Pressure, Recoil, and Newton

[rbrowning]

recoil

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]

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]

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]

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]

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]

Principles

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]

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]

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]

So we can completely take the bullet out of the system and still have the same recoil?

 

[RJ357]

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]

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]

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]

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]

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]

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]

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

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]

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]

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]

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]

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]

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]

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]

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]

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]

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?