# :confused:Figuring muzzle energy

BunnMan

January 14, 2009, 10:58 AM

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

If you enjoyed reading about ":confused:Figuring muzzle energy" here in TheHighRoad.org archive, you'll LOVE our community. Come join

TheHighRoad.org today for the full version!

TimRB

January 14, 2009, 11:08 AM

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

Tim

BunnMan

January 14, 2009, 11:35 AM

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?

The Bushmaster

January 14, 2009, 11:35 AM

See if this helps...

http://kwk.us/powley.html

TimRB

January 14, 2009, 12:14 PM

"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.

BunnMan

January 14, 2009, 02:03 PM

Soooo.... if I'm putting this together right:

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

Sounds about right.

rcmodel

January 14, 2009, 02:10 PM

Here, do it the easy foolproof way:

http://www.handloads.com/calc/quick.asp

Muzzle energy is:

bullet weigth x velocity2 / 450400

Example 150 gr bullet at 2,700 FPS

150 x 2700 squared / 450400 = 2428 ft-lbs.

rc

joneb

January 14, 2009, 02:47 PM

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

BunnMan

January 14, 2009, 02:52 PM

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

rcmodel

January 14, 2009, 03:09 PM

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

TimRB

January 14, 2009, 03:28 PM

"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

rcmodel

January 14, 2009, 03:36 PM

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

rc

TimRB

January 14, 2009, 04:14 PM

"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

BunnMan

January 14, 2009, 06:11 PM

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

rcmodel

January 14, 2009, 06:51 PM

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

RidgwayCO

January 14, 2009, 11:43 PM

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?"

1858rem

January 14, 2009, 11:59 PM

[(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?

TimRB

January 15, 2009, 12:14 AM

"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.)

joneb

January 15, 2009, 01:54 AM

255grain bullet X 940fps X 940fps= 225318000... then divide by 450240...= 500.44 ft/lb energy

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.

MCMXI

January 15, 2009, 04:34 AM

Since we're getting picky, the units are ft-lb (units of work) and NOT ft/lb ... there's a difference.

:)

rcmodel

January 15, 2009, 02:36 PM

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

SSN Vet

January 15, 2009, 03:57 PM

He thinks g=32.171 ft/sec^2

You think g=32.160 ft/sec^2

Wikipedia thinks g=32.174 ft/sec^2

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

:neener:

TimRB

January 15, 2009, 04:27 PM

"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

BunnMan

January 15, 2009, 04:32 PM

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

joneb

January 15, 2009, 06:10 PM

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

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.

TimRB

January 15, 2009, 06:33 PM

"your 150gr bullet weighs 25gr grains on the moon. This is why the Earth's gravitational force is factored into the ft-lbs equation."

The weight of the bullet is less on the moon, but its mass is the same, so if muzzle velocity is also the same, muzzle energy will be the same.

Tim

BunnMan

January 15, 2009, 06:33 PM

your 150gr bullet weighs 25gr grains on the moon

Jib,

This is the reason the gravitational force is factored in. The same "mass" bullet has a different "weight" depending upon the gravity acting upon it. We have to solve for the mass to work the equation. I think rc is right about the muzzle energy being the same. The bullet gets launched out of the muzzle with the same force regardless of gravitational pull. It would carry further if fired at the same angle in less gravity because it wouldn't fall as fast and due the the atmosphere being less dense would hold on to more velocity downrange. This is kinda makin' my head hurt a lil'. I think I'll just use that handy link.

Ponder this though...it's absolutely true:

If you lay the bore of your favorite rifle perfectly horizontal on the bench shooting over flat land and hold a bullet identical to the one loaded in your cartridge at the same height as the bore then drop the bullet at the precise instant your fire the rifle, both bullets will hit the ground at the same time ;)

-BunnMan

RidgwayCO

January 15, 2009, 08:29 PM

Not only that, but even if the bullet you shoot weighs 55gr and the bullet you drop weighs 300gr, they'll still touch the ground at the same time (tip-of-the-hat to Gallileo).

joneb

January 15, 2009, 09:41 PM

We have to solve for the mass to work the equation.

The equation I have uses slugs, just for fun I will plug in 5.36 ft/sec(2) stay tuned :)

GooseGestapo

January 15, 2009, 10:52 PM

Before I retired, I was the only officer on the 240+member force that owned or even knew what a chronograph was.

For many years my state (Georgia) had a 500ft/lb energy requirement, and at least a .357mag minimium w/6" bbl and adjustable sights. I only made a half dozen "illegal ammo/weapon" charges in over 20yrs. Most were for FMJ ammo, where hunter was known and "knew better". One involved a complete penetration of a deer, and hit landowners house (hunting w/o permission also, no license, ect...... you get the picture!!! FMJ from AKS.......)

Now, after a fellow academy graduate who has a Master's degree in Game Mgt., and was on the pistol team is the director...... Law has been changed to something enforceable....... (.22cf or larger w/ expanding type ammo....) I wrote the reg's on rimfire (.17cal or larger "RIMFIRE" for "small game"....)

Use something "recognizeable" or state/imply that it is such. If approached/checked by an officer/agent, tell him that It's (headstamp should reflect this) that its XYZ load, or facsimile and yeilds XXXft/lbs.

If everything else is compliant, he'll most likely accept your explanation.

Just have your "ducks in a row"...............(licenses, permits, tags, written permission, flourscent orange clothing worn, ect.)

Energy dosen't kill anyhow, it's bullet construction and shot placement.........

assuming sufficient velocity to do the job...........

A close friend used one of my .45Colt loads (975fps from pistol, 1,225 from Rifle w/255gr FN cast) to kill a 110lb doe in November. Frontal chest entry, and exited opposite ribcage with .45" hole through about 18" of deer. Bled out before hitting ground approx. 45yds from site of shot. Very dead deer! Only about 400ft-lbs from handgun, about 550ft-lbs from rifle. Enough to kill a steer however.....

The "gravity function" is to convert units. As others have stated, Kg is a "mass" measurement, POUNDS is a "gravity" function.

My formula is"

Velocity squared X Bullet wt in Grains / 450,240 (64.32 x 7,000) = E

Converts grains to lbs, and Newtons to lbs,...........

BunnMan

January 15, 2009, 11:41 PM

Ridgway,

That's only true in a vacuum ;)

joneb

January 16, 2009, 04:10 AM

The equation I have uses slugs, just for fun I will plug in 5.36 ft/sec(2) stay tuned

Well that doesn't work :o

We have to solve for the mass to work the equation.

Oh that :D so 150gr / 7000 = .0214 / 32.174 =.000665, so this is the 150gr bullets mass in slugs and 1/2MV2 is 2700 fps x 2700 x .000665 / 2 = 2423.925 ft-lbs

Ok the 150gr bullet @ 2700fps will have the same energy on the Moon, but the bullets trajectory will be 6x flatter :)

Thanks guys

rcmodel

January 16, 2009, 02:28 PM

SO, is gravity used to measure muzzle energy or not?

All this math gives me a headache.

rc

TimRB

January 16, 2009, 03:33 PM

"SO, is gravity used to measure muzzle energy or not?"

Gravity is used to calculate bullet mass from the bullet weight they publish in the reloading handbook.

Tim

joneb

January 16, 2009, 03:53 PM

SO, is gravity used to measure muzzle energy or not?

I don't think it is.

When the bullet's weight is converted to slugs the Earth's gravity is taken out of the equation, now the bullet's Mass should be the same everywhere, and used in the formula as M

rcmodel

January 16, 2009, 03:55 PM

As was gravity used to weigh the bullet in grains in the first place.

I understand that mass & weight are not the same, and that is where the 450400 conversion number comes into play for average earth gravity & the english measurement of energy.

But it seems I don't need rocket science math, or know the earths exact gravitational pull in eastern Kansas, to use the old muzzle energy formula that has been around forever, and works close enough.

Anywhere I'm gonna be using it at least. :D

rc

MCMXI

January 16, 2009, 05:51 PM

I understand that mass & weight are not the same

The confusion for some comes from the fact that lbf and lbm are definitions and nothing more AND many people are sloppy with their notation. lb should be written as lbf or lbm so that there's no confusion.

1 lbf = 32 lbm * (ft/sec^2)

This is the definition of 1 lbf ... the force required to give a 1 lbm an acceleration of 32 ft/sec^2 which is similar to the DEFINITION of a Newton (N).

1 N = 1kg*(m/sec^2)

When calculating lbf (pound force) from lbm (pound mass) in the -y direction (towards the center or the earth), lbf = lbm (close enough for government work). When you calculate muzzle energy, the energy of the bullet in the -y direction is ignored since it's very small compared to the energy in the x direction. If you were calculating the deflection of a rifle barrel due to its own weight, then the mass of the barrel is the same as the weight of the barrel.

Since the majority of calculations of mass, weight, force, moment, work, energy etc are applied where we live i.e. on Earth, the definition was tailored to where we live. So here on earth, the mass of a bullet is the same as the weight of a bullet IF you're only interested in forces in the -y direction such as calculating the force that 10 boxes of bullets exert on your loading bench. Once those bullets leave the barrel, mass and weight are NOT the same since we neglect gravitational force because it's much less than "chemical" force from the burning powder and expanding gasses. Eventually, as the bullet slows down and its velocity in the -y direction IS significant compared to its velocity in the x direction, then the mass and weight of the bullet would be the same.

:)

rcmodel

January 16, 2009, 06:14 PM

See, all this just confirms why I almost flunked high-school algebra.

And why I use the http://www.handloads.com/calc/quick.asp calculator to figure muzzle energy.

It may not be exactly right.

But it's close enough for GooberMint work here on earth! :D

rc

If you enjoyed reading about ":confused:Figuring muzzle energy" here in TheHighRoad.org archive, you'll LOVE our community. Come join

TheHighRoad.org today for the full version!

vBulletin® v3.8.6, Copyright ©2000-2015, Jelsoft Enterprises Ltd.