"Thrust Vector" and other ideas

[Tamara]

Well, in a locked breech gun, the slide CAN'T move based on pressure. The bullet must move and generate recoil in the slide/barrel UNIT before the slide can move.

In a straight blowback gun, I guess you could make the statement--

How does the cartridge know what kind of gun it's in?

The barrel does not recoil, unless it is being dragged rearward by the slide. The projectile does not start its forward motion as one unit with the barrel. It moves forward and engages the barrel.

Where the hell is owen when we need him anyway? He's got a slide rule. He does this stuff for a living.

 

[NMshooter]

Test fixture: put pistol in vise. Affix barrel plug to vise. Slide should still be capable of motion. Load one blank cartridge. Fire. I bet the slide cycles normally, and the blank is extracted and ejected.

 

[RJ357]

A question.

During that first 1/10" of movement of the slide, does the barrel move downward at all? In other words, have the barrel lugs begun moving out of the slide notches during this 1/10" movement?

 

[1911Tuner]

Movement?

Howdy RJ,

I see where you're going with your question. One thing to keep in mind is that barrel linkdown occurs AFTER the vector has done its work...not during.
if it happened during a signifigant part of the pressure curve, the case head would probably blow out.

The answer is yes. The distance from the centerline of the slidestop pin to the beginning of the convex radius on the lower lug marks the start of barrel linkdown...but the barrel doesn't fall into bed...The link stops the barrel's rearward motion and forces it to change directions. IF the linkdown/unlocking sequence is correctly timed, this occurs after the
chamber pressure has fallen off enough to allow primary extraction.

At this point in the cycle, the slide momentum needed to complete the cycle
is established. Unless the slide is oversprung, or a mechanical interference
occurs between the slide and barrel, the slide will make full travel and return to battery.

This distance varies from gun to gun, and can occur with as little as .080 inch of barrel travel, to as much as .125 inch. In many pistols that are timed just so...the empty case will exit the chamber and eject from the port without an extractor even being in the gun.

 

[RJ357]

1911Tuner -

OK thanks

What I was wonder was, if the barrel was stuck to the slide, as from the vector force, if the slide could still move back that small amount (1/10").

Was barrel lugs the right term for the lugs on top of the barrel?

 

[JohnKSa]

How does the cartridge know what kind of gun it's in?

Obviously it doesn't. The slide and barrel "know". In a locked breech gun the barrel and the slide can't unlock from each other until the whole slide/barrel unit is set in motion. In a straight blowback gun the slide is free to move without the barrel.

The barrel does not recoil, unless it is being dragged rearward by the slide.

Which, of course, in a locked breech gun it is. In a straight blowback gun the barrel doesn't recoil except through the force transferred to it through the frame from the slide hitting the end of travel and from the force transferred from the slide through the recoil spring to the frame before that.

 

[1911Tuner]

RJ's Question

RJ asked:

What I was wonder was, if the barrel was stuck to the slide, as from the vector force, if the slide could still move back that small amount (1/10").

No. The barrel will only move fore/aft by the amount of clearance between the locking lugs on the barrel and the slots that they fit into in the slide. In an unworn/undamaged ordnance-spec pistol this difference is very small...about the thickness of a sheet of typing paper...up to about .010 inch (ten thousandths) if the pistol is loosely fitted. Much more than that, and the gun is barely serviceable The lugs will beat themselves to death in a fairly short time.

Tightly fitted pistols have almost none. The 1/10th inch of movement in the initial unlocking phase is made by the barrel and slide together. Hand-cycle a 1911 pistol and watch the top of the barrel hood. You'll have to cycle it very slowly and watch closely to see the barrel start to move. By about .250 inch of total slide travel, the barrel is completely unlocked and linked down...so you can see that it all happens pretty fast and in a very short space.
_______________________

Was barrel lugs the right term for the lugs on top of the barrel?

Yep...Those are the locking lugs. There are three. The two forward lugs are obvious. The third one is the "shoulder" formed at the front of the hood.

 

[ryoushi]

All I know...

Ryoushi put bullets in gun, ryouchi aim gun at rotten grapefruit. Ryoushi then pull trigger and grapefruit explode in most satisfactiry manner.

 

[Thegman]

It seems this keeps going round and round, but I think it's becoming overly complicated....

Here's my (short) take (not necessarily correct, but I don't currently see were it wouldn't be):

1. Conservation of Momentum: As bullet mass moves forward, the remainder of psitol moves rearward. Barrel and slide mass can move, so they do move, rearward, proportionally to their mass WRT the mass of the bullet. If you don't agree with that, you're one of the people that apparently doesn't believe in laws of physics.

2. Friction of bullet on barrel: I think I'd more or less forget this. This can also be looked at as barrel friction on bullet. Remember, while bullet is in barrel, pistol-bullet is a single unit. Frictional forces of bullet on barrel are equal and opposite of frictional forces of barrel on bullet. Conservation of momentum, I think, is FAR more important to consider.

3. After bullet leaves barrel, slide-barrel unit have rearward momentum from #1 and at this point, the only mass left to act on recoil is the mass of the gasses from the powder. Small mass, but, after bullet leaves barrel, very high acceleration due to high pressure gradient between atmosphere and muzzle. This gas HAS to have an effect on recoil for the same reason as #1.

4. All of the other internal pulling-pushing frictional forces, ect. should essentially cancel each other out. The pistol can't pull itself "by its boot straps". Forward accellerating masses (solids and gasses), I think, are what to consider for all of the pistol's reward acceleration.

 

[1911Tuner]

Mass Acceleration

Gman said:

This gas HAS to have an effect on recoil for the same reason as #1.

Absolutely. Anything that CAN have an effect WILL have an effect,
and can't be ignored...no matter how small.
__________________________

4. All of the other internal pulling-pushing frictional forces, ect. should essentially cancel each other out. The pistol can't pull itself "by its boot straps". Forward accellerating masses (solids and gasses), I think, are what to consider for all of the pistol's reward acceleration.

Except for one consideration...The gasses are pushing (accelerating) in both directions. The larger mass (barrel and slide) is moving backward
as the bullet is moving forward. Recoil begins at the exact moment of bullet movement. It just moves (accelerates) slower because of its greater mass.

The ignition of the powder is like any other explosion...even though the powder burns progressively...It produces force in all directions. Whatever
that force happens to act on is also accelerated, but at a lower rate due to
its having mass, so the faster acceleration of the unopposed gasses can
(and will) follow the path of least resistance.

Nobody can deny that less force is required to rack a slide than push a bullet through a rifled barrel, even after the initial coefficient of friction is broken. At the point that the vector begins to apply force in the direction
of least resistance and gets the slide moving, the slide's greater mass/momentum starts pulling the barrel backward with it via the mechanical connection (the lugs) even though the bullet hasn't yet exited.

The force is pushing, and the slide is straining to pull the barrel (and bullet) backward. ZIP! The bullet exits...(One side in a tug of war lets go of the rope) and the slide's conserved momentum is suddenly loosed...and that
momentum carries it through to full travel. Watch a slow-motion video of
the firing/recoil cycle. You'll see the gun move forward just a tiny bit, followed by the slide moving...slowly at first...and suddenly speed up just
as the bullet breaks free. I had to watch it 40 or 50 times before I saw
what was happening clearly.

 

[RJ357]

1911Tuner -

Thanks

You answered, "No. The barrel will only move fore/aft by the amount of clearance between the locking lugs on the barrel and the slots that they fit into in the slide. "

Sorry, I should have asked, "if the barrel was stuck to the slide, as from the vector force, if the slide and barrel together as a unit could still move back that small amount (1/10")."

 

[1911Tuner]

Stuck Slide

I should have asked, "if the barrel was stuck to the slide, as from the vector force, if the slide and barrel together as a unit could still move back that small amount.

Yes. As long as there was also an additional force acting on the slide to pull them in the same direction AND it was stronger than the force that was pulling them forward. It would be much the same as lifting a barbell
with 200 pounds on it. If you're exerting 201 foot-pounds of lifting force,
you would lift the weight one foot per minute. If Mr. Keenan had yanked on the slide at the same instant that he pulled the trigger in his blocked-bore experiment, the slide would have moved, pulling the barrel back with it until it unlocked because the vectors of force wouldn't have been equal.

 

[RJ357]

1911Tuner -

OK, thanks.

Here is what I am thinking:

There seem to be two separate meanings to the term "locked".
One is as we normally think of it in a 1911, the barrel lugs engage slots in the slide and prevent them from separating. As long as the slide is sitting within a certain distance of it's full forward position, the barrel and slide cannot separate. The pair must move back a certain distance before the lugs will disengage.

The other meaning of "locked" due to the thrust vector. Here the friction between the barrel lugs and the slide slots does not allow the barrel to link down and disengage the lugs from the slots.

It seems that the normal locking, as in the first definition above, will guarantee that they remain locked until the bullet is gone. This is because the bullet is gone by the first 1/10" of movement. And 1/10" of movement is not enough to completely disengage the lugs.

It looks like even if Kuhnausen intended the thrust vector to work, it doesn't matter if it doesn't. The barrel and slide would be locked together anyway.

Or should I say "if Browning intended it to work".
Who originated it?

 

[Thegman]

Tuner,

I'm not sure I'm totally following what you're suggesting about the bullet-barrel getting pulled/pushed, etc. but the frictional forces between the barrel and bullet are not really doing work on the system. The friction is producing lots of heat, but probably little to no work.

Also, I'm not totally sure I know what you're saying about the gasses, but, while the bullet is still part of the system (in barrel), the gasses are adding a little to recoil, very little. If the powder charge is, say 10 grains, it's center of mass is accelerating from the cartridge case (where it is a solid) to approximately the middle of the barrel's length as it expands and fills the barrel. It's acceleration at this time cannot be any greater than the bullet in front of it. So, while the bullet is in the barrel, the gas is like an extra ~10 grain mass moving ~ 1/2 the distance of the bullet. That won't affect the system's momentum very much. Once the bullet leaves the system, the gas, with it's center of mass ~1/2 way through the barrel, will then begin to accelerate out of the system much faster than the bullet did. That momentum is now added to recoil. How much it affects the total recoil, I don't know.

As far as the slide speeding rearward due to the bullet leaving barrel; I wouldn't think that it should, not much, anyway. The final velocity of the slide/barrel, due to the bullet moving through the system cannot be more than the momentum the bullet is taking forward, and the bullet is building momentum throughout it's acceleration, so the barrel/slide are as well. I would guess that the sudden accelleration of the gas has a lot to do with what you're seeing at this time, which is where a compensator's baffles probably come into play (slowing the gasses acceleration out of the system). If one were to do some calculations on gas acceleration it would go a long way, I think, to answer that question. How about filming the sequence with various compensators?

I've never seen the video you're talking about (I'd like to), but as far as the pistol moving forward: I wouldn't say that couldn't be in part from bullet movement into the lands (I think that's what you're saying), but don't forget that the hammer just fell at that moment as well. Was the pistol in a vice during firing? Was the test also done with dryfiring to observe effects of the forward rotation of the hammer's mass? Either way, the little forward movement you're talking about isn't the recoil we're talking about, anyway, I guess.

 

[1911Tuner]

Simplified

Howdy Gman,

The frictional resistance of the bullet in the barrel is pretty signifigant. If
you've ever had a bullet lodge halfway through a barrel, and tried to drive it out...You know that it takes a lost of force to get it out.

Okay...We've spoken of the hypothetical pulling of the bullet through the barrel with a cable. Take it a step further and mount this pistol in a vise.
Attach a cable to the bullet AND to the back of the slide. Start applying equal force in opposite directions, and gradually increase the force until
the bullet moves. At that point, the bullet requires LESS force to keep it moving than it did to GET it moving, while the force rearward on the slide
is unchanged. The balanced vector of thrust isn't equal any more, and the slide will begin to move, pulling the barrel with it via the lug connection, and it will continue to move at a set pace UNTIL the bullet breaks contact with the barrel. The force on the slide remains unchanged...established momentum if you will...and the slide will now accelerate quickly and move until it's forced to stop upon impact with the frame.

Imagine the theoretical tug of war in which opposite force begins, but
as one opponent starts to get tired, the balance shifts to the stronger side.
The tired tugger begins to slowly advance toward the mudhole that awaits the loser. Just before falling into the mud, he releases the rope...and the stronger man is suddenly free to move backward unimpeded.. He will move suddenly and quickly. The applied force that he has been imposing has become momentum because he was moving....no matter how slowly...
while before he moved, it was just an application of force. (Momentum
begins with movement.)

And the hammer's momentum in the video was taken into account. Holding the pistol loosely and letting the hammer fall will move the gun forward a little...but not nearly as far as the bullet's impact into the rifling would be
if the gun was held loosely and fired. Held in a normal firing grip, the hammer's effect on the gun moving would be nearly insignifigant as compared to the bullet's effect. The video's purpose was to show the
firing/recoil sequence...not the effect of the hammer's impact on the gun,
so I'd venture a guess that the gun was held in a normal fashion....gripped
fairly solidly in order to prevent a limp-grip malfunction.

 

[Thegman]

Tuner,

Respectfully, I think you're really over-thinking it (or thinking about it in the wrong way).

If you have to do all of this imagining and use analogies, but cannot find dynamic equations to describe them, the equations probably don't exist. Notice no one seems to have an equations for the 'broken force vectors'? We can all agree, I think, and calculate, slide-barrel velocity as the bullet leaves the system using m(bullet)v(bullet) = -(m(barrel-slide)v(barrel-slide)), and if we assume constant acceleration for the bullet, we can even predict the amount of slide-barrel movement. I think that's about it.

There should be a calulatable "thrust vector" once the bullet leaves the system - From the escaping gasses, but not from the bullet.

I fully agree there's significant friction as the bullet moves through the barrel (but remember that it will be different than pushing a bullet through the barrel at almost zero velocity - I suspect the frictional coefficient changes with velocity and the heat created), but your hypothetical 'pulling' the bullet analogy is seriously flawed.

--The bullet and pistol aren't being pulled from forces from outside the system, creating 'broken force vectors' like snapping cables --

--It's being pushed from within the system.--

In your tug-o-war example, the pistol is being pulled from EXTERNAL forces, that's not what's happening. The forces are internal, and that's an important difference. The barrel is getting stretched and deforming slightly, like a spring. The only rope cutting analogy when the bullet leaves would be the barrel returing to it's original length and diameter as the pressure drops.

Maybe think of your rope as a rigid pipe, closed on one end, filled with, let's say concrete. A charge is ignited at the closed end. Is there friction between the pipe and concrete? Sure. But if the friction is less than the strength of the materials, that's were the movement (the break) will occur. Assuming the pipe is much less massive than the concrete, the pipe will go one direction, and, regardless of the friction, the concrete will move a bit in the other. The pipe, won't, indeed can't, drag the concrete around from internal pressue and friction in the system. The pipe and concrete will heat up as they're -"pushed apart"- The friction will reduce the final velocity of the pipe and concrete due to this energy loss, which produced heat. But momentum will be conserved. All of the 'tugging' is producing -internal- strain on the pipe and concrete, stretching and compressing both to some extent, especially the pipe, but nothing gets dragged anywhere from the friction (other than some atoms at the interface of the pipe and concrete); the friction creates heat and robs velocity through (energy stealing) to some extent.

Rather than thinking about pulling the pistol with cables, think about a really strong spring being released between the breech and bullet - what happens in that analogy? Regardless of the frictional resitance, the gun goes in one direction and the bullet the other, right? The friction between the bullet and the barrel heats the barrel, but doesn't drag it along by its nose. The barrel stretches behind the bullet due to internal strain in the system, but the bullet can't drag the gun along by it's nose, as it's being pushed forward! Conservation of momentum is what moves the system.

Does this change you mind? If not, give me some numbers on those "broken vectors", and explain where they're coming from (no cables pulling things around )

 

[1911Tuner]

Equations, Equations

Hiowdy gman,

There was a time that I could do the math and show the numbers...but it's been about 35 years or so since I've had to, so I called on the lads and gents who still have a handle on it.

You offered:

I fully agree there's significant friction as the bullet moves through the barrel (but remember that it will be different than pushing a bullet through the barrel at almost zero velocity - I suspect the frictional coefficient changes with velocity and the heat created), but your hypothetical 'pulling' the bullet analogy is seriously flawed.

IIRC, my first hypothetical situation was to envision pulling the bullet at the
identical rate of acceleration as it would be if it were pushed by powder ignition...Which would serve to cause the slide to recoil if the bullet's movement was all that caused it. When the bullet is blown through the barrel, the vector of force is also blowing in the opposite direction.
_______________________

And:

The bullet and pistol aren't being pulled from forces from outside the system, creating 'broken force vectors' like snapping cables --

--It's being pushed from within the system

And there is no difference between the tow except for the direction.. 500
Newtons of force is 500 Newtons...whether it's applied as a push or a pull.
The analogy is valid in showing that the vector is imposing force in opposite
directions....regardless of which direction it's imposed.

If you lift a 100 pound barbell, you've applied the same force and performed the same amount of work by lifting it as you would by pushing it up from a bench-press...assuming that the distance that the weight moved is equal.

Push equally in opposite directions or pull...The amount of force is the same. Pull the bullet to 800 fps in 4.5 inches of rifled barrel, and the effect
on the barrel is the same as pushing it...even down to the stretching. If the bullet's forward momentum is the cause of the recoil, the slide should move if the bullet is pulled through...at 800 fps.

 

[Thegman]

Tuner,

Push-pull is important. Let me give it another shot at convincing you.

Yes, a pull on one side is equivalent on the object being moved to a push in the other, but what is not equivalent are the affects on the surroundings. In your weight lifting analogy, PUSHING the weight up puts an opposite force on the floor (the breech of the gun), that's a correct analogy for the powder in the barrel. PULLING the weight up uses the same amount of force, in the same direction, and affects the weight the same, but it puts an opposite force on a cable attached to the celiling, THERE IS NO CEILING in the analogy for the gun system. That's the difference. There is a real floor (the breech), but no imaginary ceiling. That's the importance of push-pull, the affect on the surroundings, not the equivalent effect on the bullet.

In the firearm, where is the force originating? On an imiginary cable attached to a wall? No. It's between the bullet and the breech. The breech is real, the back end of the bullet is real, and the powder's pressure is real, and it's pushing the two apart. Final relative velocities of the system is from conservation of momentum. Does that make sense?

Your analogy of the tug-o-war implies the pistol is the rope, esentially. Something is pulling the bullet out of the barrel (the non-existent cable attached to the non-existent wall) while the opposite force is the shooter, hanging on, resisting the forward pull of this cable, keeping the gun's position esentially static. In your anaolgy for this 'broken force vector', when the bullet leaves the barrel (cable breaks), the shooter is falling back because they were tugging the other side of the rope (the pistol). That's absurd of course, but that's actually what you're saying. Give it some thought. Push-pull is very important, not so much on the bullet, but what else it affects, and how.

 

[1911Tuner]

Push/Pull-2

No...In my first analogy, there wasn't anything holding the gun back...I
simply intended to paint a picture of a bullet being pulled through the barrel and expecting the slide to recoil because o fthe movement of the bullet...Oversimplification at its oversimplified best.

The second analogy...the "Tug of War" I intended to paint a picture of
one cable pulling the bullet, and another pulling on the slide...under equal force. Nothing moves under these conditions until the bullet starts to move. When the bullet starts to move, the balance is lost because of
less force required to KEEP it moving/accelerating...bit the force on the slide
remains constant. Forward "thrust" is lower than rearward...and the slide moves. The faster the bullet moves, the less force is required to keep it moving. The slide, has greater mass, and accelerates more slowly, but it keeps moving because the balance can't be restored due to the changing
conditions brought on by bullet movement and falling coefficient of friction,
so there is always greater force on the slide. Even though it's accelerating,
its greater mass...and the effect of the recoil spring...it's accelerating slower
than the bullet. The bullet breaks contact, and all that gathering momentum in the slide has to go somewhere...so it goes backward toward the path opf least resistance. Of course, it all happens in a short time frame and in a short distance.

Yes...it all makes sense, but I won't be convinced that the bullet's forward momentum is the only factor in the slide's movement. It's only a part of
the whole picture. The vector of force between the bullet and the slide/barrel unit has an effect...and I believe that effect is greater than the bullet's acceleration and momentum.

 

[R.H. Lee]

The vector of force between the bullet and the slide/barrel unit has an effect...and I believe that effect is greater than the bullet's acceleration and momentum.

Could this be tested by cutting the barrel off at the nose of the bullet, or maybe just in front of the link? Does this mean that a longer barrel will increase recoil? Or conversely, do short barrel 1911's have less recoil?

 

[RJ357]

RileyMc -

Does this mean that a longer barrel will increase recoil? Or conversely, do short barrel 1911's have less recoil?

This is where conservation of momentum is a good choice for analyzing things.
If the bullet has greater velocity, then the recoil will be greater. For the same weight bullet, more velocity means greater momentum. And so the slide momentum would necessarily be greater also.

Felt recoil is another matter. A longer barrel usually means a heavier gun.

 

[R.H. Lee]

If the bullet has greater velocity, then the recoil will be greater.

And in order for the bullet to have greater velocity, it would have to be in the barrel with the expanding gas pushing on it longer. All the while that same expanding gas is pushing on the breechface/slide, causing recoil.

 

[RJ357]

Yes, both bullet and slide continue to accelerate as long as the bullet is in the barrel. When it finally pops out, the slide gets a small amount more from the rapidly falling pressure remaining.

 

[1911Tuner]

Popped Pressure Drop

The amount of residual gass pressure in the barrel/breech after bullet exit is nearly insignifigant...althougn it can't be ignored. The momentum that
was imparted to the slide during the first .100-.125 inch of travel carries it
through to full cycle, assuming that the recoil spring strength isn't off the scale. Relate to light "Softball" loads with the standard spring sometimes causing short cycle malfunctions.

 

[Jim K]

Wow! And I thought I knew how the gun works.

But I am convinced. It is all magic. Browning called the 1911 a recoil operated pistol, but what the heck did he know?

Some folks just can't accept that the laws of physics really exist and really do work. I grant the effects of pressure and friction are a lot easier to grasp than the idea of motion in one direction causing motion in the other, but I can only recommend talking to a physics professor who might be able to explain better than I.

One word on blanks. No, the gas does not need to "push" on anything. That is why a rocket works in space where there is nothing to "push" against. When a blank is fired, there is a slight recoil (not nearly enough to operate an auto pistol) caused by the mass of the forward moving gas, but that recoil will be there in a vacuum. Recoil is not dependent on the presence of air or gravity.

And there is still no such thing as a magic "thrust vector" that stops recoil from taking place until the bullet it out of the barrel.

Jim