:confused:Figuring muzzle energy

[TimRB]

"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]

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]

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]

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]

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]

Ridgway,

That's only true in a vacuum

 

[joneb]

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

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

Oh that 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]

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

All this math gives me a headache.

rc

 

[TimRB]

"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]

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]

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.

rc

 

[MCMXI]

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]

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!

rc