rifling rate of twist vs. velocity needed to stabilize

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FLORIDA KEVIN

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I was noticing while checking out the specs for different guns online that with Ruger the rate of twist in .45 colt blackhawks is 1 turn in 16 inches andthe rate in the ..45/.454 casull super redhawk is 1 turn in 24 inches ! i am guessing that this is because the higher velocity 454 would cause excessive drag in the barrel maybe even skip the rifling ? Also I would think that the slower rate of twist would require a higher velocity to spin the bullet fast enough to stabilize the bullet ! Am I on the right track ?:confused: I have noticed that my SRH shoots more precisly (smaller groups ) with 454 loads than .45 colt loads ! So is there an optimum velocity needed to stabilize the bullet for a given pitch of rifling ?
 
i am guessing that this is because the higher velocity 454 would cause excessive drag in the barrel maybe even skip the rifling ? Also I would think that the slower rate of twist would require a higher velocity to spin the bullet fast enough to stabilize the bullet !

First one, no. 5.56mm barrels are commonly 1 twist in 7", which is a whole lot tighter than you're likely to find in a pistol. Second one is the real reason.

The old Greenhill formula will get you roughly within spitting distance of the amount of spin needed to stabilize a bullet, though it only works for blunt bullets, like roundnoses.

150 * caliber^2 / bullet length
 
Thanks ! I new someone had a formula for this ! i am spending way too much time thinking about shooting than shooting lately ! Kevin
 
That looks like about 30,646.268 rpm for a 240 grain hornady xtp/mag bullet ! or about 510 revs /second ! so if the bullet leaves the barrel at 1500 ft/sec and the rifling pitch is 1turn /2 ft ,, the rotational speed is the velocity in ft/sec multiplied by pitch of rifling in ft (1turn in 2 ft )or 750 rev/second.
 
Bullet length/wieght and rifling pitch

Bullet length (and wieght) has more to do with what a given twist rate will stabilize than does velocity.The longer (and heavier) the bullet,the faster the twist rate must be.:scrutiny:
 
The longer (and heavier) the bullet,the faster the twist rate must be

That's my undestanding. So a 7x57 Mauser you plan on shootin 120-140 grain weight out of is going to need a 1 in 10, while a 7x57 you plan on shooting 140 to 190's out of is going to need more like a 1 in 9 or 1 in 8.
 
I think Kevin's got it right. By way of example, both will shoot the same 240 grain bullet, the .45LC @ 950 fps, and the .454 @ 1,900 fps. The difference in twist rate (1:16 & 1:24, respectively) gets them both spinning the same (57,000 rpm).
 
It's not quite as simple as RPMs, though. At higher velocities, bullets become less stable. IIRC, for muzzle velocities over 2,800 fps or so, the 150 becomes 180 in the Greenhill formula.

As for why the .454 has a slower twist, I don't know. I tried once, to improve on the greenhill formula, but factory twist rates in pistols just plain boggle me. There's very little rhyme or reason. Like for Glocks, all 9mm (except .357 SIG) and 10mm calibers use a twist rate of 1 in 9.84". .357 SIG is 1 in 15.98", and .45 is 1 in 15.75". .357 SIG and .45 ACP/GAP are just about as close to polar opposites as it gets, in the calibers Glock makes. So why are they both slower?

Rifles make a little more sense, though, and it is definitely bullet length, nearly independent of velocity. For instance, .22 LRs commonly have twist rates around 1 in 16". That'll stabilize bullets from 32 grains up to about 45. The 60 gr Aguilla Sniper Subsonic rounds often keyhole. 1 in 12", like in many target .223s and the M16A1, will stabilize bullets from 40 gr up to 55-60 gr or so. Massive increase in velocity going from .22 LR to .223 Rem, yet a relatively minor increase in twist rate, proportional to the increase in bullet length.
 
.357 SIG and .45 ACP/GAP are just about as close to polar opposites as it gets, in the calibers Glock makes. So why are they both slower?

Length is proportional to mass divided by the square of the radius. Do the calculation and you'll find a 125 grain .357sig has about the same length as a 230 grain .45 acp, so they have the same twist.

The .357sig has a higher muzzle velocity than the 9mm, so if a 125gr .357sig and 9mm bullet have roughly the same length, the faster one (.357sig) will have the slower twist rate to impart the same net spin at the muzzle.
 
Length is proportional to mass divided by the square of the radius. Do the calculation and you'll find a 125 grain .357sig has about the same length as a 230 grain .45 acp, so they have the same twist.

It's not that simple, and doesn't explain why the twist rates are so much faster than those for 9mm and 10mm/.40.

9mm 147 gr bullet, about .64" long at most. 1 in 29.5" minimum twist rate, by the Greenhill formula.
9mm 125 gr bullet, about .56". 1 in 34".
10mm 180 gr bullet, .64" again. 1 in 37.5"
.45 ACP 230 gr, .64" yet again. 1 in 47.5"
.380 95 gr bullet, about .45" long. 1 in 42", but .380 Glocks have the same twist rate as 9x19mm ones. That one's probably to save on money by not having to make a new barrel thing for .380s.

Makes no sense. Apparently, twist rates are not chosen based on what will stabilize the bullets to a certain degree. If that was the case, both .357 SIG and .40 S&W should be about 1 in 12", since that'd be proportional to 29.5--9.84, and 47.5--15.75.
 
Thanks guys for all your inputs ! When i used the Twist calculator from Tanks rifle shop , and plugged in the data from the bullet i am using it gives a twist rate of twist of about 1 turn in 45 inches ! the actual rifling in my srh is 1 in 24 ! The only other reason I can think of is that ruger uses a faster than calculated twist to compenstae for the loss of rotational speed as the bullet travels down range ! I t makes sense to me that the bullet has a decay of rotational speed just as its velocity decays ! so if you want the bullet to maintain stability downrange you have to start at a much higher initial rotational speed ! this is interesting !! thanks again Kevin !
 
the loss of rotational speed as the bullet travels down range ! I t makes sense to me that the bullet has a decay of rotational speed just as its velocity decays ! so if you want the bullet to maintain stability downrange you have to start at a much higher initial rotational speed !

I don't think this is true. I recall some of this being discussed on another forum some time ago, and I came away understanding that little or no rotational speed is lost in flight.

http://www.rimfirecentral.com/forums/showthread.php?t=166879&page=2
 
Hmm I wonder how it could be proved or disproved ? Probably someone who is a lot better than i am at math and physics !:) thanks for your input guys ! I am going to shutup for a while ! my head hurts !!Kevin
 
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