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Mathematical Challenge re: CE 399
#5
David Josephs Wrote:
Bob Prudhomme Wrote:
David Josephs Wrote:
Quote: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?


Hi Bob..


Isn't a full rotation defined by the barrel itself... the grooves in the barrel complete 4 rotations,or something like that? or is it a 4 groove barrel with an unknown number of rotations?

The distance between these complete rotations within the barrel will give you your distance since it does not matter whether the bullet falls out of the barrel or is shot out,
.... the bullet wouldn't rotate at all without those grooves I imagine...

So I looked up rifling and found this on wiki: do we know the Twist Rate of the MC?
DJ

Rifling is the process of making helical grooves in the barrel of a gun or firearm, which imparts a spin to a projectile around its long axis. This spin serves to gyroscopically stabilize the projectile, improving its aerodynamic stability and accuracy.

Rifling is often described by its twist rate, which indicates the distance the rifling takes to complete one full revolution, such as "1 turn in 10 inches" (1:10 inches), or "1 turn in 254 mm" (1:254 mm). A shorter distance indicates a "faster" twist, meaning that for a given velocity the projectile will be rotating at a higher spin rate.

http://en.wikipedia.org/wiki/Rifling


Yes, David, those definitions are correct. A rifle barrel that has rifling grooves with a 1:10 rate of twist means that the bullet will make one complete rotation in the space of 10 inches of the barrel's length. It will leave the distinctive rifling marks, as seen on CE 399.

However, different rifles have differing rates of twist in their riflings, as well as the "lands and grooves" being of varying widths. It is not uncommon, in sporting rifles, to see rates of twist vary from 1:7 to 1:12. Naturally, the tighter twist will leave much tighter rifling marks on a bullet. As CE 399 is the evidence presented by the WC, I am attempting to establish the rate of twist of the barrel it was fired from to see if it matches the known rifling specs for Carcano rifles.

It is my opinion that CE399 was fired from the MC planted on the 6th floor well before-hand and was given to Elmer Todd by Chief Rowley...

Bob... how can you tell from the bullet what the twist rate is... this is the exhibit that supposedly shows CE399 and a bullet fired from the C2766 are the same...
http://aarclibrary.org/publib/jfk/wc/wcv..._0142a.htm

yet unless one knows the twist rate of the barrel, i assume there is no way to reverse engineer that info from the bullet... ??

David

The riflings will leave an exact impression on the bullet of the last part of the inside of the barrel (the muzzle) to contact the bullet. The copper jacket of the bullet is quite malleable, and the riflings will imprint onto the bullet jacket just as if they were clay. If the rate of twist at the muzzle of the rifle is 1:8, an imprint, corresponding to a 1:8 rate of twist, will be left on the bullet.

I don't doubt that CE 399 was fired from the rifle found on the 6th floor of the TSBD. That is not at issue here. What I am trying to determine is the rate of twist of the rifle found on the 6th floor.

Think about this really hard for a second, Dave, and don't make assumptions without the knowledge to back them up. Why would we need to know the rate of twist of the barrel's rifling? We do not even need to know the length of the bullet in the photo we are told is CE 399. All we need is to measure the diameter of the bullet in the photo and compare that to the known diameter of a 6.5 Carcano bullet, which is .268". This will give us a ratio that we can use to establish a 1 inch length on the bullet. In that 1 inch, we need to determine how much of the circumference of the bullet the rifling mark has covered. This is where I have problems and why I am asking help from more experienced members. 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.
Mr. HILL. The right rear portion of his head was missing. It was lying in the rear seat of the car. His brain was exposed. There was blood and bits of brain all over the entire rear portion of the car. Mrs. Kennedy was completely covered with blood. There was so much blood you could not tell if there had been any other wound or not, except for the one large gaping wound in the right rear portion of the head.

Warren Commission testimony of Secret Service Agent Clinton J. Hill, 1964
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Messages In This Thread
Mathematical Challenge re: CE 399 - by Bob Prudhomme - 08-03-2014, 09:13 PM
Mathematical Challenge re: CE 399 - by Marc Ellis - 10-03-2014, 06:39 AM
Mathematical Challenge re: CE 399 - by Marc Ellis - 11-03-2014, 08:08 AM

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