Drag coefficient of 2012 Game Piece

[Ian Curtis]

If you decide to include drag in your calculations, you turn the algebraic equations of projectile motion into differential equations and differential equations are complicated. This is because while the acceleration of the ball in simple projectile motion depends only on the time since launch, if you include drag it depends upon the velocity, and velocity also depends on acceleration -- see how it might get complicated?

It is pretty easy to calculate the magnitude of the force though, and check what the magnitude is.

Drag Force=.5*density*velocity^2*Area*Cd

For a sphere, a Cd of .5 is a pretty good. Entire books have been written on drag, but FIRST robots tend to operate pretty well on the back of a napkin.

The projected area of the ball should be about .34 square feet. That is, the area of a circle with a diameter of 8".

The density of air at sea level in the standard atmosphere is .00237 slugs/ft^3.

Velocity will obviously depend, but I'd say 30 ft/s is probably a pretty decent upper bound.

If you do the math, you'll find that the force of drag on the ball at that speed would be about .36 lbs, and it'll go down fairly quickly, if the ball is traveling at 15 ft/s, it goes down to .09 lbs.