# Mean free path regarding billiards

Is there a generalization of the mean free path for macro-objects such as the distance the white billiard ball could travel before hitting an other billiard ball?

• Mean free path doesnt make much sense unless you talk about a large number of macroscopic objects with each of the object having a initial velocity. Jan 5, 2022 at 22:22
• @JunSeo-He That's what i actually meant, a more suitable example may be (as harsh as it sounds) that of a bullet which is traveling high speed through an area that is filled with N (stationary) people . What distance could the projectile travel, until hitting a person? So lets say there are N people placed arbitrary on a field of an area of P. A shooter outside the field is shooting into the field, how much could a bullet travel on mean until hitting someone? It would be interesting to know that when for example police shoots into a area full of protesters.. _
– NHSH
Jan 5, 2022 at 22:51
• The problem is that a billiard table evolution isn't well described by random motion or any such thing as Brownian motion. It is usually not-too-far from the human operator's intended path and so has a collision quite quickly. Also, the damped motion due to friction is very important. So a mean free path isn't a good abstraction.
– Dan
Jan 5, 2022 at 23:17

The mean free path in two dimensional space (surface of table) is mathematically defined by$$^1$$ $$\lambda = \frac{A}{\sqrt{8}Nd}$$ where $$d$$ is the diameter of the billiard balls, $$N$$ is the number of balls and $$A=\text{length}\times \text{width}$$ is the area of the billiard table.
Knowing the average area for a billiard table to be $$3.4\ m^2$$ (2.6m$$\times$$1.3m), and the number of balls to be 16 (15+1 cue ball) and the diameter of each ball to be $$\approx 0.05\ m$$ or 5cm, then $$\lambda\approx 1.5 \ m$$ which is the average distance a ball will travel before colliding with another.
$$^1$$ I have not included the derivation here for the sake of brevity, but will include it if asked to by people in comments.