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Dr Ian Plummer

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Technical
Lawn Speeds

This article presents an introduction to lawn speeds and a simple derivation of the distance a ball travels assuming a constant friction against grass. A more complete mathematical treatment is given in the appendices of Don Gugan's paper Experiments on the Game of Croquet; Bouncing and Rolling Balls .

Players often debate the speeds of lawns, although little can be done about them. It can be a factor in comprehending the results of a tournament as very fast lawns require a high level of skill.

The term 'fast' indicates that balls only need to be struck gently to travel a large distance and hence a delicacy of touch is required. Conversely a 'slow' lawn requires balls to be struck firmly. The speed of a lawn depends on the length, dampness and quality of the grass, and on the smoothness of the underlying surface; long, damp coarse grass will result in a very slow lawn.

Lawn speeds are measured either by rolling a unbiased bowl across the lawn and measuring the distance it rolls (similar to the stimpmeter used in golf) or by hitting a ball to the exact length of the lawn and timing how long it takes.

An unbiased bowl is launched down a ramp and the distance it travels across the lawn is measured and averaged over a number of trials. The height and angle of the ramp are set to ensure that when the bowl meets the lawn it is rolling at exactly the right rate to match its forward velocity; this means that it rolls rather than skids across the grass. (See Sliding & Rolling Balls). The bowl generally travels further each time if the same track is used as it is sensitive to the flattening of the grass in previous trials. This method give a distance as its result.

Striking a ball the length of a lawn (35 yards) so that it stops exactly at the right point is a difficult task. Fortunately the ball can be hit repeatedly to approximately that length and timed. A graph can be drawn of length versus time and interpolated or extrapolated to 35 yards. Times ranging from 7 to 14 seconds are in a normal range.

The Croquet Association has conversions for lawn speeds measured by either method.

Newton's Basics

This section attempts to use simple mechanics to fit the measured data.

One of Newton's basic equations of motion states that a body's velocity is equal to its initial velocity plus the time it is under the influence of an acceleration.

v = u + a.t   

(1)

v is the velocity, u is the initial velocity, a is the acceleration and t the time. Hence for a decelerating body which comes to rest, vfinal = 0.

tfinal = -u/a 

(2)

Integrating (1) with respect to time, assuming that acceleration is not a function of time, yields

s = u.t + ½.a.t2 + c 

(3)

s is distance (v.t) and c is the constant of integration. At time t = 0 the ball has travelled no distance, hence s = 0 therefore c = 0.
 

We can use the above to get an expression for the initial velocity given the acceleration and total distance, stotal. Rearranging (3) by substituting t = tfinal as in (2).

stotal = -u.u/a + ½.a.u2/a2

 

stotal = -u2/a + ½.u2/a

 

stotal = -½. u2/

(4)

and rearranging yields

u = √(-2.a.stotal)

(5)

If we substitute (5) in (2) we get the (final) time in terms of the (total) distance:

tfinal = -√(-2.stotal/a

(6)

Balls on lawns

The above equations can now be applied to croquet. Before going too far it has to be realised that we are making a large number of assumptions by applying simplistic equations. In this case a major factor is that moving balls have energy in two forms; kinetic (sliding motion) energy and angular (rotational) energy (this is discussed by Gugan). In the discussion below any rotational energy of the ball is ignored.

It is also assumed that the friction offered by the grass is independent of the sliding velocity of the ball and is constant. The frictional force on a moving ball depends on the length, dampness and quality of the grass, and on the smoothness of the underlying surface, so that whether or not it is reasonable to assume constant deceleration is something which needs to be tested by experiment. We however assume that the friction between the grass and the sliding ball is a negative constant acceleration, i.e. a slowing down.

a = -afriction

 

this gives:

tfinal = √(2.stotal/afriction)

(7)

Hence for a fixed distance, e.g. the length of a croquet lawn, the time for a ball to come to rest is proportional to:

tfinal is proportional to √(1/ afriction)

(8)

  • The smaller the coefficient of friction the larger the time; transit on a fast lawn is long (in real terms you give the ball a small nudge and it trickles the length of the lawn).
     
  • The larger the coefficient of friction the smaller the transit time (you give the ball a smash and it hurtles down the lawn screeching to a halt).

The graph below shows a measurements made by Samir Patel and myself at the Hurlingham Club and in Bayeux (France, where the courts were recently mown rugby pitches).

measured lawn speeds

Measured Lawn Speeds. Blue diamonds: Hurlingham Lawn 6 2001; magenta squares: Hurlingham Lawn 3; green triangles: Bayeux (Fr.) 2002.

The brown line is an indicative plot of (n.sqrt(distance)).

A square root plot has been added above to indicate the expected form of the data if equation (6) is followed. Assuming that the data follows equation (6) we can find the value of afriction from the data. We can rearrange the equation as stotal = -½.a.t2final. Taking the magenta sequence (Lawn 3) from the above graph, a plot of distance (stotal), versus t2 has been plotted as red spots (below).

time^2 vs distance

Lawn 3. red spots: time squared vs distance. Note that distances have been converted to metres.

The gradient of Least Mean Squares fitted straight line through the red points is 0.2725. The fit is good as indicated by the R2 value. The data shows close agreement with our simple model.

We can now calculate the deceleration. A straight line has the form y = m.x, where m is the gradient, here the gradient is -½.afriction, hence afriction is -0.545 ms-2. Since the acceleration is negative it is a deceleration (slowing down or resistance) which is what we would expect. We can apply the same analysis to the data given by Gugan in the appendix to his paper:

distance vs time^2
Distance versus Time2 plot with a fitted least mean squares fit

hence gradient = 0.3794 = -½a , a = -0.759ms-2.

In practice once the transit time, tfinal for a lawn is known, making the assumptions above, we know stotal = 35 yards (32.004m) and hence can calculate afriction from (7) and u, the initial velocity, from (2), directly. In the table below the Time is the time for a ball to travel 35yards and stop.

Time
(s)
afriction
(ms-2)
u
(ms-1)
u
(km/h)
u
(mph)

3

-7.112

21.34

76.81

45.36

4

-4.001

16.00

57.61

34.02

5

-2.560

12.80

46.09

27.21

6

-1.778

10.67

38.40

22.68

7

-1.306

9.14

32.92

19.44

8

-1.000

8.00

28.80

17.01

9

-0.790

7.11

25.60

15.12

10

-0.640

6.40

23.04

13.61

11

-0.529

5.82

20.95

12.37

12

-0.445

5.33

19.20

11.34

13

-0.379

4.92

17.73

10.47

14

-0.327

4.57

16.46

9.72

15

-0.284

4.27

15.36

9.07

16

-0.250

4.00

14.40

8.50

17

-0.221

3.77

13.55

8.00

18

-0.198

3.56

12.80

7.56

19

-0.177

3.37

12.13

7.16

20

-0.160

3.20

11.52

6.80

As examples of measured lawn speeds:

 

 
Time (sec)

2001

Hurlingham: lawn 3

11.2

2001

Hurlingham: lawn 6

10.6

2002

Bayeux

6.8

2003

Bayeux

4.7

2003

Cheltenham Opens

17

2003

Hurlingham: lawn 4

14

2006 April 15

Surbiton

10

2006 April 16

Surbiton

10.3

2006 April 29

Hunstanton

11

2006 May 01

Parkstone May Westerns*

10/9

* 10 sec with the lawn nap, 9 sec against it

 

 

All rights reserved © 2004-2014


Updated 1.i.11
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