**Last updated on 14 May/98**

ball |
weight (lb) |
diameter(in) |
terminal
speed(mi/hr) |
Cor at 15 mi/hr |
CoR at 55 mi/hr |

16 lb shot | 16 | 4.72 | 325 | ||

football | 0.91 | 11.1 x 6.8 | 100 | ||

baseball | .32 | 2.9 | 95 | 0.57 | 0.55 |

golf ball | .1 | 1.68 | 90 | 0.60 | 0.58 |

softball | .4 | 3.82 | 80 | 0.55 | 0.40 |

handball | .14 | 1.88 | 75 | 0.80 | 0.50 |

tennis ball | .13 | 2.56 | 70 | 0.70 | 0.50 |

squash ball | .07 | 1.77 | 55 | 0.52 | 0.40 |

soccer ball | .94 | 8.75 | 55 | 0.75 | 0.65 |

basketball | 1.31 | 9.47 | 45 | 0.75 | 0.64 |

volleyball | .59 | 8.43 | 35 | ||

ping-pong ball | .006 | 1.47 | 20 | 0.80 | 0.70 |

superball | 0.90 | 0.85 | |||

The CoRs apply to balls dropped or thrown at a rigid wooden surface. Adapted from Plangenhoef, Patterns of Human Motion.

CoR = coefficient of restititution = (speed after collision)/(speed before collision)

The CoR can be measured directly by velocity measurements but often it is handier to measure the height of rise of the ball after it bounces relative to the height that it fell. Since v^2= 2gh. Cor = v'/v = sqrt(h'/h) where h' is after and h is before. For example a regulation tennis ball is tested at low speed where a ball is dropped from about 1 meter. The relative height of the bounce should be h'/h = CoR^2 = 0.7^2 = 0.49. The selection of balls for official games includes the Cor test in tennis or baseball or any other game.

Air resistance is zero at zero velocity. However, it increases with speed and is the limiting factor in high speed travel in bikes or cars. The high speeds encountered in baseball and tennis also have a large effect on the flights of these balls. The subject of fluid effects in the general case are difficult to calculate exactly but we can approach this subject more narrowly for limited conditions. The first is to consider the speed of the ball compared to it's terminal velocity. If it is a small percentage of the terminal velocity , we can safely ignore the air effects. Below I show you some approximate calculations to take the air effects into account.

The terminal speed is the the maximum speed reached by a particular ball which is dropped from a great height. A thrown or batted ball may travel faster than the terminal speed but it would suffer a large slowing down frictional force from the fluid which is greater than it’s weight. At the terminal speed, the frictional force = gravitational force. With no net force, the acceleration = 0 and the ball falls at a constant velocity.

This varies according to whether and how much turbulence the ball leaves in its wake.

Here’s the basic formula for calculating the drag force of an object moving through a fluid.

Fd = 1/2 r v^{2} Cd A

where r is the density of the fluid and A is the area of the
ball. For air at sea level , r
= 1.2 Kg/m^{3}

The coefficient of drag, Cd ~ 0.5 for most spheres for most sports conditions..

For basketball with v = 20 m/s and area = .046
m^{2}. Fd = 1/2 *1.2*400 *.5* .046 = 5.5N

Weight = 4.5*1.31 =6 N. So this is comparable to drag force. Cd is not a fixed number but depends on Reynolds number and surface smoothness. This calculation of drag force is close enough. Note that this calculation almost replicates the terminal velocity data from the table above.

Range of a ball launched at 45 degrees. R = V0^{2}/g
where V0 is the launch velocity assuming the ball is launched
from the height of the basket.... tall man with jump shot.

For a 3 point shot, R = 24’ = 7.3m. So V0 =sqrt(Rg) =sqrt(7.3 * 10) = 8.5 m/sec.

For 3 point shot, launch velocity in vacuum is
8.5 m/s so drag force (8.5/45)^{2} = .036 = >3.6% of
weight.

Since the range varies as sqrt(K), and the drag force is proportional to the loss of energy. the range is reduced by about 1.8% or 0.13 m. The basket has an allowance of .15/2 m so the loss of range doesn’t cause a clean miss.

This is rather academic. The players obviously practice shooting at the actual target hoop to get used to the way it looks and to get the feel for the launch velocity from experience... the ONLY way.

For shorter shots, the change in range is considerably smaller because the launch velocities are much smaller than the terminal velocity.

In Denver, the air pressure is much decreased so that the air density is significantly decreased. One must be sure to practice shooting enough to get to know the feel of the reduced air density. At Mexico City [ 2340m] the air density is about 75% that of the sea level air density. At Denver [1600m] , the air density would be about 82%.

Of course, the air effects can be your friend! The flight of a frisbee or of a golf ball is much enhanced by the air effects used positively. We will study this later.

The dynamic pressure is the force/area on a flat surface with a fluid incident on it. From the collision of a unit of fluid dm of velocity,v. the momentum dp = vdm. The force produced in this collision is F = Dp/Dt = v Dm/Dt. The increment of mass that collides in Dt is dependent on its volume and mass density, r.

Dm = Dvolume*r = D(area*x)*r . Let the area be a constant and we have:

F = Dp/Dt = v*area*r*Dx/Dt where Dx/Dt = v

F/area = pressure = r v^2. Because the fluid is not usually absorbed at the surface but moves around it, we conventionally define it to be:

Dynamic pressure = 1/2*r v^2

If you double your speed on a bike, the dynamic pressure increases 4x. You pay a lot for going fast.

for a basketball released at the hoop height,
the Range of the ball to basket is R = V^{2} /g.

For a 3 - point shot, the margin of error permitted for the V0 is small.

dR/R = 2 dV/V. dR = .35’ so dR/R = .35/24 = .014 so dV/V = .007.

Your launch velocity accuracy must be within 0.7%!!!! Obviously, the velocity accuracy is less demanding for closer shots as being proportionate to the distance. The foul shot at 15’ requires 1.1% velocity accuracy which is still considerable.

**Energy and Power requirements for launching
a three pointer.**

V = sqrt(Rg) =sqrt( 7.3*10) = 8.5 m/s

K = 1/2 mv^{2} = 1/2 * 0.6 * 8.5^{2}
= 22 Joules

The time for launch is t = d/vavg = .3/4.3 =.069 s

The power is then E/t = 22/.069 = 315 watts.. a modest amt for a short time.