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2020-05-08T23:51:07.860 | <p>A fan can run at 600 - 2000 RPM, at 2000 RPM it has 50 m<sup>3</sup>/h airflow.</p>
<p>0% = 0 RPM</p>
<p>100% = 2000 RPM</p>
<p>At 50% = 1000 RPM is the airflow halved to 25 m<sup>3</sup>/h?</p>
| |fluid-mechanics|airflow| | <p>Yes it does. Check this <a href="https://www.axair-fans.co.uk/news/applications/understanding-basic-fan-laws/" rel="nofollow noreferrer">link</a>.</p>
<blockquote>
<p>The First Fan laws: Volume of Air</p>
<p>The first law of fans is a useful tool when working out the volumetric
flow rate supplied by a fan under speed control or conversely working
out what the RPM would be to deliver a required volume of air and
hence what frequency to set a variable speed drive (VSD) to.</p>
<p>Volumetric flow rate (V, m³/hr) varies directly proportional to the
ratio of the rotational speed (RPM) of the impeller.</p>
<p><span class="math-container">$V_2 = (\frac{U_2}{U_1})V_1$</span></p>
<p>Where:</p>
<p>: Volume 1, m³/hr – Original volume of air</p>
<p>: Volume 2, m³/hr – New Volume of air</p>
<p>: RPM 1 u/min – Original Speed</p>
<p>: RPM 2, u/min – New Speed</p>
</blockquote>
| 35642 | How does fan airflow relate to its RPM setting? |
2020-05-09T05:47:13.873 | <p>When making structural analysis with FEM, do we normally use plate or shell elements to represent slabs? I checked and did some reading for the difference between shell and plate elements in FEM in general but particularly I am trying to understand the differences between the two as far as representing a slab in a structure such as a building or bridge. </p>
<p>Also does the type of slab make a difference, such as RC concrete, metal deck or post-tensioned? </p>
<p>Also, the difference should also depend on the software we use right? I mean as far as how the calculations are formulated in each software, so may be there is no definite answer to this and we go case by case for each software?</p>
| |structural-engineering|structural-analysis|finite-element-method| | <blockquote>
<p>When making structural analysis with FEM, do we normally use plate or shell elements to represent slabs?</p>
</blockquote>
<p>You really need to specify what software you are using. The names "plates" and "shells" may be used for very different finite element formulations. Another name sometimes used is "membrane element". Since you have specified the "structural engineering" tag, you are probably using something like STAAD, ETABS, SAP2000, MIDAS, LS-Dyna or Oasys GSA, and your shell and plate formulations are probably both 2D elements (<strong><em>not 3D</em></strong>) - since they have only 2D geometrical connectivity, and the formulation typically is based on shape functions based on two-dimensional local coordinates. The difference between "plates" and "shells" will depend upon the software you are using, but typically there are three types of behaviour in most structural analysis software:</p>
<ol>
<li>In-plane properties only - sometimes called membrane elements as
they have no flexural stiffness. They can be used for shear wall
analysis - the benefit for design calculations being that they do
not generate minor axis shears or bending moments. This may be a
disadvantage in some cases though. They should not be used for slab design (although they may be used as diaphragm elements to tie lateral system elements together).</li>
<li>Out-of-plane properties only - sometimes called grid or plate elements,
and are used for flexural analysis (I don't think these are commonly
used these days). These may be used for design of floor slabs without significant in-plane forces,
with
the advantage being that they don't generate axial loads, which
makes design easier. However, in some cases, this can be a
disadvantage - they cannot be used to model slab and beam acting
compositely. Also they will not provide diaphragm action to tie lateral systems together (shear walls, braced frames and moment frames).</li>
<li>In-plane AND out-of-plane properties - these are more general than the first two, and may be used for wall and slab and beam and
slab design. However, the engineer must be aware of the larger
matrix of forces and moments that they carry and will need to
design for them. In their most basic formulation they do not account for through thickness shear or "drilling moments" at the nodes, although some do (for example the Allman-Cook formulation). You will need drilling stiffness if you want to directly connect 1D beam elements to your 2D mesh at a node.</li>
</ol>
<p>You may also come across "thick shell formulations" which explicitly account for through thickness shear (good for transfer plates and flat slabs). </p>
<p>For slab design you will also need to look out for formulations or tools that convert torsional moments (Mxy and Myx) into direct moments - such as by means of the Wood and Armer method. If you ignore these you may under-design your slabs for flexure. Alternatively you can look for design software that creates strips for design (they will handle the torsional moments I mentioned). </p>
<p>There are other formulations for 2D elements - depending on whether they have linear or parabolic shape functions, or even more exotic. Note that I have not covered plane stress, plane strain or axisymmetric 2D element types.</p>
<p><strong><em>STRONG REMINDER: You need to check the user manual of the software that you are using in order to fully understand what the implementation is in that particular software.</em></strong> </p>
<p>Resources:</p>
<ul>
<li><a href="https://www.oasys-software.com/help/gsa/10.0/GSA_Manual.pdf" rel="nofollow noreferrer">https://www.oasys-software.com/help/gsa/10.0/GSA_Manual.pdf</a> (section
3.6.1 from page 78 onward) </li>
<li><a href="https://www.oasys-software.com/help/gsa/10.0/GSA_Theory.pdf" rel="nofollow noreferrer">https://www.oasys-software.com/help/gsa/10.0/GSA_Theory.pdf</a> (from
page 61)</li>
<li><a href="https://ocw.mit.edu/resources/res-2-002-finite-element-procedures-for-solids-and-structures-spring-2010/nonlinear/lecture-19/MITRES2_002S10_lec19.pdf" rel="nofollow noreferrer">https://ocw.mit.edu/resources/res-2-002-finite-element-procedures-for-solids-and-structures-spring-2010/nonlinear/lecture-19/MITRES2_002S10_lec19.pdf</a></li>
</ul>
| 35644 | Representing slabs in FEM |
2020-05-09T06:13:23.657 | <p>What is the difference between a control joint and a contraction joint in concrete? I mean, a control joint is to control the cracks in concrete due to temperature or shrinkage etc... I understand it. But then what is a contraction joint? I saw this term when I was reading about dams, when they pour large concrete masses and make contraction joint to account for shrinkage. Why don't they also call it a control joint?</p>
| |structural-engineering|concrete| | <p>They're the same thing. See the link below for an article by the Portland Cement Association. Just different vocab from different regions of the country. Another common example of this in structural engineering is mat slab versus raft slab. They're synonyms. </p>
<p><a href="https://www.cement.org/learn/concrete-technology/concrete-construction/contraction-control-joints-in-concrete-flatwork" rel="nofollow noreferrer">https://www.cement.org/learn/concrete-technology/concrete-construction/contraction-control-joints-in-concrete-flatwork</a></p>
| 35645 | Control joint and contraction joint in concrete |
2020-05-09T12:40:24.107 | <p>A PC case is basically a rectangular pathaway for air to flow through but at the places where the air enters and exits there are solid plastic or metal panels commonly with meshes and little holes.</p>
<p><a href="https://i.stack.imgur.com/q7MNb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q7MNb.jpg" alt="enter image description here"></a></p>
<p>When a fan is rated at 50cfm or 80m3/h at 100% speed by how much is this figure decreased when taking into account the characteristics of the volume of the case?</p>
<p>Is it 20% or 80% for example?</p>
| |fluid-mechanics|airflow| | <blockquote>
<p>When a fan is rated at 50cfm or 80 m<sup>3</sup>/h at 100% speed by how much is this figure decreased when taking into account the characteristics of the volume of the case?</p>
<p>Is it 20% or 80% for example?</p>
</blockquote>
<p>The volume of the case will have nothing to do with the airflow. The resistance to airflow will and this will typically be determined by the minimum cross-sectional area of the airflow path.</p>
<blockquote>
<p>What is the minimum cross-sectional area of the airflow path?</p>
</blockquote>
<p>If it's not given you would have to work it out by measurement.</p>
<p>For a given airflow rate the volume of the case will determine the number of air changes per unit time.</p>
<blockquote>
<p>What is the number of air changes per minute?</p>
</blockquote>
<p>If the volume of the cabinet is, for example, 0.5 m<sup>3</sup> and the airflow is 80 m<sup>3</sup>/h then the number of air changes per hour will be given by <span class="math-container">$ n = \frac f v = \frac {80}{0.5} = 160 $</span> (where <em>f</em> is the airflow and <em>v</em> is the volume. 160 changes per hour is almost 3 changes per minute.</p>
<p>Any filters or blockages will reduce the airflow.</p>
| 35656 | At what efficiency does a fan cool a PC case? |
2020-05-09T13:36:23.357 | <p>I'm interesting in static* roll stability of submarines. I'm always tempted to compare submerged submarine to blimps and other bigger airships. For airships, roll stability is assured by a <a href="https://aviation.stackexchange.com/a/24866/3394">pendulum effect</a>, the heavier components being located low into the structure **. I have no idea of how the masses are spread inside a submarine, but I imagine they are located such that the center of gravity is as low as possible to improve stability.</p>
<p>When submerged, I imagine the center of gravity varies as ballasts can contain a significant quantity of water. <a href="https://www.reddit.com/r/WarshipPorn/comments/352emn/graphic_virginiaclass_submarine_cutawayview_and/" rel="nofollow noreferrer">Cutaways</a> <a href="https://www.dailymail.co.uk/sciencetech/article-2203110/Deadly-Hunter-Killer-submarine-capable-hearing-ship-leaving-port-New-York--sat-underwater-English-channel.html" rel="nofollow noreferrer">I found</a> show ballast tank extend over a significant height of the submarine. Filling them may thus bring the center of gravity closer to the geometrical center of the submarine. This is only an assumption I fail to confirm.</p>
<p>How does the location of the center of gravity change when filling the ballast? </p>
<p>The consequences of this variation on roll stability is out of the scope of this question as the geometry of the immersed part of the hull also change.</p>
<hr>
<p><sub>
* i.e. when the submarine is not moving<br/>
** cutaway found on <a href="https://www.airships.net/lz-130-graf-zeppelin/]" rel="nofollow noreferrer">this website</a> give a good overview of where payload was located.
</sub></p>
| |submarines| | <p>From the picture that I found from this <a href="https://books.google.com/books?id=IHC_AwAAQBAJ&dq=Variable+weights+in+a+submarine" rel="nofollow noreferrer">book</a>, the answer seems to be yes.
<a href="https://i.stack.imgur.com/1FBFH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1FBFH.png" alt="change in the different centers of a submarine"></a> </p>
<p>In regards with stability of submarines, there are two main terms involved; center of gravity and center of buoyancy. Center of gravity is the point on the body where gravitational force is acting and, Center of buoyancy is center of gravity of the volume of water the submarine displaces. As quoted from this <a href="http://web.mit.edu/12.000/www/m2005/a1/Robotics/ballast.htm" rel="nofollow noreferrer">MIT page</a> on Submersible Ballast. </p>
<blockquote>
<p>There are two very important constraints on the centers of mass and
buoyancy which a submersible vessel must obey:</p>
<ol>
<li><p>The center of buoyancy and the center of gravity of the vessel must lie on a common vertical line. This is true because were these to lie
on different vertical lines, this would produce a torque which would
rotate the vessel.</p></li>
<li><p>The center of gravity of the vessel must be below the center of buoyancy. It is common knowledge that dense objects sink in fluids
with lower densities, and this fact can be used to infer the previous
requirement. If the center of gravity of the vessel is below its
center of buoyancy, this means that the majority of the weight rests
below the center of buoyancy. This implies that the volume of the
vessel below this point is denser than the same volume of displaced
water, and the volume above this level is less dense than the same
volume of displaced water. Thus the lower portion of the vessel will
tend to sink while the upper portion will tend to rise, preventing the
vessel from rolling</p></li>
</ol>
</blockquote>
<p>To understand more about the different submersion methods of a submarine, check this <a href="https://www.quora.com/Submarine-ballast-tanks-are-open-at-the-bottom-Doesnt-this-make-their-submerged-buoyancy-unstable-Wouldnt-a-slight-downward-error-cause-increased-flooding-leading-to-further-downward-motion" rel="nofollow noreferrer">link</a>.</p>
<blockquote>
<p>Basically, there are two ways to submerge a boat: dynamic diving and
staticdiving. Many model submarines use the dynamic method while
static diving is used by all military submarines. Dynamic diving boats
are submarines that inherently float that is, they always have a
positive buoyancy. This type of boat is made to dive by using the
speed of the boat in combination with the dive planes to force the
boat under water. This is very similar to the way airplanes fly.
Static diving submarines dive by changing the buoyancy of the boat
itself by letting water into ballast tanks. The buoyancy is thereby
changed from positive to negative and the boats starts sinking. These
boats do not require speed to dive hence this method is called static
diving.</p>
<p>Modern military submarines dive use a combination of dynamic and
static diving. The boat submerges by filling the main ballast tanks
with water. After that, the buoyancy is accurately adjusted with the
trim tanks. Once underwater, the depth of the boat is controlled with
the hydroplanes.</p>
</blockquote>
<p>And this <a href="https://www.marineinsight.com/naval-architecture/submarine-design-unique-tanks-submarine/" rel="nofollow noreferrer">link</a> has great details on the different ballast tanks used in a submarine.
<a href="https://i.stack.imgur.com/cVCQe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cVCQe.png" alt="The different tanks inside a diesel submarine"></a></p>
<p>And to understand more about the stability of a submarine and the working of a ballast tank, check out this <a href="https://www.marineinsight.com/naval-architecture/understanding-stability-submarine/" rel="nofollow noreferrer">link</a>.</p>
| 35659 | How do center of gravity varies in submarine between surface and submerged? |
2020-05-09T14:08:21.213 | <p>AA batteries have about 4 watt-hours (14400 joules, 10620.9 pound-feet).</p>
<p>Ignoring losses, does that mean a AA battery could theoretically lift 10,620.9 pounds, 1 foot high?</p>
<p>You'd need an electric motor, some sort of crane or lift, the right gearing, and a fair amount of time. What's the best efficiency you could hope for? Even with 50% loss, you should still be able to lift a pickup truck, but this feels wrong (impossible).</p>
| |mechanical-engineering|electrical-engineering|energy| | <p><a href="https://i.stack.imgur.com/ipDL3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ipDL3.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Extract from <a href="https://data.energizer.com/pdfs/e91.pdf" rel="nofollow noreferrer">Energiser E91</a> datasheet.</em></p>
<p>The datasheet shows that at 100 mA you'll get about 2500 mAh (2.5 Ah) if you discharge to 0.8 V (and it doesn't get hot). Let's say that the average voltage is 1.25 V and we get the total useful energy is about 2.5 × 1.25 = 3.125 Wh = 3.125 × 3600 = 11250 Ws or 11250 J. Your calculation is the correct order of magnitude.</p>
<p>At 100 mA the time to discharge will be 25 hours.</p>
<p>I'm not going to use Imperial units so we'll go with 5000 kg. From <span class="math-container">$ E = mgh $</span> we get <span class="math-container">$h = \frac E {mg} = \frac {11250}{5000 \times 9.81} = 0.23 \ \text m $</span>.</p>
<p>So, again your calculations are about right in theory. The problem is that it will be a 0.15 W motor running for 25 h and I can't think of any gearbox that will be capable of supporting 5000 kg and will run at 0.15 W.</p>
| 35660 | Understanding the total energy in a AA battery |
2020-05-10T07:56:40.160 | <p>I have seen plenty of YouTube videos in which they use carbon fiber fabric to build various things. One such channel is Mike's Patey's channel where he makes amazing aircrafts from carbon fiber.</p>
<p>In these videos I usually see (but not always) that after they apply the resin to the fiber fabric, they put the structure in a kind of plastic bag and they vacuum pump it.</p>
<p>Why is that process? Is it hardening and giving some extra rigidity to the fiber in any way? Can this process be done by the use of pressure? (eg. a kind of mold that pressurizes the fiber from the outside so that the fiber also takes the correct shape).</p>
| |materials|material-science|carbon-fiber| | <p>Solar Mike's answer is accurate. </p>
<p>Carbon fiber has a resin to fiber ratio which provides the optimum strength. This is typically measured by weight. The amount of resin is applied to the fiber prior to enclosing it for vacuum application.</p>
<p>Once the vacuum begins, all of the air is removed from the fiber, forcing the resin into the voids, ensuring the strength aspect.</p>
<p>The pressure aspect will not remove the air from the enclosure containing the fiber. The voids within are going to be smaller but not non-existent. Another reason for vacuum bagging as a system is that it is not restricted to any specific container. Sufficient plastic and sealing will allow fiber constructions of immense size.</p>
| 35675 | Why do they vacuum pump the carbon fiber fabric? |
2020-05-10T14:18:45.617 | <p>Is it possible to make a oblique wing just by skewing a profile of a standard wing?</p>
<ol>
<li>Take a wing profile from web</li>
<li><strong>Divide it in sagittal slices</strong></li>
<li><strong>Move slices along a new axis</strong> (still lying on horizontal plane)</li>
<li><S>Scale slices around wind direction to keep wing horizontal section of same surface</S></li>
</ol>
<p>Note that the slices, stay vertical. It Is a skew, not a rotation. Skewing means cutting the figure in infinitely many slices, and the translate vertically each slice.
Like this</p>
<p><a href="https://i.stack.imgur.com/UpVAS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UpVAS.jpg" alt="pencils skewed to oblique arrangement, and straight pencils"></a></p>
<p>The drawing illustrates what I mean:
<a href="https://i.stack.imgur.com/r9Z4k.jpg!" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r9Z4k.jpg!" alt="sketch I did when trying to think on how to make a wing profile"></a></p>
<p>The 3 sketches are the wing, seen from above.</p>
<p><strong>Edit:</strong></p>
<p>I removed the scaling part, since skewing preserve surface, as long as width stay constant</p>
<p><a href="https://i.stack.imgur.com/cSKUu.png!" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cSKUu.png!" alt="sketch file"></a></p>
| |aerodynamics| | <p>Oblique winged aircraft have been investigated as demonstrated by the <a href="https://en.wikipedia.org/wiki/NASA_AD-1" rel="nofollow noreferrer">NASA AD-1</a>. That said, the whole reason for an oblique wing arrangement is to provide good performance at both low speeds and at supersonic speeds. At low speeds a normal straight wing is best. At transonic speed and supersonic speeds, swept wings have advantages. An oblique wing only makes sense if the wing can be rotated from straight at low speeds to skewed at high speeds.</p>
<p>Not to be smug, but if you are asking this question, you aren't designing an airplane that is going to approach the speed of sound. For any practical homebuilt or model airplane, an oblique wing is going to perform less well than a straight wing even if you plan on rotating the wing. If you are planning on rotating the wing, then there is no particular reason the wing ribs need to be parallel to the airflow. The only effect of the "skewing" you are proposing is to make the airfoil section proportionally thinner as the thickness stays the same but the chord length increases.</p>
<p>In any case, questions about aviation and airplane design are better asked at <a href="https://aviation.stackexchange.com/">https://aviation.stackexchange.com/</a>.</p>
| 35680 | How to make a oblique wing without areodynamics skill? |
2020-05-11T06:40:12.803 | <p>A <a href="https://www.designingbuildings.co.uk/wiki/Types_of_raft_foundation#Blanket_raft" rel="nofollow noreferrer">blanket raft</a> is a type of solid slab raft consisting of deeper ‘toe’ sections running around the edge of the foundation as well as a central deeper ‘beam-like’ section which offer rigidity to the slab. The blanket raft in question is 250 mm deep with top and bottom A393 reinforcing mesh supported by 400 mm deep 600 mm wide toes with 4 no. top and 4 no. bottom 20 mm diameter bar reinforcement and 16 mm diameter shear links (see below).</p>
<p><a href="https://i.stack.imgur.com/Z6qvt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z6qvt.png" alt="slab elevation"></a></p>
<p>This blanket raft is to be extended to accommodate additional plant, however, the average loading per square metre is to be less. Consequently, the same depth of 250 mm and reinforcement detail can be used with two layers of A393 mesh top and bottom. The primary difficult depends on where to place the deepened ‘beam-like’ sections and how to achieve fixity between the existing foundation and its extension. The proposed method to achieve this is to scabble back the edge of the existing slab (to achieve good bond with the existing concrete) and to drill and dowel the new rebar into existing slab. By way of a concept design, I have proposed the location of the deepened 400 mm deep by 600 mm wide sections of the slab extension.</p>
<p><a href="https://i.stack.imgur.com/TSuvE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TSuvE.png" alt="plan on proposed extension"></a> </p>
<p>Is this a good layout for the deepened sections of the extension to the blanket raft?</p>
<p>Is it necessary to put a deepened toe on the extension immediately adjacent to the deepened toe of the existing slab?</p>
<p>Should the dowels be epoxied in or greased in (do they only need to function in shear)?</p>
<p>Can you suggest what are the primary considerations during the concept design stage?</p>
| |reinforced-concrete|foundations| | <p>Foundation design starts from soil and geology and ends to the structure, live and dead loads.</p>
<p>First one needs to investigate the soil, its competency, stratification, any subterrainian water level, proximity to fault lines. Underlaying geological conditions, etc.
Then you can start to thing of what type of foundation best answers the particular situation your project has.</p>
<p>Many cities and town's and building authorities require an applicant for a building permit to hire a soils engineer to investigate the soils and advise the owner on the type of foundation required.</p>
<p>In some cases they have their own geology departments and will check the site and recommend the type and details of the foundation.</p>
<p>Your first step is to check with your building department. </p>
| 35689 | Concept design for an extension to a blanket raft slab |
2020-05-12T06:47:38.380 | <p>In your experience, what items that should be in geotechnical reports are sometimes missing, and what do you do in that case, can you provide example? Do we look at codes or make best guess and based on what? Or do we need to make our own soil testing? Please give example / examples from your experience or what you know, if you could...</p>
<p>Edit: This question was closed by the moderators stating that it is opinion based. However, this is the nature of this question. Missing items in geotechnical reports occur often and the only way to discuss this is through opinion of experienced engineers. See the answer below, which was given before the moderators closed the question, which provides a very informative answer. This question should stand open which would provide valuable information to readers. This question is a useful discussion for geotechnical and structural engineers. Asking for typical missing items can not be a fact based question in nature in the first place. Experiences are also valuable. </p>
| |structural-engineering|civil-engineering|structural-analysis|geotechnical-engineering|soil| | <p>A structural engineer should never conduct their own soil testing unless being supervised by a competent person. This is not your area of expertise, let the geotechnical engineer do their job. </p>
<p>Generally the geotechnical engineer should outline any risks specific to the site that way other engineers can design for them. An example of this would be hydrostatic pressure, radon in the soil, expansive soils, etc. </p>
<p>However I have come across geotechnical reports that lack content. Usually the report specifies their recommended foundation system (1-2 different types) based on the soil conditions on site. They may also understand that certain engineers tend to use nonconventional foundation systems (such as a post-tension slab on grade) and cater the report specifically to that engineer. </p>
<p>Minimum info to be included in the reports for each foundation system are as follows:</p>
<p>Spread Footings:</p>
<ol>
<li>Bearing capacity</li>
<li>Sliding friction coefficient</li>
<li>Minimum dead load </li>
<li>Frost Depth</li>
</ol>
<p>Drilled Piers or Caissons:</p>
<ol>
<li>Bearing Capacity at different depths</li>
<li>Side friction coefficients at different depths</li>
<li>soil stiffness properties at different depths (for lateral loading)</li>
<li>minimum embedment into bedrock</li>
<li>minimum diameter</li>
</ol>
<p>Mat Slab or Raft:</p>
<ol>
<li>Bearing capacity</li>
<li>Soil spring factor (k)</li>
<li>Frost Depth</li>
</ol>
<p>Foundations Walls</p>
<ol>
<li>Bearing Capacity</li>
<li>Active, at rest, and passive earth pressures</li>
<li>Sliding friction coefficients</li>
<li>recommended drainage systems behind the walls</li>
<li>Frost depth</li>
</ol>
<p><strong>If they're missing info just reach out to them and ask for it.</strong> They're only human. </p>
| 35711 | Missing items in geotechnical reports |
2020-05-13T06:28:34.307 | <p>I was watching this engineering video <a href="https://www.youtube.com/watch?v=dQf_vE7tOlw" rel="nofollow noreferrer">here</a> and it talks about a technique used to prevent the pillars of a bridge from falling over in an earthquake. The pillars of the bridge are partially submerged under water, but instead of resting on sand at the bottom, they actually rest on a bed of gravel just above the sand. The rationale for this is that gravel has a larger particle size than the sand below it and therefore the pillars are less likely to fall over when they shake and move. </p>
<p>So my question is, why is it easier for pillars to move on gravel than sand? What role does particle size play in this? Does it reduce the friction when the pillars slides and if so why? It feels like this question should be obvious but I can not seem to fully understand why.</p>
| |materials|friction|bridges| | <p>saturated silt and sand are susceptible to liquefaction under sudden shock loads such as earthquake and pile-driving, etc. And they lose their bearing strength behaving like a liquid.</p>
<p>The reason for a large part is because of small grain size theey entrap the pore-water and once shaken they don't allow pore-water content pressure to decrease easily and actually the pore-water pressure goes up decreasing the contact forces between the grains and hence reducing their bearing strength.</p>
<p>Large size aggregate doesn't have this issue, and have large openings in between that allows the water pressure buildup to drain easily. also large size aggregate has a higher level of resiliency and elasticity.</p>
<p>That is why they are used under the railroad tracks as well.</p>
| 35732 | Why Does Particle Size Matter For Bridge Pillars here? |
2020-05-13T09:42:09.757 | <p>I have a linear control system defined by the following equations:</p>
<p><span class="math-container">\begin{array}{l}\dot{x}_{1}(t)=-\frac{1}{2}\left(x_{1}(t)-x_{2}(t)\right) \\ \dot{x}_{2}(t)=\frac{1}{10}\left(2 x_{2}(t)-10 x_{1}(t)\right)+u(t) \\ y(t)=x_{2}(t)\end{array}</span></p>
<p>I would like to find the transfer function from a constant input <span class="math-container">$u(t)=\bar{u}$</span> to the output <span class="math-container">$y$</span>. </p>
<p>I know that the transfer function is equal to <span class="math-container">$G(S)=\frac{\mathcal{L}(y(s))}{\mathcal{L}(\bar{u})}=\mathcal{L}(y(s))\cdot \frac{s}{\bar{u}}$</span></p>
<p>In order to find <span class="math-container">$\mathcal{L}(y(s))$</span> we solve the following system of equations:</p>
<p><span class="math-container">\begin{array}{l}s\mathcal{L}(x_1(s))-\mathcal{L}(x_1(0))=-\frac{1}{2}\mathcal{L}(x_1(s))+\frac{1}{2}\mathcal{L}(x_2(s))\\s\mathcal{L}(x_2(s))-\mathcal{L}(x_2(0))=\frac{1}{10}(2\mathcal{L}(x_2(s))-10\mathcal{L}(x_1(s))+\frac{\bar{u}}{s}\end{array}</span></p>
<p>However, this can be tricky to solve, so we calculate the transfer function using the following formula: <span class="math-container">$G(s)=C(s I-A)^{-1} B$</span>.</p>
<p>My question is: is the reasoning correct? Is the formula true for all types of control systems? </p>
| |control-engineering|control-theory| | <p>Yes, your reasoning is right and is applicable to all control systems with a valid state space representation. The formula to go from state-space to transfer function can be easily derived like so:
<span class="math-container">$$\dot{x} = Ax +Bu$$</span>
<span class="math-container">$$y = Cx + Du$$</span></p>
<p>Taking laplace transform on both equations one by one
<span class="math-container">$$sX= AX + BU$$</span>
i.e. <span class="math-container">$$(sI-A)X = BU$$</span>
i.e. <span class="math-container">$$X = (sI-A)^{-1}BU\cdots(i)$$</span>
<span class="math-container">$$Y = CX + DU...(ii)$$</span>
put (i) in (ii) you get <span class="math-container">$Y = C(sI-A)^{-1}BU + DU$</span>.
We know transfer function is <span class="math-container">$$G(s) = \frac{Y(s)}{U(s)}$$</span>
<span class="math-container">$$G(s) = C(sI-A)^{-1}B + D$$</span></p>
<p>Now your equations are:
<span class="math-container">$$\begin{bmatrix}\dot{x_1} \\ \dot{x_2} \\ \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \\ -1 & 0.2 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix}u$$</span><br>
<span class="math-container">$$y = \begin{bmatrix}0 &1\end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \begin{bmatrix}0 \end{bmatrix}u$$</span>
using these equations in <span class="math-container">$G(s)$</span> we get
<span class="math-container">$$G(s) = \frac{s + 0.5}{s^2 + 0.3s + 0.4}$$</span></p>
| 35734 | How to find the transfer function of a control system? |
2020-05-13T12:50:17.070 | <p>I am reading an <a href="https://arxiv.org/pdf/math/0212212.pdf" rel="nofollow noreferrer">article</a> that says:</p>
<blockquote>
<p>stabilize the multi-vehicle system to one of its local minima via dissipative control</p>
</blockquote>
<p>And <a href="https://ieeexplore.ieee.org/document/6000456" rel="nofollow noreferrer">other</a> that deals with dissipative system:</p>
<blockquote>
<p>(PID) controllers is designed to make the closed-loop linear system asymptotically stable and strictly quadratic dissipative</p>
</blockquote>
<p><strong>Question</strong>: What exactly is dissipative control or quadratic dissipative? </p>
| |control-engineering|pid-control| | <p>The intuitive idea is that a dissipative system cannot store more energy than what was initially stored plus what is supplied during an experiment, which is schematically depicted below.
<a href="https://i.stack.imgur.com/oYqEn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oYqEn.png" alt="Visualization of dissipativity"></a>
This figure is adopted from: <a href="http://www.eeci-institute.eu/pdf/M012/lec2.pdf" rel="nofollow noreferrer">http://www.eeci-institute.eu/pdf/M012/lec2.pdf</a></p>
<p>So we write that a system <span class="math-container">$\dot{x} = f(x,u)$</span>, <span class="math-container">$y = g(x,u)$</span> is dissipative with respect to the supply rate <span class="math-container">$s(u,y)$</span> if there exists a storage function <span class="math-container">$V:\mathbb{R}^n\to\mathbb{R}$</span> such that the dissipation inequality
<span class="math-container">$$V\big(x(t_1)\big) \leq V\big(x(t_0)\big) + \int_{t_0}^{t_1} s\big(u(t),y(t)\big) \;\mathrm{d}t$$</span>
hold for all system trajectories and for all <span class="math-container">$t_0< t_1$</span>.</p>
<p>We call it quadratic dissipative if the storage function is a quadratic function, e.g. <span class="math-container">$V(x) = x^\top P x$</span>.</p>
<p>Thus dissipative control is a controller such that the closed-loop system is dissipative with respect to the in- and output of the closed loop system.</p>
| 35736 | What is Dissipative Control? |
2020-05-13T16:50:06.310 | <p>Is it in general possible when you have a mechanical structure (in this case I am referring to the structure in <a href="https://engineering.stackexchange.com/questions/35600/calculate-forces-torques-on-a-parallel-spring-array/35615#35615">this</a> post) and you want to determine the global stiffness matrix of this structure, to solve for the matrix based on measurements?</p>
<p>More specifically, I have a structure where parts are connected by springs. I apply a known force to one of the parts and measure the displacement. In this way I get <span class="math-container">$n$</span> force/moment - displacement/rotation pairs.
Is it possible to solve for a matrix <span class="math-container">$K$</span>, for </p>
<p><span class="math-container">$F = KX$</span>, </p>
<p>with <span class="math-container">$F =\begin{bmatrix}F_x & F_y & F_z & M_x &M_y&M_z\end{bmatrix}$</span> and <span class="math-container">$X= \begin{bmatrix}dx & dy & dz & \theta_x &\theta_y&\theta_z\end{bmatrix}$</span>?</p>
<p>Is this feasible from a mechanical point of view, because e.g. this "stiffness matrix" <span class="math-container">$K$</span> may not be unique according to the load applied?
Are there any constraints or requirements regarding the measurements?</p>
<p>P.S. Applied forces are rather small (within the elastic range of the spring) and the displacement is restricted, so no large deflections should occur. This is why I assumed a linear approximation.</p>
| |mechanical-engineering|structural-analysis|applied-mechanics|solid-mechanics|stiffness| | <p>This is feasible and can be used to modify a theoretical stiffness matrix calculated by the Finite Element method to match experimental results more accurately. The FE model can then be used to calculate things which would be impractical to measure directly.</p>
<p>The simple approach you suggest is possible but not necessarily the best practical method. It may appear paradoxical that it is more practical to create both the stiffness and mass matrix by measuring the <em>dynamic</em> response of the structure. The reason is that if you excite the structure by an impact at one point and measure the <em>time history</em> of the response at other points (which is straightforward to do using an accelerometer instead of direct measurements of displacement, and a computer to record the real-time response at a sufficiently high sampling rate), you can obtain information about the motion in many different modes of vibration from "one measurement" of the dynamic response.</p>
<p>This can be done effectively on a "lab bench" scale using simple hand-held equipment, but with more sophisticated measuring devices, for example a scanning laser doppler vibrometer, it is possible to measure the response at hundreds of points on a structure "simultaneously" without any physical contact, including measurements under real operating conditions (e.g. at high temperatures, or the behaviour of rotating machinery while it is operating).</p>
<p>Most of this is not covered in a typical first engineering degree. Google for phrases like "experimental modal analysis" "model updating", or "system identification" for more information, both practical and theoretical.</p>
| 35741 | Solving for stiffness matrix numerically by a set of measurements |
2020-05-13T20:25:47.917 | <p>I understand that if supply and demand are not in equilibrium synchronous generators make the difference by increasing or decreasing their rotational energy - thus changing the electrical frequency in the grid. It is usually noted that a frequency shift <span class="math-container">$\geq 1\%$</span> is unacceptable and might damage the grid; however, I cannot find anywhere what in particular such frequency shift can damage. Which components of the grid are the ones we are mostly scared of getting damaged?</p>
<p>I am also curious what are the usual methods of mitigation of such situations, if I understand correctly, if the generated power is much higher than the demand the power stations disconnect and a blackout occurs (i.e. on a very windy night in a country powered by wind power stations). Similarly when the demand is much higher than the supply then some areal blackouts are purposefully introduced to decrease the load.</p>
<p>Lastly, are there currently any methods - beyond classical generators - that are widely used to increase the grid inertia?</p>
| |electrical-grid| | <p>A 1% shift in grid frequency will shift the running speed of every AC induction motor connected to it by 1%. Factory machinery and AC-powered clocks would all slow down, with potentially disastrous effects in those factories and big upsets would occur in any setting where event timing relied on AC synchronous motor-driven clocks. </p>
<p>As to the condition of the grid, its components are all tuned and impedance-matched to manage the transmission of power at the line frequency. Running the grid off-frequency would knock all the tuning off, and potentially cause unmanageable current and voltage transients and surges in the transformers, lines, and switchgear. </p>
<p>Furthermore, if part of the grid goes off-frequency then there is another bad effect that arises, in that all of the generators in the grid normally tend to automatically sync themselves together so they are phase-locked. If one part of the grid goes out of phase lock, then there will be power flowing not from all the generators to all the loads, but power flowing between the on-phase generators into the off-phase generators to "motor" them back in phase. Those currents could be large enough as to overload those parts of the grid between the out-of-phase nodes and cause the voltages in the whole grid to fluctuate. </p>
<p>There are very complex automatic load-and-generator-capacity management systems in place at all generating nodes, which hold the system in balance and maintain phase lock on the correct line frequency over a broad range of loads and generating capacities- with human override in case of accidental failures. </p>
<p>Increasing the grid "inertia" means adding <em>inductance</em> to it so it strives to maintain constant current at the design frequency for short duration perturbations. This is done within the grid by using large coils or capacitors in the line, usually near locations where large amounts of power are being consumed, to "trim" it and thereby keep the correct phase relationship between voltage and current. This is known as <em>power factor adjustment</em>. </p>
| 35745 | What is the specific issue with frequency fluctuation in the grid? |
2020-05-14T03:25:07.067 | <p>I am really challenged as I need to fit a bunch of motors in a 6" box acting as linear actuators. I was looking for linear actuators (most I found were brushed DC) at first but the ones I found that would fit in 6" didn't have a suitable speed torque curve for my application. I started looking at steppers and continuous spin servos, combined with a rack and pinion to transfer the rotational movement into a linear motion.</p>
<p>I need to calculate a linear force resulting from a torque applied on a rack from the pinion gear attached to the spinning motor. Let's say I have a servo applying a torque of 29 kg-cm on the rack, how is this force going to be transferred on the linear rack? (I do not know the pitch and size of rack and pinion yet, but I want to know how big the gear could be to make the rack move faster at a given motor RPM).</p>
<p>Thank you so much!</p>
| |motors|gears|torque|servo| | <p>The linear force of the rack is equal to the tangential force on your pinion teeth (less power lost due to friction, typically 2-3% for spur tooth and racks). Tangential force = torque/(pitch diameter/2).</p>
| 35752 | Calculating force transferred from pinion gear to rack |
2020-05-14T04:07:46.487 | <p>I found <a href="https://www.digikey.ca/product-detail/en/nmb-technologies-corporation/23KM-K267-00V/P14277-ND/2417001" rel="nofollow noreferrer">this motor</a> which I believed had interesting specs, but when I checked out the data sheet, the torque-speed graph was replaced by a torque-frequency graph! I do not know what it means. How does it intuitively translate to torque at a given speed?</p>
<p>Thank you so much!</p>
| |torque|stepper-motor| | <p><a href="https://www.se.com/id/en/faqs/FA337686/" rel="nofollow noreferrer">RPM of a stepper motor</a></p>
<blockquote>
<p>Formula for calculating stepping motor speed.</p>
<p><span class="math-container">$$RPM = \dfrac{a}{360} \cdot f_z \cdot 60$$</span></p>
<p><span class="math-container">$RPM$</span> = Revolutions per minute.<br />
<span class="math-container">$a$</span> = step angle<br />
<span class="math-container">$f_z$</span> = pulse frequency in hertz</p>
</blockquote>
| 35753 | What does the Torque vs Frequency plot mean for a stepper motor |
2020-05-15T19:45:59.153 | <p>(for the record, I already attempted to ask this in the physics stackoverflow)</p>
<p>I'm attempting to create something similar to a vertical multi-siphon. The device would pull water from multiple sources, at multiple elevations (a total range of about 5 feet) up to the top source so it can redistribute down again.</p>
<p>This could be simply accomplished by (a) pump(s) or a mini-turbine, but the transfer medium should be about 4 inches diameter: about large enough for a Betta fish to move up.</p>
<p>I can't use a multi-siphon going down and just out-pump the down flow of water, this means the fish would have to fight against the current up the piping so this isn't really an option.</p>
<p>Chemical and thermal lift are also not feasible to move the fish.</p>
<p>I'll take you quickly through some research. Anyone reading this knows that (after priming) a siphon attempts to equalize pressure which equates to leveling out the water involved.
<a href="https://i.stack.imgur.com/pefq0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pefq0.jpg" alt="enter image description here"></a></p>
<p>While I can't find any examples of this, reason should stand that with a sufficient priming force, you can pull from multiple sources.
<a href="https://i.stack.imgur.com/KkQyP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KkQyP.jpg" alt="enter image description here"></a></p>
<p>You should even be able to do crazy things (to a point).
<a href="https://i.stack.imgur.com/kNEJM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kNEJM.jpg" alt="enter image description here"></a></p>
<p>So when you attempt to siphon vertically, it requires a pump.
<a href="https://i.stack.imgur.com/Krepf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Krepf.jpg" alt="enter image description here"></a></p>
<p>And it stands to reason that a (properly calibrated) pump system could be used to supply multiple sources.
<a href="https://i.stack.imgur.com/tNlHp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tNlHp.jpg" alt="enter image description here"></a></p>
<p>So the question becomes, how do we safely lift a fish that went down a siphon back up to its original level?
<a href="https://i.stack.imgur.com/K1ekE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K1ekE.jpg" alt="enter image description here"></a></p>
<p>The only thing that is currently coming to mind, is use airline tubing connected to a pump to pull the water from the point at which the "siphon" was primed.</p>
<p>Continually applying negative pressure might be enough to pull the water at a rate that isn't going to vacuum a fish. Though, I'll admit I'm concerned it might need a series of pumps like this to imitate a caterpillar propulsion system.</p>
| |pumps|siphon| | <p>A Venturi pump may do what you're looking for:</p>
<p><a href="https://i.stack.imgur.com/dDpxh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dDpxh.png" alt="venturi pump diagram"></a></p>
<p>The pictured pump is for air, but the principle works for water too.</p>
<p>There are inline types that don't change diameter from inlet to outlet, but I haven't seen one of this type for use with water. I'm sure it could work for a small elevation gain.</p>
<p>For higher head pressures the water jet could be harmful to the fish, but in that case you could use multiple Venturis?</p>
<p>Another option would be to build some kind of "airlock" mechanism, that allows the fish to get into the up tube after the pump.</p>
<p>Or you could use a ram pump in reverse, where you pump upwards and then stop suddenly. The inertia of the water would cause some brief suction at the pump end which could suck a fish through a one way valve. One issue with this is that it would be hard to make a valve that would not injure the fish if it didn't make it all the way through. Maybe some kind of soft orifice?</p>
| 35783 | How can I create a small pump that can pull fish to a higher elevation without killing them? |
2020-05-15T20:35:34.833 | <p>Hold height constant. Then a Rectangular Prism, with length and width = radius of the cylinder, has greater volume! So why aren't tank wagons (railroad vehicles that transport gas and liquid) Rectangular Prisms? </p>
<p><a href="https://i.stack.imgur.com/q2pPj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q2pPj.jpg" alt="enter image description here"></a></p>
| |rail| | <ul>
<li>A cylindrical structure will have a lower surface-area to volume ratio than any other shape that will fit in the structure gauge of railway infrastructure.</li>
<li>I can't be certain but that tank looks as though it has been formed by continuously rolling a continuous helical coil from a flat strip. This lends itself to a continuous process of coiling, welding (with the seam passing by the fixed welder) and then chopping the "pipe" into wagon sized lengths.</li>
<li>The cylinder will be very strong. It appears from the photo that the tank is self-supporting. No need for chassis so there is further weight saving.</li>
<li>Possibly the weight of the full tank results in close to the maximum axle loading. A bigger tank couldn't be filled (although a shorter wagon could be built).</li>
<li>The cylindrical shape eliminates all corner stress points and seams except for the end caps.</li>
</ul>
| 35784 | Why are tank cars cylindrical? |
2020-05-15T20:37:33.627 | <p>I'm reading the <a href="https://transitinnovation.org/wp-content/uploads/2019/12/R211%20Tech%20Spec.pdf" rel="nofollow noreferrer">procurement documents</a> for the R211 railcar contract, which is for electric multiple unit subway cars designed for rapid transit applications in New York City. The railcar's propulsion is specified to use AC induction motors powered from PWM inverters with IGBTs.</p>
<p>The specification related to propulsion motor overcurrent is given as follows:</p>
<blockquote>
<p>10.3.20.2 Over-temperature protection shall be provided as follows:</p>
<p>a) If the winding temperature rises 27°F above (15°C above) the operating class temperature limit, the control unit shall decrease the motor duty cycle by removing the dynamic braking and converting to frictional braking, until the temperature falls below the operating class temperature.</p>
<p>b) If the winding temperature rises above its design class operating temperature limit, the control unit shall remove power to the affected truck, until the temperature falls below the operating class temperature. The condition shall be reported and recorded by the MDS.</p>
<p>c) Motor temperature shall be calculated using a thermal model of the motor and the power that the motor is handling.</p>
<p>d) The measurement of winding temperatures using embedded sensors is not permitted.</p>
</blockquote>
<p>Points c) and d) stand out to me as a little bit strange, since the motor is expected to be modeled with its loads, instead of directly measuring the temperature. The only reasoning I can determine for this is that it protects against the failure of the thermal sensor.</p>
<p>However, it's not clear to me why this failure more is harder to guard against, than (for example) the failure of current sensors needed to track the motor's power, or an obstruction of the normal cooling process by dust or debris. Is there a common technical reason why motor thermal protection is specified to use only modeled heating rather than measured heating or a combination of safeguards?</p>
| |motors|rail| | <p>Yes. Along with your reasoning of failure mode, reliability and cost, there is a disadvantage of slower response time of the sensors. The thermal model will give temperature values as fast as the speed of the digital processor you use. A <a href="https://www.gegridsolutions.com/multilin/family/motors/motor_thermal_model.pdf" rel="nofollow noreferrer">Research</a> by GE states:</p>
<blockquote>
<p>An additional reason to reject such
temperature sensors as the main basis for thermal protection,
is the fact that the traditional Resistance Temperature Detector
(RTD) has a relatively slow reaction time and can’t respond
adequately to the high speed of the heating process during
motor acceleration. </p>
</blockquote>
<p>Also, this model ensures greater transparency between motor manufacturer and stakeholder (as manufacturer supplies a family of characteristic curves for the motor):</p>
<blockquote>
<p>Another important part of thermal model implementation
is “Expected values stored in Motor Protection Device”. This term implies that
information is available from the motor designer and motor
manufacturer, that is related to the thermal reserve, allowed
performance and thermodynamics of the motor in question.</p>
</blockquote>
<p>This paper goes on to further discuss the terminology like thermal capacity used and allowed performance and is a great read if you want to build a model. </p>
| 35785 | Why would a specfication for a motor require that overtemperature protection be realized by modeling heat generation and not with an embedded sensor? |
2020-05-15T21:25:13.427 | <p><a href="https://grist.org/energy/oil-demand-plunge-coronavirus-tankers-california-coast/" rel="noreferrer">About all those oil tankers off the coast of California … | Grist</a></p>
<blockquote>
<p>The giant ships burn fuel to keep lights on, power equipment, and heat the large volumes of crude oil resting in their tanks.</p>
</blockquote>
<p>I'm assuming that crude oil can't be heated in <a href="https://www.reuters.com/article/us-global-oil-storage-fracking/a-hunt-for-any-storage-space-turns-urgent-as-oil-glut-grows-idUSKBN2230I3" rel="noreferrer">steel frac tanks</a>, <a href="https://canada.constructconnect.com/dcn/news/resource/2020/05/texas-eases-underground-oil-storage-rules-raising-concerns" rel="noreferrer">places other than salt caverns</a>, and <a href="https://science.howstuffworks.com/environmental/energy/strategic-petroleum-reserve1.htm" rel="noreferrer">salt caverns</a>.</p>
<blockquote>
<p>While underground caverns may not seem like the best place to store an emergency oil supply, they're actually very secure. For one thing, since they're 2,000 to 4,000 feet (610 to 1,219 meters) underground, the extreme pressure prevents cracks from forming and leading to leaks [source: <a href="http://www.fossil.energy.gov/programs/reserves/spr/spr-sites.html" rel="noreferrer">DOE</a>]. Also, the natural temperature difference between the top and bottom of each cavern encourages the oil to circulate, which maintains its quality.</p>
</blockquote>
| |petroleum-engineering| | <h2>Because the ocean can be very cold</h2>
<p>As MartyMcGyver discusses, underground storage does not need to be heated. Tankers, however, are sitting in water, which (apart from the salt) has the highest thermal absorption of any substance; that is to say, the worst possible insulator. </p>
<p>The US Pacific coast is some of the coldest seawater in the continental U.S. (Alaska notwithstanding of course). Don't let the palm trees and 90 degree days fool you: due to ocean currents, it gets a lot of arctic water come down from Canada/Alaska. That's why L.A. surfers wear wetsuits and Florida surfers do not. </p>
<p>Different deal in the Gulf coast or Florida generally, where ocean temp is in the 70s F. I suspect even the port of Stockton CA (which is tidal) would have respectable water temps in the 110F summer heat.</p>
| 35788 | Why do oil tankers heat crude oil? |
2020-05-16T11:10:27.757 | <p>I've been working with control systems and transfer functions recently, but I've come across a format that I haven't seen before using coefficients of theta, and I can't find any information about it.
Does anyone recognize the format below, used to describe a control system. As I was writing this question I realised what it is probably saying, but I'm unable to fully understand it. My thought process is written below so could somebody help me fill in the gaps?</p>
<p><span class="math-container">$$\ddot{\Theta_0} + 4\dot{\Theta_0} + 68\Theta_0 = 34\Theta_i$$</span></p>
<p>I'm assuming that the dots above theta are the powers. 2 dots means squared etc. <span class="math-container">$\Theta_0$</span> is representing the output while <span class="math-container">$\Theta_i$</span> is representing the input. (The 'O' is actually a zero not a letter but I'm assuming thats a typo?). Since a transfer function is:
<span class="math-container">$$T(s) = \frac{Y(s)}{U(s)}= \frac{output}{input}$$</span>
we could divide by <span class="math-container">$\Theta_i$</span> to get the transfer function. But this is where I get lost. I think this has something to do with the characteristic equation - the denominator of the transfer function when equal to <span class="math-container">$0$</span> - but I dont see how?</p>
| |control-engineering|mathematics|transfer-function| | <p>In mathematics,which are very closely related and broadly applied in control engineering, the derivatives with respect to time are often notated using the so-called: <strong>dot notation</strong>. You can check this link for more information: <a href="https://en.m.wikipedia.org/wiki/Derivative" rel="nofollow noreferrer">Derivative</a>. This means that the particular equation (the one at the question) can be rewritten in the following form:</p>
<p><span class="math-container">$$ \ddot{Θ}_ο+4\dot{Θ}_ο+68Θ_ο=34Θ_i \Rightarrow \ddot{y}+4\dot{y}=68y = 34u \Rightarrow $$</span></p>
<p><span class="math-container">$$ \frac{d^2y}{dt^2}+4\frac{dy}{dt}+68y = 34u $$</span></p>
<p>This notation is used for up to the third derivative usually. For higher order derivatives, due to appearance reasons, the notation used changes and becomes: <span class="math-container">$y^{(n)}$</span> for the n-th order derivative with respect to time. </p>
<p>Now, what's left to do is to convert the equation from time-domain to <span class="math-container">$s$</span>-domain. In order to do so the following formula is used:</p>
<p><span class="math-container">$$ \mathscr{L}\{\frac{d^nf}{dt^n}\}=s^nF(s)-s^{n-1}f(0)-s^{n-2}f^{(1)}(0)- \ ... \ -sf^{(n-2)}(0)-f^{(n-1)}(0) $$</span></p>
<p>where</p>
<ul>
<li><span class="math-container">$f^{(n-a)} \ \rightarrow \ \text{(n-a)-th derivative order of the function f}$</span></li>
<li><span class="math-container">$f^{(n-a)}(0) \ \rightarrow \ \text{initial value of the (n-a)-th derivative order of the function f}$</span></li>
</ul>
<p>For this particular ordinary differential equation (ODE) this formula yields:</p>
<ul>
<li><span class="math-container">$\ddot{y}=s^2Y(s)-sy(0)-\dot{y}(0)$</span></li>
<li><span class="math-container">$\dot{y}=sY(s)-y(0)$</span></li>
</ul>
<p>By considering the usual fact of <span class="math-container">$0$</span> initial conditions it all comes down to:</p>
<p><span class="math-container">$$ \frac{d^2y}{dt^2}+4\frac{dy}{dt}+68y = 34u \Rightarrow s^2Y(s)+4sY(s)+68Y(s)=34U(s) $$</span></p>
<p><span class="math-container">$$ Y(s)\cdot (s^2+4s+68) = 34U(S) \Rightarrow \frac{Y(s)}{U(s)} = \frac{34}{s^2+4s+68} $$</span></p>
<p>The overall transfer function of this particular system is:</p>
<p><span class="math-container">$$ \frac{Y(s)}{U(s)} = T(s) = \frac{34}{s^2+4s+68} $$</span></p>
| 35804 | Help understanding a control system represented by coefficients of theta? |
2020-05-16T13:41:19.543 | <p>I was learning about the working of drum brakes, and I came to know, that each of the two brake linings of drum brake apply horizontal force in opposite directions. </p>
<p>That confused me, because, if somewhat equal forces are being put on opposite horizontal directions on the wheel, then how is it stopping the vehicle, that is bringing it instantly to rest. Considering, the wheel is rotating in anti-clockwise direction, how are the forces' directions working on the wheels, I mean how are the components/vector components of the forces applied by brake linings affecting the components of the kinetic motion of the wheels. Can anybody clarify? </p>
| |mechanical-engineering|friction| | <p>In a drum brake when you force the pads out you create a friction force which is gradually increases and creates a torque opposing the rotation of the wheel.</p>
<p>This torque is transfered to the tire and starts to decellerate the car. </p>
<p>In the old brakes you could apply so much breaking power as to lock up the wheel and make it skid and burn the tire. But not anymore, After ABS (antilock brake assistance).</p>
<p>How ever no matter how strong the grip of the brake and the degree if torque it causes, nothing can bring a car to a sudden stop. </p>
<p>Even if it collides into a concrete wall it will still keep going untill it crashes the structure of the car enough to dissipate the kinetic energy. </p>
| 35807 | Need help regarding working of drum brakes |
2020-05-17T14:59:23.210 | <p>My structural engineering friend and I passed a construction site years ago, and he called those temporary fences a technical name. Now I ask him the technical name he said those years and he doesn't even remember the moment, understandably, and is not sure there is even a technical name to begin with. I googled "what is the fence built around ongoing projects called" and 'temporary fencing" came up.</p>
<p>This is my final straw to chalk it up to a distorted memory or finally learn that elusive name (every few months I remember that moment and still don't remember the name). So is there any technical name for temporary construction fences just as we have aggregates for sand, gravel, etc? If there is, please for the love of sanity, what is it?</p>
| |structural-engineering|civil-engineering| | <p>It is called Tapume. It is an English term but more often used in Brazilian Portuguese.</p>
| 35822 | What is the technical name for temporary construction fences? |
2020-05-18T14:55:47.540 | <p>I would like to cool a small device, with a size of 45x45 mm² and a thermal load of ~40 W. Currently, I have a cooling system available, capable of cooling a thermal load of 52 W, and a cold plate with the dimensions of 60x80 mm². Due to the size difference I expect a small drop in the efficiency.</p>
<p>Is there a way to calculate that drop, and to estimate if I still can use that cooling device? As a naive approximation I would take the difference in the area size (<span class="math-container">$80\cdot60-45^2=2775$</span>), which is approximately 60 % of the original cold plate. Can I therefore expect a loss of 60 % of the original cooling capacity, i.e. I only can cool a thermal load of 30 W for the given dimensions, or is that approach wrong?</p>
| |cooling| | <h1>Summary</h1>
<p>Assume that you put the heating element directly on the cooling element. The cooling element can handle an input flux (mW/mm<span class="math-container">$^2$</span>) of </p>
<p><span class="math-container">$$ \frac{\dot{q_c}}{A_c} = 52/(80*60) = 10.8 $$</span></p>
<p>The area of the heating element is <span class="math-container">$45^2 = 2025$</span> mm<span class="math-container">$^2$</span>. You will pull a maximum of <span class="math-container">$2025*10.8 = 21.9$</span> W power from the heater.</p>
<p>Assume that you can "pipe" the heat from the heating element directly to the cooling element through a tapered conducting path (that is otherwise insulated on all sides). At that point, you can pull 52 W into the cooling element. Your heating element is only giving 40 W maximum. The input flux (mW/mm<span class="math-container">$^2$</span>) to the cooling element is</p>
<p><span class="math-container">$$ \frac{\dot{q_h}}{A_C} = 40/(80*60) = 8.33 $$</span></p>
<h1>Conclusion</h1>
<p>To improve the efficiency of the system, pipe the heat from the heating element to the cooling element using a tapered conducting path that is insulated on all sides.</p>
| 35830 | Efficiency loss for a cooling element for cooling parts smaller than the cold plate |
2020-05-19T05:00:10.150 | <p>I don't understand p 24 of <a href="https://www.rystadenergy.com/globalassets/pdfs/rystad-energy_covid-19-report_14_may_2020-public-version.pdf" rel="nofollow noreferrer">this PDF</a> from <a href="https://www.rystadenergy.com/newsevents/news/press-releases/covid-19-demand-update-oil-seen-down-10point9-jet-fuel-down-33point6-road-fuel-down-11point1-in-2020flat/" rel="nofollow noreferrer">Rystad Energy</a>. Can someone please distinguish, and ELI5, 7 ((Crude supply (current OilMarketCube, downside risk)) and 8 (Crude supply (with required shut-ins to balance))? </p>
<p><a href="https://i.stack.imgur.com/CUlhZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CUlhZ.jpg" alt="enter image description here"></a></p>
| |petroleum-engineering| | <p>The precise definition of Oilmarket cube will be explained elsewhere, earlier most likely, but it will be a unit that is used to compare sources much like kWh can be used for gas as a cubic foot of gas varies its heating power with temperature.</p>
<p>It could be 1m^3 or 1000m^3 but it will be an amount stated.</p>
| 35841 | What does OilMarketCube mean? |
2020-05-20T05:39:49.540 | <p>Here's a typical outdoor gas stove:</p>
<p><a href="https://i.stack.imgur.com/fnS57.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fnS57.jpg" alt="enter image description here"></a></p>
<p>It is fed by a low-pressure gas supply in the form of a camping cartridge or via an LPG cylinder connected to a low-pressure regulator (around 30mbar) like this one:</p>
<p><a href="https://i.stack.imgur.com/WodsU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WodsU.jpg" alt="enter image description here"></a></p>
<p>My question is: is it possible to increase the heat output of the burner by connecting two low-pressure supplies in parallel via a T-piece like this one?</p>
<p><a href="https://i.stack.imgur.com/orXrY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/orXrY.jpg" alt="enter image description here"></a></p>
<p>Would this actually result in double the gas flow/double the heat output, or would the stove itself impose some limitation on the maximum pressure of gas able to flow through it (excluding the obvious extreme cases of catastrophically high pressure where the stove or pipes might crack)? </p>
<p>How dangerous is it? Obviously I assume it's not a recommended procedure and it probably has safety considerations, but for example would doubling a low-pressure supply from 30mbar to 60mbar pose a serious risk, or are appliances in general built to withstand far higher pressures than this? I know standard gas hoses, for example, can withstand pressures of 20 bar (100x higher than low pressure canisters).</p>
<p>Is there a way of determining whether an appliance will be able to handle higher flow? </p>
<p>I've seen this <a href="https://youtu.be/VfxkrrndtMQ?t=225" rel="nofollow noreferrer">done on cheap/flimsy camping stoves</a> so I assume it's not a catastrophic risk especially with better-made burners.</p>
| |gas|welding|compressed-gases|brazing| | <p>There is a mathematical formula called "The Hanson Theorum" basically it's the calculating the area of a circle, very accurate without the use of Pi.</p>
<p>Imagine a pipe with a <span class="math-container">$10cm$</span> in diameter bore, well the calculation is <span class="math-container">$10 x 10$</span> (to get the square of the bore) and multiply it by "The Hanson Theorum" which is <span class="math-container">$0.78$</span>.</p>
<p>The calculation is: <span class="math-container">$10^{2}\cdot 0.78cm = 78cm^{2}$</span> of bore area.</p>
<p>If your gas jet is say <span class="math-container">$0.3mm$</span> we can create a higher gas flow, by boring that jet out to a slightly larger size with either a specialist drill size or some fine guitar wire... This will increase the gas flow, but at the same pressure. The regulator, should be able to cope with a <span class="math-container">$20 - 30\%$</span> greater flow.</p>
<p>So let's assume the jet is:<span class="math-container">$0.3mm \rightarrow 0.3\cdot 0.3 \cdot 0.78 = 0.07mm^{2}$</span> of jet bore area</p>
<p>Now if we want to increase the flow rate, by <span class="math-container">$20\%$</span> - that means we need to figure out what a jet bore <span class="math-container">$20\%$</span> larger is IN AREA, not diameter, the simple maths.</p>
<p><span class="math-container">$$ (10+2)\cdot 7 = 0.084 \rightarrow \text{of area} $$</span>
<span class="math-container">$$ 0.084 = 0.327\cdot 0.327 \cdot 0.78 = \dots \rightarrow \text{when converting back to diameter} $$</span></p>
<p>My brain is tired.... So you have to increase the jet size from <span class="math-container">$0.3mm$</span> to <span class="math-container">$0.327mm$</span> in diameter.</p>
<p>OK you can drill them out - with a special drill or you can scrap them out with a piece of 0.009" or <span class="math-container">$0.23mm$</span> steel guitar wire... with a diagonally cut end, and a small drilling machine etc.. and just scrape away - you will get the technique... or you can hammed the wire into a square or rectangular shape....</p>
<p>Since you really are ONLY reboring it by a hair or two's increase in diameter.... it's not much and then you can see if your stove works well. Try to make the issue of small shavings in terms of rejetting the stove, rather than way too big.</p>
<p>If you go too big first go, your regulator may not flow that amount of gas AND OR your you may need to enlargen the air mixing stage, to get a hot clean flame, instead of a sooty yellow flame...</p>
<p>Yellow flames also produce Carbon Monoxide which is lethal in significant amounts in confined spaces.... and people under some conditions - like it's freezing cold, they are in a snow storm and the only place to cook is inside the little tent...</p>
<p>My little butane can single burner portable camping stove - on cold days like 10<em>C, the burning rate which is based upon the boiling rate of the gas, which slows down as the evaporating gas, chills the liquid gas, so it cools and boils off at an incrediby slow rate, which means cooking a big meal in a pot takes 10 x as long as cooking on a 35</em>C day.... So I need to fiddle with the jet a little to increase the rate of gas flow and not fiddle around much to make the stove improvements into a major engineering project, rather than a simple improvement.</p>
| 35855 | Is it possible to increase the heat output of a gas stove by connecting two gas supply canisters to it at the same time, and is it safe to do so? |
2020-05-20T09:46:30.517 | <p>I have pipework system with an inlet pressure, <span class="math-container">$P_0$</span>, which passes from a pipeline of fixed diameter, <span class="math-container">$D_0$</span>, through a tee with reducer to a smaller diameter pipe, <span class="math-container">$D_1$</span>, before returning to another pipeline (same diameter, <span class="math-container">$D_0$</span>) of lower but unknown pressure, <span class="math-container">$P_1$</span>, through another tee and reducer arrangement.</p>
<p>I am attempting to determine the pressure drop required for natural gas flowing through the system to meet choked flow conditions (Mach 1).</p>
<p>I have used the following equation to calculate a critical pressure ratio:</p>
<p><span class="math-container">$$\dfrac{p^*}{p_0} = \left(\dfrac{2}{n+1}\right)^{n/(n-1)}$$</span></p>
<p>Which works out as 0.542 assuming ideal gas condition using <span class="math-container">$n = 1.32$</span>, natural gas ratio of specific heats (<span class="math-container">$k$</span>). Therefore a local pressure drop of approximately <span class="math-container">$P_0/2$</span> would be required to meet choked flow conditions.</p>
<p>However, I am concerned that this is overly simplifying the problem. </p>
| |gas|pipelines|compressible-flow| | <p>Here's a diagram that I believe depicts your situation.</p>
<p><a href="https://i.stack.imgur.com/prg5K.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/prg5K.png" alt="enter image description here"></a></p>
<blockquote>
<p>I am concerned that this is overly simplifying the problem.</p>
</blockquote>
<p>You're probably fine but it may be useful to review assumptions tied to that equation and your system.</p>
<h2>Ideal gas law.</h2>
<p>The equation appears in API 520 Part 1 9th Edition (2014) on page 56 under section <strong>5.6 Sizing for Gas or Vapor Relief</strong> which assumes that "the pressure-specific volume relationship along an isoentropic path is well described by the expansion relation,"</p>
<p><span class="math-container">$$PV^k={constant}$$</span></p>
<p>where</p>
<p><span class="math-container">$k$</span> : is the ideal gas specific heat ratio at the relieving temperature.</p>
<p>Or, in other words, that the gas is an ideal gas.</p>
<p>The document cautions that this assumption may not be valid if the compressibility factor (a.k.a. "Z-factor") falls outside the <span class="math-container">$0.8<Z<1.1$</span> range, as may be the case for high pressure conditions, high molecular weight gasses such as <span class="math-container">$CO_2$</span> or <span class="math-container">${H_2}{S}$</span>, or low temperatures. Compressibility is a measure of the "non-ideality" of a gas and is a function of both pressure and temperature. </p>
<p>Based on your use of the ratio of specific heat <span class="math-container">$n=1.32$</span>, it seems you are assuming the fluid is methane (the primary component in natural gas transmission lines). Use a process simulator / equation of state calculator to calculate the compressibility or calculate it <a href="https://www.youtube.com/watch?v=zU-IiBKOZHw" rel="nofollow noreferrer">manually</a> such as with the method described in the GPSA Handbook (ex: <a href="https://www.worldcat.org/oclc/57680370" rel="nofollow noreferrer">GPSA 12th edition</a>, Section 23 "Physical Properties", page 23-10, "Z-FACTOR FOR GASES") to confirm the ideal gas assumption is still valid.</p>
<p>If the gas cannot be assumed to ideal, then API 520 Part 1 provides a numerical calculation method in Annex B. The choked flow pressure drop is found as a side effect of identifying the maximum "Mass Flux" from a numerical integration process that uses a set of pressure, temperature, and specific volume values describing the state of gas at different points along an isentropic expansion (you'd use a process simulator / equation of state calculator to get these values).</p>
<p><span class="math-container">$$G^2 = \left[ \frac{- 2 \cdot \int^P_{P_1} v \cdot dP}{v^2_t}
\right]_{\text{max}}$$</span></p>
<p>where:</p>
<p><span class="math-container">$G$</span> : is the mass flux (mass flow per unit area) through the nozzle, [<span class="math-container">$\frac{kg}{s \cdot m^2}$</span>]</p>
<p><span class="math-container">$v$</span> : is the specific volume of the fluid [<span class="math-container">$\frac{m^3}{kg}$</span>]</p>
<p><span class="math-container">$P$</span> : is the stagnation pressure of the fluid [<span class="math-container">${Pa}$</span>]</p>
<p><span class="math-container">$1$</span> : is the fluid inlet condition to the nozzle</p>
<p><span class="math-container">$t$</span> : is the fluid condition at the throat of the nozzle where cross-sectional area is minimized (the integral term corresponding to maximum mass flux)</p>
<p>The pressure drop at choke flow (<span class="math-container">$P_1 - P$</span>) is found when additional terms from the <span class="math-container">$\int^P_{P_1} v \cdot dP$</span> integral slices don't cause <span class="math-container">$G$</span> to rise any more.</p>
<p>Or you could buy commercial software and trust that they perform the non-ideal gas calculations correctly.</p>
<h2>One-phase gas flow</h2>
<p>The equation assumes one-phase gas flow. If two-phase flow is present then a different set of equations are required to calculate the critical pressure ratio (see Annex C in the document).</p>
<h2>Restriction diameters</h2>
<p>I am having a hard time imagining a plausible scenario in which your system would be useful unless it contains a valve of some sort on the <span class="math-container">$D_1$</span> diameter pipe so the amount of gas flow to the second pipeline can be controlled. If a valve is present to perform this task, make sure to identify the location of the tightest flow restriction when designing the system. For example, if you are using a a ball valve, then make sure you know if it's a restricted port ("RP") or a full port ("FP") valve. If you're installing a pressure relief valve, make sure to reevaluate the sizing equations based on the actual orifice areas ("flow areas") and certified capacity values ("coefficient of discharge") provided by the manufacturer for valves they quote you (see the <a href="https://www.nationalboard.org/Index.aspx?pageID=64" rel="nofollow noreferrer">NB-18</a> "<a href="https://www.nationalboard.org/SiteDocuments/NB18/NB18.pdf" rel="nofollow noreferrer">Redbook</a>"). Beware erosion issues that can alter the pipe inner diameter.</p>
<h2>Summary</h2>
<p>If you assume the gas is an ideal single-phase gas, then yes, the calculation of the pressure drop needed to achieve choked flow conditions is simply the equation for critical pressure ratio you provided. Otherwise, things get complicated.</p>
| 35862 | Determining required pressure drop for choked flow conditions to be met |
2020-05-20T12:16:11.300 | <p>I'm sorry if the question doesn't fit the site but I didn't know where to ask. I found this at a car service where the guys used this thing to correct the painting on the car on the regions where paint flowed down. All I know is that it scratches glass and it weighs 32 grams, dimensions being 3 cm x 2.15 cm approximately.</p>
<p><img src="https://i.stack.imgur.com/5aapM.jpg" alt=""></p>
| |materials| | <p>Considering its purpose and the density being in the region of 15 g/cm3, I would say it is the replaceable blade from a carbide scraper, so it is a <a href="https://en.wikipedia.org/wiki/Cemented_carbide" rel="nofollow noreferrer">cemented carbide</a> (something like WC-Co).</p>
| 35870 | What type of material is this? |
2020-05-20T14:31:38.483 | <p>See the follow file for details: <a href="https://www.filedropper.com/filemanager/public.php?service=files&t=757744643cd3d3972ae382af4a4deb91" rel="nofollow noreferrer">https://www.filedropper.com/filemanager/public.php?service=files&t=757744643cd3d3972ae382af4a4deb91</a> . I create a spiral structure which is smooth and beautiful, then scale it up by a factor of 10. Once it's scaled up, the geometry has changed. Can anybody tell me why, and if there is a way of avoiding it? Thanks! </p>
<p><a href="https://i.stack.imgur.com/cs3qG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cs3qG.png" alt="operations"></a>
<a href="https://i.stack.imgur.com/jOkeU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jOkeU.png" alt="original_geometry"></a>
<a href="https://i.stack.imgur.com/yFzUd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yFzUd.png" alt="changed geometry"></a>
edit: <a href="https://i.stack.imgur.com/jCfbQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jCfbQ.png" alt="scale dialogue"></a></p>
| |solidworks| | <p>See below. As I said in my comment - the problem that you're having is that each body scales about its <em>own</em> centroid. If the origin is not OK for you, then you will need to add another coordinate system.</p>
<p>See the .gif below showing how to do this.</p>
<p><a href="https://i.stack.imgur.com/NcEhK.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NcEhK.gif" alt="enter image description here"></a></p>
| 35872 | solidworks 2018 part geometry changes on scaling |
2020-05-20T21:50:19.170 | <p>I would like to increase the vacuum power of my desoldering station. It’s a multifunction unit that includes a single vacuum pickup tool, hot air wand and desoldering gun. When using the desoldering tool or the pickup tool, the hose from the tool is connected to the vacuum port. Air is drawn in from the vacuum port and out through the wand via a hose connected to it. When using the wand tool, it works the same way, except heater located inside its housing are turned on. In order to prevent premature burn out of its heater, no tool should be connected to the vacuum port. </p>
<p><a href="https://i.stack.imgur.com/xM2kw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xM2kw.png" alt="enter image description here"></a></p>
<p>To increase the vacuum pressure, I am considering adding a vacuum reserve tank with a solenoid valve. It should allow the vacuum to build up negative pressure and give a stronger, momentary suction when the valve is closed for a few seconds and then opened by triggering the desoldering gun. Since I know nothing about pneumatics, I am unsure wether or not the vacuum reserve decreases the airflow when using the wand. This is only a hunch since the tank is like a capacitor which can slow down or impede the airflow. Am I correct here? In the case that it does inhibit airflow, I also added a bypass line with another valve to circumvent the tank when the wand is in operation. Will this solve the issue?
<a href="https://i.stack.imgur.com/mgT3q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mgT3q.png" alt="enter image description here"></a></p>
| |airflow|vacuum|pneumatic|vacuum-pumps| | <p>I'm not sure what the first paragraph was about (pickup tool?), but if the vacuum reserve is installed like you have drawn it, there will be no appreciable loss of airflow when the solenoid is open. More likely is there will be a slight increase in flow since the line loss of the previous piping will be reduced because it is reduced by a large opening. It's not clear to me what the function on the bypass pipe on the bottom is, either. Just make the reserve tank larger and save a solenoid.</p>
| 35878 | How do I increase negative air pressure without inhibiting airflow? |
2020-05-21T08:53:44.027 | <p>Ive been given a question and asked to calculate the kinetic energy of a gas given the temperature, pressure, diameter of pipe and velocity. I've been looking and have been unable to find any equations that can use this information. The only hint Ive been given is "STP conditions" but am unsure how that is relevant to solving the question. I cannot figure out how to solve this question. Below is the question given. </p>
<p>Air at 300 °C and 130 kPa flows through a horizontal 7-cm ID pipe at a velocity of 42.0 m/s.
a) Calculate the kinetic energy (in J/s), assuming ideal gas behaviour.</p>
<p>Any formulas or advice about how to approach this question would be greatly appreciated.</p>
| |chemical-engineering| | <p>I worked out an answer below. If you want to work it out for yourself just look at the equations that don't have any units in them.</p>
<h1>Quick and dirty guess:</h1>
<p><span class="math-container">$$\text{K.E.}=(\text{mass flow rate}) \cdot (\text{KE term from Bernoulli equation})$$</span></p>
<p><span class="math-container">$$\text{K.E.}=(\bar{V} \cdot A \cdot \rho) \cdot (\frac{\bar{V}^2}{2})$$</span></p>
<p>Use the temperature, pressure, and composition of Earth air to calculate density <span class="math-container">$\rho$</span>. Calculate <span class="math-container">$A$</span> as cross sectional flow area of the pipe. <span class="math-container">$\bar{V}$</span> is average velocity.</p>
<h1>Verbose exploration:</h1>
<p>When asked about kinetic energy or potential energy you should immediate think "energy balances".</p>
<p>Bernoulli equation</p>
<p>The "Bernoulli equation without friction" (as derived and shown in Chapter 4, page 87 of <a href="http://www.worldcat.org/oclc/835549358" rel="nofollow noreferrer">Unit Operations of Chemical Engineering by Warren L. McCabe 7th Edition</a>) is possibly a useful energy balance equation for a unit mass flowing along a streamline through a tube. The energy of fluid at point <span class="math-container">$a$</span> along the tube is balanced with energy at point <span class="math-container">$b$</span>. Here is the equation and some accompanying explanations from McCabe:</p>
<blockquote>
<p><span class="math-container">$$\frac{p_a}{\rho} + gZ_a + \frac{u^2_a}{2} = \frac{p_b}{\rho} + gZ_b + \frac{u^2_b}{2}$$</span></p>
</blockquote>
<p>Where:</p>
<p><span class="math-container">$p_a$</span> : absolute pressure of fluid in the tube at point <span class="math-container">$a$</span></p>
<p><span class="math-container">$p_b$</span> : absolute pressure of fluid in the tube at point <span class="math-container">$b$</span></p>
<p><span class="math-container">$\rho$</span> : fluid density (assumed constant in this example)</p>
<p><span class="math-container">$g$</span> : gravitational constant</p>
<p><span class="math-container">$Z_a$</span> : elevation of the tube at point <span class="math-container">$a$</span></p>
<p><span class="math-container">$Z_b$</span> : elevation of the tube at point <span class="math-container">$b$</span></p>
<p><span class="math-container">$u_a$</span> : velocity of a unit mass of fluid in the tube at point <span class="math-container">$a$</span> along a streamline</p>
<p><span class="math-container">$u_b$</span> : velocity of the same unit mass of fluid in the tube at point <span class="math-container">$a$</span> along the streamline</p>
<blockquote>
<p>Each term in [the equation] is a scalar and has the dimensions of
energy per unit mass, representing a mechanical energy effect based on
a unit mass of the flowing fluid. Terms <span class="math-container">$gZ$</span> and <span class="math-container">$u^2/2$</span> are the
potential and kinetic energy, respectively, of a unit mass of fluid;
and <span class="math-container">$p/\rho$</span> represents the mechanical work done by forces, external
to the stream, on the the fluid in pushing it into the tube or the
work recovered from the fluid leaving the tube.</p>
<p>...</p>
<p>The term <span class="math-container">$u^2/2$</span> in [the equation] is the kinetic energy of a unit
mass of fluid all of which is flowing at the same velocity <span class="math-container">$u$</span>.</p>
</blockquote>
<p>A later version of this equation takes friction and the fact that <span class="math-container">$u$</span> isn't average velocity:</p>
<blockquote>
<p><span class="math-container">$$\frac{p_a}{\rho} + gZ_a + \frac{\alpha \bar{V}^2_a}{2} = \frac{p_b}{\rho} +
gZ_b + \frac{\alpha \bar{V}^2_b}{2} + h_f$$</span></p>
</blockquote>
<p>where:</p>
<p><span class="math-container">$\alpha$</span> : kinetic energy correction factor ("<span class="math-container">$\alpha$</span> is <span class="math-container">$2.0$</span> for laminar flow and is about <span class="math-container">$1.05$</span> for highly turbulent flow.")</p>
<p><span class="math-container">$\bar{V}^2_a$</span> : average velocity of the fluid at point <span class="math-container">$a$</span></p>
<p><span class="math-container">$\bar{V}^2_b$</span> : average velocity of the fluid at point <span class="math-container">$a$</span></p>
<p><span class="math-container">$h_f$</span> : mechanical energy lost to friction</p>
<h2>Applicability to compressible flow</h2>
<p>Although the equation is first used in McCabe's chapter on incompressible flow, it is used in the McCabe's treatment of compressible flow in a subsequent chapter as a way to show how the energy balance between kinetic energy, pressure, and friction loss. In other words, that over a short enough length of pipe, all three quantities must sum to zero:</p>
<p><span class="math-container">$\frac{dp}{\rho} + d \left( \frac{\alpha \bar{V}^2}{2} \right) + gdZ + dh_f =
0$</span></p>
<p>where <span class="math-container">$d$</span> carries the same meaning as the <span class="math-container">$d$</span> in a derivative of differential calculus. The equation's applicability to compressible flow depends upon whether density changes are important. The air in our example is compressible but we are not having to account for density changes anywhere in the example as stated. Therefore, using the kinetic energy term <span class="math-container">$\frac{\alpha \bar{V}^2}{2}$</span> to answer the question should be appropriate.</p>
<h2>Strategy</h2>
<p>You were given <span class="math-container">$\bar{V}=42.0 \space \frac{\text{m}}{\text{s}}$</span> and you don't care about what happens at two points. Also, since you were not provided material of construction information about the pipe you cannot calculate <span class="math-container">$\alpha$</span> (which is a function of pipe smoothness) so therefore the easiest way to move forward is to assume <span class="math-container">$\alpha=1$</span> and <span class="math-container">$h_f=0$</span> (flow is turbulent and pipe wall is frictionless). Regarding <span class="math-container">$\alpha$</span> on page 113, McCabe indicates "In most practical situations [<span class="math-container">$\alpha$</span> is] taken as unity in turbulent flow". From personal experience, my guess is the inner pipe diameter of <span class="math-container">$D=7 \space \text{cm}$</span> coupled with air velocity of <span class="math-container">$\bar{V}=42 \space \frac{\text{m}}{\text{s}}$</span> is turbulent, but it may be useful to check that.</p>
<p>Once we have a density we can use the cross-sectional area of the pipe plus the average air velocity to calculate a mass flow rate. Then, we just have to multiply this mass flow rate with the kinetic energy term of the Bernoulli equation to calculate the kinetic energy component of the stream.</p>
<h2>Check turbulence state via Reynolds number</h2>
<p>If the Reynolds number is greater than <span class="math-container">$4,000$</span> then flow in the pipe is definitely turbulent. (from McCabe pg. 54: "Under ordinary conditions, the flow in a pipe or tube is turbulent at Reynolds numbers above about <span class="math-container">$4,000$</span>").</p>
<p>The formula for Reynolds number is:</p>
<p><span class="math-container">$$\text{Re} = \frac{D \bar{V} \rho}{\mu}$$</span></p>
<p>where:</p>
<p><span class="math-container">$D$</span> : inner diamter of tube</p>
<p><span class="math-container">$\bar{V}$</span> : average velocity of fluid</p>
<p><span class="math-container">$\rho$</span> : density of fluid</p>
<p><span class="math-container">$\mu$</span> : [dynamic] viscosity of liquid (not kinematic viscosity)</p>
<p>From the information provided I believe it should be possible to calculate <span class="math-container">$\rho$</span> since you are permitted to treat air as an ideal gas. <span class="math-container">$\mu$</span> can be looked up from a table.</p>
<h2>Calculate density</h2>
<p>From the ideal gas law:</p>
<p><span class="math-container">$$PV = \frac{m}{M} RT$$</span></p>
<p>density can be calculated</p>
<p><span class="math-container">$$\rho = \frac{m}{V} = \frac{PM}{RT}$$</span></p>
<p>where:</p>
<p><span class="math-container">$\rho$</span> : gas density</p>
<p><span class="math-container">$m$</span> : unit mass of gas</p>
<p><span class="math-container">$V$</span> : unit volume of gas</p>
<p><span class="math-container">$P$</span> : absolute pressure of gas</p>
<p><span class="math-container">$M$</span> : molecular weight of gas</p>
<p><span class="math-container">$R$</span> : ideal gas constant</p>
<p><span class="math-container">$T$</span> : absolute temperature of the gas</p>
<p>Given <span class="math-container">$P=130 \space \text{kPa}$</span> (which I am assuming is absolute pressure, not gauge pressure), <a href="https://physics.stackexchange.com/a/466096"><span class="math-container">$M_\text{air}=28.8 \space \text{mol}$</span></a>, <a href="https://web.archive.org/web/20200510082716/https://physics.nist.gov/cgi-bin/cuu/Value?r" rel="nofollow noreferrer"><span class="math-container">$R=8.314462618 \space \frac{\text{J}}{\text{mol} \cdot \text{K}}$</span></a>, and <span class="math-container">$T=(300+273.15)=573.15 \space \text{K}$</span>, then <span class="math-container">$\rho$</span> can be calculated.</p>
<p><span class="math-container">$$\rho = \frac{(130 \space \text{kPa}) \cdot ({28.8 \space \frac{\text{g}}{\text{mol}}})}{(8.314462618 \space \frac{\text{J}}{{\text{mol}} \cdot \text{K}}) \cdot (573.15 \space \text{K})} \cdot (\frac{\text{J}}{\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}}) \cdot (\frac{1000 \space \text{Pa}}{\text{kPa}}) \left( \frac{\left( \frac{\text{kg}}{\text{s}^2 \cdot \text{m}} \right)}{\text{Pa}} \right) \cdot (\frac{\text{kg}}{1000 \space \text{g}})$$</span></p>
<p><span class="math-container">$$\rho = 0.786 \space \frac{\text{kg}}{\text{m}^3}$$</span></p>
<p>This value seems somewhat reasonable since it is close to the air density calculated by <a href="https://www.omnicalculator.com/physics/air-density?advanced=1&c=USD&v=P:130!kPa,Temp:300!C,aaa:0.000000000000000" rel="nofollow noreferrer">this real air density calculator</a> which I'm guessing takes compressibility into account.</p>
<p><a href="https://i.stack.imgur.com/cg5eA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cg5eA.png" alt="enter image description here"></a></p>
<h2>Look up air viscosity</h2>
<p>Using <a href="https://www.engineeringtoolbox.com/air-absolute-kinematic-viscosity-d_601.html" rel="nofollow noreferrer">this webpage</a> for calculating air viscosity as a function of temperature, we get:</p>
<p><span class="math-container">$$\mu = 29.28 \cdot 10^{-6} \space \text{Pa} \cdot \text{s}$$</span></p>
<h2>Calculate Reynolds number to determine if turbulent state present</h2>
<p>Plugging our values into the Reynolds equation, we get:</p>
<p><span class="math-container">$$\text{Re} = \frac{\left( 7 \text{cm} \right) \left( 42.0 \frac{\text{m}}{\text{s}} \right)
\left( 0.786 \frac{\text{kg}}{\text{m}^3} \right)}{\left( 29.28 \cdot 10^{- 6}
\text{Pa} \cdot \text{s} \right)} \cdot \left( \frac{\text{m}}{100 \space \text{cm}} \right) \cdot
\left( \frac{\text{Pa}}{\left( \frac{\text{kg}}{\text{s}^2 \text{m}} \right)} \right)$$</span></p>
<p><span class="math-container">$$\text{Re} = 80,000 $$</span></p>
<p>This is well above <span class="math-container">$4,000$</span> and therefore the flow is turbulent.</p>
<h2>Calculate mass flow rate</h2>
<p>Since we're given average velocity <span class="math-container">$\bar{V}$</span>, pipe diameter <span class="math-container">$D=(7 \space \text{cm})$</span>, and density <span class="math-container">$\rho$</span>, we can calculate mass flow rate <span class="math-container">$\dot{m}$</span> like so:</p>
<p><span class="math-container">$$\dot{m} = (\text{volumetric flow rate}) \cdot (\text{density})$$</span></p>
<p><span class="math-container">$$\dot{m} = (\bar{V} \cdot A) \cdot (\rho)$$</span></p>
<p><span class="math-container">$$\dot{m} = (42.0 \frac{\text{m}}{\text{s}}) \cdot (\pi \cdot \left( \frac{7 \space \text{cm}}{2} \right)^2 ) \cdot (0.786 \frac{kg}{\text{m}^3}) \cdot \left( \frac{\text{m}}{100 \space \text{cm}} \right)^2$$</span></p>
<p><span class="math-container">$$\dot{m} = 0.127 \frac{\text{kg}}{\text{s}} $$</span></p>
<h2>Apply Bernoulli equation</h2>
<p>Since each term in the Bernoulli equation is in units of energy per unit mass (but with the actual mass factor divided out), all you need to do to calculate kinetic energy is to multiply the mass flow rate by the kinetic energy term, <span class="math-container">$\alpha \frac{\bar{V}}{2}$</span>,from one side of the Bernoulli equation.</p>
<p><span class="math-container">$$\text{K.E.} = \dot{m} \cdot \frac{\alpha \bar{V}^2}{2} $$</span></p>
<p>This makes sense from a physics perspective because kinetic energy of any object is proportional to its mass multiplied by its velocity squared. Here, the constant of proportionality is <span class="math-container">$\frac{\alpha}{2}$</span>.</p>
<p>Because flow is turbulent (<span class="math-container">$\text{Re}>4000$</span>) we can set <span class="math-container">$\alpha=1.0$</span>, simplifying our calculation.</p>
<p><span class="math-container">$$\text{K.E.} = \dot{m} \cdot \frac{\bar{V}^2}{2} $$</span></p>
<p>Plugging in our values we get:</p>
<p><span class="math-container">$$\text{K.E.} = \left( 0.127 \frac{\text{kg}}{\text{s}} \right) \cdot \left( \frac{(42.0 \space \frac{\text{m}}{\text{s}})^2}{2} \right) \cdot \left( \frac{\text{J}}{\text{kg} \cdot \frac{\text{m}^2}{\text{s}^2}} \right) $$</span></p>
<p><span class="math-container">$$\text{K.E.} = 112 \space \frac{\text{J}}{\text{s}} $$</span></p>
<h2>Summary</h2>
<p>When sizing a gas compressor to move air at an average velocity of <span class="math-container">$42.0 \space \frac{\text{m}}{\text{s}}$</span> at this point through the pipe at temperature <span class="math-container">$300^{\circ}\text{K}$</span>, and absolute pressure <span class="math-container">$130 \space \text{kPa}$</span>, <span class="math-container">$112 \space \frac{\text{J}}{\text{s}}$</span> of energy is needed in the form of kinetic energy. Other calculations would need to be performed to:</p>
<ul>
<li><p>Determine the energy required to bring air starting (presumably) from STP conditions (<span class="math-container">$0^{\circ}\text{C}$</span>, <span class="math-container">$100 \space \text{kPa}$</span>) to <span class="math-container">$130 \space \text{kPa}$</span></p></li>
<li><p>Determine the energy required to overcome and elevation increase (or energy available from elevation decreases).</p></li>
<li><p>Determine the energy required to overcome friction losses throughout the system.</p></li>
</ul>
| 35885 | How to calculate kinetic energy of gas flow in pipe given tempearture, pressure, pipe diameter and velocity |
2020-05-21T19:41:36.500 | <p>I'm struggling with the following problem:
<a href="https://i.stack.imgur.com/cIEcP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cIEcP.png" alt="enter image description here"></a></p>
<p>The boundary conditions are <span class="math-container">$w(x,y) = 0$</span> around the edges.</p>
<p>I am supposed to find the natural frequencies of the system as a whole, and find the solution for <span class="math-container">$w(x,y,t)$</span> based on the inout <span class="math-container">$p_{in}$</span>. The air can be perceived as a 1-D continuum to simplify the equations.</p>
<p>There is a body of air above the membrane with the pressure at height L above the membrane being defined as <span class="math-container">$p_{in} = p_0 e^{jwt}$</span>
I know how to solve this problem for the membrane alone with some initial conditions but I'm lost as how to work with this problem as a whole.
I don't see how could I use the known equation <span class="math-container">$\frac{\partial^2 w}{\partial t^2} = \frac{\sigma_1}{\rho}(\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2})$</span> to find the solution.</p>
<p>So my guess would be that I have to find the solution for a 2D wave equation with a source, then write a boundary condition equating the pressure in the air at the surface of the membrane to the source in the membrane equation and go from there.
Meaning <span class="math-container">$\frac{\partial^2 w}{\partial t^2} = \frac{\sigma_1}{\rho}(\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2}) - \frac{p(x,y,t)}{h \rho}$</span> and <span class="math-container">$p_{air}(x = L, t) = p(x,y,t)$</span>.</p>
<p>Any help, insight or link to relevant materials will be greatly appreaciated.</p>
| |mathematics|vibration| | <p>I wasn't able to find the analytical solution but I at least tried to make a few steps in the right direction, which was enough for a B from the professor, which likely means it was indeed a step in the correct direction.</p>
<p>The solution of a 2D wave equation with a source is a sum of homogenous + particular solution of the equation.
<span class="math-container">$$ w(x,y,t) = w_h(x,y,t) + w_p(x,y,t)$$</span>
The boundary conditions for the system at the air-membrane junction are
<span class="math-container">$$\dot{\xi}_{air}((L-w(x,y,t),t) = \frac{\partial w(x,y,t)}{\partial t}$$</span>
and
<span class="math-container">$$p(x,y,t) = p_{air}(L-w(x,y,t),t)$$</span></p>
<p>The obstacles in being able to solve this equation with given boundary conditions are</p>
<p>i) <span class="math-container">$(L-w(x,y,t))$</span> is problematic. The approximation of <span class="math-container">$p_{air}(L-w(x,y,t),t) = p_{air}(L,t)$</span> is not feasible.</p>
<p>ii) The necessity to find a particular solution of a sourced 2D wave equation satisfying these conditions.</p>
<p>Due to my background in Mechanical engineering (and lack of experience with finding solution of this type of PDEs) I wasn't able to continue any further. </p>
<p>If anyone ever faces a similar problem and manages to make any further progress, please do contact me as I'm interested in what the actual solution looks like. </p>
| 35890 | Air - rectangular membrane system vibration and natural frequency |
2020-05-23T09:48:13.257 | <p><strong>What I want to build:</strong></p>
<p>So I have an AMD Wraith cooler. And I want to put it on my desk.</p>
<p>I managed to connect the coolers power input (12v fan header) to an USB port. And it runs on 5V.
I don't want the cooler to spin so i disconnected the fans. </p>
<p><strong>Problem:</strong></p>
<p>I don't have enough voltage for the RGB of the fans, the blue color is barely visible. Ring and logo work fine.</p>
<p><strong>Question:</strong></p>
<p>Can I connect to the cooler's power input 2 USBs? That would give the cooler 10V?</p>
<p>Problem is that one USB will connect to another USB... I think this might be a problem...
I don't really know...</p>
<p>Thank you!</p>
| |electrical-engineering| | <blockquote>
<p>Can I connect to the cooler's power input 2 USBs? That would give the cooler 10V?</p>
</blockquote>
<p>No. Both ports are fed from the same 5 V power supply so their common pins are connected. Connecting the +5 V of one port to the ground of the other will short-circuit the first port.</p>
| 35911 | More voltage through USB ports |
2020-05-24T03:30:03.020 | <p>I was watching <a href="https://youtu.be/em1O8mz7sF0?t=242" rel="nofollow noreferrer">this</a> video about how steering on a vehicle. It states:</p>
<blockquote>
<p>If you track the meeting point of the left and right wheels, you can see that the meeting point <strong>always</strong> lies on the rear wheel line.</p>
</blockquote>
<p>I have a few questions about this:</p>
<ol>
<li>If the car is longer or shorter or wider or thinner, the meeting point of the left and right wheels <strong>would not</strong> be on the rear wheel line. How then is the steering setup such that the meeting point is <strong>always</strong> in line with the rear wheels? Is there a process for determining it?
Below is a photo of the meeting point not aligning with the rear wheels:
<a href="https://i.stack.imgur.com/gJMns.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gJMns.png" alt="Common point"></a></li>
<li>Why does the meeting point need to be in line with the rear wheels? Would the wheels skip or slip along otherwise?</li>
</ol>
| |manufacturing-engineering|wheels|car| | <p>Wikipedia explains it fairly well.</p>
<p><a href="https://i.stack.imgur.com/UpUXQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UpUXQ.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Image source: <a href="https://en.wikipedia.org/wiki/Ackermann_steering_geometry" rel="nofollow noreferrer">Ackermann steering geometry</a>.</em></p>
<blockquote>
<p>A simple approximation to perfect Ackermann steering geometry may be generated by moving the steering pivot points inward so as to lie on a line drawn between the steering kingpins and the centre of the rear axle. The steering pivot points are joined by a rigid bar called the tie rod which can also be part of the steering mechanism, in the form of a rack and pinion for instance. With perfect Ackermann, at any angle of steering, the centre point of all of the circles traced by all wheels will lie at a common point. Note that this may be difficult to arrange in practice with simple linkages, and designers are advised to draw or analyse their steering systems over the full range of steering angles.</p>
</blockquote>
<p>If the vehicle is stretched then the steering pivot points will move out and the angle of the solid arms (dark brown in the image) will change to suit.</p>
<p>You might enjoy the <a href="https://www.youtube.com/watch?v=ob-FUF0Sh_E" rel="nofollow noreferrer">Project Binky steering setup</a>. This is episode 11 of (currently) 31 half-hour episodes and is a fantastic series.</p>
| 35924 | Meeting point of turning wheels with rear wheels |
2020-05-25T17:54:05.187 | <p>I need to decide upon a color coating for a steel surface. This will be a grill top accessory and will be in contact with flames. I would prefer a black/brown coating color. I was considering teflon but what other suitable coatings are out there?</p>
| |steel|manufacturing-engineering| | <p>Teflon will burn off the top of grill racks; vaporizes above about 700 F. Carbon steel will turn brown with oxides, Stainless will turn black with oxides and some of mine may be carbon from burned stuff (I have not used the "self clean" burn for awhile). I suggest an oxide ( iron ,chrome) coating. It depends on what is the temperature of "flames" , a few hundred degrees or 1200 F degrees</p>
| 35945 | What are good color coatings for a steel surface? |
2020-05-26T17:17:34.283 | <p><a href="https://i.stack.imgur.com/rTGkZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rTGkZ.jpg" alt="enter image description here" /></a></p>
<p>It seems to me there is too many unknowns to solve further</p>
| |statics| | <p>Yes, it's a determinate structure and you can solve for T4 and T5 using statics.</p>
<p><strong>Method of Joints</strong></p>
<p>One approach is to use Method of Joints to progressively solve for the member forces, and it looks like this is the approach you've chosen. With two equations of equilibrium (<span class="math-container">$\sum F_x = 0$</span>, <span class="math-container">$\sum F_y = 0$</span>) we can solve for two unknown member forces at any node. From your calculations, this is what you've arrived at for the node where <span class="math-container">$F_2$</span> is applied (Node C, below). Everything is known except T4 and T5. Sometimes it is necessary to use a system of equations to achieve a solution, and that is the case here.</p>
<p>Below is a sketch of the truss with named nodes for ease of reference. I kept A and B as you had designated. Lengths of the angled members are noted in parentheses.</p>
<p><a href="https://i.stack.imgur.com/NF4le.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NF4le.jpg" alt="Truss Sketch"></a></p>
<p>The steps followed in your calculations are appropriate for the Method of Joints:</p>
<ol>
<li>Determine external support reactions</li>
<li>Solve Node B</li>
<li>Solve Node G</li>
<li>Solve Node C (where the difficulty arose)</li>
</ol>
<p>Looking at Node C:</p>
<p><a href="https://i.stack.imgur.com/J2u7Bm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J2u7Bm.jpg" alt="Node C"></a></p>
<p><span class="math-container">$F_2$</span> is known from the problem statement. <span class="math-container">$F_{CB}$</span> and <span class="math-container">$F_{CG}$</span> were solved for in Steps 2 and 3, respectively, leaving us to solve for <span class="math-container">$F_{CD}$</span> and <span class="math-container">$F_{CA}$</span>.</p>
<p>We use the known geometry to break each member force into x and y components, and then solve our two equations of equilibrium, <span class="math-container">$\sum F_x = 0$</span>, <span class="math-container">$\sum F_y = 0$</span>. Both equations will have <span class="math-container">$F_{CD}$</span> and <span class="math-container">$F_{CA}$</span> as unknowns. Solve simultaneously and, <em>voila</em>.</p>
<p><strong>Side note:</strong> I find it convenient to use the triangle geometry of each member (i.e. the member is the hypotenuse of a right triangle) along with SOH-CAH-TOA to solve for the X and Y components of any member force. It's often simpler than dealing with the angles themselves. As an example, the vertical component of the force in Member AC can be calculated as <span class="math-container">$F_{AC} \frac{2.57}{3.95}$</span>.</p>
<p><strong>Method of Sections</strong></p>
<p>Another option is to make a vertical section cut through T5, T4, and T3 and use Method of Sections to solve for those element forces. For example, take the portion of the structure from the section cut to the left. We can sum moments about Support A to determine T5 (<span class="math-container">$F_{DC}$</span>), because T4 and T3 will produce no moment about A. Sum of the forces in X and Y directions get you the rest of the way.</p>
<p><a href="https://i.stack.imgur.com/vUmCVm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vUmCVm.jpg" alt="Method of Sections"></a></p>
<p><strong>Solving the Remainder of the Truss</strong></p>
<p>Should you want or need to solve for all the member forces, you can continue with Method of Joints. By inspection we can say the only way equilibrium can be satisfied at Node F is for the forces in both members to be zero. Similarly, with no external force applied at Node D, we know that <span class="math-container">$F_{DC} = F_{DE}$</span> and thus <span class="math-container">$F_{DA} = 0$</span>.</p>
| 35956 | How do I solve this truss to get T4 and T5 force I got T3 |
2020-05-29T11:49:11.973 | <p>The Maximum induced bending stress within the beam.</p>
<p>Given that:</p>
<p><span class="math-container">$$\frac{My}{I}$$</span></p>
<p>The beam depth is 300mm, the width is 150mm. The beam is an I section beam with a 5mm wall thickness.</p>
<p><a href="https://i.stack.imgur.com/llFD5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/llFD5.jpg" alt="enter image description here"></a></p>
<p>My calculations:</p>
<p><span class="math-container">$$\begin{align}
I_{total} &= ( I_1 + A_1 \cdot d_1^2) + ( I_2 + A_2 \cdot d_2^2) + ( I_3 + A_3 \cdot d_3^2) \\
I_3 = I_1 &= \frac{1}{12}bh^3 = \frac{1}{12} \cdot 150 \cdot 5^3 = 1.562 \cdot10^3 \\
I_2 &= \frac{1}{12}bh^3 = 150 \cdot 290^3 = 1.016 \cdot 10^7 \\
A_3 &= A_1 = bh = 150 \cdot 5 = 750\text{ mm}^2 \\
A2 &= bh = 5 \cdot 290 = 1,450\text{ mm}^2 \\
I_{total} &= (1.562 \cdot 10^3 + 750^2 \cdot 150) \\
&+ (1.016 \cdot 10^7 + 1,450^2 \cdot 0) \\
&+ (1.562\cdot10^3 + 750^2 \cdot 150) \\
&= 42.797.499\text{ mm}^4 \\
&= 4.28\cdot10^{7}\text{ mm}^4 \\
\sigma_{bend,max} &= \frac{Mc}{I} \\
M &= 48\text{ kNm} = 48\cdot 10^6\text{ Nm} \\
c &= 150\text{ mm} = 0.150\text{ m} \\
I &= 4.28\cdot 10^{7}\text{ mm}^4 = 4.28 \cdot 10^4\text{ m}^4 \\
\therefore \sigma_{bend,max} &= \frac{48,000 \cdot 0.1475}{4.28 \cdot 10^4} \\
&= 168.23\text{ MPa}
\end{align}$$</span></p>
<p>The maximum induced bending stress within the beam is 168.23;MPa</p>
<p>These are my calculations. Are these correct? </p>
<p>I am not sure on the 147.5. As seen different ways to calculate.</p>
<p>Also not sure if i have done the calculations correct.</p>
| |beam|moments|homework| | <p>Two concepts are in play here:</p>
<ol>
<li>Calculation of moment of inertia for a composite section.</li>
<li>Calculation of elastic beam stresses.</li>
</ol>
<p><strong>First, the calculation of <span class="math-container">$I_{total}$</span> for the wide flange section.</strong></p>
<p>Your equation for <span class="math-container">$I_{total}$</span> via the parallel axis theorem is correct, but the execution went awry. </p>
<p><span class="math-container">$$I_{total} = \sum (I + A \cdot d^2)$$</span></p>
<p>Since the section is symmetric, the centroid is at mid-height and the <span class="math-container">$Ad^2$</span> term will be zero for the web component. Thus:</p>
<p><span class="math-container">$$I_{total} = 2*(I_{flange} + A_{flange} \cdot {d_{flange}}^2) + I_{web}$$</span></p>
<p>Since we're calculating moment of inertia about a horizontal line, let b = the width and h = the height of each component. Let d = the vertical distance from the composite centroid to the centroid of the component. Note that this is a totally distinct choice and concept from the distance used to calculate maximum stress.</p>
<p><span class="math-container">$$I_{flange} = \frac{1}{12} bh^3 = 1,562 \ mm^4$$</span>
<span class="math-container">$$A_{flange} = bh = 750 \ mm^2$$</span>
<span class="math-container">$$d_{flange} = 147.5 \ mm$$</span>
<span class="math-container">$$I_{web} = \frac{1}{12} bh^3 = 1.016 \cdot 10^7 \ mm^4$$</span></p>
<p><span class="math-container">$$I_{total} = 4.2797 \cdot 10^7 mm^4$$</span></p>
<p>In your numerical calculation of <span class="math-container">$I_{total}$</span> it looks as though you squared the A terms instead of the d terms. It also looks as though you measured d to the outer edge of the flange instead of to the centroid. Note also that to convert from <span class="math-container">$mm^4$</span> to <span class="math-container">$m^4$</span>, divide by <span class="math-container">$1000^4$</span>.</p>
<p>An easy way to check your results is to use one of the many online moment of inertia calculators. You can find ones for an array of common cross sections, including wide flange sections.</p>
<p><strong>Now the calculation of maximum bending stress</strong></p>
<p>When we're assuming that plane sections remain plane and that the section remains elastic (an Euler-Bernoulli beam), the equation for pure bending stress is as you noted.</p>
<p><span class="math-container">$$\sigma = \frac{My}{I}$$</span></p>
<p>The derivation of this equation should be available in any introductory mechanics of materials textbook. <a href="https://en.wikipedia.org/wiki/Bending" rel="nofollow noreferrer">Wikipedia</a> also contains a discussion of Euler-Bernoulli bending theory.</p>
<p>In this equation, <span class="math-container">$y$</span> is the distance from the neutral axis (our centroid). At the neutral axis, bending stress is zero. The maximum bending stress must therefore occur at the cross section height that is farthest from the neutral axis - at the extreme fiber. For the beam you sketched, the extreme fiber is located at the outer edge of the flange, <span class="math-container">$150 \ mm$</span> from the centroid. Because we are so often interested in the maximum bending stress, this particular distance of <span class="math-container">$y$</span> is assigned its own variable name - typically, <span class="math-container">$c$</span>.</p>
<p>In design of steel beams, we often talk about the "elastic <a href="https://en.wikipedia.org/wiki/Section_modulus" rel="nofollow noreferrer">section modulus</a>" which simply lumps together the equations you've looked at here, giving engineers a quick equation to calculate the maximum bending stress in the elastic section.</p>
<p><span class="math-container">$$Elastic \ Section \ Modulus, \ S = \frac{I}{c}$$</span></p>
<p>This gives us a quick way to calculate the yield Moment, <span class="math-container">$M_y$</span>.</p>
<p><span class="math-container">$$M_y = \sigma_y \cdot S$$</span></p>
<p>Where <span class="math-container">$\sigma_y$</span> is the yield stress of the material.</p>
<p><strong>Some notes on real-world steel beams</strong></p>
<p>As Leafk noted, wide flange beams are usually designed to optimize the use of material, which typically results in flanges being thicker than webs. And while it's beyond the scope of this question it's worth noting that there are several other beam design considerations beyond pure bending stress.</p>
| 35983 | Maximum induced bending stress within the beam |
2020-05-30T03:38:06.473 | <ol>
<li><p>Why exactly isn't the US mining and producing its own rare earth metals (REEs)? I'm baffled by the mixed messages below. Because China uniquely possesses some REEs? US safeguards against pollution more than China? US would take too much time or is inefficient to establish its own production?</p></li>
<li><p>Can the US gainfully do so? How long would the US need to self-rely? </p></li>
</ol>
<p>I know of <a href="https://en.wikipedia.org/wiki/Mountain_Pass_mine" rel="nofollow noreferrer">the Mountain Pass mine</a> in California, <a href="https://www.google.com/search?client=firefox-b-d&q=canada+rare+earth+metals" rel="nofollow noreferrer">Canada</a>, Australia's <a href="https://en.wikipedia.org/wiki/Lynas#Mount_Weld" rel="nofollow noreferrer">Mount Weld</a> and Lynas Corp's <a href="https://www.ft.com/content/ae16d3b2-3a28-447e-a5b4-7e8064a20f17" rel="nofollow noreferrer">processing facility in Malayasia</a> and <a href="https://old.reddit.com/r/wallstreetbets/comments/bqxqhd/xi_sends_trump_a_message_rareearth_export_ban_is/" rel="nofollow noreferrer">Brazil, Vietnam, and Russia</a>.</p>
<p><a href="https://old.reddit.com/r/geopolitics/comments/cwc0w9/pentagon_in_talks_with_australia_on_rare_earths/eybjzgi/" rel="nofollow noreferrer">Pentagon in talks with Australia on rare earths plant : geopolitics</a></p>
<blockquote>
<p>There is a large subsection of rare earth materials that are not found outside of China in any mining development.</p>
<p>There's I believe a list of 18 and Australia's Lynas only has about 5-6 iirc, and altogether the world can't cobble all 18 together without relying on China.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/geopolitics/comments/chh3f9/rare_earths_in_the_uschina_trade_war/euzlptg/" rel="nofollow noreferrer">Rare Earths in the US-China Trade War : geopolitics</a></p>
<blockquote>
<p>There's no suppliers outside China. There are potential minerals deep in some countries with no technology or existing extraction operation.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/geopolitics/comments/davucd/united_states_implements_energy_resource/f1vlie0/" rel="nofollow noreferrer">United States implements Energy Resource Governance Initiative (ERGI) with countries such as Australia, Peru, Argentina, Namibia and the Philippines to limit China's control of rare earth minerals : geopolitics</a></p>
<blockquote>
<p>It is not the source of supply that matters but the extraction and refining tech that help China dominate. The supply still have to be sent to China for processing.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/wallstreetbets/comments/gppjnu/pentagon_legislation_seeks_to_end_us_dependence/" rel="nofollow noreferrer">Pentagon legislation seeks to end US dependence on Chinese rare earth metals : wallstreetbets</a></p>
<blockquote>
<ul>
<li><p>REMs actually aren't that rare, despite their name. Japan recently found deposits that could basically supply Earth from now to infinity. The US has huge deposits as well.</p></li>
<li><p>The US already has infrastructure in place. Before China undercut the market, the US actually supplied the world with most of the REMs needed from mining at Mountain Pass. Mountain Pass still exists and has output shutdown because China didn't make it profitable, but the point is that it still exists and infrastructure is already there. In theory, we could just start it up again without needing too much more investment.</p></li>
<li><p>REMs can easily be recycled. In fact, this is what Apple does out of environmental concerns and to hopefully reach a point where REMs never have to be mined for again:</p></li>
</ul>
<p><a href="https://www.engadget.com/2019-09-18-apple-will-use-recycled-rare-earth-metals-iphone-taptic-engine.html" rel="nofollow noreferrer">https://www.engadget.com/2019-09-18-apple-will-use-recycled-rare-earth-metals-iphone-taptic-engine.html</a></p>
<ul>
<li>The US and even private companies have some stockpiles of REMs to hold out in the short term if there were disruptions in REM supplies.</li>
</ul>
<p>When you combine Mountain Pass, the ability to recycle REMs, and REM reserves, there's very little strategic gain for China to cut supplies. It'll just encourage development in other countries and shifting of sourcing to places like Japan. The US would just fire up Mountain Pass again. China would lose market share. In fact, China already tried to restrict REM supplies earlier in the 2010s; it wasn't effective at all.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/wallstreetbets/comments/ccplf6/china_accounts_for_around_80_of_us_rare_earths/etoqcb5/" rel="nofollow noreferrer">enronCoin. 11 points 10 months ago </a></p>
<blockquote>
<p>You definitely aren’t wrong, but maybe you’re underestimating the fact that China has 1/3 of the world rare earth reserves and 40x as much untapped supply as the US.</p>
<p>Who can produce at the lowest cost? Probably China, where rare earth is abundant and labor is cheap.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/wallstreetbets/comments/ccplf6/china_accounts_for_around_80_of_us_rare_earths/etopnot/" rel="nofollow noreferrer">LukeSkyWRx. 12 points 10 months ago*</a></p>
<blockquote>
<p>They are also the only ones that will separate the ores as it is a really nasty process with vats of acid and tons of toxic chemical waste. Look up lake Baotou in China, that is where lots of China’s rare earths come from and where they pump the waste when they are done. <a href="http://www.bbc.com/future/story/20150402-the-worst-place-on-earth" rel="nofollow noreferrer">http://www.bbc.com/future/story/20150402-the-worst-place-on-earth</a> it could be done ‘clean’ if there were some incentive to do so.</p>
<p>Since lots of the deposits in the US have a bunch of Thorium in them as well they are heavily regulated similar to uranium mining. With that extra burden most deposits are uneconomical to extract.</p>
</blockquote>
<p><a href="https://old.reddit.com/r/wallstreetbets/comments/brr9u9/china_mouthpiece_on_twitter_says_they_may_stop/eog5klq/" rel="nofollow noreferrer">China Mouthpiece on Twitter says they may stop Rare Earth sales to US. : wallstreetbets</a></p>
<blockquote>
<p>The reason why China and REMs are always mentioned in the same sentence is because China pulled an Amazon and undercut their competitors with government subsidies to dominate the market. You can find other sources but they cost much more, mainly because they don’t have people backing them trying to monopolize the entire industry. This is actually a blessing in disguise. We need more neodymium if we ever want to fulfill a dream of renewable energy overtaking coal and gas. Just like any business, people follow the money, and we know the US is sitting on mountains (literally) of resources waiting to be mined and used. Provided domestic industry gets some help I can see this blowing up in China’s face.</p>
</blockquote>
| |metallurgy|mining-engineering| | <p>There's an adage spoken by some in the minerals industry ... "gold is you find it". That adage is equally applicable to all mineral commodities, including <a href="https://en.wikipedia.org/wiki/Rare-earth_element" rel="nofollow noreferrer">rare earth elements</a> (<a href="https://www.scientificamerican.com/article/dont-panic-about-rare-earth-elements/" rel="nofollow noreferrer">REE</a>).</p>
<p><a href="https://www.thoughtco.com/rare-earth-metals-2340169" rel="nofollow noreferrer">Rare earth elements</a> are not particularly rare, they are found throughout the Earth's crust. The trouble is, their concentrations within the crust are very low.</p>
<p>Mineral resources are classified as either mineral deposits or ore reserves. The difference between the two is economics. Ore reserves are mineral deposits that can be mined for a profit. This includes the costs of geological exploration, evaluation, mining, subsequent processing (sometimes primary and secondary) and marketing/sales.</p>
<p>What makes REEs regarded as being rare is they are rarely found in large concentrations that would make them economical to mine. This is a quirk of <a href="https://geology.com/usgs/ree-geology/" rel="nofollow noreferrer">geology</a>.</p>
<p>As is stated, China contains the largest reserves of REEs, with smaller deposits occurring in a small number of countries. That's just mother nature!</p>
<p>The other thing that can be attributed to geology is not all deposits of REEs contain every rare earth element.</p>
<p>REEs are have been classified as being either <a href="https://www.nrcan.gc.ca/our-natural-resources/minerals-mining/minerals-metals-facts/rare-earth-elements-facts/20522" rel="nofollow noreferrer">light or heavy</a>:</p>
<blockquote>
<p>Light REEs (lanthanum, cerium, praseodymium, neodymium, promethium, samarium, europium, gadolinium and scandium) are produced in global abundance and are in surplus supply</p>
<p>Heavy REEs (terbium, dysprosium, holmium, erbium, thulium, ytterbium, lutetium and yttrium) are produced mainly in China and are in limited supply. Global efforts to bring new resources to the marketplace continue.</p>
</blockquote>
<p>American REE deposits may not contain all the REEs America may require and given that some of the REEs, particularly the heavy ones, are mostly found in China, America may be dependent on China for supplies of these metals. After all, there 17 REEs.</p>
<p>Even if American deposits contained every REE, the concentrations for some may be so low that it is uneconomic to extract all of them. Having high concentrations of a small number of metals and very low concentrations of other metals, within the same deposit, is very common. It's just another quirk of geology. Murphy's second law comes to mind - <em>Mother nature is a bitch</em>.</p>
<p>The other thing to consider is America's reserves of REEs is small and finite. Unless it finds something else, once it has mined its resources, it will have to source the metals from elsewhere.</p>
<p>The other thing that needs to be <a href="https://www.defensenews.com/opinion/commentary/2019/11/12/the-collapse-of-american-rare-earth-mining-and-lessons-learned/" rel="nofollow noreferrer">considered</a> are the health and safety and the environmental impacts of mining and processing REEs.</p>
<p>To begin with, all REE deposits contain thorium, some also contain uranium, another quirk of geology. Both thorium and uranium are radioactive. The REEs cannot be mined and initially processed separately.</p>
<p>The other aspect is the <a href="https://e360.yale.edu/features/boom_in_mining_rare_earths_poses_mounting_toxic_risks" rel="nofollow noreferrer">chemicals used</a> in the <a href="https://arstechnica.com/information-technology/2019/05/australian-rare-earth-ore-processor-wants-to-build-a-plant-in-the-us/" rel="nofollow noreferrer">initial processing</a> of <a href="https://www.911metallurgist.com/separating-heavy-rare-earth-elements/" rel="nofollow noreferrer">REEs and the metals</a> not recovered, such as cadmium, result in the production of <a href="https://e360.yale.edu/features/china-wrestles-with-the-toxic-aftermath-of-rare-earth-mining" rel="nofollow noreferrer">toxic tailings dumps</a>.</p>
<p>The US EPA has produced a <a href="http://reviewboard.ca/upload/project_document/EA1011-001_US_EPA_-_Rare_Earth_Elements_-_Associated_Environmental_Issues.PDF" rel="nofollow noreferrer">comprehensive review</a> of the environmental impact of mining and processing REEs.</p>
<hr />
<h2>Edit 22 July 2020</h2>
<p>Additional new information, as of today.</p>
<p><a href="https://arstechnica.com/science/2020/07/are-we-ready-to-recycle-the-rare-earths-behind-an-energy-revolution/" rel="nofollow noreferrer"><em>Are we ready to recycle the “rare earths” behind an energy revolution?</em></a></p>
<blockquote>
<p>Ores with high concentrations of rare earths mostly fall into two general categories: igneous rocks and weathered sediments. The igneous ores are mostly carbonatite—an unusual product of magmas rich in carbonate minerals. It’s unusual enough that there’s only one volcano in the world erupting carbonatite lavas today, although others have in the past.</p>
<p>Something like half of current global rare earth element production comes from China’s Bayan Obo mine alone, which features many carbonatites. Southern California’s Mountain Pass mine along Interstate 15 has exploited similar rocks over its on-again-off-again history.</p>
<p>Australia’s Mount Weld straddles the two categories of rock and sediment. The ultimate source of REEs is carbonatite rock, but current mining is focused on the soil and sediment on top of this rock. That soft stuff is the result of weathering that has broken down the bedrock, carrying away some of the less resilient minerals and further concentrating the rare-earth-rich ones. Similar processes are responsible for deposits of ionic clays in China and of mineral sands in India.</p>
<p>The different sources have different ratios of rare earth elements in them.</p>
<p>“In general, all carbonatites are enriched in lanthanum and cerium,” UNLV’s Simon Jowitt told Ars. “So as you go from lanthanum down towards lutetium, basically the concentrations drop off sharply. In the ionic clays, it's the other way around; you get far less lights and far more heavies. But what we actually want is some of the stuff in the middle.”</p>
</blockquote>
| 35996 | Can the US realistically mine and produce all Rare Earth Elements, without relying on China? |
2020-05-30T06:48:03.243 | <p>The context of this question is the Form, Fit and Function (FFF) rule defined in ASME. It states that form is used to define any product with reference to its size, mass, shape, dimensions, and other visually identifiable features. <a href="https://www.arenasolutions.com/resources/articles/form-fit-function/" rel="nofollow noreferrer">For example</a> "SCREW, PAN HEAD, M3 x 0.5, <strong>2mm Lg</strong>, 316 SS". I am guessing this is Lag length based on some research i.e. the length from the head where the thread starts. </p>
<p>Q) What does Lg actually mean? Please quote resources for more reading </p>
| |mechanical-engineering|fasteners| | <p><a href="https://i.stack.imgur.com/1ZhLZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1ZhLZ.png" alt="enter image description here"></a></p>
<p>*Figure 1. For McMaster-Carr 'Lg' is the length of the screw."</p>
<p>Note that they specify the threaded length separately so Lg will be the distance from the bottom of the head to the tip of the screw.</p>
| 35998 | What does "Lg" mean when defining form of screws |
2020-05-30T16:57:26.457 | <p>I’d like to know what the below movement mechanisms are called so I can better research them and incorporate them into my own designs.</p>
<p>They are both quite similar, they are from rotary electrical isolator switches found on vehicles and industrial switch gear.</p>
<p>Type 1:
<a href="https://i.stack.imgur.com/y7rNh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y7rNh.jpg" alt="Type 1"></a>
A pin on the internal shaft or key follows the path of the cut out therefore forcing it in.</p>
<p>Type 2:
<a href="https://i.stack.imgur.com/W7N2I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W7N2I.jpg" alt="Type 2 Plunger"></a>
<a href="https://i.stack.imgur.com/YrdjH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YrdjH.jpg" alt="Type 2 Rotary Shaft"></a>
A linearly fixed T shaped shaft rotates and the internal plunger has a curved ramp. The plunger is sprung against the shaft. As the shaft rotates the T shaped runs up the ramp forcing the plunger down.</p>
<p>What are the two types of mechanism called?</p>
<p>I’ve tried searching for various terms but can only find basic cams etc. If there isn’t a specific terminology maybe someone could provide some search terms or some references </p>
| |mechanical-engineering|control-engineering| | <p>This is a type of <strong>barrel cam</strong>, but also resembles a <a href="https://en.wikipedia.org/wiki/Bayonet_mount" rel="noreferrer">bayonet mount</a>. Barrel cams consist of a cylindrical surface with a slot or raised ridge that wraps around the curved surface. As the cylinder rotates, the cam follower (in your case, the arms on the T-shaft) rides through the grooves, moving up and down along the length of the cylinder.</p>
<p>Barrel cams are commonly found in camera zoom lenses, as in the photo below. Notice how the three cam followers also ride in a slot, which is what enforces the linear motion and prevents the entirety of the lens from rotating.</p>
<p><a href="https://i.stack.imgur.com/cWYdt.png" rel="noreferrer"><img src="https://i.stack.imgur.com/cWYdt.png" alt="Barrel cams inside of a parfocal zoom camera lens"></a></p>
<p>As with many physical mechanisms, there are some barrel cams that work in reverse, where driving the cam follower in an a) linear, or b) rotary motion induces the cam to move in an a) rotary or b) linear motion. One example of type b is the reverse bayonet connector, where rotating the collar (which contains two or more followers) pulls the two connectors together.</p>
<p><a href="https://i.stack.imgur.com/J2hmO.png" rel="noreferrer"><img src="https://i.stack.imgur.com/J2hmO.png" alt="Reverse bayonet connectors"></a></p>
<p>Compare that to a bayonet connector, (such as <a href="https://en.wikipedia.org/wiki/BNC_connector" rel="noreferrer">BNC connectors</a>), where the collar contains the cam slot:</p>
<p><a href="https://i.stack.imgur.com/SCWI5.png" rel="noreferrer"><img src="https://i.stack.imgur.com/SCWI5.png" alt="BNC connectors are an example of bayonet connectors"></a></p>
| 36003 | What are these rotary to linear movements called? |
2020-05-30T23:56:27.227 | <p>I am working on design of the column which contains the water tanker and the pump + electronics at the bottom.
I have tried different solutions and they failed. I would like to know your advice on this R&D problem with the goal to be able to achieve the the robustness, manufacturing easy process and low cost.</p>
<p>Objective:
User put the tube as a column in vertical position.
The tube is separated inside in 2 parts by the false bottom. The upper part is used as a water tank. The lower part contains the water pump.
The false bottom contains the hole which is used to pass the water from the upper section to the lower section with the help of the pump. The water pipe
fitting hose connector is fixed through the hole.</p>
<p><a href="https://i.stack.imgur.com/UMZJ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UMZJ9.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/0qg3a.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0qg3a.png" alt="enter image description here"></a></p>
<p>Requirements: Tube + false bottom</p>
<ol>
<li>The material of the tube shall be one of the next list:
PVC, uPVC, ABS, Acrylic/PMMA, PC/Polycarbonate</li>
<li>The material of the wall separation shall be one of the next list:
PVC, uPVC, ABS, Acrylic/PMMA, PC/Polycarbonate</li>
<li>The tube shape shall be cylindrical</li>
<li>The tube shall be separated inside in 2 sections: the
upper section to contains the water and used as a water tank the
lower section to contains the water pump</li>
<li>The false bottom shall not pass the water from the upper section
to the lower</li>
<li>The false bottom thickness shall not exceed 6 mm</li>
<li>The upper tube section shall contain minimum 4 liters of volume</li>
<li>The upper tube section shall contain the water with no leakage</li>
<li>The lower tube section shall contain maximum 1 liter of volume</li>
<li>The lower tube section height shall be maximum 100 mm</li>
<li>The tube diameter shall be >= 110 and <= 150 mm</li>
<li>The tube height shall be >= 400 and <= 700 mm</li>
<li>The tube wall thickness (the difference between outer diameter and
inner diameter) shall be >= 2 and <= 6 mm</li>
<li>The false bottom shall be resistant to the weight of 10 kg
placed on the wall separation inside the upper section</li>
<li>The false bottom shall be functional after 10 drop tests with
different orientations from the height of 1 m</li>
</ol>
<p><a href="https://i.stack.imgur.com/J8oHB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J8oHB.png" alt="enter image description here"></a></p>
<p>I have designed this custom plug and 3d printed it with different sealing rings diameters. The problem is that:</p>
<ol>
<li>It is very difficult to insert the plug with sealing in the PVC pipe (12 cm diameter).</li>
<li>The PVC pipe is not uniform and so the sealing does not works perfectly, so there is a water leakage</li>
<li>This solution doesn't seems to be a good one for production</li>
</ol>
<p>I am going to search more on:</p>
<ol>
<li>existing plumbing solutions to separate the water from the dry section</li>
<li>factories if they are able to produce the tube with the inner separation wall</li>
<li>extruded plastic disks which could be fixed to the tube walls by the heater</li>
</ol>
<p>Do you know any existing plumbing solutions or could you advice me something to solve this problem in the most efficient manner?</p>
<p>Note: The solution with a silicone sealing is acceptable for a prototype but not the good one for production</p>
| |mechanical-engineering|fluid-mechanics|applied-mechanics|pipelines|product-engineering| | <p>Consider the use of a double-lip seal.</p>
<p><a href="https://i.stack.imgur.com/Tkjbm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tkjbm.png" alt="enter image description here"></a></p>
<p><em>Figure 1. A double-lipped seal in the false base of the tube.</em></p>
<p>This can then be inserted from the top of the tube to take advantage of the tapered edge. It seems to me that increasing the pressure in the top side will increase the seal tightness but check with someone that uses these things. (I'm an electrical engineer.)
w
<a href="https://i.stack.imgur.com/qYJ6x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qYJ6x.png" alt="enter image description here"></a></p>
<p><em>Figure 2. Having three legs will simplify horizontal alignment of the base.</em></p>
<p><a href="https://i.stack.imgur.com/sEoRn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sEoRn.png" alt="enter image description here"></a></p>
<p><em>Figure 3. Doing it this way avoids the internal seal problem completely.</em></p>
<p>With the arrangement of Figure 3 the assembly, sealing and inspection becomes much simpler. With PVC, for example, a solvent can be used to fuse the plastics. This used to be quite common on plumbing waste pipes, for example. In addition, if it <em>does</em> leak it's visible and external to the electronics. I would consider dipping the end of the pipe into a flat-bottomed container of solvent at the right depth just prior to assembly. There is a fire risk and solvent risk here so you require ventilation and a flame-proof dispenser.</p>
<p><a href="https://i.stack.imgur.com/Emozi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Emozi.jpg" alt="enter image description here"></a></p>
<p><em>Figure 4. A bench can with spring loaded flame arrester plate. Image: <a href="https://www.justrite.com/bench-can-to-clean-small-parts-in-solvents-2-ltr-plated-steel-dasher-hinged-cover-steel.html" rel="nofollow noreferrer">JustRite</a>.</em></p>
<p>Read the description in the linked article for details.</p>
<hr>
<p><a href="https://i.stack.imgur.com/kb3XF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kb3XF.png" alt="enter image description here"></a></p>
<p><em>Figure 5. This design while it makes the two parts visible has the aesthetic advantage that there is no step in the outline. The disadvantage is increased difficulty in solvent bonding.</em></p>
| 36014 | How to design internal wall separation in the PVC tube to use it as a water tanker |
2020-05-31T18:02:07.433 | <p>The muzzle velocity of a firearm can, all else equal, be increased by lengthening the barrel of the weapon which can theoretically be extended up to the point that friction and air resistance would overcome the benefit of pressure buildup in the barrel.</p>
<p>I think that the range of a howitzer be increased simply by adding more explosive propellant. Why is this not used? I would think that this would be a straightforward way to increase the gun's effective range. Any ideas?</p>
| |pressure|pressure-vessel|combustion| | <p>Here are the limitations to this approach. </p>
<p>First of all a <em>howitzer</em> is specifically designed for relatively short-range use, by using a very steep (in some cases, almost vertical) launch angle which allows the howitzer projectile to be lofted high up above trees, buildings, and even hilltops and then strike the intended target by falling almost vertically downward upon it. </p>
<p>Because of this, adding range to the howitzer does not increase its utility. For hitting more distant targets, a <em>gun</em> (i.e., cannon) is used instead, which is fired on a flatter trajectory. </p>
| 36032 | Can I increase the range of a howitzer by using more propellant? |
2020-06-01T03:35:51.343 | <p><a href="https://i.stack.imgur.com/dQXoy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dQXoy.jpg" alt="enter image description here"></a></p>
<p>Source:byjus.com</p>
<p>Do these forces cancel out each other? If yes, how is forward motion initiated?</p>
<p>Also, in this situation, since friction between the road and the wheels is enabling motion, which force is opposing the motion?</p>
| |friction|bicycles| | <p>No they do not cancel.</p>
<p>The frictional force on the front wheel is that required to overcome bearing friction etc so the wheel rotates.</p>
<p>The back wheel has sufficient friction to match the input power by the rider plus bearing friction etc.</p>
| 36040 | How is forward motion initiated in this case? |
2020-06-01T21:41:26.477 | <p>Courses on automatics define usually the second order transfer function as</p>
<p><span class="math-container">$$
H(s) = \frac{1}{\frac{s^2}{w_n^2} + \frac{2z}{w_n} s + 1}
$$</span>
(sse <a href="https://www.wikimeca.org/index.php/Syst%C3%A8mes_du_second_ordre" rel="nofollow noreferrer">here</a> for instance).</p>
<p>However, it assumes that the coefficents <span class="math-container">$a,c$</span> of the corresponding second order differential equation
<span class="math-container">$$
a\ddot x(t) + b \dot x(t) + c x(t)= u(t)
$$</span>
have the same sign.</p>
<p>Why is that ? If I consider the inverted pendulum linearized at the unstable equilibrium, then these coefficients won't have the same sign since the dynamics of the angle is:
<span class="math-container">$$
m\ddot \theta(t) - mg\theta(t) = u(t)
$$</span></p>
| |transfer-function| | <p>The classical form of a second order system, as I have usually seen it, is:</p>
<p><span class="math-container">$$ T(s) =K \frac{ω_n^2}{s^2+2ζω_ns+ω_n^2} $$</span></p>
<ul>
<li><span class="math-container">$K \ \rightarrow $</span> DC Gain of the system</li>
<li><span class="math-container">$ζ \ \ \ \rightarrow $</span> Damping Ratio of the system</li>
<li><span class="math-container">$ω_n \rightarrow $</span> Natural Frequency of the system</li>
</ul>
<p>By definition the damping ratio of a system is <span class="math-container">$ζ \geq 0$</span>. As its name suggests this term is responsible for the exponential decay of any oscillation the system is subject to. Imagine it being equal to <span class="math-container">$ζ=0$</span>, this means that there is nothing to "slow down" the system when it oscillates and it will oscillate forever. It is obvious that it can't be negative since this would mean that instead of "slowing down" the system, it would enhance its oscillation. For a more physical approach you can think of the damping as being the friction, the air resistance, water resistance etc. Anything out there in the physical world which adds resistance to the movement of the system. </p>
<p>As far as the natural frequecny goes there is a very good definition at <a href="https://en.wikipedia.org/wiki/Natural_frequency" rel="nofollow noreferrer">Wikipedia</a>. Imagine that there isn't any damping force acting on a system (<span class="math-container">$ζ=0$</span>). Then the natural frequency is the frequency at which the system oscillates when no damping forces are applied to it. For physically realizable systems, this frequency can never be less than <span class="math-container">$0$</span> (I think this is obvious). </p>
<p>The second order differential <span class="math-container">$a\ddot{x(t)}+b\dot{x(t)}+cx(t) = u(t)$</span> is transformed into the <span class="math-container">$s$</span>-domain as follows (assuming <span class="math-container">$0$</span> initial conditions):</p>
<p><span class="math-container">$$ s^2aX(s)+sbX(s)+cX(s)=U(s) \Rightarrow \frac{X(s)}{U(s)} = \frac{1}{s^2+\frac{b}{a}s+\frac{c}{a}} $$</span></p>
<p>The denominator of the transfer function is the characteristic equation of the system. In order for a second order system to be stable, the coefficient of <span class="math-container">$s$</span> and the constant coefficient need to be greater than <span class="math-container">$0$</span> (derived from Routh-Hurwitz criterion). So, suppose <span class="math-container">$a>0$</span> then <span class="math-container">$b$</span> and <span class="math-container">$c$</span> should also be positive in order for this system to be stable. Same goes for the case of <span class="math-container">$a$</span> being negative. </p>
<p>Now for the inverted pendulum, as you wrote, the straightforward linearized system is unstable. This means that the coefficients of the characteristic equation don't have the same sign. Although I can't speak french I managed to understand that the link you provided assumes that the coefficients are positive. This goes for the case of the stable systems and also for the systems represented using the damping ratio and natural frequency terms, which is indeed the truth. </p>
| 36050 | second order linear filter and coefficients sign |
2020-06-02T07:43:58.280 | <p>I was comparing the wheels of the Curiosity rover and the Perseverance rover and noticed that the latter had a very flat cylindrical design as compared to the former.</p>
<p><a href="https://i.stack.imgur.com/B4dObm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/B4dObm.png" alt="Curiosity vs Perseverance wheels"></a></p>
<p>This got me thinking about which wheel profile was sturdier. That got me thinking about various things that might be different. Intuitively, I felt that the convex wheel would be better off because it would be able to transfer some of the applied force into normal components, which would keep the shear stresses lower (Aluminum 6061 has a shear strength of only 207 MPa, compared to the tensile yield strength of 276 MPa).</p>
<p>So I decided to test it on Ansys:</p>
<p><a href="https://i.stack.imgur.com/fK4Lk.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fK4Lk.png" alt="Eq. Stresses"></a>
<a href="https://i.stack.imgur.com/EkqrQ.png" rel="noreferrer"><img src="https://i.stack.imgur.com/EkqrQ.png" alt="Normal stresses"></a>
<a href="https://i.stack.imgur.com/CD5Qd.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CD5Qd.png" alt="Shear stresses"></a></p>
<p>From these, I can visibly see that the regions of maximum stress are lower in the convex wheel. But I am starting to doubt if I'm misreading something. Could someone knowledgable help?</p>
<hr>
<p><strong>TL;DR</strong></p>
<p>Among convex and flat profiles, which one is sturdier? (If you feel the word sturdy is ambiguous, please interpret it as "less metal fatigue", perhaps.)</p>
| |mechanical-engineering|stresses|wheels| | <p>I found a <a href="https://www.jpl.nasa.gov/news/news.php?feature=7631" rel="nofollow noreferrer">JPL article</a> on it, but did not locate a CAD file (let us know if you find one). From the article it sounds like both wheels are quite close to the same with the main change being the tread. I think the image is likely photoshopped to be side by side rather than rendered side by side; so the shape change is likely less drastic than it looks in the photo.</p>
<blockquote>
<p>Illustrated here, the aluminum wheels of NASA's Curiosity (left) and Perseverance rovers. Slightly larger in diameter and narrower, 20.7 inches (52.6 centimeters) versus 20 inches (50.8 centimeters), Perseverance's wheels have twice as many treads, and are gently curved instead of chevron-patterned. Credit: NASA/JPL-Caltech</p>
</blockquote>
<p>Your ANSYS FEA makes sense. A curved or dished surface is stronger than a flat surface of the same thickness. When something is rounded or dished its has a larger <a href="https://en.wikipedia.org/wiki/Second_moment_of_area" rel="nofollow noreferrer">area moment of inertia</a> compared to the flat version. But as far as representing the rover wheels you would certainly need to model the spokes to get the whole picture. And without the CAD model, you wouldn't know for sure.</p>
<p>Here are some of my thoughts on the rover wheel design change:</p>
<ol>
<li>If the wheels are in fact less spherical, the designers may have found that the rover wheels did not need as much surface angle variation than previously thought. The main reason a wheel is rounded is so it can keep tread in contact with the surface at various angles as opposed to a perfectly vertical force. That is why motorcycle tires are round, and car tires are more flat.</li>
<li>If the wheels are in fact less spherical, the designers may have found that they were not steering torque limited. Rounded tires take less torque to rotate because there is a smaller distance between the center of rotation and the furthest tread touching the surface.</li>
<li>The tread change from chevrons to straight perpendicular treads, is likely because they found that during operation the wheels were never subject to any loads perpendicular to the direction of rotation. This would be a disastrous design for a car on earth because a car is frequently subjected to cross winds or lateral gravitational loads when traversing a ramp perpendicularly (side-hilling)(think driving a NASCAR track at 5mph). However, an expensive rover on Mars that has 6 wheels and every movement planned by a team, has no need to attempt a side-hilling maneuver. Consequently all the treads should be oriented for maximum traction in the direction of rotation. This tread shape also increases the strength of the wheel by increasing the area moment of inertial in the axial direction (like lots of little angle iron reinforcements).</li>
</ol>
| 36057 | Rover Wheels: Convex or Flat? |
2020-06-03T07:58:54.807 | <p>I have two 5.1 speaker sets, one in the front and one in the back of my living room. I have connected them together with an "audio wire" (non-shielded unfortunately) that's 12 meters in length.</p>
<p>Since I did this, the speakers started buzzing/humming when turned on. This however stops when I connect the audio wire (through an Y splitter) to my laptop. This got me curious: What happens if I connect the COMM wire on from the Y audio split to the COMM on my laptop charger? The buzzing/humming stops!</p>
<p>There exists some circuit in the laptop charger that is being able to cancel/even out the noise.</p>
<p>Why does the humming/buzzing happen? And Why does it stop when I connect audio COMM to charger COMM?</p>
<p>Can I fix the buzzing/humming noise by making some circuit using passive components I have lying around from DYI kits (Arduino etc)?</p>
<p>This is how the setup looks like in diagram form:</p>
<p><a href="https://i.stack.imgur.com/wLk8U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wLk8U.png" alt="enter image description here"></a></p>
| |electrical|consumer-electronics|audio-engineering|noise| | <p>As mentioned the cable is acting like an aerial. A ferrite core to wrap the cable around a couple of times might help. The comm to the charger comm is grounding it with less resistance as the other answer said. </p>
| 36072 | Why does audio equipment buzz/humm when a long signal cable is used and how do I eliminate it? |
2020-06-03T13:13:56.673 | <p><a href="https://i.stack.imgur.com/biEYg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/biEYg.jpg" alt="enter image description here"></a></p>
<p>How does the author get that formula(d alpha formula) using the momentum equation . The book here is Mechanics and thermodynamics of propulsion by Philip Hill and Carl Peterson. </p>
| |fluid-mechanics|cfd| | <p>It involves simple differentiation operation to get there.</p>
<p>We have <span class="math-container">$\frac{dP}{\rho} = -UdU$</span> and <span class="math-container">$\frac{dP}{\rho} = -udu$</span>. These equations can be re-organised as
<span class="math-container">$$\frac{-dP}{\rho U} = dU, ~\frac{-dP}{\rho u} = du$$</span>
Given, <span class="math-container">$\alpha = \frac{u}{U}$</span>. Differentiate this relation on both sides, we get
<span class="math-container">$$d\alpha = \frac{du}{U} - \frac{u}{U^2}dU$$</span>
Now, substitute the expressions for <span class="math-container">$du$</span> and <span class="math-container">$dU$</span> in the above, we have
<span class="math-container">\begin{eqnarray*}
d\alpha &= &\frac{1}{U}\frac{-dP}{\rho u} - \frac{u}{U^2}\frac{-dP}{\rho U}\\
&= &-\frac{1}{U^2}\frac{U}{u}\frac{dP}{\rho} + \frac{1}{U^2}
\frac{u}{U}\frac{dP}{\rho} \\
&= &-\frac{1}{U^2}\frac{1}{\alpha}\frac{dP}{\rho} + \frac{1}{U^2} \alpha\frac{dP}{\rho} \\
&= & \frac{dP}{\rho U^2}\left(\alpha - \frac{1}{\alpha}\right)\\
&= & \frac{dP}{\rho U^2}\left(\frac{\alpha^2 - 1}{\alpha}\right)
\end{eqnarray*}</span></p>
<p>Hope this answers!</p>
| 36078 | Boundary layer momentum equation |
2020-06-03T17:10:34.477 | <p>I want to be able to hold a 1/2" diameter shank boring bar on my existing lathe tool post but don't know how to buy the correct one.</p>
<p>The post we have is a Phase II 250-111: <a href="https://www.mscdirect.com/product/details/09044017?item=09044017" rel="nofollow noreferrer">https://www.mscdirect.com/product/details/09044017?item=09044017</a></p>
<p>The boring bars we have a 1/2" shank diameter. I'm hoping to get a different brand for the tool holder because it is cheaper. Which of the options below would fit? I'm confused by the OXA or AXA as I don't see that specified in the post description from Phase II.
<a href="https://www.mscdirect.com/product/details/30579627" rel="nofollow noreferrer">https://www.mscdirect.com/product/details/30579627</a>
<a href="https://www.mscdirect.com/product/details/30574867" rel="nofollow noreferrer">https://www.mscdirect.com/product/details/30574867</a></p>
<p>or even
<a href="https://rads.stackoverflow.com/amzn/click/com/B07THWWQQD" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Quick-Change-Boring-Holder-250-004/dp/B07THWWQQD</a></p>
<p>or an even cheaper option:
<a href="https://rads.stackoverflow.com/amzn/click/com/B077V5HKNR" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Boring-Bar-Holder-1-2/dp/B077V5HKNR</a></p>
| |machining| | <p>Lathe quick change tool posts or QCTP are sized for the lathe your using it on, particularly, the swing.</p>
<p>0XA = UP TO ~8"
AXA = 9-12"
BAX = 12-15"
and so forth...</p>
<p>Looks like you have a phase 2, 100 series post, or an AXA sized tool post, so you would be looking for AXA tool holders. I believe a standard AXA boring bar holder is for 3/4" shank so you'd need a bushing. You can also purchase a standard tool holder that has a v grove at the bottom of the tool slot that can hold your boring bar.</p>
| 36080 | Boring bar holder for lathe tool post |
2020-06-04T21:21:07.640 | <p>This rail junction is located on a steep, short, local railroad line from the "mainline" in Reno, NV to the community of Stead.</p>
<p>The track is new/maintained and there is no evidence of an old siding. </p>
<p>As a train travels uphill, the train engineers stop and fiddle with it (I can only assume they are switching the junction), allow the train to pass, and then fiddle with it again before proceeding uphill. </p>
<p>Once the train has passed, it is in the "siding" position (a train traveling towards the photographer would turn off the main rail line into the ditch).</p>
<p>So if the same train then travels downhill and forgets to switch the junction, it would end up in the ditch.</p>
<p>What is the purpose of this junction? </p>
<p><a href="https://i.stack.imgur.com/JyrQh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JyrQh.jpg" alt="enter image description here"></a></p>
<p>A</p>
| |rail| | <p>I think it's a derail to prevent any loose rolling stock from careening down the hill.</p>
<p><a href="https://en.wikipedia.org/wiki/Derail" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Derail</a></p>
<p>The big D on the sign is one clue :)</p>
<p>See appendix A-34 - <a href="https://www.bnsf.com/ship-with-bnsf/ways-of-shipping/pdf/indytrkstds.pdf" rel="nofollow noreferrer">https://www.bnsf.com/ship-with-bnsf/ways-of-shipping/pdf/indytrkstds.pdf</a></p>
| 36098 | What is the purpose of this dead-end rail junction? |
2020-06-05T07:11:05.027 | <p>I am working on a project on detecting damage in structures using fiber optic sensors. I read the paper <em>Structural Health Monitoring in Composite Structures by Fiber-Optic Sensors</em> and found out that fiber optic sensors can not only detect strains but also locate them; how is this possible? Do they measure the intensity of the reflected light or something? </p>
| |mechanical-engineering|structural-engineering|structural-analysis|sensors| | <p>Searching for "fibre-optic strain measurement" yields plenty of results. One from <a href="https://www.hbm.com/en/6827/article-how-does-an-optical-strain-gauge-actually-work/#:~:text=Optical%20strain%20gauges%20(also%20called,in%20temperature%2C%20acceleration%20or%20displacement." rel="nofollow noreferrer">HBM</a> states:</p>
<p><a href="https://i.stack.imgur.com/fnQMm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fnQMm.jpg" alt="enter image description here"></a></p>
<blockquote>
<p>To create the actual strain sensor, the optical fiber is inscribed during production with a so-called Fiber Bragg Grating (FBG). This is basically a pattern of material interferences, which reflects the light differently from the rest of the fiber. For better understanding, you can visualize the fiber as a cylindrical length of transparent material, with a number of thin slices in it. When the light from the laser hits this pattern, certain wavelengths are reflected, while others pass through.</p>
<p>The material interferences—the “slices” —are placed at certain intervals. When the fiber is stretched or compressed—and is therefore subjected to positive or negative strain—these intervals change. When the fiber is stretched, it lengthens and the spaces get bigger and vice versa.</p>
<p>Not only does the reflected light take a little longer or shorter to travel back when the FBG is under strain, but the wavelength that is reflected also changes. In scientific terms, the FBG has a certain refractive index. The refractive index of a material describes how much light is bent or refracted when passing through the material. When the grating changes shape due to strain, its refractive index changes as well.</p>
</blockquote>
<p>The article goes on to explain the topic further including problems with temperature, etc.</p>
| 36105 | How do fiber optic sensors detect strains in a structure? |
2020-06-05T16:13:15.717 | <p><a href="https://i.stack.imgur.com/2gARi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2gARi.jpg" alt="motor"></a>I know how to connect motor to Arduino with 3 wires: 5v, digital pin and ground. As I know only digital wire need to control "speed control" of my motor. So, if my Arduino connected to computer and it is not needed to get this 5v from "speed control". So I decided to remove useless 5v and ground wires. When I remove ground wire, it is okay. But when I remove 5v wire, my motor works about 3-5 seconds on random frequency and then die. So, can I use only 1 digital pin to control my motor?</p>
| |motors|microelectronics|radio| | <p>What you have a is a DC brushless motor of type A2212/10T 1400KV. In order for the motor to work you need all three wires. The motor has 3 winding's. Below is good image describing the motor.</p>
<p><a href="https://i.stack.imgur.com/yoYq1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yoYq1.png" alt="enter image description here"></a></p>
<p>When you connect two wires example R or L1 at 12V and Y or L2 at 0V only the winding w1 and w2 are energized. So motor will turn a little bit and stop, as you indicated. </p>
<p>Below is an image of the inside of the motor. </p>
<p><a href="https://i.stack.imgur.com/uIxDT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uIxDT.png" alt="enter image description here"></a></p>
<p>So in order for motor to continue to turn you have energize the another pair of winding's. Take a look at the sequence below. </p>
<p><a href="https://i.stack.imgur.com/bixLt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bixLt.png" alt="enter image description here"></a></p>
<p>I hope I answered your question. Take a look at the references below. </p>
<p><strong>References</strong></p>
<ul>
<li><a href="https://components101.com/motors/2212-brushless-motor" rel="nofollow noreferrer">2212 Brushless Motor</a></li>
<li><a href="https://www.youtube.com/watch?v=bCEiOnuODac" rel="nofollow noreferrer">Brushless DC Motor, How it works ?</a></li>
</ul>
| 36112 | How to connect Arduino to brushless motor through only 1 digital pin |
2020-06-07T01:01:13.340 | <p>I have a motor that I have created a simple adapter for; this lets us attach any parts we might design to solve the final problem:</p>
<p>I would like to convert the motors motion to oscillating motion in two directions from one.</p>
<p><a href="https://i.stack.imgur.com/SHrJG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SHrJG.png" alt="The motion I would like"></a></p>
<p>The classic "woodcutter" solution to this is:</p>
<p><a href="https://i.stack.imgur.com/Z7kmS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z7kmS.png" alt="The woodcutter solution"></a> </p>
<p>However that does not give me the throw I would like for the final motion in either direction. Ideally this mechanism will sit on the floor - which SORELY limits the available "height" during rotation of any such "arm" </p>
<p>This motor has very high torque so it may be possible to move other structures while at a mechanical disadvantage in oblique (aka not smashing through the floor) planes?</p>
<p>In a nutshell after a couple of days thinking it idly over haven't managed a better solution yet.</p>
<p>Considered some sort of scissor motion? but couldn't firm up the details.</p>
<p>Is there a classic - obvious - solution to this problem?</p>
<p>Full disclosure: I have no formal training whatsoever in engineering.</p>
<p>Appreciate any assist.</p>
<p>EDIT: I should add that I did think about some sort of reversing screw mechanism?
but didn't manage to make that work yet in practice. perhaps That could be part of the solution but it doesn't solve the problem of throw either, well not without being really long and actually working. I might figure something out with this I guess?</p>
<p><a href="https://i.stack.imgur.com/RCdW6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RCdW6.png" alt="Reversing Screw mockup, shown on adapter, just doodled up in fusion"></a></p>
<p>This would result in motion in the other direction - which would be okay but if there is a more reliable or "better" way to get this motion would love to hear of it. Struggling with the screw solution.</p>
<p><a href="https://i.stack.imgur.com/wZxKj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wZxKj.png" alt="Other direction via a reversing screw"></a></p>
| |mechanical-engineering|gears|applied-mechanics|mechanisms|mechanical| | <p><a href="https://i.stack.imgur.com/UNbWz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UNbWz.png" alt="enter image description here"></a></p>
<p><em>Figure 1. Something like this?</em></p>
<p>It will be faster as the drive pin moves through the lower section of the slot.</p>
<p><a href="https://i.stack.imgur.com/bDQYV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bDQYV.png" alt="enter image description here"></a></p>
<p><em>Figure 2. This option should result in more even speed on the forward and reverse strokes.</em></p>
| 36130 | Is there a way to convert a short throw in one direction into a long throw in two directions |
2020-06-07T05:46:08.333 | <p>I would like to know what standard(s) may exist for formatting log data from sensors that record information such as:</p>
<ul>
<li>temperature</li>
<li>pressure</li>
<li>humidity</li>
<li>vibration</li>
<li>acceleration vector</li>
<li>magnetic field vector</li>
<li>location vector (ex: GPS)</li>
<li>etc.</li>
</ul>
<p>I have been an end-user of <a href="https://en.wikipedia.org/wiki/SCADA" rel="nofollow noreferrer">SCADA</a> software products which record various data fields for each stream of log data it receives into a MySQL database. However, I have never looked into determining how exactly that data was stored. I would like to know if there exists an industry consensus standard for storing sensor data that facilitates information interchange. For example, the <a href="https://en.wikipedia.org/wiki/Internet_Engineering_Task_Force" rel="nofollow noreferrer">Internet Engineering Task Force</a> (IETF) publishes <a href="https://en.wikipedia.org/wiki/List_of_RFCs" rel="nofollow noreferrer">documents called "RFCs"</a> for the purpose of standardizing implementations of commonly used protocols like IMAP (email), TLS (secure web browsing).</p>
<p>For example, such a standard might indicate to me the best practices for:</p>
<ul>
<li>formatting <em>units of measurement</em> (ex: <code>degF</code> vs. <code>°F</code>)</li>
<li>formatting of <em>timestamps</em> of each measurement (ex: <a href="https://en.wikipedia.org/wiki/ISO_8601" rel="nofollow noreferrer">ISO-8601</a>)</li>
<li><em>file format</em> type (<a href="https://en.wikipedia.org/wiki/Comma-separated_values" rel="nofollow noreferrer">CSV</a> files vs. <a href="https://en.wikipedia.org/wiki/SQL#Interoperability_and_standardization" rel="nofollow noreferrer">SQL</a> database)</li>
<li><em>delimiters</em> for measurement data (ex: <code>,</code> in <code>.csv</code> files)</li>
<li>indicating origin and license of data (ex: end device owner/operator, equipment tag number, <a href="https://wiki.creativecommons.org/wiki/RDFa" rel="nofollow noreferrer">CC RDFa tags</a>)</li>
<li>encrypting measurements (esp. in a way compatible with sudden power-loss, ex: <a href="https://en.wikipedia.org/wiki/Datagram_Transport_Layer_Security" rel="nofollow noreferrer">DTLS</a>)</li>
</ul>
<p>While researching this question I noticed that <a href="https://www.ietf.org/topics/iot/" rel="nofollow noreferrer">the IETF says the Internet of Things (IoT) is a topic of study for several of its working groups</a>. Its definition of IoT is</p>
<blockquote>
<p>The Internet of Things is the network of physical objects or "things"
embedded with electronics, software, <strong>sensors</strong>, actuators, and
connectivity to enable objects to exchange data with the manufacturer,
operator and/or other connected devices.</p>
</blockquote>
<p>However, I did not find any IETF standards document describing standard methods for storing log data. Is there an <a href="https://en.wikipedia.org/wiki/International_Society_of_Automation" rel="nofollow noreferrer">ISA</a> standard covering methods for logging data? I really like their ISA 5.1 standard for P&ID symbols, "<a href="https://archive.org/details/ANSIISA5.12009InstrumentationSymbolsAndIdentification" rel="nofollow noreferrer">Instrumentation Symbols and Identification</a>".</p>
<p>Thank you for any direction you may provide.</p>
<hr>
<p><strong>Edit-Update (2021-02-02)</strong></p>
<p>After several months of off-and-on research, the closest standard that defines what I was looking for is the <a href="https://www.w3.org/TR/vocab-ssn/" rel="nofollow noreferrer">Semantic Sensor Network (SSN) Ontology</a> published by the W3C. It is an expanded version of the Sensor, Observation, Sample, and Actuator (SOSA) Ontology. It is capable of associating metadata about an observation (i.e. units, time, location, site-specific names, relationships to other entities in a data collection hierarchy) in an <a href="https://en.wikipedia.org/wiki/Resource_Description_Framework" rel="nofollow noreferrer">RDF graph</a> data structure. RDF graphs can be "<a href="https://en.wikipedia.org/wiki/Serialization" rel="nofollow noreferrer">serialized</a>" into JSON (specifically, <a href="https://en.wikipedia.org/wiki/JSON-LD" rel="nofollow noreferrer">JSON-LD</a>). The JSON graph data can be formatted using the <a href="https://jsonlines.org/" rel="nofollow noreferrer">JSON lines specification</a> so each observation can be appended to a newline-delimited log file in a streaming fashion. Because the logs are machine-readable, an appropriate software package could later import the readings into a database or whatever a user requires.</p>
<p>Search results aren't clear whether SSN will become an industry-standard for general encoding of sensor information but it seems to me the best method for storing sensor data in a self-documenting way. For now, I'm happy to have discovered the method of recording measurements + metadata as an RDF graph using the SSN ontology, serializing the graph as as JSON-LD, and formatting the JSON-LD as JSON Lines that are streamed and appended to a continuously growing log file.</p>
| |sensors|instrumentation|iot| | <p>My experience is there is no particular standards for saving sensor data. Usually one just picks a format that is useful to the task at hand. If, for instance one wants to do post processing analysis in Excel then CSV formatted text files work pretty well. If you do use CSV, then just save numerical values. Don't add text indicating units to each number as this would have to be stripped to do any processing. You can provide units in the header row. Commas or tabs are fine as delimiters. Please use the extension ".csv" for comma delimited files. I've seen ".tsv" for tab delimited files. Python and Pandas provide very nice functions for handling CSV files as do many other languages.</p>
<p>I would suggest looking into using the <a href="https://www.json.org/json-en.html" rel="nofollow noreferrer">JSON file format</a>. From JSON.org:</p>
<blockquote>
<p>JSON (JavaScript Object Notation) is a lightweight data-interchange
format. It is easy for humans to read and write. It is easy for
machines to parse and generate.</p>
</blockquote>
<p>It is still human readable like CSV, but provides a greater ability to encode useful meta information with the data and more complex data structures. Many, many programming languages have libraries for reading and writing JSON files. I find JSON much less verbose than XML.</p>
<p>As for units, I highly encourage the use of metric units unless there is some particularly customary unit involved.</p>
| 36134 | Is there an industry consensus standard for storing sensor data? |
2020-06-08T14:26:15.817 | <p>I'm trying to drive an aluminium plate using a motor via a 20mm-wide 2mm-thick stainless steel tube. The tube is driven by a 6mm motor shaft via a pulley and the torque will be around 5Nm at most. Is it a good idea to attach the plate to the tube with a split collar like <a href="https://shaft-collars-couplings.staffordmfg.com/viewitems/flanged-mounting-collars/accu-flange--shaft-mounting-collars" rel="nofollow noreferrer">this</a>?</p>
<p><a href="https://i.stack.imgur.com/zTeKW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zTeKW.jpg" alt="enter image description here"></a></p>
<p>However it's not clear what the rated torque and axial force are. If this isn't a good way, what might be an alternative without welding or adhesive?</p>
| |mechanical-engineering|motors|torque|power-transmission|coupling| | <p>There are dozens of ways to connect to the shaft of a motor, A few examples:</p>
<ul>
<li>set screw/grub screw</li>
<li>through taper pin</li>
<li>interference fit</li>
<li>split ring</li>
<li>taper lock</li>
<li>welded</li>
<li>keyway & key (most common industrial use)</li>
</ul>
<p>Each of these has a torque spec that I think you can find in a mechanical engineer's handbook. There are probably decades old free ones on the internet which are fine. These are not new technologies.</p>
| 36152 | How to connect a motor shaft to a plate |
2020-06-09T12:59:01.300 | <p>Please can someone help out with the actual concept of stagnation,static and dynamic pressures . I have learnt all that stagnation pressure is when velocity of flow is 0 when isentropically stopped and dynamic pressure is the product of 0.5 times density and velocity squared and static as actuall pressure. But I am still not able to comprehend the physical significance of it. </p>
| |fluid-mechanics|aerodynamics|propulsion| | <p>The total (stagnation) pressure is the summation of the static and dynamic pressures for incompressible flow (see Bernoulli equation). The dynamic pressure is a measure of the fluid's kinetic energy (<a href="https://www.engineeringtoolbox.com/dynamic-pressure-d_1037.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/dynamic-pressure-d_1037.html</a>). The faster a fluid travels, the more kinetic energy and dynamic pressure it has. This website has a good description and diagrams of the concept (<a href="http://kb.eng-software.com/eskb/pipe-flo/general-theory-and-equations/total-and-static-pressure" rel="nofollow noreferrer">http://kb.eng-software.com/eskb/pipe-flo/general-theory-and-equations/total-and-static-pressure</a>). A more physical interpretation of this concept is flow in a convergent duct. For isentropic subsonic conditions, the flow velocity increases as the area decreases. Therefore, the dynamic pressure increases and the static pressure decreases. The opposite can be said for isentropic subsonic flow in a divergent duct; the velocity decreases as the flow area increases. Thus, the dynamic pressure falls in a divergent duct with subsonic flow. For further reading on the concept of total, static, and dynamic conditions, look into the operating principle behind Pitot tubes. I hope this was helpful. </p>
| 36171 | Stagnation,static and dynamic pressures |
2020-06-09T22:49:22.630 | <p>If I build a quad copter but angle each rotor slightly inwards, should it be self stabilizing like an aeroplane with V wings, or does a variation of the rocket pendulum fallacy apply here? I feel like it should work but I keep second guessing myself.</p>
| |aerodynamics| | <p>First of all, lets look at what the basic state vector and control vector looks like for a coaxial rotor arrangement.</p>
<p><a href="https://arxiv.org/ftp/arxiv/papers/1908/1908.07401.pdf" rel="nofollow noreferrer">https://arxiv.org/ftp/arxiv/papers/1908/1908.07401.pdf</a></p>
<p>Now trace this down to equation 24, which gives the input vector as <span class="math-container">$u^T = (U1\space U2\space U3\space U4)$</span> , where the <span class="math-container">$U$</span>s are total thrust, total pitch moment, total roll moment, and total yaw moment of the four rotors as a function of the speed of each rotor. You need a workable formula for each of those. If the rotors are coaxial, you can arrange things so that you can change just one of the <span class="math-container">$U$</span>s and leave the others alone. If the rotors are not coaxial, you have a giant mess on your hands. It is not insurmountable, but you now have pitch, roll and yaw defined on the body, but they are mathematically different for each rotor and different for CW or CCW rotation. So you now need new expressions for the <span class="math-container">$U$</span>s that reflect the geometry. And the four <span class="math-container">$U$</span>s will not be totally independent.</p>
<p>Then, when you have the equations of motion sorted, you can look at stability cases. That part is done the same way as for a flat rotor, but evaluating the math is much harder, and many of the simplifying strategies, such as separation of variables, won't work.</p>
<p>The math on aircraft dihedral isn't any simpler, but there is a lot less of it.
noncoaxial rotors have been done before. But they were always in sync.</p>
<p>Flettner Fl 282 "Kolibri"</p>
<p><a href="https://i.stack.imgur.com/5q5Vh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5q5Vh.png" alt="enter image description here"></a></p>
| 36182 | Does a V shape of quadcopter rotors give stability? |
2020-06-10T07:23:04.640 | <blockquote>
<p>A cube made of an <strong>isotropic</strong> material is subjected to a tensile stress as shown in Figure 2. What is the relationship between the strain in different directions ?<br>
<a href="https://i.stack.imgur.com/02bg4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/02bg4.jpg" alt="enter image description here"></a></p>
</blockquote>
<p>There is an elongation in <span class="math-container">$z$</span> direction and contraction in <span class="math-container">$x, y$</span> directions. So <span class="math-container">$\epsilon_x, \epsilon_y < \epsilon_z$</span>. Is my reasoning correct ? Since they say material is isotropic, can I say anything else about <span class="math-container">$\epsilon_x, \epsilon_y$</span> ? (Like <span class="math-container">$\epsilon_x = \epsilon_y$</span>). Please explain how someone can arrive at a conclusion about <span class="math-container">$\epsilon_x, \epsilon_y, \epsilon_z$</span>.</p>
| |mechanical-engineering|materials|material-science| | <p>In this case, we say that <span class="math-container">$\epsilon_{axial}$</span> (<span class="math-container">$z$</span>-direction in your diagram) is positive by convention. Then, for a normal material with a <a href="https://en.wikipedia.org/wiki/Poisson%27s_ratio" rel="nofollow noreferrer">Poisson's ratio</a> <span class="math-container">$\nu \ge 0$</span>, <span class="math-container">$\epsilon_{transverse}$</span> (<span class="math-container">$x$</span>- and <span class="math-container">$y$</span>-directions in your diagram) will be negative. The fact that the material is isotropic means we have no way of distinctly labeling one direction <span class="math-container">$x$</span> and one direction <span class="math-container">$y$</span>, so it must be the case that <span class="math-container">$\epsilon_x = \epsilon_y$</span>.</p>
<p>So in summary, yes, assuming <span class="math-container">$\nu \ge 0$</span> you are justified in saying <span class="math-container">$\epsilon_x = \epsilon_y < \epsilon_z$</span>.</p>
<p>Note, however, that even though <em>almost all</em> materials will have <span class="math-container">$\nu \ge 0$</span>, there does exist a class of materials known as <a href="https://en.wikipedia.org/wiki/Auxetics" rel="nofollow noreferrer">auxetics</a> which have <span class="math-container">$\nu \lt 0$</span>. A common naturally occurring example is <span class="math-container">$\alpha$</span>-cristobalite silica. In this case, the relationship between the strains is not so immediately obvious, and would depend on the exact value of <span class="math-container">$\nu$</span> (since the strain would be positive in all directions). These materials are not usually isotropic though, so <span class="math-container">$\nu \ge 0$</span> is probably a safe assumption.</p>
| 36187 | What is the relationship between the strain in different directions? |
2020-06-11T02:12:36.770 | <p>I'm a beginner in chemical engineering so apologies in advance if this question is very basic!</p>
<h2>The Problem</h2>
<p>You are required to measure the mass and volumetric flow rate of a slurry stream. The slurry contains only water and solids. You take a bucket and stop watch and measure the time taken to fill the bucket. From these measurements, the following information is obtained.</p>
<ul>
<li><p>The stream fills a 10 litre bucket in 13.5s.</p>
</li>
<li><p>A sample of the slurry weighing 767g is filtered and dried. The dry solids weigh 148g. </p>
</li>
<li><p>The density of the solids only is known to be 3200kg/m<sup>3</sup>.</p>
<p>TASK: <em>Calculate the mass flow rate of the slurry stream.</em></p>
</li>
</ul>
<h2>My Attempt</h2>
<ol>
<li>I converted the rate at which the slurry filled the bucket to m<sup>3</sup>/s.</li>
</ol>
<p>10L / 13.5s = 0.01m<sup>3</sup>/13.5s = <strong>0.000741 m<sup>3</sup>/s</strong></p>
<ol start="2">
<li>I calculated the density of the slurry by adding together the relative densities of its dry solids and its water.</li>
</ol>
<p>density of slurry = (ratio of dry solids:slurry stream)(density of dry solids) + (ratio of water:slurry stream)(density of water) =(0.148/0.767)(3200 kg/m<sup>3</sup>) + (0.619/0.767)(1000kg/m<sup>3</sup>) = (0.193)(3200 kg/m<sup>3</sup>) + (0.807)(1000 kg/m<sup>3</sup>) = <strong>1424.6 kg/m<sup>3</sup></strong></p>
<ol start="3">
<li>I plugged in the respective values into the equation mass flow rate = density * volumetric flow rate.</li>
</ol>
<p>mass flow rate = 1424.6 kg/m<sup>3</sup> * 0.000741 m<sup>3</sup>/s = <em><strong>1.055 kg/s</strong></em></p>
<h2>Correct Answer</h2>
<p>According to the answer key, the answer is supposed to be <em><strong>0.854 kg/s</strong></em>.</p>
<p>I'm not sure where I went wrong.</p>
| |fluid-mechanics|chemical-engineering|chemistry| | <p>It's the density calculation you have. You're almost there! This site explains it pretty well:</p>
<p><a href="https://www.engineeringtoolbox.com/slurry-density-calculate-d_1188.html" rel="nofollow noreferrer">https://www.engineeringtoolbox.com/slurry-density-calculate-d_1188.html</a> </p>
<p>Density of slurry=1/[(ratio of solid/mixture masses)/(solid density)+(ratio of liquid/mixture masses)/(liquid density)]</p>
<p>Density of slurry=1/[0.148 kg/0.767 kg /(3200 kg/m^3) +0.619 kg/0.767 kg/(1000 kg/m^3)]</p>
<p>I hope this makes some sense!</p>
| 36194 | Help with Mass Flow Rate |
2020-06-11T14:37:22.997 | <p>I'm having some trouble figuring out how to calculate the density of a lined pipe given the diameter measurements and the specific gravity of the materials.</p>
<h2>The Problem</h2>
<p>Calculate the density of an empty rubber lined steel pipe with an inside diameter of 0.075 m, and an outside diameter of 0.079 m. The rubber lining reduces the pipe inner diameter of the pipe to 0.069 m. Assume that the SG of steel is 7.9 and the SG of the rubber is 1.52.</p>
<h2>My Attempt</h2>
<p>I multiplied the ratio of the cross-sectional area of the steel part to the cross-sectional area of the whole pipe with the density of steel. I did the same for the rubber lining and then added the two values together.</p>
<p><span class="math-container">$$\begin{align}
\rho_{substance} &= SG * \rho_{water} \\
\rho_{steel} &= 7.9 * 1000 kg/m^3 = 7900\text{ kg/m}^3 \\
\rho_{rubber} &= 1.52 * 1000 kg/m^3 = 1520\text{ kg/m}^3 \\
r &= \dfrac{\phi}{2} \\
r_{outer} &= \dfrac{0.079}{2} = 0.0395\text{ m} \\
r_{inner} &= \dfrac{0.075}{2} = 0.0375\text{ m} \\
r_{lining} &= \dfrac{0.069}{2} = 0.0345\text{ m} \\
A_{pipe} &= \pi(0.03952^2 - 0.0345^2) = 0.00116\text{ m}^2 \\
A_{steel} &= \pi(0.0395^2 - 0.0375^2) = 0.000484\text{ m}^2 \\
A_{lining} &= \pi(0.0375^2 - 0.0345^2) = 0.000679\text{ m}^2 \\
\text{steel:pipe} &= \dfrac{0.000484}{0.0011} = 0.417 \\
\text{lining:pipe} &= \dfrac{0.000679}{0.00116} = 0.585 \\
\rho_{pipe} &= \text{steel:pipe} * \rho_{steel} + \text{rubber:pipe} * \rho_{rubber} \\
&= 0.417 \cdot 7900 + 0.585 \cdot 1520 = 4184\text{ kg/m}^3
\end{align}$$</span></p>
<h2>Correct Answer</h2>
<p>According to the key, the answer is supposed to be <strong><em>990 kg/m<sup>3</sup></em></strong> (neglecting the air in the tube).</p>
<p>Is my entire approach wrong? I know how to find the composite density of, say, a slurry stream (water + solids) but how do I find the composite density of a lined pipe?</p>
| |civil-engineering|chemical-engineering| | <p>What the answer key seems to mean is to neglect the <em>mass</em> of air in the pipe, not its volume. That is, we want:</p>
<p><span class="math-container">$\rho_{pipe} = \frac{m_{rubber}+m_{steel}}{V_{pipe}} = \frac{\rho_{rubber}\cdot V_{rubber} + \rho_{steel}\cdot V_{steel}}{V_{pipe}} \propto \frac{\rho_{rubber}\cdot(R_{inside}^2 - R_{lining}^2) + \rho_{steel}\cdot(R_{outside}^2 - R_{inside}^2)}{R_{outside}^2}$</span></p>
<p>Note that I canceled out <span class="math-container">$\pi$</span> and <span class="math-container">$L$</span>, the length of a section of pipe (which is arbitrary) for compactness above, but you may want to show them in your work. It depends on whether your professor is really keen on showing everything or not.</p>
<p>Doing this with the numbers you give above results in an answer of <span class="math-container">$990.2 \frac{kg}{m^3}$</span>, so this must be what they want. I think it's poorly phrased though. Simply adding the word "mass" to the answer key would clear up a lot of this confusion.</p>
| 36205 | How do I calculate the density of a lined pipe? |
2020-06-11T20:56:23.353 | <p>Under Hooke's law, we know that sigma stresses are responsible for volumetric strain, while shear stresses are responsible for the formation of form strains. In a lecture at my university it was shown as follows (drawings). However, one thing puzzles me: if our cube were subjected to very high shear stress resulting in large gammas, shouldn't the top wall of the lecture cube move downwards (red arrow)? If so, then we would have a change in volume (shortening height) without a sigma. In that case, does Hooke's law only make sense for very small taus resulting in negligible gamma? Did I make a mistake somewhere in my thinking?</p>
<p>Thank you.</p>
<p><a href="https://i.stack.imgur.com/lprSd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lprSd.png" alt="enter image description here"></a></p>
| |stresses| | <p>Think of a stack of playing cards- if you provide a shear load in one direction to slide the top card in the x-z plane, does it reduce the total stack height in the y axis?</p>
| 36210 | Hook's law - beginner's question |
2020-06-17T01:04:22.100 | <p>Here is an image from Shigley's Mechanical Engineering Design about circularity...
<a href="https://i.stack.imgur.com/eyOTP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eyOTP.png" alt="enter image description here"></a>
Now here is another image from the same textbook about cylindricity...
<a href="https://i.stack.imgur.com/6TDNQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6TDNQ.png" alt="enter image description here"></a>
If circularity applies along the entire length of the cylinder then they seem like they achieve the exact same thing. What am I missing? What is a situation where these two would not be the same?</p>
| |drafting| | <p>I'm going to present a <em>highly</em> contrived example of a shape which would conform to the circularity constraint, and not cylindricity:</p>
<p>You can see how in the front view, every cross section is a perfect circle, but it's clearly <em>not</em> a cylinder.</p>
<p>Examples and/or discussion of when you would choose one over the other are out of scope for this question, but suffice to say that circularity is the less limiting of the two, so is used in most cases where cylindricity is not a design-critical additional requirement.</p>
<p>Often the straightness of the cylinder will be controlled by other dimensional constraints or tolerances, making the use of cylindricity redundant, and/or overconstraining.</p>
<p><a href="https://i.stack.imgur.com/EzOea.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EzOea.png" alt="circular but not cylindrical example" /></a></p>
| 36282 | GD&T What is the difference between circularity and cylindricity? |
2020-06-19T18:45:36.673 | <p>Is it possible to calculate the amount of water (mol, mass...) in a gas, in a heat exchanger, if I know the partial pressure of the water? The stream through the tubes is mainly benzene and hydrocarbons. Partial pressure of the water and the temperature is assumed constant.</p>
<p>I first tried to use the ideal gas law with the compressibility factor, but most people told me this equation is meant for closed systems (like in a tank) and not open systems like the one described. (If I'm wrong and it can be used, give me a good reason why so I understand).</p>
<p>Extra info: I need to calculate how much water is in the gas that will cause a relative humidity of 10 %. After taking the coldest temperature in the heat exchanger (because the RH would be highest here) as a constant, with the formula of relative humidity I can calculate the max partial pressure of water in the tubes thats would cause RH of 10 %. But I need the amount (mass, mol) that would be in the gas.</p>
<p>If you have any more questions, please ask.</p>
| |fluid-mechanics|thermodynamics|chemical-engineering|gas|steam| | <p>To your challenge: Most people tell you ... is not a sound working principle for analysis. Molar volume (the inverse of molar density) does not care whether the gas is in an open system or a closed system. Recast your thought process in terms of either a molar volume (density) or take a basis of a certain (enclosed) volume of gas.</p>
<p>To your main question:</p>
<ul>
<li><p>Use Dalton's law <span class="math-container">$y_{H_2O} = p_{H_2O}/p_T $</span></p>
</li>
<li><p>Assume a volume of the heat exchanger tubing or TAKE A BASIS of volume <span class="math-container">$V_T$</span></p>
</li>
<li><p>Assume ideal gases: <span class="math-container">$m_{H_2O} = y_{H_2O}(p_T V_T/RT)M_{H_2O}$</span></p>
</li>
</ul>
<p>Yes, you will not be able to determine <span class="math-container">$m_{H_2O}$</span> without a basis volume. That is however not the same as saying that you cannot use a gas law in an open system.</p>
| 36326 | Can I calculate the amount (mol or mass) of water in a gas, in a heat exchanger, if I know the partial pressure of the water? |
2020-06-21T11:28:15.130 | <p>I was wondering if it'd be a good idea to attach a 10kg load to the end of a movable steel rod by fastening it with just one M5 bolt. Looking at Misumi's <a href="https://us.misumi-ec.com/pdf/tech/mech/US2010_fa_p3549_3550.pdf" rel="nofollow noreferrer">datasheet</a>, M5 (Class 10.9) has an allowable load of 111kgf. 111kgf is 10x more than my requirement for static load. However, it's not quite clear if 111kgf is for both axial and radial directions. If the load was de/accelerated in any direction, would an M5 bolt be able to support that? For a 10kg load, does that mean if the de/acceleration of the load stayed below 10*9.8m/s^2 = 98m/s^2 (by <code>F=ma</code>), then a single M5 bolt would still be OK? Perhaps if the load was allowed to swing as well, the centrifugal force wouldn't be too far off?</p>
<p>Any guidance would be greatly appreciated.</p>
<p>EDIT1: Here's a diagram:</p>
<p><a href="https://i.stack.imgur.com/FcB84m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FcB84m.png" alt="enter image description here" /></a></p>
<p>EDIT2: Here's a revised version:</p>
<p><a href="https://i.stack.imgur.com/d9HIp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d9HIp.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|stresses|strength|bolting|fatigue| | <p>Simply put this is a textbook example of what not to do.</p>
<ul>
<li><p>The geometry of the connection, a solid bar to a narrow bolt invites stress concentration at the necking where the bolt enters the rod.</p>
</li>
<li><p>The heavy disk will rattle and slowly wear the bolt threads out, allowing play at the connection</p>
</li>
<li><p>The play of the disk will cause intense momentum back and forth causing miniature fatigue cracks both on the bolt and on the end of the rod.</p>
</li>
<li><p>This mechanism will collect some grind dust lumps inside the sleeve acting like plastic constraints causing pivoting of the bolt around them leading to a sudden complete failure.</p>
</li>
</ul>
| 36343 | Strength of bolts |
2020-06-21T19:26:32.017 | <p>I've found a shaft coupler that has a steel plate in between, but it's not clear what its function is:</p>
<p><a href="https://i.stack.imgur.com/hsnQy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hsnQy.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/SLqtu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SLqtu.png" alt="enter image description here" /></a></p>
<p>If it is to add strength to the coupler, isn't the coupler as strong as the weakest link, i.e. the aluminium ends? Does anyone know what the purpose of the steel plate is?</p>
<p>EDIT: Here's another similar design:</p>
<p><a href="https://i.stack.imgur.com/c2xGf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c2xGf.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|motors|steel|aluminum|coupling| | <p>The plate allows for small amounts of angular misalignment between the input and output shafts.</p>
| 36348 | What is the purpose of the steel plate in this coupler? |
2020-06-22T18:25:52.147 | <p>Suppose that a ball bearing has axial and radial force ratings of 300N. If the bearing is to hold a 1m long shaft in its bore with a 1kg load on the end of the shaft, will the ratings be exceeded? The torque due to gravity is 10Nm, but how should torque be compared to the bearing's force ratings?</p>
<p>Thanks in advance.</p>
| |mechanical-engineering|torque|stresses|bearings| | <p>A one-meter cantilever shaft supported by a single ball bearing is not going to work regardless of the bearing's specs.
All cantilever shafts and rods when spinning go into an increasingly unbalanced whiplash motion bending the shaft like a quadratic graph rotating about the X-axis with the centripetal force constantly increasing, leading to breaking the shaft or the ball bearing.</p>
<p>it is recommended to use at least two ball bearings set apart at least 1 meter and dynamic frequency of the system be designed such that there is no resonance between the natural frequency of the system and frequency of the rotation or their harmonics.</p>
| 36363 | Assess bearing force ratings |
2020-06-23T14:54:27.050 | <p>I have a compressor with this type of outlet:</p>
<p><a href="https://i.stack.imgur.com/h4BEV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h4BEV.jpg" alt="enter image description here" /></a></p>
<p>I am not sure how it is called, but it works for my bicycle and car.</p>
<p>Then I have a blow out gun to clean electronics by a stream of compressed air. The problem is that the blow out gun has a different inlet:</p>
<p><a href="https://i.stack.imgur.com/J0jx4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J0jx4.jpg" alt="enter image description here" /></a></p>
<p>It makes an audible click when it connects.</p>
<p>I would like to connect the two, but unfortunately I cannot find any <a href="https://www.amazon.de/s?k=kompressor%20adapter" rel="nofollow noreferrer">adapter</a>. The closest thing that remotely looks like it can do the job is this <a href="https://www.amazon.de/dp/B07WD53FWV/ref=sspa_dk_detail_3" rel="nofollow noreferrer">coupling for a welding torch</a>.</p>
<p>I have already used the two together with a plastic hose that I forcibly held in place by one hand while operating the blow out gun with the other. However, the lack of an adapter in the marketplace makes me wonder if I am doing something very wrong.</p>
<p>I see that there are essentially two types of compressors:</p>
<ol>
<li>The ones that have this click-on outlet by default - generally more expensive</li>
<li>The ones that don't - generally cheaper</li>
</ol>
<p>Is there a way I can make the compressor work with the blow out gun, or do I need to buy a second one?</p>
| |compressed-air|tools| | <p>If your car has a tire infiltrator ; no. They are too low volume for any "blowing". Unless you want to build a reservoir , like a large tire. Then fill it for a minute or so and you have enough volume for a several second blow. I have a similar tire inflater , free standing, it is pretty hot after pumping a few minutes to top off four tires. So I doubt they are good to operate for hours. It is about a $ 100 Michelin brand.</p>
| 36374 | Is it possible to connect a car air compressor to a blow out gun? |
2020-06-24T00:06:44.923 | <p>Suppose I have a wheel, about 6 feet in diameter, with empty soup cans affixed around it, powered by a big windmill. This drawing only has four spokes, but there will be more:</p>
<pre><code> ___
___| <-- tin can
... \ | /
.... <--bamboo chute| <--spoke
... \ | /
<--water is sent that way /
\ | /
\ | /
__ \ | / | |
| |-------------------( )-------------------|__|
| | / | \
/ | \
/ | \
/ | \
/ | \
~~~~~~~~~~~~~~~~~~~~~~~~|~~~~~~~~~~~~~~~~~~~~~~~~~ <-- water
|__
|___ <--water is picked up here
</code></pre>
<p>The water needs to be dispensed into a chute, or aqueduct, made from bamboo, cut in half. The water is sent down the chute to another higher pool of water.</p>
<p>The first problem I'm running into with this design is that there isn't any way to position the chute in a way that catches the water pouring out of the cans. If I position the chute under the can, it falls within the radius of rotation, and blocks the turning.</p>
<p>How can I get the water from the cans to spill out into the chute, at ideally the highest point possible?</p>
| |wheels|liquid| | <p>Consider an alternative to the current placement of the cans.
<a href="https://i.stack.imgur.com/ziYl9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ziYl9.png" alt="side dump water wheel" /></a></p>
<p>The image above from the Pinterest web site, UN Food and Agricultural Organization shows a wheel in which the cans (bamboo tubes) dump the water to the side into a trough. I'm not suggesting that you duplicate this design, but if you turn your cans slightly to the side, you can have a chute or trough to collect the water without contacting the wheel.</p>
<p>There will be velocity of the water as it dumps to clear the gap necessary to keep the cans from hitting the wheel and there will be some loss, but it's not going to be consequential in the bigger picture.</p>
<p>I used the search terms "water lifting wheel" and selected images to find this wheel. Others appeared, but this provides the best representation of the concept. The trough can be placed higher as well.</p>
| 36382 | How to get a wheel pump to deposit water near the top of the wheel? |
2020-06-24T10:50:59.780 | <p><strong>Summary:</strong> In a gear train, with given target ratio and configuration (eg. number of gears and how many compound gear steps are involved), what are the considerations when trying to select the optimal size (number of teeth) of the gears?</p>
<p><strong>Full story:</strong> I have a small lathe with manually adjustable gear train for the power feed. Manual means that I have a selection of gears to choose from and I can decide which 3 or 4 of them to install to achieve the required final ratio. There are three shafts altogether, out of which the driver shaft and the driven leadscrew have fixed position and the middle one is adjustable (this can accommodate a single idler gear or two as a compound).</p>
<p>Now I basically understand how to calculate the gear ratios that are needed, but I have many gears and I can achieve the same final ratio by several different configurations. For the sake of the argument, please ignore that some of the theoretical configurations might be physically impossible to fit.</p>
<p><strong>Example:</strong> My leadscrew has a pitch of 1.5mm, and say I would like to cut a 0.75mm pitch thread, so I need a 2:1 gear ratio. With 3 gears, the size of the idler in the middle does not affect the ratio so I can use (driver/idler/driven) 20/X/40, 30/X/60, 40/X/80, the idler being anything between 20 to 80 teeth as long as I can fit it. In a similar fashion, with 4 gears where the idlers are compound on the same shaft, I can use something like 20/30/60/80 or even 20/80/60/30 to achieve required final ratio.</p>
<p>Obviously some of the configurations are more optimal than others considering stresses, wear, and maybe for other reasons. I would like to know if some basic thumb rules can be used like "more teeth are the better due to stress distribution" and "try to avoid great differences in number of teeth". Another issue is that the gears have actually been made of different materials, most being POM but some aluminum and the smallest ones with 20 teeth are from steel. I have a feeling that the differences in strength might affect this for a greater extent than the differences in teeth.</p>
| |mechanical-engineering|gears|lathe| | <p>Generally you want more than 24 teeth in a gear to avoid having an undercut tooth profile, if you're using off the shelf 20 degree gears. Spur gears you also want to have at most a 5:1 ratio between teeth. Your smaller gear (pinion) should be of a harder material than it's mating gear. If possible have the greatest common factor between the number of teeth be 1, this way every tooth on the gear will encounter every tooth on the pinion and wear will be more even than if the same pairs of teeth mesh every time.</p>
| 36388 | Optimal gear combination with the same ratio |
2020-06-27T20:21:04.820 | <p>I have seen in many data sheets of Hydraulic Gear motor a spec called : <strong>"Minimum Speed = 450 RPM"</strong> ( 450 is an example)</p>
<p>here is a capture from the linked datasheet below :
<a href="https://i.stack.imgur.com/OPrF6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OPrF6.jpg" alt="enter image description here" /></a></p>
<p>In your experience or theoretical knowledge what happens If I run the motor slightly below this speed (say 150 RPM ) by reducing the flow of Oil.</p>
<ol>
<li><p>Does it lose torque/or any other spec but stay running normally ?</p>
</li>
<li><p>Does it stop moving smoothly? ( I have seen this "discontinuity effect" in real life when flow is too low the motor speed was around 10RPM)</p>
</li>
</ol>
<p><em>note: what I mean by discontinuity is that the motor moves a push then stops as if it is gathering oil then move a push again.</em></p>
<p><a href="https://www.brevinifluidpower.ie/custom/public/files/ot200-motor-general-information.pdf" rel="nofollow noreferrer">Datasheet</a></p>
| |mechanical-engineering|fluid-mechanics|hydraulics| | <ol>
<li>Torque will drop off with speed for the Brevini external gear motors in your datasheet (OT200 series). Unfortunately Brevini doesn't publish torque / speed curves below 1000 rpm. Here is a comparable graph from Vickers for their CR-04 internal gear motor:</li>
</ol>
<p><a href="https://i.stack.imgur.com/pDJGn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pDJGn.png" alt="enter image description here" /></a></p>
<p>Without providing a bunch of sources, external gear motor performance rapidly drops off with decreasing speed in pretty much every category - torque, heat, overall efficiency, motor life, etc.. Use a different type of motor if you need to maintain high torque at low speed - gerotor / piston / rol-vane. Or if you have only two specific speeds to run (one high speed, one low speed), there are tandem "2-speed" external gear motors for that purpose.</p>
<ol start="2">
<li>Yes motor speed will become discontinuous at some point. The transition point is difficult to predict because it depends on oil viscosity, pressure, load static friction, and other factors.</li>
</ol>
| 36427 | hydraulic Motor/pump minimum speed |
2020-06-28T14:39:19.327 | <p>I am planning a series of ponds, where the first pond (near the source of electricity) has the highest water level, the water flows to the second, lower pond, via a small water fall, and to a third, lowest pond, again via a small water fall. The pump, located in the first pond, has a fountain that sprays into the first pond, but I plan to place an underground pipe to pull water from the third, final pond.</p>
<pre><code>Bird's eye view:
________________________
__________ / 2nd pond \
/ 1st pond \ | |
| | | |
| >o< <--fountain (pump's output) |
| . | | |
| . | | |
| . ~~~~~~ |
| . ~~~~~~ <-- waterfall down to |
| . ~~~~~~ 2nd pond |
| . ~~~~~~ |
| . | | |
| . | | |
\_____.____/ \__________!!!!!!!!!_____/
. !!!!!!!!! <--waterfall down to 3rd
. __________!!!!!!!!!____________
. / 3rd pond \
. | |
. | |
. <--underground PVC pipe |
. | |
. | |
. | |
. | |
. . . . . . . . . . <--water flows into pump here
| |
| |
| |
| |
\_______________________________/
</code></pre>
<p>The pump is like this:</p>
<p><a href="https://i.stack.imgur.com/4219d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4219d.jpg" alt="enter image description here" /></a></p>
<p>Water is pumped in through the "Flat Box Filter", then sprays out the fountain at the top. If I were to lengthen the bendy pipe, replacing it with 10-20' of underground PVC to pull water from the third pond, assuming the same width pipe as the bendy tube, would that reduce the power of the pump? Or because it is all submerged anyway and water pressure will equalize, would the pump impellers not have any additional strain?</p>
| |fluid-mechanics|pumps|liquid| | <p>Theoretically, yes, but practically ,no. Regular 1" PVC will have very little flow restriction compared to the strainer and flex suction tube. Suction restrictions are always more important. And to be certain of no affect of discharge pipe I used about 30 ft of 2" PVC for my pond system . It cost a couple dollars more but nothing in the total cost. PS- pumps don't "pull", they push, and rely on gravity, etc, to produce a "net suction head". at the inlet. And , you don't say what the elevation difference is between the water surface where the pump is and the discharge at the fountainhead. That is by far the most important factor. The pump info should have a table something like 600 g/hr @ 1', 300 g/hr @ 4', 100 g/hr @ 7'. Oops! reread the question; The pump must be in the third/lowest pond to have necessary net positive suction head or flow will be very low to zero.</p>
| 36435 | Will an electric water pump has slower flow if it must pump through a longer tube? |
2020-06-28T22:25:16.230 | <p>You need a higher voltage to let current flow.
So how is it that you can to charge those big battery packs from home outlets without any additional equipment?</p>
| |electric-vehicles| | <p>The additional equipment is onboard the EV in today's crop of electric cars. The charger is part of the vehicle, while the station, formally known as EVSE, Electric Vehicle Service Equipment, negotiates with the charger, "letting it know" the source voltage and confirming that no one will be electrocuted once the power is turned on.</p>
<p>My 220v EVSE is bolted to the wall and provides 40A current to the onboard vehicle charger, after the aforementioned negotiation. I also have a portable 110v EVSE carried in the cargo bay. It too, informs the onboard charger that there's electricity.</p>
<p>My recent charge took four hours and seven minutes on the 220v EVSE. The charging estimate indicated that it would be forty or fifty hours if I used the 110v EVSE. I get about 30 miles per hour on the 220v and about 7 miles per hour on the 110v unit. Somewhere my math is incorrect on the low power EVSE, but it's unimportant to the overall assessment.</p>
<p>In both cases, there is additional equipment. It's a matter of convenience and cost which is used. For people who do not have regular access at home, public EVSE units, both free and paid can be found, but this also represents additional equipment.</p>
<p>Only two of my early EVs required no additional equipment other than an electrical cord and standard 110v outlet.</p>
| 36441 | How do you charge 400v battery packs from 120v outlets for EVs? |
2020-06-29T13:18:36.270 | <p>I had come across a question on shear force and bending moment diagram where a cantilever beam with a roller support at point C is given and the portion beyond point C in overhang. Does this hinge act as a internal release point in the structure? If yes then the does this mean bending moment about point B is zero.<a href="https://i.stack.imgur.com/d25aq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d25aq.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|statics|solid-mechanics| | <p>A hinge is a point where there is no restriction on rotation. For other points on a beam, the rotation to the left of a point must be equal to the rotation to the right of that same point; that is, there can't be a discontinuity in the rotations along a beam.</p>
<p>Hinges, however, don't have this restriction and therefore allow for discontinuities of rotation. And since bending moment is generated by a beam trying to resist changes to its curvature, we can conclude that there is no internal bending moment at a hinge.</p>
<blockquote>
<p>does this mean bending moment about point B is zero</p>
</blockquote>
<p>To be clear, a stable structure will have zero bending moment about any imaginable point. We tend to perform our <span class="math-container">$\sum M = 0$</span> calculations around supports because it eliminates some variables, but that equilibrium equation is valid about any point in the universe.</p>
<p>But if that's the case, how do we obtain non-zero bending moments along a beam? Well, that's because what we're calculating there is <em>internal</em> bending moment. And internal bending moment is calculated as the sum of bending moments to one side of the relevant point.</p>
<p>So, at midspan in a beam, for example, you calculate the bending moment generated by all the loads to the left (or right) of the beam, which will (usually) result in a non-zero value, representing the beam's internal reaction to the curvature being felt at that point.</p>
<p>But hinges don't resist rotation, so we know that the internal bending moment at the hinge is zero. Therefore, the bending moment <em><strong>to either side</strong></em> of the hinge is zero. That's what makes hinges different from other points on a beam. Any other point has a guaranteed null external bending moment (sum of all loads throughout the structure), but only hinges have guaranteed null internal bending moments (sum of all loads to the left of the hinge equal to zero).</p>
| 36449 | What is the significance of hinge in the internal structure of the beam? |
2020-06-29T18:28:56.127 | <p>I'm thinking of repurposing computer fans (case fans and/or CPU/GPU fans) for a hobby steam turbine project. So the idea would be to use the fan rotors in the turbine to drive the shaft (discarding their electronics.)</p>
<p>However, this means that the fans would run in quite high temperatures; 100 degrees celsius I suppose, since it's steam.</p>
<p>Hence my question, what's the upper temperature limit of a typical, plastic computer fan? Will it survive steam?</p>
| |materials|temperature| | <p>The steam pressure in the boiler will be greater than atmospheric, so the steam temperature will be above 100C. A typical ready-built boiler boiler for steam powered models (sold with a pressure test certificate!) would have a maximum working pressure of 4 or 5 bar, which would give steam temperatures around 150C.</p>
<p>Many common plastics will start to soften at temperatures even below 100C, and there is no obvious reason why a computer fan should be designed to work at very high temperatures, so using computer fan parts is probably not going to work.</p>
<p>Temperature ratings of some common plastics are here: <a href="https://omnexus.specialchem.com/polymer-properties/properties/max-continuous-service-temperature" rel="nofollow noreferrer">https://omnexus.specialchem.com/polymer-properties/properties/max-continuous-service-temperature</a>. There are not many options for temperatures over 100C.</p>
<p>Live steam is not stuff you should play with unless you understand what you are doing. It is too easy to get third degree burns if things go wrong! A compressed air powered turbine might be a safer project, with no high temperature issues to worry about.</p>
| 36457 | What's the thermal bounds of a typical computer case fan? |
2020-06-30T04:29:39.713 | <p>For state space systems, there is a test for 'controllability' involving finding the determinant of a 'controllability' matrix. The instructions for the test is typically to see if the determinant is equal to zero. If that determinant is zero, then the system is said to be NOT controllable.</p>
<p>My question I'd like to ask is about the condition of the determinant being 'zero'. We know that 'zero' means zero obviously. But what about a relatively small value like 1E-10? And what about 0.001?</p>
<p>Would a determinant having a value of 1E-10 mean that the system is controllable ----- because the value is not zero?</p>
| |control-engineering| | <p>A clear criterium for the property concerns rank decrease, which avoids the real computation of matrix determinant. It is clearer than the latter since the characteristic matrices A and B might have ill-conditioned matrices from emerging problem.</p>
| 36463 | controllability matrix of a state space system - determinant |
2020-07-01T02:20:56.120 | <p>Given a nonlinear system <span class="math-container">\begin{equation}
\dot{x}=f(x),~x(0)=x_0 \tag{1}
\end{equation}</span>
where <span class="math-container">$f\in{\mathcal{C}^{1}}:D\to\mathbb{R}^{n}$</span>. The Lasalle invariance theorem statement goes as follows:</p>
<p>Let <span class="math-container">$\Omega\subset{D}$</span> be a compact set that is positively invariant with respect to (1). Let <span class="math-container">$V:D\to{\mathbb{R}}$</span> be a continuously differentiable function such that <span class="math-container">$\dot{V}(x)\leq{0}$</span> in <span class="math-container">$\Omega$</span>. Let <span class="math-container">$E$</span> be the set of all points in <span class="math-container">$\Omega$</span> where <span class="math-container">$\dot{V}(x)=0$</span>. Let <span class="math-container">$M$</span> be the largest invariant set in <span class="math-container">$E$</span>. Then every solution starting in <span class="math-container">$\Omega$</span> approaches <span class="math-container">$M$</span> as <span class="math-container">$t\to\infty$</span>.</p>
<p>Here <span class="math-container">$M$</span> is the largest invariant limit set (every invariant set is a subset of M) where <span class="math-container">$x(t)$</span> converges to at infinite time. My question is how is <span class="math-container">$M$</span> and <span class="math-container">$E$</span> different. Since <span class="math-container">$\dot{V}$</span> is zero at whole of <span class="math-container">$E$</span>, then isn't <span class="math-container">$E$</span> itself the largest invariant set?</p>
| |control-engineering|control-theory|stability|nonlinear-control| | <p>The above answers are great and helped me understand better but I thought I might add my intuitive/visual understanding of how this works. So using the pendulum example above by <a href="https://engineering.stackexchange.com/a/36492/42689">useless-machine</a>, the Lyapunov function represents the energy of the pendulum system. Therefore, the height of the function <span class="math-container">$V(\mathbf{x})$</span> represents the energy at a given state. Therefore, each contour at a particular height represents all the states for a given amount of energy. We know that because of damping in the pendulum the energy should decrease over time. So if we start at some point, <span class="math-container">$x_1(t_0), x_2(t_0)$</span>, and plot out a trajectory for it, over time we should see that <span class="math-container">$V(\mathbf{x}(t)) \leq V(\mathbf{x}(t_0))$</span>. You can imagine the trajectories "flowing" down the surface of <span class="math-container">$V$</span>.</p>
<p>Importantly, the slope of the surface is not <span class="math-container">$\dot{V}$</span> since we are taking a derivative with respect to time. Instead <span class="math-container">$\dot{V}$</span> is the slope of these trajectories on the surface. With that in mind now let's consider the point mentioned <span class="math-container">$\mathbf{x} = [1, \; 0]^T$</span> which corresponds to <span class="math-container">$x_1 = 1$</span> and <span class="math-container">$ x_2 = 0$</span> remembering that <span class="math-container">$x_1 = \theta$</span> and <span class="math-container">$x_2 = \dot{\theta}$</span>. So in this case, since the instantaneous angular velocity is 0, the instantaneous loss in energy due to damping is also 0. As a result, the height of <span class="math-container">$V$</span> does not change with respect to time indicating that the slope of the trajectory is briefly along one of the contours of <span class="math-container">$V$</span>. Despite this, we know that due to gravity both <span class="math-container">$\theta$</span> and <span class="math-container">$\dot{\theta}$</span> will be changing and in this case that is along the contour. Therefore, at an infinitesimally small step away, <span class="math-container">$\dot{\theta}$</span> is no longer 0 and so <span class="math-container">$\dot{V}$</span> is no longer 0 and the trajectory is pointing down the surface again. So this point we found is more like a point of inflection along a trajectory rather than an actual equilibrium point and clearly this means that this point does not form an invariant set with the other points that follow it on the trajectory since <span class="math-container">$\dot{V}$</span> is negative semi-definite (it will never climb back up again).</p>
<p>I am very new to this so if I have a mistake in my understanding here, please correct me!</p>
| 36480 | A clarifying question on Lasalle invariance principle |
2020-07-01T06:49:09.990 | <p>(Sorry, not an engineer or student, just looking for the most metallurgically relevant stack.)</p>
<p>Recently I was in someone's office and, while playing with their paper clips, it occurred to me that copper is kind of a precious material for humble paper clips ... one look at the box, they're merely copper <em>plated</em>. But why even do that though?</p>
<p>Why not just leave them bare steel or aluminium or whatever they are made of (which, curiously, the box doesn't say)?</p>
<p><a href="https://i.stack.imgur.com/pkdzx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pkdzx.png" alt="ip" /></a></p>
| |metallurgy| | <p>It makes them not rust in humid conditions while holding papers together. The required amount of copper coating on the clip is extremely small.</p>
| 36486 | What's the benefit of copper plating paper clips? |
2020-07-01T17:05:01.717 | <p>I'm using FreeCAD, and I would like to copy all the geometry in a 2D sketch to another plane. So you have a 2D drawing on XY, and you want the same geometry to be copied to XZ. I can't seem to find an good way to do this.</p>
<p>One way I have tried is to copy the whole sketch, create a cube, and then map the geometry of the new sketch copy to the cube object face. It's quite a clunky method however and I doubt it's the most efficient way to do this.</p>
| |cad|technical-drawing| | <p>Use the carbon-copy feature of Sketcher WB:</p>
<ol>
<li>Create an empty new sketch in the plane you want.</li>
<li>On the Model tab, make the sketch you want to copy visible (click and hit spacebar).</li>
<li>In the sketch, rotate if necessary to view the to-be-copied sketch.</li>
<li>Click on the carbon-copy tool in the toolbar.</li>
<li>Hold ctrl-alt and click anywhere on the to-be-copied sketch.</li>
</ol>
<p>You may find this helpful:
<a href="https://forum.freecadweb.org/viewtopic.php?f=9&t=21762" rel="nofollow noreferrer">https://forum.freecadweb.org/viewtopic.php?f=9&t=21762</a></p>
| 36495 | In freecad, how do you copy a sketch geometry from one plane to another? |
2020-07-01T21:14:31.477 | <p>Can anyone identify what the blue and orange floating item with paddles is in this picture of what appears to be a Chinese fish or turtle farm?</p>
<p><a href="https://i.stack.imgur.com/HFzi1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HFzi1.jpg" alt="enter image description here" /></a></p>
| |liquid| | <p>It is an aerator for the pond. They use them for oxygenating the water and for circulation. <a href="https://pentairaes.com/paddlewheel-aerators-for-aquaculture-waste-water.html" rel="nofollow noreferrer">Link to the Pentair model in picture</a></p>
| 36501 | What is the function of this device in the middle of the man-made lake? |
2020-07-02T15:10:43.390 | <p>I'm looking for some advice from an ME or anyone who understands heat transfer,convection, etc.</p>
<p>I'm currently laying out in Inkscape a top panel for an audio amplifier I'm building. Now, my question lies in the orientation of the exhaust fins that I will be having laser cut out of acrylic.</p>
<p>There is a need for these exhaust fins so the cold air from below the heatsink has a place to go.</p>
<p>Now, in regards to the image below, does it make more sense to have:</p>
<ol>
<li>Situation 1, where the <strong>exhaust fins (green) are perpendicular</strong> to the heatsink fins?</li>
</ol>
<p>or</p>
<ol start="2">
<li>Situation 2, where the <strong>exhaust fins are parallel</strong> to the heatsink fins?</li>
</ol>
<p><a href="https://i.stack.imgur.com/9kvtD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9kvtD.png" alt="enter image description here" /></a></p>
<p>Thank you!</p>
| |thermodynamics|heat-transfer|mechanical|heatsink| | <p>When the slots are parallel to the aluminum fins, they promote a smooth, continuous convective stream of heat rising up and accelerating without hindrance.</p>
<p>Roughly the same concept as of a chiminea. It promotes the stream of heat to pick up momentum and accelerate, among other things.</p>
<p>Or similar to the dynamics of wildfires, they seek vertical grooves and natural re-entrant corners in the hill's face to build a rising flame.</p>
| 36511 | Most efficient orientation of exhaust vents on top plate of chassis for air flow with regard to heatsink fins? |
2020-07-02T16:53:07.677 | <p>Given a ballon and only one inlet, where some gas enters at a constant rate and the control volume of the balloon changes with time, why does conservation of mass hold?</p>
<p>To give some background, this is an example problem in the Raynold's Transport Theorem section.</p>
<p>What I don't understand and looking for clarification is that, since the mass is added to the system, how is the mass of the system constant. It changes with time, is it not? So how is the conservation of mass holds <span class="math-container">$\dfrac{dm_{sys}}{dt}=0$</span>? (See "<em>comments</em>" section of the below image)</p>
<p><a href="https://i.stack.imgur.com/gAsM8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gAsM8m.png" alt="Schematic of control the volume of a balloon" /></a></p>
<p>Here is the original question:</p>
<p><a href="https://i.stack.imgur.com/gQvVf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gQvVf.png" alt="Example problem 3.2" /></a></p>
<p>Where eq. 3.1 is <span class="math-container">$\dfrac{dm}{dt}=0$</span>,</p>
<p>and the eq. 3.16 is the Raynold's Transport Theorem:</p>
<p><span class="math-container">$$\frac{d}{dt}(B_{sys})=\frac{d}{dt}\biggl(\int_{CV} \beta \rho d\mathcal{V}\biggr) + \int_{CS} \beta \rho (\mathbf{V_r} \cdot \mathbf{n})dA$$</span></p>
<p>From this book:</p>
<p><a href="https://i.stack.imgur.com/qdg9Zs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qdg9Zs.png" alt="Fluid Mechanics 7ed, Frank M. White" /></a></p>
<p>Thank you in advance!</p>
| |fluid-mechanics| | <p>The system is a collection of contents and, because of that, its mass does not change. Hence <span class="math-container">$Dm/Dt = 0$</span>.</p>
<p>So, the problem you are facing has to do with: "since the mass is added to the system". The mass is not added to the system. The system is the contents of the balloon at the instant of observation (it is like you put a label on every molecule at that instant).</p>
<p>The control volume/surface corresponds here to the balloon and, as stated, it is deformable (it inflates).</p>
<p>After some time (i.e. after the instant of observation above), there will be more air in the balloon, but your system is still the same (i.e. the initial particles/molecules). This is, inside the balloon you will have molecules with label (the system) and molecules without label (the ones that entered afterwards).</p>
<p>EDIT:</p>
<p>The other answer is, unfortunately, missing the point. The Reynolds Transport Theorem relates a Lagrangian view (the system) with an Eulerian view (the control volume) and it does that for the instant they are coincident. This is what is done in the equation with the <span class="math-container">$B_{sys}$</span>.</p>
<p>On the left side, you have the rate of change of property B in the system (for the instant the system and the cv coincide).</p>
<p>On the right side, you have two components:</p>
<p>(1) the time rate of change of that property in the control volume (for the instant the system and the cv coincide)
+
(2) net rate (in/out) of that property flow through the control surface (for the instant the system and the cv coincide)</p>
<p>And so, because it is in the instant the system and the cv coincide, the left side equals the right side of the equation.</p>
<p>Because in this case the property is mass, we know (by definition of the system - first part of my answer) that its rate of change (in the system) is zero.</p>
| 36514 | Why is conservation of mass holds on filling balloon with gas? |
2020-07-03T11:06:08.237 | <p>When welding stainless steel pipes, the welds need some treatment (e.g. ball blasting and pickling). Is there anything equivalent for carbon steel (not stainless)? I'm looking at a tender document where someone specified pickling etc. for C-steel and I'm wondering wether someone copy-pasted from the specifications for stainless steels.</p>
| |welding| | <p>agree with blacksmith37. Here are the materials science issues.</p>
<p>the welding properties of low carbon steel from the standpoint of its time-temperature-transformation curves are well-enough understood by now that the quality of a weld in it can be determined by visual inspection and by swatting it with a 5-pound hammer. if it has the wrong color or flies apart, it's no good, and you go have a chat with the welder or the supplier of the steel.</p>
| 36525 | What treatment is necessary after welding carbon steel pipes? |
2020-07-03T16:19:32.057 | <p>I want to add a mid-drive e-bike motor to a pull wagon. I have to carry a heavy load up ramps frequently and adding an ebike motor to my wagon would make the job a lot easier. I already decided to put the motor on the "back wheel" (non-directional) and use the cart in "reverse" while on electric power, so the directional wheels would be at the back, with me. The issue I have is how to transfer the torque to the wheel? I can attach a sprocket to one wheel and use a bicycle chain, but that would give me a single wheel drive. Would that work? would that make the wagon really hard to control?</p>
<p>Other ideas I got was installing a shaft and then powering both wheels, but then they would have the same speed which is bad for turning. My last idea was to add a 5th wheel in the center and power through that wheel.</p>
<p>We are talking about a speed of "walking" speed, with loads of 350-500 lbs on a cart design to take ~1200 lbs wheels of 7 inches (total diameter).</p>
<p>Any ideas?</p>
<p>Wagon <a href="https://i.stack.imgur.com/C4NRo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C4NRo.jpg" alt="enter image description here" /></a></p>
| |electric-vehicles| | <p>I would'nt put the motor on the cart, but install a winch and use that to pull the cart up the ramp. You should be able to find one with a remote control so you can walk with the cart while operating the winch.</p>
<p>Advantages are:</p>
<ul>
<li>simpler system (no battery)</li>
<li>can be bought off the shelf, with warranty</li>
</ul>
| 36530 | One Wheel Drive utility wagon |
2020-07-04T10:42:56.087 | <p>According to argument principle, if a contour encircles a number of poles and zeros of a transfer function, the number of origin encirclements can be deduced by : <code>N = Z - P</code></p>
<p>In which N can be both positive and negative which means it can either be counter clockwise or clockwise.</p>
<p>But in Nyquist theorem, most of the time we are only counting the number of counter clockwise encirclements of point <code>(-1,0)</code> with no mentions of possible clockwise encirclements. Why is this?</p>
| |control-engineering|control-theory|signal-processing| | <p>After looking through the book "Feedback control of Dynamic Systems", I should say the clockwise encirclements are as much mentioned as the counter clockwise encirclements. However, the importance of both differs whether the open loop transfer function has unstable poles. So as you might know, N the number of clockwise encirclements in the nyquist plot, equals Z the number of unstable poles in the closed loop transfer function minus P the number of open loop unstable poles. P is in these examples assumed to be known. it is desired to get Z = 0. Therefore, if the open loop has RHP poles (P>0), The nyquist plot should have P counter clockwise encirclements. However, if the open loop has no poles in RHP, there should be no encirclements of the point (-1,0) in the nyquist plot. Every encirclement (which can be assumed to be clockwise) will show that the closed loop TF is unstable. If P > 0 and you count clockwise encirclements, the closed loop has more poles in RHP than the open loop has. So in short, counting counter clockwise encirclements to achieve stability, if there is a clockwise encirclement (you dont have to count), the system is unstable anyway.</p>
<p>I hope this helped</p>
| 36535 | Nyquist Stability Theorem and Clockwise/Counter Clockwise Encirclements |
2020-07-06T13:41:49.423 | <p>Trying to build a simplified physics simulation for cars, along the lines of <a href="https://asawicki.info/Mirror/Car%20Physics%20for%20Games/Car%20Physics%20for%20Games.html" rel="nofollow noreferrer">"Car Physics" by Marco Monster</a>. I'm stuck at straight line acceleration. I don't know if I understand the processes correctly or not:</p>
<p>I'm using engine torque, gear ratio and throttle input to calculate the "potential drive torque" on the axle. From this, I directly calculate the angular acceleration of the drive wheels. Then I compare the actual speed of the vehicle with the angular velocity of the wheels to get the longitudinal slip ratio. If too much power is used, the result indicates that the tyre is spinning by a certain amount. And here's where my confusion starts. I then use the spin ratio and the load on the drive wheels to get the final drive force.</p>
<p>Am I correctly using the spin ratio curve by plugging in the spin ratio and using the resulting force directly as the final drive force?</p>
<p>This feels wrong, over time there's a discrepancy between the angular velocity of the wheel (calculated from engine torque) and the car's speed (calculated indirectly from load on the wheel, the only input that was carried over from the engine calculation before is the spin ratio). The difference between the two speed values gets larger as long as I accelerate.</p>
<p>Edit: Adding my current pseudo-code</p>
<pre><code>var engineTorque = getEngineTorque(engineRpm, throttleInput)
var loadOnRearWheels = getLoadOnRearWheels(...)
var maxTireTraction = frictionCoefficient * loadOnRearWheels
var potDriveTorque = engineTorque * gearRatio
var potDriveForce = potDriveTorque / wheelRadius
var driveWheelAngularAcceleration = potDriveTorque / rearAxleInertia
driveWheelAngularVelocity += driveWheelAngularAcceleration * dt
driveWheelRpm = driveWheelAngularVelocity * 2 * PI
var slipRatio = getSlipRatio(driveWheelRpm, wheelRadius, speed)
var driveForce = getForceFromSlipRatio(slipRatio, loadOnRearWheels)
var totalForce = driveForce - resistanceForces - brakeForce
speed += dt * totalForce / mass
</code></pre>
| |automotive-engineering|simulation|car| | <p>You need to convert the torque at the axle into force against the ground through the tire (Force = torque/radius). Then F=ma for the entire mass of the car (plus if you want to be precise the rotational mass of the wheels, but this will be only a fraction of the mass of the car).</p>
<p>Tire slip would be based on total force from the wheel versus friction available to keep the wheel from spinning. Friction = coefficient of friction times weight according to pure theory; real life shows that the width of the tires also matters. If the car is accelerating, the friction available to rear real drive wheels goes up due to weight transfer to the rear wheels, and goes down for front wheel drive wheels for the same reason. A car in a turn has less available friction for acceleration due to the lateral force due to centripetal acceleration (force of turning). A car at speed using aerodynamic devices to increase downforce will have more friction avaialble to prevent tire slip both during acceleration and cornering. F1 cars are the best example of this and is why they look like they corner on rails.</p>
| 36552 | Calculating final drive force from engine torque and tyre spin ratio |
2020-07-06T14:40:12.487 | <p>I need to invent a construction based on standard square tubes with holes placed in certain distance from each other and from ends of the tubes:</p>
<p><a href="https://i.stack.imgur.com/S2hup.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S2hup.png" alt="enter image description here" /></a></p>
<p>I also need to imagine the tube with holes as a parametric part. Therefore I want to base all the holes on alone sketch line with points, something like the following:</p>
<p><a href="https://i.stack.imgur.com/LyCgh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LyCgh.png" alt="enter image description here" /></a></p>
<p>How can I base all the tube holes on the above sketch line?</p>
| |modeling|cad|technical-drawing|autodesk-inventor| | <p>It feels a little unintuitive, because the part is a square tube, but you can use the function Circular pattern. Select the holes on one plane as elements to repeat, select 4 repetitions and as turning axis the axis along the tube, then the holes get placed in the desired fashion.</p>
| 36556 | Autodesk Inventor Make Holes on Different Edges Based on One Line |
2020-07-06T22:28:01.557 | <p>So i want to have a reel (~4cm diameter) that will pull a ~80cm string with a torque of ~8kg-cm (edit: thanks for the clarification). To pull I want to use a gear motor. E.g. 12V DC motor like this: <a href="https://i.stack.imgur.com/61vRb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/61vRb.png" alt="dc motor" /></a></p>
<p>The problem is that I want to be able to easily unwind the string with my hands when the motor is not pulling in the string. So I would need some kind of possibility to disengage the motor. The problem is something like a sprag clutch (or one-way bearing) would not work as I pull the string in the opposite direction of the way the motor would spin. I found something like the "freecoaster" designs for bikes, but I don't know how I could build this cheap and easy especially given the small form factor (and limited tooling: mostly a 3d-printer, and normal tools). My other idea would be to try and create some kind of clutch that i would engage using a servo motor, by sliding a clutch like <a href="https://www.thingiverse.com/thing:28134" rel="nofollow noreferrer">this</a> on the axle using a servo motor. But I am not sure how well that would work or if there is a better alternative.</p>
<p>So what would be your suggestion to create a system that allows me to engage the motor to a free spinning spool when I need it to wind the string.</p>
<p>This motor will only be used to pull in the string once every month or so, so it doesn't need to be able to withstand thousands of cycles. But when it is used, it should work relatively reliable. I also found <a href="https://engineering.stackexchange.com/questions/2934/how-can-i-get-a-stepper-motor-to-engage-with-a-freely-spinning-wheel">this</a> question that seems to try and solve a similar problem. But I don't know how well that would work for my usecase. Thank you for your help!</p>
| |motors|gears| | <p>A common solution to this which may or may not work for you is to use a centrifugal clutch.</p>
<p><a href="https://i.stack.imgur.com/akZ1m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/akZ1m.jpg" alt="centrifugal clutch" /></a></p>
<p>Off the shelf options would be too large for your application; and they wouldn't kick in until too high an rpm. But you may be able to 3D print something based on this principal.</p>
<p>If you make the arms very heavy, with light springs it will work at lower rpms. You could also replace the friction element with teeth or a pawl style mechanism so that once it's engaged it will never slip.</p>
| 36565 | Engage and disengage DC motor from free-spinning reel |
2020-07-07T01:19:57.493 | <p>What's the interpretation of the margins when the curve of magnitude never crosses 0 dB or the phase curve never crosses the -180º? For example:</p>
<p><span class="math-container">$$G(s) = \frac{s+20}{(s+1)(s+7)(s+50)}$$</span></p>
<p><a href="https://i.stack.imgur.com/alXzZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/alXzZ.png" alt="enter image description here" /></a></p>
<p>I know that this TF is stable looking it's pole/zero map. But how I can achieve this information just looking the bode plot, when the margins aren't explicity?</p>
| |control-engineering|control-theory| | <p>In theory, this means you can increase the gain of your controller by an infinite amount without losing stability. However, as one might notice, by increasing the gain the magnitude of the controller will at some point cross the 0dB line, which means that you will suddenly have a phase margin. See, gain margin and phase margin indicate how far you can tweak your controller until the feedback loop loses stability.
The feedback loop (GC/(1+GC)) loses stability when GC = -1, because -1/(1-1) = -1/0. in terms of magnitude and phase, -1 has a magnitude of 1 (0db) and a phase of -180 (think of the unit circle). So, as you can see, there exist no frequency at which your open loop system will destabilize the closed loop system.</p>
<p>Checking for stability using only data (suppose you have done some frequency identification on a practical model) can be done in 3 ways:</p>
<ol>
<li>I just explained above using the gain and phase margin</li>
<li>Unstable poles will have a different kind of phase compared to stable poles. this phase starts at -180 and moves towards -90 for high frequencies. So look for parts in the bode plot where the sloop of the magnitude decreases (for instance from -20db/dec to -40db/dec), but the phase increases. Those might indicate unstable poles.</li>
<li>by checking the nyquist stability criterion using the nyquist plot. Assuming the open loop is stable (which is very likely as you have performed open loop measurements on it), if the nyquist plot encircles the -1 point, the closed loop system is unstable.</li>
</ol>
<p>Also slightly unrelated note, even though having this system might look amazing as it appears to be very hard to destabilize, it must be noted the closed loop performance is terrible. try plotting the sensitivity function (1/(1+GC)). low gain means high noise / disturbance suppression.</p>
<p>Hope this clarifies your problem, if you have any questions please ask. if anyone spots a mistake or disagrees with my explanation, please let me know. I am doing most of this from my brain right now, but can get some references if desired.</p>
| 36570 | Gain and Phase Margins from Bode Plot |
2020-07-07T07:52:55.657 | <p>For all the analysis to find work done by a compressor or work done on a turbine, the book I'm reading (<em>Fundamentals of Turbomachinery</em> by Venkanna B.K) uses the Euler turbine and pump equation, <span class="math-container">$$W=\dot{m}(V_{w1}U_1\pm V_{w2}U_2) $$</span> where <span class="math-container">$V_w$</span> is the whirl velocity of fluid at inlet and exit, and <span class="math-container">$U$</span> is the mean rotational speed of the rotor blades and inlet and exit.
It is based on the conservation of angular momentum of the fluid by drawing velocity triangles.</p>
<p>While this might give the value of work done due to momentum of the fluid, what about the work done by the pressure energy in the fluid or work done to increase the pressure energy of the fluid? Especially in cases like Francis turbine and axial compressors where change in pressure energy plays a big role, how can we consider only the momentum of the fluid in our analysis? I'm guessing work needs to be done to increase the pressure energy as well/work is done by pressure energy in turbines like Francis turbine.</p>
<p>Maybe because of complicated aerofoil shapes of the blades its hard to do an analytical approach but shouldn't we at least account for a factor of change in pressure energy?</p>
| |fluid-mechanics|pumps|turbines|turbomachinery| | <p>If you examine a derivation of the Euler work equation (<a href="http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node91.html" rel="nofollow noreferrer">http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node91.html</a>), you will see that the change in angular momentum of the fluid is proportional to the change in enthalpy of the fluid. Remember the definition of enthalpy is the flow work and the internal energy. Therefore, the enthalpy contains the change in pressure via the flow work term. This is why pressure does not directly appear in the Euler work equation, but is physically accounted for. I recommend stepping through a derivation of the equation yourself to gain a better understanding.</p>
| 36579 | Accounting for pressure energy in Euler turbine/pump equation |
2020-07-07T10:11:57.557 | <p>I use the following equation to calculate a control signal of a time-discrete PID controller:</p>
<p><span class="math-container">$$ u(k) = u(k-1) + K_{R} \cdot \left(1+ \frac{T_{0}}{2 \cdot T_{I} }+ \frac{T_{D}}{T_{0}}\right) \cdot e(k) -K_{R} \cdot \left(1 -\frac{T_{0}}{2 \cdot T_{I} } +\frac{2 \cdot T_{D}}{ T_{0}} \right) \cdot e(k-1) + K_{R} \cdot \frac{T_{D}}{T_{0}} \cdot e(k-2) $$</span>
where</p>
<ul>
<li><span class="math-container">$u(k)$</span> is the control variable</li>
<li><span class="math-container">$u(k-1)$</span> is the previous control
variable</li>
<li><span class="math-container">$K_{R}$</span> is the gain factor</li>
<li><span class="math-container">$T_{0}$</span> is the sampling time</li>
<li><span class="math-container">$T_{I}$</span> is the integral time</li>
<li><span class="math-container">$T_{D}$</span> is the derivative time</li>
<li><span class="math-container">$e(k)$</span> is the error</li>
</ul>
<p>Now I want to add a parameter adaption functionality to this, since there are slow physical changes happening in the system (dirt is depositing around the valve of a water-bearing heating system).
My research so far, has revealed, that the most common approach to adaptive PID controllers is via. system identification, as shown below:</p>
<p><a href="https://i.stack.imgur.com/Fo6wb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fo6wb.png" alt="DOI: 10.1007/978-3-642-45398-4_8" /></a></p>
<p><em>From: "Runtime Evolution of Highly Dynamic Software" (2014)</em></p>
<p>A typical approach to PID tuning is rule based, e.g., something like Ziegler Nicols. However, this approach (and the other rule based ones) I follow a clear recipe:</p>
<ul>
<li>Set integral and derivative gain to 0</li>
<li>Increase P until xxx ....</li>
<li>Optimize integral gain</li>
<li>Optimize derivative gain</li>
</ul>
<p>Since in adaptive controllers, there is already a tuned controller who's parameters just have to be adapted, these methods cannot be applied, with the exception of the final step itself, <a href="https://www.belektronig.de/web/content/1292?unique=dce691b6283fa7a3c9bce3b0347aeab35a8f4049" rel="nofollow noreferrer">for example</a>:</p>
<ul>
<li>If oscillation is to high --> increase <span class="math-container">$T_{d}$</span> (view below)</li>
</ul>
<p><a href="https://i.stack.imgur.com/R8B0A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R8B0A.png" alt="enter image description here" /></a></p>
<ul>
<li>If controller is instable --> decrease <span class="math-container">$T_{d}$</span> (view below)</li>
</ul>
<p><a href="https://i.stack.imgur.com/G8Hy7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G8Hy7.png" alt="enter image description here" /></a></p>
<p>My question now is, is there a system of rules that I can apply for the adaptive control of my controller? Particularly great would be one that can translate to the formula I gave at the start ...
Advice on good Python implementations would also be of great value to me.</p>
<p>Edit:</p>
<p>The pid controller controls a 3-way valve that mixes hot supply water into the circulation of a water-bearing radiator. A schematic I found on the internet is shown below. The values in my case: supply temperature (where it says 180°F) would be 40-70°C, inflow (where it says 110°F) would be around 20-40°C, outflow (where it says 90°F) would be around 5-10°C lower than the inflow. The valve itself is operated between 0% and 100%</p>
<p><a href="https://i.stack.imgur.com/MmvZn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MmvZn.png" alt="enter image description here" /></a></p>
<p>Below I've included an actual 6 hour snapshot of the processes (current control, not implemented by me). The purple timeseries is the setpoint, yellow is the inflow temperature (the controlled temperature). The Y axis shows "temperature":</p>
<p><a href="https://i.stack.imgur.com/55eWa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/55eWa.png" alt="enter image description here" /></a></p>
| |control-engineering|pid-control| | <blockquote>
<p><strong>My question now is, is there a system of rules that I can apply for the adaptive control of my controller?</strong></p>
</blockquote>
<p>As far as I know there are no such rules. Creating an adaptive controller without checking the convergence and robustness can have an unwanted effect on the performance.
This requires knowledge about the system dynamics and more advanced tools than just using trail-and-error or rules of thumb, such as the Ziegler and Nichols method.</p>
<blockquote>
<p>I discovered follow a clear recipe:</p>
<ul>
<li>Set integral and derivative gain to 0</li>
<li>Increase P until xxx ....</li>
<li>Optimize integral gain</li>
<li>Optimize derivative gain</li>
</ul>
</blockquote>
<p>This method you describe above is called trail-and-error. With the Ziegler and Nichols method you determine the system properties. Based on that, you can determine the gains of the desired controller.</p>
<p>This brings me to a simple solution that you could use.
Determine the required system properties for the Ziegler and Nichols method during the day and than update them occasionally.
It is important that this update is much lower frequent than all the system dynamics so that you can assume that is the same as restarting the system.</p>
| 36580 | Adapting PID controller parameters using expert rules |
2020-07-07T17:13:52.193 | <p>My girlfriend will get a refrigerator from the landlord, but right now she hasn't one. But she has a freezer. So we will get ice packs (images: <a href="https://commons.wikimedia.org/wiki/File:Freezer_ice_packs.jpg" rel="nofollow noreferrer">1</a>, <a href="https://commons.wikimedia.org/wiki/File:Kompresse.jpg" rel="nofollow noreferrer">2</a>, <a href="https://commons.wikimedia.org/wiki/File:KYK_ice_pack_760g.jpg" rel="nofollow noreferrer">3</a>) and use a cooling box for the time.</p>
<p>Now I was wondering: Would a box filled with water be just as good as any of the ice packs? What is the part that distinguishes the "cooling effect" of different materials, given that the temperature of the freezer is not changeable? The space the cooling pack may take is limited, the weight is not.</p>
<h2>Ideas</h2>
<p>From school I vaguely remember that it takes a lot of energy to warm water up / melt ice. So it also cools well. Is that thermal conductivity?</p>
<p>Would a solid block of iron cool better / longer than the same volume of water? Would water with something added (e.g. salt) be better for that?</p>
| |cooling|thermal-conduction| | <ol>
<li>No water in its liquid form will not work as well as ice or ice packs for keeping your cooler cold.</li>
<li>No a block of iron will not work as well as ice or ice packs.</li>
<li>The "hard" packs in your first and last photos are made for the job. There are also soft gel packs that do the same. However the type of gel pack intended for use on muscle injuries are not intended to go in a freezer only an ice tray and may be damaged if you use them for this. Typically they are rated for -4°C or some down to -10°C.</li>
<li>Using ordinary ice works. You can use bags of ice cubes or makeshift ice packs (e.g. freezer bags or plastic water or soda bottles) just make sure the material used is elastic enough as water expands as it freezes so will burst rigid containers.</li>
</ol>
<p>More Notes:</p>
<p>The store-bought packs with the gel mix will take longer to melt than pure water ice of your ice cubes or makeshift ice packs, so will stay cold longer and will not need to be swopped out as often and is also less likely to burst in the freezer or leak in the cooler.</p>
<p>The inside of your ice box will get cooler and stay cool longer if you use ice or ice packs than if you use a similar amount of liquid water or a block of iron. Reason: The ice absorbs heat from its surroundings, while staying at the same temperature, thus cooling down the inside of the ice box and its contents. The reason it can do this is because it is undergoing a phase change (melting). If you use liquid water or a block of iron, the water or iron will slowly heat up while the surroundings cools down and an equilibrium will be reached that is warmer than the iron or water originally was. Of course once the ice in the ice packs melts, it will do the same, but it will take significantly longer.</p>
<p>You can also cool your drinks and fresh produce down by wrapping in newspaper and placing them in the freezer for a while -not long enough to freeze- before putting them in the cooler. The newspaper insulates somewhat, so the soda cans or whatever you wrap will cool more uniformly and not freeze as fast.</p>
| 36590 | Are there differences between Ice Packs? |
2020-07-08T04:26:35.010 | <p>How do you weld carbon fibre to another piece of carbon fibre? If it can't be welded together what other solutions are there for connecting carbon fibre together?</p>
| |welding|carbon-fiber| | <p>The bonding in carbon is directional which means one carbon atom must bond in a fixed way to other atoms, e.g. 4 single bonds to hydrogen or two double bonds to oxygen. In the case of carbon-carbon bonding, you have one carbon to four others in diamond and one to three others in graphite.</p>
<p>The bonding in metals is delocalised which means that the metal atoms generally do not need to be ordered in a specific way to bond to each other. They have a preference to do that (hence they form crystals) but it is not a requirement.</p>
<p>So the point being that you can weld metals (with some exceptions) because their bonds are delocalised and they will happily join up with neighboring metal atoms; you can get a polycrystalline structure in a metal sample that is still a single cohesive piece. For graphite bonding to graphite - unless you can stitch at an atomic level these directional bonds, you can not "weld" or join two separate graphite parts.</p>
| 36610 | Welding Carbon fibre |
2020-07-08T07:58:44.940 | <p>When looking at various small fans (like those found in a computer or the bathroom) you often see in their specifications the type of bearings they use. And the typical values there are either "sleeve bearing", "1 ball bearing" and "2 ball bearing".</p>
<p>Now I can imagine what a "sleeve bearing" is - it just means that the axis of the fan is stuck in a hole, maybe with a bit of lubricant, and that's that. In essence, it means "no bearing". But what are "1 ball" and "2 ball" bearings? I know what a "ball bearing" is in general, but how do you make that with just 1 or 2 balls? I think that the very minimum theoretical amount would be 3 balls but in practice I don't think I've seen anything less than 6 or so.</p>
| |bearings| | <p>A "1 ball bearing" does not mean the bearing has 1 ball. Or 2...</p>
<p>It means that the motor / fan has been designed with one ball bearing assembly - which may contain 6, 8, 10 or 12 balls depending on the size and design. For example a caged bearing will have fewer balls to allow for the cage and this reduces inter-ball friction. Otherwise, the balls can touch each other when rotating which adds to the wear rate, but this is also heavily influenced by the other loads on thee bearing.</p>
<p>So, when the fan / motor assembly says "2 ball bearings", that tends to mean that there will be a ball bearing at each end of the shaft, but in some situations there can be 2 ball bearings at one end of the shaft to support the offset load.</p>
| 36617 | What are 1 ball and 2 ball bearings? |
2020-07-09T22:49:36.130 | <p>The radiator in my car is damaged. One tube has a severe gash in it. I need to drive the car about 100 km before I can get a replacement radiator installed and coolant drains too fast to reasonably refill the cooling system every so often (with water of course, I'm not going to be leaking antifreeze all the way!)</p>
<p>Can I, e.g.</p>
<ol>
<li>rip out some of the cooling fins around the damaged tube and crimp it shut on both sides of the gash or</li>
<li>fill the gash completely with a high-temperature sealant</li>
</ol>
<p>in order to (hopefully) dramatically impede the loss of coolant?</p>
<p>Will doing one or both of these things only cause the radiator to lose efficiency due to no flow going through a single tube, or <strong>will completely blocking the flow of one tube stop all coolant from flowing through the system?</strong></p>
<p>I.e. are the tubes of a radiator run in series:
<a href="https://i.stack.imgur.com/3N4mN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3N4mN.jpg" alt="enter image description here" /></a></p>
<p>or parallel:
<a href="https://i.stack.imgur.com/88kiX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/88kiX.jpg" alt="enter image description here" /></a>
?</p>
<p>I imagine they are run in parallel and these are safe temporary fixes.</p>
| |automotive-engineering|cooling|thermal-radiation| | <p>All (car) radiators are parallel, it is the only way to get the flow rate and reduce the fluid velocity in each tube sufficiently to allow the time for heat transfer.</p>
<p>However, some radiators are series parallel where each section is in series and the section has the small tubes in parallel. Not seen that in a car though.</p>
<p>If you can carefully « extract » enough of that damaged tube to put a small clamp on it that might help. But if the tube is copper/brass you can solder it reasonably easily, but most radiators are aluminium now which is more challenging.</p>
| 36647 | Do the tubes of an automotive radiator flow in parallel or series? |
2020-07-10T20:20:43.620 | <p>Suppose a common scale using 4 load cells is placed on an inclined surface and gives a 1kg reading for a given load. Will the 'true' weight of the load be (precisely) <code>1kg / cos(theta)</code> where theta is the angle of the plane? This should be true in physics, but is it the case in practice for common bathroom scales? Being on an inclined plane, different forces will be applied to the load cells. Should that matter?</p>
| |electrical-engineering| | <p>Usually the electric scales have one load cell with a mechanism to balance the offset loading that happens due to not standing in optimal point on the scale.</p>
<p>One would assume the same applies to an scale with a four cell scale.</p>
<p>If that is the case and the scale does not have circuitry to take into account small unevenness of the support, then the scale will read <span class="math-container">$ m_{scale}= m_{real}\cdot cos\alpha .$</span></p>
| 36659 | Scale on inclined plane |
2020-07-11T10:41:17.697 | <p>I am a student trying to learn about how OXO made this cup.</p>
<p>Here are some constraints that I know: I figured out that the text on the outside can be made using screen printing, but then I don't think you can use that same tech to print on the inside since it's so tight. I know this is a high volume part so the process has to be scalable. It's also got to be low cost because the total price is under $8. The text also needs to be able to stay on for 10+ years after contact with various fluids since OXO promises durability. Lastly, the material being printed on is Tritan, a food safe plastic material.</p>
<p>Here is the product: <a href="https://www.oxo.com/1-cup-angled-measuring-cup-451.html#" rel="nofollow noreferrer">https://www.oxo.com/1-cup-angled-measuring-cup-451.html#</a></p>
| |materials|manufacturing-engineering|plastic| | <p>It turns out my mother in law has one!</p>
<p><strong>It’s printed on the underside</strong> The red numbers are printed the same way as those on the sides of the jug, and then a layer of white paint is applied on top of that in order to provide contrast when viewing from the top.</p>
<p><img src="https://i.stack.imgur.com/qM0ab.jpg" alt="enter image description here" /></p>
| 36661 | How did OXO get text into this inside feature? |
2020-07-11T15:06:21.857 | <p>Here's my guess, but I would love to hear another perspective.</p>
<p>I think this is injection molded. This colander has draft angles and the two mold halves use an intricate step-like wall pattern to create the final holes on the wall.</p>
<p>(For extra point, how would you efficiently produce the wall pattern in CAD?)</p>
<p><a href="https://i.stack.imgur.com/OaoBl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OaoBl.jpg" alt="OXO colander" /></a></p>
| |design|manufacturing-engineering|cad|plastic|computer-aided-design| | <p>This colander was definitely not CNC machined, it was injection molded. Without being able to inspect the part closely, I will guess that the <em>shutoffs</em> (the open spaces between the plastic ribs) were formed by a retractable feature in the mold which presses against the mold <em>core</em> during the injection process, and then is pulled away from the core when the mold is split, so the part will not hang up in the mold.</p>
<p>The other way to do this without a <em>mechanism</em> in the mold as described above is to cleverly shape the halves of the mold in such a way as to form the shutoffs naturally as the two halves of the mold are pressed together. This is a tricky thing to accomplish because it creates an extremely complex <em>parting line</em> between the mold halves which greatly increases the chances of creating <em>flash</em>, where small amounts of plastic squeeze inbetween the shutoff zones and leave thin webs of plastic that close off the shutoffs.</p>
| 36662 | How did OXO manufacture this colander with such intricate wall feature? |
2020-07-11T15:42:08.563 | <p>I'm interested in knowing whether or not submerged welding can be done in fluids other than water, specifically mineral oil in this case. The reason I want to know is that I'm looking into designing an extrusion 3D printer for metals that works on the same principle as stick welding, like these ones:</p>
<p><a href="https://hackaday.com/2020/03/31/3d-metal-printer-uses-welding-wire/" rel="nofollow noreferrer">https://hackaday.com/2020/03/31/3d-metal-printer-uses-welding-wire/</a></p>
<p><a href="https://3dprint.com/226829/researchers-develop-low-cost-metal-3d-printer/" rel="nofollow noreferrer">https://3dprint.com/226829/researchers-develop-low-cost-metal-3d-printer/</a></p>
<p>However, I don't want the metal to oxidize, and the machines in these examples produced very poor surface quality due to the metal overheating and melting. I know that the oxidization could be prevented by surrounding the entire print in an inert gas, like argon, carbon dioxide, or sulfur hexafluoride, but I wanted something that would be easier to recycle without getting contaminated, plus a fluid would help to keep the print cool. Water isn't an option because it rusts metal and its conductivity makes it dangerous to use with electricity, so mineral oil seemed like the best option, as it is also used to <a href="https://en.wikipedia.org/wiki/Transformer_oil" rel="nofollow noreferrer">cool transformers</a>.</p>
<p>My major concern with using oils of this kind is, obviously, that they are flammable when heated to high temperatures in the presence of oxygen, so for this question I'm assuming the welding is taking place either in an airtight box (but expandable to account for the oil evaporating), or in a swimming pool full of mineral oil big enough for the heat to dissipate into safely. I was also wondering if the oil might decompose into toxic chemicals at welding temperatures or introduce unwanted inclusions or contaminants into the weld. Anything you can tell me would be helpful. Thanks.</p>
| |heat-transfer|manufacturing-engineering|welding|3d-printing|chemistry| | <p>Mineral oil is not used to cool transformers because it is flammable. <em>Transformer oil</em> is a different compound which is not flammable and possesses high dielectric strength. In any case, the temperatures inside a welding arc are high enough to decompose almost all chemical compounds, as Blacksmith37 points out.</p>
| 36664 | Can wet welding be done in mineral oil? |
2020-07-11T18:11:28.100 | <p>I'm looking for some kind of "standard" hydrofoil profile suited for a water vehicle comparable to the <a href="http://lancet.mit.edu/decavitator/Decavitator.html" rel="nofollow noreferrer">MIT Decavitator</a> (similar weight/speed/power/etc.), so I can use it as a base to get started with my own simulations/experiments. As far as I can tell, the team behind the Decavitator didn't publish details about their foils in more detail than the following:</p>
<blockquote>
<p>The larger 60<span class="math-container">$\times$</span>2.35-in / 1520<span class="math-container">$\times$</span>60-mm
(span <span class="math-container">$\times$</span> mean chord) wing is placed about 6 in / 150 mm below the
pontoon bottoms, and the smaller 30<span class="math-container">$\times$</span>1.4-in / 760<span class="math-container">$\times$</span>35-mm
wing is placed another 6 in / 150 mm lower. [...]<br />
The wings employ a custom 14%-thick airfoil which has been tailored
for the operating Reynolds-number range of 150,000 -- 400,000,
using the design principles and numerical simulation methods
employed for the Daedalus wing airfoils [Drela_JA88,Drela_SV89].<br />
The structural merit of the relatively thick
airfoil allows smaller wing areas and less overall drag than
the 10-12%-thick airfoils more commonly employed at these
low Reynolds numbers. The thick airfoil also gives the rather
wide usable lift-coefficient range <span class="math-container">$0.2 < C_L < 1.1$</span> , which
translates to low wing drag over a wide range of speeds.</p>
</blockquote>
<p>(yes, I checked out those references, they are very general and don't have any hydrofoil example)</p>
<p>I have found some details on the design of foiling surf boards, but as far as I can tell they usually heavily restrict the wing span (for safety/maneuverability reasons I guess) to the detriment of efficiency so they aren't useful to me.</p>
| |airfoils| | <p>This is basically a list of resources gleaned from having followed the HPV boat development for the past twenty years.</p>
<p>Here's a good sketch of how to go about exploring the hydrofoil design space. You need human factors input very early on, a good sense of drive train layout and expected efficiency, and a realization that one pedaling person has a massive torque pulse problem that affects propeller craft totally differently from road bikes.</p>
<p><a href="http://people.cst.cmich.edu/yelam1k/asee/proceedings/2011/data/36-187-1-dr.pdf" rel="nofollow noreferrer">Human Powered Hydrofoil Design & Analytic Wing Optimization</a></p>
<p>The principal examples of HPV solo hydrofoils are <em>Flying Fish</em> and <em>Decavitator</em>. There were several incarnations of both craft. <em>Flying Fish</em> won in the match race, but <em>Decavitator's</em> air prop let it exploit a rule permitting up to five knots of wind during the record attempt. The record was set running downwind in 4 knots, giving the air prop an efficiency boost.</p>
<p>Tom Speer developed foils and control systems for Boeing, for land yacht record attempts, and was part of Oracle's America's Cup team. He has generously made large amounts of his research and experiences available through his web site. <a href="http://tspeer.com/" rel="nofollow noreferrer">tspeer.com</a></p>
<p>boatdesign.net has hundreds of pages of discussions on hydofoils in several of it's subforums including Hydrodynamics, Sailing, and Powerboats. These include US navy hydrofoil engineers and control system designers, as well as Tom Speer and Mark Drela, and a host of others.</p>
<p>Sailing Anarchy has also, primarily in the Dingy Anarchy forum.</p>
<p><a href="https://m-selig.ae.illinois.edu/" rel="nofollow noreferrer">https://m-selig.ae.illinois.edu/</a> has hundreds of foil offsets.</p>
<p>Xfoil and AVL are freeware analytic and design programs out of MIT, largely the work of Mark Drela. AVL presumes the user has a fairly good knowledge of vehicle stability issues.</p>
<p>Tom Speer designed the <a href="https://www.boatdesign.net/posts/134426/" rel="nofollow noreferrer">Speer H105</a> foil for this application. In the past, he has provided offsets in exchange for a copy of any performance data the user obtains.</p>
<p>Mark Drela has indicated the <em>Decavitator</em> used his MRC16 foil for part of the wing. IIRC, the main foil was in three parts with different sections (middle, intermediate, and outboard). <a href="https://www.boatdesign.net/threads/foiler-design.2447/page-45#post-249880" rel="nofollow noreferrer">MRC16 foil discussion</a>. <a href="https://m-selig.ae.illinois.edu/ads/coord/mrc-16.dat" rel="nofollow noreferrer">MRC foil offsets</a></p>
<p>One of several outstanding discussions of hydrofoil issues on boatdesign.net <a href="https://www.boatdesign.net/threads/foil-cavitation-at-lower-speeds-than-expected.53927/" rel="nofollow noreferrer">foil-cavitation-at-lower-speeds-than-expected</a></p>
<hr />
| 36667 | What foil profile(s) does the MIT Decavitator (or similar vehicle) use? |
2020-07-12T00:10:05.823 | <p>I've gathered a lot of different details online - trying to understand exactly what they mean by 'manipulated variable'. So I only decided to finally ask about it here.</p>
<p>In one source - eg. at this link <a href="https://processdesign.mccormick.northwestern.edu/index.php/Process_controls" rel="nofollow noreferrer">here</a>, it says "<em>Inputs can further be separated into disturbance inputs and <strong>manipulated inputs</strong>. <strong>Manipulated variables</strong> refer to the quantities that are directly adjusted to control the system</em>."</p>
<p>Ok ----- I can see the source started to mention 'manipulated inputs', and then magically in the next word refers to 'manipulated variables'. So I'm assuming manipulated inputs means the same thing as manipulated variables.</p>
<p>May I ask if manipulatED variables refers to ALL variables (entire set) that are 'allowed to be manipulated' (ie. manipul<strong>ABLE</strong> ?</p>
<p>Or does 'manipulated variable' mean that a system may have various variables that can all potentially be manipulated, but the 'manipulated variables' are subsets of the entire set of potentially manipulated variables?</p>
<p>For example, if the system has in <em>total</em> five potentially manipul<strong>able</strong> variables, and then I decide to 'fix' (make constant) three of those variables and then decide that the remaining two variables stay adjustable (and will be adjusted), then does this mean that there will be two manipulated variables (even though there are actually five potentially manipulable variables)?</p>
<p>My question is just about getting a proper understanding about what is a 'manipulated variable'. I so far assume it's not the entire set of 'manipulable' variables in the system, and so far assume it means only the variables that we are going to independently manipulate (and not keep fixed at the beginning).</p>
<p>Thanks very much all!</p>
| |control-engineering| | <p>Niels already gave a good answer, but to still put some focus on the difference between manipul<strong>able</strong> and manipul<strong>ated</strong> variable:</p>
<p>Imagine you have a process in which you can influence the flow rate of a fluid and the temperature of that fluid. Then you have 2 manipul<strong>able</strong> variables.</p>
<p>However, while developing your control strategy, you decide that it would be best to keep the flow rate constant and let the closed-loop controller directly adjust the temperature of the fluid. Now the temperature became a manipul<strong>ated</strong> variable, because it is literally directly manipulated, while the flow rate remains constant.</p>
<blockquote>
<p>but the 'manipulated variables' are subsets of the entire set of potentially manipulated variables?</p>
</blockquote>
<p>Correct, maybe try to imagine the manipul<strong>ated</strong> variable more as a property of the controller, not the controlled system. Everything that a controller could potentially influence is manipul<strong>able</strong>, everything that it <em>actually</em> influences as a result of your control strategy is manipul<strong>ated</strong>.</p>
| 36673 | Definition of a 'manipulated variable' in process control - what is a manipulated variable exactly? |
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