Formula for energy

[armycar01]

What is the formula for finding the energy of a load? Used to know it, been so long since I used it I forgot.

 Thanks

[John Traveler1]

Kinetic energy E= mass in pounds x velocity in feet squared

E=1/2 m/7000 v x v,

where E = kinetic energy in foot-lbs
m = bullet mass in grains
v = velocity in feet per second

Divide bullet in grains by 7000 to get the weight in pounds.

[Reed1911]

I've always used {(VxV)/450400} * W

Velocity squared divided by 450400 times bullet weight in grains.

Same equation just a little faster to 10-key on the computer.

[Castaway]

Things like this can get complicated and I just don't know why.  Formula is:

1/2m(vv) divided by 32.

Mass is grains divided by 7,000 which gives pounds.  Velocity is in f/s.  32 is the gravitational constant of 32 ft/ss.

[Graybeard]

The formula for "paper energy" of bullets is the weight of your paper bullet times the speed of it squared. It's all just a paper work exercise anyway so use heavy paper.  :-D It has no real world use.

[ricciardelli]

Same as Reed"

Ek=((Vb*Vb)*Wb)/450,240

[Reed1911]

Graybeard, I disagree. It's real world use is comparative ballistics. While its not ever going to be an exact science, it does give you an idea of where the load/bullet stands against another. For example, most people are going to look at the .35 Remington (one of my favorite deer getters) and think "Well hell, that's a mighty fine cartridge, looks big, looks fast, lets check it out." Maybe this guy is looking for a round that will touch to say 250-300yds. SO.....

.35 Remington 200g RNSP SD - .223  BC - .195 MV 2000FPS

Energy at the muzzle is a fairly impressive 1776Ft-lbs. Well over a mid size deer load. We all know that you aren't going to shoot a deer point blank, so lets look further.

At 100yds the velocity has dropped to 1632fps/1183ft-lbs (we'll assume a zero of 100yds for the example) Easily sufficient to drop a deer.

at 200yds things are beginning to look dim 1328fps/783ftlbs still enough, but only on a really good shot add to the fact that our drop is 12" and the fact that most folks cannot really judge 150-200yds.

at 300yds we see a serious problem 1112/549 Pretty low on the energy side and we can assume that the bullet will not expand too much. Add to the fact that we now have to compensate for 43.5" of drop and we can clearly see that the cartridge needs to be kept at about 150yds or less. My general theory is 1000+ Ft-lbs at the strike point is what I’m aiming at. I’ll drop that figure to about 700ish if I have a back-up shot quickly (read semi-auto or leaver action) and I’m going to be very close (i.e. less than 50-60yds).


So I feel it does have a real world application, but I also see where you are coming from. You'll hear WAY TOO many people talking about their MV and ME and never consider the down range effect. Point is, all the energy in the world is no good at the muzzle if you loose most of it before you hit the animal.

[Graybeard]

Paper energy (and I stubbornly refuse to call it anything else) just isn't a measure of anything of value to a hunter. I stand by that statement. It does not extrapolate to killing power. It's NOT a measure of work. That's pounds-feet, not foot pounds.

It's a theoritical made up number which puts the emphasis on velocity over all else. There is no fixed value in spite of what so many rag writers wish to claim that makes a bullet equal to any fixed animal killing task. So of what possible value is it?

Folks claim it takes 1000 of those ft. lbs. of paper energy to kill a deer and 1500 of them to kill an elk. The .44 Mag. doesn't even start off there but is perfectly adequate on either to 100 yards and beyond if you place it right. In a previous example the .220 Swift is the paper energy equal of the .45-70. Sorry but that don't make it a big game round.

An arrow from a bow seldom leaves the bow with more than 50 of those paper energy pounds yet kills as quickly and efficiently as a rifle with 5000 of them.

It's a meaningless exercise in futility foisted on us long ago by magazine writers and most of them refuse to forget it. But there is hope. Of late Boddington has actually admitted in writing it is meaningless. Wonder who else will some day?

You guys will never change my mind and it looks like I won't yours either.

[mountainview]

Work and energy are actually used interchangeably in engineering terminology. Also foot-pounds and pounds-feet are the same thing. Simply moving the units does not change the quantity or the mathematics of the calculation.

[Slamfire]

I'm with Greybeard, a long, long time ago I helped build a ballistic pendulum. A factory round from a .30-06 would move the pendulum a few inches. By pushing on it I could move it to it's maximum swing. I never was able to push a horse or cow that far.  :lol:

[John Traveler1]

Ah, there is a significant difference between kinetic energy and momentum!

It's true that there are observed differences in the "real world" between calculated kinetic energy, calculated momentum, and observed stopping power/killing power.  

As much as we may disagree, various state fish and game authorities use the le energy of factory-loaded ammunition to qualify a cartridge as legal or not legal to use on deer sized game.  As an example, at least one state uses 955 ft-lb as the minimum muzzle energy for a deer-legal hunting caliber.  Coincidentally, (??) this makes the .30 US M1 Carbine a legal deer hunting caliber, but not the .357 Magnum.  Go figure!  

The US military has long established that a minimum of 150 ft-lbs on target is needed to seriously disable or wound a man.  This was a major premise for adopting the 5.56x45 NATO cartridge.  As we all know, a .22 LR has 150 ft-lbs muzzle energy.  Does this mean that it is a suitable military rifle caliber cartridge?

As long as enough people use the muzzle energy figures to establish hunting or defense requirements, the ME calculations have merit.

John

[Rifle25]

Gentlemen, the way I see it, both sides have some merit. The poster who gave us all the insight on how energy works at different ranges for the 35 Rem is indeed correct. In physics, kinetic energy does equal 1/2 mv squared. That's a simple truth. Another poster stated that momentum and energy are very different things. They indeed are. Momentum equals mv, as opposed to 1/2 mv squared. both are known physical quantities that can be calculated. The problem is that we all know that there is a hell of a lot more than this involved when a bullet strikes something. The fact that someone came up with a "minimum energy" to kill certain types of game animals...well, that's completely subjective. Calculating bullet energy and momentum has to have some merit from a physical standpoint. The problem is, no one has figured out exactly what merit this is...and who's to say that an Elk can't be taken with less than a certain amount of energy? This is where the argument becomes subjective and there is no real proof for either side. Hell, a properly placed bullet from a 223 or something along those lines could take an Elk...no matter what the kinetic energy on impact is.

Bottom line, there are many more factors involved than energy and momentum.