What is rate of twist?

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I understand what the nomenclature specifies when it says 1-12" or 1-10" (one rotation every ten or twelve inches) but what does that mean to me the shooter?

I read the other day on this site (which I think is awesome) some guys talking about rate of twist and the amount of powder they were using in their handloads.

What should I consider (if anything) as a "Novice" concerning rate of twist when buying a rifle that I will be strictly shooting factory ammo out of?

The majority of rifles I am considering have 22" barrells with 1-12" rate of twist, however there is one rifle I am looking at that has a 24" barrell with
1-10". Is barrell length the key to the equation?

I'm looking forward to learning from you.

Thanks,
Novice
 
Faster twist stabilizes longer bullets; slower twist stabilizes shorter bullets.

For example, 1-9" and 1-7" are common twists in an AR-15 barrel. They stabilize the spitzer (long, skinny, pointed) bullets in the 5.56 NATO round. 1-7" is used for heavier bullets than 1-9", but it's not the weight that matters, it's the length.

A regular old .22 is usually 1-12" -- the bore is almost exactly the same size as an AR-15, but .22LR bullets are blunt and relatively short. So, a slower twist is used to best stabilize them.

WRT what twist you want, it depends on the caliber -- since the bullet length for a given bullet weight is determined by the diameter of the bullet.
 
Rate of Twist is needed to stabelize a bullet.

long bullets need short twist and short bullets need a longer twist.

many will say that it is about the weight of a bullet, but that is just a coincident, since for example a 90gr 5.56 bullet with normaly be longer then a 43gr 5.56 bullet.

without knowing the Caliber, it will be hard to tell your the best rate of twist or a 'Novice' gun

(ArmedBear was faster.)
 
If I understand correctly, after reading several materials on twist rate, the lenght of the bullet is not the only important variable...caliber plays a role (larger caliber tend to need less spin) and muzzle velocity as well.
 
caliber plays a role (larger caliber tend to need less spin) and muzzle velocity as well.

I'm sure that's true.

Bottom line, whatever rule of thumb might lead you to pick 1:10" twist in a .24 caliber barrel isn't the same as what would make you pick 1:10" twist in a .30 caliber. Each must be considered in the light of all the variables.
 
What caliber and what you want to use the rifle for are the questions that you need to answer before anyone can give a knowledgeable response.
If you want to hunt deer type game you'll get one answer. If you want to shoot long range targets, chances are you'll get something different.
Let us know.

NCsmitty
 
Pulse said:
many will say that it is about the weight of a bullet, but that is just a coincident, since for example a 90gr 5.56 bullet with normaly be longer then a 43gr 5.56 bullet.

I'd like to see some rigorous proof of that ... it just seems like semantics. Maybe the mass is important and the length is "coincident". Is it likely that the optimum twist rate for dimensionally identical bullets made from depleted uranium, lead/copper, aluminum, delrin, wood, etc would be the same?

The rotational inertia of a cylinder (kind of like a bullet) about its long axis (the center of the bore) is equal to (mass*diameter^2)/8. Mass is a function of length the same way it's a function of the diameter of the cylinder and the density of the material. In other words, mass is function of volume and density, volume is a function of cross-sectional area and length. Since the density of most bullets is kind of a constant, as the length increases the volume increases and the mass increases.

If anyone has any PROOF or logic or math or science pertaining to this then I'd sure like to see it.

:)
 
Thank you all for the info. so far. I am eating this stuff up.

The caliber I am considering is 30-06for deer hunting and I prefer to shoot a 165 gr. bullet.

Novice
 
Well, 1858, I'm not a bullet engineer, but I've had some hands-on experience that lends strong support to what saturno_v says.

This bullet is VERY long, used in .45-70. I've cast, loaded, and used them on paper and for hunting. It's a Postell 535 grain bullet, well over an inch long.

457132.jpg


A 1:18" twist in a Sharps rifle will stabilize it just fine. Nevertheless, the M4 has a 1:7" twist to stabilize "heavy" 5.56 rounds that are a fraction of the Postell .45 bullet's length, and a smaller fraction of its weight.
 
Novice-

Any normal poduction .30-06 you buy will be made to stabilize a 165 grain bullet. No worries. 165 is about the ideal theoretical bullet weight for .30-06. Some guns will, for one reason or another, group better with 150 or 180, but 165 is about the middle of the bell curve.
 
Greenhill formula

A mathmetics professor at cambridge and the woolrich military academy wrote the the formula for finding optimum bullet weight for any given twist rate was to divide 150 by the length of the bullet in hundredths of a inch then divide again by its dia. So a 5.56 in a 55 grain fmj length is .647 and its diameter is .224. So .647 divided into .224 gives you 2.89 calibers divide that by 150 and you get 51.90 calibers then multiply 51.90 by bullet diameter of .224 gives you a ideal twist rate of 11.63. The closest thing being a 1:11 twist for the 5.56. This cmae from the abc's of reloading manual 8th edition
 
If anyone has any PROOF or logic or math or science pertaining to this then I'd sure like to see it.

the standart 5.56 NATO 62 grain round is best used in 1:7 Twist rate
the Swiss GP90, our version of the 5.56 Round, is made from denser Metal and Alloys -> is therefore storter to complay with NATO regulations and for our standart 63gr round, one should use a Rifle with 1:10 Twist.
 
Originally Posted by Pulse
many will say that it is about the weight of a bullet, but that is just a coincident, since for example a 90gr 5.56 bullet with normaly be longer then a 43gr 5.56 bullet.

I'd like to see some rigorous proof of that ... it just seems like semantics. Maybe the mass is important and the length is "coincident". Is it likely that the optimum twist rate for dimensionally identical bullets made from depleted uranium, lead/copper, aluminum, delrin, wood, etc would be the same?

The rotational inertia of a cylinder (kind of like a bullet) about its long axis (the center of the bore) is equal to (mass*diameter^2)/8. Mass is a function of length the same way it's a function of the diameter of the cylinder and the density of the material. In other words, mass is function of volume and density, volume is a function of cross-sectional area and length. Since the density of most bullets is kind of a constant, as the length increases the volume increases and the mass increases.

If anyone has any PROOF or logic or math or science pertaining to this then I'd sure like to see it.

Yes, look no further than the military's 1:7 twist, the SS109, and their tracer. The SS109 with it's steel insert makes it longer than a "normal" 62gr 5.56mm bullet. As such, some can get away with using an older 1:12 AR15 barrel with lead core 62gr rounds, BUT when the SS109 is tried the bullets go sideways always. 1:9 is the best twist for SS109 bullets; however, the .mil's tracer round, although similar in weight, is much longer and requires the 1:7 rate. That's way 1:7 is the standard not 1:9.

Russian 5.45 is another example. 52gr and an inch long requires between about a 1:7-8 twist to work. Oddly enough that's about the same lenght as a 77gr 5.56 SMK which is best pared with a 1:8.

There's always exceptions, like some poeple's 1:9 AR15 barrels will stablize a 77gr SMK where other's 1:9 doesn't have a prayor.
 
To the OP's question on .30-06: 1:12 will work with almost everything under the sun that'll fit in a .30-06 case.

If you begin to handload or find some "custom" loads with 200gr+ bullets or real long bullets like the A-max, you might hit a problem then.
 
A mathmetics professor at cambridge and the woolrich military academy wrote the the formula for finding optimum bullet weight for any given twist rate was to divide 150 by the length of the bullet in hundredths of a inch then divide again by its dia. So a 5.56 in a 55 grain fmj length is .647 and its diameter is .224. So .647 divided into .224 gives you 2.89 calibers divide that by 150 and you get 51.90 calibers then multiply 51.90 by bullet diameter of .224 gives you a ideal twist rate of 11.63. The closest thing being a 1:11 twist for the 5.56. This cmae from the abc's of reloading manual 8th edition


That Professor in question, George Greenhill, suggested to replace 150 with 180 for muzzle velocities over 2800 fps in that formula.
 
jbkebert, the COMPLETE equation is as follows:

twist_rate.jpg


So let's compare the twist rates of the bullet that you mentioned having

Length = 0.647"
Diameter = 0.224"
SG = 10.9"

to bullets made from balsa wood (SG = 0.2), aluminum (SG = 2.7) and lead (SG = 11.35).

The "ideal" twist rates using the Greenhill Formula (rule of thumb and not rigorous) are as follows:

Balsa wood = 1.57
Aluminum = 5.79
Lead = 11.87

As I said before, how is twist rate a function ONLY of length and not of weight/mass/density? That was my question which should be directed to Pulse.


KW, providing examples of bullets with the same or similar mass but different densities (and hence different lengths) and then drawing a conclusion from that is meaningless.

Anyway, this discussion got me interested enough to calculate the "ideal" twist rate for a 168gr SMK HPBT and a 190gr SMK HPBT (two bullets that I shoot) in .308. I measured the lengths of a couple of each with the average for the 168gr being 1.206" and the average for the 190gr being 1.362". The 190gr bullet is 13.1% heavier and 12.9% longer than the 168gr bullet. The "ideal" twist rates are 11.8 and 10.4 respectively. Hmmm ... so in this example, the mass is increasing at a faster rate than the length.

I have to wonder what formula George Greenhill would derive today given the exotic bullets, ogive profiles, composite tips, hollow points and materials available, along with the data processing and modeling capabilities we take for granted.

P.S. And where did that mysterious "C" come from ... empirical data maybe. :confused:

:)
 
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What should I consider (if anything) as a "Novice" concerning rate of twist when buying a rifle that I will be strictly shooting factory ammo out of?
Nothing. Unless you are going to try to "stretch" a caliber by trying to extend its performance or range then standard twists will almost always work fine.
 
A final thought here, since it's inception a hundred years ago, the 30'06 has worn a 1in10 twist. It works for the full spectrum of bullet weights from 100gr plinkers to 220gr RN. It remains a favorite of many shooters for a good reason.It works.

NCsmitty
 
Thanks NCsmitty for the last word because that is where I was leaning (although in this case I understand there is no real crisis in the choice before me).

I am leaning toward the 24" barrell with 1-10" twist, not soley because of the barrell but that is one element in the rifle among others that I like.

Thanks again,
Novice
 
In lay terms , in the same caliber longer bullets weigh more, not always , but most of the time.

That mass has to go somewhere and it can't increase in diameter, so it has to get longer as it gets heavier. Thus requiring more twist to stabilize.

When in doubt over spin is better than under spin.

Some newer Green (no lead), bullets made up of sintered metal with an exterior jacket,(Barnes Varmint grenade) are exceptionally long for their weight. Which goes hand in hand that length, more than weight, dictates greater twist rates.
 
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