So where does this leave us?

The final over of the match on Saturday 5th May has prompted some deeper musings from the club’s “physicist’s corner”.   Dr Thomas Haydon writes:

Screen Shot 2018-05-07 at 14.02.04

Dr Thomas Haydon

Summary of everything below is that Appy would have needed to bowl a ball that reached a minimum of 3m height in its flight path and was bowled with max speed of 32.3 mph.  Anything given less air or more speed would not satisfy the conditions for a no ball that hit the stumps.

Farmer Appleyard is bowling the last over in a tense village game of cricket.  1 wicket is needed for victory for his side and only a few runs for Churchill.  Farmer Appleyard bowls the ball which the batsman misses and it hits the stumps without bouncing first.  The umpire calls this a no ball under the conviction that it was above waist height when passing the batsman.  Captain White wasn’t entirely convinced with this decision.  Given the following inputs provide some calculations that would help Captain White’s argument.

  • Farmer Appleyard is 6 feet tall and launches the ball from 7 feet (2.12m)
  • Stumps – popping crease = 1.22m
  • Stumps – stumps = 20.12m
  • Ball hit the stumps at 0.5m height
  • Waist height of batsmen 1m
Proposed solution to above (with help of proper theoretical physicist Dr Paul Rowland who was visiting me Sunday):
Firstly follow this link to see the parabola that fits the data points given by the inputs that can be derived from the above (screenshot below):
You can see that Farmer Appleyard would have had to give the ball enough air that it reached a height of over 3m at some point in its path to the stumps.
The above doesn’t give the full story – because even though its possible with the arc shown above it does not tell us how fast (slow) Farmer Appleyard would need to bowl the ball.  The equation below is the standard formula for the range of a projectile launched from a height y0, angle theta with velocity v falling under gravity.  The launch angle is implied from the parabola (its 14.2 degrees) and using a range of 20m we should be able to then solve for velocity (I haven’t solved for velocity but i’ve plugged a few typical velocities into the formula until I got a range of 20m….)
{\displaystyle d={\frac {v^{2}}{2g}}\left(1+{\sqrt {1+{\frac {2g\ y_{0}}{v^{2}\sin ^{2}\theta }}\ }}\right)\sin 2\theta }
And the answer suggests that to achieve the flightpath that fits the points, Farmer Appleyard would have needed to have bowled the ball at a pace of 14.45 metres/second which is around  32.3mph.  My suggestion is that Farmer Appleyard should be put through a laboratory controlled pace test to see if this is indeed his typical bowling speed.
Dr Alex Comer adds:
“Thank you Thomas – I’d obviously quickly done these calculations in my head hence my exchange of thoughts with the official.
I may point out that whilst the umpires are the arbiters of the laws of cricket they are not always conversant with the laws of Physics. I therefore think all umpires should be required to have the equation (above) on hand along with Duckworth/Lewis in order to resolve such issues. I also recommend that Dr Rowland offer his services as consulting Physicist to the Shrubbery Somerset League.
My only hesitation is the 32mph seems a bit quick for Farmer Appleyard.”