E.G.: 190 ft. lbs. of muzzle energy (ME) = ???
As in the sentence:
As loaded by Federal the .380 ACP delivers a 95 grain FMJ bullet at a muzzle velocity (MV) of 960 fps with 190 ft. lbs. of muzzle energy (ME).
Does this mean the bullet hits with the force of 190 pounds over a square foot concentrated in just the bullet diameter?
As loaded by Federal the .380 ACP delivers a 95 grain FMJ bullet at a muzzle velocity (MV) of 960 fps with 190 ft. lbs. of muzzle energy (ME).
Does this mean the bullet hits with the force of 190 pounds over a square foot concentrated in just the bullet diameter?
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No, it's not similiar to a "pressure" issue; the "foot" is not "square footage"--it is linear (ie. "distance").
"Energy" is the concept of masses having the "ability to do work." An object at rest is considered to have no energy. Another object impacting the object at rest will transfer "energy" to the object at rest. The "transfer of energy potentials" that a moving mass has is the way we measure "amounts" energy.
The amount of energy of "190 ft-lbs" is the equivalent of a one pound force being applied steadily to a "free body" over a distance of 190 feet. The "free body" would be accelerated up to a specific velocity from this constant force. The free body will then be described as having an energy amount of "190 ft-lbs."
In the case of the 95 grain bullet, the energy is applied with higher accelerating force over a much shorter distance (ie. the length of the barrel). The "power"--think horsepower--involved is much higher than "a one-pound force applied over 190 feet," but power is different than energy. The "foot-pounds" just gives us "standard" to measure--that is: to compare--different amounts of energy.
Comparison is what all measuring is about; it is taking a "known" concept and comparing that to an "unknown". If I say a tree is 30 feet tall; I'm explaining an unknown concept of "height" in terms (feet) of a known length. (I could go on and argue that all knowledge represents comparison, but that's more than I need to explain to answer your question...)0 -
Another way to think of foot pounds of energy is the amount of force needed to move a 1 pound object a distance of 1 foot under specific conditions.
If you only have time to do two things so-so, or one thing well ... do the one thing!0 -
It's kinetic energy, KE = (M*V*V)/2.
M = the mass of the bullet in slugs (English measure of mass)
V = the velocity of the bullet in ft/sec.
Conversion factor, grains to slugs: 1 grain = 0.00000444 slugs
Weight to mass conversion: mass (in slugs) = [Weight(in lbs)]/32
95 gr = 0.0004218 sl
KE = (0.0004218 sl* 960 * 960 ft2/sec2)/2 = 194.4 ft-lbs of energy
Hope this helps!
Another useful/useless formula:
Recoil Energy of gun: RE(gun) = [m(bullet)*KE(bullet)]/M(gun)
where m(bullet) = mass of bullet in slugs
M(gun) = mass of gun in slugs
KE(bullet)= muzzle-energy of bullet in ft-lbs
Example: What is the recoil energy for an 8.5 lb rifle that shoots
.45-70 ammo which drives a 405 grain bullet to a muzzle velocity of 1800 ft/sec?
m(bullet) = 405 * (.00000444) = 0.00180 sl
M(gun) = 8.5 lb/32 = .266 sl
KE(bullet) = (0.00180 sl * 1800 * 1800 ft2/sec2)/2 = 2916 ft-lb
So... RE(gun) = (0.00180 sl * 2916 ft-lb)/0.266 sl = 19.7 ft-lb recoil
Usually, you already know the muzzle-energy of the bullet, and don't have to compute it.0 -
EXCELLENT WRITING, ALL !!! THANK YOU !
trstone So now I know why my AMT 380 Backup was kind of ...painful, and my
Colt Double Eagle Commander 40S&W (one heavy chunk of stainless) shot so effortlessly![:D]0 -
quote:Originally posted by trstone
Another useful/useless formula:
Recoil Energy of gun: RE(gun) = [m(bullet)*KE(bullet)]/M(gun)
where m(bullet) = mass of bullet in slugs
M(gun) = mass of gun in slugs
KE(bullet)= muzzle-energy of bullet in ft-lbs
Example: What is the recoil energy for an 8.5 lb rifle that shoots
.45-70 ammo which drives a 405 grain bullet to a muzzle velocity of 1800 ft/sec?
m(bullet) = 405 * (.00000444) = 0.00180 sl
M(gun) = 8.5 lb/32 = .266 sl
KE(bullet) = (0.00180 sl * 1800 * 1800 ft2/sec2)/2 = 2916 ft-lb
So... RE(gun) = (0.00180 sl * 2916 ft-lb)/0.266 sl = 19.7 ft-lb recoil
Usually, you already know the muzzle-energy of the bullet, and don't have to compute it.
Ummm somthing missing there, I think. Dosn't the mass/vel of the powder enter the formulia? Vel of the powder is normaly figgured at 1.5-2 times the bullet velocity (for smokeless powders). 50gr of powder at approx 3000 fps has to add somthing to the recoil level.
Some guys like a mag full of lead, I still prefer one round to the head.0 -
Quote:
"Ummm somthing missing there, I think. Dosn't the mass/vel of the powder enter the formulia? Vel of the powder is normaly figgured at 1.5-2 times the bullet velocity (for smokeless powders). 50gr of powder at approx 3000 fps has to add somthing to the recoil level."
Consider that the bullet also has ROTATIONAL kinetic energy which is not accounted for in tabulations of the muzzle energy; add to this the fact that bullets are slightly squashed by the force of the expanding gas---a phenomenon called "bullet upset"---and that this deformation takes up energy that is unrecoverable as bullet kinetic energy, and you have two energy "sinks" which lower the recoil level, possibly taking up the energy the mass of moving propellant would otherwise contribute....0 -
It's sort of like the difference of a yugo @50 mph running into a Mack truck versus a Chevy Suburban running into a Mack at 50mph. The Yugo will maybe wrinkle the bumper of the Mack. The Suburban will take out the radiator , the hood and the bumper. Result is they scrape the occupants of the Yugo off with a spatula, the suburban's occupants are dead but recognizable and the Mack gets towed. 0 -
Tailgunner: By Ned, YOU WERE RIGHT! I consulted a book on ballistics last night, and it expressly said you DO have to account for the momentum of the powder and gases produced thereby. My apologies for the off-the-cuff hypothesizing. 0
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