# The Math and Physics of Billiards

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In this well known Page http://archive.ncsa.illinois.edu/Classes/MATH198/townsend/math.html I don't understand the R Parameter following mg in this Equation: Where did that R come from?

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eh, they use R twice in that article there...
the First time it's the radius of the sphere, the second time it's a vector (denoted by the arrow above it) for the vector extending from the center of the ball to the point of contact with the table.

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I am talking about the Capital R in the Cross Product. I understand that is a Scalar that represents the Radius of the Sphere. The R^2 comes from the Inertia Tensor. But the R inside the Cross Product is the one buging me.

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I thought I found the answer but I still don't know what the R is.
The vector r already gives the Position from the Center of Mass.
F does not need any scaling.

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Maybe it's a constant, like the R in the ideal gas law (PV = nRT).

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Yeah, I think you're right, it shouldn't be there. The only way it works is if r is a unit vector.

Look on the bright side, if you're working in inches, the radius of a standard billiard ball is only 1.125, so an extra radius in there won't hurt too much. :)

I came across this other resource that you might find useful, though it is much lengthier:
http://www.sfbilliards.com/shepard_apapp.pdf

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The cross product on the right hand side is the torque. Torque is calculated as

$\vec{\mathbf{\tau}} = \vec{\mathbf{r}}\times\vec{\mathbf{F}}$

Where F is a force being applied, and r is the position where the force is being applied - relative to the rotation center. In this case, the force would be the friction force of the ball on the table.

In order to derive this expression for friction, there was probably some assumptions made, such as the ball is rolling without slipping. This seems to be the case, because they use μs - the coefficient of static friction. Static friction only would apply if they are assuming that the point of contact is not in relative motion with the table. This may mean that putting english on the billiard will not be entirely accurate.

My guess is that the R is present due to the definition of vf. For instance - you need the friction force to be opposed to the velocity of the ball, or at least the relative velocity of the point on the ball that is in contact with the table. The velocity of this point is

$\vec{\mathbf{v}} = \vec{\mathbf{r}}_{down}\times\vec{\mathbf{\omega}} = R \left(\vec{\mathbf{n}}_{down}\times\vec{\mathbf{\omega}}\right)$

Can you find the definition of vf in the paper?

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The cross product on the right hand side is the torque. Torque is calculated as

$\vec{\mathbf{\tau}} = \vec{\mathbf{r}}\times\vec{\mathbf{F}}$

Where F is a force being applied, and r is the position where the force is being applied - relative to the rotation center. In this case, the force would be the friction force of the ball on the table.

In order to derive this expression for friction, there was probably some assumptions made, such as the ball is rolling without slipping. This seems to be the case, because they use μs - the coefficient of static friction. Static friction only would apply if they are assuming that the point of contact is not in relative motion with the table. This may mean that putting english on the billiard will not be entirely accurate.

My guess is that the R is present due to the definition of vf. For instance - you need the friction force to be opposed to the velocity of the ball, or at least the relative velocity of the point on the ball that is in contact with the table. The velocity of this point is

$\vec{\mathbf{v}} = \vec{\mathbf{r}}_{down}\times\vec{\mathbf{\omega}} = R \left(\vec{\mathbf{n}}_{down}\times\vec{\mathbf{\omega}}\right)$

Can you find the definition of vf in the paper?

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hiigara: I'd advice you not to use that page. The physics reasonings are pretty wrong. Why don't you look for a proper paper?