Pressure, Recoil, and Newton


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RJ357
August 20, 2004, 06:41 PM
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.

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JohnKSa
August 20, 2004, 09:52 PM
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?

RJ357
August 20, 2004, 10:01 PM
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?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.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?In any particular radial direction, the force is cacelled by an equal force on the opposite side. No net force.

JohnKSa
August 21, 2004, 01:04 AM
It didn't move initially because the bullet was stuck.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?the breechface moves (because) It has a net force on it not equal to zero.What caused the net force to become nonzero? The reduction in pressure inside the chamber?

RJ357
August 21, 2004, 01:15 AM
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? 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.What caused the net force to become nonzero? The reduction in pressure inside the chamber? No. When the bullet became free, the forward force on the barrel ceased.

JohnKSa
August 21, 2004, 01:50 AM
forward force on the barrel ceased.Not unless friction between the bullet and the barrel is zero.

RJ357
August 21, 2004, 02:02 AM
Not unless friction between the bullet and the barrel is zero. 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.

1911Tuner
August 21, 2004, 05:36 AM
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?

Art Eatman
August 21, 2004, 03:17 PM
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

Jim K
August 21, 2004, 03:57 PM
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

1911Tuner
August 21, 2004, 04:18 PM
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:

Jim K
August 21, 2004, 04:46 PM
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

Jonathan
August 21, 2004, 05:02 PM
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.)
...
*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.


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.

benEzra
August 21, 2004, 05:04 PM
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.

John Ross
August 21, 2004, 05:11 PM
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

R.H. Lee
August 21, 2004, 05:15 PM
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.................

RJ357
August 21, 2004, 06:37 PM
Jim Keenan - First, in terms of guns, we are talking milliseconds, not "a couple of seconds". When the primer fires, things happen very rapidly.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.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 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.The gas pressure causes bullet movement, but it in itself does NOT cause 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?Newton's second law, now made in a lemon scented, borax freshened version called the Law of Conservation of Momentum, is still in effect. 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.

firearms_instructor
August 21, 2004, 06:53 PM
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?

RJ357
August 21, 2004, 07:09 PM
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.RJ...to get back to your question on Newton's 3rd law...It occurs because there is an equal force operating in both directions. 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.

1911Tuner
August 21, 2004, 07:16 PM
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...

RJ357
August 21, 2004, 07:18 PM
benEzra -

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

RJ357
August 21, 2004, 07:28 PM
John Ross -The "Pressure on the breechface" stuff is irrelevant. 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.

Jim K
August 21, 2004, 07:33 PM
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

carpettbaggerr
August 21, 2004, 07:46 PM
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. 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

JohnKSa
August 21, 2004, 09:09 PM
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.

rbrowning
August 21, 2004, 09:16 PM
Jim I think your comment
"The gas pressure moves the bullet. But it does not cause recoil, only bullet motion does that. " causes a bit of the problem from a symantic's point of view.

Recoil is caused by the bullett's movement.
The bullet's movement is caused by the gas pressure.
Ergo the gas pressure is the root cause of the recoil. You are just talking about effects at one level and others are intuitively jumping to the root cause level. Don't burn any powder and you get no recoil either.

I absolutely agree that it all comes down to the momentiums being equal and that heavier bullets produce more recoil and that it begine as soon as the bullet starts to move. But they don't cause the recoil by them selves. Otherwise Newton's first law of motion would be thrown out the window, ie a body at rest tends to stay at rest until acted upon by an outside force, like chamber pressure.

As an aside, the burn rate of powder tends to produce it's maximum pressure after the bullet is approximately 6-8 inches down the barrel. The powder continues to burn even after the maximum peak has been reached. Even a 22LR needs about 16" of barrell to get complete burn. But just because the powder is all burnt doesn't meant the bullet has stopped accelerating. IIRC Dan Lilja's experiment with long barrels found that, with the cartridge that he tested, the maximum velocity was achieved at about a 46" long barrel. Kind of tough to get one through the brush though!

Someone asked why a bullet would not stop after the peak pressure has passed and the answer lies in the relationship between static and dynamic friction. Once a body starts sliding the coefiecient of friction drops, to about 1/2 of what it was when static. Also there is the plastic deformation that needs to be done to the bullet to get it it fit to the riflings. The lead needs to be moved, requiring work to be done. Once this is done the force required to move the bullet decreases.

Any way I think I need more range time to experiment. Care to join me?

Rick

RJ357
August 21, 2004, 09:21 PM
Hi Jim Keenan

There could be some musunderstanding of what I am trying to say and that could my fault. As you noted, I have said the laws are true, and therefore, what I am claiming is that my description is true in addition to those laws and not instead of them. Laws apply to everything in my description and none are violated by it.

One can argue the some laws make for a better explanation than others. That is in fact exactly what I am doing. Your own description of events is accurate. I believe mine are also. I think the balanced vector problem is more easily analyzed by my description, and that it was started all of this.

It probably sounded like I was claiming that the laws of mechanics do not apply or are wrong. That caused things to focus there instead of what I was trying to say.

Again, I would request seeing if you can find an error in my description of the thought experiment.

Also see next post.

RJ357
August 21, 2004, 09:23 PM
Almost everyone is overlooking a basic fact:

There is no fundamental difference between the bullet and the breech

Before you can argue that the bullet is caused to move and the breech moves as a result, you must show that the bullet has some kind of prefered status. Unless you can do this, you must also admit that it is the breech that is acted upon and bullet moves as a result.

Something else that is being overlooked:

The bullet base, breechface, barrel walls, everything is being subjected to virtually the same pressure from the same mass of hot gas. Why allow that this pressure is causing some part to move but not another? Again, why a preference for the bullet?

RJ357
August 21, 2004, 09:28 PM
Hi JohnKSa

Concerning your first few posts, You know all that stuff!
That was just a test wasn't it?
How'd I do?

More serious question:

Is your position that pressure causes recoil in a straight blowback, but not a locked breech?

JohnKSa
August 21, 2004, 10:07 PM
It's the same in either case.

In the locked breech case the entire barrel and breech/slide assembly must recoil together because they are locked together.

In the straight blowback case the breech/slide can recoil without the barrel because they are not locked together.

How about a thought experiment...

Take the imaginary gun you described in the initial post and put a valve in place of the bullet. Imagine that this valve is controlled so that it regulates the pressure in the chamber. It is programmed to exactly duplicate the pressure curve that would occur if a bullet were fired from the barrel. Now put the powder charge behind the valve and ignite it. If pressure is the key, then the gun should recoil exactly the same with either the bullet or the valve.

Will it?

1911Tuner
August 21, 2004, 10:07 PM
John Said:

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.
_________________

Exactly my point. If the forces acting on the bullet and slide are the same, the effect will be the same. The difference is that when the slide and barrel
aren't connected by a mechanical link...the lugs...the slide will move even if the bullet doesn't. If the barrel and slide aren't connected, and the bullet CAN move, the slide will STILL move. So, in this instance...which one is the projectile and which is the launching pad?

When the barrel and slide ARE connected, the slide CAN'T move unless the bullet moves because they're fighting one another in opposite directions.
An isometric lock...exactly the same as a tug of war between equally strong adversaries. Nobody moves until something happens to break the equally applied force. A foot slips, one gets tired and weakens...or one turns loose of the rope.

Standin' by with barrel and lathe...

JohnKSa
August 21, 2004, 10:15 PM
When the barrel and slide ARE connected, the slide CAN'T move unless the bullet moves because they're fighting one another in opposite directions.No, it can't move because the method for unlocking the slide is motion--backward motion generated by recoil. There is no recoil until the bullet moves and therefore they stay locked together regardless of the pressure.

In a straight blowback gun, the slide can be opened by pressure OR recoil because there is no locking arrangement to prevent pressure from opening the breech.

RJ357
August 21, 2004, 10:35 PM
JohnKSa -Take the imaginary gun you described in the initial post and put a valve in place of the bullet. Imagine that this valve is controlled so that it regulates the pressure in the chamber. It is programmed to exactly duplicate the pressure curve that would occur if a bullet were fired from the barrel. Now put the powder charge behind the valve and ignite it. If pressure is the key, then the gun should recoil exactly the same with either the bullet or the valve.

Will it?I assume you mean that the valve is opening by some certain amount and venting gas forward. In that case yes. And here is why:
When you opened the valve, you necessarily reduced the surface area at the "bullet base". The force there (pressure*area) is now less. I have only taking this to the point that it shows the breech will move.

JohnKSa
August 21, 2004, 10:38 PM
So we can completely take the bullet out of the system and still have the same recoil?

RJ357
August 21, 2004, 10:41 PM
So we can completely take the bullet out of the system and still have the same recoil? If you could produce the identical pressure on the breechface, by whatever means, without also producing a new opposing force that wouldn't be there with a bullet, then yes.

RJ357
August 21, 2004, 10:51 PM
JohnKSa -

Are you saying then, that there are two forces on the slde, 1) the recoil effect from the moving bullet, and 2) the force produced by the gas pressure on the breechface?

And I would ask everyone else that same question.

JohnKSa
August 21, 2004, 11:25 PM
Now let's cut off the barrel just in front of the valve and weld a tank in its place.

The tank is large enough to easily accomodate the gases escaping from the chamber without becoming significantly pressurized but is manufactured to weigh exactly the same as the portion of barrel we removed.

Now fire the "gun" again with the valve and tank in place. Nothing escapes from the "gun" but the pressure on the breech is still identical to what it was when the bullet was fired.

Now how does it recoil?

RJ357
August 21, 2004, 11:34 PM
Now let's cut off the barrel just in front of the valve and weld a tank in its place.

The tank is large enough to easily accomodate the gases escaping from the chamber without becoming significantly pressurized but is manufactured to weigh exactly the same as the portion of barrel we removed.

Now fire the "gun" again with the valve and tank in place. Nothing escapes from the "gun" but the pressure on the breech is still identical to what it was when the bullet was fired.

Now how does it recoil? There will be zero net recoil; there will be an initial recoil quickly followed by a forward movement. The rearward movement would occur until the pressure wave reached the front of the tank.

I think you may have missed the post that I added just after I answered your previous one, so I will add it here:

Are you saying then, that there are two forces on the slde, 1) the recoil effect from the moving bullet, and 2) the force produced by the gas pressure on the breechface?

JohnKSa
August 21, 2004, 11:38 PM
Are you saying then, that there are two forces on the slde Maybe...

Go back to the valved gun without the tank.

What happens if you put a bullet in front (on the muzzle side) of the valve.

RJ357
August 21, 2004, 11:44 PM
Go back to the valved gun without the tank.

What happens if you put a bullet in front (on the muzzle side) of the valve. it will recoil as normal. By your own description of the valve, it is going to produce a force on the breechface equal to what it normally experiences. to do that, with a bullet in place, the valve would have to open wide.

Jonathan
August 21, 2004, 11:51 PM
Quick intro for those that don't know me: I'm a PhD student in a science field. While physics isn't my focus, I have served as a lab instructor responsible for teaching these exact concepts to incoming freshmen.


Dealing with these in order:

1.
firearms_instructor said:
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?


Answer:
Yes, some part of the gun (but not necessarily the gun as a whole) will recoil immediately upon any motion of the bullet. In general, it will recoil the amount that can be attributed to the bullet's mass and velocity when it exits the gun, plus the net mass*velocity (in the forward direction) of the gasses behind the bullet. There's also the mass of the gasses in front of the bullet, but those are really negligible! Do a search for "anti-torque muzzle brakes" on other sites and you'll see that issue has been dealt with previously :D

Assuming that the bullet does not receive additional pressure from the gasses after leaving the barrel (which in actuality it probably does, a little), the only additional momentum the rifle/pistol receives is the reverse complement of any additional gasses that follow the bullet out the barrel.


2.
RJ357 said:
Are you saying then, that there are two forces on the slde, 1) the recoil effect from the moving bullet, and 2) the force produced by the gas pressure on the breechface?

And I would ask everyone else that same question.


There are three forces on the slide in our 1911 example:
1 - recoil spring, in the forward direction
2 - chamber pressure minus air pressure to obtain pressure difference on the breechface, in the rearward direction
3 - FORWARD pressure from the locking lugs

On the lugs/barrel unit:
1 - forward drag from the bullet
2 - same chamber pressure minus air pressure: however, evaluated in 3D, the forces all cancel and there is no net motion
3 - rearward pressure on the lugs from the barrel

On the bullet:
1 - net chamber pressure forward on rear area of bullet
2 - rearward drag from the barrel

There are many other little incidental forces, but none are really important. When I have some free time I'll draw in a complete diagram on the 1911 schematics.


3.
On a general note, I've seen "momentum" tossed around very freely here. While there are wonderfully simple laws to momentum, it is certainly a mistake to say that momentum "causes" anything to happen.

In a system like our 1911, the slide does NOT recoil due to momentum, although it can be said that we know that it will recoil due to our knowledge of momentum. In reality, it will take a force to CAUSE the recoil. When dealing with "causes", the only stuff that matters are forces, time, and mass. You can toss in pressure and area, also.

JohnKSa
August 22, 2004, 12:04 AM
What I'm asking is with the valve working as it is PRE-programmed to do, put a bullet in front of it and fire the gun.

It's not going to adjust to deal with the bullet's presence, it's going to operate as if the bullet isn't there.

Now what happens in terms of recoil?

R.H. Lee
August 22, 2004, 12:08 AM
ARRGGHH!

You are all saying basically the same thing.

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.

Right, and recoil does not being until the pressure of the expanding gas separates the bullet from the case.

Again, why a preference for the bullet?


Because in our 1911 system, the bullet is the first thing to become an external object.

RJ357
August 22, 2004, 12:11 AM
Hi Jonathan -

My question about 2 forces on the breechface concerned only the forces that produce recoil.

The common explanation here is that it is caused by the forward movement of the bullet. My explanation is that it is caused by gas pressure on the breechface. I was asking if anyone thought that both were present. (I don't)
In a system like our 1911, the slide does NOT recoil due to momentum, although it can be said that we know that it will recoil due to our knowledge of momentum. In reality, it will take a force to CAUSE the recoil.The movement over which it aquires momentum is very small, only a small fraction of an inch. The large part of it's movement is by the momentum aquired during the short pressure impulse. I think that is what everyone meant.

Opinions welcome!

RJ357
August 22, 2004, 12:37 AM
JohnKSa - What I'm asking is with the valve working as it is PRE-programmed to do, put a bullet in front of it and fire the gun.

It's not going to adjust to deal with the bullet's presence, it's going to operate as if the bullet isn't there.

Now what happens in terms of recoil? I would expect the bullet velocity to be somewhat less due to the restriction (valve) between it and the charge. From that, I would say the recoil would be less. It is more difficult to analyze by the explanation I have been using. I would need to know the relationships between the pressure on the forward side of the valve, rearward side of the valve, etc.

If you want to say that action/reaction is a better explanation for some cases (like this one), I would agree. For example. rocket propulsion analysis becomes unmanagable using forward force vs rearward force. For the simpler cases though, I believe my explanations are better.

RJ357
August 22, 2004, 12:44 AM
RileyMc -
Because in our 1911 system, the bullet is the first thing to become an external object Isn't that completely arbitrary? What principle of physics gives the bullet preference?

R.H. Lee
August 22, 2004, 12:49 AM
I'm sorry. It's Newton's 3rd

Newton's Third Law
Newton's Third law: All forces in the universe occur in equal but oppositely directed pairs. There are no isolated forces; for every external force that acts on an object there is a force of equal magnitude but opposite direction which acts back on the object which exerted that external force. In the case of internal forces, a force on one part of a system will be countered by a reaction force on another part of the system so that an isolated system cannot by any means exert a net force on the system as a whole. A system cannot "bootstrap" itself into motion with purely internal forces - to achieve a net force and an acceleration, it must interact with an object external to itself.

You need force on an "external object" for recoil to occur. The separation of the case an bullet makes both the bullet and the case/breechface/slide each external to one another.

http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html

misANTHrope
August 22, 2004, 02:48 AM
I think the whole point here is that there seems to be a feeling that recoil must occur either due to gas pressure on the breech face, or due to the requirements of Newton's third, or because of the law of conservation of momentum, and that if one of these is the cause of recoil, then the other two can't also be the cause. But all the principles in physics are interdependent.

As an example, when I took Dynamics a couple of semesters ago, there were three basic methods taught to analyze problems of force, acceleration, motion, etc. There was the basic "unbalanced force" method, where you analyzed all the forces in a system, figured out the net force on object(s), and from there determined their resultant motion or whatever they wanted.

Then there was the "work-energy" method- the basic example of this method is that if you hold a ball in the air, it has potential energy, and if it is dropped, its potential energy decreases as it falls. Conversely, when held in the air motionless, it has no kinetic energy, but when dropped, it steadily gains more kinetic energy as it falls. Because we know that energy is conserved, we know that at any point in the ball's fall, the amount of potential energy it's lost is equal to the amount of kinetic energy it's gained. Thus, we can determine its velocity at any time, provided we know its mass and the distance it's fallen.

Finally, there was the "impulse-momentum" method, something we've all seen an example of here- the bullet and gun are initially not moving relative to each other, and at some later time, without some external influence (like the reference to pulling the bullet out of the barrel with a rope), the bullet is moving. Since we know momentum is conserved, we know that the pistol has the same momentum as the bullet, but in an opposite direction, so we could determine the velocity of the pistol if we know the bullet's velocity and mass, as well as the pistol's velocity. Note that I'm referring to a hypothetical pistol floating in space, and assuming that the activation of the firing mechanism didn't exert external force on anything.

The point of this little missive was this: The only reason to choose one method over another was that generally, one made analysis simpler than the others- and sometimes, you might not have the necessary information to use one method. But any method could be applied to solve the problem, and the results would be the same regardless of method. They aren't mutually exclusive.

So, back to the discussion at hand- the bullet if fired, and the slide cycles. Why does the slide tend to go rearward? It's perfectly acceptable to say that it's because of conservation of momentum, or because of the third law. But it's wrong to say that because of conservation of momentum, gas pressure can't do any work. You might say that the gas pressure has to do work on the breech face in order to satisfy conservation of momentum.

The key is this: The only force that causes either the bullet or slide to move is the force resulting from gas pressure. It's the captive gas pressure, exerting a resultant force on the bullet's base, that causes the bullet to move. Since gas pressure is the only thing causing forces, and since we know that the slide has to move backward to satisfy conservation of momentum, and since we also know that an object can't move unless acted on by a force, then we know that gas pressure has to exert the force that causes the slide to move rearward. The two principles don't exclude each other, they satisfy each other.

RJ357
August 22, 2004, 04:01 AM
Hi misANTHrope -

Thanks for the information. This whole thing started out as an aspect of Jerry Kuhnausen's "balanced force vector" theory. Then it moved off on it's own.

This is centered around the question of why there is no recoil when a bullet is immobilized in the barrel and recoil exists if the bullet is free to move. My position was that the "unbalanced force" method was better suited to explain things. The "balanced force vector" theory involves the barrel and slide being locked together and unable to move backward as a result (there are 3 threads on that subject going on). I personally think that the 3rd law causes too much confusion and is too easily misunderstood by many. The problem is that I have been unable to explain why the "unbalanced force" method is valid.

If you know of a convincing way of explaining it, it would certainly be welcome.

misANTHrope
August 22, 2004, 04:58 AM
Oh, I've been in the other two threads, too... :)

The best illustration I've thought of so far is this: Imagine that you hold a short length of rope in each of your hands, and, gripping tightly, you pull outwards with both arms, so that the rope stays in the same place- you're applying identical forces with each hand, but in opposite directions. Now, say the rope breaks- both hands are going to suddenly move apart, speed dependant on whether you're a scrawny guy like me orhave the arms of Ronnie Coleman.

Now, let's apply this to the pistol with the proverbial obstruction in the barrel. Let's also assume that, compared to the sort of forces that will be generated by powder ignition, the force required to separate the bullet from the case is nonexistent, so that the bullet and case are free to move with no mutual friction. We'll also assume that their is no gap between either the bullet nose and the obstruction, nor the case head and the breech face.

Once the powder ignited and pressure in the cartridge begins to rise, this excess pressure exerts a force in all directions. The force applied to the walls of the case is perfectly balanced due to symmetry, so there is no tendency for the case to move in that plane. However, this force will deform the chamber a miniscule amount, and this deformation in turn causes the steel to act as a spring and apply force back inward, just like when you sit in your car and it sinks a bit. Along the same lines, the pressure exerts a force on the bullet base and the case head- these forces also being equal in magnitude due to identical areas of the two surfaces. Since we've assumed no gaps, these forces are transferred to the barrel obstruction, and then to the barrel (bullet) and to the breechface (case). From here on out I'll refer to these as the "bullet composite" and the "case composite."

If this were a blowback gun, where the barrel and slide/bolt were not locked together, then the only thing resisting the relative motion between the bullet and case composites would be the recoil spring, which would have to be very stiff indeed to keep motion from occuring. So the two composites would move apart, motivated by the opposing forces due to gas pressure. But now recall that as these two composites move apart, the spring is compressed, and the resisting force of a spring is directly proportional to compression, so at some point either the spring will be compressed enough to stop motion, or (in the case of the obstructed barrel) the composites will be stopped at the mechanical limits of their motion.

But we're talking about a 1911, not a blowback gun. With a round chambered, the two composites are already locked together, so there is no motion from the start. The slide has to move rearward for unlocking to occur, and there is no unbalanced force to cause this- the rearward force on the case composite is balanced by the force on the bullet composite, and the focal point of these forces will be the locking lugs (hope the steel's not brittle!).

So we back out to a fully functional, non-obstructed 1911. Once again, firing increases pressure in the case, radial forces are balanced out, and equal but opposite forces are applied to the bullet and case. The case still transfers the force directly to the slide, but now the only force the bullet can transfer to the barrel is the force of kinetic friction between the two- and since the bullet moves, we know that that force doesn't balance the force on the bullet. But all of the rearward force on the case composite is still transferred to the barrel as well via the lugs, but the fraction of the forward force transferred by friciton is not nearly enough to balance the rearward force. So, the slide begins to travel aft, with the recoil spring resisting just enough to control the motion, and eventually the barrel unlocks and the remainder of the cycle ensues.


So, in summary: the key difference between the obstructed/non-obstructed cases are the ability of the bullet to transfer its forward force to the barrel and balance the rearward force on the slide. With an obstruction, all forces are balanced. With only friction to transfer the force, rearward force on the slide exceeds forward force on the barrel, and the pistol cycles.

Clear as mud?

1911Tuner
August 22, 2004, 05:38 AM
misANTHrope said:


So, in summary: the key difference between the obstructed/non-obstructed cases are the ability of the bullet to transfer its forward force to the barrel and balance the rearward force on the slide. With an obstruction, all forces are balanced. With only friction to transfer the force, rearward force on the slide exceeds forward force on the barrel, and the pistol cycles.
______________

Precisely!
_____________

And:

Clear as mud?

It is to me...

Refer to Isometric Tension versus pushing a medicine ball.

Consider pushing that medicine ball while freestanding, and again pushing it against a wall while you are chained TO the wall. Force is acting in opposite directions in both cases, and that force vector is your arms.
The force vector is going to apply force to the ball and your body. If you are freestanding, both you and the ball will be propelled away from ground zero. When the ball is solidly blocked by a wall, and you are chained to the wall, nothing will happen except the application of force that isn't sufficient
to move either.

Mr Keenan's experiment can be duplicated without firing the gun. Silver solder a rod to the barrel. Make the rod long enough to pass through a
one-ton steel plate. Thread the rod and attach a nut. Pull on the grip of the pistol with your right hand and try to pull the slide to the rear with the other. The principle is the same. The barrel is being forcefully pulled against the locking lugs in the slide. The slide is being pulled in opposite directions.

When the gun fires, the force is then BETWEEN the two objects, pushing them in opposite directions equally while the lugs resist. The higher the force, the more tightly they lock, and the harder they resist.

Want to see one fire without the slide moving? Attach a five-pound weight to the slide. If it moves at all it won't be very much.

cracked butt
August 22, 2004, 05:59 AM
Think of a gun not as a powder burning device ejecting a bullet, but as a spring loaded device ejecting a bullet. In essence a spring and charge of powder do the same thing, but in different forms- they both store energy- one as mechanical energy and the other as chemical energy.

So you have your gun with a supercoiled 50,000lb spring held back by your trigger mechanism. You pull the triger, the spring is released, pushing both at the breachface and the ball at the same time. To make the comparison as similar to powder as possible, the spring both weighs 5 gr. and is not attached to the breachface, and the ball weighs 230 gr.

What is going to happen is that the spring is going to push back on the breechface and forward on the ball. the ball will ejct from the barrel followed by the spring- as its the path of least resistance- versus pushing rearward against a much heavier object.

I'm a little rusty on my physics- its been a few years for me too, but conservation of momentum is (correct me if I'm wrong) Mass1xvelocity1=mass2xvelocity2

lets say the gun weighs 3 lbs, the ball and spring weigh a total of 235 gr. and the ball+spring are launched at 800 fps.

Mass1= 235 gr
velocity1=800 fps
Mass2=21000 grains
velocity2=?

235grx800fps=21000grx?
so..

(235gr x 800fps)/21000= Velocity2

Velocity2 therefore =8.9 fps

In order to launch a bullet at 800 fps your gun would have to travel backwards at 8.9 fps. some of this would be mitigated by the recoil spring which will cause the gun to move backwards slower over a longer period of time, and the force not being aimed straight back into the hand as the axis of the barrel is above the hand.

1911Tuner
August 22, 2004, 02:15 PM
Bingo, cracked butt.

The force vector...be it a spring, or an explosion...or a sudden burst of compressed air is applying force in both directions equally. Initially,
nothing moves...except I have seen slow-motion videos of the bun moving sharply FORWARD for a very short distance just prior to the recoil cycle.

benEzra
August 22, 2004, 04:34 PM
(firearms_instructor)
In a firearm with a rifled barrel, does the gun recoil much, if at all, before the bullet leaves the muzzle?
Theoretically, the gun is almost in full recoil by the time the bullet leaves the barrel. It gains some added recoil from the gases still being accelerated down the barrel, but I ignored that in my example. The key to understanding this situation is that the rifle accelerates backwards at the same time the bullet accelerates forwards; just as the forward push on the bullet ceases when the bullet leaves the barrel, the backward push on the rifle also ceases (ignoring residual gas effects).

What is special about the bullet, though? Would you agree that it could just as easily be the slide that was being "fired"?
Yes. I meant by "intrinsically tied together" that the acceleration of both the bullet and the rifle begin at the same time.

BTW, anyone remember the Hedgehog antisubmarine weapon from WW2? That system had a stationary "bullet" (solid rod attached to the weapon mount) and fired the "barrel" (hollow, finned cylinder with a bomb on the end).

(Jim Keenan)
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.
Entire correct. I meant that the backward force on the gun was no longer unopposed by the forward force on the gun imposed by the imaginary locking mechanism, but my wording was misleading.

cracked butt
August 22, 2004, 05:58 PM
Here's the $64000 question- can you make a short recoil operated pistol function in the weightlessness of space?

1911Tuner
August 22, 2004, 06:20 PM
can you make a short recoil operated pistol function in the weightlessness of space?

Sure. Mass and weight aren't the same thing. An object has mass, no matter where it is...and it still takes a measure of force to move it.

RJ357
August 22, 2004, 07:15 PM
When a repair crew had to go up and work on the Hubbel telescope, they trained underwater with large heavy objects. Even though the parts would be weightless in orbit, they still had mass. The crew had to be careful with the massive parts in orbit, as they would be hard to get moving, and once moving, they would be hard to stop.

cracked butt
August 22, 2004, 08:29 PM
Sure. Mass and weight aren't the same thing. An object has mass, no matter where it is...and it still takes a measure of force to move it.
Alright my bad, I asked the wrong question...

:o

On Earth, when you fire a pistol, you are firing from a fairly solid platform- meaning the friction of the ground keeps you from sliding backward uncontrollably when you fire. If you do something as innocuous as limp wrist the gun, it can have functional failures. I'm wondering if in space if you are free floating on a space walk, would this lack of friction cause a situation similar to limpwristing? or would the amount of one's mass be enough to compensate?
If I were selected by NASA to go on the next shuttle flight as a test subject on the effects of living in space on individuals who like to lay on the couch, put down a few Leinenkugels and watch womens' beach volleyball, and I wanted to bring my 1911 to ward off space goblins on an occasional space walk, what poundage of recoil spring should I switch to assuming my mass is 112kg and I'll be wearing another 25 kg of space gear?

I used to love problems like this, but unfortunately, the gerbil wheels in that part of my brain have rusted up and need oil, some have stopped altogether, and perhaps alot of the rodents have died:eek: :confused:

RJ357
August 22, 2004, 09:25 PM
cracked butt -

You can create a similar situation on earth by sitting (or standing) on a swing hanging from very long ropes. The purpose of the long ropes is to keep all movement horizontal. That way gravity will have little effect on your movement. All resistance to sideways motion is virtually due to mass.
You can also do a similar thing by standing on the floor and leaning backward as far as possible (without falling). The friction of your feet on the floor would then have little affect in holding you.

Now if you can imagine this, you can see, I think, that your mass would probably be sufficient to make the gun cycle reliably. You would, however, go floating off backwards. And would also be rotating backwards. This is in space, of course.

cracked butt
August 22, 2004, 09:34 PM
I thought of using a pendulum to the same effect, but as you stated, it would have to be a very long rope. I guess maybe the best way to test it on earth is to skydive from an airplane and then shoot while freefalling.

RJ357
August 22, 2004, 09:38 PM
That would work. In fact, if it weren't for the air, it would be identical to being in deep space or orbit. You would be completely weightless.

Jim K
August 22, 2004, 11:33 PM
Cracked butt asked:

"On Earth, when you fire a pistol, you are firing from a fairly solid platform- meaning the friction of the ground keeps you from sliding backward uncontrollably when you fire. If you do something as innocuous as limp wrist the gun, it can have functional failures. I'm wondering if in space if you are free floating on a space walk, would this lack of friction cause a situation similar to limpwristing? or would the amount of one's mass be enough to compensate?"

Your mass here on earth is enough to resist the recoil and let the pistol operate, and the same would be true in space. The difference would be that since there is no gravity to keep you tied to the earth, firing the pistol would move you, at least a short distance. You could, in theory, do a space walk and get around by firing a pistol in the direction opposite that you want to move. But watch those bullets; they have mass too, and will do to a human target just what they would do on terra firma.

This thread has been a lot of fun.

Just for the record, Hatcher tried the blocked barrel experiment with a Model 1903 rifle and service ammunition. The gun was suspended on strings and fired remotely. There was no recoil. The pressure stayed up quite well and there was a loud bang when the bolt was opened.

I have done the same thing with a 1911A1 pistol and GI ball, except that the gun was hand held. The result was the same. The gun did not recoil, the slide did not move (I had a strain gauge on it). The case evidently did not obturate as well as the rifle case, because the pressure leaked out over a couple of minutes. But please don't try this one at home; you have to ruin a barrel and there are a couple of things you need to do to keep the pistol from being damaged.

Jim

John Ross
September 3, 2004, 11:47 AM
RJ357 said, replying to my statement that the "Pressure on the breechface" stuff is irrelevant:

"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."

The key factor is not "pressure on the breechface (or anything else)," it's MOVEMENT. The breechface and the sides of the barrel and the base of the bullet all have pressure on them, but the gun only recoils when something moves, and it recoils in the opposite direction of the something that moved.

If you have a plugged barrel and have ground the locking lugs off your bolt, the gun will recoil FORWARD, away from your shoulder, because the bolt will be going backwards through your head.

If your chamber has a flaw on the right side and ruptures, blowing out chunks of metal to the right, the gun will recoil to the left.



And to answer the guy that asked if the gun started recoiling before the bullet got out the muzzle:

YES. If you want to know how much, do some calculations based on the center of gravity staying the same (it does.)

Chamber a round.
Find the balance point of the loaded rifle on a knife edge.
Mark that spot.
Chamber an empty case.
Dump the powder in the barrel and distribute it evenly along its length.
Jam a bullet in the muzzle so it doesn't fall out.
Find the new balance point.

The distance between the two knife marks is how far the gun will have moved in free recoil at the moment the bullet exits the muzzle.

Let's not make this more complicated than necessary.

JR

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