# Mechanical engineering - Cam Laws - Ct Accel.

1. Dec 21, 2008

### sicro

1. The problem statement, all variables and given/known data

If you could please tell me the "fall" laws for displacement, velocity and acceleration for a cam with constant acceleration.

2. Relevant equations

3. The attempt at a solution

Could only manage to find the laws for the rising part, in a 1996 book.

Didnt find the laws on google either. Just mentions of the existance of the constant acceleration motion law.

Thank you! And I am sorry if my English isnt very good.. I am from Eastern Europe.

Last edited: Dec 21, 2008
2. Dec 21, 2008

### PhanthomJay

I'm not an expert on cams, so i'll defer to the experts on this. But in my opinion, i would think that the 'fall' part would be just the reverse of the rise part: During the 'rise', the acceleration is constant positive over the first half of the rise, with increasing positive velocity and displacement, and with constant negative acceleration over the 2nd half of the rise, but with decreasing positive velocity, and with positive displacement. During the 'fall', the acceleration is constant negative (downward) over the first half of the fall, with increasing negative (downward) velocity and displacement, and with constant positive (upward) acceleration over the 2nd half of the rise, but with decreasing velocity, and with downward displacement. Just my thoughts; I actually wanted to comment on your English, which, like Mozart's music, is literally perfect!

3. Dec 21, 2008

### sicro

Yes, you are right that the fall is opposite to the rise, but I dont know how to "mirror" the laws.

For ex: Displacement: Xe= (Xemax/((1-ki)*(Xi)1^2))*(-ki*(Xi)1^2+2*(Xi)1*Xi-Xi^2)

Xe=movement[rad], Xemax=max value for Xe [rad], Xi=current angle of rotation [rad], Xi1=total angle of rotation for rise/dwell or fall. (in this case Rise), k1=point where the "bend" of the curve-in the chart- changes- in my case it is Xemax/2;

So, to reverse that, I dont know what I must do...

4. Dec 21, 2008

### PhanthomJay

Sicro: Since I'm not going to be of too much help, you might want to start a new post with the same question, because responders often look for questions with zero replies, and your question in this post might therefore soon get lost in the abyss. Good luck!