How do I calculate the output curve of a crank?

[poe]

I am having a difficult time communicating about concepts without the math. And I have some very specific questions. So here we go with one of them.

Regarding a crank mechanism, say with a throw of 10 centimeters, how can I calculate what percentage of a given constant force that is applied along the vertical axis on the throw of the crank, as it moves from dead top to dead bottom, is available to the crank through this motion?

Thanks in advance.

 

[Dr.D]

Why not use mathematics? A figure would help also as your system description is not entirely clear.

 

[poe]

Why not use mathematics? A figure would help also as your system description is not entirely clear.

Dr. D, thank you for responding. I would love your input. I have read many of your posts and you have a wealth of knowledge with practical experience on the subject.

Regarding why not use math, because I don't know how to. It's been more than 10 years since I took math and I have forgotten the language since.

In the picture, the black arrow represents the force applied on the throw of the crank, which is constant and along the vertical axis of the mechanism as drawn.

Because I'm asking for percentage of force available at crank as opposed to the force applied to the throw, it makes things a bit complicated.

Conceptually, I can see at dead top and dead bottom %0 of the force is available. And when the throw is along the horizontal axis of the mechanism, %100 of the force on the throw is available to the crank. I am trying to ignore the effect the length of the throw causes... like we're talking about a 0 radius crank, which I know doesn't make sense.

And I'm not sure if it would be accurate to calculate this problem with a radius that's almost 0?

 

[Dr.D]

I don't think that there is anyway to answer your question without mathematics, so let's go there.

Take the figure you drew and add a radial line representing the crank throw. Define an angle A between the upward vertical at the crank pivot and the crank throw. The add a separate view showing just the crank throw and the forces acting on it. This is called a free body diagram (FBD).

Suppose that your crank throw is a diagonal line, upward and to the right. The forces shown should include:
1) the vertical force applied at the upper right end, say F1;
2) the vertical reaction at the lower end, say F2;
3) the moment required for equilibrium, acting counter clockwise at the lower end, say M.
Then summing the vertical forces (and assuming that the crank is in equilibrium), we get
F2 - F1 =0
which leads to F2 = F1.

Now, continuing with the assumption of equilibrium, form the sum of moments about the lower end:
M - F1*R*sin(A) = 0
from which
M = F1*R*sin(A)

The sine function is a trig function available on most calculators today. Since F1 and R are constants, this says that M varies sinusoidally with the crank angle A.

I hope that this helps. If not, please write again.

 

[poe]

Thank you very much!

First I have to fully understand your response, and then I will be back.