twist rate question

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moooose102

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i want to be sure i understand this correctly. the faster the twist rate, the longer the projectile, or the heavier the projectile? i know, usually, it is one in the same, but i was doing some thinking, around the use of other materials than lead or copper coated lead. and i want to know which is the appropriate answer.
 
longer

but actually, it would be more correct to say the longer the projectile the faster twist is necessary
 
Not necessarily. Observe flechettes, long, no twist, stable. Many variables go into determining the best twist rate, length is an important one but not the only one. BC and SD enter into the equation also.
 
Flechettes are so long they can't effectively be stabilized by spinning. Which is why they have the little wings on the back.

Longer bullets require a faster twist to stabilize. BSW
 
The rule of thumb is what you stated, longer is usually heavier.But for every rule there are exceptions.But I can't recall one.

I think the longer aspect fits the twist rate more than the heavier.

Example,.224 dia. 50 gr. Barnes Varmint Grenade flat base hollow point bullet, stated for 1-10 or faster. I cut one open, the first 1/3rd of the bullet is just a hollow jacket, the majority of the weight is in the last 2/3rds, and the bullet is exceptionally long.

The majority of the 50 gr bullets being produced in .224 dia are being shot in rifles with 1-12 and 1-14 twist rates and stabilize with no problem, but the length of the Barnes Grenade calls for faster twist to stablize.
 
Another "Longer" Example

Sneaky Pete here: Hornady makes a .224-75gr BTHP(match) and a .224-75gr A-Max (it's longer with a smoother curv and a higher BC) . The 75gr BTHP will stabilize in a 1:9 twist BUT the 75gr A-Max won't and requires a 1:8 or faster. I tried both in a 1:9 24" match bbl. and in a 1:8 20" match bbl. Nedless to say the 75gr A-Max shot better in the 1:8 20"bbl. Neither shot worth a damn in the Origional SP-1 1:12 bbl.
 
I believe the "accepted" preferred twist rate,even today is still based on George Greenhills formula.The way I understand it,bullet weight is irrelevant. Only bullet length & caliber are factors.If anyone on this thread is interested,I do have the formulas to figure both SD & Greenhill.They are usually parallel,except when "odd" bullets;all cu etc. are used.
 
I think what you may be looking for is actually sectional density.

Taking two bullets of the same length and caliber, but different materials, the lighter one will have a lower sectional density. You must spin that lower SD bullet faster in order to gyroscopically stabilize it.

briansmithwins is correct: flechettes are aerodynamically stabilized with fins, not gyroscopically stabilized like a bullet.
 
The way I understand it,bullet weight is irrelevant. Only bullet length & caliber are factors.If anyone on this thread is interested,I do have the formulas to figure both SD & Greenhill.They are usually parallel,except when "odd" bullets;all cu etc. are used.
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What he said^
 
Greenhill's formula as I recall is an inexact approximation which assumes a copper jacketed lead bullet of about 90% lead and 10% copper/gilding metal. When using that formula with it's bullet composition assumption, length is the critical factor. If using bullets of significantly different densities than assumed, the formula becomes even less accurate. When deriving a formula or computing from scratch, one has to consider mass as Greenhill did.
 
I wonder if you could make a long bullet more stable (and thus work in a slower twist) by constructing it with a forward weight bias. Maybe something like a tungsten insert in the tip and/or an aluminum insert in the base?
 
Yes, Greenhills formula is an approximation.I believe it was based on a projectile with a specific gravity of 10.9.(with pure copper being 8.96 & pure lead at 11.35.)It is,however,very accurate even by todays standards.
Of the three discussed here; Greenhill,BC,and SD,only SD is a physical fact.
BC = an estimate(usually quite accurate) of aerodynamic potential,must be "proven" for any specific situation.
SD = mass & square of diameter.
Greenhill = theory
After saying that,I believe that bbl & bullet quality will be the largest factors of all.(within reasonable limits)
 
I think it has more to do with mass than length, but as others have pointed out, the longer bullets are usually more massive.
The more mass a bullet has, the more violently it needs to be rotated by the barrel in order to give it rotational momentum. This would require a faster twist.
Lighter bullets wouldn't require as tight of a twist from the barrel in order to get them spinning.

Jason
 
speed comes into affect as well. Just not as much as length. Length is (From what I am told) the determining factor.
 
speed comes into affect as well. Just not as much as length. Length is (From what I am told) the determining factor.
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I can see how speed plays into it, but I fail to see how a longer bullet, other than causing an increase in mass, would require a faster twist. Someone care to explain this to me? I'm confroosed :confused:.

Jason
 
Jason_G said:
I can see how speed plays into it, but I fail to see how a longer bullet, other than causing an increase in mass, would require a faster twist. Someone care to explain this to me? I'm confroosed
Click to expand...

Longer bullet = longer lever around the bullet's center of mass for aerodynamic forces to act on.
 
"I think it has more to do with mass than length". Try to look at it this way: A 500 plus grain .75 cal round ball or a baseball, would have far less potential to become unstable,as in wobble(yaw)than say, a needle,toothpick,or a 3 ft long piece of re-bar.Yes,these are extremes,but the principle still applies.
 
speed comes into affect as well.
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A longer projectile requires a faster rate of spin, you could call it RPM's or RPS's rotations per second. If a longer barrel can push a bullet faster it's rifling twist rate can be slower than a shorter barrel that achieves less velocity.
 
Longer bullet = longer lever around the bullet's center of mass for aerodynamic forces to act on.
Click to expand...

Wouldn't that mean that a longer bullet would be easier to stabilize, assuming the mass is the same?

Try to look at it this way: A 500 plus grain .75 cal round ball or a baseball, would have far less potential to become unstable,as in wobble(yaw)than say, a needle,toothpick,or a 3 ft long piece of re-bar.Yes,these are extremes,but the principle still applies.
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Okay, but why is a football easier to stabilize than a round ball? Why not use round balls like muskets did instead of bullets? If shorter and rounder is easier to stabilize, then why do such long bullets like 6.5mm and 6mm target loads enjoy such success?
I know there's alot more to the equation than this, but I am still failing to see how a longer bullet would necessitate a faster twist other than bringing more mass into the equation. I'm not saying that it's wrong, but I haven't seen any examples yet that have led me to understand how it could be so.

Jason
 
Jason G, Maybe? this will help clarify.The point at which "extra spin rate"doesn't have a marked affect is at a much lower RPM for a round projectile.A round projectile is far more susceptible to outside influences & surface imperfections,because it "doesn't care" how it "cuts wind".Getting the "required" mass into the smallest possible frontal area,and "cutting the wind in a straight line" is the goal.
 
You could look at it this way, two barrels of the same length one has a 1:12 twist and the other has a 1:8 twist shooting the same projectile at the same velocity, the bullet from the 1:12 twist barrel will begin to tumble or destabilize sooner than the bullet fired from the 1:8 twist barrel. What is the only difference between the two projectiles ?
Answer; It's rate of spin.
What determines a bullets rate of spin ?
answer; The twist rate of the barrel and the velocity of the bullet.
This calculator is pretty broad but it illustrates Greenhill's formula.http://www.tanksrifleshop.com/twistcalc.htm
Does any one have a formula to calculate the decrease of the rate of rotation ? I bet that formula would be full of variables :D
I should ad, I am not an expert, my understanding of this is from a laymen's point of view :)
 
BTW, flechettes are twist stabilized. The twist is typically imparted by fins rather than rifling.
 
Okay, I think I understand why a longer bullet would need a faster twist now...

To my way of thinking earlier, I was relating it all to the law of inertia. An object at rest tends to stay at rest. The more mass an object has, the greater a force has to be in order to get the object in motion. In this light I totally understood how a more massive bullet would require a faster twist in order to get the proper rotation for stability. And I think I am correct in that a bullet of longer design but same exact mass would achieve the same rotation if fired from the same rifling.

The part I wasn't thinking about was the fact that a longer bullet might require more spin to achieve stability in the air after it leaves the barrel (duh), even if it has the same mass as a shorter bullet.
I don't know why I was having such a brain block there, but I was wrapped around the axle on what was happening in the bore and not thinking about after the bullet exits.

I apologize for my earlier brain-farts.

Jason
 
GunTech said:
BTW, flechettes are twist stabilized. The twist is typically imparted by fins rather than rifling.
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Flechettes (and arrows and darts) aren't spin stabilized, that's purely the work of the fins. The reason for the twist is so that the effects of slightly misaligned fins won't affect accuracy too badly.

Misaligned fins with no twist = curving off in unpredictable directions.

Misaligned fins with twist = corkscrewing trajectory, but still in the direction of the target.
 
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