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Recoil Operated Actions (WARNING: Involves Calculus based Physics)

Discussion in 'The Powder Keg' started by BattleRifleG3, Apr 26, 2002.

  1. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Ok, I'm trying to figure out exactly how recoil operated guns actually work and have hit a point of confusion. I assumed that the momentum of the bullet (integral of force with respect to time) was the same as the bolt, that the bolt then recoiled into a spring. I assumed that the kinetic energy of the bolt carrier was transferred into the spring over the length of the action. Calculating all this out to find the spring constant, it looks like it would take two hundred pounds of force to work the action.

    So I must be missing something somewhere. I've heard terms like "mass delayed blowback" and wonder what other elements might be involved. I'm familiar with the roller-delayed blowback system used in my HK, but it's the simpler ones that are baffling me. Doesn't look that complicated.

    Here's what I was trying to work with:

    Bullet mass * velocity = Bullet momentum = Bolt momentum
    Bolt momentum/bolt mass = bolt velocity
    (Bolt velocity)^2 * mass / 2 = bolt energy
    Bolt energy = spring energy = spring constant / 2 * (x2^2-x1^2)
    Distance between x1 and x2 is the length of the action.
    Calculating for a 45 ACP 230gr at 500m/s
    4 kg*m/s momentum
    Divided by 0.25kg bolt mass (about half a pound) = 16m/s
    m*v^2 / 2 = 32J of kinetic energy
    KE=SE (spring energy) = K / 2 * (0.1m^2 - 0.05m^2) <-- (change in compression of the spring)
    Spring constant K turned out to be 8533 N/m
    Meaning holding it at the 0.1m (10cm, ~4in) point would take 853N, or nearly 200lb.

    Anyone know where I went wrong?
    I guess the greater question is WHY???
    Well I wanna build one.
     
  2. Shaun

    Shaun G&G Evangelist

    cycling barrels

    Let me guess you recently saw the thing on the history channel regarding the new AK with the barrel that cycles thus putting two rounds out in the same burst. if I were you and wanted to learn more I would contact the people behind the history channel namely the one's that are the AK experts and they can probably get all this answered for you. When you find out tell us I would love to know myself
     

  3. Klaus

    Klaus G&G Newbie

    You are forgeting the momentum of the gas. Also missing is friction in the barrel.
     
  4. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Actually, I never saw that History Channel thingy. Never heard of that AK you're talking about. Besides, all the AKs I know of are gas operated, not recoil operated.
    Maybe the thing to do is start with an existing design, measure bolt mass and spring constant, measure the force of a bullet going through the barrel to find friction, and see where that goes.
     
  5. sadiehn

    sadiehn G&G Newbie

    well I would reweigh the slide(bolt)on the 45 as I think the 1911 is well over .5 pounds i dont know what you are useing for a guide second is their some sort of barrel link involed and you missed it or a camming action these would change your numbers a great deal.lastly the parts should be weighed on a verry accurate scale lab grade and the springs should also be tested verry closely as for the math cant do it never could but something seems ascue
     
  6. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Ok... I changed the bolt weight to a whole pound, and used the calculated bullet momentum of 3.5 kg*m/s, instead of my overestimate of 4, which actually made quite a bit of difference. This brought it down to about 70lb of force necessary to charge it. Sometimes my HK-91 feels that hard to charge... But still a bit high. So I guess this is where those other factors fit in.
    I'm wondering, is most of the bolt's energy dissipated on impact with the receiver or buffer instead of the spring? That would explain a lot.
     
  7. sadiehn

    sadiehn G&G Newbie

    well I try this one last time itryed to post 3 times yesterday and my puter froze.

    in your math "45 acp 230 grain at 500 m/s what is the m/s meters a second or mila seconds as i said I am no math major

    Another thing to look at and I cant tell if you did is the spring a progressive in weight.

    Is their a way you could set up a scale to test how hard your pistol slide is to operate.

    i dont think the frame would absorb the kind of impact you are talking about for long so it would have to be a buffering system of sorts.

    if you are useing 230 grains at 500 meters a second i think this is where you are off.that comes out to over 1500 fps and a stout load for a 45 acp is 900 FPS or about 300 meters a second
     
  8. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    500 m/s was way off, I recalculated as above. Yes, I did know that force on spring increases with distance, my numbers were for max force.
     
  9. sadiehn

    sadiehn G&G Newbie

    you have peaked my intrest what have you got now.i dont know how you would do the numbers for a progressive weight spring
     
  10. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    General formulas I've been taught since high school physics are such:
    Force on spring = K*x (K is spring constant, x is distance compressed)
    Energy of compressed spring = 1/2 * K*x^2 (x^2 = x*x, or x squared)
    I used the energy of the spring and length of action to find K, and that to find charging force. But that assumes that the spring absorbs all of the energy from the bolt, and that the bolt assumes all of the momentum of the bullet. If not, it's much much hairier. Differential equations here I come... Got a test in it tomorrow. This is the make or break test. Must go study...