Piston speed

IndustriaL

how do you calculate piston speed with a single cylinder engine no transmission what im trying to do is determine the influence of rod length in that problem

IndustriaL

how do i calculate the piston acceleration for that on both strokes

uart

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.

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uart

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.

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krab

If stroke is s and conrod length is L, max acceleration is
$$(2\pi f)^2\left(\frac{s}{2}+\frac{s^2}{4L}\right)$$
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.

uart

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).

z = sin(q) + sqrt( a^2 - cos^2(q) )

dz/dq = cos(q) + sin(q) cos(q) / sqrt( a^2 - cos^2(q) )

d^2z/dq^2 = -sin(q) + (cos^2(q) - sin^2(q)) / sqrt( a^2 - cos^2(q) ) - sin^2(q) cos^2(q) / sqrt( a^2 - cos^2(q) )^3

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uart

Here are some graphs of those function for parameter a=2, a=3 and a=4 (first, second and third attachments respectively).

In each figure the solid curve is displacement, the dashed curve is velocity and the lightly dashed curve is acceleration.

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