American Scientist

To the Editors:

In "Predicting a Baseball's Path" (May-June), A. Terry Bahill, David G. Baldwin and Jayendran Venkateswaran state that a 95-mile-per-hour pitch drops 1.7 feet between the release point and the contact point and that a 75-mile-per-hour pitch drops 5.7 feet; both drops are due to gravity. These numbers are incorrect.

The relevant relationships are quite simple: Time of flight equals distance (60 feet) divided by velocity, and vertical drop equals one-half the gravitational constant times the square of the time of flight. Using a gravitational constant of 32 feet per second squared, the vertical drop for the 95 mile-per-hour pitch is 2.97 feet and the vertical drop for the 75 mile-per-hour pitch is 4.76 feet.

George C. Messenger
Las Vegas, Nevada

The authors respond:

The following explanation should help to clear up the misunderstanding. Our article said that as soon as a pitcher releases a ball, it's in gravitational free fall, whether it's a blistering fastball or a gentle change-up. A 95-mile-per-hour fastball drops 1.7 feet between the pitcher's release point and the point of a bat-ball collision. Slower pitches fall more. A 75-mile-per-hour curveball, for instance, drops 5.7 feet. Clearly, a ball's pathway to the plate also depends on other forces.

The numbers in the article gave the drop due to both gravity and spin. The distance between the front of the rubber and the back tip of the plate is 60.5 feet. The pitcher releases the ball about 5 feet in front of the rubber, and the batter hits the ball about 1.5 feet in front of the plate. So the ball is in the air about 52.5 feet. For this distance, the drops due to gravity and spin are given in the table above.