Snooker and pool margins for error
In New Scientist dated 8th May 2010,
there is an article
about different levels of difficulty relating to shots at pool or snooker.
Some of its conclusions are perhaps obvious, but not all, and as is so often its wont,
New Scientist has not reproduced the mathematics.
Furthermore, Chris points out that their assumptions may apply to pool, but they do not apply to snooker.
So lets see if we can form our own conclusions.
To start with, consider potting in a middle pocket with the object ball on the line joining the two middle pockets.
Imagine two spots (shown red on the diagram) immediately in front of the pocket placed so that a ball that passes over a spot is
exactly on the border between going in the pocket and missing.
The green line shows the ideal shot, and the red line the worst acceptable shot.
Let us call the angle between these two lines
φ, and use x
for the perpendicular distance between the red spot and the ideal shot.
The ideal shot is achieved by the cue ball hitting the object ball opposite where the green line intersects the circle of the ball.
The worst acceptable shot is achievd by hitting the object ball opposite the red line’s intersection with the same circle.
Let us call the distance between these two impact points
ε.
ε = x r/l
where r is the radius of the ball and l the distance of the object ball from the pocket.
Let us just look at the situation where we can ignore the angle between the green line and the line joining the centres
of the object ball and cue ball, and denote the maximum player error by
θ.
If the cue ball is not close to the object ball, we can use the approximation:
θ = 2 ε / L
In the New Scientist article, the pocket, object ball and cue ball are in a straight line.
Our analysis is valid for small deviation from this.
The cue ball is considered a constant distance from the pocket
(let us call that Z),
and the analysis looks at
how the difficulty varies with the location of the object ball.
| θ = | 2 ε
L | = | 2 x r
l L | = | 2 x r
l ( Z - l ) |
The minimum value of
θ
occurs when the value of
l ( Z - l )
is maximum. I.e. when
l = ½ Z.
The New Scientist article then goes on to consider what pool players call a combo shot and snooker players call a plant.
Our previous cue ball can become the first object ball, and the new
ε'
(the distance between the ideal impact point and the worst acceptable impact point, the previous
ε)
becomes:
ε' = 2 ε r / L'
where L'
is the distance from the cue ball to this first object ball.
The maximum allowable error is:
| θ' = | 2 ε'
L' | = | 4 x r2
l L L' |
But what if the angles are not small?