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2020-11-20T17:32:35.990 | <p>Is it possible to reduce line-weight (thickness) of border in Solidworks sheet format. I want to reduce the thickness of the lines shown below as I find it too much thick.</p>
<p><a href="https://i.stack.imgur.com/y5OIc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y5OIc.png" alt="enter image description here" /></a></p>
| |solidworks| | <p><strong>Yes</strong></p>
<ol>
<li>Right click in the drawing area, and then click "Edit Sheet Format", to allow you to select the lines you want to adjust</li>
<li>Open the Lines Toolbar - you can search for it if you don't have it visible</li>
<li>Change the line weight using the relevant tool</li>
</ol>
<p>See the .gif below to show this process.</p>
<p><a href="https://i.stack.imgur.com/TCERu.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TCERu.gif" alt="enter image description here" /></a></p>
| 38733 | Reduce line weight of border in Solidworks sheet format |
2020-11-20T19:01:12.483 | <p>ACI 318 7.7 covers requirements for cover when designing rebar placement within concrete. However, they are for general components and makes no exceptions for small irregularities such as lifting hook anchors that one may wish to place. I am not an expert with concrete so I wondered if anyone can point out to me whether there are exceptions to these requirements and whether these minor irregularities are actually required to have the same cover requirements, or are they small enough that it's usually not a significant concern.</p>
<p>Example of hook lift anchors:
<a href="https://meadowburke.com/product/hook-lift-anchor/" rel="nofollow noreferrer">https://meadowburke.com/product/hook-lift-anchor/</a>
<a href="https://meadowburke.com/product/hook-lift-recess-former/" rel="nofollow noreferrer">https://meadowburke.com/product/hook-lift-recess-former/</a></p>
<hr/>
**EDIT:**
<p>So, actually, I am more curious about the recess that I am putting in as you can see within that catalogue (I added a link above). It's obvious that the recess can reduce cover for some of the rebars involved, and I wondered if that will be an issue or whether or not that will create issues later. Any advice?</p>
| |civil-engineering|reinforced-concrete| | <p>Yes, the ACI Standard 318 does specify “Concrete protection for reinforcement” (cover) for various issues, including: 1) standard placement in “principle structural members”, 2) reinforcement in extremely corrosive atmospheres, 3) fire protection, 4) bars, inserts and plates intended for future expansion, 5) precast construction.</p>
<p>In (4) it makes a statement that it must provided “adequate covering”. It states that it’s for protection against corrosion, but could imply bonding too. (See ACI-318 808 (f). Note: I have an old edition from 1963 (I’ve been doing this awhile), but I’m sure it’s similar to current editions.)</p>
<p>In your case, if the inserts cannot be placed such that they leave adequate “cover”, to the nearest reinforcing, then adjustments must be made. I’d say that includes recessed pockets that could allow “standing water”.</p>
| 38736 | Concrete rebar cover requirements - Hook lift anchors |
2020-11-20T22:04:25.050 | <p>I need help with this exercise. I 've tried many times with pen and paper but I'm stuck. I don't want a full answer but some feedback .
<a href="https://i.stack.imgur.com/NGr98.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NGr98.png" alt="enter image description here" /></a></p>
<p>Here is my progess:</p>
<p><span class="math-container">$$\frac{dx}{dt} = Ax(t) + Bu(t) \Rightarrow sX(S) = AX(S) + BU(S) \Rightarrow \\\Rightarrow X(s)=(sI-A)^{-1}BU(S) \ \ \ (1)$$</span><br />
Now let's note as <span class="math-container">$$G(t) = \frac{x(t)}{u(t)} $$</span> So : <span class="math-container">$$G(s)= (sI-A)^{-1}B $$</span>
Then we can write: <span class="math-container">$$\frac{U(s)}{R(s)} = \frac{1}{1+G(s)k} $$</span><br />
Now , we need a zero - steady state: <span class="math-container">$$\lim_{t \rightarrow \infty}u(t) = \lim_{s \rightarrow 0}sU(s) =0$$</span>
<span class="math-container">$$sU(s) = s\frac{1}{1+G(s)k}R(s)$$</span> but <span class="math-container">$$R(S) = \frac{1}{s}$$</span> so<br />
<span class="math-container">$$sU(s) = \frac{1}{1+G(s)k}$$</span><br />
This can happen , iff : <span class="math-container">$$\lim_{s \rightarrow 0}G(s)k =\infty$$</span><br />
<span class="math-container">$$\lim_{s \rightarrow 0}((sI-A)^{-1}B)k)=\infty \rightarrow \lim_{s \rightarrow 0}(\frac{adj(sI-A)}{det(sI-A)}Bk)=\infty \rightarrow \lim_{s \rightarrow 0}adj(sI-A)Bk=\infty$$</span><br />
But <span class="math-container">$$(sI-A)= \begin{bmatrix}
s+7 & -10\\
1 & s
\end{bmatrix} \rightarrow adj(sI-A) = \begin{bmatrix}
s & 10\\
-1 & s+7
\end{bmatrix}$$</span> and<br />
<span class="math-container">$$\begin{bmatrix}
s & 10\\
-1 & s+7
\end{bmatrix}\begin{bmatrix}
1\\
0
\end{bmatrix}k=\infty$$</span> ...
And here is my struggle...</p>
| |control-engineering|control-theory|transfer-function|feedback-loop| | <p>Let's start by obtaining the state space form of the closed-loop system (closed loop means that you plug in the equations the expression of the controller). The controller of this specific system has the following form:</p>
<p><span class="math-container">$$ u = -Kx+r $$</span>
This is a full state feedback controller with feedforward gain of <span class="math-container">$1$</span> (feedforward is the gain by which the input signal is being multiplied). So, the state space equations are now:</p>
<p><span class="math-container">$$ \dot{x} = (A-BK)x+r $$</span>
<span class="math-container">$$ y = Cx $$</span>
Since the system is a classic linear one, convert it into the corresponding <span class="math-container">$\text{s-domain}$</span> representation aka transfer function. The transformation from state space to transfer function is:</p>
<p><span class="math-container">$$ \frac{Y(s)}{R(s)} = c(sI-(A-BK))^{-1}B $$</span>
Plugging in the values of the matrices and doing the math (I leave them for you) but leaving the gain vector as <span class="math-container">$K =\begin{bmatrix} k_1 & k_2 \end{bmatrix}$</span> yields the following closed loop transfer function of the system:</p>
<p><span class="math-container">$$T(s) = \frac{Y(s)}{R(s)} = \frac{1}{s^2+(7+k_1)s+(10+k_2)} $$</span>
In order for the steady state error to be <span class="math-container">$e_{ss}=0$</span>, we impose the requirement for the DC gain of the system to be equal to <span class="math-container">$1$</span>. This is derived from the final value theorem:</p>
<p><span class="math-container">$$ \lim_{t \to \infty}{y(t)} = \lim_{s \to 0}sY(s) $$</span></p>
<p>and in order to achieve satisfactory tracking we want :</p>
<p><span class="math-container">$$ sY(s) \approx sR(s) \text{ as } s\rightarrow0 \Rightarrow \left. \frac{Y(s)}{R(s)}\right\vert_{s=0} = 1 $$</span></p>
<p>The DC gain of the system is the value by which the input of the system is multiplied (and produces the output) as <span class="math-container">$t \rightarrow \infty$</span> or in <span class="math-container">$\text{s-domain}$</span> notation as <span class="math-container">$s\rightarrow 0$</span>. Inserting the value <span class="math-container">$s=0$</span> into the closed loop transfer function results to:</p>
<p><span class="math-container">$$\left. \frac{Y(s)}{R(s)}\right\vert_{s=0} = \frac{1}{10+k_2} = \text{DC-Gain} $$</span>
And as stated above, we want:</p>
<p><span class="math-container">$$ \text{DC-Gain}=1 \Rightarrow \frac{1}{10+k_2} = 1 \Rightarrow k_2 = -9 $$</span>
Obviously, only the gain <span class="math-container">$k_2$</span> influences the steady state error of the system. You can verify that by noticing that at the denominator of the closed loop transfer function the gain <span class="math-container">$k_1$</span> belongs to the expression which is multiplied by <span class="math-container">$s$</span> and as a result at steady state (<span class="math-container">$s\rightarrow 0$</span>) this expression will vanish.</p>
<p>However, you can't conclude that any value of <span class="math-container">$k_1$</span> will be ok because you have to consider the stability of the system. In order for a second order system to be stable the coefficients of the characteristic polynomial (denominator of closed loop transfer function) have to be positive:</p>
<p><span class="math-container">$$ 7+k_1 > 0 \Rightarrow k_1 > -7 $$</span>
<span class="math-container">$$ 10 + k_2 > 0 \Rightarrow k_2 > -10 $$</span>
Even though, you only have one possible choice where <span class="math-container">$k_2=-9$</span>, it is always good to check everything that's essential for your system to function properly, Stability is, if not the most, among the most important characteristics.</p>
<p>To sum up, the final closed loop transfer function is:</p>
<p><span class="math-container">$$ T(s) = \frac{1}{s^2+10s+1} $$</span></p>
<p>Notice that if you put <span class="math-container">$s=0$</span> at this equation you get:</p>
<p><span class="math-container">$$\left. T(s)\right\vert_{s=0} = \frac{1}{1} = 1 = \text{DC-Gain} $$</span>
which yields <span class="math-container">$e_{ss} = 0$</span> and a stable system with negative poles at <span class="math-container">$p=\begin{bmatrix}-9.899 & -0.1010\end{bmatrix}^T$</span>. The step response of the system proves them also graphically:</p>
<p><a href="https://i.stack.imgur.com/6edWT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6edWT.png" alt="enter image description here" /></a></p>
| 38739 | Feedback loop and zero - error in steady state |
2020-11-21T14:30:50.690 | <p>I am working on trying to determine the velocity of one end of a rod that is pinned on both ends into two different tracks. One track is linear and the pinned end has known values. The other side is pinned to a circular track in which no values are known.</p>
<p><a href="https://i.stack.imgur.com/pMuWl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pMuWl.png" alt="Rod pinned into two different tracks" /></a></p>
<p>I have attempted to obtain the angular velocity first as I am using
<span class="math-container">$$ V_B=V_A+V_{B/A} $$</span></p>
<p>by using
<span class="math-container">$$ \omega = v/r $$</span></p>
<p>I have also noted to obtain <span class="math-container">$ V_{B/A} $</span> I need to find <span class="math-container">$ \dot\theta $</span> which equals <span class="math-container">$ \omega $</span></p>
<p><span class="math-container">$ V_{B/A} $</span> has the equation of
<span class="math-container">$$ (\dot\theta\hat k) \times I*(cos\theta\hat i - sin\theta\hat j) $$</span></p>
<p>I am stuck at finding <span class="math-container">$\theta$</span> or <span class="math-container">$ V_{B/A}$</span> because it seems both need the other in order to get an answer.</p>
<p>Does anyone have a suggestion of equations to look at to obtain one of those two values with the information known?</p>
<p>Could I treat point A is being the center of the circle and use the formula of <span class="math-container">$ V_A = r\omega $</span> to obtain the value for <span class="math-container">$\omega$</span></p>
| |dynamics| | <p>Calculate <span class="math-container">$\varphi_0 = \arctan(s)$</span>. Then the position of point <span class="math-container">$A$</span> is
<span class="math-container">\begin{align}
&X_A = X_A\\
&Y_A = s\,X_A + b
\end{align}</span>
The position of point <span class="math-container">$B$</span> can be expressed in terms of the angle <span class="math-container">$\theta$</span> as
<span class="math-container">\begin{align}
&X_B = X_A + l\cos(\theta + \varphi_0)\\
&Y_B = b + s\,X_A + l\sin(\theta + \varphi_0)
\end{align}</span>
However, we have a restricition for point <span class="math-container">$B$</span>, called a holonomic constrant, which is that <span class="math-container">$B$</span> always moves along the circle <span class="math-container">$(X-c)^2 + Y^2 = r^2$</span>, Therefore <span class="math-container">$(X_B-c)^2 + Y_B^2 = r^2$</span> which explicitly is
<span class="math-container">$$\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \,
+ \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2$$</span>
Thus, the position of the bar, moving so that point <span class="math-container">$A$</span> is always on the line <span class="math-container">$Y = sX + b$</span> and point <span class="math-container">$B$</span> is always on the circle <span class="math-container">$(X-c)^2 + Y^2 = r^2$</span> can be described by the three equations
<span class="math-container">\begin{align}
&X_B = X_A + l\cos(\theta + \varphi_0)\\
&Y_B = b + s\,X_A + l\sin(\theta + \varphi_0)\\
&\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \,
+ \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2
\end{align}</span>
Hence, if you know the way <span class="math-container">$X_A = X_A(t)$</span> changes with respect to time <span class="math-container">$t$</span>, then you can blug it in the third equation and solve it for <span class="math-container">$\theta = \theta(t)$</span>. After you have found <span class="math-container">$\theta$</span> you can plug it alongisde <span class="math-container">$X_A$</span> in the first two equations to find the coordinates <span class="math-container">$(X_B, \, Y_B)$</span> of <span class="math-container">$B$</span>.</p>
<p>To find the angular velocity <span class="math-container">$\frac{d\theta}{dt}$</span> of the bar, you simply differentiate the third equation with respect to <span class="math-container">$t$</span> and add the new differentiated equation to the system, as a forth equation:
<span class="math-container">\begin{align}
&X_B = X_A + l\cos(\theta + \varphi_0)\\
&Y_B = b + s\,X_A + l\sin(\theta + \varphi_0)\\
&\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \,
+ \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2\\
&\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)
\left(\,\frac{dX_A}{dt} - l\sin(\theta + \varphi_0) \frac{d\theta}{dt}\,\right) = \, \\
&+ \,\big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)
\left(\,s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0) \frac{d\theta}{dt}\,\right) \, = \, 0
\end{align}</span>
To find <span class="math-container">$\frac{d\theta}{dt}$</span> you need only the last two equations:
<span class="math-container">\begin{align}
&\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)^2 \,
+ \, \big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)^2 \, = \, r^2\\
&\big( \, X_A + l\cos(\theta + \varphi_0) - c \,\big)
\left(\,l\sin(\theta + \varphi_0) \frac{d\theta}{dt} - \frac{dX_A}{dt}\,\right) \, = \,\big(\,b + s\,X_A + l\sin(\theta + \varphi_0)\,\big)
\left(\,s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0) \frac{d\theta}{dt}\,\right)
\end{align}</span>
Given <span class="math-container">$X_A = X_A(t)$</span> and <span class="math-container">$V_A = V_A(t) = \frac{dX_A}{dt}$</span>, you can take the first equation from the latter system of two equations, plug <span class="math-container">$X_A$</span> in it and solve for <span class="math-container">$\theta = \theta(t)$</span>. This equation is the hardest to solve. After that, plug in the second equation <span class="math-container">$X_A, \, \theta,\, \frac{dX_A}{dt}$</span> and solve for the angular speed <span class="math-container">$\frac{d\theta}{dt}$</span>.</p>
<p>Finally, to find the velocity of <span class="math-container">$B$</span>, you simply take the first two equations of the four equation system above and differentiate them with respect to <span class="math-container">$t$</span>:
<span class="math-container">\begin{align}
&V_{x,B} = \frac{dX_B}{dt} = \frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\\
&V_{y,B} = \frac{dY_B}{dt} = s\,\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt}
\end{align}</span>
So, you just have to plug in this equation the already calculated <span class="math-container">$\theta, \,\frac{d\theta}{dt}$</span> and <span class="math-container">$\frac{dX_A}{dt} = V_A$</span>.</p>
<p><span class="math-container">$$$$</span></p>
<p><span class="math-container">$$$$</span></p>
<p><strong>OLD VERSION.</strong>
Let's simplify things a bit. First, perform the translation:
<span class="math-container">\begin{align}
&X = \tilde{x} + c \\
&Y = \tilde{y}
\end{align}</span>
Then the equation of the circle becomes
<span class="math-container">$$r^2 = (X - c)^2 + Y^2 = \tilde{x}^2 + \tilde{y}^2$$</span>
Then find the angle between the line <span class="math-container">$Y = sX + b$</span>, which in new coordinates is <span class="math-container">$\tilde{y} = s\,\tilde{x} + (sc + b)$</span>, and the horizontal axis: the slope is the tangent of that angle, i.e. <span class="math-container">$$\varphi_0 = \arctan(s)$$</span>
Next, perform a clock-wise rotation of angle <span class="math-container">$\varphi_0$</span> so that the line <span class="math-container">$\tilde{y} = s\,\tilde{x} + (sc + b)$</span> becomes a line <span class="math-container">$\tilde{y} = h$</span> (one can calculate the distance <span class="math-container">$h$</span> between the center of the circle (the origin) and the line in question) parallel to the horizontal <span class="math-container">$x-$</span>axis:
<span class="math-container">\begin{align}
\tilde{x} = \cos(\varphi_0)\, x \, - \, \sin(\varphi_0)\, y\\
\tilde{y} = \sin(\varphi_0)\, x \, + \, \cos(\varphi_0)\, y
\end{align}</span>
Denote by <span class="math-container">$x_A$</span> the <span class="math-container">$x-$</span>coordinate of the point <span class="math-container">$A$</span> moving along the line. The <span class="math-container">$y-$</span>coordinate is <span class="math-container">$h$</span> and is fixed. The equation of the upper half of the circle in these new rotated and translated coordinates can be written as
<span class="math-container">$$y = \sqrt{r^2 - x^2}$$</span><br />
If <span class="math-container">$\theta$</span> is the angle between the rod <span class="math-container">$AB$</span> and the line <span class="math-container">$y = h$</span>, which is parallel to the <span class="math-container">$x-$</span>axis, then the equations for the position of the other end of the rod, point <span class="math-container">$B$</span>, are
<span class="math-container">\begin{align}
&{x}_B = x_A + l\, \cos(\theta)\\
&{y}_B = \sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}
\end{align}</span> Observe, there are two free parameters for the position of <span class="math-container">$B$</span> on the circle, namely <span class="math-container">$x_A$</span> and <span class="math-container">$\theta$</span>. However, there is another restriction - the distance between <span class="math-container">$A$</span> and <span class="math-container">$B$</span> is always <span class="math-container">$l$</span>. Hence:
<span class="math-container">$$\big(x_B - x_A\big)^2 + \big(y_B - y_A\big)^2 = l^2$$</span> or after substitutions
<span class="math-container">$$l^2 \cos^2(\theta) \, + \, \Big(\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h\Big)^2 \, = \, l^2$$</span> which establishes a link between the coordinates <span class="math-container">$x_A$</span> and <span class="math-container">$\theta$</span>. You can move the first term from the lefthand side to the right one, then apply a central trigonometric identity to the right hand side, after which you can take square root on both sides, and finally obtain the simplified equation
<span class="math-container">$$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, \pm \, l\sin(\theta)$$</span> where you should have in mind the sign <span class="math-container">$\pm$</span> depends on the sign of the right hand side. On your picture, <span class="math-container">$\theta \in [0, \pi/2)$</span> so you can choose a plus sign and the equation is
<span class="math-container">$$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)$$</span><br />
Now, in this latter equation <span class="math-container">$x_A = x_A(t)$</span> and <span class="math-container">$\theta = \theta(t)$</span> are function of the time <span class="math-container">$t$</span>, so we can differentiate the equation with respect to <span class="math-container">$t$</span> and pair it with the latter equation above:
<span class="math-container">\begin{align}
&\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)\\
&\frac{\big(x_A + l\cos(\theta)\big)}{\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}}\left(l\sin(\theta)\frac{d\theta}{dt} \, - \, \frac{dx_A}{dt}\right) \, = \, l\cos(\theta) \frac{d\theta}{dt}
\end{align}</span>
You can simplify the second equation, using the first one, by solving for the square root <span class="math-container">$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}$</span> and write the system as follows:
<span class="math-container">\begin{align}
&\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2\,} - h \, = \, l\sin(\theta)\\
&\frac{\, x_A + l\cos(\theta)\,}{h \, + \, l\sin(\theta)}\left(l\sin(\theta)\frac{d\theta}{dt} \, - \, \frac{dx_A}{dt}\right) \, = \, l\cos(\theta) \frac{d\theta}{dt}
\end{align}</span>
This system of equations features four variables:
<span class="math-container">$$x_A, \, \theta, \, \frac{dx_A}{dt}, \, \frac{d\theta}{dt}$$</span>
So, if you are given any two of these, you can solve the system and find the other two. For example, if you know the position and velocity of <span class="math-container">$A$</span>, then you know <span class="math-container">$x_A$</span> and <span class="math-container">$ \frac{dx_A}{dt}$</span>. Then, you can plug <span class="math-container">$x_A$</span> in the first equation and solve that same first equation for <span class="math-container">$\theta$</span>. Then, knowing already <span class="math-container">$x_A, \, \theta, \, \frac{dx_A}{dt}$</span>, you can plug these three values in the second equation and solve it for the angular velocity <span class="math-container">$\frac{d\theta}{dt}$</span>. This second equation is easier to solve with respect to <span class="math-container">$\frac{d\theta}{dt}$</span> because it is linear with respect to <span class="math-container">$\frac{d\theta}{dt}$</span>.</p>
<p>The next step is to find the linear velocity of <span class="math-container">$B$</span>, which should be tangent to the circle. If you take the equations
<span class="math-container">\begin{align}
&{x}_B = x_A + l\, \cos(\theta)\\
&{y}_B = \sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2}
\end{align}</span>
By the first equation from the system of equations discussed above, you can express <span class="math-container">$\sqrt{r^2 - \big(x_A + l\, \cos(\theta)\big)^2} = l\sin(\theta) + h$</span>
and re-write the latter parametrization as follows:
<span class="math-container">\begin{align}
&{x}_B = x_A + l\, \cos(\theta)\\
&{y}_B = l\sin(\theta) + h
\end{align}</span>
To find the linear velocity of <span class="math-container">$B$</span>, you just have to differentiate the latter parametrization with respect to <span class="math-container">$t$</span>
<span class="math-container">\begin{align}
&\frac{dx_B}{dt} = \frac{dx_A}{dt} - l\, \sin(\theta)\frac{d\theta}{dt}\\
&\frac{dy_B}{dt} = l \, \cos(\theta)\frac{d\theta}{dt}
\end{align}</span><br />
plug the already determined values of <span class="math-container">$\frac{dx_A}{dt}, \, \theta, \, \frac{d\theta}{dt}$</span>.</p>
| 38749 | How to determine the linear and angular velocity of one end of a rod that is pinned on a track on each side? |
2020-11-21T15:30:18.287 | <p>Presumably, temperature is a function of height, so the x-axis should be the height not the temperature, but graphs like this tend to be the norm:</p>
<p><a href="https://i.stack.imgur.com/weGTh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/weGTh.png" alt="standard atmosphere" /></a></p>
| |aerospace-engineering| | <p>Because it is more intuitive to see altitude on vertical axis.</p>
<p>In many graphs they use the altitude axis, to show different strata of the atmosphere that are naturally graduated vertically.</p>
<p>like this graph.</p>
<p><a href="https://i.stack.imgur.com/k96XK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k96XK.jpg" alt="universal Atmosphere" /></a></p>
| 38751 | Why does the graph of the Standard Atmosphere have temperature on the x-axis? |
2020-11-21T17:25:47.783 | <p>I appreciate this may not be the best forum to pose this question but...</p>
<p>I've been attempting to model a copy of a Fender Telecaster electric guitar in SketchUp. I've been mainly working from a PDF document which provides the measurements for the hardware routing and the neck pocket but provides nothing regarding the measurements and radii of the actual body shape.<br />
I've had a good look around online but am struggling to find anything.</p>
| |measurements| | <p>I have worked with Autocad and ACAD3DS, but I know SketchUp has the same capability.</p>
<p>In Autocad, you can import an image and scale it, then the program will turn it into a vector file that you can block and edit or add to your drawing. You can use a photo of the model you're interested in and turn it into a vector block.</p>
<p>There are online apps that convert JPG to DWG like this:</p>
<p><a href="https://anyconv.com/jpg-to-dwg-converter/" rel="nofollow noreferrer">JPG to DWG</a></p>
| 38754 | Fender Telecaster outer body measurements |
2020-10-21T18:25:48.300 | <p>I am making a sandwich design. I designed two parts that fit well together and assembled on Solidworks.</p>
<p>Top part:
<a href="https://i.stack.imgur.com/adgDa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/adgDa.png" alt="enter image description here" /></a></p>
<p>bottom part:
<a href="https://i.stack.imgur.com/9pSGU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9pSGU.png" alt="enter image description here" /></a></p>
<p>The two parts are attached like two LEGO pieces. I want to pump water through one of the middle holes at the top part and exist through the other. My first plan is to glue the two-parts together. However, I feel that water will leak at the points where the two parts meet. Is there a smarter method in the 3D printing community ?</p>
| |seals| | <p>Without a seal or glue, you can't make two-object prints watertight.</p>
<h2><a href="https://en.wikipedia.org/wiki/Face_seal" rel="nofollow noreferrer">face seal</a></h2>
<p>A face seal is a compressible item that gets put under pressure and as a result, prevents the fluid from creeping through crevices. Rubber o-rings are a typical seal of this type, as are gaskets, but so is rubber (tube or string) in a dedicated indention or a <a href="https://en.wikipedia.org/wiki/Gasket" rel="nofollow noreferrer">gasket</a>. Another type would be a <a href="https://en.wikipedia.org/wiki/Stuffing_box" rel="nofollow noreferrer">Gland aka stuffing box</a> or grease grove, which is then stuffed with a <a href="https://en.wikipedia.org/wiki/Sealant" rel="nofollow noreferrer">sealant</a> - this could be string, hair, <a href="https://en.wikipedia.org/wiki/Grease_(lubricant)" rel="nofollow noreferrer">grease</a>, or any type of other flexible material that reduces flow through by adhering to both sides and staying in the groove. With some designs, the gland can be packed with a fluid or viscous sealant after closing the item into its final position, for example by use of a syringe through a hole or a grease gun through a nipple.</p>
<p>Note that each of these designs has maximum pressures it can hold against.</p>
<p><a href="https://i.stack.imgur.com/xQV4h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xQV4h.png" alt="enter image description here" /></a></p>
<h2><a href="https://en.wikipedia.org/wiki/Labyrinth_seal" rel="nofollow noreferrer">Labyrinth Seal</a></h2>
<p>A Labyrinth seal is a variant of a face seal, where two parts seal against one another by having many interlocking surfaces. A screw is pretty much a labyrinth seal. Most labyrinth seals are also packed with a sealant and, if around a spinning shaft, made from 2 halves that have a face seal against one another but only the (grease packed) labyrinth seal to the shaft.</p>
<h2>Permanent sealant</h2>
<p>A permanent sealant is a subtype of a sealant that can't be removed after installation. Some sealants in glands are permanent. The most common type of permanent sealant is glue. To pack them into the sealant gland from the outside, a syringe would be the best way. Depending on the pressure expected and the curing time of the sealant, just having a set of interlocking shallow grooves can be enough. Those then are coated before pressing the two halves together. The volume of the sealant needs to be taken into account in the design!</p>
| 38759 | How to prevent water from leaking? |
2020-11-21T22:04:25.600 | <p>I am building a new control system for an injection moulding machine, the last part I am trying to work out is the main pump.</p>
<p>It has two proportional valves on the output, one labelled throttle valve and one labelled pressure limiting valve.</p>
<p>The throttle valve is connected between the pump outlet and the pressure limiting valve is connected between the pump control line and the tank return.</p>
<p>I think that you use the throttle valve to control the pump pressure and use the pressure valve to control the bypass pressure which is just like an energy saving thing. Is this the correct way of thinking?</p>
<p>The main question is though was voltage is do I use to control the valves and what is the pinout for the 3 pin feedback connector on the end on each valve?</p>
<p>Pictures of the pump and valves:
<a href="https://i.stack.imgur.com/Zd6ba.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zd6ba.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/ZVGLL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZVGLL.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/lPfpj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lPfpj.jpg" alt="enter image description here" /></a></p>
| |electrical-engineering|control-engineering|hydraulics| | <p>After some research, I found out that the valve is 0-24v DC and the output from the feedback sensor is 0-20ma. They are designed to work with a Bosch controller but can be used with any PLC if an amplifier with the correct gain is used.</p>
| 38763 | What voltage are these hydraulic valves and how to I control them? |
2020-11-22T07:11:02.660 | <p>What am I measuring if I set my multimeter to AC voltage readings and take the black probe out of COM? It seems that I am getting higher readings when I go near a light or my computer screen. This measurement seems to get more sensitive (I think) when I make a circular loop with an ordinary copper wire and connect the red probe with the loop.</p>
<p>What am I measuring in the two cases? Are they measuring the same thing?</p>
| |electrical-engineering| | <p>You are measuring the electric field in the air. Your multimeter has a very high input impedance - typically 10 MΩ. If you connect a low resistance - even 1 kΩ - between the V and COM inputs you will find that the reading collapses.</p>
| 38767 | Multimeter with only one probe |
2020-11-23T05:49:14.827 | <p>I am looking at a rod that has an acting force along one end that's pinned into a frictionless track. The other end is pinned into a track that has friction. I am trying to determine the contact forces at each end of the rod, as well as the friction force on the end that's in the frictioned track.</p>
<p><span class="math-container">$$ \vec{\dot{\omega}} = 0.1 $$</span>
<a href="https://i.stack.imgur.com/uqJlN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uqJlN.png" alt="enter image description here" /></a></p>
<p>I am looking at using the equations <span class="math-container">$ M_{cm} = I_{cm} $</span> and <span class="math-container">$ F = mA_{cm} $</span></p>
<p>I've worked the force equations to equal
<span class="math-container">$$ F_x = mA_{cm} $$</span>
<span class="math-container">$$ 5-cos(\phi)F_b = sin(27.65)98.1N $$</span></p>
<p>and</p>
<p><span class="math-container">$$ F_y = mA_{cm} $$</span>
<span class="math-container">$$ N_A + sin(\gamma)N_B+cos(\phi)=cos(27.65)*98.1 $$</span></p>
<p>and the momentum equation</p>
<p><span class="math-container">$$ M_{cm} = I_cm * \vec{\dot{\omega}} $$</span>
<span class="math-container">$$ M_{cm} = 10 * 0.1 $$</span></p>
<p>I'm having difficulties figuring out the angle for <span class="math-container">$ F_B$</span> as well as determining how <span class="math-container">$M_{cm}$</span> is playing a role in here at all.</p>
<p>Any suggestions would be appreciated.</p>
<p><span class="math-container">$$ I = 1.41 $$</span>
<span class="math-container">$$ s = 0.5 $$</span>
<span class="math-container">$$ b = -0.5 $$</span>
<span class="math-container">$$r = 1$$</span>
<span class="math-container">$$c = 2$$</span>
<span class="math-container">$$X_A = 1.25$$</span>
<span class="math-container">$$V_A = .5$$</span>
<span class="math-container">$$A_A= 0.25
$$</span></p>
| |dynamics| | <p>Maybe something like this:</p>
<p><strong>Step 1:</strong> By the <a href="https://engineering.stackexchange.com/questions/38749/how-to-determine-the-linear-and-angular-velocity-of-one-end-of-a-rod-that-is-pin">previous answer</a>, the coordinate <span class="math-container">$X_A = X_A(t)$</span> and the angle <span class="math-container">$\theta = \theta(t)$</span> are connected by the equation
<span class="math-container">$$\big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)^2 \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)^2 \, = \, r^2$$</span>
Knowing <span class="math-container">$X_A$</span>, plug it in the equation and solve for <span class="math-container">$\theta$</span>.</p>
<p><strong>Step 2:</strong> Differentiate the equation from step 1 with respect to <span class="math-container">$t$</span> and obtain the equation
<span class="math-container">$$\big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)\left(\frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)\left(s\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt}\right) = 0$$</span>
Knowing <span class="math-container">$X_A, \, \theta, \, \frac{dX_A}{dt} = V_A$</span>, plug them in the equation and solve for <span class="math-container">$\frac{d\theta}{dt}$</span>.</p>
<p><strong>Step 3:</strong> Differentiate the equation from step 2 with respect to <span class="math-container">$t$</span> and obtain the equation (which you should calcualte):
<span class="math-container">$$\frac{d}{dt} \left( \, \big(\,X_A + l \cos(\theta + \varphi_0) - c\,\big)\left(\frac{dX_A}{dt} - l\sin(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, + \, \big(\,sX_A + l \sin(\theta + \varphi_0) + b\,\big)\left(s\frac{dX_A}{dt} + l\cos(\theta + \varphi_0)\frac{d\theta}{dt}\right) \, \right) = 0$$</span> for <span class="math-container">$\frac{d\theta}{dt}$</span>.
This will be an equation for <span class="math-container">$X_A, \theta, V_A = \frac{dX_A}{dt}, \, \frac{d\theta}{dt}, \, A_A = \frac{d^2X_A}{dt^2}$</span> and <span class="math-container">$\frac{d^2\theta}{dt^2}$</span>. Given <span class="math-container">$X_A, \theta, V_A = \frac{dX_A}{dt}, \, \frac{d\theta}{dt}, \, A_A = \frac{d^2X_A}{dt^2}$</span>, plug them in the equation and solve for <span class="math-container">$\frac{d^2\theta}{dt^2}$</span>.</p>
<p><strong>Step 4:</strong> Calculate the coordinates of the point <span class="math-container">$B$</span> on the circle:
<span class="math-container">\begin{align}
&X_B = X_A + l\cos(\theta + \varphi_0) \\
&Y_B = sX_A + l\sin(\theta + \varphi_0) + b
\end{align}</span></p>
<p><strong>Step 5:</strong> Calculate the coordinates of the unit vectors <span class="math-container">$\vec{n}_A$</span> and <span class="math-container">$\vec{n}_B$</span> at points <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, perpendicular to the line and the circle respectively. For <span class="math-container">$\vec{n}_A$</span>:
<span class="math-container">\begin{align}
&n_{x,A} = \frac{-s}{\sqrt{1+s^2}} \\
&n_{y,A} = \frac{1}{\sqrt{1+s^2}}
\end{align}</span>
and for <span class="math-container">$\vec{n}_B$</span>:
<span class="math-container">\begin{align}
&n_{x,B} = \frac{X_B - c}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{X_B - c}{r}\\
&n_{y,B} = \frac{Y_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{Y_B}{r}
\end{align}</span></p>
<p><strong>Step 6:</strong> Calculate the coordinates of the unit vectors <span class="math-container">$\vec{t}_A$</span> and <span class="math-container">$\vec{t}_B$</span> at points <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, tangent to the line and the circle respectively. For <span class="math-container">$\vec{t}_A$</span>:
<span class="math-container">\begin{align}
&t_{x,A} = \frac{1}{\sqrt{1+s^2}} \\
&t_{y,A} = \frac{s}{\sqrt{1+s^2}}
\end{align}</span>
and for <span class="math-container">$\vec{t}_B$</span>:
<span class="math-container">\begin{align}
&t_{x,B} = \frac{Y_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{Y_B}{r}\\
&t_{y,B} = \frac{c - X_B}{\sqrt{(X_B - c)^2 + Y_B^2}} = \frac{c - X_B}{r}
\end{align}</span></p>
<p><strong>Step 7:</strong> Calculate the acceleration of point <span class="math-container">$B$</span>, by differentiating twice with respect to time <span class="math-container">$t$</span> the equation for the coordinates <span class="math-container">$X_B, \, Y_B$</span>:
<span class="math-container">\begin{align}
&A_{x,B} = \frac{d^2X_A}{dt^2} = A_A - l\sin(\theta + \varphi_0)\frac{d^2\theta}{dt^2} - l\cos(\theta + \varphi_0)\left(\frac{d\theta}{dt}\right)^2\\
&A_{y,B} = \frac{d^2Y_A}{dt^2} =s\,A_A + l\cos(\theta + \varphi_0)\frac{d^2\theta}{dt^2} - l\sin(\theta + \varphi_0)\left(\frac{d\theta}{dt}\right)^2
\end{align}</span>
Since, up to know, you are either given or have calculated the variables <span class="math-container">$A_A, \, \theta, \, \frac{d\theta}{dt}, \, \frac{d^2\theta}{dt^2}$</span>, you can plug them in the equations and calculate the coordinates <span class="math-container">$A_{x,B}, \, A_{y,B}$</span> of the acceleration vector <span class="math-container">$\vec{A}_B$</span> of point <span class="math-container">$B$</span>. The coordinates of the acceleration vector <span class="math-container">$\vec{A}_A$</span> of point <span class="math-container">$A$</span> are <span class="math-container">$A_{A}, \, sA_{A}$</span>.</p>
<p><strong>Step 8:</strong> Finally put together the linear system of three scalar equations and three unknown force magnitudes <span class="math-container">$N_A, \, N_B, \, F_B$</span>:
<span class="math-container">\begin{align}
& \left(\frac{ml^2}{12}\right) \, \frac{d^2\theta }{dt^2} \, \hat{k} \, = \, \frac{1}{2} \, \big(l \, \cos(\theta + \varphi_0)\hat{i} + l \, \sin(\theta + \varphi_0)\hat{j}\big) \times \left(\,F_B \vec{t}_B + N_B\vec{n}_B - P_A \vec{t}_A - N_A \vec{n}_A\,\right)\\
&\frac{m}{2} \big(\vec{A}_A + \vec{A}_B\big) = F_B \vec{t}_B + N_B\vec{n}_B + P_A \vec{t}_A + N_A \vec{n}_A
\end{align}</span> and since you already know all the other parameters in this system, including <span class="math-container">$P_A$</span> which is given, solve for <span class="math-container">$N_A, \, N_B, \, F_B$</span>. Here <span class="math-container">$\hat{i}, \, \hat{j}, \, \hat{k}$</span> are the pairwise orthogonal unit vectors of the inertial coordinate system of the system. The vector <span class="math-container">$\hat{k}$</span> is perpendicular to the 2D picture.</p>
| 38782 | Determining Friction and Contact forces |
2020-11-23T15:21:01.280 | <p>I have a vector <span class="math-container">$\omega$</span> and it associated covariance, a rotation <span class="math-container">$\mathbf R$</span> and its associated covariance. What is the covariance of <span class="math-container">$\mathbf R \cdot \omega$</span> ?</p>
<p>More rigorously:
I have an estimation of a vector expressed in the <span class="math-container">$A$</span> frame: <span class="math-container">$^A \mathbf \omega \in \mathbb R^3$</span>, and its associated covariance <span class="math-container">$\Sigma_{^A\omega} \in \mathbb R^{3\times3}$</span>. I <em>also</em> have an estimate of the rotation between <span class="math-container">$A$</span> and <span class="math-container">$N$</span> expressed by the vector <span class="math-container">$\mathbf \theta \in \mathfrak{so}(3)$</span> such that the rotation matrix from <span class="math-container">$A$</span> to <span class="math-container">$N$</span> is: <span class="math-container">$^N \mathbf R _{A} = \text{expm}(\theta) \in SO(3)$</span>. This estimate of the rotation vector <span class="math-container">$\mathbf \theta$</span> has the associated covariance <span class="math-container">$\Sigma_{\theta} \in \mathbb R^{3\times3}$</span>.</p>
<p>Given that <span class="math-container">$^{N}\omega= \;^{N} \mathbf{R}_{A} \cdot \,^{A} \mathbf \omega \;$</span>, what is <span class="math-container">$\Sigma_{^N\omega}$</span> ?</p>
<p>Thank you.</p>
<p>[EDIT]: If the rotation is without covariance (fully known), then <span class="math-container">$\Sigma_{^N\omega} = \, ^N \mathbf R _{A} \cdot \Sigma_{^A\omega} \cdot ^N \mathbf R _{A}^\intercal $</span></p>
| |mechanical-engineering|control-engineering|statistics| | <p>Turns out, it is not as complicated as it seems.</p>
<p>On a general basis, given a function <span class="math-container">$ f: \mathbb{R}^m\to \mathbb{R}^n$</span> and a vector <span class="math-container">$\mathbf x \in \mathbb R^m$</span> and its associated covariance <span class="math-container">$\Sigma_x \in \mathbb R^{m \times m}$</span>, if <span class="math-container">$\mathbf y = f(\mathbf x)$</span> then:</p>
<p><span class="math-container">$$\Sigma_x \simeq \left . \frac{\partial f}{\partial \mathbf x}\right |_{\mathbf x} \Sigma_x \left . \frac{\partial f}{\partial \mathbf x}\right |_{\mathbf x} ^\intercal$$</span></p>
<p>It requires to compute the jacobians, which can be tedious.</p>
| 38786 | Covariance of an uncertain vector going through an uncertain rotation |
2020-11-24T10:42:20.240 | <p>I have been trying to solve the problem below involving a concrete anchor block subject to two equal horizontal forces.
Normally I would just apply the force as a moment acting about the centroid of the concrete block which would produce a linear bearing pressure distribution at the base. Adding the self weight of the block as a compressive bearing pressure and I would then have an idea of the complete bearing pressure distribution.
Am I missing something?
taking moments about the centroid with the forces turned into vectors via the position vector and using the cross product only gets me so far.</p>
<p><a href="https://i.stack.imgur.com/3Z0cf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Z0cf.jpg" alt="enter image description here" /></a></p>
| |statics| | <p>Usually, these kinds of problems can be handled as a footing with P load and M overturning moment. I don't quite get those assumptions though.</p>
<p>The overturning moment will cause the anchor block to triangular positive and negative stresses on the soil, which can be calculated as, call this stress Q at the maximum point the end of the triangle:</p>
<p><span class="math-container">$$ \Sigma M_y=0,\quad (1500+1500)5kNm=2(Q*7.5/2)_{total\ stress}*(15*1/3)_{lever\ arm}$$</span>
<span class="math-container">$$Q=15kNm/37.5=400N$$</span> but this is for the 5 meters base width, therefore <span class="math-container">$$400/5=80N/m^2$$</span></p>
<p>The slope of this stress surface is <span class="math-container">$80/7.5=10.66\text{ N}$</span> per meter. With 0 stress at the middle of the 15m length.</p>
<p>So the total stress distribution is <span class="math-container">$$P_{total}=(P-80+10.7*x)Nm2$$</span></p>
<p>please check my arithmetic.</p>
<p><a href="https://i.stack.imgur.com/xCTIu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xCTIu.jpg" alt="diagram" /></a></p>
| 38792 | Vector expression of bearing pressure |
2020-11-24T16:20:32.587 | <p>When we send some drawings to laser cut operation, they use a proprietary software (usually specific to the laser machine) to duplicate the drawings per request, with the spaces between parts optimized. It's a bit like "panelizing" operation in PCB manufacturing.</p>
<p>What is the name of this optimized assembly operation? (How can we find a proper software to make it by ourselves?)</p>
| |manufacturing-engineering| | <p>The operation is called usually <a href="https://en.wikipedia.org/wiki/Nesting_(process)" rel="nofollow noreferrer"><em>nesting</em></a>. It is used for almost any type of cutting shapes out of sheets of metal.</p>
<p>One piece of software I used, free (at least at the time) and with very good results (i.e. low slack material) was <a href="https://deepnest.io" rel="nofollow noreferrer">deepnest</a>.</p>
<p>There should be other though. A quick search yielded this <a href="https://all3dp.com/2/best-laser-cutter-software-for-nesting/" rel="nofollow noreferrer">page</a> with :</p>
<ul>
<li>TruNest</li>
<li>ProNest 2019</li>
<li>SVGnest</li>
<li>SigmaNEST</li>
<li>MyNesting</li>
</ul>
| 38795 | What is the name of optimizing space between laser cut parts? |
2020-11-25T20:36:47.827 | <p>Today as I was filling up a bottle of water, and I could hear the distinctive change in pitch that signals that the bottle is near filling up, I started wondering what equation determines the frequency of the sound.</p>
<p>More specifically, I suspect this is fundamentally a problem in vibration dynamics. I suspect that there is some type of equation that changes the natural frequency of the cavity but I cannot put my finger on it.</p>
<p>Does someone know how to derive this equation? I suspect it will involve a lot more variables than the the natural frequency of the harmonic oscillator <span class="math-container">$\sqrt{\frac{k}{m}}$</span>, or the pendulum frequency <span class="math-container">$\sqrt{\frac{g}{l}}$</span>.</p>
<p>This is not for a project, just plain old curiosity, so I would be very happy for pointers to books or links.</p>
| |dynamics|vibration|sound| | <p>Most acoustic vibrations of air more or less confined to a given geometry can be explained by two basic models. One is the one-dimensional treatment of the air in a tube, and the other is the lumped-parameter treatment of the air within a volume that has a small hole in its side. The former is called a quarter-wave or half-wave tube, depending on the nature of the vibration, and the latter is called a Helmholtz resonator.</p>
<p>When we fill a bottle with water, further investigation is required in order to determine which of the above models apply. Key to that determination is the frequency range of the sound we hear. The two models will predict very different frequency ranges.</p>
<p>The proper tube model in this case is the quarter-wave tube model simply because the liquid in the bottle provides a closed end and the mouth of the bottle the other end. The natural frequency of vibration of such an air column has a wavelength four times the distance from the edge of the mouth to the water. If say the water surface is a couple inches from the edge, the wavelength will be about 8 inches, giving a vibration frequency of about 1500 Hz, which our ears can readily detect.</p>
<p>The natural vibration frequency of a Helmholtz resonator is given by f = (c/2/pi)<em>(A/L/V)^0.5, where c is the speed of sound, pi = 3.14, A the area of the opening, L the approximate length of the air "slug" that vibrates in the opening, and V the volume. Let's say the mouth diameter is 0.5 inch, the length of the air slug about twice the mouth diameter, and the volume at a certain moment of filling is roughly (pi</em>D^2/4)*L', where D is the bottle diameter and L' the length of the air space in the bottle. If we say that D = 3 inch and L' = 4 inch, we get f = 164 Hz, which is also well within the range of hearing.</p>
<p>Many of us can get a feel for the Helmholtz frequency when we blow edge-on across the mouth of an empty bottle, and the 164 Hz is very reasonable. The quarter-wave frequency is much higher, at 1500 Hz. In my view, during the process of filling an empty bottle, the frequencies start quite low, due to Helmholtz resonance, and as the liquid level rises, the Helmholtz formula shows that the Helmholtz frequency will also rise (the air volume decreases). At some point, however, the Helmholtz model breaks down, because we no longer have a well defined volume and a relatively small opening. Thus, the quarter wave model then becomes more applicable, producing a higher range of frequencies.</p>
| 38810 | Why does the frequency of a bottle filling up changes? |
2020-11-26T11:18:02.860 | <p>Suppose I have a small stepper motor, and for reasons (like wiring considerations) it is advantageous for me to not actually have the motor's body fixed as the shaft rotates, but rather have the shaft fixed in place while the entire mechanism of the motor rotate around the shaft. Are the two schemes equivalent if we assume that the shaft is perpendicular to the ground?
My ultimate case is to have a stepper motor rotate while it is upside down, without using additional gear wheels or any other component other than a coupler which secures the shaft to a solid base.</p>
| |mechanical-engineering|motors| | <blockquote>
<p>Are the two schemes equivalent?</p>
</blockquote>
<p>Mostly so. Usually the motor is affixed to a sturdy chassis resting on a solid surface, sinking vast majority of vibrations and jerk coming from the motion. In your configuration, both the motor and anything attached to it will be fully exposed to these vibrations - which introduce all kinds of headaches like electric connectors losing contact, wear of joints from material fatigue, screws coming loose, and so on and so forth - not a single thing critical, just a mountain of small headaches. You're also dealing with accelerations in case the element turns fast (and especially if it stops rapidly, say, knocking against a brake), and you have to get power and control to the rotating part somehow. Control is fairly easy using short-range wireless communication like bluetooth or IRDA. Power is worse - you'll either limit your angle so that an attached cable doesn't get too twisted, or supply power through batteries. Or you can use a slip-ring for both power and data, but slip-rings are another can of worms, something that works okay in 'big industrial', but on small scale is so fault-prone and quirky you're better off redesigning everything to have a fixed stepper.</p>
| 38819 | Fixing a stepper motor by its shaft |
2020-11-26T17:18:04.733 | <p>Consider the collision of two soliton pulses propagating with identical amplitude, equal velocity
in two opposite directions on a nerve. Assume that the collision is head, reversible and adiabatic. Consider the presence of viscous friction in the system. Then the pulses pass through each other almost undisturbed with the generation of some small amplitude noise.</p>
<p>Now, How can I prove this mathematically or in any other way?</p>
<p>Forgive me if this is not the right place to ask this question. If it is not, please tell me where to ask this question.</p>
| |biomedical-engineering| | <p>I have no clue what you are referring to in biology terms.</p>
<p>But we know in general two waves passing by each other regardless of it being straight head-on or some angle, interfere with each other (with the exception of photons which pass through each other most of the time).</p>
<p>The waves magnify or cancel each other depending on the individual phase at the interaction point. They are called constructive interference and destructive interference.</p>
<h1>constructive Interference</h1>
<p><a href="https://i.stack.imgur.com/B7cO2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B7cO2.png" alt="constructive" /></a></p>
<h1>Destructive interference.</h1>
<p>.</p>
<p><a href="https://i.stack.imgur.com/rZZ2F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rZZ2F.png" alt="destructive" /></a></p>
| 38824 | Collision of soliton pulses in nerves |
2020-11-27T10:04:48.773 | <p>What does it mean if you've been told to design a compensator to ensure that the closed loop phase margin is at least 120 degrees? Isn't a compensator typically designed using the open loop transfer function? Using the matlab command <code>allmargin()</code> for both closed loop and open loop transfer functions of the same system yields different results!</p>
| |control-engineering| | <p>Phase margin is indeed read from an open-loop transfer function, but the rationale behind it is to prevent the closed-loop from going unstable. As you might recall:
<span class="math-container">$$T(s) = \frac{GC}{1+GC}$$</span>
Looking at this function from a mathematical viewpoint, this function explodes when <span class="math-container">$GC = -1$</span>. In bode-plot terms, <span class="math-container">$GC = -1$</span> if <span class="math-container">$|GC| = 0dB$</span> and <span class="math-container">$\angle GC = -180 \deg$</span>. As most dynamical systems cross this <span class="math-container">$0dB$</span> point somewhere, one should prevent the phase is anywhere close to <span class="math-container">$-180\deg$</span>. Phase margin describes the distance to <span class="math-container">$-180\deg$</span> when the magnitude crosses the <span class="math-container">$0dB$</span> line.</p>
<p>As you see, the importance of it is to save the closed-loop transfer function from exploding, but determining it can be done directly from the open-loop transfer function. Because of this, the compensator is also designed in open-loop, but only works properly if implemented in a closed-loop system.</p>
| 38833 | Closed loop compensator |
2020-11-28T14:41:09.787 | <p>I see granite blocks in ancient Egyptian temples with perfect holes drilled into them. The historians say at that time (2500BC) they did not possess iron tools or the wheel and certainly no electricity. The most capable metal tool was made of copper.</p>
<p><a href="https://i.stack.imgur.com/O9fEo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O9fEo.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/kNAtd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNAtd.png" alt="enter image description here" /></a></p>
<p>One theory suggests that a two-man saw with a flat copper blade was used to make horizontal cuts in granite by throwing sand at the cut line; the silicon quartz pushed by the blade make cutting granite possible. But this process yields something like a 1/2 inch deep cut after many hours of sawing (impractical).</p>
<p>I assume they may have employed a copper hole-saw or tube-like tool using the above method. But it still seems very tedious and impractical.</p>
<p>There is a video here: <a href="https://youtu.be/puXsFyainQU?t=118" rel="nofollow noreferrer">"Lost Ancient High Technology In Egypt: Saw Marks And Drill Holes"</a>.</p>
| |machining|engineering-history|tools| | <p>If I got a long steel cylinder with a cutting head on it and the other end like a ball joint</p>
<p>I’d have 2 holes in opposite sides of the cylinder. I can put a knot in the rope on the inside, take the rope and wrap it several times around the cylinder. Then with the other hole, I do the same but wrap it in the other direction.</p>
<p>Then a lubricated plate would fit into the ball joint where pressure was applied.</p>
<p>Now I’d have 10 people pull each rope in opposite directions and 10 people push into the plate.</p>
<p>People used teamwork to quickly do one task at a time. I’d think drilling a hole in granite could take only 5 minutes at most</p>
<p>Probably have someone chisel the start so it doesn’t slip outward</p>
<p>Also steel only takes 500 years to rust to nothing, obviously they had steel tools.</p>
<p>Also wheels have existed since the first tree fell over, don’t fool yourself.</p>
| 38852 | How to make holes in granite without electrical or iron tools? |
2020-11-28T19:56:49.073 | <p><a href="https://i.stack.imgur.com/P5QA1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P5QA1.png" alt="enter image description here" /></a></p>
<p>I'm trying to understand bolt preloading. I found this analogy on Wikipedia. I understand that as the spring is loaded to some pretension and the block is inserted, as we apply an external load to the hook the spring won't move unless the load is bigger than the preload, since the resultant force is then towards up.</p>
<p>From Wikipedia:</p>
<blockquote>
<p>When an external tension load is applied, it relieves the compressive strains induced by the preload in the clamped components, hence the preload acting on the compressed joint components provides the external tension load with a path (through the joint) other than through the bolt. In a well designed joint, perhaps 80-90% of the externally applied tension load will pass through the joint and the remainder through the bolt. This reduces the fatigue loading of the bolt.</p>
</blockquote>
<p>I still don't understand why only a part of the tension load goes through the bolt. The bolt is the only fastener in the joint, the components don't have any kind of adhesive or anything, how can the tension have "a path" through them?</p>
| |bolting| | <p>The other answers are pretty good, but allow me to throw one more analogy out there. I think the last quote in the question is poorly worded so I wouldn't get too hung up on it.</p>
<p>Imagine you're pushing a sheet of plywood up against a door frame, the plywood is larger than the door so it's essentially like pushing into a wall. Now imagine you're pressing on the plywood with 10 pounds of force, but you can actually push with 100 if you exert yourself.</p>
<p>Now imagine someone is on the opposite side of the door pushing back. If they push back with 20lbs of force. You will feel the push, you can then push back harder to keep the plywood in place. This is similar to a bolt without much preload.</p>
<p>Now imagine the same scenario again, but you press on the door with 90lbs of force. The person on the other side presses with 20 again, but nothing budges and you won't be able to feel it (assuming the plywood is stiff). Unless they push with 90lbs of force you won't feel any difference.</p>
<p>The reason the last quote says force is "traveling through the joint", is because if that second person presses with 20lbs force, then the load on the door frame simply reduces by 20lbs, so the door frame is experiencing the change in load; but you're correct in that all the force is ultimately conducted through the bolt.</p>
| 38857 | Bolt preload confusion |
2020-11-29T08:55:15.157 | <p>Say a body rotates by 90 degrees, then the yaw angle will be 90 which can be calculated using the data provided by IMU.</p>
<p>Now from that point, the body traverses some distance and is again rotated by 90 degree in the same direction. What will be yaw angle shown by the IMU? Will it be 180 degrees or 90 degrees?</p>
<p>In a nutshell what I want to ask is that, is the angle measured from the initial starting point or from the last reference point?</p>
<p><a href="https://i.stack.imgur.com/mJ87h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mJ87h.png" alt="enter image description here" /></a></p>
<p>For Roll, Pitch, Yaw, please refer to the above diagram.</p>
| |sensors|robotics|instrumentation|electronics| | <p>The yaw angle is defined as the difference between the body's orientation vector and velocity vector in the yaw plane of the body.</p>
<p>So the IMU computes the velocity vector using integrating accelerometers and computes the orientation vector using gyrocompasses. It then compares the two and extracts pitch, roll, and yaw angles. The way it does this depends on whether you are using a strapdown IMU or a gimballed inertial platform.</p>
<p>To answer your question, you have to tell us in what direction the vehicle is moving.</p>
| 38866 | How is the YAW angle calculated by an IMU sensor? |
2020-11-29T11:30:26.123 | <p>I was trying to solve this problem. But I could not figure out how to do it. Please help or provide a hint how can I solve it?</p>
<p><a href="https://i.stack.imgur.com/mOVmB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOVmB.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|vibration| | <p>Some hints:</p>
<ul>
<li>you need to consider a few constraintsdue to strength.
<ul>
<li>compession strength</li>
<li>buckling strength</li>
</ul>
</li>
</ul>
<p>Those two will give you a mininum D, d.</p>
<p>Additionally you need to consider the vibrational behaviour:</p>
<ul>
<li>you need to calculate the sdof harmonic oscillator model. Ie what is the equivalent mass and equivalent stiffness for horizontal movement</li>
<li>transmissability ratio due to base excitation. That way you'll be able to find the force induced by the vibration</li>
<li>then you d need to use the force to calculate the bending stress on the base of the column. That will give you another set of D,d constraints.</li>
</ul>
<p>Keep in mind that the equivalent stiffness will be depended on the diameters. Therefore deepening on how you solve this problem you might need to do iterative calculations.</p>
| 38869 | I need a solution to a problem. I could not figure out myself how to do it |
2020-11-29T18:01:21.817 | <p>Which is stronger, a round sign post or a square-cross-section sign post?
I will be using one or the other to hold the top corner of a shade sail. I tried an aluminum pipe but a strong wind on the sail bent the pipe. I am about to buy a 10-foot-long sign post because they look stronger than the pipe. I found online a 2 by 2 inch square-cross-section post with a lot of holes in it. The steel is 14 gauge (0.0781 in) thick. I also found a 2 inch diameter hollow cylindrical post with no holes in it. The steel is 16 gauge (.065 in) thick. Which one should I buy?</p>
| |structural-engineering|architecture| | <p>If you install the square post in a way that its face is perpendicular to the cord it has to support it will be stronger. It has a larger 2nd moment of the area (I) that way which makes it stiffer.</p>
<p>I would also try to figure out the prevailing wind and position the sail in a way it is not going to get a lift out of that angle and also the cord is tight to keep the sail flat. Make sure you minimize the sail's flutter by pulling it tight.</p>
<p>The sail, like an airfoil, gets a lift when it is laying parallel or near parallel to the wind. I would position it leaning into the wind so the wind will push it down as opposed to lifting it up.</p>
<p><strong>Edit</strong></p>
<p>I correct my error after a comment by aleph zero. the second moment of area, I, of a square section is constant regardless of the angle of the square.</p>
<p><span class="math-container">$$I_u=\frac{I_x+I_y}{2} + \frac{I_x-I_y}{2} \cos{2\phi} -I_{xy} \sin{2\phi} \\ I_v = \frac{I_x+I_y}{2} - \frac{I_x-I_y}{2} \cos{2\phi} +I_{xy} \sin{2\varphi} \\ I_{uv} = \frac{I_x-I_y}{2} \sin{2\phi} +I_{xy} \cos{2\varphi} $$</span></p>
<p>In this cas the above simplifies to
<span class="math-container">$$I_u=\frac{I_x+I_y}{2}=I_{xx}=I_{yy}$$</span>
So the parallel orientation is equally as suitable as any, except it's easier to install and easier to find fasteners for.
The rest of my answer is still valid.</p>
| 38876 | Which is stronger, a round sign post or a square-cross-section sign post? |
2020-11-29T22:41:54.710 | <p>Okay, so recently I got a Dewalt 735 Thickness Planer.</p>
<p><a href="https://i.stack.imgur.com/o2vCT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o2vCT.png" alt="Dewalt 735" /></a></p>
<p>It's great at what it does. (It even LOOKS like it's ready to consume the tender flesh of lumber, like some twisted Victorian lumberjack... though, maybe that's just me).</p>
<p><a href="https://i.stack.imgur.com/U1ROt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U1ROt.png" alt="Jack the Crosscutter" /></a></p>
<p>Anyway, the problem I've been experiencing with the tool is that, while planing, should a board exceed a certain length, such that the midpoint of the piece is no longer supported while some portion of it is still being cut, the board tips, resulting in snipe (a gouge in the board, resulting from the angle shifting mid-operation).</p>
<p><a href="https://i.stack.imgur.com/dpiCS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dpiCS.png" alt="Illustrating snipe" /></a></p>
<p>Okay, no problem: I'll just build some supports. Now, I just so happen to have some Heavy Duty 22" full-extension 150-pound-rated drawer slides - about a dozen of em - from a different project. Cool. I'll run a roller bar between em and make the whole deal extendible at need. I can even add flip down leg supports if needed (though if the slides are strong enough, I'd rather not).</p>
<p><a href="https://i.stack.imgur.com/D4Nej.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D4Nej.png" alt="Extensive support" /></a></p>
<p>As I was doodling this out, my son walks in and asks what I'm looking at. I muse aloud that it's a shame I don't have longer slides. 22 inches isn't terribly huge for infeed/outfeed.</p>
<p><strong>Then he asks me: "So why can't you just hook two together? Just stick another piece between em so you can screw into both sides of that."</strong> Heh. From the mouth... of... babes?</p>
<p><a href="https://i.stack.imgur.com/4H5VM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4H5VM.png" alt="FORM BLAZING SWORD!" /></a></p>
<p>Now, intuitively, this seems like a bad call (certainly without the leg support). But I'll be damned if I can figure out WHY or what to tell him. Some long-forgotten, niggling memory of similar suggests this is a Static Equilibrium problem, which - despite it having been decades since I've looked at any of this - I have some vague recollection on how to calculate. But it's the addition of the strut that hooks the two slides together I cannot account for.</p>
<p><a href="https://i.stack.imgur.com/s4hpU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s4hpU.png" alt="enter image description here" /></a></p>
<p>Now, although the finished version of either approach will have two slides with a bar spanning them, I'm reducing this to two dimensions for my own old brain to walk through.</p>
<ul>
<li><p>HOW do I figure out how much load would be put on the roller if I were to take his approach?</p>
</li>
<li><p>COULD a 150 lb capacity (well, two combined, for the 2d example we're using, per side - four total in actual implementation) support a portion of the mass of a single, call it 25lb board?</p>
</li>
<li><p>WOULD it need legs?</p>
</li>
</ul>
<p><strong>Edit</strong> <em>additional clarification</em>: the dimensions of the mobile base are 38" wide (the axis perpendicular to the force) by 26" deep (the axis parallel) by 30" tall (including the locking steel casters it rests atop).</p>
<p>It is constructed of the better part of a full sheet of 3/4" Baltic Birch plywood (call it 75 lbs), and contains a small cabinet containing various bits and bobs (~30 lbs). Also seated alongside the planer is a cast iron bench top jointer (~65 lbs), which itself has 2 drawers beneath it containing various sleds, jigs, and blades (~20 lbs). This is all in addition to the four 4" steel casters (~2.5 lbs ea.) the fixture rests atop, and the 105 lbs. planer itself. This accounts for an approximate total weight of ~300 lbs and change. I've been simply treating it as the proverbial "immovable object" in the equation, Archimedes be damned, but that is, I admit, simply because it hadn't occurred to me it's stability might be compromised.</p>
| |mechanical-engineering|materials|torque|stresses|beam| | <p>Your best bet is to rig the planer with a couple of carpenter's horses or something on the sides. However, if we want to go your way need to check the following.</p>
<ul>
<li><p>Overturning moment on the fully extended setting with 44" cantilevered rollers.</p>
</li>
<li><p>Torque on the joint between the two drawer slides.</p>
</li>
</ul>
<p>Let's say your machine and its base weigh 40# and the base is 24 inches wide. And half of 25#board weight is on the roller half on the blades, say 10lbs sum of the roller, and half of 4 sliders.</p>
<p>We compare the moment of the board's end with the moment due to the base and planer and the other end of the board.</p>
<p><span class="math-container">$$\text{moment due to the base side}\ (40+12.5)12" = 630lbs.inch \\ \text{moment due to the roller side}\ (12.5 +10)(44+12)=1260lbs.inch>>630$$</span></p>
<p>Therefore the planer will tip over and we need legs.</p>
<p>As for the torque between the sliders, we have a torque of <span class="math-container">$\tau= (12.5+7.5)*22=177.5lbs.inch$</span></p>
<p>say 200, do you think you slider can take that much torque?</p>
| 38880 | "Heavy-duty" or not, can my drawer slides really support this weight? |
2020-12-03T16:44:05.253 | <p>I'm reading about classification of beam cross sections <a href="http://www.steel-insdag.org/TeachingMaterial/Chapter8.pdf" rel="nofollow noreferrer">here.</a></p>
<p>It is said:</p>
<blockquote>
<p>The ratio of the ultimate rotation to the yield rotation is called the rotation capacity
of the section. The yield and the plastic moments together with the rotation capacity of
the cross-section are used to classify the sections.</p>
</blockquote>
<p>I understand the concept of <em>yield moment</em>, it is the moment at which the extreme fiber of the cross section reaches the yield. But the page also talks about <em>yield rotation</em>. Rotation of what? I understand that we normally measure the rotation of the ends of the beam, but here it is supported rigidly at the ends, so what rotation are we measuring here?</p>
<p>If we thought about a simply supported beam so we could measure the rotation of the end, isn't the yield rotation a property of the <em>beam</em>, rather than the cross section? If the beam is longer, wouldn't the required rotation (to make the critical cross section reach yield) probably be smaller, making the yield rotation dependent on the whole beam rather than just cross section?</p>
| |structural-engineering|beam| | <p>Rotation of the section over the length with the particular moment.</p>
<p>Skip forward to the end: in a theoretical section made of a material with perfect plasticity (no hardening, infinite strain capacity) at the plastic moment, there is a plastic hinge. Once you've reached the fully plastic moment, any increment of moment will result in infinite rotation of that hinge.</p>
<p>Obviously real beams made of real material don't exhibit infinite rotation in a hinge.</p>
<p>In most real sections formed of real plastic material (like steel) there are other mechanisms at play, and often you can't actually get the whole section plastic. With respect to classification of sections, the most significant other effect is local buckling. Some part of the section reaches a limiting compressive force and buckles, so limiting the compressive force that element can carry, thus limiting the moment the section can carry. So you get a sort of hinge type behaviour in the section (no greater moment can be carried) without plasticity across every part of the section.</p>
<p>Furthermore, you can get a behaviour where the full plastic moment is generated, but as rotation occurs then an element buckles (rather than squashes). As the hinge rotates, the moment the section carries drops dramatically (just like a classic strut buckling - once it has buckled the axial load drops compared to the load that caused it to buckle). In this case you don't get a long flat plateau on your moment/rotation curve, you get a buckling type behaviour and the moment plummets past the peak.</p>
<p>Thus, you get the classes of section described, illustrated in your reference at figure 4.</p>
<p>A slender section doesn't even get to first yield - you get a local buckling (typically of the flanges) before the section even gets to yield. When the flanges buckle they can't take any more compressive force, so you get a hinge-ish behaviour, there's a localised rotation.</p>
<p>A semi-compact section can carry enough moment that some part of it has started to yield before the flanges buckle, but then they do buckle before you get to the whole cross-section at yield stress.</p>
<p>Compact and plastic sections both achieve the full plastic moment, and differ in the post-yield behaviour. If you had a square or circular solid cross section with nothing to buckle you can generate the plastic hinge and then just keep bending it, getting vey high concentrated rotations (i.e. hinge-like) without any reduction in the moment carried. However, quite a few sections that can generate the full plastic moment can't actually sustain that moment with significant rotation - when they start rotating the outstands actually buckle. Then the moment carried drops dramatically. So you can achieve the plastic moment, but you don't achieve an actual plastic hinge.</p>
<p>Note that the current UK (and Europe) codes - EC3 (aka EN1993) has classes 1 to 4. Class 1 can form a plastic hinge, class 2 can generate the plastic moment but not a hinge - local buckling prevents the hinge, class 3 gets to yield moment but cannot get to plastic moment, and class 4 doesn't even get to yield stress at any point in the section before some part of it buckles - i.e. the same classification, just different labels.</p>
<p>Possibly (arguably) figure 4 could be thought of as rotation per unit length (i.e. some function of curvature), but these behaviours occur over a finite length of beam (a buckle, for example, needs some length). To be rigorous, the graph is probably the rotation (i.e. change in angle) of an elemental but finite length of beam with a constant moment along the whole length of that element, thus giving rise to a finite rotation across the element. It's not curvature, because that breaks down at a plastic hinge.</p>
<p>Mathematically, the graph (and discussion) is probably conflating (and muddling) delta-theta/delta-position with d-theta/d-position (something that engineers are somewhat prone to do, causing mathematicians to shake their heads and tut).</p>
| 38918 | What is meant by the rotation capacity of a cross section? |
2020-12-04T06:13:17.190 | <p>We all know if we use less electricity we save energy.</p>
<p>But the energy we're getting is result of <em>burning</em> (not necessarily the literal meaning) the fuel. Even if we don't use that energy that fuel is gone forever. (Except saving money on bills)</p>
<p>My questions</p>
<ol>
<li><p>Am I wrong in above statement? If yes then how does actually electricity production work so that we're able to save it?</p>
</li>
<li><p>Do power plant (or associate agencies) constantly monitor demand and reduce the production of electricity on real time thus saving fuel (hence energy)?</p>
</li>
</ol>
| |power|energy|power-engineering|fuel-economy| | <h1>Timescales on the grid</h1>
<p>Power demand fluctations can be broken into timescales from micro-seconds to decades. On the "decades" end of the scale, the power industry and utility regulators work together to plan and fund construction of power plants and the associated transmission and distribution infrastructure. When you turn your air conditioner off <em>today</em> it doesn't affect these long term processes, but if you (and enough of your neighbors) do this consistently, then there will be a measurably decline in cyclic demand, which will be factored into the planning process.</p>
<p>But even on the micro-second time-scale, turning off an air conditioner (or even just a light) does result in less fuel being burned now. To explain how, we need to build up the timescales from micro-seconds to seconds (beyond that, <a href="https://engineering.stackexchange.com/a/38922/17004">NMech</a> did a good job explaining the hourly to annual time scales).</p>
<h2>0.000001 - Inertial response at microsecond timescales</h2>
<p>Fundamentally, generators are large pieces of spinning iron. This applies to coal, nuclear, natural gas, hydro, and wind generators. All of this spinning iron has inertia -- a fact around which the system is designed (in the case of solar power, grid-connected PV inverters are actually designed to emulate the behavior of spinning iron).</p>
<p>Inertia is "the resistance of any physical object to any change in its velocity" (<a href="https://en.wikipedia.org/wiki/Inertia" rel="noreferrer">Wikipedia</a>). If you touch a huge spinning wheel, it will slow down -- but this change in speed will be imperceptible, and not instantaneous, due to the wheel's inertia.</p>
<p>On the grid, touching the wheel is akin to increasing the load -- turning on a light, toaster, hair dryer, etc. The inertia present on the grid causes a time delay between the change in load, and the change in speed.</p>
<p>This time delay is where the first level of grid control comes in: <strong>droop control.</strong></p>
<h2>0.001 - Droop control at millisecond timescales</h2>
<p><a href="https://en.wikipedia.org/wiki/Droop_speed_control" rel="noreferrer">Droop control</a> is a proportional control on the "throttle" of each generator. Each generator on the grid is programmed to increase its speed by x% for every y% reduction in the frequency detected on the grid (and decrease speed when the frequency increases -- the proportion x:y is the same in both directions). This x:y proportion is the <em>droop percent</em>. The frequency, in turn, is directly proportional to the speed of each generator.</p>
<p>Droop control is like a string connecting the needle on a speedometer to the "gas pedal" of the generator -- as the speed decreases, the string pulls the pedal to give it more gas -- once the speed increases, the tension is released and less gas is supplied. This seems a bit backwards, because normally we think of using the throttle to control the speed of a car, but in this case we're talking about dozens or more cars all welded bumper-to-bumper, so one throttle can't actually do much. But if each car has the same exact speedometer/string/throttle set-up, we start to see how this could affect the speed control of the whole system.</p>
<p>This system has two shortcomings, though:</p>
<ol>
<li>What about when the load increases or decreases outside the range of droop control (i.e. all the generators are at full throttle and load is still increasing)?</li>
<li>How is a constant frequency maintained?</li>
</ol>
<p>This is where <strong>automatic generator control</strong> comes in.</p>
<h2>1.0 - Automatic generator control at second timescales</h2>
<p>The "throttle" functions differently for each type of generator. For natural gas, it's similar to a car -- you adjust the amount of fuel burned and the power output is directly adjusted. For coal and nuclear systems, it's more complicated, because the fuel is used to produce steam, and the steam then runs the generator. Droop control directly controls the "steam valve."</p>
<p><a href="https://en.wikipedia.org/wiki/Automatic_generation_control" rel="noreferrer">Automatic generator control</a> (AGC) incorporates the diversity of generator types into a single algorithm which can be applied across the grid. Variables in AGC include:</p>
<ul>
<li><strong>Droop percent</strong>, as discussed above</li>
<li><strong>Frequency setpoint</strong>, fixed at the grid level (60 Hz or 50 Hz, depending on the continent)</li>
<li><strong>Area control error</strong> (ACE), a power output adjustment amount which depends on how much power the node where the generator is connected needs to import or export to other nodes</li>
</ul>
<p>Where droop control is simple and straightforward to implement (at it's most basic, it can be done using passive electrical components), AGC requires more complicated, usually digital controls.</p>
<h2>10 - Minute timescales and beyond</h2>
<p>After AGC things get more complicated and less standardized across the industry. At the minute timescale and beyond, variables such as fuel prices, weather forecasts, emissions rates, and load forecasts can be included in control algorithms. Design of these algorithms vary by region and utility. The Wikipedia article on <a href="https://en.wikipedia.org/wiki/Regional_transmission_organization_(North_America)" rel="noreferrer">regional transmission organizations</a> is a good starting point.</p>
<h1>tl;dr - how does this save fuel when I switch off a light?</h1>
<p>When you switch off a light, grid inertia immediately kicks in to increase the frequency (speed up). This triggers the droop control algorithm across the grid to throttle down generators. The generators which can physically respond fastest are programmed to do so -- these generators also tend to be those directly burning fossil fuels -- usually natural gas, but sometimes diesel.</p>
<p>Even in the case of coal, however, there will be decreased fuel use, but not until AGC kicks in. Droop control throttles down the steam valve directly, but as the pressure in the steam vessel responds to this change, less coal will need to be burned to maintain the set pressure.</p>
| 38920 | How does turning off electric appliances save energy |
2020-12-04T22:03:54.677 | <p>I'm looking for information pertaining to the design of an engine cooling system in a heavy duty truck -- one that's used for long haul shipping. I've found some resources that provide details on the main components -- radiator, fan, pump, reservoir, thermostat, etc. -- but, I've found little in terms of the specific design features -- pipe diameter(s), flow rates, etc. Some design information on the cooling system in a popular truck model, such as a Volo VNL series with a D13, D13TC or Cummins X15 engine would be appreciated.</p>
<p>My ultimate goal is to try to construct the cooling system, removed from the engine. Specifically, I'd like to know the exact parts that go into the cooling system of one of these trucks. I know the pump is driven from the timing belt of the engine, I'm assuming this means the rpm of the pump is equal to the rpm of the crankshaft, and therefore varies depending on the vehicle demand. I understand the mechanism for the fan might be slightly more complicated, as its necessity is only imperative for when the vehicle is stationary. I'd also appreciate some recommended resources for learning about some of these design components. I'd really appreciate some guidance and any useful information. Please let me know fi you have any questions.</p>
<p>Regards,</p>
<p>Lucas W.</p>
| |mechanical-engineering|automotive-engineering|cooling| | <p>Well, first the rpm of the water pump is not equal to the rpm of the engine crank, the pulley diameter ratio drives the water pump faster.</p>
<p>As for design parameters, start with the amount of heat that has to be removed to keep the engine within its temperature limits at max power as that is the worst condition. Remember to include things like if the turbo has extra cooling.</p>
<p>Once that point is defined, you can start to size the radiator needed, considering capacities, flow rates and issues such as loosing ram cooling while engine is at max power ie it is going slow pulling hard and relying on the cooling fan(s) only.</p>
<p>Then the control such as bypass and thermostat needs consideration as the amount of cooling needs reducing if the engine is only gently working so it is not overcooled.</p>
<p>For fans, there are many choices such as electrically driven with full electronic control to provide a speed and airflow matched to the cooling required.</p>
<p>Lots of things that I have not mentioned but it’s a start for you to expand on.</p>
| 38934 | Cooling system design of a heavy duty truck |
2020-12-05T05:11:36.717 | <p>The symbol:</p>
<p><a href="https://i.stack.imgur.com/ncGiO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ncGiO.png" alt="enter image description here" /></a></p>
<p>Here is a part of the diagram:</p>
<p><a href="https://i.stack.imgur.com/owgt1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/owgt1.png" alt="enter image description here" /></a></p>
| |control-engineering|technical-drawing|instrumentation|diagram|pi-diagram| | <p>From <a href="https://www.hydraulicspneumatics.com/technologies/other-components/article/21884775/tip-16-what-is-the-condition-of-your-diamonds" rel="nofollow noreferrer">Tip #16 — What is the condition of your diamonds?</a></p>
<p><a href="https://i.stack.imgur.com/VLqA7m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VLqA7m.png" alt="enter image description here" /></a></p>
<blockquote>
<p>Example d), as I’m sure you can easily figure out, is a heater. With all other things being equal, this one shows the inward arrows adding thermal energy to the hydraulic oil. I have left out the coolant flow arrows, which is common, meaning the coolant circuit isn’t specified.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/TUZMJm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TUZMJm.png" alt="enter image description here" /></a></p>
<p>Symbol is clearly a heater showing heating lines going in and out. Reference shows cooler coolant lines, which they were left off on heater symbol.</p>
<p>Missing horizontal line, but it follows format of filter.</p>
<p>From <a href="https://www.hydraulicspneumatics.com/technologies/other-technologies/article/21884116/chapter-4-iso-symbols" rel="nofollow noreferrer">CHAPTER 4: ISO Symbols</a>:</p>
<p><a href="https://i.stack.imgur.com/FLPZj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FLPZj.png" alt="enter image description here" /></a></p>
| 38936 | What does this P&ID symbol represent? |
2020-12-05T11:15:32.523 | <p>I am working on a personal project where I need to select a motor to rotate a system that is in static equilibrium. The motor torque specification is given in kg.cm.</p>
<p>This is what I have:
Bar : 2 meters in length, weighing 1kg (for simplicity)
Axle: 10mm (5mm radius) with a bearing to reduce friction.
F : F1 and F2 both weigh 10kg</p>
<pre><code> (F1) (F2)
10kg 10kg
-------o-------
^
T1 -T2
</code></pre>
<p>I am not entirely sure what all the parameters are to arrive at a formula to use.
Also, what formula can I use to calculate the torque required to rotate the system at a constant speed of 3rpm ?</p>
<p>Thank you in advance.</p>
| |motors|torque| | <p>I would disregard friction and drag.</p>
<p>let's say an estimated acceleration to get to 3rpm is 10 seconds.</p>
<p><span class="math-container">$$I=(1/12ml^2)_{bar}+(2*10*1^2)_{10kgmass}=\\1/12*1*4+20=1/3kg+20kg=20.33kg.m$$</span></p>
<p><strong>This line added after OP's comment</strong></p>
<p>Using Newton's second law
<span class="math-container">$$F*dt= \Delta mv\quad or\ \tau dt=\Delta L=\omega I$$</span></p>
<p><span class="math-container">$$\tau dt=I\omega=20.33*3/60=61/60\ \approx 1 kgm^2/s$$</span></p>
<p>plugging 10s for dt we get the torque.</p>
<p><span class="math-container">$$10\tau= 1, \ \tau=1/10*9.8 \approx 1Nm,\ or\\ \tau= 10kg.cm $$</span></p>
<p>Now by changing dt to up or down you get the torque you need.</p>
| 38942 | How to calculate the required torque in a static equilibrium system |
2020-12-06T17:50:01.927 | <p>So I am using a complementary filter to find the attitude in my quadcopter, doing the following</p>
<p><span class="math-container">$$\text{angle} = 0.98\cdot (\text{angle}+\text{dt}\cdot \text{angle_rate})+0.02\cdot \text{accelerometer_angle}$$</span></p>
<p>Now the problem is that the angle is wrong when my system(quadcopter) is accelerating, do you know how to solve the problem, or it is normal not to get accurate values when accelerating even if one is using the complementary filter?</p>
| |control-theory|sensors|electronic-filters|kalman-filters| | <p>I have scrolled through the data sheet of this device, and first things first: never expect accurate values from a device that costs like 5 euro. looking at you equation, I think you are calculating the angle by effectively integrating the gyro value. Since integrating a sensor value is not an accurate estimation (because noise), the angle measured from the accelerometer could compensate for this deviation. This means that if the device is hovering statically, the angle saturates to the value of the accelerometer angle. The accelerometer calculates the angle using the gravitational acceleration (which it measures). However, the accelerometer also measures any other force acting on the system and that causes your accuracy problem.</p>
<p>The solution to an issue like that is hard, increasing the weight of the accelerometer improves estimation at constant velocities. decreasing it prevents measurement offsets when accelerating.</p>
<p>However, the sensor seems to have a couple tricks for you to use (DPM and noise filtering), so if you havent already, you might to check these.</p>
| 38956 | Complementary filter in MPU_6050 is giving me the wrong answer when the system is accelerating |
2020-12-06T18:03:36.890 | <p>I am in the process of developing a robotics design on SolidWorks.</p>
<p>The goal of this design is to tilt a platform using 2 actuators.</p>
<p>My design utilizes 2 indirect actuators that are separately powered and operated to tilt the platform on each end.</p>
<p>Each actuator has a fixed rotational joint at the top, that should rotate along the hinge as it gets pushed up with it's actuator, thus lowering the other side at the same time.</p>
<p><em>(See <strong>MODEL 1 below</strong>, which contains a brief 360 and <strong>identifies the actuators</strong> on my SW model).</em></p>
<p><strong>Please note the 2 rotational joints at the top of each actuator, and notice that they have full rotational capability at any stage.</strong>
<img src="https://s8.gifyu.com/images/Model1.gif"></p>
<p>However, when I attach a metal plate onto my rotating joints and try to animate my SolidWorks design, it ends up pulling the opposite actuator, and keeps the metal plate level the entire time. It does not move the rotational joint at the top of it's actuator at all. Which does not make any sense because the rotational joint should account for the actuators push.</p>
<p><strong>As the actuator moves up one one side, both joints should rotate by the same angle, and the plate should tilt. However this does not happen at all, both actuators move together, and the plate remains flat.</strong></p>
<p><strong>Can anyone please explain why this is happening and what I can do to fix this issue?</strong></p>
<p><em>Is this a problem with SolidWorks perhaps?</em></p>
<p>Any input would be highly appreciated, I can provide more information if needed.</p>
<p>Thank you for your time</p>
<p><em>(See <strong>MODEL 2 below</strong>, which contains the problem, a brief 360, and <strong>identifies the actuators with the metal plate</strong> on my SW model).</em>
<a href="https://web.archive.org/web/20210506211923/https://s8.gifyu.com/images/Model2.gif" rel="nofollow noreferrer"><img src="https://web.archive.org/web/20210506211923/https://s8.gifyu.com/images/Model2.gif" alt="" /></a></p>
| |mechanical-engineering|design|modeling|solidworks|robotics| | <p>According to me (if I understand correctly the design and what the goal is) this happens because the distance between the two actuators is fixed. Visualize a parallelogram. It has one degree of freedom, both pairs of sides always pairwise parallel. Your actuators are like two parallel opposing sides of the parallelogram. However, when you flex it, the two actuator sides of the parallelogram come closer together, i.e. the distance between them (represented by a segment that is perpendicular to both sides simultaneously) shortens. In your mechanism, you have eliminated the possibility for the two parallel actuators to get closer to each other. This over-constrains your motion. To fix it, you should allow the two horizontal platforms at the bottom of the actuators to move closer to each other or something like this. For now they are fixed, but try to allow for a horizontal translational motion.</p>
| 38957 | Trouble with an actuator on a Robotics SolidWorks design. Any ideas why? |
2020-12-06T19:29:37.890 | <p>I always think about how there surely must be natural ways to cool/heat a house, rather than using electricity, like solar panels, heaters, and AC. Does anyone have any good examples of houses/buildings that were engineered to be really effective at cooling and heating themselves, like with lots of ventilation through the ceiling or lots of glass at certain spots to concentrate sunlight?</p>
<p>I have tried looking this up but most results I find use solar panels. I'm looking for more radical examples that really don't rely on technology and electricity.</p>
<p>I'm looking for someone who has gone down this train of thought and has interesting examples to share! Thank you!</p>
| |energy-efficiency|architecture| | <p>There are many ways that one can achieve natural cooling and heating. However, the best solutions are very dependent on the situation. Also, the problem with most of these solutions is that you'd need to plan ahead (don't expect to incorporate them easily into an existing structure). And finally almost all of them are most effective for one or two story buildings.</p>
<p><strong>Basic principle</strong>: <em>Keep the sun out in the summer months. Let the summer in during winter months.</em></p>
<p>In any project that involves natural heating/cooling, an engineer/architect in order to offer the most effective solution has to take into consideration:</p>
<ul>
<li>location (latitude and longitude): this is to consider available sunshine during the day and throughout the season.</li>
<li>climate and microclimate (temperatures and humidity): to prioritize between heating / cooling</li>
<li>Orientation of the building.</li>
</ul>
<p>If that information is available then the engineer can start to deploy a number of tools. The caveat is that most of the solutions need to be planned in the building before hand (i.e. need to be incorporated in the design of the building).</p>
<p>An outline of some of the solutions is presented in the list below.</p>
<p><strong>DISCLAIMER</strong>: this is not a complete list.</p>
<p><strong>WARNING</strong>: they will seem in all cases very low tech (that's the point).</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Trombe_wall" rel="nofollow noreferrer">Trombe wall</a>:</li>
<li>Effective use of plants.</li>
<li>creating passively airflows in the building that promote heating and cooling</li>
<li>using "heat batteries" in the house</li>
</ul>
<h1>Trombe Wall</h1>
<p>The Trombe wall is a passive heating wall arrangement which collects heat during the day and then slowly releases it/creates warm/cool air drafts. The principle of operation can be seen in the next image.</p>
<p><a href="https://i.stack.imgur.com/Q966i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q966i.png" alt="enter image description here" /></a></p>
<p>Basically is a wall behind a glass. The glass allows sun radiation to come through and hit the wall. Depending on whether cooling/heating is required the user can open up openings in the wall to allow for warm/cool drafts of air.</p>
<p><a href="https://i.stack.imgur.com/Mg4Cb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mg4Cb.png" alt="enter image description here" /></a></p>
<p>Notice, that the wall needs to be south facing in the north hemisphere, and north-facing in the south hemisphere (this is a common pattern in this post). Additionally, there is a requirement for sun and shading devices (to control the amount of sun in the summer months).</p>
<h1>Sun controlling devices</h1>
<h2>Effective use of plants.</h2>
<p>Deciduous trees can be used very effectively as sun controlling devices. Also trellis with seasonal plants on the wall with the greatest sun exposure can be used to great effect.</p>
<p><a href="https://i.stack.imgur.com/bxTof.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bxTof.png" alt="enter image description here" /></a></p>
<p>The idea here is to use a plant, to stop the sun's radiation from hitting the house and warming it in the summer, and allowing it to pass in and warm the house and warm up the house.</p>
<p>The following image shows a house in the <em>southern</em> hemisphere (Australia), with vines on the <em>north</em> side.</p>
<p><a href="https://i.stack.imgur.com/ORUVB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ORUVB.png" alt="enter image description here" /></a></p>
<p>Again the orientation is extremely important. Care needs to be taken with regard the humidity that some plants (especially in the trellis version) can impart.</p>
<h2>Louvres</h2>
<p>Another option for sun control devices are louvres.</p>
<p><a href="https://i.stack.imgur.com/dHSDa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dHSDa.png" alt="enter image description here" /></a></p>
<p>Again, the idea is:</p>
<ul>
<li>block the sun rays from entering in the midday of summer day</li>
<li>allow the sun rays to enter on the midday of a summer day.</li>
</ul>
<h2>creating passively airflows in the building that promote heating and cooling</h2>
<p>This is a matter of design (in part of the world its called 'bioclimatic'). Again the orientation of this design is of paramount importance. i.e. Left side in the image in the north hemisphere should be facing south ( while the right should be facing north).</p>
<p><a href="https://i.stack.imgur.com/FZk8S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FZk8S.png" alt="enter image description here" /></a></p>
<p>The general principle is the following. The windows in the center section (near the top) allow sunlight to pass in during the winter months, but not on the summer months). Also notice that heat collectors on the roof (facing the sun).</p>
<p>Additionally, the windows in the summer months act as a chimney. They can be opened, and as warm airs travels up, it accumulates at the top, and gets out of the building. This also creates a small draft, which creates a sense of cooling.</p>
<h2>using "heat batteries" in the house</h2>
<p>This is something that, is very interesting, however arguably it is difficult to tune it just right. I'll try to describe the principle. The idea is that you have a heat storage buried in the basement of the building. That heat storage can be used for heating and cooling.</p>
<p>The way that heat storage works is that you have heat conductive materials (e.g. steel) exposed and connected directly to the heat storage. The idea is that <strong>during the summer</strong> to open up the house in the evening so that the house cools down (and together with it the heat storage), and during the day to allow the heat storage to heat up.</p>
<p><strong>Conversely, in the winter</strong>: the idea is to heat up as much as possible the house (and the heat storage), and that heat is again release during the evening and night.</p>
<h1>Insulation: The most important thing.</h1>
<p>All of the above, don't make much sense unless the building has a <strong>adequate insulation</strong>. In order to be able to control the heat comfort in a house, with the minimum effort you need to have it properly insulated (thermally and humidity-wise). This is the most important pre-requisite for using any of the above techniques effectively.</p>
<h1>Final thoughts</h1>
<p>The best place to find technologies for passive heating and cooling, is in researching old buildings. There has been a lot engineering ingenuity that has been lost over the ages due to the simplicity and convenience of an HVAC.</p>
| 38959 | Natural ways for passively cooling/heating a home? |
2020-12-07T09:46:08.457 | <p>I bought a cheap version of this machine, because all decent ones had costs of hundreds of dollars of transport to my region. The machine seems to be made of solid parts, but is really badly assembled. I hope to reassemble it into a working state. This is how it should look like:</p>
<p><a href="https://i.stack.imgur.com/rLJMG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rLJMG.jpg" alt="enter image description here" /></a></p>
<p>And it is the photo of the worst part of what I bought. My question is about the brass (?) nut:</p>
<p><a href="https://i.stack.imgur.com/38BX3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/38BX3.jpg" alt="enter image description here" /></a></p>
<p><strong>What is the role of the nut? Why it has something which looks like a small ball inside? Should it be put in the position, in which it is in every other photo of this machine?</strong></p>
| |machine-design|machine-elements| | <p>The image shows what is often known as a zerk fitting, aka a grease fitting. The ball seals the passage from the outside to the internals which require grease.</p>
<p>A grease gun with a properly sized connector will snap onto the zerk fitting. The gun is used to pump grease under pressure to the area described in the image as "no space, grinds," likely removing the grinding.</p>
<p>Pump grease through the fitting until some appears, then wipe off the excess.</p>
| 38970 | What is this strange nut and what to do with it |
2020-12-07T09:48:53.113 | <p>I have been reading various research papers regarding IMU filters and I came to the question of why do we not need to calibrate the accelerometer values as we do for the gyroscope and magnetometer?</p>
| |mechanical-engineering|control-engineering|kalman-filters|calibration| | <p>The first thing that one needs to point out is the following that )<em>usually</em>):</p>
<ul>
<li>accelerometers report acceleration in 1, 2 or 3 orthogonal directions (i.e. second derivatives of <em>length</em>)</li>
<li>gyroscopes report angular velocity (i.e. first derivative of angular position)</li>
<li>magnetometer report amplitude and orientation of a magnetic field.</li>
</ul>
<h2>Magnetometers</h2>
<p>Magnetometers can be used to obtain the orientation using earth's magnetic field. However, magnetometers are very prone to magnetic field disturbances. It is very similar to having a magnet needle near a compass. It essentially introduces a bias/offset. So the idea is to try, and remove the offset with that calibration.</p>
<h2>Gyroscopes</h2>
<p><strong>Gyroscopes</strong> although similar in construction to accelerometers, they serve another purpose altogether. They are usually used for "attitude-sensing". I.e. they are used to find the orientation of an object. However, what they do, is they report the rate of change (i.e. the angular velocity). Given the angular velocity, it is possible to integrate to the angular position.</p>
<p>The problem is that there is always a small error in the measurement. if that error is large, then in the long run you get accumulation and significant error. Algorithms exist that can combine input from sensors (e.g. GPS) and using various techniques (e.g. kalman filtering), it is possible to correct for that error. However, for a gyroscope that remains static it is very difficult (/impossible) to make that type of correction.</p>
<h2>Accelerometers</h2>
<p>The MEMS accelerometers are based on a very simple principle. Its basically a mass on spring which oscillates.</p>
<p><a href="https://i.stack.imgur.com/U2Yt4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/U2Yt4.png" alt="enter image description here" /></a></p>
<p>Based on the oscillation which is generated by the acceleration on the object (see Base induced Vibration), there is a very clear understanding and theory which allows prediction of the acceleration from the mass oscillation.
Since the masses. are really small, there is very little that can change over the period of time (no drift). Therefore, modern MEMS accelerometers don't really require much calibration.</p>
<p>One might say, that I can also use accelerometers to get acceleration and integrate that to velocity and then integrate again over time to obtain position. However then you'd be integrating twice the error (let alone that the axis on the accelerometer change over time), and the error accumulates a lot faster. So usually, accellerometers are not used for orientation but they are used to gauge the resultant acceleration of all three accelerations and then do smart things with that (e.g. identify movement, walking etc). (There are -of course - some exceptions).</p>
<h2>why calibration is not as important on accelerometers</h2>
<p>gyroscopes and magnetometers are used to obtain orientation in space. Both measurements are prone to errors for different reasons (see <em>error accumulation</em> and <em>electromagnetic offset</em>).</p>
<p>Accelerometers are not usually used for orientation measurements (although is some cases they can be used under static conditions). Therefore, the very small error (usually less than 1%) doesn't affect much effective operation.</p>
| 38971 | Why not to calibrate accelerometer in IMU filters? |
2020-12-07T17:09:21.510 | <p>If a 6"x4" asphalt sample is cut down to smaller dimensions (let's say to prepare a DCT testing specimen), does the Gmb value change from the original bigger sample? If it does change during testing, then what might be the factors affecting this change?</p>
| |civil-engineering| | <p>Bituminous Materials and binders specific gravity is much less than the gravel/sand aggregate and is sensitive to ambient temperature. ( AASHTO T 228 and ASTM D 70: Specific Gravity and Density of Semi-Solid Bituminous Materials-) roughly around 1.03 versus 2.6 to 2.65 of aggregate.</p>
<p>So depending on the mixture of these two components that are not uniform due to the random space filled with the binders the specific gravity of smaller samples is expected to vary.</p>
| 38978 | Does bulk specific gravity (Gmb) of asphalt specimen change with sample size? |
2020-12-07T20:02:04.267 | <p>In solid state lasers a light source pumps the medium, which is a crystal. A crystal is excited and emits infrared light, until then everything is clear.</p>
<p>What I want to know: after excitement, what controls the flow of those photons emitted to oscillate in the cavity?</p>
| |optics|lasers| | <p>If a photon hits an already excited atom, it can make it to de-excite and also eject another photon. That is <em>induced emission</em>. This new photon will have the same properties (phase, direction) than the original one.</p>
<p>If more than 50% of the atoms are excited, this causes an exponential increase of the photons in the system, all in the same phase and direction.</p>
<p>The cavity length must be an integer multiple of the photon wavelength, with mirrors on both ends. (One of the mirrors can be a little bit transparent, to enable the beam to get out.) To get the laser beam, the photons must be reflected many times on the mirrors.</p>
<p>Photons whose direction was not exactly perpendicular to the mirrors, will soon leave the system (among with the other photons they made with the induced emission). This is one of the losses which make laser beams highly ineffective.</p>
| 38982 | How solid state lasers work? |
2020-12-08T09:52:26.057 | <p>So , I was reading a course book : ''Modern Control Engineering '' by Ogata and I came forward this statement that made me skeptical:<br />
''The points of root locus only satisfiy the angle - condition. Closed - loop poles are the roots of the characteristic equation and therefore satisfiy both the magnitude and angle condition. If , we want to find the closed - loop poles for a given gain value K , then we have to ...''</p>
<p>I am not sure why does not every point of root locus correspond to a closed - loop pole for some K. Is it true? And if so , these points that are not closed - loop poles for any K , do they have any meaning/ physical/geometrical/control theory ?</p>
<p>I even have an assumption to this: Maybe he means , that if K has to be integer ( K may has to be integer , in ''real world applications''(?))? Then of course , root locus would have a discrete-represantion graph , so he means that if K - is integer , in root locus we connect the dots (to have a continoous represantantion) and thefore not every point of it is closed-loop pole and everything would make sense.<br />
I even graphed, a root-locus in python : If you click an any point python returns the corresponding K for this point (which may be float) . So is this the reason? Let me know if I am wrong/right</p>
| |control-engineering|control-theory|feedback-loop| | <p>Suppose we have a system <span class="math-container">$G(s) = \frac{1}{s(s+1)}$</span> and controller <span class="math-container">$K$</span> (this is purely a gain) and we close the loop:
<span class="math-container">$$T(s) = \frac{KG(s)}{1+KG(s)} = \frac{K\frac{1}{s(s+1)}}{1+K\frac{1}{s(s+1)}}$$</span>
<span class="math-container">$$ = \frac{K}{s^2+s+K} $$</span>
As you might notice, the poles of this closed-loop equation depend on the value of <span class="math-container">$K$</span>:
<span class="math-container">$$s^2 + s + K = 0 \rightarrow s = -0.5\pm\sqrt{0.25 - K}$$</span>
This means that one can influence the behaviour by only changing this <span class="math-container">$K$</span> to any arbitrary real value (imaginary might sound cool in simulation, but its kinda hard to put an imaginary voltage to a system for instance). The root locus plot represents for how the poles shift for changing values of <span class="math-container">$K$</span> (where <span class="math-container">$K > 0$</span>). As you stated, if <span class="math-container">$K$</span> can only be an integer, you will indeed not get a continuous function, only dots at the places where <span class="math-container">$K$</span> exists. But as I stated earlier <span class="math-container">$K$</span> can by any real value, however for negative values you can quickly see the system becomes unstable.</p>
<p>EDIT: I have rewritten this part in a more elaborate proof.
Instead of looking at a specific system, suppose an arbitrary system
<span class="math-container">$$G(s) = \frac{N(s)}{D(s)}$$</span>
The angle condition is the point at which the phase of the open loop system is an odd multiple of <span class="math-container">$-180^o$</span> or in other words:
<span class="math-container">$$\mathcal{Im}\left\{\frac{KN(s)}{D(s)}\right\} = 0 ~~\text{ and }~~ \mathcal{Re}\left\{\frac{KN(s)}{D(s)}\right\} < 0$$</span>
The magnitude condition is the point where the magnitude of the open loop transfer equals 1. If both the magnitude condition and the angle condition match, the denominator of the closed loop transfer function becomes 0 (which is something we tend to avoid at all cost). Now, suppose we take a value for <span class="math-container">$K$</span> and calculate the value for <span class="math-container">$s = s_0$</span> such that the open loop transfer equals <span class="math-container">$-1$</span>:
<span class="math-container">$$\frac{KN(s_0)}{D(s_0)} = -1 \rightarrow \frac{N(s_0)}{D(s_0)} = -\frac{1}{K}$$</span>
We know this <span class="math-container">$s_0$</span> lies on the root locus. What you might also notice is that the sign does not change if I change <span class="math-container">$K$</span> to any positive, real value. Therefore, if <span class="math-container">$s$</span> lies on the root locus, the angle condition holds for any positive, real <span class="math-container">$K$</span>. You could also reflect this in the bode plot, as <span class="math-container">$K$</span> only changes the magnitude, not the phase plot. As you might expect, this also implies the inverse holds: the angle condition holds for any value of <span class="math-container">$s$</span> that lies on the root locus of <span class="math-container">$G(s)$</span>.
The magnitude condition only holds for a finite set of values of <span class="math-container">$s$</span> on the root locus (the solution at which the characteristic equation equals 0).
I hope I explained it a bit better, you can try also to just write out a couple cases and you quickly notice the same.</p>
| 38990 | Is it true that points of root locus only satisfiy the angle condition? |
2020-12-08T09:53:22.467 | <p>Lately, I've been pondering why some engine blocks are so bulky, I always thought intuitively it was because they had to last a long time whilst containing thousands of combustion cycles but the more, I look into the reality of engine design that understanding doesn't always hold up to the design of an engine.</p>
<p>It seems cylinder walls have a set thickness which makes pretty good sense, then there is a water-jacket chamber for coolant which again, makes sense but then there is an outer shell of material that is sometimes even thicker than the cylinder walls themselves. This is where I become confused (An example image below)</p>
<p><a href="https://i.stack.imgur.com/EECAz.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/EECAz.jpg" alt="enter image description here" /></a></p>
<p>Surely, this outer shell doesn't need to be as thick as it is. In my mind all it's doing is containing a pressurized coolant which can't amount to more pressure than the combustion of the engine itself. So why is it sometimes as thick as the cylinder wall?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|automotive-engineering|vibration| | <p>No one has mentioned yet that there are structural loads placed on the block by other components. The first that comes to mind are the cylinder heads, and as it (they) are torqued, the block needs to resist distorting due to those loads.</p>
| 38991 | Why are engine blocks so robust apart from containing high pressure? |
2020-12-08T12:18:31.103 | <p>In a gear you normally have a center hole for the shaft to be put into. Why do some gears have holes surrounding that center hole?</p>
<p>Is it for weight reasons? Does it add stability? Is it so you can add special parts?</p>
| |mechanical-engineering|materials|gears|statics|structures| | <p>Having been a designer of the old style parking meters (AMF)and time switches (Venner) which contained very many brass gears the unused extra holes were there because of weight saving, but more importantly cost savings. The holes cost little extra on a pierce and blanking tool, but the scrap brass was worth a lot. At that time all our scrap brass and copper could be sold as scrap for around 80%+ of the new purchase price. At times of high inflation of our raw materials we often joked that selling our stores stock of hundreds of tons of copper, brass and stainless steel for scrap would make us more profit than turning it into products.</p>
| 38994 | Why does a gear have extra holes? |
2020-12-08T12:48:59.277 | <p>I'm interested in how the steering angles of a modern car are calculated, I'm assuming it should be similar to Ackermann Steering formula.</p>
<p>However, the wikidepia page for Ackermann Steering mentions that "<em>Modern cars do not use pure Ackermann steering, partly because it ignores important dynamic and compliant effects, but the principle is sound for low-speed maneuvers.</em>"</p>
| |automotive-engineering| | <p>In my opinion, steering angle can be obtained by the information of lateral forces which are produced by the tires. Force can be calculated by chasis dynamo or any mechanism and any angle can be estimated or obtained by filtering methods or some of calculations. In bicycle model, we ignore some of important things and results from it may be far from correct values. Even so, bicyle model gives us useful informations.</p>
| 38998 | What's the exact formula for calculating steering angles of a modern car? |
2020-12-09T03:17:29.867 | <p>In an airfoil, a low-pressure region forms in the turbulent wake created when the flow separates. Wouldn't this cause an increase in the net vertical force acting on the airfoil since the pressure at the top decreases?</p>
| |fluid-mechanics|aerospace-engineering|aerodynamics| | <p>To add more to Kamran's answer, Generally for streamlined bodies like aerofoils, pressure drag force << Skin friction drag force << lift force. If the flow is separated, there will a reverse flow on the surface of the aerofoil and this will lead to a drastic increment in skin friction drag. so the efficiency <span class="math-container">$(C_L/C_D)$</span> of the airfoil also comes down.</p>
| 39024 | Why does flow separation cause stall when it creates a low pressure zone? |
2020-12-09T08:41:27.473 | <p>Could you explain me the following convertion:</p>
<p>The address range is 0x00100000 to 0x1fffffff</p>
<p>How to convert it to bytes or Mb?</p>
| |electrical-engineering| | <p><span class="math-container">\begin{array}{cc}
& 0x &\ 1 &\ F &\ F &\ F &\ F &\ F &\ F &\ F \\
- &0x &\ 0 &\ 0 &\ 1 &\ 0 &\ 0 &\ 0 &\ 0 &\ 0 \\
+ &0x &\ 0 &\ 0 &\ 0 &\ 0 &\ 0 &\ 0 &\ 0 &\ 1 \\
= &0x &\ 1 &\ F &\ F &\ 0 &\ 0 &\ 0 &\ 0 &\ 0 \\
\times & &\ 16^7 &\ 16^6 &\ 16^5 &\ 16^4 &\ 16^3 &\ 16^2 &\ 16^1 &\ 16^0 \\
\end{array}</span></p>
<p>Adding 1 goes from start to end.</p>
<p><span class="math-container">$$1 \times 16^7 + 15 \times 16^6 + 15 \times 16^5 = 535,822,336\ bytes$$</span></p>
<p>0x100000 = 1MB = 1,048,576 bytes, so the address range from 0x00100000 to 0x1fffffff is 511MB.</p>
| 39030 | Address range to Mb |
2020-12-10T05:46:40.203 | <p>Oil industry in permain basin is collapsing.</p>
<p>Also, no new technology or anything to save it this time.</p>
<p>Is this the beginning of the end for oil?</p>
| |energy|petroleum-engineering| | <p>This is an interesting subject, although its mostly an opinion rather that anything else.</p>
<p><strong>TL;DR: Oil industry is certainly facing challenges and its probably on its way out, but there still a long way until it completely phases out</strong></p>
<p>I'll start with the a quote from <a href="https://www.eia.gov/todayinenergy/detail.php?id=45096#:%7E:text=The%20share%20of%20U.S.%20total%20energy%20consumption%20that%20originated%20from,of%2086%20quads%20in%202007." rel="nofollow noreferrer">eia.gov</a></p>
<p><em><strong>The share of U.S. total energy consumption that originated from fossil fuels has fallen from its peak of 94% in 1966 to 80% in 2019.</strong></em></p>
<p>In the following graph is a graph from Monthly Energy Review</p>
<p><a href="https://i.stack.imgur.com/m9nT2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m9nT2.png" alt="enter image description here" /></a></p>
<p>Obviously, renewable energy has started to make dent. Hydro already had approximately 5-10% since the early 1950s. Also in the following graph, it can be seen that hydro and the other renewables have a steady growth (especially after early 2000 (basically they doubled energy production in the last 15 years).</p>
<p><a href="https://i.stack.imgur.com/6IwEj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6IwEj.png" alt="enter image description here" /></a></p>
<p>However, while fossil fuels (coal, gas, oil) can be used for heating/cooling, electricity generation and as transportation fuels, the renewables have only managed to penetrate convincingly electricity generation and heating/cooling application. That leaves transportation. Some might argue that there are already electrical vehicles, however IMHO there is still some headway towards that.</p>
<p>To me the largest problem is the infrastructure that has already been built around automotive fuels. In the last 100 years, there has been a steady investment in automotive fuel stations and technology. The result is that there is now almost in any place in the world a fuel station ready to supply fossil fuel. Sadly the same cannot be said for electric vehicles.</p>
<p>Of course, there is heavy legislative push worldwide, towards electric vehicles and infrastructure. However to me, the need for subsidy (of sorts) is an indication that still there are still challenges with respect to the adoption of electrical vehicles (and implicitly adoption of renewable energy generation).</p>
<p>Additionally, almost no progress has been done towards the industrialization of electrical aviation and naval transportation. And there doesn't appear to be any indication of that happening any time soon.</p>
<p>Another thing that me is very important to is that, despite all the changes, still 80% of the energy consumption is based on fossil fuels. Although coal is definitely on its way out, oil and gas are still controlling approximately 70% of the energy production.</p>
<p>Given that it took approximately 50 years to go from 94% to 80% its hard to see how in 50 years its going to be possible to become independent from oil industry. Especially given the fact, that oil industry is the base of the plastic industry.</p>
<h2>Economics</h2>
<p>There are many issues also with respect to economics, which indicate the elasticity of oil (elasticity in the 'economics' sense). This results in High volatility of oil. This high volatility is depended on many aspects geo-political in nature primarily. However, in some cases, there are other issues that affect the price, see how the COVID-19 pandemic -momentarily- made price of oil negative.</p>
<h2>Lack of an immediately available viable alternative</h2>
<p>Finally another issue that needs to be addressed is that currently world energy consumption is increasing with a steady pace. If the next image is any indication, then the increase is in the order of 15000TWh per 10 years.</p>
<p><a href="https://i.stack.imgur.com/GCRLq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GCRLq.png" alt="enter image description here" /></a></p>
<p>Renewable energy is managing marginally to keep up with the annual increase in energy (currently its at a pace of about 5000TWh per 10years). So unless there is a another viable alternative there is no way to become independent of the oil industry. As viable alternative, currently I would only consider energy generation through fusion, which however is still trying to break even and its not expected to yield any commercial results for at least 15 to 20 years.</p>
| 39049 | Is oil collapsing |
2020-12-10T13:21:45.620 | <p>I am looking for a voltage input module that measures a voltage input in the range of 0 to 10 Volts (slowly changing during a 5 seconds period). It should be connected to a local Windows computer that has a preexisting .NET (C#) software program (with its source code available) for plotting the input. I am completely inexperienced in this area of device acquisition, so I find it difficult to select the device model that would fit the purpose, and, confirm the existence of the accompanying .NET API.</p>
<p>For example, if I select a USB voltage input module from Lucid Control company, they offer these two downloads (scroll below to see the API + .Net examples) that do not seem to function. The supplied .dll file cannot be added to the project. The documentation is lacking.
<a href="https://www.lucid-control.com/downloads/" rel="nofollow noreferrer">https://www.lucid-control.com/downloads/</a></p>
<p>That shows that a solid device recommendation is needed to accommodate an easy update of the program code using the provided API.</p>
<p>Thanks.</p>
| |electrical-engineering|measurements| | <p>You can try <a href="https://labjack.com/support/software/examples/ljm/net" rel="nofollow noreferrer">Labjack</a>. They are relatively cost effective solutions, with good documentation (I've primarily used python though).</p>
<p>They come in different configurations.</p>
| 39052 | Can you suggest a 0 to 10 Volts mesuring unit with an easy .NET (C#) API? |
2020-12-10T15:54:39.580 | <p>I´m new to Siemens NX 12 coming from Fusion 360. I´m having problems with the extrusion command in NX 12. I´ve created a sketch where I want to extrude one big body and then add with the second extrusion some parts to it. When I select the lines of the sketch and then look at the output of this operation I only get walls with no thickness but not a solid body. How can I change this behavior?</p>
<p>thanks for your help :)</p>
| |cad| | <p>I had an open sketch because I used projected curves as a sketch ending.</p>
| 39054 | How can I change the type of extrusion output in Siemens NX 12? |
2020-12-10T18:55:36.110 | <p>I'm learning about composite resins, and the various filler additives, but I wondered about the relation between the two. For example....</p>
<p>An epoxy resin has poor adherence to a material such as polyethylene, but it has strong adherence to a material such as polyamide 66. If you have an epoxy resin with powdered polyethylene, vs polyamide 66, beyond the characteristics of the powdered material itself, what impact does the resins ability to adhere to the powered filler have?</p>
| |materials|composite-resin| | <p>Generally composites with longer fibers are more affected by the fiber matrix interface. The smaller the reinforcement particulate size usually the effect from bad interface is less pronounced. However again that has to do with the properties of the matrix e.g energy for crack propagation in the bulk matrix material.</p>
| 39059 | Learning about composite resins, should the resin be able to adhere to the selected filler? |
2020-12-11T02:54:02.177 | <p>My team is trying to build a mechanical cannon for a final project for a course and we are running into some problems with our motor stalling.</p>
<p>Our motor stalls when the plate hits either side of this railing (as shown with circles below) because, I think, it creates infinite tension that the motor can’t overcome. Once our DC motor is fully stalled and refuses to move in the direction of infinite tension, we try to move the motor in the opposite direction to free it up, but it stays stuck. We need to physically intervene to “unjam” the motor. Sometimes we can also press our button to move the motor backwards quickly and consecutively and it will unwind.</p>
<p>We don’t want to use a limit switch since we have to stay within a “required budget”, so we are wondering if there would be any other way to prevent this motor from becoming jammed. We were thinking that the jamming of the motor occurs because there is too high of a current draw when it stalls, so a solution would be to put a resistor in series with the motor driver + DC motor interface to reduce the amount of current that is drawn when the motor stalls. We have no idea if this is a robust solution that would allow the motor to easily unjam. Any ideas would be really appreciated !</p>
<p>Some specs for context:</p>
<p>DC motor that moves plate:</p>
<p><a href="https://rads.stackoverflow.com/amzn/click/com/B07556CZL1" rel="nofollow noreferrer" rel="nofollow noreferrer">DC Motor Threaded Output Shaft Lead Screw 6V Motor Stepper Lead Screw Actuator Gear Motor with Long M355MM Lead Screw Thread Output Shaft</a></p>
<p>Motor has a 6V rating and a stall current of 1A</p>
<p><a href="https://i.stack.imgur.com/ULj6I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ULj6I.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|motors|torque|power-electronics|embedded-systems| | <p>I think it's the effect of "tightening the screw", a lot of friction distributed over a large area. Even if the plate doesn't bend, it will still accumulate a bunch of tension, and as a result, friction. You need to stop the motor before that happens.</p>
<p>I think a mechanical stop could work: put a small "blade" of sheet metal on the shaft (squeeze it between two nuts to fix it in place) and a nail or other small bar / protrusion / obstacle on the moving plate. Angle the "blade" slightly, like a propeller, so that during the motion of the plate outwards, its leading edge will catch against the nail and stop any further rotation dead, but on motion in the opposite direction, will slide over the nail, at worst flexing slightly.</p>
<p><a href="https://i.stack.imgur.com/CIJ96.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CIJ96.png" alt="enter image description here" /></a></p>
| 39064 | Preventing a DC motor from stalling |
2020-12-11T08:57:00.097 | <p>To obtain lateral force according to slip angles, I use Pacejka Magic Formula, but my graph's style is not similar with picture which i added. I cannot find where the mistake is.</p>
<pre><code>clear all
clc
a1y=-22.1;
a2y=1011;
a3y=1078;
a4y=1.82;
a5y=0.208;
a6y=0.00;
a7y=-0.354;
a8y=0.707;
Fz=2;
Sh=-0.28;
Sv=-118;
Cy=1.5;
Dy=a1y*Fz^2+a2y*Fz;
BCDy=a3y*sind(a4y*atand(a5y*Fz));
By=BCDy/(Cy*Dy);
Ey=a6y*Fz^2+a7y*Fz+a8y;
alpha_r=-10:0.01:10;
PCKy=(Dy*sind(Cy*atand(By*rad2deg(alpha_r+Sh) - Ey*(By*rad2deg(alpha_r+Sh) - atand(By*rad2deg(alpha_r+Sh))))))+Sv;
plot(alpha_r,PCKy)
</code></pre>
<p><a href="https://i.stack.imgur.com/q2ddu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q2ddu.png" alt="Lateral force" /></a></p>
<p><a href="https://i.stack.imgur.com/PaEbh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PaEbh.png" alt="enter image description here" /></a></p>
| |automotive-engineering|wheels| | <p>The offending line was</p>
<p><code>PCKy=(Dy*sind(Cy*atand(By*rad2deg(alpha_r+Sh) - Ey*(By*rad2deg(alpha_r+Sh) - atand(By*rad2deg(alpha_r+Sh))))))+Sv;</code></p>
<p>The correct form is without the rad2deg</p>
<p><code>PCKy=(Dy*sind(Cy*atand(By*(alpha_r+Sh) - Ey*(By*(alpha_r+Sh) - atand(By*(alpha_r+Sh))))))+Sv;</code></p>
<p>if you just correct that you should get:</p>
<p><a href="https://i.stack.imgur.com/oGsHg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGsHg.png" alt="enter image description here" /></a></p>
| 39072 | Magic Formula: Calculating of lateral forces |
2020-12-12T06:49:22.157 | <p>In a college project, I was instructed to design the floor plans and the conduit layout of a duplex building. It was instructed that the duplex is 2500 square feet. No other instruction was provided. Now while drawing, should I draw 2500 square feet for each of the floors (that is 5000 sq ft altogether) or 1250 sq ft for each (2500 altogether)?</p>
| |civil-engineering|building-design| | <p>It’s total square footage not per floor square footage.</p>
<p>Btw, the AIA has established EXACTLY how to calculate square footage of a building. That is to say, overhangs, canopies, balconies, pipe chases, open terrace, etc. is considered 1/2 square footage. So, if there is a 200 square foot entry canopy, it’s considered 100 square feet in the calculations. (You can see this graphically drawn in AIA Document D101...you can Google it. This document is part of the Architect’s contract.)</p>
<p>So, your 5,000 sf building is not necessarily 2500 sf per duplex nor 1250 sf per floor in each duplex. You’ll need to subtract canopies, overhangs, etc. and then divide the balance of the square footage up with each duplex and each floor.</p>
| 39084 | How to calculate the square footage of a duplex building |
2020-12-12T14:47:16.073 | <p>I am trying to complete the following question. But feel it is not correct, so need some help.</p>
<p>Given that heat transfer via convection off a planar wall is given by:</p>
<p><a href="https://i.stack.imgur.com/opiCp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/opiCp.png" alt="enter image description here" /></a></p>
<p>Write out the combined thermal resistance for the following system (only thermal conduction and convection present)</p>
<p><a href="https://i.stack.imgur.com/WImRR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WImRR.png" alt="enter image description here" /></a></p>
<p>My attempt:</p>
<p>Combined Thermal Resistance:</p>
<p><span class="math-container">$$R1 = \frac{L1}{K1A1}$$</span>
<span class="math-container">$$R2 = \frac{L2}{K2A2}$$</span></p>
<p><span class="math-container">$$RTotal = R1 + R2$$</span>
<span class="math-container">$$RTotal = \frac{L1}{K1A1} + \frac{L2}{K2A2}$$</span></p>
<p><span class="math-container">$$RTotal = \frac{L1}{K1A} + \frac{L2}{K2A}$$</span></p>
<p><span class="math-container">$$Q= \frac{\delta T}{ER}$$</span></p>
<p>Is what i have done so far correct?</p>
| |thermodynamics|heat-transfer| | <h1>Conductive</h1>
<p>The Conductive heat transfer is given by:</p>
<p><span class="math-container">$$\dot{Q} = \frac{k}{L} A \Delta T$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$k$</span> = is the heat conductivity of the material in this case aluminimum (<span class="math-container">$\frac{kCal}{m°C}$</span>)</li>
<li><span class="math-container">$L$</span> is the thickness of the wall</li>
<li>A is the total exchange surface</li>
<li><span class="math-container">$\Delta T$</span> the temperature difference</li>
</ul>
<h1>Convective heat transfer</h1>
<p>Convective heat transfer is when a solid surface and a fluid (liquid or gas) exchange heat. The total rate of exchanged heat is:</p>
<p><span class="math-container">$$\dot{Q} = h_c A \Delta T$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$h_c$</span> = heat transfer coefficient (<span class="math-container">$\frac{kCal}{m^2h°C}$</span>)</li>
<li>A is the total exchange surface</li>
<li><span class="math-container">$\Delta T$</span> the temperature difference</li>
</ul>
<h1>Total thermal resistance:</h1>
<p>The total thermal resistance in your example will be given by</p>
<p><span class="math-container">$$R = \frac{1}{h_{1}A} + \frac{L_1}{k_1 A}+\frac{L_2}{k_2 A}+ \frac{1}{h_{2}A}$$</span></p>
<p>where:</p>
<ul>
<li>R is the thermal resistance</li>
<li><span class="math-container">$h_{1}$</span> is the convective coefficient on the left side of the problem</li>
<li><span class="math-container">$h_{2}$</span> is the convective coefficient on the right side of the problem</li>
<li><span class="math-container">$k_i$</span> is the heat conductivity coefficient of the material</li>
<li><span class="math-container">$L_i$</span> is the thickness for each wall</li>
<li><span class="math-container">$A$</span> is the area of the surface.</li>
</ul>
<p>When you calculate R, then you can use it the following way:</p>
<p><span class="math-container">$$ \dot{Q} = \frac{T_{\infty 1}-T_{\infty 2}}{R}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$T_{\infty 1}$</span> is the operating temperature inside of the box</li>
<li><span class="math-container">$T_{\infty 2}$</span> is the operating temperature outside of the box</li>
</ul>
| 39088 | Thermal Resistance |
2020-12-13T19:34:35.157 | <p><a href="https://i.stack.imgur.com/1jlaq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1jlaq.png" alt="enter image description here" /></a></p>
<p>In this question I understand what we have to do to solve it and is equate the sum of moments with the moment of inertia. However I am not sure on how the LHS became negative as I see that the angular acceleration follows the same direction as the moment, so positive.</p>
<p><a href="https://i.stack.imgur.com/Irzxj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Irzxj.png" alt="enter image description here" /></a></p>
<p>So why is the LHS negative?</p>
| |mechanical-engineering|dynamics| | <p>If im not mistaken the angle is increasing when the door is opening. That is defined as positive angular displacement.</p>
<p>On the other hand if you see the moments that are acting in the door they appear to be directed towards clouding the door.</p>
<p>Thus the -sign in the equation.</p>
| 39097 | How do I determine the angular acceleration? |
2020-12-13T22:30:10.300 | <p><a href="https://i.stack.imgur.com/mlBN3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mlBN3.png" alt="enter image description here" /></a></p>
<p>In structural engineering, a pad footing under a normal load N and a moment M is assumed to have a linear distribution of stress under it. <strong>Is there any intuition why we can make this assumption?</strong></p>
| |structural-engineering| | <p>The stress distribution must be in equilibrium with the applied load, and it is a reasonable assumption that it cannot be in tension. Notice how that creates the difference between cases (b) and (c) in your attachment, depending on the relative size of the direct load and the moment.</p>
<p>There must be at least two independent parameters in the stress distribution to match up the two independent parameters in the load (i.e. the direct load and the moment).</p>
<p>In the general situation there is not enough information to determine more than two parameters, so if the simplest assumption (a linear distribution of stress) is reasonably consistent with experiment, there is no reason to do anything more complicated.</p>
<p>A more complicated assumption with more parameters would require more assumptions about the behavior of the footing.</p>
<p>Civil engineering is not an exact science. If simple assumptions plus a safety factor are enough to make designs that don't fall down when you build them, that is usually "good enough" unless you are designing a safety-critical structure.</p>
| 39101 | Why can we assume the stress distribution under a pad footing to be linear? |
2020-12-14T10:03:51.020 | <p>I have to study the effect of low-pass filters (if any) on the outcome of the experiment. And the experiment here is to measure the vibration of a (metal) specimen by hitting the said specimen with impulse (or "knocking").</p>
<p>Parameters for measurement include:</p>
<ul>
<li>Average complex (convert from time to frequency domain with FFT)</li>
<li>Hann (or Hanning) window function (this is the next parameter study,
for now, bear with it)</li>
<li>20kHz total bandwidth with 25600 FFT lines (or
25.6kHz sampling rate)</li>
</ul>
<p>Theoretically speaking, applying the low-pass filter should lead to some differences in the frequency spectrum of the transfer function. Yet, in the image below, there is practically none. Sure, there is a drop of 1~2 dB from 350 to 500 Hz, but my instructor says that this frequency range is mostly noise, and they are not as important as the peaks (where there is no significant change).</p>
<p>Can anyone explain this to me, why there is no difference in the spectrum?</p>
<p>Further check on the components also say that there is only negligible differences in both the input (knocking force) and the output (vibration). My hunch is that for the force, because I only "knock", which means there is no frequency range, and I generate the knocking signal in time domain, so there is no frequency range to block. In other words, because of my setting for the input, the filter has no effect. However, this would not explain the situation with the output and the transfer function.</p>
<p><a href="https://i.stack.imgur.com/LtBgW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LtBgW.png" alt="enter image description here" /></a></p>
<p>Edit: Additional graph, with frequency spectrum runs from 0 to 20kHz, comparing with and without 100 Hz filter (with 3 options: No filter (black), Filter on Output only (Yellow), Filter on both input and Output (Blue))</p>
<p>Linear scale for frequency
<a href="https://i.stack.imgur.com/IaxEi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IaxEi.png" alt="enter image description here" /></a></p>
<p>Logarithmic scale for frequency
<a href="https://i.stack.imgur.com/HnCmP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HnCmP.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|vibration|signal-processing| | <p>This is more of a continuation of the comments. Maybe later with more information it will be converted to a proper answer.</p>
<h1>Low pass filter frequency response</h1>
<p>Low pass filters have a specific response with respect to frequency. A typical example for a low pass filter is presented below.</p>
<p><a href="https://i.stack.imgur.com/CFqU3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CFqU3.png" alt="enter image description here" /></a></p>
<p>In the image above, the cutoff frequency is 1.</p>
<p>Notice that <strong>the x-axis is logarithmic.</strong> One of the most important features is that the filter attenuation is linear in this log scale.</p>
<p>As you might notice, that at a frequency which is double the cutoff frequency you have only a few <em>db</em> drop.</p>
<h2>Your example</h2>
<p>In your example, the x-axis is linear. The plot is not entirely clear and I can't be sure what the cutoff frequency is used. However, I will assume that you use 200Hz as a cutoff frequency. If you notice up to 200Hz hardly anything happens. Only above 200Hz you start seeing a difference. That difference becomes greater for higher frequencies. Around 400 Hz its in the order of a couple db.</p>
<p>This is consistent with the typical response graph, presented above. However lacking more specific data on the filter parameters its difficult -for me- to hazard another guess.</p>
<p>Given the sampling frequency, you could present data up to 12.8kHz. If you present that you should see greater attenuation of the higher frequencies.</p>
<h2>Other filters</h2>
<p>There are other filters with more pronounced cutoff regions. If you used one of these you could have higher attenuation at frequencies closer to the cutoff frequency. The main problem, is that some of them could mess the signal, and the phase.</p>
<p>Below is comparison between</p>
<ul>
<li>different orders of a low pass Butterworth signal</li>
</ul>
<p><a href="https://i.stack.imgur.com/90V3g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/90V3g.png" alt="enter image description here" /></a></p>
<ul>
<li>different types of filters</li>
</ul>
<p><a href="https://i.stack.imgur.com/WaKjN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WaKjN.png" alt="enter image description here" /></a></p>
| 39109 | Non-observable effect of low-pass filter on the Transfer function (of Vibration measured with Laser scanner) |
2020-12-14T20:07:12.137 | <p>I am writing a report and need to reference every equation used. I am using the equation for shear stress in a rectangular beam of</p>
<p>tau = 3/2 * V/A</p>
<p>but am not sure where it came from or what textbook to reference it from! It just seems like common engineering knowledge so I don't know who to put down as the originator of the equation.</p>
<p>Thanks</p>
| |structural-engineering|structural-analysis|stresses|beam|shear| | <p>you can use the following link</p>
<p><a href="https://www.informit.com/articles/article.aspx?p=2982118&seqNum=7" rel="nofollow noreferrer">Advanced Mechanics of Materials and Applied Elasticity, 6th Edition</a> by By Ansel C. Ugural and Saul K. Fenster.</p>
<p>You can find this derivation in literally every handbook of mechanics of materials.</p>
| 39118 | Where does the equation for shear stress of a rectangular beam (3/2 V/A) come from? |
2020-12-15T01:12:28.060 | <p>What does it mean to say that there are "coupling effects" between some elements of an optical system? I've searched through the literature, but I cannot find a general explanation of what this means – every mention of it assumes that the reader already knows what it means.</p>
| |optics|coupling| | <p>Based on <a href="https://www.osapublishing.org/josaa/abstract.cfm?uri=josaa-16-7-1730" rel="nofollow noreferrer">this paper</a>, as well as other sources I've come across in my research, it seems to just be a technical way of saying that the elements, or the action of the elements, "depend" on each other (have "dependancy" on the state of each other in some way).</p>
| 39122 | What does "coupling effects" mean in the context of optical systems? |
2020-12-15T08:12:11.760 | <p>I have a sensor whose datasheet mentions that it should be connected to 5V Vcc.</p>
<p>A green LED glows on the sensor when it is powered from the source at 5V.</p>
<p>However, even if I power it with 3.3V, still the green LED glows.</p>
<p>Does this affect the working of the sensor in any way?</p>
<p>If somebody could please let me know about this, I would be very much thankful.</p>
| |sensors| | <p><strong>If it says 5V power it up with 5V.</strong> The fact that the LED is glowing does not necessarily mean that the sensor is operating as it should.</p>
<p>You can only safely say that the LED is receiving enough voltage to operate (nothing more nothing less), unless you have the manual of the sensor that says otherwise (e.g. if the greenlight on the LED is indicative of appropriate power supply).</p>
| 39126 | Sensor compatibility with power source |
2020-12-15T12:41:16.463 | <p>I'm studying <a href="https://structural-analyser.com/domains/SteelDesign/Buckling/" rel="nofollow noreferrer">this</a> text about buckling.</p>
<p>There a factor called <span class="math-container">$\lambda_1$</span> is defined:</p>
<p><span class="math-container">$$\lambda_1=\pi \sqrt{\frac{E}{f_y}}=93.9\sqrt{\frac{235}{f_y}}$$</span></p>
<p><strong>But how was that last step made?</strong> Where did that factor of 93.9 come from? Usually we take the elastic modulus of structural steel as around 200GPa, so how is there 235 under the square root? I feel like there is some step missing.</p>
| |structural-engineering| | <p><strong>Edit</strong></p>
<p>I have changed the units to MPa, basically cranking their numbers back to see what unit they have used for E.</p>
<p>Without looking at your source, it makes sense if they are using the unit of MPa.</p>
<p><span class="math-container">$E=210.000MPa$</span></p>
<p>then:</p>
<p><span class="math-container">$$\lambda_1=\pi\sqrt\frac{210,000}{F_y}=\sqrt\frac{893.617*235}{F_y}$$</span></p>
<p><span class="math-container">$\lambda_1=93.91\sqrt\frac{235}{F_y}$</span></p>
| 39135 | $\lambda_1$ factor in context of buckling |
2020-12-15T20:21:43.453 | <p>I have trouble understanding Virtual Work (in statics) as I understand what's happening but I do not understand the signs or directions.</p>
<p>This is what I know. If the force acts in the opposite direction to the displacement then it's taken as negative. But sometimes we don't do this and instead we take the derivative of an angle. Then, the derivative will account for the direction. (Do we also assume the direction of the derivative and plot it and that is why the following part happens?)</p>
<p><a href="https://i.stack.imgur.com/CbDoz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CbDoz.png" alt="enter image description here" /></a></p>
<p>In this exercise, the work would be the work done by the spring in the positive y and x axis (standard) and the <span class="math-container">$W$</span>, multiplied by their respective distances. In this case I took <span class="math-container">$d_y=-0.45\sin\theta\text{d}\theta$</span> and <span class="math-container">$d_x=-0.45\cos\theta\text{d}\theta$</span>.</p>
<p>What I thought was substituting <span class="math-container">$d_y$</span> and <span class="math-container">$d_x$</span> into <span class="math-container">$\text{d}U= Wd_y + F\cos\alpha\text{d}x + F\sin\alpha\text{d}y$</span>.</p>
<p>However in the solution it is said that instead is <span class="math-container">$-Wd_y - F\cos\alpha\text{d}x + F\sin\alpha\text{d}y$</span>.</p>
<p>What am I misunderstanding? I hope it is clear what I'm trying to say.</p>
| |mechanical-engineering|statics| | <p>I came up with yet another result.</p>
<p>Assuming <span class="math-container">$a$</span> is the angle of spring force with the horizontal, I end up with the following equilibrium around point B</p>
<p><a href="https://i.stack.imgur.com/NIDTw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NIDTw.png" alt="enter image description here" /></a></p>
<p>Note that:</p>
<ul>
<li><span class="math-container">$ds = 0.45\cdot d\theta$</span> , where <span class="math-container">$ds^2 = dx^2 + dy^2$</span> and the angle of <span class="math-container">$ds$</span> with the negative horizontal is <span class="math-container">$\theta$</span>.</li>
</ul>
<p>Although <span class="math-container">$ds$</span> doesn't come into the calculation it is important because it determines the direction of dx, dy. More specifically, for increasing <span class="math-container">$\theta$</span> the direction of <span class="math-container">$ds$</span> is pointing down and to the left.</p>
<ul>
<li><span class="math-container">$R_{BC}$</span> is perpendicular to <span class="math-container">$ds$</span> (therefore produces no work).</li>
</ul>
<p>So the virtual work (I'll denote it as <span class="math-container">$VW$</span> to avoid confusion with W) for each force is:</p>
<ul>
<li><span class="math-container">$VW_{R_{BC}}=0$</span> perpendicular to <span class="math-container">$ds$</span> (therefore produces no work).</li>
<li><span class="math-container">$VW_W = W\cdot dy$</span></li>
<li><span class="math-container">$VW_F =-Fx \cdot dx- Fy \cdot dy$</span></li>
</ul>
<p>The reason this is negative for weight, is because the displacement dx and dy is opposite to the direction of the force.</p>
<p>Therefore I end up with:</p>
<p><span class="math-container">$$dU = W\cdot dy -Fx \cdot dx - Fy \cdot dy$$</span></p>
| 39145 | Sign of the sum of Virtual work (Direction) |
2020-12-16T10:54:48.743 | <p>My phone is pretty smashed up and I won't be able to buy a new one for quite a while. I know you can buy a phone for ~$30, but I'm in an extremely tight situation and I just can't afford that right now. I do, however, have some Protite (clear-casting/embedding) fibreglass resin and catalyst. So my question is, if I were to cast a (temporary) new screen, how safe would it be to use? Would I even be able to attach it to the digitiser without losing functionality? And, if anyone could even approximate, how long could I expect it to last? I know an epoxy bond would be stronger but aside from that I'm not sure how it would go enduring heat from the phone, or if it might pose any other risks. I live an hour out of town and rely on my phone greatly, it's kind of essential, so any help would be greatly appreciated.</p>
<p><em>NB: I posted here because I felt this query pertained most to 'a specific engineering problem' and didn't fit the criteria for the electrical engineering site, nor do I believe it is specific to a particular type of device e.g. iPhone or Android. I apologise if that is not correct. I'm new to SE and definitely no expert in such matters.</em></p>
| |casting|composite-resin| | <p>I'll be assuming that you are talking about a "smartphone".</p>
<p><strong>To be honest, if your phone is marginally working, and you are strapped for cash, I would avoid doing so.</strong></p>
<p>Having said that, I'd be very interested to hear what the results of this experiment is.</p>
<p>To be honest, I'd expect that you would probably brick the phone if you tried that. The heat from the phone is not the problem (if its a resin, its most likely a thermosetting one, so there is no worry about it having a problem).</p>
<p>The main problem (IMHO), is that you'd be adding an very stiff layer on top of your screen. Even if its a tenth of a mm, it would be orders of magnitude stiffer than the film you use. The end result would be, that you'd be applying the force over a bigger area. So the responsiveness of the phone would degrade - a lot.</p>
<p>Having said that, you'd probably encounter other problems should you try to go down that road. e.g. Chemical reactions. you might find that the resin, bonds to the screen.</p>
<p>Also something another question is, how will a capacitative or resistive touch screen behave in this scenario?</p>
| 39153 | Casting a phone screen using fiberglass resin? |
2020-12-17T06:11:27.263 | <p>This water sewer plant is about 1/4 mile from vacant rural lot I own. With no particular weekly pattern I occasional hear what sounds like a large generator or pump faintly in the distance.</p>
<p>From this image, can somebody explain what they might be doing there? Does it appear that any waste or sludge is being reintroduced back into soil in those two dry ponds?</p>
<p>I have not tested the water yet, planning to , but how safe for contact with human skin (not drinking) can I expect a shallow well to be a 1/4 mile away from this plant?</p>
<p>It's in central Florida, please do not ask me for detail on the location.</p>
<p><a href="https://i.stack.imgur.com/fvQgW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fvQgW.jpg" alt="enter image description here" /></a></p>
<p>update: here's another picture: <a href="https://i.stack.imgur.com/o33NN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o33NN.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jp9WK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jp9WK.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/MFO1f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MFO1f.jpg" alt="enter image description here" /></a></p>
| |waste-water-treatment| | <p>This is a bit speculative on my part, But I think what's going on here is this:</p>
<p>Hard to see from the pictures.</p>
<p>The concrete tructure on the right, It could be some mechanical treatment or mechanical tgreatment plus aeration (I'm not sure if I see bubbles in the upper left basin)</p>
<p>The patch of ground on the left:</p>
<ul>
<li>At first I thought <em>constructed wetlands</em>, but then there should be more greenery and less tire tracks.</li>
<li>You write the circles are about 3m below ground, I'd assume the patches are either <em>attenuation ponds</em> who happen to be always dry (though I don't see any overflow structure to empty the ponds).</li>
<li>The next option would be groundwater infiltration, infiltration ditches for stormwater I've seen looked like approximately so.</li>
</ul>
<p>It may be that your hear noises when the level in a pump sump reaches a certain point and a pump activates. Constructed wetlands are doused with wastewater intermittently for process reasons (the distribution hose can be buried).</p>
<p>You could check the following:</p>
<ul>
<li>If wastewater treatment, is it <em>domestic</em> wastewater or something else? Fi</li>
<li>Is there an ope body of water nearby into which the WWTP discharges, or could discharge via a buried pipeline?</li>
<li>Are there any official and up -to-date statements about groundwater quality in your area availible - this could give you an indication without paying for the analysis yourself</li>
<li>Is your well upstream or downstream of the plant? (Often groundwater follows the surface morphology - but that is far from assured)</li>
</ul>
| 39170 | what goes on in this 6 acre water sewage plant? |
2020-12-17T11:52:26.030 | <p>I was referring to some Thermodynamics text books and found that their definitions of a "pure substance" seem very subjective.</p>
<p>"Air, for example, is a mixture of several gases, but it is often considered
to be a pure substance because it has a uniform chemical composition" (Fundamentals of thermal-fluid sciences, fifth edition - YUNUS A.
ÇENGEL, JOHN M.
CIMBALA, ROBERT H.
TURNER)</p>
<p>"A system consisting of air can be regarded as a pure substance as
long as it is a mixture of gases; but if a liquid phase should form on cooling, the liquid would
have a different composition from the gas phase, and the system would no longer be considered a pure substance" (Fundamentals of
Engineering Thermodynamics - Michael J. Moran, Howard N. Shapiro)</p>
| |mechanical-engineering|thermodynamics| | <p>I have seen these statements. They can be confusing. They can also be mis-leading. In trying to avoid giving the fuller details, these statements sometimes hinder the ability to make valid projections later.</p>
<p>The term substance generally means a particular kind of matter with uniform properties. In chemistry, a substance is matter that has a specific composition and specific properties.</p>
<p>Here is the first lesson. Every pure compound is a substance. But not every substance is a (single) compound.</p>
<p>Air is NOT a pure compound. So, it cannot be classed as a substance by default. Air (as a single gas phase) is ONLY a substance by the fact that it has a specific <em>uniform</em> composition and specific <em>uniform</em> properties throughout. What are the specifics? Air (as a gas phase) contains a well-defined set of chemical compounds (nitrogen, oxygen, and so on) in well-defined relative concentrations (79 mol%, 21 mol%, and so on) mixed uniformly as a single gas-phase solution (not as multi-phase system).</p>
<p>What then is air as a "pure" substance as opposed to simply calling air a substance? The inference by using the word "pure" is that, as long as the composition and properties of (gas phase) air are uniform throughout, we may as well just believe that air is composed of "air molecules”. We do not need to know that air is truly composed of nitrogen molecules and oxygen molecules (and water molecules and argon atoms and ...). We never get that far in our treatment of air to discover where it makes a difference.</p>
<p>Now the second lesson. Even when the gas of a pure compound forms a liquid, we still regard the gas as a substance. In fact, we also regard the pure liquid as a substance. Why? Because the default is that pure compounds are substances. The term "substance" for pure compounds is not tied to the phase state of the compound.</p>
<p>But, even though each phase is itself a substance, we never say is that the entire two phase system is a substance. Why not? Because anything that has more than one phase by definition does not have uniform properties throughout the entire system. In particular, at a fundamental (molecular) level, either the chemistry or the structure (or both) within each of the various phases must be different from any of the other phases in order to have more than one phase.</p>
<p>By example, we do not call a system that contains a two phase mixture of PURE water vapor + PURE liquid water a substance. With this same reasoning, we cannot call a two phase system of gas phase air + liquid phase air a substance either.</p>
<p>To be clear, in a multi-phase system, each phase is by itself a substance, so long as that phase retains its own uniform composition and properties throughout.</p>
<p>What about composition? Why is that notion even raised?</p>
<p>We need to explore what happens to air when it forms a liquid. Because air is not a pure compound, it will not form a pure compound as a liquid. In fact, the liquid state of air will naturally have a different composition of the components that it contains. The normal boiling point of nitrogen is -320 <span class="math-container">$^o$</span>F and of oxygen is -297 <span class="math-container">$^o$</span>F. As we lower the temperature on gaseous air, the oxygen will prefer to form liquid before the nitrogen. So, the liquid phase that forms will be richer in oxygen than in nitrogen. At any temperature between -297 <span class="math-container">$^o$</span>F and -320 <span class="math-container">$^o$</span>F, the equilibrium state of the two phase system will have a liquid that is richer in oxygen than the gas phase state.</p>
<p>Does this composition difference make air no longer a compound?</p>
<p>This finding does NOT make the GAS PHASE AIR no longer a substance by itself. This finding also does NOT make the LIQUID PHASE AIR no longer a substance by itself. Both phases by themselves completely follow the guidelines for uniform composition and properties throughout.</p>
<p>So, part of the confusion is due to a redundancy. Simply put, we do not call a <strong>SYSTEM</strong> of gas + liquid a substance because no SYSTEM that contains more than one phase ever has uniform properties throughout. Regardless of whether the system contains pure water, pure benzene, or ... "pure" air, a multi-phase system is never a substance.</p>
<p>Return now to the ending phrase "the liquid would have a different composition from the gas phase, and the system would (therefore) no longer be considered a pure substance". This phrase is conceptually misleading if not wrong. It projects to you that the reason to believe that a system of gas phase air + liquid phase air is no longer a substance is because the compositions in the two phases are different from each other. The real reason is simply because you have more than one phase in the system. There is absolutely no reason to have to talk about composition.</p>
<p>In summary, there is a lot that is hidden in the simplified statements that you quote. They make life so much easier to solve homework problems (oh ... this system has gas phase air ... it is a pure substance ... there is an ideal equation for this case). They make life so much harder when it comes time to solve problems to the real world (oh ... my gas phase air pump is experiencing cavitation ... why would a "pure substance" cause such a thing).</p>
| 39173 | Is air a pure substance? |
2020-12-17T17:18:58.470 | <p>Is it possible to introduce torque on a lead screw (with no magnets attached to it) with a moving magnet that can only move parallels to the screw?</p>
<p>(the lead screw is ferromagnetic)</p>
<p>Something like that:</p>
<p><a href="https://i.stack.imgur.com/jaT48.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jaT48.png" alt="enter image description here" /></a></p>
| |magnets| | <p>Yes, it is POSSIBLE but it's not very practical.</p>
<p>Take a look at how <strong>switched reluctance motors</strong> work. Essentially, torque is generated by the motion of the soft magnetic material to align itself with the magnetic field lines. You're using a hard magnetic material (ferromagnetic), but the approach is essentially the same.</p>
<p>IT would work a lot better if you can introduce some level of asymmetry to your screw cross sectional area, but if you don't want to do that, you can get an effective elongated magnetic profile by aligning your stators with the thread pitch of your screw.</p>
| 39176 | Rotating a lead screw with a moving magnet |
2020-12-17T22:37:32.553 | <p>I'm reading <a href="https://structural-analyser.com/domains/SteelDesign/Buckling/" rel="nofollow noreferrer">here</a> on buckling resistance of members according to Eurocode.</p>
<p>Non-dimensional slenderness is defined as:</p>
<p><span class="math-container">$$\bar{\lambda}=\frac{\lambda}{\lambda_1}$$</span></p>
<p>where</p>
<p><span class="math-container">$$\lambda_1=\pi \sqrt{\frac{E}{f_y}}$$</span>
<span class="math-container">$$\lambda = \frac{L_{cr}}{i}$$</span></p>
<p>The plain <span class="math-container">$\lambda$</span> is by definition the slenderness of the column, the ratio of its effective length to its radius of gyration. But what is the meaning of <span class="math-container">$\lambda_1$</span>? No explanation besides its definition is given in the text. It is some factor by which we can divide the plain lambda or slenderness to make it "non-dimensional", but why does it take the form it does? What meaning does the square root of the ratio of elastic modulus to yield stress multiplied by pi have?</p>
| |structural-engineering| | <p><span class="math-container">$\lambda_1$</span> is the minimum slenderness ratio at which buckling will be the dominant condition for a purely axially loaded element. Interestingly, it is an intrinsic property of the material, not the geometry.</p>
<p>It's derivation is simple:</p>
<p><span class="math-container">$$\begin{align}
P_{crit} &= \dfrac{\pi^2 EI}{L^2} \\
\dfrac{P_{crit}}{A} &= \pi^2 E \cdot \dfrac{I}{L^2A} \\
\sigma_{crit} &= \pi^2 E \cdot \dfrac{i^2}{L^2} \\
\sigma_{crit} \equiv f_y &= \pi^2 E \cdot \dfrac{1}{\lambda_1^2} \\
f_y &= \dfrac{\pi^2 E}{\lambda_1^2} \\
\therefore \lambda_1 &= \pi\sqrt{\dfrac{E}{f_y}}
\end{align}$$</span></p>
<p>From this we can calculate a slenderness ratio for steel and another for aluminium, above which any element made of each material will fail by buckling, and below which they will fail by simple compression.</p>
<p>And that's all that's described by <span class="math-container">$\bar\lambda$</span>: if it's greater than 1 (the element's slenderness ratio is greater than <span class="math-container">$\lambda_1$</span>), the element will fail by buckling; if lower, by compression. If exactly equal to 1, it'll do both simultaneously.</p>
<p>In reality, codes usually have "fuzzier" rules for elements with <span class="math-container">$\bar\lambda \approx 1$</span>, since there can be interactions between both failure states which lead to ugly math.</p>
| 39181 | Meaning of $\lambda_1$ factor in Eurocodes |
2020-12-18T15:11:56.287 | <p><strong>I have the following problem</strong><br>
E= 210 MPa
d = 40 mm
n = 20
y = 1.5
a =1
calculate the maximum allowed load for the system</p>
<p>I am confused on how to solve the problem, I really need advice.</p>
| |solid-mechanics|buckling| | <p>The 1m "a" bars each carry a tension load of <span class="math-container">$F=\sqrt2*P/2 \ $</span>and they have a vertical component along the vertical bar equal to <span class="math-container">$F=P/2\ $</span>compression. The horizontal reactions are canceled by the left side reactions.</p>
<p><strong>Edit.</strong> [To clarify how we get the P in vertical bars. The two vertical right-hand components cancel each other and leave the vertical bar with P/2 compression. The same thing applies to the bars on the left-hand side, again leaving the vertical bar with another P/2 compression and canceling the right-side horizontal components of reaction P.]</p>
<p>So the vertical bar is under <span class="math-container">$F=P/2+P/2= P \ $</span> compression</p>
<p>we check for two cases, buckling and yield. And we assume the <span class="math-container">$E_{steel}= 210GPa.$</span></p>
<p><span class="math-container">$I_{cylinder}=1/4 \pi R^4$</span></p>
<p><span class="math-container">$ P_{cr} = { \pi^2 \, E \, I \over L^2 }$</span></p>
<p><span class="math-container">$ P_{cr} = { \pi^2 \, 210GPa \, 1/4 \pi R^4\over L^2 } $</span></p>
<p>We just plug the R=0.02m and L=1.41m then and multiply by 1/n to get the allowable buckling load.</p>
<p>For allowable compression load, we just figure axial stress.</p>
<p><span class="math-container">$\sigma=P/A{bar} \quad 240MPa=P/ \pi*0.02^2$</span></p>
<p>and then multiply by1/1.5 SF.</p>
<p>The lower load between the two is the allowable load.</p>
| 39191 | maximum allowed load |
2020-12-18T19:06:16.613 | <p>I’m thinking some kind of torque measuring device but I can’t find confirmation.</p>
<p><a href="https://i.stack.imgur.com/hBbnM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hBbnM.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>Your picture is also on Wikipedia <a href="https://en.wikipedia.org/wiki/Planimeter" rel="noreferrer">Planimeter</a>.</p>
<p>Planimeters were used more prior to the development of computer digitizers during the 1980s. They were useful in measuring areas of cut & fill in civil engineering, such as in road works, or in determining the area mined in a open pit.</p>
<p>During the design stage, vertical sections would be taken through the region being cut or fill. If the sections were uniform they would be multiplied by the thickness to obtain volume of cut or fill.</p>
<p>If the sections weren't uniform, or in open pit work, where surveyors would provide before and after survey outlines of a region that had been mined, the volume would be found using <a href="https://www.howtoexcel.info/Civil/earthwork-volume/" rel="noreferrer">Simpson's Rule</a>. Areas measured using a planimeter would be multiplied by the distance between sections.</p>
<p>The same technique would apply to measuring the volume of fill in civil engineering or waste dumps in mining.</p>
<p>By the early 2000s, planimeters were generally not being used because computer digitization had taken over.</p>
| 39196 | What is this mechanical device and what does it measure? |
2020-12-19T16:00:45.063 | <p>My partner and I are having a friendly debate. We live in a walk-up style apartment building which has a radiator-like unit in the corner, which we believe is connected to some sort of central steam heat (ie, the steam is piped through the radiator from a central unit in the basement, which heats our house). We also have a ceramic electric space heater. We're curious which of these methods of heating our living room is more "environmentally friendly" - that is, which one will consume less energy (from the power/gas company) to increase the temperature in our living room by one degree. We're assuming that the steam is generated by a gas boiler, and the electricity by an oil-fired power plant.</p>
| |heating-systems| | <p>This is not a straight forward question to answer.</p>
<p>If for example you use solar or photovoltaic energy to power up your ceramic heater then that is definitely more environmentally cleaner than any type of fossil fuel burner that could boil water in order to produce steam.</p>
<h1>Sankey Diagram</h1>
<p>If you need to debate it more depth, it will be easier if you used Sankey Diagrams. They are a visual tool for getting an overview of how energy is redistributed within a process.</p>
<p><a href="https://i.stack.imgur.com/VH4Ox.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VH4Ox.png" alt="enter image description here" /></a></p>
<p>Sankey diagrams describe how the <strong>primary energy</strong> from any fuel gets redistributed to other types of energy (some useful some not).</p>
<h1>Electrical energy</h1>
<p>A typical Sankey for energy production from fossil fuels is presented in the following image, which shows that only approximately 35% of energy is useful for electricity:</p>
<p><a href="https://i.stack.imgur.com/MFb2g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MFb2g.png" alt="enter image description here" /></a></p>
<p>So you only get about 1/3 of the total energy for electricity. The good thing for ceramic heaters is that you have almost 100% efficiency, so all the electrical energy is converted into heating.</p>
<h1>Steam Boilers</h1>
<p>The problem with steam boilers is that they are usually old installations, with limited data. Additionally, they can be very dependent on the materials used during installation and the maintenance afterwards. Generally speaking you get about 2/3 of energy in steam that you can use for heating.</p>
<p><a href="https://i.stack.imgur.com/sXWTk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sXWTk.png" alt="enter image description here" /></a></p>
<h1>Comparison between the two.</h1>
<p>So you can see that, while you use more efficiently the primary energy with the steam boiler, if the electricity you get is from a renewable source then the ceramic might be more environmentally friendly.</p>
<p>Of course, I avoid going into the discussion about carbon emissions, or the total impact on the environment (e.g. consider if your electricity came from nuclear). That makes the problem even more complex.</p>
<h1>A Better solution</h1>
<p>If you live in a relatively warm climate (i.e. you don't get temperatures under 0 more that a few days per year), a better alternative for electrical power can be a heat pump.</p>
<p>The difference with heat pumps, is that they don't use the electrical energy directly for heating. (Simply put) What they do is they <strong>use the electrical energy to mechanically move an arrangement of pumps/compressors/condensors to pump heat energy from one side of the wall to the other</strong>. Heat energy that would not usually be inclined to flow towards that direction. More specifically, they take heat from the colder environment and pump it inside a warmer room.</p>
<p>That process, <em>in ideal conditions</em>, is <em><strong>very efficient</strong></em>. For example, for one unit of electrical energy you can nowadays move up to 4 units of heat energy. That ratio of useful energy to energy expended is the basic formula for the famous COP - <a href="https://en.wikipedia.org/wiki/Coefficient_of_performance" rel="nofollow noreferrer">Coefficient of performance</a>. In this specific example COP would be 4.</p>
<p>So in ideal conditions, if you used 100 units of fossil fuel, and got 35 units of electrical energy, in theory you could get 140 units of heat energy pumped in your room.</p>
<p>Of course, there are limitations. E.g. at cold climates using a heat pump would create ice/frost on the heat pump, which ultimately would lower significantly the COP.</p>
| 39205 | Which uses less energy to heat a room: central steam heat, or a ceramic space heater? |
2020-12-19T21:02:33.407 | <p>Please help me to identify what C5 stands for in this picture. Full drawing is available <a href="https://www.qualtekusa.com/images/AC_Receptacles/pdfs/703w0004.pdf" rel="nofollow noreferrer">here</a>.</p>
<p><a href="https://i.stack.imgur.com/taoFe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/taoFe.png" alt="Drawing" /></a></p>
| |technical-drawing| | <p>The "C5" annotation in this example refers to a "5mm Champher". It's important to note, however, that this is non-standard notation. I have included an example below with the standard annotation for the champher on the LHS, and another alternative on the RHS.
<a href="https://i.stack.imgur.com/ppIL8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ppIL8.png" alt="5mm champhers" /></a></p>
| 39210 | Connector drawing - please help find C abbreviation/symbol meaning |
2020-12-20T06:16:18.927 | <p>I am working on a new project right now and I was wondering if anyone had any ideas on how to keep a spring fixed and connected to this flat plastic piece [image shown below].</p>
<p>Our goal is to keep the spring erect and allow for an upright, linear compression.</p>
<p>I was wondering if there was a mechanical method of attaching/fixing these two pieces together without using glue, epoxy, or other such products.</p>
<p>The best solution that we have been able to come up with so far is to add a custom threading into the plastic such that the spring can just screw into it, but we are afraid that this wouldn’t be too robust. Another downside is that the spring is winded in a counterclockwise manner so I am not sure how it would even be able to “screw” it in.</p>
<p>If anyone has a clever solution to this problem, I would really appreciate it.</p>
<p>Thanks,
Ryan</p>
<p><a href="https://i.stack.imgur.com/ARY57.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ARY57.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|materials|applied-mechanics|manufacturing-engineering|springs| | <p>It might seem simplistic, but if you have the option, you can drill a hole with a diameter slightly less than the spring and then pressure fit it inside the spring.</p>
<p><a href="https://i.stack.imgur.com/3oDzU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3oDzU.png" alt="enter image description here" /></a></p>
| 39218 | How to mechanically attach a compression spring to a flat surface |
2020-12-20T09:48:06.387 | <p>I am working on a small waste treatment project where there is a small mixing vessel (about 5 liters - more or less 20cm diameter). The mixing process produces methane gas which shouldn't escape the confines of the mixing vessel (odor and flammability).</p>
<p>Some other salient parameters (as I perceive them):</p>
<ul>
<li>The shaft diameter is small (about 16mm).</li>
<li>the shaft rotates really slow (about 10-30 rpm).</li>
<li>the pressure of the gas is low (just over 1Atm ).</li>
</ul>
<p>I was wondering whether what type of seal would be most appropriate for this type of application. My first instinct was a lip sea with a garter spring. I know there are other types of seals (e.g. packing, mechanical seals, tandem, cartridge), but I have little to no experience with most of them.</p>
<p>The cost would be a primary driver in this application, so any indication about the *cost effectiveness * would be welcome. For example, I have seen some of gas seals used in the petroleum industry but the cost is prohibitive.</p>
| |process-engineering|seals| | <p>Nitrile rubber (Buna-N) o-ring seals would be the most cost effective choice for your application. <a href="https://www.parker.com/Literature/O-Ring%20Division%20Literature/ORD%205700.pdf" rel="nofollow noreferrer">This</a> is a valuable resource on o-ring seal design, for your case a dynamic rotary seal.</p>
| 39220 | Gas Seal for slowly turning shaft |
2020-12-20T13:24:23.007 | <p>To work with Magic Formula for load dependent conditions, we have to use some of additional parameters. In this <a href="https://www.mathworks.com/help/physmod/sdl/ref/tireroadinteractionmagicformula.html" rel="nofollow noreferrer">link</a>, you can find the formula about this.</p>
<p>What is the meaning of Fz0 (Nominal vertical load on tire)? Is it static load?</p>
<p>Thanks a lot,</p>
| |mechanical-engineering|wheels| | <p>The meaning of <span class="math-container">$F_{Z0}$</span> is the vertical load on the tire. Since tire-ground interaction is friction based, greater load leads to greater friction forces (longitudinal and transverse).</p>
<p>So, for example if a truck is loaded then the force on each tire is greater (compared to an unloaded one), and so are the transverse and longitudinal forces on the tire.</p>
<p>Regarding, whether that is static or dynamic load: the <span class="math-container">$F_{z0}$</span> represent a dynamic load in the sense, that when taking a turn, the wheels on the outer side of the turn are loaded more than the wheels on the inside of the turn. So most of the time, when a vehicle is turnign there is continuously redistribution of the forces on the wheels.</p>
| 39223 | Pacejka's Magic Formula:Nominal vertical load on tire |
2020-12-21T20:23:05.597 | <p>We would like a carriage mounted element (CME) on a rotating linear guide rail (LGR) that negates the LGR rotation and thusly remains rotationally stationary relative to the LGR's pivot affixation surface.</p>
<hr />
<p>Here is a diagram to illustrate what is desired for just three example angles and three example distances from the pivot superimposed (but all angles and legal distances should work - even right up to the pivot itself, though that is less important). P denotes the pivot and the CME has a notch on it to indicate constant orientation relative to the surface the pivot is affixed to:</p>
<p><a href="https://i.stack.imgur.com/Bgz25.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bgz25.png" alt="Pivot and three examples of CME position" /></a></p>
<hr />
<p>The axis of rotation is the same for both the LGR and CME in question. This axis is the axis you'd intuitively imagine when imagining a LGR rotating, which is to say the axis that pierces through the top of the carriage and into the LGR, although that is only describing the direction of the axis, the origin/pivot doesn't follow the carriage; it is in a constant position near the end of the LGR.</p>
<p>I've already attempted various solutions, but most fail because of their inability to function with a dynamic distance from the pivot due to the carriage's linear freedom of movement.</p>
<p>One solution that could account for the carriage's dynamic distance from the pivot was to use sensors to measure the rotation of the LGR and a servomotor to rotate the CME to the desired angle. I will begrudgingly use this solution if nothing better is found, but I'd much prefer something with good old gears/pulleys/chains/etc. to ensure there's no "lag" and to keep costs down, etc.</p>
<hr />
<p>Sorry for being cryptic, I am working on a prototype for something and my partner in crime is paranoid that their idea will be stolen. If you have any queries relevant to finding an answer I'll be happy to answer them.</p>
<p><strong>edit: There will be two carriages on the LGR, each with their own CME, differing to the diagram. Both needing to negate the rotation of the LGR as already described. This is one of the things my partner would have rather withheld if you are wondering why I didn't say it sooner.</strong></p>
| |mechanical-engineering| | <p>Not knowing the distances or latency, you could use a 2nd LGR in parallel with your green rail with another CME. Connect the two with an extension spring and hooks. Or mount the whole thing vertically (pivot axis is horizontal) and let gravity maintain orientation of your CME.</p>
<p><a href="https://i.stack.imgur.com/XJVGz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XJVGz.png" alt="enter image description here" /></a></p>
| 39248 | How to negate rotation of a rotating linear guide rail on a carriage? |
2020-12-22T19:43:29.450 | <p>Ok so what happend is that i started to think about Submarines and why exactly you cannot open the bottom heatch at lets say the Challenger Depth. At least i am pretty sure you cannot do that. Right ?</p>
<p>Because, i was told that if you were to open the Heatch, water would still rush in and kill you rather fast. But why ? All of the Water and thus all of the Pressure is above you. Here is a bad illustration of what i mean<a href="https://i.stack.imgur.com/YvF2D.png" rel="noreferrer"><img src="https://i.stack.imgur.com/YvF2D.png" alt="enter image description here" /></a></p>
<p>My theory: In theory you can do that, but now, since there is no floor, all the Pressure from above is resting on the Water below. Which itself is the support for the Submarine. And with no floor, the Water gets forced in. So what happens is that the Air is acting like the Floor that is missing. But while the Floor could resist the Pressure from above pusing it against the support from below, the Air cannot. And thus gets compressed.</p>
<p>Is that right ?</p>
| |pressure| | <p>I add to all the good answers my little different answer.</p>
<p>the pressure in the water increases by one atmosphere every 10 meters down and that is basically the weight of the 10 meters by 1 square cm column of the water 10times 100cm is 1000 cm3 hence 1kg additional pressure.</p>
<p>so if you open the bottom hatch of the submarine at 1000 meters depth, the high-pressure water will flood the submarine and compress the air inside to 1/100 of its volume and 100 atm extra pressure.</p>
<p>even if you are lucky and found a pocket of air to hide, the pressure of the air will crush you immediately and break your ribcage. Even if there is a leak from a cracked seal, the water will shoot in with huge pressure and can and has killed all the sailors.</p>
<p>that's why the deeper a submarine can go the stronger and thicker the haul must be to protect us from the crushing pressure.</p>
| 39267 | Understanding Pressure |
2020-12-24T12:13:40.793 | <p>I have an intermediate shaft that drives my spindle with pitch (3). On the spindle there is a slide which is moved translationally by the spindle. My intermediate shaft is rubbery and has a very low torsional stiffness. Is it possible to convert this torsional stiffness by the pitch of the spindle to a translational stiffness?</p>
<p><a href="https://i.stack.imgur.com/Ljgei.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ljgei.png" alt="r/EngineeringStudents - Convert torsional stiffness to translational stiffness
Similar to this picture except that an intermediate shaft is connected to reduce the stiffness of the system" /></a></p>
<p>I am relatively new to the subject and would appreciate any help.</p>
<p>Many thanks</p>
| |mechanical-engineering| | <p>Lets assume you have the following system:</p>
<p>A motor supplies a torque, which turns a shaft. The shaft has a gear on it (with <span class="math-container">$z_1$</span> teeth), which meshes to a second gear (with <span class="math-container">$z_2$</span> teeth) designed such that the speed of the second shaft is greater than the first.</p>
<p><a href="https://i.stack.imgur.com/bLWzL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bLWzL.png" alt="enter image description here" /></a></p>
<p>The shafts are mounted on bearings each with a torsional damping coefficient <span class="math-container">$c_t$</span>.</p>
<p>The angular displacement of the shaft 1 (directly connected to the motor) is denoted with <span class="math-container">$\theta_1$</span>. The relationship between the angular velocity of shaft 1 and shaft 2 is given by:</p>
<p><span class="math-container">$$\frac{z_1}{z_2}=\frac{\dot{\theta}_2}{\dot{\theta}_1}$$</span></p>
<h1>1. kinematic relationship</h1>
<p>Seemingly, the problem here is that you have two different angular displacements. However, in truth you only have one, because <span class="math-container">$\theta_2$</span> is proportionally dependent to <span class="math-container">$\theta_1$</span>. The exact equation is :</p>
<p><span class="math-container">$$\theta_2 = \frac{z_1}{z_2} \theta_1$$</span></p>
<h1>2. derivation of equivalent system.</h1>
<p>The main idea is that you come up with parameters <span class="math-container">$I_{eq}$</span>, <span class="math-container">$c_{eq}$</span> and <span class="math-container">$k_{eq}$</span> that for a given angular displacement have the same energy with the system.</p>
<h2>2.1 Kinetic energy of shafts.</h2>
<p>The kinetic energy of a rotating object is given by:</p>
<p>$$T_i = \frac{1}{2} I_i \dot{\theta}_i^2$</p>
<p>where:</p>
<ul>
<li><span class="math-container">$J_i$</span> is the mass moment of inertia <span class="math-container">$[kg m^2]$</span></li>
</ul>
<p>because both the gear (g) and the shaft (s) are rotating the above can be written for a shaft with a gear as:</p>
<p>$$T_i = \frac{1}{2} (I_{s,i} + I_{g,i}) \dot{\theta}_i^2$</p>
<p>so the total energy is :</p>
<p><span class="math-container">$$T_{total} = T_1 + T_2 = \frac{1}{2} (I_{s,1} + I_{g,1}) \dot{\theta}_1^2 + \frac{1}{2} (I_{s,2} + I_{g,2}) \dot{\theta}_2^2$$</span></p>
<p>this can be simplified by substituting <span class="math-container">$\dot{\theta}_2 = \frac{z_1}{z_2} \dot{\theta}_1$</span> to:</p>
<p><span class="math-container">$$T_{total} = T_1 + T_2 = \frac{1}{2} (I_{s,1} + I_{g,1}) \dot{\theta}_1^2 + \frac{1}{2}(I_{s,2} + I_{g,2})\left( \frac{z_1}{z_2} \dot{\theta}_1\right)^2$$</span></p>
<p><span class="math-container">$$T_{total} = \frac{1}{2} \color{red}{\left( (I_{s,1} + I_{g,1})+ (I_{s,2} + I_{g,2}) \left( \frac{z_1}{z_2} \right)^2 \right)}\dot{\theta}_1^2 $$</span></p>
<p>So the equivalent mass moment of inertia <span class="math-container">$I_{eq}$</span> to simulate the system which is rotating with <span class="math-container">$\dot{\theta}_1$</span> is:</p>
<p><span class="math-container">$$ I_{eq} = (I_{s,1} + I_{g,1}) + (I_{s,2} + I_{g,2}) \left( \frac{z_1}{z_2} \right)^2 $$</span></p>
<h2>2.2. Damping</h2>
<p>The work done by the viscous dampers from time a to time b is equivalent to the rotational speed.</p>
<p><span class="math-container">$$W_{ab} = -\int_{\theta_{1,a}}^{\theta_{1,b}} c_t \dot\theta_1 d\dot{\theta}_1 -\int_{\theta_{2,a}}^{\theta_{2,b}} c_t \dot\theta_2 d\theta_2 $$</span></p>
<p>Again, this can be simplified by substituting <span class="math-container">$\dot{\theta}_2 = \frac{z_1}{z_2} \dot{\theta}_1$</span> and <span class="math-container">$\theta_2 = \frac{z_1}{z_2} \theta_1$</span> to:</p>
<p><span class="math-container">$$W_{ab} = -\int_{\theta_{1,a}}^{\theta_{1,b}} \color{red}{c_t \left(1 + \left(\frac{z_1}{z_2} \right)^2\right)} \dot\theta_1 d\dot\theta_1 $$</span></p>
<p>So the equivalent damping coeffient is the red part:</p>
<p><span class="math-container">$$c_{eq} =c_t \left(1 + \left(\frac{z_1}{z_2} \right)^2\right)$$</span></p>
<h2>2.3. Dynamic energy.</h2>
<p>In this system, some energy is stored in the shafts as torsional energy. However, we won't calculate the <span class="math-container">$k_{eq}$</span> because in this system A) the angular displacement related to the stored torsional energy is orders of magnitude smaller to that of <span class="math-container">$\theta_1$</span>, B) the stored energy - usually (however probably not in your example)- is normally comparatively small, and C) because the elastic energy remains constant if the applied torque remains the constant.</p>
<h1>final comments</h1>
<p>This is the general idea of <strong>a</strong> modelling method to turn a system like this to an SDOF. In your particular problem you would have other things, like bending of the beam, etc. At the end of the day, it might not be possible to model it accurately with a SDOF, but you'd need to go to a MDOF System.</p>
<p>Additionally, there are other issues of concern (like how to calcucate the damping coefficients etc). As I said, in my comment, I don't think this is particularly useful for what you are intending to do.</p>
| 39311 | Convert torsional stiffness to translational stiffness |
2020-12-25T03:42:44.723 | <p>I am using AutoCAD and I was trying to figure out how to draw a circular arc through 2 known point and tangent to a line.</p>
<p><a href="https://i.stack.imgur.com/DwJXm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DwJXm.png" alt="Sample image" /></a></p>
<p>When I look at the circle tools I see the following options:</p>
<p><a href="https://i.stack.imgur.com/TfnqO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TfnqO.png" alt="Circles" /></a></p>
<p>When I look at the arc options I see the following:</p>
<p><a href="https://i.stack.imgur.com/G8CKu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G8CKu.png" alt="Arcs" /></a></p>
<p>I am able to do it through a convoluted use of parametric constraints, but I cant figure out how to do it from first principals. Is there a way?</p>
<p>What I figured out so far is that the center of the circle will be on the perpendicular bisector (Red Line) of the line between the known points.</p>
<p><a href="https://i.stack.imgur.com/UMxbt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UMxbt.png" alt="trial 1" /></a></p>
<p>Using algebra I can mathematically come up with the center for the two blue circles. And visually I know I want the larger circle. What am I missing to find that center point graphically?</p>
<p>I can do it through parametric constraints, but I thought one should be able to do this with essentially a straight edge and a compass (straight lines and circles).</p>
| |technical-drawing|drafting|drawings| | <p>After more than a few days with the wrong key words for a google search, I stumbled on the answer while trying to navigate to the math stack exchange...and the answer was some place completely different:</p>
<p><a href="https://www.geogebra.org/m/CgHVUJs8" rel="nofollow noreferrer">https://www.geogebra.org/m/CgHVUJs8</a></p>
<p>Great animation. Basically these are the step to figure it out graphically with the assumed initial setup below:</p>
<p><a href="https://i.stack.imgur.com/pk2pe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pk2pe.png" alt="setup" /></a></p>
<hr />
<p>Step 1</p>
<hr />
<p><a href="https://i.stack.imgur.com/9mLBQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9mLBQ.png" alt="Step 1" /></a></p>
<p>Make the line AB</p>
<hr />
<p>Step 2</p>
<hr />
<p><a href="https://i.stack.imgur.com/4GFR6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4GFR6.png" alt="Step 2" /></a></p>
<p>Draw the perpendicular bisector of AB. call the intersection point E and the new line FG.</p>
<hr />
<p>Step 3</p>
<hr />
<p><a href="https://i.stack.imgur.com/hq7Zb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hq7Zb.png" alt="Step 3" /></a></p>
<p>Extend the line AB so it intersects with the tangent line CD. The point of intersection will be point H.</p>
<hr />
<p>Step 4</p>
<hr />
<p><a href="https://i.stack.imgur.com/NszAI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NszAI.png" alt="Step 4" /></a></p>
<p>Draw a circle centered on E so it passes through point H. (Radius = |EH|)</p>
<hr />
<p>Step 5</p>
<hr />
<p><a href="https://i.stack.imgur.com/8TjcQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8TjcQ.png" alt="Step 5" /></a></p>
<p>Draw a line from point B that is perpendicular to the line AB. Where this perpendicular line intersects the the circle, call that point J.</p>
<hr />
<p>Step 6</p>
<hr />
<p><a href="https://i.stack.imgur.com/uhykf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uhykf.png" alt="Step 6" /></a></p>
<p>Draw a circle with radius |BJ| centered on point H. ( |BJ|=|HK| ) Where this circle intersects with the tangent line CD, call the points of intersection L and M.</p>
<hr />
<p>Step 7</p>
<hr />
<p><a href="https://i.stack.imgur.com/11SYk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/11SYk.png" alt="Step 7" /></a></p>
<p>Draw lines perpendicular to the tangent line CD, at points L and M. Call these perpendicular lines LN and MP. Where LN and MP intersect with the line FG mark the points of intersection Q and R.</p>
<hr />
<p>Step 8</p>
<hr />
<p><a href="https://i.stack.imgur.com/WkBS2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WkBS2.png" alt="Step 8" /></a></p>
<p>Points Q and R are the two possible centers for a circle or arc passing through points A and B and being tangent to line CD.</p>
| 39320 | How to draw an arc through 2 known points and tangent to a line |
2020-12-26T00:01:23.890 | <p>I have come across a problem that for the life of me I can't seem to figure out. I'm sharing this here in the hopes that the amazing community here can help push me in the right direction.</p>
<p>The problem goes like this, in my project I require 4 High torque BLDC Motors each having an output torque of 105 N-m or approximately 1070 kg-cm. Due to size constraints we need to go with geared motors. The problem starts here, I have come across a 12 rpm planetary gear DC motor with encoder that has the stall torque of around 584 kg-cm with a gear ratio of 721:1. On further research I can't seem to be able to find a motor with a higher torque than this. I would love to get a guidance from experts regarding what should I do next?</p>
<p>Should I increase the gear ratio or order a custom DC motor more suited to my application.</p>
<p>N.B: Keep in mind this will be used in a vehicle with a load capacity of around 400-500kg. I have verified all calculations and need to source the DC motors now for this load capacity.</p>
<p>I look forward on hearing your views regarding this problem</p>
<p>DC Motor Link: <a href="https://www.robotshop.com/en/12v-12rpm-81102oz-in-hd-planetary-gearmotor.html" rel="nofollow noreferrer">12V, 12RPM 8110.2oz-in HD Planetary Gearmotor</a></p>
<p>Edit Update: The project is a automated warehouse management system in which we are developing an AGV (Automated Guided Vehicle). 400-500 kg will be the weight on it after stacking bins on top.
Just like shown in the link: daifuku-logisticssolutions.com/image.jsp?id=3767</p>
<p>We can't go lower than 10 rpm (velocity of 0.11 m/s) since we have to cover 100 m in 15 minutes max. I'm attaching a link for related to the custom build DC motor directindustry.com/prod/electric-motor-power-pty-ltd/…</p>
| |mechanical-engineering|electrical-engineering|motors|automotive-engineering|linear-motors| | <p>Contact motor vendors. They will be happy to recommend options. Their goal is to help you ultimately buy their offering for your product.</p>
<p>I've seen where they recommend one option, and if it doesn't perform well enough, provide others at no (or reasonable) cost to help you get the right combination. This at least is my experience in big industry.</p>
| 39328 | Selecting the right DC motor for horizontal motion |
2020-12-26T02:28:30.603 | <p>My house is heated by a condensing natural gas furnace that is rated as 92% efficient. It has the option to draw combustion air directly from outside. I know the main advantage of this is avoiding negative pressure within the house to reduce drafts. However, I thought of a few questions:</p>
<ol>
<li>Outside air is colder. Does this decrease the maximum temperature of the flames?</li>
<li>As the outside air is colder, it is also denser. Does the furnace's performance increase due to increased oxygen content?</li>
<li>The furnace's exhaust temperature exiting outside is about 43 degrees Celsius, so there is still some heat in the exhaust that is not going towards heating the house. What could be done to utilize more of that heat?</li>
</ol>
| |thermodynamics|heat-transfer|heat-exchanger| | <blockquote>
<p>Outside air is colder. Does this decrease the maximum temperature of the flames?</p>
</blockquote>
<ol>
<li>Your furnace will always need air for combustion. This combustion air must ultimately come from the outside. Combustion air can either come directly from outside (say it's 30F outside) or it can leak into your house, go through the heater and be warmed up from 30F to 70F, and then go into the furnace.</li>
</ol>
<p>In the first case, your flame temperature is a little cooler than the second case, but all the heat is used for your house.</p>
<p>In the second case, the flame temperature is a little higher than the first case, but some of the heat is diverted into heating the combustion air.</p>
<p>Ultimately, it works out the same.</p>
<ol start="2">
<li><p>Yes, denser air would increase your furnace's maximum output but it is seldom that your furnace would be running at maximum output. What you always care about is your fuel bill which is affected by total BTUs required and efficiency. Denser air would not increase your furnace's efficiency.</p>
</li>
<li><p>As others have already pointed out, a coaxial heat recuperator can increase your furnace's efficiency and is also safer. My two furnaces (for a two story house) do have coaxial heat recuperators.</p>
</li>
</ol>
| 39330 | Furnace Combustion Air Properties |
2020-12-26T11:30:07.140 | <p>I'm currently researching some mechanisms for a simulation game, which also covers several electrical power plant options. As part of this, I'm currently looking at a solar thermal power plant, which in the first phase is meant to use thermal oil for heat transfer and storage.
Unfortunately I cannot find details on which exact thermal oils are usable in such a plant and how they are produced. I found that there are several options:</p>
<ul>
<li>Mineral oil based</li>
<li>Synthetic silicon based</li>
<li>Biological</li>
</ul>
<p>I checked several power plant pages, but they never go into that detail.
So I'm looking for the kind of thermal oil that can be used in this context, the chemical composition and where/how it can be produced, so I can model a rough production chain. Also it would be nice to know if that oil has to be exchanged frequently or if it can stay inside the closed cycle for a long time.</p>
<p>Thanks in advance</p>
| |solar-energy| | <p>Solar heating systems have been designed with many different fluids depending on the ambient temperatures and max desired temperature.</p>
<p>The ambient temperature is because the night-time temperature or even no sun temperature will affect the performance of the fluid.</p>
<p>Some systems are designed to be above the boiling point of water so the system pressure is increased, others go higher than that so water is no longer viable.</p>
<p>To avoid fluid temperature problems some systems are designed as drainback, where the fluid is stored in a reservoir and can then be kept at a storage temperature. Drainback is a legal requirement in some countries as no antifreeze can be used to prevent leaks and subsequent or possible environmental pollution.</p>
<p>Here is a link to one site the lists several fluids that have or are in use:
<a href="https://www.energy.gov/energysaver/solar-water-heaters/heat-transfer-fluids-solar-water-heating-systems" rel="nofollow noreferrer">types of fluids</a></p>
| 39336 | Which kind of thermal oil is used in solar thermal power plants and how are they produced? |
2020-12-27T10:23:34.120 | <p>Consider a liquid flow in which, keeping all else constant, we can vary the thermal conductance(K). Now, if we increase the thermal conductance of the fluid the Prandlt number will decrease. I read that as Prandtl number decreases the thermal boundary layer thickness becomes larger than the velocity boundary layer.</p>
<p>What if we have to compare the boundary layer thickness of liquid with lower K and higher K?Will the liquid with higher K( lower Pr) have thinner TBL than the liquid with lower K (higher Pr)?</p>
<p>I am asking this because I think higher value of K should cause heat to diffuse more easily. If heat diffuse easily then in the TBL the temperature should reach the free stream temperature in a shorter distance. So shouldn't the thermal boundary layer be thinner for higher K.</p>
| |thermal-conduction| | <p>A higher thermal conductivity ratio in the flow causes the heat diffusion to catch up with farther layers of fleeing molecules of the flowing liquid that otherwise would have washed past unaffected. If we compare k to the speed of heat diffusion, a higher k means reaching more in less time.</p>
<p>For example, if the k converges to infinity the TBL will be the complete thickness of the liquid because it will heat up immediately before getting a chance to be carried away cold by the momentum.</p>
| 39351 | Why does the thermal boundary layer(TBL) thickness increase with decreasing Prandtl number? |
2020-12-27T18:07:05.073 | <p>I have a greenhouse that get’s mold in the winter because it’s to humid.</p>
<p>I have now installed a heater and a ventilation system drawing outside air into the greenhouse. I wish to start the vent based on weather conditions however I cannot figure out which settings to use. I know the temperature and relative humidity inside and outside, and I want to know when to draw outside air in, only when then outside air is drier than the inside air. Any ideas?</p>
| |temperature| | <p>See the chart in "Algorithm for your setup" (above).
If you circulate air through a chilling set-up, you will condense excess moisture out of the air and reintroduce it to your environment drier.</p>
<p>If your air is 25C and 50%RH then there is ~10g of water per KG of dry air inside.
Draw the air into your chiller, lower the air temp. Chill the circulating air to 5C. About 4.5g of water will condense per kg of air processed (as liquid and dump it outside) and you will reintroduce the cold air back into the grow chamber, but when the temperature comes back to 25C it will only have 5.5g moisture per kg of dry air and equilabrate to ~27%RH. Because 5C air holds at most ~5.5g moisture per kg of air. (The difference between 5.5g moisture in 1kg of "dry" air versus 5.5g moisture in 1kg of air, is negligible.) Warning: The colder air may shock the plants so follow the below instead.</p>
<p>Alternatively: After drying the air (condensing the excess moisture by making it colder, to 5C, and dumping the condensed water outside) you can heat the cold air back to 25C and reintroduce it at ~27%RH. Eventually, this will become your steadystate. Since you continually water the plants, the humidity and temperature of your processed air should remain constant through this processing.</p>
| 39364 | At which temperature and humidity should outside air be drawn inside? |
2020-12-29T15:15:57.387 | <p>My understanding of MAWP in ASME BPVC Sec VIII Div 1 code is that it is the maximum pressure a pressure vessel should be subjected to. If that is the case, then why should the hydrotest pressure be kept at 1.3 times MAWP . Would'nt it damage the vessel, since it is exceeding MAWP.
Can anybody please clarify on the same.</p>
| |piping| | <p>A Mechanical engineer, or some one who works in materials/reliability would be better suited to answer this, but I believe the reasoning to be the a combination of safety factor (ensuring fittings, gaskets, etc are all going to withstand the pressure load) in addition to the difference between peak acute strength vs sustained or cyclic loads. MAWP stands for Maximum Allowable <em><strong>Working</strong></em> Pressure, and is talking about the maximum pressure the vessel can see on a sustained, or cyclic & repeating basis. Note that metals are all subject to fatigue, especially under cyclic loads. For many materials, this means listing the MAWP as some number that is significantly lower than say, absolute yield strength, knowing that over time the effective yield strength will be reduced.</p>
<p>The EXACT ratio (1.3, 1.5, 2.0 etc) and the application will depend on what materials it's holding, the type of environment it is in, and especially the actual materials of construction and that material's physical & wear characteristics. This ratio takes into account that new piping/vessels/equipment are all considerably stronger than the MAWP, which is defined as the limit the vessel will see during it's service/working life, and will benefit in terms of ensuring sound construction by going above the MAWP (I.E. if a vessels holds up against 275 PSI, you know it will suffice for 150 PSI)</p>
<p>There may be additional reasoning behind this, but from a practical perspective this is my interpretation in industry.</p>
| 39395 | Why is hydrotest pressure more than MAWP as per ASME Sec VIII Div 1 |
2020-12-29T19:16:17.687 | <p>I want to build a machine that pumps liquids (roughly density of water) out of an enclosed cannister. The cannister has for one inlet for air at the very top and has an outlet (near the top as well) that feeds a tube down to the bottom of the bottle. Both these in/out-lets are sealed with silicon and the one that allows air in has a one way valve for the air.</p>
<p>The outlet is connected to a manifold with valves that lets me choose when to allow liquid to flow out.</p>
<p>I want to use an air pump to pressurize the cannister and then control the liquid output (on/off, not flow rate) via the manifold valve.</p>
<p>In addition when the manifold is open the air pump will be constantly pumping to replace lost pressure.</p>
<p>I am considering a pump like this which <em>claims</em> to have an air flow rate of 3.5 liters/minute. Does this mean that if I keep the manifold open and pump I air at 3.5 liters/minute it will flow liquid out of the outlet at the same rate? Or is there something i'm not considering that is going to make it flow at a slower rate?</p>
<p>Such things that might be of concern is the the liquid is acting against gravity (not much, it's being pumped up maybe 6 inches to the manifold) as well as the inner diameter of the fittings on the cannister (0.095 inch).</p>
<p>Pump in question:
<a href="https://smile.amazon.com/NW-5V-6VDC-Miniature-Vacuum-100KPa/dp/B078H8V563/ref=sr_1_2?dchild=1&keywords=air%20pump%205v&qid=1609264991&sr=8-2" rel="nofollow noreferrer">https://smile.amazon.com/NW-5V-6VDC-Miniature-Vacuum-100KPa/dp/B078H8V563/ref=sr_1_2?dchild=1&keywords=air%20pump%205v&qid=1609264991&sr=8-2</a></p>
<p>And a diagram of my intended design:
<a href="https://i.stack.imgur.com/obEqG.png" rel="nofollow noreferrer">https://i.stack.imgur.com/obEqG.png</a></p>
| |fluid-mechanics|pumps|airflow| | <blockquote>
<p>Does this mean that if I keep the manifold open and pump I air at 3.5 liters/minute it will flow liquid out of the outlet at the same rate?</p>
</blockquote>
<p><strong>Short answer: No.</strong></p>
<p>The 3.5 L/min rate is a flowrate of a compressible fluid. The keyword here is <em>compressible</em>. The true flow rate of the water-like liquid being driven by the air pressure (from your 3.5 L/min of airflow) that you will see will be roughly proportional to the difference in pressure that your air pump produces, and the pressure at the outlet of the manifold valve (looks like you would be using atmospheric), minus the height that you must pump the liquid (you described as ~6"), minus the loss of friction in the pipes/tubing.</p>
<p>We could actually try to calculate this using Bernoulli's equation + friction factor + physical properties (density & viscosity) of your fluid if provided. It looks like we already have the outlet pressure of the pump (from the link you provided).</p>
| 39396 | Does air flow rate of an air pump imply the same rate when using it to pump water via air pressure? |
2020-12-29T22:20:08.457 | <p>I was wondering how to construct a hydraulic circuit to control a hydraulic double acting piston/actuator enabling the electronic control of the piston extension. I have found the following pneumatic solution. However, I am not sure if this can be applied to hydraulics too.</p>
<p><a href="https://i.stack.imgur.com/ROkmM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ROkmM.png" alt="enter image description here" /></a></p>
| |hydraulics|circuits|pneumatic|pistons| | <p>The pneumatic circuit in your question provides simple start / stop control of the piston. While the 5-port valve is opened you have full flow in either forward / reverse directions. Your control logic would monitor piston position as it approaches a target, then close the 5-port valve when position is within tolerance. This is accurate enough for some applications.</p>
<p>Your circuit also has the manually-adjustable "speed control" valves which are referred to in industry as "flow control" valves. These allow you to further improve accuracy by sacrificing some piston speed. Restricting these manual valves causes the piston to approach a target more gradually, making it easier to shut off the 5-port valve close to the target.</p>
<p>Lastly the circuit has PO check (pilot-operated check) valves. These lock cylinder position after the 5-port valve is shut off. PO check valves are common in vertical cylinder applications where locking piston position is necessary for safety reasons. PO check valves are usually avoided in applications requiring accurate control because they tend to "chatter" at low flow rates. Chatter is rapid stop/start motion of the valves which causes problems with piston positioning. So the PO checks are usually replaced with other valve options. Everything up to this point is true for hydraulics or pneumatics.</p>
<p>Your question asks for position control in a hydraulic system. Here is the typical hydraulic system design you're looking for -</p>
<p><a href="https://i.stack.imgur.com/XPojo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XPojo.png" alt="enter image description here" /></a></p>
<p>How does this position control loop work? Piston position is monitored by a feedback device (m). The feedback device communicates with a PID controller (X inside the circle). The PID controller communicates with the 4-port valve through an amplifier (triangle) to adjust valve opening size. The valve opening size controls flow rate, which controls piston position, thereby completing the control loop.</p>
<p>The key to this system design is the 4-port "proportional" hydraulic valve. Proportional means that valve opening % will follow the electrical input %. For example using a 0-10V valve, inputting 3V would result in 30% of maximum hydraulic flow. We refer to this capability as electro-proportional flow control. The advantage is you have 100% flow for the majority of the piston's travel toward the target. Then flow is reduced (100, 90, 80, ... 10%, 5%, 1%) as the piston nears the target. Keeping flow rate high during most of the cycle is necessary to maximize overall cycle time. Final positioning accuracy is improved because of dramatically reduced flow rate near the target piston position.</p>
<p>Couple of final notes -</p>
<ul>
<li>Modern electro-proportional valves are typically assembled together with the electronic amplifier. This means you simply supply the 0-10V or 4-20mA control signal. No need to worry about 600mA current draw of the valve solenoids because the manufacturer already accounted for that.</li>
<li>Highly accurate proportional valves require better hydraulic oil filtration. Typical industry recommendations would be 5 micron particulate filtration. Aerospace-grade proportional valves (such as servo-valves) may require up to 1 micron filtration.</li>
<li>In case cylinder locking is needed when the proportional valve shuts off, externally-vented counterbalances valves may work. The option for best accuracy would be solenoid-operated 2-port logic valves. Also many other locking options which would be a different topic for a different post.</li>
<li>Final positioning accuracy is affected by other factors besides the proportional valve quality. Position measurement is critical. PID controller bit rate plays a role. Electrical noise can interfere so don't run your signal wires right next to a 50kW induction motor.</li>
</ul>
| 39402 | Circuit to provide position control for double action hydraulic actuator |
2020-12-30T06:08:52.660 | <p>I have a pretty complicated part in solidworks that I need to 3D print. The part is designed to be a support, holding up a gauge on the side of a small pump.</p>
<p>On the other side of the pump, I would like to have the same part but in the opposite orientation to hold up another gauge. Rather than going through the tedious process of building the mirrored part from scratch, it would be easier to mirror the entire part about the relevant face. Is this possible in solidworks?</p>
| |solidworks| | <p>Probably Jonathan R Swift will give the best reply, however I'll give it a try.</p>
<p>It can be done in many ways.</p>
<h1>Assembly level</h1>
<p>In the assembly you have an option under Linear component Pattern.</p>
<p><a href="https://i.stack.imgur.com/8kPlU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8kPlU.png" alt="enter image description here" /></a></p>
<h1>Part level - 1</h1>
<p>If the distance is fixed and the orientation You can create another body within the same part. (This is something people sometimes forget)</p>
<p>Then you can insert that body as you would normally do</p>
<h1>Part Level - 2</h1>
<p>You can <strong>Save as</strong> a new part, then in the new part mirror (and delete the old).</p>
<p>Now you will have two mirrored parts that you can insert.</p>
| 39405 | Mirror an entire part in solidworks to create a new part |
2020-12-30T09:19:40.293 | <p>Due to the ongoing COVID pandemic, I decided to put a basketball hoop on one of our outside walls, so that me and my son can stretch a bit. The problem is that the wall I plan to put the basketball is built with hollow bricks, (see below).</p>
<p><a href="https://i.stack.imgur.com/vgTKY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vgTKY.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/SeKfu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SeKfu.png" alt="enter image description here" /></a></p>
<p>The thickness of the brick is close to 10cm.
Additionnal specs :</p>
<ul>
<li><p>the wall is getting its fair share or rain during autumn and winter and its facing the wind, so I want to make sure that no moisture will find its way in.</p>
</li>
<li><p>there may be some dunking (we are on average 100 kg each) involved so I need to make this as robust as possible.</p>
</li>
</ul>
<p>I considered but I find them lacking :</p>
<ul>
<li>screw plastic wall plugs</li>
</ul>
<p><a href="https://i.stack.imgur.com/nNd9Hm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nNd9Hm.png" alt="enter image description here" /></a></p>
<ul>
<li>screw anchors</li>
</ul>
<p><a href="https://i.stack.imgur.com/YwxkYm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YwxkYm.png" alt="enter image description here" /></a></p>
<ul>
<li>Toggle Bolts</li>
</ul>
<p><a href="https://i.stack.imgur.com/Htxcam.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Htxcam.png" alt="enter image description here" /></a></p>
<ul>
<li>expansion anchors</li>
</ul>
<p><a href="https://i.stack.imgur.com/YG5k5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YG5k5m.png" alt="enter image description here" /></a></p>
<ul>
<li>hollow wall anchors</li>
</ul>
<p><a href="https://i.stack.imgur.com/yVBuhm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yVBuhm.png" alt="enter image description here" /></a></p>
<p>In most of the cases my main worry is that because the brick is hollow the won't find enough support. So, I wonder what is the best practice for this type of brick.</p>
<p>I would greatly appreciate any guidance on the matter.</p>
| |mechanical-engineering|structural-engineering|bolting|fasteners|waterproofing| | <p>I'd recommend first making a hole to fill the channels in the brick with mortar/concrete locally. This way you will effectively have a solid brick which allows for a much stronger fastening and a much larger range of applicable anchors.</p>
<p>The anchors you show above are all mechanical anchors, which is not the optimal solution when a waterproof result is required. You could consider using a chemical anchor in the form of a threaded rod installed with an appropriate resin. (This will require filling the channels of the brick.)</p>
| 39408 | What is the best way to fasten securely a load onto a hollow brick? |
2021-01-01T23:07:03.287 | <p>I have recently bought myself a Flux Beamo laser cutter and am now in the process of setting up fume extraction for it to vent outside. The laser unit has a built in fan of which I haven’t been able to find the specs but it’s fairly powerful. It also comes with a 2 meter length of 100mm diameter ducting which attaches to the laser via a hose clamp. My plan is to use about 400mm of that ducting to go from the laser cutter that is sitting on a table to a section of 100mm PVC pipe about 1 meter long, which will sit again the wall and go up to a 90° bend and then though the wall to the outside. I now have a couple of questions. The first is that I’ve seen small fans on AliExpress that are designed to fit into 100mm pipe: <a href="https://a.aliexpress.com/_mMqOjk9" rel="nofollow noreferrer">https://a.aliexpress.com/_mMqOjk9</a></p>
<p>They claim to be 12 watt and I thought that I could add one in between the ducting and PVC pipe to increase extraction, but by doing this am I actually increasing air flow even if it is less powerful than the laser fan before it? I don’t want to restrict the flow of the existing fan. My second question is that I will need to put some type of cover over the hole on the outside to stop rain coming in, however I’m not sure wether to use one with angled slats that are always open or one the has tilting slats that only open when the fan is pushing? I feel like the second one has more resistance but I’m not sure to what degree.</p>
| |fluid-mechanics|airflow| | <ul>
<li><p>The setup you are referring to is called "fans in series" and unless the 2nd fan is much weaker than the first laser fan it improves the performance of the system moderately. But it is only better than parallel systems in high resistance ducting, like narrow windy ducting, not your case. Fans in series add to the outlet pressure (not its volume) by adding torque to the flow, if they don't create destructive turbulence.</p>
</li>
<li><p>I Suggest you use fans in parallel if you feel the need to increase the volume.</p>
</li>
<li><p>The tilting vents are designed to offer little resistance to flow by being balanced around the hinge and open easily at least the decent ones and have the additional benefit of stoping the back-draft or stray raindrops.</p>
</li>
</ul>
| 39450 | Extraction Fan Setup |
2021-01-02T10:49:07.837 | <p>I made an object in Blender (default cube), then exported it to <code>STL</code>, then imported this <code>STL</code> into Solidworks.</p>
<p>Now I see this object just exists and I can do nothing with it.</p>
<p>I can't reference any face of the cube and can't scetch on it. Also I can't use <code>Mesh Prep Wizard</code> since it is dimmed.</p>
<p><a href="https://i.stack.imgur.com/2NC1i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2NC1i.png" alt="enter image description here" /></a></p>
<p>What can be done here?</p>
| |solidworks|meshing| | <p>You can convert it from a graphic body to a mesh body (use the command search at the top right of the screen), and then you can use the Boolean tools to affect it using other mesh bodies.</p>
<p>You will not be able to "sketch on the face", since by their very definition, mesh bodies are made up of vertices, not faces. Native solidworks objects (BREP) <em>are</em> made of faces however. If you need to sketch on a face, You can use the three vertices of any triangle to define a plane, and sketch on that.</p>
<p>See below for a demonstration from the same question I previously answered elsewhere on the internet.</p>
<p><a href="https://i.imgur.com/Jx4UtTk.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/Jx4UtTk.gif" alt="enter image description here" /></a></p>
| 39458 | Unable to do anything with the mesh imported from blender |
2021-01-02T13:14:25.180 | <p>I was trying to compute the <a href="https://en.wikipedia.org/wiki/Fixed_end_moment" rel="nofollow noreferrer">fixed end moment</a> for the beam (given below) using equilibrium equation for x, y displacement and moment but there are 4 unknown variables reaction and moment at both ends but I have 3 equations only, vertical equilibrium and moment about both ends. could someone help me to get the answer <span class="math-container">$-Pab^2/L^2$</span>?</p>
<p><a href="https://i.stack.imgur.com/DsRe3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DsRe3.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|structures|moments| | <p>As Wasabi correctly pointed out, this can't be done using only equilibrium equations. To get the extra equations you need, you must take displacements and rotations into account. Enter <a href="https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory" rel="nofollow noreferrer">Euler-Bernoulli beam theory</a>. To make a long story short, your main equation is now
<span class="math-container">$$
\tag{1}
\frac{d^2v}{dx^2}=\frac{M(x)}{EI}
$$</span>
And recall that a section's rotation is given by <span class="math-container">$\theta(x)=\frac{dv}{dx}$</span>.
To calculate the bending moment diagram, we must first "get rid" of the unknown moment at the fixed supports. You can split your problem as follows
<a href="https://i.stack.imgur.com/TyNVM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TyNVM.png" alt="Splitting of the problem" /></a></p>
<p>Now you can calculate the bending moments for each case, which are
<span class="math-container">$$
\begin{aligned}
M_1(x)&=\begin{cases}
\frac{Pb}{L}x & \text{if }x\leq a \\
\frac{Pb}{L}x-P(x-a) & \text{if } x> a
\end{cases}
\\
M_2(x)&=\frac{M_1}{L}x-M_1
\\
M_3(x)&=\frac{M_2}{L}x
\end{aligned}
$$</span>
Notice we just replaced the unknown moment from the fixed support for a positive moment <span class="math-container">$M_i$</span>, which we want to determine.</p>
<p>Now let's proceed to integrate these bending moment expressions as per equation <span class="math-container">$(1)$</span>.</p>
<p><span class="math-container">$$
\begin{aligned}
v_1(x)&=\frac{1}{EI}\begin{cases}
\frac{Pb}{6L}x^3+C_ax+D_a & \text{if }x\leq a \\
\frac{Pb}{6L}x^3-\frac{P}{6}x^3+\frac{Pa}{2}x^2+C_bx+D_b & \text{if } x> a
\end{cases}
\\
v_2(x)&=\frac{1}{EI}\left(\frac{M_1}{6L}x^3-\frac{M_1}{2}x^2+C_2x+D_2\right)
\\
v_3(x)&=\frac{1}{EI}\left(\frac{M_2}{6L}x^3+C_3x+D_3\right)
\end{aligned}
$$</span>
Now we need boundary conditions to determine the integration constants, these conditions come from the pinned supports and from compatibility, they are
<span class="math-container">$$
v_1(0)=0\qquad v_1(a^-)=v_1(a^+) \qquad \theta_1(a^-)=\theta_1(a^+) \qquad v_1(L)=0 \\
v_2(0)=0 \qquad v_2(L)=0 \\
v_3(0)=0 \qquad v_3(L)=0
$$</span>
Some algebra to determine these constants and we arrive at
<span class="math-container">$$
\begin{aligned}
v_1(x)&=\frac{1}{EI}\begin{cases}
\frac{Pb}{6L}x^3+\frac{3PLa^2+PL^3-3PL^2a-PL^2b-Pa^3}{6L}x & \text{if }x\leq a \\
\frac{Pb}{6L}x^3-\frac{P}{6}x^3+\frac{Pa}{2}x^2+\frac{PL^3-3PL^2a-PL^2b-Pa^3}{6L}x+\frac{Pa^3}{6} & \text{if } x> a
\end{cases}
\\
v_2(x)&=\frac{1}{EI}\left(\frac{M_1}{6L}x^3-\frac{M_1}{2}x^2+\frac{M_1L}{3}x\right)
\\
v_3(x)&=\frac{1}{EI}\left(\frac{M_2}{6L}x^3-\frac{M_2L}{6}x\right)
\end{aligned}
$$</span></p>
<p>Now that we have the displacements, we can determine the values for <span class="math-container">$M_1$</span> and <span class="math-container">$M_2$</span>. For this we need to solve the following system
<span class="math-container">$$
\theta_1(0)+M_1\theta_2(0)+M_2\theta_3(0)=0 \\
\theta_1(L)+M_1\theta_2(L)+M_2\theta_3(L)=0
$$</span>
Again, doing some algebra leads to
<span class="math-container">$$
\begin{aligned}
M_1&=\frac{Pa(L-a)^2}{L^2} \\
M_2&=-\frac{PL^2a+PL^2b+PLa^2-PL^3-Pa^3}{L^2}
\end{aligned}
$$</span></p>
<p>And finally, using that <span class="math-container">$b=L-a$</span> and <span class="math-container">$a=L-b$</span> you can simplify this to the values you were originally looking for, which are
<span class="math-container">$$
\begin{aligned}
M_1&=\frac{Pab^2}{L^2} \\
M_2&=-\frac{Pa^2b}{L^2}
\end{aligned}
$$</span></p>
<p>And there you have it, the full derivation for the reactions of a fixed-fixed beam with an eccentric concentrated load. It's not overly hard, but the algebra is somewhat cumbersome at times.</p>
<p>Of course, you can extend this method to any loading and boundary condition, even supports at mid-span and such.</p>
| 39461 | How to determine the fixed end moment? |
2021-01-03T06:09:40.870 | <p><strong>Background</strong></p>
<p>I want to build a drip-style coffee maker for fun. My "design" features a sealable cylindrical housing for the water and heating element. <a href="https://rads.stackoverflow.com/amzn/click/com/B08BRFM416" rel="nofollow noreferrer" rel="nofollow noreferrer">Silicon tubes</a> will carry boiling water from the cylinder to the coffee grounds. I plan on using a <a href="https://rads.stackoverflow.com/amzn/click/com/B0006JLVBW" rel="nofollow noreferrer" rel="nofollow noreferrer">cheap water heater element</a> that I can power using a 120V household outlet. A valve will be installed on the side of the cylinder to allow steam to escape. By partially closing the valve, I think I can modulate how quickly boiling water is pushed through the silicone tubing to the coffee grounds.</p>
<p><a href="https://i.stack.imgur.com/7V4jjm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7V4jjm.jpg" alt="sketch" /></a></p>
<p><strong>Question</strong></p>
<p>What material can I use for the cylinder? Right now I'm looking for easy and cheap. I have access to a drill press and welding station, but not a mill/lathe.</p>
<p>Options that I'm considering:</p>
<ul>
<li>3D printing using a food-grade filament. Not a great option. It seems there aren't a lot of good filaments that can withstand boiling water for long periods of time.</li>
<li>CPVC piping. I think this would work, but I'm seeing mixed answers online. A few answers say CPVC is rated for 200 F, not 212 F. Another resource rated CPVC for much higher temperatures.</li>
<li>Copper tubing. This would be safe, but it's an expensive option. And I guess I would need to worry about welding/threading the top and bottom.</li>
</ul>
| |materials|plastic|boilers| | <p>Buy a pressure cooker, drill holes as needed for tube and electric connections .It will have an approved safety valve and a second vent that usually allows for a couple different pressures . This should protect you from yourself. Steam - boiler explosions prompted creation of ASME over 130 years ago . I believe a government agency let manufacturers know if they did not make boiler rules , the government would. So ASME was founded and wrote the boiler code , generally known as the "code". Then ASTM was formed to write specifications for materials to use to build boilers conforming to Code. Basically anything larger than a home hot water heater must conform to Code. Generally anything that can generate steam pressure follows the guide lines. The point is when you build something that can make steam pressure ,you should be careful.</p>
| 39472 | Building a coffee maker - food grade, heat resistant material? |
2021-01-03T16:15:11.200 | <p>Ok so thinking about aircraft like the 787, which have some quite drastic wing flex (upper image), I was wondering what the loads are that are placed on the skin on the the upper and lower surfaces of the wing (not considering torsion from control surfaces etc...) - considering tension and compression only.</p>
<p>So naturally, one would assume that the upper skin panels experiences compression, and the lower skin panels, tension due to the different radii of rotation they both follow. However, does, for example the lower surface of the upper wing skin experience tension, not compression, due to the material thickness (I'm thinking about the principle in the lower image).</p>
<p><a href="https://i.stack.imgur.com/whfAF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/whfAF.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/f6Dlp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f6Dlp.png" alt="enter image description here" /></a></p>
| |materials|aviation| | <p>For the purposes of this thought experiment, you can assume that wing deforms as a whole structure, or more precisely a sandwich structure. So you have a thin skin of composite material and a foam (or air) in between.</p>
<p>Regarding the loads that the wing experiences, is mainly drag and lift, which you can assume that are distributed loads (almost uniform, or triangular).</p>
<p>In that case, the strain distribution can be seen in the following image (see below <span class="math-container">$\varepsilon_c$</span>. The only difference is that the lift has the opposite direction to the load in the image. You can see that the top distribution is entirely in tension and the bottom skin is totally in compression.</p>
<p>So the <strong>strain distribution is continuous</strong>, i.e. there are no abrupt changes (even though there is no material.</p>
<p><a href="https://i.stack.imgur.com/PH0DV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PH0DV.png" alt="enter image description here" /></a></p>
<p>Additionally you can see that the <strong>stress distribution is not continuous</strong>.</p>
| 39480 | What tension/compression loads does aircraft wing skin experience? |
2021-01-04T09:36:53.727 | <p>Is it possible to constraint a line on another line by way of a perpendicular constraint? This also includes center points on other objects.</p>
<p>What I am trying to accomplish is to move a rectangular array to a set distance from another object (all in sketch) at the center points. So basically, I need to move this array to a set location from the corner or another object (inside the object which is a rectangle)</p>
<p>I have been trying to get this to work in different ways for the past 3 hours with no luck or even trying to find additional help on the Autocad inventor forums.</p>
<p>Please keep in mind, I am VERY new to Inventor (started with Fusion and, YUP, there are a ton of differences). Are there even guidelines? I will not be able to use the grid as my measurements are not a whole number and precision is 3 places.</p>
<p>Hopefully, I have explained what I am trying to accomplish, If you need any further information to assist me in answering this question, please do not hesitate to ask. :)</p>
<p>Attached is a sample of what I am dealing with. the column of objects HAS to be x, y from the top and right sides of the outer rectangle. The additional lines outside of the rectangle in the upper right are directional lines for creating arrays only.</p>
<p><a href="https://i.stack.imgur.com/vTpZb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vTpZb.png" alt="Sample" /></a></p>
<p>Thanks in advance.</p>
| |cad|autodesk-inventor| | <p>Ok, so I've re-installed Inventor...</p>
<p>It's only now become apparent to me that your slot is actually a rectangle with all four corners filleted?</p>
<p>If you instead create a 'slot' entity directly, then the relevant point will be automatically created for you, ready for dimensioning.</p>
<p>In any case, below I have shown both how to add the point I was describing before to allow you to dimension to the midpoint, and also I have shown how to make a slot entity instead, which is better practice</p>
<p><a href="https://i.stack.imgur.com/AqU8J.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AqU8J.gif" alt="enter image description here" /></a></p>
| 39493 | Intersection constraints? |
2021-01-05T01:49:45.250 | <p>EDIT: I will state my test problem description and then my solution. But before that, I would like to apprise you of the absence of the actual description given during test (which happened in Zoom and there I was provided with the link leading to tasks).</p>
<p>Problem 1. I was given a beam (that you may see below added as a picture from my textbook) and assigned to draw diagrams of internal forces (<span class="math-container">$Q$</span>, <span class="math-container">$N$</span>, <span class="math-container">$M$</span>) which are shear force, axial (normal) force and bending moment. My question is how you know immediately that these diagrams are drawn incorrectly if all subconditions such as "differential identities" and "all equilibrium equations" hold true. Otherwise, I miss something and do not realise about this.</p>
<p><img src="https://i.stack.imgur.com/Rl8wq.jpg" alt="enter image description here" /></p>
<p>Here you might see a beam whose endings are A and B. At them, there are supports (at A it is non moving, so there arises two reactions, and at B it is moving, so leaving me with one reaction). The data is as follows: <span class="math-container">$q = 12 \frac{kN}{m} $</span>, <span class="math-container">$M = 8 kNm$</span>, <span class="math-container">$F_1 = 5 kN$</span>, <span class="math-container">$F_2 = 14 kN$</span>, <span class="math-container">$l_1 = 2.6 m$</span> and <span class="math-container">$l_2 = 1.2$</span>, so <span class="math-container">$l = l_1 + l_2 = 3.8 m$</span>.</p>
<p>First step is to solve reactions, for which I must compose equations of equilibrium:</p>
<p><span class="math-container">$$ \sum_n F_n ^x = 0 \Rightarrow X_A - F_2 = 0 \Rightarrow
X_A = 14 kN $$</span></p>
<p><span class="math-container">$$\sum_n M_y ^A (F_n) = 0 \Rightarrow M + F_1 \cdot l_1 + l \cdot Z_B - q \cdot l \cdot \frac{l}{2} = 0 \Rightarrow Z_B \approx 17.27$$</span></p>
<p><span class="math-container">$$\sum_n M_y ^B (F_n) = 0 \Rightarrow M - F_1 \cdot l_2 + q \cdot l \cdot \frac{l}{2} - Z_A \cdot l = 0 \Rightarrow Z_A \approx 23.33 kN$$</span></p>
<p>Notice that:</p>
<p><span class="math-container">$$\sum_n F_n ^z = 0 \Rightarrow ql - Z_A - Z_B - F_1 \equiv 0$$</span></p>
<p>So, these reactions are correctly found and the equilibrium is maintained. Next, we are going to draw diagrams of internal forces. Here I am not going to be idle and show you my actions, so you would help me:</p>
<p><img src="https://i.stack.imgur.com/DBqYO.jpg" alt="enter image description here" /></p>
<p>Idea is shown above on the pictures. Let us begin. The cut comes to the point C or to the part of the beam at C. Thus, we must consider two parts of the beams.</p>
<p>The part AC where <span class="math-container">$0 \le x \le l_1$</span>. Internal forces are determined with equilibrium equations, so</p>
<p><span class="math-container">$$\sum_n F_n ^z = 0 \Rightarrow Q_z + ql - Z_A - Z_B - F_1 = 0 \Rightarrow Q_z = 23.33 - 12x$$</span></p>
<p>Note that <span class="math-container">$l = x$</span> along the segment according to method of cut, sections. Next,</p>
<p><span class="math-container">$$\sum_n M_y ^C (F_n) = 0 \Rightarrow M_y + M - Z_A \cdot x + q \cdot x \cdot \frac{x}{2} = 0 \Rightarrow M_y = 23.33x - 6x^2 - 8$$</span></p>
<p><span class="math-container">$$\sum_n F_n ^x = 0 \Rightarrow X_A + N_x = 0 \Rightarrow N_x = - 14 kN$$</span></p>
<p>Note that the following equations are true:</p>
<p><span class="math-container">$$\frac{\partial M_y}{\partial x} = Q_z, \\ \frac{\partial Q_z}{\partial x} = - q$$</span></p>
<p>Thus, internal forces are found correctly and the equilibrium at the cut is maintained. Let us compute the internal forces' values at endings of the cut:</p>
<p><span class="math-container">$$M_y (0) = - 8 kNm, \ M_y (2.6) = 12.098 kNm \\ Q_z (0) = 23.33 kN, \ Q_z (2.6) = - 7.87 kN$$</span></p>
<p>The second part is CB where <span class="math-container">$ 2.6 \le x \le 3.8$</span>. Internal forces:</p>
<p><span class="math-container">$$\sum_n M_y ^B (F_n) = 0 \Rightarrow M_y + M - Z_A \cdot (x + 2.6)- F_1 \cdot x + q \cdot (x + 2.6) \cdot \frac{x + 2.6}{2} = 0 \Rightarrow M_y = - 6x^2 - 2.87x + 20.098$$</span></p>
<p>Hence,</p>
<p><span class="math-container">$$Q_z = \frac{\partial M_y}{\partial x} = - 12x - 2.87$$</span></p>
<p>The axial force is the same because on the cut there is no force <span class="math-container">$F_2$</span>, because it is cut off. Again we compute values of internal forces and draw our diagrams:</p>
<p><img src="https://i.stack.imgur.com/RVGv2.jpg" alt="enter image description here" /></p>
<p>The second problem was much harder. I had to do the same things but in that time, I was to determine the distribution load <span class="math-container">$q$</span>, which they denoted by <span class="math-container">$p$</span>. I had to draw diagrams of internal forces. In addition to that, I was provided with I-beam whose dimensions were given. I guess that this I-beam was given to make me compute the polar moment of the figure and use some formulae from strength materials theory to find such values of <span class="math-container">$q$</span> for which the beam is stable. That is the point. Technically, I drew incorrect diagram. But numbers are correct, so I drew incorrectly and compute correctly :)</p>
| |beam|diagram| | <p>In such exercises, I suggest doing things by steps.</p>
<p>Your work goes from the equilibrium equations directly to the internal force equations. Instead, I suggest doing one thing at a time. The equilibrium equations give you the reactions, so let's make it clear what those are:</p>
<p><span class="math-container">$$\begin{align}
\sum F_x &= X_A - F_2 = 0 \\
\therefore X_A &= 14\text{ kN} \\
\sum F_y &= Y_A + Y_B + F_1 - q(\ell_1 + \ell_2) = 0 \\
\therefore Y_A + Y_B &= 40.6 \\
\sum M_A &= M + F_1\ell_1 + Y_B(\ell_1 + \ell_2) - \dfrac{q(\ell_1 + \ell_2)^2}{2} = 0 \\
\therefore Y_B &= \dfrac{\dfrac{q(\ell_1 + \ell_2)^2}{2} - M - F_1\ell_1}{\ell_1 + \ell_2} = 17.27\text{ kN} \\
\therefore Y_A &= 23.33\text{ kN}
\end{align}$$</span></p>
<p>So, now that that is clear we can move on to the internal force diagrams.</p>
<p>Now, we can do this directly, as you seem to have tried to do, or explicitly by integration. To keep things clear, I'll use integration.</p>
<p>For the shear diagram, it's important to remember that there's a concentrated force, so there'll be a discontinuity.</p>
<p><span class="math-container">$$\begin{align}
Q(x) &= \int q(x) = qx + C \\
Q(0) &= C_{[0,2.6]} = 23.33 \\
Q(3.8) &= -12 \cdot 3.8 + C_{[2.6, 3.8]} = -17.27 \\
\therefore C_{[2.6, 3.8]} &= 28.33 \\
\therefore Q(x) &= \begin{cases}
-12x + 23.33 & x \in [0, &2.6) \\
-12x + 28.33 & x \in (2.6, &3.8] \\
\end{cases} \\
\therefore Q(2.6^-) &= -7.87\text{ kN} \\
Q(2.6^+) &= -2.87\text{ kN} \\
\end{align}$$</span></p>
<p>So, what is different between this result and what you obtained? The values for <span class="math-container">$Q(0)$</span> and <span class="math-container">$Q(2.6^-)$</span> are right, but the first mistake we see is for <span class="math-container">$Q(2.6^+)$</span>.</p>
<p>I don't understand how you got that result, but I assume you did it "intuitively", without performing explicit calculations for each segment of the beam. The correction then is to note that, when moving from left to right, a positive concentrated load leads to a positive change in shear force.</p>
<p>The way to remember this is by remembering the sign convention:</p>
<p><img src="https://i.stack.imgur.com/TfxCv.png" alt="" /></p>
<p>When moving from left to right, we are always looking at a beam element and looking at its left face (not the right face as you drew), where an upward vertical force means positive shear.</p>
<p>But the biggest red flag in your diagram is what you got at B. When calculating the reaction, you correctly obtained 17.27 kN, but your shear diagram at that point is -48.47 kN.</p>
<p>The fact that it's negative is correct, but it should be (in magnitude) equal to the reaction at B. This is because the reaction is upward and would therefore be a positive change to the shear, generating the expected zero shear to the (infinitesimal) right of B.</p>
<p>I can't quite understand your work to obtain the results to the right of <span class="math-container">$F_1$</span>, so I can't quite point out what you did wrong,</p>
<p>And then we move to the bending moment, by integrating the shear diagram:</p>
<p><span class="math-container">$$\begin{align}
M &= \int Q(x) = \begin{cases}
-6x^2 + 23.33x + C_{[0, 2.6]} &= 0 & x \in [0, &2.6) \\
-6x^2 + 28.33x + C_{[2.6, 3.8]} &= 0 & x \in (2.6, &3.8]
\end{cases} \\
M(0) &= C_{[0, 2.6]} = -8 \\
M(3.8) &= -6 \cdot 3.8^2 + 28.33 \cdot 3.8 + C_{[2.6, 3.8]} = 0 \\
\therefore C_{[2.6, 3.8]} &= -21.01 \\
\therefore M &= \begin{cases}
-6x^2 + 23.33x - 8 & x \in [0, &2.6) \\
-6x^2 + 28.33x - 21.01 & x \in (2.6, &3.8]
\end{cases} \\
\end{align}$$</span></p>
<p>Once again, your results to the left of <span class="math-container">$F_1$</span> are correct (though the curvature of your curve is wrong, it should start a sharp drop and become flat near <span class="math-container">$F_1$</span>, since the shear force is greatest at the support).</p>
<p>But then you have a discontinuity at <span class="math-container">$F_1$</span>, which is obviously wrong: a discontinuity should only appear where there is a concentrated bending moment. A concentrated force causes a discontinuity in the shear diagram, but that simply means there's a discontinuity in the <em><strong>tangent</strong></em> of the bending moment.</p>
<p>And then, even more problematically, you have a non-zero moment at B, which is an edge simple support. As such, by definition, it should have zero bending moment.</p>
<p>And then to check our work:</p>
<p><a href="https://i.stack.imgur.com/oBsUu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oBsUu.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/k0NB2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k0NB2.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jho8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jho8m.png" alt="enter image description here" /></a></p>
<hr />
<p><sup>Diagrams obtained via <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free educational 2D frame analysis tool.</sup></p>
| 39514 | How to check if bending moment diagram is correctly drawn |
2021-01-05T15:52:59.270 | <p>I was thinking about this question from quite a while back about <a href="https://engineering.stackexchange.com/questions/18574/height-limits-of-brick-and-mortar-tower">height limits of a brick and mortar tower</a>. One <a href="https://engineering.stackexchange.com/a/37923/61">answer pointed out</a> that, due to the high compressive strength of brick, one could build a tower 7.4 km high <em>without taper</em> before individual bricks fail in compression. So it seems highly likely that a high brick structure will fail in another way. <em>One candidate</em> for a failure (the other two that come to mind would be lateral forces and accumulated errors in the masonry) is buckling pr more precise self buckling. Wikipedia gives a formula for self buckling but for the purpose of this question the simpler case of critical stress to cause buckling is more interesting (because the formula is simpler):</p>
<p><span class="math-container">$$\sigma = \frac{F}{A} = \frac{\pi^2 E}{(l/r)^2}$$</span></p>
<p>The crucial element is, I think, Youngs' modulus <span class="math-container">$E$</span>, which represents stiffness in tension. You often hear the statement that the mortar or rather the connection brick - mortar - brick has practically no strength in tension. Intuitivly, a structure build from several parts should be <em>worse</em> at resisting buckling than a monolith - So the formula given above for buckling would give an upper limit to the maximum stress. However quite slender brick structures, like catalan vaults or tall chimneys, where built and withstood the test of time.</p>
<p>My ultimate question is: <span class="math-container">$E$</span> is a material constant. Strictly speaking a brick tower is a a structure made of bricks and mortar. <em>How is the effective <span class="math-container">$E$</span> determined in this case?</em></p>
<p>*</p>
| |structural-engineering| | <p>To answer this, let us first make quick list of old and new building materials:</p>
<p>Old:
Limestone
Marble
Lime concrete
Clay bricks/ tiles
Slate</p>
<p>Modern:
Portland cement concrete
Steel
Reinforced concrete
Reconstructed stone
Pre-cast concrete
Sandlime bricks</p>
<p>As for bricks, modern bricks are are sandlime, they have smoother appearance than old bricks, but they're acceptable for masonry walls as veneer and decoration, but they suck for compressive retaining structures. The best building materials are practically inert, whereas the greatest defect of all modern materials is their high coefficient of expansion. This means that their seasonal and nighttime temperature differential expansion and contraction is such that expansion joints are essential; A modern brick wall has to have expansion joints every 30 feet. This breaks up the monolithic nature of any structure into little isolated blocks with expansion joints. The weathering and attrition at said joints is a long-term weakness, whereas a traditionally built structure has none of these defects because the matrix is oldschool lime mortar, instead of cement.</p>
<p>Think of the Pantheon in Rome, built around 120 AD; in brick and lime mortar. It's over 140 feet wide and stood for two millennia. No reinforced concrete structure could last that long because once air and moisture have penetrated to the reinforcement there is nothing which can halt its breakdown. It's why steel skyscrapers are actually better than modern highrises today which are essentially wedding cakes made of poured concrete.</p>
| 39526 | How do you account for buckling in a brick and mortar structure? |
2021-01-06T13:59:15.513 | <p>I have carried out 10 cantilever sample tests that provide a braking force as follows (in kN):</p>
<p>22.1, 20.98, 21.03, 20.7, 21.03, 21.09, 20.98, 21.08, 21.24, 21.19</p>
<p>The goal is to recommend the worst case breaking force</p>
<p>The mean is 21.14 and the STD is 0.37.</p>
<p>If I use the 3-sigma rule, the minimum force is 20.04kN (mean-3sigma). I feel like the 22.1kN sample is skewing the minimum data.</p>
<p>Whats the best way to recommending the worst case scenario for the breaking force?</p>
| |mechanical-engineering|statistics| | <p>The problem with your proposed solution of using 3 times the standard deviation, is that it doesn't take into account how accurately the standard deviation has been determined. As you only have 10 test values, it has a significant error bar, which needs to be accounted for somehow.</p>
<p>I would recommend using the method in EN1990 annex D or whatever design code is applicable in your case. That would mean using 3.04 times the standard deviation if you had performed an infinite number of tests, but 4.51 times the standard deviation when it is only known from 10 test values. The safety index is the same in the two cases.</p>
| 39546 | Usage of the three-sigma limits in calculating minimum force |
2021-01-06T23:23:19.260 | <p>So I'm new to using Arduino's in general and just want to have my code by reversed each subsequent startup using the eeprom on my Arduino uno. I've been trying to read up on how to use the eeprom but I'm kind of confused so I've posted below a picture of some sample code that I think should work (but it's probably wrong). Basically I want to on start-up, execute some code, and next startup it executes some other code (the reverse of the original. Then the next startup to execute the original code. I'm unsure of how to fit my Boolean system into the loop part though, so any help would be greatly appreciated!</p>
<p><a href="https://i.stack.imgur.com/UYXEj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UYXEj.png" alt="sample code" /></a></p>
<p>So What I want this code to do is the first time I power the Arduino to print "Code A". Then after its shutdown, and restarted again to print "Code B" But what I get at the moment is <a href="https://i.stack.imgur.com/ovEbk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ovEbk.png" alt="output of code" /></a></p>
| |mechanical-engineering|electrical-engineering| | <p>The problem is that during setup, you first write the EEPROM address to 1 and printA, and then you read the same address which is now 1 so the if condition is again true.</p>
<p>What you should use :</p>
<pre><code>if(EEPROM.read(address)==0)
{
Serial.println('A');
EEPROM.write(address)==1;
} else {
{
Serial.println('B');
EEPROM.write(address)==0;
}
</code></pre>
<p>(Hopefully I explained clear enough the problem. I'am a bit distracted by the newsfeed).</p>
| 39555 | Simple EEPROM Code for Arduino to act as a Boolean |
2021-01-07T10:33:21.910 | <p>I have imported <code>STL</code> into Solidworks, converted it, then built an intersection of this body with a plane. I got 3 objects: first half of a body, second half of a body and an intersection ellipse.</p>
<p>I can select any of these parts:</p>
<p><a href="https://i.stack.imgur.com/NDqK5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NDqK5.png" alt="enter image description here" /></a></p>
<p>On this picture I have selected one half and hovered ellipse.</p>
<p>I would like to sketch in a plane, having this ellipse as a reference.</p>
<p>Unfortunately, the halfs of the body occlude sketch for me. Hence I would like to hide them, but can't: deselecting <code>eye</code> icon hides entire body and an ellipse.</p>
<p>How to accomplish?</p>
| |solidworks| | <p>See below - expand the side bar and the solid bodies folder, to then individually hide solid bodies.</p>
<p>You can use the same side bar to quickly toggle visiblity of Sketches, Planes, Axes etc., too</p>
<p><a href="https://i.stack.imgur.com/JEgpy.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JEgpy.gif" alt="enter image description here" /></a></p>
| 39566 | How to hide part of intersection in Solidworks? |
2021-01-07T14:14:23.710 | <p>I live very near a bulding contruction site, so the "hammer beating wall" noise is a constant in my daylife for the next one year and half (when the building construction will be finished).</p>
<p>In order to cancel the hammer noise, I use a 40 mm headphone with fan noise playing in medium to high volume (I mean, not low volume, but not maximum volume). (My headset doesn't have noise-canceling feature by the way.)</p>
<p>I ask:</p>
<p>1 - Based on Acoustics Science, is this the best way to cancel a "hammer beating wall" noise?</p>
<p>2 - Is there any other frequency, or combinations of frequencies, which I could build using a software, which would be the perfect match for canceling hammer noise? I mean, even better than fan noise frequencies?</p>
<p>Hearing a loud fan noise with headphones is not that nice thing, but is way better than the hammer. It cancels the hammer noise (I hear nothing, but I feel the floor vibrating a bit). So I am searching for the ultimate sound for cancelling the hammer beating wall noise.</p>
<p>Thank you for any help.</p>
| |acoustics| | <p>The word best can be subjective and is dependent on how much money you want to spend and what you can practically do with the technology available.</p>
<p>You're current strategy of listening to white noise via headphones at a volume so that it masks the noise of construction is not a good strategy because the volume of noise emitted by your headphones could be <a href="https://en.wikipedia.org/wiki/Noise-induced_hearing_loss" rel="nofollow noreferrer">damaging your hearing system</a> and it could be causing you permanent <a href="https://www.nidcd.nih.gov/health/noise-induced-hearing-loss" rel="nofollow noreferrer">hearing loss</a>. You need a system that will <a href="https://en.wikipedia.org/wiki/Hearing_protection_device" rel="nofollow noreferrer">protect you hearing</a>.</p>
<h2>Ear Plugs and Ear Muffs - Passive Noise Reduction</h2>
<p>The first thing you could do is wear <a href="https://www.3m.com.au/3M/en_AU/company-au/all-3m-products/%7E/All-3M-Products/Personal-Protective-Equipment/Hearing-Protection/Safety/Personal-Safety/Ear-Plugs/?N=5002385%208709322%208711017%208711405%208720539%208720546%208720869%203294857444&rt=r3" rel="nofollow noreferrer">ear plugs</a> or <a href="https://www.3m.com.au/3M/en_AU/company-au/all-3m-products/%7E/All-3M-Products/Personal-Protective-Equipment/Hearing-Protection/Safety/Earmuffs/?N=5002385%208709322%208711017%208720539%208720546%208720749%203294857444&rt=r3" rel="nofollow noreferrer">ear muffs</a>. These will reduce noise but not eliminate it.</p>
<p>Some issues with ear plugs are:</p>
<ul>
<li>some people find them uncomfortable to wear for prolonged periods.</li>
<li>people with ear drum damage or ear infections cannot use them.</li>
<li>to be effective, the plugs must be inserted correctly, with most of
the plug inserted within the ear canal.</li>
<li>for hygiene reasons, most ear plugs are disposable and should only be
worn once.</li>
<li>ear plugs made of rubber and <em>fitted to you ears by a professional auditory expert</em> can be reused, but they must be washed very carefully between each time they are used.</li>
<li>they muffle all sounds, including the sounds in the room were you
are, so they make conversations and listening to music or other
sounds difficult. This also applies to ear muffs.</li>
</ul>
<p>Some issues with ear muffs are:</p>
<ul>
<li>each muff must completely cover each ear</li>
<li>the muffs must be periodically cleaned, preferably with antibacterial
wipes, for hygiene reasons and to prevent ear infections.</li>
<li>ear muffs can be uncomfortable to wear for prolonged periods, but
generally are more comfortable than ear plugs.</li>
<li>the muffs must be worn so there is no air gap between the muff and
the ear or sides of the head.</li>
<li>for <em>slightly</em> increased noise reduction, ear plugs can be worn while
using ear muffs.</li>
</ul>
<p>Some people, such as underground miners in some countries, must wear either ear plugs or muffs, or both, for the entire 12 hour duration of their working shift. Because of this, it is possible to wear either ear plugs or muffs constantly while you are at home, but it would not be an ideal situation. While sleeping it is possible to sleep with ear plugs inserted in ones ears, but it would be very difficult to sleep with ear muffs on. Which would be a problem for people who work night shift and must sleep during the day.</p>
<h2>Active Noise Reduction</h2>
<p>Noise cancelling headphones could offer a solution, but they are best at cancelling constant noise that has a limited frequency range, such as that produced by vehicle engines. They work by producing a sound with the same frequency as the disturbing noise but out of phase so the two sounds cancel each other.</p>
<p>With noise cancelling headphones it would be best to use headphones that surround each ear. The issues that I listed for ear muffs also apply to noise reduction headphones.</p>
<p>It may be possible to purchase a noise reduction system that could be <a href="https://www.nytimes.com/2020/07/11/science/windows-street-noise.html" rel="nofollow noreferrer">installed in a room</a>, but such a system would at best reduce noise in that <a href="https://www.extremetech.com/extreme/170649-sono-a-noise-cancelation-and-isolation-device-that-sticks-on-your-window" rel="nofollow noreferrer">room</a>.</p>
<h2>Alterations to the House or Apartment</h2>
<p>These <a href="https://soundproofnation.com/noise-canceling-system-for-home/" rel="nofollow noreferrer">options</a> <a href="https://soundproofliving.com/home-noise-canceling-system/" rel="nofollow noreferrer">may be expensive</a> and may or may not be practical in your situation.</p>
<ul>
<li>Seal and thicken the doors (replace the door with one made of dense
timber). Use a door sweep at the bottom of the door to disrupt the
gap between the base of the door and the floor.</li>
<li>Seal the windows and cover the windows with heavy drapes. The drapes act as a sound insulator.</li>
<li>Add mass to the walls. This can include retrofitting sound proofing
insulation, which would be very expensive to do or you can hang
drapes or carpets on the walls.</li>
<li>Soundproof the ceiling and floor by installing sound proofing
insulation above the ceiling and below the floor.</li>
<li>If possible erect a noise barrier between the source of the noise and
your dwelling.</li>
</ul>
| 39572 | Better way to cancel “hammer beating wall” noise |
2021-01-07T14:45:51.180 | <p>I am trying to make a simple dispenser without using any electricity. I just want to create a simple dispenser that dispenses flour with the turn of a knob. I am creating this because wherever I bake, I need so many different measuring cups for each item, so I want to make a dispenser that dispenses each item, but I will start with just flour. Since flour is usually used in cups, I will have the dispenser dispense flour in ¼ cups. Now to get into the problems, I have a design, but I dont know how to make it.</p>
<p>Everything has to be in the right size since I am having the measurement of ¼ cup, so the space in between the spinning disc must be equivalent to ¼ cup. The other issue is for this to work; the canister holding the flour must be just big enough for the spinning disc to fit inside, so that means I can’t buy it online. So my two most significant issues are how do I make the disc and how do I make the canister that would fit. If anybody can help, that would be much appreciated.</p>
<p>If anyone has any alternative designs or ideas, they are all welcomed. I also want to mention that I dont know how to do a lot of stuff since I am still in high school.</p>
<p><a href="https://i.stack.imgur.com/1lxLl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1lxLl.png" alt="The desgin I am using" /></a>
<a href="https://i.stack.imgur.com/GgdNX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GgdNX.jpg" alt="Knob and spinning disc desgin " /></a></p>
| |mechanical-engineering|design| | <p>If your objective is to use the design shown in the patent drawing or similar, you'll have to have a detent mechanism; something to restrict movement to carefully designed intervals. That's not likely to be particularly difficult to integrate into the design and you can research detents independently of this answer.</p>
<p>For the rotating paddle, consider that you can calculate the necessary volume to generate a quarter cup. The volume of a sphere is <span class="math-container">$\frac{4}{3}\pi.r^3$</span> and I'd suggest converting cups to cubic centimeters to keep the math easier. <span class="math-container">$\frac{1}{4}$</span>cup = 59 cc, rounded to 60 cc.</p>
<p>If you'll want to use the full turn for <span class="math-container">$\frac{1}{4}$</span>cup, each paddle should be the appropriate fraction of the sphere volume calculated previously. A six-paddle dispenser would have approximately 10 cc per segment. You can scale things up as desired, see below.</p>
<p>Also consider the exit path. The hopper would enable the material to fill all segments, but the exit should be restricted to open only one segment to gravity per detent click. Think of the mouth appearance of the Pac-Man game, or a wedge removed from a lemon. The opening allows one segment of material to fall, while the detent clicks into place providing the necessary delay for that action.</p>
<p>If six clicks per quarter cup is excessive, increase the overall volume to make it 60 cc per segment, which provides for one-quarter cup per detent click. Because baking is not a particularly exact science, one can tolerate a bit more or a bit less during the measuring process.</p>
<p>For the construction of such a mechanism, 3D printing immediately jumps to the forefront. That's a completely different question/answer sequence, I believe.</p>
| 39576 | Simple Flour Dispenser |
2021-01-07T17:18:43.533 | <p>For example in yellow:</p>
<p><a href="https://i.stack.imgur.com/5Id3q.png" rel="noreferrer"><img src="https://i.stack.imgur.com/5Id3q.png" alt="enter image description here" /></a></p>
<p>if all bars were rectangular, the entige frame would tend to skew. To prevent this, these diagonal bars are mounted.</p>
<p>What is the term for them?</p>
| |terminology|architecture|wood| | <p>The generic term is <strong>Bracing</strong>.</p>
<p>You will often find it further qualified as <strong>Cross Bracing</strong>, <strong>Diagonal Bracing</strong> or <strong>Triangular Bracing</strong>, depending on the whim of the person doing it.</p>
| 39582 | What is the term for diagonal bars which are making rectangular frame more rigid? |
2021-01-08T02:18:59.827 | <p>Calculate the area of the triangular tract of land and its most</p>
| |surveying| | <p>I'll add my answer because the question is requesting the probable error.
To calculate the area A (essentially the same as kamran's answer but in line with the OP symbols) as:</p>
<p><span class="math-container">$$A= \sqrt{k(k-a)(k_a-b)(k-c)}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$k =\frac{a+b+c}{2}$</span></li>
<li>a, b,c are respectively 180.21, 275.26, 156.31</li>
</ul>
<p>This simplifies to:
<span class="math-container">$$\sqrt{(a+b) (a+c) (b+c) (a+b+c)}$$</span></p>
<p>In order to calculate the error you can use the following relationship:</p>
<p><span class="math-container">$$ dA = \frac{\partial A}{\partial a}da+\frac{\partial A}{\partial b}db+\frac{\partial A}{\partial c}dc$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\frac{\partial A}{\partial a}= \frac{(b+c) \left(3 a^2+4 a (b+c)+b^2+3 b c+c^2\right)}{2 \sqrt{(a+b) (a+c) (b+c) (a+b+c)}}$</span></li>
<li><span class="math-container">$\frac{\partial A}{\partial b}=\frac{(a+c) \left(a^2+4 a b+3 a c+3 b^2+4 b c+c^2\right)}{2 \sqrt{(a+b) (a+c) (b+c) (a+b+c)}}$</span></li>
<li><span class="math-container">$\frac{\partial A}{\partial c}=
\frac{(a+b) \left(a^2+3 a b+4 a c+b^2+4 b c+3 c^2\right)}{2 \sqrt{(a+b) (a+c) (b+c) (a+b+c)}}$</span></li>
<li><span class="math-container">$da, db, dc$</span> are respectively <span class="math-container">$0.05, 0.02, 0.04$</span></li>
</ul>
<p>Since you probably want the relative error you'd need to divide by A:</p>
<p><span class="math-container">$$\frac{dA }{A} =\frac{1}{A} \left(\frac{\partial A}{\partial a}da+\frac{\partial A}{\partial b}db+\frac{\partial A}{\partial c}dc\right)\Rightarrow$$</span></p>
<p><span class="math-container">$$\frac{dA }{A} =\frac{1}{A}\frac{\partial A}{\partial a}da+\frac{1}{A}\frac{\partial A}{\partial b}db+\frac{1}{A}\frac{\partial A}{\partial c}dc$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\frac{1}{A}\frac{\partial A}{\partial a}=\frac{3 a^2+4 a (b+c)+b^2+3 b c+c^2}{2 (a+b) (a+c) (a+b+c)}$</span></li>
<li><span class="math-container">$\frac{1}{A}\frac{\partial A}{\partial b}=\frac {a^2 + 4 a b + 3 a c + 3 b^2 + 4 b c +
c^2} {2 (a + b) (b + c) (a + b + c)}$</span></li>
<li><span class="math-container">$\frac{1}{A}\frac{\partial A}{\partial c}=\frac{a^2+3 a b+4 a c+b^2+4 b c+3 c^2}{2 (a+c) (b+c) (a+b+c)}$</span></li>
</ul>
<p>A simpler example for the error (for the rectangle case) can be seen <a href="http://spiff.rit.edu/classes/phys311/workshops/w1b/area_example.html" rel="nofollow noreferrer">here.</a></p>
| 39594 | Probable error (Area) |
2021-01-08T22:04:46.380 | <p>There is a missile guidance algorithm called proportional navigation. Is there an opposite algorithm for missile avoidance? Something that tries something other than escaping in the direction of a homing missile that doesn't work when the escaping object has a little less power?</p>
| |aerospace-engineering| | <p>Short answer: if there was one on the target, it would basically seek to be unpredictable, i.e. random (but within bounds of wanting to close the distance)</p>
<p>Long answer:
If the target is ahead of the interceptor, and they are both flying in approximately the same direction, a faster/more agile interceptor will eventually intercept the target, unless the target does something like release chaff or the interceptor runs out of propellant.</p>
<p>If they are head-on, closing distances, then the relative agilities makes a difference. If the interceptor has a relatively small maneuverability envelope, then the target may be able to dodge the interceptor, and then fly past.</p>
<p>From these examples above, we see that the evasion strategy implemented by the algorithm must take into account the relative speed and agilities of itself, and the interceptor. I'm not aware of any air-to-surface, air-to-air, ballistic or other missile, that is capable of detecting and discriminating what interceptors are launched against it, and plan an evasion route accordingly.</p>
<p>Usually what is done is the route to the target has features that make the target hard to intercept en route, such as a terrain following, hypersonic velocities or random course changes.</p>
<p>Specific answer:
Let the:
Interceptor be I
Missile be M
Target of missile be T
Bearing of M to be BM
Bearing of I to be BI</p>
<p>Essentially you want M to close the distance with T in the most direct route possible. In 2D space without obstacles, if T is fixed then you just need a bearing, BM.</p>
<p>M may assume that I will use proportional navigation to intercept it, and will follow BI. Plotting a course for M to hit T will therefore be a compromise between the shortest path, and the path that evades interception.</p>
<p>How I would code the algorithm is to let M have a random evasion time interval, RT, specified by an average + variance. Every RT, M makes a random course change.</p>
<p>The random course change is itself defined by another Gaussian distribution with the average = BM and the variance an "evasiveness parameter".</p>
<p>Hope this helps.</p>
| 39611 | How to dodge homing missile? |
2021-01-08T22:32:46.493 | <p>I have a outdoors wooden window blinds and currently I am raising/lowering them manually - by pulling/releasing a rope that's hooked on them using pulley. It's the most basic setup but still I can't understand how the calculation of torque works.</p>
<p>I know the formula is <em>Torque = Length x Force</em> and Force is gravity pull (mass x 9.81 N*kg<sup>-1</sup>) (+ some friction). But I am not sure what "Length" is in this case. And where in this formula is considered the motor shaft diameter and step size. From what I understand the torque should be bigger when using microsteps.</p>
<p>So, what is the correct formula in this setup to estimate required torque for the step motor to be able to pull the blinds all the way up and back?</p>
| |torque|stepper-motor| | <p>Two things.</p>
<ul>
<li><p>What your 0.35cm pulley/wire is doing is just redirecting the tension in the rope. You need a separate pulley to wind the rope around.</p>
</li>
<li><p>I see something is wrong in the dimension of your pulley.</p>
</li>
</ul>
<p>let's say your blind is 180cm long and the rope touched the pulley 4cm away from its support at each end, and is lifting 5kg/ half of the weight of the blind.
A 0.35 cm wire can support the following moment.</p>
<ul>
<li><p>I= moment of inertia</p>
</li>
<li><p>s= section modulus</p>
</li>
</ul>
<p><span class="math-container">$$I = \frac{πr^4}{4}=0.00073cm^4, \quad and\ S=\frac{I}{C}=0.0042cm^3$$</span></p>
<p><span class="math-container">$$M_{max}= Fy*S=2500*0.0042=10.5kgcm $$</span></p>
<p>But your existing moment with just a safety factor of 2 for dynamic loading is</p>
<p><span class="math-container">$$5kg*4cm*2=40kgcm$$</span></p>
<p>It means your wire/pulley should have bent by now.</p>
<p>Ignoring that
Your pulley here is the pulley attached to the motor, or to the wall and say it is a 5cm radius pulley, then you need a torque</p>
<p><span class="math-container">$\tau= 0.05*10kg*9.8*2_{safety-factor}=9.8Nm$</span></p>
| 39612 | Calculating required torque of stepper motor for manipulating blinds |
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