Is this a valid formula for energy?

Buzzcook

New member
velocity(fps)^2 x bullet weight in grains / 450436 = Muzzle Energy.

ie 900(fps) x 900(fps) x 230gr / 450436 = 413.599ft/pd


I've been using E=1/2MV^2 since high school. But this seems to be much easier.
 

Shoney

New member
Energy (of motion of a bullet) - Kinetic energy or force carried by a bullet at that point in its trajectory. In common use and popular shooting literature it is expressed in foot-pounds, one ft/lbs being the amount of force required to lift a one-pound weight one foot above the ground. Formula: Energy ( in ft/lbs ) equals bullet weight ( in grains ) multiplied by the velocity ( in feet per second ) squared, divided by 450240. E=MC2/450240 Often wrongly equated with killing power, energy is not a reliable gauge of this, as it does not take into account penetration or bullet performance.
 
Buzzcook,

That is correct. The 450436 constant includes the 1/2 term in 1/2mV². It also includes dividing weight by 1G and converting grains to pounds, which has to be done in the English system because units of mass and units of force are both expressed in pounds, with a pound of force (lbf or just lb) being the force 1G exerts on a pound of mass (lbm or just lb). In the more sensible MKS and CGS forms of the metric system, mass is in kilograms or grams, while force is in Newtons. Keeping force and mass separate simplifies the math. It's the "just lb" being common to both mass and force under the old way of writing the English units that causes confusion. The old timers just figured whether it was mass or force was implicit and you had to be following the calculations well enough to know which it was?

If you ever have to do it the hard way:

1 grain = lbm/7000

1 lbf = 32.174 lbm×ft/s²

Nick
 
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Kernel

New member
Looks right. The two equations are the same. 1/450436 is just a constant to make the units come out in foot-pounds (ft-lbs). A different constant could be used if you wanted the results in Joules, or BTUs, or calories, or ergs, or .... whatever. The constant changes, but the equation up front would be the same. If you wanted, you could also lose the all those decimal places. Your equation only has three Significant Digits. 413.599 is the type of superfluous precision you get from a calculator.
 

Buzzcook

New member
Thanks for the replies. I'm not a mathlete, so when I see a novel way of calculating stuff I need reassurance.
 
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