# Single piston engine problem - position of piston

1. Nov 19, 2008

### grog

1. The problem statement, all variables and given/known data
Consider the simplified single-piston engine in the figure. The wheel rotates at a constant angular speed of 1 rad/s and the radius of the wheel is 2 m. If the piston is fully extended at time t=0, find the position of the piston at 2s. The figure depicts a wheel with a peg that draws a piston in and out of a bore as the wheel rotates (counter clockwise, with the piston horizontal to the right).

2. Relevant equations
x=A cos (omega*t)
where A is the radius of the wheel

3. The attempt at a solution

cos 2 * 2m = x.
since 2 > pi/2, I add 2meters to the resulting x.

The answer I get is reportedly wrong. what's wrong with my approach?

2. Nov 19, 2008

### Staff: Mentor

It would help to see a diagram, but are you taking into account the angle of the connecting rod? When the pin is not at 0 or PI, you lose some of the stroke distance since the pin is above or below the axis of the cylinder.

3. Nov 19, 2008

### Staff: Mentor

Actually it looks like you are trying to include that factor, but the distance x is not cos(2rad*2m). I don't think that's the angle that you should be taking the cos of...

4. Nov 19, 2008

### grog

after doing some searching, I was able to find another problem with an identical diagram.

http://faculty.ksu.edu.sa/alkurtass/Tut102/Tut102-9.pdf

problem number 7 shows the piston and the wheel. my thinking of the angle being 2 rad stems from the piece that catches on the piston assembly following the path of the wheel. Since the piston is fully extended at time t=0, I believe that puts the initial angle at 0. am I wrong in thinking so?

5. Nov 20, 2008

### Staff: Mentor

At t=0, the pin is at the far right. At 2s into rotation, the wheel is 2 radians CCW, which is most of the way to PI radians which is the farthest left. The x displacement to the left of the piston is most of the way to max displacement possible, but a little less. To calc the displacement x, draw a triangle. The theta that you take the cos of is most definitely not 2 radians. Draw the triangle on that figure, and show us your calculation...

6. Nov 20, 2008

### Staff: Mentor

And actually, I don't think it's a simple cosine. It's a simple angle calculation, though...

7. Nov 20, 2008

### livewire852

Sorry to intrude, but what is it? I have the same problem, with the same omega = 1 rad/s, but my r = 1.3m.
I am trying x = Acos(pi - 2rad) but this is not working. I have drawn a little triangle and concluded that the angle is pi minus omega(t), but no bueno! Any suggestions would be greatly appreciated

8. Nov 20, 2008