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Re:Perfectly centered break of a perfectly aligned pool ball... by rcthompson on Sunday February 02, 2014 @07:51PM (#7168280)
Can anyone explain the "rules" for a 3-way simultaneous colision? Freshman physics only covered collisions between 2 objects, and I've never figured out how to generalize it.
Re:Perfectly centered break of a perfectly aligned pool ball... by waqf on Sunday February 02, 2014 @08:37PM (#7168391)
Right, with 3 balls you can't just do an impulse calculation like you can with 2 (too many unknowns, too few equations).

The author is simulating the collision as an event which happens gradually over an (admittedly short) period of time, using a differential equation, instead of something that happens all at once. That's why he has to talk about the elasticity model he's using, F ∝ (2-d)^1.5, and how the results vary when you change it.

Re:Perfectly centered break of a perfectly aligned pool ball... by tel on Sunday February 02, 2014 @09:08PM (#7168462)
Could you get more equations by symmetry?
Re:Perfectly centered break of a perfectly aligned pool ball... by waqf on Sunday February 02, 2014 @09:53PM (#7168549)
Yes … in the symmetric case. That might give you enough to solve the first one-against-two collision. But the subsequent collisions in the pool break problem won't be symmetric (the velocity of the "incoming" ball is not perpendicular to the axis joining the centers of the other two).
Re:Perfectly centered break of a perfectly aligned pool ball... by rcthompson on Monday February 03, 2014 @02:41AM (#7169352)
Hmm. How does the elasticity model provide the additional constraint needed to give an unique solution?
Re:Perfectly centered break of a perfectly aligned pool ball... by waqf on Monday February 03, 2014 @10:58AM (#7171283)
It provides a way to apportion the impulse between the two simultaneous collisions. It behaves as a differential equation because the instantaneous force in each collision depends (via the elasticity model) on the instantaneous velocities of the balls, which in turn depend on how much momentum has already been transferred.