Talk:Coefficient of restitution
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The article states that a cannonball has a low Coefficient of Restitution. I'm not so sure about that. Certainly a ball bearing has a very high Coefficient of Restitution as demonstrated when one bounces a ball bearing on a piece of concrete. If one dropped a cannonball on a suitably non-deforming surface, might it bounce quite high? --Dumbo1 16:28, 13 January 2006 (UTC)
- I suppose it depends on the surface. That confuses things a lot and I can't be positive that the article is right. Run! 18:04, 13 January 2006 (UTC)
- If the surface deforms, then the height of the bounce will depend to some degree on the coefficient of restitution of the surface, as well as that of the object bouncing. I think the article is wrong in this respect, specifically that of the example of the cannonball. At a guess I think that the coefficient of a cannonball is greater not less than that of a rubberball. --Dumbo1 00:59, 2 February 2006 (UTC)
- the coefficient of restituion all depends on the to materials interacting. For example, a ball bearing bouncing on solid steel has a high coefficient of resitution, but that same ball bearing boucing on, say, vinyl flooring will have a much lower coefficient of restituion. (Cabin Tom 03:29, 20 April 2006 (UTC))
I've basically entirely rewritten the page, expanding it from a stub. I couldn't find a reference to the equations I listed in "Usage" for the general collision case. They are modified slightly from equations listed in Tricks of the Game Programming Gurus 2nd Edition by Andre Lamothe and the equations listed on the elastic collision page. --Numsgil 02:23, 11 February 2006 (UTC)
I've added an equation for the coefficient that uses heightยน. --Lewk_of_Serthic contrib talk 02:05, 24 February 2006 (UTC)
Someone needs to fix something here, specifically having to do with the "Failed to parse (Can't write to or create math output directory)" messages that are showing instead of the equations. I'd do it if I knew how... (Cabin Tom 03:31, 20 April 2006 (UTC))
- What "failed to parse" messages? --Numsgil 06:22, 20 April 2006 (UTC)
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- It seems to have been fixed...(Cabin Tom 23:43, 22 April 2006 (UTC))
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[edit] Equation Sign
Seems to be disagreement both on the page and on the web about what the signs of the equations need to be.
As I understand it COR is, for any realistic collision, positive and between 0 and 1. Maybe an sbsolute value would take care of this? --Numsgil 02:18, 21 April 2006 (UTC)
I've fixed the sign of the COR to be correct. I know it's correct, because it is used to derive the equations in the "use" section. The other definition, with the sign flipped, wouldn't give an equation that agreed with the equations on the elastic collision page when the COR = 1. --Numsgil 10:37, 29 October 2006 (UTC)
[edit] Landmine Example
I wonder if that is a good example. After all the collision is not generating kietic energy, rather the chemical potential of the explosive is being converted to heat, which in turn give the objects kinetic energy. Any opinions? 147.10.17.211 13:20, 31 May 2006 (UTC)
- I agree it's not a perfect example, but there really aren't any natural phenomenon I can think of which exhibit CORs outside the range [0,1], so you have to be somewhat inventive with your metaphors.
- It does get the intended point across, despite its faults. --Numsgil 02:34, 1 June 2006 (UTC)
[edit] Combining two COR's
Hi, this is a question bearing on the discussion. I am actually doing a computer simulation of collisions. If two balls with COR's c1 and c2 collide, what COR should be used? It would depend on the relative masses I guess. Also, if one has COR 0, seems that the final COR should be 0, even if the other has COR=1. A steel ball will stick to a putty ball, no matter what their relative masses.
- Alberto
Yes information about an equation that deals with two cor's would be greatly helpful
- I've written a small caveat on the page: CORs are for collisions, not objects. That said, in any sort of simulation it's not going to be practical to pick CORs for every object. I have no idea if this is correct or not, but a good half way solution would be to use the minimum of the two CORs of the colliding objects as the COR for the collision. That should take care of the case Alberto pointed out where putty and a steel ball collide. Again, I'm not sure if this is physically accurate or not. --Numsgil 10:41, 29 October 2006 (UTC)
[edit] What the hell?
How in the world are the equations for the final velocities derived?
- You mean in the "use" section? I had to derive them myself, though they're very similar to equations given by Andre Lamonthe in one of his intro to game programming books. It's been a while, I don't have my original notes for how they're derived, but I could reconstruct the method of derivation if you're interested/skeptical --Numsgil 02:50, 29 October 2006 (UTC)
- edit: Here's what I said at the top of the page when I rewrote the page: "I've basically entirely rewritten the page, expanding it from a stub. I couldn't find a reference to the equations I listed in "Usage" for the general collision case. They are modified slightly from equations listed in Tricks of the Game Programming Gurus 2nd Edition by Andre Lamothe and the equations listed on the elastic collision page. --Numsgil 02:23, 11 February 2006 (UTC)". --Numsgil 02:51, 29 October 2006 (UTC)
Yes, those ones. I'm writing up a lab report for physics where I would like to derive them, but I have been trying for some time with no success! If you could reconstruct the derivation, that would help me out a lot -- plus it would be nice to have it included in this article. Thanks!
- Posted the way to find the derivation, I've left the solving of the system of equations as an excercise for the reader. Should be managable for anyone with any sort of algebra background. --Numsgil 10:42, 29 October 2006 (UTC)
Hey Numsgil great work on the page. I was wondering how you got the the second equation of the derivation? If you could just paste the derivativation here that would be greatly appreciated.
- Thanks. The first equation is the conservation of momentum that holds true for all collisions. The second is the definition of the coefficient of restitution. Both equations are then manipulated to put it into a form that linear algebra is good at, specifically the "standard form" aX + bY = c. For the purposes of the derivation, you can treat c, v1 initial and v2 initial as constanst and v1final and v2final as your unknowns, since you know c, v1 initial and v2 initial before the collision occurs. Tell me if I've been obtuse at any point in my reasoning, I'm trying to bridge the gap between pure theoretical and the practical level of algebra most computer scientists (say, game programmers) are comfortable with. --Numsgil 22:23, 2 November 2006 (UTC)
