Figuring Energy of a round
 

A lot of times when I quote a Velocity from one of my tests I'll also quote the Energy. Some of you might wonder how I come up with these numbers( No Mike, there's no room for them in my posterior). Chuck and I use the Factory standard formula for figuring Energy in a round.

Velocity times velocity
450400
Times bullet-weight =FPE
For example; a 158 grain bullet at 1250 fps.
1250 x 1250 divided by 450400 times 158 equals 548FPE

very interesting Dave! i'm going to write this down in my reloading manual.

that is great to know. I have a chrony and always messing with loads and this is an invaluable formula...now if I only knew how to do math....

Thanks, Dave. As an engineer, I'm always wanting to know the backing info. With that in mind - what does 450400 represent?

Lane

Lane,The constant 450400 takes into account the proper unit conversions and the half mass in order to get the energy balanced correctly.

You can use different constants to figure it in Joules or Ergs. However, Americans prefer Pounds*feet or, more commonly, foot-pounds, for energy.

Interesting.

The general formulae for Kinetic Energy is E=.5(mass)(velocity*velocity) European standard uses meters/second and yields Joules.

of course the bullet is in grains and needs to be converted to pounds for consistency. Many sources when publishing kinetic energy tables for small arms ammunition, use an acceleration due to gravity of 32.163 ft/s2 rather than the standard of 32.1739 ft/s2 which is used above.

The formula then becomes
KE=.5(mass)(velocity*velocity)/(7000*32.163) with the alternate gravity number. The ICAO standard accepted for the gravitational constant is 32.1742 which gives 450,437.4. Speer uses this standard number for acceleration due to gravity. Although several online sources use the alternate value, I can not find a 'why' to the use of it. I doubt the small variance will matter much at our level of specificity regardless of your choice.

Now for a completely different approach:
1
Divide the bullet weight by 100. For example, the average weight of a 9mm full metal jacket bullet is usually either 115 grains or 124 grains. When divided by 100, the result is 1.15 and 1.24, respectively.

2
Divide the velocity by 100. Again, using an average 9mm full metal jacket bullet at 115 grains, a common velocity is 1,145 feet per second. Divided by 100 to get 11.45.

3
Multiply the adjusted weight by the adjusted velocity, then by the adjusted velocity again, then by 2.22.

Using the figures from the example, the equation is 1.15 x 11.45 x 11.45 x 2.22 = 334.7046825. Rounded off, the muzzle energy ends up as 334.7 foot-pounds.

Compare this with the actual formulae you get 334.9... close enough I'd reckon.

Don't know what happened, I just got in the physics mode this morning.

Colonel, mine is the same, just modified to use with a standard pocket calculator. Plugging your 115 grain 9mm at 1145 into mine I come up with 334.7 FPE. This way you don't have to break down the velocity to a managable size. also, for all practical purposes, .2 FPE is a non-existant problem. It's just a faster way to do it, unless you're a Math-Junkie who enjoys the minutia.

I don't if any of you noticed, but this is cut from the same cloth as the famous equation for energy from mass E=Mc2

and C is the speed of light.

Right, energy = mass times velocity squared.