## Calculating Bullet RPM — Spin Rates and Stability

Most serious shooters can tell you the muzzle velocity (MV) of their ammunition, based on measurements taken with a chronograph, or listed from a manufacturer’s data sheet. (Of course, actual speed tests conducted with YOUR gun will be more reliable.)

**Bullet RPM = MV X 720/Twist Rate (in inches)**

Photo by Werner Mehl, www.kurzzeit.com, all rights reserved.

However, if you ask a typical reloader for the rotational rate of his bullet, in revolutions per minute (RPM), chances are he can’t give you an answer. Knowing the true spin rate or RPM of your bullets is very important. First, spin rate, or RPM, will dramatically affect the performance of a bullet on a game animal. Ask any varminter and he’ll tell you that ultra-high RPM produces more dramatic hits with more “varmint hang time”. Second, RPM is important for bullet integrity. If you spin your bullets too fast, this heats up the jackets and also increases the centrifugal force acting on the jacket, pulling it outward. The combination of heat, friction, and centrifugal force can cause jacket failure and bullet “blow-ups” if you spin your bullets too fast.

**Accuracy and RPM**

Additionally, bullet RPM is very important for accuracy. Nearly all modern rifles use spin-stablized bullets. The barrel’s rifling imparts spin to the bullet as it passes through the bore. This rotation stablizes the bullet in flight. Different bullets need different spin rates to perform optimally. Generally speaking, among bullets of the same caliber, longer bullets need more RPM to stabilize than do shorter bullets–often a lot more RPM.

It is generally believed that, for match bullets, best accuracy is achieved at the minimal spin rates that will fully stabilize the particular bullet at the distances where the bullet must perform. That’s why short-range 6PPC benchrest shooters use relatively slow twist rates, such as 1:14″, to stabilize their short, flatbase bullets. They could use “fast” twist rates such as 1:8″, but this delivers more bullet RPM than necessary. Match results have demonstrated conclusively that the slower twist rates produce better accuracy with these bullets.

**Calculating Bullet RPM from MV and Twist Rate**

The lesson here is that you want to use the optimal RPM for each bullet type. So how do you calculate that? Bullet RPM is a function of two factors, barrel twist rate and velocity through the bore. With a given rifling twist rate, the quicker the bullet passes through the rifling, the faster it will be spinning when it leaves the muzzle. To a certain extent, then, if you speed up the bullet, you can use a slower twist rate, and still end up with enough RPM to stabilize the bullet. But you have to know how to calculate RPM so you can maintain sufficient revs.

**Bullet RPM Formula**

Here is a simple formula for calculating bullet RPM:

**MV x (12/twist rate in inches) x 60 = Bullet RPM**

**Quick Version: MV X 720/Twist Rate = RPM**

Example One: In a 1:12″ twist barrel the bullet will make one complete revolution for every 12″ (or 1 foot) it travels through the bore. This makes the RPM calculation very easy. With a velocity of 3000 feet per second (FPS), in a 1:12″ twist barrel, the bullet will spin 3000 revolutions per SECOND (because it is traveling exactly one foot, and thereby making one complete revolution, in 1/3000 of a second). To convert to RPM, simply multiply by 60 since there are 60 seconds in a minute. Thus, at 3000 FPS, a bullet will be spinning at 3000 x 60, or 180,000 RPM, when it leaves the barrel.

Example Two: What about a faster twist rate, say a 1:8″ twist? We know the bullet will be spinning faster than in Example One, but how much faster? Using the formula, this is simple to calculate. Assuming the same MV of 3000 FPS, the bullet makes 12/8 or 1.5 revolutions for each 12″ or one foot it travels in the bore. Accordingly, the RPM is 3000 x (12/8) x 60, or 270,000 RPM.

**Implications for Gun Builders and Reloaders**

Calculating the RPM based on twist rate and MV gives us some very important information. Number one, we can tailor the load to decrease velocity just enough to avoid jacket failure and bullet blow-up at excessive RPMs. Number two, knowing how to find bullet RPM helps us compare barrels of different twist rates. Once we find that a bullet is stable at a given RPM, that gives us a “target” to meet or exceed in other barrels with a different twist rate. Although there are other important factors to consider, if you speed up the bullet (i.e. increase MV), you MAY be able to run a slower twist-rate barrel, so long as you maintain the requisite RPM for stabilization and other factors contributing to Gyroscopic Stability are present. In fact, you may need somewhat MORE RPM as you increase velocity, because more speed puts more pressure, a destabilizing force, on the nose of the bullet. You need to compensate for that destabilizing force with somewhat more RPM. But, as a general rule, if you increase velocity you CAN decrease twist rate. What’s the benefit? The slower twist-rate barrel may, potentially, be more accurate. And barrel heat and friction may be reduced somewhat.

Just remember that as you reduce twist rate you need to increase velocity, and you may need somewhat MORE RPM than before. (As velocities climb, destabilizing forces increase somewhat, RPM being equal.) There is a formula by Don Miller that can help you calculate how much you can slow down the twist rate as you increase velocity.

CLICK HERE for Miller Formula in Excel Spreadsheet Format

That said, we note that bullet-makers provide a recommended twist rate for their bullets. This is the “safe bet” to achieve stabilization with that bullet, and it may also indicate the twist rate at which the bullet shoots best. Though the RPM number alone does not assure gyroscopic stability, an RPM-based calculation can be very useful. We’ve seen real world examples where a bullet that needs an 8-twist barrel at 2800 FPS MV, would stabilize in a 9-twist barrel at 3200 FPS MV. Consider these examples.

**MV = 2800 FPS
8-Twist RPM = 2800 x (12/8) x 60 = 252,000 RPM**

**MV = 3200 FPS
9-Twist RPM = 3200 x (12/9) x 60 = 256,000 RPM**

Of course max velocity will be limited by case capacity and pressure. You can’t switch to a slower twist-rate barrel and maintain RPM if you’ve already maxed out your MV. But the Miller Formula can help you select an optimal twist rate if you’re thinking of running the same bullet in a larger case with more potential velocity.

### Similar Posts:

- Twist Rate: Common Misconceptions about Twist and Stabilization
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- Find Optimal Barrel Twist-Rate with Berger Stability Calculator
- Are You Spinning Your Bullets Fast Enough? Twist Rate Calculator Predicts Gyroscopic Stability
- TECH TIP: How to Determine Your Barrel’s Actual Twist Rate

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Excellent! Your explanation and examples offered here are super, easy read.

So many folks never fully realize the importance of RPM or rotational moment when pondering bullet stabilization and the true effects of muzzle velocity, which your last example clearly demonstrates. I sure wish the bullet makers would advertise the required RPM for each bullet offered as opposed to rate of twist; otherwise just let folks calculate the maximum stabilized distance for their expected case capacity and MV.

Thanks again, and Godâ€™s best!

Tom Mc

Required twist rate is maximum displacement per turn. It has nothing to do with time(turns per time).

mikecr, the “time” comes into the formula by the acromun RPM where R is revolutions, P is Per, and M is minutes. The formula is correct. Think about it, take a 1:12 twist barrel, push the bullet 3000 fps, so 3000 * 60 to make a minute, and in this case, the bullet does exactly 1 turn in 1 foot, so the bullet is doing exactly 180,000 RPM.

Mike,

As Keith noted, in a given section of spiraled rifling, the end bullet rotation RPM WILL depend on how fast the bullet moved through the rifling. The shorter the time spent traversing the fixed rifling distance (as expressed by MV) the faster the bullet will be spinning on exit.

I checked this basic formula with both “Green” and “Red” bullet makers, and they both use the [MV x 720/twist rate] formula to give a basic determination of bullet RPM. There are some other fine points involved, but the basic principle is there.

In discussing the issue with Hornady regarding the .224 80gr Amax, for example, they specifically said stability should be achieved so long as about 250,000 RPM is maintained. That’s based on proven stability at 2750 fps in an 8-twist barrel (247500 RPM).

Here’s an online twist-rate calculator. You’ll note that as you change the muzzle velocity value, the recommended minimum twist changes. This assumes that bullet weight, length, and Specific Gravity are constant.

Link: http://kwk.us/twist.html

I agree that your formula accurately calculates RPM. However, it doesn’t answer the fundamental question: What is the ‘optimal’ or minimum RPM’s required for a specific bullet at a specific speed? In other words, you can’t figure out what twist is actually required for stability.

The fundamental measure of bullet stability is the Gyroscopic Stability Factor: SG, also known as the static stability factor. In order to achieve adequate stability, the SG should be greater than one. In order to ensure a margin of error, an SG of 1.4 should be achieved.

How do you calculate SG? Don Miller has worked for many years to develop an experimentally based ‘twist rule’ which very accurately calculates the SG for a given bullet based on: length, caliber, twist rate, muzzle velocity, and atmospheric conditions. Don has published his work in Precision Shooting Magazine over the last couple years. I use Don’s formula exclusively for doing stability calculations because it’s so accurate. I’ve offered on several shooting forums to email a spreadsheet with the formula programed to anyone interested, and I’m extending that offer now. Please email me if you’re interested, and I’ll send you the SG calculator.

RPM’s is more like a ‘symptom’ of stability, not the fundamental determining factor. If you want to know the ‘correct’ twist rate to use, you should be looking at SG, not RPM.

Take care everyone,

-Bryan

I though my email address would appear in the last post. It’s:

bsl135@yahoo.com

Please be a little more careful to differentiate conventional wisdom (the things that ‘everybody knows’) from fact. The lore about the tremendous effect of bullet rotation on terminal performance (the ‘buzz saw’ effect) is often repeated, but just can’t happen. For example, a 70 grain, 6mm bullet at 3000 fps has about 1400 ft-lbs of energy, and less than 1% of it is due to rotation. When a bullet comes apart in flight it is largely due to temperature from frictional heating due to passage through the bore. A rifle bullet spinning at typical rpm would never fail from centrifugal force at room temperature. It would be easier to destroy the bullet by firing it a very high velocity from a smoothbore and letting friction do the work. This is just plain physics.

JOSH— sorry to have to disagree with you but, that “buzz saw” effect exist. I’ve shot well over 10,000 varmints and let you tell fast twist of the identical caliber and bullet do make a notiable difference! (it’s the direction the energy travels when imparted into the mass at point of termination)

am pleased with your definitions thue, it needs to be clear that bullet max rpms dose require a bullet to trevel quite long distance in order to spin that many times so this is in fact a formula which is used to determain best twist rate??!! some folks i know thinks the bullet spins 237.600 rpms at the muzzle as stated in speer manuel #14 page 21. that is what i’ v been trying to get a clear answer from the ‘big boys” but still no luck. in fact no bullet can stay in flight that long it will reach next state if so again could you make it clear so all people could remember the time fraction it takes a bullet to leave the barrel so in no way it could spin that many times right at the muzzle right?? i do thank you for a proper respond.

respectfully Avi.

consider this: The .015″ thick jacket of a “frangible” bullet gets immediately reduced to an effective .010″ thick jacket the moment the rifling gets “impressed” (engraved insinuates the removal of metal) into it’s thin surface; weakening the soft copper jacket’s tensile strength by about 33%. Then, the lead core gets heated to a “flash” temperature approaching it’s melting point, from the hot gasses pushing it and the copper jacket’s friction with the barrel; further reducing it’s (their) tensile strength. So both metals, already seriously weakened, get wound up to a rotational speed of 150,000-250,000+ rpm and CAN violently burst apart in a radial (not foreword) burst pattern the moment they leave the confinement of the barrel. I extensively studied the radial burst patterns of high velocity .22’s many years ago, and the radial burst patterns they produce explain why you cannot “skip” a high velocity .22 bullet into a ground squirrel even if the bullet hits the ground only 6″ ahead of it. The moment the bullet’s thin copper jacket gets scuffed (such as a “skip” contact with the ground), the bullet violently explodes in a radial pattern; none of it hitting the squirrel.

To get a good idea of what happens when a “frangible” projectile bursts (either from excessive rpm, or on impact), read a chapter or two on “flywheels” in any good mechanical engineering book. When the centrifical forces exceed the tensile strength of the material, there is a sudden, EXPLOSIVE release of the stored energy in that material. In a Hot Rod Magazine published back in the 60’s, there was a picture of the interior of a dragstrip operated ’55 Chevrolet. Or what was left of it anyway, as when the flywheel burst, the floorboard was blown up into the dashboard, the dashboard was blown clear to the roof, the windshield was blown completely out, and the front seat was totally demolished. By a miracle, the driver survived.

This is what happens when a high rpm “frangible” bullet strikes the animal. It not only expends it’s stored energy foreword into the animal; it also VIOLENTLY releases it’s stored rotational forces in a radial burst as well. The faster it spins, the more violently it comes apart.

If you want to see what a RPM induced radial burst pattern looks like, take a piece of 24″ X 36″ construction paper (or your wife’s favorite Brad Pitt poster), roll it up into a 12″ diameter tube (stapled or taped together), lay the tube on the ground, stand up a piece of 1/2″ plywood in front of the tube opening, then get back about 10 feet and fire a high velocity frangible .22, .20, or .17 bullet thru the middle of the plywood. You will see the radial burst pattern show up as VERY tiny little holes around the circumference of the tube. You will also notice the pattern isn’t very far beyond the plywood. How far behind varies with the foreword velocity and RPM of the bullet on impact. When the bullet bursts, it bursts APART at a higher velicity than it’s foreword velocity. That’s the reason why you cannot kill 2 ground squirrels with a single frangible bullet, even if one squirrel is standing right behind the other.

Regarding the “buzz saw” effect, I’ll explain this little known fact. When a bullet is spinning at, let’s say 200,000 rpm on impact with the animal, the foreword motion of that bullet immediately slows down from the impact. But until ALL foreword motion completely stops, the bullet is still spinning (although also rapidly slowing down as well). So there IS a momentary “buzz saw” effect. Using a non-frangible deer hunting bullet, the folded back sharp petals act like an end mill; cutting on it’s front and sides at the same time. Do you remember the “shot in the stomach, but the bullet wound up in the guy’s shoulder” stories that came back from Vietnam? Guess how the little 68 fmj bullet got there? It spun it’s way up there. With retained RPM.

Anyone who had seen a very slow speed photography of a bullet hitting the weaved surface of an armor vest could readily notice the twisting effect imparted by the spinning bullet on the material it hits. Yes, there is a “buzz saw” effect, at least until all the rotational inertial energy of the bullet is used up!

What I’d like to know is whether the bullet gets heated by passing through the air at Mach 2 like a small Apollo capsule. I’m prepared to believe that the bullet gets heated to around half the

temp of molten lead by barrel friction (from the rifling) plus a small effect at the base from hot propellant. But is there an appreciable extra heating (and later presumably a cooling)from flying through the air?

Regarding post #13:

The M193 ball used in Vietnam had a 55 grain bullet. The 62 grain M855 was adopted as NATO STANAG 4172 in 1980.

Hi could you send me the SG calculator if you still have it.

This information is just what I needed.

Why don’t the T.V shows on guns and ammo

ever explain this part?

Great reading I am trying to stabilise a Nosler 90gr. Etip copper alloy bullet, 1.180″ long, in my .243 Win. with a 1 in 10 twist, can I acheave this. The other option could be a Barns 85gr. TSX copper alloy bullet. Any thoughts

Dougie H

I may have missed something but what is the optimal SG we are looking for?

I have to admit that I am having a lot of trouble swallowing this buzz saw effect. One cannot compare a tiny little bullet with only a few percent rotational energy to a huge engine flywheel with 100% rotational energy.

I’ll submit just one tiny little piece of info and then let the flack fly… LOL! Even though the bullet is spinning at several hundred thousand RPM, it is not spinning any faster than the spin rate imparted by the rifling. Ie is is spinning just one revolution in say a foot for a 12″ twist. in other words, it makes just one revolution as it travels 12″ through the carcass of the target. That said, I recognize that it slows down rather rapidly, and probably maintains some rotational speed even as it slows down, but this is likely offset to a great extent by the increasing diameter of the expanding bullet just like a spinning figure skater speeds and slows as they extend or retract their arms.

I’d put a lot more credibility into the energy imparted by forward energy of the bullet and the pressure shockwave of the impacting bullet travelling at Mach 2 than I do in any buzz saw.

Last but not least, a bullet flying apart at impact ensures that that this forward energy is dissipated with a short distance and therefore a larger radius of damage within that distance.

My two cents.

The effects of bullet spin are obviously important, but the discussion was about calculating bullet spin. It might seem like nit picking to argue about the formula since it gives a consistant indication which satisfies the needs of shooters whether it is accurate or not.

My objection is that in theory nothing can accelerate to say 2800 fps in zero time. If a bullet is at rest and reaches a muzzle velocity of 2800 fps, the average velocity within the barrel is only 1400 fps if the acceleration is linear. Therefore,the spin rates obtained using the given formula would be twice the actual spin rates. But, acceleration might not be linear and might depend on the characteristics of the propellant used.

Bullet and firearm manufacturers must have a better handle on this than I do, before I found this discussion i tried to derive my own formula and didn’t know if using the muzzle velocity was valid.

The SG calculator removed some of the doubts I had about muzzleloader rounds, but with the wide variety of bullets in use I wonder if the 1:28 twist is suitable for all of them.

I didn’t notice any difference when shooting slugs from a smooth bore shotgun vs a rifled shotgun, so I think the buzz saw effect is rather minimal. Can’t wait to try and find a difference now that I am looking for one.

Buzz saw? Foreword?

For Pete’s sake people! You all must realize that we have high speed cameras, as well as clear ballistic gelatin these days. Do you not?

How is any of this “buzz saw” garbage even a conversation?

Get Smart – So I’ll call mine the Router Effect – Late ’80s A hunting buddy and I experimented with a Ruger Mini-14 to see if it would work for Deer Hunting. To 1st sight in, we used a Mil-issued M193 round shooting into a stack of abandoned wood that was “Pithy” inside the bark. We split the log in 1/2 to find the FMJ bullet had kinked almost 90* and cut a channel into the 18″ log, bullet still intact. It had rotated,horizontally, not tumbled end for end, and veered to the right – Leaving a channel of chewed up wood, the entire path.

I’ll call it the “Router Effect” if it makes you feel better.