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If you look at the cushions (bumpers) on a pool table, you'll see that they're not vertical. They're tilted inwards. About 10 years ago, I came across a physics exam in which one of the problems stated a set of physical assumptions and then asked the examinee to determine the optimal angle so that if a ball was rolling without slipping when it hit the cushion, it would rebound such that it was again rolling without slipping. The implication seemed to be that this was the angle used on actual pool tables.

I didn't have access to the examiner's own solution, and when I analyzed the problem myself, I found that I couldn't reproduce the actual angles found on pool tables. There seems to have been some speculation that somewhere back in the mists of time, somebody set the angle empirically by maximizing the rebound distance for balls incident along the normal, and that this angle was well explained by this type of calculation. In fact, high-speed video (Mathaven 2009) shows that in snooker, a ball that hits a cushion along the normal doesn't rebound by rolling without slipping; it slips for about 0.1 s before the torque due to kinetic friction brings it into the non-slipping state. (I don't think the physical situation is vastly different between pool and snooker.)

Some data. Let $r$ be the radius of a pool ball, and let the angle of the cushion be such that the point of contact between the ball and the cushion is above the center of the ball by a height $b$. Real pool tables have $b/r\approx 0.26$. In snooker, the coefficient of kinetic friction, as measured by Mathaven, is $\mu_k \approx .18-.24$, and the deceleration due to rolling resistance is about $a/g\approx 0.0127-0.0129$.

My question: Is there any physical explanation for the angle of the cushions on a pool table? It doesn't seem to be optimal for normal incidence, but is it possible that it's optimized in the sense of an average over all possible angles of incidence? Are there models and possibly computer simulations that are sufficiently sophisticated to address this kind of question?

Mathaven, S., et al., "Application of high-speed imaging to determine the dynamics of billiards," American Journal of Physics Vol. 77, No. 9, pp. 788-794, 2009.

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Fascinating. Another possible goal would be to retain the angle of reflection equal to the angle of incidence (on the assumption of rolling straight ahead?) despite the inelastic nature of the collision. – dmckee Apr 29 '13 at 14:55
One factor that was likely considered was to minimize the likely hood of high speed balls bouncing off the pool table. Perhaps the current cushion angle was found to be optimal for this situation. – Michael Luciuk Apr 29 '13 at 19:15
This analysis claims the 'sweet spot' of a pool ball is (7/5)r . I would have guessed that height was the reason for table design, rather than a particular angle. – Mark Rovetta Apr 29 '13 at 19:21
@MarkRovetta: The angle $\theta$ of the cushion is related to the height at which it touches the ball. The result you quote would be, in my notation, $b/r=0.4$, or $\theta=\sin^{-1}(b/r)=41$ deg. As noted in the question, actual pool tables have $b/r=0.26$, or $\theta=15$ deg. Their result is derived under the assumption that the force acting on the ball is horizontal, but that's not correct for a ball hitting a cushion, as discussed in the link to my own analysis. So the result of the analysis you linked to (a) doesn't match reality, and (b) is based on assumptions that don't apply here. – Ben Crowell Apr 29 '13 at 22:23
@BenCrowell: According to this site, the 'sweet spot' of $b/r=0.4$ isn't used because the downward push of the rebounding ball onto the surface would cause too much wear. So instead a slightly lower height is used. According to BCA specifications, billiard balls have a diameter of 2 1/4", while the cushion height is usually taken as 1 7/16" (but there is variation between manufacturers). For these values, $b/r\approx 0.28$. – Pulsar May 22 '13 at 7:46

If the bumpers where vertical then the contact point would be at the center and since gravity is more than the bounce force it means there isn't going to be enough friction in change the rotation of the ball when the direction of the ball changes. If the contact point is further up, then the contact force is towards the center of the ball, and hence is pushing the ball down towards the table increasing friction allowing for the rotation to change as needed for a good bounce.

I did some quick math with inelastic collisions and here is what I found:

  1. The mass moment of inertia of a solid ball is $I=\frac{2}{5}m r^2$
  2. The contact impulse is $$ J = (\epsilon+1) \frac{\frac{3}{5} m \omega r}{\sqrt{ \frac{h}{r} \left( 2 - \frac{h}{r} \right)}} $$ where $h$ is the height above the floor the tip of the bumper is and $r$ is the ball radius. $v=\omega\,r$ is the linear speed of the ball before impact and $\epsilon = 0 \ldots 1$ is the coeffcient of restitution.
  3. The required frictional impulse to reverse the rotation of the ball is $$ R =(\epsilon+1) \frac{2}{5} m \omega r $$
  4. The vertical impulse from the table is $$ N =(\epsilon+1) \frac{ \frac{3}{5} m \omega r \left(\frac{h}{r}-1\right)}{\sqrt{\frac{h}{r} \left( 2 - \frac{h}{r} \right) }}$$
  5. The minimum coefficient of friction needed to reverse the rotation is $$ \mu \gt \sqrt{\frac{4}{9} \left(\frac{1}{\left(\frac{h}{r}-1\right)^2} -1\right)} $$

So if the bumpers are cause the contact point to be in the center of the sphere with $h=r$ then $\mu \gt \infty$ which is imposible. A more typical value of $\mu$ requires that the height is more than $$ \boxed{ h \gt r \left(1 + \sqrt{ \frac{1}{1+ \frac{9}{4} \mu^2} } \right) } \\ \mu = 0.5 \\ h \gt r \left(1 + \frac{4}{5}\right) = \frac{9}{5} r $$

On the other hand, the higher $h$ is then the forces involved increase as $J$ becomes larger and thus the potential losses and non-linearities become dominant. So the trick is the find the smallest possible height $h$ for the surface friction $\mu$.

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NOTE: Actually gravity does not play a role in impacts because they said to occur in infinitesimal small time and thus any acceletation will not significantly effect the speed. – ja72 Feb 20 '14 at 21:59
The impulse $J$ points downwards with an angle $\theta$ from vertical so the equations I used are $$ R - J \sin \theta = m \Delta v \\ N - J \cos \theta = 0 \\ r R = I \Delta \omega \\ \Delta v = r \Delta \omega \\ \Delta \omega = -(\epsilon+1) \omega \\ \mu \gt \frac{R}{N} $$ – ja72 Feb 20 '14 at 22:01

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