Question about Greenhill twist rate formula

tpcollins · April 13, 2009, 07:55 AM

Not sure where to post this but here goes. I understand the Greenhill formula but I'm not sure how to apply it. I shoot a saboted bullet in my muzzleloader that is about 1.166" long overall including the bullet that protrudes out of the sabot. The sabot alone is .850" long.

When using the Greenhill formula for a regular bullet, there is the portion from the ogive to the tip that doesn't contact the rifling. So when calculating for a muzzleloader sabot, do I include overall length or just the length of the sabot jacket? I'm a bit perplexed since the sabot falls away from the bullet shortly after leaving the muzzle but it's job is to impart spin on the projectile. Thanks.

Jim Watson · April 13, 2009, 08:40 AM

I would use the length of the bullet alone. You are not stabilizing the sabot, just the projectile. Greenhill is pretty coarse, what are you trying to determine?

**Unclenick** · April 13, 2009, 09:24 AM

Even without rifling (smooth bore), a bullet will travel further than it takes a sabot to fall off before tumbling gets a serious hold on it. So, as Jim said, you care what happens to the bullet after the sabot falls away. Use only the bullet information.

The Greenhill formula is premised on artillery shells launched at around 1500-1800 fps. If you have Excel or the free Open Office Suite, I have a bullet stability and twist estimator you can download from my file repository here. It is based on Don Miller's updated and more comprehensive version of the Greenhill formula. The calculations take bullet weight and velocity and air conditions into account, as well as its length and diameter. It is easy to use and the results window estimates both the stability factor from your existing rifling and the twist rate needed to achieve a targeted stability factor.

tpcollins · April 13, 2009, 11:21 AM

Thanks for the response. I have a Knight MK85 with a 1:28" twist. For years I've shot a rather lightweight 180gr bullet/sabot from PR Bullet. It is a .40 cal polymer tipped bullet and the Greenhill formula indicates exactly a 1:28" twist when I do the math - duh.

My step-son (shoots the exact same bullet) and I bought some of PR Bullets new Duplex sabots which is a 195 gr .357 bullet housed in a special flat base .45 cal sabot and then this is inserted into a regular .50 cal sabot - hence the term Duplex.

If I use the .3535" diameter and 1.021" length of the projectile, I come up with a recommended twist rate of 1:18" using the Greenhill formula. However, this Duplex sabot had the least amount of bullet drop and best accuracy from an independent tester. I don't recall what twist the testing was done with but I've not seen a ML with that fast of a twist. That's why I was wondering whether to include the sabot in the formula. Thanks.

dmazur · April 14, 2009, 12:36 PM

My understanding is that the formula recognizes two aspects of spin stabilization. The first is diameter, as larger bullets have more angular momemtum. The second is length, which acts as a lever arm to destabilize the bullet. Stabilization is roughly proportional to diameter squared, inversely proportional to length. (Plus a fudge factor constant for what velocity range we're playing with...)

A heavier bullet can show a need for faster twist, if it is longer or smaller in diameter than a lighter one (or both).

That's about all the insight I have to offer. I'd ignore the sabot as it isn't the stabilized projectile. The angular momentum / length is the lead projectile.

**Unclenick** · April 14, 2009, 12:55 PM

All else being equal, more bullet weight increases stability (more inertia to overcome). More air density means more turning force acting to destabilize the projectile, so higher altitude and higher temperature mean better stability as they reduce air density. Velocity's effect actually depends on the bullet profile. Most of the pointed projectiles have drag functions that cause stability to increase with velocity once you have passed beyond the transonic velocity range. That is, the increased rate of spin from increased velocity in the barrel does more to stabilize the bullet than the higher velocity air flow does to destabilize it.

The other thing that has to be considered is that the bullet's rate of spin (angular velocity) decays more slowly than its forward velocity does when flying through air. Thus, the stability tends to increase with range, not counting the transonic exaggeration of drag.

April 13, 2009, 07:55 AM	#1
tpcollins Senior Member Join Date: February 18, 2009 Location: SE Michigan Posts: 558	Question about Greenhill twist rate formula Not sure where to post this but here goes. I understand the Greenhill formula but I'm not sure how to apply it. I shoot a saboted bullet in my muzzleloader that is about 1.166" long overall including the bullet that protrudes out of the sabot. The sabot alone is .850" long. When using the Greenhill formula for a regular bullet, there is the portion from the ogive to the tip that doesn't contact the rifling. So when calculating for a muzzleloader sabot, do I include overall length or just the length of the sabot jacket? I'm a bit perplexed since the sabot falls away from the bullet shortly after leaving the muzzle but it's job is to impart spin on the projectile. Thanks. __________________ What direction did that last shot at Kennedy come from?

April 13, 2009, 09:24 AM	#3
Unclenick Staff Join Date: March 4, 2005 Location: Ohio Posts: 21,061	Even without rifling (smooth bore), a bullet will travel further than it takes a sabot to fall off before tumbling gets a serious hold on it. So, as Jim said, you care what happens to the bullet after the sabot falls away. Use only the bullet information. The Greenhill formula is premised on artillery shells launched at around 1500-1800 fps. If you have Excel or the free Open Office Suite, I have a bullet stability and twist estimator you can download from my file repository here. It is based on Don Miller's updated and more comprehensive version of the Greenhill formula. The calculations take bullet weight and velocity and air conditions into account, as well as its length and diameter. It is easy to use and the results window estimates both the stability factor from your existing rifling and the twist rate needed to achieve a targeted stability factor. __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member and Golden Eagle Last edited by Unclenick; April 13, 2009 at 09:29 AM.

April 13, 2009, 11:21 AM	#4
tpcollins Senior Member Join Date: February 18, 2009 Location: SE Michigan Posts: 558	Thanks for the response. I have a Knight MK85 with a 1:28" twist. For years I've shot a rather lightweight 180gr bullet/sabot from PR Bullet. It is a .40 cal polymer tipped bullet and the Greenhill formula indicates exactly a 1:28" twist when I do the math - duh. My step-son (shoots the exact same bullet) and I bought some of PR Bullets new Duplex sabots which is a 195 gr .357 bullet housed in a special flat base .45 cal sabot and then this is inserted into a regular .50 cal sabot - hence the term Duplex. If I use the .3535" diameter and 1.021" length of the projectile, I come up with a recommended twist rate of 1:18" using the Greenhill formula. However, this Duplex sabot had the least amount of bullet drop and best accuracy from an independent tester. I don't recall what twist the testing was done with but I've not seen a ML with that fast of a twist. That's why I was wondering whether to include the sabot in the formula. Thanks. __________________ What direction did that last shot at Kennedy come from?

April 14, 2009, 12:36 PM	#5
dmazur Senior Member Join Date: July 5, 2007 Location: Pacific NW Posts: 1,310	My understanding is that the formula recognizes two aspects of spin stabilization. The first is diameter, as larger bullets have more angular momemtum. The second is length, which acts as a lever arm to destabilize the bullet. Stabilization is roughly proportional to diameter squared, inversely proportional to length. (Plus a fudge factor constant for what velocity range we're playing with...) A heavier bullet can show a need for faster twist, if it is longer or smaller in diameter than a lighter one (or both). That's about all the insight I have to offer. I'd ignore the sabot as it isn't the stabilized projectile. The angular momentum / length is the lead projectile. __________________ .30-06 Springfield: 100 yrs + and still going strong

April 14, 2009, 12:55 PM	#6
Unclenick Staff Join Date: March 4, 2005 Location: Ohio Posts: 21,061	All else being equal, more bullet weight increases stability (more inertia to overcome). More air density means more turning force acting to destabilize the projectile, so higher altitude and higher temperature mean better stability as they reduce air density. Velocity's effect actually depends on the bullet profile. Most of the pointed projectiles have drag functions that cause stability to increase with velocity once you have passed beyond the transonic velocity range. That is, the increased rate of spin from increased velocity in the barrel does more to stabilize the bullet than the higher velocity air flow does to destabilize it. The other thing that has to be considered is that the bullet's rate of spin (angular velocity) decays more slowly than its forward velocity does when flying through air. Thus, the stability tends to increase with range, not counting the transonic exaggeration of drag. __________________ Gunsite Orange Hat Family Member CMP Certified GSM Master Instructor NRA Certified Rifle Instructor NRA Benefactor Member and Golden Eagle

April 13, 2009, 08:40 AM	#2
Jim Watson Senior Member Join Date: October 25, 2001 Location: Alabama Posts: 18,539	I would use the length of the bullet alone. You are not stabilizing the sabot, just the projectile. Greenhill is pretty coarse, what are you trying to determine?