The horizontal line extends outward from the building to a small rectangular block, and then a sloping line illustrates a 5-foot slope from the 175-foot point to the 265-foot point.
(At this point, Representative Boggs entered the hearing room.)
Mr. Frazier. The time of flight of the bullet of approximately 8/100ths of a second and, again, it was necessary to assume—the time of flight of the bullet from the window to this first location of 175 feet is approximately 8/100ths of a second, which means a 2-foot lead on the target. That is, the target would move 2 feet in that interval of time, thereby necessitating shooting slightly ahead of the target to hit your aiming point. That has been diagrammatically illustrated by a 2-foot distance laid off on this rectangular block here, and two lines, very fine lines, drawn back towards the window area.
The right-hand side of Commission's 556 shows the same rectangular block, again with two lines drawn to it, one illustrating the point of aim and the other the amount of lead which would be necessary to strike an object aimed at which was moving, according to the time of flight of the projectile.
Mr. Eisenberg. And you calculated the speed of the car by translating the figures on total time elapsed between first and third shots?
Mr. Frazier. Yes, sir. The time—the speed of the moving object was calculated on the basis of an assumed 5.5-second interval for a distance of 90 feet, which figures out mathematically to be 11.3 miles per hour.
Mr. Eisenberg. Now, you said before that in order to give this 2-foot lead, you would have to aim 2 inches—for a target going away from you, you would have to aim 2 inches above the target, or in front of the target.
Mr. Frazier. 2 feet in front of the target, which would interpolate into a much lower actual elevation change.
Mr. Eisenberg. The elevation change would be 2 inches, is that it?
Mr. Frazier. Well, no. It would be on the order of 6 to 8 inches.