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Pressure, Recoil, and Newton

Discussion in 'General Gun Discussions' started by RJ357, Aug 20, 2004.

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  1. RJ357

    RJ357 Member

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    I am starting a new topic for this because it had become too off topic on the "balanced vector" threads. Also I became too off topic on my own argument.
    And I apologize for all that.

    The question is that of why recoil occurs with a fired bullet that is free to move in relation to the breechface.

    The common explanation calls upon a law of mechanics, that every force is accompanied by an opposite and equal force. Consequently, we can say that every action has an opposite and equal reaction.
    There is no question that these laws are true.
    This can be adapted to our application by making it more specific: Every bullet fired is accompanied by an equal and opposite movement of the breechface.

    But now look:

    Q - "Why does a gun have recoil?"
    A - "Because every bullet fired is accompanied by an equal and opposite movement of the breechface."
    Q - "Yes, I know they all have recoil, but why?"
    A - "Because every bullet fired is accompanied by..."

    Do you see the problem?

    It gets even worse when we add the explanation that the breechface moves backward because the bullet moved forward. For some people it even becomes unbelievble.

    I claim that we do not need to use Newton's 3rd law, that everything can be explained through simple laws that are already familiar through common experience.

    Our test subject will be very simple. A barrel with a breechface permanently attached to one end. A powder charge is at the breechface end and a bullet sits right in front of the charge. The bullet is stuck in the barrel and will not move at first. We ignite the charge. For a couple of seconds, nothing happens. Inside the barrel, there is tremendous pressure behind the bullet.
    The pressure produces a rearward force on the breechface. It also produces a forward force on the barrel through the stuck bullet. These forces are equal and being in opposite directions, they cancel. The net force on the barrel/breechface assembly is zero.
    Suddenly the bullet breaks free. The pressure of the gas, still enormous, is still producing a force on the bullet's base, accelerating it down the barrel.The pressure is also still producing a force on the breechface, accelerating the barrel/breechface assembly backward.

    I believe nothing else is required. I also claim that this simple explanation will also predict results that agree with reality.
     
  2. JohnKSa

    JohnKSa Member

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    When the bullet starts moving, the pressure begins to decrease.

    If there wasn't enough pressure to move the breech rearward when it was high, why can it move the breech after the pressure begins to drop?

    The pressure is also pushing outward on all sides of the chamber. At the point that the pressure is high enough to move the bullet and the breech, why doesn't it also move the gun in one of those directions?
     
  3. RJ357

    RJ357 Member

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    It didn't move initially because the bullet was stuck. It didn't begin moving until a couple seconds later. Barral, breech, and bullet were one piece.

    And just added this:
    Once the bullet begins to move, the breechface moves for the same reason the bullet does. It has a net force on it not equal to zero.
    In any particular radial direction, the force is cacelled by an equal force on the opposite side. No net force.
     
  4. JohnKSa

    JohnKSa Member

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    What's the bullet got to do with it? Remember, we're not talking about the 3rd law. We're saying that pressure is moving the breech. Well the pressure was higher before the bullet began moving. So why didn't the breech move then?
    What caused the net force to become nonzero? The reduction in pressure inside the chamber?
     
  5. RJ357

    RJ357 Member

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    The bullet is stuck in the barrel and therefor transfers any force on itself to the barrel. The base of the bullet has a force on it equal to the force on the breechface. So the barrel also has a force on it equal to the force on the breechface.
    Just added:
    The barrel/breechface assembly has two equal and opposite forces on it.
    No. When the bullet became free, the forward force on the barrel ceased.
     
  6. JohnKSa

    JohnKSa Member

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    Not unless friction between the bullet and the barrel is zero.
     
  7. RJ357

    RJ357 Member

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    That's true. I meant that it ceased in a relative sense. The rearward force on the breechface is much greater than the forward drag from the bullet. The main point there was that the forward force became substancialy less than it was, and the net force was no longer zero.

    Hey, speaking of Newton, did you see my "scale problem" over in the roundtable? Four different answers so far.
     
  8. 1911Tuner

    1911Tuner Moderator Emeritus

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    Pressure Vectors

    RJ! You're sneakin' around over here tryin' to figger it out huh?:D

    John said:

    When the bullet starts moving, the pressure begins to decrease.

    Incorrect. Smokeless powders burn progressively. The pressure
    begins on ignition and rises to a peak and then begins to fall off...Like a
    torque curve. This occurs even with very fast-burning pistol powders, such as Bullseye and HP-38. The faster the powder, the steeper the curve and the quicker the peak. This explains why magnum revolver rounds produce
    the charactistic fireball at the muzzle...The bullet exits near the peak of the pressure curve while there is still a lot of unburned powder in the barrel,
    and it explains why slow powders perform best with a long barrel and/or a heavy bullet.

    Also, the slower the burn rate, the broader the curve. Peak pressure is maintained for a longer period before starting to taper off. Quick powders
    peak quickly and fall off quickly. The curve looks more like a spike than a curve. Blackpowder spikes quickly and falls off almost instantly after the peak, making it more of an explosion than a progressive burn. Even
    Bullseye burns slower than blackpowder, but it produces higher peak pressures for a given charge weight...Much higher.

    RJ...to get back to your question on Newton's 3rd law...It occurs because there is an equal force operating in both directions. Stand with your nose
    to a wall and push off the wall with your hands. The amount of force that you exert against the wall will push you away from the wall with equal force. In this example, you are the bullet and the wall is the breechface and your arms become the vector. Now...Stand toe to toe with someone of equal weight and push hard against his shoulders. You will both move
    an equal distance and speed from the center of the vector, assuming that each one of you offers equal resistance to the shove. In this example, you are the breechface and he is the bullet. He moves and you recoil...Equal and Opposite action/reaction.

    Now, do it again with somebody twice your weight. The force will still be equal, but he will move half as fast and half as far. The "kick" that you feel
    when you shoot a rifle is the reaction of the rifle to the bullet's inertial resistance AND the barrel's frictional resistance to the bullet passing through it. You can decrease the felt recoil by using a barrel in which the
    rifling is removed, and the bullet passes through easier, even with a perfect gas seal so that no pressure can escape...OR...Fire a lighter
    bullet at the same velocity as the heavy one, and recoil will likewise decrease. Less inertial resistance to action equates to reduced opposite reaction.

    Recoil occurs because of the vector. In order to produce "Equal and Opposite", there must be a driving force to initiate the action. Simple, what?
     
  9. Art Eatman

    Art Eatman Administrator Staff Member

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    Don't confuse force and motion. The pushing against the wall of a house is a good example.

    For all practical purposes as to recoil, the change in pressure as the bullet travels down the barrel is irrelevant.

    To get a handle on the amount of recoil, the most relevant number is the maximum pressure. (Always remember that the pressure from an expanding gas is equal in all directions. This means sideways as well as forwards and backwards.) The next numbers of importance are the weights of the bullet+powder and the rifle; then, some rational assumption of the velocity at the time of the peak pressure.

    Force = Pressure x Area. The dimensions which are of concern are the diameter of the bullet and the diameter of the inside of the base of the cartridge case.

    Recoil itself, you're getting into Impulse and Momentum. Since I'm some forty-plus years away from studying those, I'll leave it up to those more current on the subject. Possibly a textbook on Physics would address this; for sure an engineering text on "Statics and Dynamics".

    Art
     
  10. Jim K

    Jim K Member

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    First, in terms of guns, we are talking milliseconds, not "a couple of seconds". When the primer fires, things happen very rapidly.

    Now to the rest. Recoil is NOT caused by the gas pressure acting on the breech face. It is cause by bullet movement. Period. The gas pressure causes bullet movement, but it in itself does NOT cause recoil.

    I have said this before, but no one pays attention:

    IF THE BULLET DOES NOT MOVE, THERE IS NO RECOIL. NONE. NO WAY. NO HOW. NO MATTER HOW HIGH THE PRESSURE IS. NO MATTER WHAT POWDER IS USED. NO MATTER WHAT THE RACE, RELIGION OR NATIONALITY OF THE SHOOTER. Let me try that again. IF THE BULLET DOES NOT MOVE, THERE IS NO RECOIL. PERIOD.

    If the barrel of a gun is plugged so the bullet cannot move at all, and the gun fired, there will be no recoil. If the same thing is done to a recoil operated pistol (the 1911 for example), the pistol will not operate.

    Recoil is not caused by gas pressure; the gun is not kept closed by the bullet pushing the barrel forward; the "thrust vector" explanation in Jerry Kuhnhausen's book is nonsense. (I like the books, and use them continually; I respect the author as an expert in the 1911 pistol, but he simply does not know how the darn thing works!)

    Recoil begins at exactly the same time the bullet begins to move; not sooner, not later. In a pistol like the 1911, the barrel/slide unit begins to recoil the instant the bullet begins to move, not after the bullet exits the barrel, no matter what anyone says, and high speed photos prove it. But the barrel/slide unit weighs more than the bullet, so it travels slower than the bullet. The bullet moves about 4", while the barrel and slide are moving about 1/10 inch, so unlocking does not begin until the bullet is out of the barrel and pressure drops, which is the whole idea.

    Newton's second law, now made in a lemon scented, borax freshened version called the Law of Conservation of Momentum, is still in effect.

    Jim
     
  11. 1911Tuner

    1911Tuner Moderator Emeritus

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    Lemon-Scented

    Jim! Ease up brother...Remember your blood pressure!:D

    There's no argument that the bullet has to move to get the slide to move...and that they both start at the same time, but the force that gets'em movin' is what produces the recoil...not the bullet movement alone.
    It can't. If you yank the bullet through the muzzle at 830 fps or 8350 fps,
    the slide ain't gonna recoil....at least not in a normal fashion.

    It has made for an intersting discussion though...what?:cool:
     
  12. Jim K

    Jim K Member

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    I'll try again. The gas pressure moves the bullet. But it does not cause recoil, only bullet motion does that. (There is a small component of recoil from the forward movement of the gas particles, but it is negligible* and again, it is the forward movement, not the pressure that is involved.)

    If it were pressure, then a gun with its barrel blocked would recoil even though the bullet never moves. It does not.

    Yanking the bullet through the barrel is not the same thing, since an outside force is involved.

    *In a gun, it is negligible in comparison with the recoil component caused by bullet motion, but it is what put a man on the moon. A rocket functions from the recoil of gas at very high speed going out of the "barrel" and pushing the "gun" in the opposite direction.

    Jim
     
  13. Jonathan

    Jonathan Member

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    If I remember correctly, this started with a discussion centered around handguns, so I can understand why you state it is negligible. When dealing with hundreds of grains in bullet weight and tens of grains of powder, the mass and momentum will usually favor the bullet over the gasses.


    However, why over-simplify? In a rifle cartridge where the powder charge can approach 50% or more of the bullet weight, it is most certainly significant. With a strictly decomposition reaction you're going to be dealing with at least the powder weight sent downrange, and any oxidation would only increase the numbers.

    After all, if gasses have nothing to do with recoil, then brakes would be totally ineffective.
     
  14. benEzra

    benEzra Moderator Emeritus

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    Let's try a little thought experiment.

    Consider an imaginary gun whose chamber we can freely pressurize to any value, and a bullet that can be locked into place or released at will. Let's assume for the sake of argument that the bullet has a cross-sectional area of 1 square inch. We'll also assume a cylindrical chamber, so the breechface also has an area of 0.1 square inch. We'll also assume that friction between the bullet and barrel is negligible. To keep things simple, let's say the bullet has an inertial mass of 0.0001 slug (roughly 22.5 grains) and the gun has a mass of 0.1 slug (3.2 pounds). (I really don't like U.S. customary units, but bear with me.)

    We lock the bullet in place and pressurize the chamber to 30,000 PSIG. There is no movement and no recoil; the compressed gas pushes the bullet forward with a force of 3000 pounds. The compressed gas likewise pushes the breechface backward with a force of 3000 pounds. But the whole system is locked together so nothing happens.

    We then release the bullet, and things get interesting. The force of 3000 pounds pushing on the bullet is no longer opposed by our imaginary bullet-locking mechanism, so the bullet begins accelerating at a rate of 30,000,000 ft/sec^2 (3000 lb/0.0001 slug). The force of 3000 pounds pushing on the breechface is no longer opposed by the locked bullet pushing forward on the barrel, so the gun begins accelerating backward at 30,000 ft/sec^2 (3000 lb/0.1 slug). Recoil begins at this moment.

    Let's ignore gas pressure falloff and just assume that for barrel length reasons, the bullet attains a final velocity of 3000 ft/sec by the time it exits the barrel. The gun is accelerating backward at 1/1000 this rate, so it attains a final backward velocity of 3 ft/sec.

    Now let's examine the system from a conservation of momentum standpoint.

    Bullet momentum: 3000 ft/sec * 0.0001 slug = 0.3 ft-slug-sec.
    Recoil momentum: 3 ft/sec * 0.1 slug = 0.3 ft-slug-sec.

    So it checks out.

    Some thoughts:

    (1) Yes, recoil is intrinsically tied to bullet movement, and can even be calculated from the bullet momentum (ignoring gas-mass effects, which I've ignored here).

    (2) Recoil is transmitted to the firearm by the unopposed gas pressure upon the breechface. No surprise there.
     
  15. John Ross

    John Ross Member

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    Jim is the only one to really get it right here: The Conservation of Momentum is the relevant law here.

    Imagine a massively strong bolt action rifle with a solid "barrel" that had only a chamber and nothing else. It would chamber a cartridge but the bullet would have nowhere to go if the trigger were pulled. You pull the trigger and the primer ignites the powder. Chamber pressure soars.

    For the sake of the illustration, let's assume a cartridge that under the above circumstances produces a pressure low enough that the case doesn't rupture, action doesn't leak, whatever.

    Apart from the tiny momentum opposite the release of the firing pin, the gun has no recoil at all. It's just a container holding a small, extremely high pressure cylinder (the fired cartridge.)

    Fire the same cartidge in a gun with a normal barrel, the bullet and burning powder fly out the barrel with a momentum of some value, and the gun moves to the rear at a speed such that it has THE SAME momentum value IN THE OPPOSITE DIRECTION as the bullet and burning powder.

    The "Pressure on the breechface" stuff is irrelevant.

    An inflated balloon with the neck pinched shut has pressure on all its "breechfaces" and no recoil. Release it, and the air goes one way and the rubber balloon goes the opposite way, with the equal amount of momentum.

    The MOMENTUMS (MxV) are equal, not the energies (MxV squared). That's why heavier bullet loads of the same muzzle energy have more recoil. They have more momentum.

    JR
     
  16. R.H. Lee

    R.H. Lee Member

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    Think of it like this-the pressure of the expanding gases is what pushes the bullet out of the case and down the bore. The same pressure also pushes in all directions, including against the breechface through the casehead. So as the pressure is pushing the bullet down the bore, it is also pushing against the breechface and attached slide. The bullet moves forward, and the slide moves rearward. Action & reaction. Whatever is pushing the bullet down the bore needs something to push against. That is the slide. With regard the the 3rd law, the external object is the bullet, but only after it disengages from the case.

    FWIW, I'm an accountant, not a physics professor :) so if my explanation is lacking.................
     
  17. RJ357

    RJ357 Member

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    Jim Keenan -
    True, the above is just an experiment. The couple seconds allows us to imagine it easier since it separates the two conditions of immobilized bullet and free bullet.
    The bullet must move in relation to the barrel in order for recoil to exist. Movement alone means nothing. That's why yanking it out with a rope doesn't work. If you immobilize the bullet to a solid wall, you will get a rather large recoil.
    In my example above, explain what is different between the bullet and the breech/barrel assembly. Aside from weight, what is different? We could make the bullet the same weight as the breech/barrel assembly. Why then couldn't you say that the breech/barrel assembly was propelled and the bullet was recoiling?
    And I have said that myself - "There is no question that these laws are true."

    My position is that these laws of mechanics are not the reason or cause. They are observations and predictions. Which is not trivial. For example, one of the most significant proofs of a theory is that it predicts reality.

    Newton's law is an observation of action/reaction and predicts that it will happen in every single case. However, it is not a commandment to do so; it is a prediction that it will do so.

    I would respectfully request, and welcome, that you look at my original post above and my exchanges with JohnKSa (who understands the laws well, as do you), and see if there are any flaws in my description.
     
  18. firearms_instructor

    firearms_instructor Member

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    Intersting thread, and a question...

    In a firearm with a rifled barrel, does the gun recoil much, if at all, before the bullet leaves the muzzle?

    I suspect that the tremendous friction of the bullet being squeezed through the barrel kinda "pulls the barrel forward" until the bullet leaves the muzzle, whereas in a cylinder-choke shotgun, the gun can start recoiling immediately on pressure buildup since the projectile is not impeded in any way. Is this more or less correct?
     
  19. RJ357

    RJ357 Member

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    1911Tuner -

    Yeah, I was making way too much of a mess over in the vector threads.

    I think we agree more than we don't, actually.
    Yes, I agree. Look:
    "The pressure of the gas, still enormous, is still producing a force on the bullet's base, accelerating it down the barrel.The pressure is also still producing a force on the breechface, accelerating the barrel/breechface assembly backward."
    There are two equal and opposite forces, one pushing the bullet, and one pushing the breechface.
     
  20. 1911Tuner

    1911Tuner Moderator Emeritus

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    Experiment

    I propose an experiment if Mr. Keenan is willing. I will remove the locking lugs from a Colt barrel and mail it to him if he still has the rod, the tap and the set screw. I propose a handload of 2.5 grains of Unique pushing a 230-grain bullet, and I'm willing to bet that the slide will move with the bullet static.

    The only difference between the two experiments is that the barrel won't be locked to the slide in the second one. The cartridge is to be reduced in the interest of preventing damage to the frame and slide.

    Standin' by...
     
  21. RJ357

    RJ357 Member

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    benEzra -

    That is basically an exact description of my thought experiment with the addition of some actual values.
    What is special about the bullet, though? Would you agree that it could just as easily be the slide that was being "fired"?
     
  22. RJ357

    RJ357 Member

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    John Ross -
    If that is true, then the "pressure on the bullet stuff" is also irrelavent. There is no fundamental difference between the bullet and the breechface (and whatever is rigidly attached to the breechface). Both are masses. Both are free to move.
     
  23. Jim K

    Jim K Member

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    Hi, John Ross,

    Thanks. I wonder if we are not fighting a losing battle.

    Hi, Jonathan,

    Guilty as charged to over-simplifying; the mass of the gas is a factor in recoil when the proportion of the mass of the powder charge to the mass of the bullet is fairly large, as in a .30-'06 rifle with a 150 gr. bullet and a 50 gr. powder charge. The case in the normal .45 pistol load is different, and that is what I had in mind. Yes, the "rocket" effect of powder gas does contribute to recoil and muzzle brakes do reduce that recoil by redirecting the gas.

    Hi, RJ357,

    I am confused (even more than usual). You say there is no doubt that the laws are true, but then you say you don't believe them, because they are just observations which are not "commandments". I doubt you could get much support for that idea from physicists, who seem to believe that those laws are pretty much just that, which is why they are laws, rather than theories or hypotheses. No slam intended, but those are the kind of arguments usually heard in college dorms after a few brewskis.

    You are right in the sense that things don't happen because Newton says they must. People and things fell long before Newton was born; he observed gravity and explained its effects mathematically. But that does not mean there is no gravity or that if you jump off a building you won't fall if you consider the "law of gravity" doesn't apply to you.

    Hi, ben Ezra,

    Your initial take is correct; mass x velocity in one direction equals mass x velocity in the other. But your second conclusion, that "Recoil is transmitted to the firearm by the unopposed gas pressure upon the breechface. No surprise there", is not correct. In fact, recoil began before the gas pressure was unopposed; it began when the bullet first moved. The bullet exit causes pressure to drop almost instantly to the ambient level. The slight remaining pressure (called "residual pressure") can have some effects, but it is not enough to cause recoil.

    Jim
     
    Last edited: Aug 21, 2004
  24. carpettbaggerr

    carpettbaggerr Member

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    Try and find a case where Newton's laws don't apply. Simple mechanics are well explained -- it's not until you get down to the quantum level that there is difficulty in predicting behaviors.

    Can't wait 'till laser and plasma guns are popular. Imagine the threads we'll see then......

    :p
     
  25. JohnKSa

    JohnKSa Member

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    Removing the locking lugs and then locking the bullet in place just turns the slide into the projectile and the bullet into the breech. It doesn't change the principles involved.
     
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