Mathematical Challenge re: CE 399

[David Josephs]

Here is a challenge for those members with mathematical or geometrical talents. Using this photo of CE 399,



or any other photo of CE 399, can anyone determine how many inches this bullet would have to travel, in the barrel it was fired from, in order for it to make one complete revolution?

Thanks in advance.

Once we have that figure, either as a percentage of the circumference or as a fraction of .842" (circumference of a .268" diameter bullet), it should be a simple matter to determine the rate of twist of the rifle.

Y'know Bob, if it was "simple" you'd do it, right? I did not mean to make it more complicated, I simply did not understand the mechanics of a bullet moving thru a barrel.


Mr. EISENBERG - Yes; for the record, these cartridges were found on the sixth. floor of the School Book Depository Building. They were found near the south east corner window--that is, the easternmost window on the southern face of the sixth floor of that building.
Mr. Frazier, are these cartridge cases which have just been admitted into evidence the same type of cartridge-- from the same type of cartridge as you just examined, Commission Exhibit No. 141?
Mr. FRAZIER - Yes; they are.
Mr. EISENBERG - That is, 6.5 mm. Mannlicher-Carcano, manufactured by the Western Cartridge Co.?
Mr. FRAZIER - Yes, sir.
Mr. EISENBERG - You gave the weight of the bullet which is found in this type of cartridge. Could you give us a description of the contour of the bullet, and its length?
Mr. FRAZIER - The bullet has parallel sides, with a round nose, is fully jacketed with a copper-alloy coating or metal jacket on the outside of a lead core. Its diameter is 6.65 millimeters. The length--possibly it would be better to put it in inches rather than millimeters The diameter is .267 inches, and a length of 1.185, or approximately 1.2 inches.

So these are Centimeters

[ATTACH=CONFIG]5769[/ATTACH] [ATTACH=CONFIG]5768[/ATTACH]

Each of these ruler marks is 1/10th of a centimeter...
the length of the groove is about 21 cms or .826772 inches...
so it appears to have rotated .25 cms = 0.0984252 inches of turn over .826772 inches of bullet or an 8.4:1 ratio of twist