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Automotive Deriving an Equation for the Stroke of a Crank Slider Mechanism with an offset Crankshaft

  1. Jul 30, 2017 #1
    So I know that the relationship between the crank throw and the stroke of a non-offset CSM is S=2R. But then I realized that when offset is introduced, the stroke length changes just a little. So I spend some time drawing diagrams and then found out that the maximum piston position for any CSM is at the angle formed by the offset @ 0 degrees CA, and minimum is just 180 degrees plus that. For example, a CSM has a Connecting Rod length of 4 and an offset of 1.5. Doing some math, we come up with Angle=Sin^-1(d/L), where L is connecting Rod length and d is offset. Our answer is approximately 22.02 degrees, and thus when the angle of the crank is 22.02 or 202.02 degrees, maximum and minimum displacement occur. Let's say this mechanism has a crank throw (radius) of 2. The equation I came up with so far is S=2R(1/cos(Z)), where S is the stroke length, R is the radius, and Z is the angle formed by the offset. Plugging in our variables, we get: S = approximately 4.315 units, but when I check it by using actual displacement functions, it differs by just a little bit!: S = 4.330. Does anyone know why this occurs?

    Displacement Equation: D=Rcos(X)+sqrt(L^2-(Rsin(X)-d)^2)

    Where X is crank angle.
    Last edited: Jul 30, 2017
  2. jcsd
  3. Jul 31, 2017 #2
    The stroke is simply the distance between the extreme positions. Determining this does not require anything more than some triangle solutions. The max distance from crank to wrist pin happens when the crank and connecting rod are aligned. The min distance occurs when they are again aligned but now overlapping.
  4. Jul 31, 2017 #3
    Yes, but I was thinking in terms using a formula for a spreadsheet.
  5. Jul 31, 2017 #4
    So, what is the difficulty?
  6. Jul 31, 2017 #5
    I was thinking that there must be some sort of relationship between The crank radius and stroke, even with offset
  7. Jul 31, 2017 #6
    Let R = crank radius, L = connecting rod length, and e = offset. Further, let X = wrist pin distance from the crank axis. Then
    (L + R)^2 = e^2 + Xmax^2 Pythagoras at full extension
    (L - R)^2 = e^2 + Xmin^2 Pythagoras at minimum extension
    Solve these for Xmax and Xmin, then take the difference to get
    stroke = S = Xmax - Xmin
    You can do the algebra to roll all this into a single expression.
  8. Aug 1, 2017 #7
    Thanks man!
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