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2019-02-17T14:06:06.297 | <p>Is it possible to ( how do you ) export a series of points along a spline.
A spline has been created in 3d from various segments.
I would like to export a table of 3d points along the spline starting at one end and every mm or so along the spline.</p>
<p>The result would be a file with Z,Y,Z points that are every mm along the spline.<br>
I want to end up with lots of little line segments that approximate the spine. </p>
<p>I have seen a few tutorials on how to export intersection points but that in not what I am looking for.</p>
<p>Thanks</p>
| |solidworks| | <p>Answering my own question in hopes others will stumble upon it.
I did not find an acceptable way to directly export points off a spline.<br>
Attached is my two part solution.<br>
I was able to export the parameters that define the spline from Solidworks and then process that data with a small python script.</p>
<p>Here is the VBA script to run in Solidworks that will export the data for a spline.<br>
In my particular application it is looking for two 3d sketches (centerLineFull and rackPosFull), each of these sketches has exactly 1 spline.<br>
The parameters of the spline are then output to a file that is formated to look like a python dictionary.<br>
You will need to modify all of this for your needs.. but you get the idea..<br>
[A lot of this was lifted from various Solidworks API help files. There are a lot of goof search terms in here that will lead to the correct places in the API help files ]</p>
<pre><code> Attribute VB_Name = "Macro21"
Option Explicit
Dim swApp As SldWorks.SldWorks
Dim Part As SldWorks.ModelDoc2
Dim swSelectMgr As SldWorks.SelectionMgr
Dim swFeature As Feature
Dim swSketch As Sketch
Dim swSketchSeg As SldWorks.SketchSegment
Dim swCurveIn As SldWorks.Curve
Dim varSplineParams As Variant
Dim success As Boolean
Dim swActiveSketch As Sketch
Dim SketchSegments As Variant
Dim SketchSegment As Variant
Dim swSketchSegment As SketchSegment
Dim FilePath As String
Dim FileName As String
Dim boolstatus As Boolean
Dim longstatus As Long, longwarnings As Long
Dim myModelView As Object
' Extract two integer values out of a single
' double value, assuming that a C int has 4 bytes
Type DoubleRec
dValue As Double
End Type
Type Int2Rec
iLower As Long
iUpper As Long
End Type
' Extract two integer values out of a single
' double value, by assigning a DoubleRec to the
' double value then copying the value over an
' Int2Rec and extracting the integer values
Function ExtractFields(dValue As Double, iLower As Integer, iUpper As Integer)
Dim dr As DoubleRec
Dim i2r As Int2Rec
' Set the double value
dr.dValue = dValue
' Copy the values
LSet i2r = dr
' Extract the values
iLower = i2r.iLower
iUpper = i2r.iUpper
End Function
Sub SplineDump(varSplineParams As Variant)
Dim dTmpValue As Double
Dim iNumKnots As Integer
Dim iNumCtrlPts As Integer
Dim iDimension As Integer
Dim iOrder As Integer
Dim iPeriodicity As Integer
Dim iSplineArraySize As Integer
Dim iSplineIndex As Integer
Dim iVarIndex As Integer
Dim i As Integer
dTmpValue = varSplineParams(0)
ExtractFields dTmpValue, iDimension, iOrder
dTmpValue = varSplineParams(1)
ExtractFields dTmpValue, iNumCtrlPts, iPeriodicity
'Debug.Print "Dimension "; iDimension, " Order"; iOrder, " NumCtrlsPts"; iNumCtrlPts, "Periodicity "; iPeriodicity
Print #1, "'Dimension':"; iDimension; ","
Print #1, "'Order':"; iOrder; ","
Print #1, "'NumberOfControlPoints':"; iNumCtrlPts; ","
Print #1, "'Periodicity':", iPeriodicity; ","
Print #1, "'KnotPoints':["
iNumKnots = iOrder + iNumCtrlPts
iVarIndex = 2
For i = 0 To (iNumKnots - 1)
Print #1, varSplineParams(iVarIndex); ","
iVarIndex = iVarIndex + 1
Next i
Print #1, "],"
Print #1, "#End of Knot Points"
Print #1, ""
Print #1, "#Control Points..."
Print #1, "'ControlPoints':["
For i = 0 To (iNumCtrlPts - 1)
Dim j As Long
Dim X, Y, Z As Double
X = varSplineParams(iVarIndex) * 1000#
Y = varSplineParams(iVarIndex + 1) * 1000
Z = varSplineParams(iVarIndex + 2) * 1000
'Debug.Print X, Y, Z
Print #1, "[";
Write #1, X, Y, Z;
Print #1, "],"
iVarIndex = iVarIndex + 3
Next i
Print #1, "],"
Print #1, "#End of Control Points"
End Sub
Sub SelectStartFeature()
Dim SketchArcs As Variant
Dim SketchArc As Variant
Dim SketchPoints As Variant
Dim SketchPoint As Variant
Dim swSketchPoint As SketchPoint
Dim swArcType As SketchArc
Dim Point(2) As Double
Dim PointVariant As Variant
Dim swXForm As MathTransform
Dim swMathUtil As MathUtility
Dim swMathPt As MathPoint
Dim status As Boolean
Dim work As String
Dim i As Integer
Print #1, "#Start Features"
status = Part.Extension.SelectByID2("start", "SKETCH", 0, 0, 0, False, 0, Nothing, 0)
Debug.Print "Selected Shetch 'start' is ", status
Set swFeature = swSelectMgr.GetSelectedObject6(1, 0)
Set swSketch = swFeature.GetSpecificFeature2
Set swXForm = swSketch.ModelToSketchTransform
Set swXForm = swXForm.Inverse
Set swMathUtil = swApp.GetMathUtility
Debug.Print "Sketch is in 3d:"; swSketch.Is3D
Debug.Print "Number of arcs ", swSketch.GetArcCount
Print #1, "'NumberArcsFound':", swSketch.GetArcCount; ","
SketchArcs = swSketch.GetArcs2
Debug.Print "Arc Center on Sketch", SketchArcs(12) * 1000, SketchArcs(13) * 1000, SketchArcs(14) * 1000
'Create a Temporary point to use to transform from Sketch coor to model
Point(0) = SketchArcs(12)
Point(1) = SketchArcs(13)
Point(2) = SketchArcs(14)
PointVariant = Point
Set swMathPt = swMathUtil.CreatePoint(PointVariant)
Set swMathPt = swMathPt.MultiplyTransform(swXForm)
Debug.Print "Arc Center in Model", swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000
Print #1, "'ArcCenter':[";
Write #1, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000;
Print #1, "],"
Print #1, "'NumberPointsFound':", swSketch.GetSketchPointsCount2; ","
Print #1, ""
Debug.Print "Number of points", swSketch.GetSketchPointsCount2
i = 0
SketchPoints = swSketch.GetSketchPoints2
For Each SketchPoint In SketchPoints
work = "'Point" + CStr(i) + "':"
i = i + 1
Set swSketchPoint = SketchPoint
'Debug.Print swSketchPoint.Type, swSketchPoint.X, swSketchPoint.Y, swSketchPoint.Z
If swSketch.Is3D Then
Debug.Print swSketchPoint.Type, swSketchPoint.X, swSketchPoint.Y, swSketchPoint.Z
Else
Point(0) = swSketchPoint.X
Point(1) = swSketchPoint.Y
Point(2) = swSketchPoint.Z
PointVariant = Point
Set swXForm = swSketch.ModelToSketchTransform
Set swXForm = swXForm.Inverse
Set swMathUtil = swApp.GetMathUtility
Set swMathPt = swMathUtil.CreatePoint((PointVariant))
Set swMathPt = swMathPt.MultiplyTransform(swXForm)
Debug.Print "Point Model", swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000
Debug.Print "Point shetch", swSketchPoint.X * 1000, swSketchPoint.Y * 1000, swSketchPoint.Z * 1000
Print #1, work;
Print #1, "[";
Write #1, swMathPt.ArrayData(0) * 1000, swMathPt.ArrayData(1) * 1000, swMathPt.ArrayData(2) * 1000;
Print #1, "],"
End If
Next
Print #1, ""
'Part.ClearSelection2 True
End Sub
Sub main()
Set swApp = CreateObject("SldWorks.Application")
Set Part = swApp.ActiveDoc
Set swSelectMgr = Part.SelectionManager
FilePath = CurDir$
FilePath = Left(Part.GetPathName, InStrRev(Part.GetPathName, ".") - 1)
Debug.Print FilePath
FileName = FilePath + "_Table.txt"
Debug.Print FileName
Open FileName For Output As #1
Print #1, "#Spline data"
Print #1, "{"
Print #1, "'FileName':' "; Part.GetPathName; "',"
Print #1, "'Date':'"; Now(); "',"
Print #1, ""
SelectStartFeature
Print #1, "#Begin Center Line Spline"
Print #1, "'CenterLineSpline': {"
Part.ClearSelection2 (True)
success = Part.Extension.SelectByID2("centerLineFull", "SKETCH", 0, 0, 0, False, 0, Nothing, 0)
Part.SketchManager.Insert3DSketch (False)
Set swActiveSketch = Part.GetActiveSketch2
SketchSegments = swActiveSketch.GetSketchSegments
For Each SketchSegment In SketchSegments
Set swSketchSegment = SketchSegment
Debug.Print swSketchSegment.GetName
Debug.Print swSketchSegment.GetType
If swSketchSegment.GetType = 3 Then
Debug.Print "This is type 3"
Exit For
End If
Next
Set swCurveIn = swSketchSegment.GetCurve
Print #1, "'Spline Name':' "; swSketchSegment.GetName; "',"
Print #1, "'Length':"; swSketchSegment.GetLength * 1000#; ","
' False - do not want a cubic spline
varSplineParams = swCurveIn.GetBCurveParams(False)
SplineDump (varSplineParams)
Part.SketchManager.Insert3DSketch (True)
Print #1, "},"
Print #1, ""
Print #1, "#End Center Line Spline"
Print #1, ""
Print #1, "#Begin Rack Spline Data"
Print #1, "'RackSpline': {"
Set swActiveSketch = Nothing
Set swSketchSegment = Nothing
Set SketchSegments = Nothing
Set swCurveIn = Nothing
Part.ClearSelection2 (False)
success = Part.Extension.SelectByID2("rackPosFull", "SKETCH", 0, 0, 0, False, 0, Nothing, 0)
Part.SketchManager.Insert3DSketch (False)
Set swActiveSketch = Part.GetActiveSketch2
Set swSketchSegment = Nothing
SketchSegments = swActiveSketch.GetSketchSegments
For Each SketchSegment In SketchSegments
Set swSketchSegment = SketchSegment
Debug.Print "in rack", swSketchSegment.GetName
Debug.Print swSketchSegment.GetType
If swSketchSegment.GetType = 3 Then
Debug.Print "This is type 3"
Exit For
End If
Next
Print #1, "'Spline Name':'"; swSketchSegment.GetName; "',"
Print #1, "'Length':"; swSketchSegment.GetLength * 1000#; ","
Set swCurveIn = swSketchSegment.GetCurve
Set varSplineParams = Nothing
varSplineParams = swCurveIn.GetBCurveParams(False)
SplineDump (varSplineParams)
Part.SketchManager.Insert3DSketch (True)
Print #1, "}"
Print #1, "#End Rack Spline Data"
Print #1, "} #End of everything"
Close #1
End Sub
</code></pre>
<p>Below is a python script that will read the output of above and give you as many points as you want along that spline.</p>
<pre><code> import scipy
from geomdl import BSpline
import math
#import easygui
from math import *
import numpy as np
import math
from functools import partial
np.seterr(divide='ignore', invalid='ignore')
filename = 'my_splines_Table.txt'
filehandle = open(filename,"r")
infile = filehandle.read()
filehandle.close()
spd=eval(infile)
CenterCurve = BSpline.Curve()
CenterCurve.rawdata = spd['CenterLineSpline']
CenterCurve.degree = spd['CenterLineSpline']['Dimension']
CenterCurve.ctrlpts = spd['CenterLineSpline']['ControlPoints']
CenterCurve.knotvector = spd['CenterLineSpline']['KnotPoints']
NumPoints = 250
for i in range(NumPoints+1): #Rember to generate the last point so you really get Numpoints + 1
position = i / 250
point = CenterCurve.evaluate_single(position)
print(" Point ",point)
</code></pre>
<p>All of this is specific to my application. You will need to modify accordingly.<br>
There in no error checking or sanity checking.<br>
One thing to note is that when the part is rebuilt, the direction of the spline data may reverse. That is the start and end swap. All the data is correct but the python code my process it from end to start. For my application I have another reference point that is near the beginning of the spline. I check the start and end of the spline data to see which end is closer to the reference point. (That is what some of the extra bits are in the VBA script.)</p>
| 26026 | Solidworks export the 3d points of a spline |
2019-02-18T14:27:44.487 | <p>One end of a small insulated rod is attached to a support which is maintained at a constant temperature of 473 K,and once the rod has reached a uniform temperature in its entire length the insulation is removed. Now the surrounding air is at 298 K with convective heat transfer coefficient h.</p>
<p>I am required to find the time elasped when the temperatures at some length of rod reaches a given temperature.The radius,thermal conductivity and thermal diffusivity are all given</p>
<p>This is a unsteady state heat conduction and convection problem.I got the following pde <span class="math-container">$$k \frac{\partial^2 T }{\partial z^2} - \frac k\alpha \frac{\partial T}{\partial t}=\frac{2h}{R} (T-T_0)$$</span></p>
<p>where <span class="math-container">$k$</span> is thermal conductivity <span class="math-container">$\alpha$</span> is thermal diffusivity and <span class="math-container">$T_0$</span> is surrounding ambient temperature.I am facing trouble finding out the two boundary conditions.The initial condition is that at <span class="math-container">$t=0$</span> the rod is at a uniform temperature of 473K.The first boundary condition is at <span class="math-container">$z=0$</span> the temperature is 473K.However i am unable to find out the second boundary condition.The tip of the fin is not insulated and is neither maintained at a constant temperature.</p>
| |mechanical-engineering|heat-transfer|chemical-engineering|finite-element-method| | <p>The BC at the far end is the flux condition.</p>
<p><span class="math-container">$$ -k \left. \frac{dT}{dx}\right|_{x=L} = h(T - T_o) $$</span></p>
<p>See Chapter 17 in the book by <a href="https://www.wiley.com/en-us/Fundamentals+of+Momentum%2C+Heat%2C+and+Mass+Transfer%2C+Revised+6th+Edition-p-9781118947463" rel="nofollow noreferrer">Welty etal</a> for examples of extended surfaces and their solutions at steady state. Finally, just for fun, here is an example temperature profile of an extended beam at steady state with the heat transfer BC. Your final answer will end at a comparable profile when <span class="math-container">$m^2 = 10$</span>.</p>
<p><a href="https://i.stack.imgur.com/mjlHZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mjlHZ.png" alt="temperature profile in extend beam"></a></p>
| 26034 | Boundary conditions in a heat transfer problem |
2019-02-20T00:13:14.293 | <p><img src="https://i.stack.imgur.com/4OhNj.jpg" alt="The truss"></p>
<p>How do I calculate the reactions in this truss?I would usually cut in the joint where beams 5,6,7 and 8 are and calculate reactions in the two trusses that are created.How do I do that when there is a force acting on the joint?</p>
| |civil-engineering| | <p>You need to realize that this isn't one large truss. In fact, it's three separate trusses, each supporting the other.</p>
<p>Here's your structure with some noteworthy points added and the three "sub-trusses" shown:</p>
<p><a href="https://i.stack.imgur.com/gPJZz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gPJZz.png" alt="enter image description here"></a></p>
<p>Start by looking at the left-most truss which ends at node B. If you deleted the vertical member at B, that truss would no longer be stable. This means this truss is basically supported by the rest of the structure at A.</p>
<p>Which means we can treat this truss as having a vertical support at B and solving it independently from the rest of the structure.</p>
<p>This is trivial, and we get that <span class="math-container">$R_A = R_B = 50\text{ kN}$</span>.</p>
<p>Now, when looking at the rest of the structure, we can pretend that left-most truss doesn't exist and replace it with a concentrated downwards force of 50 kN at B.</p>
<p><a href="https://i.stack.imgur.com/iy8QU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iy8QU.png" alt="enter image description here"></a></p>
<p>However, looking at the central truss (which ends at node D), we can see the same thing applies: if it weren't for member 8 (the right-most diagonal), this truss would also be unstable. Therefore, we can once again consider node D to be a vertical support for this truss and solve for the reactions.</p>
<p>Now, you can choose whether to include the load applied at node D on this truss or on the next one. Since the load would be applied directly on the fictional support at D, that support will absorb the entirety of that load, which will then be applied when calculating the last truss. Or you can choose not to consider the load now, and then add the reaction you find for D to that load when calculating the last truss. It's exactly the same.</p>
<p>This one is still pretty straightforward though, and you get <span class="math-container">$R_C = 225\text{ kN}$</span> and <span class="math-container">$R_D = 25\text{ kN}$</span> (assuming you added the applied load now, otherwise it'd be -75 kN).</p>
<p>And then you move onto the last truss, replacing everything else with a downwards concentrated load of 25 kN at D (or with an upwards 75 kN but then adding in the applied load at the same location).</p>
<p><a href="https://i.stack.imgur.com/4t8Jq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4t8Jq.png" alt="enter image description here"></a></p>
<p>Unlike the other trusses, this one is self-standing, so you can just solve it as-is. This is also simple, and you get <span class="math-container">$R_E = 87.5\text{ kN}$</span> and <span class="math-container">$R_F = 37.5\text{ kN}$</span>.</p>
<hr>
<p>In image form, this is what we did:</p>
<p><a href="https://i.stack.imgur.com/TfPiU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TfPiU.png" alt="enter image description here"></a></p>
| 26055 | How do I calculate the reactions in this truss? |
2019-02-21T07:24:38.443 | <p>In CFD we focus on the NS equations as the governing equations (together with other equations) and solved them using Some numerical method (usually FVM). What are the governing equations (counter parts of the NS equation) in solid mechanics?. </p>
| |finite-element-method| | <p><strong>Momentum equation</strong></p>
<p>The Navier-Stokes equations represent the equations for the conservation of linear momentum. In convective form they are written as
<span class="math-container">$$
\rho\frac{D\mathbf{v}}{Dt} = - \nabla p + \nabla \cdot \boldsymbol \tau + \rho\,\mathbf{g}
$$</span>
where <span class="math-container">$\mathbf{v}$</span> is the velocity and the stress is <span class="math-container">$\boldsymbol{\sigma} = -p\,\mathbf{I} + \boldsymbol{\tau}$</span>.</p>
<p>The linear momentum equation in solids is identical
<span class="math-container">$$
\rho~\frac{D\mathbf{v}}{Dt} = \boldsymbol{\nabla} \cdot \boldsymbol{\sigma} +\rho~\mathbf{b}
$$</span></p>
<p><strong>Constitutive relation</strong></p>
<p>For compressible Newtonian fluids, the relationship between stress and velocity is
<span class="math-container">$$
\boldsymbol \sigma = \lambda (\nabla\cdot\mathbf{v}) \mathbf I + 2 \mu \dot{\boldsymbol \varepsilon}
$$</span>
where <span class="math-container">$\lambda$</span> is the bulk viscosity, <span class="math-container">$\mu$</span> is the dynamic viscosity, and
<span class="math-container">$$
\dot{\boldsymbol \varepsilon} = \tfrac{1}{2} \left(\nabla\mathbf{v} + ( \nabla\mathbf{v})^\mathrm{T}\right)
$$</span></p>
<p>For linear elastic solids, the stress-strain relation is
<span class="math-container">$$
\boldsymbol \sigma = \lambda (\nabla\cdot\mathbf{u}) \mathbf I + 2 \mu \boldsymbol \varepsilon
$$</span>
where <span class="math-container">$\lambda$</span> and <span class="math-container">$\mu$</span> are Lame parameters, <span class="math-container">$\mathbf{u}$</span> is the displacement, and
<span class="math-container">$$
\boldsymbol \varepsilon = \tfrac{1}{2} \left(\nabla\mathbf{u} + ( \nabla\mathbf{u})^\mathrm{T}\right)
$$</span></p>
<p><strong>Eulerian vs Lagrangian</strong></p>
<p>Because it typically does not make sense to solve CFD equations from the point of view of individual particles, an Eulerian FVM approach is often used. In contrast, Lagrangian methods are essential for solids because we typically need to know where each point in a structure moves to under the action of loads.</p>
| 26072 | What are the governing equations solved in FEA for structural mechanics? |
2019-02-21T16:50:37.047 | <p>I apologize in advance if this is off topic - maybe it is more of an English language query... Please feel free to migrate or close if needed :)</p>
<p>I have seen mine cages/service lifts referred to as Mary-Ann (or Maryann/Mary Anne etc.) for example <a href="https://www.sudbury.com/letters-to-the-editor/mine-cage-technology-flawed-jodi-blasutti-243919" rel="nofollow noreferrer">here</a>, <a href="http://www.canadianminingjournal.com/features/red-lake-just-hitting-its-stride/" rel="nofollow noreferrer">here</a> and <a href="http://www.meglab.ca/en/portfolio/instrumentation-and-control/automation-of-the-mine-cage-maryann-underground-mine-canada-s-mid-north-2013/" rel="nofollow noreferrer">here</a>. Does anyone know where this comes from? And is it a common term - I have heard it used in South Africa and the sources linked are Canadian, which suggests this is the case. </p>
| |mining-engineering| | <p>My understanding is that the Mary Anne (in the sense of an escape hoist) originated in Australia. A convicted horse thief was able to escape his island prison when his girl friend, Mary Anne, swam to the island towing a jury-rigged life saver. Her lover, who could not swim, donned the jacket and Mary Anne virtually towed him back to shore through shark-infested waters. Unfortunately he then continued his nefarious profession until he died in a shoot-out with a posse of "rangers."
This tale is authenticated and well detailed in several independent posts on the internet.</p>
| 26077 | Why is a mining cage called a Mary-Ann? |
2019-02-23T16:19:12.617 | <p>I'm having trouble welding the inner corners of my material. Welding material is sometimes deposited on either one side or both sides, but never really connecting the two parts.</p>
<p>Side/flat welds work fine: </p>
<p><img src="https://i.stack.imgur.com/C8x6g.jpg" alt="photo"></p>
<p>But those pesky inner corners don't: </p>
<p><img src="https://i.stack.imgur.com/du8MX.jpg" alt="photo"></p>
<p>The material I'm using is 1mm thick 16x16mm square steel tubing.
Welding settings are 20 Amps DC electrode positive and welding electrode is 1.6mm steel.</p>
<p>Am I doing something wrong? Is there a trick to this I'm missing?</p>
| |welding| | <p>It's all about angles, try a main weld at around 45 degrees down the corner then have a weld on the bottom piece of metal that half laps over the original weld, follow that up with a weld on the top piece of metal which also half laps the original weld this one should be welded at around 30 degrees. I'm not the best welder but that's how I was taught and if you can work with angles then you should be fine, also don't go too fast or too slow, and try notice patterns like almost draw ticks or C's in your weld. EDIT: also tie what your welding together, for example weld ties at each corner and then it's just a case of filling in the gap.</p>
| 26097 | Stick welding inner corner |
2019-02-23T20:47:14.787 | <p>Just began studying automata. I've already seen several examples of the applications of finite automata: pattern matching, automatic doors, etc. And so far, I understand the logical connection between those mechanisms and state diagrams and formal definitions. But I'm curious, what <em>thing</em> is implementing time steps—if that's what they're called—allowing states to transition. I know a tiny bit about computer architecture so I'm thinking it's like a processor clock, you know, something just ticking on and off very rapidly. Then the automata takes input on every "tick" of said clock. But with different devices using different current types (AC/DC) how is this actually being accomplished in an arbitrary physical device? </p>
<p>Now that I'm writing this out, it seems clear there are multiple ways to accomplish that ticking depending on what we're talking about, but I'm just looking for the most common examples found in real-life. For example what is allowing the sequence of input signals to occur in an automatic door?</p>
| |electrical-engineering|computer-engineering|consumer-electronics| | <p>A synchronous system with a clock is not necessary. The transition between states can be triggered asynchronously by inputs to the system (e.g. someone pushes a button to operate an automatic door). The finite state machine needs to have some logic elements to "remember" of what state it is currently in, but that does not have to be anything like the addressable memory used in digital computer system. It could simply be a set of mechanical switches and electrical relays. </p>
<p>For example a mechanical combination lock is an example of an asynchronous finite state machine where the "memory" is just the physical positions of the different parts of the lock.</p>
<p>To interface the real world inputs to a synchronous control system such as a digital computer, there are two basic techniques: either the computer program regularly <em>polls</em> the input devices to check what state they are in, or the input devices generate <em>interrupts</em> which force the program to jump to a different point and continue executing from there. Textbooks on computer system design explain both methods in detail.</p>
| 26101 | What is the mechanism of transition in applications of finite state automata? |
2019-02-24T17:49:52.810 | <p>I want to use a round-over bit on a CNC router after cutting a profile with a 3mm (1/8") bit. So the round-over bit needs a shank thinner than that. I could only find shanks with 6mm (1/4") diameter.</p>
<p>Do round-over bits with shanks thinner than 3mm (1/8") exist? If not, what is the physical limitation that prevents it?</p>
| |cnc| | <p>I found this CNC bit that seems to do the job:</p>
<blockquote>
<p>1pc 6mm Shank Classical Round Nose Point Cut Wood Router Bit Tungsten Cobalt Alloy 2 Flute Wood Milling Cutters Woodworking Tool</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/e8cP5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e8cP5.jpg" alt="CNC bit" /></a></p>
<p>I have not tested it because I don't have access to the CNC during the pandemic.</p>
| 26125 | CNC round-over bits with thin shank |
2019-02-25T14:45:47.863 | <p>If we wanted to open gate with maximum force that man can apply at distance of 1m from pivoted joint.
So I want torque to determine motor and gearbox that can be used to automate gate.</p>
| |torque| | <p>The maximum push force for an adult male, is 818N according to NASA - <a href="https://msis.jsc.nasa.gov/sections/section04.htm#Figure%204.9.3-6" rel="nofollow noreferrer">https://msis.jsc.nasa.gov/sections/section04.htm#Figure%204.9.3-6</a></p>
<p>Applying this at a distance of 1m from the pivot, corresponds to a torque of:</p>
<p><span class="math-container">$$T=F*d=818\text{ N} * 1\text{ m}=818\text{ Nm}$$</span></p>
<p>This is likely much more torque than is required to automate the gate, however - your calculations should be based on the mass and shape of the gate, and any speed requirements for opening time and/or acceleration, rather than on the force of a human.</p>
| 26138 | How much torque is needed? |
2019-02-25T15:08:56.310 | <p>I am trying to understand what is the difference between the wave equation ad the advection equation? Bot the equation seems to move a quantity. While in literature it is said that the advection equation is the simpler of the two I don't see how the wave equation is that much different than the advection equation except moving the quantity in both directions when talking in terms of 2d.</p>
<p>Ref :</p>
<ul>
<li><a href="https://www.rose-hulman.edu/~bryan/lottamath/wave.pdf" rel="nofollow noreferrer">https://www.rose-hulman.edu/~bryan/lottamath/wave.pdf</a> </li>
<li><a href="http://farside.ph.utexas.edu/teaching/329/lectures/node90.html" rel="nofollow noreferrer">http://farside.ph.utexas.edu/teaching/329/lectures/node90.html</a></li>
</ul>
| |convection|waves| | <p>As alephzero mentioned the difference is huge. </p>
<p>I will explain mathematically and then physically. </p>
<p>Mathematically, </p>
<p>For any given partial differential equation of the form, <span class="math-container">$$A \frac {\partial^2\phi}{ \partial x^2} + B \frac {\partial^2\phi}{ \partial x \partial y} + C \frac {\partial^2\phi}{ \partial y^2} + \rm first\ derivative \ terms\ =0 $$</span>
We get,</p>
<p><span class="math-container">$$\rm B^2 - 4AC = 0 , - Parabolic, $$</span></p>
<p><span class="math-container">$$\rm B^2 - 4AC < 0 , - Elliptic $$</span></p>
<p><span class="math-container">$$\rm B^2 - 4AC > 0 , - Hyperbolic. $$</span></p>
<p>Based on the above conditions the wave equation is a hyperbolic equation and the diffusion equation is a parabolic equation.</p>
<p>These conical conditions decides the zone of influence and zone of dependance in your domain of interest. i.e., To be get disturbed at any point of interest in the domain, you should know where and when you should pinch. </p>
<p>Physically, </p>
<ol>
<li><p>Consider a 2D square plane, apply heat on the west side edge, then
the heat will flow from west to east. if you are running from north
to south you will feel the same temerature. <span class="math-container">$->$</span> This process is
goverend by unsteady heat diffusion equation mathematically
parabolic and zone of dependance is all east side to where apply
heat.</p></li>
<li><p>Throw a stone in the water pool which is still. after some time you will find, the wave reach all the corners of the pool. So you disturb anywhere to get disturbed anywhere. These waves are goverend by Laplace equation which is elliptic. </p></li>
<li><p>Now consider a moving channel, throw a stone, then see the waves now. Do the disturbances reach upstream of the channel? Nope. The disturbances created by the stone are carried downstream by the moving stream. This is in the 2D sense. Imagine same thing for 1D also. The created disturbances are carried by the material elasticity (noted by the wave/sound speed). when and where (the disturbed region) is decided by the wave speed. These inrteresting problems are governed by hyperbolic equations. So disturb only certain location to get disturbed somewhere.</p></li>
</ol>
<p>You can google more to get pictorial explanations for these. </p>
<p>Hope this answer helps!</p>
| 26139 | What is the difference between the wave equation and the advection equation? |
2019-02-25T15:25:26.203 | <p>Got a Tech Drawing today with measurements and all that. I understand everything except for one part where they've specified a bolt with M4 P=0.35. The M part I get but what does the P specify?</p>
| |bolting|technical-drawing| | <p>I would go with Pitch of 0.35mm ie 1 turn advances by 0.35mm</p>
<p>A view of the drawing may clarify this.</p>
| 26140 | What does P0.35 stand for in a Tech drawing for a Bolt |
2019-02-27T04:50:36.950 | <p>According to uniaxial stress strain curve, specimen fails at a certain stress value since necking begins. </p>
<p>My question is that, same stress value appears before necking especially around yield point however specimen keeps its integrity and continue up to uts point and then rupture. Why ?</p>
| |mechanical-engineering|structural-engineering|materials|solid-mechanics| | <p>It just seems the way you mention because most of the books use the so called engineering graph. Which assumes the specimen surface area as constant throughout the loading which as we know is not right. Under tension the specimen extends axially and becomes narrower by the Poison's contraction.</p>
<p>So if we just plot stress versus strain curve without modifying the stress due to reduction in cross section of the specimen we get the known curve with straight elastic part and the large extension under higher stress due to hardening and the brittle rupture. </p>
<p>Because at the first point on the graph stress level the specimen is still ductile and has flexibility to accept even a little bit higher stress and keep stretching while it becomes harder and more brittle.</p>
<p>So after a large expansion it has lost almost all its ductility and it starts a sort of free fall elongation without increasing the stress and breaks at that point.
In reality there is never negative slope on the curve and it keeps rising albeit with different shapes at distinct stress levels.
The Wikipedia graph shows this, the blue curve,B, is the real and thered,A, is the engineering curve..<a href="https://i.stack.imgur.com/AVhKj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AVhKj.jpg" alt="enter image description here"></a></p>
<p>They explain this clearly here.<a href="https://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve" rel="nofollow noreferrer">Wikipedia Stress/strain</a></p>
| 26162 | Stress strain curve |
2019-02-28T00:12:58.860 | <p>Is it common to take equal values for the proportional and derivative gains of a PID controller? if so what does it mean?</p>
| |pid-control| | <p>The transfer function of a PID controller is given by</p>
<p><span class="math-container">$$
C(s) = K_p + \frac{K_i}{s} K_d\,s = \frac{K_d\,s^2 + K_p\,s + K_i}{s}.
$$</span></p>
<p>When <span class="math-container">$K=K_p=K_d$</span> and <span class="math-container">$K_i$</span> is relatively small compared to <span class="math-container">$K$</span>, so <span class="math-container">$K_i=\varepsilon\,K$</span> with <span class="math-container">$|\varepsilon|<1$</span> this can also be written as</p>
<p><span class="math-container">$$
C(s) = K\frac{s^2 + s + \varepsilon}{s},
$$</span></p>
<p>which has a zero close to <span class="math-container">$s=-1$</span>. This means that <span class="math-container">$C(s)$</span> has an asymptote of slope zero below a frequency of one rad/s (0.16 Hertz) and a slope of plus one above that frequency. With slope I am referring how many times the magnitude in the Bode of <span class="math-container">$C(s)$</span> increases by 20 dB per decade. Usually this transition frequency is placed close to or below the bandwidth of your closed loop system (assuming your plant acts like a second order system at that frequency range).</p>
<p>So the assumption that <span class="math-container">$K=K_p=K_d$</span> might imply that the closed loop bandwidth is roughly one rad/s or higher.</p>
| 26182 | Equal proportional and derivative gains of PID |
2019-02-28T13:10:45.570 | <p>I am releasing CO2 from a cylinder into a ventilated room and want to mimic a human breathing (let's ignore the in/out breath). Problem is that the gas that comes out is at room temperature (or lower due to decompression) not body temperature. How can I heat the gas that comes out of the cylinder to 37ºC?</p>
| |mechanical-engineering|fluid-mechanics| | <p>take a cheap hair dryer apart so you can control the heater element separate from the fan blower. Connect the heater element to 120VAC through a variac (variable AC transformer) so you can control the voltage it sees. Put a temperature sensor in the outlet stream of the hair dryer and build a duct that carries CO2 from the tank to the inlet of the hair dryer. turn on the CO2 so it blows through the hair dryer with the dryer's fan off. adjust the variac to get the desired discharge temperature. </p>
| 26193 | How to heat CO2 coming out of a gas canister to human breath temperature? |
2019-03-01T15:19:50.127 | <p>I'm new to plasma cutting with a hand-held plasma cutter (as opposed to CNC). Its straightforward to cut with a fence/guide and I am getting good results that way. </p>
<p>However, when I want to freehand a rough shape without setting up a complicated jig, I can't figure out a way to see a line I want to cut. I wear a welding mask on a low setting for eye safety and once it dims I can't see any lines I've drawn on the part. </p>
<p>Does anyone have a good way to safely see a cut line when plasma cutting with a dimming mask?</p>
| |machining| | <p>I use the liquid correction pens, the sort of thing used for whiting out marks on paper, Tippex etc. <a href="https://uniball.in/shop/img/p/4/3/1/431-thickbox_default.jpg" rel="nofollow noreferrer">https://uniball.in/shop/img/p/4/3/1/431-thickbox_default.jpg</a></p>
<p>They produce a reasonably thin line about the same as the kerf of a plasma cutter and work pretty well on most metals also because they are quite bright white the marks are very visible when cutting with the sort of shade of eye protection you are likely to be using for plasma cutting. </p>
<p>Another option is lightly scoring the surface with a thin slitting disk in an angle grinder which can work well for straight lines and shallow curves and has the advantage of not being burned off by the cutting process so you keep a good reference for any final grinding to shape you might need to do. </p>
| 26210 | Visible markings for plasma cutting? |
2019-03-03T00:09:53.427 | <p>Which is more reliable. For example, in setting steel posts in the ground, the pole should go under ground 1/3 of its height from the surface up. In IBC, steel post depth is made rather on a storey foundation, calling for 18" depth for 1 storey/floor, and 24" depth for 2 storey/floors. Which should be followed?</p>
| |concrete|foundations| | <p>You don't get to pick what code or standard to use for a project, the local governing authority has established a code and all applicants for a building permit have to follow it, even state when it has to construct roads and bridges, or other public projects has to follow these codes.</p>
<p>Also the examples you used don't sound correct. In the case of steel columns, there are specified load combination factors(such as LFRD) and design criteria, building types, fire protection, the utility of the column as axial or lateral load bearing member, etc. But in all the cases the columns are supported by a concrete foundation through a base plate and anchor bolts all designed according to governing codes.</p>
<p>For example, the state of California has its building code. The city of Los Angeles has adopted this code with some modifications. </p>
<p>Building industry institutes, such as AISC, American Steel Institute, or ACI work closely with current codes and upgrade to comply with the latest codes. </p>
| 26222 | International Building Code vs Industry Standards |
2019-03-03T13:47:36.740 | <p>My understanding of a typical jet engine is that when running, the turbine extracts energy from the hot exhaust gases. This drives the turbine, which is coupled to the compressor at the front of the engine. As a result, when the turbine accelerates, the compressor accelerates. This increases the mass flow rate of air into the engine, which in turn increases the mass flow rate out of the engine, producing more thrust. It also means more energy is extracted by the turbine, which drives the compressor even faster.</p>
<p>It seems as though it is a cycle that should continue to increase without limit. Therefore, my question is what stops the engine spin speed from increasing to infinity (ignoring material limits)? Why does the spin speed of the engine stabilise if the compressor and the turbine are continually driving each other at increasing rates?</p>
| |aerospace-engineering|turbines|compressors| | <p>This an oversimplification, but it should get the idea across</p>
<p>How does the turbine extract energy from the air? The turbine stator (nozzle) turns the air from axial to tangential (converting internal energy to kinetic energy), and then the turbine rotor removes the tangential velocity from the air and converts it into rotational energy of the rotor. So, the amount of energy extracted by the rotor depends on the relative tangential velocity between the air and the turbine. Put another way, <strong>as the turbine speeds up, it extracts less work from the air.</strong></p>
<p>The exact opposite is true for a compressor. The compressor rotor add tangential velocity to the air, and then the compressor stator turns the air from tangential back to axial. So <strong>as the compressor speeds up, it is doing more work on the air.</strong></p>
<p>So start at a low speed, and then dump a bunch of fuel in the combustion. Now the turbine produces a lot of torque, and the compressor is not consuming much. The excess torque accelerates the rotor. Now at the high speed, the compressor is consuming more torque, but the turbine is producing less. There's still some excess torque, but not as much. The rotor keeps accelerating. As it goes the compressor is consuming more and more and more torque, and the turbine is producing less and less and less, until some point it balances out. The turbine is producing exactly as much as the compressor is consuming. No excess torque. Therefore it stops accelerating.</p>
<p>Now what this simplified description has assumed is that the amount of fuel burned (i.e. amount of heat added) is constant with speed. That's not really true. But again this should get you the idea. </p>
| 26224 | If the turbine drives the compressor, then what limits the engine speed? |
2019-03-03T19:49:17.497 | <p>I'm trying to get an answer for something potentially quite simple, but I can't visualize it (nor do I have the parts yet to build it). It's for a little hobby project, but I'm slightly concerned I have made an assumption about what will happen, which may not be correct. I've drawn the problem below. </p>
<p><a href="https://i.stack.imgur.com/OA3y7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OA3y7.jpg" alt="enter image description here"></a></p>
<p>In essence you have two pulleys connected by a timing belt.</p>
<pre><code>Pulley A = 80 Teeth
Pulley B = 120 Teeth
</code></pre>
<p>Pulley A is static, it does not rotate at all. Pulley B is free to spin around its own axis. It also then rotates around the axis of A, keeping the timing belt taught. </p>
<p>In one revolution of pulley B around the axis of A (dotted line), how many times would pulley B have rotated?</p>
<p>My assumption would be that it would be 2/3 rotations of pulley B for every time it does an entire revolution around pulley A.</p>
| |pulleys| | <p>yes, you are right, but only partially. </p>
<p>In one clockwise revolution, the belt is wound around, pulled, by A at a rate of 80 teeth. So pully B will have turned</p>
<p><span class="math-container">$80/120 =\frac{2}{3}\quad$</span> counterclockwise.</p>
<p>But we have to add to this one clockwise rotation that we initiated.</p>
<p>So the total B rotation is 1-2/3 = 1/3 clockwise. </p>
| 26229 | Trying to visualize the resulting interaction between two pulleys |
2019-03-05T17:55:28.843 | <p>I am given these sets of equations for a gantry crane:</p>
<p>1) <span class="math-container">$M\frac{d^2x}{dt^2}+D\frac{dx}{dt}-mg\theta =u$</span></p>
<p>2) <span class="math-container">$ml\frac{d^2\theta }{dt^2}+mg\theta +m\frac{d^2x\:}{dt^2}=0$</span></p>
<p>3) <span class="math-container">$y=x+l\theta $</span></p>
<p>Where <span class="math-container">$x$</span> is the position of the trolley; <span class="math-container">$M$</span> is the mass of the trolley; <span class="math-container">$m$</span> is the mass of the payload; <span class="math-container">$\theta$</span> is the angular position of the payload; <span class="math-container">$D$</span> is the damping coefficient; <span class="math-container">$y$</span> is the payload position; <span class="math-container">$u$</span> is the force applied to the trolley; <span class="math-container">$l$</span> is the cable length.</p>
<p><a href="https://i.stack.imgur.com/8fZt3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8fZt3.png" alt="Gantry Crane"></a></p>
<p>Based on the equations above, I have found the transfer function of this system to be <span class="math-container">$G\left(s\right)=\frac{y\left(s\right)}{u\left(s\right)}=\frac{s^2\left(l-1\right)+g}{Mls^4+Dls^3+\left(M+m\right)gs^2+Dgs}$</span> </p>
<p>Now, I have to find the system response if <span class="math-container">$u$</span> is zero and the payload is shifted by an angle of <span class="math-container">$10^{\circ }$</span>. </p>
<p>What this means is that we must now consider the initial condition of a <span class="math-container">$10^{\circ }$</span> angle when we do the Laplace transform; however I can't figure out how to incorporate the initial condition in my equations. </p>
<p>Also, if <span class="math-container">$u$</span> is zero, how is it mathematically possible to find <span class="math-container">$y(s)$</span> to be not zero, given that <span class="math-container">$y(s)=u(s)G(s)$</span>?</p>
| |mechanical-engineering|electrical-engineering| | <p>When deriving the transfer function it is usually implicitly assumed that the initial conditions are zero. However the <a href="https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems" rel="nofollow noreferrer">Laplace transform</a> of the derivative of a variable with respect to time with initial conditions is defined by</p>
<p><span class="math-container">$$
\mathcal{L}\left\{\frac{d^n y}{dt^n}\right\}(s) = s\,\mathcal{L}\left\{\frac{d^{n-1} y}{dt^{n-1}}\right\}(s) - \left.\frac{d^{n-1} y}{dt^{n-1}}\right|_{t=0^+}.
$$</span></p>
<p>So for example <span class="math-container">$\mathcal{L}\{\tfrac{d}{dt}y\}(s)=s\,\mathcal{L}\{y\}(s)-y(0^+)$</span> and <span class="math-container">$\mathcal{L}\{\tfrac{d^2}{dt^2}y\}(s) = s\,\mathcal{L}\{\tfrac{d}{dt}y\}(s) - \tfrac{d}{dt}y(0^+) = s^2\,\mathcal{L}\{y\}(s) - s\,y(0^+) - \tfrac{d}{dt}y(0^+)$</span>.</p>
<p>The unforced system response of the system with initial conditions can be calculated by taking the inverse Laplace transform of the resulting expression. However it might be easier to not switch to and from the frequency domain and just solve it directly in the time domain.</p>
| 26254 | Find the system response given the initial condition |
2019-03-06T20:55:57.440 | <p>I'm trying to build a cablecam, a picture that serves only for illustration is:
<a href="https://i.stack.imgur.com/9NMGa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9NMGa.jpg" alt="enter image description here"></a></p>
<p>The DC motor is</p>
<ul>
<li>48V</li>
<li>5000W</li>
<li>6000rpm</li>
<li>6000 oz-in stall torque</li>
<li>85% effiencity</li>
<li>0.066 Ohms Resistance</li>
<li>135Kv (RPM / Volt)</li>
<li>Reduction ratio with timing belt is 1:2, so speed on 60mm output pulley is 3000rpm.</li>
</ul>
<p>Battery is Li-Ion</p>
<ul>
<li>48V</li>
<li>15Ah</li>
<li>70A constant/120A peak for 20sec.</li>
</ul>
<p>Mechanical properties </p>
<ul>
<li>Max load 40 kg (including motor, wires, frame, electronics, pulleys...) </li>
<li>Friction coefficient is 0.7</li>
</ul>
<p>I want to know if this 48V DC motor can even start this mass on the rope without burn itself. I would like to know if it could reach speed of 50km/h within 10 seconds or 50 meters of travel on rope.</p>
| |motors|torque| | <p>I think it goes something like this - the energy from the motor is just it's rated power <span class="math-container">$P$</span> <span class="math-container">$\times$</span> time <span class="math-container">$t$</span>, which produces kinetic energy (mass <span class="math-container">$m$</span>, velocity <span class="math-container">$v$</span>) in the system:
<span class="math-container">$$P t = \frac{1}{2} m v^2$$</span> solve for <span class="math-container">$t$</span>, knowing <span class="math-container">$P$</span> (5000 W), <span class="math-container">$m$</span> (40 kg) and final <span class="math-container">$v$</span> (50 km/h) results in the time to reach 50 km/h is about 0.8 s. </p>
<p>It's a simplification so I'm disregarding</p>
<ul>
<li>slipping of the drive wheel on the rope - likely to be some initial slippage with a constant power input because the initial acceleration is infinite.</li>
<li>additional mass of the motor raising the total accelerated mass over OP's 40 kg </li>
<li>any motor/battery/efficiency related power limitations</li>
<li>might need a different gearing ratio over what OP has specified (3000 rpm on 60 mm diameter pulley is 33 km/h)</li>
<li>friction in pulleys, inertia of pulleys</li>
</ul>
<p>The frictional between the drive pulley and the rope isn't just related to the mass of the cart, it's also related to the tension in the rope. </p>
| 26270 | Calculating proper specs of DC motor for rig driven on the rope |
2019-03-07T16:20:45.583 | <p>The device have to climb a pipe vertically. I have a DC motor with power rating 21.2 w, output speed 13360 rpm and maximum output torque 154.4 gcm.
My device is 1 kg and I have wheel with 0.012 m radius. I did calculations and I found that my gear box should be 1:72 but I am not sure about that. I need the device to go upwards as fast as possible. With gear ratio 1:72 is enough for the DC motor to lift the device? And what I have to do to make it go faster? Make the gear ratio smaller or bigger?</p>
| |motors|gears|torque|power| | <p>lifting a 1kg weight requires 9.8 WATT, we need to add to that the power needed to accelerate your mass from zero to the final speed.</p>
<p>So we assume an acceleration of.</p>
<p>a =1.8m/s </p>
<p>The power needed <span class="math-container">$ \ P= 9.8 + 1.8*1kg =11.8N.s =11.8W$</span></p>
<p>Say your system efficiency is 80% then you need 11.8W* 1.25 = 14.75 W which is less than your DC motor output.</p>
<p>As for the gear.</p>
<p>you need a torque of <span class="math-container">$14.75* (1.2cm/2*100cm/m) = 0.0885Nm \ < $</span></p>
<p>your DC motor torque of <span class="math-container">$0.1513Nm = 154.4*9.8/100cm/m*1000g/kg$</span></p>
<p>This means no gear needed, your motor with the 1.2cm wheel directly attached to its shaft will lift the load.</p>
<p>As for the question of benefit of increasing gear ratio, you need to have the motor's torque and performance ratio diagrams, but adding gear will add to friction.</p>
| 26280 | The torque I need to lift the device and correct gear ratio for the device to be as fast as possible |
2019-03-07T21:32:59.593 | <p>My middle schooler is building a catapult that’s supposed to hurl a ping-pong ball 5 meters:</p>
<p><a href="https://i.stack.imgur.com/t6CiT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t6CiT.jpg" alt="enter image description here"></a></p>
<p>The problem is that the pressure on the latch is so high that the kid just cannot trigger it:</p>
<p><a href="https://i.stack.imgur.com/4KiQH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4KiQH.jpg" alt="enter image description here"></a></p>
<p>The door latch used as a trigger is being under so much pressure that it just wouldn’t slide out. I doubt that adding lubricant would help enough.</p>
<p>Thus the question: what latch design could you suggest so that it would operate smoothly under lateral pressure?</p>
| |pressure|mechanisms|mechanical| | <p>My son made a model trebuchet which is a sort of medieval catapult. The trigger mechanism is quite simple using three screw eyes and a pin. In case the picture below isn't clear, the there are two screw eyes attached to the frame and a single screw eye attached to the arm with a pin through all three screw eyes. Attaching a string to the pin keeps hands away from things when launching.</p>
<p><a href="https://i.stack.imgur.com/djuQp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/djuQp.jpg" alt="enter image description here"></a></p>
| 26286 | A latch that’s easy to open under lateral pressure |
2019-03-08T19:58:13.017 | <p>I have been trying to calculate the analytical solution for the following problem but have not succeeded to reach an exact solution. Can anyone propose a way?</p>
<p>A concrete beam is to be lifted by two cranes. The beam is 20 m long and weights 900 kN. The cranes have lifting capacities of 500 kN and 400 kN. What are distances <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$c$</span> in the scheme below so the cranes are not overloaded? (position of cranes are marked with | |)</p>
<pre><code> <---- a----> <----- b -------> <--- c --->
=========================================== <-- the beam
|| ||
A = 500 kN B = 400 kN
</code></pre>
<ul>
<li>A & B: reaction forces of cranes</li>
<li>distributed self weight q = 900 kN / 20 m = 45 kN/m</li>
<li><span class="math-container">$a + b + c = L = 20\text{ m}$</span></li>
</ul>
<p>The beam is statically determinate.</p>
<p>My attempted solutions:</p>
<p>I started from the equilibrium equations of the sum of bending moments about the left and right ends which are equal to zero:</p>
<p><span class="math-container">$$\begin{align}
\sum M_{left} &= 0 = Aa + B(a + b) = \dfrac{qL^2}{2} \\
\sum M_{right} &= 0 = Bc + A(b + c) = \dfrac{qL^2}{2}
\end{align}$$</span></p>
<p>A third equation is needed; I have tried <span class="math-container">$a + b + c = L$</span>, but I arrive at the "apple = apple" situation, which implies that the three equations are not independent.</p>
<p>I also tried to add an additional constrained by putting any of the <span class="math-container">$a$</span>, <span class="math-container">$b$</span> or <span class="math-container">$c$</span> values equal to a constant value. The issue then is that each of the equilibrium equations become dependent on only one of the two remaining unknowns, as in the following case, where I have put <span class="math-container">$b = L/3$</span>:</p>
<p><span class="math-container">$$\begin{align}
\sum M_{left} &= 0 = \dfrac{qL^2}{2} = Aa + B(a + L/3) \\
\sum M_{right} &= 0 = \dfrac{qL^2}{2} = Bc + A(c + L/3)
\end{align}$$</span></p>
<p>And thus, each solved unknown will not satisfy the other equilibrium equation.</p>
<p>I have run out of tricks in my sleeve ...</p>
| |structural-engineering| | <p>Let's call the distance from the anchor to 500 kN crane from the middle of the beam <span class="math-container">$D_1$</span> and the distance to 400 kN, <span class="math-container">$D_2$</span>.</p>
<p>To do exact analysis we get the moment of cranes' hook forces about the midpoint of the beam at equilibrium configuration.</p>
<p><span class="math-container">$$P_1 D_1= P_2 D_2$$</span></p>
<p>denoting <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span> as the tributary weight of the beam on 500 kN crane and 400 kN, respectively.</p>
<p>Substituting <span class="math-container">$P_2$</span> with <span class="math-container">$4/5 P_1$</span> in above equation we get</p>
<p><span class="math-container">$$P_1 D_1 = 4/5 P_1 D_2$$</span></p>
<p>Dividing both sides by <span class="math-container">$P_1$</span>:</p>
<p><span class="math-container">$$D_1= \dfrac{4}{5}D_2$$</span></p>
<p>One of the safest ways to lift the beam is to attach the 400 kN crane to one end of concrete beam and 500 kN to a point at the other end 90% the length of the beam.</p>
| 26303 | Exact anlytical solution for beam-lifting problem constrained by crane capacity |
2019-03-09T01:52:41.777 | <p>I'm pressing glass and I want to know if the pressure that the walls of my mold feel is the same through all the internal surfaces. I'm willing to assume that glass will behave purely as a fluid (very viscous) in the temperature range I'm working it.</p>
| |fluid-mechanics|pressure|glass| | <p>Yes, in an enclosed container, with fluid at rest, the pressure is equal on all the walls and at any point inside the container. </p>
<p>The fluid at rest indicates the Kinetic energy of the fluid,</p>
<p><span class="math-container">$ \ E =\frac{1}{2}\dot{m}v^2 $</span> </p>
<p>is insignificant, there is no rapid flow of matter.</p>
<p>It is Pascal’s law.</p>
<p>This law is used in hydraulic presses as well.</p>
| 26310 | Will the pressure inside a pressing mold will be constant if I'm working with a fluid inside? |
2019-03-09T13:25:54.950 | <p>The 40 kg weight accelerates and runs on a rope from 0 to 50 km/h in 10 seconds. </p>
<p>I want to calculate the minimum required rolling friction that I need (acceleration without slipping) to achieve acceleration in a time of 10 seconds from 0-50km / h. </p>
<p>Friction adjustment (later for different speeds and accelerations) I would change by increasing /decreasing the angle between the drive pulley and the pulley 2 as shown in the figure. Currently, the angle is 15 '.</p>
<p>My question is:
How to calculate the minimum required rolling friction (or better say, minimum required angle which is 15' now) between the drive pulley and the rope of 40kg of weight moving along the rope, as shown in the drawing?</p>
<p><a href="https://i.stack.imgur.com/XC7fz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XC7fz.jpg" alt="enter image description here"></a></p>
<p>All pulleys are covered with rubber, and the rope is synthetic, Dyneema. What aditional information do I have to apply for the correct calculation?</p>
| |friction|pulleys|acceleration| | <p>let's find out acceleration first.</p>
<p><span class="math-container">$$v= at = 50000M/h * 1h/3600s *1000M/kM =13.888M/s \\ a= \frac{v}{t} =13.88/10 =1.388 m/s^2$$</span></p>
<p>the tributary weight on the two pullies is</p>
<p>weight on the right pully is <span class="math-container">$$40 *(1147.5-154.6)/1147.5= 34.6kg \ and \ on \ the \ left =40-34.6 = 5.4 kg $$</span></p>
<p>the angle of rope to left is <span class="math-container">$15 * 154.6/1147.5 = 2.02$</span> degrees. </p>
<p>say negligible, we assume the entire friction from the right wheel.</p>
<p>The friction force is </p>
<p><span class="math-container">$ F_f = \mu*40*9.8/ sin(15) \quad = 40*1.388$</span></p>
| 26314 | Minimum required rolling friction between pulley and the rope |
2019-03-10T19:32:30.883 | <p>This problem is about bending moment of a simple beam subject to a mixture of 4 or 5 UDLs and point loads. The beam is assumed to be rigid enough that deflection of the combined load is "small".</p>
<p>Solving this analytically for all load combinations simultaneously, in order to graph the result, is a tricky formula. </p>
<p>I'm wondering if, in this kind of simple case without significant distortion, one can simply use standard formulae for bending moments for single UDLs and point loads on a simple beam, and sum the results, which would be a much easier approach.</p>
<p>Intuitively, it should work. But is this actually a correct intuition? Can bending moments be summed in this way for multiple loads acting on a beam, to generate a graph of total bending moment acting at different positions on the beam?</p>
| |beam| | <h1>YES</h1>
<p>This is known as the <a href="https://en.wikipedia.org/wiki/Superposition_principle" rel="nofollow noreferrer">superposition principle</a>. For beams composed of linear-elastic materials where all loads are constant and independent from changes in beam geometry (i.e. not rain or snow loads, which increase as the beam deflects, creating an ever-deeper "pool"), the loads can all be considered independently. This means you can model each load alone, obtain the relevant results and then add them all up.</p>
<p>Just note that determining which results are "relevant" is not necessarily easy to do without modeling everything simultaneously. For example, a beam with complicated loading won't <strong><em>necessarily</em></strong> have maximum bending moment at midspan (unless all the loads are maximized at midspan, of course). In these cases, it may be hard to determine <em>a priori</em> which points to sample the bending moment, so finding the maximum bending moment may require solving all the loads simultaneously. </p>
<p>For example:</p>
<p><a href="https://i.stack.imgur.com/cTgCR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cTgCR.png" alt="simply-supported beam with asymmetric loading and resulting bending moment diagram"></a></p>
<p>This beam has a maximum bending moment 4.10 m from the left support, which couldn't be determined <em>a priori</em> since the loading can be split into three "sub-loads":</p>
<ol>
<li>UDL of 10 kN/m along the entire span, which we trivially know will cause maximum bending at midpsan;</li>
<li>UDL of 10 kN/m on the middle 2 m, which we trivially know will cause maximum bending at midpsan;</li>
<li>UDL of 30 kN/m applied 2nbsp;m to 4 m from the left support. This one is harder to determine, but has maximum bending 3.40 m from the left support.</li>
</ol>
<p>Given this information, you couldn't possibly know that the most important point along the span is actually at 4.10 m. The only way to know this is by considering all the loads simultaneously.</p>
<p>But if the loading is more trivial or if you already know which points to sample, then the superposition principle is an excellent friend to have.</p>
| 26321 | Can bending moments for a combination of loads on a simple beam, be calculated by adding the individual B.M. of each load? |
2019-03-11T21:42:47.480 | <p>I'm searching for an approach on how to calculate the adiabatic temperature loss of air when increasing its humidity by a humidifier.</p>
<p>Assumed we have the following setup:</p>
<p><strong>Air measuring position 1 (before humidifier):</strong></p>
<p><span class="math-container">$T = 30{ °C}$</span></p>
<p><span class="math-container">$x = 10\text{ g/kg (absolute humidity)}$</span></p>
<p><strong>Air measuring position 2 (after humidifier):</strong></p>
<p><span class="math-container">$T = \text{? °C}$</span></p>
<p><span class="math-container">$x = 20\text{ g/kg (absolute humidity)}$</span></p>
<hr>
<p><strong>The question now is:</strong> How can I calculate the air temperature at measuring position 2? With a <code>Mollier</code>-diagram it would be easy to figure out, but I want to solve it the analytical way.</p>
<p>I guess there is something possible with the enthalpy of air:</p>
<p><span class="math-container">$$h_1=h_2$$</span></p>
<p><span class="math-container">$$Q_1\cdot x_1\cdot T_1=Q_2\cdot x_2\cdot T_2$$</span></p>
<p><span class="math-container">$$\frac{Q_1\cdot x_1\cdot T_1}{Q_2\cdot x_2}=T_2$$</span></p>
<p>What is missing? :-)</p>
| |thermodynamics|cooling|temperature|vapor-pressure| | <p>You are missing an equation of state that can be solved analytical or preprogram SRK or PR in a TI calculator when make it solve the enthalpy for you (enthalpy 1=enthalpy 2 but the phases are different and so is the volume), because what you want will have changes in the molar Volume and it has to be solved, that is why you solve it with a Mollier Diagram or Steam tables; Someone already solved/measured those properties so any engineer (without access to a simulator) can solve it with a pen and a napkin. </p>
<p>You can try solving it yourself (and it is quite fun) but please by all means use a computer because you will be wasting time for a simple problem otherwise. </p>
| 26337 | How to calculate adiabatic air temperature loss caused by increase of humidity? |
2019-03-13T04:15:24.060 | <p>I built/assembled a small drill press recently and used a part from a Junk Drill because I couldn't figure out what a particular component I needed was called. </p>
<p>What I was looking for was a way to step up in chuck size so I could accommodate a bigger bit on a rod that was a smaller size (see: 3.17mm motor shaft driving 1/4"+ bits).</p>
<p>I now know the function of what I was looking for, it's a chamber of gears driven by a central D shaft gear the motor shaft can be inserted into. It's the internal component used to drive keyless chucks available in most retail drills. </p>
<p>Does anyone know what this would be called so I could look for it properly next time? I resorted to stripping an existing finished product for a part just because I couldn't figure it out, I was originally wanting to mount a keyed chuck</p>
<p><br> <a href="https://i.stack.imgur.com/8pBSj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8pBSj.jpg" alt="enter image description here"></a></p>
| |tools|drilling| | <p>I think it is not the chuck you want, but you've taken the <em>planetery gearbox</em>(full of planetary gears, may be called orbital gears instead) and the <em>torque limiter</em> from a battery operated drill, so it seems likely one of those words will help you. If I may ask, is there a reason you've gone this route instead of buying a cheap drill press? Mine isn't industrial grade, but it's more robust than the one in your picture and only cost $40 CAD on sale.</p>
<p>Also be advised that if you want the torque limiting function, broken battery operated hand tools can often be found cheaply on second hand sources, so you could take the gearbox from a higher quality hand drill and put yours back in service. If you go shopping for broken drills I'd go for Milwaukee(Extremely well documented, high quality) if you can find it. </p>
| 26356 | Terminology for drill component? |
2019-03-13T12:58:21.060 | <p>Part of my project is to compare linear elements and quadratic elements with a Quad mesh within Ansys workbench - static structural. The button that my classmates have to change between element orders is missing on my Ansys. Has anyone had ths issue and knows a solution or knows something I need to check to get the button to show? the option should appear under relevance in the Defualts section of the mesh. </p>
<p><a href="https://i.stack.imgur.com/RdPqm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RdPqm.png" alt="Mesh details"></a></p>
<p>Kind regards,<br>
George Packham</p>
| |ansys|ansys-workbench| | <p>Solution is based on the fact that the user has an older version of Ansys, in this case user has 16.2. There is the option for lines and quadratic nodes but it is found under automatic method as Element midside nodes. The option are either dropped (linear) or kept(quadratic). </p>
<p>The issue arose as my colleagues are using the new version whilst my licence is an older model </p>
| 26364 | Element order option missing |
2019-03-14T12:03:45.540 | <p>I'm in the process of building a hovercraft. In order to lift it, I'll need a motor directing the flow of air into a skirt. The specs for the motor I'm looking at say that it can turn a 15x5 dual-blade propeller at 6,000 RPM with 262W of power.</p>
<p>However, the motor is designed to be used by a large drone. In that case, the motor only has to work against atmospheric pressure. My hovercraft will have about .17 psi of pressure within the skirt. How do I adjust the required power for a given RPM based on the pressure below the propeller?</p>
<p>Mass is 87kg</p>
<p>1.6m x .45m</p>
<p>Airgrap: 12.7mm</p>
<p>Cushion Pressure: 1.185 kPa</p>
<p>Inward flow during static flight is 3,000 CFM (I believe that means flow out is also 3,000 CFM?)</p>
<p>These are the motors I'm looking at:
<a href="https://www.getfpv.com/tiger-motor-mn4014-330kv-antigravity-2-motors.html" rel="nofollow noreferrer">https://www.getfpv.com/tiger-motor-mn4014-330kv-antigravity-2-motors.html</a></p>
| |motors|pressure| | <p>Alright, I think I figured out the answer. Bernoulli's equation tells us about the energy density of air. Each pascal of pressure is one joule of energy per cubic meter. If we multiply the in-skirt pressure of the hovercraft by the volume of flow per second, we should get the cost in additional power.</p>
<p>1,185 Pa * 1.42 m^3/s = 1,683W</p>
| 26379 | How to adjust power requirements for a propeller working against a pressure head |
2019-03-14T17:33:07.537 | <p>I'm hoping some kind person can tell me if I have this right or not. I swear this isn't a homework problem. </p>
<p>I'll be using a heated pneumatic press to heat seal lidding to plastic containers (imagine I'm sealing yogurt containers). I need to verify that the machine I'm looking at can deliver a certain psi at the interface of the plastic and the lidding. That interface has much less area than the total area of the heated platen applying the force.</p>
<p>From what I understand I take the pressure delivered to the machine, subtract the ambient air pressure, multiply by the surface area of the cylinder and then add the total weight of the platen to get the full downforce the machine is capable of. From there I divide that by the surface area of the contact between the object under the press and the platen.</p>
<p>Please enjoy my wonderful art below (these are just random numbers):
<a href="https://i.stack.imgur.com/06uqT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/06uqT.png" alt="psi diagram"></a></p>
<p>So in this case the total downforce of the platen is:
(100psi - 14.7psi) * 1 sq in + 10 lbs = 95.3 lbs</p>
<p>And the force experienced by the red object under the press is:
95.3lbs / 2 sq in = 47.65psi</p>
<p>I feel like that's right but all my attempts to google for this answer only lead me to finding the total downforce of the press and not the force experienced by a smaller object <em>under</em> the press. Do I have it right? Thanks!</p>
| |pressure|hydraulics|compressed-air|compressors| | <p>This looks right- but note the units of <em>pressure</em> (not <em>force</em>) experienced by the red object are pounds per square inch.</p>
| 26384 | Pressure delivered to an object under a pneumatic press |
2019-03-14T19:57:40.993 | <p>The following image is an example that will help to demonstrate what i am saying and this below is a blender image version of an object with a slant surface at the top.</p>
<p><a href="https://i.stack.imgur.com/EaNI2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EaNI2.png" alt="enter image description here"></a></p>
<p>to draw the multi view projection of this view all the necessary dimensions are given, so the top and the front will be as follows</p>
<p>top:<a href="https://i.stack.imgur.com/IEfXx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IEfXx.png" alt="enter image description here"></a>
front:<a href="https://i.stack.imgur.com/mGSNw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mGSNw.png" alt="enter image description here"></a></p>
<p>i very well understand how the front and the top projections worked and here is where my question arises, as seen both from the pictorial view and the two side projection there are two thoroughly subtracted cylinders in the slant surface which means when i try to do the right side projection i will be seeing two ellipses rather than the original circles i have seen by the top view as shown below, so how could i figure out the dimensions of the major and minor axis of the ellipses.</p>
<p>right view:</p>
<p><a href="https://i.stack.imgur.com/fAynY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fAynY.png" alt="enter image description here"></a></p>
| |mechanical-engineering|technical-drawing|architecture| | <p>This is a simple trigonometric exercise. You really only need the "front" view to solve this.</p>
<p>Here's the cleaned up version of that view (I have omitted the holes for now, and I assumed the right-hand "vertical" is 1.50 m thick, which isn't explicitly stated in your diagrams):</p>
<p><a href="https://i.stack.imgur.com/cCfLG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cCfLG.png" alt="enter image description here"></a></p>
<p>So, really, the only things we need to observe here is the slope of the incline, which can be trivially found to be</p>
<p><span class="math-container">$$\dfrac{\text{d}y}{\text{d}} = \dfrac{0.99}{9.00} = 0.11$$</span></p>
<p>So, for every meter traveled horizontally, the incline has traveled 11 centimeters vertically.</p>
<p>Now, one of the cylindrical holes is shown with a diameter of 1 m<sup>[1]</sup>. I'm assuming this is valid for both.</p>
<p>Knowing the cylinder's diameter and the slope of the incline, it's easy to see that the vertical drop will be of 0.11 meters. After all, if you're at the top-most lip of the hole and move horizontally to the other end, you'll need to drop 0.11 meters to get to the incline once again.</p>
<p>So when you're drawing the holes in the "right" view, your ellipses need to have a <strong>major diameter of 1 meter</strong> (just like the actual circle, since there's no "warping" at this angle) and a <strong>minor diameter of 0.11 meters</strong>, since that's the vertical drop between the top and bottom lips of the circle.</p>
<hr>
<p><sub><sup>[1]</sup> There's a point of uncertainty here as to whether the cylinder's cross-section is circular or whether the hole on the inclined surface is circular. Only one of these can be true (if the cylinder is circular, then the inclined hole will be an ellipse; if the inclined hole is circular, then the cylinder is elliptical). I have further assumed here that the cylinder is circular since, well, that seems much more likely.</sub></p>
<p><sub>For completeness, in the unusual case where the inclined hole is circular (and the cylinder and hole at the bottom face are therefore elliptical), the solution would require a tiny bit more math: we'd need to use the Pythagorean Theorem to find the hypotenuse associated with the given slope, and then just use proportions to find the vertical step equivalent to a 1 m hyotenuse with the same slope.</sub></p>
<p><span class="math-container">$$\small\begin{align}
h &= \sqrt{0.11^2 + 1^2} = 1.00603\text{ m} \\
\dfrac{x}{1} &= \dfrac{0.11}{1.00603} \\
\therefore x &= 0.10934\text{ m}
\end{align}
$$</span></p>
| 26386 | How to calculate the dimensions of an ellipse in a multi-view projection |
2019-03-15T16:11:20.737 | <p>I am working on calculating the stiffness of a cantilever beam by applying a distributed load. I wanted to ask if the distribution of the load on the beam effects its stiffness.<br>
For eg., if I apply a point load at the free end vs a constant distributed load throughout the beam. Will the stiffness I get be different? How do I calculate the stiffness in later case?<br>
Thanks</p>
| |mechanical-engineering|beam|vibration|stiffness| | <p><strong><em>Stiffness</em></strong> is a murky term frequently used ambiguously in engineering.</p>
<p>However, the most common definition of stiffness is the product of a beam's Young's Modulus <span class="math-container">$E$</span> (which is a function of its material) and its moment of inertia <span class="math-container">$I$</span> (which is a function of its cross-section). So <span class="math-container">$\text{Stiffness} = EI$</span>.</p>
<p>Loading has nothing to do with stiffness according to this definition, which you could say describes an isolated beam's stiffness. This value allows you to say whether one beam would be stiffer than another in identical circumstances (whatever those may be).</p>
<hr>
<p>Now, another possible definition is stiffness as the deflection a beam or structure suffers under load. This would be an analogy with a spring's stiffness (which is literally measured in force needed to move the spring a unit distance).</p>
<p>This could be described as the whole structure's stiffness (even if that structure is a single cantilever beam). Obviously, a structure with pinned supports will be less stiff than one with fixed supports. Likewise, a cantilever beam with a concentrated load at midspan will deflect less (be stiffer) than one where the load is at the free end.</p>
<p>Now, as mentioned in <a href="https://engineering.stackexchange.com/a/26394/1832">@kamran's answer</a>, different loading patterns will lead to different deflections even if the total applied load is equal:</p>
<ul>
<li>A cantilever beam with a uniformly distributed load will have a deflection at the free end of
<span class="math-container">$$\delta = \dfrac{qL^4}{8EI}$$</span></li>
<li>The same beam with a concentrated force (equal to <span class="math-container">$P=qL$</span> to keep the same total force) at the free end will have
<span class="math-container">$$\delta = \dfrac{PL^3}{3EI} = \dfrac{qL^4}{3EI}$$</span></li>
<li>The same beam where the concentrated force is applied at the midspan will have
<span class="math-container">$$\delta = \dfrac{5PL^3}{48EI} = \dfrac{5qL^4}{48EI}$$</span></li>
</ul>
<p>Note that all these solutions have the constant <span class="math-container">$EI$</span> given as the more common definition of stiffness above. So a stiffer beam (one with greater <span class="math-container">$EI$</span>) will deflect less than a more flexible one in any and all cases. However, how much any given beam will deflect depends on the loading and support conditions.</p>
<hr>
<p>Given this, the answer to your question depends on what you're looking for.</p>
<p>If you're looking for the isolated beam's stiffness, then the result will be the same no matter your loading pattern (since you'll use different equations depending on the loading to get that beam's stiffness).</p>
<p>If, however, you're looking to get the stiffness as "force per deflection", then the loading will change your result. Any loading is valid, but your result will only be valid for that type of loading (i.e. this beam needs 10 kN/mm at midspan, but only 3.2 kN/mm at the free end).</p>
| 26392 | Stiffness of a cantilever beam |
2019-03-17T03:55:01.420 | <p>I recently started as designer at a making custom signage (small signs viewed up close, so there are lots of sizing limitations) and none of the other designers have any solid rules to rely on for building these signs when they are light through light diffusing acrylic. Often signs will look patchy, with individual LED's being distinguishable through the acrylic, due to the variation in lighting, but not the consistency of the acrylic.</p>
<p>Are there any established equations/relationships between the light diffusing properties of a material and the resulting scattering and intensity of the light? How can I measure or estimate the light diffusing properties?
The variables I can currently estimate or measure are: light intensity of the LED's, their spacing, thickness of material its shining through and the resulting light intensity, as well as, grading the resulting light pattern.</p>
<p>One of the issues we face is that our supplier does not send a consistent opacity of light diffusing acrylic, so a thickness that would normally look fine, will now either not let enough light through, or it will look patchy (because of point source LED's.) To be clear, the variation is batch to batch, not on the same sheet.</p>
<blockquote>
<p>Ideally, these relations/equations would help me understand the
difference in thicknesses required. Any starting point would be helpful.</p>
</blockquote>
| |optics|lighting| | <p>You seem to face a few issues.</p>
<p>First is to account for the potential to have patchiness in any one your sheets. Theoretical equations for light transmission through a system will be easiest when applied for isotropic materials, not for "patchy" systems. This problem has to be solved at the distributor end.</p>
<p>Second is to account for the transmission (opacity) for a given thickness in an isotropic sheet. For an isotropic material that is uniformly absorbing light at a constant amount per unit volume, the transmission falls off exponentially with thickness. For a given thickness, the total amount of light that is absorbed is a function of such factors as the concentration of absorber and its absorbing or scattering efficiency. The theoretical formulations are found under themes such as Beer's law for absorption of light. In essence, the light transmission equation is as below, with <span class="math-container">$I$</span> as the transmitted intensity, <span class="math-container">$I_o$</span> as the incoming intensity, <span class="math-container">$\alpha$</span> as an absorption factor, and <span class="math-container">$t$</span> as thickness.</p>
<p><span class="math-container">$$ \frac{I}{I_o} = \exp(-\alpha t) $$</span></p>
<p>This equation neglects reflection at the interfaces.</p>
<p>Third is to account for scattering to cause opacity. In polymers, light scatters from crystalline region. A perfectly glassy polymer is transparent. Uniform scattering comes from uniformly sized, uniformly distributed regions of crystallinity. A rough "off the top of my head" equation is to model the opacity as a rule of mixtures approach using the volume fraction of uniformly-sized, uniformly distributed crystalline regions <span class="math-container">$f_c$</span>.</p>
<p><span class="math-container">$$ O \equiv \frac{I}{I_o} = 1 - f_c $$</span></p>
<p>This models a polymer with a "full volume distribution" of crystalline regions <span class="math-container">$f_c = 1$</span> will be fully opaque. You are free to set <span class="math-container">$f_c$</span> as some other metric for linearity between scattering and opacity. Patchiness in any given sheet of polymer indicates uneven distribution of sizes or number density of crystalline regions. This problem has to be solved at the distributor end.</p>
<p>Fourth is to account for inconsistent stocks from sheet to sheet. When for example the concentration of absorber, type of absorber, or size/number density of scattering regions varies from sheet to sheet, <span class="math-container">$\alpha$</span> or the scattering potential will vary from sheet to sheet. The light transmission will therefore vary from sheet to sheet even when the sheet has the same thickness <span class="math-container">$t$</span>. This problem has to be solved at the distributor end.</p>
<p>Finally is account for the LEDs as point sources. Even in the best cases, LEDs are point sources not uniform sources of light when compared to tubes or bulbs. One suggestion to get around this is to invert your LEDs. Point them toward a diffuse scattering surface. Have that surface project the scattered light outward.</p>
| 26415 | Relations/Equations to do with light diffusing materials |
2019-03-17T18:19:22.263 | <p>I have been reading <a href="https://www.mdpi.com/1996-1073/12/2/272/htm#B22-energies-12-00272" rel="nofollow noreferrer">this</a> paper, which addresses the advantages of diffusers in cross flow turbines.</p>
<p>The abstract says (emphasis is mine):</p>
<blockquote>
<p>A CP’s extraordinary improving resulted when yaw increased up to 22.5° for the hydrofoil shaped and up to 30° for the <strong>symmetrical diffuser</strong>. Similar behaviour in yawed flows also occurred in case of a <strong>ducted</strong> single rotor, demonstrating that it is a characteristic of CFTs.</p>
</blockquote>
<p>I am confused as to the difference between a duct and a diffuser in terms of turbines. </p>
<p>According to Wikipedia, the diffuser sounds similar to a duct.</p>
<blockquote>
<p>a wind turbine modified with a cone-shaped wind diffuser that is used to increase the efficiency of converting wind power to electrical power. The increased efficiency is possible due to the increased wind speeds the diffuser can provide. </p>
</blockquote>
<p>My question is, what is the difference between the two? Do they have different functions?</p>
<p>I ask because it seems like the author is implying that the two are different.</p>
| |power|turbines|energy|energy-efficiency|diffusion| | <p>A definition of a diffuser is:</p>
<p>"The (divergent) diffuser is a duct so shaped that the fluid flowing through it decelerates, the pressure increasing from inlet to outlet."</p>
<p>This is from Engineering Thermodynamics Work & Heat Transfer, Rogers & Mayhew, page 38.</p>
<p>A duct is assumed to have neither a convergent or divergent shape.</p>
| 26421 | What is the difference between a diffuser and a duct in turbines? |
2019-03-17T21:02:23.833 | <p>I was looking in google, but I did not find any information on how the drilling was done, I would appreciate it if someone who knew how it was answered. As typical it could be an axis of 200 [mm] in diameter and a hole of 24 [mm]</p>
<p>Thank you!</p>
| |manufacturing-engineering| | <p>Holes or threaded holes can be done on a lathe.</p>
<p>These are bored or machined using drill bits to start then a cutting tool or boring bar to reach the final internal diameter.</p>
<p>Here is a video of a boring bar being made to give you an idea : <a href="https://www.youtube.com/watch?v=P8kmZ-9SjRA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=P8kmZ-9SjRA</a></p>
| 26422 | How is the drilling done along the axis for a typical synchronous motor shaft? |
2019-03-18T17:22:43.917 | <p>I have a standard non-linear static problem to solve with FEA. I am applying displacement boundary conditions only, no external force. The equation to solve look like these:</p>
<p><span class="math-container">$K(u)u=R(u)$</span> Eq(1)</p>
<p>The terms <span class="math-container">$K, R$</span> are the stiffness matrix and residual vector(in this case internal force vector), in any standard FEA formulation, <span class="math-container">$u$</span> is the field variable(displacement). The relationship between <span class="math-container">$K, R $</span> can be given as per standard FEM definitions, <span class="math-container">$\partial F/ \partial U=R$</span>(obtained from Taylor series expansion). Due to the non-linear nature, the equations cannot be solved directly, an incremental iterative approach is needed. </p>
<p>I am confused on how to set up the NR algorithm. All examples I see are breaking up the external load into steps, and run iterations in each step to check for the convergence <span class="math-container">$Residual=External Force-Internal Force= 0$</span>. But I don't have an external force vector, then how should I proceed with the incremental steps? Moreover, in this case, <span class="math-container">$External Force=0$</span> so how should the convergence be checked in each iteration step? </p>
| |mechanical-engineering|applied-mechanics|finite-element-method|modeling|numerical-methods| | <p>The discretized problem that you are trying to solve is
<span class="math-container">$$
\text{find}\,\, \mathbf{u}^{n+1} \,\, \text{such that} \,\, \mathbf{r}(\mathbf{u}^{n+1}) = \mathbf{0} \,\, \text{subject to the constraints} \,\, \mathbf{g}(\mathbf{u}^{n+1}) = \mathbf{0}
$$</span>
where <span class="math-container">$\mathbf{u}^{n+1}$</span> is the displacement at the <strong>end of load step</strong> <span class="math-container">$n$</span> and the residual is
<span class="math-container">$$
\mathbf{r}(\mathbf{u}^{n+1}) := \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}) - \mathbf{f}^{\text{ext}}(\mathbf{u}^{n+1}) \,.
$$</span>
In the <strong>absence of external forces</strong>, we have
<span class="math-container">$$
\mathbf{r}(\mathbf{u}^{n+1}) = \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}) \,.
$$</span>
Therefore, at the end of the load step, the displacements should be such that the internal forces are zero.</p>
<p>Newton's method can be used to find the values of <span class="math-container">$\mathbf{u}^{n+1}$</span> at which <span class="math-container">$\mathbf{r} = \mathbf{0}$</span>. </p>
<p>We start with the solution at the <strong>beginning</strong> of load step <span class="math-container">$n$</span>:
<span class="math-container">$$
\mathbf{u}^{n+1}_0 = \mathbf{u}^n
$$</span></p>
<p>(<strong>Caveat</strong>: For implicit dynamic computations with the Newmark-<span class="math-container">$\beta$</span> method, a better initial guess is <span class="math-container">$\mathbf{u}^{n+1}$</span>.)</p>
<p>As the iterations proceed, we will get better and better estimates of the quantity <span class="math-container">$\mathbf{u}^{n+1}_k$</span> where <span class="math-container">$k$</span> is the <strong>iteration number</strong>.</p>
<p>A Taylor series expansion of the residual about the current value of the displacement leads to
<span class="math-container">$$
\mathbf{0} = \mathbf{r}(\mathbf{u}^{n+1}_k) = \mathbf{r}(\mathbf{u}^{n+1}_{k-1}) + \frac{\partial \mathbf{r}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}} \cdot (\mathbf{u}^{n+1}_{k} - \mathbf{u}^{n+1}_{k-1}) + \dots
$$</span>
Reorganizing, we get
<span class="math-container">$$
\mathbf{u}^{n+1}_{k} = \mathbf{u}^{n+1}_{k-1} - \left[\frac{\partial \mathbf{r}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}}\right]^{-1} \cdot \mathbf{r}(\mathbf{u}^{n+1}_{k-1})
$$</span>
In the absence of external forces, we have
<span class="math-container">$$
\begin{align}
\mathbf{u}^{n+1}_{k} &= \mathbf{u}^{n+1}_{k-1} - \left[\frac{\partial \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1})}{\partial \mathbf{u}}\right]^{-1} \cdot \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1}) \\
& = \mathbf{u}^{n+1}_{k-1} - \left[\mathbf{K}^{\text{int}}\right]^{-1} \cdot \mathbf{f}^{\text{int}}(\mathbf{u}^{n+1}_{k-1})
\end{align}
$$</span>
The quantity <span class="math-container">$\mathbf{K}^{\text{int}}$</span> is called the <strong>tangent stiffness matrix</strong>. We continue iterating until a convergence criterion is satisfied - typically some sort of the energy norm computed from force and displacement.</p>
<p><strong>Update</strong>:</p>
<ol>
<li><strong>Constraints</strong>:</li>
</ol>
<p>I haven't talked about the process of applying the displacement constraints in the above procedure. There are several ways of applying these constraints. The most easily understood approach is to use Lagrange multipliers. A more common approach is to use a <a href="https://en.wikipedia.org/wiki/Penalty_method" rel="nofollow noreferrer">penalty method</a>.</p>
<ol start="2">
<li><strong>Computing the tangent stiffness</strong>:</li>
</ol>
<p>The tangent stiffness matrix can be derived in several ways. For example, in the Belytchko et al. book on nonlinear FE, they proceed using the convected rate of the Kirchhoff stress. Simo's book on computation inelasticity tends to focus on algorithmically consistent tangent stiffness matrices. Deriving the tangent stiffness for complex material models and implementing it without errors is one of the major difficulties experience by authors of FE codes. Very few commercial codes get everything right.</p>
| 26429 | FEA: Newton-Raphson algorithm for Dirichlet BC non-linear static analysis |
2019-03-19T13:19:50.497 | <p>I am currently working on developing a simulation for a restricted UAV, and I’m having some issues with Simulink while doing this. I have my mathematical equations which have been approved by my professor, but while I’m trying to simulate the system I am running into strange issues. Moreover, neither my professor nor my TA can figure out what is causing this issue, despite multiple private sessions to troubleshoot. If anyone can shed some light on the issue it would be highly appreciated. </p>
<p><strong>Firstly, the equation I’ve modeled on Simulink is this:</strong></p>
<p><a href="https://i.stack.imgur.com/7Fxxg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7Fxxg.png" alt="equation 1"></a></p>
<p><a href="https://i.stack.imgur.com/9ZtJh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ZtJh.png" alt="diagram"></a></p>
<p><strong>My Simulink block diagram is above. Input 1 Is Fb+Ff which, in this scenario = 6.8. Input 2 is theta which = 0.</strong> </p>
<p>I have annotated the values that should be and are at each of the nodes before the summing junction. In this section of the block diagram the theoretical value and the simulation agree.</p>
<p>The problem comes in the more complex half of the diagram. The graph I am validating should look like this.</p>
<p><a href="https://i.stack.imgur.com/UPdzV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UPdzV.png" alt="graph1"></a></p>
<p>There is an initial rise and then oscillations about a non-zero number. Obviously, my graph will be ideal so oscillations will not persist like they do in the graph above.
My final graph is this…</p>
<p><a href="https://i.stack.imgur.com/0U7j1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0U7j1.png" alt="graph2"></a></p>
<p>Where the y-axis is elevation angle in degrees and the x-axis is in seconds. Obviously these two graphs are in no way similar. In an attempt to analyze where I could have made an error in my block diagram, I analyzed the behavior at each node. </p>
<p>Immediately after the cos block to the lower right I get oscillating behavior</p>
<p><a href="https://i.stack.imgur.com/DEKuo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DEKuo.png" alt="graph 3"></a></p>
<p>Immediately after the summing junction and gain (1/0.91) I get:</p>
<p><a href="https://i.stack.imgur.com/2KcID.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2KcID.png" alt="graph 4"></a></p>
<p>The above bears a resemblance to the graph I’m validating against.</p>
<p>After the first integrator, however, I get this.</p>
<p><a href="https://i.stack.imgur.com/mWhEX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mWhEX.png" alt="graph 5"></a></p>
<p>Adding the two graphs together, should I not be getting a graph that rises and then oscillates about a non-zero number? </p>
<p>Have I made a mistake in my block diagram somewhere? </p>
<p><strong>The only value I’m allowed to change is the gain marked <em>De</em>.</strong> </p>
<p>At this point I am starting to hit a wall and really need some insight on this. Surely this cant be something wrong with the Simulink source code.</p>
<p>Nothing I do gets me anywhere near the validation graph. Can someone please point me in the right direction or let me know if you see any errors?</p>
<p>Any insight is highly appreciated, thank you for reading.</p>
<p><strong>Should you want to view the Simulink file for further analysis, it is available for download here: <a href="https://ufile.io/08aqv" rel="nofollow noreferrer">https://ufile.io/08aqv</a></strong> However please note that the attached file is my entire Simulink project. The circuit reference in the question above is contained in the subsystem "elevation".</p>
| |mechanical-engineering|electrical-engineering|control-engineering|aerospace-engineering|simulink| | <p>Your results are correct. I have solved the differential equations in Octave using the <code>ode</code> solver and I get the same results:</p>
<pre><code>function dX = elevation(t,X)
g = 9.81;
m_cw = 1.87;
l_w = 0.47;
m_uav = 1.17;
l_a = 0.66;
J_e = 0.91;
D_e = 0.04;
F_in = 0.68;
theta = 0;
dX(1) = X(2);
dX(2) = (F_in * l_a * cos(theta) + (m_cw*l_w - m_uav*l_a)*g*cos(X(1)) - D_e * X(2)) / J_e;
</code></pre>
<p>Which gets saved as <code>elevation.m</code> and solved by:</p>
<pre><code>tspan = [0 60];
X0=[0,0];
opts = odeset('MaxStep',0.1,'InitialStep',0.1);
[t,X] = ode45(@elevation,tspan,X0,opts);
plot (t,X(:,1))
grid
xlabel('time')
ylabel('elevation [rad]')
title ('elevation vs. time')
</code></pre>
<p>and which gives the following results:</p>
<p><a href="https://i.stack.imgur.com/cZUkc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cZUkc.jpg" alt="enter image description here"></a></p>
<p>So the only possible explanations are:</p>
<ol>
<li>The equations you use are incorrect (unlikely because you say they have been checked)</li>
<li>The initial conditions of the differential equations are incorrect or do not match those used to get the validated results (assumed to be zero for both elevation and elevation rate at t=0). For example, it looks like the initial condition for the elevation in the validated graph is about -30 deg or so.</li>
<li>The numerical values for the various parameters are incorrect or do not match those used to get the validated results</li>
<li>The inputs used (<code>Fb+Ff</code> and <code>theta</code>) do not match those used to get the validated results.</li>
</ol>
<p>To me, #4 seems like the most plausible explanation. You have assumed both inputs to be constant. Is that really the case? Are they not a function of time?</p>
<p>I have tried changing the value of <code>D_e</code> as you suggest and you do get some quite different results. For example, this is with <code>D_e = 0.2</code> instead of <code>0.04</code>:</p>
<p><a href="https://i.stack.imgur.com/hvYwA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hvYwA.jpg" alt="enter image description here"></a></p>
| 26442 | Strange simulink error. Professor & TA don't know. Any Ideas? |
2019-03-19T13:32:40.697 | <p>I'm a computational person, and I don't quite understand how rotational DoFs are manifested in the linear elasticity equations. I understand the translational DoFs are simply the displacement components. When specifying boundary conditions for linear elasticity simulations, we typically specify the displacement and/or traction boundary conditions at the boundaries. Where do the rotational boundary conditions come in? Are they manifested in the traction boundary conditions? </p>
<p>This is not very intuitive to me and I was hoping someone can explain.</p>
| |structural-engineering|structural-analysis|solid-mechanics|traction| | <p>Just answering the traction part of your question. A traction on the surface of a shell or beam that has a component tangent to the surface is equivalent to forces on the translational DOFs of the nodes plus moments on their rotational DOFs.
<a href="https://i.stack.imgur.com/mVIXY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mVIXY.png" alt="enter image description here"></a></p>
| 26443 | Rotational and translation DoFs in structural mechanics |
2019-03-19T18:08:50.470 | <p>So, I have a 1,000 pounds load.</p>
<p>What would be the required torque in inch/pounds to lift that load for a motor?</p>
<p>If the motor is installed at the center of the radius generating that torque, does it matter how long the radius is if the end requirement still 1,000 pounds load?</p>
<p>Would the answer be 1,000 in-lbs? </p>
| |torque| | <p>Any torque and power demand greater than your motor rating will cause it to stall.
for a 1000lbs.inch motor the pully's radius r must be, <span class="math-container">$$r < 1 \ \text{inch}$$</span></p>
<p>Otherwise you need a gear. Also the motor must have at least the power of,</p>
<p><span class="math-container">$$P= 9.8 \cdot 1000\cdot 0.45 \ lbs/kg \cdot0.305 \ ft/meters=1345 \ \text{watts}$$</span>
This is if you want to lift it in 1 second. If you want to lift it in 10 seconds then the power will be 134.5 watts, but the required torque will still be the same.</p>
| 26452 | How much torque do I need to lift 1,000 pounds? |
2019-03-20T08:33:57.367 | <p>The datasheet of the AD630 mentions that the device can be used as a Precision Phase Comparator or as a Lock-in Amplifier.</p>
<p>What's the difference? They both seem to map the amplitude of a modulated signal to around DC. That is, the output of both circuits is the demodulated signal. (Assuming the reference signal is the modulated signal)</p>
<p>Why would I want to pick one circuit over the other?</p>
| |electrical-engineering|modulation| | <p>The same circuit may be used for different purposes depending on what inputs you apply. And you pick depending on what you want to achieve. Amplifiers are very versatile components and you can achieve quite different goals with the same component.</p>
<ul>
<li><p><em>Assuming the reference signal is the modulated signal</em> is one of possible inputs and then you use the circuit as a demodulator - you get the demodulated signal on output - a simple receiver of modulated signal.</p></li>
<li><p>If you pass two identical, or very similar signals (modulated or not) which are phase-shifted, you can treat the output as measurement of the phase shift - Phase Comparator - measuring system delay, e.g. round-trip time of echo vs reference in distance measurements using ultrasound sensors.</p></li>
<li><p>If you apply a noisy signal and the reference input gets attenuated noise plus carrier, you're getting a demodulator that can deal with the noise - just like the first case but filtering high-frequency noise away.</p></li>
</ul>
| 26467 | Difference between AD630 'Phase Comparator' and 'Lock-in Amplifier' modes |
2019-03-21T08:52:35.597 | <p>I suspect I know the answer to this, but hoping someone may know something I don't here.</p>
<p>Is there a way I can determine the specific heat capacity at constant pressure (Cp) of an oil if I know the density, temperature and viscosity? </p>
<p>The oil I am interested in: <a href="http://www.bakerprecision.com/neo/75w90rhd.htm" rel="nofollow noreferrer">http://www.bakerprecision.com/neo/75w90rhd.htm</a></p>
<p>The company is only small and hasn't got a value of Cp as far as I can tell, I have contacted them. They are unlikely to have tested it in my experience of dealing with them. </p>
<p>As far as I know this could only be done by experimental correlation, and there is no universal equation. </p>
<p><strong>Is there any theory you know that can estimate the value of Cp of an oil from basic properties without experimentation?</strong> </p>
<p>I know:
Dentity, viscosity and surface tension at room temperature. Plus viscosity at various other temperatures. </p>
| |fluid-mechanics|heat-transfer|fluid| | <p>There is no practical way to do this.</p>
<p>It is possible to estimate specific heats from first principles using computer simulation and quantum mechanics. There are also some semi-empirical "laws" which apply to particular situations, such as materials whose elements have large atomic numbers, (bigger than iron, and therefore irrelevant for lubricating oil) or at very low (cryogenic) temperatures. None of those is applicable to gear oil.</p>
<p>Measuring specific heat is not difficult - you could do it yourself with a "kitchen table" experiment. Put a known mass (or volume) of oil in a thermally insulated container (e.g. a thermos flask), apply a known amount of heat (for example using an electrical resistor immersed in the oil), and measure the temperature change.</p>
| 26481 | Determine Cp of an oil from density and temperature? |
2019-03-22T14:56:01.293 | <p>I need to make PTFE caps on my test tubes. Why PTFE? The test tubes contain industrial solvent mixtures & the rubber caps that came with the test tubes are very, very slowly being dissolved.</p>
<p>I need about a dozen caps. If I were to make it out of solid PTFE, turned on a lathe creating the cap, then it would be expensive. So I'm thinking of having thick cups of PTFE. Then it would be filled by a soft elastomer, creating the cap. Seeing that most PTFE pieces are created by taking PTFE powder & thermoforming it in molds, the production method is expensive even if the cups are lightweight. So I'm thinking "what about vacuum forming?", so I'm here. Can Vacuum forming be used on PTFE?</p>
| |plastic| | <p>Don't make them, just google "ptfe stoppers", and buy them. (You've asked an XY problem.)</p>
<p><a href="https://i.stack.imgur.com/mKtGS.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/mKtGS.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/LWfTi.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/LWfTi.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/93eiI.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/93eiI.gif" alt="enter image description here"></a></p>
| 26507 | PTFE sheet vacuum forming |
2019-03-22T22:33:42.440 | <h1>The context</h1>
<p>I’ve taken an interest in the design of pressure vessels for use in submarine applications. From a cursory survey of what is used in industry, it seems the <em>de facto</em> standard housing is a thick-walled cylinder (rather than a thin-walled cylinder which would be prone to buckling when loaded in hydrostatic compression). While the equations for stress and deformation in a thick-walled cylinder subject to internal and external pressures are well established (<a href="https://www.amazon.ca/Roarks-Formulas-Stress-Strain-8th/dp/0071742476" rel="nofollow noreferrer">Young</a>, <a href="https://www.amazon.ca/Machine-Design-5th-Robert-Norton/dp/013335671X" rel="nofollow noreferrer">Norton</a>), I’ve also read that prestressed compound cylinders can result in a housing with a greater strength to weight ratio (<a href="https://nptel.ac.in/courses/112105125/pdf/module-9%20lesson-2.pdf" rel="nofollow noreferrer">Kharagpur</a>, <a href="https://www.mae.ust.hk/~meqpsun/Notes/Chapter2.pdf" rel="nofollow noreferrer">Sun</a>).</p>
<p><img src="https://i.stack.imgur.com/l7W3C.png" alt="Stress distributions in compound pressure cylinders"></p>
<p>As a result, I’ve been trying to understand the derivation of the thick-walled cylinder equations to better understand how they can be applied and optimized.</p>
<h1>The question</h1>
<p>With the help of multiple online sources (primarily <a href="http://engweb.swan.ac.uk/~c.kadapa/teaching/2017-2018/EGF316/week2/EGF316%20Thin%20and%20Thick%20Cylinders%20-%20notes.pdf" rel="nofollow noreferrer">Kadapa</a> and <a href="http://www.engr.mun.ca/~katna/5931/Thick%20Walled%20Cylinders(corrected).pdf" rel="nofollow noreferrer">Katna</a>), I’m confident that I understand the derivation of Lamé’s equations</p>
<p><span class="math-container">$$\sigma_r = C_1 + \frac{C_2}{r^2}$$</span>
<span class="math-container">$$\sigma_\theta = C_1 - \frac{C_2}{r^2}$$</span></p>
<p>This is where things get interesting. In both sources mentioned above, and indeed everywhere else I can find it, the equation for the change in radius of a differential shell is calculated from the hoop strain, rather than the radial strain as I would expect.</p>
<p><span class="math-container">$$\varepsilon_\theta = \frac{\sigma_\theta - \nu \left( \sigma_r + \sigma_z \right)}{E}$$</span>
<span class="math-container">$$u = \varepsilon_\theta r$$</span></p>
<p>While I’m satisfied this is correct, I don’t understand why. Why doesn’t the radial deformation result from the radial strain as I expect it should, according to</p>
<p><span class="math-container">$$\varepsilon_r = \frac{\sigma_r - \nu \left( \sigma_\theta + \sigma_z \right)}{E}$$</span>
<span class="math-container">$$u = \varepsilon_r r$$</span></p>
<h1>What I've tried so far</h1>
<p>I’ve tried evaluating both equations for <span class="math-container">$u$</span> for the simple case of a single cylinder subject to internal and external pressure (the boundary conditions given in every textbook).</p>
<p><img src="https://i.stack.imgur.com/B4DSt.png" alt="Differential cylinder element showing stresses and deformation"></p>
<p>While I haven’t compared the values to a more reliable source, the equation with hoop strain always produces deformations in the direction of the pressure differential (an external pressure causes the cylinder to contract). The equation using radial strain has so far always produced deformations of differing magnitude and in the direction opposite the pressure differential (an external pressure causes the cylinder to expand).</p>
<p>In both Kadapa and Katna, it is offhandedly mentioned that and <span class="math-container">$\varepsilon_r$</span> and <span class="math-container">$\varepsilon_\theta$</span> are equal on the bases that</p>
<p><span class="math-container">$$\varepsilon_r = \frac{u}{r} = \frac{2 \pi u}{2 \pi r} = \frac{\delta\ circumference}{circumference} = \varepsilon_\theta$$</span></p>
<p>While I can’t find fault with this statement, I think I have shown that it is only true for thin-walled cylinders where the hoop stress is assumed to be constant throughout the wall as below:</p>
<blockquote>
<p>Based on the methods presented in Kadapa, Katna, and others,</p>
<p><span class="math-container">$$\varepsilon_r = \frac{du}{dr}$$</span>
<span class="math-container">$$\varepsilon_\theta = \frac{u}{r}$$</span></p>
<p>Assume that</p>
<p><span class="math-container">$$\varepsilon_r = \varepsilon_\theta$$</span></p>
<p>If so, </p>
<p><span class="math-container">$$\frac{du}{dr} = \frac{u}{r}$$</span></p>
<p>This first order linear ODE has the solution</p>
<p><span class="math-container">$$u = kr$$</span></p>
<p>As a result, </p>
<p><span class="math-container">$$\varepsilon_\theta = \frac{u}{r} = k$$</span></p>
</blockquote>
| |mechanical-engineering|pressure-vessel|solid-mechanics| | <p>I think I've identified my misconception. As alephzero commented on my question, I was conflating the local strain with the overall radial deformation. The strain is defined as the normalized change in length of an infinitesimal line segment, and not as the change in length of a macroscopic dimension such as radius. As such, the results </p>
<p><span class="math-container">$$\varepsilon_r = \frac{du}{dr}$$</span>
<span class="math-container">$$\varepsilon_\theta = \frac{u}{r}$$</span></p>
<p>give me immediate access to the deformations as</p>
<p><span class="math-container">$$u = \int \varepsilon_r dr = \varepsilon_\theta r$$</span></p>
<p>Thanks for pointing me in the right direction.</p>
| 26512 | When calculating the deformed radius of a pressurized thick-walled cylinder, why is the hoop strain used rather than the radial strain? |
2019-03-22T23:34:25.323 | <p><a href="https://i.stack.imgur.com/0lfOo.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/0lfOo.jpg" alt="enter image description here"></a></p>
<p>From doing some research I've learned that internal stresses inside an object can "concentrate" on sharp edges, as shown in the above picture. For that reason we often make fillets so that the change in diameter is smoother and we get a smaller stress concentration.</p>
<p>But why does the stress concentrate on sharp edges? From Googling I've found numerous inadequate answers, such as imagining "stress flow lines" getting packed at the sharp edge. <strong>Here I can see an analogy to electromagnetism, where we talk about flow lines of an electric field, but what is "flowing" here? What do the lines represent?</strong></p>
<p>Intuitively, I can imagine this: If we press the object on the left in the above picture, there is a higher stress after the edge, since there is now a smaller area enduring the same force. Hence the stress is higher. But this does not explain why the stress concentrates <em>at</em> the edge. I mean, even with the fillet, the radius decreases anyway, so why is there not a similarly high stress on the right of the fillet?</p>
<p>Similar effect also occurs when we have a continuous piece of material with a hole in it. The stress is the highest around the hole.</p>
<p><a href="https://i.stack.imgur.com/cCo0b.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/cCo0b.gif" alt="enter image description here"></a></p>
<p>Once again, force lines, what do they represent? Again I can see the similarity to electromagnetism. But what kind of a force field are we talking about here?</p>
<p>One source I found simply explained that we can use methods like finite element analysis to confirm this concentrating of stress does happen. But this is not an explanation either. <strong>Use of FEA implies there is some kind of differential equation to be solved, so what kind of theory is there behind it?</strong></p>
<p>Thank you!</p>
| |structural-analysis|stresses| | <p>I would not overthink the lines of force explanation too much; its a concept that really only makes sense in the context of vanishing shear forces, like masonry.</p>
<p>To me, the simplest explanation of why stress concentrations occur is as follows: consider a rod with a step function in its radius, like in your initial picture. Given a fixed axial load, the most optimal distribution for carrying that load is a uniform distribution over the cross section. Every element carries the same load, and no one area will have a reason to fail before any others do. A uniform rod might reach that distribution if the force is led-in sufficiently gently. But can a rod with a sudden change in cross-section do the same thing?</p>
<p>The best case scenario would be if the thin end of the rod would still maintain its optimal uniform stress distribution. But resistance to deformation in the thick end will tend to favor the same flat stress distribution profile in the thick end, and it will settle into that at least far away from the cross sectional change. But the uniform distribution in the thick and thin end will have a different stress magnitude; they cant just meet as two uniform distributions of different magnitudes and be in equilibrium.</p>
<p>So by that argument alone, we have proven that the stress distribution near the sectional change, can and will not be the optimal uniform profile; that is it will have some local maxima; concentrations if you will.</p>
<p>Now that still leaves open the possibility that at least the thin end would have the good taste to maintain its optimal uniform distribution, gradually spreading to a wider distribution in the thin end, without unnecessarily weakening our construction. And you can engineer a situation like that, with a nicely rounded transition. But if you consider the effect of the 'extra' material that exists, from a sharp geometry relative to a rounded geometry, you realise that relative to the deformation in the rounded case, that extra material would have to experience a deformation to match the deformation of the rod. You can think of the stress concentration as the reaction force required to deform that 'extra' material along with the rest of the rod. And since the extra material is located out there near the sectional change on the perimeter, thats where you will find a stress peak. Even though the thin end of the rod would love to stay in a nice flat stress distribution, if you put some extra material squeezing it into another shape away from a flat stress profile; well, ask and you shall receive.</p>
| 26514 | Why do stress concentrations really occur? |
2019-03-23T18:59:48.207 | <p>In educational construction material as well as engineering publications, the corbel angle for significant excess loads, and also for substantial openings, in an ordinary brick wall is usually stipulated as being taken as 60 degrees to the horizontal. </p>
<p>For example:
<a href="https://i.stack.imgur.com/hyypO.png" rel="noreferrer"><img src="https://i.stack.imgur.com/hyypO.png" alt="enter image description here"></a>
<em>In the first illustration, the reduced load is spread out, in the second a distributed load is spread out, and in the third, a point load is spread out. There are no buttresses or other "out-of-plane" support. The wall can be imagined as an indefinite length ordinary 100mm thick brick wall.</em></p>
<p>I've tried to find the source of that 60 degree value, to check when it's applicable and how it's calculated, but I can find nothing whatsoever on the topic. The few mentions online of "corbel angle" are all of a form that just state/assume 60 degrees (with no reasoning). I can't find anything else.</p>
<p>Related to this, when dealing with an opening, which of these is the correct way to use the corbel angle, and why? They can't both be correct, as they contradict, but there doesn't seem to be an obvious reason why the answer should be one rather than the other.</p>
<p><a href="https://i.stack.imgur.com/3S2vF.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3S2vF.png" alt="enter image description here"></a></p>
<p><em>In the first diagram, the reduced weight is treated as causing a reduced load in nearby masonry at 60 degrees outward; in the second the heavier surrounding weight is treated as causing an increased load at 60 degrees inward.</em></p>
<p>How does the corbel angle actually work in these typical situations, where there is an excess loads + openings at some point in an ordinary brick wall, and how universally appropriate is it to always assume it's 60 degrees in ordinary brick or concrete construction?</p>
| |structural-engineering|design|structural-analysis|masonry| | <p>I recommended reading EN1996-1-1, and especially section 6.1.3, at least if you are in an area where eurocodes are applicable. This section contains the load distribution angle of 60 degrees. It has a few figures showing applications of the angle, but no formulas for calculating an alternative value.</p>
<p>I would not expect to be able to find a formula to calculate a more exact value of the angle. To me, it appears very much to be an empirical value, which is backed by experiments rather than theory.</p>
<p>As for the question on window openings: The angle of 60 degrees is for distributing concentrated over a higher length of wall. Therefore the figure on the right in your diagram is the correct one. Strictly speaking, you could use the figure on the left for distributing the weight of the window sill and the figure on the right for distributing the weight of the wall between and above the windows, but that will rarely be worth the bother. Just think of it as distributing a load, not distributing the absence of a load.</p>
| 26522 | Should the corbel angle always be estimated as 60 degrees in masonry construction, or is there a formula to use instead? |
2019-03-25T05:05:55.667 | <p>These holes are caused by bars that are put through the frame before the concrete is poured and then clamped, so the mold can't swell outward. </p>
<p>Is there a name for these holes though that is widely recognized?</p>
<p><a href="https://i.stack.imgur.com/5GpL9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5GpL9.jpg" alt="enter image description here"></a></p>
| |concrete| | <p>“Ando dots”. Named after Architect <a href="https://en.wikipedia.org/wiki/Tadao_Ando" rel="nofollow noreferrer">Tadao Ando</a> who popularized highlighting his design with the dots prominently showing.</p>
| 26543 | What are the regular holes in poured concrete walls called? |
2019-03-25T14:16:27.620 | <p>Hello everyone I am designing a parts that needs to rotate so I am planning to insert a round bushing.</p>
<p>But I am curious of how much space I need to leave between the round bushing and the parts that need to rotate? (Assuming the round bushing will be fixed by tighten the screw)</p>
<p>The outer diameter of the round bushing is 3.5mm so how many mm hole should I design for that 3.5mm round bushing? If produce by a CNC machine how accurate can it mill? </p>
<p>Thank you very much<a href="https://i.stack.imgur.com/B32Al.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B32Al.jpg" alt="enter image description here"></a></p>
| |cnc| | <p>This is one of those questions that can be answered by "you get what you pay for." Ignoring the poor grammar of that statement, one does have to spend money for quality performance. </p>
<p>In the 3D printer world, you'd have to have as much as 0.2 mm spacing to ensure the rotation you seek. In the CNC world, it's not that large. A friend who owned a machine shop created parts with his CNC with tolerances of as small as 0.0005". That's five ten-thousandths of an inch. In these circumstances, one must ensure temperatures are uniform for both parts. If the parts were machined in 65°F conditions, those parts might not fit together in colder or warmer conditions.</p>
<p>You can get a good sliding fit with room for lubrication by having a clearance of two or three thousandths of an inch. When I want a clean sliding fit for tubing, I'll purchase a "standard" size from Aircraft Spruce and Specialty, say, 0.375 outside diameter. The matching tubing, next size up would be 7/16" in the list, and I'd have to pick a wall thickness to match the 0.375, which in this case is 0.035, giving me an inside diameter of 0.368, a difference of 0.007". The next size down would cause too much slop and the next size up would result in an impossible-to-fit set of tubes.</p>
<p>Seven thousandths of an inch is tight enough that a small burr will prevent a fit. One last consideration is your bushing size. Typically a manufacturer will specify that part's tolerance as well. Is the 3.5 mm listed as +/- 0.1 or something similar? </p>
| 26547 | How much space should I leave between round bushing and the rotating parts to minimize the play? |
2019-03-25T17:32:47.160 | <p>I have a device 4cm X 5cm X 15cm that stands on asphalt/concrete/road. So, the device is taller than its base which makes contact with the asphalt.</p>
<p>I want to maximize the friction so that the device stands as well as possible without any attachment.</p>
<p>I'm considering a thin rubber add-on attached at the bottom of the device and I'm wondering what is the best design for such rubber to maximize the friction with asphalt.</p>
<p>I selected rubber because it has the highest friction coefficient with asphalt - there is a reason why tires are made of rubber...</p>
<p>But how should I design the bottom side of this rubber part that will be in contact with the asphalt? Here is an idea which I got inspiration from a bathtub mat. What is the best pattern possible? And what should be the pattern size (small or big relative to the 5cm X 4cm area)?</p>
<p><a href="https://i.stack.imgur.com/3Dq7D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Dq7D.png" alt="enter image description here"></a></p>
| |mechanical-engineering|friction|concrete|industrial-engineering|product-engineering| | <p>I would use three(triangulated) very sharp tip toes, stainless steel conical points so that they dig into the asphalt a little bit with a little pressure from the installer :-)
I don't know if they have them anymore, but track and field shoes use to have threaded conical spikes with the shoe soles having threaded inserts to receive the spikes. You could use something similar. Me thinks this will grip the ground better than a textured rubber pad.</p>
| 26550 | Which design to maximize friction for a device standing on concrete? |
2019-03-25T20:11:20.350 | <p>I am building a roll stabilized model rocket. I am beginner at controls systems. How do I find out the transfer function of the rocket. Although, I am aware of definitions and stuff of control systems. I just a need a nudge in the right direction.</p>
<p>Also what your thought on using State Space Analysis?</p>
<p>Thanks so much! :D</p>
| |control-engineering|pid-control|rocketry| | <ul>
<li>Define your inputs (i.e., control surfaces on the fins, steerable motors, etc.)</li>
<li>Define your output (roll position of the rocket)</li>
<li>Find the differential equation that describes the roll in terms of the input.</li>
<li>Linearize (note that if you're using fins then the linearized equation will be dependent on airspeed, and that if you're using steerable motors then the linearized equation will be dependent on thrust -- this will complicate your life, but get over the first hurdle first!)</li>
<li>Extract the transfer function</li>
</ul>
<p>State space is not a bad way to go, but if you're only concerned with roll control, and if you're confident that you're not going to be affecting the behavior in yaw or pitch, then this is a classic single-input, single-output system that can be handled with transfer functions.</p>
| 26554 | How to design a transfer function for a Model Rocket? |
2019-03-27T14:11:49.427 | <p>I'm trying to find the maximum bending moment and deflection for this symmetrical simple beam. The formulae I can find, deal with either cantilever beams with one fixed end, or continuous beams without cantilever ends, not this kind of situation.</p>
<p><a href="https://i.stack.imgur.com/0UhVh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0UhVh.png" alt="enter image description here"></a></p>
<p>I've tried to model the beam as two fixed-end cantilevers, on the basis that symmetry demands the middle of the beam is horizontal, and then modelled the loads as separate loads superimposed. But I'm not getting anywhere and not sure how I'd tell whether my answer was correct, if I did. I don't know if the fact that the beam can rise or fall in the middle (depending how F compares to W) is the problem, or if it's something else.</p>
<p>The beam won't be exposed to non-linear behaviour. It could bend either way, as I don't know whether the 2 point forces will be larger or smaller than the UDL, or the exact lengths involved.</p>
<p>Is there an easy way to work out the maximum deflection and B.M. of this beam, and if so, what are they?</p>
<p><em>Note, this isn't homework or coursework (I haven't had those for many years!). It's part of a real-world practical design I'm trying to work on.</em></p>
| |beam| | <p>HINT</p>
<p>After drawing Free Body, Bending moment <span class="math-container">$ M/EI $</span> diagrams, integrations by</p>
<p><a href="https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Aerospace_Structures_and_Materials_(Alderliesten)/02%3A_Analysis_of_Statically_Determinate_Structures/07%3A_Deflection_of_Beams-_Geometric_Methods/7.05%3A_Deflection_by_Moment-Area_Method" rel="nofollow noreferrer"> First & Second Moment Area Methods</a></p>
<p>help determine rotations and deflections.</p>
| 26581 | Simple beam with cantilever ends |
2019-03-27T15:31:30.517 | <p>I have a cable which dissipates <span class="math-container">$\dot Q_{total}$</span> and I need to determine the surface temperature of the insulation <span class="math-container">$T_{ins}$</span> and the surface temperature of the conductor <span class="math-container">$T_{cond}$</span>. </p>
<p>There is an insulating layer with the thermal resistance <span class="math-container">$R_{cond}$</span> and the convective thermal resistance <span class="math-container">$R_{conv}$</span>. The radiation component is neglected. From the geometry of the insulating layer I can obtain <span class="math-container">$R_{cond}$</span>, but for <span class="math-container">$R_{conv}$</span> I need to calculate the convection coefficient via the Nusselt-number. I have an empirical equation for the Nusselt-number, but the Rayleigh-number calculation requires the value of <span class="math-container">$T_{ins} - T_{ambient,air}$</span>. </p>
<p>So I have the following equation: </p>
<p><span class="math-container">$\dot Q_{total} = (1/R_{total}) \cdot (T_{cond} - T_{ambient,air})$</span> </p>
<p>with <span class="math-container">$R_{total} = R_{cond} + R_{conv} = R_{cond} + Nu \cdot \lambda/L$</span>. </p>
<p>Where <span class="math-container">$\lambda$</span> is the thermal conductivity and <span class="math-container">$L$</span> the length of the cable. There are two unknown values in this equation <span class="math-container">$T_{cond}$</span> and <span class="math-container">$T_{ins}$</span>, because <span class="math-container">$Nu$</span> depends on <span class="math-container">$T_{ins}$</span> and <span class="math-container">$\dot Q_{total}$</span> depends on <span class="math-container">$T_{cond}$</span>. Is there a way to calculate the temperatures or do I have to define a temperature? </p>
<p>Furthermore I cannot determine the film temperature because <span class="math-container">$T_{ins}$</span> is unknown. At which temperature should I determine the thermodynamic properties for the Rayleigh- and Prandtl-number calculation?</p>
<p>Summary:
Known: heat flow <span class="math-container">$\dot Q_{total}$</span>, <span class="math-container">$R_{conv}$</span> (because geometry is known), <span class="math-container">$T_{ambient,air}$</span> Unknown: <span class="math-container">$T_{ins}$</span>, <span class="math-container">$T_{cond}$</span>, film temperature <span class="math-container">$T_{f}$</span> and therefore also the exact values of Rayleigh, Prandtl and Nusselt-number.</p>
<p>I want the thermal resistance of convection to be as accurate as possible, that is why I wanted to calculate it with these dimensionless quantities </p>
| |mechanical-engineering|thermodynamics|heat-transfer| | <h1>Problem Statement</h1>
<p>You have three unknowns: the temperature at the conductor <span class="math-container">$T_c$</span>, the temperature at the insulation <span class="math-container">$T_i$</span>, and the air heat transfer coefficient <span class="math-container">$h_a$</span>. You know the air temperature <span class="math-container">$T_a$</span>, heat flow per length <span class="math-container">$\dot{q}/L \equiv q_L$</span>, the geometry of the cable + insulation (radii <span class="math-container">$r_c$</span> and <span class="math-container">$r_i$</span>), and the insulation material (thermal conductivity <span class="math-container">$k$</span>).</p>
<p>You have two energy balance equations.</p>
<p><span class="math-container">$$ q_L = \frac{2\pi\ k}{\ln(r_i/r_c)}\ (T_c - T_i)$$</span></p>
<p><span class="math-container">$$ q_L = 2\pi\ r_i\ h_a\ (T_i - T_a) $$</span></p>
<p>In the set of unknowns, you also know that <span class="math-container">$h_a = f(T_i, T_a)$</span>. The relation is solved through a correlation equation, chart, or table using the Nusselt number <span class="math-container">$Nu$</span> (and Reynolds number and Prandtl number no doubt). The flow is (ostensibly) laminar (because you mention film temperature, meaning boundary layer theory, meaning laminar flow).</p>
<h1>Proposed Approach</h1>
<p>When we know <span class="math-container">$h_a$</span>, we can solve for both unknown temperatures.</p>
<p><span class="math-container">$\Rightarrow$</span> Make a plot of <span class="math-container">$T_c$</span> and <span class="math-container">$T_i$</span> versus <span class="math-container">$h_a$</span> over a range of values for <span class="math-container">$h_a$</span> that are expected to be relevant for the physical system.</p>
<p>We do not know <span class="math-container">$h_a$</span> until we know <span class="math-container">$T_i$</span>.</p>
<p><span class="math-container">$\Rightarrow$</span> Calculate <span class="math-container">$h_a$</span> as a function of <span class="math-container">$T_i$</span>. Superimpose the values correspondingly on the above graph.</p>
<p>The answer for the three unknown values will be at the point where the line for <span class="math-container">$T_i$</span> versus <span class="math-container">$h_a$</span> made using the energy balance equations intersects with the line for <span class="math-container">$h_a$</span> versus <span class="math-container">$T_i$</span> made using the correlation expressions.</p>
<p>Here is an example of the plot that I believe you might have at the end.</p>
<p><a href="https://i.stack.imgur.com/WqkRK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WqkRK.png" alt="profile plot" /></a></p>
<p>The circles show how to obtain the three unknown values for your system.</p>
<p>You <em>could</em> also iterate through the equations. I prefer a graphical approach.</p>
| 26583 | Calculating the Surface Temperature of a Cable with known heat flow |
2019-03-28T00:21:27.540 | <p>I have a shaft inside a bushing, the shaft has a gear on one end that is supposed to provide resistance to other gears it is driven with.</p>
<p>I'm having a hard time finding the correct term for the sort of grease or fluid, or in more lay-man's terms a "goo", that will provide <em>increased</em> friction between the shaft and the bushing, rather than reducing friction, as it is with regular oil or grease.</p>
<p>Essentially a tar-like substance, that isn't as messy as actual tar. Perhaps also similar to tree-sap, except more manageable in terms of putting it into a mechanical device. Also tree-sap gets hard over time, I'd like to keep its mechanical properties, like viscosity, etc.</p>
<p>In case you're wondering, this is going to be part of a soft-open gearing, such that a small door doesn't fling open when the lock is released, instead it provides some resistance.</p>
| |mechanical|friction|fluid| | <p>They are called haptic greases. Used on knobs and levers to provide stiction but very smooth movement when actuated.</p>
<p>Here's an explanation from Nye Technologies - <a href="https://www.nyelubricants.com/motion-control" rel="nofollow noreferrer">https://www.nyelubricants.com/motion-control</a></p>
| 26595 | What are fluids or "goos" called, that increase friction? |
2019-03-28T03:05:08.820 | <p>Hello all I am drawing a hole for M3 countersunk screw(flathead) I have checked the specs online the diameter of the head is largest 5.5mm thick 1.65 90degree</p>
<p>But when I draw a 5.5 hole 90degree it's only 1.3 thick as I remember, so I measure the actual product that comes with M3 countersunk it's roughly 6.5mm diameter</p>
<p>So my question is which size should I draw for a M3 countersunk screw? they're on 3mm abs plastic and 2mm aluminum, thank you very much. </p>
| |cnc| | <p>This is what Inventor recommends for a clearance hole:</p>
<p><a href="https://i.stack.imgur.com/4nQPb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4nQPb.png" alt="enter image description here"></a></p>
<p>And for a tapped hole:</p>
<p><a href="https://i.stack.imgur.com/o5OFD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o5OFD.png" alt="enter image description here"></a></p>
| 26597 | Curious of drawing a hole for M3 countersunk screw, 5.5mm or 6.5mm? |
2019-03-28T17:48:51.347 | <p>After introducing a quantization block in the simulink model, the output of the controlled system starts to oscillate around the zero fixpoint.
I am using a lqr controller to regulate the system, which was designed based on the continuous system.
Can I tune the parameters of the lqr design to fix this problem ? Which method is recommended here ?</p>
| |control-engineering| | <p>The typical method is to choose hardware whose quantization is lower than the system's ambient noise.</p>
<p>You can't damp it out -- it's a nonlinear phenomenon, which, when linearized around a switch in the output, has infinite incremental gain. If for some reason you can't just make the quantization small enough, you can add some dither to it -- either add a sawtooth right before the worst quantization step, or implement a delta-sigma modulator (a 1st-order is enough in my experience).</p>
| 26603 | How can I damp oscillating behaviour caused by quantization in a digital control system? |
2019-03-29T02:45:19.180 | <p>I want to secure a box (roughly 20kg in total) that contains electronic equipment to the underside of a flat steel structure. We can weld onto the underside.</p>
<p>The steel structure has a high chance of receiving mechanical impacts from above (for example, large rocks falling onto it), so I'd like to find ways to dampen the mechanical shock and vibration reaching the box.</p>
<p>I have seen shock mounts/isolation mounts before (like moulded rubber mounts), but they seem to be designed for supporting objects that rest on top of the mount -- in other words, the mount is designed to be under compression.</p>
<p>What options do I have if the object is "hanging" from the mount?</p>
| |structural-engineering|structures|vibration|shock| | <p>Use U brackets with the tops of the U welded to the table underside and flatten the bottom of the U to provide the support - then the vibration mounts can be in compression also side supports can be used to minimize the sidways movement.</p>
| 26611 | Shock/Isolation mounting under tension |
2019-03-29T23:22:40.037 | <p>I want to connect this to wires , I know where the negative terminal is on this ( the spring) but I can’t figure out where the positive terminal is, I can’t find and answer online anywhere else either. It’s a blue laser. Please help<img src="https://i.stack.imgur.com/7LcbF.jpg" alt="enter image description here"></p>
| |battery| | <p>The positive terminal is the leg connected to the led diode case, which is in contact with the metallic led holder.</p>
<p>On the board this leg is soldered on the opposite side of the components.</p>
<p>More details here: <a href="https://www.datasheetarchive.com/pdf/download.php?id=3986888527796ddb5524529252bea59f4ce640&type=P&term=Kingbor%2520LED" rel="nofollow noreferrer">https://www.datasheetarchive.com/pdf/download.php?id=3986888527796ddb5524529252bea59f4ce640&type=P&term=Kingbor%2520LED</a></p>
| 26628 | Where is the positive terminal on the laser pointer |
2019-03-30T10:39:17.623 | <p>Please explain how the normal shear stress (shear stress normal to the cuve, C, defining the cross-section of a thin-walled beam) is zero. I have attached a screenshot from a book which says that the outer surfaces of the beam are stress free. This cannot be true except in very special cases. Right?</p>
<p><a href="https://i.stack.imgur.com/o4Mtr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o4Mtr.jpg" alt="Structural Analysis with Applications to Aerospace Structures by Craig"></a></p>
<p>Now consider the following image. Here axis <span class="math-container">$1$</span> is the axial direction of the beam, and axes <span class="math-container">$2$</span> and <span class="math-container">$3$</span> define the cross-section of the beam. For the normal shear stress to be zero in the member <span class="math-container">$A$</span> of the cross-section, the shear stress <span class="math-container">$\sigma_{21}$</span> should be zero on the lateral face <span class="math-container">$1$</span> which is possible only if there is no axial force on the lateral face <span class="math-container">$1$</span>. Am I right?
<a href="https://i.stack.imgur.com/GnYlE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GnYlE.jpg" alt="enter image description here"></a></p>
| |structural-analysis| | <p>We don't know what equation 1.5 says in your book, but the statement "the outer surfaces are stress free" is nonsense, unless the book is only talking about some particular <em>component</em> of the stress.</p>
<p>The stress distribution on the outer surface of any component has to be in equilibrium with the external forces on that surface. That doesn't mean the stress must be zero. The simplest example I can think of is a rod under tension. The axial stress at the surface is the same as everywhere else in the rod.</p>
<p>In the diagram in the book, the stress component <span class="math-container">$\tau_n$</span> will be zero at the surface <em>if there is no pressure load on the surface</em> at that position along the beam. Otherwise, it will not be zero.</p>
<p>The book says it is <em>assumed</em> that "<span class="math-container">$\tau_n$</span> is zero through the wall thickness." That is an <em>assumption,</em> and it is only an approximation. In fact, for a uniform beam there is a parabolic variation of <span class="math-container">$\tau_n$</span> through the thickness. This can be <em>ignored</em> for a slender beam (which is thin relative to its length) because it is small compared with the other stress components, but it is not <em>zero</em>. For a thick beam, this shear stress component <em>is</em> included in the beam formulation.</p>
| 26636 | Why/how normal shear stress is zero in thin-walled beams? |
2019-03-30T15:07:44.373 | <p>"Work (as in Thermodynamics) is said to be done by a system if sole effect on the surroundings could be the raising of a weight." So, how is non mechanical work like : Chemical work, Magnetization work or even electric current flow work be reduced to the same?</p>
| |thermodynamics|chemical-engineering| | <p>Other work is also a flow of energy between system and surroundings (across a boundary). Heat is flow of energy too. Consider these three sayings as analogous:</p>
<ul>
<li><p>Mechanical work is said to be done BY a system if the sole effect on the surroundings could be the raising of a weight (in the surroundings).</p></li>
<li><p>Heat flow is said to be EXOTHERMIC if the sole effect on the surroundings could be an increase in the temperature of the surroundings.</p></li>
<li><p>Other work is said to be done BY a system if the sole effect on the surroundings could be either the raising of a weight (in the surroundings) or an increase in the temperature of the surroundings.</p></li>
</ul>
| 26639 | equivalent deduction of Non-mechanical thermodynamic work |
2019-03-30T20:25:27.773 | <p>Surge protectors are banned on cruise ships, and so are confiscated during boarding. Extension cords and cube taps etc. are discouraged, but not banned.</p>
<p>I'm asking specifically about surge protector incompatibility, and would like an answer that includes the technical details of the ship's cabin power supply.</p>
<hr>
<p>I've asked in various places (not StackExchange) and have never been able to get a technical answer as to how a surge protector can cause damage.</p>
<p>Most responses are along the lines of "<em>they cause fires</em>", "<em>fire is really bad on a ship</em>", and "<em>they ban all extension cords</em>", so please don't give a similar answer.</p>
<p>To discourage "<em>they ban all devices</em>" responses, please look at this ad from Amazon and note the "Non Surge Protection & Ship Approved":</p>
<p><a href="https://i.stack.imgur.com/rdzls.png" rel="noreferrer"><img src="https://i.stack.imgur.com/rdzls.png" alt="enter image description here"></a></p>
<hr>
<p>(Note that I wrote this last night before the answer from @StainlessSteelRat)</p>
<p>Learning that ships use an isolated-neutral system was a great help.</p>
<p>Here is a summary (many details omitted) of where I am so far.
In particular, I run into what seems like dead ends (in bold)
where the bad effects aren't nearly as bad as they at first sound.</p>
<hr>
<p>Wiring Systems:</p>
<ul>
<li><p>Normal house wiring (North American standards) is asymmetric, having one line at 120V,
and the other shorted to ground at the main breaker panel.
The neutral side (larger slot in outlet) is, <em>in theory</em>, safe to insert a fork without shock.
The live side is, in practice, dangerous to insert a fork.</p></li>
<li><p>Ship wiring is symmetric, having two lines, each at 60V and 180 degrees out of phase.
Both sides <em>in theory</em> pass the fork test (similar to birds safely sitting on a distribution wire).</p></li>
</ul>
<p>Surge Protection:</p>
<ul>
<li><p>With normal house wiring, when a surge protector notices a voltage spike between ground and the live wire,
it diverts the excess current.</p></li>
<li><p>a surge protector has a continuous but small current (for the detection), and occasionally a brief but large current (when clamping).</p></li>
</ul>
<p>Surge Protectors on Isolated-neutral systems:</p>
<ul>
<li><p>The connection between one of the live wires and ground causes the system to behave like a grounded system.
The current is normally small, but with many cabins on the same circuit, it might become significant.
This makes the system more dangerous to users, <strong>but no more dangerous than home wiring</strong>.</p></li>
<li><p>The grounding occurs on the side of the circuit that connects to the live socket.
Devices that rely on the difference between the two sides will be compromised.
For instance, the on/off switch is normally on the live side, but would now be on the neutral side, leaving voltage on the device
(e.g. even when switched off, the threaded part of a light-bulb socket would be live).
This makes the system more dangerous to users, <strong>but no more dangerous than home wiring that has live and neutral reversed</strong>.</p>
<ul>
<li>There will be multiple connections to ground in the circuit, which can create ground-loop currents.
This can be annoying (e.g. hum in sound systems), <strong>but doesn't create any danger</strong>.</li>
</ul></li>
</ul>
<p>Speculation:</p>
<ul>
<li><p>If the two wires aren't different colours, or if the installer knows that it doesn't make any difference,
one line could end up connected to the neutral outlet in one cabin and to the live outlet in another.
That could increase the continuous current leakage through ground.
In the case of an actual spike, both sides would be briefly clamped to ground creating a large current and possibly tripping the circuit breaker.
This would be annoying, <strong>but not dangerous</strong>.</p></li>
<li><p>I'm sure that the ship's electrical system has devices to monitor and regulate the power supply.
The current leak to ground caused by surge protectors might cause this mechanism to get false readings.
This could be dangerous to users.</p></li>
<li><p>There is an effect on metals known as <em>stray current corrosion</em>.
Even without any added electricity, random currents occur in metals, and, particularly in the presence of salt,
will cause the metal to corrode at points that act as anodes.
Many marine vessels have magnesium-anode devices attached to their hulls to mitigate this effect.
I imagine this is the reason that ships use isolated-neutral wiring, to avoid introducing any electrical currents into their hulls.
This isn't immediately dangerous, but would be immensely expensive in the long run.</p></li>
</ul>
| |electrical-engineering|power-electronics|marine-engineering| | <p>Residential institutions have a history of fatal fires caused by evacuation delays. That's historically true of ships as well as hotels and hospitals.
Institutions have two options: periodically check all surge protection circuits, or ban all surge protection devices. Hospitals I've worked in ban all power boards and test everything. Building sites I've worked in have test-and-tag requirements. Passenger ships ban "surge protection".</p>
<p>Anyway, regarding the main question:</p>
<p>The USCG document referenced above concludes with the warning that load disconnection devices should disconnect both legs. This would be true for any site that doesn't have the expected line connection at the socket. That's true for badly wired homes, but for technical reasons is also true of ships.</p>
<p>There seems to be some confusion about what "surge protection" means. If the USCG doesn't recognize a difference between the terms "overload protection" and "surge protection", that is clearly an indication that both passengers and crew will fail to recognize a difference.</p>
<p>So we have several problems:</p>
<ul>
<li><p>any power board with overload protection should be excluded, regardless of whether it includes surge protection, because you can't tell by looking at it that it disconnects both legs.</p>
</li>
<li><p>any power board with surge protection connected to or after the overload circuit should be excluded (but should be anyway), because a voltage surge may trigger load disconnection.</p>
</li>
<li><p>any power board with voltage surge protection should be excluded, even if it doesn't include overload protection, because surge protection circuits are designed to protect against transient voltage conditions and then be discarded. This isn't a problem that is "worse" than it is at home, but at home you'll only kill yourself if there is a fire.</p>
</li>
<li><p>there are also fault conditions that cause the the neutral and earth to be way off to one side. This is a continuous-fault condition that will cause surge-protection overheating and may cause power-board overheating, failure and fire. This is a fault you are less likely to see at home unless you are multi-phase and independently-earthed. The USCG document referenced in the other answer seems to suggest that ships (like large institutions and factories) are susceptible to this kind of fault: (<a href="https://www.dco.uscg.mil/Portals/9/DCO%20Documents/5p/CSNCOE/Safety%20Alerts/USCG%20Marine%20Safety%20Alert%2003-13%20Surge%20Protective%20Devices%20Onboard%20Vessels.pdf?ver=2017-08-08-082206-293" rel="nofollow noreferrer">https://www.dco.uscg.mil/Portals/9/DCO%20Documents/5p/CSNCOE/Safety%20Alerts/USCG%20Marine%20Safety%20Alert%2003-13%20Surge%20Protective%20Devices%20Onboard%20Vessels.pdf?ver=2017-08-08-082206-293</a>)</p>
</li>
</ul>
| 26645 | What is it about the cabin electrical supply on cruise ships that makes surge protectors dangerous? |
2019-03-31T23:14:57.093 | <p>I saw these specifications listed for a compressed air dryer (desiccant type). I also wonder, can we extract moisture without make it liquid (dew) first when in it is in a vapor state? How would we do this?</p>
| |thermodynamics| | <p>Start with the phase diagram of water found <a href="https://en.wikipedia.org/wiki/File:Phase_diagram_of_water.svg" rel="nofollow noreferrer">at this link</a>. The solid <-> vapor transformation occurs at pressures below 1 bar. Using a temperature of -40 <span class="math-container">$^o$</span>C puts the pressure around 7 Pa. This is vacuum conditions.</p>
<p>Unlike all five other transformations (e.g. fusion is solid -> liquid), the vapor -> solid transformation has no standard definition. In some cases, it is called solid-state condensation. In others, it is simply called vapor deposition.</p>
<p>The better term for the transition vapor -> solid is the frost point. However, the use of the word dew point is a hypothetical approach. It says that, when you would try to cool the air to saturate it with water, you must cool to that temperature. Yes, in this case, when you would cool an infinitesimal amount just below that temperature, you would create solid ice.</p>
<p>Finally, you can use a desiccant to extract water vapor from air without going through a liquid state. The water vapor adsorbs on to the solid.</p>
| 26656 | What is the meaning of dew point -20 or -40 or -70 deg.C? Since below zero centigrade, water cannot dew (already freeze)? |
2019-04-01T18:45:05.730 | <p>At first, I had a particular goal, but I'm starting to doubt feasibility which has led me here.</p>
<p><strong>Original Goal:</strong> Create a small <em>(3in²)</em> electromagnet with 100% duty cycle capable of holding at least 50lbs for 1-2 hours at a time.</p>
<p>I haven't fully killed this idea off, but I'm starting to think heat may be a significant issue here; so I'm considering a different approach.</p>
<hr>
<p><strong>New Question</strong>: Can I use a permanent magnet with holding force of +50lbs and use an electromagnet to disrupt the holding ability, which would then release it?</p>
<p>The idea here being that I could possibly use a magnet capable of holding hundreds of pounds, but it'd always be "on" so to speak. If my original electromagnet idea is plagued by heat problems; maybe I could leverage a powerful electromagnet at < 5s intervals to separate a metal object from a permanent magnet.</p>
<hr>
<p>Does any of this make sense? Am I speaking jibberish? I'm not sure exactly which direction I can go, but my end goal is that I need to hold ~50lbs for 1-2 hours and then release it... with a magnet that won't become overwhelmingly hot.</p>
| |electrical-engineering|electromagnetism|magnets| | <p>An electropermanent magnet exists in theory, but in practice hard to find.</p>
<p>To meet your goal of a 50 lb load without overheating if attached to some thermally conductive angle iron. It is cheap and easy to find, which may be better than a theoretical answer that has no cheap commercial solution.</p>
<p><strong>Does any of this make sense?</strong> </p>
<p>example of random web electromagnetic solenoid 110lb/50kg @ 8W $21 (1pc)</p>
<p><a href="https://i.stack.imgur.com/3ztKo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3ztKo.jpg" alt="enter image description here"></a>
<a href="https://www.amazon.com/Electric-Lifting-Magnet-Electromagnet-Solenoid/product-reviews/B01GJWLGZU" rel="nofollow noreferrer">YXQ 50Kg 110LB 24V 50mm Electric Lifting Magnet Electromagnet Solenoid</a></p>
<ul>
<li>8Watts Thermal rise depends on heatsink to base and applied voltage Pd=V^2/R</li>
<li>Holding power depends on no air gap to steel surface.</li>
<li>if object is suspended object jarred loose to cause any separation, it falls then a swivel connection is needed.</li>
</ul>
| 26669 | Can you use an electromagnet to break the connection of a permanent magnet? |
2019-04-01T20:52:49.733 | <p>I'm stuck on this question. I got the first part right which is to work out the dimension change in the x direction but I got the y and z directions wrong. I had to submit this online and it doesn't show what you did wrong so I'm really frustrated on why I got this wrong I was hoping someone could correct me. Since I put my answers online I left the negatives in both my y and z values which might be why I got it wrong I just wanted someone to help to see if I have actually done something wrong. </p>
<p><a href="https://i.stack.imgur.com/sy7El.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sy7El.jpg" alt="enter image description here"></a></p>
<p>I forgot to add in my diagram the length of the x which is 200mm sorry. It's in my working though. </p>
| |mechanical-engineering|materials|mathematics| | <p>Did you include the stress applied to the Y direction (your drawing shows Py = 10 kN) this contributes to the strain in the Y direction.</p>
<p>Also If you take the Y direction as "up" (as you have with the load 10 kN) then your Z dimension is out of the page, when you calculate the displacement in those directions it looks like you got the signs mixed up.</p>
<p>Remember the load in the Y direction will cause expansion in Y and contraction in X and Z. You need to add up all these effects in each direction.</p>
| 26670 | Help with determining the dimension change in the xyz direction of a component under a load? |
2019-04-02T08:28:00.073 | <p>The primary purpose of a vehicle (car, locomotive, etc.) horn is to <em>warn</em> other traffic units that something potentially dangerous is about to happen. And so, others must pay attention to the signal.</p>
<p>So, why not to tune the horns to dissonant intervals or chords (for example, several notes semitone or even quarter-tone apart, e.g. C natural, C half-sharp and C sharp sounding simultaneously), which are much more unpleasant to listen and so more "attention-payable"? </p>
<p>Maybe this is not a proper place for the question, but I've unfortunately not found a better one, since is more related to engineering and design (than to, for example, music).</p>
| |design|car| | <p>Tuning car horn pairs to a major third seems to be a tradition, better suited for discussion on the automotive stack exchange than here. But it is definitely not the only option, and it boils down to cost versus benefit. </p>
<p>For example, the horn in my 1965 Cadillac hearse had three trumpets, two of which were tuned to a minor interval, which created dissonance. The sales literature described that horn as being "oddly compelling", and it was, and the extra expense of furnishing a three-trumpet horn was small in comparison to the cost of the car, which was about $12000 in 1965 money- or about twice the price of a base model Caddy at the time.</p>
<p>In comparison, my 1960 VW Bug had a single-tone horn, because it was a cheap car. </p>
| 26678 | Why vehicle horns are tuned to consonant intervals? |
2019-04-02T16:20:44.677 | <p>I'm trying to figure out how to model projectile motion which can be defined using a simple [quadratic] equation which defines projectile parabola:</p>
<p><span class="math-container">$$y = y_0 + v_{0y}t - \dfrac{1}{2} gt^2$$</span></p>
<p>I came across some <a href="http://mlrg.cs.brown.edu/dynamics-slides.pdf" rel="nofollow noreferrer">slides</a> from Brown University which model the motion as follows:</p>
<p><span class="math-container">$$\begin{align}
\text{Matrix form: }& \begin{pmatrix} 0 & 100 & -10\end{pmatrix} \\ \\
\text{state: }& \begin{matrix}y & y' & y'' \\ p & v & a\end{matrix} \\ \\
&\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
0 & 2 & 1 \\
\end{pmatrix}
\end{align}$$</span></p>
<p>But I'm not quite sure how they came up with that transition matrix. Can anyone shed some light how to pick the state variables and how to model the motion?</p>
| |dynamics|systems-engineering|systems-design| | <p>He started from <span class="math-container">$$ y = y_0 + v_0 t + \frac 1 2 a t^2$$</span> but then for some reason best known to himself decided to take <span class="math-container">$\frac 1 2 a$</span> as <span class="math-container">$y''$</span> rather than a, (and ignored the fact that his <span class="math-container">$y''(t)$</span> is no longer equal to <span class="math-container">$d y'(t) / dt$</span>, but hey, Lewis Carroll was also a mathematician, and this guy, like Humpy Dumpty, can make words mean whatever he feels like they ought to mean).</p>
<p>So he gets the approximate equations
<span class="math-container">$$\begin{matrix} y(t) \\ y'(t) \\ y''(t) \end{matrix} \approx
\begin{matrix} y(t-1) + y'(t-1) \\ y'(t-1) + 2y''(y-1) \\ y''(t-1)
\end{matrix}$$</span>
or
<span class="math-container">$$\begin{bmatrix} y(t) & y'(t) & y''(t) \end{bmatrix} \approx
\begin{bmatrix} y(t-1) & y'(t-1) &y''(t-1) \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\\
\end{bmatrix}
$$</span>
He can't seem to make up his mind whether <span class="math-container">$y''$</span> is really a variable, or just the arbitrary constant <span class="math-container">$-10$</span>, (or was that <span class="math-container">$+10$</span>, or <span class="math-container">$\pm 20$</span> - I've lost the will to care...</p>
<p>It would be more obvious what's going on if you replace the 2's with 1's in his equations, and let <span class="math-container">$y''(t)$</span> actually BE the second derivative of <span class="math-container">$y(t)$</span>, not something close.</p>
<p>Do people actually keep their jobs as university lecturers for teaching like this nowadays? It's 50 years since I was an undergrad, so I don't know....</p>
| 26686 | How to model projectile motion using dynamical system matrices |
2019-04-03T05:26:38.783 | <p>Refer to <a href="https://www.concretepipe.org/wp-content/uploads/2014/09/DD_17.pdf" rel="nofollow noreferrer">this document</a>, and particular to Figure 2:</p>
<p><a href="https://i.stack.imgur.com/bvgWf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bvgWf.png" alt="enter image description here"></a></p>
<p>I wonder what is the formula that governs the box culvert partial flow with respect to full flow? In other words, base on what formula the above graph is generated?</p>
<p>The <a href="https://www.cedengineering.com/userfiles/Partially%20Full%20Pipe%20Flow%20Calculations.pdf" rel="nofollow noreferrer">formula for pipe partial flow</a> is as thus:</p>
<p><a href="https://i.stack.imgur.com/eprDU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eprDU.png" alt="enter image description here"></a></p>
<p>So essentially I'm trying to find the analogous formula for pipe culvert.</p>
| |civil-engineering| | <p>To answer my own question:</p>
<p>The formula for hydraulic radius, velocity and flow remains the same. But what is important to note is the wetted perimeter, <span class="math-container">$P$</span>.</p>
<p>When the flow is partial flow, </p>
<p><span class="math-container">$P=2h+b$</span></p>
<p>When the flow is full flow,</p>
<p><span class="math-container">$P=2h+2b$</span>.</p>
<p>(because now the top part of the box is also "wetted").</p>
<p>This explains why the "strange" behavior how when <span class="math-container">$D:RISE$</span> approaches <span class="math-container">$1$</span>, <span class="math-container">$\frac{Q}{Q_f}$</span> can go way beyond <span class="math-container">$1$</span>, and when <span class="math-container">$D:RISE=1$</span>, <span class="math-container">$\frac{Q}{Q_f}$</span> goes back to <span class="math-container">$1$</span>. </p>
| 26701 | The formula for partial flow of Box Culvert |
2019-04-03T09:00:41.503 | <p>I have heard many times that four-wheel drive is less efficient than two-wheel drive. Many car producers have a normal drive mode where only two wheels are powered, and turn the four-wheel drive on only when needed, to "save energy".</p>
<p>I have never understood why it should be so. You need to transfer only half the energy to each wheel (which must also save the material wear), so the total energy consumption must be the same.</p>
<p>Please answer the question also for the case of electric cars with four separate small engines each attached to a single wheel.</p>
| |automotive-engineering|energy-efficiency|car| | <p>As others have pointed out, this isn't really a matter of physics so much as the choice of how to implement the 4WD system. With modern ECU governed AWD systems, the mileage penalty is much less than it used to be with the hydraulically controlled AWD units. A common system was to hang a small hydraulic pump on the front and rear shaft, piped in a loop. If the machine was FrontWD normally, and everything was hooked up, no pressure would develop in the system. If the front axle's tires started to slip, it's pump would pump more fluid than the rear and transfer some torque. If the imbalance was large, the pressure would engage a clutch and transmit power directly though the shafts. These systems had appreciable parasitic losses.
The two pumps can be eliminated now and the entire thing controlled with sensors and an ECU. They can also talk to other ECU's and as a coordinated group comprise a dynamic stability control system.</p>
<p>See this short video on the Honda CRV comparing the pre 2012 system with the post 2012 system.
<a href="https://www.youtube.com/watch?v=d2fnBxjAWso" rel="nofollow noreferrer">Honda CRV AWD systems</a></p>
| 26704 | Is it true that four-wheel drive is less efficient and if so, why? |
2019-04-03T11:44:37.393 | <p>I've always wondered if this is just a marketing thing, but all across the electronic supply spectrum I see "gold plated" listed as the best conductors for various contacts and connectors. However, silver is considered the most conductive element (6.2×107 S/m), followed by copper (5.9×107 S/m), with gold (4.5×107 S/m) being third. I get that any of those are better conductors than the most common (tin and steel) but I would think silver, being less expensive, more abundant, and a better conductor, would be preferred over gold. What am I missing here?</p>
| |electrical-engineering| | <p>Pure silver or pure copper tarnishes and gains a patena over time (oxidisation) whereas gold if pure should never develop such effects. Plating a metal with gold will preserve the metal by preventing oxidisation indefinitely so long as the gold seal remain complete. In simple terms you dont have to clean your gold tipped headphone jack because pure gold doesn't rust.</p>
| 26710 | If gold is a worse electrical conductor than silver and copper, why are gold plated contacts considered "better" by the market? |
2019-04-04T23:00:56.343 | <p>When we use Abaqus to solve the static problem, one step is to set:</p>
<p><strong>Time period, the number of increments and increments size</strong>.</p>
<p>But since the static problem is not time-dependent, there is no time, which means the above time is virtual time. Then I was wondering how we look at those parameters in the static case.</p>
<p>Can we think that way:
The Time period is the real/physical total deformation time for the static problem and is scaled to 1. And its substep which from Time period/number of increments comes from the sub deformation which is also scaled.</p>
| |statics|abaqus| | <p>While the other answers make some good points the only one really addressing the issue is Schneider's response.
The time increments in a static problem in abaqus means that the total load, say for example 10 N, is divided into chunks of say 1 N each (if you have 10 substeps), and the software adds 1 N each time the previous substep has converged. This is done because it is easier for the solver to converge with small loads.</p>
| 26741 | Physical meaning of increment in static problem in abaqus |
2019-04-05T06:27:18.590 | <p>i have Question that Mechatronics Engineering is all about Precision or intelligence , from precision i mean that to follow pattern precisely with precise time and from intelligence i mean making intelligent decisions based on input using image processing etc.
for example i google these Mechatronics keyword and i got these images </p>
<p>a) auto industry robotic arms manufacturing cars : <a href="https://i.stack.imgur.com/5NIJe.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/5NIJe.jpg</a>
so my Question is these auto industry robotic arms just follow precise pattern or they have cameras in them and they assemble or cut parts of cars based on what they see in camera added to them ( image processing ).
if no camera then it mean if car assembly is bit off from its position they could make mistake and could mess up things ?</p>
| |robotics| | <p>Modern robots are precise. given the same inputs they produce the same outputs. A signal turns on and the arm moves to a position. Every time the same signal turns on, the robot arm moves back to the same position, often with in 10's of microns. </p>
<p>The intelligence is to some degree illusionary. A computer program running on the robot controller is simply executing a program responding to various inputs such as single sensors detecting the presence of part or values from a vision camera providing an X and Y coordinate of part center in its field of view. Based on the inputs in accordance with program code it produces outputs, including moving the robot arm to various locations. </p>
<p>For some applications the programming is fairly simplistic, it other cases it maybe vary complicated. Besides the application code there will be code to actually control robot functions such as motion planing, handling emergency stop signals, and controlling and monitoring its own power supplies. </p>
<p>Mechatroincs combines knowledge of mechanical systems such as the robot arms transmission systems, electrical / electronic such as the electrical motors and electronic circuits for things like inputs and outputs, with the controls and software. it is a large skill set to master, but the career field is rewarding.</p>
| 26746 | Mechatronics Engineering is about Precision or intelligence? |
2019-04-05T17:46:22.100 | <p>We can get local coordinate system information (location and rotation angle (in degrees) relative to the global Cartesian coordinate system.) by using <code>*GET, Par, CDSY, N, LOC, X (Y or Z)</code> or <code>*GET, Par, CDSY, N, ANG, XY (YZ or ZX)</code>.</p>
<p>Is there a way to get the element coordinate system information for each element?</p>
| |ansys|ansys-workbench|ansys-apdl| | <p>Just like any other <code>*get</code> function you can use in APDL, you can loop on each element (or on each element of a group) and fill an array:</p>
<pre><code>allsel
*get,nbe,elem,all,count
wanted_SYS= 11
*del,myarray
*dim,myarray,array,nbe,6
*do,i,1,nbe,1
esel,s,,,i
*get,myarray(i,1),cdsy,wanted_SYS,loc,x
*get,myarray(i,2),cdsy,wanted_SYS,loc,y
*get,myarray(i,3),cdsy,wanted_SYS,loc,z
*get,myarray(i,4),cdsy,wanted_SYS,ang,xy
*get,myarray(i,5),cdsy,wanted_SYS,ang,yz
*get,myarray(i,6),cdsy,wanted_SYS,ang,zx
*enddo
</code></pre>
| 26754 | How to get element coordinate system information (location, angle) for each element using ANSYS Mechanical APDL |
2019-04-05T19:27:21.353 | <p><a href="https://i.stack.imgur.com/KFnDw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KFnDw.png" alt="enter image description here"></a></p>
<p>(a)Z=(B+C')A+B(C+D')+BD</p>
<p>Z=AB+AC'+BC+BD'+BD</p>
<p>Z=AB+A'C+B</p>
<p>Z=B+AC'</p>
<p>(b) A B C|Z</p>
<pre><code>0 0 0|0
0 0 1|0
0 1 0|1
0 1 1|1
1 0 0|1
1 0 1|0
1 1 0|1
1 1 1|1
</code></pre>
<p>(c) BC 00 01 11 10</p>
<pre><code> A 0 |0 |1 |1 |1|
1 |1 |0 |1 |1|
</code></pre>
<p><strong>I was able to complete the 1st 3 Q's i.e.(a),(b),(c).But i didn't understand the last 2 Q's i.e. (d),(e)</strong>.Can someone give me a hint or push me in the right direction? Thanks in advance</p>
| |electrical-engineering|electrical| | <p>Question D refers to the fact you can use NAND gates in different combinations to replicate different types of gates. IE to replicate a NOT gate, simply tie the 2 inputs of the NAND gate together, and there you go. This <a href="https://www.instructables.com/id/NOT-AND-OR-Gates-From-NAND-Gates/" rel="nofollow noreferrer">Link</a> details the different connections of NAND gates to replicate your gates.
Question E refers to your schematic you'll have made in question D once you've removed any redundant NAND gates, if there are any. Every gate has a propogation delay, or the time it takes for the signal to go from the input pins the the output pin. This is usually in nanoseconds. As the question states it'll take 20ns to go from the input of a gate to the output, simply add up the gates in series to come to the answer.</p>
| 26757 | Implementing logic expression and the truth table of logic function |
2019-04-05T21:01:36.230 | <p>I have rope on a spool that is on a cart that is powered and moving at a constant velocity of 0.5 m/s. The end of the rope is passed thru a break and fixed to the wall. one break has a compression spring with an unknown force to create friction to maintain a constant tension on the rope. The spool has a friction of .1 and is .5 m in diameter. </p>
<p>I want to calculate the force the spring needs to put on to maintain a constant tension as a result of the movement of the cart but do not know where to start</p>
<p><a href="https://i.stack.imgur.com/6ykC9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ykC9.jpg" alt="enter image description here"></a></p>
| |statics|dynamics|springs| | <p>You have not specified the diameter of the spool nor the friction constant. Neither have you specified the friction constant in the pulleys. I’m guessing you’re trying to find the required spring force in the pulleys so that an appropriate force is subtracted from 38 N such that the remaining force cancels out the rotation friction in the spool.</p>
<p>Edit:
I am unable to comment so I will write here. You want to calculate the required spring force to maintain a constant tension in the rope. However, since the rope is fixed to a wall it cannot move between the pulleys and therefore the tension results purely from radius, friction and moment of inertia (if there is angular acceleration) of the spool. The friction in the spool causes a moment <span class="math-container">$M_{friction}$</span> (opposing uncoiling of the spool). This is balanced by tension in the rope so</p>
<p><span class="math-container">$ M_{total} = M_{friction}+M_{inertia} = Tr \:\: => \:\:T=\frac{M_{total}}{r}$</span></p>
<p>Therefore, if you wish to maintain a constant tension in the rope, you must maintain a constant <span class="math-container">$\frac{M}{r}$</span>.</p>
| 26761 | Dynamic spring tension |
2019-04-08T04:30:26.057 | <p>I've read that desert sand is too round for concrete mixes. But, I'm not building a dam, I'm building a 3 foot landscaping wall. I'm interested in using desert sand for the color, so this is an aesthetic choice. How will using desert sand affect the strength of my mix? Are there ways to counteract the effect? I've been reading everything. Is desert sand ok to use, just uneconomical for large projects?</p>
| |materials|civil-engineering|concrete| | <p>You can use desert sand ( or use the approved concrete coloring admixtures which will give the tint you need) with the following considerations:</p>
<ul>
<li>Wash the sand thoroughly to remove any possible dirt or clay mixed with it, roundness is not a big issue.</li>
<li>Add more cement to compensate for possible reduction in strength.</li>
<li>Add concrete acrylic glue as per manufacturer's recommendation.</li>
<li>Generally the following is recommended proportion of concrete materials: </li>
</ul>
<p>Add 6 bags of 94lbs of cement, water not more than 40% of cement weight, to a cubic yard of sand and gravel. The more water the weaker your concrete. so consider the moisture of your sand as part of water ratio.</p>
<p>The aggregate should be roughly half less than 3/8" and half between 3/8 to 1.25" for your application and lack of slump tests.</p>
<p>Don't forget curing the wall, watering it for at least a week.</p>
<p>One could seek advice from local trades-people as to the ways to make your concrete stronger and as far as plasticity workable. </p>
| 26781 | How will using desert sand instead of concrete sand affect my concrete recipe? |
2019-04-10T01:18:26.863 | <p>This is an off-level rigging problem. When solving for the tension forces in the red slings (lifting from point A), can the assumption be made that the y force vector at points B and C are each half of the weight of the object? Or must one calculate the force vectors at point A? In which case the y vectors are not symmetrical. </p>
<p><a href="https://i.stack.imgur.com/AWCMW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWCMW.jpg" alt="Off-level pick points"></a></p>
| |statics| | <p>No. The issue here is that the problem appears to be overconstrained at first glance. So you can't even assume that there is a physically valid solution at all. How many equations do you have? How many unknowns? Has it been constructed so that two of the equations are linear multiples of each other and one can be dropped?</p>
<p>Trying to take shortcuts like this <em>will</em> bite you in the butt someday.</p>
| 26801 | Can you assume that the forces in the vertical direction are evenly distributed for an off level rigging problem? |
2019-04-10T22:54:01.307 | <p>Why there are different dimensions for the same measurements on this drawing?
For example 0.127mils and 0.097 mils pin to pin. Or highlighted 405 and 385 mils.
I can't find anything on the datasheet indicating the difference?? Is this typical and max? Or typical and min? (some actually say max)</p>
<p>Need to know highlighted distance so I know where to place this part on the PCB. I could just make it 405mils to be safe, but I still want to know why they did this on the drawing?</p>
<p>Complete datasheet link below drawing. Thanks.</p>
<p><a href="https://i.stack.imgur.com/4twdp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4twdp.png" alt="enter image description here"></a>
<a href="https://www.mouser.com/datasheet/2/418/NG_CD_1218434_D1-1234416.pdf" rel="nofollow noreferrer">https://www.mouser.com/datasheet/2/418/NG_CD_1218434_D1-1234416.pdf</a></p>
| |cad|drafting|technical-drawing|drawings| | <p>The link you've provided has a pretty comprehensive set of dimensions. One bit of text that jumped out to me is worded "See sheet 2 for recommended PC board layout."</p>
<p>I would read the reference of 0.127 / 0.097 as a difference of 0.030, which would then translate to 0.112 +/- 0.015 which appears consistent on many of the measurements. A couple appear a bit tighter at 0.020 or +/1 0.010 here and there.</p>
<p>The reference to Max is not so much to determine hole placement as it is to provide you with a dimension that the part will not exceed. If you allocate 0.225 for the shell, you'd have 0.001 clearance for the housing/components/accessories placed adjacent to that portion of the connector. It's not impossible to consider that the 0.224 Max could be measured at 0.220 which would still not cause problems for a design that allocates 0.225 in the housing/case.</p>
<p>Another location on the document specifies that the tolerance is +/- 0.010 unless otherwise specified. That means that the diagram on sheet two which has no tolerance listed individually for the PCB hole layout would be assigned that figure.</p>
<p>The above is a layman's interpretation mixed with a bit of mechanical drawing instruction and a good bit of tradesman's osmotic absorption from a career machinist (60+ years). It could be incorrect.</p>
| 26821 | Help interpreting dimensions from this drawing |
2019-04-11T02:15:44.207 | <p>I am creating a technical drawing of a system I created. Let's call my system, system X. System X is a mounting mechanism for already invented system Y. System Y is customizable, however, system X is a general mechanism that can be fitted to any system Y. </p>
<p>My question is, when labelling my technical drawing, which includes a physical outline of system Y (otherwise the paper would have a lot of blank space), how can I articulate that my general depiction of system Y is replaceable with any customized version of system Y?</p>
<p>I am creating other drawings that detail the mechanisms of X up close without Y, however, in the case of my question, I am trying to create an overview drawing of what system X would look like when applied.</p>
<p>Thank you.</p>
| |technical-drawing|drawings| | <p>If the paper has a lot of blank space without drawing Y, then you can draw X at a bigger scale to fill it up, or use smaller paper!</p>
<p>If Y is physically bigger than X (which seems possible, if X is just a mounting system) you might want to add a reduced scale "sketch" drawing showing how X and Y are assembled.</p>
<p>If Y could be any component with a standard type of mount (e.g. any make of TV with a VESA standard mounting) explain that in a text comment, and just draw a "generic" version of Y in your assembly sketch drawing.</p>
| 26824 | Technical Drawing: Labelling a component that's not part of my design but integral to my system |
2019-04-12T10:19:03.953 | <p>I've just started to study reinforced concrete design and I came across this</p>
<p><a href="https://i.stack.imgur.com/PjxWU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PjxWU.png" alt=""></a> in the "Design of Concrete Structures" by Dr.Nilson.
and I was wondering why do we increase the minimum thickness if we are using stell grade more than 60 ksi instead of decreasing it?</p>
| |deflection| | <p>The principle behind that table is that you will be doing a strength calculation in any case and need the table for a guideline of whether deflections or bending strength will be the design driver. High strength reinforcement does not contribute any more to the stiffness than low strength reinforcement, as the elasticity modulus of steel is the same for reinforcement of different strength grades. (If you used a more accurate approach and actually calculated the deflection, the yield strength of the reinforcement would not change the deflection.)</p>
<p>But if you are using low strength reinforcement you will need relatively more or larger rebars to satisfy the strength requirement, resulting in a stiffer beam or slab, which in turn will make it less likely that deflections become the design driver.</p>
| 26839 | Why do we increase the minimum beam/slab depth if we increase the grade of the reinforcement to control the deflection |
2019-04-13T12:27:40.907 | <p>Of course, most 2-stroke engines require oil to be mixed with the fuel (I appreciate there are 2-stroke engines that don't require this). Therefore, it makes sense to me that oil will be burnt during the combustion process, given it is mixed with the fuel.</p>
<p>However, what doesn't make sense to me is why 4-stroke engines don't burn just as much oil. In both engines, my understanding is that the oil has to be used within the cylinders to keep them from seizing, so surely a 4-stroke engine would also be burning oil too, with the oil just being added at a different stage?</p>
<p>I would like to understand why 2-stroke engines burn more oil than 4-stroke engines?</p>
| |mechanical-engineering|automotive-engineering|mechanical| | <blockquote>
<p>In both engines, my understanding is that the oil has to be used within the cylinders to keep them from seizing, so surely a 4-stroke engine would also be burning oil too, with the oil just being added at a different stage?</p>
</blockquote>
<p>Oil is not added because piston rings separate it from fuel. Rings keep the oil at one side of the piston (the crankcase side) and the burning fuel-air mixture at the piston head side. Ideally, the oil is not supposed to get to the head side at all: no mixing and no burning.</p>
<p>Of course, some is oil is left on the inner cylinder wall in the form of a film, some of it get washed away by fuel and burned by the combusting mix. But this amount is much, much smaller than the amount of oil that has to be mixed into fuel in cheap 2-stroke engines. While the mixed-in oil passes through the crankcase just once yet it still has to provide similar level of lubrication as the oil that's (almost) permanently contained in the oil pan of a 4-stroke engine (or a proper2-stroke engine with a dedicated scavenging blower, as you've mentioned). </p>
<p>Basically: 4-stroke keeps reusing oil and loses only as much oil as slips past piston rings. 2-stroke uses oil once and throws it overboard. That's why so much more oil is necessary when it's mixed into fuel.</p>
| 26858 | Why do 2-stroke engines burn more oil? |
2019-04-13T14:24:12.007 | <p>I need to build a steel frame with a steel web in it, as illustrated below.
Flanges of the frame will be welded, as well as the web, along its whole perimeter.</p>
<p>Would adding a diagonal fold to the web allow this structure to whistand a bigger load? Or, for an equivalent loading, would the fold allow me to use a thinner web?</p>
<p><a href="https://i.stack.imgur.com/lXcHH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lXcHH.jpg" alt="no fold"></a>
<a href="https://i.stack.imgur.com/8Ug8u.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8Ug8u.jpg" alt="fold"></a></p>
<p>Addendum:</p>
<p>Considering the doubly folded strap of the web material welded on the ends to the frame, is there a critical angle for this double fold, where, at some point it is stronger to have no folds at all, rather than two folds. </p>
<p>Intuitively and I may be wrong about this, concentrating matter close to the diagonal tends to make the web behave as a truss. However shear flow must be disrupted as well with two folds of very obtuse angle. See illustration below. </p>
<p>Is this case can it be assumed that one single fold, whatever the angle of the fold, weakens the structure, but two diagonal folds, whatever the angle, makes it stronger?</p>
<p><a href="https://i.stack.imgur.com/jD5LI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jD5LI.jpg" alt="two fold"></a></p>
| |structural-analysis|steel| | <p>There is potential value to adding stiffener plates that are transverse to the plane of the web. For tall welded plate wide flange girders, it is very common to provide a stiffener element in order to reduce the web plate thickness.</p>
<p>As mentioned, the shear wants to flow through the plate analogous to a truss, this same analogy is used by AISC in Chapter G they call it Tension Field Action.</p>
<p>Whether or not this is the most economical solution depends on your actual configuration, loads, and fabricator capabilities. </p>
| 26861 | Influence of a diagonal fold on a web + double fold situation |
2019-04-13T23:43:18.920 | <p>For example, the step response of <span class="math-container">$\frac{s+1}{s+2}$</span> is decreasing although its gain is +1, and this system has higher gain at high frequencies than low frequencies based on its Bode plot. But if I make the transfer function strictly proper, no matter how transient the pole I added is, such as <span class="math-container">$\frac{s+1}{(s+2)(s+1000000000000)}$</span>, the system tends to respond "nicely" like those responses shown in textbooks and professor's examples.</p>
<ol>
<li>Mathematically, <span class="math-container">$\frac{s+1}{s+2}=1-\frac{1}{s+2}$</span>, so its response to a step input is an exponential decay, is there a more physical/intuitive way to explain this? </li>
<li>Why systems with higher gain at high frequencies have decreasing response to step inputs?</li>
<li>Is there a reason that we prefer increasing(nice) step response (or do we prefer it at all)?</li>
</ol>
<p>By nice, I mean something like this</p>
<p><img src="https://i.stack.imgur.com/Ime0l.png" width="400" /></p>
<p>By not nice, I mean something like this</p>
<p><img src="https://i.stack.imgur.com/yuRkz.png" width="400" /></p>
| |mechanical-engineering|electrical-engineering|control-engineering|control-theory| | <p>Normally a lead compensator is used to increase the phase of the open loop transfer function in a certain frequency range, while trying to keep the magnitude close to constant. So one would normally not encounter a lead compensator all by itself.</p>
<p>But to also answer the question regarding the shape of the step response you can use that</p>
<p><span class="math-container">$$
\lim_{s\to0}
\frac{s+1}{s+2} = \frac{1}{2}
$$</span></p>
<p><span class="math-container">$$
\lim_{s\to\infty}
\frac{s+1}{s+2} = 1
$$</span></p>
<p>So low frequencies get halved, while high frequencies pass straight through. The moment the step starts the input has an infinite slope which one could also view as a "high frequency", so passes straight through the lead compensator. But the steady state response to the step would be a "low frequency", so that magnitude would get halved. So initially the step would pass straight through the lead compensator, but as time progresses its magnitude would drop to a half.</p>
| 26870 | Why does a lead-compensator type system decrease to a step response |
2019-04-15T04:59:00.140 | <p>Is it true that computer chips "float" inside of the computers ? Can someone give me references(books, articles, papers...) about that fact ?</p>
| |fluid-mechanics|rf-electronics| | <p>No, they are securely fixed (soldered) to the circuit boards.</p>
<p>If you think you have heard the term "float" with reference to a chip, then you are probably hearing about "floating point arithmetic" which is the process used inside the chip to solve the mathematical equations and calculations we want.</p>
| 26885 | A naïve question about computer chips |
2019-04-15T14:35:34.867 | <p>I am trying to calculate the <span class="math-container">$I_{sp}$</span> of a rocket with a thrust of 500 kN. I know that you can calculate <span class="math-container">$I_{sp}$</span> with <span class="math-container">$I_{sp} = \dfrac{v_e}{g_0}$</span>. However, to calculate <span class="math-container">$v_e$</span>, you need the <span class="math-container">$I_{sp}$</span>: <span class="math-container">$v_e = g_0 \cdot I_{sp}$</span>. How do I calculate both without having to set one value (Other than <span class="math-container">$g_0$</span>)?</p>
| |aerospace-engineering|rocketry| | <p>Easiest: from thrust.</p>
<p><span class="math-container">$T = v_e {dm\over dt}$</span></p>
<p>Both fuel mass flow and thrust are simply, directly measurable quantities - measure the force exerted by the engine on a test stand and fuel flow rate.</p>
<p>That way you're getting the exhaust speed, and as result, the specific impulse.</p>
<p>An alternative is using the rocket equation - <span class="math-container">$\Delta v = {I_{sp} \over g_0} \ln {m_0 \over m_f}$</span> - measure rocket mass change (fuel expenditure) over time, and change in velocity, find the specific impulse.</p>
<p>But since <span class="math-container">$I_{sp}$</span> is directly proportional to <span class="math-container">$v_e$</span>, and nothing else (<span class="math-container">$g_0$</span> is a constant), you can't change one without changing the other.</p>
| 26893 | Is there an easier way to calculate rocket engine Isp? |
2019-04-15T17:49:29.620 | <p>The type of hose involved is the thinnest one on the right in this <a href="http://www.flexiblerubberhoses.com/photo/pl6863074-id_8mm_to_22mm_epdm_air_conditioning_hose_for_conveying_r134a_r12_etc.jpg" rel="nofollow noreferrer">picture</a>. This hose contains an external rubber layer, a textile reinforcement layer, an inner rubber layer and a nylon lining.</p>
<p>The hose has been ziptied to a plastic electrical conduit and it serves as a discharge hose of an automotive air conditioner, so it gets warm during the system operation, and internal pressures are high (the reason for the reinforcement layer).
Also, it's located not very far from the exhaust plumbing.</p>
<p>Can the zip tie and the conduit possibly restrict the hose internally due to thermal expansion as the hose warms up, or would the zip tie expand together with the hose? Can the reinforcement layer prevent the hose from developing a restriction except in the case of a lining delamination?</p>
<p>Here's an exemplificative picture of the arrangement:</p>
<p><a href="https://i.stack.imgur.com/KUiYo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KUiYo.jpg" alt="enter image description here"></a></p>
| |reinforcement|thermal-expansion| | <p>No, the zip tie and conduit won't cause the pipe to develop an internal restriction.</p>
<p>Even squeezing the outer cover will not compress the inner sleeve... There are metal pipe clamps that will compress the inner but they have serious leverage that you won't get from a nylon zip tie.</p>
| 26901 | Can a reinforced hose develop an internal restriction due to thermal expansion? |
2019-04-15T21:36:12.807 | <p><a href="https://i.stack.imgur.com/XSKMh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XSKMh.png" alt="Parshall Flume Outflow"></a>
How do I calculate the discharge from a Parshall flume that flow over a bulkhead (XY) into three partitions (A, B and C) divided 25%/50%/25%. There is no restriction of flow when the pipe gates are open (the incoming volume does not exceed the capacity in any of the partitions). The partitions A & C feed 8" pipe, while the partition B feeds a 10" pipe. When either A or B partitions are closed (using gates A1 and C1), a downstream gate (A2 and C2) can be opened to dump that partition's portion into the 10" pipe (Section C).</p>
<ol>
<li><p>If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is closed, what percentage of the water goes into Pipe A and B? (My guess is A 33% and B 66%)</p></li>
<li><p>If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is open, what percentage of the water goes into Pipe A and B? (My guess is A 25% and B 75%)</p></li>
</ol>
<p>Thanks, Chris</p>
| |fluid-mechanics| | <p>Hydraulics get complicated fast, the beaty of weirs (that's the function of your bulkhead hear) is that you simplify the hydraulics, as you separate the flow upstream of the weir from the flow conditions downstream the weir: As long as water level downstream is a bit below the crest, flow conditions upstream can be calculated backwards from the weir withour regarding downstream conditions. This is why you can build flow distributing structures as per your drawing: Flow into one compartment depends only on weir length (again, only so long as the water level downstream is below the weir crest). You "pay" for this with hydraulic losses.</p>
<p>Your guesses is correct, provided two things:</p>
<ul>
<li>the wall partinioning B & C is higher than the resulting water level in those compartments.</li>
<li>There's enough distance between upstream of the bulkhead for the flow to evenly distribute, or the flow velocity is low (how much distance? How low? I don't know. You want even distribution along the bulkhead). Given the shape of your flume upstream, I'd assume that you are good.</li>
</ul>
<blockquote>
<p>If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is closed, what percentage of the water goes into Pipe A and B? (My guess is A 33% and B 66%)</p>
</blockquote>
<p>Reservoir C will fill up to the water level upstream of the bulkhead, then flow over the bulkhead into the compartments A and B will be proportional to bulkhead lengths. IF the bulkhead between B and C is lower than that level, you will have an additional flow into compartment B, to be calculated by the <a href="https://engineering.stackexchange.com/a/8039/61">weir formula</a>. You need to check the hydraulic losses for the pipes from reservoir A and B and calculate the water levels in A and B at the relevant flow rates.</p>
<blockquote>
<p>If gate C1 is closed (effectively shutting off the water to that pipe), and the downstream gate C2 is open, what percentage of the water goes into Pipe A and B? (My guess is A 25% and B 75%)</p>
</blockquote>
<p>IF the hydraulic losses from the pipes A or B are so high that the water level rises above weir crest, you'd need to account for a lower discharge coefficient and the math gets hariy fast.</p>
| 26906 | How to calculate divided irrigation discharge from a Parshall flume |
2019-04-16T15:44:53.420 | <p>Yesterday, Notre Dame cathedral caught fire. The roof and the central spire were destroyed and some of the stonework was subjected to high temperatures for an extended period.</p>
<p>What are the likely implications for the strength of the individual stones, the walls and other structures?</p>
| |structural-engineering|structural-analysis|fire| | <p>There is a specific temperature range that causes calcination of limestone.</p>
<p><span class="math-container">$$\text{CaCO}_3 + \text{Heat} = \text{CaO} + \text{CO}_2$$</span></p>
<p>The temperature range of reaction is 840°C-900°C and is used for creating cement in a kiln. Adding oxygen and sulphur dioxide causes limestone to become gypsum... which is structurally significantly weaker than limestone from which the old cathedral is built. Above 900°C the carbon dioxide emissions accelerate beyond permissible emissions levels.</p>
<p>Effectively, superheating limestone causes chemical reactions that result in physical changes of the structural material blocks. Exfoliation is a likely result with degradation of surfaces reaction to other chemical compounds in the fire itself. If sulphur compounds were present then there is a risk that the stone may have been changed into another type of rock to varying depths.</p>
<p>Add water to carbon dioxide and it creates carbolic acid which may attack limestone.</p>
<blockquote>
<p>Carbonic acid, hydrochloric acid and acetic acid are some acids that react with limestone, causing it to dissolve. Each of these acids reacts with limestone in different ways. Limestone is made up mainly of calcite that is the chemical compound calcium carbonate.</p>
<p>Carbonic acid, which is a weak acid, forms when rainwater and carbon dioxide in the air react with one another. When carbonic acid comes into contact with limestone or calcium carbonate, it can cause it to dissolve over a long period of time. In nature, this reaction leads to the formation of caves.</p>
</blockquote>
<p>Source: <a href="https://www.reference.com/science/acids-dissolve-limestone-8485a52922d6ebe1" rel="nofollow noreferrer">https://www.reference.com/science/acids-dissolve-limestone-8485a52922d6ebe1</a></p>
| 26913 | What are the possible implications of the heating of the structural stonework of Notre-Dame cathedral during the fire? |
2019-04-16T16:06:20.717 | <p>Is it possible to power an AC motor, using alternating positive and negative DC current? I would assume toggling between each at 60hz would work.</p>
| |power-electronics| | <p>This is possible! The AC motors are designed to run with AC currents so this is not a very good idea. A very popular way to do it is PWM, it basically mimics a sin wave by switching between two states, you need a filter and proper timing to run a single phase AC motor. </p>
<p>Some spacial applications require to run at desired frequencies, so we design a rectifier to gain a DC signal and then turn the rectified signal to an AC signal with desired frequenties(this is not possible without the presence of an AC grid). </p>
<p>A multiphase AC motor needs a rotating field, three sin waves make a very smooth rotating field, however a modulated DC signal can't generate a rotating field as smooth as sin wave no matter how well it mimics the sin wave (think about the harmonics). The results are extra vibrations, arising of recoils between mechanical parts of the motor and introducing harmonics to the AC grid. The motor won't work efficient at all.</p>
<p>I have seen many researches and papers coming out each day, DC drive is yet in development stage. </p>
| 26914 | Simulating AC using DC |
2019-04-17T00:29:55.287 | <p>If two surfaces rub together with a normal force between them, friction is generated. You could add lubrication between those two surfaces and reduce that friction. </p>
<p>Bearings essentially consist of two surfaces rubbing together and, as before, under load, friction will occur. As before, you could lubricate the bearings and reduce that friction.</p>
<p>So absent a better understanding, I can't see how bearings add much of anything to efficiency. This must be wrong or all of industry would be spending money frivolously on unnecessary parts.</p>
<p>So my question is, in what way is my argument wrong?</p>
| |friction|bearings|energy-efficiency| | <p>Your initial assumption is a bit flawed. You can not just draw the conclusion that all friction is the same. It is like saying all birds i know of can fly thus all birds can fly. Which is obviously untrue. Moreover, continuing this similie, you also imply that there is only one kind of flight.</p>
<p>In reality friction is a highly complicated phenomena that includes a lot of the things we don't deem worth thinking about in our day to day life. In fact if you have ever seen the formula:</p>
<p><span class="math-container">$$
F_f= \mu F_n
$$</span></p>
<p>You have seen a physicist saying i have no idea of how it works but seen a simple property. Everything in your question is hidden in the symbol <span class="math-container">$\mu$</span> and the formula is approximate. Also by the way the formula is actually <span class="math-container">$F_f <= \mu F_n$</span> which is even less useful to answer the question asked.</p>
<p>See, the friction coefficient is highly dependent on not just the material pair, but also the quality of surface finish, surfaces capability to carry the load, how the materials react on the interface, temperature, lubrication and geometry. But it gets better, rolling is way different it has much lower friction so a roller bearing has just the fraction of the friction a conventional sliding contact pair. Even better you can have no solid contact at all. Typically this is achieved by lubrication, but there are even bearings that do this naturally like airbearings or magnetic bearings that don't even have touch anything.</p>
<p>It is not always even about the efficiency increase. Rather about controlling the wear, or some other characteristic. See by having specialized pieces you can change them if damaged, you can spread the load so they last longer, and you can get these cheaper than manufacturing them yourself. The bearing can even have a function in aligning the shaft properly and thereby reduce vibration.</p>
| 26925 | Refute my assumption: Bearings Don't Improve Efficiency |
2019-04-17T15:33:40.040 | <p><img src="https://i.stack.imgur.com/hP4rg.png" alt="enter image description here"></p>
<p>Engineers often use Finite Element Analysis software to solve for stresses inside complicated structures to generate pictures like the one above.</p>
<p>I am familiar with FEA as a method to solve differential equations. So what (differential) equation are we using to describe forces/stresses inside objects?</p>
<p>When it comes to analyzing forces, I know only Newton's laws. But how can we use the laws in here, when the internal stresses are dependent on the geometry of the object?</p>
| |structural-analysis| | <p>For structural FEA, there are a few approaches. Most commonly differential equations are developed that relate the deformation state and the stress state of each element in the structure in order to satisfy equilibrium.</p>
<p>A good resource for the details of structural FEA is a book by Ed Wilson Called
Three Dimensional Static and Dynamic Analysis Of Structures.</p>
<p>Here is a chapter that shows the basis of equilibrium and compatibility.</p>
<p><a href="http://www.edwilson.org/Book/02-equi.pdf" rel="nofollow noreferrer">http://www.edwilson.org/Book/02-equi.pdf</a></p>
<p>**To be more descriptive: For linear elasticity problems that use FEA to find internal forces and deformations, most FEA solvers do not directly solve the the partial differential equations. Instead FEA is used to approximate the solutions to the PDEs. I would like to note that using this method, solutions can be found to problems which are nearly impossible (if not impossible) to write closed formed elastic solutions.</p>
| 26937 | What equation are we using to describe forces in FEA? |
2019-04-17T23:23:41.450 | <p>I have a Ford F350 cab & chassis and a gooseneck trailer. For the gooseneck hitch I want to use a steel plate.
The steel plate will be BOLTED to the frame, above the rear axle And the gooseneck ball will be welded to the steel plate.</p>
<p>The distance between the frames is about 34". So the plate Is going to be 34"long x 8" wide.</p>
<p>The trailer is 16,000 lbs GVWR. And a maximum tongue weight of 4,000 lbs (at 25%)</p>
<p>A regular A36 steel plate, 3/8" thick, will bend under load. (A36 mechanical properties: Brinell=112, Tensile=58,000 psi, Yield=36,000 psi)</p>
<p>So I'm thinking to use a 1/8" thick, high strength steel ASTM 514 / AR400F (Tensile Strength=184,000 psi, Yield Point=150,000 psi, Elongation = 23% in 2",Brinell Hardness = 360/444)</p>
<p>Is a 1/8" AR400F steel plate strong enough, for maximum 4,000 lbs?</p>
<p>If not, what do you think about AR400F 3/16" thick ?</p>
<p><a href="https://i.stack.imgur.com/HvqLm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HvqLm.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/MT76b.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MT76b.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/R7mzS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R7mzS.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/J8Wrc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J8Wrc.jpg" alt="enter image description here"></a></p>
| |structural-engineering|materials|steel|strength| | <p>I could see a moment when an overloaded trailer with defective brakes is rushing into the cabin.</p>
<p>You need to consult the trailer's hitch manual, or a professional engineer.</p>
<p>My estimate is you need a minimum 5/8" thick plate with manufacturer recommended geometry and anchored at a the correct distance. </p>
<p>This is a safety critical motion-dynamics problem which needs due deliberation as to the possible forces which may be encountered during acceleration, turning, hitting a pothole, braking, jackknife possiblity, limping due to unbalanced load, and many more issues. It's not just a motorcycle trailer.</p>
| 26951 | Is a 1/8" AR400F steel plate strong enough, for 4,000 lbs? |
2019-04-18T13:17:03.173 | <p>Risers are added reservoirs used to feed liquid metal to solidifying casting. In the case of alloys that shrink during solidification (this is not the case of all alloys), one uses such reservoirs to make liquid metal available throughout solidification. Risers should be kept liquid throughout the solidification, which means the solidification must be directed toward the risers, which must solidify last.</p>
<p><strong>What are the principles used to design such risers?</strong></p>
<p>What I already know:</p>
<ul>
<li>Their volume should be sufficient to compensate the solidification shrinkage</li>
<li>They should remain liquid last, so maximizing the volume/surface ratio seems appropriate</li>
</ul>
<p>What I am looking for:</p>
<ul>
<li>How to ensure the riser can efficiently feed the rest of the cast part?</li>
<li>Are there other rules (some kind of general algorithm maybe) for efficiently designing risers?</li>
</ul>
| |materials|manufacturing-engineering|casting| | <p>I am a metal casting process researcher who has several past and ongoing collaborations with commercial foundry and tooling experts. In commercial steel sand casting, I have observed the terms "riser" and "feeder" to be used more-or-less interchangeably. In the literature, British commonwealth authors tend to prefer feeder, while American authors tend to prefer riser. I will use feeder because it is a more descriptive term. When I use the term "section" I mean a contiguous region of the casting. The term is left vague because it is not a particularly well-defined term in practice. When I use the term "hotspot" I mean the last part of a section or feeder to solidify.</p>
<h2>The Seven Feeding Rules</h2>
<p>John Campbell defines seven rules for feeder design in <em>Complete Casting Handbook</em> (located on p. 661-688 of 2004 edition).</p>
<ol>
<li><strong>Do not feed unless necessary</strong> — only attach feeders to locations that have demonstrated shrinkage problems. Demonstration could take the form of simulation results. Prefer to use thin sections which taper away from the gates so feeding is done by the filling system. Ignoring this requirement increases the cost of production by consuming more liquid metal per casting. It is worth noting that this is not always possible nor the most economical route.</li>
<li><strong>Heat transfer requirement</strong> — The feeder must solidify later than the section it is intended to feed. Influenced by feeder size and shape, and feeder-mold interfacial heat transfer properties. Ignoring this requirement can result in a feeder freezing before the section, leaving the section without feed metal.</li>
<li><strong>Mass transfer requirement</strong> — The feeder must have sufficient <em>useful</em> volume to meet or exceed the contraction of the section it is intended to feed. Useful volume means the amount of liquid available to feed the section that isn't consumed by direct solidification of the feeder itself, or by thermal contraction. Influenced by all of the same factors as (2.). Ignoring this section can result in the feeder freezing before it has supplied sufficient feed metal to the section. The geometry of the section determines which of rules (2.) and (3.) is a stricter requirement. Heat transfer (2.) tends to be more important for thick sections, and mass transfer (3.) for thin sections. The reason is that thin sections will rapidly solidify, well before attached risers, so that useful feeder volume becomes the dominant factor.</li>
<li><strong>Junction requirement</strong> — The point of attachment between feeder and section can become a hotspot. If this occurs, porosity will form at the attachment point. What results is referred to as "under-riser shrinkage porosity". Increasing the size of the feeder, or changing its location relative to the section, can help resolve this issue. A common rule is to use a feeder whose diameter (via Heuvers' circles or spheres) or section modulus (care of Chvorinov, Wlodawer, et al.) is 1.0 to 1.2 times that of the section. For sufficiently thick sections as much as 2.0 times may be necessary to avoid junction issues.</li>
<li><strong>Feed path requirement</strong> — There must be a clear feeding path available from the feeder to every point in the section intended to be fed. If there is a narrow feature between the feeder and some part of the overall section, then that narrow feature will solidify before the section, cutting off feeding. The rest of the section will be isolated. To achieve this, ensure there is a monotonically increasing gradient of modulus or section thickness from most extreme points back to the feeder. Attaching excess metal in the form of feedpads can assist with this requirement. Feeding distance is limited by both feeder geometry and section geometry. Tapering sections can increase the feeding distance. Attempting to feed too far can result in microscopic shrinkage called center-line shrink or microshrink.</li>
<li><strong>Pressure gradient requirement</strong> The feeder must have higher metallostatic pressure than the section intended to be fed, or the porosity will form in the section by the action of gravity. Ensuring the feeder is located at a higher point than the attached section can help, but is not sufficient. Vented feeders exert more pressure because they are not working against atmospheric pressure – atmosphere supports up to 1.5 meters of steel. The metallostatic pressure gradients change over the course of solidification as sections become cut off from one another. All of these changes must be accounted for.</li>
<li><strong>Pressure requirement</strong> The feeder must have sufficient metallostatic pressure to ensure that points in the attached section are above points in the feeder. If the liquid level in the feeder drops below an un-solidified part of the attached section, that part will experience shrinkage porosity. Whereas the pressure gradient requirement considers the entire course of solidification, the pressure requirement is concerned with a single feeder-section system once it has become isolated from the rest of the casting.</li>
</ol>
<h2>Additional considerations</h2>
<p>Other literature sources give additional criteria. Some of these are obvious, some follow from the seven rules above, and a couple are less obvious.</p>
<ul>
<li>Locate feeders near thick sections</li>
<li>Place side feeders on gates</li>
<li>Place top feeders away from gates</li>
<li>Do not cluster feeders (minimum distance)</li>
<li>Blind feeders height:diameter in range of 1:1 and 3:1</li>
<li>Feeders located on flat, accessible surfaces to ease removal and minimize risk of damage to the casting</li>
<li>Select feeder shapes from a list of standards</li>
</ul>
<p>These are sourced from a Master's Thesis by Jensen, 1998, who cited:
- S Guleypoglu and J L Hill AFS Transactions, 1995
- J L Hill, et al., AFS Transactions, 1991
- J L Hill, et al., AFS Transactions, 1993
- B Ravi and M N Srinivasan, Computer-Aided Design 22(1), 1990.</p>
<h2>Numerical models</h2>
<p>A few mathematical and geometric models exist that can guide engineers toward better solutions. These models tend to center around requirements (2.) and (3.), with a little bit into (4.) and (5.).</p>
<p>Two models that are still widely used 70 years after their invention are <a href="https://www.giessereilexikon.com/en/foundry-lexicon/Encyclopedia/show/heuvers-circle-method-4649/?cHash=8b1a894ab6cde93d18d8ce7070c9b2c7" rel="nofollow noreferrer">Heuvers' circles</a> and <a href="https://www.giessereilexikon.com/en/foundry-lexicon/Encyclopedia/show/chvorinovs-rule-4627/?cHash=2d0751fb7828d4549443fd324577ef4f" rel="nofollow noreferrer">Chvorinov's rule</a>. Heuvers' circles are used to ensure that the diameter of the largest inscribed circle or sphere in a feeder is larger than that of the attached section. Chvorinov's rule is based on experimental work that shows solidification time is proportional to <span class="math-container">$\left(\frac{V}{A}\right)^{n}$</span>, where <span class="math-container">$n$</span> is often taken to be <span class="math-container">$2$</span>. If the modulus of a feeder is larger than its attached section, then it should solidify later.</p>
<p>Wlodawer and Heine independently published work extending Chvorinov's rule to individual parallelpiped sections of castings. They each noticed that castings can be broken up into sections with shared boundary segments. The value of <span class="math-container">$A$</span> for each section can then be discounted by shared area since very little heat is transferred across shared boundaries.</p>
<p>Sorelmetal's Ductile Iron book <a href="http://www.sorelmetal.com/en/publi/Gating-risering/Gating-Risering.pdf" rel="nofollow noreferrer">(pdf link)</a> and the Redbook Rules from Steel Founders' Society of America <a href="http://user.engineering.uiowa.edu/~becker/documents.dir/redbook.pdf" rel="nofollow noreferrer">(pdf link)</a> both deal directly with sizing feeders based on geometric considerations. The models are too involved to share here, but are based on practical and empirical considerations from experts in the field.</p>
<h2>What about computer models?</h2>
<p>I am unaware of any commercially available computational models that can accurately guide an engineer through every aspect of feeder design. For what it's worth, <a href="https://www.magmasoft.com" rel="nofollow noreferrer">MAGMASoft</a> is a casting process simulation software capable of both solidification simulation and parametric geometry optimization based on simulation results. So if you have a "close-enough" design, it can optimize the geometry to minimize defects and maximize process efficiency.</p>
<p>Getting that "close-enough" design is the part that still largely relies on human judgement and experiential knowledge. While there are a number of mathematical and geometric models (described above), none of them are perfect or cover all of the above requirements. Furthermore, manual calculations associated with these methods are time-consuming. They can also be error-prone if used by engineers unfamiliar with the effects of geometry on feedpaths. There is at least one open-source program for automatically placing feeder geometries based on (5.) <a href="https://github.com/wwarriner/casting_geometric_toolsuite" rel="nofollow noreferrer">(GitHub link)</a>. Full disclosure: I am the author of that program. The original intent with the software is to divide the casting into feed-able sections using a watershed approach and then determine feeder size based on section geometry. It works reasonably well, and when we have an open source solidification solver it will be that much more accurate.</p>
| 26956 | What are the design principles for risers in sand casting? |
2019-04-19T13:18:19.163 | <p>I am writing my first 3D cad-file and I want to place into it a <a href="https://en.wikipedia.org/wiki/Turnstile" rel="nofollow noreferrer">turnstile mechanism</a> (a mechanism that allows a fixed axis to rotate only one way, and not in the opposite). Are there cad-templates for devices I can find in some online-platform for engineers? I found <a href="https://grabcad.com/library/turnstile-mechanism/details?folder_id=130515" rel="nofollow noreferrer">this cad-file</a>, though it looks too large and complicated, I want to place it in a device that is 20cm*20cm*20cm. As an engineer, where would you look for a template of a device you want to create physically? Or any suggestions how to create my own simple turnstile cad-file?</p>
| |mechanical-engineering|cad|bearings|3d-printing| | <p>I found what I was looking for in this <a href="https://www.youtube.com/watch?v=UIhCPl8eb7s" rel="nofollow noreferrer">video</a> - apparently what I need is a Sprag Clutch or a Roller Clutch. The authour of the video provides his 3d-cad files to a Roller Clutch <a href="https://makersmuse.podia.com/roller-clutch" rel="nofollow noreferrer">here</a>, although they cost all together 4$. Anyways, I purchased them, because it would save me a lot of time and it looks like a decently working clutch.</p>
| 26976 | Cad file for turnstile mechanism/ one-way clutch/ one-way rotation? |
2019-04-22T00:26:31.603 | <p>In the engineering material book. It shows the following figure
<a href="https://i.stack.imgur.com/B5ht0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B5ht0.png" alt="enter image description here"></a></p>
<p>The following is the text to depict the picture
"The origin of rubber elasticity is more difficult to picture than that of a crystal or glass. The long molecules, intertwined like a jar of exceptionally long worms, from entanglements—points where molecules, because of their length and flexibility, become knotted together (Figure 25.6)."</p>
<p>But I still don't know the point and line.</p>
| |structural-engineering|materials| | <p>The <strong>dots</strong> represent <strong>atoms</strong> (as Dibujo de Croquis pointed out, they can be C, H, O, F, Cl, etc.), and the <strong>lines</strong> represent that there's a <strong>chemical bond</strong> between two atoms, as exemplified by this picture: <a href="https://i.stack.imgur.com/X2c90.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X2c90.png" alt="enter image description here"></a>
Your picture can be particularly confusing because:</p>
<ol>
<li><p>It just draws dots for the atoms for the branches of each monomer, and skips the atoms that make the chain itself, this is standard practice (but you may imagine an atom the end of each line segment, usually carbon).</p></li>
<li><p>It decides to suddenly stop drawing those atoms at the branches near the center of the picture. They do that because they want to show how it is tangling in the middle, so in order to get a cleaner picture, they just decided to spot drawing those atoms.</p></li>
<li><p>It includes some dash lines between the atoms in the branches. This is to show that, in spite of not having a strong chemical bond, there's some weak interaction between this atoms.</p></li>
</ol>
| 27011 | What is the meaning of line and points in the schematic of a polymer |
2019-04-22T21:40:13.983 | <p>So I'm given this question with the following diagram:</p>
<p>Two sections of a pipe are joined in a flanged joint as shown in the Figure below. The difference in pressure from interior to exterior is 2 MPa.</p>
<p>The joint is connected with 12 off bolts, equally pitched around a pitch circle diameter of 350 mm. If the maximum allowable tensile stress on the bolts is 300 MPa, what diameter do the bolts need to be?</p>
<p><a href="https://i.stack.imgur.com/lIh8H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lIh8H.png" alt="enter image description here"></a></p>
<p><strong>My working out</strong></p>
<p><a href="https://i.stack.imgur.com/qdYTc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qdYTc.png" alt="enter image description here"></a></p>
<p>First I find the area of the section that the bolt's diameter takes up:</p>
<p><span class="math-container">$A = \pi (0.175+r)^2 - \pi(0.175-r)^2 $</span></p>
<p><span class="math-container">$A = 0.7\pi r\ $</span></p>
<p>Then I find out the force acting on that area of section and divide by twelve to obtain the force for a section that has one bolt.</p>
<p><span class="math-container">$F_{12} = (2\times 10^6)(0.7\pi r) = 14 \times 10^5 \pi r$</span></p>
<p><span class="math-container">$F_1 = 366519.1429r$</span></p>
<p>I then sub this into the stress equations since I know the maximum tensile stress for 1 bolt:</p>
<p><span class="math-container">$A = {F_1 \over \sigma_{max}}$</span></p>
<p><span class="math-container">$\pi r^2 = {366519.1429r \over 300 \times 10^6}$</span></p>
<p><span class="math-container">$Diameter = 0.039 m$</span></p>
<p>But the answer I'm given says it is supposed to be 5.9mm. Is there anything wrong in my thinking?</p>
| |structural-engineering|structural-analysis|stresses|bolting| | <p>Your calculation for the maximum forces to the bolt appears flawed.</p>
<p>The forces resisted by the bolts in tension is equal to the total force of the area on either side of the tube.</p>
<p>You have the tube diameter and thickness, you can calculate the interior area.</p>
<p>Multiple the pressure by this area to find the total force trying to pull these two sides apart.</p>
<p>Divide that by 12 and you have your force in one bolt. Then divide by the allowable bolt stress to get the minimum required area and then diameter.</p>
<p>When I do this I get 5.8mm this is the root area (doesn't include the threads).</p>
| 27037 | How to find the diameter of bolts in a flanged joint? |
2019-04-23T01:57:18.333 | <p>I am trained as a seismic geophysicist but really enjoy getting to translate what I know into what other people can understand and use for their applications, if necessary.</p>
<p>For those with working in and/or retain practical knowledge of Oil/Gas E&P in terms of the engineering side, what is a usable or comprehensive list of terms to know in order to <em>read</em> (and understand) data from my engineer colleagues? For example, I know terms such as Water Cut, GOR, etc. I want to learn more and expand this list.</p>
| |mechanical-engineering|petroleum-engineering|drilling| | <p>There are a number of online reference. I entered "dictionary oil terms" into a search engine & came up with these as an example:</p>
<p><a href="https://www.glossary.oilfield.slb.com/" rel="nofollow noreferrer">The Schlumberger Oilfield Glossary</a></p>
<p><a href="https://cogcc.state.co.us/COGIS_Help/glossary.htm" rel="nofollow noreferrer">Glossary of Oil and Gas Terms - </a></p>
<p><a href="http://bruinep.com/glossary-of-oil-and-natural-gas-terms/" rel="nofollow noreferrer">Bruin Glossary of Oil and Natural Gas Terms</a></p>
<p><a href="https://bfs-usa.com/oil-gas-glossary-of-terms/" rel="nofollow noreferrer">Oil and Gas Glossary of Terms for Production Financing</a></p>
<p><a href="https://www.opisnet.com/resources/glossary-of-terms/" rel="nofollow noreferrer">OPIS Glossary of Terms</a></p>
<p><a href="http://www.oil150.com/about-oil/oil-gas-dictionary/" rel="nofollow noreferrer">Oil & Gas Dictionary of Historical Terminology</a></p>
<p>There are many more references.</p>
| 27041 | Basic Drilling/Production/Reservoir Engineering Terms |
2019-04-25T08:24:35.203 | <p>I have a top level project directory with about 40 Assembly files and several hundred SLDPRT files and they are all highly dependent on each other.</p>
<p>The original version of this works fine and has no errors that we know of. Now I wish to <strong>safely</strong> clone this entire set of design files to form the basis of the next iteration of our design.</p>
<p>(The originals are of course backed up off-site and are entirely safe from any accidental corruption by stray references in the new cloned version.)</p>
<p>The question is, given the complexity of the project and the huge number of cross-references in the design, how can I make a cloned copy that has absolutely no residual links to the old design files, so that I can be 100% confident that the new cloned directory is going to be self-contained and free from broken reference errors?</p>
<p>For a single assembly or part it would be a simple case of using the "Pack and Go" feature of solidworks, but how can I reliably do this for multiple assemblies that are themselves cross-linked? I'm worried that I might make a mistake somewhere and not realise it until a year later by which time it would be almost impossible to fix.</p>
| |solidworks|computer-aided-design|project-management| | <p>You hinted at the answer in your original question - Pack and Go.</p>
<p>If this were my project, I would create a 'Package Master' top level assembly that contains your 40 assembly files as sub-assemblies. No need to mate them to each other, just have them there contained within the file.</p>
<p>Then use the Pack and Go feature that you're familiar with. I always use the Select/Replace tool to remove the previous 'version number', and update with the latest one, so that you never get in trouble with two different parts having the same filename, and either windows or a user getting confused between these.</p>
<p>Make sure you tick the 'Include Drawings', 'Include Suppressed Components' etc. boxes at the top left of the Pack and Go dialogue according to your needs.</p>
<p>You can delete the 'Package Master' from the target directory once everything has gone through, if you don't need it any more.</p>
| 28078 | Solidworks - How to safely clone a large multi-assembly project? |
2019-04-25T15:57:31.020 | <p>Suppose we have a water reservoir at some elevation say z and the flow is frictionless then by Bernoulli's equation we have the exit velocity (2gz)^0.5 . This means that the static pressure in the middle of horizontal pipe would always be atmospheric.
Now what if we split the pipe into two at the exit, would the velocity at each exit be the same (2gz)^0.5 ? If so, then this implies that the pressure in the middle of the same horizontal pipe is negative (gauge). Does this make sense ?</p>
| |fluid-mechanics|fluid| | <p>Yes.</p>
<p>Conservation of mass dictates that the velocity at an exit with double the area of the upstream pipe is 1/2 the velocity in the upstream pipe. For an ideal fluid in an ideal (horizontal) expansion transition, there is no change in entropy. So Bernoulli's equation gives you the relationship for calculating the pipe's static pressure given that the exit is at atmospheric pressure. This pressure will be the same throughout the pipe, and will be less than atmospheric pressure. We are also assuming an ideal bell transition from the reservoir to the pipe.</p>
| 28093 | Frictionless Flow |
2019-04-26T01:28:02.830 | <p>What are the non-practical features of a sterling engine that cause the internal combustion engine to be used instead (in a car)? (e.g. price, size, the power to weight ratio, etc.)</p>
<p>Can you name any that aren't as obvious as those stated above?
Please provide answers that aren't obvious but are still relevant.</p>
| |mechanical-engineering|motors|automotive-engineering|combustion|engines| | <p>I found out this for one of the other features but would still like more.
City driving where there is stop-start activity is not beneficial for the usage of sterling engines as well as the fact that you are not able to link the cylinders of multiple sterling engines to one another like you can with a combustion engine.</p>
| 28100 | What are the non-practical features of a sterling engine that cause the internal combustion engine to be used instead? |
2019-04-27T03:21:54.803 | <p>Say I am trying to create a metal composite and have a mixture of 3 different metal powders each with varying weight %'s. I'm going to compact them into the shape of a ring by applying a force P. The inner and outer radius of the ring are given and I am asked to determine the thickness of this ring after compaction.</p>
<p>My first question is, each of these metals have a true density (ie. 8900 kg/m3 for Al). Looking at this powder alone, if density is reached, can it be further compressed or does adding any more force P do nothing because it physically cannot obtain a greater density (due to atoms themselves etc.). </p>
<p>Second question, assuming you cannot compress more than the true density, is there a formula to determine how much force corresponds to the density achieved?</p>
<p>Thanks in advance</p>
| |materials|metallurgy|composite| | <blockquote>
<p>assuming you cannot compress more than the true density, is there a
formula to determine how much force corresponds to the density
achieved?</p>
</blockquote>
<p>I scanned through some literature and found <a href="http://www.worldcat.org/oclc/961217538" rel="nofollow noreferrer">this book</a> to lead to useful information. Chapter 2 "Bulk Solid Characterization", section 4 "Compressibility", references several sets of equations, each set from a different paper about compressibility of solids:</p>
<h2>Heckel Equation</h2>
<p>See: Heckel, R.W., 1961. An analysis of powder compaction phenomena. Transactions of the Metallurgical Society of AIME 221, 1001e1008</p>
<blockquote>
<p><span class="math-container">$$ln \left( \frac{1}{1-D} \right)=kP+A$$</span></p>
<p>where <span class="math-container">$D$</span> is the relative density of the bulk solid at the applied pressure
<span class="math-container">$P$</span> and is equal to <span class="math-container">$1-\varepsilon$</span>, and <span class="math-container">$k$</span> is a material-related
parameter and is inversely related to the mean yield pressure of the
bulk solid. A larger value of <span class="math-container">$k$</span> indicates a smaller yield pressure
and hence a greater degree of plasticity. <span class="math-container">$A$</span> is a parameter related
to the densification due to die filling (e.g. initial powder packing)
and rearrangement of particles.</p>
</blockquote>
<p>Note: <span class="math-container">$\varepsilon$</span> is void fraction, or "proportion of the total volume not occupied by particles".</p>
<p><a href="https://doi.org/10.1016/S0032-5910(02)00111-0" rel="nofollow noreferrer">P.J. Denny states in Powder Technology 127 (2002)</a> ("Compaction equations: a comparison of the Heckel and Kawakita equations") that:</p>
<blockquote>
<p>The powder metallurgy area tends to use the Heckel equation because
metal all compact by the same mechanism (plastic deformation), and
Heckel did his work using metal powders. The ceramics area tends to
use that first used by Brusch, which plots the relative density (<span class="math-container">$D$</span>)
against the logarithm of applied pressure as shown in the following:</p>
<p><span class="math-container">$$D=\frac{1}{V}=a_2+K_2\ln{P_a}$$</span></p>
</blockquote>
<p>Also, regarding the Heckel Equation:</p>
<p><span class="math-container">$$k=\frac{1}{3\sigma_0}$$</span></p>
<p><span class="math-container">$$A=\ln \left(\frac{1}{\varepsilon} \right)+B$$</span></p>
<blockquote>
<p>where <span class="math-container">$\sigma_0$</span> is the yield strength of the material being studied.
<span class="math-container">$K$</span> is thus inversely related to the ability of the material to deform
plastically. Heckel studied mainly powders and the equation was only
meant to apply to materials that compact by plastic deformation. The
term <span class="math-container">$3\sigma_0 \space (=1/K)$</span> is often called the yield pressure.
Heckel found that, for metal powders, the value of the yield pressure
was increased considerably by the presence of oxide surface layers
(which are especially likely to be significant with ultrafine metal
powders).</p>
</blockquote>
<p><span class="math-container">$B$</span> appears to be another material-specific empirical constant.</p>
<p>Denny also indicates the significant variance in Heckel plots can appear for low pressures especially when the powder consists of an aggregate of a various particle sizes.</p>
<h2>Kawakita Equation</h2>
<p>See: Kawakita, K., Lüdde, K.H., 1971. Some considerations on powder compression equations. Powder Technology 4, 61-68. <a href="https://doi.org/10.1016/0032-5910(71)80001-3" rel="nofollow noreferrer">https://doi.org/10.1016/0032-5910(71)80001-3</a></p>
<blockquote>
<p><span class="math-container">$$C=\frac{V_0 - V}{V_0}=\frac{abP}{1+bP}$$</span></p>
<p>where <span class="math-container">$C$</span> is the degree of volume reduction, <span class="math-container">$V_0$</span> is the initial
volume of the bulk solid, <span class="math-container">$V$</span> is the volume of the bulk solid at
pressure <span class="math-container">$P$</span>, and <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are two material constants with <span class="math-container">$a$</span>
indicating the initial powder porosity before compression (i.e., the
total proportion of reducible volume at maximum pressure), and <span class="math-container">$b$</span>
being a constant related to the yield stress of particles.</p>
<p>[The equation] can be expressed as</p>
<p><span class="math-container">$$\frac{P}{C}=\frac{P}{a}+\frac{1}{ab}$$</span></p>
<p>For given compression data, one can plot <span class="math-container">$P/C$</span> as a function of the
compression pressure, which is often referred to as the Kawakita plot.</p>
</blockquote>
<h2>Adams Equation</h2>
<p>See: Adams, M.J., Mullier, M.A., Seville, J.P.K., 1994. Agglomerate strength measurement using a uniaxial confined compression test. Powder Technology 78, 5e13.</p>
<blockquote>
<p><span class="math-container">$$\ln \left( \frac{\tau_0}{c_0} \right) + c_0 \epsilon + \ln (1 - e^{-c_0 \epsilon}) \space \space \space \space \text{(for large strains)}$$</span></p>
<p>where <span class="math-container">$\tau_0$</span> is the apparent single particle strength and <span class="math-container">$c_0$</span> is a
constant related to the friction between particles.</p>
<p>... the natural strain <span class="math-container">$\epsilon$</span> is defined as</p>
<p><span class="math-container">$$\epsilon=\ln \left( \frac{h_0}{h} \right)$$</span></p>
</blockquote>
<p>where <span class="math-container">$h$</span> is bed height of a load-bearing column.</p>
<h2>Summary</h2>
<p>The increase in pressure required to compress metal powders is roughly proportional to the <span class="math-container">$\ln\left(\frac{1}{\varepsilon}\right)$</span> where <span class="math-container">$\varepsilon$</span> is the void coefficient (percentage of space in the mixture that isn't occupied by metal). Different empirical correlations attempt to model nonlinear deviations for low compaction pressures but most Heckel plots (<span class="math-container">$\ln\frac{1}{1-D}$</span> versus <span class="math-container">$P$</span>) for metal powders eventually become linear for sufficiently high <span class="math-container">$P$</span>. The papers I skimmed show that metals powders do not reach what you call the "true density" (<span class="math-container">$\varepsilon=1-D=0$</span>) for pressures tested. The <a href="https://doi.org/10.1016/S0032-5910(02)00111-0" rel="nofollow noreferrer">Denny paper</a> has several plots for different groups of metals for compaction pressure up to <span class="math-container">$700 \space\text{MPa}$</span>. I would recommend reading it.</p>
| 28117 | What are equations related to the compaction of composite metal powders? |
2019-04-28T20:08:08.697 | <p>I often find these metal cylinders in the streets of London: a silver-colored tube with a round tapering at one end and a hole with no thread at the other.</p>
<p><a href="https://i.stack.imgur.com/J65Ea.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J65Ea.jpg" alt="Top view of the metal cylinder"></a>
<a href="https://i.stack.imgur.com/8KQMx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8KQMx.jpg" alt="Front view of the metal cylinder with hole"></a></p>
<p>What are they?</p>
| |metals| | <p>These containers are <em>nitrous oxide canisters</em> </p>
<p><a href="https://www.talktofrank.com/drug/nitrous-oxide" rel="nofollow noreferrer">https://www.talktofrank.com/drug/nitrous-oxide</a></p>
<p><a href="https://en.wikipedia.org/wiki/Recreational_use_of_nitrous_oxide" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Recreational_use_of_nitrous_oxide</a></p>
| 28136 | What are these metal cylinders often found in the streets of London? |
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