T17:12:27.117 | <p>This is a stress-strain diagram
<a href="https://i.stack.imgur.com/405ro.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/405ro.gif" alt="enter image description here"></a></p>
<p>I have noticed that the curve decreases between points C & D and between E & F.
Why does that happen? I expected the curve to always increase since the tension force in a tensile test is always increasing and the cross sectional area of the specimen decreases, which means that the stress is always increasing.</p>
| |mechanical-engineering|materials|civil-engineering|industrial-engineering|strength| | <p>Such an explanation of the voltage drop at the beginning of the yielding plateau is generally accepted. However, the same voltage drops occur in other materials without solid carbon solution (doi:10.1016/j.ijsolstr.2009.09.018). In addition, the same short voltage drops are found at the end of the yielding plateau (doi:10.1016/j.ijsolstr.2011.02.019).</p>
| 13583 | Why does the stress-strain curve decrease? |
2017-02-04T17:55:45.560 | <p>A spec characterizes the vibration stability by acceleration noise density</p>
<p>$$Na < \dfrac{1\text{ μg}}{\sqrt{Hz}}$$</p>
<p>for 1-100 Hz, where g=9.8 m/s<sup>2</sup>. But usually what I see is something like 3.5 nm pk-pk.</p>
<p>My question is, how to convert these two?</p>
<p>I tried this as below, but not sure if it's correct, any comment is appreciated.</p>
<hr>
<p>Displacement noise density (is this correct?)</p>
<p>$$Nx = \frac{Na}{4\pi^2f^2}$$</p>
<p>Thus</p>
<p>$$\begin{align}
x^2 &= \int (Nx)^2 df = \int \frac{(Na)^2}{16\pi^4f^4} df \\
&= \frac{(1\mu g)^2}{16\pi^4} \int_{1}^{100} \frac{1}{f^4} df \\
&\approx 2 \times 10^{-14} m^2
\end{align}$$</p>
| |mechanical-engineering| | <p>Conversion between $g/\sqrt{hz} $ and $m/\sqrt {hz} $ is possible. However, converting from that to m pk-pk is usually not possible. The reason is that spectral density is a unit used to quantify <em>random</em> vibration, and m pk-pk is a unit used to quantify <em>sinusoidal</em> or periodic vibration. Random and periodic vibration simply are not the same thing. For specific applications, there are ways to come up with equivalences between them. But in the most general sense it is apples and oranges. I have some tutorials on my webite that may help explain this concept in more detail <a href="http://mechanicalvibration.com/Power_spectral_density.html" rel="nofollow noreferrer">http://mechanicalvibration.com/Power_spectral_density.html</a></p>
| 13601 | Vibration characterization: how to convert acceleration noise density to displacement pk-pk value? |
2017-02-04T18:25:14.397 | <p>In the old days there was a type of sign/light you would find in bars and restaurants which was animated.</p>
<p>It would be a lit scene usually of water, like a lake with a waterfall, and by some means it would animate and look like the water was flowing.</p>
<p>They would be used to advertise cigarettes or beer.</p>
<p>What was this technology called? How did it work?</p>
| |lighting| | <p>At least some of such signs were animated using polarized light. The water or whatever would be made as a mosaic of little pieces of polarizing plastic, and the whole thing would be illuminated from behind by a lamp shining through a rotating polarizer. This causes different segments to brighten and dim at different times.</p>
| 13602 | Animated light box signs |
2017-02-05T04:46:13.427 | <p>I know that the 4G network is only for data and doesn't support voice calls. Most of phones switch to 2G/3G when receiving a voice call. </p>
<p>But my problem is how does the phone gets to know about the reception of the voice call when it is in 4G network. We can think, a signal is sent over 4G data to the phone to switch from 4G to 2G/3G. But what will happen if the mobile data is switched off in the device, when receiving a phone call.</p>
| |telecommunication| | <p>First things first. 4G is not only for data, it is a only -packet switched network as opposed to both circuit switched and packet switched in 3G.</p>
<p>Now as per your question as to how does a phone(user equipment or better UE) knows the reception of a voice call, it is done by control plane signalling over the network. So, yes, a signal is sent before the reception of call. But what is important to note that this procedure is not exclusive to 4G, 3G and lower networks also employ similar mechanism.</p>
<p>Now, as per your doubt if call would be received with data switched off, yes it would be received. This is because signalling messages are transferred over LTE and not data( they are two separate things). In fact event before you can access data on LTE networks a process similar to voice call is carried out, and it is technically known as data calls.</p>
<p>As I had mentioned in my comment, VoLTE is used to carry voice calls. This is a new standard for transmitting voice messages. The need for this method arose because LTE supports only Packet Switching, while conventionally voice was carried through Circuit Switched networks.</p>
| 13607 | How does 4G know the reception of a voice call? |
2017-02-06T02:31:14.130 | <p>Why do some high speed trains like the Shinkansen E5 and E6, have <em>really</em> long noses while other trains, like the Eurostar and Javelin, have shorter noses?</p>
<p>How does the shape of the nose affect the train?</p>
| |mechanical-engineering|aerodynamics|rail| | <p>It all depends on the level of efficiency you seek. </p>
<p>A train, given its size, has a ridiculously small cross section. This small frontal area footprint is being 'pushed' by the inertia of hundreds of tonnes of metal.</p>
<p>All high speed train have aerodynamic noses, but some will tend to be pedantic about how many thousandths of a percent efficiency they can squeeze out of an aerodynamic profile. </p>
<p>There are a couple of other factors that come into play with the design of the nose and that is the pressure pulse that is generated as a train passes another object, generally another train going in the opposite direction or a tunnel. It's why passengers are told to stay away from the platform edge as fast trains pass by, its not that you'll be blown away, but may sucked towards the train by the low pressure. </p>
<p>Two high speed trains passing each other can generate a huge amount of suction between them at the nose, because of the low pressures there, so having a longer nose helps spread this out a bit and reduce the area of minimum pressure. If not the trains could be pulled towards each other, or some instability set up by the rapid sideways pressure pulse.
The nose of the E5 Series, at 15 metres, is a massive 9 metres longer than the previous incarnation of the bullet train (Shinkansen), the E2 Series. This, according to its designers at JR East, will help eliminate the phenomena of "tunnel boom".</p>
<p>Japan's rail tunnels are somewhat narrower than their European counterparts, so when the Shinkansen enters a tunnel at speeds above 200 kilometres per hour, the sudden increase in air pressure can cause a loud "boom" at the other end of the tunnel. In some cases, such shock waves are thought to have damaged tunnels in Japan, ripping chunks of material from tunnel ceilings.</p>
<p>The shape of the front car has evolved gradually to combat this danger, and the striking "Long Nose" design of the E5 Series is the result.</p>
<p>Reference <a href="http://www.jreast.co.jp/e/development/tech/pdf_1/22_28tecrev.pdf" rel="noreferrer">1</a>,
<a href="https://www.newscientist.com/blogs/onepercent/2011/03/japans-long-nosed-bullet-train-1.html" rel="noreferrer">2</a></p>
| 13621 | Why do some high speed trains have long noses, while others do not? |
2017-02-06T15:25:31.517 | <p>I'm having some trouble figuring out how to design the corresponding internal gear for any particular spur gear. With spur gears, meshing is accomplished by having two gears with the same modulus (MOD) or, in the US, the same diametral pitch. The spacing between the two gears is such that their pitch circles touch.</p>
<p>Looking at an internal gear the tooth profile for an internal involute gear is the inverse of the external involute. However if I try to create such a system (in Fusion 360), I get some interference issues:</p>
<p><a href="https://i.stack.imgur.com/ETImA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ETImA.png" alt="Internal to Spur gear"></a></p>
<p>What I've done is create two spur gears, both MOD = 1, one with 24 teeth which I've just simply inverted to create an internal ring gear and the other with 12 teeth. My questions:</p>
<ul>
<li><p>The pitch circles are marked in green. Is the pitch circle for the internal gear the same as the spur gear that I used in inverse to create it? Or is it moved somewhat. The proper relative gear positions should be where the pitch circles touch so that's a main issue.</p></li>
<li><p>Although it doesn't look too bad, the left and right teeth in the image are interfering. I think if the addendum of the internal gear's teeth was reduced this would help matters. It seems that the larger the ratios (eg: if the spur gear was 8 teeth instead) the more the gears would crash and perhaps the more the addendum should be reduced.</p></li>
<li><p>In addition I could add backslash, would this help the interference issue?</p></li>
</ul>
<p>Are there specific rules to design an internal involute gear for a particular MOD?</p>
| |gears| | <p>First of all, yes the pitch circle is the same for an internal gear as it is for the equivalent external gear. Second, note that indeed the addendum refers to how much the teeth extend inward of the pitch circle for an internal gear (instead of outward, as is the case for external gears). Vice versa for the dedendum. For standard gears, both internal and external, the addendum $a$ and dedendum $d$ are as follows:</p>
<p>$$a=m=\frac{1}{D_p}$$</p>
<p>$$d=1.25m=\frac{1.25}{D_p}$$</p>
<p>$m$ is the module (in millimetres) and $D_p$ is the diametral pitch (in inches$^{-1}$). The units of the addendum and dedendum are consistent with that of the module or diametral pitch, depending what you chose.</p>
<p>Therefore, it looks like you will need to increase the dedendum for the internal gear to $1.25m$, and make sure the addendum is only $m$. This will then prevent these inference issues, and will generally be fine as long as any external gears are not significantly larger than half the diameter of the internal gear they are inside of.</p>
<p><a href="https://i.stack.imgur.com/k3HeO.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/k3HeO.gif" alt="enter image description here"></a></p>
| 13633 | Internal teeth involute gear pitch circles |
2017-02-06T23:41:12.447 | <p>I am aware that it is easier to run a static timing analysis on a CPLD than a FPGA, but why is this?</p>
<p>Any information would be greatly appreciated.</p>
| |electrical-engineering|computer-engineering| | <p>Basically, a CPLD has far fewer choices in terms of the number of different <em>kinds</em> of paths through which a signal can flow from one register to the next, which means that the job of analyzing all of the different <em>combinations</em> of paths is reduced by many orders of magnitude.</p>
| 13639 | Why is it easier to run an STA on a CPLD than an FPGA? |
2017-02-07T16:25:09.080 | <p>I'm playing around with a radio receiver on 433MHz and I am receiving signals from a temperature / humidity sensor.</p>
<p>I already figured out what bytes represent humidity, but I'm stuck with the temperature as I don't know how it is represented in the signal.
Here are some example messages I receive:</p>
<pre><code>C1 | 75.2F | 70%
0001010001010011010000000100100000000100011011010111000011
C1 | 24.1C | 68%
0001010001010100100000000100100000010100010000000001000011
C1 | 24.1C | 56%
0001010001010100100000000100100000010011100001010110000011
</code></pre>
<p>The temperature sensor has the following readouts:</p>
<ul>
<li>Temperature</li>
<li>Humidity</li>
<li>Unit (Celsius / Fahrenheit)</li>
<li>Channel (1-8)</li>
</ul>
<p>I already figured out what the <strong>humidity</strong> is as well as what represents the <strong>channel</strong>. I also figured out a <strong>temperature range(?)</strong> [1 = ..-11 | 2= -1..-10 | 3= 0..15 | 4= 15..30 | 5=31..40]. But I have a hard time figuring out where I can find the exact temperature or the unit (F/C).</p>
<p>I prepared this public Google Sheet with my findings and raw data so far (sorry I don't know a better way to share on Stackexchange):</p>
<p><a href="https://docs.google.com/spreadsheets/d/1q2sk5bK6Bv3gi8ahTLIUM4DaMR7AvnTVPY-NvmGMHoQ/edit?usp=sharing" rel="nofollow noreferrer">https://docs.google.com/spreadsheets/d/1q2sk5bK6Bv3gi8ahTLIUM4DaMR7AvnTVPY-NvmGMHoQ/edit?usp=sharing</a>
<a href="https://i.stack.imgur.com/vrhIM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vrhIM.png" alt="enter image description here"></a>
(the channel with a zero based index I already figured out is violet, the humidity is in blue, some bits always stay the same hence I made them grey)</p>
<p>What is my best course of action now? I have been trying around with all the bits and bytes, adding them together, adding the one to the other, multiplying the temperature times 10 to get a decimal... nothing.</p>
<hr>
<p>Yeah... Plotting is an idea, but it doesn't really help me:
<a href="https://i.stack.imgur.com/UE91o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UE91o.png" alt="enter image description here"></a></p>
<p>Column <strong>I</strong> seems to follow the temperature in a way (negative temperature/high temperatures and the values just don't relate, look to the end of the graph), but looking at the decimal representation of I can see values such as 7,8,9 and wildly changing temperatures. I am assuming there is some kind of correlation between the temperature range and this byte but I haven't figured out yet how to achieve the calculation.</p>
<hr>
<p>Alright I found something:</p>
<p><a href="https://i.stack.imgur.com/w0bdj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w0bdj.png" alt="enter image description here"></a></p>
<p>The temperature ranges in correlation with I & J seem to be the key. I & J corresponds to a value between 0-255 (I guess) for each range and represents a particular temperature value. The I & J scale seems to be logarithmic looking at the difference between 17,8C (17) and 20,8C (71) and the difference between 20,8C (71) and 22,8C (106). But that's just a guess.</p>
<p>So I somehow need to correlate the range (15 degrees C wide) with the 0-255 index to get a temperature. Crazy.</p>
<hr>
<p>Converted H & I & J 12 bit number to decimal and get the following:</p>
<p><a href="https://i.stack.imgur.com/Lnjtn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lnjtn.png" alt="enter image description here"></a></p>
<p>It does look very related, but I can't figure out a formula.</p>
| |wireless-communication| | <p>Figured it out!</p>
<p>The temperature is always being send in Fahrenheit not in Celsius, see the following table:</p>
<p><a href="https://i.stack.imgur.com/ygrGR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ygrGR.png" alt="enter image description here"></a></p>
<p>An increase of one in the 12bit (columns H&I&J together) column leads to 0,1 degree Fahrenheit increase.</p>
<p>Converting the 12bit to decimal, subtracting 400 and dividing by 10 actually leads to the exact value in Fahrenheit.</p>
<p>The formula is as follows:</p>
<pre><code>(H&I&J-400)/10 = Temperature in Fahrenheit
((H&I&J-400/10)-32)/1,8 = Temperature in Celsius
</code></pre>
<p>The formulas hold for all values!</p>
| 13654 | How to decode radio thermometer / hygrometer bytes to floats? |
2017-02-07T18:59:17.533 | <p>As an assignment for my Automotive Engineering study, I have to practice manual FEM calculations. I'm free to choose whichever existing construction, and with that I have to make FEM analyses of both the original and the improved construction. I chose the chassis of the '57 MGA i'm restoring, which is shown below: </p>
<p><a href="https://i.stack.imgur.com/J3GWa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J3GWa.jpg" alt="mga chassis"></a></p>
<p>I suppose there are several ways to improve the driveability, but I expect increased torsional stiffness to have the most impact. The rear bridge seems rather wobbly to me, but I can't really tell; I haven't driven an MGA before.<br>
(If you have better ideas how to improve the chassis please do suggest.)</p>
<p>What would be a good approach to improve the torsional stiffness of the chassis?<br>
With that, I mean what kind of construction/structure would be effective; a triangular structure between the two side beams for instance. </p>
<p><strong>Please note:</strong> it doesn't have to be practically feasible or cost effective. It's just for practicing.</p>
| |structural-engineering|finite-element-method|reinforcement| | <p>The overall stiffness of the frame is determined by the stiffness of the side beams augmented primarily by the three tubular cross members.</p>
<p>An interesting academic exercise would be to evaluate how changing the diameter of the cross members affects the overall stiffness.</p>
| 13656 | How could an MGA's frame/chassis best be improved? |
2017-02-07T22:02:25.827 | <p>I am fascinated by construction. My university is building a new facility, and they are using 2 tower cranes in the construction.</p>
<p>Why would a contractor choose to use 2 cranes on one construction project? Is it simply a matter of saving time by having the two cranes work simultaneously? What are some other considerations that might be taken into account? </p>
<p>EDIT: after checking your answers, I agree that it is most likely that using two cranes in this situation was mostly done to increase productivity/throughput, and the positioning of the cranes depends on the building design and ease of loading materials to be transported by the crane. <img src="https://i.stack.imgur.com/sSpAw.jpg" alt="enter image description here"><img src="https://i.stack.imgur.com/2N1u9.jpg" alt="enter image description here"></p>
| |civil-engineering|construction-management| | <p>Tower cranes are a very significant cost on any construction site, and so builders will usually only use as many cranes as is necessary. </p>
<p>I work for a construction scheduling company, and the one of the biggest factors for deciding on the number of cranes and their location is access. Obviously the crane needs to be able to access all parts of the building during construction, but the placement of the crane is also often determined by access to roads and materials on the ground. Other big restrictions such as site boundaries and roads also often come into play in urban areas. If the access limitations mean you need two cranes, then two cranes will be used. </p>
<p>Secondly, as you noted, is time. Time is money on a site, and if having a second crane can save you a couple of months off your finish date, then it is usually a cost effective idea. Other times, there may be a hard deadline for a project finish, in these cases more than one crane may be used to accelerate the program at the request of the client despite it being more expensive. </p>
| 13658 | Why use two tower cranes on one construction site? |
2017-02-08T07:55:57.670 | <p>I am planning on buying a kitchen stand mixer with planetary action.
<br/>One important factor to consider is the type of gears the mixer uses, they can be plastic, composite or metal.
<br/>As far as I know metal gears are the best and plastic the worst in terms of durability and handling stress.
<br/>However I don't know <em>what</em> are composite gears and how they fare, compared to metal or plastic.</p>
| |materials|gears| | <p>Composite is a very broad term in engineering - it's essentially any material that is made of two or more substances merged at macroscopic level - so not a mixture or alloy. This may be just a mix of larger grains, or it can be a complex structure.</p>
<p>One most common composite is reinforced concrete, where structure of rebar is embedded in concrete. Another is carbon fiber reinforced polymer (CFRP), where a texture/fabric of carbon fiber is embedded in a resin. Even plain concrete, even if rarely defined as such, is a composite - cement binder plus gravel. </p>
<p>In mechanisms, composite is usually the middle ground between plain polymer and metal. This may be metal-reinforced polymer (thin sheet metal gears embedded in plastic for stability), or it may be a CFRP cut-out, or a gear with metal axis mount and plastic teeth. Or it may be plain plastic with some waste organic fiber (cotton) added, increasing its strength insignificantly. </p>
<p>It's hard to say without knowing the exact device in question, and while that may be something of performance quite comparable to metal, the marketing department can get away with the buzzword to peddle a total BS that way - any additive that won't dissolve in plastic will make the plastic "a composite" regardless of any performance gains - or losses.</p>
| 13665 | Composite vs metal gearing |
2017-02-08T08:22:38.177 | <p>Ohms law defines R = V / I, so according to this equation, when resistance increases, voltage and current decreases.
Now i have two questions.</p>
<ol>
<li>Ain't there somekind of "pressure" buildup in front of a resistor? because the resistor limits the current flow, I would expect something like a "traffic jam".</li>
<li>How is it that a resistor reduces the voltage? Is it converted to heat by electrons "bumping" into more massive particles or does something else happens to the voltage?</li>
</ol>
<p><strong>Edit</strong></p>
<p><em>Wasabi</em>:
Sorry if this question doesn't have the correct tag, didn't know which tag to give. If you have any suggestions for tags i'll put that in.</p>
<p><em>hazzey</em>:</p>
<p>I'm trying to visualize for myself how Volt\Amp\Resistance work togather. </p>
<p>I understand that Voltage is a measurement of potential energy expressed in Jouls / Coulomb and the Current is the number of Coulombs per second passing a certain point, the only part that isn't quite clear to me is the resistance. How resistance physically works to reduce current and in the process uses some energy (voltage).</p>
<p>If I understand the answer correctly, the following is true.
The voltage that is used by a resistor is always converted to heat.</p>
<p>So if I have a circuit with a single resistor of 1 Ohms with 1 Amp of current, the resulting 1 volt is converted to heat?</p>
| |electrical-engineering| | <p>No. If resistance increases, voltage increases and current decreases. The extra energy is spent thermally, which is why the resistor gets hot when current flows through it, and the higher the resistance, the hotter it becomes for the same current. </p>
<p>You have to remember that voltage is a differential measurement, like pressure or altitude: you always measure a voltage relative to something or a reference (like earth). Current, on the other hand, is measured when flowing through something, like flow rate. </p>
<p>A good analogy would be water flowing through a dam or similar flow restriction. The more you restrict the water flow (i.e. the higher the resistance becomes), the lower the water flow through the dam, but the higher the pressure upstream of the dam.</p>
| 13666 | What happens to the voltage? |
2017-02-09T10:29:48.567 | <p>Hi there (from Europe),</p>
<p>I am completing what should be a relatively quick assessment for the client, but I have started to second guess my self and would like some reassurance. </p>
<p>Essentially the job involves removing a portion of an existing 2.6m x 1.22m concrete slab in order to install a manhole (1.2m x 0.75m). I've taken the fairly rudimentary (and conservative imo) approach of simply reducing the existing slabs moment capacity by reducing it's width. For example, for assessing the moment capacity of the current slab I took a width of 1.22m, but it's reduced capacity would be based on a width of 1.22-0.75=0.47m. </p>
<p>With an original moment capacity (89kNm), and reduced moment capacity(37kNm), we know the difference between these two (<strong>52kNm</strong>). My idea was to simply take two Square Hollow Sections and attach them to the underside of the slab, on either side of the manhole, and size them such that each was capable of holding half the difference in moment, <strong>26kNm</strong>.</p>
<p><a href="https://i.stack.imgur.com/gaBkq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gaBkq.png" alt="Sketch of situation and proposed solution"></a></p>
<p>This is a fairly simple approach but the more I think about it the more I'm beginning to reconsider and instead consider assessing the steel and concrete as a single composite section to Eurocode 4.</p>
<p>This is a small scheme with a low budget and I lack experience of Eurocode 4 to quickly assess this.</p>
<p>My question is thus:
<strong>Is my approach conservative enough to safely size steel members or is there something fundamental that I am missing in composite design that makes my solution inaccurate and dangerous?</strong> </p>
| |civil-engineering|steel|concrete|composite|eurocodes| | <p>Regarding composite action, it will be minimal therefore you are doing well to ignore it. </p>
<p>Probably the design is adequate (except maybe for point 4 below)but i want to mention some missing elements from your analysis:</p>
<ol>
<li>You are restoring the slab strength for sagging moments but not for hogging ones.</li>
<li>It is important to know when the shs members are installed. If after the hole is opened, then a verification is needed for the reduced slab under the dead loads plus any live before the shs installation.</li>
<li>Verification of deflections of the reduced slab.</li>
<li>The hole surface should probably get strengthened, particularly its corners, due to stress concentrations.</li>
</ol>
| 13679 | Assessment of a concrete beam being strengthened by an additional steel member |
2017-02-11T09:06:43.680 | <p>I'm a software engineer whose main concentration has been data science with some minor electrical engineering on the side.</p>
<p>I want to delve into some mechanical engineering by building a proper mini vending machine but am having trouble knowing which materials to use as a consequence of not knowing where to find good documentation.</p>
<p>Is corrosion resistance important for a vending machine? Do I use aluminum or steel? What factors would I need to take into account if it were a 2 meters tall, 1 meter deep machine, for example? Or another size?</p>
<p>Apologies if this is mistagged.</p>
<p>Thank you.</p>
| |mechanical-engineering| | <p>Before you start thinking about technical solutions like materials and mechanisms you really need to come up with a reasonably detailed design specification, this should address the problem in terms of what you want to achieve rather than how you will achieve it. This also involves precisely setting out the problem that you want to solve. </p>
<p>Clearly one crucial element will be the product you are dispensing for example :</p>
<ul>
<li>Shelf life : how will you ensure efficient stock rotation ?</li>
<li>Volume : how much of the product should the machine hold ?</li>
<li>Storage requirements : temperature, humidity etc </li>
<li>Is the product individually packaged or does it need to be dispensed into containers at the point of delivery ? Does it have specific handling requirements ?</li>
<li>How many different types of product will the machine stock ?</li>
</ul>
<p>Other important elements might include :</p>
<ul>
<li>Security </li>
<li>Safety </li>
<li>Branding and aesthetic appearance </li>
<li>How payment will be taken </li>
<li>User interface </li>
<li>Reliability </li>
</ul>
| 13703 | What factors are needed to be taken into account when building a vending machine? |
2017-02-11T16:43:42.620 | <p>Suppose that there is a piston cylinder device . It contains a gas with pressure P. The external pressure is Pext . In order for the gas to expand , I think that P must be greater than P ext , and it will expand until the pressure of the gas is equal to the external pressure.
The thing that I cannot understand is that when we calculate the boundary work we multiply the external pressure by the change in volume. Since the gas is expanding , then the gas is DOING the work ON the surrounding , then I expect that when calculating the boundary work is to consider the gas pressure. What is wrong with that ?</p>
| |mechanical-engineering|thermodynamics| | <p>In case of a Quasi-static expansion, the internal pressure ($P_{in}$) would at all times be just infinitesimally greater than the external pressure($P_{ex}$). Mathematically, $P_{in} = P_{ex} +
\text{d}p$. Now work done as you said is, $\int P_{in} \text{d}v$. Therefore, using $P_{in} = P_{ex} + \text{d}p$, then after multiplication inside the integral you would have $\int P_{ex} \text{d}v + \text{d}p\text{d}v$. Since, $\text{d}p$ and $\text{d}v$ are both infinitesimally small, therefore their product can be neglected and you end up with an approximate value of work, which is $\int P_{ex} \text{d}v$.</p>
| 13708 | why do we consider the external pressure when calculating the boundary work |
2017-02-12T01:05:31.573 | <p>I am trying to create a simulation for car suspension and tyre physics and to do the tyre physics I need to know how much weight is over the friction point that is under the spring.</p>
| |springs| | <p>The weight supported by each spring will depend on how the weight is distributed by the vehicle. Most vehicles are slightly heavier at the front so, a 60:40% or 55:45% distribution would be a good approximation for most cars. Sports cars get closer to an even split. Then, you would subtract the weight of the tyre and supporting axle, brake, etc. since that weight is not supported by the spring. </p>
| 13714 | How would I calculate a force on a spring? Or more specifically the weight of an object that spring is supporting |
2017-02-12T01:25:01.190 | <p>How to finish part (a) and (b) of this dynamics problem?</p>
<p><a href="https://i.stack.imgur.com/aIQGb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aIQGb.png" alt="enter image description here"></a></p>
<p>There are no solution or answer in the book...</p>
<p>My professor say that this problem is very tricky, don't even try by yourself.</p>
| |mechanical-engineering|dynamics| | <p>Starting with equating the total energies at the starting and max velocity point. The energy at the starting point comes from the potential energies of the mass and spring. The energy at the other point also has the kinetic energy component of the mass.</p>
<p>$$
m g (R \tan (\alpha )+R)+\frac{1}{2} k \left(\sqrt{R^2+(R \tan (\alpha
)+R)^2}-\ell \right)^2 = \\
m g (R-R \sin (\beta ))+\frac{1}{2} k \left(\sqrt{(R-R
\sin (\beta ))^2+(2 R-R \cos (\beta ))^2}-\ell \right)^2+\frac{1}{2}m v^2
$$</p>
<p>($\ell$ is the unstretched spring length)</p>
<p>Now we can solve for $v^2$ (or $v$).</p>
<p>$$
\frac{1}{m}(-4. \sin ^2(\beta )-4. \cos ^2(\beta )+16. \cos (\beta )+40. \sqrt{-0.08 \sin
(\beta )-0.16 \cos (\beta )+0.24}+(3.92 m+8.) \sin (\beta )-12 +\\
4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan (\alpha )+0.08}+8.
\tan (\alpha )+3.92 m \tan (\alpha ))
$$</p>
<p>Looking at this expression we see that part of it depends on $\beta$ and the other part on $\alpha$, and there are no cross variables. The other variable is the mass $m$. Given the information that the max velocity is at $\beta=34 {}^{\circ}$, we can equate the derivative of $v^2$ to 0 and solve for $m$. This gives $m = 0.143 kg$. </p>
<p>I think this is as far as we can get with the information provided. The actual velocity is a function of $\alpha$ and as expected higher values produce larger velocities.</p>
<p>$$
2.64447 \sqrt{4. \tan ^2(\alpha )-40. \sqrt{0.04 \tan ^2(\alpha )+0.08 \tan
(\alpha )+0.08}+8.56054 \tan (\alpha )+12.0611}
$$</p>
<p><a href="https://i.stack.imgur.com/487UA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/487UA.png" alt="enter image description here"></a></p>
<p><strong>[Update]</strong> If $\alpha=45{}^{\circ}$ then the velocity is $6.86192\ m/s$.</p>
<p>(All the computations were done in Mathematica and I am attaching the screenshot).</p>
<p><a href="https://i.stack.imgur.com/NtN37.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NtN37.png" alt="enter image description here"></a></p>
| 13715 | How to compute the mass and max speed of a ball moving on a circular track under the influence of gravity and a spring |
2017-02-12T05:19:31.233 | <p>I'm trying to find PSIG from other variables that I have values for. I'll try my best to explain.</p>
<p>I have Helium flowing through a 1" copper pipe at 65 psig and at 40 deg F to a 1" ball valve and power control valve. There is a flow meter connected after the power control valve, with the ball valve throttled to achieve 40 CFM. I need to find the PSIG after the throttled ball valve.</p>
<p>I'm looking for the formula or a web calculator to solve this equation. I don't need a precise answer, just a rounded.</p>
<p>Any help will be appreciated.
Thanks, Steve.</p>
<p><a href="https://i.stack.imgur.com/P3WEE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P3WEE.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>Unfortunately for all of us who have to deal with fluid mechanics, pressure drop through a valve from a flow rate can't be computed without knowing the Cv of the valve. You actually need the Cv of both valves here. From that, and knowing your target flow rate, you can compute how much the pressure will need to drop to achieve the flow rate. </p>
<p>You can work the problem the other way, too: have a fixed pressure at the outlet and work backwards to find what your flow rate will be. Either way, you need the Cv or the pressure.</p>
<p>There are a million different references out there on valve sizing and the relevant equations: my personal favorite is "Fluid Flow Basics of Throttling Valves": <a href="http://www.controlglobal.com/assets/Media/MediaManager/RefBook_Cashco_Fluid.pdf" rel="nofollow noreferrer">http://www.controlglobal.com/assets/Media/MediaManager/RefBook_Cashco_Fluid.pdf</a></p>
<p>The equation you want, for gas flow, is: </p>
<pre><code>Cv = Qdot / (N7 * Fp * P1 * Y) * sqrt(Mw * T1 * Z1 / X)
</code></pre>
<p>With:</p>
<pre><code>Qdot - volumetric flow rate in SCFH
N7 - Unit correlation constant = 1360
Fp - Piping geometry factor (can be taken as 1 for your case)
P1 - Absolute upstream pressure in psia
Y - Expansion factor (maximum of 1, minimum of 0.667 - plug both in to get a range of possible pressures)
Mw - molecular weight in lb/lb mole
T1 - Temperature in Rankine
Z1 - compressibility factor
X - Pressure drop ratio (i.e., (UpstreamP - DownstreamP) / UpstreamP)
</code></pre>
<p>So clearly here you need to either solve for X given Cv and Qdot, or Cv given X and Qdot...and probably now you can see why this can end up being someone's full time job.</p>
| 13716 | Need help solving for PSIG |
2017-02-12T20:51:06.747 | <p>I am trying to make a device that can go up or down a screw by twisting a knob, kind of like a wing nut, but the wing would be sideways (I suppose). To clarify what I mean by this, please see a simplified version in a diagram below.</p>
<p><a href="https://i.stack.imgur.com/Dcw10.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dcw10.jpg" alt="enter image description here"></a></p>
<p>I think this type of engineering has already been used in various devices, such as a microscope, where the course adjustment knob will raise or lower the "stage" of the microscope.</p>
<p>Does anyone know where a part/device like this could be found, or how I would go about making something similar?</p>
<p>Thank you so much for your time and kindness,</p>
<p>Kelsey Nealon</p>
| |mechanical-engineering| | <p>I can't see a worm gear working in this instance as it would be trying to reverse drive the worm - there is too much friction to achieve this. A rack and pinion would not stay in place after releasing the knob. </p>
<p>You could use a 90 degree bevel gear with an internal thread in the output gear like the one in the photo. </p>
<p><a href="https://i.stack.imgur.com/DZozr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DZozr.jpg" alt="enter image description here"></a></p>
| 13724 | How to Make an Object Able to be Screwed In/Out With a Sideways Knob |
2017-02-13T04:30:07.313 | <p>With folding barriers like the one below, how does the orange bar get turned by the green mechanism? What's inside the yellow triangular part?</p>
<p><a href="https://i.stack.imgur.com/2gqpQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2gqpQ.png" alt="enter image description here"></a>
<a href="https://www.youtube.com/watch?v=LF8kSTCZIxw" rel="nofollow noreferrer">https://www.youtube.com/watch?v=LF8kSTCZIxw</a></p>
| |mechanical-engineering|gears|linkage| | <p>Since the pink linkage sits above the blue arm in the post, it has no problem reaching up to the trangular part when blue arm is raised. When blue arm is lowered, the pink linkage doesn't reach as far up the blue arm and starts to pull back on the triangular part, erecting the orange arm</p>
| 13729 | How do folding barriers work? |
2017-02-14T00:19:56.200 | <p>I would appreciate a quick peer check on the following.</p>
<p>I am currently enrolled in a class where, I think, the professor is repeatedly confusing <em>percent increase</em> and factor by which something increases.</p>
<p>There <em>is</em> a difference isn't there? I'm not going crazy here am I? </p>
<p>As I understand it, to calculate the percent increase I would do:</p>
<p>$$
IV = initial \; value\\
FV = final \; value\\
Assume \; FV > IV\\
\% \, Inc. = \left( \frac{FV - IV}{IV} \right) * 100 = \left( \frac{FV}{IV} - 1 \right) * 100
$$</p>
<p>If I want the factor by which the value increased with respect to the initial value I would take the ratio of the final value to the initial:</p>
<p>$$
Inc. \; Factor = \frac{FV}{IV}
$$</p>
<p>To reiterate, these are not the same things. I just want to make sure I'm not confusing anything before I point it out to him and ask for clarification.</p>
| |heat-transfer| | <p>Yes, what you show is correct.</p>
<p>For example, a value going from 8 to 12 is a 50% increase. You can also say it was increased by a factor of 1.5.</p>
<p>People sometimes get a little sloppy with this. If it's clear enough from context what the professor really means, I would let it go. If you're the professor, it's OK to insist your students use the terms correctly. When you're the student, it's better to reserve picking a fight for something more important.</p>
| 13750 | Percent Increase vs. Factor By Which Something Increases |
2017-02-14T14:40:20.667 | <p>As far as I know, propane with oxygen will burn at 2800 °C and steel will melt at ~1500 °C. Brass is even less.</p>
<p>What I am trying to do is a rocket nozzle.</p>
<p>Is there a way I can built this with steel or is it simply too hot? And for the burning temperature of propane, 2800 °C, is it the temperature <strong>IN</strong> the nozzle, <strong>OUT</strong> of it, or <strong>IN</strong> the combustion chamber?</p>
<p><a href="https://i.stack.imgur.com/3nDU4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3nDU4.png" alt="enter image description here"></a></p>
| |cad|propulsion|autodesk-inventor|rocketry| | <p>To answer the other part of your question, the temperature is that in the chamber, but note that that figure assumes perfect combustion, which may not be the optimum point for specific impulse in your geometry. Many rockets run fuel rich as a reducing atmosphere in the hot bits is easier to cope with then an oxidising one. </p>
<p>The largest thermal flux is usually at the nozzle throat, as that is where the combination of high pressure, high temperature and high mass flow come together. </p>
<p>The nozzle is a complex thermodynamic machine, which trades hot and high pressure, for cooler, a pressure ideally just above external atmospheric, and high velocity. </p>
<p>For ablative nozzles, graphite is a magic material. </p>
| 13758 | Propane rocket nozzle material |
2017-02-14T15:57:32.287 | <p>Ok, I wish to plot the speed time curve of the my urban service . The travel includes the curvature ( as shown in image ( just for reference )).
<a href="https://i.stack.imgur.com/BPMSw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BPMSw.jpg" alt="curvature in the track"></a>
The train speed is decreased while it runs along the arc ( i am pretty sure ) . The question is whether this decrease in train speed is considered while plotting a speed time curve or is it considered to be negiligible. I have to plot one . <a href="https://i.stack.imgur.com/KNjF3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KNjF3.jpg" alt="urban service speed time curve"></a></p>
| |rail|transportation| | <p>As is often the case the answer is that it depends. Urban train services tend to operate at fairly modest speeds and it may well be that they are run at an optimal constant safe speed for a given stretch of track, bearing in mind also that there are a lot of energy losses associated with accelerating and braking a heavy train.</p>
<p>Equally there may be no real advantage in running at the maximum possible speed if there are other bottlenecks in the system associated with signalling and spacing of trains, especially in an urban system where there are likely to be lots of stops and large volumes of passengers getting on and off. For example there is little point expending fuel getting to between stations 5 minutes faster if that just means you need to wait an extra five minutes for the previous train to clear that section of track. </p>
| 13763 | Does the decrease in speed of the train at the curvature changes the properties of Speed Time Curve Of Urban Service |
2017-02-14T23:56:14.687 | <p>Is there a device that acts like a spring (i.e. $F=kx$) but the spring constant, $k$, has different values in compression and tension?</p>
<p>For example, if $k_c > k_t$ (where c and t refer to compression and tension), then the spring would quickly compress und a compressive force but would slowly expand under a tensile force.</p>
<p>I imagine you could design a hydraulic system like this with some check valves and control valves. Is there a single device that does this? Or alternatively, are there practical engineered systems that do this?</p>
| |springs| | <p><a href="https://i.stack.imgur.com/vgRc5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vgRc5.png" alt="enter image description here"></a></p>
<p>Essentially a bi-modulus or bilinear characteristic? In this suggestion two strips are pinned together and one end is fixed at the other. When compressed the buckling resistance is low. In tension strip stresses in metal are high, so acts as a stiff spring.</p>
| 13770 | Spring-like device with anisotropic k value |
2017-02-15T10:08:39.507 | <p>I want to engrave some letters from a 2D dxf file with single stroke fonts.</p>
<p>Is this easy to do in most CAM software?</p>
<p>I have not decided what CAM software to use yet, because the ability to do single stroke fonts is very important to me. </p>
<p>I have read on forums of different CAM software, that people often find this difficult. Maybe it is because single stroke fonts are not included in the software.</p>
<p>However I already have the font as a dxf, so all I am asking is for the machine to follow a single line, where the centre of the cutting tool follows the line.</p>
| |cnc| | <p>Yes this is a simple operation. The only real consideration is the shape/size of the cutter. Obviously a tapered cutter will carve a path where the width is proportional to the depth. Therefore some understanding of the toolpath and interference with other objects or cuts is necessary.</p>
<p>This is easily visualized and controlled by most CAM applications once the machine setup is understood. The DXF files may also need to be cleaned up if they were produced from a raster conversion.</p>
| 13772 | CNC engraving single line fonts |
2017-02-16T15:11:21.757 | <p>I need to understand some basic concepts about the cooling characteristics of a black "polymer"-and-zinc-alloy handgun (specifically, <a href="http://www.hi-pointfirearms.com/Hi-Point-handguns/9MM_handgun.html" rel="nofollow noreferrer">this one</a>).</p>
<p>If this gun was left on the ground under daytime sunlight, I assume that both the ground and the gun would get warmer. I assume, also, that after the sun went down, both the ground and the gun would begin to cool. My questions are:</p>
<p>1) would the gun warm under sunlight to a higher temperature than the ground, because it is black and/or because of what it's made of?</p>
<p>2) after sundown, would the gun cool at the same rate as the ground? Or, would the rate of cooling be different?</p>
<p>Edit: I should clarify that the surrounding ground is essentially brown dirt/mud.</p>
| |heat-transfer| | <p>There are three main heat transfer mechanisms: conductive, convective, and radiative.</p>
<p>Conductive transfer means direct transfer from one item to another it's in contact with. The transfer rate depends on the empirically determined thermal conductivity of both materials. </p>
<p>Convective refers to heat transfer between an object and a moving fluid such as wind. This depends heavily on the total surface area available (such as fins on a radiator).</p>
<p>Radiative means absorption or emission of photons. The energy transfer rate depends on the surface properties, called "emissivity" or equivalently "absorptivity" and is wavelength-dependent for any given material. </p>
<p>OK, so with that in mind, yes a black object generally will absorb more sunlight than a lighter shade of ground/dirt. This is a generality, as the absorptivity in the infrared may be different than in the visible. At the same time, the conductive heat loss/gain to the ground depends on the total contact area, and further will be much greater if the ground is damp, because water is good at conducting a lot of heat 'fast.' After dark, the radiative processes are sort of reversed. It's not an exact reversal because the spectral radiation rates depend on the material AND on the temperature of the object. (actually the radiation rates apply all the time; it's just that in daylight each object absorbs more light than it radiates until it reaches a thermal equilibrium). </p>
<p>In sum, the black object probably will get cold faster, and get colder than the ground. As a practical example, come to New England in the late Fall. On cool nights, black stuff such as roof shingles often frost over when other things like white walls don't, because of radiative cooling. </p>
| 13780 | Rate of Cooling of a Black "Polymer" Handgun |
2017-02-17T04:13:21.307 | <p>I am working on a DIY project, I am trying to replicate a bar compressor which for some reason is only sold in Europe so I thought it would be fun to make my own without having to waste so much material. I simply want to make sure the force going down with the plate somewhat replaces a person to jump in the trashcan. Somewhere above 80+ lbs, since I am no engineer I was wondering how would one go about finding the force applied going down to the trash if about 50lbs is pulling down the lever. I understand that materials are key to this but as an amateur I just wanted to know how to get started. Sorry for posting here but I thought it would suit this page the most for this question. Thank you.</p>
<p><a href="https://i.stack.imgur.com/8CRBu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8CRBu.jpg" alt="Drawing of beam"></a></p>
<p><a href="https://i.stack.imgur.com/aZwty.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aZwty.png" alt="Actual use of bar"></a></p>
| |mechanical-engineering|civil-engineering|applied-mechanics|statics|dynamics| | <p>Balance the moments acting.</p>
<p>So,</p>
<p>EffortForce x 39 x cos 30 = 18.5 cos 30 x resulting force.</p>
<p>Or more accurately,</p>
<p><a href="https://i.stack.imgur.com/pBX46.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pBX46.jpg" alt="enter image description here"></a></p>
<p>Sin(90+ theta - alpha ) x 39 x effortForce = cos(theta) x 18.5 x requiredForce</p>
<p>This equation will give you answer for any angle of level. </p>
| 13792 | Finding downward force of lever bar |
2017-02-17T09:55:55.440 | <p>I have a rectangular shape with uniform mass, approx 6m x 1m. It rotates around an axis in the centre of its width but 1m from one end. The axis has some friction but this is not known.</p>
<p>What I need to work out is how much force is required to start rotating this object from different positions from the axis.</p>
<p>I fear I may be plunged out of my depth but if anyone could point me in the right direction it would be really appreciated!</p>
| |mechanical-engineering|applied-mechanics|moments| | <p>If we neglect friction and the object is rotating about its centre of mass then it will experience an <a href="https://en.wikipedia.org/wiki/Angular_acceleration" rel="nofollow noreferrer">angular acceleration</a> proportional to the torque applied and its moment of inertia. </p>
<p>Note that this is a direct analogue of the linear acceleration equation <em>F = ma</em> </p>
<p>But instead of force we use torque (force x radius) and instead of mass we use <a href="https://en.wikipedia.org/wiki/Moment_of_inertia" rel="nofollow noreferrer">moment of inertia</a>. </p>
<p>So as with linear acceleration, if we ingnore friction any force greater than zero will produce some acceleration. </p>
<p>The situation is slightly different if the axis of rotation is offset from the centre of mass. In this case the weight of the body will exert a torque on it ie it will behave as a <a href="https://en.wikipedia.org/wiki/Pendulum" rel="nofollow noreferrer">pendulum</a>. </p>
<p>So for your example with a 1m x 6m plank rotating around a point 1m from one end the centre of mass will lie 3m from each end so it can be simplified to a point mass connected to a pivot by a mass-less rigid bar 2m long. </p>
<p>There will be a constant force acting on the point mass of Mg acting vertically downwards. This will create a torque when it is displaced at some angle x from the horizontal so, again any torque greater than zero will produce some displacement but in order to get it horizontal you need to support the its full wieght so :</p>
<p>the minimum constant torque to rotate it in a full circle is given by :</p>
<p>T > M.g.r</p>
<p>Where M is the mass of the object, g is acceleration due to gravity and r is the distance from centre of mass to the pivot point. </p>
<p>Strictly speaking this is the torque required to hold it stationary in the horizontal position. </p>
<p>To work out the holding torque for other positions you need to use trigonometry to work out the resultant toque from the vertical force due gravity. </p>
| 13798 | Force required to rotate an object on a plane - moment of inertia? |
2017-02-18T12:35:42.947 | <p>I've heard that in the past, spray cans (aerosol cans) contained Freon. I found that the Montreal Protocol made the usage of Freon illegal due to its effect on the ozone layer, but I can't find the reason for why Freon was used in the first place. I understand its role in refrigeration systems, but I fail to see a functionality for spray cans.</p>
<p>So why would anyone use Freon in spray cans, what was its contribution? I assume that it affected the pressure inside somehow, but it's just a guess.</p>
| |thermodynamics|pressure|compressed-air|refrigeration|compressed-gases| | <p>Spray cans work by containing a propellant which has a boiling point a small amount below room temperature. When the can is sealed the liquid and vapour reach equilibrium at a relatively modest pressure. When the spray valve is opened this pressure forces out the contents as an aerosol and more of the propellant evaporates as the internal pressure drops. </p>
<p>This has two advantages firstly the can doesn't need to be anywhere near as strong as if the propellant was a fully pressurised gas and it provides a fairly constant pressure untill the propellant is exhausted without needing mechanical regulators which would add cost and complexity. </p>
<p>This is pretty much the same principal as a butane lighter. </p>
<p>So what you need in a propellant is something with a very specific boiling point which provides the required vapour pressure at the expected operating temperature. </p>
<hr>
<p>As mentioned in the comments butane and/or propane are now often used as propellants in aerosol type cans indeed 'canned air' dusters used for cleaning computers etc is often just butane. </p>
<p>One downside of butane is that there is an inherent cooling effect as the liquid evaporates and the gas expends which can cause a significant cooling effect and, as the boiling point of butane is a bit on the low side this can cause significant cooling of the contents of the can and a consequent drop-off in pressure. </p>
| 13811 | Why was Freon used in spray cans before it was declared a threat to ozone? |
2017-02-19T03:10:11.253 | <p>I encountered a problem in measuring electrical energy using a multi ratio current transformer (CT). The catalogue of CTs is unavailable.</p>
<p>I have little information about the type of CT I am using. I have several questions about this matter. Firstly, the CT I work with has 7 connections:
X7,100,200,300,400,500&600.</p>
<p>The connection for ratio 100/5 was on X7 & 100. Something that is confusing me is that X7 a common connector and if connection X7 to 100 measures 100/5, are other windings from 200 to 600 in circuit or open? If they are in circuit how does it produce a 100/5 ratio, and if they are open how it that possible for a CT?</p>
<p>Secondly, the power transformer has delta connections and the CT connection is a Star. After I have connected the meter to the CT I get an error in energy metering. Is it possible that this error is due to the star connection of the CT and should it be a delta connection, as for the power transformer?</p>
<p>Its impossible to change the CT connection to delta. Can I define any coefficient factor to remove this error with out changing the circuits?</p>
<p>The power transformer ratio is 14400/3300, with a delta connection and the
CT's ratio is 200/5, with star connection.
<img src="https://i.stack.imgur.com/jKvz5.jpg" alt="enter image description here"></p>
| |electrical-engineering|sensors|measurements|circuits| | <p>I've encountered this problem before.</p>
<p>This problem didn't depend on the connection of your power transformer.</p>
<p>This problem happens when meter coefficients (depend on your CT's and PT's ratios)are inappropriate.</p>
<p>The note is this:Check your circuits node and use KCL law to calculate the powers in node.</p>
<p>Best regards</p>
| 13826 | Multi Ratio Current Transformers |
2017-02-19T19:00:17.193 | <p>I am building a free flight helicopter for a Science Olympiad competition, and the general rules are as such: <a href="https://www.soinc.org/sites/default/files/uploaded_files/Helicopters_2017_Diagram.pdf" rel="nofollow noreferrer">https://www.soinc.org/sites/default/files/uploaded_files/Helicopters_2017_Diagram.pdf</a></p>
<p>This leaves a lot of opportunity for math that is well beyond my ability, so my questions are as such:</p>
<ol>
<li><p>You get scored based on how long the craft stays in the air, but you can get 25% added (up to 75% total) for each single bladed rotor, is a single bladed rotor a worthwhile investment?</p></li>
<li><p>I have tried using an airfoil shaped wing at the regional competition, and it failed because of how heavy it was, but people have been suggesting a helical blade shape, what are the mathematical advantages of this over a flat rotor.</p></li>
<li><p>The central shaft has no given dimensional restriction, what is the optimum length of the core? </p></li>
</ol>
<p>Thanks in advance.</p>
| |mathematics|building-physics| | <ol>
<li><p>Yes, even through a simple parallelization of a single-rotor design. If there's no other payload than the helicopter itself, and the challenge is for staying airborne, you gain about nothing performance-wise by adding more blades, but hey, +75% score. Go with quadcopter and four bands. Driving four rotors through a single shaft in line would be a hell of mechanics.</p>
</li>
<li><p>Single-bladed airfoil will mean the blade is way off-center, introducing hopeless torque. If your single blade is a full turn of a helix, its center of mass is dead in the axis of rotation. It's not really beneficial aerodynamically but it's a way to cheat center of mass into rotation axis with the contrived "single blade" restriction through making the blade big.</p>
<p>The airfoil gives a decent lift to mass ratio at higher airspeeds, but you won't get high airspeeds from a rubber band. You must buy your lift with wing surface area - and the helical blade will give you that. The blade doesn't need to be THAT big for that, but that's one way to make center of mass and axis the same.</p>
</li>
<li><p>It will likely be dictated by many factors - keeping the rubber band tension, minimizing mass, keeping center of mass low enough. I don't think you can find a good <em>formula</em> for it. It's something that should be found experimentally.</p>
</li>
</ol>
| 13832 | Rubber band Powered Free Flight Helicopter |
2017-02-19T19:45:36.843 | <h1>Background</h1>
<p>Looking to build a table for a dryer to fit inside a laundry closet. The table must be assembled inside the closet, as such it has three parts: two legs and a metal sheet for the table top. Here's a picture of the table outside the closet:</p>
<p><a href="https://i.stack.imgur.com/vzyFI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vzyFI.png" alt="laundry closet washer" /></a></p>
<p>Here's a picture of the dryer on top of the table:</p>
<p><a href="https://i.stack.imgur.com/JZbG9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JZbG9.png" alt="laundry closet washer and dryer" /></a></p>
<h1>Weights, Measures, and Materials</h1>
<h2>Dryer</h2>
<ul>
<li>Dimensions: 65.5cm x 60cm</li>
<li>Weight: 50kg</li>
</ul>
<h2>Table</h2>
<ul>
<li>Dimensions: 106cm x 63.5cm</li>
<li>Material: Steel alloy</li>
</ul>
<p><a href="https://i.stack.imgur.com/XdBHr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XdBHr.png" alt="Table Dimensions Wireframe" /></a></p>
<p>Same, but without the wireframe:</p>
<p><a href="https://i.stack.imgur.com/JR9qI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JR9qI.png" alt="Table Dimensions" /></a></p>
<p>Without the table top to show the supporting flaps:</p>
<p><a href="https://i.stack.imgur.com/nS1UI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nS1UI.png" alt="Table Flaps" /></a></p>
<h1>Problem</h1>
<p>I'm concerned that the table thickness will be insufficient to support the load. A <a href="https://www.engineersedge.com/calculators/flat-plate-deflection.htm" rel="nofollow noreferrer">flat plane deflector</a> calculator is useful for calculating the total load applied to center of plate, but doesn't quite tell me if the distributed load of the dryer (not a point load) will be structurally sound.</p>
<p>In comparison, commercial <a href="http://www.globalindustrial.ca/g/work-benches/tables-stands/machine-tables/welded-12-gauge-machine-tables" rel="nofollow noreferrer">heavy gauge machine tables</a> support over 900kg, so it looks like the design is sound.</p>
<h1>Question</h1>
<p>Are these calculations reasonable? (Especially pertaining to the forces imparted by the dryer?) That is, will a 1.25cm (approx. ½") steel alloy table top outlast the dryer?</p>
<h1>Related Links</h1>
<ul>
<li><a href="https://www.engineersedge.com/beam_bending/beam_bending3.htm" rel="nofollow noreferrer">https://www.engineersedge.com/beam_bending/beam_bending3.htm</a></li>
<li><a href="https://www.engineersedge.com/calculators/flat-plate-deflection.htm" rel="nofollow noreferrer">https://www.engineersedge.com/calculators/flat-plate-deflection.htm</a></li>
<li><a href="https://www.engineeringtoolbox.com/young-modulus-d_417.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/young-modulus-d_417.html</a></li>
<li><a href="https://www.engineeringtoolbox.com/poissons-ratio-d_1224.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/poissons-ratio-d_1224.html</a></li>
</ul>
| |structural-engineering|materials|stresses|mathematics| | <p>Based on the numerous helpful comments, I've settled on the following design.</p>
<p>After researching structural integrity of steel alloys, my understanding is that fatigue failure due to cyclic stresses can be eliminated by ensuring that the endurance limit of the alloy is higher than the cyclic forces the dryer will impose. I've also read that some steel alloys will not fatigue regardless of the number of cycles.</p>
<p>Here are my assumptions:</p>
<ul>
<li>Point load = 35kN (estimated)</li>
<li>(x, y) offset to the right of the table top</li>
<li>Dimensions = 64cm x 106.5cm x 1cm</li>
<li>Young's modulus = 200GPa</li>
<li>Poisson's ratio = 0.287</li>
</ul>
<p>The (x, y) stresses are therefore (193MPa, 113MPa).</p>
<p>Structural steel (ASTM-A36) has a yield strength of 250MPa and tensile strength of 400MPa. AISI 8620 is 385MPa/530MPa, respectively. At 2cm thickness, the point load would have to exceed 140kN for the same stresses as 1cm, thus a ½" thickness for the table top should not have its endurance limit exceeded by the dryer's cyclic stresses.</p>
<p><a href="https://www.homedepot.ca/en/home/p.laundry-anti-vibration-pads.1000672766.html" rel="nofollow noreferrer">Laundry pads</a> prevent walking and reduce vibration. The end result will resemble (pads not rendered):</p>
<p><a href="https://i.stack.imgur.com/weEZw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/weEZw.jpg" alt="Table w/Dryer" /></a></p>
<p>Instead of an X brace, I've opted for two horizontal braces.</p>
<p><a href="https://i.stack.imgur.com/dQOQr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQOQr.jpg" alt="Table without Dryer" /></a></p>
<p>Without the top, it'll look like:</p>
<p><a href="https://i.stack.imgur.com/uQuuU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uQuuU.jpg" alt="Table without Top" /></a></p>
<hr />
<p>The metalworker decided simple was best and basically welded a table together inside the closet:</p>
<p><a href="https://i.stack.imgur.com/ZM8Bq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZM8Bq.jpg" alt="Finished table" /></a></p>
| 13833 | Custom heavy gauge machine table to support clothes dryer |
2017-02-19T22:45:38.343 | <p>I'm building a server type computer consisting of integrated motherboard, CPU, RAM and hard disk. There will also be a standard power supply unit. This is to be a Steam Punked computer, so...</p>
<p>I want to know the feasibility of passively cooling the server using only a chimney effect. So the electronics would be enclosed, with a chimney at the top and air intake(s) at the bottom. All the components will be custom mounted so there is no typical computer case and therefore no particular size constraint. How would I begin to evaluate /design this? I believe that there are relationships between chimney proportions, temperatures and exhaust velocities. Perhaps like this...</p>
<p><a href="https://i.stack.imgur.com/LiGi9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LiGi9.png" alt="Computer cooled by chimney effect"></a></p>
<p>Mini computers (ARM /Celeron) exist that have no fans, so perhaps it isn't really a big deal and anything at all will work?</p>
| |thermodynamics|cooling| | <p>The key technical term you're looking for here is convective cooling. Either you allow the air to rise through the chimney on its own accord (natural convection cooling) or you pump it through with a fan (forced convection cooling). The second is obviously more effective at pulling heat out of the system.</p>
<p>The amount of heat removed will be a factor of the flow rate of air through the system as well as the heat transfer coefficient (HTC) between the air and the CPU. You can get a good estimate by grabbing an HTC function from engineering toolbox or a text book and using Newton's Law of Cooling:</p>
<pre><code>q = hc A dT
</code></pre>
<p>where q is heat transfer (in watts), hc is the heat transfer coefficient (W/m^2 K), A is the surface area to be cooled (m^2), and dT is the temperature difference between the air and the surface (in K).</p>
<p>The HTC for air is (for v between 2-20 m/s):</p>
<pre><code>hc = 10.45 - v + 10 v^1/2
</code></pre>
<p>where v is your air flow speed in m/s. The faster the air is moving, the higher your HTC (and the more heat you can pull out of the system). </p>
<pre><code>Broad estimates of HTCs:
Free convection: 0.5-1000 W/(m^2 K)
Forced convection: 10-1000 W/(m^2 K)
</code></pre>
<p>You should know from the specs of the machine roughly how much heat is produced in watts. Then your air flow will dictate your HTC - and try to maximize it so that you can pull more heat out of the system.</p>
<p>The concept of convective cooling is pretty widely discussed; any general engineering thermodynamics textbook will have more detail ("Thermodynamics: An Engineering Approach" by Cengel or "Fundamentals of Heat and Mass Transfer" by Incropera are good resources). There's quite a lot of stuff about general engineering approaches online, too: where to locate the object to be cooled in your enclosure, what sort of fans to use, etc. These are the sorts of problems that can be analyzed at a really detailed level with computational fluid dynamics. To check the feasibility, I would start with Newton's Law of Cooling, though.</p>
| 13837 | How to design a cooling chimney for a computer server? |
2017-02-20T04:23:03.620 | <p><a href="https://en.wikipedia.org/wiki/Combined_cycle" rel="nofollow noreferrer">Combined Cycle Turbines</a> are a gas turbine whose exhaust heat powers a second turbine running on steam. These can get thermal efficiencies greater than 50%, whereas a plain turbine would get 30 to 35%. Large power plants tend to use combined cycle for obvious reasons.</p>
<p>But I want to know if this can be small enough to power a train?</p>
<p><a href="http://www.dreamworksrc.com/catalog/Jetcat-P200Rx" rel="nofollow noreferrer">Example</a> of a very small jet engine, which is very similar to a gas turbine.</p>
<p>Desired specs:</p>
<ul>
<li><p>Fuel: either natural gas or kerosene</p></li>
<li><p>Horsepower: ~4500 (taken from a typical diesel train engine (<a href="https://en.wikipedia.org/wiki/GE_Dash_9-44CW" rel="nofollow noreferrer">GE Dash 9</a>))</p></li>
<li><p>Max volume: 25 x 3 x 6 meters (about the volume of a train car)</p></li>
<li><p>Max weight: 100 metric tons</p></li>
</ul>
<p>I'm assuming a fuel car can follow behind the engine car. The turbines would generate electric power sent to the car wheels, which is what a modern diesel-electric train does. I was imagining combined cycle turbo trains for freight and passenger services, but at "normal" speeds (100 - 150 kmh).</p>
<p>So can this be done with today's technology? The engine would still need to be insulated enough and efficient enough to retain 50+ % thermal efficiency.</p>
<p>BTW, I'm aware of <a href="https://en.wikipedia.org/wiki/Turboliner" rel="nofollow noreferrer">this failed example</a> from Amtrack starting in the 1970's. I don't think those were combined cycle turbines, and I could not find an explanation for why "The high cost of operating the trains led to their withdrawal from the Midwest in 1981." But I did notice those were tiny trains, 5 cars total (2 engine cars, 2 coaches, and a food car). That is probably far too small to get economy of scale.</p>
| |civil-engineering|turbines|energy-efficiency| | <p>We can try to answer this question by looking at existing equipment in terms of size and weight and make an educated guess from there if all this fits on a train. Our CC will consist of:</p>
<ul>
<li>Gas turbine</li>
<li>Exhaust HX/Boiler</li>
<li>Steam turbine with condensor and other aux. parts</li>
<li>Transmission</li>
</ul>
<p>For the following, I will not match components exactly, but pick stuff that sits in the right power range, just to get a feel if the parts could fit into the design envelope given above.</p>
<p><strong>Basic sizing</strong><br>
You give 4500hp, that's a bit more than 3MW. Looking at wikipedia, the gas turbine is typically twice the power of the steam turbine. Soe we look for a 2MW gas turbine and a 1MW steam plant. </p>
<p><strong>Gas turbine</strong><br>
What are some turbine around 2MW? </p>
<ul>
<li><a href="http://www.kawasaki-gasturbine.de/files/Gas_Turbine_Generator_Set_GPB17D_englisch_rev2.pdf" rel="nofollow noreferrer">Kawasaki has one at 1.7 MW</a>, it seems the 'naked' turbine fits into a 6.5m container + there's another one stacked on top for air vents. No mass given, does not seem to include exhaust HX.</li>
<li><a href="http://www.opraturbines.com/en/OPRA-WEB---Products---Products/" rel="nofollow noreferrer">This 2MW Opra engine</a> again fits into a 6.5 m container (without air supply), the naked engine and weighs maybe 13.5 t.</li>
</ul>
<p>For the following, I'll take the stats of the Kawasaki engine and assume the weight of the Opra engine is not too far off.</p>
<p>Kawasaki:<br>
Electrical Power 1.690 kW<br>
Fuel input 6.360 kW<br>
Efficiency 26,9 %<br>
Exhaust gas mass flow 8,00 kg/s<br>
Exhaust gas temperature 521 °C </p>
<p><strong>Boiler/Exhaust HX</strong><br>
If we take the Kawasaki engine ebove, we have about 480kg/min exhaust gas. Looking at <a href="http://www.ejbowman.co.uk/pdf/Exhaust%20Gas%20R-angle%20Iss%201.pdf" rel="nofollow noreferrer">this table of exhaust HX</a>, we would need 7 of the largest, each about 600kg and length x diameter 2.6 x 0.7 m. Roundabouts. At up to 700kW heat recovery each, we should get north of the 2MW worth of steam we need for our steam turbine.</p>
<p><strong>Steam turbine</strong><br>
<a href="http://www.energy.siemens.com/hq/pool/hq/power-generation/steam-turbines/Pre-Designed_Steam_Turbines_en.pdf" rel="nofollow noreferrer">Siemens has one that covers the 800kW-6MW range</a>. Without condensor it fits into a LxWxH evelope of 1.5 x 2.5 x 2.5 m. No weight is given, I guess a few tons.</p>
<p><strong>Condensor</strong><br>
We need to shed about 1MW of heat for our 1MW steam engine. Let's take some measurements from this <a href="http://www.guentner.eu/fileadmin/literature/europe/condensers_drycoolers/GVW/Guentner_GVW_Data_sheet_DE_EN.pdf" rel="nofollow noreferrer">product sheet from Güntner</a>. Note that these a condensors for refridgerant. I guess the bottleneck is rather the airflow and the heat transfer towards air, then these should be ok.</p>
<p>The largest have a nominal capacity of 500kW, weigh 2t and are about LxWxH 10x1.2x1.7 m. Two of those side by side.</p>
<p><strong>Transmission</strong>
My guess is that an electrical trasnmission would be used, then you'd have to add a 3MW motor and a 1MW and 2MW alternator. Because by now I'm lazy I'd say that these add maybe 2-3 m to our 'length budget'. </p>
<p><strong>Conclusion</strong><br>
We don't seem to really reach the mass budget of 100t (which seems way large for just the engine, OTOH).</p>
<p>Our train could look like, from from to back: </p>
<ul>
<li>6m length dedicted to air intake, filters etc.</li>
<li>6.5 m for the gas turbine (13t)</li>
<li>3m for a stack of HX (3.5t)</li>
<li>3 m for the steam engine (3t?)</li>
<li>10m for the condensors (4t)</li>
<li>2-3 m for alternators and motor (10t?)</li>
</ul>
<p>For a total of 31m. This is larger than the size envelope given by in the OP, but not by much. So my conclusion is: <em>You can fit 3MW worth of combined cycle plant on a train car.</em> </p>
<p>There's lots of room to improve the whole thing at the HX and condensors, OTOH we need to fit large electrical systems on the train. In summary I believe you could build a combined cycle powered train & you would not have to invent too many components to do it. </p>
| 13838 | Can combined cycle turbines be small enough for trains? |
2017-02-20T06:29:44.617 | <p>I recently purchased a <a href="http://www.ebay.com/itm/282070870420" rel="nofollow noreferrer">digital scale from eBay</a>.</p>
<p>According to the listing it can weigh in the following units:</p>
<blockquote>
<p>Multiple Weighting Units: g, oz, tl, ct.</p>
</blockquote>
<p>Screenshot of that page:</p>
<p><a href="https://i.stack.imgur.com/mkdxa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mkdxa.png" alt="Units of weighing"></a></p>
<p>Now I know that "g" is grams, "oz" is ounces, and "ct" is carats (and the figures displayed agree with that) however what is "tl"?</p>
<p>If I measure a 10g reference weight it shows as 0.853 tl, thus 1g is 0.0853 tl. Allowing for slight measurement errors (the last significant figure changes a bit with different measurements) I can't find any unit of weight that "tl" could correspond to.</p>
<p>The closest in name I could find is "tael" but the conversion factor doesn't agree. (10g is 0.264555 tael).</p>
<p><a href="https://i.stack.imgur.com/jRd7O.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jRd7O.jpg" alt="10g in tl"></a></p>
<hr>
<p>The unit also measures in grains (gn) although that is not advertised.</p>
<p>The device only measure up to 200g, so I imagine that whatever "tl" is will be some sort of measurement of jewellery or similar.</p>
<p>So, my question is: <strong>What is tl?</strong></p>
| |measurements| | <p>tl is a measure of weight in Asia and is the tael. 1 tael is equal to 1.5 oz or 50 grams.</p>
| 13840 | What is the "tl" unit of weight measurement? |
2017-02-20T11:18:24.717 | <p>I have asked this question in 3DPrinting.stackexchange forum but got no answer answer. Please take a minute to see if you can help. </p>
<p>I am building a simple RWD vehicle using 3d printer parts. See the picture below. <a href="https://i.stack.imgur.com/FQIfX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FQIfX.jpg" alt="RWD vehicle with steering"></a></p>
<p>As you can see from the picture, I am using 3d printed linkages for steering mechanism and I used regular bolt and lock nuts to secure them together (thats what I had laying around the house at the moment). The problem here is that, holes in the linkages are not exactly the size of the bolt, and because the bolts are threaded, it generates a lot of friction. Other than that, the bolt seems to sit loose inside the hole and creates an unstable pivot.
<a href="https://i.stack.imgur.com/pTxK4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pTxK4.jpg" alt="enter image description here"></a></p>
<p>If I tighten the nut more, it just presses them together making it hard for the servo to operate. I am looking for a solution that will use simple bolts, nuts and washers which can be bought easily off the shelf (or even with the bolts and screws everyone has around there house). What can be done to make this joint stable in xy plane, rotating freely around the bolt. By the way, even though I used nylon lock nuts, it seems to loosen after couple of repetitive rotations. </p>
<p>Thanks for your help in advance and I hope ya'll have a great day. </p>
| |3d-printing|bolting|linkage| | <p>Simple, easy but lower-quality approach: two nuts and a tension washer between them. Turn the nut until the linkages are secured in position with correct amount of play. Put the tension washer and a second nut, then holding the first nut immobile with a wrench, tighten the second nut very firmly. That way the pair of nuts will remain in place on the bolt without moving either way.
Put a flat washer between the linkages; friction of 3d-printed plastic against 3d-printed plastic is much higher than plastic against metal.</p>
<p>Better quality, harder approach: nut-based bearings. Drill in the holes of the linkages, to size a little smaller than the nut. Heat the nut to temperature sufficient to melt the plastic, and place it in the hole, fusing it with the linkage. Screw the bolt in; leave some room between the linkages - about 2 turns worth. Secure the bolt by tightening it to one (top-most) nut, so that it's immobile relative to one of the linkages; the nut in the other acts as a bearing; it will introduce some half a millimeter of travel up or down as it turns, but that should be well within your tolerances.</p>
<p>Of course the best approach would involve actual ball bearings, but it seems for your application that might be an overkill. </p>
| 13845 | How to build a pivot linkage for a robot. |
2017-02-20T13:03:40.970 | <p>The answer to a question on <a href="https://engineering.stackexchange.com/a/12951/2555">involute gear</a> curve calculations shows how the gear tooth curve proceeds as an involute calculation from the base diameter.</p>
<p>If the <em>pitch diameter</em> is D = m.N where N is the number of teeth and m is the modulus (for metric gears), then the <em>base circle diameter</em> for a pressure angle α, most usually 20 degrees, is D.cos α and the <em>root diameter</em> is given by the amount the tooth falls below the pitch circle. This amount is the dedendum and is usually 1.25m. Thus the root circle diameter is D - 2.5m.</p>
<p>The involute is calculated in a standard way and looks like the unwinding of a string from a circle, the base circle:</p>
<p><a href="https://i.stack.imgur.com/YafhH.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YafhH.png" alt="enter image description here"></a></p>
<p>The gear tooth is then described by the intersection of the involute with the root diameter (the bottom of the tooth) and the outer diameter (the top of the tooth). An issue arises if the root diameter is less than the base diameter as the involute is not defined below the base diameter (you get a math domain error in Python if you try it!).</p>
<p><a href="https://i.stack.imgur.com/pJUHq.png" rel="noreferrer"><img src="https://i.stack.imgur.com/pJUHq.png" alt="enter image description here"></a></p>
<p>These tooth are described as "undercut". I cannot find a description of how the curve proceeds once you "run out of involute" - ie cross the base diameter.</p>
<p>Is there a standard way of describing the rest of the gear tooth curve?</p>
| |design|gears| | <p>In general, each tooth of an involute gear is made up of two curves: the involute curve and the root fillet. The involute curve is essential for gears to operate properly, and allows power to be smoothly transmitted from one gear to the other. The root fillet on the other hand is not involved directly with power transmission or the kinematics of the gears (i.e. how the motion of one gear influences the motions of a meshing gear.). This means that the root fillet curve's shape is reasonably arbitrary, provided, of course, that the root fillet on one gear does not at any point make contact with the teeth of the other gear. In addition, the root fillet also plays a roll in determining how much bending stress develops at the root of the teeth during operation: if the root fillet has a low radius of curvature, this will result in a large stress concentration at the root.</p>
<p>Despite the root fillet curve having the potential to be almost any shape, there are a few shapes that occur commonly, the two which I believe to be most common are as below:</p>
<p><strong>1. Connecting involute curve and root circle with radial lines</strong></p>
<p>The simplest root fillet to determine mathematically is simply a straight radial line that connects the start of the involute curve to the root circle. This is shown below:</p>
<p><a href="https://i.stack.imgur.com/RdPj3.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RdPj3.png" alt="enter image description here"></a></p>
<p>The advantages of the root "fillet" (a sharp corner is not much of a fillet I guess) is the geometry is very simple. However, there are two big disadvantages: 1. The sharp corner produces a very large stress concentration at the root of the gear, meaning larger root bending stresses. This means that these gears are just not suitable for applications that subject them to strong torques. 2. In cases where <em>undercut interference</em> becomes an issue, these radial lines are going to get in the way of the teeth, causing gears to jam. For gears with pressure angles at the standard value of 20°, undercut interference usually occurs for gears with teeth less than 18 (if gear is meshing with a rack gear) or 13 (if gear is meshing with another gear of equal number of teeth). For these reasons, you will rarely see gears with this kind of root fillet in industrial application, but are typically fine in areas where teeth numbers are quite high and the gears are not subjected to high torques.</p>
<p><strong>2. Trochoidal root fillet</strong></p>
<p>One of the most common processes for generating spur gears and helical gears is <a href="https://en.wikipedia.org/wiki/Hobbing" rel="noreferrer">hobbing</a>, and the root fillet that is naturally produced as a result of hobbing is the trochoidal curve. An example of a gear with trochoidal root fillets is shown below:</p>
<p><a href="https://i.stack.imgur.com/9H7cD.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9H7cD.png" alt="enter image description here"></a></p>
<p>The individual trochoidal curve and involute curve are displayed along with the root circle and outer circle as below:</p>
<p><a href="https://i.stack.imgur.com/ueewg.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ueewg.png" alt="enter image description here"></a></p>
<p>The first thing to notice is that the trochoidal fillet is considerably more rounded than the previous case, so less bending stress will develop at the root of the tooth and so the gear is more suitable in high-torque scenarios. Also, notice how the trochoidal curve intersects with the involute curve at a point slightly to the right of the start of the involute: some of the involute curve has been inadvertently lost by the hobbing process! The smaller the number of teeth, the greater this involute loss will be, and it becomes apparent as undercut interference starts to occur: in such a case, the gears are said to be <em>undercut</em>. While for radial-line fillet curves the undercut interference would have caused the gears to jam, gears with trochoidal fillets will not jam in such circumstances.</p>
<p>Whenever gear teeth are generated by hobbing, it is as though a rack gear is rolling along a gear blank (circle with diameter equal to outer diameter of gear tooth), material being subtracted wherever the rack gear has been. The below animation should illustrate how the rack gear can generate the shape of a spur gear with trochoidal fillets:</p>
<p><a href="https://i.stack.imgur.com/3wnsT.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/3wnsT.gif" alt="enter image description here"></a></p>
<p>Notice how the teeth on the rack gear are trapezia shapes (and also that the rack gear here, being a cutter, has an addendum of $1.25m$ and a dedendum of $m$, unlike usual rack gears), and the cutting of these trapezium teeth into a gear blank manages to magically create two curves, the involute and the trochoidal curves. More precisely, the involute curve is generated by either of the diagonal lines of the trapezium, and the trochoidal curve is generated by either of the two points on the end of the trapezium. With this information, it is possible to express the trochoidal fillet as a parametric equation.</p>
<p>The next section of this question will look at the mathematics behind deriving such a parametric equation.</p>
<p><strong>Deriving the trochoidal curve</strong></p>
<p>First of all, let's consider a single trapezium of the rack gear, and see how a point on the end of a trapezium moves relative to the gear being cut. The curve that this point follows will be the trochoidal curve. The pitch point of the trapezium, $P$, is the point on the trapezium that will initially lie on the gear's pitch circle, as shown below.</p>
<p><a href="https://i.stack.imgur.com/wUrQj.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/wUrQj.jpg" alt="enter image description here"></a></p>
<p>Note that the pitch circle radius is $R$, and that there is a general point $Q$ that lies on the trapezium, offset from $P$ by $x_0$ and $y_0$ in the $x$ and $y$ directions (later, we will set this point to be one of the points on the end of the trapezium). Also, note the solid black triangle: this is to help identify the orientation of the gear. The current position of $Q$ can be expressed as follows:</p>
<p>$$
\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} R+x_0 \\ y_0 \end{array} \right]
$$</p>
<p>Now, let's rotate the gear anticlockwise by an angle $t$, which then pushes the trapezium up by distance $Rt$ (no slip between pitch circle and pitch line). Illustration:</p>
<p><a href="https://i.stack.imgur.com/E9wP1.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/E9wP1.jpg" alt="enter image description here"></a></p>
<p>Note how the solid black triangle has moved in accordance with the gear rotation. The position of $Q$ is now:</p>
<p>$$
\left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} R+x_0 \\ Rt+y_0 \end{array} \right]
$$</p>
<p>The trapezium has moved, but so has the gear: we are interested in the motion of $Q$ <em>relative</em> to the gear. Therefore, we need to rotate the entire system about the origin by angle $t$ clockwise. This simply rotates the perspective so that it returns the gear to its default position while retaining the new movement of the trapezium, i.e. we get the motion of a trapezium as the rack rolls around the stationary gear, similar to the animation above.</p>
<p><a href="https://i.stack.imgur.com/uyWKN.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/uyWKN.jpg" alt="enter image description here"></a></p>
<p>To get the position of point $Q$, we need to multiply the previous position by a clockwise rotation matrix, so the position is expressed as follows:</p>
<p>$$
\left[ \begin{array}{c} x \\ y \end{array} \right] = \begin{bmatrix} \cos{t} & \sin{t} \\ -\sin{t} & \cos{t} \end{bmatrix}\left[ \begin{array}{c} R+x_0 \\ Rt+y_0 \end{array} \right]= \left[ \begin{array}{c} (R+x_0)\cos{t}+(Rt+y_0)\sin{t} \\ -(R+x_0)\sin{t}+(Rt+y_0)\cos{t} \end{array} \right]
$$</p>
<p>Finally, by setting $Q$ to be the "upper left" point on the trapezium, i.e. $x_0 = -1.25m$ and $y_0 = \frac{1}{4}\pi m - 1.25m\tan{\phi}$, where $\phi$ is pressure angle, and also noting that $R=\frac{1}{2}Nm$, where $N$ is number of gear teeth, you get the parametric equation for the trochoidal fillet curve:</p>
<p>$$
\left[ \begin{array}{c} x(t) \\ y(t) \end{array} \right] = \left[ \begin{array}{c} (\frac{1}{2}Nm-1.25m)\cos{t}+(\frac{1}{2}Nmt+\frac{1}{4}\pi m - 1.25m\tan{\phi})\sin{t} \\ -(\frac{1}{2}Nm-1.25m)\sin{t}+(\frac{1}{2}Nmt+\frac{1}{4}\pi m - 1.25m\tan{\phi})\cos{t} \end{array} \right]
$$</p>
<p>Note that the parameter $t$ will start at $t=-\frac{y_0}{R}=-\frac{\frac{1}{4}\pi - 1.25\tan{\phi}}{\frac{1}{2}N}$ and increases up until a particular value that represents the intersection of the trochoid with the involute.</p>
<p>Hope that answers your question :)</p>
| 13852 | Involute gear curve when root diameter falls below base diameter |
2017-02-20T20:33:13.083 | <p>How are these springed ball connectors called? Can not find them without right name.</p>
<p><a href="https://i.stack.imgur.com/OJmBV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OJmBV.jpg" alt="enter image description here"></a></p>
| |connections| | <p>Spring pin connector<br>
Spring loaded connector<br>
Spring-loaded connector<br>
Spring loaded pin connector<br>
Magnetic spring connector (obviously only if magnetic)</p>
<p><a href="https://www.google.com/search?q=spring%20pin%20connector&tbm=isch" rel="nofollow noreferrer">Google image search "spring pin connector"</a></p>
<p><a href="https://www.mill-max.com/new_products/detail/74" rel="nofollow noreferrer">Mill-Max spring-loaded connector</a></p>
| 13861 | Name or type of these connectors |
2017-02-20T22:49:07.947 | <p>After some research, I found that diesel is my choice of fuel.</p>
<p>Now I need a way to pump it to the engine. My main constraint is price.</p>
<p>So my question is can I use a water pump like one of <a href="https://www.amazon.ca/SODIAL-Priming-Diaphragm-Spray-Dispenser/dp/B00U8N07HU/ref=sr_1_8?ie=UTF8&qid=1487630446&sr=8-8&keywords=water%20pump%20sodial" rel="noreferrer">these</a>, to pump the diesel?</p>
<p><a href="https://i.stack.imgur.com/VXuZu.png" rel="noreferrer"><img src="https://i.stack.imgur.com/VXuZu.png" alt="enter image description here"></a></p>
| |pumps|diesel|rocketry| | <p>With pumping diesel, your main constraint should be safety, not cost. There are very good reasons diesel pumps aren't for sale for $5 on amazon. </p>
<p>The main things you want to look at are 1) whether the wetted materials are <a href="https://www.coleparmer.com/Chemical-Resistance" rel="nofollow noreferrer">compatible with diesel</a> (EDPM and silicone for example will dissolve) and 2) how it is grounded. Many pumps can build up static charge which can cause the diesel to explode in the pump. You'll also need to be careful to avoid cavitation which could easily lead to explosions also.</p>
<p>Edit: Explosion proof motors (and intermediate ratings like totally enclosed) are also worth investigating. If your plumbing leaks, is the vapor going to find its way to the motor and explode?</p>
<p>I would also recommend reading more about explosion-proof pumps and reading up more generally on what types of pumps are used for diesel.</p>
| 13865 | Can a water-pump pump diesel? |
2017-02-20T23:28:46.850 | <p>Most specs for light bulbs list luminous efficiency (or, more accurately sometimes, luminous efficacy) but not energy efficiency. Luminous efficacy tells you how bright the light is as observed by a human eye per unit of input electricity. Unfortunately, the human eye is terrible at measuring power and its very hard to convert between lumens/spectrum/human eye/radiant power.</p>
<p>I'm curious as to which types of lights have the highest energy efficiency, which I'm defining the amount of energy that would be absorbed by a perfect absorber surrounding the light bulb divided by the electric input. In this metric, the light is just a one form that the energy takes between electricity and its absorbed form (eg, heat). Obviously the exergy is degraded in this process, but I'm not worried about that here.</p>
<p>I would expect that incandescents are near 100% efficient by this metric since the only losses should be conduction/convection from the element through the near-vacuum and wires. <a href="https://www.dial.de/en/blog/article/efficiency-of-ledsthe-highest-luminous-efficacy-of-a-white-led/" rel="nofollow noreferrer">This weakly scientific analysis</a> suggests that LEDs are probably only 40% efficient by this metric because all their luminous inefficiencies are thermal losses in the cells. I think fluorescent lights might be extremely efficient by this metric but I can't really find any data.</p>
<p>I found the phrase "radiant efficiency" somewhat useful but still can't find much in manufacturers data sheets. Is there a better name for the metric I've described or a good database for what values might be?</p>
<p>In response to comments: I am interested in the power radiated away from the light bulb in the form of electromagnetic radiation divided by the electric power that goes in to the lightbulb. I don't care about the wavelength of the light (even though I understand that it changes both how much you can see and how much can be recaptured in another form later). I do care about thermal losses inside the bulb that don't radiate power across a distance.</p>
| |energy|lighting| | <p>By your metric, a electrically-heated black body radiator in vacuum would be very efficient.</p>
<p>Incandescent bulbs are close, but run at high temperatures to make (at least some) visible light. However, the filament is not in vacuum, so incandescent bulbs have significant heat conduction losses. A typical bulb gets too hot to touch after running for a few minutes. The filament is run so hot that if it were in a vacuum, it would evaporate too quickly and coat the inside of the bulb with the filament material.</p>
<p>A modified incandescent that runs a bit cooler with the filament in vacuum should work nicely. At dimly glowing red temperatures, the evaporation of the filament is greatly reduced, and you should get useful run times in vacuum. The spectrum would be shifted even lower than typical incandescent light bulbs, but your spec says nothing about wavelength restrictions.</p>
<p>The only losses would be a little heat conduction thru the filament connections to the outside, and transmission losses thru the bulb at the wavelengths actually emitted.</p>
<p>Another way to do this is to start with a standard incandescent light bulb and place it within another bulb that is evacuated. Now there can't be conduction losses from the incandescent bulb, except the little bit thru the feed wires. The problem with this setup is that the inner bulb will likely overheat at full power, since it was probably designed with some air cooling assumptions. You could compensate by running the inner bulb at lower than rated power. That will shift the light more toward longer wavelength, but probably doesn't need to be as far as in the first example.</p>
| 13866 | Light bulbs with highest energy efficiency |
2017-02-22T07:13:50.710 | <p>I am working at a construction site and came accross this issue. a borehole needs to be expanded in length inside a container-office. Working space is as little as 6m x 3m (19ft by 9.5ft). Height of the container is about 2m (6'6"). Heigt can somehow be altered up to 5 meters (16'5"). Borehole diameter is 150mm (6"). Current depth of the borehole is 30 m (98 ft). The depth should be increased to 42m (138 ft). See the attached sketch for details. </p>
<p><a href="https://i.stack.imgur.com/ddjJf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ddjJf.jpg" alt="enter image description here"></a></p>
<p>The borehole was bored before the containers were placed. It is used to measure water levels with piezometric sensors inside them. Water levels have dropped lower than expected and the borehole needs to be expanded in depth to reach lower groundwater deposits. Containers cannot be moved, and the location cannot be changed since this specific borehole is in a sensitive location, so please do not offer boring a new borehole outside the container. </p>
<p>What method or which equipment can be used to work in a small container space and can reach such depth. </p>
| |civil-engineering|geotechnical-engineering| | <p>I would recommend you consider a low profile diamond drill rig; something similar to the <a href="http://www.miningandconstruction.sandvik.com/sandvik/1181/Internet/CIS/S000924.nsf/Alldocs/Products*5CDrill*rigs*and*rock*drills*5CUnderground*exploration*drill*rigs*2ADE140/$File/DE140_080814_low.pdf" rel="nofollow noreferrer">Sandvik DE140</a> (I'm not advertising). </p>
<p>Such rigs would definitely be able to drill the length of hole you require. Your space restrictions are the complicating factor.</p>
<p>Given you space restrictions, I also recommend you consider the underground drill rig version, as opposed to the surface drill rig version, as equipment designed to operate underground tends to be smaller than equipment designed to operate on the surface.</p>
<p>You will most likely have to raise the ceiling of the container where the hole is to be extended.</p>
<p>The diagram of the rig on page 7 shows it be 4250 mm tall when drilling vertical holes. It is advisable you have about 1 metre clearance above the rig, when it is operating; the more clearance, the better. For such a rig you will need to raise the ceiling of the container to the 5 metres mentioned in your question.</p>
<p>Depending on the type of access and the dimensions of the access available into the container, where the hole is located, it may be possible to used a vehicle mounted rig similar to the one in the first picture on <a href="http://www.resemin.com/en/pdf/agxyzq-drillmeters.pdf" rel="nofollow noreferrer">this site</a>. However, re-positioning the drill directly over the hole every time the rig has to be moved to add more drill steels, as the hole gets longer, will be a problem. </p>
<p>For more expert advice, contact an underground mining drilling contractor who has low profile drilling rigs.</p>
| 13883 | Equipment or method for borehole |
2017-02-22T22:07:40.227 | <p>According to the <a href="http://voyager.jpl.nasa.gov/spacecraft/spacecraftlife.html" rel="nofollow noreferrer">NASA Website</a>, the Voyager probes will begin powering down electrical components soon with all electronic functions ceasing some time around 2025. </p>
<p>With no ability to transmit back to earth and thus of 0 value in the Interstellar Medium...</p>
<p>...could either one (or both) of the Voyagers be <strong>directed</strong> back to Earth at all before electronic shut down?</p>
<p>If not; do aerospace engineers have any idea where either (or both) of the probes may end up?</p>
| |aerospace-engineering| | <blockquote>
<p>...could either one (or both) of the Voyagers be directed back to Earth at all before electronic shut down?</p>
</blockquote>
<p>No. It would be impossible since these satellites do not carry any type of course changing propellant. According to Wikipedia the Voyager satellites carry 16 hydrazine thrusters that are only used to reorient the satellite. This satellite was made to be like a bullet. Once it left the rocket that got it off earth it was only affected by gravity.</p>
<blockquote>
<p>do aerospace engineers have any idea where either (or both) of the probes may end up?</p>
</blockquote>
<p>Yes we do know where the probes are going. The short answer is nowhere in particular. Space is much more empty than many people can even fathom. The long answer is that in 300 years it will reach the Oort cloud which itself is theoretical and after that 40,000 years till it passes relatively close to a star. A mere 1.6 light years will separate them!</p>
<p>These probes were not made to return to earth. They were designed to capture and return data to earth as they traveled out of our solar system.</p>
| 13896 | Can Voyager 1 or 2, theoretically, return to earth if given instructions before their electronic instruments shut down in 2025? |
2017-02-23T09:07:35.813 | <p>I made a mathematical model (transfer function) of a rlc circuit. I simulated it in matlab simulink with step input and square wave input and observed the result. I can also calculate this system with step input since its transfer function is simple 1/s. But i do not know how to calculated square wave transfer function. Is there a way to find square wave transfer function or just i have to break the signal in to high or low form.
THanks </p>
| |control-theory|mathematics|transfer-function|systems-engineering|circuit-design| | <p>I thought I would expand a little on the answer offered by Karlo.</p>
<p><strong>Long story short, I would not try to calculate the analytical time response of a system to a square wave. That would be a serious pure-maths exercise, and not necessary for most engineering applications.</strong></p>
<p>Instead, I would suggest using the step response analytical vs. simulated to validate your model, and then trust that your MATLAB Simulink model is working properly for the square wave case.</p>
<p>But if you feel like trying it out, here is what you need to know:</p>
<p>The transfer function $G(s)$ is a description of the relationship between the input $U(s)$ and output $Y(s)$ of your system.</p>
<p>$$G(s) = \frac{Y(s)}{U(s)}$$</p>
<p>When you want to determine the time response of your system to a specified input analyticalally, you rearrange for $Y(s)$ and then use the inverse Laplace Transform:</p>
<p>$$Y(s) = G(s)U(s)$$</p>
<p>$$y(t) = \mathcal{L}^{-1}\{G(s)U(s)\}$$</p>
<p>As stated in the Question, for a simple step input $U(s) = \frac{1}{s}$ the time response is then:</p>
<p>$$y(t) = \mathcal{L}^{-1}\{\frac{G(s)}{s}\}$$</p>
<p>It gets a lot more difficult for a square wave. A square wave is a series of time-shifted step functions (or Heaviside functions) $H(t-T)$ where $T$ is the time at which the step occurs. The derivation for the Laplace transform of a square wave is given in the answer to <a href="https://math.stackexchange.com/q/1782998" title="this question">this question</a> by <a href="https://math.stackexchange.com/users/103399/alexjo">alexjo</a>:</p>
<blockquote>
<p>$$u(t) = A\sum_{k=0}^{\infty} \left[H\left(t-kT\right)-2H\left(t-\frac{2k+1}{2}T\right)+H\left(t-(k+1)T\right)\right]$$
$$U(s) =\frac{A}{s}\frac{1-\mathrm e^{-\frac{Ts}{2}}}{1+\mathrm e^{-\frac{Ts}{2}}}$$</p>
</blockquote>
<p>Where $A$ is the amplitude of the square wave. You would then multiply this expression by your transfer function and then take the inverse Laplace transform of the entire mess. I am not aware of any simple way of doing this. Good luck.</p>
| 13901 | Square wave transfer function? |
2017-02-23T20:44:36.997 | <p>What is the best way to design a circuit based on the following transfer function: (0.0364*s)/(0.0002*s + 1) ? I am trying to design the derivative stage of a PIDF controller using op-amps</p>
<p>EDIT: This is part of a bigger problem where i have to design a PIDF controller to achieve the minimum possible settling time for the plant function: $$\ G(s)= \frac{100*20*100}{(s+10)(s+20)(s+100)}$$</p>
| |pid-control|matlab| | <p>To my knowledge the simplest way to accomplish the transfer function...</p>
<p>$$ G(s) = \frac{E_o(s)}{E_i(s)} = \frac{0.0364s}{0.0002s + 1} = \frac{KTs}{Ts +1} $$</p>
<p>...is the following high-pass filter circuit:
<a href="https://i.stack.imgur.com/Xc24a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xc24a.png" alt="High pass filter with opamp gain"></a></p>
<p>The transfer function of this circuit is:</p>
<p>$$ \frac{E_o(s)}{E_i(s)} = \frac{R_2+R_3}{R_3}\frac{R_1 C s}{R_1 C s + 1} $$</p>
<p>So let $R_1C = T = 0.0002$ and $\frac{R_2+R_3}{R_3} = K = 0.0364/0.0002$. Then solve for $R_1$, $R_2$, $R_3$ and $C$. You'll have some wiggle room as to what values you pick, because you have two equations and four parameters. There are multiple solutions and you will have to decide what combination of values work best for your application.</p>
| 13909 | Design op-amp circuit from transfer function? |
2017-02-24T08:06:26.997 | <p>In Ansys workbench , structural module, in modelling, what is the exact difference between pressure and surface force application? Is surface force applied in same manner as that of pressure application or different?
In more detail, I want to know how the total force is applied on the surface equally?</p>
| |ansys|ansys-workbench| | <p>As you point out, the underlying elements on the surface will take the applied surface load and the FE element formulation allocates, consistently, to the nodes as nodal forces. Hope this explanation helps.</p>
<p>CR</p>
| 13922 | ansys workbech structural- difference between pressure and surface force application |
2017-02-26T00:16:07.463 | <p>I have a small flat surface (~80mm x 80mm) that needs to adhere to a round surface (radius = 100mm). Does anyone know how to mate the two surfaces together using tape to give a good enough bond?</p>
<p>Any thought appreciated.</p>
<p>EDIT: The part with the flat surface (~250g) is supposed to be hung on the round part (a lamp shade). Both parts are plastic. I'm hoping to avoid causing damage to either part. The whole assembly will be kept indoor.</p>
| |adhesive| | <p>Without knowing more about your requirements here are some options:</p>
<p>"Permanent" options:
Try a two-part epoxy designated for plastics. Using sandpaper, like 120 grit, to roughen the surfaces helps with adhesion. </p>
<p>Alternatively, maybe a double-sided foam tape would work. Specifically VHB tape. </p>
<p>If you're using acrylic, you could use the special acrylic glue solvent which makes a strong bond.</p>
<p>"Non-permanent" options:
zip ties/cable ties or a hose clamp to hold the parts together</p>
| 13939 | Mating a round surface with a flat surface |
2017-02-26T01:27:21.390 | <p>Arch bridges operate according to the following (I am assuming arch dams are the same):</p>
<blockquote>
<p>Tensional force in arch bridges are virtually negligible. The natural curve of the arch and its ability to dissipate the force outward greatly reduces the effects of tension on the underside of the arch.</p>
<p>The greater the degree of curvature (the larger the semicircle of the arch), the greater the effects of tension on the underside of the bridge.
(<a href="http://science.howstuffworks.com/engineering/civil/bridge5.htm" rel="nofollow noreferrer">Source</a>)</p>
</blockquote>
<p><img src="https://i.stack.imgur.com/7cuBs.png" alt="enter image description here" /></p>
<ol>
<li><strong>WHY does a greater degree of curvature results in more tension?</strong> A diagram would be useful if possible.</li>
</ol>
<p><a href="https://i.stack.imgur.com/wvRJN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wvRJN.png" alt="enter image description here" /></a></p>
<p><strong>2. How does the changing the radius of curvature affect the tension and compression in the dam, and ultimately the strength of it.</strong>
<a href="https://i.stack.imgur.com/GhRjD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GhRjD.jpg" alt="enter image description here" /></a></p>
| |civil-engineering|bridges|hydroelectric-dams| | <p>The presence of tension in an arch is not really dependent on the curvature but on how well the arch matches a <a href="https://en.wikipedia.org/wiki/Catenary" rel="noreferrer">catenary shape</a>.</p>
<p>Robert Hooke famously stated:</p>
<blockquote>
<p>As hangs a flexible cable so, inverted, stand the touching pieces of an arch.</p>
</blockquote>
<p>So a catenary arch will have only compression stresses (since a flexible cable can only have tension stresses). </p>
<p><a href="https://i.stack.imgur.com/GKhAq.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/GKhAq.jpg" alt="Catenary Arch"></a></p>
<p>Notice how the line of action of the compression in the arch runs through the centre of the arch. If the arch shape is not a catenary, the line of action will still follow a catenary shape but will not follow the centre line of the arch:</p>
<p><a href="https://i.stack.imgur.com/CLcKTm.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/CLcKTm.jpg" alt="Non-catenary"></a></p>
<p>We can see how this can cause tension by examining the distribution of stress as a result of the position of the line of action (green is tension):</p>
<p><a href="https://i.stack.imgur.com/aZWXF.png" rel="noreferrer"><img src="https://i.stack.imgur.com/aZWXF.png" alt="Stress Distribution"></a></p>
<p>In general the rule for arches is the cantenary line must be within the middle $\frac{1}{3}$ of the arch to avoid tension.</p>
<p>Form finding can be used to find the catenary shape for a particular loading. This can be done computationally, or even by the use of <a href="http://dataphys.org/list/gaudis-hanging-chain-models/" rel="noreferrer">hanging models</a>.</p>
<p>Arch dams behave in the same way. However, since the main loading on the dam (from water) is a pressure normal to the dam surface the equivalent of the catenary shape is a circle. In this case the compressive stress is calculated in the same way as the <a href="https://en.wikipedia.org/wiki/Cylinder_stress#Hoop_stress" rel="noreferrer">hoop stress</a> of a pressure vessel (where I have assumed r >> t):</p>
<p>$$\sigma = \frac{Pr}{t}$$</p>
<p>where $\sigma$ is the stress P is the applied pressure, r is the radius and t is the thickness. Since the pressure P increases with the depth of water dams sometimes decrease radius with depth (variable-radius dams) or change the thickness (constant-radius dams).</p>
<p>Naturally the design of a dam has many other complicating factors which affect the design.</p>
| 13941 | How and WHY does the tension in an arch dam or bridge change as the degree of curvature and radius of curvature change? |
2017-02-26T03:05:44.277 | <p>Could someone tell me if there's a general name for the kind of spring clips that slide onto a rod and set a position for some other attached element. I've seen them on sinks and toilet tanks -- here are pictures:</p>
<p><a href="https://i.stack.imgur.com/vvz3K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vvz3K.jpg" alt="toilet float level setter"></a></p>
<p><a href="https://i.stack.imgur.com/6Burj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Burj.jpg" alt="Sink stopper lever position adjustment"></a></p>
| |mechanical-engineering| | <p>After doing some searching it seems that style of clip is limited to toilet hardware like you have shown.</p>
<p>A google reverse image search found the <a href="http://www.prettyhandygirl.com/repairing-pop-up-sink-drain/" rel="nofollow noreferrer">originating site</a> of your first image. They call it a "joint clamp clip" but a <a href="https://en.wikipedia.org/wiki/Joint_clip" rel="nofollow noreferrer">join clamp clip on wiki</a> says "Joint clips are used to prevent a joint from separating during a reaction process" so that name does not fit.</p>
<p>Many niche industries and unique devices do not have names even if the device is very old. I'm sure various plumbers refer to this clip in many different ways. From my experience in construction, the names are often unintelligent, named after a specific brand of that product, racist, sexist, or overlap with another industry.</p>
<p>When I am creating engineering documentation and I see a new device that does not have an obvious, appropriate, or searchable name; I will often create a name for it. If the name is logical and valuable, searchable documentation is spread with that name; it may very well become the industry standard name for it.</p>
<p>Here potential names for it. If you like one of them set it as the alt text for your image; and you may have just named it ;-)</p>
<ol>
<li>rod retention clasp</li>
<li>adjustable rod retention clasp </li>
<li>rod retention clip</li>
<li>adjustable rod retention clip</li>
<li>float retention clip</li>
<li>adjustable float retention clip</li>
<li>float retention clasp</li>
<li>adjustable float retention clasp</li>
</ol>
| 13943 | What's the name of position-setting spring clips used in plumbing? |
2017-02-26T13:24:25.177 | <p>If one searches for the different levels of CO2 emissions in cars, one can find two very different levels. For instance:</p>
<p>1: <a href="https://ec.europa.eu/clima/policies/transport/vehicles/cars_en" rel="nofollow noreferrer">Here</a> it is written that:</p>
<blockquote>
<p>The law requires that the new cars registered in the EU do not emit more than an average of <strong>130 grams of CO2 per kilometer</strong> (g CO2/km) by 2015.</p>
</blockquote>
<p>So it can be seen that the mass of CO2 is described as being over 100 grams/km. </p>
<p>2: <a href="https://www.dieselnet.com/standards/eu/ld.php" rel="nofollow noreferrer">Here</a>, one can see that the level of <strong>CO is around 1 g/km</strong>. So way less than 100 g/km, which is a huge difference. </p>
<hr>
<p>Now, in the second reference, the table shows that those are the values for CO and not CO2, as in the first case. But in the second one, I can't find the CO2 levels and I assumed that the CO is a typo, and it's missing the "2". However, something tells me that it is not a typo and that they read the levels of CO and not CO2. Now, I have two questions. </p>
<p>Is there a typo in the second reference? </p>
<p>If Yes, <strong>Q1:</strong> Why is the difference so huge?</p>
<p>If No, <strong>Q2:</strong> Why do they give the values in CO and not in CO$_2$?</p>
| |manufacturing-engineering|combustion|emissions| | <p>Q1.
The CO is Carbon Monoxide Level, and CO2 is the Carbon Dioxide Level.</p>
<p>The difference is huge because the potent levels for both of them are very different. Generally for clean fresh air there will be between 0-0.5ppm CO, and between 250-350ppm CO2.</p>
<p>Exposing ourselves to around 800ppm of CO will kill us in 2-3 hours, where as we would need to be exposed to over 40,000ppm of CO2 for it to become fatal for an average person dur to oxygen deprivation.</p>
<p>Q2.
The reason for some datasets using CO while other use CO2 is usually down to standards within the Governments, Councils, Companies parameters. Certain datasets require CO levels where as others use CO2 levels. Theis could also be due to the questions that are being asked.</p>
| 13951 | What is the real CO2 emission for cars? 1g/km or >100g/km? |
2017-02-26T13:30:14.400 | <p>This question is similar to a question I am stuck on, but I have changed it to get an understanding how to works.</p>
<p>A spring is being made by pulling a 12cm long cylinder of material in tension with a desired stiffness of 6200kN/m. </p>
<p>It has a density of 2.9 g/cm^3, an ultimate tensile strength of 375 MPa and a Young's modulus of 70 GPa.</p>
<p>What should the diameter (in mm) be for the metal "spring"?</p>
<p>So, my question is how can I go about solving this question?</p>
<p>I have looked at the Young's Modulus and Hooke Law's formulas. But keep getting stuck with material displacement. Do I need to know the amount of length the object changes to work out this solution, or am I on the wrong track?</p>
<p>Can I assume: $\Delta L = 1$ if I used the following formula to find the $A_0$, which I can then calculate the diameter from:
$$
F = \frac{E A_0 \Delta L}{L_0}
$$</p>
<p>Any help will be most appreciated.</p>
| |springs| | <p>The axial stiffness of an isotropic material with a uniform cross-section is a fundamental part of most engineering mechanics of materials concepts. We can derive it as follows:</p>
<p>The force-displacement relation of a spring is described by Hooke's Law,</p>
<p>$$F = k \Delta L$$</p>
<p>where $F$ is the force exerted on the spring, $\Delta L$ is the change in length or displacement, and $k$ is the stiffness or spring constant.</p>
<p>We can rearrange to get an expression of the stiffness</p>
<p>$$k = \frac{F}{\Delta L}$$</p>
<p>Now we want to express $k$ exclusively in terms of the geometry and material properties. For that, we need to make use of the engineering stress $\sigma = F/A_0$ where $A_0$ is the cross-sectional area of the material before deformation. We can rearrange and substitute this back into the previous equation</p>
<p>$$k = \frac{\sigma A_0}{\Delta L}$$</p>
<p>We still need to get rid of $\sigma$, and to do that we can use $\sigma = E \epsilon$ where $E$ is the Young's modulus of the material and $\epsilon$ is the engineering strain. Substituting that and using the definition of engineering strain $\epsilon = \Delta L/L_0$ where $L_0$ is the length of the material before deformation and we get</p>
<p>$$ k = \frac{E(\Delta L/L_0)A_0}{\Delta L}$$</p>
<p>$$ k = \frac{E A_0}{L_0}$$</p>
<p>Note that $\Delta L$ cancels, so we don't need to know the change in length.</p>
<p>Now if you want to figure out the necessary dimensions for a block of material to have a given stiffness, just substitute the appropriate equation for $A_0$ and rearrange. For example, a cylinder has cross-sectional area $A_0 = \frac{\pi}{4} D_0^2$, sub that in and you get</p>
<p>$$ D_0 = \sqrt{\frac{4 k L_0}{\pi E}}$$</p>
| 13952 | Finding the diameter of a material using tension stiffness and Young's Modulus |
2017-02-26T18:36:51.930 | <p>I've been working on simulating drone flight in Gazebo (an open source robotics simulator).</p>
<p>Currently, I'm trying to control the drone's pitch / roll rotation using PIDs. It's important to remember that a drone moves linearly only by angular rotation (forward / backward requires pitch rotation, left / right requires roll).</p>
<p>Right now, I'm testing PID pitch control. The goal is to get the drone to rotate to a specified angle and hold that position - perfect job for a PID.</p>
<p>But the physics implementation complicates that. Essentially, the PID takes the setpoint - the desired pitch rotation angle (in radians), and the current pitch rotation angle (which I get directly form the simulation state).</p>
<p>The PID output is then used to calculate the resultant torque. The torque equation is t = I * a (torque = Intertia * angular acceleration). So, essentially, I'm using a PID to control the angular acceleration of the drone given the current and desired angles of rotation.</p>
<p>The problem, of course, is that a PID output scales linearly with time, so although it decreases to zero as the drone approaches the setpoint, it has been applying positive accelerations through the entire time. The result is that the drone has a lot of angular momentum, and so it naturally overshoots. Then, of course, the PID responds by reversing the rotation, but with the same problem.</p>
<p>In the end, what happens is that the drone starts with a horizontal attitude, rotates to the setpoint, then rotates back to horizontal, then back to the setpoint. If I add any I or D gain it becomes unstable.</p>
<p>The problem is really simple. It's like controlling a vehicle in space. Thrust is applied in pulses in order to give fine control over position / rotation. The PID however, provides a continuous output. Which means I need some sort of physical system that behaves as depicted here:</p>
<p><a href="http://vignette4.wikia.nocookie.net...imation.gif/revision/latest?cb=20141124024926" rel="nofollow noreferrer">http://vignette4.wikia.nocookie.net...imation.gif/revision/latest?cb=20141124024926</a></p>
<p>or I need to rewrite the physics so that the PID output scales appropriately to bring angular velocity to zero by the time it reaches the desired angle of rotation.</p>
<p>Solving this is proving more difficult than I thought - probably because I'm not a strong physics or controls guy.</p>
<p>I've posted this problem already in another forum. You can read the thread here:</p>
<p><a href="https://www.physicsforums.com/threads/pid-control-for-drone-rotations.905431/#post-5701849" rel="nofollow noreferrer">PID control for drone rotations</a> </p>
<p>where I get really specific. I'm afraid I wasn't able to articulate the problem well enough, partly because the guy who tried to help wasn't really familiar with PIDs (or C++).</p>
<p>Anyway, Any thoughts from someone who has dealt with this sort of thing (or thinks they know more about it than I do...) would be really welcome. :)</p>
<p>Edit: You can see the behavior in the following animated gif:</p>
<p><a href="http://gph.is/2lKf9aT" rel="nofollow noreferrer">http://gph.is/2lKf9aT</a></p>
| |control-engineering|aerospace-engineering|pid-control| | <p>So, I'm going to base my answer off a simulation I just finished in a flight dynamics class I'm taking. The book is called Small Unmanned Aircraft Theory and Practice by Beard and McLain. There are slides available at <a href="http://uavbook.byu.edu/doku.php?id=lecture" rel="nofollow noreferrer">http://uavbook.byu.edu/doku.php?id=lecture</a> (check out chapter 6). While we are simulating fixed-wing UAVs, I think the concepts should be similar. For our pitch and roll loops (which are kind of the same thing in a quadrotor), we use a basic PD loop (no integral term because each is an inner loop). The equation for a commanded control input is $y=k_pE - k_d\dot{E}$. For convenience, we used $q$ (pitch rate) instead of $\dot{E}$ because we have measurements of it. </p>
<p>In your PID loop, you add $k_d\dot{E}$ instead of subtracting it. By subtracting, it reduces the input as you increase angular velocity. For instance, if I am $\frac{\pi}{2}$ radians off and not rotating, my input will be large. As I increase in angular velocity, my input decreases from both the decrease in error and the increase in angular velocity. Eventually, there must be 0 input and it will happen at some point before I get to my position. The UAV keeps rotating and error decreases. However, $\dot{E}$ remains at its previous level and the input becomes negative, decreasing my angular velocity. If, somehow, the angular velocity, got to 0 before achieving the setpoint, this whole process would be back at the beginning. This means you should only have to tune your gains. </p>
| 13960 | Using a PID to control rotation |
2017-02-27T08:46:03.323 | <p>Apparently, during a world fair (presumably Paris in 1889<sup>[Don't trust that fact]</sup>), an engineer showed a perpetual machine that could not be stopped. For months, he challenged visitors to stop it, but that machine always recovered the energy that people tried to take from it. Everyone was convinced it worked!</p>
<p>The "trick" was so simple: the action of trying to stop it would give the machine extra energy.</p>
<p>Now, I could have heard that from a highly unreliable source, but whether the story is true or not doesn't change my question:</p>
<p>What would that machine look like and how exactly would it work?</p>
| |mechanical-engineering| | <p>There are many possible approaches - let's take a simple one; electrostatic. A static charge based <a href="http://physics.kenyon.edu/EarlyApparatus/Static_Electricity/Electric_Whirl/Electric_Whirl.html" rel="nofollow noreferrer">whirl</a> connected to a capacitor for storing more charge, and a fluffy carpet around. Stopping the whirl with your fingers charges it (and the capacitor) with new static charge picked through your feet from the carpet.</p>
| 13970 | How does this obviously fake but convincing perpetual machine work? |
2017-02-27T09:42:47.523 | <p>Note - This is a duplicate of the question I asked over on <a href="https://physics.stackexchange.com/questions/314474/two-dimensional-plasticity-with-radial-return">Physics</a> which hasn't received any attention.</p>
<p>What is the proper method to scale deviatoric stresses back to the yield surface in plane stress 2D?</p>
<p>I cannot get the <a href="http://www.dynasupport.com/tutorial/computational-plasticity/radial-return" rel="nofollow noreferrer">radial return</a> method to produce the desired results under plain stress conditions in 2D. I'll show this using an example in both 3D and 2D in the hope that my mistake might become apparent.</p>
<p><strong>Three Dimensions</strong></p>
<p>Assume an arbitrary trial <a href="https://en.wikipedia.org/wiki/Cauchy_stress_tensor" rel="nofollow noreferrer">Cauchy Stress tensor</a> of:</p>
<p>$\sigma_{tr} = \begin{bmatrix}
90 & 20 & 0 \\
20 & 90& 0 \\
0 & 0& 0
\end{bmatrix}$</p>
<p>Then the <a href="http://www.continuummechanics.org/hydrodeviatoricstress.html" rel="nofollow noreferrer">hydrostatic component</a> is given by:
$\sigma_{tr,hyd} = \begin{bmatrix}
50 & 0 & 0 \\
0 & 50& 0 \\
0 & 0& 50
\end{bmatrix}$</p>
<p>And the <a href="https://en.wikipedia.org/wiki/Cauchy_stress_tensor#Stress_deviator_tensor" rel="nofollow noreferrer">deviatoric component</a> by:
$\sigma_{tr,dev} = \begin{bmatrix}
40 & 20 & 0 \\
20 & 10& 0 \\
0 & 0& -50
\end{bmatrix}$</p>
<p>We will use the <a href="https://en.wikipedia.org/wiki/Von_Mises_yield_criterion" rel="nofollow noreferrer">Von Mises criterion</a> to determine whether the stresses are still in the yield surface. This calculated as (for general plane stress): </p>
<p>$\sigma_v = \sqrt{\sigma_{11}^2-\sigma_{11}\sigma_{22}+\sigma_{22}^2+3\sigma_{12}^2} = 86.6025$</p>
<p>For the sake of argument let's assume a yield stress of $\sigma_y=80$. In which case the deviatoric stresses need to be scaled back by a factor of $\alpha={\sigma_y}/{\sigma_v}=0.9238$. So the new deviatoric stress is given by:</p>
<p>$\sigma_{dev}=\alpha*\sigma_{tr,dev}
= \begin{bmatrix}
36.95 & 18.48 & 0 \\
18.48 & 9.24& 0 \\
0 & 0& -46.19
\end{bmatrix}$</p>
<p>And the scaled back </p>
<p>$\sigma=\sigma_{tr,hyd}+\sigma_{dev}
= \begin{bmatrix}
86.95& 18.48 & 0 \\
18.48 & 59.24& 0 \\
0 & 0& 3.81
\end{bmatrix}$</p>
<p>If we calculate the corresponding Von Mises stress we get $\sigma_{v}=80$ indicating that the scaling was correct. </p>
<p><strong>Two Dimensions</strong></p>
<p>I haven't repeated a lot of the text here for brevity. Check page 157 <a href="https://www.wou.edu/las/physci/taylor/es406_structure/TM_chap8_stress.pdf" rel="nofollow noreferrer">here</a> for confirmation of hydrostatic stress in 2D, it's just $(\sigma_{11}+\sigma_{22})/2$. Assuming plane stress then:</p>
<p>$\sigma_{tr} = \begin{bmatrix}
90 & 20 \\
20 & 90 \\
\end{bmatrix}$
$\sigma_{tr,hyd} = \begin{bmatrix}
75 & 0 \\
0 & 75 \\
\end{bmatrix}$
$\sigma_{tr,dev} = \begin{bmatrix}
15 & 20 \\
20 & -15 \\
\end{bmatrix}$
$\sigma_v = 86.6025$</p>
<p>The problem is that scaling the deviatoric stresses ($\alpha$ is the same as before) results in:</p>
<p>$\sigma_{dev}=\alpha*\sigma_{tr,dev}
= \begin{bmatrix}
13.86 & 18.48 \\
18.48 & -13.86 \\
\end{bmatrix}$</p>
<p>And then:</p>
<p>$\sigma=\sigma_{tr,hyd}+\sigma_{dev}
= \begin{bmatrix}
88.86& 18.48 \\
18.48 & 61.14 \\
\end{bmatrix}$</p>
<p>Calculating the Von Mises stress gives a value of $\sigma_{v}=85$, which is outside the yield surface. What is the problem here? What is the proper method to scale the deviatoric stresses back to the yield surface in 2D?</p>
<p><strong>Update</strong>
I am slowly coming around to the ideal that this radial return method is only appropriate for full 3D or plane strain (in which case the $\sigma_{33}\neq0$). Page 25 of <a href="http://publications.jrc.ec.europa.eu/repository/bitstream/JRC102336/vpjcfin.pdf" rel="nofollow noreferrer">this</a> seems to also indicate that this is the case, but I am still unsure. </p>
| |modeling|stresses| | <p>Let's look at this problem mathematically a la Simo.</p>
<p>Let stress tensor ($\boldsymbol{\sigma}$) and its deviator ($\boldsymbol{s}$) in plane stress space be
$$
\boldsymbol{\sigma} = \left[\sigma_{11}\, \sigma_{22}\, \sigma_{12}\right]^T
\quad \text{and} \quad
\boldsymbol{s} = \left[s_{11}\, s_{22}\, s_{12}\right]^T
$$
Note that $\sigma_{33} = \sigma_{13} = \sigma_{23} = s_{23} = s_{13} = 0$. Even though $s_{33} \ne 0$, we can ignore $s_{33}$ because $s_{11} + s_{22} + s_{33} = 0$.</p>
<p>Let $\bar{\mathsf{T}}$ be the matrix that takes $\boldsymbol{\sigma}$ to $\boldsymbol{s}$, i.e., $\boldsymbol{s} = \bar{\mathsf{T}}\,\boldsymbol{\sigma}$.
Also recall the convention for writing the plane stress strains ($\boldsymbol{\varepsilon}$) and their deviators ($\boldsymbol{e}$):
$$
\boldsymbol{\varepsilon} = \left[\varepsilon_{11}\,\varepsilon_{22}\, 2\varepsilon_{12}\right]^T \quad \text{and} \quad
\boldsymbol{e} = \left[e_{11}\, e_{22}\, 2\varepsilon_{12}\right]^T
$$
You can show, after some algebra and making sure that the convention of the strain tensors is respected, that if we use a slightly modified version of $\bar{\mathsf{T}}$ (let's call it $\mathsf{T}$):
$$
\mathsf{T} = \frac{1}{3}\begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 6 \end{bmatrix}
$$
we can get relatively simple expressions for the yield condition and flow rule for plane stress.</p>
<p>The yield function for von Mises plasticity in 3D is (in tensor notation)
$$
f = \sqrt{\frac{3\, \boldsymbol{s}:\boldsymbol{s}}{2}} - \sigma_Y
$$
In plane stress, and using the matrices we have defined earlier (and a bit of algebra), you can show that the yield function can be written as
$$
f = \sqrt{\frac{3\, \boldsymbol{\sigma}^T\,\mathsf{T}\,\boldsymbol{\sigma}}{2}} - \sigma_Y
$$
You will get yielding at your trial stress values.</p>
<p>You will have to compute the return using the plane stress flow rule:
$$
\dot{\boldsymbol{\varepsilon}}^p = \dot{\lambda}\,\mathsf{T}\,\boldsymbol{\sigma}
\quad \text{where} \quad
\boldsymbol{\varepsilon}^p = \left[\varepsilon_{11}^p\,\varepsilon_{22}^p\, 2\varepsilon_{12}^p\right]^T
$$
These modifications are needed for you to reach the right point on the yield surface in your return process.</p>
<p><em>Update</em> </p>
<p>If the linear elastic relation is
$$
\boldsymbol{\sigma} = \mathsf{C}\,\boldsymbol{\varepsilon}^e
$$
where $\mathsf{C}$ is the plane-stress $3\times 3$ stiffness matrix, we can use the flow rule to show that
$$
\boldsymbol{\sigma_{n+1}} = \left[\mathsf{C}^{-1} + \Delta\lambda\mathsf{T}\right]^{-1} \mathsf{C}^{-1} \boldsymbol{\sigma_{n+1}}^{\text{trial}}
$$
We solve for $\Delta\lambda = \dot{\lambda} \Delta t$ by using the consistency condition $f(\sigma_{n+1}) = 0$. The resulting equation is a quadratic in $\Delta\lambda$ and therefore a simple scaling relation cannot be used for the plane stress radial return.</p>
| 13971 | Two dimensional plasticity with radial return |
2017-02-27T10:23:38.380 | <p>I'm being annoyed by a pulsating hum which comes from my neighbor's HVAC system. I read up a little on the design of HVAC systems and I came to a tentative conclusion that what I'm hearing is a compressor being driven by an AC induction motor. A frequency spectrum shows a spike at 120 Hz and another at 117 Hz; which of course produces a 3 Hz beat. The 117 Hz spike actually moves around between say 117 and 118 Hz. I guess what I'm hearing is called a "slip frequency beat"? I'm in the US, where the mains frequency is 60 Hz; so 120 Hz is the audible hum produced from magnetic fields in motors, transformers, and so on.</p>
<p>According to <a href="https://www.bksv.com/doc/BO0269.pdf" rel="noreferrer">this document</a> (p. 3), the "slip frequency", defined as "per-unit slip times line frequency" is the "actual frequency of current in the rotor conductors". Per-unit slip is proportional to torque and is commonly 0.03 - 0.05. This means 3 Hz could be a typical slip frequency, and I guess 117 Hz could be a slip-frequency sideband of 120 Hz. But there is no 123 Hz spike, and in fact the rest of the spectrum is pretty quiet except for spikes at 59 Hz and 29 Hz. There are harmonics at 234 Hz and 240 Hz, but the harmonic at 352 Hz is stronger than that at 234 Hz. All the harmonics have much less energy than the 117 Hz and 120 Hz fundamentals (say >20 times less). The 29 Hz and 59 Hz spikes don't seem to vary much when the compressor turns on and off, and neither does the 120 Hz spike. Only the 117 Hz spike and its harmonics show an "on-off" square wave in a time-series plot, corresponding to compressor activation at 10-minute intervals.</p>
<p>I can't provide decibel-calibrated sound pressure levels (and since I don't have access to the equipment, I can't measure the actual vibration intensity). I would say the sound is not loud enough to be annoying unless you have to live next to it.</p>
<p>I have calculated that the sound amplitude in the 117 Hz "spike" is 3 times <em>higher</em> than that in the 120 Hz spike. I don't know if this reveals anything significant.</p>
<p>I've read that the speed of an AC motor is 120 Hz / (number of poles), and that the minimum number of poles is 2, so I'm concluding that the motor cannot be turning at just under 120 Hz (7200 RPM) but rather must be going at some slower speed like 60 Hz (3600 RPM). This seems to imply that the 117 Hz vibration is not mechanical vibration from the rotor - but I'm not sure what else it could be. The linked document mentions certain faults producing vibrations at 2 x RPM, but it seems these should be concurrent with vibrations at 1 x RPM. But, as I said, the 59 Hz spike doesn't vary with compressor activation.</p>
<p>Here's a time-series plot of various spectral amplitudes, taken over 6 hours at night. The spectral bands shown here are 58-59 Hz, 116-118 Hz, 174-177 Hz, etc. You can see that compressor activation doesn't seem to affect the 59 Hz spike, but there is a clear signal at 117 Hz and 3x117 = 351 Hz. (I also appended a spectrum plot)</p>
<p><a href="https://i.stack.imgur.com/tYhpV.png" rel="noreferrer"><img src="https://i.stack.imgur.com/tYhpV.png" alt="Harmonics of 59 Hz, 6-hour time-series, plus spectrum"></a></p>
<p>According to the linked document, some amount of vibration is normal and the surest way to diagnose a developing fault is to take spectral readings over time and see if anything is getting worse. The neighbor's HVAC system is 20 years old, but I only have readings for the past six months, and there seems to be a lot of fluctuation just from weather conditions and diurnal temperature variations influencing the propagation of sound across the property line. Because of this I can't say that there has been an obvious trend upwards in sound amplitude over this time period.</p>
| |vibration|hvac|compressors|ac|frequency-response| | <p>A 3Hz pulsating hum-- noise that is getting louder and softer at 3Hz. Yes, I agree that would be quite annoying.</p>
<p>Since this is a residential unit in the US, both the compressor and the fan are run with single phase asynchronous motors. Almost certainly, they are four pole with an 1800 RPM synchronous speed, running about 1750 RPM or so, depending upon load. The peak you are seeing at 116.8 Hz is consistent with this since that corresponds to the second harmonic of a 1752 RPM rotation rate. The big peak you see at 120Hz is interesting. I wonder if this is due to a loose winding in one of the motors.</p>
<p>The compressor is inside the A/C unit, in this case, on top of the house. The fan has to be exposed to the outside. My experience is the fan is what you hear most of the time.</p>
<p>From the graphs, it looks like everything is working correctly. The compressor or fan is cycling in and out every 20 minutes which is to be expected. The duty cycle is less than 50% which means it is properly cooling so I guess the coolant levels are adequate. The sounds are probably caused by a combination of loose sheet metal and maybe a loose winding. If it were my unit, I would put my hand here and there to try to guess where the noises are coming from and make sure everything I can get to was bolted down securely. Recognizing that it is not your unit and you don’t have access to it, much less permission to go monkeying around, this won’t do you much good. If it is not running in heat pump mode and it is running at close to 50% duty cycle in February, (when this question was originally posted), it won’t be able to keep the house cool when it starts to get hot and your neighbor will get it fixed, hopefully with a replacement unit.</p>
<p>After reading the linked document, I think you should go over there in the dark of night and do a “power trip test” if for no other reason than to get some peace and quiet.</p>
| 13972 | Is a 3 Hz beat frequency from my neighbor's HVAC a sign of a malfunction? |
2017-02-27T14:30:28.327 | <p><a href="https://i.stack.imgur.com/yqMly.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yqMly.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/1QX6o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1QX6o.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/lmZxe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lmZxe.png" alt="enter image description here"></a></p>
<p>I know that the closed-loop transfer function is equal to
Y/Ysp = (P controller * DC motor) /(1+(P controller * DC motor)) then I am struck. Can anyone give me some hints on how to proceed?</p>
| |control-engineering| | <p>You don't need to use the transfer function for this problem, just substitute the known equations in the time domain to get a differential equation for the closed loop. You have</p>
<p><span class="math-container">$$ V(t) = K_p e(t) \qquad e = Y_{sp} - Y(t) \qquad \tau_L = 0$$</span>
<span class="math-container">$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K V(t) - K_B \tau_L(t)$$</span></p>
<p>Just insert the first three equations into the fourth and you have your closed loop differential equation.</p>
<p><span class="math-container">$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} = K K_p (Y_{sp} - Y(t))$$</span></p>
<p>Rearrange to get into the standard differential equation form with your states on the left-hand-side and your inputs on the right-hand-side:</p>
<p><span class="math-container">$$ T \frac{d^2Y(t)}{dt^2} + \frac{dY(t)}{dt} + K K_p Y(t) = K K_p Y_{sp}$$</span></p>
<p>Now you can analyze the control system as a simple second-order DE with a constant input.</p>
<h1>Edit: Steady state value</h1>
<p>The steady state value can easily be derived from the differential equation. We know that when the system reaches steady state, by definition, <span class="math-container">$\frac{d^2Y(t)}{dt^2} = 0$</span> and <span class="math-container">$\frac{dY(t)}{dt} = 0$</span>. Therefore, substituting those values in the DE and setting <span class="math-container">$Y(t)=Y_{ss}$</span>:</p>
<p><span class="math-container">$$ T \cdot 0 + 0 + K K_p Y_{ss} = K K_p Y_{sp}$$</span>
<span class="math-container">$$Y_{ss} = Y_{sp}$$</span></p>
| 13978 | Find the differential equation to the closed-loop system |
2017-02-28T02:12:42.857 | <p>Internal Combustion Engine powered cars need multi-speed gear shifts since their power is highly dependent on speed and engines have narrow usable power/speed bands. </p>
<p>Some electric vehicles are also coming out with gear shifts. We have Antonov already working on a 3-speed gearbox for electric motor drive, and the new Exagon Furtive eGT is said to be coming with a 4-speed gearbox. Why multiple speeds on an electric vehicle? The reasoning is the same as for an engine. It is to allow lots of torque in first gear to start but also to eliminate the need for the motor to run insanely high RPM's when you're going 80 mph on the freeway. </p>
<p>Metal band saws also come with some form of mechanical speed control (CVT or* gear shift) since you will want to cut softer stuff using high gear, but you will need low gear to cut iron, steel, etc. There's Clausing Industrial who have put two xor THREE speed controls: a belt-drive CVT, 4-speed gearbox, and possibly an infinitely-variable** inverter motor, in their V4014H vertical band saw! </p>
<p>But what's odd is that many tower cranes also have gear shifts (on the hoist winch) even though they're electric! </p>
<p>For example, the hoist winch on the Liebherr 357 HC-L 12/24 luffing boom tower crane appears to have a 3-speed gearbox immediately after the motor. Using a 215-horsepower (160 kW) motor and 1-fall operation, the maximum no-load lifting speeds are (capacities based on halving the capacities of 2-fall operation): </p>
<ul>
<li>330 m/min (3rd gear) maintainable up to 1,500(?) kgf </li>
<li>185 m/min (2nd gear) </li>
<li>113 m/min (1st gear) </li>
</ul>
<p>And the maximum full-load line speed at 12,000 kgf (even on third gear, I presume) is 68 m/min. </p>
<p>The smaller 147-horsepower (110 kW) hoist winch also, apparently, has 3 speeds (these are again on 1-fall operation): </p>
<ul>
<li>261 m/min (3rd gear) maintainable up to 1,100(?) kgf </li>
<li>157 m/min (2nd gear) </li>
<li>91 m/min (1st gear) </li>
</ul>
<p>And the maximum full-load lifting speed (again, on any gear) is 45 m/min. </p>
<p>NOTES: </p>
<ul>
<li>"No-Load Speed" is at zero hook load (when the crane is lifting only the hook-block) </li>
<li>kgf = Kilogram-force</li>
<li>*Logical OR</li>
<li>** "Infinitely-variable" = "Continuously variable down to zero speed" </li>
</ul>
<p>My questions are: </p>
<ol>
<li><p>Despite the fast lifting speeds often encountered and the high load capacities, shouldn't the working range of a tower crane be smaller than that of a car? </p></li>
<li><p>If electric motors do so well on cars without a gear shift, why would you need such a thing on a winch? In this case, a crane hoist) </p></li>
</ol>
| |mechanical-engineering|gears| | <p>You need very slow speeds to make precise movements of the load. At high speed, the cables will stretch as the load accelerates and decelerates and then oscillate longitudinally, "bouncing" the load up and down (and bouncing the whole crane boom as well). That is not a good idea if you want to raise or lower the load accurately through a small distance (say 10mm).</p>
<p>Also, moving a large load a small precise distance <em>vertically</em> is different problem from moving a car a small precise distance <em>horizontally</em>, because the crane motor has to provide the force to counter-balance the full weight of the load all the time that the brakes are off, but the car engine does not. A lower gear ratio means the torque on the motor is less, and therefore the motor generates less heat while it is effectively stalled holding the load stationary. </p>
<p>You can easily move a car 10mm horizontally by pushing it with one hand, but you can't do the same thing to a 12-tonne weight hanging from a cable!</p>
<p>Just to avoid any misunderstanding, you don't "change gear" on the crane to accelerate up to the maximum speed <em>while the load is moving</em>. You select the appropriate gear for the next step in the procedure while the load is stationary and the crane brakes are on.</p>
| 13992 | Why do tower cranes have gear shifts? |
2017-02-28T13:12:05.403 | <p>None of my fellow commuters know what this white panel with, what appears to be a red slider attached, is called and what it is for.</p>
<p>This particular one is on a platform edge which happens to be over a bridge (Maidenhead in Berkshire).</p>
<p>Does anyone know? The track is a straight section at this point. So I don't think it's any indication of cant.</p>
<p><a href="https://i.stack.imgur.com/nGKzR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nGKzR.jpg" alt="enter image description here"></a></p>
| |rail| | <p><a href="https://www.google.co.uk/search?q=indicator%20track%20side%20at%20station&sourceid=ie7&rls=com.microsoft:en-GB:IE-Address&ie=&oe=&gfe_rd=cr&ei=Zoi1WKKHF4Om8we5mK2gBA&gws_rd=ssl" rel="nofollow noreferrer">Google</a> has the answer (<a href="https://groups.google.com/forum/#!topic/uk.railway/_z2B3F_8jyM" rel="nofollow noreferrer">second result</a>):</p>
<blockquote>
<ul>
<li>They mark the height of the nearest rail. </li>
<li>Height with respect to the little slider thing, so the track position is
usually an amount below the indicator. They are also found on some pieces
of trackside equipment such as OHLE or signal gantries - but they aren't so
readily seen, so mostly go without comment. Anywhere where the track
position cannot vary too much away from the 'as installed' position
basically. </li>
<li>There are datum plates, when track is relaid or worked on it allows
measurement to ensure the vehicle clearance is maintained </li>
</ul>
</blockquote>
<p>So basically it's there to mark what height the rail should be installed to, presumably in your case to ensure sufficient clearance to the bridge above.</p>
| 14001 | What is this indicator called? |
2017-02-28T15:44:59.690 | <p>I'm using airfoil data that is generated by a program. The generated airfoil has a chord length of 1 unit. I want to know how to scale up this airfoil to required chord length. My airfoil series number 638xx. NACA five digit series.</p>
| |aerospace-engineering|aerodynamics|airfoils| | <p>MountainClimberi is correct. Multiplying both x and y coordinates by the same factor will scale your airfoil correctly. Using a different factor for each will result in a different aspect ratio and a different(stretched) version of that airfoil that will not match its published properties.</p>
<p>Also remember that after scaling the data, your reference location will have moved if it was not (0,0). This may require a translation operation if you need to adjust it.</p>
<p>Per <a href="https://en.wikipedia.org/wiki/Scaling_(geometry)" rel="nofollow noreferrer">Scaling(geometry) wiki</a>: "The scaling is uniform if and only if the scaling factors are equal (vx = vy = vz)"
<br>Willamate.edu has a good article on <a href="https://www.willamette.edu/~gorr/classes/GeneralGraphics/Transforms/transforms2d.htm" rel="nofollow noreferrer">2D transformations</a> including scaling.
<br>Python would be a good language to manipulate this data with. Here is an <a href="https://stackoverflow.com/questions/2450035/scale-2d-coordinates-and-keep-their-relative-euclidean-distances-intact">SE python scaling question</a>.</p>
| 14005 | How to scale airfoil coordinates to achieve the required size? |
2017-03-01T08:53:36.003 | <p>Internet is full of twisted belt examples (30, 60, 90 degrees), however, the allowed misalignment tolerances claimed by the manufacturers are less than several degrees. </p>
<p>The very term misalignment seems to impose the two pulleys are out of single plane. </p>
<p>So, would it be correct stating that?:</p>
<ul>
<li>180 / 360 degree twist is ok</li>
<li>any other twist is a misalignment and therefore is a bad practice</li>
</ul>
<p>Apart from timing belts, are there other belt types which would enable me driving i.e. 30kW load at 30 degree belt twist?</p>
<p><a href="https://i.stack.imgur.com/92mQk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/92mQk.jpg" alt="enter image description here"></a></p>
| |torque|pulleys| | <p>If the motive power and the load are great enough the twist can gain an ability to avoid tearing out ribs on the belt from continuous start/stop circumstances.</p>
<p>A need to reverse belt direction many times per minute or hour is another consideration.</p>
<p>So to is the need to transmit power 90 degrees to drive equipment on multiple levels as was done in my grandfather's broom factory. A one cylinder natural gas engine in basement of a hillside factory used ribless belting that could slip on a large drive wheel until engine speed was great enough to engage belt with a long lever to tighten slack and start to 'smoke' from slippage as central belts got up to operating speed and belt slipping stopped.</p>
| 14015 | Is twisted belt configuration a good practice? |
2017-03-01T12:17:43.237 | <p><a href="https://i.stack.imgur.com/NpVA8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NpVA8.jpg" alt="enter image description here"></a></p>
<p>The above is my daughter's tricycle that I plan to make run off some spare Lithium-ion 18650s. I was just thinking of mounting a small DC motor to where I am pointing, and it will use a small tyre to make contact with the wheel, and rotate it.</p>
<p>I have a few questions / concerns:
1) How do I calculate the Power (Watt) and/or Torque required from the motor, given the total weight of the tri-cycle, motor, and my daughter... in order to make it go at 5Kmph on a flat surface?</p>
<p>2) Both the rear wheels are free rotating independent of each other. So, if I rotate only one of them using the motor, will that push the cycle straight and forward, or in circles/curved path?</p>
<p>3) If she pedals instead of using the electric mode, I am guessing it will charge the battery, although I doubt she'd pedal fast and enough to overcharge it. Would I need to do something to prevent this?</p>
<p>4) Is there anything else that I haven't considered?</p>
<p>Considering the tricycle will not be used for more than an hour in a day, I can make a relatively small battery pack with less mAh capacity.</p>
| |torque|electric-vehicles| | <ol>
<li>To calculate required torque, we are going to need a lot more information. Probably easiest to just pick a power like say 100 Watts. Then calculate the motor rpm you need based on the tire diameter you select.</li>
<li>Powering one is actually better. If you powered both wheels you would need a differential. The steering of the front wheel should easily overpower the offset driving wheel.</li>
<li>Don't mess around with charging the battery. Just install a double pull double throw (DPDT) switch (if using a dc motor) to give you forward and reverse options. Then install a on/off momentary push button.</li>
<li>Design the motor interface such that fingers wont be pinched (remember kids are always trying to get injured). A tire type drive might not be your best option. Getting a gearbox that can direct drive the wheel (might have to modify its axle) might be a better option.</li>
</ol>
<p>It is going to be a pretty large project when you get into it, so feel free to post new questions specific to gearing, battery calculations, etc. Good Luck!</p>
| 14019 | Evaluate my electric tricycle project |
2017-03-01T20:30:22.070 | <p>The sentence is </p>
<blockquote>
<p>In a closed loop system when the loop gain grows typically we have an instability</p>
</blockquote>
| |control-engineering| | <p>I can explain the very basic sentence with a very basic idea what happens.</p>
<p>If your gain grows, that means you're amplifying the signal <strong>more</strong>.</p>
<p>By loop gain I assume they mean the total gain from the loop. If that is growing constantly, and the loop is closed, every time you go around the loop your signal strength should increase. A constantly increasing signal isn't stable behaviour.</p>
<p><a href="https://en.wikipedia.org/wiki/Positive_feedback" rel="nofollow noreferrer">Wikipedia's article on positive feedback</a> has a lot of information and more links to more information (where I'm sure you could find actual academic sources).</p>
| 14035 | Can someone explain my this sentence about feedback systems? |
2017-03-01T23:06:29.800 | <p>The high-pressure oxygen in the cabin is supplied through a system of regulators and valves from a high-pressure oxygen tank. There are two pairs of the sub-systems of check valves, shut-off valves and nonreturn automatic shut-off valves in the system. The function of these valves is to stop the reverse flow of air from the cabin to the gas tank in the case of pressure drop and to close the line supply if there is some sudden pressure drop in header line or cabin in order to avoid the waste of the gas. Each pair of the valve systems consists of two alternative branches. One consists of a non-return automatic emergency shut-off valve, and the other consists of a check valve and a shut-off valve in series. Any branch of the two pairs is capable of supplying sufficient gas to the cabin. There are three alternative paths between the oxygen tank and the pair of valves. Oxygen can be transmitted to the cabin through either of the two regulators and the pair of valves connected to the regulator. It can also be transmitted to the cabin through a selector valve and either of the two pairs of valves.</p>
<p><a href="https://i.stack.imgur.com/DHUx4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DHUx4.png" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>From Oxygen tanks or H2O electrolysis (back to fuel cell as H2 and O2)</p>
| 14037 | Where is the source of the high pressure oxygen supply system in this spacecraft life support system? |
2017-03-02T10:22:36.677 | <p>Does work behave the same way for all temperature range while in a refrigeration setup. When quality of heat changes with the temperature, Does it apply to work also. IF we consider refrigerator Source-sink of 1000-500k and 300-250k, Does 1KJ of work input have the same effect or the effect is different. My doubt may be a very silly, i might be missing something. Any input is appreciated. Thank you.</p>
| |thermodynamics|refrigeration| | <p>In theoretical, ideal setup, work will be approximately the same for the same temperature deltas between original temperature of the body, target temperature, and heat sink temperature. Cooling something from 200K to 100K given 300K heatsink should take the same work as cooling it from 1000K to 900K given 1100K heatsink.</p>
<p>In reality, obtaining efficient heat sinks for low temperatures is difficult. You can just leave something sit in open air for a couple minutes to bring it from 1000K to 900K while you need to jump through hoops and spend a lot of energy to get something from 200 to 100K - simply because your heatsink will be at ~300K in both cases. If you can provide a heat sink of temperature that is the same offset from your target temperature, you have comparable work - but usually, you can't.</p>
<p>There are also operational temperatures of the coolant. Most heat is stored/extracted at phase change, gas<->liquid. This needs to correspond to boiling point of the coolant, so you'll use e.g. sodium to cool a nuclear reactor, but helium to cool superconducting magnets. Since they have different enthalpy of evaporation, they will provide different performance - not due to difference in temperature, but due to being different substances, that happen to boil at different temperatures needed for their respective processes - affecting performance of the refrigerators.</p>
| 14041 | How does Work behave in two different refrigeration setup? |
2017-03-03T07:46:01.813 | <p>What will be the best thermal insulation material for insulation of pipe lines having temperature of 500℃. Mineral wool and Calcium Silicate can be used but here we have a constraint of both weight and space.</p>
| |thermal-insulation| | <p>There is a well known but counter-intuitive result that for a given temperature, below a certain diameter, any insulation at all is worse than no insulation, because it conducts heat (poorly but better than air) to a larger external radiating surface. </p>
<p>Details should be in any good book on heat flow, but for common steam pipe temperatures, the critical diameter is of the order of 5mm.</p>
<p>So, below a certain pipe diameter, leave the pipe exposed to keep the radiating area small. We don't know from your question if your pipe dia can be small enough to allow this.</p>
| 14058 | Best insulation material for pipes @ 500℃ |
2017-03-03T19:12:57.023 | <p>I am on the task of analysing the load capacity of a warren-ish truss bridge similar to <a href="https://images.duckduckgo.com/iu/?u=https%3A%2F%2Fs-media-cache-ak0.pinimg.com%2F736x%2F7f%2Ff1%2Fa1%2F7ff1a1719eadfdc3567338d9b659284b.jpg&f=1" rel="nofollow noreferrer">this one</a>. I do not, however, have the software or skill to model and analyse a 3D structure.</p>
<p>I was wondering wether or not i could analyse a <a href="https://images.duckduckgo.com/iu/?u=http%3A%2F%2Fwww.mathalino.com%2Fsites%2Fdefault%2Ffiles%2Freviewer-mechanics%2F418-warren-truss.gif&f=1" rel="nofollow noreferrer">2D</a> simplification and still come to a somewhat credible conclusion. What considerations should be made when approximating a 3D truss as a 2D truss? What would be the limitations of such an approach? Thank you.</p>
<p><strong>Edit:</strong>
I do not have to know the exact maximum load. My only goal is to figure out where the truss will fail first.</p>
| |structural-engineering|statics| | <p>It depends on the loads applied to it. If there are only vertical loads (or the lateral loads are insignificant in comparison) then of you can assume each side to behave in similar manner to a 2D truss. Of course you have to consider that the vertical loads in such a bridge reach to the side trusses through bending of the transverse members, so you must consider the possibility of failure of these members too. Furthermore even if you analyse each truss as a 2D structure you should propably consider the presence of the the transverse members when you calculate the resistances of the truss, for lateral buckling failure. </p>
<p>In other circumstances, such as when severe lateral loads exist, or the dynamic response is wanted (which is typical true for bridges) then 3D modeling is required. </p>
| 14062 | Drawing conclusions about 3d truss from 2d model |
2017-03-04T13:16:34.493 | <p>I have a choice of two hydraulic power units with specs as follows:</p>
<p>A. 2.2kW motor, 8lpm flow rate, 120 Bar pressure rating</p>
<p>B. 1.1kW motor, 2lpm flow rate, 200 Bar pressure rating</p>
<p>Using F=PA I assume pump B is capable of producing more force in the same system (albeit a lot slower!)</p>
<p>However, hearing the phrase "resistance to flow produces the pressure not the pump" many times gives me a nagging feeling.</p>
<p>Am I correct or is there something I'm missing?</p>
<p>Thanks</p>
| |fluid-mechanics|hydraulics| | <p>For a given cylinder, B has higher pressure and will produce a larger force.</p>
<p>You could select a larger diameter cylinder for A such that it would produce the same or larger force.</p>
<p>If a cylinder is moving with zero resistance, the pressure in the cylinder will be very low (just enough to overcome its own friction).</p>
<p>You are likely looking for maximum force (pump max pressure x cylinder head area) which occurs when the cylinder blocked or fully extended (zero flow).</p>
<p>At flowrates and pressures in between full flow zero pressure and full pressure zero flow will require a pump curve, plumbing information, and more involved calculations.</p>
| 14070 | Hydraulic pump choice - which produces most force |
2017-03-05T23:33:33.563 | <p>I have been looking at the location of shear flow centers for profiles for my mechanics of materials course.</p>
<p>I am trying to figure out how to determine the direction of the shear flow in a given beam section with a shear force which is either directed up or down (aligned with y-axis). In my lectures it has been said that 'the shear flow in the web of the beam section must always follow the same direction as the shear force'.</p>
<p>I have found some examples where it appears this rule does not work though:</p>
<ol>
<li><p><a href="https://www.youtube.com/watch?v=zs_dEtwgxSM" rel="nofollow noreferrer">https://www.youtube.com/watch?v=zs_dEtwgxSM</a></p></li>
<li><p><a href="https://engineering.stackexchange.com/questions/6601/interpretation-of-shear-flow-shear-center-of-l-profile">Interpretation of shear flow/shear center of L-profile</a></p></li>
</ol>
<p>In these 2 examples the shear flow direction in the web of the beam section is drawn in opposite direction of the shear force. I do not know why the aforementioned rule does not apply to these examples. Is this rule (shear flow direction in web must follow shear force direction) correct?</p>
<p>How do I determine the direction of shear flow for these 2 examples/in general?</p>
| |beam|shear|cross-section| | <p>Yes, shear <em>flow</em> in a cross sectional element will always be in the same direction as the shear <em>force</em> in that same cross sectional element because the shear flow is essentially the distribution of the shear force.</p>
<p>Confusion may arise due to the distinction between the "shear force" in an element and the externally applied load that is producing said shear. The net shear force in the cross section will be in the opposite direction of the applied shear-producing load. (At least, that's the sign convention I use, and that appears to be what is happening in the two examples you provided.)</p>
<p>A couple sketches based on the video example you provided:</p>
<p><a href="https://i.stack.imgur.com/BVSCjl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BVSCjl.jpg" alt="shear flow"></a></p>
<p>Note that the net horizontal shear force will be zero and the net vertical shear force will be equal and opposite to the applied force. (Congruent with basic statics)</p>
| 14087 | Determine shear flow direction in profile/beam section |
2017-03-07T02:32:54.750 | <p>I have been trying to find out the significance of the usual streetlight design - Why is there curvature of the arm holding the bulb of the light? I have been unable to any reason as to why it may be. Here is an picture of the "usual" of which I am referring to: <a href="https://i.stack.imgur.com/KsFqw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KsFqw.jpg" alt="enter image description here"></a></p>
| |design|lighting| | <p>okay heres a thought. </p>
<p>I have had similar thoughts about sign boards and their life.
We know what happens when there is a cantilever beam. There is more stress at the ends (SFD)
So think of the deflection to be maximum at the ends with respect to the original reference or starting point of the pipe.</p>
<p>If we keep a curved pipe, the deflection with respect to the starting point will be lesser than the first case. </p>
<p>The life and the amount of stress the pipe can handle would be more by using the curved pipe</p>
<p>Think of this example of a camber in rollers, we change the shape a bit to accommodate the pressure the rollers face.</p>
<p><a href="https://i.stack.imgur.com/qPX3E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qPX3E.png" alt="Roll Bending"></a></p>
| 14102 | What is the significance of curvature in the arm of a streetlight? |
2017-03-07T18:23:14.700 | <p>If you were aiming for 300mph then what would need to change and would it be possible. There is less heat so less drag which must be a huge advantage.</p>
<p>What kWh battery pack would be needed?</p>
| |automotive-engineering|motors|electric-vehicles| | <p>Maximum speed in a straight line is largely determined by the vehicles maximum power output compared to its drag although once you get past about 200 mph dynamic stability can start to become a serious issue as well. </p>
<p>The current Land speed Record is held by <a href="https://en.wikipedia.org/wiki/ThrustSSC" rel="nofollow noreferrer">Thrust SSC</a> which was powered by two jet engines, reached supersonic speeds and has as much in common with a plane as a conventional car. </p>
<p>A crucial point is that aerodynamic drag is proportional to the square of speed so all else being equal to double the maximum speed you need to quadruple the thrust/torque available and 8x the power. </p>
<p>In any car, even a high performance one there is a trade off between maximum speed and handling and even very high end performance car manufacturers have pretty much given up chasing headline top speed as if becomes a bit meaningless once you get cars which can top 200mph and acceleration, braking and handling become more meaningful measures of performance. Even current F1 cars don't get anywhere near 300 mph simply because adding aerodynamic down-force increased overall average speeds around a track at the expense of increased drag. </p>
<p>I suppose it is possible that you could use the electric drive train from a Tesla to attempt to get to 300mph but it would for all practical purposed be a completely different vehicle as you would need totally different chassis, body shape and gear ratios. </p>
<p>As a point of reference the current <a href="http://www.autocar.co.uk/car-news/motorsport-formula-e/electric-land-speed-record-set-venturi" rel="nofollow noreferrer">Land Speed Record for an electric car</a> is 341 mph. As you can see from the article this is a very different vehicle from a Tesla. </p>
| 14117 | What would be needed to make a tesla fastest car in the world? |
2017-03-07T23:57:58.887 | <p>I have a dynamical model of a planetary gearbox of the form:</p>
<p>$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{[K_b+K_m-\Omega_c^2K_\Omega]q}=T(t)+F(t)} $</p>
<p>$K_b$: bearing stiffness symmetrical matrix
$K_m$: mess stiffness matrix
$K_\Omega$: diagonal centripetal stiffness matrix
$T$: applied external torque
$F$: static transmission error excitation
$q$: the vector with the traslation and rotational coordinates for the sun,ring and carrier</p>
<p>For the free vibrations I have the equation:</p>
<p>$\mathbf{M{\ddot q}+{[K_b+K_m]q}=0} $</p>
<p>$\mathbf{-\omega_i^2M{\phi_i}+{[K_b+K_m]\phi_i}=0} $</p>
<p>Where $\phi_i$ is the vibration modes.</p>
<p>$\mathbf{\omega_i^2M{\phi_i}={[K_b+K_m]\phi_i}} $</p>
<p>I know that the $\Omega_c$ term for the gyroscopic effect is neglected at first and there's no excitation input because it's free vibration. But then I don't understand where the $\phi_i$ comes from.</p>
<p>I have basic knowledge of vibration theory for a single degree system, but not for multi DOF which is this case.</p>
<p>I need to determine the natural frequencies and vibrational modes of the system from this equation but I don't know how or where to start. I have initial numerical values for Stiffness, mass, preassure angle of the carrier, sun and planet gears.</p>
<p>Numerically how would I know how many modes of vibration I have and how to determine them with their natural frequencies? Like I said this is the first time I deal with a MDOF system and I've read a lot but this model is far to complicated and the math is confusing me.</p>
| |mechanical-engineering|gears|dynamics|vibration|frequency-response| | <p>To obtain the modes shapes and resonant frequencies, you start from your equation of motion with no externally applied forces, which is indeed as you've stated.</p>
<p>$$\mathbf M \mathbf{\ddot q} + \mathbf K \mathbf q = \mathbf 0 \qquad (1)$$</p>
<p>For brevity, I've let $\mathbf K = \mathbf K_b + \mathbf K_m$.</p>
<p>Currently, $\mathbf q(t)$ is a function of time. If the system is vibrating in the $i$th mode, everywhere will be vibrating at frequency $\omega_i$ in the same phase, assuming there is no damping. Therefore, we can assume the following:
$$\mathbf q(t) = \mathbf u^{(i)}\exp{(j\omega_i t)}\qquad(2)$$
, where $j$ is the imaginary number defined as $\sqrt{-1}$ and $\mathbf u^{(i)}$ is the $i$th mode shape, which is independent of time. (This assumption may seem bizarre as $\exp{(j\omega_i t)}$ is a complex number, but this does not matter in this scenario, and is used just to show that $\mathbf q(t)$ is sinusoidal in time with frequency $\omega_i$. Equivalently, you could instead use $\mathbf q(t) = \mathbf u^{(i)} (A \sin{\omega_i t} + B \cos{\omega_i t})$.)</p>
<p>Note that:</p>
<p>$$\mathbf {\ddot q}(t) = \frac{d}{dt}\left(\mathbf u^{(i)}\exp{(j\omega_i t)}\right)=-\omega_i^2\mathbf u^{(i)}\exp{(j\omega_i t)}\qquad (3)$$</p>
<p>Substituting equations (2) and (3) into (1) will give you the following, noting that the exponential functions cancel out:</p>
<p>$$\left(-\omega_i^2 \mathbf M + \mathbf K \right) \mathbf u^{(i)}=0 \qquad (4)$$</p>
<p>Equation 4 is a type of problem called the <em>generalised</em> eigenvalue problem, which involves solving equations of the form $\mathbf A \mathbf x = \lambda \mathbf B \mathbf x$ (compare with the <em>standard</em> eigenvalue problem, which involves solving $\mathbf A \mathbf x = \lambda \mathbf x$). In this case, the eigenvalue is $\omega_i^2$ and the eigenvector is $\mathbf u^{(i)}$. To obtain the eigenvalues and eigenvectors involves solving the generalised eigenvalue problem. There are two main approaches: hand calculation or numerical computation. For your problem, the latter will definitely be the recommended option, however, the former should help with smaller problems and lends some insight to how generalised eigenvalue problems are solved.</p>
<p><strong>HAND CALCULATION</strong></p>
<p>Note how equation (4) also has the form $\mathbf A \mathbf u^{(i)} = \mathbf 0$. Obviously, $\mathbf u^{(i)}=\mathbf 0$ would satisfy such an equation, but this is a trivial result that corresponds to no actual vibration. Instead, we are interested in values of $\mathbf u^{(i)}\ne\mathbf 0$ that solve the equation. In order for such a non-zero answer to exist, $\mathbf A$ must have no inverse, i.e. $\mathbf A$ is a singular matrix. (If an inverse did exist, then you could premultiply both sides of the equation with the inverse matrix, getting $\mathbf u^{(i)} = \mathbf A^{-1} \mathbf 0 = \mathbf 0$.) In order for a matrix to be singular, the <a href="https://en.wikipedia.org/wiki/Determinant" rel="noreferrer">determinant</a> of the matrix must be zero. i.e.</p>
<p>$$\left| \mathbf A \right| = \left| -\omega_i^2 \mathbf M + \mathbf K \right| = 0$$</p>
<p>This will produce a polynomial in $\omega_i^2$, the roots of which will be each of the resonant frequencies squared. Take care to note that some modes will have the same resonant frequencies, in the form of repeated roots.</p>
<p>For example, say:</p>
<p>$$\mathbf M = \left[\begin{matrix}
m & 0\\
0 & m\\
\end{matrix}\right]
\qquad
\mathbf K = \left[\begin{matrix}
2k & -k\\
-k & 2k\\
\end{matrix}\right]
$$</p>
<p>The determinant of interest would be:</p>
<p>$$
\left|\begin{matrix}
2k-\omega_i^2 m & -k\\
-k & 2k-\omega_i^2 m\\
\end{matrix}\right|
=0
$$</p>
<p>By expanding out the determinant expression, we obtain the polynomial:</p>
<p>$$(2k-\omega_i^2 m)^2-k^2=0$$</p>
<p>The two roots of which are:</p>
<p>$$\omega_i^2 = \frac{k}{m},\frac{3k}{m}$$</p>
<p>So the two resonant frequencies are:</p>
<p>$$\omega_1 = \sqrt{\frac{k}{m}} \qquad \omega_2 = \sqrt{\frac{3k}{m}}$$</p>
<p>Note that you should obtain a number of (not necessarily distinct) eigenvalues equal to the number of rows or columns of your $\mathbf M$ or $\mathbf K$ matrices.</p>
<p>To obtain the eigenvector corresponding to a particular eigenvalue, substitute back into equation (4), to obtain a series of simultaneous equation, the solving of which should determine the eigenvector.</p>
<p>In our example, finding the eigenvector corresponding to the eigenvalue $\omega_1^2$, substitute $\omega_i=\omega_1$ into equation (4)...</p>
<p>$$
\left[\begin{matrix}
2k-k & -k\\
-k & 2k-k\\
\end{matrix}\right]
\left[\begin{matrix}
u^{(1)}_1\\
u^{(1)}_2\\
\end{matrix}\right]
=
\left[\begin{matrix}
k u^{(1)}_1-k u^{(1)}_2\\
-k u^{(1)}_1+k u^{(1)}_2\\
\end{matrix}\right]
=
\left[\begin{matrix}
0\\
0\\
\end{matrix}\right]
$$</p>
<p>From this, we can see that $u^{(1)}_1 = u^{(1)}_2$, and so the eigenvector is therefore:</p>
<p>$$\mathbf u^{(1)} =
\left[\begin{matrix}
1\\
1\\
\end{matrix}\right]
$$</p>
<p><strong>NUMERICAL COMPUTATION</strong></p>
<p>Your problem does not seem practical to do by hand. Thankfully, there are a variety of different eigen-solver algorithms out there that are dedicated to solving this very problem.</p>
<p>A couple of these include:</p>
<ul>
<li>Power Iteration method</li>
<li>Lanczos method</li>
<li>Krylov-Schur method</li>
<li>Jacobi-Davidson method</li>
</ul>
<p>To explain how each of these algorithms work would take more time than I currently have, however I would definitely recommend tracking down these algorithms, reading how they work and using them to solver the generalised eigenvalue problem.</p>
<p>Hope this helps.</p>
| 14120 | How to simulate the Natural Frequencies of a Planetary Gearbox? |
2017-03-08T00:59:43.463 | <p>I am trying to build a low power output generator with 12 pole pairs and 9 stator coils. The design will be similar to the one described in this link <a href="http://www.sparweb.ca/2_Gen_Ax/AXIAL_FLUX_HowItWorks_V4.pdf" rel="nofollow noreferrer">Basic Principles Of The Homebrew Axial Flux Alternator</a> , but at a different scale/size. I am trying to find some way of estimating the number of turns I will need in the coils to produce an output voltage of 6.5V. The generator should produce this voltage at 100rpm. I am getting very confused about how to use Faraday's law to do my calculations.</p>
<p>$\varepsilon = N\frac{d(BA)}{dt}$</p>
<p><a href="http://www.6pie.com/faradayslaw.php" rel="nofollow noreferrer">Faraday's Law</a> as explained in this example where the area used is that of the magnet, while <a href="https://www.electrical4u.com/faraday-law-of-electromagnetic-induction/" rel="nofollow noreferrer">Faraday Law of Electromagnetic Induction</a> states at the bottom that A is the area of the coil. Does this difference have to do with when the magnetic field is stationary while the coil is moving versus when the magnetic field is moving relative to a stationary coil? Also, should I be using $$\varepsilon = N\frac{d(BAcos(\theta))}{dt}$$ instead? Thanks very much in advance, and apologies if I am missing something very obvious here.</p>
| |generator| | <p>No expert, but the problem is interesting.</p>
<p>Assumptions:</p>
<ul>
<li>Generate $6.5V_{RMS}$ at 100 rpm with 9 stator coils.
<ul>
<li>$V_{MAX} = 9.2V$</li>
<li>100rpm = 1.67rps</li>
<li>if radius to center of magnets/coils = 0.1m</li>
<li>$C = 2 \pi r = 2 \pi \times 0.1m = 0.628m$</li>
<li>at 1.67 rps, v = 1.05m/s</li>
</ul></li>
<li>N48 Neodymium Bar Magnets 1 in x 1/2 in x 1/4 in
<ul>
<li>L = 0.0254m, W = 0.0127m, D = 0.00635m.</li>
</ul></li>
<li>Separation z = 0.00635m.
<ul>
<li>Biggest problem is separation between magnets and windings. Closer they are the better.</li>
</ul></li>
</ul>
<p>From: <a href="https://www.supermagnete.de/eng/faq/How-do-you-calculate-the-magnetic-flux-density" rel="nofollow noreferrer">How do you calculate the magnetic flux density?</a>
<a href="https://www.supermagnete.de/eng/faq/How-do-you-calculate-the-magnetic-flux-density" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KMaqh.png" alt="enter image description here"></a></p>
<p>From: <a href="http://www.cy-magnetics.com/Mag_Pros_NdFeB.htm" rel="nofollow noreferrer">Magnetic Properties of Sintered NdFeB Magnets</a></p>
<p>N48 has a Remanence field of $B_r$ = 1.48T, which gives a flux density B = 0.164T, when you process the above equation.</p>
<p>$$V_{Ind} = N B l v$$</p>
<p>$$N = \frac {V_{Ind}} {B l v} = \frac {9.2V} {0.164T \times 0.0254m \times 1.05m/s} = 2108 turns$$</p>
<p>So that assumes you have one coil. Divide that by 3 (in series) gives 703 turns/coil (which is a lot). This coil has to be within length of magnet of 0.0254m. Whole area of coil has to fit within area of magnet for greatest effect.</p>
<p>The math is a bit distorted, but it should give you a ballpark to do your initial calculations. I'd guess, by the second or third prototype, you will have what you want. Not all three coils in series will generate maximum voltage at the same time, so some experimentation will have to come into play (i.e. more turns).</p>
<p>Realize that the coil will not experience a constant magnetic field. The further away windings are from the magnet the less the voltage induced. As already stated, biggest problem is air gap between windings and magnets. Air is a poor conductor of flux. I'd use a steel plate to attach your magnets.</p>
| 14122 | What area should I use when calculating induced voltage? |
2017-03-08T08:21:39.617 | <p>Is it possible to define any camera without "flash" (sensible to any band of radio wave) as a passive radar?</p>
<p>If so, where could I find references to that? I've been looking all over the place and nothing has come up. </p>
<p>If anyone has any sources or comments, please let me know.
Thanks!</p>
<p>Edit: The idea of camera is merely a device that captures light in any range (visible light, infrared, ultraviolet).
This is close in meaning to a passive radar, which is sensible to the electromagnetic waves coming from a target.</p>
| |electromagnetism|waves| | <p>Not really. </p>
<p>Firstly the RA in RADAR stands for RAdio which covers significantly longer wavelengths than the visible light (an in some cases near infra-red) which cameras can detect. </p>
<p>While light and radio are both part of the electromagnetism spectrum they are <em>different</em> parts of it and not interchangeable. While they are all part of a spectrum in terms of physics, from an engineering perspective they might as well not be. </p>
<p>Equally there are laser based systems analogous to radar, usuallly called <a href="https://en.wikipedia.org/wiki/Lidar" rel="nofollow noreferrer">LIDAR</a> </p>
<p>Also radar implies a device which is capable of deducing information about the range and possibly speed and direction of a target object, most cameras can't do this, or if they can it is through a separate system. In fact the auto-focus on most cameras is more akin to the <a href="https://en.wikipedia.org/wiki/Coincidence_rangefinder" rel="nofollow noreferrer">optical rangefinders</a> widely used for military purposes before the advent of radar. </p>
<p>A radar system consists not just of an emitter and receiver but also the processing equipment to deduce the required information while a camera is intended to capture and store a 2D image. </p>
<p>The closest thing to a 'radio camera' is a <a href="http://abyss.uoregon.edu/~js/glossary/radio_telescope.html" rel="nofollow noreferrer">radio telescope</a>. A key point here is that the differences in wavelength between radio and visible light means that a radio telescope is very different from any camera in terms of both scale and structure. </p>
<p>While radar and cameras could both be described as devices for detecting electromagnetic radiation that about as far as the similarity goes and is not a complete definition of either. </p>
| 14128 | Defining camera as a passive radar |
2017-03-08T14:25:30.510 | <p>Today, I was thinking about seat belts. </p>
<p>On that note, why do the ones in cars and planes differ? Planes have only belts on the waist, but cars have waist and diagonal. Is there a reason that this is so, or are the difference just... [Insert suitable word]?</p>
| |automotive-engineering|aerospace-engineering|stresses|safety| | <p>Originally car seat belts were like the airline <a href="https://en.wikipedia.org/wiki/Seat_belt" rel="nofollow noreferrer">seat belts</a>. In 1959 the Swedish car marker <a href="http://jalopnik.com/volvo-gave-away-their-most-important-invention-to-save-1069825878" rel="nofollow noreferrer">Volvo invented the three point seat belt</a> and determined it to be the safest type of seat belt, by the way it restrains vehicle occupants. Volvo let other car makers use the belt for a fee of just one dollar ($1) to save lives around the world.</p>
<p>Airline type seat belts are simple to use & simpler to get out of in an emergency in a bulk passenger situation. The seat belt ends stay on the seat. If airlines used the triple point seat belts that are used in cars some passengers may get injured by the belt buckles as they fly back up to there resting positions.</p>
| 14133 | Difference between seat belts |
2017-03-09T15:33:03.303 | <p>Is it possible to build a two storey building just using bricks (no wooden beams, steel beams or steel reinforcement) to suspend the floor of the second storey?</p>
<p>I'm aware that using wooden beams is an option, however because of the environment of the build (tropical rains, humidity, termites, ants, beetles and millions of other insects), I would much prefer to use only brick - if this is possible.</p>
<p>Ideally there will be no intermediate supports; the floor structure will span from one external wall to the other.</p>
<p>I have an intuitive feeling that an efficient structure can be built using just bricks - but I can't quite articulate it. </p>
<p>Has this been done before - is this even possible?</p>
| |structural-engineering|civil-engineering|structures| | <p>Of course it can be done - <a href="https://www.google.com/search?site=&tbm=isch&source=hp&biw=784&bih=797&q=masonry%20bridge&oq=masonry%20bridge&gs_l=img.3..0j0i8i30k1j0i24k1l2.187.3369.0.3672.16.14.1.1.1.0.157.1193.11j3.14.0....0...1ac.1.64.img..0.16.1201.2LIkfpgDGSE#tbm=isch&q=masonry%20arch%20bridge&*" rel="nofollow noreferrer">masonry bridges</a> suspend a bridge deck above a void, you could use the same technique to support the floor for your second storey.</p>
<p>Of course, in order to avoid tension in your masonry (masonry is not good at tension) you need an arch that is close to a catenary in shape. This means that the crown (top) of your arch is going to be a fairly big height above your springing (bottom of the arch). My intuition is that your "two storey" house will be the height of a normal three storey house, with the arch taking up the height of a storey.</p>
<p>If you want the floor of your second storey to be flat, you'll need fill over the arch. This adds weight and hence causes more thrust.</p>
<p>With a masonry arch bridge there's a big thrust resisted at the abutments. This is resisted by the mass of soil. For your house to resist it you're going to need very thick walls.</p>
<p>So: technically feasible but not very practical.</p>
<p>Of course, there are also plenty of other materials out there which aren't steel or wood. Fibre reinforced polymers would work, but they're not going to be cheap!</p>
| 14162 | Two storey brick building without using wood or steel |
2017-03-09T16:11:49.880 | <p>I'm working on a seaplane and find myself doubting about the specific nautical term for a keel placed between the <em>bilge keel</em> and the <em>main keel</em>. This might be a bit academic in nature since I have a working way of naming it, but I'd rather take the chance to find if it's correct or just a fossilized mistake.</p>
<p>My working term is "sister keelson", which is commonplace in seaplane design, but something tells me it's not really right, as in naval engineering a keelson is the <strong><em>inner</em></strong> structural element that holds ribs against a keel. </p>
<p>The second nearest term would be "docking keel", because of its location, but that implies it can be used to beach the craft, which it may not be necessarily designed for.</p>
<p>I can't add images right now, but a google image search for "sister keelson seaplane" will give you pictures of what this is, while a search for "sister keelson" will point you to the inner elements that would be used in a wooden hull boat.</p>
| |aerospace-engineering|terminology|aircraft-design|naval-engineering| | <p>Sister keelson definitely does not translate to conventional marine usage. There are a couple of terms that apply to similar structural elements.</p>
<ul>
<li><p>Strake. A longitudinal hull element that may be backed by an internal longitidinal structural element. In planing hulls the strake at the turn of the hard chine may be referred to as a lifting strake. Lifting strakes may also be fitted on the flat of bottom to improve hydrodynamic efficiency.</p></li>
<li><p>Which leads to the other possible term. <a href="https://en.m.wikipedia.org/wiki/Chine_(boating)" rel="nofollow noreferrer">Chine</a>. In a multi chine vessel the hard angle between strakes is referred to as a chine. There will typically be a chine log or composite chine log backing the chine longitidinally.</p></li>
</ul>
<p>There is a reasonable comparative illustration <a href="https://www.google.com.au/search?q=strakes%20on%20a%20boat&client=ms-android-samsung&sa=X&prmd=ivsn&tbm=isch&tbo=u&source=univ&fir=ITCNuPhUtqLjfM%253A%252CYvGOzp8PjEj-eM%252C_%253B3ziv3qLqc8OgzM%253A%252CtoQs-4812V7NFM%252C_%253BftJ94Ujy71allM%253A%252C3EhS_rE2FDYqtM%252C_%253Ba0Cpu-pY-NvReM%253A%252CvWfBJMOajLkyYM%252C_%253BLla2SrsZht-k2M%253A%252CeS5eMb-SYdn4VM%252C_%253By2HpFf6GcfqsCM%253A%252CnEW36Lbel-ALkM%252C_%253BSTg5Hs9ZasXUuM%253A%252CkDr7cxOdGcNgFM%252C_%253BnJsghWHEciCu-M%253A%252CSexDWV_dTQ8ZZM%252C_%253B4ld3NXuttglZxM%253A%252CBL47r_DZrW1jpM%252C_%253BwST5lZoHWdr6bM%253A%252Co1vxoqgWu_tAFM%252C_&usg=__oKx9GkfEpZ6QQ_WDPd3Z5tm2ydw%3D&ved=0ahUKEwiSuf-m_cvSAhVE3mMKHSpnDDUQsAQIGA&biw=360&bih=560#imgrc=nJsghWHEciCu-M:" rel="nofollow noreferrer">here</a>.</p>
<p>This usage is primarily restricted to planing hulls (there are no chines on faired hulls) and on a faired hull a strake is a longitudinal run of planking / plating extending from the garboard strake adjacent to the keel to the sheer strake at the hull/deck join.</p>
<p>However I guess your seaplane float is a planing hull & so the terminology applies!</p>
| 14164 | Name of a keel between the bilge keel and the main keel |
2017-03-10T10:56:39.777 | <p>I am finding different opinion, but I can not find a mathematical demonstration about this topic.
for example here is described as non homokinetic:
<a href="https://engineering.stackexchange.com/questions/7898/oldham-coupling-vs-universal-joint/7935?newreg=9adda9c4d4144e99af298e1434ee12d2">Oldham coupling vs Universal joint</a></p>
<p>here is homokinetic:
<a href="https://books.google.it/books?id=X0AHKxwWTsYC&pg=PA120&lpg=PA120&dq=constant%20velocity%20oldham&source=bl&ots=KtiVJpXE_U&sig=5SYjn7HEgm2Yr5_mKvx1Y0H7i3Y&hl=it&sa=X&ved=0ahUKEwiDtYCV4svSAhWKXBoKHWBfAzQ4ChDoAQg_MAY#v=onepage&q=constant%20velocity%20oldham&f=false" rel="nofollow noreferrer">Book</a></p>
| |mechanical-engineering|mechanisms| | <p>It's kinda trivial, the middle disk is homokinetically and rotationally coupled to the outer disks. Because the slot and tab in the connections are always parallel (not counting play in the fit).</p>
| 14180 | Is the oldham coupling homokinetic? |
2017-03-11T19:38:19.000 | <p>I read that a human</p>
<p><a href="https://www.google.co.uk/search?client=ms-android-samsung&ei=l0_EWNjsHKXEgAafuIUg&q=john+stapp+45+g&oq=john+stapp+45&gs_l=mobile-gws-serp.1.0.0.6460.9902.0.10829.15.14.1.8.8.0.185.2085.0j13.13.0....0...1c.1j4.64.mobile-gws-serp..1.14.1112.3..35i39k1j0i67k1j0i10k1.tC-NnmcrLLw" rel="nofollow noreferrer">https://www.google.co.uk/search?client=ms-android-samsung&ei=l0_EWNjsHKXEgAafuIUg&q=john+stapp+45+g&oq=john+stapp+45&gs_l=mobile-gws-serp.1.0.0.6460.9902.0.10829.15.14.1.8.8.0.185.2085.0j13.13.0....0...1c.1j4.64.mobile-gws-serp..1.14.1112.3..35i39k1j0i67k1j0i10k1.tC-NnmcrLLw</a></p>
<p>Can survive 46g force so what type of rocket and how much fuel would be best to stop a human falling at terminal velocity without exceeding this g force limit.</p>
<p>I also read spacex use rockets not parashutes for the weight advantage so would a short (i think 1 second) burst rocket be better in anyway to a parashute (despite such deceleration causing injury(yet non permanent)).</p>
<p>Or would the rocket and fuel be far too big.</p>
| |aerospace-engineering|rocketry| | <p>There are human portable <a href="https://en.wikipedia.org/wiki/Jet_pack" rel="nofollow noreferrer">rocket packs</a> which do work, albeit to a fairly limited extent. </p>
<p>The minimum terminal velocity in a stable 'belly to earth' position is about 50 metres per second so even at a fairly modest 10 g that is only about 5 seconds to slow down to a stop. So it's certainly not impossible. </p>
<p>However parachutes work pretty well at what they do. They are very simple and reliable and light and compact enough that they don't encumber the wearer much and you can carry a second one as a reserve which further improves their reliability. They don't require much maintenance beyond proper storage and for a safety-critical item they aren't really that expensive. </p>
<p>Equally importantly they are pretty easy to use. If you jumped out of a plane wearing one you would have a pretty good chance of surviving, even with minimal instruction. Obviously a moderate amount of training is desirable for dealing with emergencies and reducing the risk of injury on landing but even so there are a significant number of competent amateur parachutists. </p>
<p>Also once a parachute has deployed successfully there is little left to go mechanically wrong </p>
<p>On the other hand rocket packs require significant skill to fly at all they usually launch from a standing position and the pilot stays upright throughout the flight whereas free-fall parachutes are usually deployed from a belly down spreadeagled position which is inherently stable and works nicely with a backpack type parachute harness. </p>
<p>With a limited fuel supply you need to be able to time the deployment of thrust pretty accurately so you have a fairly narrow window to get it right, as opposed to a parachute where any altitude above the minimum safe deployment is basically fine (obviously jumping from very high altitudes has a whole separate set of problems). </p>
<p>Equally a rocket pack is mechanically complex, heavy and bulky and most rockets use fuel which is hazardous and requires special handling. </p>
<p>So really you are talking about replacing a simple, inexpensive and proven system with something which has never really got beyond the prototype stage, doesn't offer any obvious advantages and has several inherent drawbacks. </p>
| 14200 | Rocket to slow freefall |
2017-03-11T20:25:35.910 | <p>For a cantilever beam of circular cross section, loaded at the end, how does the deflection of the free end vary as a function of beam length and diameter?</p>
<p>I expect the answer will be something like:</p>
<p>$$\dfrac{L^2}{D^3}$$</p>
| |beam|deflection| | <p>For a cantilever with a point load at the free end, the deflection at that free end is:</p>
<p>$$\dfrac{WL^3}{3EI}$$</p>
<p><sup>Handbook of Structural Steelwork - 3rd edition</sup></p>
<p>where</p>
<ul>
<li>$W$ = point load</li>
<li>$L$ = beam length</li>
<li>$E$ = Young's modulus</li>
<li>$I$ = second moment of area</li>
</ul>
<p>The Young's modulus depends on the material, see <a href="https://en.wikipedia.org/wiki/Young's_modulus#Approximate_values" rel="nofollow noreferrer">Wikipedia</a> for various examples.</p>
<p>The second moment of area for a circle is:</p>
<p>$$\dfrac{\pi r^4}{4} = \dfrac{\pi d^4}{64}$$</p>
<p>So your overall equation is:</p>
<p>$$\dfrac{64WL^3}{3\pi Ed^4}$$</p>
| 14203 | What is the formula for deflection of an end-loaded cantilever rod beam? |
2017-03-13T16:53:57.200 | <p>So I got a new belt, and it was the right length but the wrong thickness. I needed a 3/8 belt but it was a 1/2. I did go to get it but the guy looked it up and told me I needed this one. So I got it. But when I installed, as I figured, there was a small issue. </p>
<p>So the belt is running up on the "brake" as I call it. It is along the idler and when not engaged this flat metal holds it in place, so the belt does not run when I do not want it to I guess. But the belt is so thick that it is rubbing against this metal when engaged, when the metal is far from the pulley (I think it is called). So it has been burning the belt from friction.</p>
<p>So this is my thought to solving...</p>
<ol>
<li>Let it burn. I am sure it wouldn't take time to burn down to the right thickness. My only problem is that the 'strings' could show. If this is a good idea in general. </li>
<li>Cut that metal off. All I see it doing is keeping the belt from spinning when not engaged. But the belt would be loose were I do not think that would be a problem to begin with. So unless it does have another purpose than to just keep the belt from moving. Couldn't I just cut that off? Then it would not contact the belt and problem solved. Thing I do not like on this is when I do get the proper belt will it be needed then. Side thought: Does this keep the belt from falling off?</li>
</ol>
<p>2nd Update: Another thought is, since the belt is too thick, how could I thin it while it is running? I would have to remove a whole 1/8 inch.</p>
<p>I do not want to return the belt, it was only $7 anyway. So I hope this was understandable and there is a answer to this. </p>
<p>Note: I again, am not sure were to ask a question like this so I wold hope this is fine.</p>
| |automotive-engineering| | <p>A big item to note is a belt is not just a piece of rubber. In fact, typical v-belts and other engineered belting systems have steel wires running through them:</p>
<p><a href="https://i.stack.imgur.com/H7pGE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H7pGE.png" alt="enter image description here"></a></p>
<p>These steel wires do all the real work, but the steel has a low coefficient of friction. It would take a lot of area of wire allow the belt to operate with just the wire alone, without welding the wire in place. The rubber has a high coefficient of friction, and allows the sheaves to grip the belt and let the steel do the work.</p>
<p>Bottom line is, don't modify a belt because you could damage the wires inside. That's why they're sized differently instead of simply being modifiable.</p>
| 14226 | Belt is to thick for my snowblower, how to thin or leave? |
2017-03-13T17:32:27.717 | <p>I have mounted a piezo-ceramic ultrasonic transducer (MODEL # SMD10T2R111WL, Vendor Website: <a href="https://www.steminc.com/PZT/en/piezo-ceramic-disc-10x2mm-r-215-khz-wire-leads-smd10t2r111wl" rel="nofollow noreferrer">PIEZO CERAMIC DISC 10X2MM R 215 KHZ WIRE LEADS</a>) under a plastic tank using epoxy. Please see the below picture. </p>
<p><a href="https://i.stack.imgur.com/FlGC6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FlGC6.jpg" alt="[1]: https://i.stack.imgur.com/sN7Sj.jpg"></a></p>
<p>I am trying to confirm that the wire did not get detached from the transducer while mounting. Is there a way to test it. The two wires that are soldered to the transducer are not connected so there is no way to check the continuity of the wires using multi meter short circuit test. </p>
| |electrical-engineering|sensors| | <p>Apply a small voltage to the piezo from a signal generator, maybe around 1 kHz. If you can hear something, then the wires are connected. A few Volts should be good enough, and won't damage the piezo, especially one that size.</p>
<p>Conversely, you can connect the wires to a scope and see if there are any voltage spikes as you jar the device the piezo is attached to. Be careful that this doesn't cause the piezo to touch anything other what it is already mounted on. Otherwise, the voltage spikes could be due to the leads picking up static charges from different objects. You want to make sure you are getting voltage spikes only from mechanical stress on the piezo.</p>
| 14227 | How to test connectivity between wire and piezo-ceramic transducer |
2017-03-14T10:35:12.160 | <p>I am struggling with the ASME B16.5 code and the specified stud-bolt length that is taken into account in the tables (7, 10, 13, 15, 17, 19 and 21).</p>
<p>If you calculate the stud-bolt length according the "guidelines for stud-bolt calculation" for a NPS 24 Class 150 welding neck flange, the stud-bolt length should be <strong>7.45in</strong>.</p>
<p>When I look in the ASME B16.5 Table7 the length is <strong>6.75in</strong> this is <strong>0.7in</strong> difference with the ASME B16.5.</p>
<hr>
<p><a href="https://i.stack.imgur.com/rL8wX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rL8wX.png" alt="Studbolt length image"></a></p>
<hr>
<p>What are the required number of free threads? As far as I am aware from bolt size 1in and up the number of threads per inch is the same and size <code>S</code> is fixed. If you keep calculating with <code>S = 1/3 bolt diameter</code> the number of free threads will only increase with no use?</p>
| |mechanical-engineering|bolting| | <p>If hydraulic tensioning is being completed you need 3 bolts not 2. Hence the extra length.</p>
| 14236 | Flanged connection studbolt length calculation |
2017-03-15T14:36:50.733 | <p>Is there a way to use a radioactive elements (Americium in my case, but for demo, any radioactive material) to create a usable amount of energy (Power small LED?) AT HOME! Using materials that can be bought from eBay/amazon and the store?</p>
| |power|energy-efficiency|energy|thermal-conduction|radiation| | <p>Take the americium as one lump ball. Surround them with a conductor with a decent (1-2 cm) air gap. Connect the conductor to one side of a capacitor, and the Americium to the the other side of the capacitor. Then connect to a <a href="https://electronics.stackexchange.com/questions/103930/automatic-capacitor-charge-discharge-circuit">circuit</a>. Each alpha particle that decays will increase the negative charge on the Americium side, and will increase the positive charge on the conductor's side.</p>
<p>Once the charge has a sufficient voltage, it should discharge through the capacitor to generate the experiment. Connect to a device that emits a <a href="https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=creepy%20sound%20youtube" rel="nofollow noreferrer">creepy sound</a>, and leave in the bedroom of your worst enemy. Of course, if you really wanted measurable nuclear power at home, I'd suggest <a href="https://www.youtube.com/watch?v=KKdzhPiOqqg" rel="nofollow noreferrer">tritium</a></p>
| 14268 | Radioactive Materials to Energy AT HOME |
2017-03-15T17:19:01.940 | <p>I have begun to study control engineering by several books and awesome <a href="https://www.youtube.com/playlist?list=PLUMWjy5jgHK1NC52DXXrriwihVrYZKqjk" rel="nofollow noreferrer">youtube videos by Brian Douglas</a>. I love them because they give me new understanding of math that I learned before but I still can not get the point of what actually control engineers do. </p>
<p>What is meant by systems here? I mean all examples always are just physical systems like springs and etc, but all of them are governed by physical laws and no matter how dissatisfied with output we are, we are not able to modify physical laws. And in the case of, for instance, mobile robots what do control engineers do? Deriving PID coefficients? I feel like missing the most crucial part of it. </p>
| |control-engineering|control-theory| | <p>In short, 'systems' refers to a combination of components that act together and perform a certain objective. A system can span across different physical and virtual domains.</p>
<p>Controls engineers primarily seek to change these inputs to achieve an optimal response based on their objectives/requirements through means of some process (usually electrical or mechanical actuation).</p>
<p>On a side note, in a commercial setting due to inherent complexities associated with non-isolated systems, engineers usually specialize in a specific process or task (eg. Tuning PID coefficients for different plants). Also controls engineers are sometimes called systems engineers in their designation.</p>
<p>Sources: Modern Control Engineering (K. Ogata) and personal experience. </p>
| 14273 | What do actually control engineers do? |
2017-03-16T10:50:39.227 | <p>I am making a vehicle simulation, but just noticed a big issue - there are a lot of cars with no clutch in them, but I haven't driven a car with automatic/clutchless-manual gearbox, nor been even a passenger in such a car. Consequently, I have no idea how such vehicle drive or operate.</p>
<p>For a vehicle with a clutch, I know how it drives and I also know the physics behind it operation - torque from engine gets transmitted through the clutch to the transmission, then to differential and then it reaches wheels, the wheels generate torque from ground friction which goes the other way around and slows down the engine. If the clutch is disengaged, then the engine won't transfer torque. If the clutch is partially engaged, then some of the engine torque will be transmitted and the rest of it will be lost in some form of heat during clutch slippage.</p>
<p>However, I have no idea how vehicles with an automatic gearbox or a semi-automatic gearbox (e.g. sequential) work. As far as I know, they don't have a clutch. However, I've seen that such vehicles can have their engine revving while not transmitting any torque to the differential.</p>
<p>Is it that they're directly connected to the transmission or is there something else there? What I am asking for is some sort of guidelines for me to figure out how to calculate their maths (plus, if possible, some equations would help).</p>
| |automotive-engineering|motors|torque|simulation| | <p>Conventional automatic transmissions have a torque converter which uses a viscous rather than a rigid coupling to connect input and output. This performs a similar function to a friction clutch but doesn't require manual engagement. Also automatic gearboxes are based on epicyclic gear trains so all the gears are permanently meshed with each other and ratios are selected by locking different combinations of the sun, planet and annular gears so there is no need to match different shaft speeds to each other. </p>
<p>Sequential gearboxes use a combination of straight cut (as opposed to helical) gears and dog clutches with gears engaging by sliding along the input and/or output shaft so the gears never go fully out of mesh with each other but rather are either locked to or freely spinning on their respective shaft. Again this means that it is not necessary to match input and output speeds, however a clutch is still required to start from a standstill. Motorcycles, for example typically have sequential gearboxes and hand operated clutches. </p>
<p>The more modern type of auto/semi-auto gearbox typically uses twin wet plate clutches to change between pre-selected gears in this case the gear selection and clutch engagement is done by actuators which are controlled by a computer. In many cases they can be operated as fully automatic or semi automatics where the driver shifts up or down by operating a paddle or lever. So while they do have a clutch is it not manually operated to change gear. Although, as in a sequential gearbox, manual clutch operation <em>may</em> still be required to start from a standstill often this will be an additional paddle or paddles. Although many semi-auto transmission will also incorporate an anti-stall system which automatically releases the clutch if the revs drop below a certain level at low speed. </p>
<hr>
<p>So from the perspective of the driver there are 3 main cases. </p>
<ul>
<li><p>Full auto (either conventional auto or modern semi auto in full auto mode) : press accelerator pedal to go and the brake to stop and the clutch (or equivalent) sorts itself out automatically eg with automatic transmission if you brake to a stop the engine won't stall so no manual clutch operation required. </p></li>
<li><p>Sequential/semi auto : manual clutch operation is required to start/stop but not to change gear. Often this will be a hand clutch. </p></li>
<li><p>Manual : manual (usually foot pedal) clutch required to start and to change gear. </p></li>
</ul>
<p>These cover the most common systems, although there are various other transmission types around. </p>
<p>Sequential gearboxes are most common on motorcycles and some classes of competition cars (especially those heavily adapted from road cars (eg rallying). </p>
<p>Most conventional road cars have either conventional automatic or manual transmissions. </p>
<p>Computer controlled semi-auto/auto transmissions are mostly found on high-end sports and racing cars eg F1 and GT formulae. </p>
<p>There are also automatic transmissions which allow for an approximation of manual gear selection eg the Porche Tiptronic system and it is also <em>possible</em> to automate a conventional manual transmission, although this is a lot of complexity for no real gain. </p>
<p>For the sake of completeness there are also centrifugal clutches which only engage above a certain engine rpm but these are only really found on things like chainsaws and self propelled mowers. Also steam piston engines don't <em>require</em> a clutch as they can develop torque from zero rpm. </p>
<hr>
<p>It is also worth noting that for a driving simulation, unless you have really good haptic feedback in the control hardware it will be very difficult to simulate the feel of a manual clutch.</p>
| 14283 | Non-Clutch Vehicle Drivetrain - How Does it Work? |
2017-03-17T10:58:26.410 | <p>I have a general question regarding energy coupling in multi phase flow. My question comes is based on the text from this book:</p>
<p><a href="https://books.google.co.uk/books?id=CioXotlGMiYC&pg=PA33&lpg=PA33&dq=thermal+coupling+parameter+continuous+dispersed&source=bl&ots=s9uWpatRmM&sig=-aRSlKoQXHc7gTlZTsl30ROncfo&hl=en&sa=X&ved=0ahUKEwjh6-DfhdzSAhVELMAKHfKOBHAQ6AEIGjAA#v=onepage&q=thermal" rel="noreferrer">https://books.google.co.uk/books?id=CioXotlGMiYC&pg=PA33&lpg=PA33&dq=thermal+coupling+parameter+continuous+dispersed&source=bl&ots=s9uWpatRmM&sig=-aRSlKoQXHc7gTlZTsl30ROncfo&hl=en&sa=X&ved=0ahUKEwjh6-DfhdzSAhVELMAKHfKOBHAQ6AEIGjAA#v=onepage&q=thermal</a> coupling parameter continuous dispersed&f=false</p>
<p>(page 33)</p>
<p>This is kind of obscure so I'll provide a quick background:</p>
<p>Basically, the premise of the question is that there is a flow of fluid, the fluid has a continuous phase, and within the continuous there exists a dispersed phase, so something like water flowing with little droplets of oil inside the water. The idea is that, suppose the water is flowing through a pipe and into a control volume. The coupling parameter will compare the effect of the dispersed phase on the control volume to the continuous phases effect. So, a flow of fluid with a continuous phase containing n drops per volume flows into a cubic control volume of length L (a volume of $L^3$).</p>
<p>Energy coupling is related to temperature. So, in energy coupling the two things that are to be compared are:
The energy (heat) released by the droplets (the convective heat transfer from droplets to water)
The energy provided by the continuous phase (the heat energy of the water)
Thus, the coupling parameter will be:
$$
\Pi_{\text{Energy}} = \frac{\text{Heat Released by Droplets}}{\text{Enthalpy Flux from Water}}
$$
The heat released by the droplets comes from Newtons law of cooling:
$$
mC_d\frac{dT_d}{dt} = Nu k_c \pi D (T_c - T_d)
$$
D is the diameter of the droplet, Tc and Td are the temperatures of the continuous/dispersed phases. Nu is the Nusselt number. Again, there are n droplets per unit volume therefore the total heat released from the droplets is:
$$
nL^3Nu k_c \pi D (T_c - T_d)
$$
The heat flux due to the continuous will just be:
$$
\rho_c u C_c T_c L^2
$$
Which is the density of the continuous phase multiplied by the volume flowing per unit time multiplied by specific heat capacity and it's temperature. The coupling parameter is therefore:
$$
\Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d)}{\rho_c u C_c T_c L^2}
$$</p>
<p>Ok, so that's all fine and dandy. The issue is when they simplify the above formula. So there exists a thermal response time constant (I do not believe it matters where it comes from, this is basically an algebraic exercise at this point):
$$
\tau_T = \frac{\rho_dC_dD^2}{12k_c}
$$
This can be included in the above equation such that :
$$
\Pi_{\text{Energy}} = \frac{nL^3Nu k_c \pi D (T_c - T_d) \rho_d C_d D^2 \tau_T}{\rho_c u C_c T_c L^2 \tau_T }
$$
This can be cleaned up, Nu/2 can be approximated to 1 at low Reynolds numbers. piD^2/6 multiplied by density of a droplet (spherical) gives the mass of that droplet, multiply that by n and you end up with the total droplet density per unit volume. The above equation simplifies to:
$$
\Pi_{\text{Energy}}= \frac{T_dCLC_d}{uC_c T_C \tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)
$$
Where C is the ratio of total dispersed phase density and continuous phase density.</p>
<p>Here is the problem, the book quotes the final result as:
$$
\Pi_{\text{Energy}}= \frac{CL}{u\tau_T } \bigg(1-\frac{T_c}{T_d}\bigg)
$$
Which implies that:
$$
\frac{C_dT_d}{C_c T_c} = 1
$$</p>
<p>I don't see any justification for that though...does anyone know where the extra terms went?</p>
<p>Thanks.</p>
| |fluid-mechanics|heat-transfer|multiphase-flow| | <p>No answers, I agree with Carlton in the comments, some very large liberties have been taken.</p>
| 14291 | Multiphase Flow - Energy Coupling Parameter |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.