The Energy of a Bullet

The rotational speed of the bullet depends on the length of the barrel also.

OKAY,calculate this:A barrel has a 1 in 8 twist,that is every 8 inches it rotates once.

The muzzle velocity is 4000 Feet Per Second,the barrel length is 24 inches.

How fast is the bullet rotating when it exits the barrel?

How much energy if the bullet weighs 60grams?

The heavier the round the more twist is required to maintain stability.

MV X 720/Twist Rate = RPM

MV=4000x720/8=360,000 rpm

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 etc,etc.

Okay,we have the rpm,now calculate the energy of the rotating mass,at the muzzle, if the bullet is .556"Diameter and weight is 60 grams.

Show kinetic and potential energy.

link:

https://www.accurateshooter.com/technical-articles/calculating-bullet-rpm-spin-rates-stability/

I believe the projectile weight is in grains, not grams. Typically a 5.56 bullet weighs 60 GRAINS.

A 60 gram bullet (projectile) would weigh 925.94150 grains -- A pretty large round that I certainly wouldn't want to hit me -- spinning or not.

You are right! My bad.

Thanks for pointing that out.

I stand corrected...(or sitting in this case).

Intuitively it makes sense that the rotational energy is small compared to the linear energy, likewise for the angular and linear momentums, because in firing any given load in any weapon, there is a linear kick, but not torsional kick. At least not in anything that I've fired.

The muzzle does tend to kick up however.Why up,and not any other direction?

I think that it comes down to the fact that the axis of the barrel of the gun does not pass through the center of mass of the gun. In a conventional upright firing position, in any gun I've seen, the axis of the barrel passes above the center of mass. So upon firing, a torque is created that rotates the gun, and raising the end of the barrel upwardly.

Maybe I'll go on YouTube and look for some Dirty Harry videos that demonstrate the principle. Do you feel lucky?

GA

The reaction is also above the "average" point of contact with the shoulder.

Potential or Kinetic?

Thanks guys, I appreciate the response. I had wondered for a long time. I chocked it up to an increase in inertia, and not noticed because we use static weight for calculations. And that meant little effect, which I doubted. I hadn't considered the K V drop. And the m to V ratio. Fit's perfectly.

But I was surprised at the angular to linear ratio. Much smaller than I assumed. It's there, but means little.

Thanks you all.

Assuming that the slug - of mass m, exits the rifle muzzle with forward (linear) velocity v, whilst spinning with angular velocity ω, and moment of inertia I about its axis of rotation...

Then KE_total = KE_linear + KE_rotational = ½mv² + ½Iω²

A train is moving north at 100 ft/sec. There is a wall parallel to the track at 100 ft east of track. We fire a bullet due east from train, at 100 ft/sec. How far does the bullet travel in 1 sec?

Assuming that the shooter is not "compensating" for the movement of the train:

141.4 feet

Also ignoring gravity effects,which adds another curve to the calculations.The bullet also travels Downward at 32fps/ps as well a North at 100fps and East at 100 fps.

Needs to be a tall wall if he intends to hit it unless the rifle is 32 feet above the ground.

Depending on how precise you want to measure it,there is also the Coriolis effect of the Earth's rotation and the latitude and angle of the rifle barrel in reference to the ground.

AW Shoot! I forgot to include the effects of the orbital velocity of the Earth around the sun,the orbital velocity of the solar system's rotation around the galaxy,and the galaxy's rotation in the local group,the local groups motion toward the

Great Attractor,and a few other variables too minor to mention.

At what angle is the spin axis of the bullet, when the bullet hits the wall?

Ignoring air resistance.

At right angles to the wall.

Imagine that the shooter shoots from the far side of the carriage, and, consider the state of the bullet before it exits the train. Why should anything change when it passes through the window (assuming it's open and ignoring air resistance).

Let's ignore the air resistance, gravity and Coriolis Effect. So actuality, the bullet has a 140 ft/sec muzzles velocity. And the bullet slides. This would produce a larger impact area. And the offset in spin, might cause the bullet to arc up or down, after impact, depending on the twist direction. Sounds much more damaging.

Let's replace the train with a counterclockwise rotating wheel, holding the gun. The muzzle has an angular V of 100 ft/sec. The barrel points due E when bullet leaves barrel.(at 100 ft/sec)

The train added 2 linear velocities. The wheel added an angular V to a linear V.

Where does the bullet go? And it's speed and spin alignment?

Would there be any difference at all? Would it slide as before? Or would it be normal to the wall?