Bullet RPM Per Inch Math Problem

[Tumbleweed]

Hey all. I've been interested in calculating a hunting projectile's rpm per inch of travel. In other words, I'd find it very interesting when comparing twist rates to know how many times a bullet will rotate in say 12" of animal. Many have seen the benefits of faster bullet rpm's creating more devastating wounds and fragmentation. I.E. 215 Berger out of 10 twist vs. 215 Berger out of 9 twist. Being able to calculate this and compare different twist rates and their effect is useful to me.

Coming up with bullet rpm at the target is easy, it's the distance of travel that I struggle to make the connection on...mathematical shortcomings. Ha ha! Also, some suggest only a 10% decline in bullet rpm at the target but there's no real data available on this. Can someone who is much better at math than me come up with this formula and spell out the process for me? Thank you!
Jesse

 

[Eric_F]

No clue if this is a worthwhile calculation, but I'm a sucker for ballistic math. In 12" of animal, are you assuming a pass through or coming to a stop? For both you would have to assume a velocity drop (and as you stated a decrease in spin rate). I can tell you I think a linear decrease in velocity and spin are very unlikely, so the math is going to get complicated, like calculus complicated. Further, you're either going to need a good estimate of the drag force of a bullet inside an animal (good luck) or an estimate of the exit velocity (shrug emoji). I typed up a longer response going through what would need to be integrated but there are so many guesstimate terms that I'm not sure it's even useful. If anyone knows of estimates of exit velocity for what you're looking for then I'll go through the math. It's simpler if you assume after 12" it stops, is that what you meant?

 

[Tumbleweed]

No clue if this is a worthwhile calculation, but I'm a sucker for ballistic math. In 12" of animal, are you assuming a pass through or coming to a stop? For both you would have to assume a velocity drop (and as you stated a decrease in spin rate). I can tell you I think a linear decrease in velocity and spin are very unlikely, so the math is going to get complicated, like calculus complicated. Further, you're either going to need a good estimate of the drag force of a bullet inside an animal (good luck) or an estimate of the exit velocity (shrug emoji). I typed up a longer response going through what would need to be integrated but there are so many guesstimate terms that I'm not sure it's even useful. If anyone knows of estimates of exit velocity for what you're looking for then I'll go through the math. It's simpler if you assume after 12" it stops, is that what you meant?

Maybe I can simplify this greatly Eric. Instead of trying to factor the forces that would happen inside the animal, what if we just calculate bullet revolutions within 12" of atmosphere at the target distance? For me it would still be useful information to just know how many more revolutions one twist rate will make vs. another in that given distance. Thanks for your response

 

[ShtrRdy]

It's going to depend on the bullet velocity at the distance you're wondering about.

The bullet will have some RPM rate when it leaves the muzzle. It will still be spinning at about the same rate but it will not be going as fast.

 

[Eric_F]

Hmm ok that's easier, but will be a minimum limit as the bullet will definitely slow and spend more time into he animal.

Time in animal t (sec) = 1 ft / velocity (fps)
Rpm / 60 = rev/sec
So
Revolutions in animal = rpm/(60*velocity)

There are calculators for rpm at the muzzle, and I know Brian Litz has calculated how the spin slows over flight but I don't have his book so I'd use the muzzle rpm as your maximum. That's just muzzle velocity (fps) / twist rate (in per rev) / 12 in per foot * 60 sec/min.

If you're only using muzzle rpm, that simplifies to:
Rev in animal = muzzle velocity (fps) *12 in per foot / (impact velocity (fps) * twist (inches per rev))

 

[Tumbleweed]

Hmm ok that's easier, but will be a minimum limit as the bullet will definitely slow and spend more time into he animal.

Time in animal t (sec) = 1 ft / velocity (fps)
Rpm / 60 = rev/sec
So
Revolutions in animal = rpm/(60*velocity)

There are calculators for rpm at the muzzle, and I know Brian Litz has calculated how the spin slows over flight but I don't have his book so I'd use the muzzle rpm as your maximum. That's just muzzle velocity (fps) / twist rate (in per rev) / 12 in per foot * 60 sec/min.

If you're only using muzzle rpm, that simplifies to:
Rev in animal = muzzle velocity (fps) *12 in per foot / (impact velocity (fps) * twist (inches per rev))

It works, thanks! With some of the numbers I ran with my equipment, typical revs/foot is between 1.6 and 2.8 depending on impact velocity. Very interesting. Not as much difference as I would have thought between twist rates either, although we know from field experience it does make a difference in terminal performance.

 

[longestrange]

If a barrel has a one in 12" twist, the bullet will rotate exactly once in 12" as it leaves the muzzle. If it has a 1 in 6" twist it will rotate exactly twice in 12" as it leaves the muzzle.
If that is true, then using 1 in 12 twist as an example:
At 3000 fps, the bullet will spin at 3000 revolutions per second (rps) as it starts out.
The energy involved is much more complicated. A larger diameter bullet will have more rotational energy and rotational momentum than a smaller diameter bullet of the same mass if they spin at the same rps. Also, as the bullet mushrooms it will increase in diameter and thus the rps will decrease due to something known as conservation of momentum. You can see this very clearly watching a figure skater. They start a spin with their legs and arms sticking out, but when they pull their legs and arms in they become a blur.
I'm too lazy to calculate the rotational energy, but very skeptical, given the small diameters of hunting bullets, that the spin contributes much to the damage - even if it is still rotating at the initial 3000 rps when it hits.

 

[Tiny Tim]

Here is a previous thread I started that may help a little.

Effect of Bullet Spin on Terminal Performance

I was wondering if anyone has any insight into this. According to a test done at the Aberdeen Proving Grounds, a bullets spin decay rate is approximately 2% per 100 yds. Bullets shed velocity at a much greater rate than spin. I understand that a bullet may only make 1-2 revolutions while...

www.longrangehunting.com

 

[lancetkenyon]

Simple method: Muzzle velocity × 720 ÷ twist rate=RPM

Actual method: Muzzle velocity × 12/twist rate × 60 = RPM

 

[longestrange]

Here is a previous thread I started that may help a little.

Effect of Bullet Spin on Terminal Performance

I was wondering if anyone has any insight into this. According to a test done at the Aberdeen Proving Grounds, a bullets spin decay rate is approximately 2% per 100 yds. Bullets shed velocity at a much greater rate than spin. I understand that a bullet may only make 1-2 revolutions while...

www.longrangehunting.com

Thanks. Funny how several of the posters used the figure skater example.
After reading through that thread it appears that, although the rotational energy is minimal compared to the translational energy, the centrifugal force acts to expand a bullet more rapidly, thus causing more damage. It can even disintegrate a bullet mid-flight.

 

[orkan]

RPM is the secret sauce to most explosive prairie dogs.

Though there is no "RPM per inch of animal" in that equation. There is simply spinning the bullets right up against their failure point, so as soon as they hit anything, they grenade. Excellent acrobatics ensue.

 

[Tiny Tim]

A quick and simple, albeit rough, calculation for turns prr inch of travel at the target would be:

twist rate ÷ % of remaining spin

For example 94%@300 yards with a 10 tw

10 ÷ .94 = 10.638 or 1 turn in 10.638"

 

[jasonco]

External ballistics be like this sometimes...

https://media1.tenor.com/images/e5875add5503b4bf2174c59f146b16d3/tenor.gif?itemid=15394332

 

[Tiny Tim]

External ballistics be like this sometimes...

https://media1.tenor.com/images/e5875add5503b4bf2174c59f146b16d3/tenor.gif?itemid=15394332

Now that's quality entertainment! My wife laughed at that one too.