OxfordCroquet Logo

Dr Ian Plummer

Donation Button

http://oxfordcroquet.com/tech/rush/index.asp
Technical
Derivations for Straight Rush Shot
Straight Rush - variables
Variables used in the calculations. Click on image for a larger version

The aim is to produce a formula which gives the direction of a rush for given separations of the striker's ball (s) and the roqueted ball (r) when the strikers' mallet is swung at an angle (α) to the intended target direction (T). 

The separation of the striker's ball to the roqueted ball (centre-to-centre) is given by the distance C. In a rush shot the direction of a rush (R) can be found by imagining a hypothetical ball (h) travelling in the stroke direction (S) which just makes contact with the roqueted ball. The roqueted ball moves away at an angle (β) set by the line passing through the centres of the hypothetical ball and roqueted ball with the target direction.

Our initial task is to produce the relationship between α, C and β.

The balls are assumed to be incompressible frictionless spheres.

The first thing to notice is that the separation from the centre to the hypothetical ball (h) to the roqueted ball (r) will always be 1 ball diameter.  This is the distance A on the diagram and if we use the ball diameter as the unit of length then:

1). A = 1

First we use the Sine Rule.  For a triangle of sides A, B and C with opposite angles α, β and γ there is the following relation:

2). Sin Rule:  A / sin(α) = B / sin(β)  = C / sin(γ).

Note from the diagram that for any fixed A, α and C, there are two possible results: triangle shr and triangle sh'r

Given we do not know B, which varies depending on which triangle is used, we start with a simple rearrangement of the Sine Rule to get θ or γ:

3). sin(γ) = C / A * sin(α), hence

4). γ = arcsin[ C / A *sin(α) ]

Quick sanity check, if C = A then the striker's ball is in contact with the roqueted ball, C / A = 1, hence α = θ and we have an equilateral triangle A = B = C, α = β = θ.

If we try a sample calculation at this stage with A = 1 and C = 3 and α = 3°.

γ = arcsin[ 3 / 1 *sin(3°) ] = 9° hence this is working for the triangle sh'r

Given the acute angle it is not difficult to get the complimentary angle θ = 180° - γ  

Since the internal angles within a triangle add up to 180° then for shr:

5). 180° = α + β + θ

6). β  = 180° - α - θ

7). β = 180° - α - 180° + γ = γ - α

For our example case α = 3° and C = 3, γ = 9 hence β= 6° and θ = 171°.

We can now produce the relationship between α, β and C: substitute (4) for γ

8). β = arcsin[ C / A * sin (α) ] - α,    where A = 1.

Examples

The table below gives the calculated values of β (the angle at which the rushed ball leaves) from (8), for various C and α (the angle off-target that the striker's ball is struck). Remember that C is the distance from the centre of the striker's ball to the centre of the roqueted ball, hence a value of 1.5 represents a gap of half a ball.

C =>
ball diameters

1.25

1.50

1.75

2.00

2.50

10.00

α = 0°

α = 1°

0.25°

0.50°

0.75°

1.50°

9.05

α = 2°

0.50°

1.00°

1.50°

3.01°

18.43

α = 3°

0.75°

1.50°

2.25°

4.52°

28.56

α = 5°

1.25°

2.51°

3.77°

7.59°

55.64

α = 10°

2.54°

5.10°

7.69°

10°

15.73°

miss

The figures support the idea that if the gap separating balls in a rush is less than a ball diameter (green area), then for a straight rush (!), the rushed ball will be 'more on target' than if the striker's ball alone had been hit at the target with a slight error in direction.

In descriptive terms: for small separations the effect of the geometry of two balls colliding almost head-on causes the forward ball to be cut towards the intended aiming line. If the gap increases beyond one ball diameter the geometry works against this and the forward ball deviates at a greater angle to the intended aiming line.

All rights reserved © 2011-2017


Updated 28.i.16
About, Feedback
oxfordcroquet.com/tech/rush/index.asp
on www.oxfordcroquet.com
Hits: 4672