The physics of six byes

Cricket abounds with tales of a ball going for six byes, with Charles Kortright, Roy Gilchrist and Jeff Thomson among those to which the feat has been attributed. Leaving aside for the moment the fact that the laws of cricket state that a ball only scores six runs if it clears the boundary on the full after having been struck by the bat, just how likely is it that a ball can bounce in the middle of the pitch and then clear the boundary without bouncing again? Clearly such a ball must be extremely fast - but how fast?

Assume that the ball bounces approximately in the middle of the pitch. Most international grounds have a distance of around 80m from there to the boundary in any direction, so let's say that the ball travels exactly 80m before landing. Even the fastest bouncer does not pass very far over the batsman's head - say a height of 2.5m, 10m after pitching. The path of a projectile in a gravitational field is a parabola, and identifying three points on it is enough to determine its equation. The general equation of a parabola is y = ax^2 + bx + c and the three known points on it are (0, 0), (10, 2.5) and (80, 0), so the coefficients a, b and c must satisfy the simultaneous equations

0 = 0a + 0b + c

2.5 = 100a + 10b + c

0 = 6400a + 80b + c

Clearly the first equation implies that c = 0, so the second and third can be reduced to 2.5 = 100a + 10b and 0 = 6400a + 80b. Multiplying the second equation by 8 gives 20 = 800a + 80b, and subtracting this from the third -20 = 5600a, so a = -1/280, and substituting this value into either of the equations gives b = 2/7. Thus the height y of the ball at a horizontal distance x from its point of pitching is given by the equation y = 2x/7 - x^2/280. Differentiating this gives the equation of the tangent to the graph - the direction in which the ball is travelling at a given moment: dy/dx = 2/7 - x/140. At x = 0 this has the value 2/7, the direction at the moment of pitching; this is equivalent to an angle of about 16 degrees from the horizontal.

A parabola is symmetrical, so the highest point in the ball's trajectory is reached at half the horizontal distance between pitching and landing: x = 40m, y = 80/7 - 1600/280 = 40/7m. This information enables us to use the equations of motion:

s(horizontal) = 40

u(horizontal) = 7x/sqrt53

v(horizontal) = 7x/sqrt53

a(horizontal) = 0

t = ?

s(vertical) = 40/7

u(vertical) = 2x/sqrt53

v(vertical) = 0

a(vertical) = -9.8

t = ?

where s = the distance travelled in that direction between the ball pitching and reaching its highest point, u the component of the initial velocity in that direction, v the component of the velocity when the ball reaches its highest point, a the acceleration in that direction, t the time taken for it to reach that point (which is independent of the direction), and x the magnitude of the initial velocity (including both components. The horizontal acceleration is zero if air resistance is disregarded (and thus the horizontal speed is constant); the vertical acceleration is that caused by the Earth's gravitational field (9.8m/s^2), and the ball has an (instantaneous) vertical velocity of zero at its highest point. The horizontal and vertical components of the initial velocity as functions of its overall magnitude are given by applying Pythagoras's Theorem to the known direction of the trajectory at that point.

The equation s = ut + at^2/2 applied to the horizontal motion gives 40 = (7x/sqrt53)t + 0t^2/2, so 40 = (7x/sqrt53)t and thus 40/t = 7x/sqrt53.

The equation v = u + at applied to the vertical motion gives 0 = 2x/sqrt53 - 9.8t, so 9.8t = 2x/sqrt53. Multiplying this equation by 7/2 gives 34.3t = 7x/sqrt53.

Putting the resulting two equations together gives the result 40/t = 34.3t, so t^2 = 40/34.3 and t = 1.08 seconds. Then 9.8t = 2x/sqrt53, so 9.8(1.08) = 2x/sqrt53 and x = 10.6(sqrt53)/2 = 38.6m/s. This is equal to 86.8mph, which is fairly fast in itself - but this is the speed at which the ball must be travelling after pitching. The next factor which needs to be taken into consideration is the coefficient of restitution - the ratio of the ball's speed after pitching to its speed before. Even on the bounciest of pitches this will be no more than 0.7 (on most pitches probably less), which would mean that in order to achieve a speed of 86.8mph after pitching, the ball would have to be released at a minimum of 124mph - a quarter as much again as the highest recorded speed of any ball bowled in an international match (and that's even if you believe the speed guns). The actual figure would be more, because the above calculation does not take into account the effect of air resistance

In conclusion: 'six byes' (even though they wouldn't be counted as such) might just about be possible at a very small club ground - although even if the boundary is 60m from the centre of the pitch rather than 80m, applying the same method shows that the ball would still have to be bowled at a minimum of 105mph. It is certainly impossible at an international-sized ground.

Manish Yadav/Michael Jones