Muzzle Energy = Firearm Power?
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 a firearm with a higher muzzle energy is a more powerful firearm? Providing there isn't horrible overpenatration, wouldnt the muzzle energy stat be a good one for figuring how effective the round itself is as a defensive man (or woman :knockedout: ) stopper?
I understand that I am very safe with anything 9mm and above, but I'd like to understand the science behind whats going on as much as possible.
Thank in advance, everybody.
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.