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Old April 3, 2010, 05:37 PM   #2
InigoMontoya
Senior Member
 
Join Date: January 14, 2007
Location: Ridgecrest, CA
Posts: 229
Pretty easy to determine *IF* you have a drag coefficient (because it WILL differ from bullet to bullet)...

Back of the envelope (I'll work in metric because it really is an easier system):

MuzzEnergy = KineticEnergyMeasured + WorkLost

WorkLost = Force * Distance

Distance = 10' = 3.0 meters

Force = Drag = 0.5 * rho * Vel^2 * Cd * RefArea

rho = DensityOfAir = 1.225 kg/m^3 (adjust according to local conditions)
Vel = 810 fps = 247 m/s
Cd = ?? ...Bah, we'll call it 0.5 (feel free to get "real" numbers)
RefArea = pi/4*Diam^2 = 3.14/4*0.45^2 = 0.159 in^2 = 102.6e-6 m^2

Drag = 1.92 Newtons.

So... WorkLost = 1.92 * 3 = 5.75 Joules

Going back to the beginning....

KineticEnergyMeasured = 0.5 * mass * Vel^2

mass = 230 gr = 0.0723 kg

KineticEnergyMeasured = 2.205 kJ

MuzzEnergy = 2205 J + 5.75 J = 2211 J

MuzzEnergy = 0.5 * mass * MuzzVel^2

=> sqrt(4422 / 0.0723) = MuzzVel

MuzzVel = 247.3 m/s = 811 fps


Now, that's a pretty sloppy methodology that could easily be improved with the use of a spreadsheet and integration methods, but it's close enough to give a pretty good idea of what's going on in those 10 feet.... And it looks like the answer is "not much."
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