Mechanical engineering - Cam Laws - Ct Accel.

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Discussion Overview

The discussion revolves around the "fall" laws for displacement, velocity, and acceleration for a cam with constant acceleration, particularly in the context of mechanical engineering. Participants are exploring the relationship between the rising and falling phases of cam motion, seeking to understand how to derive the equations for the falling part based on the known rising laws.

Discussion Character

  • Homework-related
  • Exploratory

Main Points Raised

  • One participant requests the "fall" laws for displacement, velocity, and acceleration for a cam with constant acceleration, noting difficulty in finding this information.
  • Another participant suggests that the fall laws would be the reverse of the rise laws, describing the acceleration and velocity behavior during both phases, but acknowledges uncertainty in how to mathematically mirror the laws.
  • A third participant provides a specific equation for displacement during the rise phase and expresses confusion about how to adapt this equation for the fall phase.
  • A fourth participant advises starting a new post for better visibility, indicating that the current post may not attract responses due to its existing replies.

Areas of Agreement / Disagreement

Participants generally agree that the fall laws should mirror the rise laws, but there is no consensus on how to derive or express these mirrored laws mathematically. The discussion remains unresolved regarding the specific equations for the fall phase.

Contextual Notes

Participants express uncertainty about the mathematical transformations needed to adapt the rise laws for the fall phase, indicating potential limitations in their understanding of the underlying principles.

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.
 
Last edited:
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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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