# Piston speed

1. Jun 14, 2005

### 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

2. Jun 14, 2005

### IndustriaL

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

3. Jun 15, 2005

### 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.

#### Attached Files:

• ###### Image1.gif
File size:
2.4 KB
Views:
112
4. Jun 16, 2005

### 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.

Last edited: Jun 16, 2005
5. Jun 16, 2005

### 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.

6. Jun 18, 2005

### 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

Last edited: Jun 18, 2005
7. Jun 18, 2005

### 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.

#### Attached Files:

File size:
4.6 KB
Views:
99
File size:
4.1 KB
Views:
91
• ###### piston4.gif
File size:
4.1 KB
Views:
98
Last edited: Jun 18, 2005