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2021-05-18T20:04:02.020 | <p>A mechanism popped up in my mind when I was studying kinematics, see picture below<a href="https://i.stack.imgur.com/HdOZP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HdOZP.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$J1,\ J2,\ J3$</span> are rotational joints and there are two translational joints <span class="math-container">$J4$</span> and <span class="math-container">$J5$</span> between two sliders and the cross hatched rail. Sliders are free to move horizontally towards or away from one another. Now when I imagine a motion in which a force is applied vertically at <span class="math-container">$J1$</span>, pushing or pulling on the said joint, thus sliders move away or come closer, I think it is a one degree of freedom mechanism. However the famous formula <span class="math-container">$3n-2j-3$</span> where <span class="math-container">$n$</span> is the number of links and <span class="math-container">$j$</span> is the number of lower pair joints, gives two degrees of freedom. Considering the rail as a ground link, with two sliders and two arms (<span class="math-container">$J1J2$</span> and <span class="math-container">$J1J3$</span>) we have <span class="math-container">$n=5$</span> and we have <span class="math-container">$j=5$</span></p>
<p><span class="math-container">$3*5-2*5-3=2$</span></p>
<p>So, what is with that?</p>
| |mechanical-engineering|applied-mechanics|kinematics| | <p>This is a 2 degree of freedom mechanism.</p>
<p>For me its easier to think of it by translation of the horizontal sliders. When you know the position of the <strong>both</strong> the horizontal sliders then you know the position of the mechanism.</p>
<p>Another way to look at it: If you know the position of J1 (which has two degrees of freedom, then you automatically know the position of the horizontal sliders, and the angles.</p>
| 43241 | Degree of freedom of a mechanism |
2021-05-19T04:58:44.760 | <p>I have a DC motor with a rated rpm of 2750. At max rpm, the rubber wheel that I've attached to it is pulling itself apart (literally, the rubber is being pulled away from the central hub due to the centrifugal force). How can I know before purchasing a new wheel weather it will be able to handle this rpm? Is there a way to estimate this or do I need a spec sheet? I haven't seen any wheels that have a spec sheet.</p>
| |mechanical-engineering|structural-engineering|wheels| | <p>Since you don't need the wheel to spin that fast, what you could do is use a gear with a very high gear ratio.</p>
<p>You can easily get a gear ratio of 10 for any size of commercially available DC motors, which would reduce the rotational velocity to the wheel to about 300 rpm. Than would reduce the forces by approximately 100 times (centrifugal forces are proportional to <span class="math-container">$\omega^2$</span>)</p>
| 43246 | How to calculate maximum rpm for a wheel? |
2021-05-19T06:50:00.097 | <p><a href="https://i.stack.imgur.com/DluTB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DluTB.png" alt="Problem Diagram" /></a></p>
<p>I'm very new to engineering and thought I'd start with some basic problems and test them out with my own constructions! I've come up with this fun problem and would love some help solving it.</p>
<p>Say you have two beams of wood with a thickness of 10cm.</p>
<p>Say you then set up a series of 13 3m tarpaulin sheets that are attached to the beams via screwed-in hooks, with 10cm between them, and placed 300kg worth of soil in each one, with a max depth of 5cm in the lowest point of the tarp.</p>
<p>What opposing force, in the form of grounding wire, will be needed to stop the beams from toppling over?</p>
<p>I have not formally studied physics since high-school, but my thinking was that there would be a torque force that needs to be opposed, my thinking is as follows:</p>
<p>The torque force at each point would be represented by the formula τ=rF, as the hooks would be perpendicular to the ground so sin(θ) would be 1.</p>
<p>We can therefore work out total torque applied to each beam with the sigma:</p>
<p><a href="https://i.stack.imgur.com/Z894y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z894y.png" alt="Sigma notation" /></a></p>
<p>The problem I now face is trying to calculate the force needed for H, as well as θ. I don't understand how the torque is distributed across the beam and how that impacts the force needed to counteract it.</p>
<p>Is the presence of a wire like H even necessary, or can the beams be grounded with foundation to support the weight?</p>
| |structural-engineering|torque|beam| | <p>To follow up on answer by NMech:</p>
<p>I suspect if this were built, it would be necessary to have tension adjustment, due to elasticity of the tarp. Then, the sag and any 3D "catenary type" shape (<a href="https://mathoverflow.net/questions/69817/surface-equivalent-of-catenary-curve">a gnarly math problem of its own</a>, and that link doesn't include there being more load in the center in our case), is simply a byproduct of however much the builder tensions the tarp. Noting here that to get any reasonably flat angle, tarp-tension >> load.</p>
<p>With the pre-set tension being maintained approximately constant by the elasticity, AND, if the angle of the tarp edges from the horizontal is small, this more complicated geometrical aspect of the problem could be neglected. Due to the small-angle approximation (cos(small_angle) approximately 1), the horizontal projection of the tarp tensions will be approximately the same as their magnitude.</p>
<p>Then you proceed per the other answers, balancing moments on the pole to get the support cable tension.</p>
| 43250 | Fun Force Problem - Basic Structure of Two beams supporting weight |
2021-05-19T07:44:23.280 | <p>In this drawing 4 of the holes have the top left quarter filled in black.</p>
<p>What is the meaning of that? It it some special type of hole?</p>
<p>The drawing is from a Waveguide and in the specs it says</p>
<pre><code>Flange type FDP32 (Cover)
</code></pre>
<p>So I'm thinking it might be some way of mounting the cover.</p>
<p>Is there a standarized meaning for the <a href="https://i.stack.imgur.com/4ztfD.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4ztfD.png" alt="enter image description here" /></a> symbol?</p>
<p><a href="https://i.stack.imgur.com/7VwdT.png" rel="noreferrer"><img src="https://i.stack.imgur.com/7VwdT.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|technical-drawing| | <p>In this particular case it appears to have no meaning, I think someone just didn't notice or care when making the drawing. I'm guessing they made two hole patterns in the software they were using and it automatically labelled them differently.</p>
<p>After doing the digging to figure out that FDP32 seems to be a name used by Chinese manufacturers for CPR284F flanges (using <a href="http://www.dolphmicrowave.com/upload/Flange-Types-Designations-Information.pdf" rel="nofollow noreferrer">this table</a> to see that FDP32 is used for WR284 waveguide, and the column is labelled "FDP (CPRF)"), I found <a href="https://www.atmmicrowave.com/wp-content/uploads/flg_cpr284g.pdf" rel="nofollow noreferrer">this drawing</a> c/o ATM microwave:<a href="https://i.stack.imgur.com/UOPGV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UOPGV.png" alt="CPR284F flange drawing" /></a></p>
<p>Some of the dimensions don't match exactly when you convert to mm, interestingly, but more importantly the holes are all part of the same pattern. Note about waveguide sizing: WRxx or WGxx refers to the dimensions of the waveguide itself (so determines which frequencies it can carry), but there are many different types of flanges which a given waveguide size can be used with.</p>
<p>Here's another Chinese manufacturer's drawing calling it FDP32 but explicitly calling the 10 holes part of the same group (from <a href="http://www.ainfoinc.com.cn/en/p_flange_s.asp" rel="nofollow noreferrer">here</a>)</p>
<p><a href="https://i.stack.imgur.com/EU4YN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EU4YN.jpg" alt="FDP32 flange drawing" /></a></p>
| 43251 | Meaning of quarter hole color fill in drawing |
2021-05-19T10:21:08.523 | <p>I need to create a series of free-standing perspex panels to add height to an existing fence for wind-proofing purposes.</p>
<p>My idea is that each panel would be fixed to two supporting poles that are mounted on flat horizontal 'feet'. In this diagram, the wind would be directly onto the unprotected side of the fence, so the new panels are on the 'inside' of the fence:</p>
<p><a href="https://i.stack.imgur.com/GLl0F.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GLl0F.jpg" alt="enter image description here" /></a></p>
<p>Dimensions:</p>
<ul>
<li>Height of existing fence: 1 m</li>
<li>Height from ground to base of perspex
panel: 1 m</li>
<li>Height from ground to top of perspex panel: 2.75 m</li>
<li>Width of each panel: 1.4 m</li>
</ul>
<p>-> Surface area of each panel: 2.45 <span class="math-container">$m^2$</span></p>
<p>Feet:</p>
<ul>
<li>Made of iron (7.86 <span class="math-container">$g/m^3$</span>) with cross-section 10 mm x 100 mm</li>
</ul>
<p>-> Mass of each foot (per meter length): 1000 g/m</p>
<p>How long / heavy do those feet have to be to make the panels stable in a wind of (say) 6 m/s?</p>
| |pressure|applied-mechanics|statics|torque|wind| | <p>This is a side view of the panels</p>
<p><a href="https://i.stack.imgur.com/6JrKH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6JrKH.png" alt="enter image description here" /></a></p>
<p>where:</p>
<ul>
<li><p><span class="math-container">$H$</span> is the height of the resultant concentrated wind force <span class="math-container">$P_W$</span> from the ground. Given the parameters of the problem, H should be equal to :
<span class="math-container">$$ H= \frac{1+2.75}{2}= 1.875 [m]$$</span></p>
</li>
<li><p><span class="math-container">$L_f$</span> is the total length of the horizontal part of the feet. The length of the feet will affect the weight, and the total weight of the feet <span class="math-container">$W_f$</span> will be equal to :</p>
</li>
</ul>
<p><span class="math-container">$$W_f = 2 L_F \cdot q\cdot g$$</span></p>
<p>where:</p>
<ul>
<li><p>q is the mass per meter (1 kg/m)</p>
</li>
<li><p>g is the acceleration of gravity (~10 <span class="math-container">$m/s^2$</span>)</p>
</li>
<li><p>2 is because there are two (2) feet per stand
therefore:
<span class="math-container">$$W_f = 20 \frac{N}{m} L_f $$</span></p>
</li>
<li><p><span class="math-container">$W_p$</span>: is the weight of the vertical part of the panel (glass and metallic part). I will be assuming here that the weight is about 10 kg (i.e. 100 N).</p>
</li>
<li><p><span class="math-container">$P_W$</span> is the total force of each panel (assuming that there is a uniform pressure on the panel (this is a semi valid assumption, but its simple enough to present here). This is the most involved part so I will break it up.</p>
</li>
</ul>
<h2>wind panel</h2>
<p>In order to calculate the wind force on the panel:</p>
<p>The nominal wind pressure is <span class="math-container">$q_p$</span>
<span class="math-container">$$q_n = \frac{1}{2} \rho v^2$$</span>
where:</p>
<ul>
<li><span class="math-container">$\rho = 1.225[kg/m^3]$</span> is the air density</li>
<li><span class="math-container">$v$</span> is the air velocity (6[m/s])</li>
</ul>
<p>The wind pressure <span class="math-container">$q_p $</span> on the flat panel is
<span class="math-container">$$q_p = C_d \cdot q_n = C_d \cdot \frac{1}{2} \rho v^2$$</span>
where:</p>
<ul>
<li><span class="math-container">$C_d$</span>: is the drag coefficient.</li>
</ul>
<p>The total wind force <span class="math-container">$P_w$</span> on the panel is:</p>
<p><span class="math-container">$$P_w= q_p \cdot A = A\cdot C_d \cdot \frac{1}{2} \rho v^2$$</span></p>
<h2>Moment equilibrium.</h2>
<p>If the wind blows in the direction of the initial image, then the panel will start to rotate/pivot around the rightmost part of the feet in the image (lets call that point A). In that case the moment equilibrium about A, will be:</p>
<p><span class="math-container">$$\sum M_A = -P_w H + W_p \cdot L_f + P_f\cdot \frac{L_f}{2}$$</span></p>
<p>This need to be positive in order for the panel not to rotate about A. Therefore</p>
<p><span class="math-container">$$\sum M_A \ge 0 \Rightarrow -P_w H + W_p \cdot L_f + P_f\cdot \frac{L_f}{2} \ge 0 $$</span></p>
<p>I will substitute the parameters that are depended on <span class="math-container">$L_f$</span>:</p>
<p><span class="math-container">$$-P_w H + W_p \cdot L_f + 20 \frac{N}{m} L_f \cdot \frac{L_f}{2} \ge 0 $$</span>
<span class="math-container">$$ \frac{20 \frac{N}{m}}{2}L_f^2 + W_p \cdot L_f - P_w H \ge 0$$</span></p>
<p>You can then solve this equation and obtain two solutions. One will be positive and one will be negative. Values greater than the positive value, and values smaller than the negative value will satisfy the above requirement. (The negative value makes sense, because negative <span class="math-container">$L_f$</span>, just means that the feet extend to the left).</p>
<hr />
<p><strong>Note</strong> that you should investigate also the other way overturning (wind blowing from the right.</p>
<hr />
<h2>another solution instead of really long feet.</h2>
<p>Provided the existing fence has adequate structural capacity, you could try the following solution.</p>
<p><a href="https://i.stack.imgur.com/VjQ7X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VjQ7X.png" alt="enter image description here" /></a></p>
<p>where:</p>
<ul>
<li><strong>red</strong> is the old fence</li>
<li><strong>green</strong> are the new panels.</li>
</ul>
| 43255 | What size feet for this wind-protection panel? |
2021-05-20T04:26:12.843 | <p>I'm using a torque wrench whose unit is <code>cN.m</code>. I wonder what unit this is, and how to convert <code>N.m</code> (Newton meter) to <code>cN.m</code>.</p>
| |experimental-physics|tools|unit| | <p><strong>cN.m</strong> = <strong>N.cm</strong> = 0.01 N.m</p>
<p>N.cm is more intuitive than cN.m.</p>
<p>'c' is a standard SI prefix indication one-hundredth as in centimeter, 0.01 m.</p>
| 43273 | Unit of "cN.m" used on torque wrench |
2021-05-20T13:24:01.417 | <p>If you look at any launch it appears as though rockets do not take the shortest path. Their trajectory bends. Why do rockets not fly straight up after launching; wouldn’t that be less atmosphere to traverse?</p>
| |aerospace-engineering|rocketry| | <p>Yes, but atmospheric drag is nearly negligable.</p>
<p>To get into orbit, the rocket has to expend energy doing several different things, and they each have performance constraints that affect the structural weight of the rocket.</p>
<p>Roughly in order of fuel consumption, the rocket has to</p>
<ol>
<li>Gain orbital velocity (a reversible kinetic energy term)</li>
<li>Overcome gravity loss (a non-reversible term for lost impulse)</li>
<li>Gain orbital altitude (a reversible potential energy term)</li>
<li>Overcome atmospheric friction without getting too hot, overstressing the structure, or sliding sideways through the air (because that increases the drag).</li>
</ol>
<p>If you look at only the first and third, they are both reversible and you can run any linear combination, including launching straight up and then kicking over 90 degrees and gaining transverse velocity. (Essentially, all paths would have the same fuel requirement.)</p>
<p>The trouble with doing that is that the gravity loss is path dependent. If a rocket was hovering just above the ground, it is still burning fuel, but not gaining kinetic energy and not gaining potential energy. The rocket has to overcome all of local gravity to stay in the air. But if we begin to traverse, the ground is falling away from us and we only need to overcome, say 90% of gravity and let the rocket accelerate towards the ground which it will never hit. The total gravity loss is the time integral of gravity force - centrifugal force. The ideal solution (with gravity, but without atmosphere) is a very high acceleration rate and short burn time, followed by a coast into orbit. Then you need a second burn near apogee to not crash back down. If you can accelerate hard enough, you can just launch sideways. But we can't accelerate that hard.</p>
<p>The atmosphere puts all kinds of practical constraints on the coupled problem of vehicle design and launch trajectory design. The way it shakes out, the atmospheric drag ends up being nearly negligible. Here are some Delta V numbers from <a href="https://web.stanford.edu/%7Ecantwell/AA284A_Course_Material/Karabeyoglu%20AA%20284A%20Lectures/AA284a_Lecture7.pdf" rel="nofollow noreferrer">Stanford course material</a> (Delta V is the change in velocity, where all of the rocket engine's efforts have been converted to what would happen in space.)</p>
<p><a href="https://i.stack.imgur.com/1HH3h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1HH3h.png" alt="Delta V budget from Stanford AA284a" /></a></p>
<p>So even with a well executed launch, the atmospheric drag loss is only 5% of the gravity loss and only 1% of the total Delta V budget. A poor trajectory can increase the gravity loss by more than the total atmospheric drag. The trajectory is driven largely by attempting to minimize the gravity loss figure. The atmosphere's real nuisance value is the structural loads that are very peaky around mach 1.4. So we have to go kind of slow until the air is thin enough to be able to push though it with a light upper structure.</p>
| 43278 | Why do rockets not fly straight up after launching; wouldn’t that be less atmosphere to traverse? |
2021-05-20T13:30:54.243 | <p>Not sure if this is the right place to ask, but it is for a physics experiment.</p>
<p>I'm trying to find out how the internal diameter of a copper pipe affects the time it takes for a magnet to drop through. Dropping the magnet in the centre would allow me to ensure consistency for each drop.</p>
| |experimental-physics| | <p>Illustration following up from the comments</p>
<p>We used vacuum to hold the ball bearing being dropped, from above. The ball was hand placed up against the output port of a manifold. Around it was open air. The port holding the ball connected to COMMON port of a 3/2 solenoid valve. NC to vacuum regulator, NO vent to atmosphere. When valve vented the output, part went into free fall.</p>
<p>There wasn't much to it, really, although we had the benefit of symmetry, and the sphere-circle contact made great repeatability in horizontal position.</p>
<p><img src="https://i.stack.imgur.com/YKjok.jpg" alt="fine-art" /></p>
| 43279 | How would I drop something directly in the middle of a vertical pipe? |
2021-05-20T15:11:14.317 | <p><a href="https://i.stack.imgur.com/zblGu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zblGu.jpg" alt="enter image description here" /></a></p>
<p>Here , in case of a death well situation as an example.</p>
<p>Q How is direction of friction going to be opposite to mg or in upwards direction ?</p>
<p>Why not:</p>
<ol>
<li><p>In case of like circular turning , direction of friction is toward the centre. Why is it not the same here ?</p>
</li>
<li><p>Online it says that when f=mg , then only net force is towards the centre. How is that possible ? Normal is pushing the bike. How is it still on the track then if f and mg cancel out.</p>
</li>
<li><p>Friction should be in a direction opposite to motion I.e since the bike is moving forward I,e towards you in picture. Then , friction should be backwards.</p>
</li>
</ol>
<p>Similarly , there are many situation in which it becomes difficult to guess the direction of friction. For example , banking of road , vertical circular motion of a car or bike.</p>
<p>Especially in cases of banked raids, direction of friction changes as magnitude of velocity changes .</p>
<p>Q:What is the way with which I can always and I’m almost every situation , guess the direction of friction or how can you find the direction of friction in an intuitive way.</p>
| |mechanical-engineering|civil-engineering|applied-mechanics|friction| | <p>Friction is the resistance of instantaneous bonds between the molecules of the two contact surfaces against their sliding.</p>
<p>Any effort or force trying to slide one of the surfaces against the other will cause a reaction exactly equal and in the opposite direction of that force. (static friction)</p>
<p>This static friction will keep increasing equal to applied force until it becomes greater than the bond between the two surfaces.</p>
<p>When this force exceeds this level the bond will break with a sudden jerk and the two surfaces will slide against each other with less resistance, kinetic friction.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/MDf9J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MDf9J.png" alt="static vs kinetic friction." /></a></p>
| 43283 | How to guess direction of friction in any case |
2021-05-21T07:09:04.100 | <p><span class="math-container">$a_t$</span> = <span class="math-container">$\frac{dv}{dt}$</span> = <span class="math-container">$\frac{d(r*\omega)}{dt}$</span> = <span class="math-container">$r $</span>* <span class="math-container">$\alpha$</span>.</p>
<p><span class="math-container">$a_c$</span> = r*<span class="math-container">$\omega^2$</span> = <span class="math-container">$\frac{v^2}{r}$</span></p>
<p>These are almost all the variables and values associated with centripetal and tangential acceleration in case of circular motion.</p>
<p>The Q 1 : is to find angular acceleration of particle when it’s speed changes from <span class="math-container">$2$</span>m/s to<span class="math-container">$ 4 $</span>m/s in 4 seconds where r of the circle =0.5m.</p>
<p><span class="math-container">$a_t$</span> = <span class="math-container">$\frac{dv}{dt}$</span> where <span class="math-container">$dv=4-2$</span> and <span class="math-container">$dt=2$</span>.Therefore , <span class="math-container">$ a_t$</span>=<span class="math-container">$0.5m/$</span> <span class="math-container">$s^2$</span>.</p>
<p>Then used <span class="math-container">$r $</span>* <span class="math-container">$\alpha$</span> = <span class="math-container">$a_t$</span>= 1 rad /<span class="math-container">$ s^2$</span></p>
<p>My Q’s are that why did we not need to use the formula of centripetal acceleration here( in Q:1)to find angular acceleration. But instead used formula of tangential acceleration.</p>
<p>If I take a situation in which there are 3 different values of centripetal acceleration. For example , 4 , 5 ,6 m/<span class="math-container">$s^2$</span> and take the same values of the above Q 1 only for dv,dt and r. Then , still Angular acceleration would be having same result since there is no mention of centripetal acceleration.</p>
<p>Why do I think it is not possible :</p>
<ol>
<li>Doesn’t centripetal acceleration also affect the angular acceleration. We have <span class="math-container">$\omega$</span> in the formula of centripetal acceleration which is stated as amount of <span class="math-container">$\theta$</span> covered per unit time. Then , it means there has to be centripetal acceleration relation with <span class="math-container">$\alpha$</span> or angular acceleration also.</li>
</ol>
<p>My confusion in brief is that , we know net acc = <span class="math-container">$a_t$</span> + <span class="math-container">$a_c$</span> . Right. So , I am thinking that both formulas when combined together should give angular acceleration</p>
<p>EDIT: Regarding a= <span class="math-container">$\frac{dv}{dt}$</span>.</p>
<p><a href="https://i.stack.imgur.com/Rf8hM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rf8hM.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|applied-mechanics|acceleration| | <p><span class="math-container">$a_t = \frac{dv}{dt} = \frac{d(r∗ω)}{dt} = r* α$</span>. Here, <span class="math-container">$v$</span> is a vector.</p>
<p><span class="math-container">$a_c = r*ω^2 = \frac{v^2}{r}$</span>. Here <span class="math-container">$v$</span> is a scaler.</p>
<p>In your numerical example, both changes in velocity and time are given, so you go ahead to calculate <span class="math-container">$a_t$</span> using the information provided. However, there is no information for the <span class="math-container">$v$</span> and <span class="math-container">$\omega$</span>, so, you need to figure them out by:</p>
<ol>
<li><p>Find the centripetal force <span class="math-container">$F_c$</span>.</p>
</li>
<li><p>Solve <span class="math-container">$a_c = F_c/mass$</span>.</p>
</li>
<li><p>Then solving <span class="math-container">$v$</span> and <span class="math-container">$\omega$</span> by the equation <span class="math-container">$a_c = r*ω^2 = \frac{v^2}{r}$</span>.</p>
</li>
</ol>
<p>Now you can find the acceleration of the particle by combining <span class="math-container">$a_t$</span> & <span class="math-container">$a_c$</span></p>
<p>Please review the linked article, which may help you to better understand the current problem, and the other similar question of yours. <a href="http://getyournotes.blogspot.com/2011/07/calculating-circular-motion-questions.html" rel="nofollow noreferrer">http://getyournotes.blogspot.com/2011/07/calculating-circular-motion-questions.html</a></p>
| 43298 | Effect of centripetal acceleration on angular acceleration |
2021-05-21T12:29:30.257 | <p>I have an airtight rigid chamber of volume <strong>V</strong> (~500 cu.ft.)</p>
<p>I want to remove the air from the chamber such that the internal pressure is equivalent to some <strong>P</strong> absolute atmospheric pressure (eg 3.46psia)</p>
<p>I would want to evacuate the chamber from ambient pressure <strong>A</strong> (eg 14psia) to <strong>P</strong> within <strong>T</strong> milliseconds (< 1 second)</p>
<p>How would I go about determining the specs for the vacuum pump needed to perform this function? Assume ambient pressure is "on land" and the output of the vacuum is wide open into the environment (minimal resistance).</p>
| |vacuum|vacuum-pumps| | <p>Evacuating 10 psi of pressure from a 500 cu ft chamber in under a second is pretty insane. <a href="https://www.engineeringtoolbox.com/vacuum-evacuation-time-d_844.html" rel="nofollow noreferrer">This</a> is a good calculator, as you will probably need to change your parameters (most likely time) in order to find a suitable pump.</p>
| 43301 | Vacuum Pump Specs |
2021-05-21T13:44:19.443 | <p><a href="https://i.stack.imgur.com/taMO4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/taMO4.jpg" alt="enter image description here" /></a></p>
<p>For this Q , the author of the textbook writes that T cos 45 =mg</p>
<p>but it also states that mg cos 45 is not equal to T. I got to know about this because this is the way , I solved it first.</p>
<p>My Q :1 is that why mg cos 45 is not equal to tension or T.</p>
<p>The main point to noice is that Tcos 45 = mg doesn’t give same value of T as T=mg cos 45.</p>
<p>According to me why I think mg cos 45 should be equal to T :</p>
<ol>
<li><p>There is no acceleration along T. Therefore acc=0.</p>
</li>
<li><p>T and mg cos 45 also come at the same vertical line.</p>
</li>
</ol>
<p>Also as Q2: can we say there is <span class="math-container">$a_t$</span> is along z axis for this question ?</p>
<p>Name of author: Dc pandey</p>
| |mechanical-engineering|applied-mechanics| | <p>Very short answer: The tension in the cable must be greater than the weight since it's supporting both the weight and the centripetal force. Cosines are always less than 1.</p>
| 43304 | Why do we write T cos theta = mg for this Q instead of mg cos theta = T |
2021-05-21T14:38:55.097 | <p>I have to inspect a sphere on a CMM, and the standard states that the "Sphericity must be within 0.02mm". I am using a Zeiss CMM, and it does not have a 'sphericity' inspection method, but it does have a 'profile' inspection method. Since profile is essentially ensuring a tolerance band that would cover the whole sphere, would this suffice?</p>
| |drafting|drawings| | <p>One could argue that sphericity is a specific type of profile tolerance. You would just need to make sure perfectly opposing points on the surface (or as close to it as possible) are inspected to properly calculate the diameter.</p>
| 43305 | Is sphericity the same as profile? |
2021-05-21T18:11:15.853 | <p>I was watching <a href="https://youtu.be/iSid1qz70vw?t=623" rel="nofollow noreferrer">this video - at 10:23 mark</a> where the person was using a hook mechanism for sliding the metal sheets together and noted that sometimes the hook would get caught on a bad angle. To solve this, he used magnets, but I'm not sure how this mechanism works.</p>
<p>If anyone could provide a drawing with explanations where the magnets are placed and how it would affect the sliding mechanism that would be great.</p>
| |metals|magnets| | <p>I suspect that he has just placed some magnets as spacers / sliding surfaces that keep the faces apart.</p>
<p>What has magnets got to do with it? I guess that magnets have the advantage that no adhesive or fastener is required but it's not clear which surface it will stick to.</p>
<p>You could try experimenting with magnetic plastic strip such as found in fridge door seals or fridge magnets. The smooth surface would be low friction.</p>
| 43310 | How does the magnet sheet sliding metal mechanism work? |
2021-05-22T05:36:18.353 | <p>I've been playing around with structural modelling on the SkyCiv platform and have created this structure, related to a project I am working on to educate myself.</p>
<p><a href="https://i.stack.imgur.com/aIsF4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aIsF4.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/KaIiL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KaIiL.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/0W5BX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0W5BX.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/Q4D09.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q4D09.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/ctGd4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ctGd4.png" alt="enter image description here" /></a></p>
<p>It is 2m wide, 3m long and 1.5m high and is made from malabar ironwood, which has a young's modulus of about 13000, a density of about 1000kg/m3 and a poisson's ratio of about 0.27 (I think).</p>
<p>Every 10cm, there is a woven PE sheet of 2m x 3m x 0.4mm dimensions, there are 14 in total. Each sheet has 3kN force acting on it and therefore a pressure of 0.5kPa, which I'm assuming is distributed uniformly. It's quite hard to be sure, but I think the PE sheet has a young's modulus of 0.3, a density of 0.91, a poisson's ratio of 0.029 and an ultimate strength of 20MPa.</p>
<p>Whenever I solve these loads using linear static + buckling simulations, one side is buckling. What causes asymmetrical buckling? Is this a known phenomenon? I have triple-checked the model and it is symmetrical.</p>
<p><a href="https://i.stack.imgur.com/wTUEu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wTUEu.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/Jy4me.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jy4me.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/dYj8x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dYj8x.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|structures|buckling| | <p>First of all, buckling is an instability failure mode. I.e. the structure resists up to a critical load, and beyond that the failure is very rapid.</p>
<p>In most textbooks, buckling is considered in single column members. (Obviously, buckling can occur in structures like the one you have.) In an example with two columns (see below),</p>
<p><a href="https://i.stack.imgur.com/fvcuN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fvcuN.png" alt="enter image description here" /></a></p>
<p>if you had two columns with one(e.g the left) having a smaller EI, then only one side would collapse, and the other would remain relatively intact.</p>
<p><a href="https://i.stack.imgur.com/fJAwE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fJAwE.png" alt="enter image description here" /></a></p>
<p><strong>Figure: source <a href="https://www.scia.net/en/support/faq/steel/buckling-coefficients" rel="nofollow noreferrer">scia.net</a></strong></p>
<p>In real life, when the properties are similar (cross-section/material etc) then the column that will start to buckle first will be determined by small imperfections, in either column.</p>
<h2>Perfection and Numerical Simulations</h2>
<p>A numerical simulation with buckling of two columns, <em><strong>theoretically</strong></em> two columns with identical characteristics will buckle at exactly the same load because the <em>physical</em> imperfections are not there. So if you set out to calculate the behaviour numerically with pen and paper you'd result into an identical load <em><strong>which doesn't explain what you are seeing</strong></em>.</p>
<p>In the extreme scenario, for a perfect column, buckling should not occur, because there is not an initial imperfection. (see below) (in a way its similar to the no</p>
<p><a href="https://i.stack.imgur.com/CXDfE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CXDfE.png" alt="enter image description here" /></a></p>
<p><em>Figure: initial imperfection in column buckling and resulting forces source: <a href="https://advantage.graitec.com/en-US/knowledgebase/article-details/?code=KA-03157" rel="nofollow noreferrer">Graitec Advantage</a></em></p>
<h2>Possible explanation Causes</h2>
<p>Some possible causes can be the following:</p>
<ol>
<li><p>Sometimes for buckling applications in order to avoid the "perfect beam" problem, they introduce intentionally random jitter to the position of the nodes, to promote buckling. This is probably not very common nowadays.</p>
</li>
<li><p>There is some artificial noise in the placement of the nodes and it has to do with the 32-bit representation of a decimal number. If you never tried it start up Calculator in Windows and perform the following calculation:</p>
</li>
</ol>
<p><span class="math-container">$$\sqrt{2}\cdot \sqrt{2} - 2$$</span></p>
<p>Any person would expect the solution to this problem to be 0, however you'd be surprised to see something like the following (left is Windows Calculator, and to the right is Octave - a Matlab clone. Click to see in more detail) .</p>
<p><a href="https://i.stack.imgur.com/MoBcl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MoBcl.png" alt="enter image description here" /></a></p>
<p>So, <strong>internally the way a decimal number (as in the XYZ coordinates of a node) has tiny discrepancies, which can act as imperfections for buckling</strong>. This is a much more plausible/common place.</p>
<ol start="3">
<li>A final reason (that I can think of at least), is the implementation of the analysis code. Some solver codes, use implicit which other use explicit schemes. Although sometimes the borderlines are blurred,</li>
</ol>
<ul>
<li><strong>implicit</strong> means that there is an inversion of a matrix and the structure is considered all at once (in which case it is less likely to explain the asymmetric buckling).</li>
<li><strong>explicit</strong> (which is very common for dynamic analyses) means that each node in the structure is updated incrementally, considering either the state of the previous timestep, or the most current state at each node (Depending on the implementation.</li>
</ul>
<p><strong>Given the type of the problem, I would expect reasons 2 and 3 are most likely explaining the behavior you are observing.</strong></p>
| 43320 | What Causes Asymmetrical Buckling in this structure? |
2021-05-22T10:23:11.783 | <p>This is a homework for my university course control theory.</p>
<p>The task is simple: Test the P, the I and the D controllers separately to see the effect.
We don't have to connect the controllers to a system it means IN -> controller - out, but the instruction doesn't say if we have to use a unity gain feedback loop.</p>
<p>Does it make sense ( practical use ) to test a PID controller without a feedback loop?</p>
<p>The inputs are step function for P and I controllers and a ramp function for the D controller.</p>
<p>And I do everything in Simulink.</p>
| |control-theory|pid-control|matlab| | <p>Testing physical controller hardware in "open loop" isn't uncommon, but it would just be part of a test, and the feedback line can be split at a point and treated as both input and output of a test setup.</p>
<p>For example you can use the feedback sensor to measure what the system is doing and inject a fake sensor signal along with a desired input to observe whether the system is behaving as you would expect it to. Whether such a system is considered closed or open loop depends on what part of the system you consider. A simple decision to abort the test based on behavior can be considered feedback. Naturally, when you have the fake sensor match the real, that becomes something close to the normal operation closed loop system.</p>
| 43323 | Do I need a feedback loop while testing PID controller? |
2021-05-24T03:54:16.347 | <p>I am looking for specific terminology around a half wall that acts as a divider to split a space into two separate uses. The top half of this features a series of wooden posts. It essentially resembles a balustrade except is not for stairs.</p>
<p>Is there a specific name for this?</p>
<p>Here are a few examples:</p>
<p><a href="https://i.stack.imgur.com/VEc2N.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VEc2N.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/BXt0n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BXt0n.jpg" alt="enter image description here" /></a></p>
| |terminology|architecture|wood| | <p>A non jargon terminology (i.e. non scientific architectural) would be <em>permanent room divider</em> or <em>wall divider</em> or plain divider.</p>
<p>Probably the proper architectural terminology is <strong><a href="https://en.wikipedia.org/wiki/Mullion" rel="nofollow noreferrer">mullion</a></strong> is a vertical element that forms a division between units of a window or screen, or is used decoratively</p>
<p><a href="https://i.stack.imgur.com/YMDtS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YMDtS.png" alt="enter image description here" /></a></p>
<p><em>Figure: Mullion (image source: Wikipedia)</em></p>
<hr />
<p>In French you mint find it as <em>paravent</em>, which comes from the latin <em>paravento</em> (which translates for the wind).</p>
<p>The use of <em>permanent</em> is to separate what you are looking for from room dividers like</p>
<p><a href="https://i.stack.imgur.com/A1pcv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A1pcv.png" alt="enter image description here" /></a></p>
<p><em>Figure: room divider source:<a href="https://www.vkf-renzel.com/room-divider-lea-with-wooden-slats-80-0720-5.html" rel="nofollow noreferrer">vkf-renzel.com</a></em></p>
<p>Some architectural terms:</p>
<ul>
<li><p><strong>Cancellus (plural: Cancelli)</strong> Barriers which correspond to the modern balustrade or railing, especially the screen dividing the body of a church from the part occupied by the ministers hence chancel. The Romans employed cancelli to partition off portions of the courts of law.</p>
</li>
<li><p><strong>Muntin</strong> A vertical or horizontal piece that divides a pane of glass into two or more panes or lites in a window.</p>
</li>
<li><p><strong>railing</strong> (also related handrail and banister)</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/40phd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/40phd.png" alt="enter image description here" /></a></p>
<p><em>Figure: Railing (source : <a href="https://www.houzz.in/photos/horizontal-deck-railing-in-bloomfield-hills-phvw-vp%7E54828788" rel="nofollow noreferrer">www.houzz.in</a>)</em></p>
<ul>
<li>(JimClark also suggested) <em><strong>Partition</strong></em> is another word that can refer to the vertical wall in a room. (This is something that I also thought of, but -<em>initially</em>- I decided not to include because in my mind partition refers to the horizontal spaces that are created by the vertical column).</li>
</ul>
| 43343 | What is the name of a half wall with posts on top that divides a space into two? |
2021-05-25T03:04:14.737 | <p>So, I just saw <a href="https://stats.stackexchange.com/questions/525697/why-is-it-that-my-colleagues-and-i-learned-opposite-definitions-for-test-and-val">this question</a> on the Statistics SE site, about how in artificial intelligence/machine learning some people use a particular set of terms in one way, and other people use the same set of terms in the exact opposite way. It got me thinking that this confusion might be resolved if there was an official standard by a body like the ISO or IEEE regulating things.</p>
<p>So, what's the official process for starting the process of creating an IEEE or ISO standard? If you wanted one of these big engineering organizations to make an official standard on the topic, what would you need to do?</p>
| |standards| | <p>The ISO committee has the following process for developing Technical committees (TC) and subcommittees (SC).</p>
<ul>
<li>Stage 1: Proposal stage</li>
<li>Stage 2: Preparatory stage</li>
<li>Stage 3: Committee stage</li>
<li>Stage 4: Enquiry stage</li>
<li>Stage 5: Approval stage</li>
<li>Stage 6: Publication stage</li>
</ul>
<p>There is a lot of detail at <a href="https://en.wikipedia.org/wiki/International_Organization_for_Standardization#Standardization_process" rel="nofollow noreferrer">this wikipedia article under Standardization process</a>. (Quite interesting read is the Criticism section, it mentions that the normal process is long,boring and tedious, and how artificially it can be speed up by pumping money into it).</p>
| 43357 | What's the process for starting an ISO or IEEE standardization? |
2021-05-25T09:33:35.150 | <p>Consider a rod with a length of 150 [mm] and a diameter of 44 [mm]. At the left boundary the rod is fixed to a wall. At the right end a torque of 1000 [N m] is applied. The remaining boundaries are free to move. As a material a S275 steel is used. Thus Young's modulus is given as <span class="math-container">$Y=210000$</span> [N/<span class="math-container">$mm^2$</span>] and Poisson's ratio is <span class="math-container">$\nu=0.3$</span>.</p>
<p><a href="https://i.stack.imgur.com/zKrmz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zKrmz.png" alt="enter image description here" /></a></p>
<p>A reference solution for the maximal displacement suggests it should be around 0.1103 [mm] and the maximal von Mises stress should be about 103.6 [N/mm^2]. In this <a href="https://www.youtube.com/watch?v=N_Oo1J2eGv4" rel="nofollow noreferrer">Solidworks tutorial</a> the same thing is done, but with different values than the reference I used (which is also based on a Solidworks example).</p>
<p>The problem I have is that the finite element simulation tool I use does not have a direct way to specify a torque as a boundary condition. One can, however, specify a boundary surface pressure. I'd like to convert the given torque into a surface pressure. Here is how I did the conversion. I created a vector field that gives the direction of the pressure load in the <span class="math-container">$x$</span>, <span class="math-container">$y$</span> and <span class="math-container">$z$</span> direction and converted the torque value into a pressure. As a vector field I used:</p>
<pre><code>{0, Sin[ArcTan[y, z]], -Cos[ArcTan[y, z]]}
</code></pre>
<p>A plot of the vector field looks like this:</p>
<pre><code>VectorPlot[{Sin[ArcTan[y, z]], -Cos[ArcTan[y, z]]}, {y, -1, 1}, {z, -1, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/etPah.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/etPah.png" alt="enter image description here" /></a></p>
<p>Then I converted the <span class="math-container">$1000$</span> [N m] into <span class="math-container">$1000*10^3$</span> [N mm]. To get a pressure I think I'd need to use something of this sort:</p>
<p><span class="math-container">$$M_t / q * 1 / (\pi r^2)$$</span></p>
<p>where <span class="math-container">$M_t$</span> is the given torque value. I would like to convert the torque into a force by dividing the force by a length. The <span class="math-container">$q$</span> is the part I am uncertain about what to use. I tried <span class="math-container">$q=\sqrt{y^2 + z^2}$</span> such that it gives the radius to the axis but that does not give correct values. Then I divide by the area of the surface to get a pressure.</p>
<p>I think I am making a mistake in the conversion. If someone could point me in the right direction...</p>
<p><strong>Update:</strong></p>
<p>Here is a different approach: Based on <a href="https://en.wikipedia.org/wiki/Torsion_(mechanics)" rel="nofollow noreferrer">this</a> we start from</p>
<p><span class="math-container">$$M_t = \frac{J_T}{r} \tau$$</span></p>
<p>where <span class="math-container">$J_T$</span> is the torsion constant, <span class="math-container">$r$</span> the radius and <span class="math-container">$\tau$</span> the maximum shear stress. We solve for <span class="math-container">$\tau$</span> to get</p>
<p><span class="math-container">$$\tau = \frac{M_t r}{J_T}$$</span></p>
<p>The <a href="https://en.wikipedia.org/wiki/Torsion_constant" rel="nofollow noreferrer">torsion constant</a> can be computed with</p>
<p><span class="math-container">$$J_{zz} = J_{xx} + J_{yy} = \frac{\pi r^4}{2}$$</span></p>
<p>With this approach I get a maximum displacement of <span class="math-container">$0.111035$</span> [mm] and a maximum von Mises stress of <span class="math-container">$104.277$</span> [N/mm^2], both close enough for government work.</p>
<p>A plot of the von Mises stress:</p>
<p><a href="https://i.stack.imgur.com/EYpYc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EYpYc.png" alt="enter image description here" /></a></p>
<p>In principal, I think, this approach could be used for other surfaces, then one would need to compute the <a href="https://mathworld.wolfram.com/AreaMomentofInertia.html" rel="nofollow noreferrer">area moments of inertia</a> for those surfaces.</p>
| |finite-element-method|modeling|simulation| | <p>I'll hazard a guess. From what I undestand what you are after is given a specific torque - I'll call it <span class="math-container">$M_t$</span> (instead of <span class="math-container">$F_t$</span>)- you want to calculate what is the <strong>uniform</strong> pressure that you need to apply on the face of a cylinder to obtain as a result <span class="math-container">$M_t$</span>.</p>
<p>Assume a thin annular stripe at distance r with with dr.</p>
<p><a href="https://i.stack.imgur.com/ZTBhp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZTBhp.png" alt="enter image description here" /></a></p>
<p>The area of that stripe will be:
<span class="math-container">$$dA = 2\pi r dr$$</span></p>
<p>Assuming a constant pressure q [<span class="math-container">$N/m^2$</span>]is applied, the total force on that stripe is</p>
<p><span class="math-container">$$dF = q dA = 2\pi qr dr$$</span></p>
<p>The torque of that thin stripe <span class="math-container">$dM_t$</span> will be</p>
<p><span class="math-container">$$dM_t = r\cdot dF = r q dA = 2\pi q r^2 dr$$</span></p>
<p>By integrating from 0 to <span class="math-container">$R$</span> (: external radius).</p>
<p><span class="math-container">$$M_t = \int_0^{R} 2\pi q r^2 dr$$</span>
<span class="math-container">$$M_t = 2\pi q \int_0^{R} r^2 dr$$</span>
<span class="math-container">$$M_t = 2\pi q \left[ \frac{r^3}{3} \right]_0^{R}$$</span>
<span class="math-container">$$M_t = \frac{2}{3}\pi q R^3$$</span></p>
<p>Then the magnitude of the constant pressure <span class="math-container">$q$</span> should be:</p>
<p><span class="math-container">$$q= \frac{3}{2}\frac{M_t}{\pi R^3}$$</span></p>
| 43362 | Modeling torque as surface pressure |
2021-05-25T10:36:26.480 | <p>Does using multiple shear pins under a limiting torque case reduce the minimum required shear pin diameter for the coupling or is the complete load distributed over the shear pins regardless of the number of shear pins that are used?</p>
<p><a href="https://i.stack.imgur.com/Fzjab.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fzjab.png" alt="Pinned sleeve shaft coupling formula (Mechanical Engineer Data handbook - James Carvill)" /></a></p>
| |torque|mechanical|shear|coupling| | <p>Multiple pins would share the torque if they are radialy positioned and the sleeve bore holes are machined snug.</p>
<p>For example in your case if we have two perpendicular pins the shear on each end is 1/4 of total shear.</p>
<p>In many precisision applications such as aviation or heavy construction that they require multiple fasteners, even in complex geometry like the empenage of an airplane, the load is divded between the rivets. So each rivet takes a contributory load as a fraction of the total joint load.</p>
| 43363 | Effect of multiple shear pins used in combination in a coupling |
2021-05-25T22:42:55.823 | <p>I was reading documents about energy importation and came across the following:</p>
<blockquote>
<p>The import was of 5,915.87 MMtoe</p>
</blockquote>
<p>I wonder if MMtoe = <span class="math-container">$10^6*10^6$</span> toe = <span class="math-container">$10^{12}$</span> toe.</p>
<p>First time coming across a unit expressed like that.</p>
| |energy|unit|fuel-economy| | <p><strong>TL;DR: the MM in MMtoe is <em>probably</em> related to the Latin Numeral M (for 1000). In that context MM means one thousand thousands (i.e. one million)</strong></p>
<hr />
<p>Although this is very confusing, the bottom line is that to (my understanding)</p>
<p><span class="math-container">$$ 1 \text{ MMtoe} = 1 \text{ Mtoe} = 10^6\text{ toe}$$</span></p>
<p>Where:</p>
<ul>
<li>1 toe = 11.63 megawatt-hours (MWh)</li>
<li>1 Mtoe = 1 MMtoe= 11.63 terawatt-hours (TWh)</li>
</ul>
<h2>Metric Ton or Tonne interpretation</h2>
<p>To my (initial) understanding (I am willing to bet to a lot of other people accustomed to SI units):</p>
<ul>
<li><strong>MMtoe</strong> stood for "Million <strong>Metric Ton</strong> of oil equivalent".</li>
<li><strong>Mtoe</strong> stands for "Million <strong>Tonne</strong> of oil equivalent".</li>
</ul>
<p>So, I thought that the confusion arose because:</p>
<ul>
<li>There are many <em>ton</em> units (see below)</li>
<li>"<strong>M</strong>etric <strong>T</strong>on" is a "Ton<strong>ne</strong>",</li>
<li>1 toe is defined strictly as <em><strong>Tonne</strong> of oil equivalent</em></li>
</ul>
<hr />
<h3>Brief description of <a href="https://en.wikipedia.org/wiki/Ton" rel="nofollow noreferrer">Ton units</a></h3>
<p>This is a brief explanation of the ton units.</p>
<ul>
<li><p>the <a href="https://en.wikipedia.org/wiki/Tonne" rel="nofollow noreferrer">metric ton</a> (<em>which is relevant in the context of <strong>toe</strong></em>), is equal to 1000 kilograms, or approximately 2204 pounds. The <strong>metric ton</strong> is officially called <em><strong>tonne</strong></em>. The SI standard calls it <strong>tonne</strong> or <strong>Megagram</strong> (<strong>Mg</strong>), but <em>the U.S. Government</em> recommends calling it metric ton.</p>
</li>
<li><p>The <a href="https://en.wikipedia.org/wiki/Long_ton" rel="nofollow noreferrer">British ton</a> is the <em>long</em> ton( or displacement ton), which is 2240 pounds, and</p>
</li>
<li><p>the <a href="https://en.wikipedia.org/wiki/Short_ton" rel="nofollow noreferrer">U.S. ton</a> is the <em>short</em> ton which is 2000 pounds.</p>
</li>
</ul>
<p>Both the long and short tons are actually defined in the same way. 1 ton is equal to 20 hundredweight. It is just the definition of the hundredweight that differs between countries.</p>
<ul>
<li>In the U.S. there are 100 pounds in the hundredweight, and</li>
<li>in Britain there are 112 pounds in the hundredweight. This causes the actual weight of the ton to differ between countries.</li>
</ul>
<hr />
<h2>Latin Numerals Interpretation</h2>
<p>Then after some prodding from PeteW comments, I looked around a bit, and found the following <a href="https://www.investopedia.com/terms/m/mcf.asp" rel="nofollow noreferrer">link</a>.
And it actually gets even more confusing.</p>
<p>It actually goes says that the <span class="math-container">$M$</span> is the Latin Numeral for 1000 (and not from Mega = 1000000). And that <span class="math-container">$MM$</span> stands for one thousand thousands (ie. one million).</p>
<p>That explains why, McF, and MMBtu are used in the manner they are used (and probably even MMtoe is the same because it seems to apply to energy units).</p>
<h2>Bottom line</h2>
<p>Despite all this (<em>inconsistency, arbitrariness and -at the end of the day- just plain craziness</em>), the bottom line is that MMtoe (irrespectively to the interpretation you use) is equal to 1 million tonnes of oil equivalent.</p>
| 43369 | What does MM prefix mean in MMtoe? |
2021-05-26T03:42:26.943 | <p>I would like to know if a lifting force will be created by the Coanda effect from high velocity air that is directed underneath a half-toroid shaped saucer by a centrifugal impeller.</p>
<p>I am seeking an answer to this question to make sure that I fully understand how the Coanda effect creates a lifting force so I do not waste time and materials building a prototype of it.</p>
<p>To illustrate how such a saucer and a centrifugal impeller could be put together to generate a lifting force via the Coanda effect, I created a 3D CAD drawing of it and I have displayed this drawing below in three different viewing perspectives.</p>
<p>The first drawing shows a cross-sectional view with its main components labeled, the second drawing shows a top side perspective view, and the third drawing shows a bottom side perspective view.</p>
<p><a href="https://i.stack.imgur.com/lj9Z8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lj9Z8.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/sawv2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sawv2.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/ly6bF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ly6bF.png" alt="enter image description here" /></a></p>
<p>I want to point out that this saucer is not a true half-toroid shape because one side of this toroid goes up at a 45 degree angle and the other side slopes downward in an arc. The reason for this 45 degree angle is to have the airflow go across a flat surface which should minimize the Coanda effect across this particular surface area of the saucer.</p>
<p>If I understand the Coanda effect correctly, the high velocity air exiting the centrifugal impeller should be pushed up into the half-toroid area of the saucer by ambient air pressure. Some of the kinetic energy of the high velocity air molecules pushing against the bottom surface of the saucer's toroid area should be transferred into the saucer which should cause the saucer to move in an upward direction.</p>
<p>There is a circular cutout in the center of the saucer to allow air to flow into the centrifugal impeller. (Inflowing air should rapidly rotate in order to enter into the rotating impeller creating an air vortex above the top surface of the saucer. This vortex should aid in lifting up the saucer, although this is secondary to the answer that I am seeking.)</p>
<p>Will high velocity air directed underneath a half-toroid shaped Coanda saucer by a centrifugal impeller create a lifting force?</p>
<p><strong>EDIT</strong></p>
<p>I am thinking that the half-toroid area of the saucer should be larger in size so that the high velocity air will exert itself against a larger surface area resulting in a stronger lifting force. I also think that the air coming out of the centrifugal impeller should be as close as possible to the bottom surface of the saucer in order to keep any pockets of low static air pressure from forming there.</p>
<p>I have revised the original drawing to show this larger size and to show the repositioned centrifugal impeller.</p>
<p><a href="https://i.stack.imgur.com/jja0f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jja0f.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/8a8oe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8a8oe.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|fluid-mechanics|design|aerospace-engineering| | <p>You have it completely reversed! Your saucer may fly but only because of the momentum of the air flow directed downward (action-reaction). Also maybe because of a possible in-ground effect triggered by a trapped toroidal (more exactly, poloidal) vortex that may arise under the saucer, exactly like for the Avrocar (which never took off beyond this low in-ground effect!)</p>
<p>Coanda radial (also called "conical") jet air streams use a semi-guided nozzle along a free stream, i.e. a curved solid surface in air, that acts as a DIVERGENT to lower the pressure onto top part of the fuselage, inducing lift, like a traditional wing.</p>
<p>The second trick however, and actually the one that produces the main lift in the case of a TRUE Coanda effect (NOT all these pseudo "Coanda saucers" with anemic low-pressure impellers and propellers that have polluted the whole web for a decade now, instead of the REAL THING i.e. Henri Coandă's 1935 original patent and experiments, which all dealt with high pressure supersonic radial superthin nozzles – this has been completely forgotten sadly) is the <em>strong radial expansion of the jet outward</em>, inducing a great pressure reduction, and the main lift.</p>
<p>Forget impellers, PC fans and propellers. They are not adapted at all, their limited ability to move only thick air at slow speed only scratch the surface of the real Coanda effect, and never really trigger it, almost a hundred years after the real deal. A true Coanda saucer would expel a supersonic ultrathin jet stream of air from the top center part of the saucer, along the whole upper part of the fuselage… but admittedly no compressor exists even today that could supply such a tremendous throughput in order to induce the appropriate lift to fly a real-world device. Maybe the air around the disc would have to be ionized and the boundary layer accelerated by the means of electromagnetic Lorentz forces over the whole wet surface instead, thanks to powerful superconducting electromagnets and electrodes, but as Rudyard Kipling said: that's another story.</p>
| 43374 | Will high velocity air directed underneath a half-toroid shaped Coanda saucer by a centrifugal impeller create a lifting force? |
2021-05-26T09:38:56.460 | <p>Suppose I have a concrete wall in front of me and I push on it, now it seems that I do work and the energy has magically disappeared.</p>
<p>I am trying to figure out how this works, the explanation I thought for it is that the wall compresses just a little (similar to how a spring does) and my energy gets stored in increasing the energy of the structure of molecules of the wall. Now, the molecules want to go back but I simply can not observe them because the magnitude of the oscillations are that low.</p>
<p>For the above analysis to be correct, I need that that all bodies are necessarily compressed by some amount when put under a load. Is this statement true? In sense has there been studies on this?</p>
| |materials|structural-analysis|applied-mechanics|solid-mechanics|elastic-modulus| | <h1>When you push the concrete wall the following happens.</h1>
<ul>
<li><p>your body starts to firm up from the major muscle groups around your back and waist core and gets ready for heavy lifting. Some muscles just tighten your waist and push the disks on your lower backbone tight together causing realignment on them.</p>
</li>
<li><p>from there a complex hierarchy of neurons shoots orders down to your feet and shoulder and hand to exert force while balancing the entire action.</p>
</li>
<li><p>your feet will push the floor material down and your hand will push the concrete wall in. The muscle near your waist are now loaded to the max and the path of force travels from your feet carrying the reaction of the floor all the way up to your fist. Creating a lot of heat.</p>
</li>
<li><p>This will cause small deflection in the cement of the wall and the floor and compresses all your joints closer.</p>
</li>
<li><p>while you increase the pressure, the wall and floor even your body will vibrate.</p>
</li>
<li><p>you can easily verify this vibration by slapping the concrete wall, the vibration of the cement surface will issue a loud crashing sound.</p>
</li>
</ul>
<h1>Conclusion</h1>
<p>Nothing is absolutely rigid and any material no matter hough hard under any small load will deform.</p>
| 43379 | Do all structures deform if put a load on them? |
2021-05-26T14:33:03.440 | <p>When using an external .txt file to define global variables for SolidWorks 2021, can one add comments to this file, and what is the syntax?</p>
| |solidworks| | <p>It works the same way as VBA - just add an apostrophe, so, for example:</p>
<pre><code>"Length" = 5mm 'Comment
</code></pre>
<p>This will then be visible in the "Comments" column within the Equation Manager, too.</p>
| 43383 | How to add comments to solidworks variable file? |
2021-05-28T00:25:13.910 | <p>So, as the question states, on steam locomotives with six driving wheels, the rear wheels are farther from the middle wheels than the front wheels are. You can see it in these pictures:
<a href="https://i.stack.imgur.com/i5VF3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i5VF3.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/bFWCV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bFWCV.jpg" alt="enter image description here" /></a></p>
<p>It's sometimes less obvious, but this detail always appears to be present, even if it's really subtle and hard to see. In this picture, if you look closely, you will see that the rear pair of wheels is slightly farther back than the distance between the first two pairs of driving wheels.
<a href="https://i.stack.imgur.com/hGwmL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hGwmL.jpg" alt="enter image description here" /></a></p>
<p>I've noticed that this doesn't appear to be the case for locomotives with eight driving wheels. (Alternatively, it is present, but it's super subtle and I haven't noticed it yet.)
<a href="https://i.stack.imgur.com/t7GXs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t7GXs.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/c1cdp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c1cdp.jpg" alt="enter image description here" /></a></p>
<p>So, what is the reason for this and why does it only seem to apply to steam locomotives with six driving wheels?</p>
| |mechanical-engineering|design|steam| | <p>In the ones pictured, there are no trailing axles - they are a 2-6-0 and a "tenwheeler" 4-6-0 respectively. The ones shown have the firebox between the two rear drive axles. .That was an evolution of the earlier 4-4-0 arrangement. In later 4-6-0, "Prairie" 2-6-2 and "pacific" 4-6-2s and tank engines with those configurations, the firebox was behind all of the drive wheels and they were as close together as possible.</p>
<p><a href="https://www.google.com/imgres?imgurl=https%3A%2F%2Fwww.nps.gov%2Fparkhistory%2Fonline_books%2Fsteamtown%2Fshs2t4.jpg&imgrefurl=https%3A%2F%2Fwww.nps.gov%2Fparkhistory%2Fonline_books%2Fsteamtown%2Fshs2t.htm&tbnid=MOjWtlp0DQCwKM&vet=12ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygIegUIARDLAQ..i&docid=IbivyJhndthhzM&w=500&h=266&q=4-6-0%20steam%20engine%20schematic&ved=2ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygIegUIARDLAQ" rel="noreferrer">Early 4-4-0 showing firebox location.</a></p>
<p>And a <a href="https://www.google.com/imgres?imgurl=https%3A%2F%2Fi.pinimg.com%2Foriginals%2F6d%2F0a%2Fdf%2F6d0adf022a42f67ca4ebd387f9055066.jpg&imgrefurl=https%3A%2F%2Fwww.pinterest.com%2Fwarwickannear%2Flocomotive-cutaways-cross-sections%2F&tbnid=iIWqMPLzjTtcfM&vet=12ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygEegUIARDDAQ..i&docid=iZw83DTCkQMRgM&w=3456&h=2152&q=4-6-0%20steam%20engine%20schematic&ved=2ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygEegUIARDDAQ" rel="noreferrer">4-6-0 with mid firebox</a></p>
<p>And a <a href="https://www.google.com/imgres?imgurl=http%3A%2F%2Fwww.traintesting.com%2Fimages%2FMN%2520diagram.jpg&imgrefurl=http%3A%2F%2Fwww.steamlocomotive.com%2Flocobase.php%3Fid%3D1435&tbnid=QQYgwckArOwFKM&vet=12ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygXegUIARDrAQ..i&docid=eNx58Dcu1uLh4M&w=1276&h=759&q=4-6-0%20steam%20engine%20schematic&ved=2ahUKEwjXv5GFsuvwAhUR8VMKHYQDB6IQMygXegUIARDrAQ" rel="noreferrer">4-6-2 with aft firebox.</a></p>
| 43394 | On steam locomotives with six driving wheels, why are the rear set of wheels placed further apart from the other two? |
2021-05-28T00:41:28.947 | <p>I have an assembly file in Solidworks 2019 with two components, the green, and yellow rectangular prisms.
<a href="https://i.stack.imgur.com/x4OBl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x4OBl.png" alt="front" /></a>
<a href="https://i.stack.imgur.com/1noeP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1noeP.png" alt="left" /></a></p>
<p>I'm having trouble understanding the instructions I found online about making an intersection with these two components. Can someone help me with this?</p>
<p>They form a subassembly in a larger file, and I'm really just interested in the volume that they share.</p>
<p><strong>Edit</strong>
I just found out how to make a 3D sketch outlining the intersection. Tools > Sketch tools > Intersection curve. But I'm not sure how to take this further and extract the intersected volume.
<a href="https://i.stack.imgur.com/yj0ir.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yj0ir.png" alt="intersection" /></a></p>
<p><strong>Edit 2</strong>
I am looking to take these two parts from the subassembly and make a new single part where they intersect.</p>
<p><strong>Yes</strong>, they ultimately do need to be an assembly, but if it can be converted temporarily that's fine. It can be saved under a new name first as a file with a multibody if necessary.</p>
<p>The only <strong>remaining part should be the intersection</strong> of yellow and green. (<em>Although if I think about it carefully, it might be important <strong>not to remove</strong> the portion of green below the intersection in the final part.</em>) The leftovers can be removed as long as it's in a separate file. I still need the subassembly for other things in the larger assembly I mentioned.</p>
| |solidworks| | <p>This is surprisingly tricky, even in SW2020.</p>
<p>If you don't need to operate on individual bodies and your parts are "simple", you can use the <strong>cavity tool within an assembly</strong> to completely remove the volume of GreenPart from YellowPart. Cavity doesn't do an intersection, but you can achieve that effect by modeling one of the parts as a "mold" of the solid you actually want.</p>
<p>If your assembly is "simple", and you just need the final result without keeping live parametrics, you can <strong>save-as the assembly as a multi-body part</strong>, and then use conventional multi-body operations to perform the intersection. Combine, intersect, etc.</p>
<p>However, if you are in the full general case of multibody parts, need to operate just on individual bodies within these, and want to preserve the parametrics for living updates, it is harder. There are two ways I've found, and what I use depends on the parts in question.</p>
<ul>
<li><p>Option A: Edit the YellowPart in-context in the assembly and use the
<strong>offset surface command with an offset of 0</strong> to make surface copies to get the geometry of a body in GreenPart into YellowPart. Then you can knit, and make solid. And then use conventional multi-body operations in YellowPart.</p>
</li>
<li><p>Option B: Within your assembly, create an <strong>incontext 3D sketch</strong> in YellowPart that contains three points that are located at easy-to-find vertices of GreenPart. These will lock down the 6 degrees of freedom. Then do an <strong>Insert Part</strong> in YellowPart to bring in the bodies of GreenPart. Use the <strong>constraints</strong> option to locate the selected verticies of GreenPart where it was in the assembly via the sketch points in the 3d sketch. Then you can again use conventional multi-body operations in the YellowPart.</p>
</li>
</ul>
| 43395 | How to use the intersect tool for two components in an assembly with solidworks? |
2021-05-31T05:18:57.787 | <p>Three blocks of 1N 2N and 3N are placed one over the other respectively. If the friction between the topmost 1N and the middle 2N block is twice the friction between the lowest 3N and floor then determine the minimum horizontal force to disturb the equilibrium. <em>(I tried to solve, the photo contains my Solution.)</em>
<a href="https://i.stack.imgur.com/LKNi6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LKNi6.jpg" alt="enter image description here" /></a></p>
| |dynamics|friction| | <p>Your answer is correct. But I don't think acceleration needs to be involved for this case, rather it can be solved by statics as depicted in the figure below. From the figure, it is clear that in order to mobilize the middle block, the force must be equal to the larger friction force on the top and bottom faces, <span class="math-container">$F \geq 6\mu N$</span>.</p>
<p><a href="https://i.stack.imgur.com/b3PFF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b3PFF.png" alt="enter image description here" /></a></p>
| 43426 | Determine the minimum horizontal force to disturb the equilibrium |
2021-05-31T06:04:16.367 | <p>I'm trying to find the forces in every member. But having a hard time resolving it. The Problem is to be solved by the Method of Sections of Truss.<a href="https://i.stack.imgur.com/mFfy9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mFfy9.png" alt="enter image description here" /></a></p>
| |civil-engineering|beam| | <p>The forces on section b-b are calculated below.</p>
<ol>
<li><p><span class="math-container">$\sum F_J = 0, R_I= +8.33$</span></p>
</li>
<li><p>Draw section diagram with member forces indicated as shown in graph below.
<a href="https://i.stack.imgur.com/hckr1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hckr1.png" alt="enter image description here" /></a></p>
</li>
<li><p>Write the external equilibrium equations for the section:</p>
</li>
</ol>
<p><span class="math-container">$\sum F_X = 0, F_{BA} + 0.8F_{EG} + F_{HG} = 0$</span> -----(1)</p>
<p><span class="math-container">$\sum F_Y = 0, F_{BD} + 0.6F_{EG} - 8.33 = 0$</span> -----(2)</p>
<p><span class="math-container">$\sum M_I = 0, 6F_{BA} + 8F_{BD} + 0.8_F{EG}*3 + 0.6F_{EG}*4 = 0$</span></p>
<p>-----> <span class="math-container">$6F_{BA} + 8F_{BD} +4.8F_{EG} = 0$</span> -----(3)</p>
<p>From (2), <span class="math-container">$F_{BD} = 8.33 - 0.6F_{EG}$</span> -----(2')</p>
<p>(2')->(3), <span class="math-container">$6F_{BA} + 66.64 = 0$</span></p>
<p>-----> <span class="math-container">$F_{BA} = -11.11$</span> (direction in reverse of direction assumed)</p>
<p>Plug <span class="math-container">$F_{BA}$</span> into (1), <span class="math-container">$-11.11 + 0.8F_{EG} + F_{HG} = 0$</span></p>
<p>-----> <span class="math-container">$ F_{EG} = 13.89 - 1.25F_{HG}$</span> ----->(1')</p>
<p>(1')->(2), <span class="math-container">$F_{BD} + 0.6*(13.89 - 1.25F_{HG} = 8.33$</span></p>
<p>-----> <span class="math-container">$F_{BD} = 0.75F_{HG}$</span> ----->(1")</p>
<p>As there are more unknowns than the external equilibrium equations available, we need to investigate an internal joint to get additional information. Let's select joint "E" (see figure below for direction of forces assumed),
<a href="https://i.stack.imgur.com/jvLii.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jvLii.png" alt="enter image description here" /></a></p>
<p>At joint E, <span class="math-container">$\sum F_X = 0$</span> ------> <span class="math-container">$F_{EB} = F_{EG} = 13.89 -1.25F_{HG}$</span>.</p>
<p>And, along member BE, <span class="math-container">$F_{BE} = F_{EB}$</span>, thus,</p>
<p>at joint B, <span class="math-container">$F_{BE} = 13.89 - 1.25F_{HG}$</span> -----> (4)</p>
<p>and, <span class="math-container">$0.6F_{BE} - F_{BD} = 0$</span> ------>(5)</p>
<p>Plug (1") and (4) -> (5)</p>
<p>-----> <span class="math-container">$0.6*(13.89 - 1.25F_{HG}) - 0.75_F{HG} = 0$</span></p>
<p>-----> <span class="math-container">$F_{HG} = 5.56$</span></p>
<p>Finally, plug <span class="math-container">$F_{BA} = -11.11$</span> and <span class="math-container">$F_{HG} = 5.56$</span> into (1)</p>
<p>-----> <span class="math-container">$F_{EG} = 6.94$</span>,</p>
<p>and <span class="math-container">$F_{HG}$</span> into (1')</p>
<p>-----> <span class="math-container">$F_{BD} = 4.17$</span></p>
<p><a href="https://i.stack.imgur.com/QIIO9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QIIO9.png" alt="enter image description here" /></a>
<strong>Please check and advise for mistakes.</strong></p>
| 43428 | Find out the forces in members (as mentioned in the figure) of the truss by method of sections |
2021-05-25T00:17:53.593 | <p>I would like to know if a lift force will be created by the Coanda effect from high velocity air that is directed under a half-toroid shaped saucer by a centrifugal impeller.</p>
<p>To illustrate how such a saucer and a centrifugal impeller could be put together to generate a lift force via the Coanda effect, I have created a 3D CAD drawing using Autodesk Fusion 360 and I have displayed this drawing below in three different viewing perspectives.</p>
<p>The first drawing shows a cross-sectional view with the main components labeled, the second drawing shows a top side perspective view, and the third drawing shows a bottom side perspective view.</p>
<p><a href="https://i.stack.imgur.com/VJTTy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VJTTy.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/uxsnh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uxsnh.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/OT0jS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OT0jS.png" alt="enter image description here" /></a></p>
<p>Will a lift force be created by the Coanda effect if high velocity air flows under a half-toroid shaped saucer?</p>
<p><strong>EDIT</strong></p>
<p>I just want to point out that this question was migrated over from Aviation SE which is showing an older design that will most likely create very little or no lifting force. I have requested that this particular question be closed or deleted.</p>
<p>Please see a newer question that I asked on Engineering SE which shows a revised design that should create a lifting force: <a href="https://engineering.stackexchange.com/questions/43374/will-high-velocity-air-directed-underneath-a-half-toroid-shaped-coanda-saucer-by">Will high velocity air directed underneath a half-toroid shaped Coanda saucer by a centrifugal impeller create a lifting force?</a></p>
| |aircraft-design|aerodynamics| | <p>No. Lift will not be created.</p>
<p>The split toroid shape is not of any use. The airflow turning into the "donut" creates turbulence (and hence losses) <em>and</em> a downwards force since the airmass is accelerated upwards.</p>
<p>Further degrading any desired lift would be the fact that the air would mostly circulate out of and into the impeller. As mentioned in Robert's answer, the air should be pulled into the impeller from above, but it would still be by far less efficient than a simple propeller.</p>
<p>Google "coanda propeller" to find a plethora of evidence that coanda devices are not more efficient than simpler propeller arrangements.</p>
| 43438 | Will a lift force be created by the Coanda effect if high velocity air flows under a half-toroid shaped saucer? |
2021-06-01T08:16:50.817 | <p>we were trying to find one <strong><a href="https://www.google.com/search?q=site%3Agrabcad.com%20Storage%20Shed%20industrial%20&tbm=isch&ved=2ahUKEwjugP2I-PXwAhX067sIHezBADMQ2-cCegQIABAA&oq=site%3Agrabcad.com%20Storage%20Shed%20industrial%20&gs_lcp=CgNpbWcQA1DLIFjLIGDNJmgAcAB4AIAB4AGIAeABkgEDMi0xmAEAoAEBqgELZ3dzLXdpei1pbWfAAQE&sclient=img&ei=POW1YO7yIfTX7_UP7IODmAM&bih=757&biw=1332&hl=fa#imgrc=ocH4uVfBToNBOM" rel="nofollow noreferrer">Dal Mill Storage Shed</a></strong>, CAD file form the <a href="https://grabcad.com" rel="nofollow noreferrer"><strong>grabcad site</strong></a>.</p>
<p><a href="https://i.stack.imgur.com/WUVXy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WUVXy.png" alt="enter image description here" /></a></p>
<p>If Possible i like to understand can we use the shared cad file of the grabcad site for <strong>commercial usage</strong>.</p>
<p>I have seen <a href="https://grabcad.com/questions/license-for-free-models" rel="nofollow noreferrer">this post</a> about the <strong>grabcad</strong> shared file licence, but I don't understand finally is it secure to use the shared files form grabcad for commercial project. if not, I like to know, <strong>How much change</strong> (percentage) on one cad files is necessary for removing the copyright of one close licence CAD file?</p>
<ul>
<li>List item</li>
</ul>
<p>Thanks.</p>
| |cad| | <p>The Terms are pretty clear on the website: <a href="https://grabcad.com/terms" rel="nofollow noreferrer">https://grabcad.com/terms</a></p>
<blockquote>
<p>6.2. each user of the Site a worldwide, non-exclusive, royalty-free, fully-paid, perpetual, irrevocable, non-sublicensable, non-assignable,
and non-transferable license to use, reproduce, copy, modify, adapt,
arrange, translate, and create derivative works of such User
Submissions for such user's own benefit and <strong>non-commercial</strong>, internal
use (the "Cross License")</p>
</blockquote>
<p>The license you get from downloading anything from GrabCAD is strictly non-commercial. This includes derivative works - there is no threshold above which you can 'remove the copyright'.</p>
<p>If you wish to use a model for a commercial project, you must contact the original author directly (<a href="https://grabcad.com/home/inbox/new?to=zulfiqar.islam-1" rel="nofollow noreferrer">https://grabcad.com/home/inbox/new?to=zulfiqar.islam-1</a> for the image highlighted in your post) and agree this with them.</p>
| 43442 | About the License of GrabCad's shared models |
2021-06-01T12:04:09.760 | <p>I want to make a vinyl stencil on a curved surface. I already use vinyl stencils with a vector cutter like Silhouette to make designs on flat surfaces. I also use heat-shrinking tape for insulating a repair of electric cables. So I suppose a combination of both exists and I can cut a heat-shrinking vinyl, glue on a curved surface, and apply heat for it to adhere seamlessly to the surface.</p>
<p>What is the name of such material?</p>
| |plastic|heat-treatment| | <p>Does it have to be vinyl? Model airplanes use this process to apply skins to wood framed wings. It's called "covering film", like Monokote. Just any sheet of shrink wrap with a pre-applied adhesive really.</p>
| 43446 | Heat-shrinking vinyl |
2021-06-01T20:53:00.383 | <p><a href="https://i.stack.imgur.com/77yjn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/77yjn.jpg" alt="a particle moves in a straight line from a constant point on straight line starting from rest such a=3/8 x^2 find velocity when x = 2" /></a></p>
<p>the question is clear and straight forward but when i get the velocity it should be ±√2</p>
<p>what i don't understand is the guide answer refusing the -ve value because acc is always +ve
<a href="https://i.stack.imgur.com/mxb3p.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mxb3p.jpg" alt="the negative is refused because a > 0" /></a></p>
<p>thank you in advance</p>
| |dynamics|acceleration| | <p>I will rephrase DKNGuyen's answer (which is correct - I upvoted ).</p>
<p>I assume that the positive (+) sign is for position, velocities and acceleration is to the right.</p>
<p>The object is starting from rest. Therefore its velocity is zero. We don't explicitly know the initial position, but we can assume that its a very small value different that zero (any value however small would do and whichever direction <span class="math-container">$\pm$</span>).</p>
<p>The acceleration is given by:</p>
<p><span class="math-container">$$a(x)= \frac{3}{8} x^2 >0 $$</span></p>
<p>Therefore, because a is greater that zero, the object will accelerate towards the right. There is no position where acceleration can become negative (and therefore at any given time the acceleration can be positive or zero), therefore, the equation:</p>
<p><span class="math-container">$$V = V_0 + a\cdot t$$</span></p>
<p>means that the velocity is always greater than zero (since the initial velocity is <span class="math-container">$V_0$</span>) in the parameters of this problem.</p>
<hr />
<p>The other solution (-ve) the post mentions is reserved for the case when:</p>
<p><span class="math-container">$$a = \color{red}{\mathbf{-}}\frac 3 8 x^2$$</span></p>
| 43454 | is a positive acceleration indication that velocity must be positive? |
2021-06-02T07:11:23.163 | <p>I am doing an experiment testing different types of gears, the gears are fairly small, with gear 1 being approx. 35mm in diameter and gear 2 being 45mm in diameter, and both are meshed together, there are also around 15mm wide. the gears will be rotating at a max speed of 2000 RPM also. I want to test a lot of the same gears with different constrains, but want the lubrication to remain the same, since these gears tangential velocity are not that fast, many suggest Grease as lubrication, but I want to use the exact same amount of grease each time I test the gears so it does not affect the results, another maybe the oil bath method, which will allow me to put an exact amount of oil each time however might seem a bit overkill for gears which are so small.</p>
<p>does anyone know Which method of lubrication will be better and if it is grease, what are the suggestions to make sure I use the exact same amount each time so It does not affect the results.</p>
<p>thanks</p>
| |mechanical-engineering|gears| | <p>one solution would be to spray the lubricant (provided it's fluid enough) on the gears</p>
<p><a href="https://i.stack.imgur.com/bZhKJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bZhKJ.png" alt="enter image description here" /></a></p>
<p><em>Figure 1: source <a href="https://www.machinerylubrication.com/Read/614/airless-spray-gear-lubrication" rel="nofollow noreferrer">machinery lubrication</a></em></p>
<p>By spraying you can make sure that you control the flow of the lubricant, and therefore you can control parameters like:</p>
<ul>
<li>heat abduction</li>
<li>estimated thickness of film on the gears</li>
<li>etc</li>
</ul>
<hr />
<h1>Splash lubrication</h1>
<p>Another solution is <strong>Splash lubrication</strong> (but not optimal for the parameters you have provided).</p>
<p><a href="https://i.stack.imgur.com/zg5eY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zg5eY.png" alt="enter image description here" /></a></p>
<p><em>Figure 2: Splash lubrication <a href="https://addinol.de/en/service-en/expert-tip/splash-lubrication/" rel="nofollow noreferrer">addinol</a></em></p>
<p>The advantage is that it is more cost effective. The only problem is that with the dimensions of the gear only at the maximum rpm you are at the low end of the recommended velocity.</p>
<p>I.e. Assuming that the 45[mm] is rotating at 2000[rpm], then the velocity will be about 4.7[m/s]. (if the 35 mm is @2000[rpm] then the velocity is about 3.67[m/s]).</p>
<p>The recommended range velocities for the different types of lubrication can be seen in the following image</p>
<p><a href="https://i.stack.imgur.com/RJ298.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RJ298.png" alt="enter image description here" /></a></p>
<p><em>Figure 3: Ranges of tangential speed (m-s) for spur gears and bevel gears source <a href="https://khkgears.net/new/gear_knowledge/gear_technical_reference/lubrication-of-gears.html" rel="nofollow noreferrer">KHK gears</a></em></p>
| 43460 | How should I lubricate my gears? |
2021-06-02T13:17:21.920 | <p>Compare to PID. The part that rejects disturbance is the I-term which integrated from error signal. On the other hand, in full state feedback there are no integral part. How does it reject the disturbance? or How we can make it to?</p>
| |control-theory| | <p>As you already stated at your question, PID-Controllers incorporate the I-Term in order to able to reject any external disturbance (unmodeled dynamics, linearizations, etc) and as a result achieve the desired output performance by driving the steady state error to <span class="math-container">$0$</span> (<span class="math-container">$e_{ss} \rightarrow 0$</span>).</p>
<p>Full state feedback control is like having only the P and D term of the PID-Controller and produce the control signal:</p>
<p><span class="math-container">$$u = -k_1x_1 - k_2x_2,$$</span></p>
<p>where <span class="math-container">$x_1$</span> is the position and <span class="math-container">$x_2$</span> is the velocity of the plant. Full state feedback relies mainly on the model you have obtained for the system, which will NEVER be perfect. So, any unmodeled dynamics and model disturbances will be present inside your <span class="math-container">$A$</span> matrix (<span class="math-container">$\dot{x} = Ax+Bu$</span>) and furthermore you will have to suffer the steady state error which will certainly exist when trying to implement such a controller to a physical system because at simulations everything can look perfect. So, with fun state feedback you can't compensate the external disturbances.</p>
<p><strong>SOLUTION:</strong> There is the extension of full state feedback, called <strong>dynamic state feedback</strong>. The whole idea is to increase the system order by <span class="math-container">$1$</span> and introduce an integral term. The new state-space representation will look like this (for a second order system):</p>
<p><span class="math-container">$$\begin{gather} \dot{x} = Ax + Bu \\
y = Cx \\
\dot{z} = Cx - r = y - r \\
u = -k_1x_1 -k_2x_2 -k_iz \end{gather}$$</span>
where <span class="math-container">$x$</span> are the states of the system, <span class="math-container">$y$</span> is the output of the system, <span class="math-container">$z$</span> is the integral term, <span class="math-container">$u$</span> is the control signal and <span class="math-container">$r$</span> is the reference point. The procedure in order to obtain the values of the gains <span class="math-container">$k_1,k_2,k_i$</span> is the usual, by obtaining the characteristic polynomial, finding its roots and so on.</p>
<p>One thing to notice is that by increasing the order of the system (from <span class="math-container">$2$</span> to <span class="math-container">$3$</span>) you have an extra pole. This pole should be considered as the third pole and the desired characteristic polynomial will look like this (one of the two forms):</p>
<p><span class="math-container">$$ p_d(s) = (s+p_3)\cdot(s+p_2)\cdot(s+p_1) = (s+p_3)\cdot(s^2+(p_1+p_2)s+p_1p_2) $$</span></p>
<p><span class="math-container">$$ p_d(s) = (s+p_3)\cdot(s^2+2\zeta \omega_ns+\omega_n^2) $$</span></p>
<p>So, you should choose the third pole far enough in order for the system's transient and steady state response to be defined by the two dominant poles. However, choosing poles far to the left-half plane will produce large gains, which most likely will not be able to be implemented by the system's actuators. So, this is a trade-off which you, as a control engineer, should try to figure out and achieve the best result.</p>
| 43467 | Can fullstate feedback control reject a disturbance? |
2021-06-02T16:46:05.753 | <p>I am experimenting to analyse the liquid wicking(radial) behaviour of various liquids. I have performed initial studies and experiments on household paper towels but they tend to deform and stick to the surface they are lying on, when wet, as shown below.</p>
<p><a href="https://i.stack.imgur.com/SbNKq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SbNKq.jpg" alt="Liquid drop spreading on paper via wicking and wetting" /></a></p>
<p>I aim to find another paper with a high absorption speed. Wicking is apparent and produces an observable difference in speed and radius of wicking different liquids.</p>
<p>I am thinking of trying out with filter papers next as some similar studies have used filter papers.</p>
<p>Are there any other kinds of papers with a high absorption speed?</p>
<p>Also, if filter papers are a good choice, which grade or quality or any other attribute should I care to maximise absorption speed?</p>
<p>PS: I checked out chromatography paper also but they are costlier than filter papers. Unless the absorption speed difference is large I would rather use filter papers.</p>
<p>This happens on filter papers on the other hand for some liquids. The spread is minimal and a large drop of liquid just stays for a long time.
<a href="https://i.stack.imgur.com/nSNDT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nSNDT.jpg" alt="enter image description here" /></a></p>
| |fluid-mechanics|fluid| | <p>If you want to gage how wetable a liquid is, I would think, wetting angle (and possibly surface tension) measurement(s) would better differentiate between liquids. Low contact angles means more wetting. Capillary force was used to dial-in ink properties to optimize the Ball Point Pen's flow to the paper. Too high a force and the ink would skip and the meniscus at the ball would break. Too low and the ink blobbed.</p>
<p>I don't know if any paper product would exhibit equivalent adsorbent properties as fabricated from one day to the next, fiber distribution density, seasons of the year, season of fiber harvest or location of manufacture (as variables), might all have an effect. There is a nose tissue that is meant for kids with colds. It has "lotion" pre-applied. It is hydrophobic. You want hydrophilic. Any treatment applied to the media other than a wetting agent probably would degrade performance. You can probably create your own highly absobent media by taking your choice of product (filter paper maybe) and pre-treating it with wetting agents (or surfactants) like sodium lauryl sulfate, to increase absorption speed. Search on the terms above.</p>
<p>Also, any contact of the "paper" media and the support surface below, will create their own capillaries and adhesion of the paper to the support. I would say, place the paper over a hole in the support surface so it is just the paper product making contact with the liquid. Use a consistent drop size, consistent temperature of fluids and media, consistent test humidity and water fraction of the media, and maybe gage with an automated optical system (for wetted surface area maybe even area versus time).</p>
| 43470 | Which kind of paper has highest absorption speed? |
2021-06-02T16:59:06.027 | <p>If I understand correctly (history shows I probably don't), then the only reason why the absorption chiller evaporators are restricted (I hope that's the right word) to extract just enough heat so the chilled water exits at around 3 Celsius is so that the danger of the chilled water becoming frozen, and thus not being able to be pumped, is avoided by a safe margin.</p>
<p>Other than that I understand that most of the heat transfer at the evaporator takes place not because of the temperature gradient, but because the refrigerant water is kinda forced to evaporate due to the vacuum headspace (okay, your eyes bleed, but you're the physicist, not me). So I suppose if the flow rate in the chilled water pipes is lower, and every molecule of water passing through can lose more heat, the chilled water could essentially freeze?</p>
<p>This leads me to my question: If water in the chilled water pipe, is replaced by air, can't we use it to freeze things in a freezer? Is it possible?</p>
| |fluid-mechanics|thermodynamics|heat-transfer|pressure| | <p>Of course we could go below freezing using absorption cooling. The question is who would want to.</p>
<p>Lithium Bromide chillers is what I'm familiar with, and the temperatures they can produce is limited less by the chilled medium and more the properties of the LiBr solution. Using brine as the heat transfer medium is common when we need to use chillers below freezing. But I can't get "more vacuum" in the evaporator since the vacuum is really low (high? I'm unclear what the usage really is, lol) already. Theoretically there should be other absorption solutions that will go to lower temperatures to take your evaporator to below-freezing temps.</p>
<p>The issue with these units is that they are a maintenance nightmare. It is really hard in the field to maintain complex equipment that maintains a very low pressure vacuum. I'm guessing the reason we don't see these in use for ice making is that ammonia refrigerant plants are cheaper to build and maintain.</p>
| 43471 | Absorption chillers: can they achieve temperatures below zero Celsius? |
2021-06-03T16:23:01.343 | <p>I read about absorption chillers and found out that the water is evaporated/boiled off at the generator.</p>
<p>Water boils over at about 3.7 Celsius in the near-vacuum conditions in the generator. Now the boiling point is also the condensing point! But when this same water condenses at 35 Celsius in the condenser despite the pressure being the same near-vacuum (or is it?), it sends my brain to physics hell.</p>
<p>The only explanation my pea-brain can think of is that the steam itself creates pressure that sort of increases... I hate to say it... its own boiling (thus condensing) point. And so now it condenses at 35 rather than 3 Celsius. Does this even make sense to you?</p>
| |fluid-mechanics|thermodynamics|heat-transfer|pressure| | <p>The whole point of an absorption chiller is that it doesn't use condensation: it uses absorption.</p>
<p>The steam doesn't 'condense'. It is 'absorbed'.</p>
<p>It is, as it happens, absorbed into a liquid phase, but that liquid is not water, and the steam doesn't become liquid by condensation: it becomes liquid by absorption.</p>
<p>Although the words 'condenser' and 'condensation' are often used by well meaning people when describing an absorption chiller that works between gas and liquid phases, that can be confusing and misleading.</p>
| 43488 | How does steam in almost vacuum condense at 35 Celsius? |
2021-06-04T06:09:11.140 | <p>Assume two buildings, one which generates a lot of air heat inside due to electrical equipment running etc. The buildings are 20-25 meters apart. Some questions:</p>
<ul>
<li>How can one easily transfer heater air from the one which produces the heat to the other?</li>
<li>Is a fan with a duct sufficient or would an improvement be seen with some sort of heat pump or exchanger or compresses or similar?</li>
<li>What is the most likely product to help here?</li>
<li>How do I search for a product to buy, i.e. what sort of terminology is used to describe such items?</li>
<li>For the duct, can I just use a standard airco duct or?</li>
</ul>
<p>I am looking for generic answers to enable further research. I am mostly stuck on the concept of "heat transfer" from one building to the next and do not know what is commonly used to do this and/or how effec</p>
| |heat-transfer|airflow|heating-systems|heat-exchanger| | <p>you should look into <a href="https://en.wikipedia.org/wiki/District_heating" rel="nofollow noreferrer">district heating</a> solutions. For your application the lengths are quite small so you probably can get away with using an air duct and a fan to push the air through.</p>
<hr />
<p>However, a better way is (<em>probably</em> optimal) to actually use a air/liquid heat pump which extracts the heat (cools down the air in one building), and heats up water which is used to transfer the heat to the other building.</p>
<p>The benefits of the second solution are:</p>
<ul>
<li>smaller diameter ducts (heat capacity of water is significantly greater)</li>
<li>less cost on insulation (the diameter of the ducts will be limited)</li>
<li>you can distribute the heat transfer medium (water) easier to different location of the building without worrying too much about losses, and then use fan coils to distribute the heat.</li>
<li>Some minor safety concerns, e.g.: in case of a fire into building A (hot), the smoke will not be driven into building B.</li>
</ul>
<p>The main drawback of the second solution, is the higher initial cost.</p>
<hr />
<p>A few additional question clarifications:</p>
<ul>
<li><strong>Would the first solution also require insulation around the air duct, however large?</strong></li>
</ul>
<p>Yes, both solutions would require significant insulation. The problem with the airduct, due to its larger cross-section it would require more insulation.</p>
<ul>
<li><strong>Looking at solutions online, there are a lot of liquid-to-air pumps, though I guess I need two pumps to 1) capture the heat in building A, 2) convert to liquid, 3) deconvert back to air, 4) heat building two?</strong></li>
</ul>
<p>I am not sure what you mean exactly here, but you'd need at least one. However if you use fan coils you can just divert the hot water directly to them.</p>
<ul>
<li><strong>Is there a way to prevent a duct from transferring smoke (and possibly fire, though that would seem like a long stretch) in the case that there is a fire issue with the first building in the first solution.</strong></li>
</ul>
<p>You could use filters (probably HEPA but I am not an expert), but the type of filters and the frequency that they would need replacing -IMHO- are going to be too costly over the long run.</p>
| 44494 | How to transfer air heat easily? |
2021-06-04T15:07:55.073 | <p><a href="https://i.stack.imgur.com/oFVrL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oFVrL.jpg" alt="enter image description here" /></a></p>
<p>We have standard profiles such as this type of tube with outer diameter of 60.3mm, used for example in railing. <strong>Why is it 60.3mm rather than exactly 60mm? In what application does the extra 0.3mm make a significant difference?</strong> I've also seen profiles ending in .7mm. These sound inconvenient to produce compared to full millimiters.</p>
| |civil-engineering|steel| | <p>Circular hollow sections were standardised a long time ago and in inches for historical reasons. This should be obvious if you look at a complete list and convert the units.</p>
<p>The question is better asked the other way around: Why should we change the standard diameter from 60.3mm to 60.0mm, which would an additional cost for the manufactorers, when the 0.3mm difference doesn't make a significant difference anyway?</p>
| 44503 | Why do we have profiles with fraction millimeter dimensions? |
2021-06-05T13:39:14.040 | <p>I'm looking for a simple way (that doesn't require high pressure chambers or state of the art tools) to permanently bind a set of carbon fibre pieces in order to make the frame of a chassis that could be used in a go kart or dune buggy or even a boat? (watercraft is just a bonus, it just needs to support the weight and impacts of a 300kg vehicle, including the driver).</p>
<p>The raw pieces of carbon fibre are as follows:</p>
<p><a href="https://i.stack.imgur.com/TLcxC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TLcxC.png" alt="enter image description here" /></a></p>
<ul>
<li><p><strong>2</strong> x 10mm square/round tubes (up to 1500mm length) For the outer frame's longitudinal beams (shown in <strong>blue</strong>)</p>
</li>
<li><p><strong>6</strong> x 10mm-by-5mm strips (up to 600mm length) For the outer frame's lateral beams and cross-diagonal support beams (shown in <strong>green</strong>)</p>
</li>
<li><p><strong>2</strong> x 10mm rods (up to 800mm length) For a longitudinal reinforcement, also to be sticking out as an axle mounting point (shown in <strong>black</strong>)</p>
</li>
<li><p><strong>1</strong> 6mm thick (very large) sheet of carbon fibre cut to shape and cover the chassis, walls and crevices (not shown)</p>
</li>
</ul>
<p>The methods I'm aware/capable of are:</p>
<ul>
<li>Drilling/tapping screw holes</li>
<li>Using epoxies and other chemical adhesives</li>
<li>using carbon fibre vinyl as a tightly wrapped joint</li>
<li>Using ropes/cables where a long piece protrudes from the body</li>
<li>Welding...</li>
</ul>
<p>That last method I believe is useless in a carbon fibre build, however I have plenty of access to steel materials and the tools to cut/bend/drill them into virtually anything, would this serve any use for my build?</p>
<p>I'm thinking about drilling holes to make mounting points which I can then glue together and seal off with a washer or bung or a locking stopper.</p>
<p>My real questions are (TLDR):</p>
<ol>
<li><p>How does drilling holes into carbon fibre affect its structural integrity?</p>
</li>
<li><p>Which bonding agents are ideal for carbon fibre in a structure that needs to withstand impact from many directions (especially up-and-down)?</p>
</li>
<li><p>Is it beneficial to use both steel and carbon fibre together in the same structural component?</p>
</li>
<li><p>Are there any composite material alternatives to Carbon Fibre that are just as, or possibly more applicable to this type of structure? I'm looking for something with a light weight, high tensile stress, resistance to abrasion and of course, a decent price.</p>
</li>
</ol>
<p>Thank you. Please forgive if this question is elementary, I've not been researching composites for long but can't wait to build something.</p>
| |structural-engineering|automotive-engineering|composite|adhesive|frame| | <p>Cool that you're considering CF, it's a fun material.</p>
<p>So first, to answer your questions.</p>
<ol>
<li>If done right it can be fine, but even then it's the least preferable (in my opinion) way to bond FRPs (Fiber Reinforced Plastics). It's hard to do well without special tools, especially in curved and/or hollow profiles, since you need to support the material around the hole on both the entry and exit side.
If not done right you will most likely get de-lamination and open up to a broad range of ways for your structure to fail.</li>
<li>General answer, Epoxy. Try a one-component epoxy adhesive for high impact resistance.</li>
<li>It can be, depends on application. Aluminum is often used due to low weight and price, tough very sensitive to corrosion and galvanically a terrible match for carbon. Titanium works well but is expensive to process.</li>
<li>Wood :) Glassfiber. Carbon is preferable for your structure though.
<strong>Note:</strong> Carbon and glass FRPs are <strong>not</strong> resistant to abrasion, in fact this is the preferred way to cut them. Because the fibers and matrix (plastic )are brittle. Composite structures often utilize metals and engineering plastics like Acetal to deal with abrasion. If you need abrasion resistance during a crash you could incorporate some aramid fibers (like Kevlar).</li>
</ol>
<hr />
<p>About your actual question, how should you bond your structure together?
Since you asked for a permanent bond I would recommend you to steer clear from nuts and bolts.</p>
<p>I would suggest you try the following.</p>
<ol>
<li>Place all your parts in their desired positions, a simple wooden jig with some clamps will make it easier.</li>
<li>Bond the parts together with a non-sagging epoxy adhesive, give it time to cure (harden).</li>
<li>Wrap the joints tightly with carbon fiber weave or NCF soaked in epoxy resin, let cure.</li>
</ol>
<p>Tools needed. Buckets to mix in. Sticks for stirring. Gloves for protection. Scissors or knife to cut carbon mats.</p>
<p>Important!</p>
<ul>
<li><strong>Always use protective gloves!</strong>
Contact with uncured epoxy can make you allergic to epoxy. Oil from your skin will interfere with the bonding. Cut (or sanded) glass and carbon fibers are very itchy and annoying ;)</li>
<li><strong>Don't mix to much epoxy resin at once!</strong> The reaction is exothermic and can cause a thermal runaway that might result in a toxic fire.</li>
<li>Always sand down any surfaces where you plan to use epoxy, don't apply epoxy to glossy surfaces.</li>
<li>Practice on some scrap pieces, the bonds are permanent.</li>
</ul>
<hr />
<p>Regarding your design.</p>
<p>I feel like your structure is severely underdimensioned if the plan is to survive a crash, sorry if I sound harsh.</p>
<ul>
<li>I think you want something closer to 40+ mm tubes for your blue and black members, skip the rods and go for tubes.</li>
<li>Maybe 30+ mm tubes for the green members, strips are terrible for all loads except tension.</li>
<li>Do not even consider using vinyl wrap for load bearing joints.</li>
</ul>
<p>Cheers :)</p>
| 44520 | What Are The Recommended Binding Methods For A Carbon Fibre Chassis Frame? |
2021-06-05T21:39:27.160 | <p>I have a style spline that when extruded have sharp coroners inside of a smooth curve</p>
<p>Is there a way to increase its resolution?</p>
<ul>
<li>Note that its not solidworks graphics setting, it does in fact actually 3d print like shown in the picture</li>
</ul>
<p><a href="https://i.stack.imgur.com/gFan3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gFan3.png" alt="enter image description here" /></a></p>
<p>When I try to split with an offset line, this happens
<a href="https://i.stack.imgur.com/e1UQ8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e1UQ8.png" alt="enter image description here" /></a></p>
<p>There is a thin wall that is cut as well, which shows that there is something going</p>
| |solidworks| | <p>Your problem is in your STL export settings - the spline <em>is</em> perfectly smooth mathematically, and there are separate settings for the SW graphic window and STL export which both approximate the curve as a series of straight lines.</p>
<p><a href="http://help.solidworks.com/2020/english/SolidWorks/sldworks/hidd_stl.htm?verRedirect=1" rel="nofollow noreferrer">http://help.solidworks.com/2020/english/SolidWorks/sldworks/hidd_stl.htm?verRedirect=1</a></p>
| 44525 | Increasing the style spline resolution |
2021-06-05T22:13:23.170 | <p>Hello I had a quick question regarding moments on hinged gates on submerged curves. Is there a general rule that we follow when we draw moments on hinges, do we draw the hinges in the direction the gate would open if enough force acts on it? Do we draw the moments in the direction the gate is current pushing against water? Please help me, I included an image as an example, where would I draw my moments on this hinge C? If its the former would this be the correct direction of moments?
<a href="https://i.stack.imgur.com/ALigi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ALigi.png" alt="Correct moments on C or no?" /></a></p>
| |moments|hydrostatics| | <p>When the gate is drawn fully closed, it is understood that the system is in structural equilibrium with the condition <span class="math-container">$\sum M_C = 0$</span> is true, so there is no moment to show on the diagram.</p>
<p>If a mechanical device is required to assist the hinge to resist the rotation due to the hydrostatic pressure, then its presence must be indicated on the hinge point "C" with the direction in agreement with its purpose. In this case, the same direction as rotation produced by the force "P".</p>
<p><a href="https://i.stack.imgur.com/VkE8L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VkE8L.png" alt="enter image description here" /></a></p>
<p>On the diagram above, it is understood that the rotational equilibrium is maintained thru <span class="math-container">$M_P + M_R - M_W = 0$</span>. (<span class="math-container">$M_P$</span> & <span class="math-container">$M_W$</span> need not shown though.)</p>
| 44526 | How do I draw moments on hinged gates? |
2021-06-07T05:55:41.493 | <p>It is becoming very difficult for me to understand the difference between temperature and heat.</p>
<p>About heat all I have read :</p>
<ol>
<li><p>Heat exchange is related to as exchange of matter + energy.</p>
</li>
<li><p>Heat is related as total kinetic energy of molecules. Translational + vibrational + rotational.</p>
</li>
</ol>
<p>Now, I can kind of relate energy here. For exchange of matter, it’s like if put a ice cube (less kinetic energy ) into fire (high kinetic energy ). Then there is exchange of kinetic energy or both will try to make their kinetic energy equal by one reducing it and the other increasing it.</p>
<p>About temperature:</p>
<ol>
<li>Temperature is related to as average kinetic energy of moles from the equipartition theorem.</li>
</ol>
<p>Now, we know as temperature of room increases, heat of the room also increases.</p>
<p>I am not able to think or find any difference between the two.</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heat| | <p>Temperature is an intensity , heat is a quantity.</p>
| 44546 | Difference between temperature and heat |
2021-06-07T08:57:34.423 | <p>What is the technical difference between a hydraulic cylinder and a spring? For sure, I understand their principles but I wonder about physical-technical differences.
For example, the most obvious aspect (for me), is that they both can deflect. A spring, on the contrary, can store potential energy, is this also valid for the cylinder? It is also able to oscillate. Are there more differences?</p>
| |mechanical-engineering|springs| | <p>@Ben This is to answer the comment whether there are configurations that a hydraulic cylinder acts as a spring.</p>
<p>Please Note that with the term hydraulic cylinder, I consider any cylinder with a hydraulic fluid that can act as an actuator.</p>
<h2>Automotive shock absorber:</h2>
<p>The following is an automotive shock absorbed. Notice the use of oil and gas (usually nitrogen). This type of shock absorved acts like a spring (the gas compresses) and a dashpot (the piston in the oil).</p>
<p><a href="https://i.stack.imgur.com/JYL0Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JYL0Z.png" alt="enter image description here" /></a></p>
<p><em>Figure 1: This is an automotive shock absorber (source: <a href="https://aresmotorsports.com/product/godspeedmmxi-7013-maxx-sports-high-performance-inverted-coilovers-kit-for-subaru-impreza-wrx-sti-gr-gv-2008-14/" rel="nofollow noreferrer">aresmotorsports</a>)</em></p>
<h2>Connected hydraulic cylinders</h2>
<p>The following image are two connected hydraulic cylinders.</p>
<p><a href="https://i.stack.imgur.com/y89wm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y89wm.png" alt="enter image description here" /></a></p>
<p><em>Figure (2): connected hydraulic cylinders (source: <a href="http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/DennisJackson211web/Howtheyworkpage.html" rel="nofollow noreferrer">University of Alaska Fairbanks</a></em></p>
<p>In this configuration you can place a dead weight on one of the two cylinders. Then by applying a force on the other end, the vertical motion of the cylinder effectively stores potential energy.</p>
| 44554 | Difference between Hydraulic cylinder and spring |
2021-06-07T10:35:02.833 | <p><a href="https://i.stack.imgur.com/fnMSv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fnMSv.jpg" alt="enter image description here" /></a></p>
<p><strong>ΔQ = ΔU + Δ(PV)</strong> (First law of thermodynamics) which has the same statement as the 1st equation.</p>
<p><strong><span class="math-container">$Q_p$</span> or ΔH= ΔU + PΔV</strong> which is the heat exchanged at constant pressure.</p>
<p>Then why in the image above in my textbook . ΔH has two equations?</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heat| | <p>I understood the answer to my Q.</p>
<p>ΔH=ΔU+Δ(pV)is the general statement for Δ q and ΔH</p>
<p>Whereas , ΔH=ΔU+pΔ(V)
is a subcase when pressure is constant.</p>
| 44558 | Why ΔH has two separate equation in my textbook? |
2021-06-07T10:56:52.630 | <p>From a particular topic : Enthalpy changes during phase transformations.</p>
<p>It says in my textbook that standard state of a substance at a specified temperature is it’s pure form at 1 bar pressure.</p>
<p>Let’s consider the phase change for <span class="math-container">$H_2O(s)$</span> to <span class="math-container">$H_2O$</span> (L). <span class="math-container">$\delta H$</span> of fusion with a subscript I.e - . It’s value is 6 kJ/mol is value needed for such a change.</p>
<p>This phase change is happening at a standard state according to my textbook.</p>
<p><a href="https://i.stack.imgur.com/AWblr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWblr.jpg" alt="enter image description here" /></a></p>
<p>Here . I have drawn is the image of how is the phase change happening. Mostly , I have just combined all the definition and statements.</p>
<p>The <span class="math-container">$\delta H$</span> of fusion is at standard state. I.e it has a pressure of 1 bar of its own and a specified temperature. It is the amount of energy needed to melt <span class="math-container">$H_2O$</span> (s).</p>
<p>The surrounding are at a temperature = 273K and pressure = 1 atm.</p>
<p>When ice melts at 273K. It is converted to water at 273K. Same temperature. Pressure in scenario is always 1 atm.</p>
<p>I wish to confirm my scenario.</p>
| |mechanical-engineering|thermodynamics|heat-transfer|heat| | <p><a href="https://i.stack.imgur.com/qVcsS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qVcsS.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. The phase diagram for water. The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties. Image source: <a href="https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_122/Chapter_2%3A_Phase_Equilibria/2.4%3A_Phase_Diagrams" rel="nofollow noreferrer">Chem.LibreTexts.org</a>.</em></p>
<p>The situation you are describing is circled in Figure 1. You are travelling along the horizont line through the melting point of water at 1 bar.</p>
<p>In engineering disciplines we don't use kJ/mol but rather work with the latent heat of fusion of water which is 334 kJ/kg.</p>
<p>Using that figure you can work out that a 1 kW (1 kJ/s) heating source will take 334 s (6.5 minutes) to convert 1 kg of ice at 0°C to 1 kg (1 L) of water at 0°C. If you leave the heater on for another 334 s the water temperature will rise to 79.8°C which illustrates how significant the cooling effect of melting ice is.</p>
<p>Now, how much ice do you need to add to a 330 ml can of drink to cool it from 20°C to 4°C?</p>
| 44559 | Can I consider this scenario for the phase change of $H_2O_{(s)}$ to $H_2O_{(L)}$ |
2021-06-07T14:30:51.433 | <p>I am looking for the proper piece of hardware to do what I'm looking for on my current prototype machine build.</p>
<p>I want a timer to begin counting upwards in seconds when a switch is activated, and reset back to zero when the switch is deactivated.</p>
<p>I know I can probably make some kind of Arduino based set up for this, and I know for a fact how to do this with a PLC/HMI, but I am having a hard time finding a cost effective, easily solution for this as its only a prototype at this stage. Everything I search keeps pointing me towards what looks like timer relays, but I'm not sure if that's what I need. I don't need to throw any outputs at any set points, basically just a cycle time readout for the operator.</p>
<p>EDIT: I suppose I could also use a 1 second pulse generator wired to the input on a basic counter, but lets see if we can find the right stuff..</p>
<p>Thanks!</p>
| |electrical-engineering|control-engineering|process-engineering|machine-design| | <p>Omron makes a wide variety of industrial <em>timer modules</em> which can count down a preset number of tenths of a second all the way to hundreds of hours and then close or open a set of switch contacts and reset themselves. A new one from one of the on-line industrial supply houses might cost about 150 dollars new, or 20 to 40 dollars for a used one at an industrial surplus house.</p>
<p>They can be configured to actuate a switch when power is first applied, count down for a preprogrammed time, and then reverse the switch action when the internal clock counts down to zero.</p>
<p>They contain multiple sets of switch contacts and can be connected up to perform a number of useful tasks, and it sounds like one of these would be what you want.</p>
| 44564 | What type of electrical machine component for counting time? |
2021-06-08T00:07:36.383 | <p>Given an orifice rated 1.8 GPM of water at 90 PSI, how do I calculate the flow rate at 30 PSI and what rating I would need to purchase to achieve 1.8 GPM at 30 PSI?</p>
<p>Below added to give some context, based on responses. Sorry, as a programmer I always reduce problems to the simplest form. Apparently, that's a bad thing on this site.</p>
<p>I'm trying to fix the shower in my new apartment in Chicago. The building owner says that as long as warm water comes out she has met her legal requirement. Google seems to agree with her. I've added "check shower, sinks, etc." to my apt inspection checklist for my next move, but that doesn't help me now. I should probably Google "apt inspection checklist" next time to see what else I should have looked at.</p>
<p>I don't know if the problem is in the pipes, values, showerhead, or what. The sink works a lot better now that I removed the 2.2 GPM aerator. Changing the showerhead would be the equivalent for a shower, but simply removing it doesn't let you see if that fixes the problem like removing an aerator does. I'm guessing flow rate is proportional to pressure, but I don't know if it is a major or minor effect.</p>
<p>The showerhead is labeled 1.8GPM, which seems to be fairly normal for a showerhead. I used a Watts gauge with an aerator adapter on the sink to find my water pressure. I spoke with the showerhead manufacturer (Waterpik) and they stated that they rate their showerheads based on 90 PSI. They were unable to tell me how well they would work at 30 PSI. Seems like a scam to me. Who has 90 PSI water pressure?!?!</p>
<p>All the formulas I find on the Internet require all kinds of other information I don't have and are just too complex for me to figure out.</p>
| |fluid-mechanics|pressure|piping|flow-control| | <p>Flow rate is directly proportional to the square root of pressure. So for any given orifice, all else being equal, <span class="math-container">$\frac{GPM}{\sqrt{PSI}}$</span> is constant. Technically PSI here is the pressure difference across the orifice, but with no backpressure only the input pressure matters.</p>
<p>For an orifice rated 1.8 GPM at 90 PSI, <span class="math-container">$\frac{GPM}{\sqrt{PSI}} = 0.1897$</span>. To calculate GPM for the same orifice, just multiply <span class="math-container">$0.1897$</span> by <span class="math-container">$\sqrt{PSI}$</span>. So for 30 PSI: <span class="math-container">$0.1897*\sqrt{30}=0.1897*5.477 = 1.039$</span> GPM.</p>
<p>For an orifice rated 1.8 GPM at 30 PSI, <span class="math-container">$\frac{1.8}{\sqrt{30}} = 0.3286$</span>. To calculate GPM for the same orifice, just multiply <span class="math-container">$0.3286$</span> by <span class="math-container">$\sqrt{PSI}$</span>. So for 90 PSI: <span class="math-container">$0.3286*\sqrt{90}=0.3286*9.487 = 3.118$</span> GPM.</p>
<p>The above is based upon the answers by NMech and Pete W, I just removed all the non-relevant parts and focused on the original question.</p>
<p>Unfortunately, the United States Energy Policy Act of 1992 limited showerheads manufactured after 1993 to 2.5 GPM at 80 PSI. <span class="math-container">$\frac{2.5}{\sqrt{80}} = 0.2795$</span> yielding a maximum at 30 PSI of <span class="math-container">$0.2795*\sqrt{30} = 0.2795*5.477 = 1.531$</span> GPM.</p>
| 44572 | Given flow rate of an orifice at one pressure, how do I calculate flow rate for a different pressure? |
2021-06-09T02:29:02.687 | <p>Does anyone know how I would find the horizontal neutral axis (<em>yc</em>) and 2nd moment of area of this cross section? (R = 210 mm) I know that <a href="https://i.stack.imgur.com/DvL7l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DvL7l.png" alt="enter image description here" /></a> and that there needs to be the same amount of area on either side of the neutral axis, but I don't know how I'm supposed to put that into practice here.</p>
<p>Any help is much appreciated. Thanks.</p>
<p><a href="https://i.stack.imgur.com/EmuvJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EmuvJ.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|civil-engineering|solid-mechanics| | <p>This problem is most easily handled using the "table" form as shown below.</p>
<p><a href="https://i.stack.imgur.com/DtcQX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DtcQX.png" alt="enter image description here" /></a></p>
<p>The result can be verified by selecting a more convenient reference line, to eliminate the potential mistake made in sign convention, as shown below.</p>
<p>Final checks:</p>
<ul>
<li><p><span class="math-container">$y_{c(A)} = y_{c(B)}$</span></p>
</li>
<li><p>Due to effect of substraction, the neutral axis must shifting above the centerline of the circle, so <span class="math-container">$y_c > R$</span>.</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/sWfTA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sWfTA.png" alt="enter image description here" /></a></p>
| 44584 | Neutral axis and 2nd moment of area |
2021-06-10T12:53:03.433 | <p>I need to solve for the mass of block C that is in the verge of slipping. All surfaces have a coefficient of static friction of 0.11. The cable connecting the wedge B and C is parallel with the incline. I have given for the mass of the wedges A and B.
<a href="https://i.stack.imgur.com/n8UX5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n8UX5.png" alt="enter image description here" /></a></p>
<p>I have established the FBD and its initial equations but I am not sure if it is correct.</p>
<p>For wedge A,</p>
<p><a href="https://i.stack.imgur.com/JSP9a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JSP9a.png" alt="enter image description here" /></a></p>
<p>The initial equations are
<span class="math-container">$\sum F_x = F_1(cos20) + N_1(sin20) - F_2(cos40) - N_2(sin40) = 0$</span> and <span class="math-container">$\sum F_y = W_A - F_1(sin20) + N_1(cos20) - F_2(sin40) + N_2(cos40) = 0$</span></p>
<p>For wedge B,</p>
<p><a href="https://i.stack.imgur.com/V44kz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V44kz.png" alt="enter image description here" /></a></p>
<p>The initial equations are <span class="math-container">$\sum F_x = F_3 + F_2(cos40) + N_2(sin40) + T(cos11) = 0$</span> and <span class="math-container">$\sum F_y = -W_B - T(sin11) + F_2(sin40) - N_2(cos40) = 0$</span></p>
<p>For block C,</p>
<p><a href="https://i.stack.imgur.com/yRI78.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yRI78.png" alt="enter image description here" /></a></p>
<p>My initial equations are <span class="math-container">$\sum F_x = -T - W_C(sin11) - F_4 = 0$</span> and <span class="math-container">$\sum F_y = N_4 - W_C(cos11) = 0$</span></p>
<p>I will edit my illustration for my tension here. I think it is wrong.</p>
<p>Did I miss something in the cases of friction? Are those all I need to solve for the mass of block C?</p>
| |statics|friction| | <p>Your work (free body diagrams and equilibrium equations) contents mistakes. The sketch below shows the force diagram of block "C" as an example:</p>
<p><a href="https://i.stack.imgur.com/bd2xc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bd2xc.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\sum Fy = 0$</span>, the normal force, <span class="math-container">$N_4 = W_Ccos\theta$</span></p>
<p>The shear friction force, <span class="math-container">$S_4 = \mu N$</span></p>
<p><span class="math-container">$\sum Fx = 0$</span>, the required tension, <span class="math-container">$T = W_Csin\theta - S_4$</span></p>
<p>You shall get the tension produced by the blocks "A" and "B", in the same manner as shown above, for which the magnitude shall be the same as <span class="math-container">$"T"$</span>.</p>
<p><em>Note that the forces <span class="math-container">$T, W,$</span> and <span class="math-container">$N$</span> must act thru the centroid of mass "C". The shear friction force <span class="math-container">$S$</span> is a reactive force, its magnitude depends on the normal force and friction coefficient only, and always in the direction against the motion.</em></p>
| 44605 | Friction in Wedges: finding the mass of the block C on the verge of slipping |
2021-06-10T14:36:03.493 | <p>We ordered a really nice light for our dinner table and ordered a special "lightswing" seperatly.
The light that we ordered however is heavier than the 'lightswing' is rated for.</p>
<p>Please see the picture for some clarification.</p>
<p>The grey part is our ceiling. The light swing is rated to handle 3.5 kg at a 80cm extension.
My question is how do I calculate the maximum weight when the light swing is only extended lets say 50 cm?</p>
<p>I tried to Google but could not figure out the right way to calculate this.</p>
<p><a href="https://i.stack.imgur.com/6T5Kk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6T5Kk.jpg" alt="enter image description here" /></a></p>
<p>Thanks in advance!</p>
| |mechanical-engineering|beam|deflection| | <p>Basic rule of thumb would say,</p>
<p><span class="math-container">$$80* 3.5 =P*50 \quad P= 80*3.5/50 = 1.6*3.5= 5.6 kg$$</span></p>
<ul>
<li>P is the light fixture weight</li>
</ul>
<p>but the fastners attaching the hanger to the ceiling need to be strong enough to support this load too.</p>
| 44607 | Calculating the load on the end of a beam |
2021-06-08T14:05:26.273 | <p>I am trying to select what stepper motor size (nema17,23,34...) to use for my project and quantity of motors needed. But am not sure how to apply the motors "torque" to real scenarios. The scenario I am interested in is moving a table, with some weight, a precise distance in the z direction using an 8mm lead screw (2mm pitch and 8 starts).</p>
<p>This may be a tall order but I have 3 scenarios below that I could really benefit in understanding how to apply stepper torque rating to calculate the max vertical lifting abilities for a <strong>typical Nema17</strong>. Basically what I am looking for is what is the maximum mass I can vertically lift at a reasonable velocity/acceleration (whichever is limiting) without taking the motor beyond its torque rating. I am not sure what assumptions are needed here.</p>
<p>Also I'd like to ask if I ensure that the stepper's acceleration and velocity is infinitely slow will my vertical lifting abilities be much higher?</p>
<p>Scenario1</p>
<p><a href="https://i.stack.imgur.com/NKHSP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NKHSP.jpg" alt="scenario1" /></a></p>
<p>A single stepper with a circular pedestal. Assuming the weights added to the pedestal were doughnut shaped and the pedestal itself is weightless, what is the max mass one could lift at reasonable speeds without experiencing missed steps or taking the motor beyond its torque rating?
Scenario2</p>
<p><a href="https://i.stack.imgur.com/MJ7wW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MJ7wW.jpg" alt="scenario2" /></a></p>
<p>Same question, assuming a weightless platform 600mmx50mm with 2 steppers on each end what is the maximum mass we can lift? Assuming the weight is precisely in the center may be necessary.</p>
<p>Scenario3</p>
<p><a href="https://i.stack.imgur.com/iMcxF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iMcxF.jpg" alt="scenario3" /></a></p>
<p>And finally with a weightless platform approx. 600mmx800mm what is the maximum weight we can lift with 4 steppers.</p>
| |stepper-motor| | <p>There are many parameters that you need to consider here, so I will just focus on the first scenario (i.e. A single stepper with a circular pedestal.</p>
<blockquote>
<p>Assuming the weights added to the pedestal were doughnut shaped and the pedestal itself is weightless, what is the max mass one could lift at reasonable speeds without experiencing missed steps or taking the motor beyond its torque rating?</p>
</blockquote>
<h2>Torque calculation as a function of rpm and supply voltage</h2>
<p>First of all, the rpm and the supply voltage are two critical factors. In the following image you can see that depending on the rpm and the Voltage you get significantly different torque.</p>
<p><a href="https://i.stack.imgur.com/lq4Pi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lq4Pi.png" alt="enter image description here" /></a></p>
<p><em>Figure 1: Nema 17 torque vs rpm curves (source: <a href="https://www.moonsindustries.com/series/nema-17-standard-hybrid-stepper-motors-b020105" rel="nofollow noreferrer">moons motoros</a>)</em></p>
<p>What is important to note here is the significant drop in Torque after 100rpm (for this particular model).</p>
<h2>Calculating the maximum weight</h2>
<p>In order to calculate the maximum weight the next thing that's required is the thread pitch (e.g. p = 1 mm/revolution). Then the most convenient thing is to use the following equations:</p>
<p><span class="math-container">$$P = M\cdot \omega = F\cdot v \tag{eq.1}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$P$</span>: power</li>
<li><span class="math-container">$M$</span>: torque in [Nm]</li>
<li><span class="math-container">$\omega$</span>: angular velocity [rad/s] (<span class="math-container">$=\frac{2\pi n}{60}$</span>)</li>
<li><span class="math-container">$F$</span>: weight on [N]</li>
<li><span class="math-container">$v$</span>: velocity [m/s] (the velocity <span class="math-container">$v=\frac{n}{60}\cdot p$</span>)</li>
</ul>
<p>In the above equations I am assumming.</p>
<ul>
<li><span class="math-container">$n$</span>: revolutions per minute of the shaft units:[rpm]</li>
<li><span class="math-container">$p$</span>: travel per revolution of the shaft (preferred unit [m/rev])</li>
</ul>
<p>As a result you can rewrite Eq.1 as
<span class="math-container">$$ M\cdot \frac{2\pi n}{60}= F\cdot \left(\frac{n}{60}\cdot p\right)$$</span></p>
<p><span class="math-container">$$ F = \frac{2\pi }{p} \cdot M$$</span></p>
<p>This is the theoretical maximum weight if there is no acceleration. However, in reality it is lower for a number of reason, eg:</p>
<ul>
<li>during the acceleration stage some of the torque is expended in the rotational energy of the rotational masses. usually this can be neglected in most application. (if you want to include that its easy assuming you know the acceleration profile, but IMHO its not worth the effort, because low accel will have almost no effect, high accels will only last a few seconds).</li>
<li>friction on the setup</li>
<li>cogging torque</li>
<li>...</li>
</ul>
<p>In general, the implementation parameters (e.g. alignment) are the most difficult to estimate.</p>
| 44615 | Stepper motors and vertically lifting a platform |
2021-06-11T08:53:50.117 | <p>I was designing a turbine to be utilized in order to measure the flow rate of a river system. I can measure the angular velocity of the shaft by utilizing a magnet and hall sensor, however, in order to relate the angular velocity with flow rate, I was unable to find an equation.</p>
<p>A rough sketch of the design is demonstrated:
<a href="https://i.stack.imgur.com/txuEq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/txuEq.jpg" alt="Sketch Of Turbine" /></a>.</p>
<p>The paddles are essentially rectangular in geometry with cross sectional area A. The flow of water causes some rotation of the turbine, thus causing the rotation of the shaft. At this point, I can consider mechanical losses in the shaft or bearings (not fluid mechanical) as negligible. Clearly the final flow velocity (<span class="math-container">$v_2$</span>) locally would be lower than the initial (<span class="math-container">$v_1$</span>) due to some energy being converted into angular (<span class="math-container">$\omega$</span>) kinetic energy.</p>
<p>I would greatly appreciate it if you could please provide some guidance on how I can relate the angular velocity with the flow rate. Approximations or appropriate simplifications are fine, but I am interested in verifying if I can relate the two quantities so I can program it into Arduino.</p>
| |mechanical-engineering|fluid-mechanics|turbines| | <p>In general, v=rω</p>
<p>Where v is the velocity of the water, r is the radius of the turbine, and ω is the angular velocity. This is the simplest that I know of that could be quickly programmed</p>
| 44625 | Angular Velocity And Flow Rate Relationship Of Turbine |
2021-06-11T17:09:20.793 | <p>Here, let’s say I have a tea cup.</p>
<p><a href="https://i.stack.imgur.com/r2TaL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r2TaL.jpg" alt="enter image description here" /></a>In that cup , let us say there is a ball inside the cup which is temporarily fixed to the surface of cup. Now , this ball is fixed very loose but strong enough that if I tilt the cup downwards , it doesn’t fall.</p>
<p>Now , what I noticed is that if I tilt my cup face in downward direction and then I move my cup down with a haste while holding the cups handle. The ball falls down.</p>
<p>I did not apply any pressure or force on the ball from the top of cup or from below the cup. How is it that the ball falls down ?.</p>
<p><a href="https://i.stack.imgur.com/ETThT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ETThT.jpg" alt="enter image description here" /></a></p>
<p>What I think is that if I Look at this movement from an inertial frame.</p>
<p>1)I can notice if the cup moves down with 5m/s , then so does the ball moves down with same speed. Also , there is a <span class="math-container">$9.8m/s^2$</span> acceleration acting all the time. Due to these two factors , it contributes to more acceleration I.e more Force since F=m*a.</p>
<ol start="2">
<li>There should be a force from the air molecules in upward direction . Due to Newton’s third law , the ball pushes the air down. Therefore , also forces itself to fall down.</li>
</ol>
<p>This is all what I think should happen.</p>
<p>Please let me know if you have any other suggestion or somewhere I went wrong.</p>
| |mechanical-engineering|pressure| | <p>The ball gathers velocity and when you decelerate it the inertia results in the development of forces that separate it from the cup.</p>
<p>Basically the inertia force which is required to develop the deceleration exceeds the maximum force of the bond between the ball and the cup.</p>
| 44631 | Why does the ball detach from the surface of cup? |
2021-06-11T17:20:45.877 | <p><a href="https://i.stack.imgur.com/WRoD9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WRoD9.jpg" alt="enter image description here" /></a></p>
<p>Here , we have two water filters and buckets .</p>
<p>Now , we know that as the water falls down , it’s speed increases because of acceleration due to gravity.</p>
<p>Let us say I started both the filters and once the water reaches the bottom of both the buckets.of course , water from 2nd filter would have had reached first at the surface of 2nd bucket.</p>
<p>Assume that exactly when the water touches the bucket form both the filters. Only then , I start the timer. Now , I am recording or noting the time it is taken by both filters to fill the bucket.</p>
<p>Volume is same of both buckets.</p>
<p>The Q is : Which bucket will fill up first ?</p>
<p>I think both the buckets would fill up at the same time.</p>
<p>Now , I also noticed as the water keeps on dripping at longer heights. There is a separation between water molecules present. You can see most of the water at below the tap and it’s amount kind of like looks fo reduce as you go down.</p>
<p>This is what confused me. From this observation , I think 2nd bucket should fill up faster than the 1st one.</p>
<p>Why ? Because if I imagine the separation between the bucket and water at an extreme level. Then , I can see like drops filling up the bucket and not like a stream of flowing water.</p>
<p>Is this correct ? Also , I’m not getting how should I write an equation for this.</p>
<p>I did research about it. It says that :</p>
<p>A decrease in temperature caused the water molecules to lose energy and slow down, which results in water molecules that are closer together and a decrease in water volume. When water is heated, it expands, or increases in volume. When water increases in volume, it becomes less dense.</p>
<p>Therefore , can I say my observation is true only depending on the temperature?</p>
| |mechanical-engineering|fluid-mechanics| | <p>The top bucket will fill up first because the water needs less time to fall from the reservoir to the bucket.</p>
| 44632 | In which bucket , should the water fill it up first? |
2021-06-12T10:54:42.953 | <p>In moment distribution method, the bending stiffness of a beam is taken as <span class="math-container">$EI/L$</span>. But in these kinds of stiffness tables, the flexural rigidity is usually divided by some power of <span class="math-container">$L$</span>:
<a href="https://i.stack.imgur.com/HJCM7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HJCM7.jpg" alt="enter image description here" /></a></p>
<p>So why is <span class="math-container">$EI$</span> divided by simply <span class="math-container">$L$</span> to get the bending stiffness, instead of some of its powers as in the tables? Could somebody provide a quick derivation? Thank you!</p>
| |structural-engineering| | <p>I/L is relative stiffness. For the same flexural member material and cross sectional characteristics, the longer the flexural member, the more it displaces/deflects. Therefore, I/L provides a relative stiffness value depending on member's length.</p>
| 44646 | Why is the beam bending stiffness taken as $EI/L$? |
2021-06-13T14:08:18.743 | <p>Standard enthalpy of formation is when one mole of compound is formed from its constituent elements.</p>
<p>Example : <span class="math-container">$H_2$</span>(g) + <span class="math-container">$O_2$</span>(g) —><span class="math-container">$ H_2O$</span>(L).</p>
<p>But we don’t say the same for <span class="math-container">$CaO + CO_2$</span> —> <span class="math-container">$CaCO_3$</span>.</p>
<p>It is because <span class="math-container">$CaCO_3$</span> is not formed it’s constituent elements but from other compounds.</p>
<p>Q1: From other compounds , does it mean the most simplest state I.e it is not formed from Ca + <span class="math-container">$O_2$</span>.</p>
<p>I want to confirm if that’s all the meaning of standard enthalpy of formation. I am getting a lot confused over it.</p>
<p>Q2 Also , for the equation above. We can never find standard enthalpy of formation but can find the standard enthalpy of reaction I.e addition of <span class="math-container">$\delta$</span>H of <span class="math-container">$CaO $</span> in solid state + <span class="math-container">$CO_2$</span> in gaseous state ?</p>
<p>Edit:
<a href="https://i.stack.imgur.com/h7Q7X.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h7Q7X.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|thermodynamics|heat-transfer|chemistry|heat| | <p>I prefer this definition for Hf: "The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states."</p>
<p>Simply put though heat of formation is a tool we use to quickly calculate heat of reaction for a compound. If I know the heat of formation for each reactant and product I can use the formula below (Hess's Law) to calculate heat of reaction.</p>
<p><a href="https://i.stack.imgur.com/kfcIm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kfcIm.jpg" alt="heat of reaction" /></a></p>
<p>Once the calculation is done we will know if the chemical reaction is going to be endothermic or exothermic and to what magnitude.</p>
<p>For your calcium carbonate example you wouldn't need to tell me the heat of reaction I could simply look up the heat of formation for <span class="math-container">$CaO, CO_2, CaCO_3$</span> and do the math in the formula above.</p>
<p><span class="math-container">$CaO: Ca(s) + (1/2) O_2(g) --> CaO (s) Hf = -635.5 $</span>KJ/mol</p>
<p><span class="math-container">$CO_2: C(s) + O_2(g) --> CO_2(g) Hf = -393.5 KJ/mol$</span></p>
<p><span class="math-container">$CaCO_3:Ca(s)+C(s)+(3/2)O_2→CaCO_3(s) Hf = -1207.0 KJ/mol$</span></p>
<p><strong>Given the reaction:</strong></p>
<p><span class="math-container">$CaO(s) + CO_2(g) --> CaCO_3(s)$</span></p>
<p><span class="math-container">$Heat of reaction = (-1207.0)-(-635.5-393.5) = -178 KJ/mol$</span></p>
| 44664 | Difficulty in understand the meaning of standard enthalpy of formation |
2021-06-14T07:36:01.597 | <p>We say kinetic of molecules as a function of temperature from the equipartition theorem. This means that more is the velocity or mass of the body , more the is the temperature.</p>
<p>My Q is why does increased motion or mass of molecules contribute to heat , increased temperature ?</p>
<p>From this. Is it also possible to say that when velocity of molecules = 0. Then , temperature is at its coldest measure ? Since even at -20K or so. There must be some speed of the molecules where I can say that they are just slow.</p>
<p>I kind of also feel fire and ice also confuse with how they can be compared here ?</p>
<p><strong>To conclude</strong>. All these few Q are a subtopic of my major Q i.e why does increased motion or mass of molecules contribute to heat , increased temperature.</p>
| |mechanical-engineering|thermodynamics|temperature|heat| | <p>IMHO you are still confusing heat with temperature (despite the answer to <a href="https://engineering.stackexchange.com/questions/44546/difference-between-temperature-and-heat">this question</a> or <a href="https://engineering.stackexchange.com/questions/44532/difficulty-in-understanding-the-scenario-for-this-q">this </a> ).</p>
<blockquote>
<p>My Q is why does increased motion or mass of molecules contribute to heat , increased temperature.</p>
</blockquote>
<p>I will try to answer in a way that address that fundamental difference.</p>
<p>Consider 1 kg of water in a well insulated vessel which is travelling with 10[m/s]. However the temperature of the water is:</p>
<ul>
<li>200 Kelvin: So water is in solid (ice) state</li>
<li>300 Kelvin (approx 27 <span class="math-container">$^oC$</span>): so its mostly in liquid state</li>
<li>400 Kelvin (approx 127 <span class="math-container">$^oC$</span>): so its mostly in gaseous state.</li>
</ul>
<p>The speed of the vessel does not have an immediate effect on the temperature.</p>
<p>It <strong>does however contribute</strong> to the overall energy of the system (this is to reply to your question) (NOTE: please edit your punctuation in the question).</p>
<blockquote>
<p>Is it also possible to say that when velocity of molecules = 0. Then , temperature is at its coldest measure?</p>
</blockquote>
<p>Regarding the part of your question:</p>
<blockquote>
<p>Since even at -20K or so. There must be some speed of the molecules where I can say that they are just slow.</p>
</blockquote>
<p>There can be no negative temperatures. Even, if you could reach absolute Zero, then the energy of the system would never be entirely zero due to quantum effects ( I've had a theoretical Laser physicist try to explain it to me, but my brain shut down after hearing a few times the word quantum).</p>
<hr />
<p><strong>Update for comment</strong></p>
<blockquote>
<p>If I put ice on my skin. Is it that the molecules of ice are trying to slow down the velocity of the molecules of my skin.</p>
</blockquote>
<p>When you put ice on your skin, the molecules of water don't have much internal energy, so they are barely moving in the lattice. When you put them in contact with your skin (which has a lot of water molecules at higher temperature), what happens is that they exchange heat .</p>
<p>I am not an expect on the exact mechanism that heat is exchanged, but I tend to think about it the classical sense of two balls of water molecules hitting each other. The result is that energy is transferred in the impact and some energy is imparted (statistically from the fast moving molecule to the slow).</p>
<p>Because of this exchange, the hotter becomes cooler, and the cold becomes warmer (until eventually the exchange statistically is the same).</p>
<hr />
<p><strong>UPDATE 2: Fan example</strong></p>
<blockquote>
<p>Another example , AC or fan. Do they try to reduce the velocity of molecules of your body.</p>
</blockquote>
<p>The reason you get cooling with the fan against your skin, is that the fan pushes to your skin air which is slightly cooler than your temperature.</p>
<p>The reason why the exchange is more severe, is that the more cooler molecules will hit your body and exchange heat in a unit of time (see forced convection.</p>
<p>The fan by itself will actually increase the temperature of the air and the room.</p>
| 44675 | What does it mean by average kinetic energy of molecules as temperature? |
2021-06-15T05:40:44.570 | <p>I was conducting a FEA analysis of a symmetric isotropic structure under a loading condition. I will not mention the material I was using. Under that loading, one part of the structure was in tension and other in compression. When I reversed the loading condition, the part that was in compression previously was now in tension and vice versa. However, the displacements and the Von-Mises stresses that were seen on the part of the structure that was in tension in the initial loading condition, and again which was in tension in the reversed loading condition, I saw a difference in it and I couldn't understand why it was happening. So I thought maybe it is possible for the material to have different elastic modulus and Poisson's ratio under compression and tension.
Is this true? If yes, then in most of the FEA softwares, there is option to add only one elastic modulus. Why is that? Is there any specific class of materials where this happens, or we have to conduct experiments to individually check each and every material if their elastic modulus is the same of different under tension and compression?</p>
| |elastic-modulus| | <p>For small strains of stable materials, <a href="https://physics.stackexchange.com/a/408992/146039">the tensile and compressive elastic moduli are equal</a>. This is equivalent to saying that a smooth energy minimum looks like a (symmetric) parabola up close; an energy well in the shape of a parabola characterizes an ideal spring with equal elongation and contraction spring constants.</p>
<p>This approximation works well for metals, ceramics, and crosslinked polymers, for example, whose elastic strain is small (1%, say). This is why you have the option to enter only one Young's modulus, for example. However, the approximation typically does not work for large elastic strains of hyperelastic materials such as elastomers, which may stretch their own length (100% strain) and much more. As shown below (<a href="https://en.wikipedia.org/wiki/Hyperelastic_material" rel="nofollow noreferrer">source</a>), the stress–strain slopes are visually identical for slight positive and negative strains but differ for larger strains.</p>
<p>Since the same argument holds for the bulk and shear modulus, it must hold for Poisson's ratio, which is not independent if Young's modulus and either the bulk or shear modulus are specified.</p>
<img src="https://i.stack.imgur.com/BuPhg.png" width="400">
| 44687 | Can the Elastic Modulus of an isotropic material be different in Tension and in compression? |
2021-06-15T19:02:01.067 | <p><a href="https://i.stack.imgur.com/t0kml.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t0kml.jpg" alt="enter image description here" /></a></p>
<p>Eurocode: design of steel connections mentions something called "bolt elogation length", defined as the grip length of the bolt plus half of the sum of bolt head height and nut height. <strong>But what is the significance of this length?</strong></p>
| |civil-engineering|bolting| | <p>The bolt elongation length is (in all likelihood ) what is referred to as <strong>Effective length</strong> in the following image</p>
<p><a href="https://i.stack.imgur.com/dvFrq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dvFrq.png" alt="enter image description here" /></a></p>
<p><em>Figure 1: Effective bolt length (source: <a href="https://www.resonic-instruments.com/wp-content/uploads/2018/03/Ultrasonic-Measurement-of-Fasteners-Rev-6.pdf" rel="nofollow noreferrer">Resonic</a>)</em></p>
<p>The physical significance, is the part of the bolt that is in tension, and effectively generates the clamping force. Additionally that length can be used to estimate the spring constant of the bolt:</p>
<p><span class="math-container">$$K = \frac{E A }{L}$$</span></p>
<p>The following image shows the axial forces (N diagram) along the axis of the bolt.</p>
<p><a href="https://i.stack.imgur.com/PigON.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PigON.png" alt="enter image description here" /></a></p>
<p><strong>Update</strong></p>
<p>There is a mistake in Figure 2, which ingenord pointed out. The distribution of the force on the nut is along the distance of the bolt (see image below)</p>
<p><a href="https://i.stack.imgur.com/AaxAe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AaxAe.png" alt="enter image description here" /></a></p>
<p><em>Figure 3: Distribution of load along threads (source <a href="http://pubs.sciepub.com/ajme/1/7/50/figure/8" rel="nofollow noreferrer">sciepub František Trebuňa</a>)</em></p>
<p>The graph shows that approximately 34% of the load is transferred on the 1st thread, about 70% by the first 3, and about 100% by all 6 threads.</p>
| 44694 | What is the physical significance of bolt elongation length? |
2021-06-16T05:31:27.230 | <p>I want to know ,can I use the empirical formula time period T=0.1*N, where N = number of floors, to find the fundamental frequency of vibration of Tesla's laboratory building in which he performed experiment on the mechanical oscillator and the result of experiment was EARTHQUAKE!.. Through research I have found that his laboratory building was 7 story in the year1898 in New York city. I could not get any other details of the building. So I found time period using above formula which I found on Research gate site and took reciprocal of it to get the frequency. Is it correct?...Please reply.
Here's the link showing image of Tesla's laboratory building. <a href="https://images.app.goo.gl/BTqpupmGp23s2BXB6" rel="nofollow noreferrer">https://images.app.goo.gl/BTqpupmGp23s2BXB6</a></p>
<p>Best Regards.</p>
| |civil-engineering|earthquake-engineering| | <p>Yes, per the design code (ASCE 7 & IBC) the formula could be used for estimating the fundamental period of buildings during a seismic event if the construction of the building meets the criteria indicated in the code provisions.</p>
<p><a href="https://i.stack.imgur.com/SJV9u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SJV9u.png" alt="enter image description here" /></a></p>
<p>See P 1-56, <a href="https://www.seismicreview.com/January_2017-Errata-1.pdf" rel="nofollow noreferrer">https://www.seismicreview.com/January_2017-Errata-1.pdf</a></p>
<p>However, as the seismic design and evaluation are quite involved, please check the latest IBC code and ASCE 7 for the "<strong>correctness</strong>" on using this approximate method to determine the building period for a specific application.</p>
| 44696 | Can I use empirical formula ,time period T=0.1*N, (N= number of floors) to find fundamental frequency of vibration of Tesla's lab building in1890? |
2021-06-16T10:08:27.060 | <p>The last couple of weeks, I've been noticing this (wifi like) symbol next to some of my contacts in my Android phone. The symbol I am interested about is in the red circle that has the outline of the wifi symbol.</p>
<p><a href="https://i.stack.imgur.com/YTJ20.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YTJ20.png" alt="enter image description here" /></a></p>
<p>I don't remember having enabled anything, so I am curious as to what it is and what does it mean (and whether I should disable it).</p>
<p>I tried looking, and I have a hunch it might be related to <strong>wifi calling</strong> but I can't be certain.</p>
<p>Apologies in advance, I know this does not fall strictly within the Engineering SE's jurisdiction, however, I know that I'll get an valid answer faster here that anywhere else.</p>
| |telecommunication| | <p>This symbol means that the contact number you have called was connected via <strong>Voice over WiFi (VoWiFi)</strong> technology instead of VoLTE. Your service automatically shifts to VoWiFi whenever there is a WiFi network connection available if you turn the option of WiFi calling on.</p>
| 44697 | Does anybody know what this symbol is? |
2021-06-16T14:42:01.293 | <p><a href="https://i.stack.imgur.com/17wZQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/17wZQ.jpg" alt="enter image description here" /></a></p>
<p>Usually the reason for having a washer under a bolt head is stated to be that it helps to evenly distribute the stress to the clamped material surface. <strong>But why is this?</strong></p>
<p>I would understand if the washer was significantly larger than the bolt head. Then there would understandably be more surface area. But for a bolt like in the picture, the washer is only slightly larger than the bolt head, so why would it make much difference?</p>
<p>EDIT: Thank you everyone for good discussion and answers! Didn't expect this to spark such a discussion.</p>
| |mechanical-engineering|civil-engineering|stresses|fasteners|machine-elements| | <p>Another important part of the answer is the symmetry of the stress pattern. The stress caused by a bolt head varies greatly between the points of the bolt head and the straight sides. As a result local stresses, which are what you really care about because those are what the materials have to withstand, can be much higher than the average stress. A washer's circular symmetry smooths those out.</p>
<p><a href="https://i.stack.imgur.com/d1ITJ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/d1ITJ.png" alt="enter image description here" /></a></p>
<p><a href="https://www.researchgate.net/figure/Contours-of-warping-displacement-a-shear-stress-b-and-components-of-shear-stress_fig5_267129617" rel="noreferrer">https://www.researchgate.net/figure/Contours-of-warping-displacement-a-shear-stress-b-and-components-of-shear-stress_fig5_267129617</a></p>
| 44701 | Why exactly does a washer help distribute the stress around a bolt? |
2021-06-16T17:15:34.307 | <p>I am currently pondering modifying my laptop's heatsink and am curious whether I stand to gain any benefit at all.
The CPU and GPU (similar heatsink design) reach high temperatures (90°C +) under load and the exhaust air is warm, but far from hot.
So far, I have tried several thermal compounds and did some experiments regarding the way the heatsinks are attached to CPU / GPU.
Unfortunately, the exhaust air is still not hot which leads me to believe that the heatsink design could be improved.</p>
<p>For context: the laptop uses a usual heatsink design, where a copper plate sits on top of the CPU / GPU.
This copper plate is connected to a few heatpipes which lead to the cooling fins.
The heatpipes are mostly round and the fins as well as the plate are flat / rectangular.
Right now, the heatpipes seem to be fixed to the fins and the plate on individual spots, what looks like spot-welding.</p>
<p>The current design looks something like this (plate is green, heatpipe is blue, spots with connection are black):</p>
<p><a href="https://i.stack.imgur.com/hdlJ3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hdlJ3.png" alt="Draft of the current design" /></a></p>
<p><strong>UPDATE:</strong> I have taken the laptop apart and taken a picture of the actual heatsink, it looks like this (green arrows show where there is some kind of connection).
<a href="https://i.stack.imgur.com/sI50M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sI50M.jpg" alt="photo of heatsink" /></a>
I did use compressed air to clean the space between the individual fins afterwards.</p>
<p>My reasoning is that all of the thermal conductance is achieved via those small dots where the heatpipes are connected to the other parts.
As thermal conductivity relies on surface area, I was thinking about filling in the gaps to the side of the heatpipes.
Hopefully, by doing so the heatsink is able to transfer heat more efficiently, leading to lower temperatures.</p>
<p>Unfortunately, there are very few materials or ideas that seem suitable.
Under normal circumstances (e.g. heatsink application) I would rely on thermal paste - but this can't be used here as it requires pressure.
While searching for some kind of epoxy or glue that is thermally conductive, I came up empty.</p>
<p>My best idea right now is to use Indium (can be bought easily, melts at a reasonably low temperature) for the gaps.
However, as handling of molten metal (at ~156°C) is tricky, I would love to know whether my idea has any merit.</p>
<p>A draft of my idea is as follows:</p>
<p><a href="https://i.stack.imgur.com/nWaaB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nWaaB.png" alt="Draft of my idea" /></a></p>
<p>Where green is the plate / fins, blue are the heatpipes and yellow (on the right side) are the filled in gaps.</p>
<p>Thank you very much for any insights or hints!
(and sorry for probably mixing several terms, like conductivity and conductance)</p>
<p><strong>UPDATE:</strong> I have since taken apart the laptop and tried to fill the gaps with solder.
Neither Indium (melts at ~ 156°C) nor regular solder would work, as both did harden as expected but could easily be pushed off with a Q-tip.
My assumption was that some kind of material on top of the surface is causing the solder to not stick - however, even when using a metal brush with a electric drill, I was not able to make the solder stick (neither on the aluminum nor the copper part).
This might be due to the inidividual parts not being heated to the melting point of the solder, but trying to heating them with the soldering iron did not even make a sensible difference.
Additionally, I am too careful to heat the complete assembly up all the way to the melting point of solder, if I risk that the few existing connections might break while doing so.</p>
<p>In the end, I did sand down my CPU's IHS to make it even flatter and have accepted that this seems to be the limit that I can hope to achieve with this device.</p>
<p>Mixing some kind of adhesive / glue with thermally conductive particles might be an option but this was an avenue that I did not pursue further due to lack of experience with this process.</p>
| |thermal-conduction|computer-hardware|heatsink| | <p>Your are correct that the smallest surface area will be a limit on the thermal conductivity. Before we dive into that lets look at the over all approach and some other limits.</p>
<p>Are your temperatures typical for your laptop on forums or per manufacture? Heatsinks are just lumps of metal so you can easily inspect the quality, but <a href="https://en.wikipedia.org/wiki/Heat_pipe" rel="nofollow noreferrer">heat pipes</a> are full of a working fluid like ethanol. If they are incorrectly filled or punctured their conductivity will be significantly lower. It takes some lab work to determine if they are working or not, but just be aware of this.</p>
<p>My recommendation would be to purchase a replacement heat sink/pipe system for your laptop. This is two fold;</p>
<ol>
<li>If the heat pipes were damaged on your first unit then your problem is solved. Also possible that even the Chinese company selling the aftermarket heatsink will have already improved it.</li>
<li>It is a good practice to have a backup before you modify an item that is critical to your primary computer. I speak from experience on this one ;-)</li>
</ol>
<p>Another potential limit is the airflow around the fins. If you can increase fan speed from the software or upgrade the fan this will help. Obviously clean fins and larger areas for air passage help. Keep wires out of the air flow path when possible.</p>
<p>Ok to your question. Brazing/soldering <strong>may</strong> be a good solution. Indium is used in different lead-free solders to lower the melting point. I would just use plumbing solder though for cost and availability. Electrical solder would work too and may be available in a lower temperature version.</p>
<p>The types of metal you are brazing matter. Copper to copper will work great. Aluminum can be brazed, but it is much more difficult requiring acid preparation and aluminum specific brazing rod. Lots of good brazing videos on youtube once you know your metal types.</p>
<p>The biggest risk with doing this is that it is possible to <strong>damage and or rupture the heat pipes</strong>. They are little pressure vessels and will pop under high temperature. Might be okay at that temperature, might not. It is a <strong>safety issue</strong>; use face shield, glasses, gloves, and bolt it down before heating.</p>
<p>Epoxy with fillers has pretty terrible conductivity compared to metal, but its much better than air if you aren't willing to risk failure of the unit. As long as you don't contaminate the chip mating surfaces; it will either be better or the same. Buy some <a href="https://www.ebay.com/itm/122512722016?hash=item1c8653c860:g:iecAAOSwurZZHfKO" rel="nofollow noreferrer">silver powder</a> and two part <a href="https://rads.stackoverflow.com/amzn/click/com/B001MUM62C" rel="nofollow noreferrer" rel="nofollow noreferrer">60 min epoxy</a>. Use as much silver powder as the epoxy will wet. Put all the ingredients in the corner of a ziplock bag, evacuate the air, and mix by squeezing back an fourth. This minimizes the amount of air (insulator) in the mix.</p>
<p>Most of the heat pipe units I have seen use a compression assembly (see photo below). Another bit more extreme option, would be to buy a heat pipe/sink system that is close to what you need, then bend the heat pipes and mechanical modify it as need. On the plus side you don't have to damage your currently working unit. Good luck!</p>
<p><a href="https://i.stack.imgur.com/HfaPfm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HfaPfm.png" alt="enter image description here" /></a></p>
| 44704 | Filling gaps in heatsink design to improve thermal conductance? |
2021-06-16T19:45:11.503 | <p>Peace and Blessings. While doing research about the Civil Engineering profession, I see that they typically do a lot of computer work, desk work, meetings, and fieldwork in the sense of overseeing their project. I think this is cool but I was wondering do they ever physically partake in the construction (or demolition) of said project. I then came across this video on <a href="https://youtu.be/6xrs1oIv5VI" rel="nofollow noreferrer">youtube</a>. Under the description it stated:</p>
<blockquote>
<p>In this video I discuss whether jobs in engineering are hands-on, more
computer work, or in the field. Some people might have the wrong image
of an engineering job which may cause confusion down the road. The
good news is that engineers do everything listed. They make computer
models, they work with hardware, and they can go to sites where their
projects are being constructed. However, you will notice that a large
majority of engineering jobs are desk jobs because that's where a lot
of design work and engineering drawings are done. The hands on work is
often left to technicians, operators, machinists, and construction
workers who physically build (and sometimes test) the designs made by
the engineers.</p>
</blockquote>
<p>The video gave more insight into my inquiry but did not fully answer my question (in the video it showed Civil Engineers just taking samples from a site to test which I know isn't indicative of the full extent of what Civil Engineers do "hands-on"). He also made mention of depending on the size of the company may determine how much "hands-on" work a Civil Engineer will do. Do Civil Engineers ever really physically get involved with the project and if so how likely or often will they? As always thank you for your time and assistance.</p>
| |civil-engineering| | <p>Of course engineers <em>can</em> do the physical trade labor, but the reality is that they likely have other jobs to do (supervision, quality inspection, etc.) instead of operating tools or equipment. It's unlikely that the average engineer has the skills necessary to perform most trade tasks, and the skills the engineer does have (working a shovel) are probably a poor labor choice.</p>
| 44707 | Can/do Civil Engineers do "hands-on work" such as physically partake in building the infrastructure as well? |
2021-06-20T22:13:36.790 | <p>In class we're learning about moments, and I'm not entirely sure what is the difference between using an equation of equilibrium for the moments about a point (in this case C), and the actual <em>moment at C</em>.</p>
<p>From what I understand, taking the "sum" of moments about C is needed to be equal to 0, since at that point, C, if the "sum" is greater than 0, then we are no longer in static equilibrium, and therefore things will move? So, to counteract this we need an "internal moment" provided by the beam at C to counter act this.</p>
<p>Edit: Also, was my professor incorrect here in his direction scheme? It seems like he's taking the moment applied to C from B as negative even though he set the positive moment direction to be in the CW direction.
<a href="https://i.stack.imgur.com/VSA1t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VSA1t.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/1QoCS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1QoCS.png" alt="enter image description here" /></a></p>
| |statics|moments| | <p>They mean the same thing to me. Although I suppose you could interpret "moment at a point" to be a specific moment due to a particular cause which composes part of the sum.</p>
<p>As for the sign convention. I think your professor did goof. He carried over the negative sign from the Vc = -6 from the Y-axis linear calculations. Not to mention he also defined his rotation axis (the z-axis) with respect to the X and Y axis contrary to the typical definition of the cross product.</p>
| 44749 | What is the difference between a sum of moments about a point versus the moment at a point? |
2021-06-21T02:58:08.033 | <p>I was looking through some course notes and realized that I never fully understood this concept. So, looking at the diagram below we can see that there is a 20kN axial force being applied in-between sections (1) and (2).</p>
<p><a href="https://i.stack.imgur.com/M4hnO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M4hnO.png" alt="enter image description here" /></a></p>
<p>Looking at the force from section (1)'s point of view, this would be a compressive force, as it it acting inwards towards the section. However, from section (2)'s point of view it is tensile, as it is acting away from the center of the section.</p>
<p><a href="https://i.stack.imgur.com/kUFjS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kUFjS.png" alt="enter image description here" /></a></p>
<p>From what I can think, in this case I believe that when drawing the axial force diagram we simply go up (+'ve compressive axial force) when drawing from left to right. Meanwhile, if we were to draw this axial force diagram from right to left we would go down (-'ve tensile force).</p>
<p>Am I interpreting this correctly? Any help/insight would be really appreciated!</p>
<p>Edit: Another way I was interpreting this was, if I were to scan, starting from the leftmost side towards the right. I would initially see a compressive force onto the beam from R, then, I would see compressive forces (by using the section method at point 1). However once I am coming up to the 20kN force, I would, from my perspective, see a 20kN force being applied towards me, and from my section's perspective this would be compressive (-'ve force), so shouldn't I go down in the Force/distance chart? Since -'ve forces are compressive and +'ve ones are tensile.</p>
<p>Edit 2: Looking at some more Youtube videos: <a href="https://www.youtube.com/watch?v=Q7kYdDNKE8E" rel="nofollow noreferrer">https://www.youtube.com/watch?v=Q7kYdDNKE8E</a>, it looks like what happens at the point of force application isn't exactly necessary. Mostly we just calculate the force <em>in-between</em> everything, and just simply connect the lines.</p>
| |statics|beam|diagram| | <p>I think your big misunderstanding is in this paragraph:</p>
<blockquote>
<p>[...] if I were to scan, starting from the leftmost side towards the right. I would initially see a compressive force onto the beam from R, then, I would see compressive forces (by using the section method at point 1). However once I am coming up to the 20kN force, I would, from my perspective, see a 20kN force being applied towards me, and from my section's perspective this would be compressive (-'ve force) [...]</p>
</blockquote>
<p>The big problem here is that you are changing your perspective between these two moments. You start by "looking to the left" and then end by "looking to the right".</p>
<p>It's important to remember that the sign convention for internal stresses depends on the face you're looking at: a force is tensile if it's to the right on the right-hand face or if it's to the left on the left-hand face.</p>
<p><a href="https://i.stack.imgur.com/TfxCv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TfxCv.png" alt="enter image description here" /></a></p>
<p>When you are scanning along a beam, you need to choose which face you're going to look at. Or, well, you'll choose the direction in which you'll move and that'll define the face you're going to think about.</p>
<p>If you choose to go from left to right (as one usually does), then you obviously must only look at the left-hand face of the element. After all, you don't know what's going on to the right of the element, only to its left.</p>
<p>So, we start at the support, where we see the left-hand face receiving a rightward force. Looking at our sign convention, that's negative (compressive). So far, so good.</p>
<p>We then keep moving to the right, until we reach the point of application of those 20 kN. Where is that force applied? Well, we're only looking at the left-hand face, so we'll only notice that force once we've "moved past it" such that the force is applied on the left-hand face*. At which point we'll have a leftward force on the left-hand face, which in our sign convention is positive (tensile). So now we'll have a net 10 kN leftward force on the left-hand face, which means we're now under tension.</p>
<p>We then keep moving right until we reach the free end with the 10 kN force applied. Once again, we'll only notice that force once it's reached the left-hand face, at which point we'll once again have a rightward force on the left-hand face, which means it's negative (compression). Adding that to what we'd already accumulated, we get the expected zero force at the free end.</p>
<p>Going from right to left will give the same result, so long as you remember to always look at the right-hand face instead.</p>
<hr />
<p>* <sub>The term "moved past it" is merely metaphorical: remember, the element of the sign convention is infinitesimal!</sub></p>
| 44751 | Is an axial force applied at the centre of a beam considered a compressive force, tensile force, or both? |
2021-06-21T10:05:49.077 | <p>If there is a body moving with a constant velocity of <span class="math-container">$1000m/s$</span> and it’s mass =<span class="math-container">$ 10kg$</span>.</p>
<p>Then , how is it possible that the force applied by this body<span class="math-container">$ = 0 N $</span>since it will definitely make an object in front of it move.</p>
<p><span class="math-container">$F_{applied}$</span> <span class="math-container">$= 10kg * acc(0) = 0N. $</span></p>
<p>EDIT:</p>
<p><a href="https://i.stack.imgur.com/Jmmck.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jmmck.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>With constant velocity, the force on the moving object is constant at the interval from the initial position to the final position, so the net change in force is zero. The graph below explains how is so.</p>
<p><a href="https://i.stack.imgur.com/LVudS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LVudS.png" alt="enter image description here" /></a></p>
<p>Note the object needs to accelerate from point 1 to point 2 to reach the velocity of 1000 m/s, thus the net change in force is 100N. The velocity from point 2 to point 3 is constant, a = 0, thus the net force change is 0N, but the force at point 3 = 100 + 0 = 100N (assume zero friction and air resistance, nothing to stop the motion). The force is mostly realized when the object travels with the same velocity from point 3 and hitting the barrier wall at point 4. Depending on the deformation characteristics of the object and the wall, the wall will exert a reactive force on the object to stop the motion, thus, F >= 100N.</p>
| 44755 | Does constant velocity means no force applied? |
2021-06-21T10:53:25.513 | <p>I'm trying to project a very thin line at a low angle over some surface defects (peaks and troughs). The scale is ~20mm square, but the defects can be sub-mm (0.2 - 1.0mm).</p>
<p>I am currently using a line generated by a laser pointer. The $4 laser diodes work well enough, but the line spreads out due to the low angle and poor focus(?).</p>
<p>Are there any better / safer alternatives, such as some LEDs + lens that could also focus the light into a line?</p>
<p>I'm new to engineering and optics so any advice would be much appreciated.</p>
| |sensors|optics|lasers|lighting| | <p>A few points that don't fit into a comment but could help you:</p>
<ul>
<li>Get a better laser source than a 4 \$ laser diode. "Better" means a laser with higher beam quality. When you look at a laser's data sheet, it should specify a value <span class="math-container">$M^2$</span> or "BPP" (beam parameter product). Both these numbers are related, the smaller they are, the better you can focus the laser beam.</li>
<li>Use a cylindrical lens (if you don't already), they are meant exactly for your application, focusing a laser onto a line instead of onto a spot.</li>
<li>From how laser diodes are built, compared to other laser sources, their beam quality differs in the two lateral directions. That means, a diode laser beam diverges for example faster in its height than in its width. It's only an assumption derived from this, but rotating the diode by 90 degree around the beam axis could possibly improve the focus of the laser line in this particular application.</li>
</ul>
<p>I think that with a laser you are already on a better way than with an LED, it's just about a few tweaks to improve your laser line.</p>
| 44756 | Alternatives to laser line? |
2021-06-22T15:56:49.243 | <p>I have a question regarding a not-so-common construction setup for a house.
In this situation, the concrete slab is "floating"/insulated with a rigid insulation underneath.
I understand the slab is not grounded and there's a risk for static charge build up.
Here's a picture of this setup:</p>
<p><a href="https://i.stack.imgur.com/Y2L5p.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y2L5p.jpg" alt="enter image description here" /></a>
(Source: https://www.buildingscience.com/documents/insights/bsi-059-slab-happy)</p>
<p>If my understanding is correct, what would be the best way to ground this floating slab?
One idea I had was to have a wire mesh inside the slab (if standard reinforcement is not enough) and ground that mesh on the perimeter of the slab (to the ground wire(s) from electric wiring already setup).</p>
| |electrical-engineering|electrical| | <p>I have used a similar construction method and have never seen any issues with static build up. It was just for thermal storage in a residential home. Like Carl mentioned in the comments the rug and pad would either isolate it or conduct to the surroundings. However, if this design is to be used in an electronics or pyrotechnics manufacturing facility you will indeed want to take some extra precautions.</p>
<p>To ground the internal floor to the foundation I would recommend running 6 gauge copper wire (mainly for flexibility and durability) from a <a href="https://rads.stackoverflow.com/amzn/click/com/B005GDFXR6" rel="nofollow noreferrer" rel="nofollow noreferrer">ground clamp</a> on rebar in the foundation to a ground clamp on the floating floor structural <a href="https://www.homedepot.com/p/42-in-x-7-ft-Remesh-Sheet-5901028/311174548" rel="nofollow noreferrer">remesh</a>. The structural remesh will be sufficient. That will certainly address any static issues, but it is always a good idea to check the <a href="https://www.nfpa.org/codes-and-standards/all-codes-and-standards/list-of-codes-and-standards/detail?code=70" rel="nofollow noreferrer">NEC (NFPA 70)</a> and your local electrical codes before a construction project. Even if grounding was not required by code, they may require any grounding you add to be installed by an electrician (if not your personal home).</p>
<p>Another option is to install a dedicated <a href="https://www.homedepot.com/p/ERICO-5-8-in-x-8-ft-Copper-Ground-Rod-615880UPC/202195738" rel="nofollow noreferrer">ground rod</a> or rods in or below the floating floor. Might be easier than messing with connections to the foundation.</p>
<p>If this is for a static sensitive application you will need to consult with a company that specializes in this. They will help specify conductive flooring, facility grounding, low static generating materials, HVAC systems and controls, garment specifications, etc.</p>
| 44774 | Electrostatic grounding of an electrically isolated concrete floor |
2021-06-22T17:31:46.160 | <p>Title says it all. I can’t find the Bluetooth version, only that it supports Bluetooth. Also, forgive me if this question is off topic here.</p>
| |consumer-electronics| | <p>Bluetooth 4.2 + EDR.</p>
<p>From Bestbuy: <a href="https://www.bestbuy.ca/en-ca/product/amazon-echo-show-5-smart-display-with-alexa-charcoal/13619574" rel="nofollow noreferrer">Amazon Echo Show 5 Smart Display with Alexa - Charcoal</a></p>
<p>EDR is Enhanced Data Rate which means it is backwards compatible to other versions that implement EDR (2.1 + EDR & 2.0 + EDR). <a href="https://en.wikipedia.org/wiki/Bluetooth#Bluetooth_4.2" rel="nofollow noreferrer">Wikipedia</a></p>
| 44778 | What version of Bluetooth does the Echo Show 5 support? |
2021-06-23T18:50:30.157 | <p>I thought I understood the ground concept but now I'm a lot confused.</p>
<p>From: Pg.67 Practical Electronics for Inventors - Scherz Paul</p>
<p><a href="https://i.stack.imgur.com/Pwtli.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pwtli.jpg" alt="Img in: Pg.67 Practical Electronics for Inventors - Scherz Paul" /></a></p>
<p>As it's explained in the book, the first circuity is the one that's dangerous if someone would put a hand on the chassis while this person is grounded(or touching something that is). But the second one is safe.
But grounding this circuity before the load wouldn't make the current go strait to the earth ground?</p>
<p>In the way I see, the second circuity is the same as the first when someone put a hand on the chassis, and as such, whatever the load is supposed to do, it won't work because there is no current passing through there.</p>
<p>Did I get something wrong???</p>
| |circuits|circuit-design|electronics| | <p>The issue is certainly NOT that the chassis ground is at -100V, it's absurd. The chassis ground by definition is at equal potential for power supply and load, you can treat it like one big wire that's large enough to take all the current you could throw into it. If you put a voltmeter between a car's chassis & earth, I promise it will never read -12V.</p>
<p>The reason the chassis needs to be grounded is 1) leakage to ground (R-sub-leak) can form a capacitor and you don't want to be the thing that lets its current flow. The second is that a fault to the chassis can result in the chassis being live at 100V, which is a problem if you touch it just like a bare wire (electricians call this getting bit, depending on how it goes through the body you can call it the end of you). This is prevented in modern electrical systems via a dedicated and properly wired ground wire, which is directly connected to true earth. It would be unusual to have a high voltage setup like shown using just a chassis ground because a fault could kill you.</p>
| 44794 | Electrical potential and grounding in a circuity, why it works? |
2021-06-23T20:20:35.947 | <p>So I've got a support system shown here in the attached image. Comparing configuration "A" to "B" it seems intuitive that "A" could support a larger load. The stubs are locking the two beams above and below the tube steel to eachother to increase resistance to deflection. How can I show this mathematically? A colleague suggested I would add the deflections of the beam above and below the tube steel to get a total deflection and you could show that this total deflection would be less than the deflection of the support beam in config. "B". This doesn't seem quite right to me, but I'm not sure. I should note I am assuming the taller base supports are 100% rigid.</p>
<p><a href="https://i.stack.imgur.com/LHqTN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LHqTN.jpg" alt="beam loading" /></a></p>
| |mechanical-engineering|statics|beam|deflection| | <p>Let's label the upper beam as "A", the lower beam as "B". Assume rigid vertical studs, the deflection at beam A at points 1 & 2 must equal to beam B at the respective locations, that is <span class="math-container">$\Delta_{A1} = \Delta_{B1}$</span> and <span class="math-container">$\Delta_{A2} = \Delta_{B2}$</span>.</p>
<p><a href="https://i.stack.imgur.com/fPSD7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fPSD7.png" alt="enter image description here" /></a></p>
<p>Now, release the studs and obtain the deflections at points 1 and 2 due to concentrate load <span class="math-container">$P_B$</span> at the mid-span of beam B.</p>
<p><span class="math-container">$\Delta_{B1} = \Delta_{B2} = \frac {P_Ba}{48EI}(3L^2 - 4a^2)$</span></p>
<p><a href="https://i.stack.imgur.com/GcBB0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GcBB0.png" alt="enter image description here" /></a></p>
<p>Next, apply a force <span class="math-container">$P_A$</span> at points 1 and 2 on beam A, and get the deflections at points 1 and 2.</p>
<p><a href="https://i.stack.imgur.com/BISVd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BISVd.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\Delta_{A1} = \Delta_{A2} = \frac {P_Aa}{6EI}(3aL - 4a^2)$</span></p>
<p>Now you have two equations and two unknowns. By equation <span class="math-container">$\Delta_{A1}$</span> and <span class="math-container">$\Delta_ {B1}$</span> you can get <span class="math-container">$P_A$</span> in proportion to <span class="math-container">$P_B$</span>. And based on the equivalent forces concept, <span class="math-container">$2P_A = P_B = Applied Load$</span>, you can get the exact/actual mangitude of <span class="math-container">$P_A$</span> and <span class="math-container">$P_B$</span>.</p>
<p>Finally, compute the beam deflections using the actual <span class="math-container">$P_A and P_B$</span> using the appropriate equations as shown above. Note the difference in deflection in segments between points 1 and 2 of beams A and B, it implies the beams suffer different internal forces.</p>
<p><em>Note, this solution is valid for the case that the load is directly attached to beam B; beam A shares the load through force transfer from the rigid studs, not through connection to the load.</em></p>
| 44796 | How to determine beam deflection with this unusual configuration (diagram attached) |
2021-06-23T20:44:17.423 | <p>When my car shifts gears, the car's acceleration never goes to 0. I feel that there is constant torque to the wheels throughout the entire shift.</p>
<p>How is this possible during a gear change?</p>
<p>Transmission is Ford's 10R80</p>
| |mechanical-engineering|automotive-engineering|car|power-transmission|transmission| | <h1>Multiple Clutches</h1>
<p>Many manual transmissions feature just a single clutch between the engine and the transmission. This requires that the clutch be disengaged before switching gears, during the switch, and until the final selected gear is fully engaged. When a human is performing this operation, the clutch may be engaged for many hundreds, even thousands of milliseconds, which is why you feel a lag in the acceleration.</p>
<p>An automatic transmission, partly due to using planetary gears, will usually have multiple clutches. Below is an older diagram, but modern high-gear transmissions are so complicated that any actual picture will be more confusing than enlightening.</p>
<p><a href="https://i.stack.imgur.com/HmlOR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HmlOR.jpg" alt="enter image description here" /></a></p>
<p>This picture is clearly more than complicated, even though the model shown is nearly 60 years old! Since each gear is defined by a particular set of clutches being engaged or disengaged, switching gears is as simple as engaging one clutch and releasing another. This can happen nearly simultaneously (< 100 ms up to a few hundred ms). In dual-clutch transmissions, it is often possible for the next gear to be pre-engaged, and only activated via a clutch. Thus, the first clutch disengages the current gear, and the second engages the next gear.</p>
<p>Naturally, there will, in fact, be a dip in power going to the wheels during this interval, but it is difficult for humans to detect changes much faster than 100 ms or so, given how slow our neurons are.</p>
| 44797 | How does my automatic car shift without complete temporary loss of acceleration |
2021-06-24T05:20:06.597 | <p>How does the (pulley based) CVT achieve the following?</p>
<ul>
<li>no requirement for a clutch</li>
<li>fuel economy</li>
<li>smooth gear transition</li>
</ul>
| |mechanical-engineering|gears|transmission| | <h2>Review of geared transmissions</h2>
<p>In general, the idea of a geared transmission is depended on the gear ratio. The gear ratio <span class="math-container">$i$</span> affects many aspects of a gear transmission, and the following equations apply:</p>
<p><span class="math-container">$$i=\frac{n_1}{n_2} = \frac{z_2}{z_1}= \frac{d_2}{d_1}= \frac{M_2}{M_1}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$n_i$</span> is the rpm in gear i</li>
<li><span class="math-container">$z_i$</span> is the number of teeth in gear i</li>
<li><span class="math-container">$d_i$</span> is the diameter of gear i</li>
<li><span class="math-container">$M_i$</span> is the torque in gear i</li>
</ul>
<p>Basically, what this gear ratio mandates is that a gear with a greater number of teeth will rotate slower than another with a less teeth.</p>
<p>The fact that the ratio of teeth is the same with the ratio of diameters,
<span class="math-container">$$i=\frac{z_2}{z_1}= \frac{d_2}{d_1}$$</span></p>
<p>essentially allows us to abstract the geared transmission with the idea of two circles that are rotating in contact (without slipping).</p>
<p><a href="https://i.stack.imgur.com/Qqt1b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qqt1b.png" alt="Enter image description here" /></a></p>
<p><strong>Figure 1: abstraction of operating pitch circles (source:<a href="https://www.tec-science.com/mechanical-power-transmission/basics/operating-principle/" rel="nofollow noreferrer">Tec-science</a></strong></p>
<p>A manual shift gearbox works by having (typically 4 to 6) different pair combination of gear diameters, thus changing the gear ratio by "shifting" to another pair.</p>
<p>However, during shifting, you need to use the clutch in order to decouple gear transmission. The activation of the clutch is what reduces the acceleration.</p>
<h2>How the concept of pulley based continuous variable transmission (CVT) works.</h2>
<p>The idea behind the continuous variable transmission is very simple, yet powerful because it achieves the following things.</p>
<ul>
<li>there is an infinite number of gear ratios (which enable the car engine to operate at its optimal point)</li>
<li>there is no need for a clutch (no need to "shift" gears)</li>
</ul>
<p>The simplest way ( Evans Friction Cone - circa 1903) to achieve that is the following:</p>
<p><a href="https://i.stack.imgur.com/lHUi3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lHUi3.png" alt="Enter image description here" /></a></p>
<p><strong>Figure 2: CVT concept (source:<a href="https://www.autotrader.ca/newsfeatures/20200317/cvt-transmission-pros-and-cons/" rel="nofollow noreferrer">autotrader.ca</a></strong></p>
<p>Basically, what happens is that the position of the pulley in the image above can be moved to any position, thus changing the diameters involved on each rotating shaft. Further the pulleys are designed to move according to the power demand of the car, always ensuring that the engine operates at its optimal range. Technically, this is possible because, CVTs have an infinite number of drive ratios.</p>
<h2>Modern implementations of pulley based CVTs</h2>
<p>The first pulley-belt CVTs is probably the <a href="https://en.wikipedia.org/wiki/Variomatic" rel="nofollow noreferrer">Variomatic</a>. It was developed by <a href="https://en.wikipedia.org/wiki/Hub_van_Doorne" rel="nofollow noreferrer">Hub van Doorne</a> for DAF and introduced around 1958.</p>
<p>Currently, the prevailing implementation of the CVT is different than the above illustration.</p>
<p><a href="https://i.stack.imgur.com/YDK0u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YDK0u.png" alt="Enter image description here" /></a></p>
<p><strong>Figure 3: CVT concept (source:<a href="http://www.rapidrepairautocenter.com/cvt-transmissions-the-pros-the-cons/" rel="nofollow noreferrer">rapidrepairautocenter</a></strong></p>
<p>In this implementation, each pulley (primary and secondary) concurrently change their effective diameter. This is achieved by moving their side walls in opposite direction (see above).</p>
<p>This effectively changes the gear ratio, without the need to decouple the two shafts.</p>
<h2>Other types of CVTs</h2>
<p>Apart from pulley based CVT's there are other implementations. Namely:</p>
<ul>
<li>Toroidal CVTs (which replace pulleys and belts with discs and power rollers) (e.g. used by the 1999 Nissan Cedric (Y34),</li>
<li>Hydrostatic CVTs (which involve converting fluid flow to rotational motion).</li>
<li>Electrical (e.g. used in Toyota e-CVT in 1997 Toyota Prius)</li>
<li>Epicyclic (or planetary CVT), uses planetary gears. (e.g. used in Toyota e-CVT in 1997 Toyota Prius)</li>
<li>Ratcheting</li>
<li>Magnetic</li>
<li>Friction-disk transmissions (e.g. used in <a href="https://en.wikipedia.org/wiki/Plymouth_Locomotive_Works" rel="nofollow noreferrer">Plymouth locomotive</a>)</li>
</ul>
| 44804 | How does a (pulley based) continuous variable transmission (CVT) work? |
2021-06-24T12:27:36.563 | <p><a href="https://i.stack.imgur.com/HPoaj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HPoaj.jpg" alt="a circuity" /></a></p>
<p>Considering that the ground is a a real and connected to earth ground, if the electrons have a low resistance path to it from the negative pole, why would the circuity still work? Maybe you could consider that the source of the voltage is a low electron density on the positive side or something similar, but just for theoretic purposes, if the source is a high density on the negative side, it wouldn't. Or i'm overlooking something?</p>
| |circuits|electronics| | <p>It is not about <em>density</em> as such. It is about potential or electric field.</p>
<p>There is an electric field which applies a force on the electrons from the negative pole to the positive. When connected to ground, at steady state there is no field forcing the electrons from the negative towards the ground.</p>
<p>Alternately, to use the potential analogy, electrons at the negative pole are sitting on a surface which <em>level</em> with <em>ground</em>. Electrons would have a tendency to roll towards the positive pole which is <em>below</em> the level or the negative pole and the ground. (positive pole is <em>below</em> the level of negative pole since electrons have negative charge). They won't have a tendency to roll from the negative pole to the ground which is at the same level.</p>
| 44813 | Why wouldn't grounding before a load in DC circuitry affect it? |
2021-06-24T12:36:37.200 | <p>I was trying to get a value for elastic modulus for 316L stainless steel material from internet, and I saw a brochure online which was giving two different values for elastic modulus, one in tension and one in torsion. I was confused why this is the case?</p>
| |solid-mechanics|elastic-modulus| | <p>Probably the value for the torsion related modulus for 316L should be close to 79[GPa], compared to 200-210 [GPa] for tension.</p>
<p>The shear modulus for isotropic materials can be analytically calculated as</p>
<p><span class="math-container">$$G = \frac{E}{2\cdot(1+\nu)}$$</span></p>
<p>where:</p>
<ul>
<li>G is the shear modulus</li>
<li>E is the tensile modulus</li>
<li><span class="math-container">$\nu$</span> is the Poisson's ratio.</li>
</ul>
<p>You can see a derivation in the following <a href="http://homepages.engineering.auckland.ac.nz/%7Epkel015/SolidMechanicsBooks/Part_I/BookSM_Part_I/06_LinearElasticity/06_Linear_Elasticity_Complete.pdf" rel="nofollow noreferrer">link (University of Auckland)</a></p>
| 44814 | Why do we have two different Elastic Modulus in Tension and Torsion? |
2021-06-25T09:18:10.850 | <p>I'm trying to create a smooth blended fillet on the outer edge of a part with fillets around it's corners in Solidworks. The results I have are pretty much what I want it to look like but when I check out the curvature the surfaces look pretty patchy - does anyone know of a good way to solve this? Or has faced similar issues?
<a href="https://i.stack.imgur.com/61LcQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/61LcQ.png" alt="desired edge fillet shape" /></a> <a href="https://i.stack.imgur.com/RJKco.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RJKco.png" alt="patchy surface curvature" /></a></p>
<p>I've tried building these corners several different ways including lofts and boundary surfaces, but the way I have settled on uses a couple of filled surfaces just in the corners (areas shown in the image below). I've played around a lot with the edge settings and the ones shown are the best I can come up with. If anyone has any ideas on a different method of building the fillet that would also be appreciated.
<a href="https://i.stack.imgur.com/cQMFn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQMFn.png" alt="patches filled with filled surface" /></a></p>
<p>Files: STEP of current surfaced fillet: <a href="http://www.filedropper.com/2021-06-25-filletedpart" rel="nofollow noreferrer">http://www.filedropper.com/2021-06-25-filletedpart</a>
STEP of unfilleted block: <a href="http://www.filedropper.com/2021-06-25-unfilletedpart" rel="nofollow noreferrer">http://www.filedropper.com/2021-06-25-unfilletedpart</a></p>
| |solidworks|cad|computer-aided-design|surface-modelling| | <p>Ok, looking at the "Filleted" file, it looks like you are trying to achieve the following:</p>
<ol>
<li>Yellow = 1mm</li>
<li>Orange = 2.3mm</li>
<li>Red = 5mm</li>
</ol>
<p><a href="https://i.stack.imgur.com/2jJjZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2jJjZ.png" alt="enter image description here" /></a></p>
<p>Your issue is primarily with order of operations. You need to do the yellow fillet <em>first</em>, in order to avoid the "sharp point where the two fillets meet" which you mentioned in your comment.</p>
<p>The file that you supplied described as "unfilleted" in fact had the yellow and red fillets already applied, so, adding the orange one after the fact does indeed look pretty rubbish:</p>
<p><a href="https://i.stack.imgur.com/J5waD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J5waD.png" alt="enter image description here" /></a></p>
<p>I removed both of these using the "delete face" command to give a true 'blank slate', and then added the yellow radius first, before adding both red and orange together using a variable radius fillet.</p>
<p><a href="https://i.stack.imgur.com/XKmE1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XKmE1.png" alt="enter image description here" /></a></p>
<p>P.S. The "Grey face" in the curvature evaluation <em>almost</em> runs 'curvature continuously' into the flat face next to it, but not quite. If you are worried about these fillets, it might be worth looking into that, too.</p>
| 44829 | Is there a way to fix 'bad' surfaces in Solidworks? |
2021-06-25T00:15:18.277 | <p>I'm thinking about using a pulley system that would allow heavy loads (e.g. a rock) to be lifted with fair ease with the intended goal of harvesting the load's falling energy.</p>
<p>Something that could be lifted over and over.</p>
<p>I think most hydroelectric dams use gravity and the water flow (like a windmill but with multiple turbines instead).</p>
<p>Here's my question:</p>
<p>Would the energy generated from the falling rock be far greater than what is consumed from lifting, and would that be a viable source of energy, theoretically (provided it can be stored efficiently)?</p>
| |energy-storage| | <p><strong>If you are thinking about "free" energy, then that's definitely not a way to achieve it</strong>. Although with some types of pulleys you might get half, 1/3, or even less force than the weight you are pulling, you'd be needing to pull the rope, twice, 3 or ever more times.</p>
<p>Basically the work is constant to raise a mass at height h, and its</p>
<p><span class="math-container">$$W = m \cdot g \cdot h$$</span></p>
<p>if you use a pulley that you need to apply half the force (see configuration 2).</p>
<p><a href="https://i.stack.imgur.com/KA8Xn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KA8Xn.png" alt="enter image description here" /></a>
<em>Figure 1: pulley configurations</em></p>
<p>then you need to pull a distance of 2h.</p>
<hr />
<p>However, if your question is</p>
<blockquote>
<p>can I use a low power input and slowly raise the weight (using a fraction of the weight of the mass that is being raised), so that I can later on obtain the energy at a much greater rate back?</p>
</blockquote>
<p>Then yes. You can use for example a small photovoltaic cell and power a small motor to slowly raise a weight. Then after the weight reached a height you can release the weight and obtain the potential energy that has been accumulated.</p>
| 44831 | Using a manual pulley system to repeatedly lift a rock and gather its falling energy to store in a power bank |
2021-06-27T06:37:23.920 | <p>I was reading the paper "Full-Speed Range Self-Balancing Electric Motorcycles Without the Handlebar" released in March 2016 by Professor Yang and Murakami in Keio University.</p>
<p>I couldn't figure out what the meaning of the following block is :</p>
<p><a href="https://i.stack.imgur.com/a5Vcq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a5Vcq.png" alt="Unkown Block" /></a></p>
<p>from the following control system diagram on page 1915 of IEEE Transactions on Industrial Electronics, Vol. 63, No. 3, March 2016.</p>
<p><a href="https://i.stack.imgur.com/Q2Bxq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q2Bxq.png" alt="Control System Diagram" /></a></p>
<p>I would greatly appreciate if someone could explain the meaning of the mentioned block symbol.</p>
| |control-engineering|control-theory|pid-control|optimal-control|diagram| | <p>If <span class="math-container">$\alpha$</span> has units of angle and <span class="math-container">$\tau$</span> has units of torque, then <span class="math-container">$I$</span> has units of moment of inertia. It is likely to be the moment of inertia about the axis which is being controlled.</p>
<p>The block is very likely a plain <code>gain</code> block which multiplies the angular acceleration signal with the value of the moment of inertia.</p>
<p>Published papers usually have a nomenclature section. Even otherwise, they will describe all symbols used. Check if the paper describes the symbol.</p>
| 44849 | What is the meaning of this symbol in Control System Diagram |
2021-06-27T14:30:48.473 | <p>Two-mass rotational system has the following form and is represented in following structural diagram.</p>
<p><a href="https://i.stack.imgur.com/Gd5L4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gd5L4.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$\tau_e$</span>, <span class="math-container">$\omega_1$</span> and <span class="math-container">$J_m$</span> - motor torque, angular velocity and moment of inertia</p>
<p><span class="math-container">$\tau_s$</span>, <span class="math-container">$\tau_s$</span>, <span class="math-container">$\omega_2$</span> and <span class="math-container">$J_d$</span> - shaft torque, load torque, angular velocity and load moment of inertia;</p>
<p><span class="math-container">$K_{md}$</span> - shaft stiffness</p>
<p><strong>Problem:</strong> how to include a gear ratio <span class="math-container">$N=\frac{\omega_1}{\omega_2}$</span> in equation of motion and in in a block diagram respectively?</p>
<p><a href="https://i.stack.imgur.com/UI6gD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UI6gD.png" alt="enter image description here" /></a></p>
<p>I wrote down the Lagrangian for each mass in terms of the gear ratio, but in the usual case, I just compiled a system of differential equations according to the Lagrange equation, but I cannot understand how now using the TWO Lagrangian to get one block diagram with a gear ratio.</p>
| |mechanical-engineering|electrical-engineering|mechanisms|modeling|mechanical| | <p>Assuming that the gear box is on the left end of the shaft (i.e. no flexible shaft between motor and gearbox).</p>
<ol>
<li>The angular velocity on the left end of the gear box is <span class="math-container">$\omega_1$</span>.</li>
<li>The angular velocity of the shaft side of the gear box is assumed as <span class="math-container">$\omega_1' = \frac{\omega_1}{N}$</span>.</li>
<li>The angular velocity on the right end of the shaft is <span class="math-container">$\omega_2$</span>. So the torque on the shaft is <span class="math-container">$\pm K_m (\frac{\phi_1}{N} - \phi_2)$</span>. (sign to be checked).</li>
<li>Because of the way I described the gearbox, <span class="math-container">$\omega_1' < \omega_1 $</span>. so the torque on the shaft when acting on the motor through the gearbox is <span class="math-container">$\frac{1}{N}$</span>. This can be seen in the below derivation.</li>
<li>Since I have assumed that shaft is directly connected to the load, the torque in the shaft is made available 1:1. This can also be seen in below derivation.</li>
</ol>
<p>(Below derivation to be verified independently by OP)
<span class="math-container">$$
L = \frac{J_m \omega_1^2}{2}
+ \frac{J_d \omega_2^2}{2}
+ \frac{Km (\frac{1}{N} \phi_1 - \phi_2)^2}{2}
$$</span></p>
<p><span class="math-container">$$
\frac{d}{dt} \frac{\partial L}{\partial \omega_1}
= \frac{d}{dt} J_m \omega_1
= J_m \frac{d \omega_1}{dt}
$$</span></p>
<p><span class="math-container">$$
\frac{\partial L}{\partial \phi_1}
= \frac{K_m}{\color{red}{N}} (\frac{1}{N} \phi_1 - \phi_2)
$$</span></p>
<p>Similarly for the other body also (exercise left to you).</p>
<p><span class="math-container">$$
\frac{\partial L}{\partial \phi_2}
= -K_m (\frac{1}{N} \phi_1 - \phi_2)
$$</span></p>
<p>I have not considered the input torque. It can be added to this result.</p>
| 44852 | Dynamic model of a two-mass electric drive taking into account the Gear Ratio |
2021-06-28T12:46:00.360 | <p>Reading the <a href="https://en.wikipedia.org/wiki/Ultrasonic_horn" rel="nofollow noreferrer">Ultrasonic Horn Wikipedia article</a>, it states:</p>
<blockquote>
<p>An ultrasonic horn is a tapering metal bar commonly used for augmenting the oscillation displacement amplitude provided by an ultrasonic transducer...</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/k6qtrm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k6qtrm.png" alt="enter image description here" /></a></p>
<p>I can see how a smaller area would increase the applied pressure, but I don't understand how it would increase the displacement. Is it stretching and compressing the bar like a spring? Anyone with experience in this area? References on how to calculate this would be very helpful. Thanks!</p>
| |vibration|acoustics| | <p><sub>Disclaimer: Even though I posted a relevant <a href="https://www.sciencedirect.com/science/article/pii/S135041771500142X" rel="nofollow noreferrer">link</a> (in a comment to the question ) which can answer the exact question asked, I am not familiar enough with continuum mechanics to actually explain it. Having said that ...</sub></p>
<p>The behavior seen in continuous elastic material may have discrete spring mass analogues.</p>
<h2>Qualitative / by Analogy</h2>
<p>Consider a <a href="https://en.wikipedia.org/wiki/Newton%27s_cradle" rel="nofollow noreferrer">Newton's cradle</a>. All the balls are identical. You input a specific displacement to a ball at one end. That ball then drops and collides with the rest. The <strong>energy is transferred</strong> to the ball at the other end which is identical and so moves by the same amount.</p>
<p>Now consider a case where the balls are non identical. The displacement can be larger or smaller depending on mass of the last (and perhaps intermediate) ball. If the last ball is smaller, it will under go a larger displacement than the ball on the <em>input</em> side.</p>
<ol>
<li><a href="https://en.wikipedia.org/wiki/Galilean_cannon" rel="nofollow noreferrer">Galilean cannon</a>.</li>
<li><a href="https://youtu.be/2UHS883_P60?t=17" rel="nofollow noreferrer">A video showing different ball sizes (dur : 3:33)</a>.</li>
<li><a href="https://youtu.be/ikDgvxAw9wM?t=71" rel="nofollow noreferrer">Newton's cradle with varying ball size (dur : 2:21)</a></li>
</ol>
<p><a href="https://i.stack.imgur.com/WTU8i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WTU8i.png" alt="newton's cradle with different ball sizes. video screenshot" /></a></p>
<p>The oscillations in an ultrasonic horn are linear in nature and the above examples involves collisions (non linear ?); so the logic may not carry over one-to-one. The basic ideas of energy transfer and energy conservation still apply.
<strong>If</strong> this logic carries over to the ultrasonic horn, then a smaller end would be expected to under go larger displacement since a smaller cross section would need to compress/expand by larger amount when forced to store the same (elastic) energy that was supplied at the input.</p>
<p><strong>But</strong>, A short search of literature clearly shows that ultrasonic horns can have diverging sections also; apart from converging sections. Examples seen in literature show horns having multiple converging-cylindrical-diverging section. This may be due to other reasons and may not invalidate the above explanation. So the discrete analog of the system may not be a perfect analogy.</p>
<h2>(kind of) Quantitative</h2>
<p>The one dimensional partial differential equation can be discretised in spatial domain. The equation given in the linked reference literature is</p>
<p><span class="math-container">$$
\frac{\partial }{\partial x}(S\cdot E \frac{\partial u}{\partial x}) dx
= S\cdot \rho \frac{\partial^2 u}{\partial t^2} dx
$$</span></p>
<p>On discretising, the RHS becomes <span class="math-container">$m_i \frac{d^2 u_i}{d t^2}$</span> where <span class="math-container">$m_i$</span> is the ith discrete mass.</p>
<p>On the LHS, the <span class="math-container">$\frac{\partial^2 u}{\partial x^2}$</span> will take the form <span class="math-container">$u_{i+1}+u_{i-1}-2u_{i}$</span> and <span class="math-container">$ \frac{\partial S}{\partial x} \frac{\partial u}{\partial x}$</span> will take the form <span class="math-container">$p\cdot (u_{i+1}-u_{i})$</span> where <span class="math-container">$p$</span> depends on the taper and <span class="math-container">$p=0$</span> for the cylindrical section.</p>
<p>So the discretised equation will be of the form
<span class="math-container">$$
a u_{i+1} -b u_{i} + c u_{i-1} = \frac{d^2 u_{i}}{d t^2}
$$</span></p>
<p>The eigen vectors of this discretised system will provide the deflection at resonance.</p>
<p>An Octave code which <strong>attempts</strong> to construct the system and then find its eigen vectors is given below. (code is not verified; doesn't follow physical units)</p>
<pre><code> clc;
% number of elements / sections
n = 100;
choice = 3;
switch(choice)
case 1
% cylindrical section
si = ones(1, n);
tstr = 'cylinder';
case 2
% linear taper
si = linspace(10, 1, n);
tstr = 'linear taper';
case 3
% step at half length
si = [10*ones(1, n/2), 1*ones(1, n/2)];
tstr = 'step';
end
% normalize area so that
% the net mass is the same for each case
si = si/sum(si);
% mass proportional to area. density = 1
mi = si;
sum(mi) % check mass
% spring constant of spring from ith to i+1th section
% spring constant proportional to average area of neighboring sections
ki_ip1 = (si(1:end-1) + si(2:end))/2;
%% differential equation
% M xdd = K x
% xdd = M^-1 K x;
M = diag(mi);
K = diag(ki_ip1, 1) ... connection to m_i+1
+diag(ki_ip1, -1) ... connection to m_i-1
-diag([ki_ip1, 0]) ... connection of other end of spring m_i
-diag([0, ki_ip1]);... connection of other end of spring m_i
% attach the first node "rigidly" to wall
% if boundary condition requires it
% but the paper states the "free-free" assumption instead.
% K(1,1) = K(1,1) - 100;
% find the mode shapes of M^-1 K
A = -M\K;
[V, lambda] = eig(A);
% normalize the input end to +-1
% this allows us to see the "amplification" at the free end
% for various rod shapes
V = V ./ repmat(V(1, :), [n, 1]);
% normalize the "free" output end to positive value
V = V .* repmat(sign(V(end, :)), [n, 1]);
% extract the square of the frequency
lambda = diag(lambda);
% sort by frequency
[l, iii] = sort(lambda);
% plot mode shape low freq modes
% the second mode is the important mode to note.
figure(1);
select = 1:4;
plot(V(:, iii(select)), 'linewidth', 1.5);
legend(num2str(sqrt(l(select))), 'location', 'northwest');
title(tstr, 'fontsize', 14);
set(gca, 'fontsize', 14);
xlabel('node number');
ylabel('normalised amplitude');
</code></pre>
<p>The resulting mode shapes (first 4 modes including 0 frequency mode) from above code is shown below. As mentioned in the linked paper, mode shapes for the cylindrical horn are sinusoids. Mode shapes for the stepped horn are sinusoids of differing amplitudes.</p>
<p>Both the converging shapes show amplification at the free end when compared to the cylindrical horn.</p>
<p><a href="https://i.stack.imgur.com/jQ6bI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jQ6bI.png" alt="mode shapes for different configurations of the ultrasonic horn" /></a></p>
<h2>Quantitative</h2>
<p>Extracting the mode shapes from a proper FEM model will give the free end amplification for complicated shapes.</p>
| 44867 | Does an ultrasonic horn increase displacement? |
2021-06-29T13:08:59.820 | <p>The Ogata and Banks analytical solution of the convection-diffusion equation for a continuous source of infinite duration and a 1D domain:</p>
<p><a href="https://i.stack.imgur.com/1y9G1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1y9G1.png" alt="enter image description here" /></a></p>
<p>where <code>C [mol/L]</code> is the concentration, <code>x [m]</code> is the distance, <code>R</code> is the retardation factor, <code>D [m2/day]</code> is the effective dispersion/diffusion, <code>v [m/day]</code> is the flow velocity, <code>Ci [mol/L]</code> is the initial concentration in the column, and <code>Co [mol/L]</code> is the influent (or injected) concentration.</p>
<p>The time-evolution plot below shows two cases of specie production and degradation obtained from the numerical solution of equation (1) with the following parameters:</p>
<p>Production (red line): <code>Ci = 0 mol/L, Co = 1.2 mol/L, R = 1, D = 0.00048 m2/day, v = 0.24 m/day</code>.</p>
<p>Degradation (green line): <code>Ci = 1 mol/L, Co = 0 mol/L, R = 1, D = 0.00048 m2/day, v = 0.24 m/day</code>.</p>
<p><a href="https://i.stack.imgur.com/2Gepf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Gepf.png" alt="enter image description here" /></a></p>
<p>The analytical equation above matches (this is not shown in the plot) the production but not the degradation, whose curve is clearly an inversion of the production curve. So, I have been looking for the equivalent analytical expression that describes the degradation process (meaning that Ci would be contained therein). I am thus hoping that there is someone who may know this equation or could point me in the right direction. Thank you in advance.</p>
| |fluid-mechanics|cfd|numerical-methods|convection|diffusion| | <p>To laymen's , R in electrical theory could describe a downward OSC reading, hence the retardation of blackbox and Weiner's whitebox, circular ambularity. Apologies for my unspecific brevity.</p>
| 44883 | Analytical solution for the 1D convection-diffusion equation |
2021-06-29T15:44:22.370 | <p>This question might come across as a newbie one but are pressure suits and anti-g suits are two names for the same thing? Google didn't turn up differences between them specifically (if there is any), and Wikipedia has separate pages for both of them. I found an <a href="https://patentimages.storage.googleapis.com/b7/0d/1b/48829c22759f49/US5007893.pdf" rel="nofollow noreferrer">old patent paper</a> which says it combines both pressure suits and anti-g suits though I couldn't glean much from it. Their working and effects seemed similar but I couldn't find any confirmation whether they are different. So are they different or not?</p>
| |pressure|aerospace-engineering|terminology| | <p>A <a href="https://en.wikipedia.org/wiki/Pressure_suit" rel="nofollow noreferrer">pressure suit</a> is <strong>a protective suit worn by high-altitude pilots</strong> who may fly at altitudes where the air pressure is too low for an unprotected person to survive, even breathing pure oxygen at positive pressure. Such suits may be either full-pressure (i.e. a space suit) or partial-pressure (as used by aircrew). Partial-pressure suits work by providing mechanical counter-pressure to assist breathing at altitude.</p>
<p>A <a href="https://en.wikipedia.org/wiki/G-suit" rel="nofollow noreferrer">g-suit</a>, or anti-g suit, is <strong>a flight suit worn by aviators and astronauts who are subject to high levels of acceleration force (g)</strong>. It is designed to prevent a black-out and g-LOC (g-induced loss of consciousness) caused by the blood pooling in the lower part of the body when under acceleration, thus depriving the brain of blood.<a href="https://en.wikipedia.org/wiki/Pressure_suit" rel="nofollow noreferrer">1</a> Black-out and g-LOC have caused a number of fatal aircraft accidents</p>
| 44886 | Difference between pressure suits and anti-g suits |
2021-06-29T16:12:16.907 | <p>I had a quick question regarding this homework problem, I can't seem to understand why they chose to use the conservation of momentum in vertical direction and not the horizontal direction as you would get a different end result if you chose the horizontal direction to work with. Would anyone kindly explain this to me ? Is this because the question states that there can be no tangential force on the plate surface so that means horizontal forces and thus working in the horizontal direction is neglected ??
Thank you very much in advance!</p>
<p><a href="https://i.stack.imgur.com/XnyWl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XnyWl.png" alt="Problem + Solution" /></a></p>
| |fluid-mechanics| | <p>The water hits the surface and the moving directions of the water particles get changed. Their speeds are assumed to stay.</p>
<p>Let's follow what happens when a particle with mass M hits the surface and splits to pieces. Piece 1 with mass M1 continues to the +X direction in the solution drawing and piece 2 which has the rest of the mass M2 = (M-M1) turns to the negative X direction.</p>
<p>Piece 1 has momentum V * M1 and it's directed to positive X.</p>
<p>Piece 2 has momentum -V(M - M1) if we calculate it to the +X direction.</p>
<p>The total momentum of the incoming particle is not conserved. The Y direction momentum is used to make the normal force to the surface. The sum of the X-direction momentums of the particles stay because the plate causes no force to the particles in X-direction (=frictionless)</p>
<p>The X-direction momentum of the incoming particle is V * M *sinθ</p>
<p>The X-direction conservation equation:</p>
<p>V *M * sinθ = V * M1 -V(M - M1)</p>
<p>V can be dropped off. After reordering we get M1 = M(1 + sinθ)/2 or equivalently</p>
<p>(M1/M) = (1 + sinθ)/2</p>
<p>If we assume the water is non-compressible the jet thicknesses must have the same ratio and that's the same result as presented in the original solution.</p>
<p>Your "working in the horizontal direction -idea is a misconception". The conservation of the momentum is only true for the "along the surface component". The rest of the momentum is lost because the surface stops the motion in the surface normal direction.</p>
| 44887 | Need help on conservation of momentum (liquid jet) Problem! |
2021-06-29T18:28:07.377 | <p>Consider a "beam". In order to perform modal analysis using the finite element method, the beam is meshed with 2nd order hexa elements. The total count of nodes is 1280. The same beam can be either clamped on one side or both sides. Clamped ends are modelled by zero displacements (dx=dy=dz=0) for all nodes on the clamped surface. There are 80 nodes per clamped surface.</p>
<p>The mass and stiffness matrices are generated using code_aster.</p>
<p>The clamped-clamped beam matrices are 4800x4800, meaning that the system has 4800 degrees of freedom (dof).</p>
<p>The simply-clamped beam matrices are 4320x4320, meaning that the system has 4320 degrees of freedom (dof).</p>
<p>I don't understand where the number of dof comes from. I would expect 6 dof per node which gives:
1280 • 6 = 7680</p>
<p>Worst than that, if constraints are already included, I would expect the clamped-clamped beam to have less dof than the simply-clamped.</p>
<p>Can anyone help me?</p>
| |mechanical-engineering|finite-element-method|modal-analysis| | <p>I found the solution. First of all, I made a mistake, in the example I presented I was using 1st order hexa elements.</p>
<p>Code_aster uses lagrange multipliers to apply constraints (in this case clamped faces).</p>
<p>In order to use the <span class="math-container">$LDL^T-SP$</span> algorithm, it doubles the lagrange multipliers.</p>
<p>The additional equations are added into the matrices (mass, stiffness,...).</p>
<p>Conclusion:</p>
<ul>
<li><p>Each unconstrained node has 3 dof.</p>
</li>
<li><p>Each node in the clamped faces has 3 dof, plus 2 lagrange multipliers for each dof (because dx=dy=dz=0). Total = 3 + 2x3 = 9.</p>
</li>
</ul>
<p>Finally, the total count:</p>
<ul>
<li><p>Clamped clamped beam: 1280x3 + 80x3x2x2 = 4800</p>
</li>
<li><p>Clamped beam: 1280x3 + 80x3x2 = 4320</p>
</li>
</ul>
<p>For more details, please refer to the <a href="https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.code-aster.org/V2/doc/default/en/man_r/r3/r3.03.01.pdf&ved=2ahUKEwjBrr2LtsvyAhWPDuwKHZoODrAQFnoECAMQAQ&usg=AOvVaw2-4fZDqvk2cgrzcLT4UN0L" rel="nofollow noreferrer">documentation</a>.</p>
| 44891 | FEM degrees of freedom |
2021-06-29T18:33:19.977 | <p>I'm not an expert in mechanics, so this is my question.</p>
<p>I will buy the motor of the image below to move my door.</p>
<p>In the motor shaft, there is a gear with 50 mm of diameter which will move the door.</p>
<p>The power you does to move the door is similar to lift 12 kg.</p>
<p>The speed needed is about <strong>0,3 m/s.</strong></p>
<p>So, can this motor move the door?</p>
<p>The text are in Portuguese.</p>
<p>Corrente = Current.</p>
<p><a href="https://i.stack.imgur.com/aycai.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aycai.jpg" alt="Motor Details" /></a></p>
<p>And here is more details of the mechanics, is the idea:</p>
<p><a href="https://i.stack.imgur.com/Wbpe8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wbpe8.png" alt="enter image description here" /></a></p>
<p>Thanks for any help.</p>
| |mechanical-engineering|motors|torque| | <p>The relation of torque with force is,</p>
<p><span class="math-container">$$\tau= F*r \quad 1.16Nm =12*9.8*r \\ r= 1.16/117.6=0.00986m=0.986cm$$</span></p>
<ul>
<li><p><span class="math-container">$\tau$</span> = torque</p>
</li>
<li><p>f=12kg(g)= 12*9.8=117.6N</p>
</li>
<li><p>r= radius of the pully= 0.986cm</p>
</li>
</ul>
<p>So your 5cm=2.5cm radius pulley can not move your door. Either you need to replace it with the small size pulley or you have to use a gear with <span class="math-container">$r= 2.5/0.986* 2.5=6.33cm \text{ or a diameter of 12.67cm } \ $</span> engaged with your 50mm pulley.</p>
| 44893 | Will a motor with 1,16 Nm rotate a 50 mm gear moving about 12 kg |
2021-06-30T00:47:45.233 | <p>I know that PLA does not bond at all, just peel right off
But i'm wondering about PETG or ABS</p>
<p>Will they bond just as well as wood?</p>
| |plastic|wood|epoxy|abs| | <p>I have had good luck with plastic specific consumer grade epoxies. Both <a href="https://www.homedepot.com/p/J-B-Weld-0-85-oz-Plastic-Bonder-50133H/303710788" rel="nofollow noreferrer">JB Weld Plastic Bonder</a> and <a href="https://www.lowes.com/pd/LOCTITE-Plastic-Bonder-Yellow-Epoxy-Adhesive/3220051" rel="nofollow noreferrer">Loctite Plastics</a> have worked well. I have not performed any structural tests with them, but in my experience they are ~25% stronger than just standard 5 minute epoxy for plastic. Perhaps they addressed some of the surface chemistry Jim Clark mentioned.</p>
<p><a href="https://en.wikipedia.org/wiki/Cyanoacrylate" rel="nofollow noreferrer">Cyanoacrylate</a> like Carl Witthoft mentioned may also work well. In my experience it only works where the surfaces are very rough and where the bond area will not flex. Note that Cyanoacrylate reacts with water (moisture in the air) to cure. This can make it difficult to use on large plastic to plastic bonds because plastic and cured Cyanoacrylate restrict the movement of water to the uncured portion of the joint.</p>
<p>ABS can be bonded to itself with Acetone and other solvent based glues. Common drain pipe is ABS and <a href="https://www.homedepot.com/p/Oatey-4-oz-Medium-Black-ABS-Cement-309993/100136815" rel="nofollow noreferrer">ABS pipe cement</a> will work. It looks like PETG can be dissolved with toluene and MEK, so you have similar options there. Lots of experimentation to get these adhesives to stick to other materials though.</p>
<p>ABS and PETG are both thermoplastics and can be <a href="https://en.wikipedia.org/wiki/Plastic_welding" rel="nofollow noreferrer">melted together</a> with heat or ultrasonic welding.</p>
<p>Anytime you are relying on an adhesive for attachment, a roughened surface helps improve the bond. And even better if you can change the geometry of your design to put the adhesive joint in <a href="https://www.3m.com/3M/en_US/bonding-and-assembly-us/resources/science-of-adhesion/common-stress-types-adhesive-joints/" rel="nofollow noreferrer">shear instead of tension or peeling stresses.</a></p>
| 44905 | How well does epoxy bond to PETG / ABS vs Wood? |
2021-06-30T12:29:05.383 | <p>I'd like to visualise/measure the flow of air in sauna to optimise the ventilation.</p>
<p>Is there a way to colour the air (non-toxic!!!) so I may <strong>watch</strong> the flow?
Or is there any other accessible technology applicable in the home conditions?</p>
<hr />
<p>EDIT: Finally I asked a DJ on a wedding for a small amount of the fog-machine liquid and I poured it directly to the sauna stove. It produced some smoke and then it (or maybe rather the gas produced by the heat) started burning, but not with long flames. Apparently the heat wires in the stove are too hot. The produced smoke visualised the air flow sufficiently. The only problem was that the smoke dissolved in the air so farther from the stove it wasn't so distinctive, but it was good enough. Thanks to @dknguyen for the tip how to do it without a fog machine.</p>
| |heat-transfer|measurements|airflow|convection| | <p>You can buy an incense stick, which burns slowly and produces perfumed smoke. you then position the incense stick at different points in the sauna and make a video of the resulting smoke trails.</p>
| 44909 | How to visualise/measure airflow in sauna? |
2021-07-01T02:31:39.187 | <p>I was just wondering whether or not the standard beam bending equations worked for all units. E.g. imperial and metric etc.</p>
<p>From what I'm guessing it shouldn't really, since once we plug everything in the units should cancel out properly.</p>
<p><a href="https://i.stack.imgur.com/0bt4V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0bt4V.png" alt="enter image description here" /></a></p>
| |statics|beam|bending| | <h1>What's an equation?</h1>
<p>In life there are two types of equations:</p>
<ul>
<li><em><strong>theoretical equations</strong></em> are obtained from first principles: make some assumptions and then play around with variables until you get a useful equation</li>
<li><em><strong>empirical equations</strong></em> are obtained by experiment and then finding an equation that adequately describes the observed behavior.</li>
</ul>
<p>Theoretical equations are always unit-agnostic. This is because the coefficients originated from your mathematical manipulations (probably integration or derivation, etc) and don't actually represent any physical quantity. For this reason, dimensional analysis must, by definition, always work out for theoretical equations. So if you have a theoretical equation <span class="math-container">$y = 3x^2$</span>, you can be sure that those coefficients popped out from some numerical manipulation, and that therefore the unit of <span class="math-container">$y$</span> will be equal to the square of <span class="math-container">$x$</span>'s unit (so if <span class="math-container">$x$</span> is a length <span class="math-container">$[L]$</span>, <span class="math-container">$y$</span> is an area <span class="math-container">$[L^2]$</span>, for example).</p>
<p>Empirical equations are another matter entirely and are usually unit-system-dependent. This is because the equations are effectively improvised with no mathematical backing to them, so the coefficients (including exponents) which are adopted may need to be entirely different. So much so that dimensional analysis will more often than not fail for empirical equations. This means you may be dealing with an empirical <span class="math-container">$y = 3x^2$</span> where <span class="math-container">$x$</span> is a length <span class="math-container">$[L]$</span> and <span class="math-container">$y$</span> is a mass <span class="math-container">$[M]$</span>, which of course makes no sense analytically.</p>
<p>Indeed, it only makes sense if you know that the 3 coefficient is actually 3 kg/m<sup>2</sup>: therefore, this equation only works if you use meters for <span class="math-container">$x$</span> and expect kilograms for <span class="math-container">$y$</span>.</p>
<p>If you know the coefficient's units, however, you could convert this equation to other units of measurement pretty easily: <span class="math-container">$3\text{ kg/m}^2 = 3000 \text{ g/m}^2 = 0.3 \text{ g/cm}^2 \approx 0.048 \text{ g/in}^2$</span>. But still, each equation (using 3, 3000, 0.3, 0.048 or any other coefficient you can think of) will only be valid for the specific units used to define that coefficient. And more complex cases can make this far more complicated to figure out.</p>
<h1>And what about bending equations?</h1>
<p>Well, now answering your question is very straightforward: the beam bending equations are 100% theoretical. Euler and Bernoulli made assumptions (parallel sections remain parallel, etc), played around with variables and -- poof! -- out came the fundamental beam equation, from which we can then derive the deflections for different boundary conditions.</p>
<p>Being theoretical equations, we can be sure that they are, by definition, unit-agnostic.</p>
| 44922 | What are the units used in beam bending equations? Do they matter? |
2021-07-01T06:25:06.913 | <p>Given some desired release angle, could someone please suggest a way that a machine could alter the angle of a launch tube per feed back from an ultrasonic sensor (I've derived the relation between range and desired angle separately, but is not really necessary here). I'm not looking for an extremely detailed description, but rather some broad outline on a method.</p>
| |mechanical-engineering| | <p>A screw thread would work, use the sensor to drive a motor with suitable gearing.</p>
| 44936 | Methods of altering the angle of a projectile launch tube |
2021-07-02T09:30:22.780 | <p>Related to the question: <em><a href="https://engineering.stackexchange.com/questions/44946/why-would-anyone-pull-the-stem-out-and-stop-the-wristwatch-at-night-or-when-not">Why would anyone “pull the stem out and stop the wristwatch at night or when not wearing it” in the context of mechanical watches?</a></em></p>
<p>When the stem of either a mechanical wrist or pocket watch is <strong>fully</strong> pulled out and the watch appears to stop, what happens inside the watch; are the hands just disengaged, is the watch stopped and what happens to the main spring, does it continue unwinding or is it prevented from unwinding?</p>
| |mechanisms|watch| | <p>It disengages the winding wheels and eventually stops the movement after the energy stored in the spring has exhausted.</p>
<p><a href="https://i.stack.imgur.com/lb46D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lb46D.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/LFulD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LFulD.png" alt="enter image description here" /></a></p>
| 44957 | What happens to a mechanical watch when the stem is fully pulled out? |
2021-07-03T14:29:50.723 | <p>Recently I have been reading about the <a href="https://en.wikipedia.org/wiki/Costa_Concordia_disaster" rel="nofollow noreferrer">shipwreck</a> of the Costa Concordia, unfortunately what I read is not online, so I have no links, but it raised a lot of questions.</p>
<p>A rock hit by the ship tore a 50-metre gash which cause the flooding of 3 water tight compartments. The flooding disabled all the electrical systems, only the emergency generator kept working, but it could only power the communication systems and the lights, it didn't even have enough power to turn the rudder.
Apart from that none of the descriptions of the events I read went into the technical details, I don't even know whether they could power a water pump to slow down the sinking.</p>
<p>I read the description of the ship on <a href="https://en.wikipedia.org/wiki/Costa_Concordia" rel="nofollow noreferrer">Wikipedia</a> which told that the ship had a diesel-electric propulsion powered by 6 generators. Since they were all damaged by the flooding they had to be in the same compartment or in two adjacent water tight compartments. Actually I am still wondering how the compartments were arranged. Did they have only bulkheads going from side to side? Were there longitudinal bulkheads?</p>
<p>But actually my question is why put all the generators close together? I guess that putting by design two engine rooms in different parts of the ship would have increased the cost and the complexity of the systems, but it would have added some redundancy. Actually there could be two ways of adding redundancy, another solution could be to add some longitudinal bulkheads at the cost of extra weight.</p>
<p>What could be the cost increase and the technical issues due to a more redundant design?</p>
<h2>Update</h2>
<p>Reading the @StainlessSteelRat post which does not answer my question I thought that maybe I put too many question marks that might have created some confusion. I'll rephrase my question, but I'll leave the old one above to leave the context, please ignore the question marks above.</p>
<p>From a 290 meters long ship with two engines and 6 generators designed to carry almost 5000 people counting crew and passengers you might expect some resilience. Instead all the systems were knocked out by a single breach. It seems that the designers didn't consider the possibility that a breach might have been that long, but also the designers of the Titanic called it unsinkable because they didn't consider the possibility of a long breach.</p>
<p>Now, to improve the resilience maybe they could have put one of the generators rooms close to the bow (far from the other one) and they could have added a longitudinal bulkhead to split the engines room.
<br>What could be the cost in term of additional weight and complexity of such design or another design with as much resilience?</p>
| |safety|marine-engineering|ships| | <p>Engineering is a knowledge of physics and a game of economic tradeoffs. I have not designed any watercraft, and have less knowledge this specific wreck than the OP, but I can explain some of the engineering design process and business management that results in situations like this.</p>
<p>A cruise ship is basically a small city and this massive project is subdivided into specialties; electrical, mechanical, structures, HVAC, plumbing, architectural, drainage, wastewater treatment, information technology, etc.</p>
<p>The individual engineers do their specific job functions to best meet the requirements they have been given by business managers that have business specific economic and functionality requirements. The engineers at this level are usually highly ethical and do the best they can within the financial and engineering constraints of their task.</p>
<p>The requirements for the power plant group was likely:</p>
<ol>
<li>Provide sufficient power to all ship systems.</li>
<li>Provide redundant generators so the ship can operate at normal capacity 24/7 year round with one or more of the generators offline for maintenance.</li>
<li>Reduce capital cost of the ship: shortest power transfer distances, small footprint, simple controls.</li>
<li>Reduce the operating costs of the ship: again short power transfer distances, keep everything in one area so only one generator operator is needed instead of two, all the same maintenance in same location, large passages and gantry cranes to unmounted equipment and move replacement equipment into place.</li>
<li>Safety, keep all the fuel in one spot to reduce the risk of fire. Reduce power transmission distances and voltages. Reduce the walking distance for preventative maintenance and safety rounds (employee patrolling relevant equipment).</li>
</ol>
<p>It is easy to see in hindsight that this specific scenario could have benefited by having two separated generator compartments, but survival of the vessel after a wreck was not their priority.</p>
<p>In contrast, a military ship like an aircraft carrier has military contractor oversight rather than business contractor oversight guiding the higher level design priority processes.</p>
<p>The backup plan on a military vessel is to have redundant redundant systems to limp back to port; because in wartime there might not be anyone coming to save you. Operating costs are considered, but they have a taxpayer budget and no competition to worry about, so they focus on reliability and functionality.</p>
<p>The backup plan on a vacation cruise ship is to have basic safety systems including life rafts and have redundant communication systems to radio the friendly outside world for help.</p>
<p>Supposing the cruise ship manufacturer's salesman did propose a redundant power plant safety feature (or a similar safety feature) to a customer at one time it would go like this: <br></p>
<blockquote>
<p><strong>Vendor:</strong> Would you like us to add a partition to the generator room for power plant redundancy?<br>
<strong>Customer:</strong> How much will it cost? and how much will it increase my operating cost? <br>
<strong>(3 months of engineering later) Vendor:</strong> 1.3M USD, estimated operating cost will be 230K USD/year in operator labor and maintenance.<br>
<strong>Customer:</strong> Says "okay", sends it to the accountants and insurance providers to crunch the numbers.<br>
<strong>(6 months later) Accountants:</strong> Based on the statistical likeliness of the generator room flooding from a collision, the insurance company offers a 3.6M USD annual premium collision insurance WITH this feature and a 3.7M USD annual
premium collision insurance WITHOUT for a savings of $100K USD/year.
As such, the finance team unanimously advises to forgo this redundant
safety feature.</p>
</blockquote>
<p>All of that said, in light of this wreck, you will likely see future ships with new fail-safes and redundancies in place for this and other failure modes.
This is because:</p>
<ol>
<li>The insurance statistics changed as a result of the wreck.</li>
<li>Everyone's financial and personal liability just went up; the captain, the crew, the passengers, the investors, the manufacture, the lawyers, etc.</li>
<li>Sinking a ship and killing people is obviously bad PR and bad marketing hurts all businesses involved.</li>
<li>I do not know if <a href="https://cruising.org/en/about-the-industry/policy-priorities/cruise-industry-regulation" rel="nofollow noreferrer">this organization</a> has legal teeth or not since cruises operate in international water, but engineering codes can be made more strict by governments much like building codes. In the interest of preserving human life, such a code could require new provisions that would reduce the risk of a similar wreck in the future.</li>
</ol>
| 44973 | Disposition of the generators on the shipwrecked Costa Concordia |
2021-07-06T19:17:25.877 | <p>I am trying to create a line made from light projected from a lens / guide of some sort onto a surface from a few mm away.</p>
<p>I have used a laser, but comes with safety considerations, space to mount the optics, and a lot of fine tuning.</p>
<p>I was testing alternatives, and had some success using the fibre-optic effect of the edge of an acrylic sheet and some cheap LEDs.</p>
<p>Does a lens or light guide of the type shown below exist?</p>
<p>I am not worried about the "shape" or "geometry" of the acrylic light guide and open to suggestions based on what exists. I am new to optics and need to know where to start Googling! Thanks!</p>
<p><a href="https://i.stack.imgur.com/NuMOQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NuMOQ.png" alt="Example" /></a></p>
| |optics| | <p>The term to search for is 'light pipe'. You may have luck with this question on Electronic Engineering SE (such a part would be part on an EE design).</p>
<p>eg <a href="https://www.digikey.com.au/products/en/optoelectronics/optics-light-pipes/102" rel="nofollow noreferrer">Search for light pipes on Digi-Key.com</a></p>
<p>While nothing appears to be an exact match to what you're trying to achieve, some of the light pipes shown with wide configuration would come close, although personally, I would use the laser cut acrylic approach suggested in another answer.</p>
| 44999 | Does this type of lens or light guide exist? |
2021-07-07T08:31:49.667 | <p>In <a href="https://www.youtube.com/watch?v=grIDv-edAOY" rel="nofollow noreferrer">this video</a> and in many others I've seen, smiths use 2 different "hammers" to forge metal together. One seems to hit it repetitively with less force (0:30 in the video) while the other "squishes" the metals together with larger force in a slower motion (1:10 in the video).</p>
<p>My question is, what is the difference? why not use one or the other?</p>
| |metal-folding|metalwork| | <p>One is called a press, the other is called a hammer.</p>
<p>At this point I am just regurgitating what I have read.</p>
<p>You can see that the press is able to shape the metal more which is is good in that it is faster but worse in that you have less control.</p>
<p>The hammer takes longer to shape the metal but that means you have more control.</p>
<p>In addition, the press is in contact with the metal for longer and over a larger area which cools down the metal faster so you don't have as long to work something on it which you might need such as when fine tuning something.</p>
<p>Does a table saw, a chop saw, a bandsaw, and a scroll saw do the same thing? Or a a rasp and sandpaper? Maybe, but there are certain tasks that are far more convenient with better results on one piece of equipment than another. Sometimes you even have a big hammer and an identical small hammer in your toolbox.</p>
| 45009 | Why do blacksmiths use these 2 tools which seem to do the same thing |
2021-07-08T07:44:51.870 | <p>I need to make a single piece (therefore automation not necessary). The PVC pipe in question is approximately 490mm long and has a 20mm diameter. I would like to cut four 60° arcs, the location of the cuts along the circumference of the cross section must be within a ±20° tolerance and the cuts must be within a 70° ± 10° tolerance. Cuts cannot extend beyond the corners, if the cut goes to far then the entire hole must become larger too. The width of the cuts (measurement on the long axis of the pipe) can be of any length longer than 3mm. How would I cut out the arcs and what tools would I need? I have provided diagrams below. Thanks in advance.</p>
<p>(I stretched the diameter of the pipe in the diagrams 10x both to make them easier on the eyes and to provide space for annotating)</p>
<p><img src="https://cdn.discordapp.com/attachments/744101506177040406/862637686656204800/Topdown-view.png" alt="Valid XHTML" />
<img src="https://cdn.discordapp.com/attachments/744101506177040406/862826321868488734/unknown.png" alt="Valid XHTML" />
Note: I am unexperienced in this sort of thing but I am sure that there is a better solution than using a handheld grinder and hoping for the best so I am asking.</p>
| |tools|cutting| | <p>Put a hex-head plug in each end so you have a flat.</p>
<p><a href="https://i.stack.imgur.com/O7xrn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7xrn.png" alt="enter image description here" /></a></p>
| 45023 | I would like to cut out arcs of the circumference of a cross section of PVC pipe so the arcs are precisely 60 degrees. How would I do this? |
2021-07-08T08:08:05.267 | <p>I have one <strong>ds18b20 temperature sensor</strong> for <a href="https://github.com/Startup-Data/UN-Hakim-PCR/tree/main/Arduino/Temperature" rel="nofollow noreferrer">this project</a>, but <strong>I don't have the metal cover</strong> for it like shown below:
<a href="https://i.stack.imgur.com/5YTnq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5YTnq.png" alt="enter image description here" /></a></p>
<p>So if possible I asked here to have your intelligence for finding some homemade alternative as metal cover for covering this sensor (Metal Covers are better).</p>
<p><strong>Update:</strong></p>
<p>If the required tools not have high price I can buy or borrow it, but I have <a href="https://www.google.com/search?q=basic%20mechanical%20toools%20at%20ordinary%20home&tbm=isch&ved=2ahUKEwjtgMytjtPxAhW7ybsIHZIpBWYQ2-cCegQIABAA&oq=basic%20mechanical%20toools%20at%20ordinary%20home&gs_lcp=CgNpbWcQA1C4SFjTkQFgg5QBaARwAHgAgAGEAogB-RGSAQUwLjcuNJgBAKABAaoBC2d3cy13aXotaW1nwAEB&sclient=img&ei=2r7mYO2rJLuT7_UPktOUsAY&bih=888&biw=1853#imgrc=nN-7UzpnWyFHyM" rel="nofollow noreferrer">the basic mechanical tools</a> in my home, like shown at below:</p>
<p><a href="https://i.stack.imgur.com/0mnA5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0mnA5.png" alt="enter image description here" /></a></p>
<p>Thanks.</p>
| |sensors| | <p>Because you don't have much fabrication equipment at home you will need to find the metal casing. There are many miscellaneous objects that might work, for example a fired .22 case, or maybe a piece of copper tube with one end soldered shut.</p>
<p>Once you've got that, solder a cable onto your sensor, and then pot it in the metal case using 2 part epoxy. Thermal epoxy would be ideal, but it's expensive and it probably will work fine with regular epoxy.</p>
| 45025 | Homemade metal cover for Arduino temperature sensor |
2021-07-08T20:20:19.873 | <p>Peltier modules are said to be less efficient than conventional air conditioners. Can a medium like moving cold water over the heat sink increase the efficiency of a peltier module? By how much? Another ideas is to bury the heat sink deep in the ground to cool the hot side through conduction instead of convection.</p>
| |electrical-engineering|motors|cooling| | <p>Be careful: "efficiency" usually means the ratio of energy applied to energy removed from the "cold side" object. Since a Peltier junction has a fixed delta-T , you won't get any colder by increasing the heat-sink capability. You will get to the cold limit faster, or even get to a lower temperature if you use cold water instead of room-temp air.<br />
The only way to get a greater delta-T is to stack junctions.</p>
<p>BTW, there's a pretty good discussion of Peltier coolers <a href="https://electronics.stackexchange.com/questions/162393/6w-12v-5a-peltier-only-drawing-1a-i-am-supplying-more-current-but-its-only-draw">here</a> , where the practical limits of drive current are explained.</p>
| 45037 | Can peltier modules be more efficiant with a medium other than air? |
2021-07-09T00:18:52.617 | <p>I'm currently working on a project in which I have to control 2 solenoid valves with arduino to switch the suction of a vacuum pump from one tube to another one (tried to explain it on the photo), I've tried one of those pneumatic 3/2 valves but it didn't work, and after some research I realized that those kind of valves work under pressure, so, what kind of solenoid valve should I use?<a href="https://i.stack.imgur.com/qt36M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qt36M.png" alt="enter image description here" /></a></p>
| |valves|vacuum|vacuum-pumps| | <p>A <a href="https://en.wikipedia.org/wiki/Solenoid_valve#Internally_piloted" rel="nofollow noreferrer">pilot operated valve</a> (often unfortunately generalized as a "solenoid valve") uses a small control solenoid valve and the pressure differential across the valve to open a diaphragm that allows flow to pass. This is a very common design because it allows for a large flow to be controlled with very little energy. This <a href="https://youtu.be/zlu4HSHycOE?t=39" rel="nofollow noreferrer">YouTube video</a> has a good diagram. Pilot operated valves have the limitation of only being able to shut off flow one direction and like you noticed they require a pressure above vacuum to stop flow.</p>
<p><a href="http://www.connexion-developments.com/solenoid-valve-for-vacuum.html" rel="nofollow noreferrer">This site on vacuum valves</a> mentions that they can be used for high flow vacuum applications, but there are some limitations. So depending on what vacuum pressures you are trying to achieve, these valves may be an option. In your testing you may have just had the direction backwards. It may also not have worked because your applications requires blocking flow in both directions, this could be mitigated with a check valve if needed.</p>
<p>What you are looking for is called a "<a href="https://airinc.net/product/310-120vac/" rel="nofollow noreferrer">direct acting solenoid valve</a>". This valve simply uses an electric coil called the <a href="https://en.wikipedia.org/wiki/Solenoid#Applications" rel="nofollow noreferrer">solenoid</a>, a ferrous plunger, and rubber seal around a hole. Most also use a spring to hold it in an off position, and the magnetic force overcomes the spring when powered giving two discrete states. The smaller orifice limits flow more than the pilot operated valves, but it is a simply on/off with less mechanical complexities. They can also operate in full vacuum. Some may still have a preferential flow direction, but most will be rated for full differential holding pressure in either direction. They can be Normally Closed NC, Normally Open NO, or some fancy ones have latching functionality like Mahendra Gunawardena mentioned.</p>
| 45042 | What kind of solenoid valve should I use for vacuum? |
2021-07-09T13:08:43.793 | <p>We take Nacl and dissolve it in water. There is formation of <span class="math-container">$Na^+$</span> and <span class="math-container">$Cl^-$</span> gaseous ions along with 2H<sup>+</sup> and <span class="math-container">$Cl^{2-}$</span> aqueous ions.</p>
<p>Then , Na<sup>+</sup> is attracted towards <span class="math-container">$Cl^{2-}$</span> and 2H<sup>+</sup> is attracted towards <span class="math-container">$Cl^-$</span> ions.</p>
<p>Energy required by water molecules to separate the NaCl molecules is knows as Lattice enthalpy.</p>
<p>Energy required to form aqueous NaCl molecule is called hydration enthalpy.</p>
<p>My questions are:</p>
<ol>
<li><p>Who is the one that converts gaseous ions of NaCl to aqueous ions?</p>
</li>
<li><p>During the dissolution of NaCl, why is there no formation of solid ions first but directly, it states that gaseous ions of NaCl are formed?</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/C73YJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C73YJ.jpg" alt="enter image description here" /></a></p>
<p>Also</p>
<p><a href="https://i.stack.imgur.com/v8A8s.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v8A8s.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|thermodynamics|heat-transfer|chemistry|heat| | <p>We define reactions uniquely depending on the reactants and products. Here are examples related to your question.</p>
<p>Formation: Na(s) + (1/2)Cl<span class="math-container">$_2$</span>(g) <span class="math-container">$\rightarrow$</span> NaCl(s)</p>
<p>Lattice Formation: Na<span class="math-container">$^+$</span>(g) + Cl<span class="math-container">$^-$</span>(g) <span class="math-container">$\rightarrow$</span> NaCl(s)</p>
<p>Solution: NaCl(s) <span class="math-container">$\rightarrow$</span> NaCl(aq)</p>
<p>Hydration: Na<span class="math-container">$^+$</span>(g) <span class="math-container">$\rightarrow$</span> Na<span class="math-container">$^+$</span>(aq)</p>
<p>Atomization: Na(s) <span class="math-container">$\rightarrow$</span> Na(g)</p>
<p>A significant observation here is that the STATE of the reactants and products is also required because it is a unique identifier.</p>
<p>As to your questions ...</p>
<blockquote>
<ol>
<li>Who is the one that converts ...</li>
</ol>
</blockquote>
<p>We can convert NaCl(s) to gaseous ions experimentally through processes that may involve multiple steps, including vaporization and ionization.</p>
<blockquote>
<ol start="2">
<li>During the dissolution of NaCl(s), why is there no formation of solid ions first but directly, it states that gaseous ions of NaCl are formed?</li>
</ol>
</blockquote>
<p>The <em>experimental</em> dissolution of NaCl(s) does not involve formation of gaseous ions. It involves formation and solvation of aqueous ions. The two steps (dissolution and solvation) are often not easily separated.</p>
<p>What you refer to is the case where we <em>imagine</em> the dissolution of NaCl(s) happening through a set of steps that include the formation of gaseous ions. The process to combine reaction steps that go from a given set of reactants to a given set of products is formalized in what is called a Born-Haber cycle. <a href="https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/Lattice_Enthalpies_and_Born_Haber_Cycles" rel="noreferrer">This example</a> demonstrates how some of the above reactions fit into a Born-Haber cycle diagram for NaCl(s).</p>
<p>Just because we imagine a reaction path on the Born-Haber cycle does not mean that the reaction occurs experimentally. Indeed, one power to the Born-Haber cycle is to back-calculate an energy for a reaction that does not occur experimentally but that has significant theoretical utility as a building block to obtain other values.</p>
| 45051 | In case of solvation , Who converts gaseous ions to aqueous ? Why do we even have gaseous but not solid ions? |
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