Start by looking at just the mininum information that you need, ignore stuff like "no transmission" and "single cylinder" as that makes no difference. Start with a simple diagram like that below and see how far you can get from there.
How are you going with this one Industrial, do you have a starting point yet?
Let P1 be the attachment point on the circle and P2 be the attachment point on the piston (big-end bearing and gudeon pin respectively).
Take P1 = ( r cos(q), r sin(q) ),
and P2 = ( 0, y).
Since you know the distance between P1 and P2 is constrainted to be the lenght l of the connecting rod then you can easily write an expression for the unknown vertical displacement y.
I recommend using "normalized" variables for y and l by expressing everything as multiples of the radius r. For example if you let l=ar and y=zr then the expression you get in terms of these normalized parameters is,
cos^2(q) + ( z - sin(q) )^2 = a^2.
Just solve this for z as a function of q and you're set up to work out the things you require.
BTW, Note that I've used q to represent the crank angle because I was too lazy to use latex and write the more conventional theta.
If stroke is s and conrod length is L, max acceleration is
where f is the rotation frequency of the crank. So if f is in rpm, and s and L are in feet, this will give you feet per minute squared.
Here are the normalized values that I got for the vertical displacement (z=y/r) and it's first and second derivatives as a function of crank angle (q) and normalized connecting rod length (a=L/r). I think they're correct, though I'm sure that the expression for second derivative could be tidied up a bit. Obviously for constant crank angular velocity (omega) you just need to multiply the first derivative by omega and the second by omega^2 to get the time derivatives (by the chain rule).
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