Muzzle Energy = Firearm Power?
This is a discussion on Muzzle Energy = Firearm Power? within the General Firearm Discussion forums, part of the Related Topics category; I am still very much in the learning stages of firearms, but I do have a question about muzzle energy... Is it save to say ...
December 10th, 2006 03:52 AM
December 10th, 2006 08:25 AM
The whole ME thing is very debateable.
I say,figureing the ME of bullets is OK if you consider other aspects of terminal ballistics.
If a bullet passes through a target,you loose some of the potential. I have been told that the "perfect" bullet will pass completely through, and lodge just under the skin on the far side.
With out being long winded on the subject,I would rather consider TKO than ME.
TKO being Taylor Knock-Out Formula. It is the only formula that I know of that takes into consideration the bullets diameter.(too bad that it doesn't consider meplate also)
Try this on your loads/bullets.
TKO=caliber in thousands X bullet weight in grains X velocity in FPS, divided by 7000. (Google for more info) -------
December 10th, 2006 08:58 AM
I don't believe that the Taylor KO really applies that much to expanding bullets, only hardcast big game type bullets.
Same thing with muzzle energy. Unless we're talking rifle bullets, that really doesn't apply.
What DOES apply is how much damage the bullet in question will do to the target. This is directly proportional to bleedout time.
Unless the bullet hits the CNS, bleedout or psychological effects are all you have. You want 1) the person to know he's been shot and 2) that shot to drop the blood pressure below the threshold at which he can maintain consciousness as quickly as possible.
Sooo... I think it's safe to say that a hypothetical 9mm that expands to .65 caliber almost immediately and drills all the way through the body is superior to a .40 or a .45 which expands to only .55" and goes 6 inches into the body.
December 10th, 2006 10:02 AM
It's a long running debate, and the science angle is inconclusive. Given the advances in medical technology, it's only a matter of probabilities. Want more damage? Make a bigger hole, strike a more-vital area, strike more often, send a faster or more highly-expanding bullet, etc. There are only a few ways to change the formula. It's damage that disrupts a body's function. The answer: cause more of it.
Sectional density, speed, expansion, energy. All play a role, though to what degree is unknown as impact and disruption dynamics are complicated.
For me, I'm willing to accept that a handgun is a pitifully weak device, hence my ability to control multiple shots accurately and quickly weighs much more heavily on the outcome. All things equal, I'd prefer a faster bullet, a bigger bullet, more expansion, greater energy dump in the BG's body ... more often in a shorter period of time. A quality bullet in a reliable, good-handling firearm works. That's enough, for me.
Your best weapon is your brain. Don't leave home without it.
self defense (A.O.J.).
How does disarming
the number of victims?
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December 11th, 2006 12:31 AM
Thanks for your input, RSSZ, Joshua M. Smith, and ccw9mm.
December 11th, 2006 02:23 PM
Depends on your definition of "powerful." What is safe to say is that a cartridge with more muzzle energy has more muzzle energy. Beyond that, it is all speculation.
Originally Posted by Tros
Energy is one thing to consider. Momentum is another. The Taylor KO factor is yet one more. Penetration, velocity, whatever. They all tell us something. None of them, however, no matter how much some people would like to believe they do, directly translate into an answer to the really important question, which is "in this particular case, against this particular assailant, will this round save my ass?"
December 11th, 2006 04:19 PM
IPSC definition of power may be useful...
"Power" is one of those nouns with multiple meanings. I looked in the dictionary and it listed about 10 different definitions.
In physics class a long time ago, the professor said that power was the rate at which work was done, and showed us an equation for calculating power. Typical units for power were horsepower or foot-pounds per second, as I recall.
In handgun shooting sports, the IPSC defines the "power factor" of a gun and ammunition as the product of bullet weight (grains) times muzzle velocity (feet per second) divided by 1,000. Using this definition, the power factor of some common ammo would be:
A 230 grain .45 ACP bullet at 850 fps.......power factor 196
A 155 grain .40 cal. bullet at 1205 fps......power factor 187
A 115 grain 9mm bullet at 1225 fps.........power factor 141
A 240 grain .44 mag. bullet at 1180 fps....power factor 283
As you can see, this power factor is just momentum (mass times velocity) divided by a constant. Personally, I am willing to use this definition whenever I wonder if .44 magnum is more powerful than 9mm, etc. Based on these numbers, I'd say the .44 magnum is about twice as powerful as the 9mm.
December 11th, 2006 09:11 PM
That makes plenty of sense, pogo2.
Thanks for all of the responses :)
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