THE CURVE BALL

As the force acts on the ball, it will be deflected along it's flight path. If we neglect the viscous forces on the ball (which will slow it down and change the magnitude and direction of the force), we have a constant force always acting perpendicular (at a right angle) to the flow direction.

The resulting flight path is a circular arc. On the figure, we see the trajectory of the baseball as it moves from the pitching mound to the plate (right to left). Each white dot is the location of the ball at .05 second intervals (it gets to the plate pretty fast!). The radius of curvature of the flight path depends on the velocity of the pitch and the acceleration produced by the force. We can solve for this acceleration from Newton's second law of motion using the force for a spinning ball and the mass of the baseball (5 oz.).

Since the radius of curvature depends on the force, all the factors that affect the force will also affect the trajectory. The higher the spin, the sharper the curve; the higher the velocity, the straighter the curve. At higher altitudes, the density is lower and the curve will straighten out. (It's very hard to make a ball curve on Mt. Everest, no matter how much it spins.)

Knowing the radius of curvature and the distance from the pitching mound to home plate, we can also calculate the distance that the ball is deflected (Yd) along the flight path. The pitcher can release the ball at different distances from the center of the plate (Yp). The difference between Yp and Yd will give the final location of the pitch relative to the center of the plate.

All that is necessary to create lift is to turn a flow of air. An airfoil of a wing will turn a flow, and so will a rotating cylinder. A spinning ball will also turn a flow and create a force.

The exact details are fairly complex. Next to any surface, the molecules of the air will stick to the surface, as discussed in the properties of air slide. This thin layer of molecules will entrain or pull the surrounding flow for a spinning ball in the direction of the spin. If the ball were not moving, we would have a spinning, vortex-like flow set up around the spinning ball (if we neglect three-dimensional and viscous effects in the outer flow).

If the ball is moving through the air at some velocity, on one side of the ball the entrained flow will oppose the free stream flow, while on the other side of the ball, the entrained and free stream flows will be in the same direction. If we add the components of velocity for the entrained flow to the free stream flow, on one side of the ball the net velocity will be less than free stream; while on the other, the net velocity will be greater than free stream.

The flow will then be turned by the spinning ball, and a force will be generated. Because of the change to the velocity field, the pressure field will also be altered around the ball. The magnitude of the force can be computed by integrating the surface pressure times the area around the ball. The direction of the force is perpendicular (at a right angle) to the flow direction.

The right part of the slide shows a view of the flow as if we were moving with the ball looking down from above. The ball appears stationary and the flow moves from left to right. On this figure the ball spins clockwise, so the free stream flow over the top of the ball is assisted by the circular flow; the free stream flow below the ball is opposed by the circular flow. In the figure we can see that the net streamlines around the ball are distorted because of the spinning. The net turning of the flow has produced an upward force.

We can consider the spinning ball to be a three-dimensional version of the two-dimensional rotating cylinder. We can then use the Kutta-Joukowski lift theorem for cylinders to approximate the magnitude of the force generated by a spinning ball. On the left part of the figure we have surrounded the spinning ball by a cylinder whose axis of rotation is the same as the ball.

The Kutta-Joukowski theorem tells us how the lift per unit length of the cylinder is related to the velocity, the density, and the strength of the vortex. The length of cylinder used in this example is equal to twice the radius (diameter) of the ball (2b). This force would act over a cross-sectional area, which would appear as a square of side 2b if viewed along the force vector. For the same view, the ball would have a projected area of pi times the radius (b) squared. So we can correct the force generated by the cylinder by the ratio of these areas to get the final approximate force equation;    F = r GV*2b*pi/4