Understanding CSM Displacement with Offset

[Jason Louison]

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.

 

[Dr.D]

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.

 

[Jason Louison]

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.

Yes, but I was thinking in terms using a formula for a spreadsheet.

 

[Dr.D]

So, what is the difficulty?

 

[Jason Louison]

So, what is the difficulty?

I was thinking that there must be some sort of relationship between The crank radius and stroke, even with offset

 

[Dr.D]

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.

 

[Jason Louison]

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.

Thanks man!