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. Doglips

    Doglips Guest

    You think in a pool way too deeb for me to tred in... The high points use a blow back design with the bolt delayed by weight and spring tension....So you got to factor the sprin weight (?) into your calculations....Maybe the gun smith form or Wolff Gun Springs could help you out..I figure you need only about 1/100th os a second delay on the action for the bullet to exit the barrel...maybe the mass and spring resistance keeps the body at rest..since it prefers to remain at rest long enough for the bullet to exit before the force causes the mass (bolt and spring) to move.

  3. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Way I've been taught, every action has an equal and opposite reaction. So if a bullet is forced forward, the bolt is likewise forced back. If the bolt locks, like in every other kind of gun, it pushes the whole bolt back. But if it doesn't lock, such as in a recoil operated semi auto, every bit of force that pushes the bullet forward will push the bolt back. Now maybe some of the energy/pressure/momentum or other stuff goes somewhere else, I dunno. Or maybe the momentum of the bullet itsn't the thing to go by.
    Maybe the thing to do is to buy a Hi-point and do some measurements, then maybe fudge it from there. Potential problem there is with patent rights.
  4. Jack O

    Jack O Guest

    BattleRifleG3 that gives new meaning to higher education. I must apoligize I thought you were going somewhere else yesterday had my chain yanked at work way to much. As for the project the math is beyond me. I'd have to stick with the tried and true method of trail and error. Good luck to you in your quest.
  5. Klaus

    Klaus Guest

    He left out a few items, such as the friction of the barrel, the friction of the chamber, and the momemtum of the gas itself.
    Last edited: Apr 27, 2002
  6. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Quite right, Klaus. Don't we all wish we lived in a perfect frictionless world?
  7. jerry

    jerry Since 03-15- 2002 Forum Contributor

    I don't know if John Browning knew anything about calculus.
  8. m14nut

    m14nut Guest

    Pull trigger, listen for bang, check intended target....come on guys!
  9. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Unless someone screwed up the gun... Then it's the prosthetic engineers' turn.
  10. jerry

    jerry Since 03-15- 2002 Forum Contributor

    Engineers! gotta love 'em. I work with several, very book smart but honestly worry me when I see one crossing the street. Not trying to fuel a silly fire but this made sense to me. Source: Ruger 10/22 owners manual.

    When the trigger is pulled the hammer is released and strikes the rear of the firing pin. The front end of the firing pin indents the cartrige case rim and ignites the priming compound inside the rim. The flame from the priming compound ignites the powder in the cartrige case and a great volume of gas is instantly generated. the high pressure gas acts on the interior of the cartige case in all directions.
    Because the bullet offers the least resistance it begins to move out of the cartrige case and down the bore. The bolt, which has also been subjected to the "push" of the gas acting on the inside of the cartrige case head, moves rearward, extracting and ejecting the fired cartrige case, at the same time cocking the hammer and compressing the recoil spring, etc. as described above.
    now, that my coffee is cold,,,,,,,,,,,,,:smash:
  11. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Yep, but the big question is "how fast" and "How heavy". "Real fast" and "real heavy" don't work out well. The problem I'm dealing with is that it looks like we're not strong enough to work the actions. But hopefully I'll learn how this all works out.
  12. azeeb

    azeeb Guest

    i think your main problem is that a 230 grain 45 won't go anywhere near 500 m/s. That would be about 1650 fps. A typical 230 grain 45 only has about half this velocity. 800 fps or so.
  13. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    I checked my numbers, and found I'd been using the energy in Joules instead of the velocity in m/s. Duhhh... Using the right numbers, I brought it down to about 70lb necessary to cock it.
    Now granted, the spring doesn't absorb all the energy, the bolt does hit a buffer... But how much of the energy is absorbed by the buffer and by the spring? And how much of the bullet's momentum is really imparted into the bolt? That is the question...
  14. azeeb

    azeeb Guest

    also i think there may be a mistake in your spring equation. Should it be (x2-x1)^2 instead of x2^2-x1^2? A typical 45 spring is about 16-17 lb. That should give you something to shoot for. Just a little help from a fellow engineering student who likes to play with guns ;)
  15. azeeb

    azeeb Guest

    Actually idealy i don't think the bolt should hit a buffer. The spring alone should stop it. A true 1911 does not have a buffer. Some people add them as aftermarket parts, but they should not really be needed if the proper spring is being used.
  16. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    The energy of a compressed spring is 0.5*K*x^2. The change in energy would be 0.5*K*x2^2 - 0.5*K*x1^2.
  17. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Ok, the spring roughly accounts for about a third of the energy I calculated for the bolt carrier.

    I think I've figured something else out...
    Maybe this is what Klaus was saying, but how about the energy lost to friction as the case slides back through the chamber?
    It would throw everything off though, as it would allow less of the bullet's momentum to be imparted to the bolt.
    And to calculate this, I'd need a friction constant.
    With chamber pressure, I'd figure out the normal force of the case on the chamber, then use the friction constant to find the frictional force...
    But the pressure would change as the the powder burned up and the bullet moved down the barrel. I can smell the Calculus already... A differential equation... Which is what I need to be studying presently.
    Last edited: Apr 28, 2002
  18. Rave

    Rave G&G Evangelist

    Been a while since I checked this forum out,must say it has shrunken a bit!
    As far as this discussion seems to be going I suggest that it is inertia that one must consider when dealing with blow-back firearms.Enjoy all!:nod: :nod: :nod: :cool:
  19. BattleRifleG3

    BattleRifleG3 G&G Evangelist

    Also known as mass. Though some people apply it to momentum too. Energy is absorbed and transferred. I think the big problem is finding where all the energy goes.
  20. JohnD

    JohnD Guest

    Well what I think is most of it is held back by friction. And the bullets get out the barrel before the preasure is right for the action to open. I have seen pictures of a 1911 type pistole being fired and the bullet was 4-6" out of the muzzle, and the action was just starting to open. In a 22 auto loader I know that the ruger mark 2 is far more efficiant than a revolver with same barrel length.