CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2022-10-04T08:24:04.620 | <p>I am trying to find the angle theta when the system is in static equilibrium but I have some difficulties in setting up my equation due to the extra arm for F1.
Can someone help me to set up the equilibrium equation to find the angle?</p>
<p>I know the length from the pivot point O to F1 and F2 and the angle between the wall and the horizontal line is 52 degrees.</p>
<p>Thanks</p>
<p><a href="https://i.stack.imgur.com/0I2pU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0I2pU.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|torque| | <p><a href="https://i.stack.imgur.com/kNZdb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNZdb.png" alt="angles marked" /></a></p>
<p>Let <span class="math-container">$LL1 \triangleq \sqrt{L1^2 + L1x^2}$</span>. This quantity is fixed.</p>
<p>Lets work with the vertical as our reference (and forget about the 52 deg wall till the end). The angle between the mass and the plate is
<span class="math-container">$
\tan^{-1}(L1x/L1)
$</span>.</p>
<p>So <span class="math-container">$\phi = \tan^{-1}(L1x/L1) - \psi$</span>.</p>
<p>Moment about <span class="math-container">$O$</span> due to <span class="math-container">$m_1$</span> is <span class="math-container">$F1 \cdot LL1 \sin(\phi)$</span>. Substitute for <span class="math-container">$\phi$</span>. Here, the only unknown is <span class="math-container">$\psi$</span>.</p>
<p>After writing down expressions for moment for the plate and <span class="math-container">$m_2$</span>, equate them, and solve for <span class="math-container">$\psi$</span>. Convert the result to <span class="math-container">$\theta$</span>.</p>
| 52732 | Static equilibrium |
2022-10-04T09:06:35.327 | <p>I want to know the flow through a vertical pipe when a valve in the bottom is opened (a disk is lowered as illustrated in the drawing)
I know the heights in the drawing and the diameter of the pipe.
<a href="https://i.stack.imgur.com/rGRz5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rGRz5.png" alt="enter image description here" /></a></p>
<p>I have tried to use Poiseuille's law (assuming laminar flow). Where I get the following:</p>
<p>Flow: <<span class="math-container">$Q_{pipe} = \frac{(p_1-p_2-\rho \cdot g \cdot h_{submerged}) \cdot \pi \cdot D_{pipe}^4}{128 \cdot \mu \cdot h_{pipe}} $</span>>
Velocity: <<span class="math-container">$Q_{pipe} = \frac{(p_1-p_2-\rho \cdot g \cdot h_{submerged}) \cdot \pi \cdot D_{pipe}^2}{32 \cdot \mu \cdot h_{pipe}} $</span>></p>
<p>where: <<span class="math-container">$p_1 = \rho \cdot g \cdot h_{submerged} \quad p_2 = p_{atm}$</span>></p>
<p>However, using this I end up with an average fluid velocity of approximately 9000 m/s.</p>
<p>Can anyone help me set up the equations needed?</p>
<p>parameters:</p>
<p><<span class="math-container">$\rho 1000$</span>>
<<span class="math-container">$g = 9.82$</span>>
<<span class="math-container">$h_{pipe} = 0.152 m$</span>>
<<span class="math-container">$D_{pipe} = 0.102 m$</span>>
<<span class="math-container">$\mu = 0.00141$</span>></p>
<p>the submerged height can vary in the range of 0.1 to 0.45 m.</p>
| |fluid-mechanics| | <p>Since the pipe is vertical, the flow will have to overcome its height as well (if <span class="math-container">$h_{submerged} = h_{pipe}$</span>, there should be no flow). So the proper pressure difference driving the flow will be:
<span class="math-container">$$\Delta p = \left(h_{submerged} - h_{pipe}\right)\cdot \rho\cdot g$$</span></p>
<p>If calculated flow is not laminar, you can use the approach from <a href="https://engineering.stackexchange.com/a/51869/21214">this answer</a>.</p>
| 52736 | Vertical pipe flow |
2022-10-04T15:12:45.833 | <p>This is a pretty basic question. So in the below figure is the representation of prismatic column. The forces are represented as P1 and P2. let x be an arbitrary cross section between A and B. The stress in the arbitrary (not considering self weight) is given as P1/A Where 'A' is the area of cross section of the arbitrary cross section between A,B. Similarly for an arbitrary cross section between B and C stress is given as (P1+P2)/A.<br />
<a href="https://i.stack.imgur.com/FFEG5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FFEG5.jpg" alt="prismatic column with loads P1 and P2" /></a></p>
<p>Now my question is the P2 will also induce some force on the cross section in AB region. So the stress should have been (P1+P2)/A for the cross section of X also. (Imagine this cross section as a trampoline. where a metal ball pushes it down from above and another metal ball is tied below, also pulls it down). So the net stress induced should be same (P1+P2)/A and also the net strain should be uniform in AB and BC cross section?</p>
| |structural-engineering|stresses|structures|solid-mechanics|strength| | <p>Is your setup required to be in static equilibrium (i.e., there is no movement of the rod at all) ? If yes, there's a very simple answer to why P2 won't affect the section x-x.</p>
<p>Let's take a section cut slightly above the load application point of P2:
<a href="https://i.stack.imgur.com/qlRo0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qlRo0.png" alt="enter image description here" /></a></p>
<p>Ask yourself - what's the force needed to keep this rod section in place ? The answer is P1:
<a href="https://i.stack.imgur.com/7WWGi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7WWGi.png" alt="enter image description here" /></a></p>
<p>So P2 doesn't affect P1 in this instance because of the requirement of static equilibrium.</p>
| 52748 | Resultant Forces in a vertical rod |
2022-10-04T19:28:38.110 | <p>How can I calculate the height of the submerged part of a partly filled bucket?
(The liquid both inside and outside is freshwater).
The bucket is trapezoidal such that the volume of the submerged can be described as:</p>
<p><<span class="math-container">$V_{bucket} = L_{bottom}*L_{bottom}*h_{submerged} + 2 \cdot (\frac{1}{2} \cdot h_{submerged}^2 \cdot (\frac{0.5 \cdot (L_{top}-L_{bottom})}{h_{bucket}}))$</span>></p>
<p>Can I calculate this using the gravitational force alone (such that I use the mass of the water inside the bucket) as:</p>
<p><<span class="math-container">$g \cdot (m_{bucket}+m_{water}) = \rho \cdot g \cdot V_{Liquid} $</span>></p>
<p>Or should I calculate the force from the water inside using the hydrostatical pressure as:</p>
<p><<span class="math-container">$m_{bucket} \cdot g + P_{atm} + \rho \cdot g \cdot h1 = \rho \cdot g \cdot V_{Liquid}$</span>></p>
<p>Also in relation to the buoyancy force, should I take the angle at the sides into account or do I just use the center of buoyancy?</p>
<p>Thanks</p>
<p><a href="https://i.stack.imgur.com/7zthi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7zthi.png" alt="enter image description here" /></a></p>
| |fluid-mechanics|fluid|hydrostatics| | <p><a href="https://en.wikipedia.org/wiki/Archimedes%27_principle" rel="nofollow noreferrer">Archimedes' principle</a> states that:</p>
<blockquote>
<p>Any object, totally or partially immersed in a fluid or liquid, is buoyed up by a force equal to the weight of the fluid displaced by the object.</p>
</blockquote>
<p>In other words; for a body partially submerged, the buoyancy force is equal to the total weight of the body. For a body fully submerged, the buoyancy force is equal to the volume of the body times the density of the fluid it displaces.</p>
<hr />
<p>If applying Archimedes' principle to your problem (a floating/partially submerged body), we know that the weight of the water displaced by the bucket is equal to the total weight of the bucket (including both the actual bucket and the liquid inside). Knowing the weight of the displaced water allows us to calculate the volume of the displaced water, and at last solve for the submerged height.</p>
<p>That is, <span class="math-container">$$m_{bucket\,including\,liquid}= V_{displaced\,water} * \rho_{displaced\,water}$$</span>
where the unknown is <span class="math-container">$V_{displaced\,water}$</span>. Then, since the shape of the displaced water volume is given by the shape of the bucket, we have: <span class="math-container">$$V_{displaced\,water} = V_{bucket}$$</span>
which can be solved for <span class="math-container">$h_{submerged}$</span> using the equation described by the question.</p>
| 52755 | Submerged fraction buoancy force |
2022-10-04T21:21:09.823 | <p>I am trying to get the:</p>
<ul>
<li>maximum deflection</li>
<li>location of maximum deflection</li>
<li>deflection at any point</li>
</ul>
<p>For a beam under an arbitrary triangular load (i.e. <em>a</em>, <em>b</em>, <em>c</em>, and <em>w</em> can vary). I tried to solve this via double integration by splitting it up into 3 piecewise functions but get stuck with the constants where I can't seem to get enough boundary conditions. I know that deflection at x = 0 is 0, deflection at x = L is 0, and the slope and deflection must be compatible at x = a and x = (a+b) at the cut points for the piecewise functions. (Where x = 0 at the left end of the beam, and positive going to the right).</p>
<p>Is there a smarter way to do this? I'm hoping to get solutions in terms of formulas, where I can just solve this once and plug and play in the future for a small coding project I'm doing.</p>
<p><a href="https://i.stack.imgur.com/1VWBA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1VWBA.png" alt="Arbitrary Triangular Load" /></a></p>
| |structural-engineering|beam|mathematics|deflection|deformation| | <p>If you are doing this for a coding product, a good possibility is doing it by numerically integrating the applied load <span class="math-container">$w$</span>. The advantage of this is that it can be easily extended to general forms of <span class="math-container">$w$</span> for which closed expressions may be hard to get, the disadvantage is that it's more computationally expensive than evaluating a known formula.</p>
<p>The relevant equations are as follows, for the simply supported case shown in the image.</p>
<p>Reactions
<span class="math-container">$$
\begin{aligned}
P_2&=\frac{1}{L}\int_0^Lw(x)x\,dx \\
P_1&=\int_0^Lw(x)\,dx -P_2
\end{aligned}
$$</span></p>
<p>Internal shear and bending moment
<span class="math-container">$$
\begin{aligned}
V(x)&=\int_0^xw(x)\,dx-P_1 \\
M(x)&=-\int_0^xV(x)\,dx
\end{aligned}
$$</span></p>
<p>And, assuming Euler-Bernoulli beam theory,
<span class="math-container">$$
\begin{aligned}
\phi(x)&=\frac{M(x)}{EI}\\
\theta(x)&=\int\phi(x)\,dx +C \\
\delta(x)&=\int\theta(x)\,dx +Cx+D
\end{aligned}
$$</span>
For the simply supported case <span class="math-container">$D=0$</span> and <span class="math-container">$C=-\left[\int\theta(x)\,dx\right](L)$</span></p>
<p>A MATLAB implementation of this is shown below</p>
<pre><code>clear variables
close all
clc
%Inputs
%--Beam properties--
L=10; %Length, meters
E=200; %Modulus of elasticity, GPa
I=1; %Moment of inertia about axis considered, m^4
%--Load properties--
a=3; %Start of triangular load, m
b=4; %Length of triangular load,m
w=1; %Maximum magnitude, kN/m
%Boundary conditions (Only isostatic cases considered in this code)
fixed_slope=[0 0]; %0 indicates free dof, 1 means fixed dof.
fixed_deflection=[1 1];
%Simulation parameters
delta=0.01; %Max element length, m
%Calculations
% ADD A BIT HERE CHECKING THAT fixed_slope+fixed_deflection = 2 AND
% fixed_deflection has at leas one element or the problem is ill-posed
% ADD A BIT HERE CHECKING THAT a+b<=L and other validations
%
n=ceil(L/delta);
x=linspace(0,L,n)';
delta=x(2)-x(1); %Recalculated because original delta may have changed
q=(x>a).*(x<=a+b).*(w/b*(x-a)); %Conditional definition of applied load
% Equivalently using for loops and conditionals
% q=zeros(length(x),1);
% for i=1:length(x)
% if x(i)>=a && x(i)<=a+b
% q(i)=w/b*(x(i)-a);
% end
% end
figure
plot(x,q); xlabel('x [m]'); ylabel('Applied load [kN/m]')
%Determine reactions
qV=delta*trapz(q); %Total force applied by w
qM=delta*trapz(q.*x); %Moment generated by w around the leftmost end
P1=0; %Leftmost reaction, kN
P2=0; %Rightmost reaction, kN
M1=0; %Leftmost moment, kN-m
M2=0; %Rightmost moment, kN-m
% Five possibilities for reactions
if fixed_slope(1)==1 && fixed_deflection(1)==1
P1=qV; M1=qM;
end
if fixed_slope(1)==1 && fixed_deflection(2)==1
P2=qV; M1=qM-qV*L;
end
if fixed_slope(2)==1 && fixed_deflection(1)==1
P1=qV; M2=qM;
end
if fixed_slope(2)==1 && fixed_deflection(2)==1
P2=qV; M2=qM-qV/L;
end
if fixed_deflection(1)==1 && fixed_deflection(2)==1
P2=qM/L; P1=qV-P2;
end
%Calculate internal shear and bending moment
V=zeros(length(x),1); %Shear, kN
M=zeros(length(x),1); %Bending moment, kN-m
for i=1:length(x)
V(i)=delta*trapz(q(1:i))-P1;
end
figure
plot(x,V); xlabel('x [m]'); ylabel('Shear [kN]')
for i=1:length(x)
M(i)=-delta*trapz(V(1:i))-M1;
end
figure
plot(x,-M) %I plot moments on tensioned fibre side, just a convention
xlabel('x [m]'); ylabel('Bending moment [kN-m]')
%Some unit transformation, due to the way inputs were stated.
E=E*1e6; %kPa, or kN/m2
%Euler-Bernoulli beam theory calculation
phi=M./(E.*I);
theta=zeros(length(phi),1);
def=zeros(length(phi),1);
%Integrate radius of curvature to get slope
for i=1:length(theta)
theta(i)=delta*trapz(phi(1:i));
end
%Integrate slope to get deflection
for i=1:length(def)
def(i)=delta*trapz(theta(1:i));
end
% Five cases for boundary conditions
if fixed_slope(1)==1 && fixed_deflection(1)==1
C=-theta(1); D=-def(1);
end
if fixed_slope(1)==1 && fixed_deflection(2)==1
C=-theta(1); D=-def(end)-C*L;
end
if fixed_slope(2)==1 && fixed_deflection(1)==1
C=-theta(end); D=-def(1);
end
if fixed_slope(2)==1 && fixed_deflection(2)==1
C=-theta(end); D=-def(end)-C*L;
end
if fixed_deflection(1)==1 && fixed_deflection(2)==1
D=-def(1); C=-def(end)/L;
end
%Recall that we integrated twice, so we have a constant of integration and
%a constant of integration multiplied by x.
def=def+C*x+D;
figure
plot(x,def); xlabel('x [m]'); ylabel('deflection [m]')
%Plot a black line showing the original shape
hold on
plot([0 L],[0 0],'k','linewidth',2)
</code></pre>
<p>This produces the following results:
Applied load
<a href="https://i.stack.imgur.com/KTxf1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KTxf1.png" alt="Applied load" /></a></p>
<p>Internal shear
<a href="https://i.stack.imgur.com/yM9dS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yM9dS.png" alt="Shear on beam" /></a></p>
<p>Bending moment
<a href="https://i.stack.imgur.com/yMfVC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yMfVC.png" alt="Bending moment" /></a></p>
<p>Deflection
<a href="https://i.stack.imgur.com/cKD5V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cKD5V.png" alt="Deflection of beam" /></a></p>
<p>If you have any questions, just leave a comment and I'll get to it.</p>
| 52757 | Maximum Deflection for simply supported beam under arbitrary triangular loading |
2022-10-04T21:37:55.003 | <p>I read there are multiple models for it,
Plate theory, euler-bernoulli, timoshenko beam theory etc</p>
<p>Which is the most accurate for modelling the bending of a stressed cantilever beam?</p>
| |materials|civil-engineering|structural-analysis|stresses|material-science| | <p>Your question is quite broad. Accuracy depends a lot on the context, so I'll caveat my response by noting that the below is valid only for completely linear-elastic materials (no plastic behaviour whatsoever).</p>
<p>For the most part, Euler-Bernoulli is less accurate than timoshenko since it doesn't account for shear deformations. However, with timoshenko formulations most implementations will approximate the shear factor (which is a complicated thing to solve for on it's own). A finely meshed plate model using standard FE formulations (Kirchoff-Love, etc) which is not susceptible to shear locking will probably be more accurate than timoshenko beam results ... but even here, the difference will be negligible for long memebers.</p>
<p>So for the most part and assuming my conditions noted above apply,</p>
<p>Euler-Bernoulli < Timoshenko < Plate Model</p>
| 52758 | What is the most accurate mathematical model of a cantilever beam's bending? |
2022-10-05T12:40:36.243 | <p>Given a composite material composed of two materials each having different thermal conductivities and heat flux vectors. This leads us to write the two properties as position dependent. One thing I don't understand is taking the temperatures of both materials to be equal at the interface when writing the boundary condition. And also, the normal heat flux vectors to be equal at the interface between the two materials. The writer in the paper <code>An Introduction to Periodic Homogenization</code> mentions that it's for the sake of continuity. It isn't clear to me.
<a href="https://i.stack.imgur.com/HiZCW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HiZCW.png" alt="enter image description here" /></a></p>
| |heat-transfer|composite| | <p>Temperatures must be equal at the interface, otherwise the gradient <span class="math-container">$\frac{\partial u}{\partial x}$</span> would be infinite which would mean infinite heat flux. Regarding heat flux: the heat that goes from one material through the interface must end up in the other material (interface by itself cannot generate or consume any heat).</p>
| 52765 | Confusion about boundary condition (steady heat conduction) |
2022-10-06T12:15:26.397 | <p>Let me preface this by saying that I did my research and have r̶e̶a̶d̶ skimmed a lot of papers on new blade profiles for vertical-axis wind turbines (VAWTs).</p>
<p>What I have yet to read about is a blade profile that is optimized with the recognition that in a VAWT (in contrast to airplanes and HAWTs) the blade is constantly rotating around its vertical axis.</p>
<p>To be fair, my intuition might be wrong here, but a blade that rotates a tiny bit along its circular path is still subject to basically the same conditions it was a mere moment before (ie the air molecules have not had time to move far across the profile). Surely this must have an influence (?), but simulation programs (like xfoil) only simulate independent moments in time and do not capture this continuity.</p>
<p>So ... are there blade profiles that are optimized to constantly travel on a circular path?</p>
| |turbines|wind-power|airfoils| | <p>I am not aware of a publicly available VAWT airfoil library. Yes there are likely optimizations that can be made, but they would be harder to generalize than the lift/drag plots of standard airfoils.</p>
<p>There are two main regimes of vertical access wind turbines:</p>
<p><a href="https://en.wikipedia.org/wiki/Darrieus_wind_turbine" rel="nofollow noreferrer">Darrieous wind turbines</a> have blades that operate at a velocity in excess of the wind speed. These blades use symmetrical airfoils which are well documented in various <a href="https://m-selig.ae.illinois.edu/ads/coord_database.html" rel="nofollow noreferrer">airfoil databases</a>. When the cord is 1/30th or less of the circumference of travel there would likely not be any optimization advantages. However, as the circumference of travel is reduced (if there are any design advantages for doing this) the rotation may become appreciable and CFD optimization would be beneficial.</p>
<p><a href="https://en.wikipedia.org/wiki/Savonius_wind_turbine" rel="nofollow noreferrer">Savonius wind turbines</a> use asymmetrical drag and are typically not considered an airfoil. Hybrids do exist however and CFD would be required to optimize.</p>
<p>I modeled a proprietary VAWT design in <a href="https://www.openfoam.com/documentation/guides/latest/man/rotateMesh.html" rel="nofollow noreferrer">OpenFOAM</a> for a client back in 2010. OpenFOAM has come a long way since then and would easily be able to handle the analysis if you are looking to investigate further.</p>
| 52775 | Is standard airfoil theory suitable to VAWT blades? |
2022-10-08T04:18:13.133 | <p>I'm trying to determine the most structurally sound way to construct a support beam for a serious treehouse. There are four options I've mocked up below - the first two are individual knee brace variants (i.e. knee braces which are not connected to each other) and the second two are tri-beam / yoke / double-knee brace variants where the two knee braces share a connection point at the bottom. The center support bolt for the top beam and the bottom of the knee braces is something called a TAB (treehouse attachment bolt) or GL (Garnier limb). It is a 1-1/4 steel bolt with a 3 inch thick collar to provide more surface contact area with the tree (oak in this case). The fact that the tree is leaning is what causes there to not be an easy answer of what to do (i.e. option C but close to the tree).</p>
<p>What would be the best way to proceed? There will be thousands of pounds supported by this. The angle in A and B is approximately 12-15 degrees if that's helpful to know.</p>
<p><a href="https://i.stack.imgur.com/Foobd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Foobd.jpg" alt="enter image description here" /></a></p>
<hr />
<p><a href="https://i.stack.imgur.com/HTqem.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HTqem.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|beam|structures| | <p>On the basis of what I think your sketches show, option C or D is probably most favorable. Treehouse design will generally be controlled by gravity (weight). Depending upon the internal forces accumulated in the members, member size and possibly even species will be important. In general however, the connections will be most critical for long-term strength of the treehouse. Forces acting through members eccentric (offset) to the connections will present more difficult strength considerations.</p>
<p>Connection eccentricity presents problems with the timber members as well as the connectors. Tear-out, bearing and grain cracking are some of the member connection limit states to consider. Determination of the forces acting on eccentrically loaded connectors and the transfer of these forces from the members is more complex than for concentric connections. For eccentric connections several forces can act simultaneously including withdrawal (tension), bi-axial flexure and of course shear. Therefore, it is desirable to keep connections as concentric to the connected parts as possible.</p>
<p>The diagonal braces in options A and B appear to have the most eccentric connections. Options C and D are more concentric to the tree and the timber members. The TAB bolts however are eccentrically loaded and will be required to resist shear, bending and possibly withdrawal forces. Your intuition is correct in that keeping these TAB bolts as short as possible is desirable. However, TAB bolts are proprietary products. As such they have a specified strength. This strength can vary depending upon the species of the tree. They are specifically manufactured for the construction of treehouses with the apparent purpose of reducing or eliminating eccentricity by acting as an artificial limb wherever you need one.</p>
<p>So, if you are able to determine the forces acting on the TAB bolt, it can simply be compared to the manufacturer specified strength. Use of these products has an added advantage in that companies that manufacture them often provide technical support at no charge.</p>
| 52798 | Compound angle for support tri-beam / double knee brace |
2022-10-08T14:17:12.970 | <p>How can a travelator (i.e. a moving walkway) such as those used in airports be mechanically made to increase speed incrementally when travelling towards the middle and to decrease speed incrementally when travelling towards the end to shorten the travel time from point A to point B?</p>
<p>Here is a video of this kind of <a href="https://www.youtube.com/watch?v=GvfF4TeXz7U" rel="nofollow noreferrer">fast moving walkway at an airport.</a></p>
| |mechanical-engineering|civil-engineering| | <p>The belt is a combination of plates together that are capable of sliding into and expanding out of each other through the comb blades.</p>
<p>Say the length of the stretched belt is <span class="math-container">$X$</span> time its contracted state.</p>
<p>Then the fast speed of the belt</p>
<p><span class="math-container">$$V_{fast}=X\cdot V_{slow}$$</span></p>
<p>The belt is driven at this speed for most of its course. Near the beginning though, it will gradually expand from a slow speed to fast and near the end, it will contract to a slower speed.</p>
<p>To reach from the slow speed, <span class="math-container">$V_{slow}$</span> to the fast speed the first N plates (the speed transition length) need to stretch <span class="math-container">$\frac{X}{N}$</span> each one by one until at the end of the Nth plate it is expanded fully and is moving at the speed of <span class="math-container">$V_{Nth-plate}=X\cdot V_{slow}=V_{fast}$</span>.</p>
| 52806 | How does a variable speed moving walkway work? |
2022-10-10T19:19:30.623 | <p>How the wedge action produce significant holding force in a Jacobs taper for holding a twist drill chuck.
(Even just sliding the tapered plug inside the sleeve by hand makes a rigid connection that it's difficult to remove again by hand)</p>
| |mechanical-engineering|solid-mechanics|machining|machine-design| | <p>It's just friction due to high pressure from the wedge effect. It's not like the pull out force that a Jacobs taper must withstand is very high; It just needs to stop the drill chuck from dropping out under its own weight. It won't withstand machine forces trying to pull the taper out.</p>
<p>The real force that the taper must withstand is the drilling force which pushes into the taper and that is pushing the wedge in so it's no surprise why that works.</p>
| 52833 | How holding force is produced in Jacobs taper for holding drill chucks? |
2022-10-10T21:33:18.747 | <p>I have been testing acrylic in and out of a N<sub>2</sub> bath. It is a very thick piece so it is holding up surprisingly well.
I need to bond pieces of acrylic together for my application that will be submerged in N<sub>2</sub>.</p>
<p>In your experience have you used or know of any epoxy's that are cheaper that work at 77 K (-420 °F)?</p>
<p>I found this [epoxy] <a href="https://www.lakeshore.com/products/categories/overview/temperature-products/cryogenic-accessories/epoxy" rel="nofollow noreferrer">https://www.lakeshore.com/products/categories/overview/temperature-products/cryogenic-accessories/epoxy</a>
but it seems a little expensive.</p>
| |cryogenics|epoxy| | <p>Why not use acrylic cement? It solvent welds the pieces together and leaves no "glue" behind, so it should work at the same temperatures the acrylic does.</p>
| 52835 | Cheap epoxy for Liquid Nitrogen temperatures? |
2022-10-11T10:26:40.883 | <p>I am learning fluid dynamics, and noticing an odd problem in my computations which I understand, but wouldn't expect, so I would like to know what I'm missing. I feel like it's something silly.</p>
<p>At time zero, I create a fluid with constant density 1 and constant velocity 1 to the right. (In other words, mass and momentum is automatically conserved without solving any equations, for this simple starter example.) There is also a non-diffusing concentration of something (say, smoke) included on the the left side.</p>
<p><a href="https://i.stack.imgur.com/eq1xv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eq1xv.png" alt="a" /></a></p>
<p>As time evolves, I expect this gradient to smoothly move to the right by convection. And it does, but something else happens, too:</p>
<p><a href="https://i.stack.imgur.com/oVfJr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oVfJr.png" alt="b" /></a></p>
<p>It starts to segment into bands of higher concentration surrounded by troughs of lower ones:</p>
<p><a href="https://i.stack.imgur.com/XVC9d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XVC9d.png" alt="c" /></a></p>
<p>And this behaviour is self-reinforcing, so after some time it goes really bad:</p>
<p><a href="https://i.stack.imgur.com/iF7aH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iF7aH.png" alt="d" /></a></p>
<hr />
<p>I'm evolving the concentration (since there's no diffusion, there's only the unstable and convection terms) according to</p>
<p><span class="math-container">$$\frac{\partial (\rho \phi)}{\partial t} + \nabla \cdot (\rho \phi \vec{v}) = 0$$</span></p>
<p>where <span class="math-container">$\rho = 1$</span> and <span class="math-container">$\vec{v} = (1, 0)$</span> which makes this very easy to compute. In my Cartesian grid I'm computing the above as</p>
<p><span class="math-container">$$\frac{\partial \rho}{\partial t} = - \frac{\partial \phi}{\partial x}$$</span></p>
<p>(with the derivative in the y axis disappearing due to the zero velocity in that direction.)</p>
<p>It's clear to me that the solution to that equation will amplify troughs in the concentration:</p>
<ul>
<li>Just before the trough, the approximated derivative in the horizontal direction will be negative. This means the divergence will be negative, and the concentration will <em>increase</em> just before the trough.</li>
<li>In the middle of the trough, the approximated derivative in the horizontal direction will be near zero. This means the trough will stay at low concentration.</li>
<li>Just after the trough, the approximated derivative in the horizontal direction will be positive. This means the trough widens.</li>
<li>In addition to the above, there's also a slight movement to the right, but for the purposes of looking at local behaviour I'm ignoring it.</li>
</ul>
<p>I get all that. As a consequence of the equation stating convection as the divergence of concentration times velocity, any troughs will "push aside" concentration and amplify themselves.</p>
<p>But that is really unsatisfying to me. What am I missing?</p>
<p>It also seems unrealistic to me that the troughs naturally form from a completely smooth gradient. Again, I get why it happens (this is in the interaction between the rightward velocity and the same trough-amplifying effect we already saw that creates "tears" in the concentration) – but that shouldn't be the case, I don't think! What am I missing?</p>
<p>I'm open to the suggestion that this might actually be the realistic behaviour of concentration in a fluid when convection is the sole driving force, and that it's just unintuitive to me because I'm used to diffusion smoothing this out.</p>
<p>However, I don't just want to add a diffusion term to fix this in case I am actually missing a critical insight that could cause problems down the line.</p>
<hr />
<p>Edit: I guess this does remind me a little of a phenomenon I am aware of: waves. But would waves really spontaneously arise in one dimension from a concentration gradient interacting with itself in a moving fluid?</p>
| |fluid-mechanics|cad|simulation| | <p>I'm starting to suspect the problem may have been a silly one: discretisation error in the time dimension.</p>
<p>When I run the same simulation with 50× higher temporal resolution (pracitcally: divide velocities by 50 and render out only every 50th timestep), the problem appears to go away.</p>
<p>I will do a few more sanity checks before I can say the above confidently, though.</p>
<hr />
<p>Edit: Yup, that was it, confirmed both by testing and reading a book.</p>
| 52838 | Steep gradients spontaneously emerge during convection (discrete approximation) |
2022-10-11T12:42:31.453 | <p>I'm new to using autocad, so I'll open up a document, look at it and close it without changing anything, but if I scrolled around it flags the files as modified and asks if I want to save them. It's especially annoying on a close all for a bunch of tabs. I imagine that it wants to save viewport position or somesuch.</p>
<p>Is there a way to disable that? I'd like to scroll, zoom, move and just close the file without it asking me to save. I only want to be asked to save if I've modified the drawing somehow.</p>
<p>Thanks a bunch</p>
| |electrical-engineering|cad|autocad| | <p>There does not seem to be a way to disable that.
It has annoyed me for years.
<br>The suggestion that there should be an "off" button for this "feature" provoked religious ire in a "Stop Save Prompt" thread on a forum hosted by AutoCAD.
<br>A post from one of the AutoCAD faithful: "This is AutoCAD. It's always been this way and many before you have made the suggestion in forums designated for feature requests."
<br>It seems to have been inspired by posts that agreed with a heretic who dared to say: "I'm amazed at how completely one-sided, and single minded these answers are."
<br>The first response in the thread was: "Use the free viewer from Autodesk if all you are doing is viewing and plotting"
<br>Another zealot piled on with, "What's confusing is using the wrong tool to view files."
<br>It seems it is anathema to many software developers to include "off" buttons for "features" they think everyone should love as much as they do.
Never mind "personal" computing, this is the age of enforced conformity. <br>Once upon a time, 16 million colors were available to customize the appearances of dialog boxes in Windows. Not now.
<br>AutoCAD is the 800-pound gorilla. They don't need to adapt to users who expect 21st century behavior.</p>
| 52844 | Autocad Electrical 2022 flags a file as modified even if I just scroll around in it, is there a way to disable that? |
2022-10-12T09:04:07.287 | <p>Given is a shaft powered by a steam turbine that rotates unloaded (i.e without generator) at <span class="math-container">$ \omega $</span> rad/s and a torque of <span class="math-container">$T$</span> Nm. How would one match a generator to extract power requiring a certain torque/rpm such that the shaft does not slow down or better operate at maximum efficiency?</p>
<p>I imagine that when choosing a generator that matches rpm and torque exactly the shaft will not rotate at all.</p>
<p>Assume all relevant parameters are known.</p>
| |torque|energy|generator|power-generation| | <p>The efficiency of a steam turbine is driven by steam inle tenergy and outlet temp/pressure. I have never heard of a steam turbine that will run unloaded without overspeeding without speed control. Clearly, taking power out of the system via a generator will slow the turbine down. Generally, this is covered by a governor which introduces more steam when it does.</p>
<p>Your "given" condition isn't realistic unless you found a turbine running out in the wild, which seems unlikely. Generators are matched to turbines based on the turbine nameplate data and the steam available to it.</p>
| 52855 | How to match a synchronous generator with a powered shaft? |
2022-10-07T12:07:38.777 | <h3>Context</h3>
<p>Even major bicycle corporations have the occasional
<a href="https://www.cpsc.gov/Recalls/2022/Trek-Bicycle-Corporation-Recalls-Road-Bikes-and-Bicycle-Handlebar-Stems-Due-to-Fall-and-Crash-Hazards" rel="nofollow noreferrer">product recall on handlebars</a>.</p>
<p>It's easy to see why. The bike itself is a gentle, light, and nimble ~10 kg (for a road bike) object, whereas the weight of the brute riding it is anywhere from 50 kg to 100+ kg, and if the cyclist <a href="https://physics.stackexchange.com/q/541997/259693">doesn't rise before hitting bumps</a>, the frame is subjected to considerable stresses. The stresses are most worrisome on the handlebar (and the seatpost).</p>
<p>The handlebar can be <em>roughly</em> modeled as a beam with two 20 kg weights at each end, and a 40 kg supporting load at the center.</p>
<p>But that's without any impacts. Either way the stresses spike at the stem-handlebar interface, and their starting values are already nontrivial because the handlebar must be firmly clamped at the stem.</p>
<p>Carbon on bicycles is, as cyclists like to repeat often and in unison, not a spontaneously explosive material. Still, anyone who has followed a serious bicycle race will have witnessed live the spectacular ways that carbon frames can fail. You don't want that to happen to <em>your</em> handlebar.</p>
<p>This is one of the reasons why the bicycle industry has moved on mountain bikes from a 31.8 mm diameter for the stem-to-handlebar connection. On mountain bikes that diameter is now used only for lower-market bicycles. The trend is moving towards 35 mm diameter at the stem-handlebar interface.</p>
<h3>Statement</h3>
<p>The stems on the market are of two designs.</p>
<p>The dominant one is an open-front design.</p>
<p><a href="https://i.stack.imgur.com/UCrid.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UCrid.jpg" alt="open-front stem" /></a></p>
<p>The alternative is a closed-front design.</p>
<p><a href="https://i.stack.imgur.com/nakuU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nakuU.jpg" alt="closed-front stem" /></a></p>
<p>Either way, the backside of the stem (the one on the side of the cyclist) is always open.</p>
<p>Especially for a carbon handlebar since the recommended torque of stem bolts is often much lower than that for an alloy handlebar, it would seem that the closed-front design is superior. It distributes the grasping stresses on a larger area. (A special gritty paste is usually required to make a firm connection possible using lower bolt torques.) Indeed, it would appear that a closed-from-both-sides design would be even better.</p>
<p>Why is the open-front bike stem design so widespread despite that it may be inferior to the closed-front design?</p>
<p>The only explanation I gather why the open-front design suffices is that
<a href="https://engineering.stackexchange.com/a/48869/36266">nothing is rigid</a>.
Even steel ball bearings, which we imagine are a rigid object, do not generate <a href="https://engineering.stackexchange.com/q/48864/36266">stresses that shoot towards infinity</a>.
When loaded, they behave just like a rubber ball. They can be squished.</p>
<p>Here also, the two pairs of bolts will generate the dominant stresses on the two narrow vertical bands—regardless of whether the center section is present in the stem. If it's present, it will take relatively little load, and it can hence be discarded. The presence of the mid-section may simply be cosmetic.</p>
<p>Can you confirm/refute and/or elaborate?</p>
<h3>Future Questions</h3>
<ul>
<li>Normally <a href="https://youtu.be/iE1Si-ndgE4?t=80" rel="nofollow noreferrer">shims</a> are only needed to use a larger-diameter stem. Might shims be used to better distribute the load and enable rougher riding with less concern of using dental insurance?</li>
</ul>
| |stresses|friction|bicycles|carbon-fiber| | <p>For starters, I think you have to differentiate between "theoretically inferior" and "practically inferior". Judging by the hundreds of thousands of bikes with open-faced stems out there which are safely zipping along, I don't believe it's really an issue in practice. Stem/bar failures generally are caused by corrosion or grossly incorrect torquing procedure in my experience.</p>
<p>Regarding the actual stress distribution, I would imagine the removed material doesn't play a huge role in clamping the bars. As you noted, the stem side of the bar clamp is universally open. Hence, you're going to be limited by the maximum contact stress on the backside of the stem, regardless of whether the faceplate is open or closed. Stems made by Bontrager for example have two separate "face plate" bands, which is a design that seems to support this theory. Also, as Michael noted, the stress risers on the edges of the stem are of far greater concern.</p>
<p>I also don't think it's necessarily "safer" to reduce the clamping pressure on carbon components. So long as you're within the safe zone so to speak, it doesn't really make you extra safe to lower it even more. In fact, high preload force tends to make bolted joints more fatigue-resistant, so there may actually be some benefit in keeping clamping pressure at a moderate level.</p>
<p>As a point of interest, I have the same stem as the one in your first picture (albeit in black) on my bike right now. No issues.</p>
| 52859 | Why is the open-front bike stem design so widespread despite that it may be inferior to the closed-front design? |
2022-10-13T17:54:09.153 | <p>Would it be feasible to design a passive CVT gearbox that converted input rpm into an output of always N by splitting the input (such as is split in a helicopter transmission) then inverting the speed from one split-torque gear to merge with the input from the other (both inputs undergoing several transformations) for instance in the ring of a planetary gearset?</p>
| |mechanical-engineering|gears|applied-mechanics| | <p>If I understand correctly what you are looking for (and provided that I translated it correctly), then a <strong>power-split transmission</strong> or <strong>power-distributed transmission</strong> should get pretty close to it.</p>
<p>It's a common transmission in tractors. Basically the rotation from the engine gets split into a mechanical part that ends in a planetary gearset and a hydraulic CVT that drives the ring gear of the planetary gearset. Via appropriate settings of the hydraulic CVT, you can control the RPM of the mechanical driveshaft continuously down to 0 RPM, that is, the ring gear turns against the sun gear and exactly cancels it out.</p>
<p>Your application sounds to me like looking for the reversed process, but maybe the concept is helpful.</p>
<p>I don't have a picture at hand right now, but can add one later.</p>
| 52866 | Is splitting torque a possible solution for a passive CVT? |
2022-10-14T01:40:47.940 | <p>I am working on a wire rope pulley at rest, shown below. The wire rope wraps once around the pulley (180°) and has tension T. The pulley axle is on a sliding track that is up against its hard stop, thus a reaction force R balances the free body diagram. If I introduce a force F that pulls on the axle to the left, I believe the free body diagram becomes statically indeterminate? When F is 0, R=2T. But when 0<F<2T then F+R=2T. Does the reaction diminish whilst the tension remains constant? Or does the reaction remain constant whilst the wire rope tension increases? Or something in-between? I know there are methods to solve indeterminate beams, but not sure how to apply here...
<a href="https://i.stack.imgur.com/HyOYm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HyOYm.png" alt="enter image description here" /></a></p>
| |pulleys|tension| | <p>This really depends on how the tension is applied at the end of the rope. More specifically, there are two (basic) cases:</p>
<ul>
<li>rope is inextensible</li>
<li>rope is extensible (acts like a spring).</li>
</ul>
<p>In most cases, if there is no translation of the pulley (ie. the reaction is not zero), then the tension of the rope remains the same.</p>
<h1>rope as inextensible</h1>
<p>In this assumption, (and also assuming there is no spring element at the end of the sides of the rope) then in order to have a tension T you'd need some sort of dead weight (that is the simplest way to think about it).</p>
<p>In the case described above, when you apply the force F then there is no change in rope tension (it remains T).</p>
<p>(Of course is the tension is adjusted/applied with a spring like element this changes).</p>
<h1>rope is extensible</h1>
<p>If the rope is extensible then you can have two different situations depending on how the tension is applied:</p>
<ol>
<li><p>the tension is applied (again) by a dead weight on one end of the rope and fixed on the other end. In this case, in the quasi static case the situation is the same as above (i.e. no change in the rope).</p>
</li>
<li><p>The tension is applied by fixing both ends of the rope to the wall, and adjusting the tension with a turn-buckle (See below)</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/oJtnv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oJtnv.png" alt="turn-buckle" /></a></p>
<p>In that case the wire tension will change, <strong>but only when the reaction R reaches 0</strong>. How much it will change, it will depend on the "spring constant" of the rope.</p>
<h2>to sum up</h2>
<p>As a conclusion, if you know the force F, and the initial force T is also known, then your only unknown is the reaction. So the system is <strong>not statically indeterminate</strong>.</p>
| 52870 | How to solve statically indeterminate pulley? |
2022-10-14T06:55:26.130 | <p>I am having some issues with a PVC coated steel cable. The PVC often strips when used with a pulley.</p>
<p>I wanted to know why this may be happening and possible solutions for it.</p>
| |design|machine-design|pulleys|maintenance| | <p>The pvc coating is designed for cables in the open as a protection from dust, the elements etc.</p>
<p>If the pvc coated cable is going over a pulley then the pvc coating is subjected to crushing and tearing forces that exceed its design parameters.</p>
<p>You likely are using the wrong cable or using a cable outside of its designed use.</p>
<p>Consider other solutions ie different cables or a different design of mechanism, difficult to have any suggestions as your question lacks any useful detail.</p>
| 52873 | PVC Stripping issue on steel wire with pulley |
2022-10-14T11:45:53.597 | <p>When having a tapped hole close to an edge in metal, how do you determine the minimum edge thickness around the tapped hole? Obviously the metal strength is a critical factor, but are there any rules of thumbs for common materials?</p>
<p>I could only find <a href="https://www.aboveboardelectronics.com/recoil-designs.html" rel="nofollow noreferrer">a rule-of-thumb for tapped inserts</a> that specifies:</p>
<blockquote>
<p>The minimum edge distance recommended is the maximum diameter of the STI tap measured from the edge of the material to the centre line of the hole.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/Mq1D8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mq1D8.png" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>The European Standard EN 1993-1-8 <em>Eurocode 3: Design of steel structures –
Part 1-8: Design of joints</em> specifies a distance from the edge of at least 1.2 * d</p>
<p>For slot holes it specifies 1.5 * d.</p>
| 52875 | How to determine minimum edge thickness for a tapped hole? |
2022-10-18T14:24:48.767 | <p>I’m basically looking for a bomb proof way to lock a ≈80lb/40kg metal frame into place, with an electronic release.</p>
<p>Alternatively, looking for alternate ideas for the entire project below.</p>
<hr />
<p>I’m building a solar panel rig to put in the roof of a van that can rotate to angle in any direction (360° rotational and +-60° tilt anyway)</p>
<p>My idea is fairly simple: build a metal frame, make it titleable with some linear actuators, and then stick this onto a lazy susan (a mechanical turn table)</p>
<p>The only snag I’m running into is how to lock the whole thing into place when it’s not moving. I’m sure this is a solved problem and I’m just not aware of the solution. I ideally want the lock to be electronically releasable, and allow 180° (Well 179°) rotation in either direction, though 359° in one direction would also work. And it needs to be stable in 100mph/160kph winds while driving.</p>
<p>Any thoughts? Thanks!!</p>
| |mechanical-engineering| | <p>You can use a ring of holes in the base with vertical pin(s) being driven by a solenoid to lock it in place at different angles of yaw.</p>
<p>If the base is rectangle with ends far from the lazy susan, to prevent wind from blowing it up in driving position you can have the ends slide it hold down brackets mounted on the roof of the car at each end.</p>
<p>A few ways to do this. Some are:</p>
<ol>
<li>Mount brackets like an upside down 'L' on the car roof which the baseplate slides under. Will require the plate to have circular ends for proper clearance so it can rotate.</li>
<li>Have brackets both on the roof and under the plate that slide into each other. Does not require a rounding the ends of the plate. Optimally, those L-brackets would be curved throats and ends when looking at them from above. That would allow a tight horizontal fit. Otherwise the throat will need lateral clearance so they can rotate through each other. I suppose the much easier compromise would be to round just the ends of both brackets and not curve the throat.</li>
<li>Fancy ass arched/curved dovetail</li>
</ol>
<p>In all cases, you want to slope the entryway to those brackets in the direction tangential to the rotational circle to guide things if there is vertical play. Otherwise it may crash. Will also probably want to treat those brackets as bearing surfaces.</p>
| 52906 | Looking for a very solid way to lock a metal frame into place |
2022-10-18T18:59:55.900 | <p>Can somebody help me how to calculate the moment of inertia of a plate with two attached boxes around a rotation point?</p>
<p>I know the lengths to each center of mass and the mass of each object but I am not sure how to calculate the moment of Inertia around point O.<br />
If the length to each center is called r can I calculate the inertia as:</p>
<p><<span class="math-container">$I = m1 \cdot r1 + m2 \cdot r2 + m3 \cdot r3$</span>></p>
<p><a href="https://i.stack.imgur.com/4z5C4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4z5C4.png" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>Try</p>
<p><span class="math-container">$I = m_1 \cdot r_1^2 + m_2 \cdot r_2^2 + m_3 \cdot r_3^2$</span></p>
| 52913 | Moment of inertia around rotation point |
2022-10-19T01:08:49.643 | <p>Fleeces typically exhibit low thermal conductivity, and are often used in winter clothing for their insulating properties.
I am seeking a material for a radiant heat application that is soft to the touch like fleece, but transmits heat well.</p>
<p>Measuring thermal conductivity in Watts per meter-Kelvin.</p>
<p><a href="https://inpressco.com/wp-content/uploads/2019/01/Paper241675-1678.pdf" rel="nofollow noreferrer">This study</a> found 15 fleece samples to all have thermal conductivity below 0.004 W/mK. Textiles with higher thermal conductivity include <a href="http://lib.buet.ac.bd:8080/xmlui/bitstream/handle/123456789/2996/Full%20Thesis.pdf?sequence=1&isAllowed=y" rel="nofollow noreferrer">leather</a> (0.1 to 0.15 W/mK) and <a href="https://thermtest.com/how-thermal-conductivity-decides-what-you-put-in-your-closet" rel="nofollow noreferrer">linen</a> (0.188 W/mK).</p>
<p>I understand that the air trapped within fleece will impede heat flow. So a heat-transmitting fleece might have very dense 'hairs' to shrink air pockets. Alternatively, the hairs may be shorter.</p>
<p>Heated blankets often incorporate fleeces. Do these blankets use particular fabrics to reduce heat loss across the fleece? Or do they just put out enough heat to compensate for the loss through poor conductivity?</p>
| |thermodynamics|thermal-conduction|thermal-insulation| | <p>We add thermal conductivity to textiles and fibres based materials via coating their fibres with graphene and other 2D nano materials. For glass fiber fabric, for ex. we increase thermal conductivity in 100 times (from 0.02 to 2 W/mK). But keep in mind - it is still lower than aluminium for ex (10-30 W/mK). Talk to us if you want to have a unique sample. grafren.se</p>
| 52917 | High thermal conductivity in soft, fleece-like textiles |
2022-10-23T07:33:44.510 | <p>As far as I know, there are two possible uses of nuclear bombs: for mass destruction and for deflecting asteroids. For example, <a href="https://en.wikipedia.org/wiki/Asteroid_impact_avoidance#Nuclear_explosive_device" rel="nofollow noreferrer">Wikipedia</a> describes how nuclear bombs can be used for deflecting asteroids. Can those two things be achieved with the same bombs? Or must they be designed in different ways? Also, can nuclear bombs which are intended for mass destruction, be used to deflect an asteroid without making large adjustments?</p>
| |aerospace-engineering|nuclear-engineering| | <p>Deflecting asteroids is far from the only proposed <a href="https://en.wikipedia.org/wiki/Peaceful_nuclear_explosion" rel="nofollow noreferrer">peaceful use of nuclear explosions</a>, but few (if any) of them made economic, environmental, or political sense.</p>
<p>The answer is yes to your specific question as to whether existing nuclear weapons could be used for deflecting asteroids.</p>
<p>Specially designed <a href="https://en.wikipedia.org/wiki/Nuclear_shaped_charge" rel="nofollow noreferrer">"nuclear shaped charges"</a> were proposed for <a href="https://en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion)" rel="nofollow noreferrer">Project Orion</a> nuclear propulsion which could have been repurposed for asteroid deflection, but these were never built and do not exist (unless there is a government project that has been magically kept secret for decades :)).
<a href="https://en.wikipedia.org/wiki/Neutron_bomb#History_and_deployment_to_present" rel="nofollow noreferrer">Neutron bombs</a> were also proposed as better than regular nukes for <a href="https://doi.org/10.1126/science.255.5049.1205" rel="nofollow noreferrer">"nudging asteroids"</a> without blowing them up, but there has been no drive to develop dedicated neutron bomb asteroid deflectors.</p>
<p>The 2007 NASA report on <a href="https://web.archive.org/web/20150701020407/http://www.nss.org/resources/library/planetarydefense/2007-NearEarthObjectMitigationOptionsUsingExplorationTechnologies.pdf" rel="nofollow noreferrer">"Near Earth Object (NEO) Mitigation Options Using Exploration Technologies"</a> proposed using existing <a href="https://en.wikipedia.org/wiki/B83_nuclear_bomb#Novel_uses" rel="nofollow noreferrer">B83 bombs</a> in a particular sequence.
As @Tiger-Guy says, most nuclear asteroid deflection research assumes "bombs go boom" and works on figuring what sizes and sequences of standoff, surface, or subsurface explosions of existing Nuclear Explosive Devices (NED) will be most effective. As noted in the November 2020 <a href="https://www.ascend.events/" rel="nofollow noreferrer">AIAA ASCEND</a> conference discussion of <a href="https://ntrs.nasa.gov/api/citations/20205008370/downloads/Nuclear_Devices_for_Planetary_Defense_ASCEND_2020_FINAL_2020-10-02.pdf" rel="nofollow noreferrer">“Nuclear Devices for Planetary Defense”</a> (PANEL-17):</p>
<blockquote>
<p>The lack of need for new NED designs to handle the most probable future NEO threats is an important finding of the work to date on this topic</p>
</blockquote>
| 52950 | Can nuclear bombs intended for mass destruction also be used to deflect asteroids? |
2022-10-23T08:36:42.370 | <p>Here is the situation: In a shell and tube condenser, the hot shell side is Pentane from 115°C to 36°C and the cold tube side is LNG at the temperature of -161℃ at the inlet of the heat exchanger. The mass flow rate of Pentane is 1kg/s.</p>
<p>So based on energy conservation, with the outlet temperature of LNG I can calculate the mass flow rate of the LNG.</p>
<p>But I find that at the case when the temperature change of LNG is small(less than 1°C), the LNG mass flow rate is very big(more than 200kg/s).
I wonder If 200kg/s is feasible in practice?</p>
| |heat-exchanger| | <h3>Possible but not likely</h3>
<p>At temperature changes less than 1 degree C you are dividing by a fraction. Make that fraction small enough and you get a huge flow rate from the equation.</p>
<p>In the real world, the flow rates of the LNG will be limited by the physical characteristics of the exchanger, and it is extemely unlikely anyone would build a heat exchanger big enough to pass such a flow rate to change temperature of the LNG by a fraction of a degree. We can build a heat exchanger to pass almost any conceivable flow rate with enough money, but the question would be why.</p>
| 52951 | If 200kg/s mass flow rate of LNG is feasible in a heat exchanger? |
2022-10-23T15:41:58.457 | <p>Consider a system where: <span class="math-container">$ \frac{Y(s)}{U(s)}=\frac{k}{s} $</span>. Is the gain k, or infinity? Given that to a step response, steady state is never reached, but it increases linearly with no end. Thanks!</p>
| |mechanical-engineering|control-engineering|chemical-engineering|pid-control|process-engineering| | <p>There are multiple definitions for the gain of a transfer function <span class="math-container">$G(s)$</span>. The two most common are the <strong>zero-pole-gain</strong> and the <strong>(static) DC-gain</strong>.</p>
<p>The <strong>DC-gain</strong> is obtained by <span class="math-container">$$K_{DC} = G(s) \mid_0 \quad \text{or} \quad K_{DC} = \lim_{s\to 0} G(s),$$</span> depending on the definition you follow.</p>
<p>The <strong>zero-pole-gain</strong> follows from writing the transfer function in the general zero-pole-gain model
<span class="math-container">$$G(s) = K_{zp} \frac{(s-z_1)(s-z_2)\ldots(s-z_m)}{(s-p_1)(s-p_2)\ldots(s-p_m)},$$</span>
where <span class="math-container">$z_i$</span> are the zero's, <span class="math-container">$p_j$</span> the poles and <span class="math-container">$K_{zp}$</span> zero-pole-gain, with <span class="math-container">$i = 1 \ldots, m$</span> and <span class="math-container">$j=1\ldots n$</span>.</p>
<p>For example
<span class="math-container">$$ G(s) = \frac{2s^2+10s+8}{s^2+5s+6} = 2\frac{(s+1)(s+4)}{(s+2)(s+3)},$$</span> has a <span class="math-container">$K_{DC} = \frac{8}{6} = \frac{4}{3} $</span> and <span class="math-container">$K_{zp} = 2$</span>.</p>
<p>For your system, the integrator <span class="math-container">$G(s)=\frac{k}{s}$</span>, the zero-pole-gain is <span class="math-container">$K_{zp}= k$</span>. Depending on the definition, the DC-gain does not exists as you cannot divide by zero, or if you use the limit results in <span class="math-container">$K_{DC} = \infty$</span>.</p>
| 52955 | What is the gain of a purely integrating process? |
2022-10-24T12:18:54.623 | <p>Given an inlet wind speed of <span class="math-container">$v_1$</span> and nozzle inlet area of <span class="math-container">$A_1$</span> and an outlet area <span class="math-container">$A_2$</span> what would be the outlet velocity. The nozzle/concentrator is positioned in an open field where the wind is blowing. The shape of the concentrator is conical.</p>
<p>Obviously, the equation <span class="math-container">$A_1v_1=A_2v_2$</span> does not apply given the compressibility of the fluid. What way can it be calculated?</p>
<p>The motivation of this question is accesing the possibility of using concentrated air in turbines/wind mills. Air velocities are low in certain places so increasing its speed may be of interest.</p>
| |fluid-mechanics|aerodynamics| | <p>If you check out where some wind turbines have been located, valleys have been used as they are, in effect, a tapered concentrator.</p>
<p>Also wind turbines have been located on top of hills - some distance back from the edge as that improves the performance - the "compression" caused by the ground slope also improves the performance.</p>
<p>As I mentioned in the comment, mass flow between inlet and outlet is constant unless you have a leak. So the change in velocity will be according to the change between the diameters and there will be a discharge coefficient as there will be lost air.</p>
<p>There have been ventilation fans using shrouds to reduce the edge losses, but that becomes impractical for wind turbines.</p>
<p>You could set up an experiment using cones to evaluate the velocity increase based on the diameter change. One thing is to then use a pitot tube to measure the velocity profile across the outlet and inlet if you really want to find out what happens. Then you can use that velocity profile to check the mass flow etc and find out which parameters are mostly responsible for the errors. This was a really good lab experiment which was 3 hours work... The analysis and write-up took much longer.</p>
| 52962 | How to calculate the increase in air speed from a wind nozzle / concentrator? |
2022-10-26T04:26:30.590 | <p>Generally speaking cutters such as end-mills and often lead screws are ground vs shaped using a sharp cutter.</p>
<p>I am curious how are high accuracy grinding wheels made? Their very nature seems to be fundamentally not precise yet they are used to create very precise tools.</p>
| |manufacturing-engineering|machining|grinding| | <p>Is there such thing as a "high-accuracy" grinding wheel? What does "high-accuracy" even mean in this context? If you mean of a precise dimension, then no. Grinding wheels wear and with the small amounts of material they are intended to remove, they are used to aim for tighter tolerances so small amounts of wear matters proportionally more than with other tools. Therefore you dress the to size with a diamond. The more precise you need the dimension the more often you dress them. There are finer grinding wheels but that doesn't mean high accuracy; It just means they leave a finer finish.</p>
| 52977 | How are high accuracy grinding stones and disks made? |
2022-10-27T11:31:25.873 | <p>I have a rod that can rotate about a point, the rotation is caused by the buoyancy of 3 balls connected to the rod as in the picture:</p>
<p><a href="https://i.stack.imgur.com/bl9Nv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bl9Nv.png" alt="enter image description here" /></a></p>
<p>I need to calculate the angle at different heights and the submerged fraction of the second ball from the left for each angle.</p>
<p>I tried to calculate the submerged fraction of each ball individually using the angle (and thereby neglecting the gravitational forces).</p>
<p>But by doing the angle does not reduce when the water level falls since I only include the buoyancy forces. This is not right as the rod should follow the water level.</p>
<p>I don't know how to include the downwards forces for each ball.
This is a problem because (looking at the second ball from the left) the ball has the forces <span class="math-container">$F_{g2}$</span> and <span class="math-container">$F_{B2}$</span> acting on it but there will also be some force acting on this from the rest of the systems right?</p>
<p>So my question is how do I set up the forces acting on the middle ball?</p>
<p>I think the moment equation should be something like
<span class="math-container">$F_{g1} \cdot L_1 \cdot angle + F_{g2} \cdot L_2 \cdot angle + F_{g3} \cdot L_3 \cdot angle- F_{B1} \cdot L_1 \cdot angle - F_{B2} \cdot L_2 \cdot angle - F_{B3} \cdot L_3 \cdot angle$</span></p>
<p>thanks</p>
| |fluid|forces| | <p>For solving the problem where the system is constantly changing, you are solving the differential equation (Newton's second law)</p>
<p><span class="math-container">$$
I \frac{d^2\ \alpha}{dt^2} = \sum_i{(M_{gi} + M_{bi})}
= \sum_i{\left((F_{gi} + F_{bi\ (\alpha, level)})\cdot L_i \cdot \sin(\alpha)\right)}
$$</span></p>
<p>where <span class="math-container">$I$</span> is moment of inertia of the total system about the hinge point, <span class="math-container">$\alpha$</span> is the angle, <span class="math-container">$M_i$</span> are the torques, <span class="math-container">$F_i$</span> are the (gravitational and buoyancy) forces, <span class="math-container">$L_i$</span> are the distances to the balls from the hinge point.</p>
<p>Assumptions include (but not limited to)</p>
<ul>
<li>rigid mass less rod</li>
<li>only forces acting are gravity and buoyancy</li>
</ul>
| 52987 | Movement caused by buoyancy |
2022-10-28T11:05:16.910 | <p>There's a video of assembling a physician's scale <a href="https://www.youtube.com/watch?v=k0psOeSMfK0&t=125s" rel="nofollow noreferrer">here</a>.</p>
<p>And here's a photo of the bottom:</p>
<p><a href="https://i.stack.imgur.com/ZWL3Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZWL3Y.png" alt="bottom of Detecto scale" /></a></p>
<p>I estimate that the poise on the scale's weigh beam have a mass or less than <span class="math-container">$1/1000th$</span> of the scale's capacity, and there's no minimum weight for the scale, so the scale must have an mechanical advantage of at least <span class="math-container">$1000$</span> between the end of the weigh beam and the platform.</p>
<p>This could be achieved with three levers in series, each with a mechanical advantage of <span class="math-container">$10,$</span> or two levers each with a mechanical advantage a little over <span class="math-container">$30,$</span> or some other combination.</p>
<p>I see that the weigh beam is one lever, and that it's in series with the long lever under the scale. The load from the platform seems to come in at these places (green arrows):</p>
<p><a href="https://i.stack.imgur.com/rLt9H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rLt9H.png" alt="scale bottom with load locations marked" /></a></p>
<p>It looks like the long lever has a mechanical advantage of about 18; I measured 47 pixels and 828 pixels for these two distances (green):</p>
<p><a href="https://i.stack.imgur.com/u8Nql.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u8Nql.png" alt="scale bottom with distances marked" /></a></p>
<p>If the mechanical advantage is just the long lever in series with the weigh beam, that would leave the weigh beam to have a mechanical advantage of more than <span class="math-container">$50.$</span> Is that right? Perhaps there is some mechanism inside the horizontal crossbar beneath the weigh beam giving part of that mechanical advantage?</p>
<p>My guess was that the short lever is not adding mechanical advantage, but instead positioned so that it doesn't matter whether the patient stands near the front of the scale or near the back. Is that right, or is the short lever needed for additional mechanical advantage?</p>
<p><a href="https://www.rehabmart.com/pdfs/dtc-475_mechanical_chair_scale_by_detecto_troubleshooting_guide.pdf" rel="nofollow noreferrer">Photo source</a></p>
| |mechanical-engineering| | <p>Here's an animated GIF I made from a <a href="https://youtu.be/0mZAjWJ1Q5M" rel="nofollow noreferrer">YouTube video</a> by
thang010146. I don't know enough to answer your questions, but the illustration may help.</p>
<p><a href="https://i.stack.imgur.com/WmHKC.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WmHKC.gif" alt="enter image description here" /></a></p>
<blockquote>
<p>My guess was that the short lever is not adding mechanical advantage, but instead positioned so that it doesn't matter whether the patient stands near the front of the scale or near the back. Is that right, or is the short lever needed for additional mechanical advantage?</p>
</blockquote>
<p>I think it's performing several functions:</p>
<ul>
<li>It handles 1/4 of the total weight. (There are two blue levers and two grey levers).</li>
<li>It provides stability for the platform.</li>
<li>As you point out, it provides an accurate weighing no matter where on the platform the load is placed.</li>
</ul>
| 52998 | Where does the mechanical advantage come from in a physician's scale? |
2022-10-28T19:37:18.367 | <p>Let’s say I had a mesh with 10000 nodes, then made another with 13000 nodes, why would the finer one give me slightly different results to the other even if it appears like results converged?</p>
<p>Also, why do results vary lots before they converge? Why don’t they converge straight away?</p>
| |mechanical-engineering|ansys|meshing| | <p>Remember that you are trying to approximate real solution using simple shape functions. Depending on the problem, you may need to use lots of elements to be able to describe the solution with sufficient accuracy.</p>
<p>It is a feature of FEA that if your model is correct, the value you are trying to calculate should converge to the exact solution when you are refining the mesh. Just try "mesh convergence". When you right click on a result in the tree, you can add "convergence" to it and ANSYS will try to refine the mesh automatically.</p>
<p>Regarding varying results before convergence in nonlinear analysis, ANSYS uses Newton method and especially in cases where the load step is big, the method may make a lot of wrong guesses all over the place. After a certain number of iterations, where solution has not been found, this is stopped, the load step is halved (bisection) and Newton method starts again.</p>
| 53000 | Mesh convergence and its affect on results |
2022-10-29T13:10:28.443 | <p>I have a system with a rot connected to a motor and I need to determine the center of mass to calculate the moment caused by gravitation.</p>
<p>To illustrate:</p>
<p><a href="https://i.stack.imgur.com/0C5rl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0C5rl.png" alt="enter image description here" /></a></p>
<p>I am uncertain about how to do this, is it a poor assumption to only consider the length along the rod? such that I get the length from the rod end (where the motor is attached) to the center of mass to be:</p>
<p><span class="math-container">$L_{cm} = \frac{L_{1,y} \cdot m_1 + L_{2,y} \cdot m_2 + L_{3,y} \cdot m_3}{m_1+m_2+m_3}$</span></p>
<p>Where <span class="math-container">$L_{i,y}$</span> is the length from the end to i'th mass along the rod (y direction).</p>
<p>Can I do this or do I need to consider the direct lengths? If this is the case how can I determine the gravitation Moment if the center of mass is not located at the system (meaning that if I use direct length the center of mass might be located beside the rod)?</p>
| |forces|centre-of-gravity| | <p>The moment experienced by the drives will be the sum of the three moments. Yes, you have to calculate to the centre of each mass, not the distance along the rod.</p>
<p><a href="https://i.stack.imgur.com/yDVVd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yDVVd.png" alt="enter image description here" /></a></p>
<p>Let <span class="math-container">$\alpha$</span> be the angle of each radius from the horizontal. The total moment can then be calculated by,</p>
<p><span class="math-container">$$ M = L_1m_1cos\alpha_1 + L_2m_2cos\alpha_2 + L_3m_3cos\alpha_3$$</span></p>
| 53004 | center of mass of rotating rod system |
2022-10-29T13:59:10.463 | <p>At the large hotel where I work in the engineering department the big Halloween event is the annual Pumpkin Derby...and we've lost every year to the culinary department. It's embarrasing!</p>
<p>This year, though, we caught them in the act. Since they're in charge of distributing the pumpkins, they kept a half dozen or so for themselves, tricked them out, and we caught them testing the racers to select the best one. We, on the other hand, are limited to a single pumpkin. So I'm soliciting advice for building a hot-rod pumpkin!</p>
<p>The rules are that you may use any kind of wheels, from ball bearing casters up to wheelchair wheels. The catch is that they must be attached only to the pumpkin. You can use axles which go through the body of the pumpkin, but you can't have any kind of a chassis or even a tie wire to align them. That's what killed us in the past; our pumpkin wouldn't travel in a straight line and dissipated its kinetic energy in a turn. The pumpkin is let go from the top of a ramp to roll across the lobby floor, and the prize goes to the pumpkin which travels the longest distance from the base of the ramp. The floor is low-pile carpet.</p>
<p>My preliminary idea this time is to use casters such as you might find in an office chair, three of them. They have 5/16" threaded stems, and we have some threaded anchors intended for concrete. My idea is to use three of them, drilling mounting holes vertically into the base of the pumpkin, and to use a pane of glass to level them into the same plane and then use RTV or similar to glue them in place.</p>
<p>One challenge I'm looking at is how to keep the pumpkin from spinning as it rolls down the ramp. I had thought that by using a triangle of casters and starting it down the ramp heaviest end first we might have the most luck. But I'm open to suggestions.</p>
<p>Help the engineers beat out the cooks!</p>
<p>Edit To Add: Thinking about it, possibly the biggest challenge is keeping the pumpkin from spinning when it leaves the ramp. While accelerating the net force is directed forward, while when it's on the carpet net force is to the rear. We plan on using swiveling casters; perhaps a touch of hot glue on the one in the lead to hold it straight?</p>
<p>Here are the casters we plan to use:</p>
<p><a href="https://i.stack.imgur.com/RqC6f.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RqC6f.jpg" alt="(Casters we plan to use)" /></a></p>
<p>We also have these wheels available (only two, but we could use a caster before or behind):</p>
<p><a href="https://i.stack.imgur.com/07ps0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/07ps0.jpg" alt="4.10/3.50 wheels" /></a></p>
<p>Here's the ramp. Pumpkin needs to fit between first two lines. Long wheelbase is a problem:</p>
<p><a href="https://i.stack.imgur.com/SmZnG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SmZnG.jpg" alt="Pumpkin Derby ramp" /></a></p>
<p>Preliminary components:</p>
<p><a href="https://i.stack.imgur.com/JITUs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JITUs.jpg" alt="Derby racer components" /></a></p>
| |mechanical-engineering|design|centre-of-gravity|rolling| | <p>Here's what we eventually came up with:</p>
<p><a href="https://i.stack.imgur.com/vEaD2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vEaD2.jpg" alt="Pumpkin Derby Racer" /></a></p>
<p>I know it looks off-kilter, but first priority was mounting the 'tailwheel'. Once it was in and the glue set we drilled it for the allthread axle, then mounted the disc wheels. A few test runs, then we used some hot glue to align the tailwheel straight.</p>
<p>Keeping it in the refrigerator till Monday. I feel somewhat confident. Thanks for the help!</p>
<p>Edit To Add: Almost race time. The stylists didn't consult with the engineers, so it'll have to go down the ramp backwards!</p>
<p><a href="https://i.stack.imgur.com/EiMDt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EiMDt.jpg" alt="Pumpkin Derby Racer decorated" /></a></p>
| 53005 | How to build a winning Pumpkin Derby racer? |
2022-10-30T17:01:13.153 | <p><a href="https://i.stack.imgur.com/xfc9e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xfc9e.png" alt="" /></a>
Consider a 1D <strong>adiabatic</strong> room (of volume <strong>V</strong>) with a wall conducting heat on its <strong>left side</strong>.
The wall is solid and heat conduction in it is driven by the common <strong>heat equation</strong>.
There is <strong>convection</strong> at the interface between the wall and the room.</p>
<ol>
<li><p>How to follow the room temperature over time using <em>finite difference method</em>?</p>
</li>
<li><p>Does it make sense to consider the temperature uniform in the room?</p>
</li>
<li><p>If so, how to use the convection coefficient as the difference <span class="math-container">$ T_{interface}(t) - T_{room}(t) $</span> would be <strong>zero</strong>?</p>
</li>
<li><p>Using <strong>Tomáš Létal</strong> (asnwer) method and discretising it, would it be:
<span class="math-container">$\frac{dT_r(t)}{dt}\cdot c_p = \left(T_w-T_r(t)\right)\cdot \alpha\cdot A \rightarrow T_r^{t+1} = T_r^{t} + \frac{dt\cdot\alpha\cdot A}{c_p}\cdot(T_N^t - T_r^t)$</span></p>
</li>
</ol>
| |heat-transfer|temperature|thermal-conduction|convection| | <p>You can simplify this, so you don't need finite difference method. If you start with room temperature <span class="math-container">$T_{r0}$</span>, constant wall temperature <span class="math-container">$T_w$</span> and constant convection coefficient <span class="math-container">$\alpha$</span>, the instant heat flux from room to wall <span class="math-container">$\dot{Q}_{r\rightarrow w}$</span> should be:</p>
<p><span class="math-container">$$\dot{Q}_{r\rightarrow w} = \left(T_r(t)-T_w\right)\cdot \alpha\cdot A$$</span></p>
<p>The room has to have heat capacity <span class="math-container">$c_p$</span>, which will make its temperature dependent on the transferred heat:</p>
<p><span class="math-container">$$\frac{dT_r}{dt}\cdot c_p = -\dot{Q}_{r\rightarrow w}$$</span></p>
<p>Combining these 2 equations:
<span class="math-container">$$\frac{dT_r(t)}{dt}\cdot c_p = \left(T_w-T_r(t)\right)\cdot \alpha\cdot A$$</span></p>
<p>This is simple differential equation, you can separate the variables and integrate:</p>
<p><span class="math-container">$$\int\limits_{T_{r0}}^{T_r(t)} \frac{1}{T_w-T_r(t)} dT_{room} = \frac{\alpha\cdot A}{c_p}\cdot \int\limits_{0}^t 1 dt$$</span></p>
<p>Resulting in:</p>
<p><span class="math-container">$$\ln\left(\frac{T_w-T_r(t)}{T_w-T_{r0}}\right) = -\frac{\alpha\cdot A}{c_p}\cdot \left(t-t_0\right)$$</span></p>
<p>Function for temperature:</p>
<p><span class="math-container">$$T_r(t) = T_w+\left(T_{r0}-T_w\right)\cdot \exp\left(-\frac{\alpha\cdot A}{c_p}\cdot t\right)$$</span></p>
| 53013 | Modeling 1D Transient Heat Transfer: Wall/Room interface |
2022-10-31T19:20:41.103 | <p>To calculate the torque caused by the buoyancy force I need the length and the angle but what angle should I use? Should I use the angle of from the center at the bottom of the angle from the center of mass?</p>
<p><a href="https://i.stack.imgur.com/lQVpy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lQVpy.png" alt="enter image description here" /></a></p>
| |fluid-mechanics|torque|forces| | <p>you should use angle 2. The buoyancy in this case is due to pressure from 3 sides. the 2 lateral sides undergo triangular pressure distribution that opposes each other and cancels.</p>
<p>Therefore the only effective force is the buoyancy pressure times the surface of the bottom, which is the applied force and is imparted at the center of the bottom of your submerged object.</p>
| 53022 | Angle of attack for torque calculation from buoyancy force |
2022-10-31T21:26:24.670 | <p>Is there a way to solve truss member forces of a complex truss like fink roof truss where the {Edit: external} forces and reactions are known, the angles of all the members and the total base length?</p>
<p>I have a text book problem in trying to solve. My difficulty is in how to resolve member forces, albeit with standard mathematical calculations. So this is why I'm wondering whether there a graphical method to estimate the individual member forces using just the mentioned givens and reactions?</p>
<p>Where the heck does the "ij" and "hi" come from? There is no labelled i or j joint.</p>
<p>Not a homework question but an educational experience.
<a href="https://i.stack.imgur.com/8moqI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8moqI.jpg" alt="original problem in book" /></a>
<a href="https://i.stack.imgur.com/31IZi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/31IZi.jpg" alt="given answer" /></a></p>
| |structural-engineering|structural-analysis| | <p>So in order to figure this out graphically I first needed to figure out the reactions. This can be done either by taking the sum of the forces about A then Figuring out the Forces at E. Alternatively, since this is symmetrical a quick short cut is simply sum the vertical applied loads (30 kN), and dividing by the number of reactions (2) to get the reaction A both A and E since they will be equal due to the symmetry. Rya = Rye = 15 kN.</p>
<p>The next thing to consider with a truss is that there are only axial forces. So this means the geometry of the truss will control the direction of the force vector in each member. In this example you will note that the same coloured members are will carry the same vector magnitude because of the symmetrical loading and geometry.</p>
<p><a href="https://i.stack.imgur.com/t5xtO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t5xtO.png" alt="enter image description here" /></a></p>
<p>The next step is to draw you vectors to a an arbitrary scale. Whatever the scale you pick you will need to multiply the results by it to get your final force. In my example I used 1 kN as the same 1 m I used in the drawing the truss. Because I was in cad and had unlimited space this was not an issue for me. If you were doing it on paper you may need to use a smaller scale or get larger paper.</p>
<p><a href="https://i.stack.imgur.com/E3baY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E3baY.png" alt="enter image description here" /></a></p>
<p>The next I did was start at the reaction A. I did this as it had the fewest members at the joint. Since the sum of the forces at a node have to sum to zero we can use the sum of vectors to solve the unknown magnitude of the vector GA and BA. I started by drawing the reaction vector RyA to scale. You can then proceed clockwise or counterclockwise around the node and draw in the direction of unknown vectors.</p>
<p><a href="https://i.stack.imgur.com/uDh1c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uDh1c.png" alt="enter image description here" /></a></p>
<p>Note that each unknown vector is placed at either end of the known vector. Since we know that the vectors have to sum to zero, we know that they unknown vectors will need to intersect. Based on the direction of the RyA vector, the direction of the unknown vectors is determined.</p>
<p><a href="https://i.stack.imgur.com/ouHkC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ouHkC.png" alt="enter image description here" /></a></p>
<p>This process is repeated at node B. It is important to note that the direction of the vector at one member member end will be opposite at the other. For node B, The 2 of the 4 vectors are known. Draw in the FyB vector to scale and draw the same vector FAB that were determined at the previous at the same scale. You will know FCB will be attached to Fyb, and you know that FGB will be attached to FAB. Extend FCB and FGB until they intersect. When you have done this successfully you will wind up with something that looks like the following:</p>
<p><a href="https://i.stack.imgur.com/iNN8R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iNN8R.png" alt="enter image description here" /></a></p>
<p>Repeat this for node C and G.</p>
<p><a href="https://i.stack.imgur.com/7Xr5a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Xr5a.png" alt="enter image description here" /></a></p>
<p>Note that in the next image vectors GF and GA are both horizontal and on top of one another.</p>
<p><a href="https://i.stack.imgur.com/9qRHt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9qRHt.png" alt="enter image description here" /></a></p>
<p>To get the magnitudes, measure the length of your vectors and multiply them by the scale you chose to use. Since this is a symmetrical system you can apply the magnitude to the corresponding members.</p>
<p>In my case I got the following results:</p>
<ul>
<li>AB=ED=27.0416 kN</li>
<li>AG=EF=22.5 kN</li>
<li>BC=DC=21.4946 kN</li>
<li>GB=FD=8.3205 kN</li>
<li>GC=FC=7.5 kN</li>
<li>GH=15 kN</li>
</ul>
<p>Vectors that point towards the node are in compression and the vectors that point away from the nodes are tension.</p>
<p>Alternatively normally loaded trusses like this have the top chord in compression and the bottom chord in tension. You can derive a convention from this to determine the interior member forces.</p>
<p>These values were also mathematically verified with a simple 2D online truss solver.</p>
| 53025 | Solving truss member forces graphically |
2022-11-01T10:21:24.460 | <p>Pushing a cylinder through the water with an angular velocity of <span class="math-container">$\omega$</span>.</p>
<p><a href="https://i.stack.imgur.com/9eI8a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9eI8a.png" alt="enter image description here" /></a></p>
<p>The cylinder diameter and length are <span class="math-container">$D$</span> and <span class="math-container">$L$</span> respectively and the length from the rotation point to the center of the cylinder is <span class="math-container">$L_{arm}$</span>.</p>
<p><span class="math-container">$F_D = \frac{1}{2} \cdot \rho \cdot C_D \cdot A \cdot v^2 $</span></p>
<p>But how do I find the drag coefficient?</p>
| |drag| | <p>There are tables for common shapes, e.g. in my fluid mechanics formulary I find for a long cylinder <span class="math-container">$C_\text{D}=0.3 \dots 1.3$</span> depending on the Re-Number.</p>
| 53035 | Drag force and coefficient for cylinder through water |
2022-11-05T04:32:51.100 | <p>Compressed air energy storage is looked down upon because of the inefficiencies it brings along. Every article I've found till today, makes use of the stored energy by converting it into mechanical torque of a shaft, and more often than not, converting the mechanical to electrical energy.</p>
<p>Is there a reason why, on a domestic scale, we cannot use the cooling of the air as it expands for air conditioning, along with whatever energy it provides of the mechanical kind? Please provide some sample calculations too if you can.</p>
<p>My own thinking would be to adiabatically pressurize a cylinder buried in a bunch of sand, which would absorb whatever low grade heat is given off, and then to decompress through a turbine to turn a shaft, while the exhaust cool air can be used for air conditioning?</p>
<p>What aspect will be uneconomical or unrealistic? Space required, efficiency, safety standards, what?</p>
<p>EDIT: Leakages from valves when decompressing are always counted as an inefficiency. If we simply pass the leakages along with exhaust cool air from the components to come (turbine etc.) through a heat exchanger, that can be used for air conditioning! Why is it a loss?</p>
<p>Similarly, heat loss when compressing can be captured by the aforementioned sand (or water etc) which will be preheated for heating applications elsewhere. I understand the compressed air will never give us back that energy when decompressing, but the capturing media will!</p>
| |mechanical-engineering|structural-engineering|fluid-mechanics|thermodynamics|pressure| | <p>You could certainly make a compressed air storage system as you describe, using the waste air for cooling. One big source of inefficiency however would be that you need some kind of mechanical work to compress the air in the first place. If the main goal is to generate mechanical work, why go through all the extra steps of compressing and storing the air? Sometimes it makes sense to do this, like to run air tools such as an impact wrench or air grinder, but that's only because an air turbine is preferable to an electric motor for those applications.</p>
<p>As for refrigeration, you could technically use air as a working fluid by just compressing and expanding it to move heat around, but it wouldn't be very efficient without undergoing a phase change between liquid and gas. Refrigerants are chosen because they change phase at conditions that are close to those of the working environment. Air (gaseous nitrogen) will liquify only at relatively cold temperatures, at least below -150°C. So, it wouldn't work for typical domestic applications.</p>
<p>As an example, say we have a tank of compressed air at 300 Kelvin being used to drive a turbine. The inlet pressure to the turbine is regulated to be a constant 500kPa, and the air expands to 200 kPa in the process. Assuming no heat transfer between the surroundings and the expanding air (probably not a very good assumption), the exiting air temperature can be calculated using the isentropic equation of state:</p>
<p><span class="math-container">$$
T_{exit} = T_{inlet}\left( \frac{P_{exit}}{P_{inlet}} \right) ^ \frac{\gamma-1}{\gamma} = 300K\cdot\left( \frac{200 kPa}{500 kPa} \right) ^ \frac{1.4-1}{1.4} = 231 K
$$</span></p>
<p>The work done per unit mass of air would be equal to the change in enthalpy of the air, times the turbine efficiency (assume ~50%):</p>
<p><span class="math-container">$$
w=-\eta \Delta h=-\eta c_p \Delta T=-\eta c_p \left( T_{exit} - T_{inlet}\right) = -0.5\cdot 1000 \frac{J}{kg\cdot K} \cdot \left( 231K - 300K\right) = 35.5\frac{kJ}{kg}
$$</span></p>
<p>Obviously I've made some simplifying assumptions here, but this should be enough to get you started with the basic concepts.</p>
| 53073 | Compressed air efficiency if expansion cooling is used |
2022-11-06T16:06:05.037 | <p>I have a CVT transmission, which consists of a <strong>metal belt</strong> and four <strong>circular toroid discs</strong>, that form <strong>two pulleys</strong>. I need to be able to move the two toroids <strong>further and closer together</strong>. This movement has to be <strong>separate</strong> for the two pulleys, so when the drive toroids (connected to the engine) get further apart, decreasing the diameter, the driven toroids (connected to the driveshaft) have to get closer together as to <strong>tense the belt</strong>. This all has to be controlled by a raspberry pi. I thought about maybe using a hydraulic piston, but I'm not very familiar with hydraulics.</p>
<p>How should I design the system to move the toroids closer and further apart?</p>
| |hydraulics|transmission| | <p>Several methods come to mind:</p>
<ol>
<li><p>mechanical : perhaps threads - would need to be fine and you need to think about the disadvantages, but stepper motors may give suitable control</p>
</li>
<li><p>pneumatic pistons, however the inherent "flexibility" due to air compressibility may not give sufficiently fine control</p>
</li>
<li><p>hydraulics: used in many control systems as liquids are only slightly compressible, and at the pressures you will use they are assumed incompressible. Hydraulics are used from small agricultural implements up to controlling the legs of drilling platforms in the North Sea - and they can do that to a tolerance of better than 10mm for a leg that supports many tons.</p>
</li>
</ol>
| 53085 | What kind of system should I use to move the toroids of my CVT transmission? |
2022-11-07T09:56:29.280 | <p>I am currently studying How excavators work? and came to know that "In excavators diesel engine is used to drive the hydraulic pump. The high pressure fluid from hydraulic pump is used to run the hydraulic motors which rotates the wheel." Is there any specific reason for using hydraulic motors instead of directly using diesel engine?</p>
| |mechanical-engineering|automotive-engineering|hydraulics| | <p>Having spent over 30 years working in the world of heavy equipement, marine industry, farm industry, construction industry, logging industry etc., my experience has been that electric motors and systems always seem to have a have a way of succumbing to environmental corrosion over time faster than other types of systems, no matter what the engineers or theories might suggest.
Sitting in a nice office someplace far away from actual construction world doesn't always cut it. There's a Canadian company (Letourne) that made log stackers using electric drives in each motor...they were a pain to work on and we always wished they were hydraulically driven rather than electric, as it seemed to us that the electric motors were more failure prone and you couldn't easy get repair parts for them or fix them easily in the field. If a hydraulic drive broke down, it was usually as simple as replacing a worn out hydraulic hose, which was cheap and easy and we could fix ourselves on site. Simplicy and realiability seemed to be the benefit of hydraulics, even if they are not as "efficient" as other drive systems, as no one cared about the cost of "efficiency" when two men and a manchine were costing the customer or the company hundreds of dollars per hour if they weren't working... Thus the extra cost of fuel was a mute point. Even the electric drive Letourne had to have a big diesel motor to generate the electricity for the electrical system in the wheels...so I'm not sure they were all that efficient, and thus like the Wankle rotary motor, it just never caught on (for good reason)... Theories and engineering only gets you so far... My best advice is to get out and get some on-site/real world work experience on Tug Boats, farming, logging or construction.</p>
| 53089 | Why excavators use Hydraulic motors for rolling or moving forward instead of diesel engine? |
2022-11-09T09:23:35.360 | <p>So from what I could understand from my Electrical Engineering lectures the electric potential at a particular point charge <span class="math-container">$A$</span> is:</p>
<p><span class="math-container">$u_A=-\int_{\infty}^{r_A}\overrightarrow{E}\overrightarrow{dr}$</span></p>
<p>The minus here might be a mistake on my side because I think since they're still written in vector form the minus should not be there. Not sure though.</p>
<p>But how is this formula derived? Why is it particularly the integral of <span class="math-container">$\overrightarrow{E}\overrightarrow{dr}$</span>? Btw I'm a beginner in this field so please go easy with the electrical engineering jargon.</p>
| |electrical-engineering| | <p>Formula for (small amount of) work done (<span class="math-container">$dW$</span>) on a body to move it over a (small) distance <span class="math-container">$\vec{dr}$</span> using force <span class="math-container">$\vec{F}$</span> is given by the <em>dot product</em>
<span class="math-container">$$dW = \vec{F}\cdot\vec{dr}$$</span></p>
<p>In electrostatics, the force acting on a charged particle is proportional to its charge <span class="math-container">$q$</span> and the electric field <span class="math-container">$\vec{E}_{(\vec{r})}$</span> at <em>that</em> point <span class="math-container">$\vec{r}$</span> where the charge is situated. So work done <em>by the electric field</em> is
<span class="math-container">$$dW = q\ \vec{E}_{(\vec{r})}\cdot\vec{dr}$$</span></p>
<p>If we want to move a charge against the force from the electric field (very slowly), we need to apply an equal and opposite force on the charge (we don't have to apply a force greater than the electric field since we don't intend to accelerate the charged particle. Remember, this is only a thought process used for understanding the formula. More details at <a href="https://en.wikipedia.org/wiki/Electric_potential" rel="nofollow noreferrer">Wikipedia</a>). Then, the work <em>done against the electric field</em> is
<span class="math-container">$$dW = (-1) q\ \vec{E}_{(\vec{r})}\cdot\vec{dr}$$</span></p>
<p>With suitable assumptions, the work that would have been done on the particle to bring it from an infinite distance away to the location <span class="math-container">$r_A$</span> <em>against the force from the electric field</em> is then (by "<em>summing</em>" up the tiny work done over tiny distances from infinity to <span class="math-container">$r_A$</span>)</p>
<p><span class="math-container">$$W = q\ \int_{\infty}^{\vec{r}_A} (-1)\vec{E}_{(\vec{r})}\cdot\vec{dr}$$</span></p>
<p>Potential is now <em>defined as</em> the work done <em>per unit charge</em>.
From <a href="https://en.wikipedia.org/wiki/Electric_potential" rel="nofollow noreferrer">Wikipedia</a></p>
<blockquote>
<p>The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. ... By definition, the electric potential at the reference point is zero units. Typically, the reference point is earth <strong>or a point at infinity</strong>, although any point can be used.</p>
</blockquote>
<p>So potential becomes</p>
<p><span class="math-container">$$u_A = - \int_{\infty}^{\vec{r}_A} \vec{E}_{(\vec{r})}\cdot\vec{dr}$$</span></p>
| 53113 | Why is the electric potential formula like this and how is it derived? |
2022-11-09T11:51:34.353 | <p>So, during my last class the teacher asked if we could go from:</p>
<p>MO = e(-π<em>ξ / √1-ξ²) to ξ= (-ln</em>(MO)) / (√π²+ln²(MO))</p>
<p>MO = Max overshoot
ξ = zeta
e = exp</p>
<p>Does anyone understand what he meant by that? Did I just misunderstand?
Tried a few times but I just can't make sense of how to go from the MO equation to the zeta one.</p>
| |electrical-engineering| | <p>It would be good to see what you tried and I'm sure your teacher would too, but does this make it any clearer?</p>
<p><span class="math-container">$$
MO = e^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}}
$$</span>
<span class="math-container">$$
ln(MO) = -\pi\frac{\zeta}{\sqrt{1-\zeta^2}}
$$</span></p>
<p><span class="math-container">$$
ln(MO)^2 = (ln(MO)^2 +\pi^2)\zeta^2
$$</span></p>
<p><span class="math-container">$$
\zeta^2 = \frac{ln(MO)^2}{ln(MO)^2 +\pi^2}
$$</span></p>
| 53117 | Control theory - overshoot max |
2022-11-10T00:07:11.050 | <p>I want to build a DIY case with the connector shown in the pictures below.</p>
<p><a href="https://i.stack.imgur.com/7ZZkb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7ZZkb.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/iN4Hc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iN4Hc.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/NhJ7d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NhJ7d.png" alt="enter image description here" /></a></p>
<p>As you can see, it consists of a circular plastic that extrudes from the main enclosure. Inside there is something that can be described as "cut_circle", that again extrudes from the main enclosure.</p>
<p>Inside the "cut_cirle" are two holes where power connectors are inserted.</p>
<p>I am not sure if this could be achieved with injection moulding - since i am complete new to the topic.</p>
<p>Even if the two holes inside the "cut_cirle" cannot be made with injection moulding (i am not stating, i do not know, i am also asking), then i guess they could be cut with a
drill and connectors could be installed inside (with a press or something).</p>
<p>So, my question is, is this design doable with injection moulding?</p>
| |plastic|injection|injection-molding| | <p>Yes it is, in fact the part in your picture was almost certainly made this way, as follows.</p>
<p>An injection mold can be made to accept those metal pins, which are positioned in the mold so that when the plastic is shot in, it envelops the pins with plastic which then become an integral component of the finished part. The part is then stripped out of the mold, pins and all, and another set of pins are loaded into the mold and the process is repeated. These pins are called <em>molded-in</em> parts.</p>
<p>Another way to do this is to mold deep holes called <em>pass-thrus</em> into the part. The inside surface of the pass-thru is formed in such a way that the pin can be pressed into the hole with a snap engagement to hold the pin tightly in place.</p>
<p>In fact, it is possible to have the pins soldered onto a piece of connecting wire and the whole assembly then loaded into the mold, so that the injected plastic flows around and encloses both the pins and a length of the wire to serve as a <em>strain relief</em> that helps prevent the wire and pins from being pulled free of the plastic plug.</p>
| 53124 | Is this shape possible with injection moulding? |
2022-11-11T19:05:10.503 | <p>I have a half cylinder which is significantly thick but not very long. In the half cylinder, a force <span class="math-container">$F_{app}$</span> is being applied over a certain area as shown in the figure below:</p>
<p><a href="https://i.stack.imgur.com/Wk44M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wk44M.jpg" alt="enter image description here" /></a></p>
<p>I have to find the hoop stress in the free edges.</p>
<p>I understand how the hoop stress is calculated over cylindrical vessels when there is internal pressure all over the cylindrical surface. Can anyone provide any insight how to start when there is force acting over a partial area. And finally how will the results change when the thickness is significantly large.</p>
| |applied-mechanics|statics|solid-mechanics|forces| | <p><a href="https://i.stack.imgur.com/Cutar.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Cutar.png" alt="enter image description here" /></a></p>
<p>The sketch above forms the base for solving reactions at "a" and "b". Note that if the loads are acting on the near edge, then there is a twisting need to be considered.</p>
<p>Without twisting, find the reactions that contribute to normal stress at "a" and "b":</p>
<p><span class="math-container">$\sum Ma = 0$</span></p>
<p><span class="math-container">$Rb = \dfrac{Fx*x + Fy*y}{D}$</span>, pointing down. "D" is the diameter measured from center to center.</p>
<p><span class="math-container">$\sum Fy = 0$</span></p>
<p><span class="math-container">$Fa = Fy - Fb$</span>, assume "positive force" pointing up</p>
| 53143 | Finding hoop stress in a half-cylinder |
2022-11-13T00:07:49.360 | <p>Have already asked this on physics-site but has been <a href="https://physics.stackexchange.com/q/736454/31729">closed</a>.
As such trying here!</p>
<p>This question is basically a thought experiment.</p>
<p>Let's say there are two pulley's <b>P<sub>1</sub></b> and <b>P<sub>2</sub></b> having equal radius <strong>R</strong> and negligible mass <b>M<sub>P</sub></b>, arranged side-by-side (parallel axis of rotation).</p>
<p>They are connected by a gears <b>G<sub>1</sub></b> and <b>G<sub>2</sub></b> with an unknown ratio. The presence of gears effectively results in pulley <b>P<sub>1</sub></b> having a higher RPM than <b>P<sub>2</sub></b>.</p>
<p>If the mass <b>M<sub>2</sub></b> is 20 times the mass of <b>M<sub>1</sub></b>, then what is the ratio of the gears that would result in the system being in equilibrium?</p>
<p>If additional values would help, then we can have the pulley radius <strong>R</strong> to be 5m and Mass <b>M<sub>1</sub></b> to be 5kg (and therefore we have <b>M<sub>2</sub></b> to be 100kg).</p>
<p>To elaborate further, I have tried to map a system involving rotation to a system that is more linear in nature. Specifically in a rotational system, I would have to consider moment-of-inertia. However the equivalent of moment-of-inertia is not as clear in a scenario involving linear motion.</p>
<p>Below is an illustration with pulley's (yellow), gears (orange) and the masses(green) for clarity.</p>
<p><a href="https://i.stack.imgur.com/xewQF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xewQF.png" alt="pulley system under equilibrium" /></a></p>
| |mechanical-engineering|experimental-physics|linear-motion| | <p>Let's assume the size of the two pulleys is the same ( if not then we have to multiply the answer by their ratio) and call the radius of <span class="math-container">$G_1, R_1, and \ G_2, R_2.$</span></p>
<p>Then for the system to be in equilibrium we have:</p>
<p><span class="math-container">$$M_2\cdot R_2 = M_1\cdot R_1 \rightarrow 20M_1*R_2 =M_1*R_1 $$</span></p>
<p><span class="math-container">$$R_1=20R_2$$</span></p>
| 53153 | Determining the balancing force for a system of unequal masses for given scenario (angular-momentum, moment-of-inertia, thought-experiment) |
2022-11-13T23:02:42.520 | <p>Reposting from <a href="https://physics.stackexchange.com/questions/736554/how-to-calculate-voltage-change-across-electrodes-in-tank-of-water">https://physics.stackexchange.com/questions/736554/how-to-calculate-voltage-change-across-electrodes-in-tank-of-water</a> because apparently this is an engineering question and not an experimental physics question.</p>
<p>Apologies for the basic question. I am a neuroscience student and not that familiar with physics.</p>
<p>I am delivering a 1 V bipolar sinusoidal stimulus at 842Hz to a tank of deionized water at 26C with two electrodes for one minute. I am also recording from two electrodes with a sampling rate of 20kHz inside that tank approximately 0.3 m away. I would like to calculate the voltage recorded by the recording electrodes that is due to the sinusoidal stimulus as a function of time. However, I am not sure how to go about doing this. I would greatly appreciate any help in calculating this.</p>
<p>My purpose for this is to separate my electrical stimulus from the electric organ discharge produced by a black ghost knifefish in the tank. My recording electrodes pick up both the discharge produced by the fish and the output of my stimulating electrodes. I have chosen the frequency of my electrical stimulus to be within 1-2Hz of the frequency produced by the fish, in hopes of observing a jamming avoidance response by the fish (i.e. the fish shifting its discharge frequency away from the stimulus frequency). This makes it hard to distinguish my stimulus from the fish's discharge in the spectrum view of my data (i.e. after Fourier transforming). Therefore it would be very helpful if I could calculate the voltage recorded by the recording electrodes in response to the stimulus voltage, given the stimulus voltage as a function of time. Then I would be able to subtract out this voltage from my data to get the discharge from the fish. I have recordings from when there is no stimulus, and by merely eyeballing my data I can see that the voltage change in the recording electrodes in response to the stimulus is somewhere around 3-4% of the stimulus voltage. But it would be great if there were a more methodical way to do this.</p>
<p>If more information is needed, let me know and I can try to provide it.</p>
| |electrical-engineering|electromagnetism| | <p>My suggestion is to first measure the response of the tank without the fish. In this way you can determine the transfer function <span class="math-container">$G(\omega)$</span> of the tank from the electrodes (input) <span class="math-container">$U(\omega)$</span> to your sensors (output) <span class="math-container">$Y(\omega)$</span>:
<span class="math-container">$$
G(\omega) = \frac{Y(\omega)}{U(\omega)}.
$$</span></p>
<p>If you then measure the response of the tank with the fish, you can substract the (expected) response from the tank and you get the response of the fish <span class="math-container">$Y_\mathrm{fish} = Y(\omega)-G(\omega)U(\omega)$</span>.</p>
<p>However, in practice your measurements will be polluted with a transient response and noise. Moreover, if you do not use the correct number of samples for the Fourier transform you will also introduce spectral leakage.</p>
<p>To avoid spectral leakage, choose a window length that fits an integer number of periods of your stimulus frequency (this is important as being of by one sample will already introduce spectral leakage).
To reduce the effect of the transient, start your window after the transient term has diminished.</p>
<p>I hope this will help designing/performing your experiment.</p>
| 53162 | How to calculate voltage change across electrodes in tank of water? |
2022-11-14T15:16:08.003 | <p>I took apart this hose faucet timer:</p>
<p><a href="https://i.stack.imgur.com/K67L1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K67L1.png" alt="1 Outlet Hose Faucet Timer from Orbit" /></a></p>
<p>When unpowered the valve is closed. When powered (either manually or by timer) it opens allowing water to flow through. I am attempting to learn how the open/close valve mechanism works. From some research I believe it is called a solenoid valve, but the youtube videos I've watched on the matter, don't seem to cover this particular design. Pictured is the valve assembly, minus the circuit board.</p>
<p><a href="https://i.stack.imgur.com/SUpxw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SUpxw.jpg" alt="Valve Assembly" /></a></p>
<p>From some research I believe the silver device on the right is a solenoid. I could tell this because when it was unpowered, a piston sat inside, and when it received power, the piston/rod would pop out (I lost the actual piston before taking pictures). The solenoid screws into a hard plastic harness with a single hole in it, which I dont believe the piston even touches or covers. On the other side of that plastic housing is what I believe is referred to as a diaphragm which depresses.
<a href="https://i.stack.imgur.com/5izzD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5izzD.jpg" alt="Harness for solenoid" /></a>
<a href="https://i.stack.imgur.com/ZKrQR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZKrQR.jpg" alt="Other side of harness plus diaphragm" /></a>
<a href="https://i.stack.imgur.com/fWZH0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fWZH0.jpg" alt="Diaphragm" /></a>
<a href="https://i.stack.imgur.com/Sqb55.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Sqb55.jpg" alt="Diaphragm back" /></a></p>
<p>The water flows from the top inlet, through a larger outer ring (top arrow in the picture below). My presumption is that when the diaphragm is pushed back the water can then continue to flow through the middle ring (bottom arrow) and out of the bottom.</p>
<p><a href="https://i.stack.imgur.com/by9do.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/by9do.jpg" alt="Inlet and outlet" /></a></p>
<p>What I am trying to understand is two fold:</p>
<ol>
<li>Why isn't this valve normally open? If the water from the inlet is directly pushing against the diaphragm, wouldn't that pressure be enough to push the diaphragm back and allow the water to flow through?</li>
<li>How does the solenoid "pull" the diaphragm back? Wouldn't the solenoid pushing out, cause positive pressure, directly pushing the diaphragm away from the solenoid, further sealing the valve?</li>
</ol>
<p>I put together <a href="https://photos.app.goo.gl/d2vuyFuLkx5zgTMA8" rel="nofollow noreferrer">this video showing a breakdown of the valve</a>.</p>
<p>p.s. Incase its not clear, I am a laymen when it comes to mechanical/electrical engineering, so please dumb down your explanation.</p>
| |valves|diagram|water-pressure| | <p>A <a href="https://www.youtube.com/watch?v=5rokXHhxRSw" rel="nofollow noreferrer">fairly clear video</a> explaining the operation of the valve is provided by Rain Bird Irrigation™. It shows how the inlet water is providing pressure against the diaphragm, sealing the exit opening. That inlet water also travels to the solenoid valve, which when closed, keeps the pressure on the diaphragm.</p>
<p>The diaphragm has a greater surface area on the inlet side than on the exit side. When the solenoid is activated, some of the inlet pressure is released, allowing the diaphragm to open.</p>
<p>The solenoid force is minimal and is only that which is required to seal the transfer opening.</p>
<p>This type of valve is common in sprinkler systems, which use low voltages for safety reasons. I have seen modifications of this valve used in massive air dump systems such as those used for launching projectiles (pumpkins) as it provides for great volumes of air to be released in a short time.</p>
| 53166 | ELI5: How does this water (solenoid) valve work? |
2022-11-14T19:01:48.093 | <p>I have been researching about the factors that affect natural frequencies, particularly about the effects of stresses.</p>
<p>Most of what I have found only discusses compressive and tensile stresses though, that they decrease and increase the natural frequency respectively</p>
<p>However, I cannot find anything about how bending/flexural stress affects the natural frequency. If I had to assume though, it probably does not affect it, as bending is just a combination of compression and tension occurring at the two sides, and they just cancel out.</p>
<p>Is my guess correct? And if not, can you link additional readings about it. Thank you!</p>
| |stresses|vibration|bending|flexures| | <p>The oscillation comes about as the interaction between a position dependent 'restoring force' and an inertia. I put force between brackets as it is a force in a mass-spring system, but can also a torque.</p>
<p>This interaction can be described by the differential equation
<span class="math-container">$m*a = -k*x$</span> so <span class="math-container">$m*x"+k*x = 0$</span>
For a linear spring, <span class="math-container">$x = \sin(\sqrt(k/m)*t)$</span> is a solution, which is harmonic and with frequency determined by the inertia term 'm' and the spring stiffness term 'k'.</p>
<p>In your bending beam, the 'restoring force' is actually a torque that is the result of the combination of compression on one side of the neutral line and tension on the other side.</p>
<p>Its hard to imagine how to add bending stress without altering either the shape or otherwise the physics of the beam. For strings (under tension) or columns (under compression) this is easily done.</p>
<p>One note (pun somewhat intended): the frequency can change with changing amplitude of oscillation, when the stiffness is not constant. Many materials exhibit spring stiffening. They are thus non-linear springs. This means that the restoring force depends on amplitude (larger deflection, larger-than-proportionally-larger restoring force. Also, the response will not be a simple harmonic anymore.</p>
| 53170 | Does Bending Stress Affect Natural Frequency? |
2022-11-14T19:17:36.110 | <p>Somebody is insisting on this design to me, and while I cant put my finger on what exactly makes this impossible, I know it is. I need your help.</p>
<p>I believe it's to do with the pressure of the headspace, and losses in friction and heat causing motion to cease. But how can it cease when the pressure of the water at its deepest is so much higher than the headspace? It just messes me up.</p>
<p><a href="https://i.stack.imgur.com/0Lrfh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0Lrfh.png" alt="enter image description here" /></a></p>
<p>This is a closed (yes, closed, no air/vapor in/out) system, with a shaft connected to the turbine inside sticking out of it. Of course, it exchanges heat with the ambient, so keeping that in mind, what will make this stop, or will it work at all?</p>
| |fluid-mechanics|thermodynamics|pressure| | <p>A siphon only works when the exit of the siphon is below the level of the supplying water. This thing wouldn't do anything except sit there. Figure out a way to make water run uphill and we'll really have something.</p>
| 53171 | Debunk closed Boyle flask for perpetual energy |
2022-11-15T16:50:24.083 | <p>I read that sodium sulphate crystals are used to make sponges. The crystals are mixed in with some other stuff, then put into a mold. The mold is heated, causing the sodium sulphate crystals to melt and leak out through holes in the bottom, leaving the pores.</p>
<p>Is there a reason sodium sulphate specifically is used and not some other substance?
Thanks</p>
| |chemical-engineering|manufacturing-engineering|product-engineering| | <p>Other chemicals may be used but sodium hydroxide is preferred since it is an effective swelling agent.</p>
<p>Source: <a href="http://www.madehow.com/Volume-5/Sponge.html" rel="nofollow noreferrer">http://www.madehow.com/Volume-5/Sponge.html</a></p>
<blockquote>
<ul>
<li>"As it cooks, the sodium sulphate crystals melt, and drain away through openings in the bottom of the mold. It is their melting that leaves the characteristic pores in the finished sponge. The size of the pores is determined by the size of the sodium sulphate crystals. A rough sponge used for washing a car, for instance, is made with coarse crystals, while a fine sponge of the type used for applying makeup is made with very fine crystals. As the celluolose mix cooks, then cools, it becomes a hard, porous block.".</li>
</ul>
</blockquote>
<p>Source: <a href="https://patents.google.com/patent/US3476844A/en" rel="nofollow noreferrer">https://patents.google.com/patent/US3476844A/en</a></p>
<blockquote>
<p>"... <strong>sodium sulphate decahydrate, which is very soluble in water and easily fusible</strong></p>
</blockquote>
<blockquote>
<p>"The porosity of the sponges obtained depends on the size of the hydrated sodium sulphate crystals used as pore producing material. Large crystals are obtained by slow crystallization of solutions of sodium sulphate and are very often composed of an agglomeration of relatively small crystals, thus producing cavities and irregularities which make the resultant mass rather fragile.".</p>
</blockquote>
<blockquote>
<p>"... Other dimensions and shapes could be used for the agglomerates, for example cubic crystals could be produced, although the tetrahedron gives better orientation effects, In place of hydrated sodium sulphate, any other fusible or easily soluble product can be used which is capable of producing under pressure, masses which are sufficiently solid to survive without breaking the various processes in the preparation of the paste used for the manufacture of regenerated cellulose sponges.".</p>
</blockquote>
<p>Source: <a href="https://patents.google.com/patent/US3131076A/en" rel="nofollow noreferrer">https://patents.google.com/patent/US3131076A/en</a></p>
<blockquote>
<p>"The pore forming agent is conventionally sodium sulfate. Glaubers salt (Na<sub>2</sub>SO<sub>4</sub>.10H<sub>2</sub>O) is normally used, though the anhydrous salt may also be used if desired. There is, however, no advantage in using the anhydrous material. <strong>In place of sodium sulfate, other fusible, water soluble salts that do not readily regenerate viscose may be used</strong>, as may gas forming substances such as aluminum powder.<br />
The cellulose swelling agent may be any of a large variety or materials including the <strong>alkali metal hydroxides, the alkali metal salts of weak acids, e.g. sodium zincate, salts of weak bases and strong acids such as zinc chloride and zinc nitrate and the like. Sodium hydroxide is, however, much preferred</strong> since it is a very effective swelling agent and introduces no new elements into the sponge making mixture.".</p>
</blockquote>
| 53182 | Why Sodium Sulphate is used in Sponge making |
2022-11-16T01:46:21.777 | <p>I'm looking for guidance to define how hot is too hot when related to human safety. I have heard of a general limit from OSAH of 60 °C (140 °F) being ok for contact up to 5 sec. Are there better sources for defining this limit for different temperatures and time limits?</p>
| |temperature|safety| | <p>Since this is a safety issue, you'll need to carefully follow relevant standards such as:</p>
<ul>
<li><a href="https://www.astm.org/c1055-20.html" rel="nofollow noreferrer">ASTM C1055-20</a>: Standard Guide for Heated System Surface Conditions that Produce Contact Burn Injuries.</li>
<li><a href="https://www.iso.org/standard/43558.html" rel="nofollow noreferrer">ISO 13732-1:2006</a> Ergonomics of the thermal environment — Methods for the assessment of human responses to contact with surfaces — Part 1: Hot surfaces</li>
</ul>
<p>These can be traced back to a seminal series of papers in 1947 on <a href="https://scholar.google.ca/scholar?q=%22Studies+of+Thermal+Injury%22+%22Moritz%22+%22american+journal+of+pathology%22&hl=en&as_sdt=0%2C5&as_ylo=1947&as_yhi=1947" rel="nofollow noreferrer">"Studies of Thermal Injury"</a> by Moritz and Henriques. <a href="https://pmj.bmj.com/content/39/458/717" rel="nofollow noreferrer">Figure 1 in an 1963 article</a> by J.P. Bull in the British Medical Journal summarizes their temperature-time threshold curve for injury and you can see where the "60°C for 5 seconds" threshold criterion for partial thickness burns comes from. According to the figure, the damage takes only about 1 second at 65°C, and about 100 seconds at 50°C.</p>
<p>For a more recent review, check out the 2017 article <a href="https://doi.org/10.1016/j.burns.2017.04.003" rel="nofollow noreferrer" title="N.A. Martin, S. Falder, Burns, Volume 43, Issue 8, 2017, Pages 1624-1639">"A review of the evidence for threshold of burn injury"</a> by Martin and Falder in the journal <em>Burns</em>. According to that article, burn injury occurs when the basal epidermis reaches 44°C and the rate of tissue damage increases logarithmically with temperature, until by 70°C the damage is so rapid that "interpretation can be difficult". This is roughly consistent with the figure mentioned above.</p>
| 53187 | What standards or industry guidance define a top temperature for human contact to surfaces? |
2022-11-16T02:20:11.437 | <p>On the boiler plate of a turbo charger (Type seems to be TPS57-F32, probably an ABB model, but manufacturer is not displayed) I found the following temperatures:</p>
<p><span class="math-container">$$
t_{M_{max}} = 650°C
$$</span>
<span class="math-container">$$
t_{B_{max}} = 620°C
$$</span></p>
<p>Obviously these temperatures are limits for temperatures at the turbine side. But what do they mean in particular?</p>
| |turbomachinery| | <p>From <a href="https://docplayer.net/25247598-Operation-manual-tps57-f32.html" rel="nofollow noreferrer">ABB Operation Manual TPS57-F32</a></p>
<blockquote>
<p>Operation above the indicated values <span class="math-container">$n_{B_{max}}$</span>, <span class="math-container">$t_{B_{max}}$</span> can considerably shorten the recommended replacement intervals. In such cases, we recommend that you contact the nearest official ABB Turbocharging service station.</p>
</blockquote>
<blockquote>
<p><span class="math-container">$n_{M_{max}}$</span> and <span class="math-container">$t_{M_{max}}$</span> normally apply only when running at overload (110%) during <strong>trials</strong> on the engine test bed. These limits can also be permitted during operation for special applications. Operation above <span class="math-container">$n_{M_{max}}$</span> and <span class="math-container">$t_{M_{max}}$</span> is not permitted.</p>
</blockquote>
<p><span class="math-container">$B_{max}$</span> is normal, not to be exceeded, operational limits, while <span class="math-container">$M_{max}$</span> are 110% one time, not to be exceeded, commisioning limits for the turbocharger (TC). As in: is everything working correctly if the TC is subjected to 110% of full-load.</p>
<p>From <a href="https://www.dieselduck.info/machine/01%20prime%20movers/2010%20-%20MAN%20Project%20Guide%206S90ME.pdf" rel="nofollow noreferrer">MAN B&W S90ME-C8-TII</a> 2-stroke Diesel Engine.</p>
<blockquote>
<p>Minimum delivery test</p>
</blockquote>
<blockquote>
<p>The minimum <strong>delivery test</strong>, EoD: 4 14 001, involves:
• Starting and manoeuvring test at no load
• Load test
Engine to be started and run up to 50% of
Specified MCR (M) in 1 hour</p>
</blockquote>
<blockquote>
<p>Followed by:
• 0.50 hour running at 25% of specified MCR
• 0.50 hour running at 50% of specified MCR
• 0.50 hour running at 75% of specified MCR
• 1.00 hour running at 100% of specified MCR
• <strong>0.50 hour running at 110% of specified MCR</strong></p>
</blockquote>
<p>During the commisioning process for large marine diesels, the diesel is run at 110% of Manufacturers Continuous Rating (MCR) for 30 minutes. The diesel and it's components can be measured to ensure ratings stabalize below specific maximums. In the case of the TC, ABB states that frequency and temperature should be below these <span class="math-container">$M_{max}$</span> limits (<span class="math-container">$t_{M_{max}} = 650°C$</span>).</p>
<p>Similarily, normal operation limits <span class="math-container">$B_{max}$</span> allow periodic testing.</p>
<p>Both are easy to verify if written on the device.</p>
| 53189 | What is the meaning of different temperature limits printed on turbo charger's boiler plate? |
2022-11-17T03:48:01.430 | <p>I searched here and even asked at McMaster-Carr, but even they couldn't provide an answer for me--</p>
<p>So, say you have either a ball-screw, or a timing belt-- Of course, in either case, having the right screw or belt for the geared purpose is 'quite important'.</p>
<p>However, I wondered if, in the end, the 'pitch' actually, at all, affected 'precision' in any reasonable way.</p>
<p>Abstractly, the easiest way for me to think of this is in the case of a 'ball screw'.</p>
<p>Lets say you have a 'really high pitch'-- Obviously for each turn of the screw, you are going to 'get where you are going' <em>much</em> faster than a very tight pitch.</p>
<p>That would require ever more turns of the screw--</p>
<p>Yet, between the two, it seems to me, between the high pitch, and low pitch screw, you should always be able to reach the 'same degree of position accuracy'-- <em>at least</em> if you are able to turn the screw slow enough (smaller, rather than bigger steps).</p>
<p>Am I wrong in this ?</p>
<p>With regard to 'timing belts', it is harder for me to think about. Would the principle be the same ?</p>
| |linear-motion| | <p>The ball screw converts the rotational position of the shaft to a linear position of the carriage. In a typical application of ball screws, like a servo linear axis or the cross slide of a lathe, the rotation is controlled, not the actual linear motion.</p>
<p>Example: on a lathe, you turn the dial which has gradation that represent linear displacement, but as you turn the wheel to a certain rotational position, you control the rotation - not the linear displacement directly. A servo controlling a linear motion often has a rotary encoder as its position feedback mechanism. Same story applies.</p>
<p>This control of rotational angle will have a limited resolution, repeatability and absolute accuracy. Same as the ball screw converts a rotational position to linear position, with conversion proportional to the pitch of the screw, it converts a rotational angle error to a linear position error that is proportional to the pitch.</p>
<p>Example:</p>
<ul>
<li>1/100 of revolution error at 5 mm pitch: linear position error = 0.01*5=0.05 mm</li>
<li>1/100 of revolution error at 25 mm pitch: linear position error = 0.01*25=0.25 mm</li>
</ul>
| 53198 | Understanding the importance of 'pitch': For ball screws and timing belts |
2022-11-17T21:46:14.320 | <p><strong>Background</strong></p>
<p>For my mechanical engineering senior design project, we are designing a cargo transporter electric vehicle for off-road conditions. The vehicle needs to transport loads weighing up to about 22,000 N or about 5000 lbf.</p>
<p>The transporter will have a very slow speed (about 6 in/s). Using a coefficient of rolling resistance of 0.35, I calculated that the total power requirements would be about 5 hp. To calculate this, I took the rolling resistance force times the target speed of 6 in/s (P=FV).</p>
<p>Assuming that we use four electric motors, one for each wheel, I found a per-motor horsepower requirement of about 1.25 hp. For large wheels of about 3 ft diameter at the speed of 6 in/s, the wheel rotation rate is about 2.5 RPM. This yields a large torque requirement of around 3700 N-m or 2700 ft-lb.</p>
<p>I'm having trouble finding the right combination of DC motor and gearbox to fit this application.</p>
<p><strong>Question</strong></p>
<p>How can I size a DC motor for a high torque, low speed application? The torque load per motor is about 3700 N-m and the nominal rotation speed is about 2.5 RPM, yielding a power requirement of about 1.25 hp per motor.</p>
<p>I can find DC motors rated for this power, but definitely not for this torque requirement or rotational speed. Would I need a special gearbox? How do industry people normally solve this problem?</p>
| |motors|torque|electric-vehicles| | <p>First, make sure you know all load cases. You need to know which torque at which RPM you need to have.
Then, find a motor + gearbox combination, check their rated torque, RPM and power. Consideration here are: cost, efficiency, type of controlling/driving and in this case especially, power supply. I can imagine that on a lunar base you might not have your 60 Hz 120 V AC?</p>
<p>Just a quick example of a process you'll need to iteratively go through.
If start-up torque is not huge compared to steady state power/torque requirements, and let's make the wild guess you'll have DC system on the lunar base, and the application requires frequent starting, stopping, positioning, you could have a look at servo systems with bolt on planetary gearbox. Let's say a 5000 rpm motor and a 2.5 RPM load, requires a 1:2000 gear. Let's divide that by a worm 1:40 and a planetary gear box in the motor 1:50. Check torques, check other assumptions. Rinse and repeat.</p>
<p>By the way, are you sure about the 0.35 rolling friction coefficient? Seems high for rolling.</p>
| 53205 | Electric motor for low speed, high torque application? |
2022-11-18T03:49:16.557 | <p>Is there some sort of general guidance or load rating for wood screws (lag bolts) going into framing wood where the force is pulling directly inline with the screw orientation? More context and specifics:</p>
<p>Trying to install a ceiling attachment for a hoist in a garage. There is a structural beam going across the ceiling. Looking at the engineering plans I see the following:</p>
<ul>
<li>Size: 7x14</li>
<li>Type: PSL (Parallel strand lumber)</li>
<li>Bearing: 5-2x6 (not sure what this means?)</li>
</ul>
<p>I can't access the top of the beam (drywall, floor joists, etc) so hoping to use lag bolts (drilled vertically up into the beam) to attach a lashing ring (to which I would attach a block-and-tackle or chain hoist). Here's a little diagram:</p>
<p><a href="https://i.stack.imgur.com/S5TPl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S5TPl.png" alt="enter image description here" /></a></p>
<p>I'm assuming there is some engineering reference table where you can see the rated load depending on the screw type, shank diameter, thread pitch, wood type, and length of screw inserted in the material?</p>
<p>I tried to search the web but I'll admit I don't even know what terms to search for.</p>
<p>My intuition tells me to use a lashing ring plate with 4 holes and use bigger (3/8 and longer) lag screws, but I'll admit I just even have an idea. Is vertical pull on a 1/4 lag bolt 50 lbs or 500 lbs?</p>
<p>Also to qualify: I'm not lifting anything insane. Most likely only lifting about 250 lbs, but would be nice to know if it's possible to lift a car engine for example.</p>
<p>I was also told to use GRK Fasteners 'RSS' screws instead of traditional lag screws. I'm guessing they have more holding power because of the aggressive thread?</p>
<p>I'm a software engineer not a civil engineer so completely a fish out of water, but also don't want to just take the advice from the ol' Home Depot boys "oh she'll be fine, ya, just use a few screws"...</p>
| |fasteners|wood| | <p>If you would make a U-shape clamp around the beam, you could bolt it to the sides of the beam. You can put as many bolts as you need (or a threaded bar right through) at the neutral line of the beam.</p>
<p>The vertical load would load the bolts or bar by a shear force. I expect the permissible shear load of a bolt to be listed somewhere, or otherwise to be calculated by cross-section area * bolt class yield strength * 0.55</p>
| 53209 | Inline pull load rating of lag bolts in wood? |
2022-11-18T12:02:52.727 | <p>Are these the same? if so, why there are two equations given? is there a specific purpose? like, combining the equations for the ease of calculation etc?</p>
<p>Why there are two equations to calculate effective depth for concrete beams?</p>
<p><a href="https://i.stack.imgur.com/cAseE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cAseE.png" alt="enter image description here" /></a></p>
| |structural-engineering|concrete|eurocodes| | <p>The referenced Eurocode equations 7.16(a) & 7.16(b) are part of a “simplified” procedure for limiting deflection of reinforced concrete beams to the limits specified in sections 7.4.1(4) and 7.4.1(5) of the code. The applicability of the equations is based on the relative values of the required tension reinforcement ratio p of beams under design loads to a reference reinforcement ratio p0. These two values are used in inequalities for each equation that define which equation is applicable. Another observable difference between the two equations that is that 7.1.6(b) appears to be applicable to doubly-reinforced concrete. There is apparently some confusion among Eurocode users surrounding the use of this procedure.</p>
| 53216 | Why there are two equations to calculate effective depth for concrete beams? Eurocode |
2022-11-18T15:44:48.430 | <p>I'm reading the <a href="https://en.wikipedia.org/wiki/Vapor-compression_refrigeration" rel="nofollow noreferrer">Wikipedia</a> article on vapor-compression refrigeration.</p>
<p>In the cycle, first the refrigerant absorbs heat from the source to be cooled, which causes it to be vaporized. Then this vapor goes through a compressor, which increases the pressure and temperature of the vapor. From Wikipedia:</p>
<blockquote>
<p>Circulating refrigerant enters the compressor in the thermodynamic state known as a saturated vapor<a href="https://en.wikipedia.org/wiki/Vapor-compression_refrigeration#Thermodynamic_analysis_of_the_system" rel="nofollow noreferrer">2</a> and is compressed to a higher pressure, resulting in a higher temperature as well. The hot, compressed vapor is then in the thermodynamic state known as a superheated vapor and it is at a temperature and pressure at which it can be condensed with either cooling water or cooling air flowing across the coil or tubes.</p>
</blockquote>
<p>I'm slightly confused about this stage. <strong>Why do we compress the vapor to increase its temperature, only to cool it down again in the condenser?</strong> The refrigerant has first absorbed heat from the heat source. Why don't we just conduct this vapor into the condenser, causing it to reject heat to cool down? It seems strange that we are artificially adding heat to the vapor, when removing heat from the heat source to be cooled is our goal.</p>
<p>In <a href="https://en.wikipedia.org/wiki/Vapor-compression_refrigeration#Thermodynamic_analysis_of_the_system" rel="nofollow noreferrer">this</a> part it is said:</p>
<blockquote>
<p>From point 1 to point 2, the vapor is isentropically compressed (compressed at constant entropy) and exits the compressor as a superheated vapor. Superheat is the amount of heat added above the boiling point.</p>
</blockquote>
<blockquote>
<p>From point 2 to point 3, the vapor travels through part of the condenser which removes the superheat by cooling the vapor.</p>
</blockquote>
<p>So we are adding superheat, only to remove it in the next step? The diagram from the section also confuses me:</p>
<p><a href="https://i.stack.imgur.com/T5MGy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T5MGy.png" alt="enter image description here" /></a></p>
<p>The text explains that the compression at 1-2 is isentropic, which can also be seen from the T-S diagram. But the text also says that <em>superheat</em> is added at this point. But doesn't isentropic mean that no heat is added or removed from the system?</p>
<p>So to summarize: are we adding heat to the refrigerant during the compression stage or not? And if we are, why are doing it only to remove it in the next step, rather than just conducting the saturated vapor through the condenser to give up heat?</p>
| |thermodynamics|refrigeration|vaporization| | <p>The compressor simply compresses the refrigerant gas. Compression itself increases the temperature. In order to "pump" heat, we need the refrigerant to be above the ambient temperature around the condensor (heat can only flow from hot to cold). So the hot gas leaving the compressor releases its heat to the hot region of the circuit and "condenses" into a liquid in the process.</p>
<p>The superheat portion has to do with the amount of heat the gas must give up at a given pressure before it begins to condense, reducing in volume and temperature as it goes. Once it reaches zero superheat, it remains at the same temperature and pressure and condenses to a liquid, after which it may continue to cool (subcooling) as a liquid. Superheat has nothing to do with whether the compression is isentropic - the paths taken by the refrigerant in a heat-pump cycle are constrained by thermodynamic limits - the gas can't be compressed into any desired state. The best (most efficient) compression possible is isentropic, everything worse will require even more work and push point 2 up and to the right, resulting in greater superheat, but it's not possible to push point 2 down and to the left beyond the isentropic limit. Note that refrigerant cycles do not require a phase change - it's possible to construct a heat pump where the refrigerant remains a gas at all times. For example, a Stirling engine can theoretically* achieve Carnot efficiency with no phase changes (*in practice not so much).</p>
| 53221 | Why do we need a compressor to heat vapor in vapor-compression refrigeration? |
2022-11-08T14:37:15.540 | <p>The bicycle component manufacturer Shimano makes crankarms named "Hollowtech".</p>
<p>Hollowtech crankarms are meant to be both strong and light. As the diagram below (source: <a href="https://bike.shimano.com/en-EU/technologies/component/details/hollowtech.html" rel="nofollow noreferrer">Shimano</a>) shows, the outer shell is made from an aluminum alloy and the inside is hollow.</p>
<p><a href="https://i.stack.imgur.com/G36pF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G36pF.jpg" alt="hollowtech" /></a></p>
<p>On closer look it turns out that the cross-section of the crankarm is not made from a box section, but from two C-sections. (Source: <a href="https://www.bikeradar.com/features/shimano-crank-failure/" rel="nofollow noreferrer">bikeradar</a>)</p>
<p><a href="https://i.stack.imgur.com/eN7cs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eN7cs.jpg" alt="two c-sections" /></a></p>
<p>Another (older) style of Hollowtech consisted of a C-section closed off by a plate.</p>
<p><a href="https://i.stack.imgur.com/diwcc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/diwcc.jpg" alt="failure mode of hollowtech" /></a></p>
<p>The two parts are glued using epoxy. It's understood that epoxy, once it sets, is more than adequate for the task of carrying the weight of a cyclist on the crankarm, but even if epoxy does bind well with a (perhaps roughened) alloy, wouldn't the difference in the stress-strain curves of the two materials mean that when the alloy bends, it is the epoxy that will take <em>all</em> of the load? Since it's used here only as a binding agent, it would appear this is a recipe for failure. Do you understand how epoxy can reliably bind two parts of a crank arm, and can you explain it?</p>
| |friction|aluminum|welding|bicycles|epoxy| | <p>Agree with Vladimir. Note that the join between the two C-section halves can be created with a deliberate interference (force) fit involving the roughened surfaces in such a way that when they are driven together in a press, the pieces gall and seize. This means the parts are in a sense friction-welded together and thus the epoxy does not have to carry all the imposed loads.</p>
<p>Another thing to consider is that if the interference fit creates residual compressive stresses in the assembly, then all you have to do to prevent the stress on the glue line from ever becoming tensile (which is how it would fail in service) is to ensure that the magnitude of the residual compressive stress exceeds the maximum tensile service stress.</p>
| 53235 | Can epoxy bond two Aluminum alloy beams? |
2022-11-20T13:00:38.900 | <p>I am a math lecturer and in my teaching of second-order linear differential equations I present, as an application, the classical mass-spring-dashpot system (and its RLC analogue). According to my understanding, if the overdamping case modelled an automobile suspension system, the ride would be uncomfortable. Furthermore, critical damping represents the minimum damping that can be applied to the physical system without causing oscillation.</p>
<p>One textbook I use (Ordinary Differential Equations and Applications by Weighfer and Lindsay) says:
<em>(Critical damping) is often the desired configuration for practical aplications since it represents the weakest damping before oscillatory becomes possible.</em></p>
<p>Nevertheless, another textbook I use (Differential Equations for Engineers by Xie) states that:
<em>Most engineering structures fall in this category (i.e. underdamping
case) with (the dimensionless) damping coeffient <span class="math-container">$\zeta$</span> usually less
than <span class="math-container">$10\%$</span>.</em>
[<span class="math-container">$\zeta=\frac{\gamma}{2\sqrt{mk}}$</span> for the mechanical system; <span class="math-container">$\zeta=\frac{2}{R}\sqrt{\frac{C}{L}}$</span> for the electrical analogue].</p>
<p>Question 1: Which author is right?</p>
<p>Question 2: In general What are some practical applications where each case (heavy, critical and light damping) is desirable or not desirable?</p>
<p>Thank you very much.</p>
| |mechanical-engineering|applied-mechanics| | <p>Question 1 constitutes a false dichotomy, since the two books might be talking about completely different applications. They might both be true or both false. To me, they both seem to be way overgeneralizing.</p>
<p>In many systems, the overall desire is for the system to achieve a new equilibrium state quickly following any change to its inputs. Critical damping is the fastest response you can get without any overshoot at all. But sometimes a different measure is used — minimum settling time, which is defined as the system getting to a point that is within some delta (error window) of its final state in the least amount of time. In second order systems, it turns out that this is achieved by allowing a certain amount of overshoot — in other words, slightly under-damped relative to critical damping.</p>
<p>Vehicle suspensions are one example of such a system. But unfortunately, they must deal with a wide range of input conditions with limited ability to modify their own parameters. It's relatively easy to make a passenger car ride comfortably under most conditions, because the change in mass due to passengers, etc. is a small fraction of the total mass. But have you ever ridden in a lightly-loaded truck? It's very uncomfortable, because the springs and shocks are designed to handle a full load. When the mass is much lower than that, the system is heavily over-damped. You feel every bump in the road.</p>
| 53250 | Applications of overdamping, critical damping and underdamping |
2022-11-20T16:34:31.270 | <p>A friend of mine who is (completely) red/green color blind once told me that he can distinguish the green traffic light from the yellow and red one despite his color blindness and that he thinks to be able to remember having heard that this is not an accident. Is there some law or guideline for it or is it just niceness on the traffic light manufacturer's part and what exact green is it?</p>
<p>In case it matters, I'm living in Germany.</p>
<p>(Sorry, if this is the wrong SE; it seemed like the most appropriate one.)</p>
| |traffic-light| | <p>I am <a href="https://en.wikipedia.org/wiki/Congenital_red%E2%80%93green_color_blindness" rel="nofollow noreferrer">red-green color blind</a>. I researched and <a href="https://www.leppik.net/david/blog/are-green-traffic-lights-really-turquoise/" rel="nofollow noreferrer">blogged</a> about this years ago. Since then, the primary sources have disappeared off the internet, except for the <a href="https://cie.co.at/publications/calculation-and-measurement-luminance-and-illuminance-road-lighting" rel="nofollow noreferrer">official CIE standard</a>, which is expensive. So unfortunately you may have to take my word for it.</p>
<p>Green traffic lights are are slightly toward the blue end of the spectrum, which lets people like me differentiate them from red lights. There is some variation in the allowed color range, but they are all distinct from the pure green that's hardest to tell from pure red.</p>
| 53255 | What color do green traffic lights have? |
2022-11-21T00:13:37.357 | <p>At standard room pressure, the highest temperature superconductor is "the cuprate of mercury, barium, and calcium, at around 133 K" (-140 °C).</p>
<p>If pressure can be increased, however, the record belongs to LaH<span class="math-container">$_{10}$</span> at -23 °C requiring 170 GPa, or possibly <a href="https://en.wikipedia.org/wiki/Carbonaceous_sulfur_hydride" rel="nofollow noreferrer">carbonaceous sulfur hydride</a> at +15 °C requiring 270 GPa.</p>
<p>(<a href="https://en.wikipedia.org/wiki/High-temperature_superconductivity" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/High-temperature_superconductivity</a>)</p>
<p>Since -23 °C and 15 °C are commonly encountered conditions in many parts of the Earth, what possible applications might there be for a superconductor at these non-extreme temperatures, but which requires 170-270 gigapascals of pressure to operate?</p>
| |pressure|solid-mechanics|electromagnetism| | <p>Such applications would be extremely rare. Cooling with liquid nitrogen is easy and cheap, so using regular high temperature superconductors would almost always be preferred.</p>
<p>The paper claiming +15C superconductivity has actually <a href="https://www.nature.com/articles/s41586-022-05294-9" rel="nofollow noreferrer">just been retracted</a> by the editors of Nature, but for the sake of this question we'll assume it is true. In that case, superconductivity would only require a cool day or very good air conditioning.</p>
<p>To reach the necessary static pressures requires a diamond anvil, so any high-pressure superconducting component would have to be very small. One possibility might be tiny <a href="https://en.wikipedia.org/wiki/Josephson_effect" rel="nofollow noreferrer">Josephson Junctions</a>, but these would only really be useful technology if you could also figure out how to miniaturize the anvil to create a small sensor. In the unlikely event you could overcome the very extreme engineering challenges, these might useful for <a href="https://en.wikipedia.org/wiki/SQUID" rel="nofollow noreferrer">SQUID magnetometers</a> and their <a href="https://en.wikipedia.org/wiki/Diamond_anvil_cell#Two_opposing_diamond_anvils" rel="nofollow noreferrer">many applications</a>, as well as <a href="https://en.wikipedia.org/wiki/SQUID#Uses" rel="nofollow noreferrer">Quantum Computing</a>.</p>
<p>There are multiple reasons why it is almost certainly impossible to scale up <a href="https://en.wikipedia.org/wiki/Diamond_anvil_cell#Two_opposing_diamond_anvils" rel="nofollow noreferrer">diamond anvil</a> technology to macroscopic (e.g. <span class="math-container">$\sim 1$</span> m) sizes:</p>
<ul>
<li>Metre-sized flawless diamonds are hard to come by:)</li>
<li>Applying 270 GPa over an area of a square metre would require 270 GN = 28 MTf (Mega Tonne force). The world's largest press is 10 stories tall and can only achieve <span class="math-container">$\sim 0.1$</span> MTf.</li>
<li>Most importantly, diamond anvils fail when they become larger than about <span class="math-container">$0.1$</span> mm, with <a href="https://doi.org/10.1063/1.5049720" rel="nofollow noreferrer">one estimate for the maximum pressure</a> being
<span class="math-container">$$P_{max}\,\textrm{(GPa)} \approx \frac{2000\,\textrm{GPa}\, \mu\textrm{m}^{0.5}}{\sqrt{d}}$$</span>
where <span class="math-container">$d$</span> is the diameter of the anvil face in microns.
The +15C result used an anvil smaller than 50 microns to achieve <span class="math-container">$270$</span> GPa, which is about the expected upper size limit.</li>
</ul>
| 53259 | What are possible applications for room temperature superconductor requiring extreme presssure? |
2022-11-21T16:38:54.683 | <p>So, the objective is to generate a water flow similar to a household water system (60 PSI) inside copper pipes that have 30 coil turns that are inside a stove that burns wood.</p>
<p><a href="https://i.stack.imgur.com/dQqRV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQqRV.png" alt="Illustration of a monotube pumpless boiler" /></a></p>
<p>A <a href="https://en.wikipedia.org/wiki/Monotube_steam_generator" rel="nofollow noreferrer">monotube steam generator</a> is basically a "boiler" without a <a href="https://www.globalspec.com/ImageRepository/LearnMore/20134/300px-Fire-tube_Boiler951b8ba0fb47443ca0edb2627de26786.png" rel="nofollow noreferrer">pressure vessel</a>/steam drum, generating steam <strong>pumping</strong> water through a coiled tubing above a heat source, transforming the water in steam in the process.</p>
<p>The idea here is to make a pumpless monotube boiler, the water will eventually evaporate and push the hydro generator, eventually emptying the copper coil, which will make the water come back into the tubing.</p>
<hr />
<p><strong>The Problem:</strong></p>
<p>The problem is that this seems too much like the <a href="https://youtu.be/3AXupc7oE-g" rel="nofollow noreferrer">putt putt boat</a> engine, because I'm assuming the water will only flow into the coil once the water inside it turns into steam.</p>
<p>And I found <a href="https://sciencetoymaker.org/wp-content/uploads/2018/04/Article-Efficiency-of-a-pop-pop-engine.pdf" rel="nofollow noreferrer">this PDF</a> that affirms that the putt-putt boat only achieves 0,02% of efficiency. And the closest thing I found to this contraption was this <a href="https://www.researchgate.net/figure/Model-of-a-water-pump-driven-by-a-putt-putt-heat-engine-which-can-raise-water-from-a_fig6_305881432" rel="nofollow noreferrer">putt putt heat engine water pump</a>, but they didn't specify the efficiency of the pump.</p>
<p><a href="https://i.stack.imgur.com/Ambzi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ambzi.png" alt="putt putt pump explanation illustration" /></a></p>
<p>There are also Water-tube Boilers, but I don't know if this system is the same or something else. Simply because the tubes in the water tube are separated and connected to a somewhat pressure vessel.
And they also need a water pump.</p>
<p><a href="https://i.stack.imgur.com/fqOC1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fqOC1.png" alt="Water tube boiler explanation illustration" /></a></p>
<hr />
<p><strong>The question:</strong></p>
<p>How efficient would this system be in generating steam pressure (and thus, work)?</p>
<p>Here are the measurements of the stove and copper tubing:</p>
<ul>
<li>Height of stove: 23 cm</li>
<li>Diameter of stove: 8cm</li>
</ul>
<p>Copper pipe has the following measurements:</p>
<p><a href="https://i.stack.imgur.com/rYsgq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rYsgq.png" alt="Copper pipe diagram" /></a></p>
| |fluid-mechanics|thermodynamics|energy-efficiency|steam|boilers| | <p>You need a pump to make steam because otherwise the water will never flow into the boiler. In this case, you would only get a little water flow when all water went to steam, and pressure in the coil dropped. As soon as any water enters the coil, pressure will instantly rise and flow will stop.</p>
<p>You need a pump.</p>
| 53278 | How efficient would a monotube steam generator without a pump in generating steam pressure? |
2022-11-22T19:31:23.980 | <p>I'm wondering how surface roughness specimens that are used to calibrate surface roughness machines are created. Specifically those in the 125-200 µin range. Is this typically done with vertical milling, electroforming, etc?</p>
<p>Here is an example of what I'm referring to: <a href="https://www.grainger.com/product/INSIZE-Surface-Roughness-Specimen-463T67" rel="nofollow noreferrer">https://www.grainger.com/product/INSIZE-Surface-Roughness-Specimen-463T67</a></p>
| |mechanical-engineering|materials|measurements|metrology| | <p><a href="https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4924890/" rel="nofollow noreferrer">https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4924890/</a> indicates that at least some methods used are:</p>
<ul>
<li>Chemical etching or oxidation and at the same time states that for 6,
20, 60, and 200 μm.</li>
<li>CNC diamond turning for 0.3 to 3um.</li>
</ul>
<p>100µin = 2.54µm</p>
<p>Skim through the article though... They have µm referencing both Ra values and wavelengths and I am not sure what the difference is so the numbers I posted above might be mixed up between the two.</p>
| 53294 | How are surface roughness specimens created? |
2022-11-23T13:45:38.200 | <p>I am using a stepper motor with a holding torque of x N.m driving a mass with inertia y. will adding a 2:1 gear reduction at the output cause effective holding torque to be 2x AND the inertia of the load to be y/2?</p>
<p>If this is the case, when calculating acceleration torque can I assume the inertia is halved and the torque available is double?</p>
| |gears|stepper-motor| | <p>Moment of inertia, like mass, does not change just because you have more torque or force available. Ultimately, it is acceleration that changes. The 2:1 gearbox causes the load to see double the torque from the motor, and it causes the motor to see half the moment of inertia from the load (as well as other load torques if there are there are any).</p>
<p>If you treat a 2:1 gearbox as doubling the torque AND reducing the moment of inertia you are double-counting the mechanical advantage being provided by the gearbox.</p>
| 53302 | Do gear ratios affect both inertia and holding torque in stepper motors |
2022-11-23T16:38:37.107 | <p>Hi I am an engineering student.</p>
<p>I have a question about oblique shock and expansion fans. In my understanding, oblique shock and expansion fans could happen on supersonic airplane foil. But how often are they happening?</p>
<p>Lets say if there is an airplane moving at constant speed at 3 times sound speed, how do we know how much energy is lost due to expansion fans and oblique shocks if this airplane travels for 3 km? How many shocks should we consider?</p>
| |mechanical-engineering|fluid-mechanics|compressed-air| | <p>First of all You can use the simulator to study the flow past a wedge.There is more complete shock simulation program that is avaliable on web. The various programs solves for flow past a wedge and for flow past a cone, including the detached normal shock conditions. Another simulation, called ShockModeler, describes the intersection and reflection of multiple shock waves.</p>
<p>Use simulation softwares to study different flows it will definitely help.
Further explanation I have mentioned below;</p>
<p>Oblique shocks are generated by the nose and by the leading edge of the wing and tail of a supersonic aircraft. Oblique shocks are also generated at the trailing edges of the aircraft as the flow is brought back to free stream conditions. Oblique shocks also occur downstream of a nozzle if the expanded pressure is different from free stream conditions. In high speed inlets, oblique shocks are used to compress the air going into the engine. The air pressure is increased without using any rotating machinery.</p>
<p>For compressible flows with little or small flow turning, the flow process is reversible and the entropy is constant. The change in flow properties are then given by the isentropic relations (isentropic means "constant entropy"). But when an object moves faster than the speed of sound, and there is an abrupt decrease in the flow area, shock waves are generated in the flow. Shock waves are very small regions in the gas where the gas properties change by a large amount. Across a shock wave, the static pressure, temperature, and gas density increases almost instantaneously. The changes in the flow properties are irreversible and the entropy of the entire system increases. Because a shock wave does no work, and there is no heat addition, the total enthalpy and the total temperature are constant. But because the flow is non-isentropic, the total pressure downstream of the shock is always less than the total pressure upstream of the shock. There is a loss of total pressure associated with a shock wave as shown on the slide. Because total pressure changes across the shock, we can not use the usual (incompressible) form of Bernoulli's equation across the shock. The Mach number and speed of the flow also decrease across a shock wave.</p>
<p>On the slide, a supersonic flow at Mach number M approaches a shock wave which is inclined at angle s. The flow is deflected through the shock by an amount specified as the deflection angle - a. The deflection angle is determined by resolving the incoming flow velocity into components parallel and perpendicular to the shock wave. The component parallel to the shock is assumed to remain constant across the shock, the component perpendicular is assumed to decrease by the normal shock relations. Combining the components downstream of the shock determines the delflection angle. Then:</p>
<p>cot(a) = tan(s) * [{((gam+1) * M^2)/(2 * M^2 * sin^2(s) - 1)} - 1]</p>
<p>where tan is the trigonometric tangent function, cot is the co-tangent function:</p>
<p>cot(a) = tan(90 degrees - a)</p>
<p>and sin^2 is the square of the sine. Gam is the ratio of specific heats. Across the shock wave the Mach number decreases to a value specified as M1:</p>
<p>M1^2 * sin^2(s -a) = [(gam-1)M^2 sin^2(s) + 2] / [2 * gam * M^2 * sin^2(s) - (gam - 1)]</p>
<p>The total temperature across the shock is constant, but the static temperature T increases in zone 1 to become:</p>
<p>T1 / T0 = [2 * gam * M^2 * sin^2(s) - (gam - 1)] * [(gam -1) * M^2 * sin^2(s) + 2] / [(gam + 1)^2 * M^2 * sin^2(s)]</p>
<p>The total pressure pt decreases according to:</p>
<p>pt1 / pt0 = {[(gam + 1) * M^2 * sin^2(s)]/[(gam-1)*M^2 * sin^2(s) + 2]}^[gam/((gam-1)] * {(gam+1)/[2 * gam * M^2 * sin^2(s)-(gam-1)]}^[1/(gam-1)]</p>
<p>The static pressure p increases to:</p>
<p>p1 / p0 = [2 * gam * M^2 * sin^2(s)-(gam -1)] / (gam + 1)</p>
<p>And the density r changes by:</p>
<p>r1 / r0 = [(gam + 1) * M^2 * sin^2(s)] / [(gam -1) * M^2 * sin^2(s) + 2]</p>
<p>The right hand side of all these equations depend only on the free stream Mach number and the shock angle. The shock angle depends in a complex way on the free stream Mach number and the wedge angle. So knowing the Mach number and the wedge angle, we can determine all the conditions associated with the oblique shock.</p>
<p>[The equations describing oblique shocks were published in NACA report (NACA-1135) in 1951.]</p>
| 53304 | Question about oblique shock and expansion fans |
2022-11-25T08:45:45.637 | <p>I'm currently reading Olson's Elements of Acoustical Engineering, in which formulae are given for the directional characteristics of various arrays of sound sources, all of which are stated in terms of ratios <span class="math-container">$R_\alpha$</span> of pressure at a given angle <span class="math-container">$\alpha$</span> to pressure normal to the array.</p>
<p>For example, the ratio given for an array of point sources is:</p>
<p><span class="math-container">$$R_\alpha=\frac{sin(\frac{n\pi d}{\lambda}sin\alpha)}{nsin(\frac{\pi d}{\lambda}sin\alpha)}$$</span></p>
<p>where: <span class="math-container">$n$</span> is the number of sources;
<span class="math-container">$d$</span> is the distance between the sources; and
<span class="math-container">$\lambda$</span> is the wavelength</p>
<p>I've <a href="https://www.desmos.com/calculator/t2i3k21yy2" rel="nofollow noreferrer">plotted this function in desmos</a> (substituting <span class="math-container">$343/x$</span> for <span class="math-container">$\lambda$</span> where <span class="math-container">$x$</span> is frequency and 343m/s is the speed of sound in air at room temperature).</p>
<p>The resultant graph shows a periodic function with both positive and negative values.</p>
<p>However, if such an array were approximated by real sound sources and sound pressures measured with a microphone at a given angle <span class="math-container">$\alpha$</span>, you'd expect to measure something resembling the absolute value of the same function, i.e. no negative values (see the second function in the Desmos link, above).</p>
<p>What I'm interested in understanding is: What does a negative pressure ratio mean, physically, in this context?</p>
<p>And, if it has physical meaning, how would you measure it?</p>
<p>Would anyone be able to point me in the right direction?</p>
| |fluid-mechanics|acoustics|sound| | <p>The sound pressure does go negative, that is out of phase with the reference signal.</p>
<p>Incidentally your plot is confusing to me, as x axis seems to be either frequency or wavelength(?), so you are looking at the response in a particular direction at all frequencies. Here's a more typical way to plot it for <span class="math-container">$n = 5$</span>, <span class="math-container">$\lambda = d = 1$</span>, at just one frequency but for all directions.</p>
<p><a href="https://i.stack.imgur.com/PtSl0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PtSl0.png" alt="Plot of R vs azimuth" /></a></p>
<p><a href="https://i.stack.imgur.com/sxUEW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sxUEW.png" alt="Polar of R+1" /></a></p>
<p>Here's a comparison for <span class="math-container">$n = 2$</span> <span class="math-container">$\lambda = 1$</span> or <span class="math-container">$2$</span> and <span class="math-container">$d = 1$</span></p>
<p><a href="https://i.stack.imgur.com/ucu4b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ucu4b.png" alt="Polar for dipole, lambda=1 or 2" /></a></p>
<p>"if it has physical meaning, how would you measure it?"</p>
<p>A reference microphone at <span class="math-container">$0$</span> degrees, and a second microphone at the same distance from the centre of the array at the angle of interest. You could then use a scope or an FFT analyser to compare the two waveforms.</p>
| 53312 | Confusion over textbook formulae for directional characteristics of sound sources |
2022-11-25T20:19:20.413 | <p>I was searching for questions on this stack and I found <a href="https://engineering.stackexchange.com/questions/49528/why-arent-pneumatic-hydraulic-artificial-muscle-actuated-humanoid-robots-more-c">this one asking why pneumatic/hydraulic artificial muscle robots aren't common</a>, TLDR: the reason is that it is incredible hard to control them.</p>
<p>However, one thing that the asnwer didn't talked about was vacuum actuated artificial muscles.</p>
<p>My gut feeling says that they are as difficult as pneumatic ones, since they are simply the opposite, but even then I want to know.</p>
| |robotics|vacuum|actuator|vacuum-pumps| | <p>With a vacuum you are limited to the pressure differential between zero and atmospheric (~100kPa or ~15PSI). With pneumatic you are limited to the pressure differential between atmospheric and whatever pressure you design your compressor and muscles to. A normal shop compressor easily exceeds 100PSI.</p>
<p>Also, with a compressor you can do things like store the compressed air at higher than the pressure you need and use a valve to regulate it which lets you use a smaller reservoir. You can't "store a more extreme vacuum" and then regulate the vacuum to be less. The only thing you can do is use a larger vacuum reservoir tank.</p>
| 53321 | Are vacuum actuated artificial muscles as hard to control as the pneumatic ones? |
2022-11-26T13:37:48.887 | <p>As posed above...</p>
<p>(sorry if not asked correctly - my first time on SE)</p>
| |gears|generator| | <p>A pair of gears.</p>
<p>The ratio of diameters will control the speed relationship as well as the torque.</p>
<p>Or a pair of pulleys but then they may be some slip between driven and driver, however the ratio of diameters still applies.</p>
<p>Or, if the locations of the two shafts are separate then a hydraulic motor and pump could be used.</p>
| 53327 | What device/system would be used to increase rpm and decrease torque of some shaft? |
2022-11-26T17:51:27.153 | <p>I often come across glass panels that are attached to the edge of the ceilings of public interior areas.</p>
<p>What purpose do they serve?</p>
<p><a href="https://i.stack.imgur.com/ciO8G.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ciO8G.png" alt="Glass panels attached to the edge of a ceiling in a shop" /></a>
<a href="https://i.stack.imgur.com/dePHb.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dePHb.png" alt="Glass panels attached to the edge of a ceiling in a train station" /></a></p>
| |civil-engineering|architecture| | <p>These are fire traps required by code. flames rising up from the initial fire don't get a chance to spread to neighboring areas.</p>
<p>The automated fire sprinklers will deploy in a smaller area thus eliminating the shortage of water.</p>
<p>Also, emergency partitions deploy in a smaller area, with less crowd to use the exits, and many other advantages!</p>
| 53330 | What use are those glass panels? |
2022-11-27T18:06:32.973 | <p>All descriptions I have been able to find mention some variation of "due to their geometrical design single point load cells are immune to load placement" with no explanation on how is it achieved. How does the geometry separate the perpendicular load measured from the moment caused by its' placement?</p>
<p><a href="https://i.stack.imgur.com/rUtBK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rUtBK.png" alt="Single point load cell illustration" /></a></p>
| |mechanical-engineering|solid-mechanics|moments|force-measurement| | <p>Notice the cut out in the middle? That turns it into a 4 bar linkage. Moments are reacted by opposing strains in the upper and lower arms. The strain gages are crosswired so that moments produce a cancelling effect whereas vertical loads add.</p>
| 53346 | How does a single point load cell eliminate the inluence of moment/load placement |
2022-11-28T19:49:22.863 | <p>I read that to make a dipped tantalum capacitor, they make tantalum powdered stick together with a binder and then sinter, which makes it porous.</p>
<p>But why do the pores form? It seems that if they are compressed then it would remove pores. Is it air bubbles?</p>
<p>Thanks</p>
| |chemical-engineering|manufacturing-engineering| | <p>Anything sintered has a sponge like structure with pores. That's what sintering is. For example, oil-impregnated bronze for use as self-lubricating bushings.</p>
<p>You're compressing the metal powder so that adjacent particles are applying pressure against each other, but they are not liquified. More like a putty, if anything I believe. They only deform slightly and fuse at the contact boundaries between particles. The empty space between the particles are still there.</p>
<p>Also, do not mislead yourself to believe it necessarily needs to look like a sponge to the naked eye. The particles can be powders after all when they are, the pores may be so small that it might just look like a solid mass. For example, sintered bronze bushings looks just like a solid piece of bronze.</p>
| 53361 | Why Sintered Tantalum Powder Forms a Sponge-Like Structure? |
2022-11-28T21:10:24.350 | <p>I read that for Ball-Grid Arrays, the BGA's act like a thermal dissipator because there are more channels, so it is more heat resistant as opposed to Thin-Quad Flat Pack which uses pins.</p>
<p>Is this because the BGA's use balls which have more surface area per high since it is a hemisphere as opposed to a thin rectangular prism? Thanks.</p>
| |thermodynamics|heat-transfer|thermal-conduction|computer-engineering|computer-hardware| | <p>The article seems to be referring to how, for the same dimensions, a QFP only has pins around the periphery whereas a BGA has pins throughout the entire underside of the package. BGAs, by design, have more pins per square inch packed into it than a QFP.</p>
<p>Since pins act as little heatsinks where heat developed on the die can travel outside the package to the PCB where it can be dissipated, a BGA supposedly can handle more heat than a QFP since not only are there more balls, there balls are also more evenly distributed across the package closer to the source of heat.</p>
<p>It's not that balls have more surface area than pins because they do not. For the same volume, a sphere has the minimum amount of surface area and a flat, thin pin has the most. But if you're going to take the total contact area of electrical connections at the surface of the body of the package (not the area of the pin or ball itself), the BGA will have more per area than a <em>regular</em> QFP, and they will be more evenly distributed with more of them closer to the heat generating die.</p>
<p>However, that article ignores the fact that many high power QFPs have giant thermal pads at the center of the package. It's difficult to imagine a BGA of the same size beating that in terms of heat dissipation, though the BGA still offers far more electrical connections.</p>
<p><a href="https://i.stack.imgur.com/GdZ4X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GdZ4X.png" alt="enter image description here" /></a></p>
<p><a href="https://www.digikey.ca/en/products/detail/texas-instruments/ADS1274IPAPT/1666706?utm_adgroup=Texas%20Instruments&utm_source=google&utm_medium=cpc&utm_campaign=PMax:%20Smart%20Shopping_Supplier_Texas%20Instruments&utm_term=&productid=1666706&gclid=Cj0KCQiA-JacBhC0ARIsAIxybyPQckeXwbW_LNtcVt-aoSRlm5kcGQjnaXTPo1q6mw8yXZmCs4dNm6QaAqATEALw_wcB" rel="nofollow noreferrer">https://www.digikey.ca/en/products/detail/texas-instruments/ADS1274IPAPT/1666706?utm_adgroup=Texas%20Instruments&utm_source=google&utm_medium=cpc&utm_campaign=PMax:%20Smart%20Shopping_Supplier_Texas%20Instruments&utm_term=&productid=1666706&gclid=Cj0KCQiA-JacBhC0ARIsAIxybyPQckeXwbW_LNtcVt-aoSRlm5kcGQjnaXTPo1q6mw8yXZmCs4dNm6QaAqATEALw_wcB</a></p>
| 53363 | Why Ball-Grid Arrays are More Heat Resistant |
2022-11-29T12:53:23.917 | <p>The equation for indicated power is ̇ = ̅ . Where L is the piston stroke (m), A is the piston area (m2), N is the number of mechanical cycles per cylinder per second, and ̅ is the indicated mean effective pressure (imep).</p>
<p>How would you calculate the indicated mean effective pressure ̅ if you were given Net indicated mean effective pressure and Pumping indicated mean effective pressure for a 4-stroke engine?</p>
<p>Thanks in advance</p>
| |thermodynamics| | <p>Noob, can't comment, so here's an answer.</p>
<p>It's not the terminology I'm used to, but apparently GrossIMEP is just the work done in the two important strokes (squeeze bang)of a 4 stroke, NetIMEP is GIMEP-PMEP, and BMEP=NIMEP-FMEP.</p>
| 53368 | Indicated mean effective pressure |
2022-11-30T15:14:08.887 | <p>Can someone tell me what influence an angled wall has on the flow over a triangular weir compared to a "normal" vertical one if the wall is angled against the upstream as in the picture?
<a href="https://i.stack.imgur.com/ElFRZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ElFRZ.png" alt="enter image description here" /></a></p>
| |open-channel-flow| | <p>I would think that at high flow rates the flow over an angled weir would tend to push more water over the weir. At lower rates there would be no difference. I imagine you owuld need to determine this emperically if this were the only way to build your weir.</p>
| 53385 | Tipped triangular weir |
2022-12-01T12:33:35.557 | <p>I know the title is quite a mouthful, but let me explain (also, I couldn't find anything on the subject, maybe because I'm using the wrong words).</p>
<p>I saw some time ago <a href="https://youtu.be/8JspshhWruQ?t=299" rel="nofollow noreferrer">a video about a variable suspension system</a> that used a variable viscosity fluid that changes its viscosity based on a electromagnetic field applied to it (it is just a fluid with ferromagnetic particles on it [<em>I think</em>]).</p>
<p>(The link redirects to the video in the exact moment its working mechanism is explained).</p>
<p>To me it seems quite an interesting system that could allow for solenoid/active valves with no moving parts that can instantaneously and progressively change the flow of its fluids by changing its viscosity until nothing can flow.</p>
<p>But I couldn't find any papers on the subject, this makes me wonder if such system would even work...</p>
<p>Of course, I don't think it would be able to withstand the absurd pressures on industrial hydraulic systems, but smaller (maybe hobbyist) things.</p>
| |fluid-mechanics|hydraulics|electromagnetism|hydrodynamics| | <p>Yes magnetorheological dampers (MRDs) are practical and available on cars today. I get about a zillion google hits on them</p>
<p>An overview is here</p>
<p><a href="https://link.springer.com/chapter/10.1007/978-1-84800-231-9_6" rel="nofollow noreferrer">https://link.springer.com/chapter/10.1007/978-1-84800-231-9_6</a></p>
<p>This looks interesting</p>
<p><a href="https://ieeexplore.ieee.org/document/9633028" rel="nofollow noreferrer">https://ieeexplore.ieee.org/document/9633028</a></p>
<p>Pressure wise automotive dampers don't work at much more than 1700 psi, which is low for a hydraulic system, but in the same ballpark (4000 psi). I can't think of any intrinsic reason why MRD wouldn't work at a the higher pressure.</p>
| 53397 | Any scientific research papers on electromagnetic active valves using variable viscosity electromagnetic fluids? |
2022-12-02T09:49:43.783 | <p>I was looking at two cars that friends have, and I was comparing the specs, and don't understand what is going on.</p>
<p>One is a 2017 7 seater Citroën C4, which has a 1.2L 3-cylinder petrol engine, producing 131hp (97kW). The car weighs 1372kg with a claimed top speed of 125mph</p>
<p>Another is a 2022 Yaris Hybrid, which has a 1.5L 4-cylinder petrol engine + electric, but only produces 100hp (74kW). The car is lighter at 1160kg, with a claimed top speed of 100mph.</p>
<p>What gives with the power? How can a 5 year newer engine that is 300cc bigger and has an extra cylinder produce LESS power than the older engine? Given high school/degree level combustion engine physics it should just get more powerful the bigger you make the engine? If the age was the other way around I'd just put it down to quality/design, but this feels like there is a missing variable here, but what is it? Is it something to do with fuel efficiency trade off?</p>
<p>If the idea is that we increase efficiency but reduce power, why does this result in a bigger engine. My understanding was that to maximise efficiency we want to compress the air-fuel mix as much as possible, so that it does maximum work when it expands? Or looking at the Carnot diagram, so that we get the biggest area we can - so does this mean the idea is to burn the same amount of fuel in a bigger cylinder will make a more efficient engine? Or is the idea to simply run an engine at lower revs, getting more power from each cycle?</p>
<p>Is there some significance in 3-cylinder vs 4-cylinder? Again, I would have assumed more means more power.</p>
| |thermodynamics|automotive-engineering| | <p>The main difference is the combustion cycle that is used.</p>
<p>Presumably the Citroen uses the <a href="https://en.wikipedia.org/wiki/Otto_cycle" rel="noreferrer">Otto cycle</a> which is the most common for intenal combustion engines. The main benefit is that for the same cylinder volume (L) produces more power.</p>
<p>On the other hand, the Hybrid Yaris uses the <a href="https://en.wikipedia.org/wiki/Atkinson_cycle" rel="noreferrer">Atkinson cycle</a>, (which is another 4 stroke cycle) that improves efficiency (i.e. manages to convert more thermal energy to energy transfer to the wheel - <strong>ultimately</strong>).</p>
<p>So basically, its a tradeoff between maximum power output (i.e. the acceleration of the car) and efficiency (mileage per gallon).</p>
<hr />
<p><a href="https://i.stack.imgur.com/adxml.png" rel="noreferrer"><img src="https://i.stack.imgur.com/adxml.png" alt="PV diagram of Otto and atkinson" /></a>
<strong>Figure: P-V diagram for Otto and atkinson combustion cycles <a href="https://car-moto.info/archives/909" rel="noreferrer">source:car-moteo.info</a></strong></p>
<p>In the diagram above, it is possible to get a feel of the difference. Essentially the</p>
<ul>
<li>the area enclosed by each polygon (black and red correspondingly) is the total energy that is retrieved as energy for motion (either rotational and/or electrical in the case of the Hybrid yaris).</li>
<li>The area below the bottom line is the energy wasted.</li>
</ul>
<h2>This diagram shows that a higher percentage of the power is retrieved</h2>
<p>The simplest example, to highlight that difference is race cars, which basically want to maximize output, and the efficiency optimization is a secondary priority.</p>
| 53408 | Why does a newer car have a bigger engine and produce less power? |
2022-12-04T12:47:43.537 | <p>I’ve got total gear ratio of 3.428 but it’s a double step up gearbox. So I have to sq root it get the ratio of the pairs which is 1.851. For the small cog I have to use a min of 18 teeth. So teeth x g = 33.326. How can I get a real gear ratio? I’m not sure on if I round the 33 down or up to 34. As I don’t have 33 on my Lewis form factor. For the assignment I have 30 and 34 on form factor. I’m not sure what to use. Thanks for any help.</p>
| |dynamics| | <p>if its for an assignment then, you shouldn't worry too much.
One solution is to use the 34 (which has a Lewis factor)</p>
<p>Another is to do an <strong>interpolation</strong> between the lewis factor for 30 and 34</p>
<p>Yet Another is to find out other tables/equation that allow you calculate the factor for 33. E.g. there is an online calculator at <a href="https://www.engineersedge.com/gears/lewis-factor.htm" rel="nofollow noreferrer">engineersedge</a></p>
| 53428 | Lewis form factor in between teeth |
2022-12-04T17:59:20.343 | <p>In many places, including but not limited to <a href="https://en.wikipedia.org/wiki/PID_controller#Ziegler%E2%80%93Nichols_method" rel="nofollow noreferrer">the Wikipedia page about PID controllers</a>, I see the following PID coefficients: <span class="math-container">$K_P = 0.6K_U$</span>, <span class="math-container">$K_I = 1.2 K_U / T_U$</span>, and <span class="math-container">$K_D = 0.075 K_UT_U$</span>. When were these constants first introduced? The sources point to the papers <em><a href="http://davidr.no/iiav3017/papers/Ziegler_Nichols_%201942.pdf" rel="nofollow noreferrer">Optimum Settings for Automatic Controllers</a></em> and <em>Rule-Based Autotuning Based on Frequency Domain Identification</em>, but I cannot find these values there.</p>
<p>Table I in the second mentioned paper introduces <span class="math-container">$K_C = 0.6K_U$</span>, <span class="math-container">$T_i = 0.5T_U$</span>, and <span class="math-container">$T_D = 0.125T_U$</span>, but I do not understand what those coefficients mean and how I go to <span class="math-container">$K_P$</span>, <span class="math-container">$K_I$</span>, and <span class="math-container">$K_D$</span> from there.</p>
| |control-theory|pid-control| | <p>For PID, Ziegler, Nichols (<a href="http://davidr.no/iiav3017/papers/Ziegler_Nichols_%201942.pdf" rel="nofollow noreferrer">Optimum Settings for Automatic Controllers</a>) cites:</p>
<p><span class="math-container">$K_p = 0.6K_u$</span>, <span class="math-container">$T_i = 0.5 T_u$</span>, and <span class="math-container">$T_d = 0.125 T_u$</span>.</p>
<p><a href="https://i.stack.imgur.com/pjyDBm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pjyDBm.png" alt="enter image description here" /></a></p>
<p>Which McCormack (<a href="https://www.researchgate.net/publication/3331995_Rule-based_autotuning_based_on_frequency_domain_identification" rel="nofollow noreferrer">Rule-based autotuning based on frequency domain identification</a>) and <a href="https://www.mstarlabs.com/control/znrule.html" rel="nofollow noreferrer">Ziegler-Nichols Tuning Rules for PID</a> agree with.</p>
<p><a href="https://i.stack.imgur.com/WfG0S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WfG0S.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/eynQp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eynQp.png" alt="enter image description here" /></a></p>
<p>OP cites <a href="https://en.wikipedia.org/wiki/Ziegler%E2%80%93Nichols_method" rel="nofollow noreferrer">Wikipedia PID coefficients</a>: <span class="math-container">$K_p = 0.6K_u$</span>, <span class="math-container">$K_i = 1.2 K_u / T_u$</span>, and <span class="math-container">$K_d = 0.075 K_u T_u$</span></p>
<p>The appropriate math:</p>
<p><span class="math-container">$$K_i = \frac{K_p}{T_i} = \frac{0.6 K_u}{0.5 T_u} = 1.2 \frac{K_u}{T_u}$$</span>
<span class="math-container">$$K_d = K_p T_d = (0.6 K_u) (0.125 T_u) = 0.075 K_u T_u$$</span></p>
<blockquote>
<p>Wikipedia: The Ziegler–Nichols tuning method is a heuristic method of tuning a PID controller. It was developed by John G. Ziegler and Nathaniel B. Nichols. It is performed by setting the I (integral) and D (derivative) gains to zero. The "P" (proportional) gain, <span class="math-container">$K_{p}$</span> is then increased (from zero) until it reaches the ultimate gain <span class="math-container">$K_{u}$</span>, at which the output of the control loop has stable and consistent oscillations. <span class="math-container">$K_{u}$</span> and the oscillation period <span class="math-container">$T_{u}$</span> are then used to set the P, I, and D gains depending on the type of controller used and behaviour desired:</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/83c7t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/83c7t.png" alt="enter image description here" /></a></p>
| 53431 | What is the original source of the common Ziegler-Nichols PID tuning coefficients? |
2022-12-07T11:24:05.950 | <p>Ok, don't know if such simple questions get asked here, but I have a plastic PP container and I'd like to drill a couple of 1/2 inch holes into it, both on the base and the wall. Was wondering how to do it and whether the thin material would be able to take it or crack.</p>
<p>Rectangle with slight slanting walls, something like this
<a href="https://i.stack.imgur.com/WE2Dt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WE2Dt.png" alt="enter image description here" /></a>
<a href="https://www.indiamart.com/proddetail/clear-reusable-container-13464112062.html" rel="nofollow noreferrer">https://www.indiamart.com/proddetail/clear-reusable-container-13464112062.html</a></p>
<p>Haven't gotten to measuring its thickness, haven't purchased it yet, just thinking whether its a good idea.</p>
| |drilling| | <p>Ok, so I found the most practical and easy solution. Take a metallic pipe of same OD, heat it up, run it through, file off the blunt edges.</p>
| 53455 | How to drill hole into thin plastic sheet? |
2022-12-07T17:42:52.060 | <p>Learning to use ANSYS Fluent as part of my computational engineering course I have previously modelled a 2D wing and been abled to XY plot for the pressure coefficient against x coordinate.</p>
<p>my struggle is when within a 3D case, how do you plot pressure coefficient against at various points across the span line?</p>
<p>the wing is an Onera M6 wing.</p>
<p><a href="https://i.stack.imgur.com/qjfGA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qjfGA.png" alt="enter image description here" /></a></p>
<p>where i want to plot 6 separate XY plots of C_p against X/m at each solid line, moving up the dotted span line.</p>
| |ansys|cfd|fluent|airfoils| | <p>For Anyone that may come across this post, the answer is: to create an iso surface using the wall as the 'from surface' and setting it to Mesh -> Z-coordinate and entering the desired value of Z.</p>
<p>this creates a surface along the span line, of which you select when you plot XY chart of C_p against X.</p>
<p>repeat the surface creation for as many points across the span as you desire</p>
| 53463 | Display Pressure coefficient for multiple points at across the span line of an airfoil in fluent |
2022-12-08T08:43:08.210 | <p>Can someone help me calculate the hydrostatic pressure on an immersed inclined body like the one in the figure?</p>
<p><a href="https://i.stack.imgur.com/hjFbf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hjFbf.png" alt="enter image description here" /></a></p>
<p>I know the force acting on the triangle is changing with height thus I can use an integral. Something like</p>
<p><span class="math-container">$F = \int_0^{h1} \rho \cdot g \cdot y \cdot \frac{L}{sin(\alpha)} dy$</span></p>
<p>Where L is the length of the triangle on the screen. However, I don't know how to include the height above the triangle (h-h1). Can I just add this into the integral as this is constant?</p>
| |fluid| | <p><a href="https://i.stack.imgur.com/hruiX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hruiX.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$p$</span> represents the hydrostatic pressure at the boundary point. The force per unit width is simply <span class="math-container">$\dfrac{(p_1 + p_0)*L}{2}$</span>. Wish this helps.</p>
| 53469 | Hydrostatic pressure on submerged surface |
2022-12-09T14:48:42.333 | <p>I have a small tube with outer diameter (OD) <span class="math-container">$R_1$</span> mm and inner diameter (ID) <span class="math-container">$r_1$</span> mm. The tube is bent using a moment <span class="math-container">$M_1$</span> to create an angle <span class="math-container">$\theta_1$</span> with its plane of origin. The stress can then be calculated as <span class="math-container">$\sigma_1 = M_1/W_b$</span>, where <span class="math-container">$W_b$</span> is the bend resistance <span class="math-container">$W_b = I/R$</span>, where <span class="math-container">$I$</span> is the <em>second moment of area</em> and <span class="math-container">$r$</span> the distance from the centre axis to the edge of the tube.</p>
<p>Now consider that I have another tube with another OD and ID, <span class="math-container">$R_2$</span> and <span class="math-container">$r_2$</span> respectively. Is it possible to get the same amount of stress (or strain) for the same amount of bend/deformation? Considering that the two tubes are made out of the same material.</p>
<p>Mathematically speaking:
<span class="math-container">$\sigma_1 = \sigma_2 \rightarrow \frac{M_1}{W_{b1}}=\frac{M_2}{W_{b2}}\tag{1}$</span></p>
<p>Since <span class="math-container">$W_b = I/R$</span> and <span class="math-container">$I=\frac{\pi(R^4-r^4)}{4}$</span>, equation (1) can be written as:</p>
<p><span class="math-container">$\frac{M_1R_1}{\frac{\pi(R_1^4-r_1^4)}{4}}= \frac{M_2R_2}{\frac{\pi(R_2^4-r_2^4)}{4}} \rightarrow \frac{4M_1R_1}{R_1^4-r_1^4} = \frac{4M_2R_2}{R_2^4-r_2^4} \rightarrow M_1R_1(R_2^4-r_2^4) = M_2R_2(R_1^4-r_1^4)\tag{2}$</span></p>
<p>Using Euler-Bernoulli beam theory, the angle <span class="math-container">$\theta$</span> can be expressed as:</p>
<p><span class="math-container">$\theta = \frac{Ml}{EI}$</span> where <span class="math-container">$l$</span> is the lenght of the tube and <span class="math-container">$E$</span> is Young's modulus. Same angle results in:</p>
<p><span class="math-container">$\theta_1 =\theta_2 \rightarrow \frac{M_1l}{EI_1} = \frac{M_2l}{EI_2} \tag{3}$</span></p>
<p>Since the lenght of the tube is kept the same and the material is the same, <span class="math-container">$E$</span> and <span class="math-container">$l$</span> are same and can be removed to yield:</p>
<p><span class="math-container">$\frac{M_1}{I_1} = \frac{M_2}{I_2} \rightarrow \frac{M_1}{\frac{\pi(R_1^4-r_1^4)}{4}} = \frac{M_2}{\frac{\pi(R_2^4-r_2^4)}{4}} \rightarrow \frac{M_1}{R_1^4-r_1^4} = \frac{M_2}{R_2^4-r_2^4} \tag{4}$</span></p>
<p>Solving (4) for <span class="math-container">$M_2$</span> yields:</p>
<p><span class="math-container">$M_2 = M_1\frac{R_2^4-r_2^4}{R_1^4-r_1^4}\tag{5}$</span></p>
<p>If the solution from equation (5) is inserted in (2) then the following result is reached:</p>
<p><span class="math-container">$M_1R_1(R_2^4-r_2^4) = M_1\frac{R_2^4-r_2^4}{R_1^4-r_1^4}R_2(R_1^4-r_1^4) \implies (R_2^4-r_2^4)R_1 = (R_2^4-r_2^4)R_2 \tag{6}$</span></p>
<p>The result in (6) basically says that <span class="math-container">$R_1=R_2$</span>, which implies that there is only <em>one</em> unique combination of cross section and stress which yields a specific deformation. Can this be true? This means that there is only one dimension of tube which will yield a specific stress for a certain bend.</p>
| |mechanical-engineering|stresses|solid-mechanics| | <p>That is right. For the same material and the same deformation, the stress will only depend on distance distance from the center. So if you want to have same maximal stresses in that situation, the tubes have to have the same outside diameters. And when outside diameters are the same and the deformation is the same, the tubes have to have also same inside diameters.</p>
| 53489 | Is it possible to achieve the same amount of stress for an equal amount of deformation but two different cross sections? |
2022-12-09T22:32:03.517 | <p>What does this dimension (see image), and particularly the divide symbol in it, mean please? I'm not a mechanical engineer and couldn't find the answer by searching for terms with the word "divide" in them.</p>
<p>If it makes any difference, I guess the drawing was done by a German engineer, as it's one of the dimensions given for a worktop cut-out for a Bosch gas hob.</p>
<p>Obviously it doesn't actually mean "divide" as that would result in <1 mm. Could it mean a range? So that this dimension of the cut-out can be between 480 and 492 mm after the tolerance is included?</p>
<p><a href="https://i.stack.imgur.com/KYRCZ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/KYRCZ.png" alt="Dimension with "divide" symbol" /></a></p>
| |mechanical-engineering|tolerance| | <p>The "<a href="https://en.wikipedia.org/wiki/Division_sign" rel="nofollow noreferrer">division sign</a>" is a form of <a href="https://en.wikipedia.org/wiki/Obelus#In_mathematics" rel="nofollow noreferrer">obelus</a> that in Scandinavia sometimes meant subtraction, not division, and in many European countries (e.g. Russia, Poland, Italy) it is used in engineering to denote a range of values.</p>
<p>Germany is not specifically mentioned, but given the context you are almost certainly correct that in this case it means the range of possible values is from 480 to 490 (plus the tolerance).</p>
<p>The use of the "division sign" is specifically recommended against in ISO math standards exactly because it has different meanings in non-anglophone countries.</p>
| 53496 | What does a divide symbol on a drawing mean? |
2022-12-11T03:06:45.687 | <p>So I have to create a control system simulation from scratch for the following scenario. For a bicycle-like system, there is a motor attached directly to the rear wheel. From here, the goal of the control system is to use the motor to maintain a constant speed, rejecting a small disturbance force pushing against the system from the front.</p>
<p>With the context out of the way, I've been struggling to create the transfer function that allows me to get the linear velocity from the rest of the system. So far, I've tried the process vaguely outlined in the picture below, where I've tried to rearrange a DC motor system to output torque instead of angular position, and then scaled that torque into a linear force before applying that force to the mass and air resistance to get the speed output.<a href="https://i.stack.imgur.com/hjYly.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hjYly.png" alt="Handwritten TF block diagram" /></a></p>
<blockquote>
<p>(R and L are the internal resistance and inductance of the motor,
respectively, Rw is the wheel radius, m is the total mass of the
system J is the inertia of the rear wheel, b is the air resistance
drag, Ke and Kt are motor coefficients and K is the constant for a
tiny spring-like component. 1/(Js^2+K) is supposed to be the inertial
load of the motor, and although I'm starting to suspect the spring
component isn't necessary the system acts worse without it.)</p>
</blockquote>
<p>The issue is that, when I try to simulate this open-loop system in Matlab, I get a rise/settling time that's almost 100 times larger than I'd consider acceptable for this kind of system. The poles and zeros aren't exactly confidence-inspiring either, with unstable zeros and poles that hang a fraction of a unit from the imaginary axes. None of my experimenting with controllers, closing the loop or tweaking my imperfect input variables has done anything to change the overall structure of the output.<a href="https://i.stack.imgur.com/TGZYu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TGZYu.png" alt="Matlab plots" /></a></p>
<p>So is there an easier or well established method to create the kind of transfer function I'm looking for? I've been hammering away at this for long enough that I'm starting to think the process I tried isn't fixable.</p>
<p>With everything said, I apologize if my terminology is incorrect or I'm missing something super simple. Control systems are not my forte, and I've only started with them a couple of months ago. Thank you so much for your time!</p>
<p>Edit: Here are the plots requested. These definitely don't look right.
<a href="https://i.stack.imgur.com/FuXOz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FuXOz.png" alt="enter image description here" /></a>
Also, here's a diagram that represents what I'm trying to refer to in my second comment.
<a href="https://i.stack.imgur.com/XkvB2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XkvB2.png" alt="enter image description here" /></a></p>
| |electrical-engineering|control-engineering|dynamics|modeling| | <p>Too long for a comment.</p>
<ol>
<li>As you suspected, a spring component in the wheel transfer function is not justified (while a viscous force <em>can</em> be justified). If you cannot identify a physical mechanism for a restoring force / torque (i.e. spring force), it will have to be removed.</li>
<li>From the orange rectangle, it can be seen that the angular position and angular velocity for a fixed torque depends on the wheel inertia only. One would assume that for a bicycle like device, the over all mass of the system would also be involved. Why would a larger <span class="math-container">$m$</span> of the vehicle <em>not</em> affect the motor angular position ?</li>
<li>Continuing the above point, if the wheel does <strong>not</strong> slip (which might be a reasonable assumption to make for many systems), the relation between motor angular position and angular velocity and linear distance and linear velocity are a fixed ratio. One would model it as a system with <strong>gears</strong> using <em>reflected</em> mass or inertia etc. Are you familiar with modeling systems involving gears ?</li>
<li>Can't comment on the rise time etc. as it depends on the values used for the variables also.</li>
<li>Can you plot the angular position and velocity of the wheel also along with the linear velocity ? Please clearly label the plots.</li>
</ol>
<h2>edit</h2>
<p>Under assumptions like no-slip between tyre and road, no <em>flexing</em> of the body or drive train, we can write equations as follows.</p>
<p><a href="https://i.stack.imgur.com/Xp1Lq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xp1Lq.png" alt="tyre and symbols" /></a></p>
<p><span class="math-container">$$
x = R_w \cdot \theta \qquad \text{no slipping condition}\\
\implies \color{blue}{\frac{d^2 x}{dt^2} = R_w \frac{d^2 \theta}{dt^2}}\\
\frac{d^2 x}{dt^2} = f/m\\
\frac{d^2 \theta}{dt^2} = \sum{T}/J =
\frac{T_{\text{electrical}}}{J}
\color{red}{\pm} \frac{f\cdot R_w}{J}\\
\implies
\frac{d^2 \theta}{dt^2} =
\frac{T_{\text{electrical}}}{J}
\color{red}{\pm} \frac{m \cdot R_w \cdot R_w}{J}\cdot \frac{d^2 \theta}{dt^2}\\
\implies
\frac{d^2 \theta}{dt^2} \color{red}{\mp} \frac{m \cdot R_w \cdot R_w}{J}\cdot \frac{d^2 \theta}{dt^2}
=
\frac{T_{\text{electrical}}}{J}\\
$$</span></p>
<p>Continue rearranging the equation till you get <span class="math-container">$\frac{d^2 \theta}{dt^2}$</span> on LHS. The complicated stuff on the RHS denominator should be the effective inertia which includes effect of mass.</p>
<p>You might have to re derive along similar lines to be able to include viscous forces or any other forces into the equation.</p>
<h2>edit 2</h2>
<p>Continuing the rearrangement,
<span class="math-container">$$
\frac{d^2 \theta}{dt^2} =
\frac{T_{\text{electrical}}}{J + m \cdot R^2_w}\\
\implies J_{\mathrm{eff}} = J + m \cdot R^2_w \qquad \text{looks quite reasonable}
$$</span></p>
<p>The instantaneous centre of rotation is the point of contact between the wheel and the road. The inertia about this point is given by the parallel axis theorem. The effective inertia derived above also looks to be the same expression.</p>
| 53503 | Transfer function to get linear velocity from a dc motor system |
2022-12-11T08:55:30.237 | <p>So I am thinking of purchasing a <a href="https://geekguider.com/product/agriansh-12v-dc-water-pump-battery-sprayer-motor-double-high-performance-12v-dc-220-psi/" rel="nofollow noreferrer">double motor pump</a> for higher flow rates. This pump is available in two pressure ratings, 150PSI and 220PSI. What does it mean? I read that it means how much resistance it can face. I am not expecting to provide any resistance while it is on except the resistance in piping walls. Does it matter whether I buy 150PSI or 220PSI? I just need a pump that would fill my 1 Ltr bottle in less than a 10 seconds.</p>
<p>It also says cutoff pressure is 10.3bar. What does that mean?</p>
| |pressure|pumps| | <p>Usually meaning is that if you were to pressurize a sealed container, it would reach that pressure (as long as the seal does not break). Similarly a maximum flow rate is how much would flow if there were no resistance (pretty much never).</p>
<p>Where there is flow, there is resistance. The faster you need something to flow (against the walls of something), the greater this resistance. It is also affected by other factors like the viscosity of the fluid and the diameter of the hose/pipe you are flowing it through. In other words if it doesn't get the rate you need, make changes to things downstream from the pump. The literal bottleneck may be that of your target container. It should be reasonably easy to test the resistance for your setup using gravity instead of a pump, and a stopwatch. How high (compared to where you'd connect the pump) does your source fed through your pipe need to be to get your 1L filled in 10s?</p>
<p>Whether you need the higher pressure pump will depend on the pressure vs flow curve of the pump - usually it's more than safe enough to assume that you can draw a line from 0 flow and max pressure to 0 head and max flow. This is because of the shape usually being convex to the 0,0 point. Take your head from the height experiment (0.0361 psi per inch water), and if your flow on that line is greater than 6L/min you can consider it OK.</p>
| 53509 | What does pressure rating of pumps mean? |
2022-12-11T15:54:31.683 | <p>Here this is a Load-Displacement curve from a compression test I did earlier with my colleges, Below is the resulted stress strain curve from the test from the lab computer why did the curve not start from zero is this just a zero error from the testing equipment or is there something else behind it?</p>
<p><a href="https://i.stack.imgur.com/aDbQs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aDbQs.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|materials|stresses|material-science| | <h2>In general</h2>
<p>In most cases, the norm is that you don't get zero starting values.</p>
<p>Normally the test procedure has the following steps:</p>
<ol>
<li>move the testing machine crosshead to a place that it can accommodate the specimen</li>
<li>place the specimen between the jaws</li>
<li>secure the specimen either by : a) tightening it or b) in the case of self locking jaws applying small tension</li>
<li>perform zeroing for displacement and load</li>
<li>start the test</li>
</ol>
<p>Normally in you need to make sure that both displacement is set to zero and that there is no inherent tension while you tighten the specimen to the testing jig (this is for not damaging the specimen). During that stage, normally you set to zero the displacement</p>
<p>However, because in most cases that (especially displacement) can be zeroed during the analysis stage, that can be easily forgotten/omitted.</p>
<p>(Removing the displacement is as easy are subtracting the first value from all the rest)</p>
<hr />
<h2>Particular case</h2>
<p>Sometimes the testing machines have an autozeroing feature. If that is the case, and the fixture is enabled, then you would expect that there might be a problem with the testing.</p>
<p>However, for this particular case seeing the shape of the curve (I don't know what type of material it is but I will guess some type of aluminium or other metallic alloy), I would guess that the specimen was not tightened properly, and there was some slippage in the grips between 1.5 and 1.7 mm.</p>
| 53513 | Why does the stress strain curve sometimes not start from zero? |
2022-12-11T22:43:08.057 | <p>Why does this barbell use two half washers?</p>
<p>For example: see this <a href="https://powerliftingtechnique.com/barbell-sleeve-replacement/" rel="nofollow noreferrer">link</a></p>
<p>Navigate to the “Taking Apart A Barbell Sleeve” section.
<a href="https://i.stack.imgur.com/6Z0vz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Z0vz.png" alt="enter image description here" /></a></p>
<p>They use a washer, two half washers, a shim washer, another washer and then retaining rings.</p>
<p>The split washer is not the traditional split washer that acts as a positive locking mechanism.</p>
<p>It looks like <a href="https://tippmannarms.com/m4-22-barrel-split-washer/" rel="nofollow noreferrer">this</a>.</p>
<p>My question is this: why did the designers use this type of washer? I do not see the advantage over a normal washer.</p>
| |mechanical-engineering| | <p>I'm looking at a video and I don't have a completely clear view:</p>
<p><a href="https://i.stack.imgur.com/8K2h2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8K2h2.png" alt="enter image description here" /></a></p>
<p><a href="https://youtu.be/zUFIJQqsoQE?t=90" rel="nofollow noreferrer">https://youtu.be/zUFIJQqsoQE?t=90</a></p>
<p>But from what I can see, I think those half-washers are primary load bearing components that hold everything on the shaft in the axial direction, not the retaining rings. That would make sense since retaining rings aren't very strong. Retaining rings are meant to be springy pieces of metal that can be removed which means they are limited in thickness.</p>
<p>From the glimpses I can see, there is a circular groove cut into the central shaft and the half washers sit into that groove from opposite sides. Sliding the sleeve over them keeps them in place by preventing motion in the radial direction. The sleeve must have a lip or shelf at the bottom of the hole inside it that hits the half washers if anything tries to drag the sleeve off the barbell.</p>
<p>Obviously, a full washer that must slide on from the end can't sit snugly in a groove cut into the shaft.</p>
<p>That means they're the most important piece out of everything you listed...I think? But then what stops the sleeve from sliding too deep onto the barbell? It seems like it would need to be the retaining rings but it doesn't seem like they would be strong enough, and obviously the sleeve itself slides back by design when you remove the retaining rings so you can access the half washers.</p>
| 53517 | Why does this barbell use two split barrel washers? |
2022-12-12T07:44:59.017 | <p>In Discrete-time Control Engineering by Katsuhiko Ogata, it is stated that <em><strong>"A necessary and sufficient condition for complete state controllability is that no cancellation occurs in the pulse transfer function. If cancellation occurs, the system cannot be controlled in the direction of the cancelled mode."</strong></em></p>
<p><em>The pulse transfer function relates a sequence of samples at the output of a system to the sequence producing it.</em> In other words, it is the discrete-time equivalent of the transfer function in the continuous domain.</p>
<p>It is stated that, if we have a transfer function in which pole-zero cancellation occurs, converting the same to state space representation, the system can not be fully controllable and observable (at least one of them will be lost). <em><strong>It is also stated that the system cannot be controlled in the direction of the cancelled mode.</strong></em></p>
<p>What does controlling in a direction mean?</p>
<p>Also, how is direction defined in state space?</p>
| |electrical-engineering|control-engineering|control-theory|transfer-function| | <p>I'll try to explain using examples.</p>
<h2>Basic setup</h2>
<p>Discrete time system in state space.</p>
<p><span class="math-container">$$
\begin{align}
x_{n+1} &= A x_n + B u_n\\
x_1 &= A x_0 + B u_0\\
x_2 &= A^2 x_0 + AB u_0 + Bu_1\\
x_3 &= A^3 x_0 + A^2B u_0 + AB u_1 + Bu_2\\
x_n &= A^n x_0 + \sum_k^{n-1}{A^{n-k-1}B u_k}
\end{align}
$$</span><a href="https://ocw.mit.edu/courses/6-241j-dynamic-systems-and-control-spring-2011/bab3ecdb64baf945acd206d8f370167f_MIT6_241JS11_chap10.pdf" rel="nofollow noreferrer">[1]</a></p>
<h2>Example 1</h2>
<p><span class="math-container">$$
x_{n+1} =
\begin{bmatrix}
0.9 & 0\\
0 & 0.8
\end{bmatrix}
x_n +
\begin{bmatrix}
1\\
0
\end{bmatrix}
u_n
$$</span></p>
<p>For simplicity, assume zero initial condition. Then
<span class="math-container">$$
\begin{align}
x_1 &=
\begin{bmatrix}
u_1\\
0
\end{bmatrix}\\
x_2 &=
\begin{bmatrix}
0.9 & 0\\
0 & 0.8
\end{bmatrix}
\begin{bmatrix}
u_1\\
0
\end{bmatrix}
+
\begin{bmatrix}
u_2\\
0
\end{bmatrix}
&&=
\begin{bmatrix}
0.9 u_1 + u_2\\
0
\end{bmatrix}\\
x_n
&=
\begin{bmatrix}
\sum{0.9^{n-k-1}u_k}\\
0
\end{bmatrix}
&&=
\left(\sum{0.9^{n-k-1}u_k}\right)
\begin{bmatrix}
1\\
0
\end{bmatrix}
+ 0
\begin{bmatrix}
0\\
1
\end{bmatrix}
\end{align}
$$</span></p>
<p>It can be seen that, whatever we chose as the input sequence <span class="math-container">$u_n$</span>, we can control the state only in the <span class="math-container">$[1,\quad 0]^T$</span> direction; i.e. we can influence only the first state variable and not the second one. The direction <span class="math-container">$[0,\quad 1]^T$</span> is not controllable.</p>
<p>Also note that, to reach this conclusion, we utilized the values of the column vectors <span class="math-container">$B,\ AB,\ A^2B, \dots$</span>. This is why the controllability matrix is made from those column matrices.</p>
<h2>Example 2</h2>
<p><span class="math-container">$$
x_{n+1} =
\begin{bmatrix}
0.9 & 0\\
0 & 0.9
\end{bmatrix}
x_n +
\begin{bmatrix}
1\\
1
\end{bmatrix}
u_n
$$</span>
Doing the same procedure as above,
<span class="math-container">$$
\begin{align}
x_1 &=
\begin{bmatrix}
u_1\\
u_1
\end{bmatrix}\\
x_2 &=
\begin{bmatrix}
0.9 & 0\\
0 & 0.9
\end{bmatrix}
\begin{bmatrix}
u_1\\
u_1
\end{bmatrix}
+
\begin{bmatrix}
u_2\\
u_2
\end{bmatrix}
&&=
\left(0.9 u_1 + u_2\right)
\begin{bmatrix}
1\\
1
\end{bmatrix}\\
x_n
&=
\begin{bmatrix}
\sum{0.9^{n-k-1}u_k}\\
\sum{0.9^{n-k-1}u_k}
\end{bmatrix}
&&=
\left(\sum{0.9^{n-k-1}u_k}\right)
\begin{bmatrix}
1\\
1
\end{bmatrix}
+ 0
\begin{bmatrix}
1\\
-1
\end{bmatrix}
\end{align}
$$</span>
It can be seen that, whatever we chose as the input sequence <span class="math-container">$u_n$</span>, we can control the state only in the <span class="math-container">$[1,\quad 1]^T$</span> direction; i.e. we can influence the first state variable and second state variable by the same amount. We cannot increase first state variable while decreasing the second state variable; i.e. The direction <span class="math-container">$[1,\quad -1]^T$</span> is not controllable.</p>
<h2>Example 3</h2>
<p><span class="math-container">$$
x_{n+1} =
\begin{bmatrix}
0.9 & 0\\
1 & 0.8
\end{bmatrix}
x_n +
\begin{bmatrix}
1\\
0
\end{bmatrix}
u_n
$$</span>
Doing the same procedure as above,
<span class="math-container">$$
\begin{align}
x_1 &=
\begin{bmatrix}
u_1\\
0
\end{bmatrix}\\
x_2 &=
\begin{bmatrix}
0.9 & 0\\
1 & 0.8
\end{bmatrix}
\begin{bmatrix}
u_1\\
0
\end{bmatrix}
+
\begin{bmatrix}
u_2\\
0
\end{bmatrix}
&&=
u_1
\begin{bmatrix}
0.9\\
1
\end{bmatrix}
+ u_2
\begin{bmatrix}
1\\
0
\end{bmatrix}\\
&=
u_1
\begin{bmatrix}
0\\
1
\end{bmatrix}
+ \left(u_2 + 0.9 u_1\right)
\begin{bmatrix}
1\\
0
\end{bmatrix}\\
\end{align}
$$</span></p>
<p>Here we can see that by suitably choosing <span class="math-container">$u_1$</span> and <span class="math-container">$u_2$</span>, we can change the state of the system in the direction <span class="math-container">$[1,\quad 0]^T$</span> as well as <span class="math-container">$[0,\quad 1]^T$</span> and in any direction which is a linear combination of these. Since this is a two dimensional state space, and we have two independent directions, their linear combination can actually generate any and all directions in the state space. So this system is fully controllable.</p>
<p>One more thing to note here is that, by looking at the original system equation, we see that the input directly influences only the first state variable due to <span class="math-container">$B$</span> matrix being of the form <span class="math-container">$[1,\quad 0]^T$</span>. But by looking at the <em>second</em> time step, we realise that input can influence the second state variable also (but indirectly). This is why this system is fully controllable; the input signal can directly and indirectly influence all the state variables.</p>
<h2>How to identify the directions ?</h2>
<p>Left as an exercise to the reader. Look at the eigen vectors of the <span class="math-container">$A$</span> matrices in the examples above.</p>
<h2>What if the system being analysed is different from the examples?</h2>
<p>The examples above were chosen to be easy to explain. But, we can transform <em>all</em> systems to the format of one of the above examples (diagonal or nearly diagonal) by using eigen decomposition of the <span class="math-container">$A$</span> matrix. This transformation is also called change of variable or change of coordinates.</p>
<h2>Transfer functions, cancellation of modes and controllability</h2>
<p>Continuing from the example 1, and assuming that its <span class="math-container">$C$</span> matrix is <span class="math-container">$[1,\quad 1]$</span>,
<span class="math-container">$$
\begin{align}
C(zI-A)^{-1}B &=
\begin{pmatrix}
1 & 1
\end{pmatrix}
\frac{1}{(z-0.8)(z-0.9)}
\begin{pmatrix}
z-0.8 & 0\\
0 & z-0.9
\end{pmatrix}
\begin{pmatrix}
1\\
0
\end{pmatrix}\\
&=
\frac{(z-0.8)}{(z-0.8)(z-0.9)}\\
&=
\frac{1}{(z-0.9)}\\
\end{align}
$$</span></p>
<p>We notice that the modes correspond to the <em>poles</em> of the system and that one of the poles (the one at <span class="math-container">$z=0.8$</span>) cancelled out. This is the mode that turns out to be uncontrollable. <strong>But</strong>, just by looking at the transfer function, we can't tell the direction in which the mode <span class="math-container">$z=0.9$</span> acts along, and the direction in which mode <span class="math-container">$z=0.8$</span> acts along.</p>
| 53521 | Controlling a system in a particular direction |
2022-12-12T13:06:04.163 | <p>As shown in <a href="https://youtu.be/XnYO4TnpTCo" rel="nofollow noreferrer">this video</a>, the simulated knee cap creates a fulcrum that allows for a better mechanical advantage for the muscles to pull the leg.</p>
<p>So I was wondering if the <strong>reverse</strong> would also work, or if it wouldn't change much at all.</p>
<p>For example, in this illustration I made on paint, it is shown a mechanical lever <strong>type 3</strong> with <strong>20 cm of length</strong>, where the load is on a <strong>20 cm</strong> distance from the <strong>fulcrum</strong> and the effort is at <strong>10 cm</strong> distance from the <strong>fulcrum</strong> and the effort is being done by a rope pulled by a motor.</p>
<p><a href="https://i.stack.imgur.com/bmFl8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bmFl8.png" alt="Illustration of a reverse knee lever" /></a>
(I know the illustration part is drawn like it is counting 20cm from the effort point, but it is because I'm bad at drawing)</p>
<p>Accordingly to <a href="https://www.translatorscafe.com/unit-converter/en-US/calculator/lever-mechanical-advantage/" rel="nofollow noreferrer">this online mechanical lever calculator</a>, inputing these numbers would give me 0.5 of mechanical advantage, so in order to lift 1 kg, I would need to input twice the value in order to achieve equilibrium.</p>
<p>However, if I insert a "reverse knee cap" over the effort at the same 10 cm distance and attached the effort rope directly to the point of the load arm, as shown in the illustration, would the mechanical advantage stay the same, or it would change?</p>
| |mechanical-engineering|mechanisms|mechanical| | <p>Well, I tested it on the program "Interactive Physics", and it seems like it doesn't make much difference.</p>
<p>I don't know how to explain it very well, but the pulley that acts as the "inverse knee cap" applies force the same way if the effort arm was directly connected to the 10cm distance.</p>
<p><a href="https://i.stack.imgur.com/YdId3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YdId3.png" alt="Screenshot of the Interactive Physics program showing the results" /></a></p>
<p>For example, i imitated the kneecap joint in this attempt at equal distances of 20 cm from the fulcrum and it kept everything under equilibrium:</p>
<p><a href="https://i.stack.imgur.com/zOqPU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zOqPU.png" alt="Screenshot of the interactive physics program" /></a></p>
<p>Even if I try to make the pulley to be really close to the load:</p>
<p><a href="https://i.stack.imgur.com/92qYs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/92qYs.png" alt="Screenshot of interactive physics" /></a></p>
| 53523 | The knee gives a mechanical advantage point for the leg, would a "reverse knee lever" also work? |
2022-12-12T22:57:31.587 | <p>I'd like to produce prototype parts by molding a silicon sheet (0.6mm thickness) to a shape that looks roughly like below:</p>
<p><a href="https://i.stack.imgur.com/sd25X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sd25X.png" alt="Hollow silicon part" /></a></p>
<p>If split in two, this part looks like this:</p>
<p><a href="https://i.stack.imgur.com/sD00g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sD00g.png" alt="enter image description here" /></a></p>
<p>I have attempted to print this part with a couple of 3d printers using flex resins, but without much success at maintaining a decent accuracy on the thickness. I have also tried to make 3d-printed molds for liquid silicone but getting working parts out of these molds proved a real challenge (I suck a degasing, and these parts have really thin walls).</p>
<p>So, I am looking in buying a flat 0.6mm silicone sheet with fairly accurate constant thickness and then mold it with a vacuum bag under controlled temperature. A couple of questions before I go down that path:</p>
<ol>
<li><p>is the general idea sound ? (shaping an existing flat sheet of silicone). I am asking because I can't find any youtube video of people who produce parts like this and everyone knows that if it does not exist on youtube, it does not exist. i.e., I can't make myself believe that no one has done that or documented that before.</p>
</li>
<li><p>Is temperature needed ? (would a vacuum bag be sufficient to mold a sheet ?) If not, how much temperature (compared to the specified max temperature of the silicon sheet) ?</p>
</li>
<li><p>Should I try an outer mold or an inner mold ?</p>
</li>
</ol>
| |mechanical-engineering|materials|prototyping|molding| | <p>Based on the comments by @DKNguyen, the answer is that the general idea is unsound.</p>
<p>The reason is that silicon is a thermoset plastic so, it is not possible to thermoform it.</p>
<p>Now, for the record, I looked into alternatives:</p>
<ol>
<li><p>thermoforming thin sheets of another type of thermoplastic plastic. This did not work because the sheets ended being sheared during forming.</p>
</li>
<li><p>Changing the shape to make it more friendly to thermoforming so that shearing is minimized. This did not work because, either the sheet still sheared or the result was not functional anymore (the shape could not do what it was supposed to do)</p>
</li>
</ol>
<p>In the end, I:</p>
<ol>
<li>redesigned the part so that ot could be molded with a two-part mold</li>
<li>built 2-part molds from 3d-printed resin</li>
<li>built an aluminium overmold to hold the resin mold</li>
<li>did plastic injection of with a desktop plastic injection <a href="https://holimaker.fr/holipress/" rel="nofollow noreferrer">Holipress</a> and <a href="https://technologieservices.fr/ts_fr/456442.html" rel="nofollow noreferrer">flex pellets</a></li>
</ol>
<p>The final product works and, in can anyone is interested, source code can be found <a href="https://github.com/mathieu-lacage/goggles" rel="nofollow noreferrer">here</a></p>
| 53528 | How to mold a silicone sheet? |
2022-12-15T22:10:15.173 | <p>I am designing a device that requires a gas (air) valve that takes the same volume of gas from two different input ports IP1 and IP2.</p>
<p>The output port OP will have lower pressure than both of the inputs, however, the pressures at the inputs are not not guaranteed to be constant and the same.</p>
<p>What is the simplest design that can guarantee that roughly the same volume of gas is been taken from both inputs despite the pressure fluctuations?</p>
<p>Additional criteria, the valve must not use any electronics or external power. It has to be a purely mechanical design.</p>
<p>Follow up questions:</p>
<ul>
<li>Can this be designed without using moving parts?</li>
<li>Can we modify the design such that the gas volume taken from IP1 and IP2 is a different fixed ration e.g. 1:3</li>
<li>How would you design it different if you were to guarantee that gas taken from IP1 is at least 1/3 of the total volume in the output?</li>
</ul>
| |mechanical-engineering|fluid-mechanics|gas|valves|pneumatic| | <p>Such a thing exists for hydraulic system. It is not simple or cheap, but passive.</p>
<p>You need two positive displacement machines for the airflows, like sliding vane or rotary lobe pumps. At least the first also exist as pneumatic actutators, I thinkt this is how you source the parts. Then you link the shafts so that airflows are in the correct ratio. So if both are the same size and you want a ration of 1:2 you would need a linkage so the the rpm are in this same ratio. If the ratio is 1:1 ist's simply the same shaft.</p>
<p>The idea is that the pressure loss across the whole system drives both machines, with the linkage they are forced to turn in the correct ratio. This does no regulate pressure! There will be some pressure loss, depending on friction in the system etc.</p>
| 53555 | Passive gas mixing valve that takes fixed ratio of gas from two different inputs |
2022-12-16T01:57:53.393 | <p>I'm trying to build a sort of 3 dimensional robotic arm that mimics the movement of a virtual robotic arm inside Unreal Engine. I'm using Arduino Mega for the hardware side of the arm.</p>
<p>I have a MPU6050 gyroscope on the robotic arm and I wrote a simple code to mimic the virtual arm's rotation in the real world.</p>
<p>For example, when the virtual arm's yaw angle is at 0, the hardware arm's angle is also 0, which is set by the MPU6050 gyro.</p>
<p>So when the virtual arm moves 90 degrees left, the hardware arm rotates left till the MPU6050 gives a reading of around 90 degrees then stops.</p>
<p>The problem is, Unreal Engine forwards the arm's rotational value in a range of -180 to 180 degrees and the hardware is hard-coded to either go left if the desired rotation is below the current rotation. Aka if the desired rotation is -90 degrees, and current rotation is 0, for left and vice versa, the robotic arm goes right for a +90 desired rotation. This becomes an issue as I reach the end limits till -179 or 179 as the virtual arm will go to -178, -179 then +179, +178, +177 and so on and in order to replicate that desired rotation, the hardware arm would turn left till -179, then start spinning right until it reaches +179.</p>
<p>I'd like to implement some sort of shortest path algorithm that would take the distance between the current and desired rotation and choose whether to go left or right based on which part is the shortest.</p>
<p>Help appreciated. I hope I was able to clarify what exactly it is I'm trying to achieve.</p>
| |mechanical-engineering| | <p>So after posting this question, I had a brain-fart and came up with the solution myself.</p>
<p>The solution is to take the current and desired rotation, and then loop twice from the current to desired rotation. The first time, you increment the loop and second time you decrement the loop. Inside the loops, you increment a "cost" variable for each iteration. You also add in an if statement in the increment loop with the condition that if i >= 179, set i to -179, and vice versa for the decrement i.e: if i <= -179, set i to 179. This will ensure the loops stay between the range -179 to 179 at all times. Then just compare both of the cost variables to see which cost is less and if the decrement loop has lower cost, you go left, and if the increment loop has lower cost, you go right.</p>
<p>Pasted below, is the entire arduino code for this. Upload to your board, set Serial baud rate according to your board and enter a desired rotational value in the serial monitor and the code should show you cost of both rotational loops as well as the final decision.</p>
<p>void setup()
{
// put your setup code here, to run once:
Serial.begin(2000000);
Serial.setTimeout(5);
}</p>
<p>int CurrentRot,DesiredRot;</p>
<p>void loop()
{
String Command;
Command = Serial.readString();</p>
<p>if (Command != "") //Cut Command String into Substrings
{
DesiredRot = Command.toFloat();</p>
<pre><code>String Result;
Result = FindShortestPath(CurrentRot,DesiredRot);
Serial.println(Result);
</code></pre>
<p>}</p>
<p>}</p>
<p>String FindShortestPath(int CurrentRotation, int DesiredRotation)
{</p>
<p>Serial.print("CurrentRot: ");
Serial.println(CurrentRotation);
Serial.print("DesiredRot: ");
Serial.println(DesiredRotation);</p>
<p>int CostLeft=0,CostRight=0,iLeft=0,iRight=0;
String Decision;</p>
<p>iLeft = CurrentRotation ;
while( iLeft != DesiredRotation) // Check Cost going left per iteration
{
iLeft++;
Serial.print("iLeft: ");
Serial.println(iLeft);</p>
<pre><code>delay(10);
CostLeft++; // Add cost per iteration
if(iLeft >= 179) // If reached rotational limit of 179
{
iLeft = -179; // Jump to -179 and continue going left
}
</code></pre>
<p>}</p>
<p>iRight = CurrentRotation ;
while( iRight != DesiredRotation) // Check Cost going right per iteration
{
iRight--;
Serial.print("iRight: ");
Serial.println(iRight);
delay(10);
CostRight++; // Add cost per iteration
if(iRight <= -179) // If reached rotational limit of -179
{
iRight = 179; // Jump to +179 and continue going right
}
}</p>
<p>Serial.print("CostLeft: ");
Serial.println(CostLeft);
Serial.print("CostRight: ");
Serial.println(CostRight);</p>
<p>if(CostLeft < CostRight) // if left iteration cost is less than right, return left
{
Decision = "left";
return Decision;
}
else if(CostRight < CostLeft) // if right iteration cost is less than left, return right
{
Decision = "right";
return Decision;
}</p>
<p>}</p>
| 53562 | Best way to find the shortest rotational path between current and desired rotation |
2022-12-17T10:20:07.590 | <p>In this article
<a href="https://www.sciencedirect.com/science/article/pii/S266711312200002X" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/S266711312200002X</a></p>
<p>the authors show a result of prediction on energy reserves (in particular solar ones) as known in 2015 (Figure 1) and as known in 2022 (Figure 2).</p>
<p>In the first one (Figure 1), they quote result in TWyr/y. For solar, they quote 23 000 TWyr/y.</p>
<p>In the second one (Figure 2), they quote result in TWyr<sub>30</sub>. For solar, they quote 8300 TWy<sub>30</sub>.
They state "we will refer to TWyr over 30 years as TWyr<sub>30</sub>".</p>
<p>Since results on estimation of solar energy reserves should not change by a huge factor between an estimation of 2015 and 2022, what mathematical operation should I do in order to translate from one to the other ?</p>
<p>(If I do 23 000 * 30, it will not decrease the result, so it is not the good operation.)</p>
<p>Figure 1 :</p>
<p><a href="https://i.stack.imgur.com/SqdFg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SqdFg.png" alt="enter image description here" /></a></p>
<p>Figure 2 :</p>
<p><a href="https://i.stack.imgur.com/hikS6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hikS6.png" alt="enter image description here" /></a></p>
| |energy|solar-energy|renewable-energy|unit| | <p><strong>To convert from TWyr/y to TWyr<sub>30</sub>, multiply by 30.</strong> This is confirmed <a href="https://doi.org/10.1016/j.seja.2022.100014" rel="nofollow noreferrer">in the text</a> under "total world energy consumption":</p>
<blockquote>
<p>Depending on the reference, estimates for 2021 global annual TFEC [total final energy consumption] ranges from 18.53 TWyr to 20.13 TWyr. The IEA estimates a middle-range value of 19.11 TWyr that we have retained for this article. Projecting expected [business as usual] 40–50% TFEC growth over the next 30 years, cumulative energy consumption over the period should amount to 660 TWyr<sub>30</sub>.</p>
</blockquote>
<p>Note that <code>660/30 = 22</code>, in line with the 2015 estimate.</p>
<p>When looking at solar the conversion is not as simple:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Year</th>
<th style="text-align: right;">TWyr/y</th>
<th style="text-align: right;">TWyr<sub>30</sub></th>
</tr>
</thead>
<tbody>
<tr>
<td>2015</td>
<td style="text-align: right;">23,000</td>
<td style="text-align: right;">690,000</td>
</tr>
<tr>
<td>2022</td>
<td style="text-align: right;">277</td>
<td style="text-align: right;">8,300</td>
</tr>
</tbody>
</table>
</div>
<p>The reason for the drastic change in the estimate from 2015 to 2022 is several important changes in assumptions, which are explained in the section on solar energy (emphasis added):</p>
<p><strong>It's only feasible to cover about 6% of the land with solar:</strong></p>
<blockquote>
<p>An important consideration we apply here to properly quantify renewable reserves is an extension of the term ‘reasonable’ underlying RARs [Reasonably Assured Recoverable Reserves] beyond pure extraction/exploitation economics. <strong>This consideration is especially relevant to solar since we have in the past reported the total amount of energy impinging annually on the earth's continents. Of course, whereas it would now be economically reasonable to deploy solar (PV) over most of the continental surface area, it would be unreasonable to assume that the entire space would be covered by PV.</strong> [...] Retaining the PV deployment assumptions from this article and its associated land-use web app we assume a reasonable upper limit of deployment to each land-use category including an implementation of emerging PV deployment strategies such as floating PV and agrivoltaic. <strong>We estimate this upper limit at 6% of the land mass.</strong> We did not account for any prospective offshore deployment at this time (a total resource twice as large as the continents’), although the practice may gain acceptability in the future.</p>
</blockquote>
<p><strong>Solar panels are only about 20% efficient:</strong></p>
<blockquote>
<p>In focusing on the energy available, we had also not accounted for the energy conversion efficiency at hand. While it is likely to increase in the future, <strong>we assume here a current state-of-the-art footprint conversion of sunlight to electricity of 20%</strong>, an assumption we justified and applied in a recently published article on large scale PV deployment.</p>
</blockquote>
<p>Taking this information, we can apply the calculation to the 2015 estimate and confirm it roughly equals the 2022 estimate:</p>
<pre><code>23,000 * 0.06 * 0.20 = 276
</code></pre>
<hr />
<p><em>Note that in the 2022 chart the units on the finite resources are incorrectly labeled with units of TWyr<sub>30</sub>. In the body of the paper the correct units, TWyr/y, are used in the discussion.</em></p>
| 53578 | What mathematical operation should I do in order to translate from TWyr/y to TWyr_30 in this article? |
2022-12-17T13:42:51.987 | <p>What is the best way to plot motor power output graphs? Does anyone know how we can plot this graph for motors and engines?</p>
| |motors|power|engines| | <p>The classic way is to make a chart with RPM on the x-axis and torque on the y-axis. Often the experiment is set up in such a way as to ensure an impedance match between the engine being tested and the load being imposed upon it to furnish a baseline "operating point" for the engine, in the following way:</p>
<p>First, remember that (power) = (torque x RPM) in all cases.</p>
<p>The engine is disconnected from the load and the throttle is opened all the way. at the resulting (high!) RPM, all the engine power output is being dissipated as internal friction, noise, heat, etc. This is the <em>no-load</em> case.</p>
<p>Then we apply a brake to the output shaft to load the engine down to the point where the engine's RPM has been cut in half, and measure the torque applied by the brake. This is the nominal operating point of the engine and represents the peak of the torque vs. RPM curve.</p>
<p>To get RPMs greater than this, the engine has to be partly unloaded and its torque output will fall until you get all the way over to the no-load case, where the engine is at maximum RPM but there is no power output at all and the engine is dissipating all its power output in internal friction.</p>
<p>Back to the nominal operating point. To get RPM's less than this, you have to haul harder on the brake which will slow the engine RPM's down. Remember in the case of 0 RPM, the power output is identically zero and so the power output of the engine will fall as you lug the engine down progressively towards zero RPM.</p>
<p>After an hour's worth of light, pleasant work, you will have enough data to plot the full curve.</p>
| 53581 | how to plot power output graph for motors |
2022-12-19T08:06:19.257 | <p>For isotropic material an elastic modulus, like Young's modulus and in combination with a Poisson ratio can be used to convert to any other elastic modulus, like for example the bulk modulus. Tables for these conversions exist, for example, on the <a href="https://en.wikipedia.org/wiki/Bulk_modulus" rel="nofollow noreferrer">wiki page of bulk modulus</a>.</p>
<p>I'd like to use the conversions for FEM material models that are plane strain, plane stress or axisymmetric. What confuses me on the wiki page is that there is a mention of 3D material and 2D materials and these have different conversion formulae. It's unclear to me what 2D material means in this context. My questions:</p>
<ol>
<li><p>Is it correct to use the 3D material conversion formulae for the plane stress case? My current thinking is that the conversions are valid for the plane strain and axisymmetric case as they model objects with 3D extend. But the plane stress case a models a thin object. Which formula should be used in that case?</p>
</li>
<li><p>Can anyone share a reference to where these conversions are derived, possibly including the 2D material conversions.</p>
</li>
</ol>
| |materials|finite-element-method|solid-mechanics|material-science| | <h1>Simplified 2D analyses of 3D situations</h1>
<p>Plain stress, plain strain and axisymmetric allow describing a domain in 2D, but it still represents a full 3D object. Therefore the material properties like Young's modulus and Poisson ratio are still the same as in 3D.</p>
<p>These 2D simplifications are "coded" in stress and strain tensors, which have some components equal to zero (or constant).</p>
<h1>Relation between <span class="math-container">$E$</span>, <span class="math-container">$\nu$</span> and <span class="math-container">$K$</span></h1>
<p>Relation between Young's modulus <span class="math-container">$E$</span>, Poisson ratio <span class="math-container">$\nu$</span> and bulk modulus <span class="math-container">$K$</span> in 3D comes from Hooke's law:</p>
<p><span class="math-container">$$\epsilon_x = \frac{1}{E}\left(\sigma_x-\nu\sigma_y-\nu\sigma_z\right)$$</span>
<span class="math-container">$$\epsilon_y = \frac{1}{E}\left(-\nu\sigma_x+\sigma_y-\nu\sigma_z\right)$$</span>
<span class="math-container">$$\epsilon_z = \frac{1}{E}\left(-\nu\sigma_x-\nu\sigma_y+\sigma_z\right)$$</span></p>
<p>Now the volumetric change per volume unit <span class="math-container">$dV/V$</span> for small deformations is just a sum of normal strains, which may be simplified using average normal stress <span class="math-container">$\bar{\sigma}$</span> (~pressure):</p>
<p><span class="math-container">$$\frac{dV}{V} = \epsilon_x+\epsilon_y+\epsilon_z = \left(\sigma_x+\sigma_y+\sigma_z\right)\cdot\frac{1-2\nu}{E} = 3\bar{\sigma}\frac{1-2\nu}{E}$$</span></p>
<p>Lastly, the bulk modulus as a ratio between pressure and volumetric change per unit is:</p>
<p><span class="math-container">$$K = \frac{\bar{\sigma}}{dV/V} = \frac{E}{3\left(1-2\nu\right)}$$</span></p>
<h1>2D materials</h1>
<p>"2D material" relations from table at the end of <a href="https://en.wikipedia.org/wiki/Bulk_modulus" rel="nofollow noreferrer">wiki/Bulk_modulus</a> may be useful for very special applications like <a href="https://en.wikipedia.org/wiki/Single-layer_materials" rel="nofollow noreferrer">single atom layers</a>, where tangential stresses should not cause contraction in normal direction to the layer.</p>
| 53599 | Conversion of elastic moduli for plane stress case |
2022-12-20T21:41:42.100 | <p>I am needing to find the gap between two pieces of wood on a conveyor belt. I am new to this and am having trouble knowing what to Google to find the right formulas. I found one answer that helped a little, but it was dealing with two conveyors, and I am not sure if I adjusted the formula correctly. I will include what I did as a reference and hopefully someone can let me know if I did this right.</p>
<p>L (product length) = 37 inches</p>
<p>Vx (Upstream Speed) = 145 FPM</p>
<p>G (gap) = ?</p>
<p>Time = 37/145 = .25517</p>
<p>Distance = 145 * .25517 = 45.93101</p>
<p>Gap = 45.93101 - 37 = 8.9 inches</p>
| |mechanical-engineering| | <p>Judging by what you stated in your comments, your formula would be</p>
<p><span class="math-container">$ \frac{speed}{(length+gap)} = f $</span></p>
<p>Where <span class="math-container">$f$</span> is your frequency, or pieces per minute.</p>
<p>If you assume (as you did in your comment) that <span class="math-container">$speed/length = f$</span>, you are assuming a gap between each piece of wood of 0.</p>
<p>P.s. there is a mistake in the calculations, as NMech is noted. you are now saying:</p>
<p><span class="math-container">$\frac{37}{145}*145 = 45.9 \\ 37*\frac{145}{145} = 45.9\\ 37* \frac{1}{1} = 45.9\\ 37 = 45.9$</span></p>
<p>p.p.s. In the formula above, be careful to fill in the numbers with corresponding units.</p>
| 53607 | How do I find the gap between two objects on a moving conveyor? |
2022-12-27T23:38:18.587 | <p>I am a Computer Science student that has a personal project to do. I decided to try and take on a project for a company, and one of the things that I need to measure is how perpendicular a laser is to a surface. The surface is non-reflective, and I need to adjust it until it is perpendicular to a 5 degree error. The adjustment is not an issue, but I am having trouble figuring out how I could check for how perpendicular it is. The surface is spherical in nature, but it is not a perfect sphere. Any alternative methods of checking it without having equipment too close to the surface (due to the destructive nature of what happens later). I'm sorry if I'm not very accurate, I have never taken any engineering courses and I'm not sure of the terminology. Any help would be appreciated</p>
| |lasers|surface-modelling| | <p>If, as you mention in a comment, you can apply a small reflective sticker, and the laser isn't too powerful (at least during this alignment procedure), and the radius of curvature of the surface isn't too small, then I think it should be doable. (You don't specify the range of possible radius-of-curvatures of the surface, but if they are comparable to or smaller than the laser beam diameter, "perpendicular" becomes ill defined.)</p>
<p>When aligning mirrors, I often put a transparent plastic ruler between the laser and the mirror so I can see a bright spot on the ruler where the laser passes through the plastic on the way to the mirror, and a fainter (but still very visible) spot where the beam reflected from the mirror hits the ruler. I adjust the mirror until the two spots are on top of each other, which tells me the mirror is perpendicular to the beam. With a camera observing the ruler, some image processing software to determine the position of the spots, and control over either the object or the laser position/angle, you should be able to automate this.</p>
<p>A possible method that does not require <a href="https://en.wikipedia.org/wiki/Specular_reflection" rel="nofollow noreferrer">specular reflection</a> is to use <a href="https://en.wikipedia.org/wiki/Photometric_stereo" rel="nofollow noreferrer">photometric stereo</a> imaging to determine the normal to the surface, and then use that information to set the angle of the laser. There is a large <a href="https://scholar.google.ca/scholar?hl=en&as_sdt=0%2C5&q=%22photometric+stereo%22&btnG=" rel="nofollow noreferrer">research literature</a>, public code (e.g. <a href="https://github.com/search?q=%22photometric+stereo%22" rel="nofollow noreferrer">on github</a>), and commercial photometric stereo systems are available.</p>
| 53676 | How can I measure how perpendicular a laser is to a surface across a distance |
2022-12-28T15:48:22.617 | <p>I have three conveyors. I am trying to find the minimum speed needed for the second (middle) conveyor in order to avoid collisions with the pieces of wood. The conveyors 1 and 2 are perpendicular to one another, as are conveyor 2 and 3.</p>
<p>Information I have:</p>
<p>*Vx = 145 feet per minute; Conveyor 1 speed/
*W = 5 inches; Width of wood/
*G = 4 inches; Gap between pieces of wood/
*Vy = 180 FPM; Conveyor 3 speed/</p>
<p>Information I am looking for:
*Speed needed for Conveyor 2 in order to avoid collisions</p>
<p>If anyone knows a formula or a resource they can point me to, that'd be awesome! I feel like I just can't find what I'm looking for online when I don't know exactly what I'm looking for.</p>
<p><strong><strong>EDIT</strong></strong> Apparently my question was closed. I will add this diagram and see if someone can help now that I've added a visual representation of my words. I want to know how fast conveyor 2 needs to run in order for the pieces not to hit each other. All pieces of wood are the same size.</p>
<p><a href="https://i.stack.imgur.com/tFmjT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tFmjT.png" alt="Diagram of conveyor set up with variables" /></a></p>
| |mechanical-engineering|conveyor| | <p>I'm going to estimate some numbers here that you haven't provided, but you should be able to adjust the math as needed.</p>
<p>Right off the bat we can forget about conveyor 3, since the rate at which staves are moved on to it is set by conveyor 1, and the speed of that is fixed.</p>
<p>So we need to figure out how fast conveyor 2 needs to move to put 4 inches between staves coming from conveyor 1.</p>
<p>The actual spacing of the staves on conveyor 2 will be 9in (4in gap + 5in stave width).</p>
<p>Now we need to know the rate (in staves per minute) that staves will be delivered onto this belt, from conveyor 1. Unfortunately we can't determine that from the numbers you have provided, so I'll make up some additional numbers.</p>
<p>Lets say there is 1 stave every 5 feet on conveyor 1, that equates to 29 staves per minute. Or 2.07s per stave.</p>
<p>You want conveyor 2 to travel 9in during this time, so the resulting speed is 0.75ft/.0345minutes = <strong>21.7ft/min</strong></p>
<p>Now lets determine whether the staves will collide during the transition.</p>
<p>Starting from the point where a stave has been moved onto conveyor 2 and has just begun moving to the right. It will have to move 5 inches (the width of 1 stave) before another stave collides with it. If the staves are 24 inches long then the gap on conveyor 1 is 36in.The staves cross this gap in 3ft / 145ft/min = 1.24seconds. The stave on conveyor 2 moves 21.7*12/60 = 4.34 inches/sec. So in 1.24 seconds it moves 5.3816 inches. So they will <em>barely</em> not collide. In practice they probably will. You could run conveyor 2 faster and increase the gap between staves to solve the problem, or leave larger gaps between staves on conveyor 1.</p>
| 53689 | How would I find the minimum speed for a conveyor transfer to avoid collision? |
2022-12-28T16:17:16.013 | <p>I have designed and 3D printed a gear system with three gears. Two identical gears at each end with a smaller gear in the center. One of the larger end gears is to be rotated by a stepper motor, the middle gear transfers the motion to the other gear that carries a device.</p>
<p>The issue is that there is a wiggle in the end gears even when one is fixed/held in place. If I hold down the gear controlled by the stepper motor, I can still create a small bit of rotation in the other end gear (carrying the device). In the video, I show this 'wiggle' by holding down one gear and still being able to move the other.</p>
<p>What can I edit/change or add to the design to prevent this movement from happening e.g. design ideas or additional components I could add? I need the gears to be precise so that they move exactly with each other and no wiggling in one of them without the other taking place.</p>
<p><a href="https://www.youtube.com/shorts/fhQGQIOtLeA" rel="nofollow noreferrer">YouTube video showing the issue</a></p>
| |mechanical-engineering|control-engineering|gears|cad|3d-printing| | <ol>
<li><p>If the load is light you may be able to make the gears with zero clearance teeth (or even negative). The gears will be very tight but you can force them to turn manually until they break in just the right amount to have low backlash. If you lubricate them at this point and use them with very light loads, this condition might last decently long. Once the gears wear the backlash will come back.</p>
</li>
<li><p>Make the teeth finer. Generally speaking finer gear teeth will have less backlash (scale of everything is smaller).</p>
</li>
<li><p>You can make zero backlash split gears. The outer gears would be split into 2, and the halves adjusted or spring loaded so that teeth are always in full contact with the center gear.</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/xfbND.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xfbND.jpg" alt="zero backlash split gear" /></a></p>
| 53691 | How to stop gear from wiggling despite connected gear being fixed in place? |
2022-12-29T06:44:01.260 | <p><a href="https://engineering.stackexchange.com/questions/45119/spherical-parallel-manipulator-in-solidworks">Spherical Parallel Manipulator in SolidWorks</a></p>
<p>I am drawing in SolidWorks a 3D arc with two holes on the edges. An axis of rotation passes through the center of one of these holes, around which the arc rotates according to an arbitrary law.</p>
<p>I want to learn how to display nice and informative reference geometry. In this case, I need to depict: the axis of rotation, a vector (for example, U) co-directional with it, the center of mass, etc. Something like this:</p>
<p><a href="https://i.stack.imgur.com/rQMMQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rQMMQ.png" alt="enter image description here" /></a></p>
<p>How is it done in SolidWorks?</p>
| |cad|solidworks|solid-mechanics|software| | <p>I'm assuming you want to display these things in the part editor?</p>
<p>If you make sketches (2D or 3D) with these lines, arcs and points, you can just leave them in your part and they'll be displayed while you work.</p>
<p>One caveat, there will be no labels. I have solidworks 2016, and maybe it's better in the future versions, but I find that adding annotations inside sketches does not work well. I wouldn't bother.</p>
<p>If I really needed a label I might insert some text in the sketch (as you would do for engraving), not elegant but it might work ok.</p>
<p>As far as I know you can't dimension to the center of mass, but you can show the coordinates, or a crude and simple way to locate relative to your part would be to zoom way in and place a point on it.</p>
<p>The editor isn't really designed to create illustrations, so although I think you'll be able to come up with something workable, I likely will not be "beautiful" without some post processing in GIMP or similar.</p>
<p>Crude example:
<a href="https://i.stack.imgur.com/SWVwD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SWVwD.png" alt="example of adding reference sketches to part" /></a></p>
<p>If I wanted this to look nicer I would draw the arrows on the ends of the lines by hand, and add labels. That should look decent.</p>
| 53697 | Reference geometry for a 3D arc in SolidWorks |