Today I was skimming over Rob Eastaway’s excellent book, “Why do buses come in threes?” and found an equation for a situation in snooker which wasn’t derived in the book. With the white ball and object ball lined up to the pocket, a formula was given for the vertical distance M from the centre of the pocket.

The formula is based on the player being professional which means that the angle α, between the horizontal and the direction of the white ball, is very small :

M = α(P – B)(B – W)/W

This makes sense on a number of levels:

1) When α = 0, the distance M = 0.

2) When B = W, the white ball will be touching the object ball and so M = 0.

3) When P = B, the object ball is over the pocket so M = 0.

The problem, which would be the same for any mathematically interested person, is that it isn’t good enough for me to see the formula – it has to be derived. I’ve had a stab at it and was hoping that other maths folk could confirm my method. Here it is:

For small α, sin(α)≈α and l ≈ B – W

Using the sine rule on the left triangle: sin(α)/W = sin(β)/l (vertically opposite angles)

Hence: α/W = sin(β)/(B – W)

sin(β) = α(B – W)/W (1)

From the triangle on the right: tan(β) = M/(P – B) (2)

Since tan(β) = sin(β)/cos(β), combining (1) and (2)

M/(P – B) = α(B – W)/cos(β)W

Since for small angles, cos(β)≈1,

M = α(P – B)(B – W)/W