Geometry of Snooker: Problem in Rob Eastaway’s Book, “Why do Buses come in threes?”

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.

Image2

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:

Image4

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

Advertisements
This entry was posted in Uncategorized. Bookmark the permalink.

2 Responses to Geometry of Snooker: Problem in Rob Eastaway’s Book, “Why do Buses come in threes?”

  1. Ben Keeping says:

    Looks good! I thought about this problem myself a while ago (I haven’t seen the book you refer to, so apologies if I’m repeating the point it made here!). I concluded (as you have) that the difficulty of any close-to-straight snooker shot is proportional to the product of the distances between white ball and object ball and between object ball and (centre of) pocket. This explains a real life fact, which is that good players can knock in balls which are close to the cue ball but six feet from the pocket almost as easily as knocking in balls which are near the pocket but 6 feet from the cue ball. The really hard long shots are the ones where the two distances are close to equal.

    I only really got as far as concludng that your beta is proportional to your (B-W)alpha and that M is proportional to your (P-B)beta. Your approach is more complete.

  2. Pingback: Mathematical Reflections after the SGIS Education Conference | Teaching Mathematics

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s