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2018-06-14T11:03:43.353 | <p>I have a high speed roller bearing rated at 3000rpm.</p>
<p>I know in the electrical industry that tolerances of around 10-15% are built into components so that they continue working in fluctuating electrical environments. </p>
<p>Is there a similar tolerance in the bearing industry ?</p>
| |bearings| | <p>The maximum speed for a bearing recommended by the manufacturer can be due to one of two limiting factors, Thermal, or Mechanical.</p>
<p>The thermal speed rating (often called "reference speed") is calculated using ISO 15312 standardised operating and cooling conditions, and as such, if you are able to provide improved cooling in your application, then, you will be able to exceed this rating. </p>
<p>The mechanical speed limit of a particular bearing should not be exceeded except in very rare cases. If you need a higher speed, then you should look to find a more appropriate bearing specification.</p>
<p>The <a href="http://www.skf.com/uk/products/bearings-units-housings/principles/bearing-selection-process/operating-temperature-and-speed/speed-limitations/index.html" rel="nofollow noreferrer">SKF website</a> says this about operating above the published speeds:</p>
<blockquote>
<p>It is possible to operate a bearing at speeds above its reference speed, its adjusted reference speed, or even the limiting speed. Before doing so, first make a detailed thermal analysis, and take whatever further measures may be required, such as use of special cage executions, or consider using high precision bearings. Regarding management of the effects of increased speed, consider the following options:</p>
<ul>
<li>Control the resulting increase in bearing temperature by additional cooling.</li>
<li>Compensate for any reduction in bearing clearance resulting from increased bearing temperature.</li>
<li>Revise the housing fitting tolerance choice to ensure that the influence of increased bearing temperature does not impair the
axial displaceability of non-locating bearing outer rings.</li>
<li>Revise the bearing tolerance class, together with the geometrical precision of the shaft and housing seats, to ensure
these are sufficient to avoid excessive vibration</li>
<li>Consider using an alternative cage execution that is suitable for higher speed operation, in particular when approaching or
exceeding the limiting speed.</li>
<li>Ensure that the lubricant and lubrication method used are compatible with the higher operating temperature and the cage<br>
execution.</li>
<li>Check that the relubrication interval is still acceptable,
particularly for grease lubricated bearings. Oil lubrication may be
required.</li>
</ul>
</blockquote>
| 22264 | What is the typical rpm variance-tolerance of a roller bearing? |
2018-06-15T21:24:04.333 | <p>I understand that there is such a metric as the Rayleigh number which governs convective cooling in a medium. Assume for example, a fist sized heatsink like that below in free air:-</p>
<p><a href="https://i.stack.imgur.com/tnvsI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tnvsI.png" alt="heatsink"></a></p>
<p>Does the Rayleigh number mean that convection might not start between the fins? So in effect, the finned heatsink might as well be a large solid lump? And insofar as thermal resistance goes, the thermal conductivity would only depend on the gross overall dimensions (height x width x length) and not the actual finned area?</p>
| |thermodynamics| | <p>The Reynolds number is irrelevant in these situations because it is very small and does not deal with buoyancy and natural convection. There is instead a similitude parameter called the Grashof number that deals with balances between buoyant forces and viscuous forces which is relevant here. </p>
| 22281 | What is the practical application of Rayleigh number to heat sinks? |
2018-06-15T21:35:55.147 | <p>Given:</p>
<p>$$
\sigma_{ij}=
\left[ {\begin{array}{cc}
-20 & 60 \\
60 & 90 \\
\end{array} } \right],\quad i,j=x,y
$$</p>
<p>I want to find the principle stress tensor $\sigma_{ij}^{pr}$.
Using the <a href="https://en.wikipedia.org/wiki/Mohr%27s_circle" rel="nofollow noreferrer">Mohr's Cirlce</a>, I get:</p>
<p>$$\sigma_{max}=116.39,\sigma_{min}=-46.39$$</p>
<p>the points where the circle intersects with the $x(\sigma_{xx},\sigma_{yy})$ axis.</p>
<p>From there, how do these points make up the principle stress tensor?</p>
<p>Is </p>
<p>$$
\sigma_{ij}^{pr}=
\left[ {\begin{array}{cc}
116.39 & 0 \\
0 & -46.39 \\
\end{array} } \right],\quad i,j=x,y
$$
correct?</p>
| |stresses| | <p>The values you determined for $\sigma_{\mathit{max}}$ and $\sigma_\mathit{min}$ are correct. However, your answer is only partially correct. </p>
<p>The value of $\sigma^\mathrm{pr}$ has to be expressed in the principal coordinate system. You can calculate the angle of the principal system to your original coordinate system using:
$$
tan(2\varphi) = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2 * 60}{-20-90}
$$
$$
\varphi_1 \approx 23.74°
$$</p>
<p>So yes the values are correct, but your coordinate system $x^\mathrm{pr}, y^\mathrm{pr}$ is rotated by $\varphi_1$ degrees, compared to your original coordinate system $x, y$.</p>
| 22282 | Mohr's circle: maximum normal stress tensor |
2018-06-16T10:58:11.857 | <p>My professor says, the internal hinge can not transfer moment:</p>
<p><a href="https://i.stack.imgur.com/UNZM7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UNZM7.png" alt="enter image description here"></a></p>
<p>But can we calculate $\Sigma M_A=0$ for whole beam ( without cutting it or something) this way?:</p>
<p>$\Sigma M_A=0 ===\Rightarrow M_A+(D_y \times 4m ) + 3KN.M =0$</p>
<p>with this equation we literally transferred the 3KN.M moment to point A while my professor says the internal hinge can not transfer moment.</p>
<p>Something is wrong, and I don't know what it is.</p>
| |structural-engineering|structural-analysis|beam|moments| | <p>The moment 3KN creates two equal and opposite forces at supports B and D.</p>
<p>Each is equal to $ 3KN/(Length- of-BD). $ With Dy pushing up and By pushing down.</p>
<p>D-y is countered by the D upward reaction and cancels.</p>
<p>The correct moment at A is $$ M_a = (\frac {3KN}{length-of-BD} \times AB) $$ which is clockwise.
$$ M_a = force_B \times (length AB) $$</p>
<p>And your teacher is right, no moment is transferred, but shear has been transferred.</p>
| 22288 | Transfering moment through internal hinge |
2018-06-17T03:32:50.713 | <p>What is the difference between a state observer and a Kalman filter? Having implemented various types of Kalman filters, I'm still a bit confused, mainly because state observers require the selection of poles similar to a control system setup, whereas Kalman filters don't. Any clarification on the two and their differences would be great.</p>
| |control-engineering|kalman-filters| | <p>In observer-based state estimation, the focus is on providing a mathematical proof that the error of state estimation goes to zero (exponentially or in a finite time). For that, the proof is usually based on characterizing of dynamic of error, and therefore the formulation is quite close to stability proof of control systems (with pole placement or through Lyaponuv function). However, in most cases, the proof of the stability of estimation error dynamic is developed with "deterministic" assumption where measurement and process noise are not considered.</p>
<p>On the other hand, the Kalman filter approach focuses on the state estimation for stochastic processes (measurement and process noise exist and are addressed). In the Kalman filter, the problem is expressed as an optimization problem where is the existence of process and measurement noise, obtain the best estimation of state and characterize uncertainty for that and we are not looking for converging the estimation error to zero anymore since, due to stochastic nature, it is not feasible. Actually, it can be shown if the assumptions of the Kalman filter hold (Gaussian measurement and process noises and linear dynamic), its performance is better than any other state estimator. For nonlinear systems, variations of Kalman filters such as Extended Kalman Filter (EKF) and Unscented Kalman Filter(UKF) are available.</p>
| 22295 | Kalman filters vs. state observers |
2018-06-17T04:17:05.247 | <p>If you were to use a centrifugal axial fan instead, won't you be able to achieve higher flow rates and better pressures? It seems like the efficiency of these motors are very low.</p>
<p><a href="https://i.stack.imgur.com/wzZvA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wzZvA.jpg" alt="Typical centrifugal fan"></a></p>
| |mechanical-engineering|fluid-mechanics| | <p>Lower pressure differential and lower speed per blade translates to lower noise levels. </p>
<p>Additionally, for this sort of applications, higher speed would mean higher torque (scales with square of airspeed) and that converts to higher power usage; many blades allow to keep a decent flow while keeping the pressure, and resulting torque low. </p>
<p>Both of these factors are important in laptops. And while better flow rate would be desirable, higher pressure is absolutely not needed. </p>
| 22296 | Why do most centrifugal fans found in laptops have tiny blades? |
2018-06-18T10:33:14.667 | <p>I am trying to calculate the running cost of my clothes dryer and have bought myself an energy meter to check its usage.</p>
<p>I plugged the meter in, reset it, and then turned it over to the aggregate kWh mode and did a single drying session with wet clothes from the washing machine.</p>
<p>After the load had finished the meter read 1.828 kWh and the total time was 1h 13m.</p>
<p>So as I understand it, a single load used 1.828 kWh, and as we are doing about 7 loads per week that would be a total of 665 kWh / year. At the current rate of 33c per kWh the annual running cost is therefore about $219.</p>
<p>Now this to me is a bit strange as just about every website I have seen says that dryers should be costing upwards of $400 per annum for this kind of usage. In addition, my dryer is only a 2 out of 5 stars for energy efficiency.</p>
<p>Can anyone see a problem with my math or logic?</p>
| |energy|cost-estimation| | <p>You measured one load to take 1.8 kWh. That implies that 7 loads would take 12.6 kWh. If these 7 loads are repeated 52 times per year, then a whole year of operating takes 655 kWh.</p>
<p>As a sanity check, 1.828 kWh over 1 hour and 13 minutes comes out to 1.5 kW, which is plausible for some type of clothes dryer.</p>
<p>\$0.33 per kWh sounds very high, but going with that, 655 kWh would cost $216.</p>
<p>It seems your calculations are correct.</p>
<p>Web sites that claim some appliance costs some amount per year to operate are a very rough guide at best, complete fabrication at worst. These sites usually have some agenda, so push the number towards whatever makes their argument look better. They also rarely state assumptions. Even then, usage patterns for something like a clothes dryer vary so widely, as do electricity prices, that they can pick from a wide range of values and then claim a case that justifies what they picked.</p>
<p>Keep monitoring the electricity usage of your dryer. It probably varies quite a bit from load to load. There will be inevitable variations in amount of clothes that are being dried, the ambient humidity, the wetness of the clothes, and other factors that are hard to predict. Over a whole year, you'll get a much better idea of the expected power usage of the dryer.</p>
<p>Are you really sure that you pay $0.33 incrementally per kWh? That's probably correct for some part of the world at least some of the time, but sounds rather high. If your area has time of day metering, it might be worth your trouble to arrange to run the dryer over night when the rates are lower.</p>
<p>If you want to save money, try using a plain old clothes line, at least during the warm parts of the year.</p>
| 22310 | Trying to understand how much it costs to run my clothes dryer |
2018-06-18T16:46:07.800 | <p>I am planning a prop for a marching band show and I want to determine how to best address the challenge of the wind blowing it over. This is definitely not my area of study and I am hoping for some guidance here.</p>
<p>The panel is 10' high and 4' wide. My working assumption is that we won't be dealing with winds any stronger than 15 mph. That may be wishful thinking but I needed to throw out a number.</p>
<p>The panel will be mounted on a small platform (5' x 3.5') and angled at roughly 80 degrees. Something like this:</p>
<p><a href="https://i.stack.imgur.com/91zgd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/91zgd.png" alt="Side View of prop"></a></p>
<p>The panel will be mounted top and bottom to the support and base.</p>
<p>Trying to get an idea about how much weight it would take to keep it from flipping over, I found a generic formula for calculating wind load:</p>
<p>F=AxPxCd</p>
<p>Using this formula and plugging in 0.00256xV^2 for P
0.00256 x 15^2 = 0.576</p>
<p>And based on some very cursory reading, I am guessing at the Cd value for a rectangular panel, setting it to 2.0.</p>
<p>I arrive at F = 40ft^2 x 0.576 psf x 2.0
= 46.08 pounds</p>
<p>But I'm not sure what this is telling me or if it is even correct for my situation. I'm primarily concerned with the wind coming from that support side and leading to it lifting or flipping the panel. I'm limited on the footprint but could add weights to the base and/or behind the base attached to the support poles and/or the base.</p>
<p>Thoughts?</p>
| |structural-engineering| | <p><strong>I am not a qualified civil engineer.</strong> <em>This advice is theoretical and not to be used for any actual design in the real world.</em> It is for informational purposes only.</p>
<p>In civil engineering we always design things for worst case scenarios for good reason. Were that panel to fall over and injure a child as happened with <a href="https://www.mirror.co.uk/news/uk-news/hugo-boss-fined-12-million-6382709" rel="nofollow noreferrer">a Hugo Boss store</a> in Bicester (pronounced Bister) Village, England in 2015, there may be a criminal case to answer and also legal issues that cost vast sums of money. No design of any structure is taken on without a full assessment of its possible environmental impact which means how that structure is going to perform in its position and the effects it might have in failure situations.</p>
<p>To formuate any coherent design we must make educated assumptions that take into account safety factors, environmental factors, mitigation of loading extremeties, alternative design issues to overcome any unforseen issues.</p>
<ul>
<li>In this case we need to assume that the maximum wind velocity is not only possible, but is likely and then use something like the Eurocodes safety factor margins to improve the safety of the structure.</li>
<li>The panel must be able to withstand a unifomly distributed load from wind at a far higher load than 15 mph for good reason: Wind has a tendency to NOT be uniform and GUSTS may easily be up to 80% times higher than median velocity. This means that 15 mph is possibly gusted up to 27 mph.</li>
<li>A safety factor must be incorporated to withstand the moment of lateral forces and usually it is an additional 5 to 30% additional force tolerance.</li>
<li>Altitude also plays a major factor in designing for wind load. Although wind is a dynamic load, all building structures are usually designed for a uniform load on account of safety and for simplified material strength calculations. The higher the board above ground level, the greater will be the unrestricted wind verlocity. There is often a vast difference in wind velocity between 4' and 40' depending on surrounding environmental fators.</li>
<li>Depending on the location of the panel, the force of the wind may be substantially greater in certain sitings such as between buildings 2' to 12' above the ground compared to open area of a playing field.</li>
<li>Aerodynamics also must be accounted for since it is flat and to be used specifically as an aerofoil or fro want of a better description, a wing.</li>
<li>There is no indication as to whether this wing is going to be fixed or in motion on a float in a parade.</li>
<li>According to the diagram provided, it is not going to be pinned (fixed) at all. It has two rollers meaning it can be moved mechanically. If this is true, then we need to consider using cable stays for a fixed device an mass-moment reaction balances for motion designs.</li>
<li>There is no indication as to what the material is to be used because we do not know the magnitude of forces that may be exerted upon the wing for now.</li>
<li>Self-weight of the entire system must be accounted for to calculate accurately what is a safe working design.</li>
</ul>
<hr>
<p><strong>Calculations</strong></p>
<p><strong>Imperial (Metric) vertical Panel Dimensions</strong>
$$
10'(3.048m)
\times
4'(1.2192m)
=40'^2(3.7161216m^2)
$$</p>
<p>Apparent inclined panel dimensions:
$$
10'\sin 80^\circ \times 4'
\approx 39.39231012 \text{ ft}^2 \text{ or } 3.659665363 \text{ m}^2
$$</p>
<p><em>Wind Velocity:</em></p>
<p>Client estimated maximum velocity: 15 mph</p>
<p>Feasible velocity: $30 \text{mph}$ with gust velocity estimate: $\approx <60\%$
$$
30 \text{mph}+(30 \times 6)=48 \text{mph}
$$</p>
<p>Safety factor: 25%
$$
48 \text{mph}\times 1.25 = 60 mph
$$</p>
<p>So now we have a design velocity for the wind force of 48 mph for which <a href="https://www.physicsforums.com/threads/what-psi-can-be-achieved-from-a-60mph-wind.698861/" rel="nofollow noreferrer">another website has calculations</a> we can use to determine pressure and therfore a lateral force measure that may be exerted on the panel with a safety margin for error of 60 mph.</p>
<p>The problem is that air density differs according to its thermal state which means we also need to account for the temperature of the air as more dense air has greater inertia and hence a greater pressure against the panel. Wind pressure is proportional to air density multiplied by the square of the velocity:</p>
<p>$$
\rho = \frac{p}{R_{specific} \times K}
$$</p>
<p>$
R_{specific}=287.058 J\cdot(kg·K)^{-1} \text { or } 53.35 ft·lbf(lb·°R)^{-1}
$</p>
<p>When taking account of density, it is unlikely that anyone will sensibly march their band in temperatures below $5^\circ$C ($41^\circ$F) for which the density of air at sea level is going to be <a href="https://en.wikipedia.org/wiki/Density_of_air#Temperature_and_pressure" rel="nofollow noreferrer">$ \rho=1.269 kg.m^3@100$KPa</a>.</p>
<p>We know that our surface area is $3.659665363 m $^2$ and therefore the wind force will calculates as 1671N according to <a href="https://www.engineeringtoolbox.com/wind-load-d_1775.html" rel="nofollow noreferrer">Engineering Toolbox</a></p>
<p>You can see from the result that that board would have to withstand a dead load of 170.4 Kg.f or 375.66 lb.f at the top vertical edge of the panel.</p>
<hr>
<p><strong>Design Brief and Recommendations</strong></p>
<p>We can use a few different ways to prevent the board being blown over as enumerated below:</p>
<ol>
<li>Deadweight welded to frame attached platform base;</li>
<li>Cable type stays like tent rope & peg systems;</li>
<li>Ground spike systems used to hold down groundsheets;</li>
<li>Equilateral triangle systems with internal deadweight.</li>
</ol>
<p>Looking at your design, a single sheet of wood would theoretically be strong enough as long as it was able to withstand a minimum weight of 9.536 lb.ft$^2$</p>
<p>Considering that it is only a wood sheet, you might want to consider using <a href="https://en.wikipedia.org/wiki/Voile" rel="nofollow noreferrer">voile</a> over a framed support as an alternative or canvas with eyes. These options are to save on the weight of the device and work as a windbreak also. This is an important considerations since your wooden support's bending moments are going to require a great mass in a small area to prevent the system from toppling over.</p>
<p>If you consider that the total lateral force is effectively applied as a combined moment of force at the top edge of the panel in a lateral direction, you will require an equal but opposing force to counter that force in two opposing directions. This is a major concern because wind arrives from 540$^\circ$ of direction at ground level. To understand this, you need to consider a circular direction in plan and add 180$^\circ$ in elevation which is effectively an hemisphere of wind action towards the centroid mass of the force system which needs to have a very low centre of gravity.</p>
<p>A centroid is where the centre of mass combined with gravitation and wind inertia is situated. An example is that the entire construction is going to be in motion so we cannot use cable stays that are not also mobile in tandem with the panel and supporting structure.</p>
<p>Ideally, if the platform is going to be transported, use of ratchet straps that pull the panel downwards against the base of the support would be a good idea. Four straps of one inch width that wrap around the panel should hold it in position as long as the base and the apex are securely fixed to each other.
In reference to previous ideas:</p>
<p><strong>1. Deadweight welded frame and base</strong>
While this is probably not a bad idea if transporting heavy loads above human carriage is not an issue, it requires financial expenditure if you are not a welder and do not have an encapsulating frame. On cost efficients grounds this may have to be a forgone conclusion and obsolete.</p>
<p><strong>2. Cable type stays</strong>
Feasible if using a vehicle like a pickup but will require some lateral moment opposition of a wider base.</p>
<p><strong>3. Ground spike system</strong>
This is clearly not going to help at all. Dropped into <em>File 13</em> for trash.</p>
<p><strong>4. Equilateral triangle system with deadweight.</strong>
This is a modification of option 1 but with a design change to your initial proposal in the next section because frankly, my recomendation.</p>
<hr>
<p><strong>Design considerations</strong></p>
<p><em>Moments of force</em></p>
<p>The basic formula for calculating moments around a point is $M_O = Fd$ where F is the total force multiplied by the combined distance from the point of rotation. Since the greatest force is accumulated at the top of the system, we combine the uniformly distributed load into a single point of force at the maximum point of rotation to calculate the rotation forces that need to be opposed.</p>
<p>$
M_O = 1671 \times 3.048
$</p>
<p>$
M_O = 5093.208 N.m \text { (1144 lbf)}
$</p>
<p>The platform base should be cut into 4 lengths of equal width. Your proposed vertical panel is going to become the new platfrom base. Your vertical support will be supported by the 4 platform pieces as diagonals attached to the new base and upright support at angles of $60^\circ$ to the horizontal and $90^\circ$ to each other being pinned to the vertical support edge.</p>
<p>What we need to ensure is that the wind pressure does not blow the system over and the height is the issue because The force at the top edge multiplied by the height of the system will explain what magnitude of downward force we need to use to compensate for the turning moment. Fortunately, this design is quite simple and we can compensate for lifting forces whilst remembering that the design load is for 48 mph winds as a maximum and the extra 14 mph safety is to prevent anyone being hurt by the construction.</p>
| 22313 | Wind load of panel |
2018-06-18T21:33:42.697 | <p>The pads are used in air coolers to cool the air but make it more humid.
Is it possible to use copper / aluminum coils instead of pads to get cold and dry air?</p>
| |cooling|air| | <p>Reverse cycle air conditioners, heat pumps by another name, are closed system coolers whereby heat is transferred form one side of a divide to another, ie from inside a room to outside the building, when cooling.</p>
<p>Evaporative air coolers operate on a totally different principle. Water is evaporated and dispersed into the air to cool the air by absorbing heat. In doing so they increase the humidity of the air inside the room. Because of this, such coolers are not suited to environments with humid air, such as tropical regions and coastal regions. They are best suited to areas with dry air, such as deserts.</p>
<p>If the pads of an evaporative air cooler were to be replaced with a coil, as you ask, there is nothing to evaporate. The only way for such a system to work is to turn the evaporative air cooler into a refrigerative air conditioner (also known as a reverse cycle air condition or a heat pump). This defeats the purpose of having an evaporative air cooler. You may as well just acquire a reverse cycle air conditioner.</p>
| 22317 | is it possible to use copper coil instead of pads in air cooler |
2018-06-19T16:54:30.687 | <p>I know that heat affects material permeation (increasing it); but I have not been able to find much in the way of data other than biologically, on how it effects materials such as plastics. </p>
<p>Does heat increase permeation exponentially, linearly, etc..? </p>
| |materials|heat-transfer|temperature| | <p>Because diffusion processes go like e^kT, the apparent permeabililty will rise exponentially. The picture is a bit more complex than this for plastics, because they expand significantly with temperature, thereby lowering their bulk density and enhancing their permeability.</p>
| 22324 | Heats affect on material permeation? |
2018-06-19T21:48:39.510 | <p>I am working on a project where I must find the maximum principal stresses on an object subjected to impact loading condition. I know the velocity of the object (the one which will hit the test object) and its weight. My questions are:</p>
<ol>
<li>For analysis should I apply force or pressure to the area of impact?</li>
<li>How do I calculate the force/pressure?</li>
</ol>
| |mechanical-engineering|structural-engineering|structural-analysis|finite-element-method|ansys| | <p>Your problem is ill-posed as stated. </p>
<p>However, a zeroth-order estimate of the impact force can be calculated using a one-dimensional balance of energy. The kinetic energy of the projectile is
$$
\text{KE} = \tfrac{1}{2} m v^2
$$
where $m$ is the mass of the projectile and $v$ is its velocity.</p>
<p>The strain energy of the target, assuming linear elasticity, is
$$
\text{U} = \tfrac{1}{2} \sigma \varepsilon V
$$
where $\sigma$ is the stress in the target, $\varepsilon$ is the strain in the target, and $V$ is the volume of the target.</p>
<p>Equation the two, we have
$$
m v^2 = \sigma \varepsilon V = \frac{F}{A} \varepsilon A L = F \varepsilon L
$$
where $F$ is the impact force, $A$ is the impact area and $L$ is the length of the target normal to the impact direction.</p>
<p>The impact force estimate is, therefore,
$$
F = \frac{m v^2}{\varepsilon L} = \frac{m v^2}{d}
$$
where $d$ is the depth of penetration.</p>
<p>If, instead, you would like to use the pressure,
$$
p = \sigma = \frac{m v^2}{\varepsilon A L} = \frac{m v^2}{A d}
$$
Either can be used, but note that the force has to be applied over the contact area.</p>
<p>The problem is ill-posed because, even if you know $A$, you don't know $\varepsilon$ and $L$. People usually assume that $L$ is the thickness of the object (which may be infinite in some cases) and that $\varepsilon$ is small (e.g., 0.001). On the other hand, you may know a value of $d$ and get a reasonable estimate that way.</p>
<p>The only accurate way to compute the impact forces is to do a full-physics dynamic simulation.</p>
| 22329 | Should I apply Force or Pressure, Ansys Static Structural |
2018-06-20T09:50:00.463 | <p>i'm doing a simulation in matlab which uses a 3D matrix as input, i.e. the matrix has the form (xDim, yDim, zDim).
With a simple STL export and a voxelization of this STL file i can get only a matrix with all elements be either one or zero.
But my part has four different materials which i need to distinguish in matlab.
One single slice in the matrix should look like:</p>
<pre><code>0 0 0 0 1 1 1 1 2 2 2 0 ...
0 0 0 1 1 1 1 2 2 2 0 0 ...
0 0 0 1 1 1 1 1 2 2 2 2 ...
.
.
.
</code></pre>
<p>As a first step: in which format should i export my part to preserve the materials as some value?</p>
| |solidworks| | <p>If you are stuck with using STL, here's a crude way to do it:</p>
<p><a href="https://i.stack.imgur.com/rpuS1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rpuS1.png" alt="enter image description here"></a></p>
<p>Steps:</p>
<ul>
<li><p>Create two references which extend beyond the object in the x, y, z space (I used the cylinders shown on the upper right and lower left corners)</p></li>
<li><p>For each material (body), export STL along with the references</p></li>
<li><p>Voxelise using the same parameters</p></li>
</ul>
<p>The references are used to ensure that the binary voxel arrays are generated in a consistent space between the different bodies.</p>
<p>You can multiply each separate voxel array using a different number (2,3,4 ...) and then add them together. If you want to get rid of the references, you can use Matlab's <code>bwlabel</code> <a href="https://au.mathworks.com/help/images/ref/bwlabel.html" rel="nofollow noreferrer">function</a> to detect unconnected regions and remove them.</p>
| 22339 | What is a method to export a SolidWorks part (with distinguishing between materials) to a voxel cloud? |
2018-06-20T13:47:18.067 | <p>As <strong>roadway</strong> <em>consists of a width of road on which a vehicle is not restricted by any physical barriers or separation to move laterally</em> <sup><a href="https://en.wikipedia.org/wiki/Carriageway" rel="nofollow noreferrer">source</a></sup>, what is a similar term which encompasses the width of the entire road's infrastructure - barriers, drainage ditches, sound screens, generally the entirety of terrain allocated for construction as a road is built?</p>
| |civil-engineering|terminology|highway-engineering| | <p>In the US this is commonly referred to as the Right-of-Way (property line to property line on either side of the roadway). Or, if the roadway right-of-way is wider than needed for construction, it may be referred to as the limits of construction.</p>
| 22345 | What is the entirety of breadth of road called? |
2018-06-21T05:56:53.667 | <p>How do "engineering leaders" work with large-scale "trends"?</p>
<p>Like matching against predicted requirements in renewable energy etc.</p>
<p>These seem like abstractly unknowable future states. Yet, "engineering leaders" seem to be able to ride the trends and collect the requirements needed to work on these trends.</p>
<p>Also, nowadays the projects are quite large-scale already, because they may touch even millions of people, or take tens of years to fully implement. It seems like black magic, because humans and human organizations have a lot of variables. One cannot e.g. control the supply of labour market.</p>
<p>So how do they progress in such?</p>
| |engineering-managment| | <p>The strategic planning of long term projects isn't solely an engineering problem but also involves business, financial and political planning. </p>
<p>To take the example of renewable energy and to generalise somewhat. </p>
<p>First the technology needs to be developed and proven. This is usually an iterative process, initially on a relatively small scale. You first have prototypes then pilot schemes then full commercial roll-out Infrastructure technology often involves a degree of collaboration between business, technology and government. </p>
<p>For example government may stimulate demand by offering grants, tax breaks etc or by legislating to reduce the attractiveness of alternatives. For example if renewable energy is taxed at a lower rate than coal power, then that makes renewable a more attractive investment proposition. </p>
<p>Any new technology is a business risk and investors will calculate whether the potential return on investment justifies that risk. This calculation will involve both engineering feasibility studies and economic modelling to try to predict future economic conditions. </p>
<p>Equally while you can't completely control future markets you can trade in them, energy wholesale contracts may be agreed decades in advance and often energy infrastructure projects will include complex contracts which set pricing structures. </p>
| 22363 | How do "engineering leaders" work with large-scale "trends"? |
2018-06-22T00:56:20.247 | <p>Convention says to measure at the bottom of the meniscus, but I think the cohesion/adhesion balance of mercury results in it "bulging out"</p>
<p>Do you measure at the top of the bulge (near the center) or the bottom (near the edges)?</p>
| |measurements|liquid| | <p>You need to measure from the top of mercury column, in tubes that are wide enough to allow a flat area of mercury on top and ignore the lower level along the perimeter where mercury is repelled by glass. </p>
<p>This page has enough detailed instruction on calibration of barometers. <a href="http://www.instruction.greenriver.edu/kmarr/Chem%20161/Instrument-%20Glassware%20Videos/How%20to%20Zero%20and%20Read%20a%20Barometer/How%20to%20zero%20and%20read%20a%20barometer.htm" rel="nofollow noreferrer">Green river U link.</a></p>
| 22387 | How do I measure mercury in a graduated cylinder? |
2018-06-22T18:32:40.187 | <p>I have a half cylinder shape, and another half cylinder sketch perpendicular to that first cylinder. I wish to create a "half-dome" between these two surfaces (picture what should look like almost a pill sliced in half, but hollow). Here is my current design:</p>
<p><a href="https://i.stack.imgur.com/mqUb2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mqUb2.png" alt="enter image description here"></a></p>
<p>I want to create the dome between the sketch pictured and the top face of the half-cylinder. I've tried using dome and loft, but am having trouble doing so. I tried using revolve, but I cannot seem to pick a good axis so that it does what I want.</p>
<p>I am new to using Solidworks, but I've used Inventor in a class before and a lot of the features seem fairly similar. Any help would be appreciated. Thanks!</p>
<p>Edit:</p>
<p>I tried adding a center axis to the cylinder and then revolving 180$^\circ$:</p>
<p><a href="https://i.stack.imgur.com/3Bidd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Bidd.png" alt="enter image description here"></a></p>
<p>Clearly something isn't right; the error says the axis cannot intersect the drawing, but the shape of the revolve seems to be in the right spot. I created an axis on the back of the half-cylinder so it didn't intersect, and the revolve command called it a not acceptable axis.</p>
| |cad|solidworks| | <p>Whenever you are sketching a profile to be revolved in any CAD package, you should start by drawing your axis of revolution into that sketch as a construction line. Then, keep <em>every</em> other line to one side of that axis. In SolidWorks, this gives you the added benefit of being able to dimension diameters, rather than radii, which may be more appropriate. I've shown a 'best practice' sketch below, showing how you can generate the object you're looking for, with just four dimensions, and one revolve feature in the tree. Simply revolve this profile by 180°
<a href="https://i.stack.imgur.com/X4L55.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X4L55.png" alt="Best Practice Revolve Sketch"></a></p>
<p>That said, there are of course several other ways to generate the 'half dome', which you are after. I'll go through one of these below. Please note that this is not recommended in this situation but I've included it to give an idea of how to use the tools involved more generally, and to show how many options are present in all situations.</p>
<p>The original question included a question on how to define the axis. If you wanted to revolve your dome in the other direction, then you need to create a 'master' sketch to not only define the axis, but also the plane on which the sketch should be created.</p>
<p>Start by extruding the half-cylinder:</p>
<p><a href="https://i.stack.imgur.com/5Zjy0m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Zjy0m.png" alt="Half Cylinder Extrude"></a></p>
<p>Then on the Right Plane, create a master sketch, which defines the shape of the dome that you would like. This should only contain one driving dimension, You should also add driven dimensions that show the radius of the dome, and the angle of the revolve to come</p>
<p><a href="https://i.stack.imgur.com/OK9m2m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OK9m2m.png" alt="Master Sketch Demo"></a></p>
<p>Now define a new plane for your sketch. It needs to intersect the centerpoint of the arc in your master sketch. I used the line from that sketch (coincident) and the Right Plane (perpendicular) as my references</p>
<p><a href="https://i.stack.imgur.com/lBY8fm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lBY8fm.png" alt="Plane Definition"></a></p>
<p>Now, start your sketch for the dome on the new plane. All the dimensions in this sketch should be equation driven, since you've already defined all the geometry elsewhere. Again, <strong>remember to put a centreline into your sketch and keep everything else on one side of it</strong>. The diameter of the circle should be twice the radius you measured in the master sketch, and the thickness should be equal to that set in your original extrude sketch. This means that if you ever wish to modify the height of the dome, or the diameter of the tube etc., this sketch will automatically rebuild to the correct dimensions.</p>
<p><a href="https://i.stack.imgur.com/VGjUFm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VGjUFm.png" alt="Equation Driven Sketch"></a></p>
<p>Now you are ready to revolve your dome. Use the angle dimension from your master sketch to define the revolve angle</p>
<p><a href="https://i.stack.imgur.com/scSOKm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/scSOKm.png" alt="Equation controlled revolve"></a></p>
<p>You will now be left with a shape which is correct on the inside, but has some undesired overhangs on the outside. </p>
<p><a href="https://i.stack.imgur.com/4sEAsm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4sEAsm.png" alt="Dome Overhangs"></a></p>
<p>In this case, you could simply run a cut-extrude around the outside, but another way to achieve this (in the spirit of explaining more complex tools for future use) is the 'delete face' tool. Simply delete the face underneath the overhang, with "delete and patch" selected, and SolidWorks will extend the Cylinder face up to meet the dome. You will have to do this once for each overhang. Deleting both at once will not work due to the way both overhang faces meet at a single vertex causing the maths to fall over.</p>
<p><a href="https://i.stack.imgur.com/aTgOZm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aTgOZm.png" alt="Finished Item"></a></p>
<p>The part has now been completed, and is dimensionally identical to what was achieved with a single revolve previously.</p>
<p>The fundamental concept of master modelling (defining geometry in a sketch that isn't actually used to build the model, and referring to that using equations in order to allow automatic updating of the built geometry) is extremely powerful, however, and in other more complicated examples, it is absolutely the best way to work.</p>
| 22403 | Creating a "half dome" in solidworks |
2018-06-25T02:01:06.970 | <p>Anyone who's spent much time in a machine shop will be familiar with the hardware commonly used for work-holding and fixtures on a machine tool. In particular, the <em>flange nut</em> is a common item used with strap clamps and other work-holding devices. TeCo is a well-known manufacturer of these items.</p>
<p><a href="https://i.stack.imgur.com/wgfON.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wgfON.jpg" alt="enter image description here"></a></p>
<p>I just happened to be shopping for some of these flange nuts for a fixture and noticed that <em>some</em> of these are specified as being made from 12L14 (leaded steel). This detail is not buried in the data sheet, but appears where I would interpret it as a <em>feature</em>, meaning a purchaser might have reasons for choosing one in this material, as opposed to one in a stronger steel for example.</p>
<p>My question is why 12L14 might be a <em>preferred</em> material in this application. I've understood leaded steels to be free-machining, which would make manufacturing easier (although making the part non-weldable), but I can't imagine that would be a driving factor in high-volume parts like this and in any case wouldn't rise to the level of a feature to be advertised if that were all it was.</p>
<p>I have two theories:</p>
<ol>
<li>The lead in 12L14 somehow increases the lubricity of the fastener, making it less likely to bind (or perhaps gall) in a frequently used fastener like one in a manufacturing fixture.</li>
<li>Lower tensile strength (although not <em>that</em> much lower) is a plus, because if over-tensioned, it's better for the fastener to give than the t-slots machined into an expensive machine table.</li>
</ol>
<p>Can anyone enlighten me?</p>
| |mechanical-engineering|materials| | <p>It is likely made from 12L14 for machine-ability of the threaded/tapped hole in the nut. Manufacturers can make them faster and probably have better tool life. Reduced cost. Simply that.</p>
<p>Those nuts should be case hardened. That protects from wear from friction and compression when spinning the nut down when clamping.</p>
<p>I’m going to disagree with your theories. There’s no galling or anti-friction happening from the leaded steel. Lead primarily works as a lubricant during cutting. This works because of lead’s lower melting temperature and heat generated at the cutting contact point. Liquid Lead acts as a lubricant to the cut. </p>
<p>Your machine table is likely some heat treated cast iron. It will likely take a few dings and dents from impacts, but your clamps will almost always yield before your table does. The t slots have much more stiffness and material than your t-slot clamping System. If your clamps damage the t-slot, something else is wrong. </p>
| 22440 | Why would 12L14 material be a "feature" of a work-holding flange nut? |
2018-06-25T09:10:31.213 | <p>The history of computers seems somewhat pervasive since there are so many steps in modern manufacturing and so many changes in trends with transistors over many years. Grasping how we got to our current level of processing power seems somewhat like a chicken/egg situation. How exactly were we able to create such precise processors when we only could have had more primitive tools to craft them with? For instance, take the UV stencil used on a silicon wafer to make a CPU.</p>
<p>How could engineers have designed a machine capable of etching out so many millions of transistors in the stencil given that they didn't originally have access to that level of processing that they are trying to achieve? Or in other words, how is it possible for computers to evolve in the process of making more precise machines from less precise machines? </p>
| |manufacturing-engineering|computer-engineering| | <p>Incrementally.</p>
<p>The first computers were made in discrete parts. At first vacuum tubes, then transistors. Then the first integrated circuits that had only a couple dozen transistors and were based on manually projected stencils. The computers would help a bit, but most of the work was done with pen on paper. What is now done by a single tiny chip was occupying a large board with a hundred integrated circuits - each designed by hand. Never mind precision of the process wouldn't allow more elements in an IC.</p>
<p>At certain point, computers were advanced enough to implement CAD software. This allowed moving the design from paper to circuitry, applying common cut&paste to stuff several chips into one, then plot a large, complex mask on paper to be shrunk by photo-techniques and made into more complex chips. As these entered use, increasingly less of the process was done "by hand" and increasingly more was automated. Specialized software that allowed use of a high-level programming language to "write" the IC architecture, that compiled to paths and transistors without human interaction, robotic assembly lines that could prepare increasingly more complex masks etc.</p>
<p>Think of it that way: you think the mask can't be made using anything less than it produces... are you sure? Can't it be made using something 5% less than it produces? And that one can't be made with something another 5% less complex?</p>
<p>ps. you refer to 'cutting'. That's not how these stencils were made. They were about universally made using photo-etching.</p>
<ul>
<li>Make a (big) printout of a layer of the IC.</li>
<li>Take a photo. Now your printout is the size of a microfiche.</li>
<li>Spray silicone wafer with conductor, copper vapor or such.</li>
<li>Spray that with photo-hardening emulsion.</li>
<li>Shine UV light through the microfiche stencil with optics focusing it to the size of the IC. It hardens the emulsion where paths are to go.</li>
<li>Apply acid bath, etching the conductor and emulsion away where they aren't needed. </li>
<li>Wash the emulsion off. </li>
</ul>
<p>...and so on. Different substances, different emulsions, different masks, layer by layer the chip was built in an entirely analog process. The only digital part was design and plotting of the first stage on huge sheets of paper.</p>
| 22444 | How did scientists develop the stencils that made even more precise processors than they had previously? |
2018-06-25T14:49:00.813 | <p><img src="https://i.stack.imgur.com/GHBz0.jpg" alt="enter image description here"></p>
<p>So I'm a hobbiest who loves doing and learning about different subjects. Often I run into difficult problems that I normally can find solutions to however this one stumped me, I have reviewed many physics journals on the problem but either they are two complex or tackle the much easier version of 'viscous' friction.(p 2/54)
Essentially when I even begin a solution I yield to this:</p>
<p>Ma = sign(v) * mg*mu - kx</p>
<p>This would be simple if the friction didn't change direction. Does anyone have a suggestion on how to approach this problem? I'm not looking for someone to solve it I just want an idea on how to start it. Thank you</p>
| |dynamics| | <h2>Method 1</h2>
<blockquote>
<p>This would be simple if the friction didn't change direction</p>
</blockquote>
<p>Indeed. So here is one way to deal with it. First, determine the initial velocity at t=0. E.g. let's just say v(0) is positive. Then instead of</p>
<p>$ma =sign(v) * mg*\mu - kx$</p>
<p>Just write</p>
<p>$ma =mg*\mu - kx$</p>
<p>Now, solve that equation, which should be pretty easy, and plot velocity versus time. You'll note that at some point, the velocity is no longer positive, and has changed to negative. Let's just call that time $t_1$. i.e. $v(t_1)=0$. So at that point, "stop" that solution, and write a new equation</p>
<p>$ma =-mg*\mu - kx$</p>
<p>this equation "starts" at time $t=t_1$ with initial conditions $v(t_1)=0$ and $x(t_1)$ determined from the ending position of the first solution. This equation should also be easy to solve. Continue the process, switching back and forth as necessary. At each point, you will also need to check if the spring force is enough to overcome the static friction. At some point it will not be, and then you stop. At the end, stitch together all of the partial solutions to each segment into one big piecewise solution. </p>
<h2>Method 2</h2>
<p>Use numerical integration. E.g. apply Euler's method for solving differential equations. Over each time step, you can assume that the friction force acts in a constant direction. Assuming the time steps are small enough, you'll get about the same answer as above. </p>
| 22448 | Damped harmonic motion with dry friction |
2018-06-26T19:17:15.697 | <p>I'm hoping this is the correct StackExchange for this question/request for advice.</p>
<p>I'm looking to buy a welder to introduce me a little to welding, my main use-case right now is the hobby projects I do with chainmail and soldering is just not good enough in my book. I'm working with stainless steel jump rings with a diameter ranging from 0.8mm to around 2mm.</p>
<p>I've seen on <a href="http://theringlord.com/cart/shopdisplayproducts.asp?page=1&id=164" rel="nofollow noreferrer">TheRingLord</a> and <a href="https://www.ganoksin.com/article/pulse-arc-welder-closing-chains/" rel="nofollow noreferrer">Ganoksin</a> that this can be done with an arc welder. As far as I've seen arc welders have the lowest prices and to me also seem like the least maintenance heavy since you don't need to stock up on gasses. So arc welders have my focus for now. My budged is around €200,- tops preferably from a dutch reseller, and I have no clue what to look for in a welder (my only clue so far is probably something that can be activated for short amounts). Would anyone be able to give me some advice on what would be a good welder and if I'm even looking in the right direction?</p>
<p>please do note that I'm not just looking for something fast and easy, I really don't mind getting my hands dirty learning how it works and getting some terrible welds in the process.</p>
<p>Some sites that are available for me (you might not be able to read the descriptions, but the welder device-names should still be the same):</p>
<ul>
<li><a href="https://www.beslist.nl/products/klussen/klussen_486173_3621539/?sortby=price_min" rel="nofollow noreferrer">CoolBlue</a></li>
<li><a href="https://www.allesvoorlassen.nl/lasapparatuur/elektrode-lassen" rel="nofollow noreferrer">AllesVoorLassen.nl</a></li>
<li><a href="https://www.beslist.nl/products/klussen/klussen_486173_3621539/?sortby=price_min" rel="nofollow noreferrer">Beslist.nl</a></li>
</ul>
| |materials|manufacturing-engineering|metallurgy|welding|strength| | <p>So from a fellow chain mail enthusiast (who uses split rings due to this), the spot welder is still the premier choice for this, and the ones on ring lord really are ideal for most carbon steels and even stainless steels.</p>
<p>However, if you want to get into aluminum, silver, or beyond, you'll want to invest into TIG welding. TIG welding setups can be found on say McMaster Carr for around $300 - $400, not sure on the conversion rates or if the british government allows you to have one of those without a license (you will also need a grinding wheel to grind down the tungsten electrode, plus a supply of inert gas). You'll find for ring welds, this will simply consist of heating up the ring at the gap, then tapping with weld rod like soldering. Unlike soldering, the inert gas keeps it sealed from oxidation, and the temperatures are hot enough to melt the base materials, so as a result you get a TIG weld.</p>
<p>When you get more advanced, you can use this method to weld large steel plates - simply chaining dots like this in a line until you get a full weld. It becomes quite satisfying to weld quite precious metals like aluminum and titanium.</p>
| 22469 | Arc welder? Cheap welding device for chainmail/chainmail jewelry |
2018-06-28T02:42:48.293 | <p>I am planning to build a large scale 3D printer that will utilise 2 1000mm linear motion shafts that each have a young modulus of 210 GPA. The ends of the shafts will be secured in a block of plastic and will not be able to move and the distance from one usable end to the other will total 900mm. Due to the large size of the parts being used I have concerns over possible deflection that could occur. Therefore I need to calculate the minimum required diameter necessary for 2 shafts to support a 2kg weight in the centre while deflecting less than 0.05mm (my layer height is going to be 0.1mm). Thanks for any help I am not an engineer so I have not the slightest clue how to do the calculations to determine this.
<a href="https://i.stack.imgur.com/GPLvh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GPLvh.jpg" alt="Diagram"></a></p>
| |deflection| | <p>I think that @Donald Gibson answered well. There is no such thing as "will not be able to move" in engineering because everything moves; we just cannot humanly observe those movements in most cases without specialist equipment. Plastic deforms substantially but its elastic modulus returns it to near-original shape whilst marginally deforming permanently under each repetitive stress action.</p>
<p>Cylindrical bars may seem a good idea but your problem is the variable average 19.6133N static force. When that carriage stops dead centre, the inertia will cause a moment of force to develop in excess of the gravitational force of a static object. To counter that, an I beam support would be better and must be in permanent tension to counter deflection in dynamic operation or the supports must be designed as curved rails with some sort of kinetic dampers to absorb inertial acceleration & deceleration forces.</p>
<p>It is not unusual for printer heads to travel up to 30 m.s. if that is combined with gravity, then maybe you need to consider a vertical arrangement instead of horizontal to negate deflection issues better. Use of a third rail would also help somewhat if it acted as a moment energy deflection rail to transfer inertial energy laterally outside the scope of the two cylindrical support rails. Vibration is another issue that may cause harmonic distortions which is why a tubular section under compression may actually be better than a solid cylinder shape. Elliptical bars may actually be the best answer in the application of printer design since they are likely to resist deformation far more readily than cylindrical shapes while being able to withstand greater deflection over the same distance. </p>
<p>This kind of precision engineering is not going to be cheap which is why large printers are so expensive. </p>
| 22493 | What is the minimum diameter necessary for 2 linear motion shafts to deflect less than 0.05mm when a 2kg weight is in the centre? |
2018-06-28T05:30:45.403 | <p>I have a drive way which has knick at point ot smoothen is out i leveled it by putting loose soil <a href="https://i.stack.imgur.com/ZQPNX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZQPNX.png" alt="enter image description here"></a></p>
<p>But its monsoon season and rain washed away all my soil as water comes running down slope.
In pic red is soil i am putting to level road, IS there way to stabilize this loose soil , so that water does not wash away soil </p>
| |civil-engineering| | <p>In Zimbabwe we experience flash flooding due to tropical thunderstorms. To prevent soil erosion, a major problem to farmers on sloping ground, channels are dug on roads diagonally across the surface to channel water away to drainage ditches that open up into fields of crops.</p>
<p>This method was proven most successful by Rhodesian farmers during the flood decade of the 1970's. The specific angle depends on the vertical gradient. </p>
<p>The channel begins as a narrow 10cm deep v in the road centre and expands into a much deeper 1m wide V about 30cm deep to the roadside. The trailing edge angle of the cut is usually 30° to the horizontal. This method of alternating cuts in the surface slows down the surface flew of water and diverts it to specific drainage channels calculated to carry maximum capacity storm water+30%.</p>
<p>Hope this method helps you solve your problem.</p>
<p>Am alternative drainage method is to create large diagonal speed humps that are backfilled with 12-15mm gravel stone across the drive way. The water flows downhill and then slows while being diverted away from the drive slope. This method is employed in many African game reserves on steep gradient dirt roads and is longer term more effective.</p>
| 22494 | Save loose soil from erosion on a slope |
2018-06-28T10:48:24.123 | <p>I need a check valve ( one way valve ) for my project which has less actuation force (IDK real term. shouldn't require high pressure for inflow of fluid) and I searched in market for the valve but couldn't find one. Also I am not in place where I can easily order online.</p>
<p>I saw an hydraulic (can hold 2 tons ~ 7k psi. I need about 500 psi only though.) with ball valve that lifts when inflow and gets down to its seat due to gravity. It is really simple with no spring and also has no rubber or anything like that to prevent seepage of oil but it works. So I was thinking if I can make one in lathe?</p>
<p>There are two problems for that:</p>
<ul>
<li>how to make a spherical hole.</li>
<li>despite making the structure how can I be sure that it won't leak.
Like theoretically it is possible but mostly somehow the products
don't turn out okay as planned.</li>
</ul>
<p>I am also thinking instead of ball, a tapered one will be easier to manufacture but not sure if it will work.</p>
<p>Any suggestions?</p>
| |mechanical-engineering|fluid-mechanics|hydraulics|valves| | <p><a href="https://i.stack.imgur.com/cpYaa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cpYaa.png" alt="enter image description here"></a></p>
<p><em>Figure 1. A sprung check-valve using a ball and a taper. Source: <a href="http://crosscontrols.com/product/walworth-api-602-ball-check-valve/" rel="nofollow noreferrer">Cross Controls</a></em></p>
<p>I'm no machinist but it seems to me that a purchased ball (a bearing, for example) should make a good fit into a well machined tapered hole. If operating with gravity assist the spring may be omitted.</p>
| 22499 | Can I make check valve in simple workshop? |
2018-06-29T01:09:07.460 | <p>I’m a electrical engineer but an amateur in mechanics. I’ve been checking some tools and machines in the market for small works which would allow me to manufacture small but precise pieces, such as a 5mm H6 tolerance shaft to fit a bearing bore under interference. The thing is this shaft should have a tolerance of +-4um, maximum. Ok, so usual lathe setups allow you to position the cutting tool in a scale of 50um marks, with god knows which accuracy. I’d guess no less than +- 50um. But even if it was perfectly match, 50um is rather large near H6 tolerances for my shaft.
Ok, let’s assume I could iteratively remove material using a lathe and use a top precision micrometer to check if my shaft is within required dimension. Still it seems to be a very non-professional way to do it, basically a try-and-error approach, moreover I could accidentally remove more material than I should and end up losing all the work and also money.
Could anyone tell me if there’s a more efficient method to produce such precise pieces?</p>
| |tolerance|metrology|lathe| | <p>If you truly need that type of tolerance and repeatability, then you are way out of the hobby equipment territory. You’ll need high precision calibrated measuring equipment, temperature stability, and specialty manufacturing equipment. </p>
<p>Likely an O.D. cylindrical grinding machine will get you there. If the bars are long, they need to be supported while being processed. </p>
<p>If you want to get close at the hobby scale, patience, scrap material and some decent micrometers will get you close. You can turn(machine) your diameter oversize and then you could use some abrasives(Emory cloth and scotch bright to polish the diameter to size. You could also make a homemade toolpost grinder. Essentially mount a grinding wheel and motor to your carriage. Trial and error. Cut and measure. Adjust, re-cut. Possible, but seems like an academic effort. </p>
<p>To start, I suggest you purchase pre-made shafts to that tolerance and focus on the lower precision items you can actually produce. </p>
| 22510 | H6 tolerance shaft manufacturing |
2018-06-29T10:44:22.740 | <p>Solar furnaces using mirrors to concentrate sunlight on a small area do not produce much power - perhaps 1kW per square metre of mirror - but can reach very high temperatures. From the earliest metalworking through to the beginnings of the industrial revolution, metal refining and working was limited by the difficulty of supplying enough heat to the furnace. For a small scale operation - the sort of activity carried out before the industrial revolution - could a solar furnace have been used instead of one using a fuel (usually charcoal) and air mixture?</p>
<p>I'm imagining a sunny hillside with perhaps 10 to 100 polished metal mirrors, perhaps a square metre apiece, aligned by humans running about, all pointing at a small crucible or forge inside as much insulation as could be wrapped around it.</p>
<p>Would it be possible to use that for refining or forging?</p>
<p>I want to ask, over on history stack exchange, whether such an arrangement was ever used, but I thought I'd make a fool of myself asking a question over here before I make a fool of myself asking one over there. Thanks!</p>
<p><strong>Edit</strong>: I did find these:</p>
<ul>
<li><a href="https://books.google.co.uk/books?id=-bQ9DwAAQBAJ&pg=PT106&lpg=PT106&dq=small%20solar%20forge&source=bl&ots=z5BeB-pO3K&sig=rVtQ_MdLaQ-O5GcIEdohsG7g0sY&hl=en&sa=X&ved=0ahUKEwiOp5zb-vjbAhXQJVAKHYZ3AUUQ6AEIrAEwFg#v=onepage&q=small%20solar%20forge&f=false" rel="nofollow noreferrer">Fresnel solar forge</a></li>
<li><a href="http://woodwaredesigns.com/energy/solarforge.html" rel="nofollow noreferrer">Woodward solar forge</a></li>
</ul>
<p>I'm guessing it would have been hard to make a large Fresnel lens until relatively recently. Mirrors are much easier, they have lovely shiny steel armour in many museums, and child labour to polish the metal and align the mirrors would have been cheap in the past. The Woodward forge looks good but the blog says that it hasn't been used for metalwork. Has anyone made this work on a small scale with mirrors?</p>
| |heat-transfer|metallurgy|solar-energy|forging| | <p>It sure is possible but be prepared to face a lot of engineering challenges especially if you are planning to install all the equipment, mirrors etc by yourself. A lot of metalworking tasks used in the 16th & 17th centuries are no longer in use by firms but you can find hobbyist videos on YouTube doing that kind of thing. Common household mirrors have less than the required reflectivity index and all concentrated solar projects make use of heliostats that are designed to withstand harsh environments.</p>
| 22520 | Is it practical to make a solar furnace for small scale metal production? |
2018-06-29T12:19:14.420 | <p>Roughness measurement instruments output those two values, where Rt is the distance between the top of the highest peak and the bottom of the lowest valley, and Ra is the average distance of every point to the mean line.</p>
<p>In almost all technical specifications I see the Ra value being reported, so when is the Rt value used/why is it measured?</p>
| |mechanical-engineering|materials|measurements|surface-preparation| | <p>Theoretically talking, the peak to peak value let us to estimate the worst case scenarios. Here is an example, A shaft fixed inside a conical coupling.<a href="https://i.stack.imgur.com/dtA6b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dtA6b.png" alt="enter image description here" /></a></p>
<p>How we make sure if the coupling can transmit the required torque? We need to calculate it using this pk-pk value. I'm not going further into details.</p>
<p>In production series of a specific component, people are less interested about how much really this peak to peak value is, but rather they are interested in the average. This average is not sensitive to strong deviations, and is an indicator of your machining tools health and lifetime. Strong deviations in average value means worn cutters ... .</p>
<p>But that's not all, sometimes the average value plays very important role in theoretical calculations as well, for instance, designing the thrust bearing requires always the average value, and the peak-to-peak value here doesn't make sense.</p>
| 22524 | In what cases are the roughness values Rt and Ra used? |
2018-06-30T14:29:51.960 | <p>During the era of the typewriter there were a few kinds of basic designs. The (afaik) most well known is the type bar design, where each key is connected to an individual type bar that hits the paper when the key is pressed, each type bar carries the uppercase and lowercase shapes of a single letter. The other variant is the type wheel or type ball design, where a single ball or wheel shaped element contains all the letters. The mechanism rotates and/or tilts the wheel/ball before it strikes the paper so that the right letter is imprinted. </p>
<p>Type wheel/balls have some advantages over type bars: They cannot get stuck, and the entire wheel/ball can be replaced to change the font. The <a href="https://en.wikipedia.org/wiki/Blickensderfer_typewriter" rel="nofollow noreferrer">Blickensderfer</a>, one of the early wheel operated typewriters, is often described as being much simpler than other contemporary (often type bar operated) machines, containing only a quarter of the moving parts of other contemporary typtewriters, so the type wheel mechanism appears to not necessarily be more complex than type bars. </p>
<p>During the last decades of the typewriter era, the electric type ball operated <a href="https://en.wikipedia.org/wiki/IBM_Selectric_typewriter" rel="nofollow noreferrer">Selectric</a> machines where very popular as higher end typewriters, but I have the impression that type bars were more common in manually operated lower end machines. </p>
<p>Were there technical reasons for type bars being more popular in simpler typewriters? If so what were they?</p>
<p>(This question is based on a few assumptions, which may be wrong. Please tell me if that is the case. And it may be that nontechnical reasons were more important in why type bar designs were more popular. I'm also interested in those, but in that case this is not the right site.)</p>
| |engineering-history| | <p>The reason turns out to be that type bar typewriters are faster. A type wheel is much heavier than a single type bar and needs to both rotate and move forward, so operating it with a finger requires more effort. For an electric typewriter that is not a problem as the motor provides the additional force. See <a href="https://history.stackexchange.com/a/46955/32441">this answer</a> for more details and references.</p>
| 22532 | What were the technical reasons for type bar designs being more common than type wheels for manual typewriters? |
2018-07-02T09:47:31.137 | <p>I am trying to make this
<a href="https://i.stack.imgur.com/hQRWp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hQRWp.png" alt="Goal"></a></p>
<p>I was able to make the base feature but I am facing problems with the secondary features. So could anyone shed some light on how to make the datum planes on cones. </p>
<p>I made one feature using this <a href="https://i.stack.imgur.com/0KnhH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0KnhH.png" alt="Plane"></a>
as was instructed from where I am learning.
However, I am unable to make the plane for the other feature.</p>
<p>So far I have made this <a href="https://i.stack.imgur.com/TZrGq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TZrGq.png" alt="Model"></a></p>
<p>Thanks in advance</p>
| |mechanical-engineering|design|modeling|cad| | <p>You dont actually need a datum plane, you can do it with a revolve. But yes you can do this much more easily than you think. </p>
<ol>
<li><p>When creating a plane click on he cone surface quilt, then change the option from <em>Through</em> to <em>Tangent</em>. Yeah the GUI is a bit sleazy the option is a drop down even if it has no affordance. Bad UX designer!</p>
<p><a href="https://i.stack.imgur.com/2mjxC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2mjxC.jpg" alt="enter image description here"></a></p></li>
<li><p>Then add another constraint and choose a plane or line that defines the rotation around the central axis.</p>
<p><a href="https://i.stack.imgur.com/TnKvN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TnKvN.jpg" alt="enter image description here"></a></p></li>
<li><p>And finally</p>
<p><a href="https://i.stack.imgur.com/Op9w2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Op9w2.jpg" alt="enter image description here"></a></p></li>
</ol>
<p>Also note extrusions can define different extrusion values to both directions so no need to create the second set of planes offset (or the extra axes for that matter). Personally I would make the planes internal to the feature creation as it clutters my view and model tree less.</p>
| 22544 | How to make datum planes on a cone in Creo? |
2018-07-03T13:54:18.537 | <p>My lab is evaluating the liquid, plastic, and shrinkage limits of some soil samples and we are following <a href="https://youtu.be/EcXJ961qjGA?t=3m31s" rel="nofollow noreferrer">this video provided by the Missouri University of Science and Technology</a>. The video indicates we should conduct at least 3 tests where it takes 15-20, 20-25, and 25-30 hits of the Casagrande Apparatus to close the gap that we make via the grooving tool. And this video also explains that the liquid limit is defined as the moisture content in which it takes 25 hits of the Casagrande Apparatus to close the divide made by the grooving tool. We have been doing our best to turn our soil sample into a "uniform paste" through adding water as the video indicates, and even when that is the case, on a consistent basis, it takes fewer than 10 hits for the gap to be closed. <strong>Bottom line: My question is could there be soil samples where their liquid limits are lower than what the Atterberg Limit and Casagrande Apparatus method could detect?</strong> I do recognize the possibility of user error on our part, that we can continue to refine our technique on making the soil sample a "uniform paste." Thanks in advance!</p>
| |geotechnical-engineering|soil|soil-mechanics| | <p>According to BS EN ISO 17892-12:2018 there are three methods to determine liquid limit and plasticity index of soils. Previously, BS 1377 PART 2:1990 used the definitive method of a cone penetrometer. This method superceded the Cassagrande method after 1975 (Smith's Elements of Soil Mechanics, 9th Ed, 2014, pp11-13)
<a href="https://i.stack.imgur.com/AKgpu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AKgpu.jpg" alt="Liquid limit apparatus diagram"></a></p>
<p>The problem with the Cassagrande method is that it is highly dependant on the person doing the test and therfore far less reliable than the cone penetrometer test.</p>
<p>Considering the variable results of the Cassagrande method, I would hazard a statement of "yes" because there are soils that cannot release their water content completely due to their molecular alignment and ionic bonds which are also affect by Van der Waal's (pronounced fun-dur-Varl's) forces too.</p>
| 22562 | Liquid Limit of Soil through Atterberg Tests |
2018-07-03T14:33:09.767 | <p>I am interested in the challenges that Solar Balloons face when flying. For example, the high altitude could cause drastic changes in:</p>
<ul>
<li>Wind speed</li>
<li>Temperature</li>
<li>Moisture</li>
</ul>
<p>How does the unique design of a solar balloon address these challenges?</p>
| |mechanical-engineering|thermodynamics|measurements|aerospace-engineering|instrumentation| | <p>I think you are asking about how the blimp is maneuvered or functions? Wind currents are widely used and is not that turbulent not to cause damage unless you guide it into a lightning storm or hurricane even then it could get high enough to be safe. Here is a picture of the altitude and wind direction:
<a href="https://i.stack.imgur.com/1sB9M.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1sB9M.jpg" alt="enter image description here"></a></p>
<p><a href="https://earth.nullschool.net/#current/wind/isobaric/250hPa/orthographic" rel="nofollow noreferrer">https://earth.nullschool.net/#current/wind/isobaric/250hPa/orthographic</a></p>
<p>To change direction the blimp in many cases the blimp would only have to change altitude. </p>
<p><a href="https://i.stack.imgur.com/X02KD.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X02KD.gif" alt="enter image description here"></a></p>
<p>Not Photo shopped Airship Yuanmeng.
<a href="https://i.stack.imgur.com/MhUYl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MhUYl.jpg" alt="enter image description here"></a></p>
<p>Related: The helium in the balloon only needs to be heated to raise the altitude in some models. This model not built yet suggest to do this by having a transparent side of the balloon/blimp facing the sun.</p>
<p><a href="https://i.stack.imgur.com/1B0Uo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1B0Uo.jpg" alt="enter image description here"></a>
<a href="https://commons.wikimedia.org/wiki/File:Stratobus_artiste.jpg" rel="nofollow noreferrer">https://commons.wikimedia.org/wiki/File:Stratobus_artiste.jpg</a></p>
<p>I deduct that a parabolic shape could be designed to heat an inner dome filled with a breathable oxygen/helium atmosphere supplying warm air to the passengers. </p>
<p>Instead of heating the hydrogen have an air compressor to compress the air to land and regulate pressure and use heat to gain more height.</p>
<p>Make the whole balloon transparent an put everything outside it that can be inside and make those part like the framework on the outside on the inside with a wider surface area that gets hot.</p>
<p><a href="https://i.stack.imgur.com/zvPLZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zvPLZ.jpg" alt="enter image description here"></a></p>
<p>Related: <a href="https://space.stackexchange.com/questions/26429/can-air-pressure-be-accumulated-this-way-for-a-biodome-or-spacesuit">https://space.stackexchange.com/questions/26429/can-air-pressure-be-accumulated-this-way-for-a-biodome-or-spacesuit</a></p>
| 22563 | How do solar balloons work at high altitudes? |
2018-07-03T14:38:44.373 | <p>I have a typical small electric induction motor given to me in pieces.
It has two <strong>wave washers</strong> which I am told should sit between the bearing and the assembly. </p>
<p>I am not sure which bearing needs the wave washers.</p>
<ol>
<li>Do I place on one wave washer before the front and rear bearings ?</li>
<li>Both wave washers on the rear bearing? </li>
<li>Both wave washers on the front bearing?</li>
</ol>
<p>Please advise.</p>
| |motors|bearings| | <p>Wave washers allow axial expansion of shaft due to heat and keep preload, takes care of misalignment of shaft wrt to connecting load shaft, allows the vibration coming to move the shaft slightly and thus reducing axial load on the bearing because of vibration and external load .</p>
<p>Hence they are to be provided on both sides</p>
| 22565 | Do wave washers need to be on either end of bearings in double bearing induction motor |
2018-07-03T14:43:39.323 | <p>Can someone help identifying the electrical connectors in this picture? These connectors are from an aftermarket igniter. Also where can one purchase replacement connectors. I have looked <a href="https://www.digikey.com/en/resources/connectors/index" rel="nofollow noreferrer">connectors from digikey</a>. Any help towards looking for replacement connectors is appreciated. </p>
<p><a href="https://i.stack.imgur.com/kmCXX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kmCXX.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|electrical-engineering| | <p>These connectors are also called Quick-Disconnect Terminals. Here is a link from Master Carr. <a href="https://www.mcmaster.com/#spade-connectors/=1djynbs" rel="nofollow noreferrer">Quick-Disconnect Terminals</a></p>
<p><a href="https://i.stack.imgur.com/PknAm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PknAm.png" alt="Quick Connects"></a></p>
| 22566 | How to identify electrical connectors? |
2018-07-03T17:32:17.880 | <p>I want to work out how much I can expand a piece of rubber tubing before it ruptures. The wall is 0.75mm and the ID 5mm. Hence I am working out the stress/strain from thick-walled cylinder theory using <a href="http://www.roymech.co.uk/Useful_Tables/Mechanics/Cylinders.html" rel="nofollow noreferrer">these</a> <a href="http://courses.washington.edu/me354a/Thick%20Walled%20Cylinders.pdf" rel="nofollow noreferrer">pages</a>.</p>
<p>In Lame's equations (derived on those pages), are the radiuses in the formula the original radiuses before expansion, or those resulting from expansion? Does the same go for the radius under investigation? </p>
<p>many thanks for your help</p>
| |stresses|elastic-modulus|strength| | <p>According to J. E. Gordon's book "Structures or why things don't fall down" he explains The concept of Work of Fracture. During his exposition he explains that cylindrical shapes have a specific relationship of tensile strength to the thickness of the cylinder weigh is what a tube is. The general rule is that the tensile strength of any material is determined by the surface area perpindicular to the plane in which the strain exists multiplied by the length of the cylinder. On page 77 he states rubber has about 300% working strain. On p122 a diagram displays an image of circumferential stress in a cylinder with the formula of $s_2=\frac{rp}t$ where s=stress, r=radius, p=pressure and t=cylinder thickness. </p>
<p>In almost all cases where internal pressure exceeds the strain capacity of the cylinder, a longitudinal rupture occurs. Since your question asks about rubber specifically there is another issue in that the cylindrical thickness decreases with pressure increases proportionally to the radial expansion. In simple terms, the rubber is thinnest where the diameter is greatest but the strain energy is identical throughout the tube because the internal pressure is constant.</p>
<p>In your example, the thickness of the tube is 0.75mm with an internal diameter of 5mm. Since we know rubber had approximately 300% working strain, the pressure required to break it must therefore be
$$ s=\frac {rp}t$$
$$p=\frac {st}{r}$$
$$p=\frac {3 \cdot 0.75}{5}$$
$$p=0.45$$</p>
<p>So pressure must exceed $0.45 units of pressure per {mm^2}$ to rupture the tube by which the length of the unexpanded section to be expanded must be multiplied.</p>
<p><img src="https://i.stack.imgur.com/gqDNT.jpg" alt="book cover"></p>
<p>The book is a great read and humorous in many places. As well as being an invaluable source of education, it is a recommended course book by the university of Wolverhampton for civil engineering honours students.</p>
| 22572 | Thick walled cylinder theory - Stress/strain in ballooned out tube |
2018-07-03T22:54:05.970 | <p>Sulfur concrete only uses sulfur as a binding agent then an aggregate. Why does it not use cement?</p>
| |civil-engineering| | <p>The molten sulfur is the "cement". I understand that hundreds of years ago sulfur concrete was occasionally used. In the 70's , Amoco experimented with sulfur + asphalt concretes. They had excellent properties but were too expensive.</p>
| 22577 | Why does sulfur concrete require no cement? |
2018-07-04T08:02:05.350 | <p>I need to duct a PVC pipe (DN80) through a stainless steel sheet metal covering. The hole thing needs to be tight against stormwater. There's tons of products to duct pipes through thick walls, but I'm at loss on how to do that here. The cover is horizontal, upside checkered. Ideas I've considered:</p>
<ul>
<li>Welding a short stainless steel pipe stub onto the sheet and use a ring seal</li>
<li>Hole with ID of pipe in steel sheet, flange the pipe through the sheet</li>
</ul>
<p>Both of these appear overblown to me.</p>
| |civil-engineering|pipelines| | <p>The way this would typically be achieved, is with an off-the-shelf "Tank Connector". This essentially replaces the welding operation in your first idea. Simply cut the appropriately sized hole, and install the connector. There are face seals on both sides of the sheet metal, which are clamped tight.</p>
<p><a href="https://i.stack.imgur.com/n1Tzw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n1Tzw.png" alt="Tank Connector"></a></p>
<p>I've shown a brass connector, since I thought it was the clearest image - but needless to say, these are available in PVC, Stainless Steel, and with any number of diameters or Male/Female/Etc. configurations, according to your exact needs.</p>
| 22590 | Simple way to duct a PVC pipe through a stainless steel sheet cover |
2018-07-04T09:54:50.260 | <p>I have a large and unwieldy Solidworks assembly and I would like to save it as an "inert" geometry file such as an STL or STEP format. Then I want to import this into another project assembly and move it into a specific location.</p>
<p>It seems that I am unable to move the STL version from its default insertion point in the new project. I have tried opening the STL file and saving it as a SLDPRT file and bringing this into the assembly but I am unable to move that either.</p>
<p>The reason I want to do this is so that there are no dependency links between one project and the other, in order to avoid any possibility of design corruption while I'm experimenting with the new assembly. </p>
<p><strong>In a nutshell, how can I import and position an inert STL (or STEP or whatever) into an existing SLDASM file and be able to move it around?</strong></p>
| |solidworks| | <p>In order to insert a 'dumb' copy of an assembly into another higher level assembly, you can save the assembly as a .SLDPRT File. Make sure to select the correct amount of geometry to save, according to your needs from the options in the 'Save As' dialogue. </p>
<p><a href="https://i.stack.imgur.com/4g7hW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4g7hW.png" alt="Save as Part Options"></a></p>
<p>This .SLDPRT will be completely independent from the original assembly, being made up of internally saved .IGES components.</p>
<p>You can then simply insert this new part into your high level assembly, and mate it as normal.</p>
<p>N.B., if you ever use the 'Edit Component' Function inside the Assembly environment, you can toggle "No External References", which will prevent you from accidentally creating references to the 'dumb assembly', e.g. by accidentally snapping a sketch point to it.</p>
<p><a href="https://i.stack.imgur.com/NVN4a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NVN4a.png" alt="enter image description here"></a></p>
| 22591 | Solidworks 2014 - Convert a complex assembly into a single "inert" geometry file and import it |
2018-07-05T09:34:29.580 | <p>I have 2 files in prt, and I need to render this parts but my software can't open this.</p>
<p>Could someone export this files for me to step?</p>
<p><a href="https://www.dropbox.com/s/m9ysajdp6vo89od/pokretlo1.prt.1?dl=0" rel="nofollow noreferrer">https://www.dropbox.com/s/m9ysajdp6vo89od/pokretlo1.prt.1?dl=0</a></p>
<p><a href="https://www.dropbox.com/s/nkgqe9scxhjqsw0/pokretlo2.prt.1?dl=0" rel="nofollow noreferrer">https://www.dropbox.com/s/nkgqe9scxhjqsw0/pokretlo2.prt.1?dl=0</a></p>
| |software| | <p>You can remove the ".1" from the extension, so they are just .prt files - the .1 is to do with sequential numbering of backups, I believe. They can then be opened / converted using Autodesk Fusion 360, which is free for non-commercial use.</p>
| 22605 | Can someone export file from prt.1 to step? |
2018-07-05T10:06:28.777 | <p>I hope this is the right place to ask since I couldn't find a control systems SE.</p>
<p>Lets say we have the transfer function $P(s) = \frac{1}{s(s+15)(s+10)} = \frac{1}{s(s^2+25s+150)}$(completely made up numbers) of some sort of device. This is without any controller or feedback attached.</p>
<p>How should I look at the $\frac{1}{s}$ that's in there? Do I treat it as a separate multiplier or do I handle this as a third order system? </p>
<p>Without the integrator this is a simple 2nd order system with clearly defined relative damping and natural frequency because of $\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$. </p>
<p>Technically I could treat the 1/s as an I controller with the value of $P_i$ value of 1. Would this system even have a steady state error with a simple unity P feedback? Technically it would automatically be a PI controlled system right?</p>
<p>If the questions don't seem clear it's because I don't really know what to ask here.</p>
<p>edit: I managed to read something about types of a transfer function(0, 1 or 2) which only modifies the error you will have. The transfer function shown here would be considered a type 1 because it has 1 pole in the origin(2 means type 2 and 0 means type 0 I guess). Type 1 means that the steady state error will be 0 for a step input, which confirms my speculation about the error. I still don't know if it has any other effect.</p>
<p>edit2: <a href="https://i.stack.imgur.com/XoEhU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XoEhU.png" alt="equivalent"></a></p>
<p>So this picture shows the exact same thing in 2 forms. They're are functionally the same. The PID controller is C(s) in this case. Well it's a PI controller with K = 1, without me actually adding anything to the system. I've just added a negative feedback loop.</p>
| |control-engineering|transfer-function| | <p>Okay, something indeed does not seem very clear to you.
Let me try to clarify that ;)</p>
<p>Let $P_s = \frac{1}{(s+15)(s+20)}$ denote the second order part you do understand. I don't know what you mean by type 2 system, but this can represent any mass-spring-damper system as you already stated.</p>
<p>Where you go wrong is that you view the integrator $I = \frac{1}{s}$ as a controller. You recognized that $P = IP_s$, which is true. A controller, however, would introduce (negative) feedback, which is not present here. </p>
<p><strong>Controller interpretation - wrong!</strong></p>
<p>Let $Y$ denote the Fourier transform of the output of your plant and $E$ that of the error. (Without loss of generality we set the reference $R=0$).
Then: $$Y = P E = P(R-Y),$$ from which we can derive the well-known expression for the complementary sensitivity: $$T = \frac{Y}{R} = \frac{P}{1+P}.$$ (In literature, often $L$ is used instead to denote the open-loop transfer function $CP$, where $C$ is the controller, but let's keep using your notation instead.)</p>
<p>$T = \frac{1}{s^3 + 25s^2 + 150s+1}$, is the real transfer function of your second order system with your integrator as negative feedback controller from input $R$ to output $Y$. Note that this system indeed has no steady state error as you correctly noted.</p>
<p><strong>Correct interpretation</strong></p>
<p>Back to your system and the correct interpretation: the integrator is nothing more than a ''multiplier'' as you stated. A physical interpreation could be a valve controlling a water flow into a vat on an old-fashioned mechanical scale. </p>
<p>Your input corresponds to the valve setting $u$. The output of the integrator, $Us$ or $\int u \mathrm dt$ in the time domain corresponds to the amount of water in a vat. Then your second order transfer function $P_s$ takes the amount of water as input (force induced by the weight in this case) and outputs movement of the scale $Y$. Where this analogy goes wrong is that $m$ obviously changes as the vat fills, and inertia of the water flow is ignored, but hopefully you get my point.</p>
<p>Another analogy would be an ideal capacitor in-between your input and what actually goes into your second-order system.</p>
| 22606 | Order of a transfer function and its properties |
2018-07-05T10:57:49.583 | <p><a href="https://en.wikipedia.org/wiki/Tham_Luang_cave_rescue" rel="nofollow noreferrer">Tham Luang cave rescue</a> incident is taking combined effort of Engineers throughout the world. Various news media is telling currently it is not possible to pass more than one person through some very narrow portions of the cave tunnel (so the boys have to learn diving within cave). </p>
<p><a href="https://i.stack.imgur.com/XBVwx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XBVwx.png" alt="BBC MAP"></a></p>
<p>This is a Map by BBC, perhaps from <a href="https://www.bbc.com/news/world-asia-44695232" rel="nofollow noreferrer">here of BBC</a>; showing the narrow places in cross sections. </p>
<p><strong>Now; my question is</strong>; why the engineers are <strong>not attempting</strong> to use <a href="https://en.wikipedia.org/wiki/Abrasion_(mechanical)" rel="nofollow noreferrer">abrasion</a> (with a <a href="https://en.wikipedia.org/wiki/Grindstone" rel="nofollow noreferrer">grinding wheel</a> or such thing) and rock cutting wheels to broaden these "too narrow" parts so that later on 2 people (a diver carrying a boy) can come out at a time? </p>
<p>Is there any hidden risk of attempting so? Is that the cause for not atempting so? </p>
| |civil-engineering|geotechnical-engineering| | <p>Any underground work requires special research which takes time. Modifying the tunnel would mean increasing flow rate of water and therefore current strength. With larger volumes comes additional erosion issues such as deposition. As the diameter of an opening increases the flow velocity decreases causing deposition of material and narrowing of the vertical space. </p>
<p>Risk of collapse is probably high in wet ground which is already causing problems. </p>
<p>Removing material means it has to be manually transported a very long distance underground and delay rescue before flooding expected widely at any moment. Rather than remove cave material, remove the problem: the children. </p>
| 22608 | Thai Cave Rescue- Why the narrow portions are not being broadened? |
2018-07-06T10:47:44.280 | <p>I’ve been working on an electric Dirtsurfer project that has many parts to be made from folded sheet aluminium. <a href="https://i.stack.imgur.com/LyYjI.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/LyYjI.jpg" alt="screenshot"></a></p>
<p>Obviously the sheet metal needs holes in it for bolts. The sheet aluminium will be either CNC laser cut or CNC water jet cut. </p>
<p>I’ve got three options:</p>
<ol>
<li><p>Add holes to my design at the exact dimensions that I need them to be. I am placing myself at the mercy of the accuracy of the CNC cutter to ensure I have no slop which might let the bolts move. But no drilling!</p></li>
<li><p>Leave the holes out of the design. I measure the cut patterns to find the right places for the holes and drill them myself. Using a drill will give me a clean hole that I can fully trust the dimensions of. This is what I’ve done in the past, but I haven’t been accurate enough with my placement. </p></li>
<li><p>Undersize the holes in my design, and drill them out to the correct size later. I don’t have to measure anything, but still get a clean accurate drill hole.</p></li>
</ol>
<p>Any practical advice from someone who’s done this before?</p>
| |mechanical-engineering|cutting| | <p>What I have done before from a very short period of time, I have used a CNC laser cutter to cut 12mm sheet of stainless steel, which I'm sure that you are going to use a thinner sheat metal. My design was something about 135 x 18 mm and the final result was completely nice. But you have to take in consideration that laser has its own cutting diameter "pierce" ( some thing near 0.2 mm - according to the machine power and the sheet metal thickness ) . Think of it as the tool used to cut metal in CNC router. Ensure that the size of this pierce is taken in consideration while cutting to get your desired hole diamter. Machine operators can handle this issue without any problem. You just have to manage this issue with them.</p>
| 22622 | How should I go about adding holes to laser or water jet cut sheet metal? |
2018-07-08T09:02:53.757 | <p><a href="https://i.stack.imgur.com/bh7XQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bh7XQ.jpg" alt="Image 1 - what type of nut it is? what tool I have to use to open it?"></a></p>
<p><a href="https://i.stack.imgur.com/UgbNb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UgbNb.jpg" alt="Image2 - what type of lock it is? what tool I have to use to open it?"></a></p>
<p>Please see the two images. I marked a nut and locking part. what type of nut and lock mechanism it is? and what is the tool I have to use to open it?</p>
| |torque| | <p>You need to hold the shaft (in a vise with protective jaws) and turn that nut with a "C" spanner - which looks like a C with a single protruding tooth.</p>
<p><a href="https://i.stack.imgur.com/EVGd9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EVGd9.png" alt="enter image description here"></a></p>
<p>A poor method is to use a drift or punch, but this always damages the nut - perhaps fine for a temporary repair...</p>
<p>Edit as Fred says that is a circlip, actually in internal circlip which means that the nut above needs to be removed first then that circlip will come out.</p>
| 22652 | What is the type of nut and tool to be used to unscrew? |
2018-07-09T18:01:31.450 | <p>Suppose I have a rough 3d model of a building created it some CAD software. Now I want to change some portion of the model based on the following factors</p>
<ul>
<li>Lighting</li>
<li>Energy consumption</li>
<li>Structural Load</li>
<li>Fire Safety</li>
<li>Heating & cooling load</li>
<li>etc.</li>
</ul>
<p>Is there any Software that can provide these feature. I know there are some good software for each of the feature mentioned above, but how can we integrate their results into the 3d model. I am from a Computer science background so don't have much idea about this, Any suggestion would be of great help.</p>
| |structural-engineering|cad|building-design| | <p>A good Starting point for what you need is REVIT:</p>
<p>Our structural engineers do not use REVIT. They use RAM Steel to perform their structural analysis (Gravity only) and then give the work to CAD operators to document in AutoCAD or Revit.</p>
<p>A lot of the building Components can be stress tested with multiple scenarios in INVENTOR per individual components, but they also can be done in revit if you know the know how. (this is much more complex)</p>
| 22666 | How buildings 3D model is created |
2018-07-11T06:11:09.273 | <p>There is a gate which is to be controlled by 2 motors, but one should work and when one fails the other should work. AND THE CONTROLLER USED IS A PLC</p>
| |electrical-engineering|motors| | <p>Probably the simplest, most efficient solution would be powering the mechanism by both motors simultaneously, connected on the low-torque side of any mechanics - e.g. both propelling the same linear actuator, or the same first gear of a gearbox. Have them with enough surplus power that each can move the gate and idle second motor simultaneously. The gate will open slower with one motor, but it will open just fine.</p>
<p>You can expand on that idea by having the PLC read the current drawn by either of the motors when active, and switch it off (to idle, open contact) if detecting either the current is zero (burnt) or way over the top (short circuit). If the current in both is way higher than normal, but still not at short circuit level, the PLC should switch both motors off and signal gate blocked / mechanism jammed. (similarly, current way low; gearbox broken, motor running without load.) </p>
<p>That way you're keeping the mechanics simple and move most of complex decisions into software.</p>
<p>Now, if you want to save up on electricity, and use cheaper, weaker motors, you'll need either electromagnetic clutches (...and there go all your savings on cheaper motors), or unidirectional (free-wheel) clutches (so that an idle motor isn't pulled by the active one), and forgo simple switching of direction by reversing polarity, in favor of electromechanical gearbox that allows switching direction of the output rotation.</p>
| 22683 | How to control a roller gate using 2 motors - when one fails the other works |
2018-07-11T22:39:15.283 | <p>With 9 blades that are in a slanting axis so that they roughly follows a spiral around the center, would this device be similar to a pump, and turn (rotate) and gently pump the water down along the central axis as it contracts and squeezes itself together?</p>
<p><a href="https://i.stack.imgur.com/SULld.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SULld.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>If it is primed so there is a stablished flow of water down with enough momentum, yes it will rotate while contracting and expanding acting like a fish swimming and will generate either pressure down to pump a central flux of water or will start to move up if free to move.</p>
<p>A fish swinging its tail to left and right, or up and down, creates propulsion and moves forward, by riding its own wake, as opposed to steering. </p>
| 22695 | If this is submerged in water, and contracts and squeezes itself together, does it rotate? |
2018-07-12T12:04:47.890 | <p><img src="https://i.stack.imgur.com/kishF.png" alt="illustration"></p>
<p>My dad has a tub by a lake.
He has a water pump in the lake to pump up water into a tub.</p>
<p>He claims that if he use a flexible hose, the pressure in the hose at the top will be lower than if he use a rigid hose. The pump has to keep a flexible hose stretched, causing a pressure drop, he says.</p>
<p>I say that if the diameter of the hose doesn’t change while the pump runs, there will be no pressure drop, so once the pump is fully running, it doesn’t matter if the hose is flexible or not.</p>
<p>Who is right?</p>
| |pressure|statics| | <p>In order for the steady state comparison assumptions to be valid (same length, same diameter, same DP per ft, etc.), the flexible hose has to be fully expanded for the full length of the hose. Hence, the supply pressure and flow rate capacity of the source has to be high enough to provide the flow at a pressure drop high enough to keep it expanded. Otherwise, the not fully expanded hose will have a smaller diameter and different flow characteristics. A more rigid hose type will not have the same issue (or at least not nearly to the same extent). For a high pressure city water supply, there may not be a big difference... where city water pressure is lower, it could be noticeable... for a pump situation, it depends on the pump specs.</p>
| 22704 | Do flexible water hose cause pressure drop with constant flow? |
2018-07-13T07:42:06.403 | <p>I am going to store used li-ion batteries for some long time <code>>1</code> year, and <a href="http://batteryuniversity.com/index.php/learn/article/how_to_store_batteries" rel="nofollow noreferrer">I have read</a> that optimal storage conditions are:</p>
<ul>
<li>cool ambient temperatures, like <em>2-5 degrees celsium</em></li>
<li>3.82V voltage per cell.</li>
</ul>
<p>What I cannot find is what is correct operation sequence to prepare a <em>used</em> battery for cold storage (which is initially in <em>random</em> state)?</p>
<ul>
<li>Should I charge it to 4.2V then <strong>discharge</strong> to 3.82V? </li>
<li>Should I discharge it to 3.6V then <strong>charge</strong> to 3.82V? </li>
</ul>
| |electrical-engineering|battery|energy-storage| | <p>You should first "balance charge" the battery to full capacity using a suitable charger. Batteries do vary between manufacturers, and between battery types so check the manufacturer's rated voltage. Most modern chargers are smart enough to recognize the change of resistance as the battery charges to prevent over charging.</p>
<p>Once the battery cells maintain the same voltage, the battery is considered to be balanced and fully charged.</p>
<p>Now discharge the battery gently so it does not warm up significantly, until the rated cell "storage voltage" is reached. Older batteries may take several cycles to achieve a stable and even cell voltage for each cell or may not balance at all. </p>
<p>If an individual cell voltage/internal resistance is significantly different than the other cells, the battery can still be stored, but of course not at the manufacturer's optimum voltages. The cell to cell difference should be less than 0.1v.</p>
<p>You can keep the battery in a cool space at the storage voltage for more than 12 months.</p>
| 22719 | Li-ion batteries storage |
2018-07-13T15:06:59.147 | <p>I'm studying Euler's work on structural engineering from a book out of curiosity and it is mentioned that he developed a mathematical theory describing the buckling of columns under a parallel load (the weight-force of the load is directed down along the column). The theory is covered quickly without much motivation.</p>
<p>But this got me thinking; why does a column "buckle" in the first place? If the load presses the column down, why does the column even start deflecting sideways? I know this happens in real life since this fact is easily confirmeable with household objects, but theoretically, why do objects start deflecting sideways instead of just compressing under loads? This might be something obvious and maybe I'm just overthinking but I find this curious nonetheless.</p>
| |structural-engineering|civil-engineering| | <p>It buckles because its simply easier for it to buckle then to get crushed, take example of a thin guitar string, lets say you try to apply compressive force on it from ends, it will obviously buckle and not compress. This is the most basic nature's law, nature will always take the easiest path to do things, for example a lightning will strike through air which offers it the least resistance to come down. Similarly the load required to buckle a ling column is less than the load required to crush it (in other words a column will offer less resistance to buckling then to crushing) and hence the buckling takes place.</p>
| 22725 | Why does column buckling occur when the load is parallel to the column? |
2018-07-13T17:01:33.743 | <p>Lets say you want to build a bridge out of carbon fiber. This has never been never been done before, and it's super expensive stuff. So your goal is to use as little material as possible, but obviously your bridge still needs to have enough strength.</p>
<p>To start you look up how strong is carbon fiber, and the online data sheets tell you that a 10 oz cloth has a tensile strength of 800KSI. </p>
<p>After running the math this sounds perfect and should give your bridge enough strength for it's uses.</p>
<p>However, before you start a super expensive project that's never been done, based on nothing more than datasheets you found online and equations that might not fully apply to your material you run a sanity check.</p>
<p>Before you start building your full sized bridge, you build a small one that's 1/10th the size.</p>
<p>Assuming this small bridge is exactly the same as the large one, just using 1oz fiber instead of 10 oz fiber would your small bridge endure 1/10th the load of the full sized one?</p>
<p>I don't mean to apply this question to carbon fiber alone. With anything in engineering if you scale up a model perfectly using the same materials would the strength scale up proportionally, or is there some kind of real life nuance that blows all this mathematical beauty out of the water?</p>
| |structural-engineering|structural-analysis|stresses| | <p>There has been some valid comments on the other answers, at the expense of being repetitive I mention just a few more.</p>
<p>In your analogy of the bridge for example. The cables' strength scales to the surface area of the cable, L^2 but the seismic forces are related to mass of the structure which is volume L^3.</p>
<p>Wind testing of a scaled model is very complex and depending on geometry and openings has to be individually element by element analyzed and scaled back up</p>
<p>In lab tests, say to test the shear strength of a manufacturer's proprietary assembly of a shear wall they do have scaled down models occasionally, but they are aware and calculate all these disparities.</p>
<p>City of Los Angeles accepts sometimes theses test results for performance of some products( LA RR reports) but they expect careful analysis and logic used to justify scaling.</p>
<p>However thanks to powerful simulation algorithms many projects can be directly simulated for analysis from the digital drawings.</p>
| 22731 | Do scaled models reflect strength characteristics of full size |
2018-07-14T20:17:31.717 | <p>I was going through the wikipedia page on polymer concretes. It seems like a good type of building material. I was wondering which types of polymers are used in the concrete, since I cannot find any online besides epoxy.</p>
| |civil-engineering| | <p>There are many categories of binders and more are being added due to research and development in high strength binders. The two categories of binders or resins are thermosetting and thermoplastic. Each has sub categories as shown below. </p>
<p>Thermosetting resins: </p>
<ul>
<li>Epoxy</li>
<li>Polyester</li>
<li>Acetone formaldehyde</li>
<li>Phenol formaldehyde</li>
</ul>
<p>Thermoplastic resins</p>
<ul>
<li><p>Methyl methacrylate</p></li>
<li><p>Indencumaron</p></li>
</ul>
<p><a href="https://books.google.com/books?id=NiQtAgAAQBAJ&lpg=PA3&ots=BLCaU29OOc&dq=Indencumaron&pg=PA3#v=onepage&q=Indencumaron&f=false" rel="nofollow noreferrer">Google book research, click here.</a></p>
| 22740 | Which polymers are used to make polymer concrete? |
2018-07-15T12:23:10.210 | <p>Are electron photographs showing electrical charges? Do electrically charged parts show up more (darker in the photo) ?</p>
| |optics| | <p>If you read the excellent discussions on the wikipedia page, you'll see that a transmission electron microscope works on the same principle as an x-ray machine: the image shows how many particles/photons made it through the subject. </p>
<p>Other techniques analyze the various secondary products produced by electron collisions with the subject, such as heat or X-ray emission. </p>
<p>The charge or lack thereof on the subject material has essentially zero bearing on the final image.</p>
| 22752 | Electron micrographs, is it showing electrical charges? |
2018-07-16T11:57:08.640 | <p>The problem is illustrated in the picture below. A viewer looks through a slit and into a room. The problem is to find the maximum height of the wall in the room that the viewer can see if they are able to move 'right in front' of the slit and up to 166 cm away from the slit 'in the right'. (If it makes things harder or anything.. the slit is 3.5 cm wide)</p>
<p><strong>In other words, what distance must the viewer be from the slit too see most of the wall which is 280 cm tall</strong>
<a href="https://i.stack.imgur.com/PMgsb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PMgsb.png" alt="enter image description here"></a></p>
<p>This is just a problem for fun. I am not sure how one might go about finding this optimal distance from the slit. While it can be done by experimental trial and error, I have a guess that one can see more of the wall if the viewer is closer (right next) to the slit.</p>
| |measurements| | <p>You have it essentially correct. The farther away from the slit, the smaller your angular field of view and thus the less of the far wall you can see. When your eye is in the aperture (that's the optics name for the slit in the near wall), your angular field of view is limited only by the eyeball itself. </p>
<p>But when you move away, draw the lines from the center of your eye to the top and bottom edges of the slit, then continue those lines to the far wall. You will quickly see that, the farther away you are, the smaller angle of view is made by those two limiting lines. </p>
| 22761 | Maximum height view-able in the box when looking through slit |
2018-07-18T15:13:01.413 | <p>Given a system dynamics
\begin{equation}
\dot{x}=A(p)x+Bu,
\end{equation}</p>
<p>where $A(p)$ is the uncertain system state matrix with the nominal matrix being $A(p_0)$. The uncertainty in $A(p)$ satisfies the matching condition, i.e.
\begin{equation}
A(p)-A(p_0)=B\phi(p)
\end{equation}</p>
<p>Now from the book "Robust control design (an optimal control approach)" I encountered an exercise problem which goes as follows:</p>
<p>If $(A(p_0),B)$ is controllable, show that $(A(p),B)$ is also controllable for any $\phi(p)$ where $A(p)=A(p_0)+B\phi(p)$.</p>
<p>My approach was to use the rank condition of the controllability matrix for the nominal case and extend that result to the original system but it does not give me the result I expected. Any hint will be much appreciated.</p>
| |control-engineering|control-theory|optimal-control| | <p>You can prove it using the PBH test for controllability. It states that $rank \left(
\begin{array}{cc}
\lambda I -A & B \\
\end{array}
\right) = n$ for all values of $\lambda$.</p>
<p>For the uncertain system this becomes $\left(
\begin{array}{cc}
\lambda I -A -B \phi & B \\
\end{array}
\right) = \left(
\begin{array}{cc}
\lambda I -A & B \\
\end{array}
\right).\left(
\begin{array}{cc}
I_n & 0 \\
-\phi & I_m \\
\end{array}
\right)$ which also has full rank.</p>
<p>(Here $n$ is the number of states, and $m$ the number of inputs.)</p>
| 22783 | Does the controllability of nominal system imply the controllability of the actual uncertain system? |
2018-07-19T05:07:34.963 | <p>Let us assume that the an object collides into me in one of two scenarios, either I’m standing on shore, or else we’re submerged underwater. The object doesn’t change it shape, and possesses the same velocity prior to the collision. Would the consequences be the same? more severe? less severe?</p>
<p>Points to take into account:</p>
<ol>
<li><p>The "wet" weight is larger than "dry" weight, thus the hitting body has more momentum prior to collision. This is true especially if for instance the hitting body has some cavity which may be filled with water thus giving it larger effective mass during collision.</p></li>
<li><p>My another concern is the effect of the medium. Since water are incompressible, I should start "feeling" the body "pushing" water towards me earlier (prior to impact), thus the effective duration of the collision is longer, which should imply smaller forces.</p></li>
<li><p>Aside from that, since the collision in water cannot be thought of as sudden, how should I acount for drag forces during the process.</p></li>
</ol>
| |mechanical-engineering|fluid-mechanics| | <p>Let us not consider the wet weight, because we don't at this stage have enough info on the size and location of cavities and how fast the can shed the water.</p>
<p>As for the duration of the impact it can be very short, crashing impact. Consider an sledge hammer hitting a glass panel under the water, assuming the same impact velocity, it will impart same force. I don't think there will be cushion of pressure wave in front of the hammer to break the blow, not significantly, water is incompressible.</p>
<p>An obvious difference is the way our body responds. On the air our body will roll with punch and take the blow in a more flexible way. If you hit me with a fast ball in my torso my rib cage will bend in under the impact and my longs will dampen the blow, may be by bulging out around and acting like shock absorber, but if they are pre-pressured with the hydrostatic pressure they will resist the punch in a less flexible way and chances of tissue injury are higher. </p>
<p>If we consider the viscosity of water added as a shear friction to the entire surface of our body, it will be that much harder to recoil, hence it will receive the impact more forcefully. Although if wear some wetsuit with lubricated skin it may help.</p>
<p>All these, of course if we assume we somehow manage to move the ball or projectile with the same speed below the water as above.</p>
<p>** EDIT **</p>
<p>After OP's adding more detail:</p>
<p>Something that is semi autonomic and follows you has to have fault safety algorithm built into it. Such as it will default to a fast left turn and stop, or suddenly dives down and slows in a circle ready to be towed back. You may be able to attach a soft cone to the tip as a sacrifice soft structure to break the impact, etc., etc.</p>
<p>We my google about design of torpedoes and submarines. </p>
| 22793 | Collision of submerged bodies |
2018-07-20T03:58:10.097 | <p>I am designing a system to have solar panels track the sun, but only with one axis of rotation that will tilt the panels from east to west. </p>
<p>I want to connect the panels together so that one motor could rotate multiple panels. I was thinking of a something like a bicycle roller chain and sprockets. Where a long chain goes around, say, 5 sprockets that each rotation a solar panel, and wraps around again forming a loop. </p>
<p>I am not sure if the chain will have enough contact with the sprockets, especially on panels that are in the middle of the chain system. The chain would basically be resting on those sprockets from the weight of gravity.</p>
<p>Is this a good design choice for such a system? Is there more cost effective way to accomplish the same thing? Should I supply tension to system so it has good contact with all the sprockets? </p>
| |gears|photovoltaics| | <p>I might go with a set-up use an offset arm to translate a long beam. Said beam has a link at each solar panel to a pole which push/pulls the panel to the desired angle. Kind of like the strings in a window's Venetian blind.<br>
The drawback here may be in the amount of mechanical parts required at each panel.</p>
| 22809 | Rotating Multiple Solar Panels with One Motor |
2018-07-20T08:50:44.043 | <p>This is what I'm talking about:</p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Ampyx_Power" rel="nofollow noreferrer">Wikipedia</a></li>
<li><a href="https://www.youtube.com/watch?v=lykCHdMNvQM" rel="nofollow noreferrer">Youtube video 1</a></li>
<li><a href="https://www.youtube.com/watch?v=uCHk4TRr5hE" rel="nofollow noreferrer">Youtube video 2</a></li>
<li><a href="https://www.ampyxpower.com/technology" rel="nofollow noreferrer">Official site</a></li>
</ul>
<p>Can you please explain me on how this plane is generating energy? Reeling back the aircraft should consume the energy produced and maybe more, as now you also fight the wind.</p>
<p>Some energy may be from gravity when the plane descends, but I don't see how that will be enough.</p>
| |energy-efficiency|wind-power| | <p>You've pretty much worked it out: when the wind is lower, or when the plane's controls reduce the plane's lift, it drops, and the winch is wound in. Then the lift increases again, and that reels out the winch, which is where the generator sits.</p>
<p>Its generation is therefore very similar system to Kite Power Systems, which is explained in <a href="https://www.youtube.com/watch?v=Ap_B-9o7uV8" rel="nofollow noreferrer">this video</a></p>
| 22812 | How is the Ampyx Power plain generating energy? |
2018-07-20T20:59:49.720 | <p>So lets compare my dirt bike and my daily driver:</p>
<p>My dirtbike is a single 350cc cylinder 4 stroke engine. It has a redline at 12.1k RPM. </p>
<p>My daily driver is a six cylinder 3.6L V6. The redline on it is 6.5k RPM.</p>
<p>On my daily driver, and without much mechanical knowledge, I understand that at 6000 RPM, each cylinder would be moving at 1/6th of the total RPM as its a "team effort"
My motorcycle piston would be working alone, so it has to reach the 12.1k on its own.
Is the bottleneck not the piston, but in the crankshaft? How can a single cylinder work faster alone as opposed to working with other cylinders?</p>
| |mechanical-engineering|automotive-engineering| | <p>Go back to first principles. Engine horsepower depends upon:</p>
<p>i.) Cubic capacity.</p>
<p>ii.) Rpm.</p>
<p>iii.) Compression ratio.</p>
<p>There is also the inevitable ‘ proportionality constant ‘ , which tidies up all other constants relating to interchanging units.</p>
<p>i. & ii. together give you the amount of air passing through the engine in 1 minute, & hence the amount of fuel that can be burnt to generate power.</p>
<p>If you want a small & compact, lightweight, engine, keep the cubic capacity small, & hence the rpm will need to increase to achieve the same total ‘flow’.</p>
<p>The more you can compress the charge prior to ignition, the greater the relative expansion, & the more power that can be absorbed from the process. There is, of course, a practical limit. Due to the heating effect of compressing gases, petrol will tend to ignite prematurely, before the ideal point in the cylinder cycle - detonation, also known as ‘pinking’. This limits the available power, and damages the engine into the bargain. The upper limit is about 8.5 -9 :1. Above this, you have to put additives into the fuel, to deter pre-ignition. The most famous of which was tetra-ethyl lead, which is now banned.</p>
<p>Because diesel oil is basically a much heavier, less volatile hydrocarbon than petrol [ gasoline], and it is only injected directly into the cylinder at the point of almost maximum compression, much higher compression ratios are possible, a factor of maybe 2–3.</p>
<p>Of course, the diesel engine has to be much more robust, & hence heavier to survive these stresses. I’ve never heard of a practical diesel motorbike!!</p>
<p>They can also be slower revving, & longer stroke, which gives greater torque. I’ve driven trucks with 13–14 litre [ 780 -840 ] that produce 450 bhp, in a power band that runs from 1000–1500 rpm. The torque is truly awesome. My road car has a 1.9 litre twin turbo diesel, connected to a 6 speed gearbox, & redlined at 4500rpm, to yield 180bhp. It does not have the same long stroke, hence low[ish] torque, & cannot really pull away in 2nd gear.</p>
| 22819 | Why do small displacement engines have higher RPM ranges? |
2018-07-21T17:51:00.383 | <p>How to represent numbers 10 to 15 in 2421 bcd?
The original question is this:
<a href="https://i.stack.imgur.com/cnKcC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cnKcC.png" alt="enter image description here"></a></p>
| |computer-engineering|computer| | <p>To Convert a hexadecimal to BCD first you have to convert the hexadecimal to decimal then we can convert decimal to BCD.</p>
<p>Now first take a look at how to convert decimal to BCD:</p>
<p>There is fix set of rule for this it is like</p>
<ul>
<li>0 in decimal = 0000 BCD(2421), </li>
<li>1 in decimal = 0001 BCD(2421),</li>
<li>2 = 0010</li>
<li>3 = 0011</li>
<li>4 = 0100</li>
</ul>
<p>Until here it is same as binary but it changes from here:</p>
<ul>
<li>5 = 1011, </li>
<li>6 = 1100, </li>
<li>7 = 1101, </li>
<li>8 = 1110, </li>
<li>9 = 1111</li>
</ul>
<p>So how BCD(2421) is converted lets see:</p>
<p>If we take an example of:</p>
<ul>
<li>$8 = 1110 = 1\times2+1\times4+1\times2+1\times0 = 8$</li>
<li>$9 = 1111 = 1\times2+1\times4+1\times2+1\times1 = 9$</li>
</ul>
<p>So the numbers here are multiplied by (2421) that's why it is called 2421 BCD.</p>
<p>Now then how to write 10?</p>
<ul>
<li>10 = combination of 1&0 i.e. 00010000, </li>
<li>11 = 00010001 likewise.</li>
</ul>
<p>So now to convert hexadecimal to binary:</p>
<p>$$\begin{align}
\text{(AF36) hexa} &= 10\times16^3+15\times16^2+3\times16^1+6\times16^0 \\
&= 40960+3840+48+6 \\
&=44854
\end{align}$$</p>
<p>Now convert this decimal no. 44854 into 2421 BCD</p>
<ul>
<li>4=0100, </li>
<li>4=0100, </li>
<li>8=1110, </li>
<li>5=1011, </li>
<li>4=0100, </li>
</ul>
<p>Combine them: 01000100111010110100 (option c)</p>
| 22826 | How can I convert (AF36) hexadecimal to 2421 BCD? |
2018-07-23T16:03:34.507 | <p><a href="https://i.stack.imgur.com/Pu91N.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/Pu91N.jpg" alt="Enter image description here"></a></p>
<p>Just by taking a train ride across my home city I can see truss bridges like the one in above picture everywhere. There are numerous variations, but the most common design seems to be this. But why are they built specifically this way?</p>
<p>I can intuitively kind-of see why such a design probably is strong, but is there any kind of in-depth reason? I would be interested to know the answer as much from the physics side of things as possible. Googling didn't help much; I could find information on the different variations and many examples, but none really covered what is it about this design specifically that makes it so popular.</p>
| |structural-engineering|civil-engineering| | <p>That is a steel truss design from the recent era when good engineering analysis of simple designs was possible. The accurate analysis allowed the use of longer, thinner elements, as you see here. But only for simple designs, and even then only with considerable effort.</p>
<p>Originally, truss bridges were almost always built to a standard design. An individual Engineer would have a standard design, and an individual Rail Company would have a standard design, and an individual agency would have a standard design.</p>
<p>Your state might have paid for all of the important road bridges in your state: your rail company might have put up all the original rail bridges in your state. When looking at the history of a collection of bridges, you find that they were all built in a 20 year period, all built by the same agency, all with the same chief engineer. Then he retired, and a different design was adopted.</p>
<p>When better analysis became possible, it was used to build cheaper bridges, but the effort involved still meant that nobody wanted to use a different truss structure for every cheap bridge. They just choose one truss design, and worked with that.</p>
<p>For some of the truss designs, there are clear advantages: you see that in the posted picture, both sides of the truss are identical, there are common profiles, verticals are interchangeable, and vertical-plan diagonals are interchangeable. Truss designs where that is not true can save on weight and material cost -- but at the cost of construction, design, and manufacture cost. The advantage of standard elements continues when more bridges are built -- all the bridges you saw were probably built to the same height and length, with repeating sections where required.</p>
<p>Generally, the reason you see repeated bridges with identical truss design is institutional costs, not because that design is clearly superior to other similar truss designs.</p>
| 22848 | Why are truss bridges the way they are? |
2018-07-25T05:57:04.503 | <p>I am a 3d designer trying to build a simple device for my project and I have a problem. As you can see in schema below, point G has weight on it and it needs to be a controlled, vertical rotation. I am afraid that when I build this, it will go faster or slower on sides. Rotation needs to be stable and I need rotation to stop where it is, as soon as motor stops.</p>
<ul>
<li>Can a mini motor achieve this and what kind?</li>
<li>Do I need to add extra parts to the device and what are they?</li>
</ul>
<p>Thanks in advance and I am looking forward to answers.</p>
<p><a href="https://i.stack.imgur.com/JyrPX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JyrPX.jpg" alt="schema"></a></p>
| |mechanical-engineering|motors|gears| | <p>The easiest, to implement would be to use a stepper motor with enough hold torque or possibly detent torque if you need the thing to hold even unenergized. The benefit of this is also that you do not need a closed loop controller as the stepper either moves a step or skips it. Which means as long as your stepper is strong enough its not a issue.</p>
<p>Now your post mentions gears. While i think involving gears is a bit premature as you havent done motor selection, remember that gears change the effective torque of your motor. So it also affects the effective hold and detent torque.</p>
| 22869 | Establish a controlled rotation on gears |
2018-07-25T14:43:49.543 | <p>Before the yield point, the deformation is elastic, and then it is plastic. But what happens to the elastic 'part' of strain after the load is removed? Is it retracted fully or partly, and how does this relate to the extent of plastic deformation?</p>
| |materials|strength|deformation| | <p>In ductile materials there is a well defined straight area at the beginning of the stress/strain curve which is purely elastic, the slope of this part of the curve is Young's modulus. If you remove the load the sample will completely relax to its original length and can give back all the work put into it.</p>
<p>If you keep loading the specimen beyond this elastic region and then remove the load it will not fully recover to its original length, but it will shrink with the same slope as Young's modulus. The area under this loading and unloading is the work gone into the system and never recovered.</p>
<p>This could be a huge amount of energy in structural frames and connections. It is employed to provide the last precious extra minutes of safety in case of a big earthquake, by absorbing the energy of earthquake.</p>
| 22875 | Springback after a plastic deformation |
2018-07-25T17:01:42.310 | <p>I am trying to optimize the heat control of a test bench at our facility. The test bench is basically a pipe with an air flow that can be heated up to desired temperature. To heat the air flow up, three heat zones are available as shown in the sketch below. Measurements of the temperature are available for <span class="math-container">$T_1, T_2, T_3$</span> and <span class="math-container">$T_4$</span>. The three zones are each equipped with there own PID controller, so it is possible to set target temperatures for each zone. However the single temperature states obviously affect the values of each other, especially in the direction of the flow, but also slightly in upstream direction via conduction in the pipe material.</p>
<p><a href="https://i.stack.imgur.com/QUlhe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QUlhe.png" alt="Sketch of test bench" /></a></p>
<p>The ultimate goal is to set a target temperature for <span class="math-container">$T_4$</span> and control the three heat zones accordingly.</p>
<p>Right now, basically, only the PID controller of the second heating zone is used for the temperature control. Ideally I would like to use all three heating zones. The PID controllers of the heating zones could either be used to do so or just set to a static gain and building a control on top of this.</p>
<blockquote>
<p>What is the general strategy to get such a MISO control set up?</p>
</blockquote>
| |control-engineering|control-theory|airflow|pid-control|temperature| | <p>Let us take a <a href="https://en.wikipedia.org/wiki/Lumped-element_model" rel="nofollow noreferrer">lumped mass</a> approach to model the thermal dynamics of the region across two zone, including the convection and conduction and approximate the state <span class="math-container">$T_4 \approx \frac{T_2 + T_3}{2}$</span>. Further, assuming that the heat flow is dominated by convection, we can apply the <a href="https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling" rel="nofollow noreferrer">Newton's law of cooling </a> (or heating) to obtain the heat flow rates and therefore the lumped-mass zone temperature dynamics are <span class="math-container">$\dot{T_{i}} = {k}_{i} (T_i - \frac{T_{iup}+ T_{idown}}{2})$</span> where <span class="math-container">$0 < {k}_{i} = const.$</span> and the subscripts <span class="math-container">$up$</span>, <span class="math-container">$down$</span> denote the upstream and downstream temperatures. Then, assuming that <span class="math-container">$k := k_i = k_j$</span> for all <span class="math-container">$i, j$</span> and denoting <span class="math-container">$T_{12}:=T_{1down} =T_{2up}$</span>, we have <span class="math-container">$$\dot{T_{1}} = {k} (T_1 - \frac{T_{1up}+ T_{1down}}{2}) \approx {k} (T_1 - \frac{T_{1up}+ T_{12}}{2}),$$</span> and similarly <span class="math-container">$$\dot{T_{2}} \approx {k} (T_{2} - \frac{T_{12}+ T_{4}}{2}),$$</span> <span class="math-container">$$\dot{T_{3}} \approx {k} (T_3 - \frac{T_{4}+ T_{3down}}{2}),$$</span> so that the approximation <span class="math-container">$T_4 \approx \frac{T_2 + T_3}{2}$</span> implies that <span class="math-container">$$\dot{T_4} = k (\frac{T_4}{2} - \frac{T_{12} + 2 T_4 + T_{3down}}{4}) = - k \frac{T_{1} + T_{2} + 2 T_{3down}}{8} < 0.$$</span> This implies that <span class="math-container">$\dot{T_4} \rightarrow 0$</span> as <span class="math-container">$t \rightarrow 0$</span> if <span class="math-container">$T_2$</span> and <span class="math-container">$T_3$</span> are stable. Considering that temperature PID control of zones <span class="math-container">$2$</span> and <span class="math-container">$3$</span> is available, a good first approach is to apply individual PID controllers for zones <span class="math-container">$1$</span> and <span class="math-container">$2$</span> to stabilize their temperatures and set <span class="math-container">$T_2^{ref}$</span> and <span class="math-container">$T_3^{ref}$</span> such that <span class="math-container">$T_4^{ref} = \frac{T_2^{ref} + T_3^{ref}}{2}$</span>.</p>
| 22879 | How can we control the output temperature of a multi-input single-output pipe air-flow system? |
2018-07-25T17:31:48.700 | <p>How can I quickly and accurately form batches (10 to 100) of parts from mild steel wire about 3mm diameter ? </p>
<p>The parts are 3D with bends of several radii, about 200mm in the longest direction with a 0.2"diam 270 degree loop in one end for a bolt.</p>
| |manufacturing-engineering| | <ul>
<li>CNC - commercial
<ul>
<li>Huge machines such as YF-CNC502. On Alibaba.com, you can order custom-bent wire.</li>
<li>Pensa Labs (2D and <a href="https://www.youtube.com/watch?v=X0J-kcLrWqc" rel="nofollow noreferrer">certian types of 3D shapes</a>). You can <a href="https://www.pensalabs.com/request-sample" rel="nofollow noreferrer">order free samples</a></li>
</ul>
</li>
<li>CNC - DYI (mostly 2D shapes): <a href="https://www.youtube.com/watch?v=HPQbKTJPsU4" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=HpuprEfqoEk" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=uMI7usyHyOQ" rel="nofollow noreferrer">link</a> (each has build guides and source files)</li>
<li>manual - one joint - commercial: Du-Bro E/Z Bender, Almhs 11-0043A Mini Universal Bender, Falon-Tech bending device, KAKA Industrial Mini Universal Bender MUB 1, ...</li>
<li>manual - one joint - DIY
<ul>
<li>2 perforated rails, 3 screws: <a href="https://www.youtube.com/watch?v=vb5r2WXo4Og" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=YR7FspY-sgw" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=NnLZp6ai8YY" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=K1Q2FnU535s" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=aMb8MAPKcU4" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=oHER2qtoOWY" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=7OpmlFAH28Q" rel="nofollow noreferrer">link</a> <a href="https://www.youtube.com/watch?v=l9RAEA7-98E" rel="nofollow noreferrer">link</a></li>
<li>2 angle irons: <a href="https://www.youtube.com/watch?v=Dr4Uasi0lLo" rel="nofollow noreferrer">link</a></li>
<li>3D printed: <a href="https://www.youtube.com/watch?v=n9Gtrzx7yJc" rel="nofollow noreferrer">link</a></li>
<li>wood: <a href="https://www.instructables.com/Quick-and-Easy-tube-or-wire-bender/" rel="nofollow noreferrer">https://www.instructables.com/Quick-and-Easy-tube-or-wire-bender/</a></li>
</ul>
</li>
<li>manual - no joints: wire bending jig; bending gauge</li>
<li>manual - pliers</li>
<li>manual - various types - DYI: 3D printed etc. <a href="https://grabcad.com/library?page=1&time=all_time&sort=recent&query=wire%20bending%20tool" rel="nofollow noreferrer">link</a> <a href="https://www.thingiverse.com/search?q=wire%20bending%20tool&type=things&sort=relevant" rel="nofollow noreferrer">link</a></li>
</ul>
<p>Round nose concave pliers have too little leverage.</p>
<p>The Du-Bro E/Z Bender's great.</p>
| 22880 | CNC or other method for Wire Forming |
2018-07-26T03:47:19.990 | <p>Let us consider a ceramic material, $\mathrm{Mg_2SiO_4}$ - Forsterite, with a sphere shape.</p>
<p>At room temperature, the ceramic sphere is "trapped" in a matrix which does not expand (null thermal expansion coefficient).</p>
<p>If I raise the temperature of the system (matrix + ceramic), the sphere will not be able to expand and internal stress will grow due to the resulting compression.</p>
<p>How would you estimate the maximum temperature, as well as the maximum internal stress associated before the ceramic sphere fails ?</p>
| |mechanical-engineering|materials|failure-analysis| | <p>I assume here, the linear coefficient of thermal expansion of the matrix is lower than that of the ceramic so we land in compression, otherwise the sphere tries to expand in all directions.</p>
<p>The fracture criterium of isotrope materials and anisotrope materials are not the same but in general, fractures occurs: </p>
<p>if the greatest principal stress is equal to the fracture stress as measured in axial tensile test. This is true for <strong>isotrope</strong> materials.</p>
<p>So what can we do, theoretically, first we should construct the stress tensor and then find the eigen values of this matrix, the greatest eigen values indicates the greatest principal stress.
Again all this is true if the material is isotope. For anisotrope materials you need a comprehensive knowledge of fracture mechanics, and it ain't as easy as decibed above. </p>
| 22884 | How to determine the maximum internal stress for a material under compression before brittle failure? |
2018-07-26T15:37:13.853 | <p>I'm talking about IEC motors-gearbox alignement. Here are an example:
<a href="https://i.stack.imgur.com/LkW3o.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LkW3o.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/YNJ8P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YNJ8P.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/H9vZT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H9vZT.png" alt="enter image description here"></a></p>
<p>I found the drawing in cross hydraulic website, so it's quite reliable.
What i don't understand, is, how this design makes sure the perfect alignement of gearbox coupling and the motor? </p>
| |mechanical-engineering|pumps| | <p>Bell housings are often fixed with bolts or studs, these are convenient and take the required loads but do not generally provide adequate precision in location. Threaded fasteners really need a certain amount of clearance otherwise there is a significant risk of damaging the threads or the through hole during assembly or at least making assembly much more difficult. </p>
<p>Instead additional matching sets of features will be machined onto both mating parts which are designed to provide alignment to the required tolerances. In some cases fine adjustment may be provided by shims or eccentric pins etc. </p>
<p>A fairly generic method is to use dowel pins, unlike bolts or studs, which generally require a certain amount of clearance to avoid damaging the threads these will be machined with a slight interference fit. </p>
<p>Other options include machining matching rings, steps or tapers onto both parts. Taper fits tend to be especially good at ensuring concentrically on standardised parts as they tend to be self aligning. </p>
<p>In some cases final machining with be carried out with both parts assembled together making them essentially a matched pair. This tends to give the best possible accuracy but makes parts non-interchangeable. </p>
| 22896 | How does bell housing guarantee the concentricity of the shafts? |
2018-07-26T17:52:38.427 | <p>What does it mean with respect to calculating stresses on a beam undergoing loading and when should it be applied?</p>
| |structures|stresses|beam| | <p>The thin-wall assumption is relevant when calculating stresses due to shear and torsion.</p>
<p>This is because shear and torsion generate <a href="https://en.wikipedia.org/wiki/Shear_flow" rel="nofollow noreferrer">shear flow</a> through the cross-section, as seen below:</p>
<p><a href="https://i.stack.imgur.com/sgKdd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sgKdd.png" alt="enter image description here"></a></p>
<p><sup><a href="https://slideplayer.com/slide/9308724/" rel="nofollow noreferrer">Source</a></sup></p>
<p>This shear flow is a representation of how the reaction to the applied shear force (or shear force binary pair is equivalent to the applied torsion) "flows" through the section.</p>
<p>For any given segment of the section (for example, each of the sides of the box section above), it's obvious that the total shear stress "going in" is the same as the stress "going out". However, how much of that stress is found near the outer face, near the inner face and around the middle?</p>
<p>Honestly, that gets kind of messy to calculate, so instead the "thin-walled assumption" is made instead. We just assume that the thickness of each of the walls is so small that whatever variance there is through the wall can be ignored. Put another way, it assumes that the shear flow is uniform throughout each of the walls.</p>
<p>As far as I know, there is no formal definition of when a wall can be considered thick or thin, but a common one is $d/t>20$ (seen <a href="https://en.wikipedia.org/wiki/Cylinder_stress#Thin-walled_assumption" rel="nofollow noreferrer">here on Wikipedia</a>), where $d$ is the section's width or diameter, and $t$ is the wall's thickness. That being said, I've seen the thin-wall assumption used in sections with $d/t=10$ as well.</p>
<p>That <a href="https://en.wikipedia.org/wiki/Cylinder_stress#Thin-walled_assumption" rel="nofollow noreferrer">Wikipedia article</a> actually has a good description of how much easier thin-walled sections are to calculate than thick-walled ones. Thin-walled sections have closed-form solutions, while thick-walled ones require integration to solve. Fun times!</p>
<p>The example I gave above is of a closed thin-walled section, but the assumption can also be made with open sections (I, L and T sections, for example). Obviously the solution for the shear flow is dependent on the cross-section, but closed sections perform drastically better under torsion than open ones.</p>
| 22900 | What exactly is the thin wall assumption in regards to beam cross sections? |
2018-07-27T15:44:20.223 | <p><a href="https://i.stack.imgur.com/cltOJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cltOJ.jpg" alt="enter image description here"></a>I am trying to apply an existing blower curve in a hydraulic model for a single stage blower with variable inlet diffuser vanes. The variety of curves shown on the plot does not seem to make sense to me. In particular, I don't know which of the blower curves I should be looking at. The legend at the top-middle of the plot shows IGV: 10 etc. but then those same line-types are used in the plot with 12, 10, 08 etc. and no units are shown anywhere.</p>
<p>I looked up the ASME document referenced (PTC-10) but did not find anything in it to shed any light on how this curve was developed or how it is meant to be read.</p>
<p>Does anybody know how to interpret this or know of a resource that explains this style curve in detail?</p>
| |design|modeling|compressed-air| | <p>Old topic but... to be sure you have the correct info : The IGV positions are given by the seperate lines as indicated at the top (continious, dashed, dotted and dash-dot) At these IGV positoins, tests are done at several outlet diffuser positions (the numbers at the bottom of these lines in the curves) Example; There are several dashed lines (IGV pos 04) and these have different numbers at the bottom (outlet diffuser vane position). Basically, IGV pivots the curve around the 0 flow max head point and outlet diffusers move the whole curve horizontally.</p>
<p>The numbers are simply the numbers of the actuator arm position (one for the IGV and one for the diffuser), that is on the outside of the machine. You can see them in the left os the picture in the link for the IGV
<a href="https://www.howden.com/Howden/media/Howden/img/products/compressors/single-stage-turbo-compressor.jpg" rel="nofollow noreferrer">https://www.howden.com/Howden/media/Howden/img/products/compressors/single-stage-turbo-compressor.jpg</a></p>
<p>ASME PTC-10 is the test protocol used at this specific test for warranty on flow/pressure and power consumption.</p>
<p><a href="https://www.asme.org/codes-standards/find-codes-standards/ptc-10-performance-test-code-compressors-exhausters" rel="nofollow noreferrer">https://www.asme.org/codes-standards/find-codes-standards/ptc-10-performance-test-code-compressors-exhausters</a></p>
<p>The curves you see here is a test curve of the machine, a KA44SV where KA44 is the size, S is for outlet diffuser vanes, V for IGV's, from testing (which was done for every machine before it left the factory) in Helsingor, Denmark. The IGV's were used to compensate isentropic head for off-design conditions (determined by pressure measurement, inlet temp measurement and diffuser position), the diffuser vanes for flow control. This way, they got a pretty constant high efficiency over the whole flow range, all the way down to ~45% from the max flow at a constant pressure, at a constant rpm.</p>
<p>You can not really determine flow/pressure from these to find a duty point in practice. These curves are used to program IGV position control in the PLC so when installed, the IGV goes to the optimal position automatically.</p>
<p>There is also another curve that comes with testing, that shows you the efficiency of the machine over the complete working range.</p>
<p>HV-Turbo used to be the world leader in centrifugal compressors for wastewater aeration. They made them from the KA2 (starting at ~ 45 kW) to the KA100 (max. 2.7 MW). They were taken over by Siemens (approx. 2005) and they sold of the whole package to Howden in 2018 I believe. Factory in Helsingor is gone sadly.</p>
| 22911 | How do you read a blower/compressor curve with multi-curves for inlet guide vanes? |
2018-07-27T16:33:54.843 | <p><a href="https://i.stack.imgur.com/xlny8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xlny8.jpg" alt="load disp curves comparison results between FEM and Experimental results"></a></p>
<p>According to my results, I can have this curves.
And somehow in this figure, it is not pretty much similar,
What can I say if it is asked why I still want to use the FEM results?
Or how can I validate the FEM results from this figure? </p>
<p>Any help?
I really appreciate it
Thank you very much in advance</p>
| |structural-engineering|abaqus| | <p>The tail end of the curve is probably not a massive issue as that is well into the region which would normally be considered failure in any case and FE systems often don't really look at failure mode in detail. </p>
<p>Of more concern is the elastic region where the FEM results suggest it is a lot stiffer than it is in the experiment. The experimental curve also doesn't follow a straight line at the start of the experiment which suggest that there is some slack being taken up somewhere. This could either be an issue with the experimental apparatus or the sample itself. </p>
<p>Without knowing the details of the part its is hard to speculate in detail but it could be say that you have bolted joints with inadequate preload and they are slipping under the initial loading. </p>
<p>The difference in Youngs modulus implies a more fundamental issue with the FE model though, perhaps an incorrect value for E or invalid assumptions about things like joint stiffness. </p>
<p>A likely possibility is that you have some elements of the model whcih are constrained in too many degrees of freedom compared to the experiment eg you have a rigid joint where you should have a pinned joint. </p>
| 22913 | What is the good reason if the curves between two results are not similar? |
2018-07-28T14:13:42.653 | <p>I have a fan on a stand, to circulate air in my home. The three fab blades are somewhat "D"-shaped, with one edge being rounded and one being straight-ish or even curving inward.</p>
<p>Now, from my sailing, as well as my understanding of wing and propeller (edit: and low-noise computer fan) design, I would expect the outward-curving edge to be the leading one, but it's not. In the photo*, the fan spins clockwise to push air towards me.
<a href="https://i.stack.imgur.com/pi4u6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pi4u6.jpg" alt="enter image description here"></a>
*excuse potato quality, the fan was spinning so this is with reduced shutter time and high ISO.</p>
<p>In contrast, <a href="https://images-na.ssl-images-amazon.com/images/I/61XFMzZLFkL._SY355_.jpg" rel="nofollow noreferrer">this low noise pc fan</a> also spins clockwise, yet its blades <em>do</em> curve "backwards" in relation to their direction of spin (as I'd expect).</p>
<p>Wouldn't my fan be less noisy if it spun the other way, <strong>at least theoretically</strong>? Obviously, including inverting the blade tilt, in order to keep airflow direction.</p>
| |airflow| | <p>I believe the main reason for the different blade shapes is ducted vs non-ducted fan design and tip vorticity. PC fans are essentially ducted fans with a fan-blade shape that produces tip vortices. These vortices hit the edge of the duct, which dissipates (or weakens) the vortex. OTOH, the portable fan in your picture has no duct, so the shape of the fan blade (forward swept) is designed to direct the span-wise flow towards the inside of the fan rather than the edge. This allows the blade to produce more airflow, since less energy is used in producing a tip vortex.</p>
| 22922 | Doesn't this fan spin the wrong way? |
2018-07-28T21:02:28.223 | <p>I was playing with a spring, and realized that I've never wondered why a spring gets harder and harder to compress the more you try to compress it.</p>
<p>It seems easy at first though. What is the scientific explanation behind this?</p>
| |springs| | <p>The question is answered, but not demonstrated by example. The clue surely, is in the fact that compression spring loading is given as force/unit travel. So if it is loaded with 1lbf and the spring compresses 1 inch then the spring has a linear “rate” 1lb/inch. To compress the spring further, say another inch, then you would need to add a further 1lb weight or apply a further 1lbf.
Ergo, the further the spring is compressed the more ‘force’ is needed to achieve it. “It becomes harder to compress”. RC (retd C Eng)</p>
| 22925 | Why do springs get harder to compress? |
2018-07-29T03:05:04.187 | <p>I am comparing the stress results between original or non-corroded model to corroded model.
I found that the stress value in non corroded model is lot more bigger than the corroded model,
is it correct?</p>
| |structural-analysis|stresses|abaqus| | <p>Maybe it is correct, maybe it isn't. </p>
<p>Suppose you "corrode away" a small radius that was causing a stress concentration, and it becomes a bigger radius. Quite likely the maximum stress in the structure (at that concentration) will <em>decrease</em> because of the corrosion. For example an "average" stress of 1.0 and a stress concentration of 10.0 might become an "average" stress of 1.1 and a stress concentration of 9.0. Whether you count that as a "decrease from 10 to 9" or an "increase from 1.0 to 1.1" is your choice.</p>
<p>On the other hand if you reduce the cross section of a structural member like a beam, or the thickness of a plate, the stress will probably increase - i.e. stress = load/area, and you decreased the area.</p>
| 22931 | Is it true that if the structure damaged or corroded, the stress will be lower? |
2018-07-30T08:14:26.323 | <p>Load sequence in phases can be considered applying <a href="https://forums.autodesk.com/t5/robot-structural-analysis-forum/load-sequencing-construction-stages/m-p/4570879/highlight/true#M18259" rel="nofollow noreferrer">the self-weight load only to the story added in the appropriate phase</a>. </p>
<p><a href="https://i.stack.imgur.com/YEMyz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YEMyz.png" alt="enter image description here"></a></p>
<p>Assuming that the material will not go into yield stage ( will always be at the elastic stage), is there any need to consider load sequence in construction stage for structural analysis?</p>
<p>I think there is <strong>no need</strong>, because no matter what, the final finished structure should always have the maximum loads at each floor/element, whereas during the construction, each floor/element loading is only partial. So if we consider the final structure analysis, we already consider everything adequate for design. No?</p>
<p>But why code of practice still recommend it?</p>
| |structural-engineering|structural-analysis| | <p>There is definitely a need to look at loading sequence through staged construction. Modern analysis usually has a section for staged construction. I normally run across this in bridge construction. </p>
<ul>
<li>The piers can be constructed before the tops are stabilized by the bridge deck. </li>
<li>The bridge decks themselves can be simply supported between piers before the concrete over the piers cures and the deck can subsequently be treated as a continuous member</li>
<li>Construction equipment and materials can generate abnormal loading conditions</li>
<li>Due to the size of your bridge deck and location, you may have limits on how much deck you can place and finish in a given time frame. The wet concrete will provide dead load in the poured sections, and there will be no strength from the deck in the incomplete sections (or the wet concrete sections) </li>
</ul>
| 22949 | Is there any need to consider loading sequence in construction stage for structure analysis? |
2018-07-30T17:14:50.977 | <p>I'm about to start learning about control engineering. But am I advised to learn first Thermodynamics? (Since control engineering has applications in Thermodynamics)</p>
| |thermodynamics|control-engineering| | <p>It is difficult to understand the aspect at play on the control side of processes before first understanding how the process behaves. Thermodynamics, fluid mechanics/dynamics, unit operations, kinetics, etc. will all improve your process knowledge and make what you are learning in controls much easier to implement appropriately. The first two listed would be very important in the learning phase in order to understand the why behind most control operations.</p>
| 22962 | Learn thermodynamics before beginning with control engineering ? |
2018-07-30T18:28:48.523 | <p>According to wikipedia there are 3 types of steel based on the crystal structure: austenitic, ferritic and martensitic. Austenitic steels are non-magnetic where 300 series stainless steels belong to. Ferritic steels have magnetic properties. But I didn't manage to find any information about martensitic steel magnetic properties. </p>
<p>In particular I am interested in stainless steel alloy 455. According to the data I have this steel is magnetic (though didn't manage to find the source of this data). But when I obtained a sample of this steel It appeared to be non-magnetic. This brings to possibilities: either my source is wrong or the sample I got is not steel 455.</p>
<hr>
<p>Upd.
Turned out that the samples I got were “accidentally” made of austenitic steel 304 instead of requested 455.</p>
| |steel|magnets| | <p>This is a precipitation hardening martensitic so will be ferromagnetic unless it has been mistreated. To anneal, it is quenched from high temperature and makes a soft martensite = ferromagnetic. It is then machined and/or cold formed . And then final heat-treatment is an age at roughly 1000 F when it will harden. There may be minor changes in ferromagnetic properties depending on condition but I expect it is always similar to conventional steel. I never worked with it but the high nickel - 9%, makes it prone to have retained austenitic ( very low ferromagentism) structure after quenching: If so, a sub zero cooling ( like dry ice and alcohol) may get it to transform to martensite. </p>
| 22966 | Is martensitic steel 455 magnetic? |
2018-07-30T18:55:49.123 | <p>On <a href="https://youtu.be/mqPMANOEG7Y?t=3m26s" rel="nofollow noreferrer">this video</a> from Captain Joe, you can see that the landing gear of the airplane has got some sort of v-shaped strut connecting the two moving parts.
<a href="https://i.stack.imgur.com/0tiNr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0tiNr.jpg" alt="v-shaped strut"></a></p>
<p>I think that I saw the same thing on some vespas with a suspension in the front wheel.</p>
<p>What are they for ?</p>
| |mechanical-engineering| | <p>This is the mechanism that allows the piston to move so absorbing the landing forces, while making sure that there is no rotation of the wheel assembly which would create a huge amount of side loading and possibly cause the landing gear to fail.</p>
<p>I have had the opportunity to visit Airbus and stand next to some of the units used to test these... serious engineering, but then 230 tonnes does take some controlling...</p>
| 22967 | What is this V-shaped strut on the landing gear for? |
2018-07-30T20:48:42.617 | <p>I am thinking of building a house with basement next to some existing structures. Are there rules of thumb on how closely/deeply you can excavate next to an existing structure? By moving farther away from the existing structure do I reduce the potential need for temporary retaining?</p>
| |structural-engineering|civil-engineering| | <p>Without a competent geotechnical analysis by a qualified civil engineer, you are likely to face difficulties and lawsuits. Generally speaking, forces exerted by buildings act at 33° to the horizontal in soil. This does not account for hogging in soil or loss of compression against neighbouring structures of the soil. Fluid content aints for much of the rigidity or fluidity of soil strength. Opening it up to air will change its characteristics and behaviour messing something unexpected could happen.</p>
| 22974 | Excavation near existing structure |
2018-08-01T15:30:55.077 | <p>I have been trying to understand the torque curve and power curve of an engine. From what I have read and understood I know that the torque curve and power curve is has a peak at a particular RPMs(RPM is different for torque curve and power curve). I also understand that when the engine is running at a particular RPM it will be producing a particular torque(drive torque) and if we increase the load on the engine(increase the load torque) then the RPM will start to decrease. To keep the RPM from decreasing, we can open up the throttle( which will increase the fuel burnt and so the energy produced at the power stroke) and so, increase the drive torque being produced to match the increased load torque.</p>
<p>What I am not able to understand is if the torque value at a particular RPM shown in the torque curve is the maximum possible torque at that RPM (i.e throttle valve fully open at that RPM) or if that torque curve is given for a particular external load(i.e we can still increase the load and to keep the RPM constant we can still open up the throttle valve). I am guessing that the curve shows the maximum possible torque at a particular RPM but I am not sure. </p>
| |mechanical-engineering|automotive-engineering|torque| | <p>Yes the engine would have its throttle wide-open (or whatever position gives the highest torque reading). Each rpm point on the torque curve shows the maximum torque resistance that the engine can maintain and NOT result in a drop in rpm. Once rpm starts to drop, that indicates the engine cannot overcome the amount of torque resistance that is being applied at that time.</p>
<p>The torque curve does not show how much torque the engine is producing at a particular rpm in normal driving situations (or in neutral). The curve shows how much torque can be produced at each rpm point when the engine is running under maximum sustainable load.</p>
<p>Maximum vehicle speed would be obtained by having the final drive ratio adjusted so the rpm where the torque is at the maximum can be reached but not exceeded. Speed is mostly limited by the ability of the engine to overcome the 'drag' on the vehicle, and drag increases with speed. So the top speed is obtained when the engine can produce both 1) its maximum torque and maximum rpm that produces that torque.
If the drive ratio permits rpm to increase beyond the 'max torque rpm', that means that less than max torque is actually being produced.
if the drive ratio prevents rpm from reaching the 'max torque rpm', that also means that less than max torque is being produced.</p>
| 23003 | What does the torque curve of an engine mean? |
2018-08-01T17:02:53.997 | <p>I am solving problems through Fox and McDonald.Here is the problem </p>
<p><img src="https://i.stack.imgur.com/nPj2G.jpg" alt="enter image description here"></p>
<p>I tried by finding out the torque due to viscous forces which is $$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}$$ and the torque due to mass m1 as $$\tau_2=m_1gR$$ Writing into equation $$\tau_2- \tau_1=m_2R^2 \frac{d\omega}{dt}$$ integrating this and using boundary condition $\omega=0$ at $t=0$ I got $$\omega=\frac{m_1ga}{2\pi Rh\mu}[1-exp(\frac{-2\pi \mu htR}{am_2})]$$ . However I am missing $m_1+m_2$ instead of $m_2$ in the exponential part.Any ideas? Thanks.</p>
| |mechanical-engineering|fluid-mechanics|dynamics| | <p>Mark has already pointed out what you are missing, and this is the Lagrangian approach which also should give the same result - and it does.</p>
<p>The position of $m_1$ is $x_1 = R \ \theta(t)$</p>
<p>The kinetic energy of $m_1$: $KE_1=\frac{1}{2}m_1x_1'(t)^2 =\frac{1}{2} m_1 R^2 \theta '(t)^2$</p>
<p>The potential energy of $m_1$: $PE_1=-m_1 g x_1=-m_1 g R \ \theta (t)$</p>
<p>The kinetic energy of the $m_2$: $KE_2=\frac{1}{2}I\theta '(t)^2=\frac{1}{2} m_2 R^2 \theta '(t)^2$</p>
<p>The Lagrangian: $L=KE_1+KE_2-PE_1=\frac{1}{2} m_1 R^2 \theta
'(t)^2+\frac{1}{2} m_2 R^2 \theta '(t)^2+m_1 g R \theta (t)$</p>
<p>The generalized force is the torque $\tau_2$: $Q=-\mu(2\pi Rh)\frac{R\omega}{a}R$</p>
<p>The equations of motion are $\frac{d }{d t}\frac{d L}{d
\theta'(t)}-\frac{d L}{d \theta(t)}=Q$ which turn out as</p>
<p>$$ m_1 R^2 \theta ''(t)+m_2 R^2 \theta ''(t)- m_1 g R=-\frac{2 \pi h \mu R^3 \theta '(t)}{a} $$</p>
<p>Substituting $\theta '(t)=\omega (t)$ and simplifying we get</p>
<p>$$\left(m_1+m_2\right) R \omega '(t)+ \frac{2 \pi h \mu R^2 \omega (t)}{a}-m_1 g=0$$</p>
<p>and the solution with $\omega(0) = 0$ is $$\omega(t) =\frac{ m_1 g a}{2 \pi h \mu
R^2} \left(1-e^{-\frac{2 \pi h \mu R t}{a \left(m_1+m_2\right)}}\right)$$</p>
<p>which agrees with the free-body diagram approach.</p>
| 23006 | Finding out torque in a viscosimeter |
2018-08-01T21:07:28.703 | <p>I'm new to part design/mechanical engineering and I have the following question.</p>
<p><a href="https://i.stack.imgur.com/HRMUW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HRMUW.png" alt="enter image description here"></a></p>
<ul>
<li>Blue: steel shaft, 5/8" diameter, connected to a motor</li>
<li>Green: aluminum disk, 10" diameter, 1/2" thick, 3" hole</li>
<li>Goal: the shaft should rotate the disk</li>
</ul>
<p><strong>The Question:</strong> I'm wondering how can the shaft be connected to the disk?</p>
<p>I tried searching for bushings, various forms of couplings, etc, but couldn't find anything that can connect the two.</p>
<p>Does this mean I would have to machine my own coupling, and there are no standard fittings/parts that would do the job?</p>
<p>FYI, the project needs high torque, so I'm guessing (please correct me if I'm wrong) that a coupling needs to have a large diameter as to apply the torque at a greater distance from the center.</p>
| |mechanical-engineering|motors| | <p><a href="http://www.masterdrives.com/" rel="nofollow noreferrer">Masterdrives.com</a> (.PDF - Catalog: <a href="http://www.masterdrives.com/bushings/QD%2520Bushing%2520Pages.pdf" rel="nofollow noreferrer">QD Bushings</a>) sells bushings of a suitable size but you may or may not want to do some machining to reduce the "E" dimension. Apparently you want bushing type SF, the E length with that taper is bound to fit. 5/8" bore size is available.</p>
<p>Catalog page:</p>
<p><a href="https://i.stack.imgur.com/PpCgA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PpCgA.jpg" alt="QD Bushings"></a></p>
<p>Another view of one of those T-bearings:</p>
<p><a href="https://i.stack.imgur.com/yHrCu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yHrCu.jpg" alt="QD Bushing"></a></p>
<p>Alternatively make your own or have one made for you:</p>
<p><a href="https://i.stack.imgur.com/mOSkt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mOSkt.jpg" alt="Homemade #1"></a></p>
<p>Another design:</p>
<p><a href="https://i.stack.imgur.com/P8OjF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P8OjF.jpg" alt="Homemade #2"></a></p>
| 23007 | How to couple/connect shaft to disk? |
2018-08-02T15:49:06.600 | <p>I recently started working with a new team doing detailed energy modeling for all flavors of building projects -- commercial, industrial, residential, new construction, renovations, additions. As the team has been growing, we've been discussing three particular problems that I believe version control could help with.</p>
<h3>Our challenges</h3>
<ol>
<li><strong>Coordinating multiple people working on a single energy model.</strong> Each building model involves different components -- geometry, envelope, HVAC, lighting, etc. For time-crunched projects different people may work on different components simultaneously. To bring each person's work together at the end can be complicated and time-consuming.</li>
<li><strong>Keeping track of "known good versions" for a particular application.</strong> To streamline our work we create templates for various building components. Each template may only be ready to use for certain building types (say, <code>HVAC template #n</code> is working for commercial buildings but not tested for residential). When starting a new model we review release notes to make sure we apply the appropriate template, until our periodic review when we test/update templates for all use-cases. Both the tracking of historic versions, and the periodic integration of updates, is complicated and time-consuming. </li>
<li><strong>Reproducing results from a report sent to a client.</strong> During model development we periodically prepare reports for clients. Each report is tied to a version of the model and templates which is archived on our server. This way we can re-open the old model to address any questions the client has, even as model development has continued. At times we also need to change aspects of the old model to answer specific client questions, before a model update is ready. At this point, the task of integrating two separate streams of model changes becomes... complicated and time consuming.</li>
</ol>
<p><strong>All of these processes can be improved with version control -- but nobody here has ever used version control!</strong> </p>
<p>I'm wondering if others here have been in a similar position, and implemented a version control system. What did you use, and how did it go? What best practices can you share?</p>
<h3>Some details about our team and our work</h3>
<ul>
<li>All engineers use Windows 10 </li>
<li>All of our modeling tools are Win32 applications (eQuest, Open Studio, TRNSYS), but modeling source files are stored as text-based files (not binary)</li>
<li>We're considering Git, Mercurial, and Bazaar</li>
<li>We do not currently have a server where we could run something like SVN, so we'd prefer a distributed system which could store files on a shared drive (such as a networked drive, SharePoint, DropBox, etc).</li>
</ul>
| |modeling|tools| | <p>To close the loop on this, and in case it's helpful for anyone else in the future... Here's the solution we ended up picking, and a few of the reasons and resources that moved us in that direction. </p>
<p>Let me preface by saying, as several have pointed out in the comments and answers, that <strong>the key to implementing version control is the mindset change and consistency of following a new process</strong>. We <em>had</em> a process that emulated what we do now, but it was complicated, time consuming, and prone to errors.</p>
<p>Here's what we ended up doing.</p>
<ul>
<li><strong>git</strong>
<ul>
<li>While looking into SVN I found a helpful article, but even more helpful was this comment on the article: <a href="http://disq.us/p/ru9xu7" rel="nofollow noreferrer">"5 reasons, coming from someone who has moved from subversion to git twice voluntarily"</a>. The ability to seamlessly work off-line or outside of our internal network were big selling points. </li>
<li>A comment I came across a few times (see <a href="https://softwareengineering.stackexchange.com/a/35080">this answer on softwareengineering.se</a>, for example) is that merging in SVN is complicated, and for those less comfortable with SVN can lead to a reluctance to commit. This defeats the purpose.</li>
<li>In contrast, I found <a href="https://www.atlassian.com/git/tutorials/saving-changes/git-commit" rel="nofollow noreferrer">this article from Atlassian</a> explaining that in git, "commits are cheap." Version control is most useful when frequent commits are made, so I wanted to lower the bar as much as possible for the team.</li>
<li>Because I knew I needed to get a certain level of buy-in before devoting significant resources to the project, setting up a server (such as would be needed for SVN) was off the table. I don't have the skills to do this, and convincing our small team to put time/money into something that they aren't convinced they need (when there are other things we all agree we need but that we can't afford yet) was a non-starter. Of course, this is a Catch-22: If we had a server with SVN set up, we likely could have gotten a satisfactory SVN process going.</li>
<li>I set up TortoiseSVN on my PC and used it for a couple of weeks. It was functional, but not exceptional.</li>
<li>Once I decided to move away from SVN to DCVC, the choice between Mercurial and git was decided by the fact that I had used git in the past. </li>
</ul></li>
<li><strong>GitLab</strong>
<ul>
<li>Free and open source.</li>
<li>We did consider GitHub, but it is limited to three collaborators on private repos.</li>
<li>Bitbucket is limited to five users for free teams.</li>
<li>As we (and our resources) grow, we may revisit hosting. This part of the process would be relatively easy to change.</li>
</ul></li>
<li><strong>Sourcetree</strong>
<ul>
<li>Based on <a href="https://unmethours.com/question/38/revision-control-for-text-based-modeling-programs/" rel="nofollow noreferrer">comments on a forum for energy modeling</a> (which is our use case), I decided to give Sourcetree a try. I found the interface helpful and intuitive.</li>
</ul></li>
</ul>
<p>Some other helpful resources:</p>
<ul>
<li>Oliver Steele's <a href="https://blog.osteele.com/2008/05/my-git-workflow/" rel="nofollow noreferrer">"My git workflow"</a>. The graphics were particularly useful for explaining git to our team.</li>
<li><a href="https://mindfulmodeller.ca/version-control-for-energy-models/" rel="nofollow noreferrer">"Version control for energy models"</a>. Helped us think through developing/implementing a workflow for our particular application.</li>
</ul>
| 23016 | Git, Mercurial, others -- what's the best system for an engineering team new to version control? |
2018-08-02T23:24:45.010 | <p>I was originally planning to as this on Physics Stack exchange, but found <a href="https://physics.stackexchange.com/questions/243555/focus-correcting-screen/243583#243583">this question</a> while searching, which directed the asker to here.</p>
<p>While my question is not a duplicate of that, it still falls in the same classification, so I believe it's on topic here.</p>
<p>That said, I recently had a <a href="https://en.wikipedia.org/wiki/Fluorescein_angiography" rel="nofollow noreferrer">Fluorescein angiography</a> and the following thought occurred to me about it.</p>
<p>Assume there's a piece of graph paper in front of me, the job of the lens in my eye is to project a sharp image of the paper on my retina. I have astigmatism, in both eyes as it happens, making glasses with a cylindrical correction necessary to read anything.</p>
<p>So, during the angiogram, the exact process is reversed. The camera involved projects a sharp image of my retina on a photographic sensor. Except I don't have my glasses on, so I would expect the astigmatism would lead to an imperfect image on the sensor.</p>
<p>How then is the camera able to get a sharp image? When reviewing the results afterwards with the doctor, I was quite impressed with the detail and clarity of the images.</p>
| |optics| | <blockquote>
<p>I recently had a Fluorescein angiography and the following thought occurred to me about it.</p>
<p>I have astigmatism, in both eyes as it happens, making glasses with a cylindrical correction necessary to read anything.</p>
<p>The camera involved projects a sharp image of my retina on a photographic sensor. Except I don't have my glasses on, so I would expect the astigmatism would lead to an imperfect image on the sensor.</p>
<p>How then is the camera able to get a sharp image? </p>
</blockquote>
<p>From the American Academy of Ophthalmology: <a href="https://www.aao.org/eye-health/diseases/what-do-astigmatism-measurements-mean" rel="nofollow noreferrer">Astigmatism</a></p>
<blockquote>
<p>"Astigmatism is measured in diopters. <strong>A perfect eye with no astigmatism has 0 diopters. Most people have between 0.5 to 0.75 diopters of astigmatism</strong>. People with a measurement of 1.5 or more typically need contacts or eyeglasses to correct their astigmatism in order to maintain clear vision.".</p>
</blockquote>
<p>Most people have some astigmatism.</p>
<p>The Topcon Medical Systems "<a href="http://www.topconmedical.com/products/trc50dx.htm" rel="nofollow noreferrer">TRC-50DX Mydriatic Retinal Camera</a>" webpage lists their <strong>TRC-50DX Datasheet</strong> (<a href="http://www.topconmedical.com//_assets/cf/downloadFile.cfm?section=productLit&file=TRC-50DX%20Datasheet.pdf" rel="nofollow noreferrer">.PDF</a>), which says the device has these settings (and autofocus):</p>
<blockquote>
<p>"<strong>Diopter Compensation Range For Patient’s Eye</strong></p>
<pre><code> 0 Setting: -10 to +6 Diopter
- Setting: -9 to -23 Diopter
+ Setting: +5 to +23 Diopter
+ Setting: +22 to +41 Diopter (Ocular Anterior Photography)".
</code></pre>
</blockquote>
<p>The machine has its own <em>glasses</em> built-in, which the doctor can adjust to match the eyes of the patient (whether they own glasses or not). Glasses are not worn since there could be smudges, scratches or dirt; it would be more glass in-between the sensor and your eye, and more to adjust for.</p>
<p><a href="https://i.stack.imgur.com/9PRT9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9PRT9.jpg" alt="TRC-50DX Mydriatic Retinal Camera"></a></p>
| 23023 | What lens is needed for a retina scan? |
2018-08-03T10:50:48.880 | <p>A 3 m deep basement has a two story timber frame house above.</p>
<p>They share a side, however both have parts "outside" of the other, as can be seen in the sketch below. The red dash is the basement, the blue is the house.</p>
<p><a href="https://i.stack.imgur.com/BFae7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BFae7.png" alt="rough diagram"></a></p>
<p>Can a basement form part of the foundations for the house in conjunction with other foundations not part of the basement?</p>
<p>Are there any risks or known issues with this concept?</p>
| |structural-engineering| | <p>There are some problems in your sketch that can be fixed. Other than that there is no fundamental issues.</p>
<p>A house or structure can be fully or partially supported by the walls of a basement, provided the bearing walls of the house are supported by bearing walls in the basement and the parts of the house that are outside of the perimeter of the basement have their own foundation.</p>
<p>In your sketch you have not shown the houses bearing walls, but you have to make sure all bearing walls especially the house's exterior walls are directly supported by the walls in the basement. Of course you can have interruptions in those walls such as doors or partial open wall but they have to be engineered.</p>
<p>In most of the US. these are called vertical discontinuities and there are special "penalty" load factors that apply to the structure, sometimes as large as 300% over load.</p>
<p>If it is not practical to have bearing walls under all the house walls in the basement You can have a solid structural concrete slab that runs under the entire house and also covers the roof of basement where it projects out of the foot print of the house, designed to carry the load of the walls. This is actually structurally the preferred method.</p>
<p>The other Issue of concern is the detailed design of decking, roof of basement, water proofing, slope of it and practical flashing, specially at the joints of the house with the roof of basement.</p>
| 23031 | irregular shape basement as part foundation for two story timber frame house |
2018-08-04T11:30:02.343 | <p>I am working on an assignment and I have a doubt in a step in my problem. Could someone please help me understand it?</p>
<p>Question and its solution is:<a href="https://i.stack.imgur.com/91xFV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/91xFV.jpg" alt="enter image description here"></a></p>
<p>.
My doubt is in the part which is underlined. How did the 2nd term become zero? Also. how did they calculate external moment vector. If someone can explain these steps it would be very helpful.</p>
<p>P.S: If this is not the correct forum to ask such question, I apologize.</p>
| |mechanical-engineering|vibration| | <blockquote>
<p><strong>How did the second term become zero?</strong></p>
</blockquote>
<p>Recall that the cross product of vectors $\vec{a}$ and $\vec{b}$ is a vector representing the area of the parallelogram spanned by $\vec{a}$ and $\vec{b}$ in the direction perpendicular on the parallelogram.</p>
<p>$$
\begin{align}
\vec{\omega}(t) \times \mathbf{J}\vec\omega(t) &=
\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}
\times
\begin{pmatrix} J_x & 0 & 0 \\ 0 & J_y & 0 \\ 0 & 0 & J_z \end{pmatrix}
\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix} \\
&=
\underbrace{\begin{pmatrix} 0 \\ \omega_y \\ 0 \end{pmatrix}}_{\vec{a}}
\times
\underbrace{\begin{pmatrix} 0 \\ J_y\omega_y \\ 0 \end{pmatrix}}_{\vec{b}}
\\
&= \vec{0}
\end{align}
$$</p>
<p>Note that within this case the vectors $\vec{a}$ and $\vec{b}$ are parallel, hence the area of the parallelogram they span is $0$ and therefore the cross product is also zero.</p>
<blockquote>
<p><strong>How did they calculate the external moment vector?</strong></p>
</blockquote>
<p>There is only one external force present in this system, the bearing force
$$\vec{F} = \begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}.$$
The moment this forces causes on the center of gravity is
$$\begin{align}
\vec{M}_C
&= \vec{r}_C \times \vec{F}\\
&= \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} \times
\begin{pmatrix} A_x \\ 0 \\ A_z \end{pmatrix}\\
&=
\begin{pmatrix} A_z r_y - 0 r_z \\ A_x r_z - A_z r_x \\ 0 r_x - A_x r_y \end{pmatrix}
\end{align}
$$</p>
<p>Assuming that $r_y = 0$ result in
$$ \vec{M}_C =
\begin{pmatrix} 0 \\ A_x r_z - A_z r_x \\ 0 \end{pmatrix}
$$</p>
<p>Note that the rotation $\varphi_y$ is clockwise and not counter clockwise as it would have been by using a right handed coordinate system.
Therefore the external moment should be defined in the opposite direction
$$ \vec{\mathbf{I}} = -\vec{M}_C =
\begin{pmatrix} 0 \\ -A_x r_z + A_z r_x \\ 0 \end{pmatrix}.
$$</p>
| 23044 | Rigid circular disc equation of motion |
2018-08-05T03:25:15.547 | <p>For a small 30 square meter home (8 x 3.75) is it feasible to have plinth beam 2 feet below ground? Technically, is it better than or worse off than standard procedure of plinth beam at ground level?
Also, please consider the fact that it is an earthquake prone zone.</p>
| |civil-engineering| | <p>It is better to have the plinth beam above the ground level.</p>
<p>For two reasons:</p>
<p>You keep the floor above rain run off or subterranean water and you reduce the height of your building or the length of the columns which helps them better perform in an earthquake.</p>
| 23053 | Can we have plinth beam below ground level for small homes? |
2018-08-05T09:37:17.800 | <p>Patrols have higher self-ignition temperature than diesel, but still, petrol engines have lower compression ratios than diesel engines. As the self-ignition temperature of petrol is shouldn't we be able to compress it more. I am not saying compress petrol to the point of auto-ignition which can cause knocking, but just have a higher compression ratio than diesel. </p>
<p>Is it because when the combustion of petrol-air mixture starts the moving flame front starts to compress the remaining air-fuel mixture which could reach a point of self-ignition when using higher CR?</p>
| |automotive-engineering| | <p>It's not because of the fuel, but because of the process.
The diesel process differs fundamentally from the otto(petrol/gasoline) cycle. </p>
<p>In the otto cycle, where fuel is present in the cylinder while compressing, the compression is limited by the auto ignition temperature of the fuel, whatever fuel is used. The compression warms up the mixture, ideally just before auto-ignition. The spark plug adds the needed flame source to start the combustion.</p>
<p>In the diesel cycle, fuel is added only when compression has already taken place, the temperature in the combustion chamber is way higher than the auto-ignition point of the fuel, which is why the fuel combusts as soon as it's injected into the cylinder. This removes the limit set by the auto-ignition temperature when choosing a compression ratio. Thus, the ratio can be higher, up to where materials begin to form a problem. </p>
<p>Compression is the very reason diesels are more efficient; the compression ratio can be higher, and they also always run at 'wide open throttle' giving high compression.</p>
| 23056 | Why do petrol engines have lower compression ratio than diesel engine? |
2018-08-06T02:44:07.903 | <p>I am attempting to design a "flying saucer." The saucer is an ellipsoidal-ish body with a RC drone inside. I plan to allow for air access to the rotors as shown in the rudimentary diagram below:</p>
<p><a href="https://i.stack.imgur.com/0481D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0481D.png" alt="enter image description here"></a></p>
<p>I am pretty sure that the drag on the air flowing through the intake and through the body of the flying saucer will reduce the thrust capacity of the drone. Can anyone confirm this? Are there any other considerations I should keep in mind (maybe there's a reason helicopters don't have a big tube surrounding the lift rotor)? I want the flying saucer to behave like a slow, mellow RC drone (I don't really want high speeds or elevations).</p>
| |aerodynamics|propulsion|thrust|drag| | <p>In the 50's the DARPA designed such a craft but due to aerodynamic instability it was a complete failure and abandoned. The problem you have is maintaining 4 separate intakes which have a greater volumetric capacity than each propeller can handle at maximum power in coldest conditions. You cannot combine the intakes and expect stability, it's just not possible. Each airscrew creates a pressure differential with the air cone immediately above and below it. This creates a spinning torus vortex when stationary as part of the stabilising effect of air wash. When that wash is excluded by ducting you introduce other aerodynamic effects. One of these effects is what causes a Dyson fan with a hollow centre to blow a larger volume of air. Your idea will cause the same effect external to the structure which will destabilise it.</p>
<p>An image is available here: <a href="https://www.nationalmuseum.af.mil/Upcoming/Press-Room/News/Article-Display/Article/198892/usafs-flying-saucer-on-display-at-national-museum-of-the-us-air-force/" rel="nofollow noreferrer">USAF's 'flying saucer' on display at National Museum of the U.S. Air Force</a></p>
| 23062 | RC drone rotor air access |
2018-08-06T03:35:19.333 | <p>I'm designing a ducted fan, and I'm deciding on the airfoil profile. NACA 16-series profiles are well-suited to M > .5 or .6, but they're not particularly efficient at lower Mach numbers. On the other hand, older profiles, like Clark YM, work quite well at lower speeds. </p>
<p>My question is, why do propeller designers generally only use one profile (with scaled chord and thickness) throughout the propeller? Is there some reason not to use, for example, a Clark YM for 2/3 of the blade and a NACA 16-709 for the outer 1/3?</p>
<p>Note: this being a ducted fan, laminar profiles will not work, so they aren't under consideration.</p>
| |aerospace-engineering|aerodynamics|propulsion|airfoils| | <p>There are propellers with different aerofoil sections.
For example, the propeller attached to the Rolls Royce engine kept in my graduate college had symmetric aerofoil sections roughly about 25 % and smoothly varying to bottom flat airfoil sections at the rest. </p>
<p>The aerofoil profile selections are based on the performance requirement, structural rigidity, and modes of operation.
You can go for any aerofoil sections with high ($C_l / C_d$) but, in the end, it should be structurally rigid to produce high thrust at high RPM. When you have variable pitch propellers you don't need different aerofoil sections since the required performance achieved by adequately rotating the blade.</p>
<p>In my point of view having the same aerofoil sections will be advantageous in the below-given ways. </p>
<ul>
<li>Matching the aerodynamic centres of the same aerofoils sections would be easier </li>
</ul>
<p>for example, NACA 0006 have the $a.c$ at $\sim c/4$ and NACA 2412 have at 0.3$c$, then locations of $a.c$'s will not be smooth along the span of the propeller. This would be critical in structural stress distribution on the propeller. </p>
<ul>
<li>Downwash of the propeller sections will have smooth variations. </li>
</ul>
<p>For similar sections, the downwash induced velocity component will be predictable and the twist of the blades could be compensated easily while designing. And the twist will be smooth for similar profiles from the structural point of view.</p>
<p>Hope this helps.</p>
| 23063 | Why don't propeller designers use multiple profiles? |
2018-08-07T08:47:32.210 | <p>I am attempting to write a user model in Fortran that can be connected to the Aspen Plus process simulator. Can Fortran 90 be used, or does Aspen only work with the older Fortran 77? I have checked the Aspen Plus documentation, but can find nothing about which versions of Fortran it is compatible with.</p>
| |chemical-engineering|simulation|process-engineering| | <p>Nat was right that Aspen doesn't care as long as your compiler can produce a DLL from the Fortran 90 code (which most do). However, I found that the Aspen files that need to be included in most user models are written in fixed format (the formatting style used back with Fortran 77). So when using Fortran 90 code in free format you get a bunch of compiler errors because of the mix of fixed and free formatting. There is a <a href="https://software.intel.com/en-us/forums/intel-fortran-compiler-for-linux-and-mac-os-x/topic/268667" rel="nofollow noreferrer">solution</a> though: put directives around the include lines that switch to fixed format only for the included files.</p>
<p>Here is an example I got from the Aspen support center. It is a user model that is supposed to calculate the viscosity of a compound.</p>
<pre><code>SUBROUTINE MUL2U (T, P, Z, N, IDX, &
IRW, IIW, KCALC, KOP, NDS, &
KDIAG, QMX, DQMX, KER)
IMPLICIT REAL*8 (A-H,O-Z)
#include "dms_plex.cmn"
#include "ppexec_user.cmn"
integer b(1)
equivalence (b(1), ib(1))
integer dms_ifcmnc
integer limuusr
limuusr=dms_ifcmnc('MUUSR')
! The following input language is required to define A and B
! (just define them for the 1st component in the list)
! PROPERTIES RKS-MU SOLU-WATER=2
! PROPERTIES RK-SOAVE
! PROP-REPLACE RKS-MU RK-SOAVE
! PROP MULMX MULMXUSR
! USER-PROPS MUUSR 1 1 2
!
! PROP-DATA MUUSR-1
! PROP-LIST MUUSR
! PVAL WATER -5 20
! To check that parameters are correctly transferred, set global diagnostics:
! HISTORY PROP-LEVEL=5
AA = B(limuusr + 1)
BB = B(limuusr + 2)
if (KDIAG .GE. 5) then
WRITE (USER_NHSTRY, *)'A = ', AA, ' B = ', BB
endif
qmx = exp(AA + BB / T)
dqmx = - BB * qmx / (T * T)
RETURN
END subroutine mul2u
</code></pre>
<p>The example was originally written for Fortran 77 (so it used fixed formatting), and I made the few changes necessary for it to be compiled using free formatting. However, when compiling you get a string of errors that come from the Aspen files included in the subroutine (so dms_plex.cmn and ppexec_user.cmn). Here is an example of one of the errors:</p>
<pre><code>c:\Program Files (x86)\AspenTech\AprSystem V8.8\Engine\commons/ppexec_user.cmn(7): error #5082: Syntax error, found END-OF-STATEMENT when expecting one of: %FILL <IDENTIFIER>
INTEGER USER_IUMISS, USER_NGBAL, USER_IPASS,
-------------------------------------------------------^
</code></pre>
<p>Again, it seems that the compiler gives the error because the Aspen files are written in fixed format and we are compiling with free format. To get around this use directives that switch to fixed format and back again.</p>
<pre><code>!DEC<span class="math-container">$ NOFREEFORM
#include "dms_plex.cmn"
#include "ppexec_user.cmn"
!DEC$</span> FREEFORM
</code></pre>
<p>When I use these directives around the include statement then the compiler works without errors and I can use a Fortran 90 user model in Aspen.</p>
| 23084 | Can Fortran 90 be used with Aspen Plus, or only Fortran 77? |
2018-08-07T20:34:54.473 | <p><strong>Question:</strong> I am currently encountering Hard Drive Memory space issue to store file on a National Instruments based CDAQ Tester. I am using Labview 2013 SP1 and have to all the software load in the CDAQ Controller. Some of the basic specifications for the CDAQ are as follows</p>
<ul>
<li>Processor: Intel Atom 1.33GHz Dual Core</li>
<li>Controller OS: Windows Embedded Standard 7</li>
<li>Hard Drive Memory Size: 16 GB</li>
<li>Number of Slots: 8</li>
</ul>
<p><a href="https://i.stack.imgur.com/OKYrP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OKYrP.jpg" alt="enter image description here"></a></p>
<p><strong>Background:</strong> I am trying to control and acquire electrical measurement data from a Keysight 34461A digital multimeter via GBIP using Labview and CDAQ. </p>
<p><a href="https://i.stack.imgur.com/ZQl2a.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZQl2a.jpg" alt="enter image description here"></a></p>
| |electrical-engineering|product-testing|labview| | <p>@Mahendra, I had run into Hard Drive Memory Space issue few months ago on my NI cDAQ. I am using cDAQ-9132 with a 16 GB hard drive, similar to your cDAQ. I was running LabVIEW Professional 2017 SP1 with all softwares installed. The memory space got full when I needed to install some additional drivers. </p>
<p>The cDAQ-9132 is not meant to have a full LabVIEW system running on it. The best practice is to only have LabVIEW run-time engine on the cDAQ and run LabVIEW application which is developed on a full computer with complete LabVIEW system. </p>
<p>To avoid memory space issue, I developed my code on the laptop which has over 200 GB memory space and has a full-fledged LabVIEW system. Once I developed the code I created a standalone application (EXE file) using instruction from the following tutorial: <a href="http://www.ni.com/tutorial/3303/en/#toc2" rel="nofollow noreferrer">Distributing Applications with the LabVIEW Application Builder: Stand-Alone Applications</a> . After that I created an installer for my application which I used to install the LabVIEW runtime engine and the drivers I needed on the cDAQ. You can find step by step instructions to create installer in the following link:<a href="http://www.ni.com/tutorial/3303/en/#toc5" rel="nofollow noreferrer">Distributing Applications with the LabVIEW Application Builder: Installers (Windows)</a>. </p>
<p>Now to cleanup space on the cDAQ, you need to uninstall LabVIEW completely. Go to the control panel in Windows 7 and follow this tutorial: <a href="http://www.ni.com/product-documentation/54335/en/" rel="nofollow noreferrer">Uninstall or Repair National Instruments Software or Drivers in Windows</a>. I have also used Temp File Cleaner which cleaned about 2 GB of space on cDAQ.</p>
<p>I hope it helps!</p>
| 23094 | How to resolve file space issue on CDAQ Controller with Labview Test System? |
2018-08-08T01:47:08.597 | <p>I am currently involved with a project using a mechanism similar to that seen in 'table tilters' and we are struggling to work out how to begin resolving the forces.</p>
<p>The image shows the state we are considering in equilibrium to resolve the forces. The plate (blue) is in a horizontal position and is 350mm in length</p>
<p>The linkages (red) are 270mm long and are at an angle of 20 degrees to the horizontal at this position.</p>
<p>The green component is restricted to motion in the vertical direction and is being pulled downwards with a force of 3000N. </p>
<p>We want to work out the downward force the linkages exert on the plate.</p>
<p><a href="https://i.stack.imgur.com/Oaw8H.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Oaw8H.jpg" alt="ISO view"></a></p>
<p><a href="https://i.stack.imgur.com/VATO7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VATO7.jpg" alt="Side View"></a></p>
<p>I started to try and break the system down into free body diagrams (below), but am struggling to make a start resolving the forces. Any advice would be greatly appreciated.</p>
<p><a href="https://i.stack.imgur.com/XfcP8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XfcP8.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/vdWBW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vdWBW.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering|statics|linkage| | <p>your sketches misses a bit of information. where is the resisting element?</p>
<p>Is it a post, or leg/legs under the table. Assuming the center of resisting forces falls right under the red bar on the point of contact. then the force on plate is same as 3 kN downward. If the resisting forces are not directly under the red bar then the downward force becomes bigger or smaller, depending on the location of resisting force.</p>
<p><strong>Edit</strong></p>
<p>After reviewing additional info, your table is receiving 3kN downward force. But you need to be mindful of horizontal forces at points green and yellow on the vertical; black bar. Those are $Fh = cos(20)/ Sin(20)\times 3kN = approx.\space 9 kN$</p>
<p>These two work as a couple trying to topple your table.</p>
| 23096 | Calculating force exerted on plate by link |
2018-08-08T04:09:57.077 | <p>I am trying to calculate the increase in pressure caused by liquid nitrogen at the moment it changes from liquid to vapor within a closed, constant volume at atmospheric pressure. Do I need to include the heat of vaporization? How can this be done?</p>
| |pressure|liquid|vaporization| | <p>You have some assumptions, so it's important to clarify the mechanism you are describing. I'll be assuming that you are describing a container filled with LN2 at the boiling point of LN2 under 1 atmosphere of pressure. Then this container is then left out in a room temperature environment while sealed off. The end result is - your container explodes. Proving it is complicated, but you wind up at pressures that far exceed most container's withholding pressure. While I know some thermodynamics, it's not my strongest area, so this may not be technically correct, but it gets a good ballpark estimate.</p>
<p>We begin by considering the phase diagram of nitrogen:</p>
<p><a href="https://i.stack.imgur.com/KMQuy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KMQuy.jpg" alt="Phase Diagram of Nitrogen showing critical point at 5 MPa and -150C"></a>
<a href="https://i.stack.imgur.com/ZPMTT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZPMTT.jpg" alt="ST Diagram of LN2"></a></p>
<p>Link <a href="https://www.researchgate.net/figure/The-temperature-pressure-phase-diagram-for-nitrogen_fig1_315888614" rel="nofollow noreferrer">here</a> and <a href="https://en.wikipedia.org/wiki/Enthalpy#Diagrams" rel="nofollow noreferrer">here</a> because images are broken as I write this. </p>
<p>We would begin our journey at point "g" on the ST diagram. The two phases can be considered by considering a fraction x at point "h" on the diagram, at gas phase. We put a permeable barrier between the two phases and begin a set amount of heat flow into the system.</p>
<p>As the internal energy increases, some portion at "g" crosses the barrier into "h". The result increases in pressure and entropy. As such, we move further up the dark red lines, since each phase can cross the barrier. After enough energy, the system reaches the critical point, at top of the dark red peak. I'm not sure how to write the equations to calculate how much energy or what the time it would take, but once we reach this point, we've reached supercritical nitrogen between 20-50 bars. This difficulty is resolved easily in practical engineering via the use of a <a href="https://en.wikipedia.org/wiki/Steam_drum" rel="nofollow noreferrer">steam drum</a>, a piece of process equipment that safely separates the phases while preventing <a href="https://en.wikipedia.org/wiki/Bumping_(chemistry)" rel="nofollow noreferrer">bumping</a>. However, advanced techniques today allow the resolution of multi-phase flow and multi-phase heat transfer.</p>
<p>From this point, the process would follow an <a href="https://en.wikipedia.org/wiki/Isochoric_process" rel="nofollow noreferrer">isochoric</a> process. While hard to describe using a TS diagram, a <a href="https://upload.wikimedia.org/wikipedia/commons/6/63/Temperature-entropy_chart_for_steam%2C_US_units.svg" rel="nofollow noreferrer">similar diagram</a> for steam will confirm that isochoric processes gain relatively little entropy compared to the pressure they gain with a rise in temperature. Doubling the temperature nearly has a 10-fold increase in pressure. With such a significant rise, the only conclusion I can arrive at is that you would be at a pressure that would be best measured in units of <a href="https://en.wikipedia.org/wiki/Orders_of_magnitude_(pressure)" rel="nofollow noreferrer">GigaPascals (GPa)</a>. In short, this would most likely destroy your container.</p>
| 23098 | How to calculate change in pressure from LN2 state change (constant volume) |
2018-08-08T10:23:01.693 | <p>I have a TEC-12706 thermoelectric peltier module. I am making a mini refrigerator (simple box refrigerator). I'm using this peltier and heatsink of equal size as compared with the peltier. I have given the picture of the heatsink attached with the hot side of the peltier. I'm facing a very big problem. While I connect this peltier with a DC source, it works properly for some time, i.e., the cold side of the peltier gets cool and the heatsink starts becoming hot. But just after a few time the peltier starts working in a different way. That is, the cold side of the peltier also gets some heated and the heatsink stops or reduces absorbing heat from the hot side of the peltier. I think the heatsink compound (or thermal paste) which I'm using is not proper. What do you think, what may be the problem? Please help me, I'm struck and am not able to make my refrigerator.!See this picture </p>
<p><a href="https://i.stack.imgur.com/FMpEh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FMpEh.jpg" alt="enter image description here"></a></p>
| |mechanical-engineering| | <p>Peltier modules are extremely inefficient and therefore need excessive heat dissipation. What you are experiencing is that the peltier module cannot dissipate all of the heat that it is moving (either from a heat load or from itself) and as a result it cannot maintain an appropriate delta-Temp.</p>
<p>I suggest adding a larger heat sink, a fan blowing over the heat sink (good 'ol convective cooling), or a water block. </p>
<p>See some options below:
Fan: <a href="https://www.newegg.com/Product/Product.aspx?Item=N82E16835230010&ignorebbr=1&nm_mc=KNC-GoogleAdwords-PC&cm_mmc=KNC-GoogleAdwords-PC-_-pla-_-CPU%20Cooling-_-N82E16835230010&gclid=EAIaIQobChMIgJCd8L7j3AIVDttkCh3MTw2lEAYYAiABEgJRHvD_BwE&gclsrc=aw.ds" rel="nofollow noreferrer">Click me!</a>
Water Block: <a href="https://customthermoelectric.com/water-block-3-0-x-3-0-x-0-85.html" rel="nofollow noreferrer">Click me!</a></p>
| 23102 | Thermoelectric Module Not Working Properly! |
2018-08-08T12:09:59.653 | <p>I'm looking for a name of a part. It looks like a cable inside another cable. The idea is that you can just pull on the inside cable and get the force at the other end. You can just attach the outside cable somewhere, and don't have to bother about pulleys and such. </p>
<p>Does this exists, and what is it called?</p>
| |mechanical-engineering| | <p>Also known as a "Push-Pull Cable" or "(mechanical) Control Cable"</p>
| 23103 | Looking for the name of a part |
2018-08-08T12:51:23.300 | <p>I already got great help on this topic, but I'd like to make sure that I get it.</p>
<p>For example, take a beam of length 6 meters, left side fixed support and right side rolled support. There is a load of 30 kN-m that starts at 3 meters and ends at 6 meters with a magnitude of 60 kN-m. </p>
<p>The function of the load is easy, f(x)=10x. The centroid is given by
$$\frac{\int_3^6 10x² dx}{\int_3^6 10x dx} = 4.67m$$</p>
<p>And the force translated into a point load is
$$ \int_3^6 10x dx = 135 kN $$</p>
<p>So the sum of forces as seen from a is 0, and can be used to calculate Mb
$$ \sum Ma =0 = 4.67*135 -6*Mb \Rightarrow Mb = 105 kN \Rightarrow Ma=35kN$$</p>
<p>Then to find the shear forces I integrate the equation from 3 to x taking into account Ma
$$ \int_3^x 10x dx -Ma = 5x² -75 $$</p>
<p>And to find the Bending moments I just integrate again, this time for the whole equation</p>
<p>$$ \int_0^x (5x²-75 )dx = \frac{5x³}{3} -75x $$</p>
<p>Alas this does not give the complete answer, I have many uncertainties, for the shear forces I notice that it only gives the correct answer for x between 3 and 6; for x under 3 it becomes Ma, and I imagine if the length of the beam would be greater than the distributed load's length, it would be Mb for x>6. If I would add a point load of 3kN at l=2, do I need to replace Ma with Ma'- 3kN in the shear force equation? What would happen with a point load after 6 meters (eg lenght of beam = 9 meters, point load at 7m)?</p>
<p>As for the bending moments, I kinda get the right answer for x between 3 and 6, but it is 90 kN too much, which is the BM at the start of the distributed load. For x=2 for example I get completely wrong answers, and I need to revert to calculating BM by using trigonometry to calculate the actual area under the shear force curve. </p>
<p>I mean I can solve them, but I don't get the why - why can't I use the BM function to calculate all of x, considering the boundary condition should have been solved in the first integral (from 3 to 6 ). Why do I need to substract the BM at the beginning of the distributed load,... </p>
<p>I would greatly appreciate if someone would take the time to enlighten me - as it is I'm not really confident for the exam (I can't afford to fail) </p>
| |mechanical-engineering|moments|shear|diagram| | <p>I am a bit confused here, so I went under the assumption you have a statically undetermined beam with a fixed support (moment resisting) on the left and a roller support (non-moment resisting) on the right. In either case, this method still applied, but the constants become significantly easier in the case where it is a pinned support (non-moment resisting, but fixed in x-axis) on the left side. I'll describe the more complicated case and you can see how it can simplify to the less complicated case.</p>
<p>Seeing how you want a method that gives the complete picture for all of x, I will suggest what I personally use, and what books such as <a href="https://rads.stackoverflow.com/amzn/click/0071742476" rel="nofollow noreferrer">Roark's formulas for stresses and strains</a> uses, the <a href="https://en.wikipedia.org/wiki/Macaulay_brackets" rel="nofollow noreferrer">Macaulay Brackets</a>. These are defined as follows:</p>
<p>$$\langle x-a \rangle^n = \begin{cases} 0, & x < a \\ (x-a)^n, & x \ge a. \end{cases}(n \ge 0)$$</p>
<p>Essentially, you can ignore what is in the bracket when x is less than a, and then start from 0 when x is greater than a. It is even defined at n=0 to be equal to the <a href="https://en.wikipedia.org/wiki/Heaviside_step_function" rel="nofollow noreferrer">heaviside step function.</a> This forms <a href="https://en.wikipedia.org/wiki/Macaulay%27s_method" rel="nofollow noreferrer">Macaulay's method.</a></p>
<p>Let's redefine your load function:</p>
<p>$$ w(x) = 30 \frac{kN}{m} \langle x-3 \rangle^0 + 10 \frac{kN}{m^2} \langle x-3 \rangle^1$$</p>
<p>This looks a lot more complicated then it did before. But the same theory applies:</p>
<p>The centroid of the load is still at $\frac{\int w(x)*x dx}{\int w(x) dx} $ and has a value equal to $\int w(x) dx$. We then need to work on the remaining equations. The reaction forces at each end are - actually unknown if this is a statically indeterminate beam as described above.</p>
<p>We can determine the shear and bending moment in the beam throughout the section simply by integrating w(x):</p>
<p>$$V(x) = 30\langle x-3 \rangle^1 + 5\langle x-3 \rangle^2 + F_a$$
$$M(x) = 15\langle x-3 \rangle^2 + \frac{5}{3}\langle x-3 \rangle^3 + F_a*x + M_a $$</p>
<p>We then can divide by $EI$, the beam property, and continue to integrate to determine the slope and deflection:</p>
<p>$$EI\theta(x) = 5\langle x-3 \rangle^3 + \frac{5}{12}\langle x-3 \rangle^4 + \frac{F_a}{2}x^2 + M_ax + C_1 $$
$$EIy(x) = \frac{5}{4}\langle x-3 \rangle^4 + \frac{1}{12}\langle x-3 \rangle^5 + \frac{F_a}{6}x^3 + \frac{M_a}{2}x^2 + C_1x + C_2 $$</p>
<p>We see immediately that $C_2 = 0$ because the deflection at a is 0.
We then see that $C_1 = 0$ because the slope at a is 0. Finally, we can solve for the reactions. We can use the fact that the moment at b is 0 on M(x) to get a single equation:</p>
<p>$$M(6) = 0 = 15\langle 6-3 \rangle^2 + \frac{5}{3}\langle 6-3 \rangle^3 + F_a*6 + M_a $$
$$ 0 = 180 + 6*F_a + M_a $$</p>
<p>We then use the deflection equation for the last equation:</p>
<p>$$EIy(6) = 0 = \frac{5}{4}\langle 6-3 \rangle^4 + \frac{1}{12}\langle 6-3 \rangle^5 + \frac{F_a}{6}6^3 + \frac{M_a}{2}6^2$$</p>
<p>$$0 = 101.25 + 20.25 + 36F_a + 18M_a$$
$$0 = 121.5 + 36F_a + 18M_a$$</p>
<p>This can then easily be resolved into $F_a = -43.3125$, $M_a = 79.875$. $F_b$ follows shortly thereafter.</p>
| 23110 | Shear force and bending diagram un-uniformly distributed load |
2018-08-08T20:43:38.307 | <p>I have a 2D lidar mounted to a plate on a robot. I am looking for a method to adjust the pitch and roll of the plate so the plane of the laser is parallel to the ground.</p>
<p>Currently, I am using thin metal shims, but I want to find an <em>easier</em> method that will last. I was thinking about using set screws, but I've been told that it's a bad idea because they will inevitably come loose due to vibrations.</p>
<p>What other methods exist for adjusting the roll and pitch of a mounting plate? Are there any specially engineered solutions for this?</p>
<p>For reference, I'm using a Hokuyo 20Lx lidar.
<a href="https://i.stack.imgur.com/VpSfn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VpSfn.png" alt="enter image description here"></a></p>
| |mechanical-engineering|calibration| | <p>A simple solution would be :</p>
<ul>
<li><p>Two matching square plates with an M8 hole in each corner.</p></li>
<li><p>Sandwich a bit of moderately thick rubber, 10mm thick maybe, with matching holes between the two plates </p></li>
<li><p>arrange whatever mounting points you need for the lidar and vehicle in the top and bottom plates </p></li>
<li><p>Bolt through the holes with bolts any nyloc nuts </p></li>
<li><p>adjust the bolt tension to level the top plate, the rubber will take up the slack and </p></li>
</ul>
<p>A slightly more sophisticated version is to drill a hole in the centre of the plates whcih you use to locate a ball bearing to act as a pivot point. </p>
<p>You could also drill through some (soft) ball bearings and use them as sort of spherical washers in conjunction with countersunk holes on the top plate to get better location for the bolt heads. </p>
<p>If you need a very large range of adjustment you could use something like this (made from modified pipe flanges) possibly substituting a bolt any nyloc for the Bristol lever, in fact with a smaller ball bearing (or no ball and a robber block or springs that is similar to what I described above. :</p>
<p><a href="https://i.stack.imgur.com/Leolc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Leolc.jpg" alt="enter image description here"></a></p>
| 23130 | Method for permanent adjustment of pitch and roll of a mounting plate |
2018-08-09T01:03:32.013 | <p>I was looking at some papers related to moving bed Biofilm reactor and it says that it can be used both in anoxic and oxic condition. So, I want to know what are the advantages of treatment plants that can work in both anoxic and oxic conditions?</p>
| |waste-water-treatment| | <p><strong>TLDR: Treatment technologies that can function in oxic and anoxic conditions are more flexible and can be applied in more circumstances.</strong></p>
<p>First, I think clearing up some terminology would be helpful. These terms are not super specific, but this is how they are generally used in my experience.</p>
<p>A <strong>Treatment plant</strong> is a facility which receives wastewater, treats it, and discharges it somewhere. It is not oxic or anoxic, but it has <strong>treatment processes</strong> which can be oxic or anoxic. Most plants have numerous processes, including both oxic and anoxic (and anaerobic) processes for treatment of different things. A treatment process uses a specific <strong>treatment technology or technologies</strong> to accomplish its goals. A <strong>moving bed biofilm reactor (MBBR)</strong> is a specific technology involving suspended media to host biofilm growth.</p>
<p>MBBR tanks can be aerated (oxic conditions) or un-aerated (anoxic conditions) depending on the goals of the treatment process. The fact that the technology can be applied in both kinds of processes doesn't usually* benefit the process itself. It does benefit the manufacturer because their potential market is larger however. This does not mean this is just marketing hype; it serves to clarify for customers the potential applications.</p>
<p>*I say doesn't usually benefit the process, but there are some plants that have "swing" zones that can be operated either oxically or anoxically depending on the influent conditions. I don't know of an MBBR that is operated this way, but I don't see why it couldn't be. In this case, the flexibility helps the plant adapt to the current conditions.</p>
| 23133 | What are the advantages of wastewater treatment technology that can work in both anoxic and oxic condition? |
2018-08-09T08:30:34.147 | <p>There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?</p>
<p>On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the "working" walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?</p>
<p>More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:</p>
<p><a href="https://i.stack.imgur.com/ewW7N.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ewW7N.jpg" alt="enter image description here"></a></p>
<p>I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?</p>
| |structural-engineering|materials|steel|strength| | <p>Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.</p>
| 23135 | How does the strength of a square tube depend on the side size? |
2018-08-10T06:31:40.357 | <p><a href="https://www.multitran.ru/c/m.exe?a=4&MessNum=349503&l1=1&l2=2" rel="nofollow noreferrer">A fellow translator is translating a list of parts into Russian</a>, and can't understand the meaning of "female eye". Here's the table: </p>
<blockquote>
<p><a href="https://i.stack.imgur.com/2mWhw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2mWhw.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>Maybe someone here could come up with some ideas about the possible meaning of this "female eye". </p>
| |terminology| | <p>The operative engineering term is "screw" (no pun intended).</p>
<p>The screw itself is the "male" part, and the shaft into which a screw is inserted and rotated is a female part. The "eyes" referred to are "receivers" for screws, and are therefore categorized as "female".</p>
<p>The terminology is derived, of course, from animal anatomy, in which females are the recipients of "screw"-like actions. </p>
| 23149 | Meaning of "female eye" in a list of parts |
2018-08-10T08:43:41.383 | <p><a href="https://i.stack.imgur.com/Hzopt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hzopt.jpg" alt="enter image description here"></a>A homogeneous rigid circular disc (radius R, mass ms, center point S) rolls without a slip on a plain with a small angle of inclination alpha against the horizontal. A point mass MG is fixed on the disc in a distance e to the center point S. The motion of the disc can be described by the angle phi between the vertical and the connection line SMG. </p>
<p>Please see the figure for reference. </p>
<p>I am trying to calculate Kinetic Energy and Potential energy of the given problem (pic #2) and I have considered both masses as one system but I am not sure if that's correct or not. Also, I am not sure if the h term in potential energy (mgh) is correct here. So I would be glad if someone could throw some light on this. Thanks in advance!</p>
| |mechanical-engineering|applied-mechanics|vibration|mechanical|linear-motion| | <p>The calculations (in Mathematica) with minimal explanations. Hopefully you can pick it up from here.</p>
<p><a href="https://i.stack.imgur.com/D0c2E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D0c2E.png" alt="enter image description here"></a></p>
| 23154 | Analytic expressions of the kinetic and potential energy |
2018-08-10T17:46:18.170 | <p>I need some help with a statics design problem. I am tasked with designing a solar panel frame to mount on a building and am the only engineer on staff so I don't have anyone else to ask the simple dumb questions. Since I have no real experience doing wind load calculations I figured I would assume a load normal to the solar panel face. My idea is to use two square aluminum tubes mounted along the long side of the three solar panels stacked vertically along the tube. My analysis involves the support structure for one side, of one array column. As of right now I am hoping the solar panels themselves will suffice as cross-member connections. The details really don't matter, in fact the wind loading really doesn't matter either. It is the method I am having trouble with. </p>
<p>Want it to be able to withstand 125 MPH gusts. That appears to equate to 30 lbf/square foot. Given 18.47 sq. ft./ panel * 30PSF * 3 Panels + 135 lb panel wt = 1797 lbf. That's 1800 lbf total distributed load accross all three solar panels in one array column. Since there will be two support structures per array column that's 900 lbf distributed load per support structure or 4.48 lbf/inch. (Not sure where I got 6.2 on my drawing)</p>
<p>I know I need to determine the reactions at each pin joint, do shear and moment diagrams and then I will do a buckling failure analysis on each support strut member. Any guidance would be appreciated, started doing some calcs on my own and things just weren't coming out right.</p>
<p><a href="https://i.stack.imgur.com/DFSVZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DFSVZ.jpg" alt="enter image description here"></a></p>
| |structural-engineering|structural-analysis|statics| | <p>Just FYI I was able to solve my design using RISA 3D. This is an excellent software! Ended up designing for a 125 MPH / 40 PSF wind load. Tested the frame in both horizontal and 50 degrees from horizontal orientations. Test loads were applied normal to the panel face and parallel to the ground in two different scenario groups. Loads were applied downward on all panels at 40 PSF, upward on all panels at 40 PSF, combination load: upward on the top panel / downward on the two lower panels, and the reverse of that combination load. </p>
<p>For the moment I am neglecting snow loads based on the fact that as you move north the required solar panel tilt angle becomes greater up to about 45 degree slope. And where the solar panel array would be most vulnerable (near horizontal) is in regions with zero snow load. But I will have to revisit this topic later.</p>
<p>Ended up using 6061-T6 3x2.03 I-beams for the long side members. Was even able to eliminate two support struts in this configuration.</p>
<p>Vortex shedding may be a problem however depending on how the assembly of frames is arranged. Structure frequency is 7 to 77 Hz for the first five modes. Based on the Strouhal number vortex shedding frequencies could range from 11 to 26 Hz. However when I analyze the shedding frequency across the assembly of all six planned frames the range drops to 2.2 to 2.7 Hz. Some type of vortex suppression devices may be required on the long struts.</p>
<p><a href="https://i.stack.imgur.com/KwVqq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KwVqq.png" alt="enter image description here"></a></p>
| 23160 | Solar Array Frame Design / Statics Design Problem |
2018-08-10T18:28:44.153 | <p>I understand that the lead ions can be rendered into a precipitate with chloride ions in water treatment, but why on earth does the U.S. still use lead pipes at all when contamination poses an enormous risk? </p>
| |waste-water-treatment| | <p>Lead pipes have not been used in new construction in the USA for many decades. The problem is with old buildings and old water lines which may still be equipped with lead pipes if they have not been upgraded since new. Copper pipes for residential and commercial construction were common for the last 50 years or so and since they were assembled on-site using lead solder, it is possible to produce lead contamination of water flowing through them even if the copper pipes themselves contain no lead. </p>
| 23161 | Why does the U.S. still use lead pipes? |
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