Hey Heloman.
One suggestion I have is to formulate the problem in terms of arc-length as a function of some time parameter and then differentiate twice to get the acceleration (or once to get the velocity).
so...
I have a rotating cylinder. there is a path plotted out on the cylinder for a cam follower to move on.
I "unrolled" the cylinder about its axis. It's 1 inches in diameter. So now I have a 2D plot of X axis = 1*pi inches wide and Y axis is lets say 4 inches high.
I know what my rotational speed is lets say 2*pi rad per second.
I need to find the instantaneous acceleration and velocity of the cam follower on a plot defined on the 2D (x,y) plot
It's just I have no idea how to "scale" the 2*pi rad/sec to the 2D plot??? How is that done?
Thanks!
I think I figured it out. Went home for the day... I have a proof in mind I will try tomorrow or Monday. But I had cam position data for each degree, figured out time to travel per degree based on RPM, then the distance traveled per degree, simply did Pythagoras theorem for the x dis traveled each degree and the vertical distance per degree, then divided that by the time per degree to get velocity, then again for acceleration... it looks right but won't be sure till I check it. Thanks!