I am new to the AR and have a question concerning twist rates. First off, ya’ll are to blame for my new obsession. All of the good advice and info. here has made me an AR junky. My wife thanks you. Here is what I have calculated(could be wrong): 1/7" twist rate in an 11.5" barrel means a bullet would twist 1.623 times upon exiting the muzzle. 1/9" twist rate in a 14.5" barrel–>1.438X 1/9" twist rate in a 16" barrel–>1.611X Does the twisting of the bullet accelerate immediately upon leaving the barrel for stabilization? Or, is my thinking all wrong and a bullet twists many times before exiting the barrel? Does the shorter barrel have a higher twist rate because there is less distance for the bullet to be twisted for stabilization? If this is so, my prior calculations show a bullet from a 11.5" with 1/7" twist at 1.623 twists for bullet before leaving muzzle does not equal the total “twisting” in a 16" with 1/9" twist rate at 1.611. Maybe too slight to make a diff. Yes, thanks to ya’ll this is what keeps me up at night.
I also realize my thinking may be way off mark. Just trying to think about the mechanics of the AR so I can become a better and more knowledable shooter. Any insight, thoughts, or help greatly appreciated.
Twist rate is a function, technically, of bullet length (or bullet bearing length), and practically of bullet weight. In general, the heavier the bullet, the slower the velocity in the forward direction, and therefore the faster the twist needed to achieve the desired rotational velocity. However, the list of factors that affects the consistency of exterior ballistics is long, and therefore one doesn’t calculate “optimal” twist with any degree of precision; bracketing is the norm.
Correct
no it continues at the same speed that it exits the barrel, only slowing by natural forces.
@3000fps a bullet from a 1:7 twist is spining at 5143 revolutions per second
same round from a 1:9 is spinning at 4000 revolutions per second
you need faster rifling for longer (heavier/slower) bullets, and slower rifling for shorter (lighter/faster) bullets… Hope this helps
You might want to check your calcs:
If the slug is spinning at 5143 rps, and
If the slug has to travel 7 feet forward for each revolution; then
The implied velocity in the forward direction is 36000 fps.
Rev/Sec * Feet/Rev = Feet/Sec.
Note also that your results imply rotational rates of 240,000 RPM and 309,000 RPM; I doubt the slugs could hold together at those rates.
I get values of 333 and 429 RPS, respectively.
the slug only travels 7 inches for each revolution not 7 feet. correct? thats how I based my calcs
I believe rifling is 1:7" and 1:9" not feet
Of course! (Slaps forehead.)
But the resulting numbers (RPM) are a bit startling, no?
I rechecked I belive they are right
WIKI:
Bullet RPM
Bullets leaving a rifled barrel can spin at over 300,000rpm depending on the muzzle velocity of the bullet and the twist rate. The rotational speed of the bullet can be calculated by using the formula below where MV is muzzle velocity. A bullet fired from the M16A2 rifle mentioned above with at twist rate of 1 in 7-inch (180 mm) and a muzzle velocity of 3050 ft/s would spin at ~315,000 RPM.[9]
MV x (12/twist rate in inches) x 60 = Bullet RPM
Excessive rotational speed can exceed the bullet’s designed limits and the resulting centrifugal force can cause the bullet to disintegrate in a radial fashion
http://en.wikipedia.org/wiki/Rifling
Yes, I corrected my calcs and (now) get the same numbers you do.
I’m just surprised that a jacketed slug can hold together at a rotational speed of 250,000 - 300,000 RPM, particularly a hollow point. I’m too foggy (obviously) to calculate the force of centripetal acceleration on the jacket, but intuitively it seems it would be huge.
here you go
centripetal acceleration (a) = 990324909.74729 foot/second^2
Centrifugal force = 0.278778 Gs
Damn we have gotten way off topic
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