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Dr Ian Plummer

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Technical
Where a Ball Ends Up after a Collision

Simple Theory

If a moving ball collides with a stationary one, then the moving ball should be replaced along its original trajectory at the point corresponding to the sum of the distances travelled by both balls from the point of collision.

Restoring balls after a collision

Restoring balls after a collision (Simple Theory).  The red ball collides with the blue at point C and the balls are scattered. The red ball should be restored to position F along the original trajectory (black line) where the distance C..F =  C..A + C..B

The above result incorporates the assumptions that there is a constant deceleration of the ball1, the collision is elastic (energy is conserved) and the balls have the same weight. A simple proof of why the distance travelled by a ball is proportional to its energy is given in the Appendix.

Under those conditions the distance travelled by a ball is directly proportional to the energy it carries. This is reassuring – hit the ball with twice the energy and it goes twice as far. When the balls collide the energy is shared between them and they each move a distance corresponding to the fraction of the energy they received.  Add the distances travelled by the separate balls and it gives you the distance which would have been travelled by the unhindered ball.

Energy Losses

It has been noted by Max Hooper who used a mallet-swinging rig, that the method above underestimates the final position of the ball. This is not surprising given the assumptions. The assumption that the collision of the ball is elastic (energy is conserved on collision) is patently false.

The coefficient of restitution (bouncyness) of a croquet ball is defined in the Croquet Laws to lie between 50-75% (6th Edition Laws Appendix 2; a ball must bounce between 30-45" when dropped from 60" on to a defined rigid surface). Consequently 25-50% of the energy is lost during a head-on collision. We know that distance scales with energy hence we should need to multiply the sum of the scattered distances by up to 2 times. The amount of energy available to be transfered between balls during a collision also depends on the geometry.

There is no real problem with the assumptions that both balls weight the same. The final assumption that the balls undergo a constant deceleration (which justifies the 'distance scales with energy') is a rough approximation. A curve fitting was used on experimental data to produce a relationship1. The data is sparse for when the ball is moving slowly (i.e. for short total travel distances = short transit times). It may be that a slowly moving ball decelerates less than a quickly moving ball. Further measurements are needed but the empirical data indicates that the estimate of constant deceleration is good and not likely to have as large an effect on the Simple Theory as the assumption that energy is conserved.

There are additional factors which affect the forces on the balls after collision, some may be non-linear with ball velocity. These include:

  • rolling friction and sliding friction against the grass
  • any bounce against the earth
  • air resistance
  • vertical components of ball rotation
  • ...

All of these will act as 'energy leaks' and hence may make the deceleration dependent on velocity (hence not constant). It is, for example, well known that air resistance scales as the square of velocity. More experimentation is needed.

An Australian System

A system is proposed in the Australian Croquet Association (ACA) Referees Manual (2003, Page B16). I hope I have paraphrased their solution accurately below (thanks to Owen Edwards for the recipe).

Australian Solution to Collisions

The Red ball collides with the Brown at point C and Red is scattered to R1 and Brown to B1. The distance C..R1 is x and the distance C..B1 is y. The Red ball should be restored to position R along the original trajectory where the distance C..R = x + (2.5 * y)

As indicated in the caption the Red ball is replaced the distance it was scattered from the point of collision plus two and a half time the distance the Brown ball was scattered. From the above sections in this document it is clear that the multiplication factor (2.5) is an attempt to scale for the energy losses on collision and for what happens subsequently. For certain collision configurations this is plausible.

Consider however the two circumstance below: A) where Red hits Brown almost full on - as in a rush, and B) where Red gives Brown only a slight glancing blow. The situtations are approximatey symmetrical with the balls interchanged.

Extrems of Collision geometry

Case A) is the hardest to reconcile. In the rush-like situation, Brown is boosted almost as far as Red might have travelled. Yet Red gets replaced two and a half times y1 up the lawn. Given the approximate diagrams above it seems unlikely that x1 + (2.5 * y1) = x2+ (2.5 * y2). Experiments have yet to be repeated.

Appendix - Distance Travelled is Linearly Proportional to Energy

The almost mathematics-free version is:

Energy = force x distance, the force being the resistance of the grass (constant, by assumption). If you start two balls with say half the energy, they each will go half the distance; add the two distances together and it is the same as a single ball with all the energy.

We can attempt to make it appear more thorough by using Newton’s Equations of Motion:

v = u + a.t

(1)

s = u.t + ½ .a.t2

(2)

where

s = the distance travelled from the initial state to the final state
u = the initial velocity (speed in a given direction)
v = the final velocity
a = the constant acceleration, if a is negative it is a decelleration
t = the time taken to move from the initial state to the final state
m = is the mass of the object

Consider a ball rolling along the grass; it will stop when its velocity becomes zero after a time tf – the ‘final’ time.  Putting v = 0 into Eqn. 1 and rearranging;

tf = -u/a

(3)

substitute t= tf in Eqn. 2 and we get the total distance, sf, travelled by the ball:

sf = u.tf +.a.(tf)2

 

sf = -u2/a + ½ .u2/a

 

sf = -½.u2/a

(4)

A little bit of deviousness now in the calculation.  We use the standard equation for the initial kinetic energy:

E = ½ .m.u2

(5)

This is the simplification; the ball will hold its energy in both kinetic and rotational energy but those forms are readily inter-convertible. We would be surprised if a ball came to rest but was still spinning!

We can rearrange Eqn. 5 to get u, the initial velocity, in terms of E:

u2 = 2.E/m

(6)

We can now substitute for u2 in Eqn. 4

sf = -E/(m.a)

(7)

This result shows that the total distance, sf, is a simple multiple of the initial energy, E.  Note that the 'acceleration', a, is negative designating a deceleration due to the friction, etc. against the grass and ground.  This yields a positive distance travelled.  Another way to phrase it is that the energy decreases linearly with distance.


Footnotes

1. Lawn Speeds: http://www.oxfordcroquet.com/tech/lawnspeed/index.asp

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Updated 1.i.11
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