04-02-2016, 10:13 PM
Mr. SHANEYFELT. Yes; because we were able to determine the speed of the camera, and thereby accurately determine the length of time it takes for a specific number of frames to run through the camera at this 18.3 frames per second, and having located these frame positions in the street, we took the farthest distance point we had in the Zapruder film which was frame 161 through frame 313.
This was found to run elapsed time from the film standpoint which runs at 18.3 frames a second, runs for a total of 8.3 seconds.
This distance is 136.1 feet, and this can be calculated then to 11.2 miles per hour.
Mr. SPECTER. Is that a constant average speed or does that speed reflect any variations in the movement of the car?
Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.
A distance of .9ft per frame at 18.3 frames per sec = 16.47ft traveled/1.47(1mph) = 11.2mph.
Total distance traveled for initial entries(z161-166 & z168-171) in previous graphic = .9ft
chris
This was found to run elapsed time from the film standpoint which runs at 18.3 frames a second, runs for a total of 8.3 seconds.
This distance is 136.1 feet, and this can be calculated then to 11.2 miles per hour.
Mr. SPECTER. Is that a constant average speed or does that speed reflect any variations in the movement of the car?
Mr. SHANEYFELT. That is the overall average from 161 to 313. It does not mean that it was traveling constantly at 11.2, because it was more than likely going faster in some areas and slightly slower in some areas. It is only an average speed over the entire run.
A distance of .9ft per frame at 18.3 frames per sec = 16.47ft traveled/1.47(1mph) = 11.2mph.
Total distance traveled for initial entries(z161-166 & z168-171) in previous graphic = .9ft
chris