:confused:Figuring muzzle energy

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BunnMan

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I just got through reading Lee's second edition and love the additional information in the load tables. There was no explanation of how to figure muzzle energy in Ft. lbs. from the provided load data. My assumption would be that this would be a factor of bullet weight in pounds and velocity in FPS but I cannot get the math to work to a realistic answer. I get closer to what I expect the numbers to work out to using bullet weight and chamber pressure but I don't think that is the right way to reach the answer. The reason for all this is my great state places a minumum muzzle energy on handgun hunting ammunition. I want to be able to work out the math to be certain I'm legal.

Thanks and God bless,

-Keith
 
So how are you calculating it? Muzzle energy generally is some form of 1/2 MV^2, with applicable conversions for units.

Tim
 
Tim,

I was looking for the right formula, I'm certain my calculations are in error. I'm not sure what the characters in your formula represent. It looks to me like your saying that muzzle energy is one half of muzzle velocity squared. It seems to me bullet weight HAS to come into play as a lighter bullet certainly cannot impart as much energy as a heavier bullet traveling the same speed.

It's been a long time since physics class but as I remember the formula it was F=ma. Where F is force (muzzle energy), m is mass (bullet wieght), a is acceleration (muzzle velocity). In this formula for a .44mag 240gr. XTP the NE load is 24gr. of ACC XMP5744 producing 35,000 CUP and 1413 FPS. Taking 240 divided by 7,000 to get bullet weight in pounds I get .0343 lbs. x 1413 FPS = 48.45 which is certainly not the ft. lbs. of energy this load is making. Maybe velocity doesn't equte directly to the acceleration called for in the equation? Maybe I have to solve for that first? Or maybe I'm using the wrong equation entirely?
 
"It looks to me like your saying that muzzle energy is one half of muzzle velocity squared."

Sorry, it *does* look like that, and that's not what I meant to say.

What I meant to say is that muzzle energy is the kinetic energy of the bullet as it's leaving the firearm, and kinetic energy is

(1/2) * (mass of bullet) * (muzzle velocity squared)

Tim

Edit: The "grains" shown in the reloading books for bullets are bullet weight (a force), not mass. Since F=ma, mass is weight divided by the acceleration due to gravity, about 32 ft/sec^2.
 
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Soooo.... if I'm putting this together right:

1/2 * (240/7000)/32 * 1413^2 = 1070 ft. lbs.?

Sounds about right.
 
Divide the Velocity by 1000 and square it, multiply that number by the bullet weight and that number by 2.219 this is the energy in foot/pounds.
Here's a example using rc's example,
2700 divided by 1000=2.7
2.7 squared = 7.29
7.29 x 150= 1,093.5
1,093.5 x 2.219= 2426.5
 
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rcmodel,

Excellent link! That sure is alot easier than all the calculations I was doing and the results are very close to my math as well. Which either means that the chart and I are both right or both wrong :)

your nick...you into flying rc model aircraft?

Thanks bro,

-BunnMan
 
Used to fly r/c but no longer. It's a long story!

I got to wonder about using handloads to meet game law muzzle energy figures though?

Say the law says 400 ft/lbs minimum.

Say you had a .44 Special giving 310 fp/lbs with factory loads.

But say, like me, you shoot Elmer Keiths old .44 Spl hunting load giving about 650 ft/lbs, or over twice that factory load figure?

How does a game warden determine our .44 Special handload is legal in the field?

rcmodel
 
"How does a game warden determine our .44 Special handload is legal in the field?"

What? You mean some legislator enacted a law concerning firearms that is impractical and unenforceable? How could that possibly be?

Tim
 
I'm just say'n, it might be worth asking some game warden type questions before you get arrested.

rc
 
"I'm just say'n, it might be worth asking some game warden type questions before you get arrested."

Oh, don't take it the wrong way! I'm not trying to give you a hard time, but rather the goofball politicians who don't think things through before they write the laws. I certainly agree that it would be a good idea to ask the fish and game folks what they do about that law.

Tim
 
You wanna get me started on firearms laws??? When I live in Maryland??? I'd get kicked off the forum for overloading the server! :fire:

Seriously, the handgun hunting laws are rather redundantly stupid. We have a 6" minimum barrel length and a 700 ft. lb. energy minimum. Seems to me if you can make the engery in say, 4.6" of barrel (like my G20 10mm) that should be good enough. Should have nothing to do with barrel length. Anyway, fairly any factory .44mag ammo makes the grade. I don't plan on shooting factory stuff and being the concensious law abiding citizen I am, I want to be able to back up my work with sound numbers. :rolleyes:

Really the restriction is terribly silly. I understand the idea of placing limits to keep folks from pursuing game with firearms not lethal enough to kill but how would any warden check you muzzle energy in the field? Even if you had the box the ammo came in I've never seen a manufacturer put muzzle energy on the label.


Thanks for the help,

-BunnMan
 
Energy figures are probably better then the law in Kansas.
We go by case length!

So, a 32-20 WCF revolver is legal, but a .38 Super auto isn't.

A .30 Carbine revolver is legal, but a 10mm auto isn't.

A 38-40 Cowboy load is legal, but a +P .45 Auto or .357 SIG isn't.

:banghead: :cuss:

rc
 
Colorado requires a minimum of 550 ft-lbs at 50 yards for handguns, with a minimum 4" barrel length. Ironically, the state also requires rifles to be no smaller than .24 caliber, and have a minimum energy level of 1000 ft-lbs at 100 yards.

So you could easily have a situation where a load out of a .44 Magnum handgun would be legal, but the same load out a .44 Magnum lever action rifle would be illegal, even though it was generating MORE energy...

I once asked a game warden about this situation. He just laughed and said "What do you expect from those lawyers sitting on their butts in Denver?"
 
[(bullet weight in grains)X(velocity)X(velocity)] << all that divided by 450240

ex.
my 45 colt load
255grain bullet X 940fps X 940fps= 225318000... then divide by 450240...= 500.44 ft/lb energy

RC, where did you find 450400 at?
 
"RC, where did you find 450400 at?"

He thinks g=32.171 ft/sec^2
You think g=32.160 ft/sec^2
Wikipedia thinks g=32.174 ft/sec^2

Tim (Yes, I know I need to get a life.)
 
255grain bullet X 940fps X 940fps= 225318000... then divide by 450240...= 500.44 ft/lb energy
Click to expand...
I get 499.98 ft/lbs :) standard gravity(Earth's gravitational field) is not constant at sea level everywhere.
the 450,400 formula has been around for along time.
 
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Since we're getting picky, the units are ft-lb (units of work) and NOT ft/lb ... there's a difference.

:)
 
Whats standard gravity (Earth's gravitational field) got to do with it?

Seems to me a bullets muzzle energy would be the same on the moon as it is on earth.

rc
 
"silly wabbits, every fudd knows that g=32.2ft/s^2"

I hear ya. We went to the moon in spacecraft that were largely designed by guys using sliderules, yet somehow a .04 percent difference in some conversion constants used in handloading seems important. Go figure.

Tim
 
rcmodel,

I think the standard gravity enters into the equation to solve for the mass of the bullet from it's given weight.

Don't tell the Govenor I said this but I don't think Marylands regulations apply on the moon :)

-BunnMan
 
Seems to me a bullets muzzle energy would be the same on the moon as it is on earth.
Click to expand...
ft-lbs are a measurement (a unit) of work or force required to move a object. If you lift a 10 lbs bag of bullets on Earth you would expend X amount of energy, to lift that same bag on the moon you would expend 1/6 the energy, your 150gr bullet weighs 25gr grains on the moon. This is why the Earth's gravitational force is factored into the ft-lbs equation.
 
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