(Testimony of Robert A. Frazier)
Mr. Eisenberg.
which is 60 feet above the ground, and several more feet above the position at which the car was apparently located when the shots were fired.
Second, that the length of the trajectory of the first shot was 175 feet, and that the length of the trajectory of the third shot was 265 feet.
And third, that the elapsed time between the firing of the first and third shots was 5 1/2 seconds.
Based on those assumptions, Mr. Frazier, approximately what lead would the assassin have had to give his target to compensate for its movement--and here I would disregard any possible defect in the scope.
Mr. Frazier.
I would say he would have to lead approximately 2 feet under both such situations. The lead would, of course, be dependent upon the direction in which the object was moving primarily. If it is moving away from you, then, of course, the actual lead of, say, 2 feet which he would have to lead would be interpreted as a considerably less lead in elevation above the target, because the target will move the 2 feet in a direction away from the shooter, and the apparent lead then would be cut to one foot or 12 inches or 8 inches or something of that nature, due to the movement of the individual.
Mr. Eisenberg.
Have you made calculations to achieve the figures you gave?
Mr. Frazier.
I made the calculations, but I don't have them with me.
Mr. Eisenberg.
Could you supply these to us, either in further testimony or by letter, Mr. Frazier?
Mr. Frazier.
I have one object here, a diagram which will illustrate that lead, if you would like to use that. This is drawn to scale from these figures which you quoted as building height, and distances of 175 feet and 265 feet.
Mr. Eisenberg.
For the record, these figures are approximations of the figures believed to be involved in the assassination. Will you supply the data at a later date?
Mr. Frazier.
Yes; I can furnish that.
Mr. Eisenberg.
May I have permission to introduce this as 556?
Mr. Mccloy.
That will he admitted.
(The document referred to was marked Commission Exhibit No. 556, and received in evidence.)
Mr. Eisenberg.
Could you show the lead in that diagram, Mr. Frazier?
Mr. Frazier.
In Commission Exhibit 556, it shows a triangular diagram with the vertical line on the left-hand side illustrating the height of the building. The figures of a 60-foot building height plus--
Mr. Eisenberg.
That is height of the muzzle above the ground?
Mr. Frazier.
No--window sill--60-foot window sill height above the ground, with an assumed 2- foot height in addition to accommodate the height of the rifle above the possible. the possible height of the rifle above the window sill.
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.
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