Home Forums Hunting & Shooting Ask The Gunwriters Rotational velocity

The reason you KNOW that rotational velocity doesn't effect penetration and expansion much is that the centrepidal force is a very small percentage of the energy that a bullet delivers, less than 1%. I don't have the equations to hand at the moment but I remember that it was a very small part of the total energy. When a gun recoils it doesn't Twist out of your hands does it? It kick straight back - that's the equal and opposite reaction from muzzle energy, not twisting which would be the energy from the bullets rotation......................DJ

dj: Sounds nice in theory;what happens to the 220,000 rpm's?...I'm trying to figure what the rifle not twisting out of your hands has to do with it?

As we know, the bullet stops. Usually, it stops within about a foot inside the animal (or therabouts - the exact distance isn't important to the discussion). When it stops its forward motion, it doesn't sit there spinning. It STOPS. So all the energy it had is now dissipated. Okay, that's foundation. Now here are the answers to the two most recent questions:

Centripetal force (CF) does have an influence on expansion, because the CF works in the same direction as the expansion.

CF does not have an effect on penetration because it can only act in a direction ninety degrees from the path of penetration.

Add: the "equal and opposite" law applies to all the energy imparted. So the rifle recoils rearward equal to the bullet's acceleration. It also has to recoil opposite to the rotational acceleration. We can hardly feel that torque in a rifle, compared to the rearward kick. So, since the reaction MUST be equal and opposite, we can deduce from the effects that the linear energy given to the bullet is much greater than the rotational energy. Ergo, there's very little spin energy there.

Does that help, Bob?

Yup

Bob,
That 7mm bullet is doing 4400 revolutions per second!

RinB,
Theory and conjecture, of course!
As to the expansion side of it, think about kids on tire swings that spin around. Even when rotated VERY slowly (in comparison to a bullet!) there is great force pulling the child out of the swing. The faster you rotate, the greater that force is. The child is maybe rotating at 1 revolution per second, the bullet is doing 4000-4400 revolutions per second!

The speed of the outer edge of a rotating bullet is tangential velocity, which is expressed in the same terms as translational velocity (ft/sec and etc.). Rotational velocity is expressed in terms of revolutions per unit of time (e.g. rpm).

Actually, the term "velocity" is not the correct term for bullets. The correct term is "speed." "Velocity" is a vector quantity which includes a measure of direction or angle.

Okay, here's one I don't know: How do we express the path or direction of a single point on the bullet's surface? It must contain both the rotary component and the translational component. Would it be a rotary vector?

-

Actually the equal and opposite reaction business applies to momentum. If it applied to energy we would all have a big problem when we touch one off.

Amen to that!

An extra 453.59 grams of mass on a D.G. rifle slows the recoil velocity considerably. Been there done that.

Wayne

Forgive my imprecision. Ya got yer basic English Lit major here. Making that substitution, however, is my argument not correct?

Quick edit:

Add: the "equal and opposite" law applies to all the energy momentum imparted. So the rifle recoils rearward equal to the bullet's acceleration forward momentum of the bullet and other ejecta. It also has to recoil opposite to the rotational acceleration. We can hardly feel that torque in a rifle, compared to the rearward kick. So, since the reaction MUST be equal and opposite, we can deduce from the effects that the linear energy given to the bullet is much greater than the rotational energy. Ergo, there's very little spin energy there.

Correction and addition gratefully accepted. This wordman defers to a mathman in all things numerical.

You are a gentleman, and you're welcome. Please feel free to straighten me out when it comes to writing, I wasn't the best student in that subject.

Couple of benchrest shooters I know are working with rifles that have the barrel offset from the centerline of the stock to counter torque. Extreme example, they consider the torque to be significant enough to modify their rifle.

Gentlemen,
The devil does ride on some bullets or should I say in some bullets..It is a little known fact that this is the reason the hollow point was designed, to give him a safer ride!

My good buddy, the late George Hoffman, Tanzania PH and designer of the 416 Hoffman, later the 416 Rem but thats another story, was a student of bullet twist and its effect on trajectory but mostly on its killing power..He drove me crazy with facts and therory on this subject, supported by many, many kills on game , he had it perfected to a science and I am sure he was right, I just could not understand nor absorb it all!

To me much to his disgust or amusement (not sure which)it was a exercise in futility, inasmuch as they both performed the act of killing the intended target as long a the bullet would stablilize in the particular rifle it was shot in.

Most gun nuts, at some point in time, get et up with therory!

I have never lost a big game animal because my bullet was under-rotating or over-rotating when it struck his heart.

I have, however, missed a few when a particularly tenacious demon stayed aboard to misdirect its flight path... .

Ray, I did not know that about hollowpoint bullets, but I believe it must be particularly true about plastic-tipped bullets, which are clearly the work of the Devil anyway... .

Dennis

Torque?

Shoot a light weight .458 with a 500 grain bullet and you will definately feel the torque.

This is all a guy needs to know....


http://www.youtube.com/watch?v=bRDGBpEqLfs

Rotational velocity and centripetal force....

When a M.193 ball projecticle hits meat, inside about 300 yards, it breaks at the cannelure. The two pieces go off in different directions, on a tangent from the line of impact. That is an effect of centripetal force. We cannot predict the direction of the two wounds, because they are a factor of grain of fracture in the bullet, which we cannot predict, and of point of impact with reference to that grain, which would require range predicition to the inch, or less, even if we could predict grain of fracture.

When those pieces go off at a tangent, is that not the effect of gyroscopic forces, not centripetal?

OK for all of you who want to play with mathmatics here are the formula's:

Rotational Speed R=(12/t)*V
R=Revolutions per second (second NOT minute)
T=Twist of Rifling
V=Muzzle velocity in fps

Surface Speed S=C*R
S=Surface speed in Feet per second circular
C=Circumference in Feet
R=Revolutions per second

Spin is integral with forward velocity

RS= (square root of)(V(squared)+S(squared)
V=muzzle velocity in fps
S=surface speed in ft. per sec
RS=Resultant Speed

Energy of Rotation (spin)
E=1/2Iw(squared)
E=Rotation(spin) energy
I=moment of inertia about bullets axis
w=angular velocity

Bullet Kinetic Energy Et=WV(squared)/450400
W=Weight of bullet in Grains
V=Velocity in FPS
Et=Translation energy in foot pounds (muzzle energy not accounting for the small amount of energy required to rotate the bullet)

Energy to Rotate bullet
ER=IS(squared)/2
ER=Rotational energy in foot lbs
I=axial moment of inertia in slug feet(squared)
S=rotation speen in radians per sec

for I the moment of inertia is generally in gr.-in(squared) (grain-inches squared). To convert to slug-feet(squared) take 3.084 x 10(-8) times the gr.-in(squared)


OK so FINALLY:

ER=(8.766*10(-5)AV(squared))/T(squared)
A=axial moment of inertia in grain inches squared
V=Velocity in fps
T=Twist of rifling in inches per turn
ER=rotational kinetic energy

or E=1/2IA(squared)
E=Spin energy in ft. pounds
A=Angular Velocity
I=Moment of inertia about the long axis


That is a lot of mathmatics and work to show and PROVE that the energy required in rotating the bullet is very small. Extremely small! If an example is worked out, it will fall in the general area, depending on cartridge, twist, etc., of between 1/4% and 1% of the available energy.

from: "Understanding Firearm Ballistics" by Robert A. Rinker
pg 139-141



Knock yourselves out!



notes:

For every action there is an equal and opposite reaction, that's why a gun doesn't torque out of your hands from rotational energy. You are welcome to try out both of my 458 Lott's, 416 Remington etc. to prove to yourself that even the heavier calibers don't twist out of your hands, you generally DO notice the recoil straight back into your shoulder =] .

1/4% to 1% of the total energy isn't enough to signifigantly effect terminal ballistics but then again it doesn't take a ton of effort for a sharp edge to cut through flesh either......................................DJ

Rocky, do you mean the Devil DOES NOT ride ride on every bullet???