Talk:Coefficient of restitution

Some correction:
I changed this...


 * $$C_R = \frac{V_{2f} - V_{1f}}{V_{1} - V_{2}}$$

for two colliding objects, where


 * $$V_{1f}$$ is the scalar final velocity of the first object after impact
 * $$V_{2f}$$ is the scalar final velocity of the second object after impact
 * $$V_{1}$$ is the scalar initial velocity of the first object before impact
 * $$V_{2}$$ is the scalar initial velocity of the second object before impact
 * $$V_(2)$$ is the scalar initial velocity of the second object before impact — Preceding unsigned comment added by 121.223.78.45 (talk) 13:48, 11 November 2012 (UTC)

to...


 * $$C_R = modulus( \frac{V_{2f} - V_{1f}}{V_{1} - V_{2}})$$

for two colliding objects, where


 * $$V_{1f}$$ is the vector final velocity of the first object after impact
 * $$V_{2f}$$ is the vector final velocity of the second object after impact
 * $$V_{1}$$ is the vector initial velocity of the first object before impact
 * $$V_{2}$$ is the vector initial velocity of the second object before impact

Reason: e.g. imagine after impact one object moves at 10m/s in positive direction and other object moves at 5m/s in oppsite (negative) direction. One would want the vector velocity difference of 15m/s rather than the scalar velocity difference of 5m/s. Since the coefficient of restitution is required to be between 0 and 1 one would want to take the modulus of the fraction incase of a negative result. —Preceding unsigned comment added by Mehtajigar (talk • contribs) 12:36, 15 May 2009 (UTC)

e (COR) is not a constant
I added some text on this subject based on experiments performed at Florida State University in the summer of 1955. What these experiments showed was that, for a variety of metals, e appeared to approach 1.0 at impact speeds approaching 0. As the impact speed increased the value of e dropped, then recovered to a high level, then dropped again to a steady state measure.

The paper is at www.burgy.50megs.com and also at

User John Burgeson —Preceding unsigned comment added by Burgeson (talk • contribs) 19:41, 15 January 2009 (UTC)

Not a property of objects
There is something very important in this article, and in everyone's understanding that needs to be cleared up. The coefficient of restitution is not a property of an object; it is not defined for any one object. Rather, it pertains only to interactions (collisions) between two objects, just as the coefficient of friction is only defined for two surfaces that interface. Hence the statement "A golf ball has a COR of 0.78" is incorrect and nonsensical. 141.213.247.14 06:45, 24 October 2007 (UTC)


 * The article does say this. Quote: "An important point: the COR is a property of a collision, not necessarily an object. For example, if you had 5 different types of objects colliding, you would have $${5 \choose 2} = 10$$ different CORs (ignoring the possible ways and orientations in which the objects collide), one for each possible collision between any two object types."  --Numsgil 16:04, 24 October 2007 (UTC)

Older Discussions
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 height¹. --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)


 * It seems to have been fixed...(Cabin Tom 23:43, 22 April 2006 (UTC))

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)

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)

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)

Would a tennis ball sitting on a basketball and dropped result in an apparent cor of >1 for the tennis ball? I don't think if it did it would contravene any laws of physics - the tennis ball would get it's extra energy from the basket ball which would, as a consequence not bounce so high. It's a bit like Newton's cradle with 2 balls hitting 1 ball - the one ball has a greater velocity after impact and travels further than its incident path. Any thoughts? Anyone? Thanks ColColi.white (talk) 17:44, 15 March 2008 (UTC)


 * The COR is defined as a velocity relative to what the thing hit. However, in this case, it's actually a 3 body system (floor, basketball, tennisball).  You're right, that it's like a Newton's cradle, but I don't think you'd ever see a COR > 1.  --Numsgil (talk) 04:16, 4 April 2008 (UTC)

A-Hs. I tried it and, by golly, you are wrong! I put a table-tennis ball on top of a superball and dropped then together down a cardboard tube about 30cm long. The table tennis ball shoots out of the tube like a mortar! COR between 2 and 3. I acquired a glass tube so I could do the expmt in front of class and it would look better than the old poster tube. Amazingly that did not work at all - the friction of the superball on the glass tube just caused it to wobble its way slowly to the bottom of the tube. —Preceding unsigned comment added by 91.108.120.119 (talk) 23:33, 18 May 2008 (UTC)


 * No, the CORs are still less than one. Remember that you are dropping both balls.  Meaning that you cannot treat the basketball as part of the table (the two never (well, that's an oversimplification.  The two never have the same instantaneous velocity unless the basketball is at its peak compression, and thus has stored potential energy waiting to be released)  have the same instantaneous velocity for one thing).  The basketball hits the table first, and the table causes it to reverse direction.  Then the basketball/superball collision occurs, and a great deal of energy from the basketball gets transferred to the superball.  Well, that's a simplification too, the collisions actually occur at the same time, so again it's very similar to newton's balls.  But the point here is that a COR must be less than one unless you are generating energy.  Dropping balls does not generate energy.  Dropping landmines on the other hand...  Hmm, I think that would be a more interesting classroom demonstration. :)  --Numsgil (talk) 03:02, 19 May 2008 (UTC)

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)

"COR" Redirect
Currently, "COR" redirects to Heart. Perhaps a redirect here or disambiguation page should be considered. --Half Cat 18:10, 2 April 2007 (UTC)

Could the COR be abbreviated as "e"?
I have seen that McGraw Hill's Sci-Tech Dictionary claims that e can be used for the restitution coefficient. Is this correct? I apologize for linking to answers.com, but it seemed most convenient at the time. —Preceding unsigned comment added by 69.124.113.50 (talk) 01:58, 10 January 2008 (UTC)


 * fwiw, this notation is employed in STEP. Mr Dactyl (talk) 15:05, 3 December 2008 (UTC)

"NEL": redirect?
Shouldn't things like "Newton's Experimental Law" and "Newton's Law of Impact" redirect to this page? PT (talk) 21:58, 6 December 2008 (UTC)

..Uhm absolutely albeit a very late response! Our French colleagues ascribe the work to a 1687 idea of Newton in the opening sentence. That is also how I learnt it at school. I intend to put a redirect in here from Newton's coefficient of restitution. Please feel free to politely discuss/object as per wp:BRD JRPG (talk) 17:10, 12 March 2013 (UTC)

Velocity shouldnt be scalar
Velocities —Preceding unsigned comment added by 173.2.20.96 (talk) 01:09, 15 December 2008 (UTC)


 * Velocities are in fact scalar along the direction of collision. If two objects are colliding, there's a normal vector for the collision.  Take the velocities of either body at the point of collision and dot product it with this collision normal.  That's the scalar velocity term to use in the equations.  --Numsgil (talk) 23:47, 17 May 2009 (UTC)


 * What does "direction of collision" even mean? In the center-of-mass frame, there is a still a vector relative velocity for each particle, and it can change direction before and after the collision. Yes, you can always define a scalar from a vector, but that's not always relevant. What's wrong with "relative speed"? — Preceding unsigned comment added by 2.249.139.10 (talk) 20:35, 14 April 2018 (UTC)

Velocity & Speed
A major defect with this article, and one which, for me, makes it unintelligible is that the terms 'velocity' and 'speed' are used interchangebly, COR being defined in terms of speed and then in terms of velocity in different parts. There is no consistency here. In English, 'velocity' is always a vector, and 'speed' is the corresponding scalar quantity; 'scalar velocity' is an oxymoron, which might reduce to 'speed' or 'the magnitude of a velocity'. If this is what is intended the term 'speed' should be used. 'Velocities are in fact scalar along the direction of collision' is absurd for several reasons; not only are velocities vectors, and therefore not scalar, but what possible meaning can be given to '...along the direction of collision' ? I feel that the entire article should be revised by someone who is capable of using appropriate language. I have abandoned the attempt to make sense of it, and have reached for a textbook written by a competent author. Wikki fails again. Dr Andrew Smith. — Preceding unsigned comment added by 82.32.54.67 (talk) 07:35, 26 October 2012 (UTC)


 * Don't like it? Fix it.  --38.110.159.138 (talk) 17:51, 28 February 2013 (UTC)


 * I'm with the good doctor. Sometimes you find yourself thinking: Life's too short to mess with this . . . 86.183.83.105 (talk) 04:39, 4 October 2013 (UTC)

height equation question
For the equation $$C_R = \sqrt{\frac{h}{H}}$$, is the height measured from the lowest point (i.e. where the object collides with the floor), the midpoint (i.e. center of mass for most sport balls) or highest point of the bouncing object? Should a note of this also be made below the equation? 121.222.251.141 (talk) 05:28, 16 October 2011 (UTC)


 * Assuming the ball is not rotating, and we drop it perfectly, we should be able to pick any point on the surface or inside the ball and measure the displacement of the point under gravity between the moment of dropping to the moment of collision with the ground. This displacement is the drop height. At the exact moment it bounces, we could pick a new point in the ball (or not, and simply stick with our old one) and measure the displacement between the moment of bouncing and the moment the ball reaches the apex (net v=0.) This is the bounce height.
 * TL;DR it shouldn't matter, just pick a consistent point inside the ball to measure the displacement of between critical timepoints. Measuring displacement from the top of the ball, middle of the ball, and bottom of the ball should all give the same answer here (again, assuming the ball is not rotating, and is approximately a rigid body.) 2603:6011:3703:141:AC61:A2BC:BBD1:6A86 (talk) 18:36, 11 December 2023 (UTC)

Expert Needed
I feel an expert is needed here to clean this article up. I found contradictions, highlighted by the Talk Section "Not a property of objects", and in the introduction "The coefficient of restitution (COR) of two colliding objects is a positive real number between 0.0 and 1.0" followed by "A COR greater than one is theoretically possible."

I've corrected those, but if I can find two within the first read of the article, there's probably a heap of them waiting to be found. It probably needs a rewrite.

Richard n 14:14, 12 June 2014 (UTC)

done
I've now given this a clean-up as required and removed the expert-needed flag. I added the Newton's Experimental Law redirect, changed the symbol to "e" (which is common), and added some more introductory context, details of the values of e, and a simple example of how it can be used.

For reference, the derivation of "speed of separation = e x speed of approach" from conservation of momentum and energy is given here: http://www.solitaryroad.com/c1013.html RichardNeill (talk) 19:28, 6 July 2015 (UTC)

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