# Homework Help: Mechanical engineering - Cam Laws - Ct Accel.

1. Dec 25, 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.

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.

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...

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

2. Jan 4, 2009

### Dr.D

This can definitely be tricky. What you need to do is to ask yourself what is the role of each of the variables in this expression, and then find comparable expressions in the new situation.

One of those variables was the distance from the beginning of the event to the current point. How will that value be described for the return (falling) curve? Another of those variables is the event length, the total angle turned during the motion. How is that to be expressed for the return (falling) motion?

Once you have identified these new expressions, then try to put together the equivalent functional forms that will describe the motion. If possible, use a computer to evaluate your expression and plot the motion, both rising and falling, and see if things look right. If things look correct for several arbitrary choices of the parameters, then there is a pretty good chance that you have worked it out correctly.

3. Jan 10, 2009

### sicro

Thank you, Ill try to do that...

4. Jan 10, 2009

### nvn

sicro: I might not understand, but I can think of different ways to reverse the equation you posted. See if any of these are what you want for reversed displacement.

(1) Xe = Xemax*[-ki*Xi1^2 + 2*Xi1*(Xemax - Xi) - (Xemax - Xi)^2]/[(1 - ki)*Xi1^2].
(2) Xe = {-Xemax*[-ki*Xi1^2 + 2*Xi1*(Xemax - Xi) - (Xemax - Xi)^2]/[(1 - ki)*Xi1^2]} + 2*Xemax.
(3) Xe = {-Xemax*(-ki*Xi1^2 + 2*Xi1*Xi - Xi^2)/[(1 - ki)*Xi1^2]} + 2*Xemax.

5. Jan 15, 2009

### sicro

Ok, i did it by mirroring the chart on paper. Wasnt that precise, but it worked.

Thank you all for your time and im sorry for the disturbance..

6. Jan 15, 2009

### Dr.D

It is not disturbance, sicro, but you must realize that simply drawing the graph will not get the job done with sufficient accuracy to be useful. You really must learn to manipulate the functions if you are going to master this material.

7. Jan 15, 2009

### sicro

Yes I know, but I didnt have anymore time left.. :(