Mechanical engineering - Cam Laws - Ct Accel.

AI Thread Summary
The discussion focuses on the "fall" laws for displacement, velocity, and acceleration in a cam with constant acceleration, seeking to understand how these laws mirror the "rise" laws. The original poster found information on the rising part but struggled to locate corresponding fall laws, noting a lack of resources online. Participants agree that the fall laws should be the inverse of the rise laws but express uncertainty about how to mathematically represent this mirroring. One user suggests starting a new post to attract more attention to the question. The conversation emphasizes the need for clarity in understanding cam motion dynamics.
sicro
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Homework Statement



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

Homework Equations





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 existence of the constant acceleration motion law.

Thank you! And I am sorry if my English isn't very good.. I am from Eastern Europe.
 
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sicro said:

Homework Statement



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

Homework Equations





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 existence of the constant acceleration motion law.

Thank you! And I am sorry if my English isn't very good.. I am from Eastern Europe.
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!
 
Thank you very much for your reply!:)

Yes, you are right that the fall is opposite to the rise, but I don't 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 don't know what I must do...
 
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!
 
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