Multiple Simultaneous Rigid Body Collisions?

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2 comments, last by narfidoo 22 years, 2 months ago
How does one simulate multiple simultaneous collisions of a rigid body? One would think that the mass of the rigid body would be distributed throughout each collision point, but this does not seem to work. Any suggestions/pointers? Thanks, Graham
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They are distributed through out the system

You need to convert the force from each "collision" to a force first. From here you can find the horizontal and vertical forces and the moment/torque that each creates. This will give you the reaction forces you are looking for.


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What you say is basically what I do, although I am using an 2d impulsed-based model.
Say I have a square fall flat onto another infinitely massive square. There will be two collision points. If I resolve both of these collisions as one normally would for a single collision, the resultant velocity has an excessive magnitude.
How do I solve this?
quote:Original post by narfidoo
What you say is basically what I do, although I am using an 2d impulsed-based model.
Say I have a square fall flat onto another infinitely massive square. There will be two collision points. If I resolve both of these collisions as one normally would for a single collision, the resultant velocity has an excessive magnitude.
How do I solve this?


Resolve one then the other. As long as both collisions are correct the result is correct.

E.g. if I drop a nearly horizontal pencil onto a flat surface it might bounce off one end then the other end. If so the first collision greatly reduces it''s velocity and transfer''s some of the linear kinetic energy into angular KE. The second collision generally reduces both the linear and angular KE: it may cause it to speed up as some of the angular KE becomes linear KE, depending on the impact.

As long as you work out each collision correctly the final motion should be correct. In general you will loose some energy at each collision, so the final KE should be less and so the final speed less. As some energy may now be stored as rotational KE the linear KE and so the speed may be much less. The only exception is an object spinning before the collisions, as it has rotational KE which might cause it to rebound faster with extra linear KE.
John BlackburneProgrammer, The Pitbull Syndicate

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