Formula 4 average bowling speed and percentage speed lost of the bounce
hi,im 15 nd hit 60-65 with no effort and im a bit lazy so thats alright
i've always used this to work out speed, this is only average speed through delivery obviously its slower of the bounce:
get a stop watch:
you start the watch as you release the ball and stop it when it reaches the batsmen
d/t=s
19.5(m)/time in secs=Ans
Ans x 60 x 60 ( into hours)= Ans in metre/hours
ans in metre/hours/ 1609(turns it from metric to imperial i.e km-miles) = average speed in miles per hour
so for example 19.5/0.5 secs= 39 so 39 x 60 x 60 = 140400 140400/1609 = 87
if it takes 0.5 secods for the ball to reach the batter your average speed is 87 mph
I also have this one that wrecks your head lol:
OK, in summary, the percentage change in speed from release to piching is approximately:
Change in Speed = 0.766 x D
where D is the distance between release and pitching in metres. So for a yorker D is about 18m and the ball slows about 14%(thats quite simple). (on the otherhand this will hurt your head) The physics behind this equation is as follows:
First you need to calcuate the aerodynamic drag of the ball which is given by:
Drag = Cd x 0.5 x Rho x A x V x V
Where Cd is the drag coefficient of a sphere (about 0.5)
Rho is air density, about 1.225 kg/m3 at sea level and 25C
A is the cross sectional area of the cricket ball, 0.004 m2
V is the speed at release in m/s
Once you have the drag force, you can can then calculate the deceleration d=drag/m , where m is the mass of the cricket ball (0.16kg). When you have the deceleration, you can calculate the change in speed over the time the ball is in flight, which is approximately t=D/V. Interesting V cancels out of the final equation, and it becomes the simple formular above when you plug the numbers in.
if this works please comment on it also if you fully understand the drag/air resistance part well done. I kind of understand it but i dislike physics so i think i skipped parts in school
thanks phil
hi,im 15 nd hit 60-65 with no effort and im a bit lazy so thats alright
i've always used this to work out speed, this is only average speed through delivery obviously its slower of the bounce:get a stop watch:
you start the watch as you release the ball and stop it when it reaches the batsmen
d/t=s
19.5(m)/time in secs=Ans
Ans x 60 x 60 ( into hours)= Ans in metre/hours
ans in metre/hours/ 1609(turns it from metric to imperial i.e km-miles) = average speed in miles per hour
so for example 19.5/0.5 secs= 39 so 39 x 60 x 60 = 140400 140400/1609 = 87
if it takes 0.5 secods for the ball to reach the batter your average speed is 87 mph
I also have this one that wrecks your head lol:
OK, in summary, the percentage change in speed from release to piching is approximately:
Change in Speed = 0.766 x D
where D is the distance between release and pitching in metres. So for a yorker D is about 18m and the ball slows about 14%(thats quite simple). (on the otherhand this will hurt your head) The physics behind this equation is as follows:
First you need to calcuate the aerodynamic drag of the ball which is given by:
Drag = Cd x 0.5 x Rho x A x V x V
Where Cd is the drag coefficient of a sphere (about 0.5)
Rho is air density, about 1.225 kg/m3 at sea level and 25C
A is the cross sectional area of the cricket ball, 0.004 m2
V is the speed at release in m/s
Once you have the drag force, you can can then calculate the deceleration d=drag/m , where m is the mass of the cricket ball (0.16kg). When you have the deceleration, you can calculate the change in speed over the time the ball is in flight, which is approximately t=D/V. Interesting V cancels out of the final equation, and it becomes the simple formular above when you plug the numbers in.
if this works please comment on it also if you fully understand the drag/air resistance part well done. I kind of understand it but i dislike physics so i think i skipped parts in school
thanks phil