How much energy?
How much energy does a 102 gr bullet traveling at 975 fps have?
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If you are going to look at a stopping capability of a cartridge, bypass energy, and go straight to the Taylor factor. The Taylor factor is less biased toward velocity, and relies on a combination of weight, diameter, and velocity, not just the mass(weight) multiplied by the velocity squared.
The Taylor Factor=(MASS xDIAMETER xVELOCITY)/7000gr./pound
In your case =(102gr. x0.355" x975FPS)/7000gr./pound= 5.04
The higher Taylor factor the better.
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As Charlie said, the answer to the question is 215 ft-lbs, or 291 Joules (if you like SI units).
Presumably you're asking about a .380 round here (in particular the Remington Golden Saber).
Even if you're using the absolute best possible round available in .380, its still a .380, and its a relatively weak caliber. Even the WORST possible 9mm luger (for example the 115 grain Walmart Winchester White box ball ammo) is still going to blow away any .380 in terms of energy and momentum. As another example, the numbers you cite still make the .380 quite a bit less powerful than say a 110 grain .38+P.
Also, speaking from experience here, the specific .380 I think you're asking about may not be entirely reliable through certain pistols (eg the Kel-Tec). Make sure you test it thoroughly before you rely on it for carry use.
On what the numbers actually mean, as TSR says, the formula that measures kinetic energy is skewed towards lightweight fast-moving projectiles, that may not actually be the best in terms of terminal ballistic effect from a handgun.
Taylor factor is basically a product of momentum and caliber, and although that formula was really intended specifically to compare RIFLE rounds using monolithic ammo against large game, it still probably is a better measure of ballistic effect than kinetic energy.
If you want to simplify this a bit, since .38 special, .357 magnum, 9mm luger, and every other 9mm round including .380 (aka 9mm short) .357 SIG, 9x23, etc, all use effectively the same caliber bullets you can simply compare bullet momentums (ie "power factor", or just mass x velocity) as a rough measure of terminal ballistic effectiveness of those rounds.
But note that no matter what you do, all these mathematical models are still ROUGH. Pure numbers don't take into account bullet design (ie expansion), how momentum affects recoil (and therefore ability to quickly place followup shots), placement, etc.
A bullet with 2.5x the momentum (or energy. . .or Taylor factor, etc), for example, probably isn't going to be 2.5x as effective.0 -
Relatively little What is the highest Taylor score? 0 -
Thanks!! 0
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