Impulse based physics: Derivation of Impulse

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14 comments, last by chand81 16 years, 7 months ago
Quote:Original post by chand81
Quote:Original post by Numsgil
wether the collision is dead center or barely glancing, the change in linear momentum is the same.


I'm not sure I can agree on that statement.
Although the equation for change in linear velocity is the same even with angular effects considered, the equation for impulse J changes drastically. It now depends on the collision point. So for a dead center collision J will be large and for a barely glancing collision J will be very small unless there is friction (in which case the direction of J changes). And when J changes, so does the change in velocity/momentum.


Hmm, I see your point. I made that statement remembering my college physics instead of looking at the equation being used. I'm pretty sure I'm right-- the only thing that changes the corrective impulse is the velocity of the contact points. I'm guessing maybe the position terms cancel in some way, but I don't see how. Either that, or I've been wrong for a long time ;)

Also, if the point of contact mattered beyond determining linear velocity caused by rotation (and maybe the collision normal), wouldn't Hecker have a different equation for 3.6? So I'm not sure. And I'm a little too tired for a formal examination.

As far as conservation of angular momentum... This page might be useful. It's talking about basically the same sort of collision response system. The notation is a little dense, but they show how angular momentum is conserved.

I'm afraid this might be a case where I know a subject well enough for my own purposes, but not well enough to teach it :P
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Numsgil,
I can understand.
Thank you for your willingness to help.
Will take a look a the page you have linked.
Quote:They both do effectively the same thing.


That was all I wanted to say as well :-) The problem with the pairwise impulse formulation is that you usually get stuck when you want to deal with several simultaineous collisons/contacts. I find it hard to extend the method. It is indeed the best method to understand the problem. You are right, I should have started with this...

I digged out a nice paper for the derivation of the angular impulse. (No Jacobian and the like). I found it very helpful when I began learning this stuff:

http://www.essentialmath.com/AngularImpulse.pdf


I also recommend their book and you can find a lot of other useful stuff on their side - also on physics. See their page and browse around...

http://www.essentialmath.com


HTH,
-Dirk
Quote:Original post by DonDickieD
I digged out a nice paper for the derivation of the angular impulse. (No Jacobian and the like). I found it very helpful when I began learning this stuff:

http://www.essentialmath.com/AngularImpulse.pdf
-Dirk


This is the same derivation as outlined in Chris Heckers articles.

From the paper
=======================
Since momentum is conserved, we can relate incoming and outgoing linear momentum
by:
=======================

What about the angular momentum? Wouldn't that need to be conserved as well?

If a body has linear as well as angular velocity, its total momentum would be linear momentum + angular momentum, isn't it?
Don't nail me on this, but from the back of my had this goes similar as described above:

a) We search a (non-centric and frictionless) impulse P such that linear and angular momentum are conserved...

(1) v1' = v1 + M1^-1 * P1
(2) w1' = w1 + I1^-1 * L1 = w1 + I1^-1 * (r1 x P1)

(3) v2' = v2 + M2^-1 * P2
(4) w2' = w2 + I2^-1 * L2 = w2 + I2^-1 * (r2 x P2)

b) ... such that the post-velocities satsify some kinematic constraint, e.g. here I choose the relative post-velocity at the contact along the contact normal n to be zero (aka plastic collision)...

(5) ( v2 + w2 x r2 - ( v1 + w2 x r2 ) ) * n = 0

c) ... also knowing the impulse direction (in the direction of contact normal n)
(6) P1 = -j * n
(7) P2 = j * n


Plugging all the equations together should give you the desired result. Basically we have 7 unknowns (v1', w1', v2', w2', P1, P2, j) and 7 linear indepedant equations. This should be a nice exercise to solve.


HTH,
-Dirk
Quote:Original post by DonDickieD
a) We search a (non-centric and frictionless) impulse P such that linear and angular momentum are conserved...

(1) v1' = v1 + M1^-1 * P1
(2) w1' = w1 + I1^-1 * L1 = w1 + I1^-1 * (r1 x P1)


Thats it, I see how it works now!
At collision we add momentum (linear P1 & angular L1) to the colliding bodies in order to keep them from penetrating. By adding equal and opposite momentums to the two bodies, we make sure that we do not change the total momentum (linear as well as angular) of the system.

Thanks Dirk, the L1 (angular impulse) in your 2nd equation made me think correctly :)

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