In need of a math guru

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    Ok the talon ss on co2 at approx 850psi yields 700 fps. and the talon ss on hpa at 3000psi yields approx 1000 fps. Then what would i get on a 1000psi nitro bottle? I know that is alot of guess work, but i am having a hard time trying to get a good guess.

    Thanks in advance


    Sirjio is getting 930’ish fps with .177 7.9 grainers if I remember correctly? this with his HPA bottle shimmed up to 1,350 psi…this is all from memory

    This is purely a WAG(wild assed guess) with NO science involved but I divided 700fps/850psi=.823529… multiply .823529 x 1,000psi = 823fps you can bet the farm on that one 😉

    Let us know or maybe Sirjio posted his initial results before the shim?



    thanks for the WAG 🙂 that is what i was looking for. Hope to do alot of testing once my package arrives. Still waiting on Fedex




    Need a physics student for that one. The match shown is assuming a linear relationship which with gas it is not. It is going to be some sort of curve like a torque curve of an engine.


    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.


    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.


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