Math / Physics Help - Barrel Pressure and Velocity Back in 1993 I derived the following equations to calculate projectile velocity and barrel pressure. Recently, I have noticed that I need to double the calculated results in order to obtain real word results.
I need some help to find out why I have to double the derived results to make the calculation work. The doubled results appear at the bottom of the image.
A Microsoft Excel Spreadsheet Download using the doubled (2x) results can be obtained here: question.xls (removed)
Thanks in Advance.

 A: The values you used for barrel pressure (55 kpsi for .50BMG, 60 kpsi for 7.62mm) are all PEAK chamber pressures not $\bar P$. 
$v_p=\sqrt\frac{2\bar P A L}{m}$
this is the last equation you wrote that was correct.
I looked at your excel sheet and found that you did $\sqrt\frac{2\bar P A L}{m}\over 12$ to obtain a $v_p$ in ft/s. This is a mistake. length dimension for $P$, $A$ and $L$ must all be converted from in to ft for $v_p$ to come out as $ft/s$. So if you want to enter pressure area and length in terms of inches, you must do $\sqrt\frac{2\bar P A L}{12m}$ for velocity in $ft/s$.
Once you apply this correction, you will obtain $v_p\approx 5000 ft/s$ then it will be clear that using 55,000 psi as average pressure is wrong--it is definitely lower than 55k psi from your graph. A rough estimate of 30 kpsi would give $v_p\approx 3700 ft/s$.
Still too high from the real world value of 2900. Here enters more facts:


*

*when the combusted gas expands, it pushes the gun and bullet in opposite directions thereby doing work on the gun (and whatever is holding on to the gun)

*the bullet is forced to rotate because of rifling thereby gaining rotational kinetic energy from the work done by the gas
these are two unaccounted forms of energy, so the work done by the expanding gas cannot all go into the translational KE of the bullet. This is why when we make that assumption in $v_p=\sqrt\frac{2\bar P A L}{m}  $, velocity is much higher than what is observed in real life--even if you used the correct $\bar P$.
