Let
<br>v = velocity in feet per second
<br>w = bullet weight in grains
<br>e = energy in foot-pounds
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<br>e = w times v times v, divided by 450,436
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<br>1. Multiply the velocity by the velocity ("v squared").
<br>2. Multiply "v squared" by the bullet weight.
<br>3. Divide the result by 450,436 (conversion factor for grains to pounds and weight to mass)
<br>4. Round-off to four figures.
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<br>Here and there, you'll see the conversion factor given as 450,200, 450,250, etc, indicating that the gravity element of the conversion had been rounded-off individually before the single, combined conversion factor was calculated. You can even use the conversion factor rounded-off to 450,000 and produce an energy figure that's very close -- after you round it off -- to any figure you derive with a more-accurate conversion factor.
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<br>The conversion multiplier 1/450,436 is a combination of 1/2g (1/2 times gravity -- 1/2 x 32.174) and the conversion of grains to pounds, 1/7,000 (one grain is 1/7,000 pound). Some writers round the g figure off first -- to 32.17 or 32.2, or even 32.0, for example. Where you round-off doesn't matter enough to the final four significant figures to fuss or fret about.
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<br>For example, the kinetic energy (potential energy by virtue of its mass and speed) of a 150-grain bullet striking at 2,300 ft/sec --
<br>v squared (v times v) = 5,290,000
<br>5,290,000 times w = 793,500,000
<br>793,500,000 divided by 450,436 = 1,761.626513 (most-accurate e)
<br>1,761.626513 rounded-off to four significant figures = 1,762 ft-lb
<br>or
<br>793,500,000 divided by 450,000 = 1,763.33333333333333 ft-lb (least-accurate e), which rounds-off to 1,763 ft-lb
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<br>The difference is only 2 ft-lb, only about 1/10 of 1% of the more-accurately calculated figure.
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<br>As matter of practical fact, the impact velocity of 2,300 ft/sec is usually (a) an approximation, a guess, and (b) probably not the actual impact velocity anyway. So either figure for the energy is as good a guess as the other, for what the actual impact energy is. Also, not all of the potential (kinetic) energy of the bullet is usefully imparted anyway. Some is wasted on making the thump or splat sound of impact. So as a practical matter, the last three figures in the conversion factor -- the xxx in 450,xxx -- don't matter much.
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