Everything else being equal, does twist rate impact pressure?

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duck911

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Well, that's the question!

If the same .223 rifle shoots the same load and everything else is theoretically the same except the twist, will a 1-9" twist create the same pressure as a 1-12" twist?

thanks!

--Duck911
 
No, a 1-in-9 inch twist will not have the same pressure curve as a 1-in-12 inch twist with all other variables held equal. Pressures increase as the twist tightens. There is a lag time with tighter twists because more work is required to move the bullet down the barrel because of the increase rotational kinetic energy as the bullet follows the rifling, therefore, the pressure from the burning powder has more time to build up behind the bullet as it propelled down the barrel.
 
Sure it will, but not enough to worry about. If you are following good load data, you will not have a problem.
 
Anytime you increase the twist rate, all else being the equal, pressures will increase. Will the increase be a problem, not unless you are on the edge of maximum pressure already.

Jimmy K
 
In theory, sure. But, in fact, so little as to be irrelivant. And that little only for the time it takes for the bullet's bearing surface to travel a small part of it's total length...a couple of nano-seconds maybe?
 
In some guns, like the older TC Contenders, the pressure created by 1-9" may be too much. Bullberry wouldn't make me a fast twist barrel saying it would spring the frame. The newer Contenders are much stronger, though.
 
I could calculate the change in pressure, but with .223, why bother?

The 223 is registered at 55,000 psi.
The load books max load is much less.
The 223 brass and guns are happy at 75,000 psi.

The little bit added by twist will not matter.

If you want to ask about 270, where is is registered at 65,000 psi, the old books have real loads, and the brass is only good for 70,000 psi, then maybe, just maybe, we could concern ourselves.
 
Yes. I know that in sixguns, John Linebaugh uses a slower twist for his five-shot .45Colt's than Ruger because it does in fact decrease pressure slightly. Although they use heavier bullets, the increased velocity means that you do not need near as fast a twist rate to stabilize the bullet. Which is why the twist rate of the .454 is 1-24", while the .45Colt is 1-16".
 
I think with the 45 colt, it's much like putting spin on a bowling ball - a little doesn't hurt, a lot won't matter.
 
Are they saying the torque will spring the frame then?

No, they said it created additional backthrust and they had broken several frames that way.
 
I think with the 45 colt, it's much like putting spin on a bowling ball - a little doesn't hurt, a lot won't matter.
Apparently it matters enough for Linebaugh to replace a Ruger 1-16" barrel with a 1-20". We're not talking 14,000psi cowboy loads but 50-55,000psi.
 
As an aside, many of the 6.8 SPC barrel makers have both adopted the SPCII chamber spec and also moved from a 1:9.5" or 1:10" twist to a 1:11" twist as a means of reducing peak pressures. Nobody has been able to quantify (at least that I've seen) the actual pressure curve differences attributable to the change in twist rate, but SOMEBODY has been convinced that it's a valid issue.
 
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