In need of a math guru

I’ll take a stab at it.

First off, sirjeigo’s gun is regulated to a constant 1350psi.
CO2, if there’s a mix of liquid and gas and some space for
expansion, acts sort of like it was regulated to the constant
850psi quoted. The AirForce guns are not regulated, and who
knows what a tank pressure of 3000psi means by the time
the air reaches the breech and barrel. You also didn’t say
whether that 1000psi in the AirForce gun was regulated HPA or
just coming from a bare tank.

So let’s just compare apples to apples here and use
the CO2 and sirjeigo’s numbers to try to figure out what
to expect from regulated HPA at 1000psi.

This ignores a few things, and some of them may be significant.
However, the basic idea is that energy available in the
compressed gas expanding out of the gun is going to go into
moving the pellet.

Energy stored in a gas is something like ln(P1/P)
where P1 is the initial pressure and P is atmospheric, taken
here as 14.5 psi.

Energy in the pellet is 1/2(m)(v)(v)

So ln(P1/P) = 1/2(m)(v)(v), assuming all of the stored energy
goes into the pellet, no loss from heat transfer, no friction, etc., etc.

For the CO2 numbers quoted:
ln(850psi/14.5psi) = 1/2(m)(700fps)(700fps)

For the AirForce gun at 1000psi:
ln(1000psi/14.5psi) = 1/2(m)(v)(v)

Assuming the same pellet each time:
ln(1000/14.5) / (1/2(m)(v)(v)) = ln(850/14.5) / (1/2)(m)(700)(700)
4.23 / (v)(v) = 4.07 / 490000
v = sqrt(((4.23)(490000)) / 4.07)) = 714 fps

Seems a little low, no? Let’s try it again using sirjeigo’s numbers:

ln(1000/14.5) / (1/2(m)(v)(v) = ln(1350/14.5) / (1/2)(m)(930)(930)
4.23 / (v)(v) = 4.53 / 864900
v = sqrt(((4.23)(864900) / 4.53)) = 899 fps

Hmmm. Seems a bit on the high side.

0850psi => 700fps
1000psi => ??
1350psi => 930fps

Either the fps and/or pressure quoted for CO2 is low, the fps and/or
pressure quoted for sirjeigo’s gun is high, the CO2 fps is for .22 or …

…CO2 is different from air. You might get a better idea from numbers
for HPA at 850psi.

William

Edit: Someone pointed out that the units aren’t right for ln(P1/P) in
the discussion above, so technically I can’t say ln(P1/P) = 1/2(m)(v)(v).

Quite true. Energy in an ideal gas is (n)(R)(T)(ln(P1/P), but I left off the
(n)(R)(T) intentionally. Strictly speaking, do they cancel out? For
T (temperature), yes, but for n and R, no. Basically, I chose
to pretend that (n)(R) was close enough for gov’t work, mainly
because I didn’t feel like looking them up.

It’s quite possible that plugging in the right values for
CO2 and air will make the calculation come out right. However,
this is just theory for amusement’s sake. You’d really have to
try it, or maybe look at numbers people are getting for 1000psi
HPA with the Hammerli 850, Crosman, and Chinese guns to get
an idea.

wvr