Sign in to follow this  

Question about impulse equation

This topic is 4826 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

I am having trouble understanding the derivation of the equation for impulse between two rigid bodies. I have yet to come across a source that doesn't leave some aspect of the derivation of the formula unexplained. Firstly, I have derived the equation on my own, and I might actually be correct...meaning the final answer is the same, but how we got there, and the layout of the actual final equations, are different. Here is my derivation of the equation for an impulse when the relative velocity along the collision normal and coefficient of restitution are known...it takes into account both linear and angular velocity. For right now, you can just assume one dimension, but otherwise it's the inertia tensor matrix in 3D. Note that DistT means 'perpendicular distance', under the assumption that the magnitude of a torque is the magnitude of the force times the perpendicular distance from the center of mass to the force vector (the projection of the center of mass onto the force vector's direction). Equation I came up with: J = -VRel(e+1) / (1/m1 + 1/m2 + (DistT1 / I1) + (DistT2 / I2) ) Now, the equation given by every other source: J = -VRel(e+1) / (1/m1 + 1/m2 + n * ((r1 x n)/I1) x r1 + n * ((r2 x n)/I2) x r2) So, basically, I want to know if this equation is true: DistT / I = n*((r x n) / I) x r Is there anyway to prove or disprove this based on other pre-existing theorems?

Share this post


Link to post
Share on other sites
Quote:
Original post by shadow12345
So, basically, I want to know if this equation is true:

DistT / I = n*((r x n) / I) x r



Mathematically it cannot be true as I is a second-rank tensor, i.e. a 3x3 matrix. You can calculate DistT/I using as DistT * I-1 but that's also a matrix, and you want a scalar in your equation as e.g. 1/m1 is a scalar.

The quoted equation from 'every other source' is the correct one as it's easy to prove and it works. Yours looks like the 2D version of it, which is valid but limited to 2D. the vector and matrix math is needed in 3D because of the greater degrees of freedom and anti-symmetry of 3D rotational mathematics.

Share this post


Link to post
Share on other sites
In both versions I substituted:
Something / I

instead of using:
Something * I^-1

In 2D they are the same, but in 3D the Something / I means the same thing as Something * I^-1, but doing that is valid and that is not the discrepancy (remember, in this case, both DistT / I and Something * I ^ -1 both end up being scalars, because the equation is for the magnitude of the
impulse along the collision normal anyway).

Could you just show me where this part of the equation comes from:

n*((r x n) / I) x r

Basically, I am making the claim that this equation is true:

DistT / I = n*((r x n) / I) x r

but I have no way to prove it unless I implement both methods and produce exactly the same results (which, I may have to do).

This really isn't an easy equation to derive by the way...it's easy enough to implement, but I'm trying to actually understand it because I might be making a little presentation to my physics professor, and she'd like me to be able to derive everything I've used.

Share this post


Link to post
Share on other sites
Quote:
Original post by shadow12345
In 2D they are the same, but in 3D the Something / I means the same thing as Something * I^-1, but doing that is valid and that is not the discrepancy (remember, in this case, both DistT / I and Something * I ^ -1 both end up being scalars, because the equation is for the magnitude of the
impulse along the collision normal anyway).


I repeat that DistT / I is not a scalar. Or at least not in the mathematics that I am familar with - if you know how to invert a 3x3 matrix and get a scalar you need to explain it, not just claim it to be true.

Quote:
Original post by shadow12345
Could you just show me where this part of the equation comes from:

n*((r x n) / I) x r


The proof is long and I don't have it to hand. It's straightforward as it arises from the definitions and relationships between the quantities and the laws of physics. In the end it's the only equation that will do as it's the only one that satisfies the initial and final conditions in a collision (collsions are all about initial and final conditions, as the outcome is largely determined by conservation of linear and angular momentum + the restitution in the collision).

Quote:
Original post by shadow12345
Basically, I am making the claim that this equation is true:

DistT / I = n*((r x n) / I) x r

but I have no way to prove it unless I implement both methods and produce exactly the same results (which, I may have to do).


I recommend you do this, or at least try. If you try and derive the full 3D collsion response equation you will hopefully find where it differs from yours and so where you've made a mistake, probably by incorrectly assuming you can simplify something when you can't.

Share this post


Link to post
Share on other sites
there is no page 46 but I will check this out. I have other papers by the same man but not this one. Again, all I *really* want to know is where that specific part of the equation came from, I don't want the whole thing derived for me.

EDIT:
and, I have gathered an army of papers on the subject, i.e the mirth (mitch?) thesis (the guy from ucla berkely) and even in that paper it's not exactly described why the equations can work...like, it shows the steps but not in a way that answers the questions I have been asking here.

[Edited by - shadow12345 on October 25, 2004 1:24:58 PM]

Share this post


Link to post
Share on other sites
The 16th page of the PDF, labelled D46: I was looking at the bound set of notes before writing my last post so pulled the page number off them. The PDF is only one of a number of sections of the full course notes so the page numbers don't match. You can grab the other sections of the course from here.

As for a more detailed explaination I don't think one exists. The proof given in the course is complete and fully justified, with many preceding pages of explaination of the concepts and issues and diagrams. The maths is at times quite long-winded but that's unavoidable in this problem as the result itself is very long.

Share this post


Link to post
Share on other sites
it's brian mirtich if i recall correctly :) still reading though

Share this post


Link to post
Share on other sites
I don't mean to bump as it's looked down upon, but I wanted to come back and say that I understand the general form of how this works now, and I wanted to thank you guys for the help. You were right, it isn't really that tough of an equation, I was just getting tied up on a stupid thing.

I also wanted to ask one more question that relates to this topic without having to start a new thread. Note it's not imperative that I understand this, but I don't understand how the mirtich thesis makes this simplification (on page 61 at the top):

Quote:

dU(y) = p(y)/m1 + (I^-1r x p(t)) x r
= (1/m1 - r~ I^-1 r~) p(y)


Firstly, there are two different parameters used to denote the impulse force: the y and then there is t.
p(y) is the impulse force 'when generality is required', and is used for the linear part of the equation, but the angular part of the equation uses t.

Also note that r~ is the 'cross product matrix' of r, and that the two equations are the same...I don't understand how they are the same (some theorem I don't know?) and if I don't ever find out it won't kill me :)

If you happen to know and can answer quickly I'd like to know, otherwise it's cool and I'm good to go

Share this post


Link to post
Share on other sites

This topic is 4826 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this