Technical
MidPoint Aiming
Nick Furse has indicated that the midpoint aiming method (as opposed to the half angle) method is only an approximation. One reason for this is that the aiming line is set by a vector sum of velocities, whereas the distances travelled by the balls are proportional to the square of their velocities. This page is solely to illustrate that a better approximation to the true aiming line lies in taking the square roots of the expected distances of travel to work out an aiming point. The following diagram may help, apologies for all the colours. Note there is a whole bundle of simplifying assumptions; the target here is to illustrate what the argument is based on. For convenience in drawing I have chosen a perpendicular set of directions, but the general case works similarly. A croquet stroke is played at position D and the balls travel North and East as indicated by the black lines. The distances are given in black (9 & 4 units). The midpoint aiming method would say aim at the midpoint of the (pale blue) line joining the final resting places. This midpoint aiming line is the dark blue line DM. However to strike the balls at the correct angle to make them travel in the correct directions we must consider the vector sum of velocities not distances. If we are happy assuming that the distance travelled is proportional to energy, (hit with twice the energy and it goes twice as far), then distance ∝ Energy = ½mv^{2}. Thus the velocity is proportional to the square root of the energy or square root of the distance. The pink arrowed lines represent the velocity vectors and are correspondingly 3 & 2 units long (being the square roots of 9 & 4). Their vector sum, and the correct direction for the stroke, is given by the orange vector DV. The takehome points are:
How Big is this Effect?Answer: a maxiumum aiming difference of around 13°. If we continue investigating the 90° case above we can plot how the deviation angle, δ (see diagram opposite), between the blue midpoint aiming line and the orange vectorderived aiming direction, varies with the ratio of the distances travelled by the balls in the stroke. We can use the reannotated diagram opposite. The trigonometry is straight forward. tan(α) = b / a ; tan β = B / A where a = √ A and b = √ B. Proportionality constants have been set to 1. Rearranging: α = arctan( b / a ) and β = arctan( B / A ), hence the deviation angle δ = β  α = arctan( B / A )  arctan (√ B / A ). ( B / A ) is the ratio of the distances travelled by the balls; let's call it x, so δ = arctan(x)  arctan (√ x). Plotting this out gives: So does this look right?When x = 1, i.e. both balls travelling the same distance ... there is no deviation as expected.The part of the curve between x = 0..1 represents distance A > B (as x = B / A) and is a squashed inverted mirror of the curve from x = 1..∞. The negative part of the curve covers a ball travelling further East than North, the positive is vice versa. The maximum and minimum arise from the form of the equation for x. Whilst 12° does not sound much, for a 10" (25cm) mallet it equates to swinging the tip of the aiming line through ~28mm, pivoting about the axis of the shaft. With thanks to Nick Furse for corrections. All rights reserved © 20092017
