“Spin” the often forgotten factor in terminal ballistics.

K

king9

Inactive
This is not new science, but I want to awaken you good folk to the ancient phenomena that spin kills. How often do we hear about the mysterious killing power of this or that chambering? Most experts list many factors affecting lethality, but they often don’t mention the bullet spin. We know Kinetic energy ½mv² sooo well, but how many of us are familiar with centripetal energy, namely ½mr²ω² where r is the radius of the bullet and ω the radial velocity. So 2 identical bullets traveling at the same linear velocity, but different rates of spin do not have the same energy, and very obviously the increased hydraulic shock of the faster spin . Just like linear velocity centripetal velocity has an exponential function. So are we still baffled by the mysterious slaying power of 6.5 X55 SE with an average spin of 1/8! Moreover the loss of centripetal velocity over a distance does not follow the same rules as linear velocity, long range energy retention can be much higher despite poor BC. But that is not the end of the story, think carefully about “r”, more on that if there is interest
 
Hello, This has been hashed out before on other sites such as the Cast Bullet Association, and others. It was found to be a non-issue.
 
...centripetal energy, namely ½mr²ω² where r is the radius of the bullet and ω the radial velocity.
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The formula is not correct. That is the formula for a thin-walled, hollow spinning cylinder. Bullets are not thin-walled and hollow, they are solid.

If we approximate a bullet as a solid spinning cylinder then the formula is:

(mr²ω²)/4
...ω the radial velocity.
Click to expand...
ω is 2 times pi times the frequency of rotation in revolutions per second. It is the angular velocity. When multiplied by r it is the tangential velocity. It's not correct to call either quantity "radial velocity". Radial velocity is velocity along the direction of the radius. No part of the bullet has radial velocity unless the bullet is in the process of deforming.
...how many of us are familiar with centripetal energy...
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What you're talking about is the kinetic energy of a spinning object due to the spin. "Centripetal energy" is not really a conventional term used in physics.

Ok, so with our terminology correct and using the proper formula, let's do a calculation with some real values. Let's take a .308 bullet with a weight of 150grains, shot from a 1:10 twist barrel at a muzzle velocity of 2700fps.

Spin rate will be 194,400rpm or 3240 revolutions per second--so ω is about 20,358 radians per second.

To find r for the equation we start with 0.308"/2 to give us the radius in inches. Then we need to convert it to feet. That gives us a value for r of about 0.0128 feet

The bullet weight is 150 grains but we need to scale it to mass by dividing by 7,000 to get pounds and then by 32.174 to get slugs, the unit of mass in the unit system we're working in. The mass of the bullet is about 0.00067 slugs

Now we can run the equation.

( 0.00067 x 20,358² x 0.0128² )/4

That gives us about 11.4 ft/lbs

If we recalculate for a 1:8 twist barrel it works out to about 17.8 ft/lbs of kinetic energy due to spin.

Let's contrast those values with the muzzle energy of the fired round due to muzzle velocity alone: 2427ft/lbs.

So the spin energy increases the overall energy of the round by about 0.5% with a 1:10 twist and by about 0.7% using a 1:8 twist.

If we compare that to the amount of energy lost due to air resistance the 0.7% increase in energy due to spin energy is less energy than would be lost due to air resistance (decrease in velocity/energy due to drag) after traveling about 5 yards.

So the bullet loses more energy due to air resistance in the process of traveling 5 yards downrange than the total amount of spin energy that the round possesses.

If the spin energy is having any significant effect at all on the killing power of the round, I would agree with you on at least one point. It would be very mysterious indeed for such tiny energy numbers to have any noticeable effect on the terminal performance of a bullet. It would be like saying that getting 3 or 4 yards closer to the target is going to make a signficant difference in the killing power of the round.
 
So, at 2700fps. the Spin rate will be 194,400rpm or 3240 revolutions per second.

On a day with no wind, how much does a sniper have to compensate windage for the spin at a target 2700 feet away?

:D
 
Another way of looking at it:
If a bullet passes through a 1 in 8 twist barrel,then passes through a 16 inch ribcage,there would be two revolutions of the bullet.
I realize there would be many other variables.
 
On a day with no wind, how much does a sniper have to compensate windage for the spin at a target 2700 feet away?
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In theory I'd say yes, in reality there is no such thing as a "no wind" day. There is always air movement and it wont be the same between you and your target 2700 feet away.

You can check your conditions, but by the time you do the math the conditions will change more the the effects of spin drift.

If you were to take the formula Bryan Litz gives us in his "Applied Ballistics for Long range Shooting" and compute your spin drift from 100 to 900 yards (2700 feet) you'll see its pretty much cumulative, much like MOA. It should be built into your zero for all PRACTICAL purposes.

If you are shooting artillery then yes it would matter, if you are shooting normal rifle ranges, I don't believe it does.

This is my opinion, there are others who believe its critical.

As far as the orginal topic........it's way beyond me, but I do believe we can OVER MATH the shooting game.
 
As the former CEO of Cast Performance Bullet Company, and as an avid shooter in both professional and sporting circles I have a background in these kind of issues that goes back over 30 years.
I can assure you that in most sporting arms "spin" is of no value or importance in the killing effect of the projectile.
There are rare cases where it has an effect, but those effects are not positive. They are negative.

An example is the 22-250 firing SX or Blitz bullets
When fired at varmints such bullets can and do come apart very quickly and give spectacular effects. When shooting an animal of 15 pound or less the effects are explosive and dramatic.

However if you hit a 150 -300 pound animal the same effect is happening to the bullet, but you sure can’t see it in the animal the same way.
Such bullets come apart 100% in the first 2-3 inches, and all the wounding effect is then just a “left over” from the explosive effect of the bullet’s velocity and shattering.

In other words, a frangible bullet that is almost at the edge of it’s integrity when fired is NOT going to penetrate as well as a slow bullet. (like a muzzleloading flintlock with a 62 caliber round ball turning 1 time every 6 feet at the muzzle. With a hard ball you would be amazed at how well there penitrate, even when only fired at 1500 FPS and impacting at 1000FPS)

If we try to fire a SX or blitz bullet in an AR-15 with a 1-6 1/2" or 1-7" twist at full velocity, some times the bullet will come apart in mid air only 5 feet to 30 feet in front of the rifle and never hit the target at all.

As a bullet slows in velocity it’s spin does not slow at the same rate. In other words, a bullet turning 1 -10” at the muzzle of a 30-06 is turning about 1 -6” at 400 yards. The spin does slow, but not at the same rate.
Depending on the BC of the bullet the spin will slow down about 20% to 35% before the bullet goes subsonic. But the velocity can slow down 66% to 71%

Spin can effect how fast a bullet opens up, but that opening is going to slow or stop the spin within the first 2-3 inches of penetration. Even a 240 grain .308” VLD bullet drops 80%-100% of it’s spin within only 4” of penetration in saturated news paper when that paper is hit at 1000 yards. The spin stops faster the closer the target is.

So it’s safe to say that the idea of the bullet spinning faster increases it’s lethality is 100% wrong.

It’s a theory that has been soundly disproven in small arms.
As KraigWy said, it is important in shooting artillery at several miles, but in small arms it’s positive effect is “somewhere between insignificant and non-existent.”
 
Crazy enough,I read an article in "The Accurate Rifle" or "Precision Shooting",shooting north or south at 1000 yds,the rotation of the earth makes a few inches difference,like 1/2 moa,iirc
 
The Marine Sniper Programs on Discovery and History Channels include compensation for "projectile spin" for long range shots. I used "No Wind" above to keep it simple.

For right-handed spin bullets, the bullet's axis of symmetry points to the right and a little bit upward with respect to the direction of the velocity vector as the projectile rotates through its ballistic arc on a long range trajectory. As an effect of this small inclination, there is a continuous air stream, which tends to deflect the bullet to the right.

Bullets from some firearms will undergo a sideways curve of almost 2 feet in a shot that goes 1/2 mile.

US Military M193 Ball, 55 grain, .223 inch diameter(5.56 mm) experiences drift of 23 inches (584 mm) at a range of 1000 yards

The 30-'06 cartridge drift about 3/8" and 300 yards, 1.5" at 500 yards, and 13" at 1,000 yards. This is why the 1903 Springfield rifle has sights that move a small amount to the left as you raise the elevation. The right hand twist results in gyroscopic drift to the right and the compensation for this was built into the sights.

Source(s):
Understanding Firearm Ballistics by Robert A. Rinker
6th Edition, 2005 Mulberry House Publishing
 
More spin might be better on a hollow point... something that has jagged edges as it passes through an object. But for a bullet that doesn't deform, i can't see it really mattering. Stab an orange with an ice pick, and then rotate it. It doesn't take much effort. If you stab an orange with a spoon, and then try to rotate it, thats a different story.

And the extra energy stored in a bullets spin was given to it by the propellant. So to compare two bullets with the same amount of powder, one would be going slightly slower with more spin. I'd take the extra velocity over the extra spin.
 
SpeedyJerry:

Could you post the formula you used to come up with those numbers?

Those numbers don't seem to compute. Close at 500 but a bit off at 1000.

Bryan Litz (Applied Ballistics for Long Range Shooting) gives the formula for the Gyroscopic (spin) drift as:

Drift = 1.25(Sg+1.2)tof^1.83

Sg = Gyroscope stability factor (for the example I use the 175 SMK with a Sg of 1.5)

tof = time of flight

Spreadsheet format: Drift=1.25(1.5+1.2)*tof^1.83

This gives the drift of the 30 cal 175 grn SMK @2800 FPS a drift of 7.19 inches at 1000 yards.

My contention that errors in wind estimation is much more critical. Using the same bullet, speed, etc. and a 1 MPH miss calculation of wind would drift the bullet 7.4 inches.

Not many around here, (me included), can estimate the wind to within one MPH.

Again, its cumulative. Starting when the bullet leaves the barrel.

At 100 yards=.06, 200=.2, 300=.5, 400=.9, 500=1.44, 600= 2.14, etc. etc, until the 7.19 @ 1000.

It's been my experience that this is pretty much taken care of when we sight in our rifles.

As I've mentioned several times, shooting is over 90% mental, the more you have to worry about the more its going to affect your shooting, The gyroscopic drift, Coriolis effect, wind, temp, humidity, light, etc etc is going to get to you mentally. Some (wind) are much more critical then others (Coriolis).

In small arms (rifles) this crap is so minor that worrying about it creates more error then the effect itself.

Yes this math, slide rules, computers, etc. are fun to play with on a cold winter day sitting by the fire, but when the rubber meets the road ( or the bullet meets the wind) its a complete different matter. Too many other things to worry about.

Concentrate on your fundamentals, get a good zero and learn to shoot in the wind.
 
In regards to the original post about terminal ballistics, how much damage could the torque (rotational power) of the bullet actually do inside a wound channel? For the bullet to maintain the highest torque it would have to not expand or open up. Yet these types of wounds from FMJ bullets are usually the least deadly. In fact, the wound channels usually close up, seal themselves off, and are difficult to find. For a bullet to do the most damage it has to open up to a larger diameter and cut tissue with sharp edges. As soon as this happens the resistance to the torque increases and causes spin (and forward momentum) to slow down and stop. In my un-scientific view there is no magical killing power possible from bullet spin in terminal ballistics. I would be interested to hear someone with an opposing opinion or facts.
 
The only effect I can see with spin rate is that a marginally stabilized bullet(less spin rate) MAY upset in the target and do bad things. Like the difference between a 1-12 and a 1-7 AR rifle.
 
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