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?

  • $\begingroup$ 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. $\endgroup$
    – Jun Seo-He
    Jan 5, 2022 at 22:22
  • $\begingroup$ @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.. _ $\endgroup$
    – NHSH
    Jan 5, 2022 at 22:51
  • $\begingroup$ 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. $\endgroup$
    – Dan
    Jan 5, 2022 at 23:17

1 Answer 1


The mean free path is defined as the average distance between collisions. There does not seem to be a valid reason as to why this cannot be applied to the case of billiard balls on a billiard table, only if we assume the motion of the balls is fairly random and not “too guided” by the players. Let’s assume the player who “breaks” hits the cue ball in a random direction very hard that when it hits the other balls they all keep moving indefinitely.

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.


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