[jee7s]

Air is a fluid, so there isn't "friction" per se as there aren't two distinct surfaces. Instead, the question is whether the resistance to motion from viscosity or drag is significant.

To determine that you first need a value called the Reynolds number:

Re = (v * L)/(k.v.)

Where v is velocity of the object, L is the characteristic dimension of the object, and k.v. is the kinematic viscosity of the fluid (which is dynamic viscosity normalized by density). For air at 70 F, the kinematic viscosity is 1.64 x 10E4 ft^2/s. For a sphere, the characteristic dimension is the diameter.

If the Re is much less than 1, then the viscous forces dominate. If it's much larger than 1, then the turbulent and drag forces dominate. While it depends on velocity, we can solidly say that the flow is always turbulent for the ball in this year's game, as the velocity where Re = 1 corresponds to something in the 10E-4 ft/s range. If the ball were travelling that slowly, then it's either at the top of an arc or on the ground. If it's at the top of an arc it would only be in that speed range for a very brief time and we can reasonably ignore the change in the dominant force. If it's on the ground, then carpet friction is the dominant force.

For a turbulent flow, the drag is modeled as:

Drag = (1/2) * (p) * (v^2) * (Cd) * (A)

Where p is the density, v is the velocity, Cd is the drag coefficient, and A is the cross sectional area. Density of air at sea level at 70F is about 0.15 lbm/cu ft. The Cd is about 0.5 for a sphere with Reynolds number of 10E5 (you need to look a a VERY non-linear chart to determine this). The cross sectional area is the area of the circle that is the maximum cross section of the sphere, or roughly 3.14 sq ft. Pick a velocity and plug in the numbers and you'll get a result with units of (lbm * ft)/(s^2). Divide by 32 (that's g in ft/s^2) to get pounds of force.

Using these nominal numbers:

Drag (in lbf) = 0.0036797 * v^2 (v in ft/s)

Taking a typical example using 20ft/s as the speed, the drag force on the ball is about 1.47 pounds in a direction opposite the motion of the ball. Whether that is significant is a judgement for you to make. Also note that this drag is non-linearly dependent on velocity. So, that 1.47 pound number at 20ft/s is roughly 0.36 pounds at 10ft/s. If you want to model the path of the ball in the air, you need to simulate it or use some seriously sophisticated math.

If you think in terms of launching the ball toward a goal, the drag is largest (and possibly significant) right after launch when speed is high, and then drops quickly as the speed drops.

Keep in mind that this is idealized, as it doesn't account for surface texture or spin. Spin typically increases drag, and surface texture can either increase or decrease it. But, as an estimate, the drag from the air at a 20 ft/s launch is about a pound and a half.

Hope that helps.