SpudFiles

I was sceptical at the 6.3bar figure due to friction and such.A pressure of 6.3bar is by no means hard to achieve,but as the pressure decreases along the barrel I will need a higher pressure and thus a higher pressure at the start of the combustion to equal out a pressure higher then 6.3bar at the muzzle.But a 3meter barrel won't be used,most probably a 4meter or 5meter.Could you do some calculations on that?I would do them myself but I am relatively mathematically illiterate.Took me 40 minutes to do a simple algebra equation that are for Grade 8 students and im in Grade 11 .

The 6.3 bar was an average pressure, not a pressure at the muzzle.
(on that tangent... it's a much easier calculation using Hamilton's approach.)

Anyways, his analysis assumed several things which just aren't true.
1) That pressure throughout the bore is constant.
2) That the mass of the air originally in the barrel doesn't matter.
3) That the mass of the propellant gasses doesn't matter.

As such, it's probably not the most useful number... you would probably be better off calculating it using EVBEC.

Another thing you might want to take into account in the design is how much energy you'll need in the chamber to get the projectile to Mach 1.

Figure the efficiency of the gun is ~10%. The energy in the chamber needs to be 10x the kinetic energy (0.5mv²) of the projectile.

The energy in the chamber can be calculated based on the volume of the chamber and the heat of combustion of your fuel.