Slug ft. lbs. of energy question

M

MikeS

New member
I'm new to this area but am interested in the study of ballistics.

I recently read where the .50cal BMG cartridge with a bullet weight of 750gr. at 2,700 fps has a ft. lb. energy rating, at the
muzzle of 12,143.

My question is : Why is it that the Federal 10ga. 1.75oz slug at a
slightly larger 766gr. and with a fps at a little less than one half (1,280), rates only at 2,785 ft. lbs. of energy or less than one quarter of the .50BMG. Does 2X the speed equal 4X the impact?

I'd appreciate any clarification. Thanks.

Mike
 
Let's haul out the ol'slide rule...

Bullet energy is calculated by the following equation:

KE=(Mx V^2 )/450400

where

KE= kinetic energy( in foot/ pounds of energy)
M=mass of the bullet (in grains)
V=Velocity of the bullet (in feet/ second)

As you can see a small change in the velocity will produce a much larger change in energy than a small change in the mass of the bullet.

I've found slight variations in the 450400 depending on who's book/website you get the formula from, but they are all close to that number.

After proofreading this post I realize that as I get older I'm starting to sound like one of my college physics professors. I gotta get away from the computer and get some trigger time
 
You've got it Rottweiler. Kinetic energy increases as the square of velocity. So, doubling the velocity will yield a 4x increase in energy.
 
Rottweiler,

Thanks for your clarification on determining bullet energy.

I was confused because the only formula I had seen was the
Taylor Knockout Value use on the Garrett Cartridge and other
large bore manufacturers' sites: Bullet Weight x Bullet Diameter
x Max. Velocity/7000. The TKV apparently puts a premium on weight
rather than speed.

This formula has the 10 slug at a rating 0f 105. vs. 144. for the
50BMG. Of course, the 50cal still retains 2.5 tons of impact at
a 1000 meters (s).

Mike
 
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