Hey Tuner, xray of 1911 firing

[peterotte]

I have been looking on the 'net for a value of the rate of pressure rise in the 45ACP chamber. Couldn't find it. Can someone help?

Thanks
Peter

 

[1911Tuner]

The one thing that puzzles me is that a normally loaded pistol will run for thousands of rounds with 'normal' wear on the barrel lugs but an overloaded pistol will hammer those same lugs to the point of barrel replacement.

Exactly so. The pistol was designed to operate at 17,500 CUP...which translates roughly into 20,000 pounds psi...and within certain parameters for recoil energy and bullet and slide momentum. While it's over-engineered to a certain point...it'll only stand so much foolishness before problems start to show up. The design is amply strong to handle pressures well in excess of standard USGI hardball... but pressure is only a part of the picture...and while modern metallurgy has rendered the gun stronger and more durable...it's still the same basic design as it was nearly a hundred years ago. The locking lugs aren't massive, and their maximum vertical engagement rarely exceeds .052 inch....and if all three aren't bearing an equal share of the load...it doesn't take long to do irreversible damage to the barrel with heavy loads. The heavier the bullet for a given pressure...the faster the damage occurs. Yes. Bullet obturation...known as "Bump Up" to we who cast our own...increases bullet drag, but probably not enough to warrant much concern unless the ammunition is already overpressured...and the higher the pressure...the more it bumps...or the more solidly it plugs the bore. Obturation of the bullet helps to seal the bore against gas cutting along the sides of the bullet. The result...if the bullet is sized correctly...is zero or near zero lead fouling in the bore, so it would have to have some effect in driving up the forces involved.

 

[1911Tuner]

Odd Job...You can make a 1911 function as a straight blowback without adding slide mass...but the recoil/action spring would probably have to be in the 40+ pound range. If you fired the gun without making some change to delay the slide...you'd blow the case before peak pressure was ever a reality.

If you ground the lugs off and welded the slide, you'd convert it into a closed breech weapon...much like a muzzle-loader. Same deal.
Force forward is force backward. No loss in velocity due to escaping gasses, and all the rearward force against the breech would result in...recoil. Remember. All that happens is that the bullet and the breechblock are simply being forced apart at the same time.

 

[presspuller]

To Tuner, Owen, Peterotte, THANK YOU again.

SD, you are a lost ball in high weeds.

 

[1911Tuner]

Now presspuller...Behave. He's tryin'. Maybe he'll come away with fresh perspective.

I'd like to remind him that a lot of my experience has been in metalworking.
First a journeyman machinist...a toolmaker's apprentice...and finally a toolmaker. Additionally, during my aborted tenure in the M.E.T program 15 years ago...we had three quarters devoted to metallurgy, in which we spent a lot of lab hours studying and experimenting with shear, tensile, and impact stresses and their effects on various grades of steel...so it ain't like I'm runnin' into this discussion blind.

I feel confident that SD and Grump will learn somethin' by the time peterotte and Owen are done. Whether or not they believe me is incidental and of no consequence. As long as they learn.

 

[SDC]

Great picture of the Schwarzlose 1908 pistol. I had never heard of it before. (I couldn't but help notice the bottle-neck cartridge. I would certainly have assumed that to be the source of a large portion of the 'operating' force).

It's part of the advertising that Schwarzlose used in its advertising, but it's an "artist's conception" and has at least two things wrong with it that I can see; one, it's chambered in .32 Auto, so the round isn't bottlenecked the way the artist has portrayed it here, and two, the standing breechface isn't shown tapered from bottom to top the way it is in real life. As this pistol cycles, the barrel moves forward and ejects the empty out the left side, then it picks up a fresh round as the recoil spring forces the barrel back against the breechface.

 

[SDC]

Re: Odd Job's drawings, I'd also have to say that these bullets would not emerge at the same velocities. In barrel one, both the bullet and ju-jube would move, but the ju-jube (being lighter) would travel considerably faster. In barrel two, the bullet would be propelled at the expected velocity for any other firearm of that calibre with that barrel length. In barrel three, both bullet A and bullet B would be propelled out the opposite ends of the barrel, at identical velocities to EACH OTHER, but at different velocities from the previous two examples.

 

[SDC]

Odd Job...You can make a 1911 function as a straight blowback without adding slide mass...but the recoil/action spring would probably have to be in the 40+ pound range. If you fired the gun without making some change to delay the slide...you'd blow the case before peak pressure was ever a reality.

I might have said the same thing about the Hi-Power, before I saw it being done, stock slide, stock spring, and all.

 

[SDC]

I feel confident that SD and Grump will learn somethin' by the time peterotte and Owen are done.

It's certainly an interesting discussion of an interesting THEORY, but the existence of working blow-forward firearms that actually USE "bullet drag" to function shows that this force is marginal at best, and that it plays no practical part in locking the slide to the barrel.

 

[1911Tuner]

Well...Believe what you will...but peterotte and Owen are fixin' to shoot holes in it. The calculations are up to almost 2,000 pounds just to accelerate the bullet...before it even hits the rifling.

 

[peterotte]

Ummm..... Tuner? Are you sitting down?

I just did some tests I thought up. I do not know what to make of the results. Maybe I should just go back and keep redoing them untill the results come out 'right'.

I made a bullet jacket from a spent 22lr case (In a die I had previously made). I loaded up a hornet round, filled to the base of the neck, inserted the jacket, loaded it into my rifle and fired it into my firing tube. Phuut! (Does that sound about right)? SDC, I hope you are reading this! OK, back one step. I put a piece of folded up nylon sun screen on top of the medium. This had a nice, melted hole in it from the still burning powder. So far so good. I loaded up a air rifle pellet the same way with a paper sabot and same thing. Lets go back some more. I also put a piece of 16mm fibre board under the nylon. Both air rifle pellet and empty bullet jacket made holes bigger than a 22lr hv. Zero pressure signs at this point. No way of gauging that level of pressure anyway.
Next, I filled some AAA shot onto a jacket, weighed it - 3.2 grams - loaded it and fired it. THUDD! Good pressure signs on the primer - almost full pressure.
Still not good enough, so I took a piece of turned brass that I had used to guage the bore of my Remmie 22lr and cut it to length. (You can see where this is going)? Weighed it - 2.3 grams - a bit light but what the heck. Made a cardboad wad, placed that over the powder charge, seated the brass 'bullet' using a paper sabot and fired it into the firing tube. Now the sound effects come again. THUDD! Near full pressure signs on the primer! And no melted nylon.

I don't know what the Heck to make of that! That's all wrong?:banghead: Isn't it? But it's not over yet. The lady person still has to sing!
More test still to do.

This is fun! I just had to tell you guys.

Regards
Peter

 

[Grump]

Go and push a bullet through a barrel with a rod and understand just how much force is required, and then consider how much would be required to not only push it through...but to push it through in the time frame involved.

Came back in after a long break. Not sure I'll bother coming back.

What's missing from this thread:
*Pressure/time/velocity curve for any .45 ACP load, or anything else chambered in the 1911. All I can remember is that we're dealing with a primer ignition to bullet exit time of much more than a millisecond.

*Consistent acknowledgment of the role of inertia on the bullet AND the slide AND the barrel. Both sides in the disagreement seem to have done this.

*Hard quantification of the masses involved. My scale capacities have a big gap between bullet weights and slide weights and aren't that precise for the heavier weights anyway. A nice graph of Slide-XXX grams/oz, Barrel-YYY grams/oz, and bullet-ZZZ grams/oz would be nice.


Tuner: What frosts me in this discussion is your one-sided cherry-picking of ideas. Your quote above illustrates my point and is the reason for my "thought experiment" with the Yugos. To "push it through in the time frame involved" unavoidably includes inertia. As the rough-and-ready quick calculations posted recently show, there's a certain component of all that gas pressure that is NOT used to overcome barrel friction. It's used to accelerate the bullet against its own inertia. That's constant, whether through a tight barrel or from an oversize one through which the bullet literally rattles--the bored-out barrel proposed by one of us THR people. Then you explain the low slide velocity with its own inertia resisting the acceleration, and talked for days about barrel pull as if that were the only large/significant/whatever adjective force resisting the bullet's forward motion.

I apologize for not citing names--it's late and I'm tired and I want to get this over with. The bored-out barrel idea is great. So are the no-locking-lug experiments, but I see Tuner and others ignoring the report(s?) of others who have seen at least one John Moses-by-God Browning-designed tilting barrel locked breech design cycle with no locking lugs and no case blowout--at higher pressure than .45 ACP, by the way.

NOW, KIDS!!! This inertia effect applies to EVERYTHING--the bullet, the barrel, the slide, and eventually to the frame some time after bullet departure. I say the barrel and slide are locked together by geometry, physical interference. A slide/barrel can fire just fine and stay locked just fine without a frame attached. The barrel pull by bullet friction contributes a part of the pressure on the lugs, but not ALL of it. Part of that gas pressure is pushing against a fairly static breechface--agreed, are we, that the entire gas pressure is pushing back full strength? Whether the slide has begun to accelerate or not, that force is there.

The "forward" component of the pressure is being relieved by bullet movement. Part of that is used to swage the bullet to the lands and grooves, part to overcoming friction, and part of it is devoted to accelerating the bullet. Inertia resists the slide's movement back, resists the barrel's movement backwards as pulled by the lugs, and similarly resists the bullet's movement forward.

The gases do all the work, just like my Crown Vic hitting the Yugo. My point was what about the bullet's inertia? For the acceleration rates involved, inertia has to be a larger factor than friction--to my educated but uncalibrated guesses.

To say nothing is everything and everything is nothing or however it turns out, then to ignore others' arguments about the forces involved has continued to annoy me and entice me to jump back in...

Anyway, the SAME inertia arguments about why the slide doesn't go so fast backwards also apply to the Swartslose pistol's barrel unit. The force imparted to it is also applied over a very short time period, and its mass is far greater than the bullet, so whatever thousands of pounds of barrel pull (whatever the number) exist will be just as unavoidably LESS than the 3 lbs needed to cock the thing, as the 22 lbs or whatever it takes to cycle the slide on a 1911 pistol.

Problem for Tuner is, there's 100% of the gas pressure pushing backwards against the slide, but less than 100% of the gas pressure being used to swage the bullet and overcome friction.

Problem for CBC or whoever is, the less-than-full chamber pressure force which ends up being applied to pull the barrel on a Swartslose pistol is being applied over a very short time period as well.

HOWEVER, I believe that comparing the mass ratios of the various parts involved will be truly instructive. IF the Swartslose pistol barrel unit weighs the same as a 1911 slide, one would expect it to reach the same velocity as a 1911 slide during the firing cycle, AND the 'operating' spring would need to be the same draw weight. But what if that part is lighter? Betcha it is. How about a firm report of the spring weight? IF that is also lighter, AND the reciprocating part velocities are similar, then there's a huge problem with Tuner's theory--bullet pull would have to be something less than what you could call "casehead press".

I don't believe that bullet pull is "needed" for "proper" functioning of a 1911 or any locked-breech pistol. It's there, but not needed. I'd even venture to say that somewhere on this Board, we might have a person who fired (or witnessed firing of) a 9mm in a .40 S&W gun, in which the gun at least partially cycled. Betcha a "zero bullet pull" 1911 firing a .452 bullet down a .460 bore would definitely cycle. The slide is going to be pushed backwards by the gases against the inertial resistance of the bullet, it will get accelerated fast enough to pull the barrel backwards against the link, swivel the barrel down, and continue the slide's rearward travel as the barrel stops going backwards and down.

I don't have the facilities to check this. Since Tuner has so many boxes of .45 parts laying around, I will throw down the gauntlet and ask that he test it for us. Now, it might be necessary to use a heavier bullet to get the proper chamber pressure to get the proper bullet speed to produce the proper RECOIL for the RECOIL-operated pistol, but once those parameters are met (someone loan the man a chronograph if he doesn't have one), the pistol will cycle just fine. Remember, without the bullet speed, the gas pressure will not produce the recoil force on the slide which is "proper" for "proper functioning".

Based on the numbers posted above, it appears that I was wrong in my position that barrel pull is only a tiny (tiny being "small fraction" or whatever--I was thinking less than 10%) part of the forces applied by the expanding gases. It's apparently bigger than I thought but still not important to functioning of a 1911 pistol. If anyone can check or refine those calculations, please do. Since force forward has to equal force backwards, Tuner, only that force which exists from gas pressure but which is "missing" from the bullet acceleration is available to be barrel pull. Just don't ignore the role of inertia on all sides of the equation, because even the tiniest movement of the slide before bullet exit is going to add to the lug engagement pressure because of the barrel's own resistance to being pulled backwards by the slide.

I'll stand back and take another peek in a couple of days.

 

[peterotte]

Grump , don't go away yet. I suspect you are dead on! SDC too.

.....barrel pull is only a tiny (tiny being "small fraction" or whatever--I was thinking less than 10%) ......

That was my original thought. Then I thought it must be more like 60%.

I have done some more tests.

Test #1
Drill out the core of a hollow point match bullet and load it as usual.
Jacket mass - 0.7 gram
Charge mass - 0.68 gram AR2205
Minimal flattening of primer
Jacket recovered from sand medium. (Pretty ripped up).
Muzzle blast minimum

Test #2
Drive lead bullet down bore with brass drift untill it can be pushed by rod. Charged case chambered with cardboard wad over powder. Bullet seated into case from muzzle end.
Bullet mass - 3.63 gram
Charge mass - 0.68 gram AR2205
Normal flattening of primer
Muzzle blast normal
Bullet not recovered. Only lump of dry sand found. (In damp sand).

Test#3
Lead bullet seated onto cardboard wad over powder as normal.
Bullet mass - 3.63 gram
Charge mass - 0.68 gram AR2205
Normal flattening of primer
Muzzle blast normal. (Could be a shade louder than in test#2)?
Bullet not recovered. Only lump of dry sand found. (In damp sand).

Folks, this was from a hornet rifle, not a 1911, I know. I do not know how to interpret my observations but it sure looks a lot like the pressure generated by bullet drag & swaging forces is closer to 10% rather than the 60% or so I was thinking. But make no mistake, driving that bullet into the bore required a mallet and brass drift. Also, it does appear that the primer flattening is slightly more with the normal bullet. (Primer flattening is not a accurate pressure indicator).

Betcha a "zero bullet pull" 1911 firing a .452 bullet down a .460 bore would definitely cycle

Grump, someone definately needs to try that one.

Tuner, what do you make of all this? Would you be able to do some tests with a 1911?

SDC, you should be enjoying this . It's sure been fun doing the tests so far.

More tests to be done. (It's not over ......).

Regards All
Peter

 

[1911Tuner]

Peterotte...10% may be pretty close...but 10% of a peak operating pressure is still about 2,000 pounds of force required to drive the bullet...which would exert a healthy forward drag on the barrel...and it seems to correlate with Owen's first calculation.

I'd have figured it a bit higher, but so far, so good.

SD and Grump...The description of the locking/lug engagement mechanism is described in Browning's original patents. All you need to do is read it through a few times and let it sink in. I hate that I've apparently alienated a couple of folks here. That truly wasn't my intent...and all ribbing was done in good nature. hard to see it via the printed word. Face-to-face would make it more plain...but here we seem to be anyway.

Keep testing, and we'll call Owen back into it later today...hopefully. The only explanation that I can find for your findings is that the resistance to the projectiles wasn't high enough.

Grump...Of course a bullet with zero pull would cycle the action. Force is still required to overcome the bullet's inertial mass. Owen showed us that with his calculations...that it would require nearly 2,000 psi to accelerate the bullet's mass to 700 fps without any other resisting factor. So...We appear to be in agreement at least on that point.

 

[SDC]

Thanks for the testing, peterotte. My only contribution to this discussion is that I don't believe that "bullet drag" plays any sort of practical role in the operation of a 1911 pistol. Does the fact that aerodynamic engineers have "proven" that a bumblebee can't fly stop that bumblebee from flying? No, so they're obviously missing something somewhere in their calculations. Ditto for "bullet pull"; since actual working firearms that actually operate on this principle EXIST, and they don't require anything near what the math claims they should in order to operate, that tells me something is missing in those calculations.

Problem for CBC or whoever is, the less-than-full chamber pressure force which ends up being applied to pull the barrel on a Swartslose pistol is being applied over a very short time period as well.

Exactly; that's what makes this talk of "thousands of pounds of bullet-pulling pressure" null and void, because if there actually WAS thousands of pounds of bullet-pulling pressure being exerted in this design, the Schwarzlose would not only not work, it would be a single-shot because the barrel would go downrange with the bullet.

 

[1911Tuner]

someone definately needs to try that one.

ALready did it...and as expected, it will cycle. There's ample recoil force to do that, even under very low pressures. That's why squibs will cycle the action...provided that the slide moves far enough for the barrel to reach the linkdown point.

applied over a very short time period as well

No argument there, either. As I've tried to point out...peak pressure occurs early and quickly, and drops quickly...and pressure falls off as the bullet accelerates and increases the volume of the cylinder behind the bullet. It naturally follows that whatever forces are at work at peak pressure will also drop...so the high forces against the barrel...and the lugs...only lasts for as long as the peak. It's variable...falling rapidly following the pressure peak.
And...so goes the horizontal lockup and slide delay. And...since about 90% of the total muzzle velocity occurs within the first half-inch of bullet travel...the same applies to the force.

It's a little like the small-radius firing pin stop's effect in delaying the slide. It only lasts until the hammer is cocked...but it occurs at a time when that's most effective...before the slide has had a chance to gain most of its momentum.

SD and Grump...We may be getting closer than you realize.

 

[peterotte]

.....but 10% of a peak operating pressure is still about 2,000 pounds of force required to drive the bullet...which would exert a healthy forward drag on the barrel...and it seems to correlate with Owen's first calculation.

Quite so. It's hard to 'picture' the magnitude of the pressures and forces involved. (And the short time durations). The force I used to drive a lead bullet into the muzzle of a hornet is large relative to my strength but quite small relative to the forces in the chamber. A copper jacket must require a heap more force to engrave. Certainly, the copper jacket produced a lot more pressure than did the spent 22lr jacket, going by the primer and the degree of still burning powder exiting the muzzle. With the copper jacket, there was no melting of the nylon.

By the way, that jacket was moving pretty fast! If my 22 air rifle pellet loads are anything to go by, muzzle velocities of very light projectiles are very supersonic even at miniscule pressures. Funny that the powder actually burns up in the barrel! - With the air rifle pellet I mean.


Regards,
Peter

 

[ClarkEMyers]

off topic flight of the bumble bee

Most of my own references on this are internal at the Boeing Company and I no longer have access but the information is public - the following is as I recall it and much simplifed but it is reasonably correct so far as making the point.

Does the fact that aerodynamic engineers have "proven" that a bumblebee can't fly stop that bumblebee from flying? No, so they're obviously missing something somewhere in their calculations.

There was and is absolutely nothing wrong with the calculations from their first appearance about 1934. At that time theories of flight were less developed than they are now. Grossly oversimplified and omitting references - an exercise for the reader try googling bumble bee and Boeing a full complete and long winded discussion is out there - a theory of flight by flapping the wings was proposed and it was promptly remarked that by that specific model or theory the bumble bee couldn't fly - in fact the flight of the bumble bee is not by flapping its wings but by a sculling motion.

That difficulty has even made its way into an urban legend of science, typically recounted as "a scientist 'proved' that a bumblebee can't fly" and often cited as an inspiring example for persevering in the face of overbearing dogma. The bumblebee story can be traced back to a 1934 book by entomologist Antoine Magnan, who refers to a calculation by his assistant Andr¿ Sainte-Lagu¿, who was an engineer. ....

Solving the Mystery of Insect Flight By Michael Dickinson Insects use a combination of aerodynamic effects to remain aloft JUNE 2001 Scientific American

The theory was proposed, the math was done and the theory was invalidated. In the present case I'm inclined to believe Tuner on the theory of the 1911 and I haven't seen the math to invalidate it or to prove it.

For another point to ponder consider taking a bullet from a standing start to supersonic as a question of horsepower - imagine for instance a straight up shot, raising the weight fairly quickly by every day standards - takes a fair horse to do things that fast even with a .22 and then perhaps show some respect for the hand as a dyno and the pistol for containing a lot of horses.

Given that slide failures on the 1911 mostly happen forward of the recoil face I do rather wish somebody would make a reinforced perhaps a small port slide so I could have my .460 Rowland in comfort.

 

[1911Tuner]

Clark...Most slide failures occur at the top, left corner of the port..adjacent to the breechface guide blocks. A reinforced port wouldn't do a lot to help with the Rowland. If somebody would machine a slide and leave a small fillet in those corners, it would be a major shot in the arm for the souped-up conversions.

 

[Iron bottom]

www.guns.connect.fi/gow/QA17.html

Check out the slide on that thing. Was the rotary bolt locked to the barrel during firing?

 

[brickeyee]

"When the breech-bolt was blown back after unlocking.."

It looks like the bottom left detail is the actuator track that rotates the bolt, but you really need to see the entire patent (and maybe have it translated).

Wildey uses a similar curved track machined on the inside of the slide to rotate the bolt for locking.

 

[Owen]

I wish there was a professional journal for this stuff.

The problem is that the companies employing the people that would care would never allow them to publish.

 

[DBR]

I have read and tried to follow this thread from the beginning. I am getting more confused by the page.

The way I see it the barrel and slide are a "closed system" until the barrel links down. The bullet drag in the barrel is resisted by the barrel lugs against the slide lugs as is the gas pressure accelerating the bullet. The rest of the recoil action/reaction is simply momentum, friction/force (of the guns parts) and inertia. The bullet drag might have a minute effect on the recoil motion of the slide until the barrel lugs and slide lugs lock up. After that time the barrel and slide are for all intents one piece where the pressure of the case head pressing on the breach face equals the total of forces acting on the bullet/barrel combo in effect trying to force the barrel forward from the breach. Because this is a closed system, the only recoil effect is action/reaction of the bullet mass/gas/slide.

Other than acting as a resisting force in keeping the slide and barrel tightly locked together until the bullet leaves the barrel - which the mechanics of the swinging link does anyway I don't see how bullet drag is relevant to the 1911s fundamental operation.

The recoil cycle will not be directly influenced by what is happening in the "closed system". As Tuner said, it does act sort of like a blow back system but the blow back is not what is delayed. The mass of the barrel is added to the slide mass and the release of breach pressure is what is delayed - not slide recoil motion. Mostly by the delay of the vertical motion of the barrel vs the slide before the barrel tilts down and unlocks but also by adding the inertia of the barrel to the slide in the initial rearward motion.

Where am I wrong?

 

[Owen]

DBR, I agree with you.

There are three ways to analyze motion. You can use momentum methods, energy methods, or Newtonian motion equations. Generally what happens is is you switch from one method to the next, to get pieces of information you need but don't have.

While trying to determine the velocity of the slide, the momentum method is simple and gets you what you are looking for right away. What it doesn't tell you is the magnitude of the forces involved. For instance, if I wanted to know what the load on the lugs was, I couldn't get that from just using conservation of momentum.

Analysis of the force on the lugs requires using Newtonian Equations (f=ma, x=vt+1/2at^2), and then the whole problem becomes very complex. Some big nasty integrals and all that.

To get that information, I would have to start drawing freebody diagrams of each component.

Tuner's position is that the friction in the barrel is a major force, and the implications of that force carry though to the rest of the gun. Some of the effects are going to be, a much slower slide velocity than if there was no friction, much higher friction forces on the locking lugs.

I agree that the friction in the barrel is a major factor in determining slide velocity of short recoil operation system, but I also don't believe there is a whole lot that can be done to manipulate that factor.

 

[1911Tuner]

I agree that the friction in the barrel is a major factor in determining slide velocity of short recoil operation system, but I also don't believe there is a whole lot that can be done to manipulate that factor.

Exactly. It is...and will remain...largely an unknown.


Owen...et al...A brief, slow-motion description of the best explanation I can come up with, based on several observations and a few discussions with engineers who are also familiar with firearms and ballistics.

Bang! Powder lights...gasses expand, and the driving forces start to overcome the bullet's inertial mass and frictional resistance imposed by the case.. Bullet goes from static friction to kinetic friction, which...working in conjunction with the bullet's fledgling momentum...reduces the magnitude of the force required to keep it moving.

At the same instant, the slide also starts to move...but because of its greater mass...accelerates sluggishly by comparison. The slide is gaining momentum. At this point, there is no delay imposed by the bullet's resistance, because...as CA noted...the momentum end of the equation essentially cancels it out.

Now...The resistance of pushing a .451 diameter plug into a .451 diameter hole requires quite a bit of force alone, and ...since the bullet's inertial mass is still fighting...the two become pretty formidable. The fly in the ointment is the land diamater...which is .010-.012 inch smaller than the bore, or groove diameter...and all of a sudden...we have a NEW force to contend with, and with it comes a new equation.

For those who don't understand what a "New Force" is...An object that is set in motion by a force can't accelerate any further after that force is removed. Only by application of a new force can it be compelled to move faster, and only by applying a new force can it be brought to a stop. Thus...A bullet that exits the muzzle can only slow down unless a new force is applied at its base...and in the absence of a restraining force...the bullet will theoretically continue at its present velocity forever. Fire a gun in the weightless void of outer space. It'll never stop, and it'll never slow down unless it's compelled to do so by the application of an outside force.

Because the bullet's initial resistance was overcome by the rising pressures and the force generated by that factor...the increasing pressure hits a short, but still rising plateau. Because the bullet has encountered this new resistant force...and because nothing happens instantly, and everything takes time...the bullet stops for a tick. This may only last a half a millisecond. Maybe more...Maybe less. We have no way of calculating it...but we can see evidence of it on a pressure graph. There's a slight, steep rise...a short, shallow rise...then a sudden, violent climb to peak.

During this plateau phase, the bullet has stopped. When the bullet stops...so does the slide. When everything stops...all momentum is lost, and it all has to start over again. But because the bullet has moved forward, and can't move backward...it has a slight head start toward its destination. By the time the bullet begins to overcome the new resistance, it's moved maybe a quarter-inch. That's 5% of its total distance.

I've caught this slide hesitation after watching extremely slow-motion/stop-action videos. You have to watch closely and you have to watch it several times to be sure that's what you're actually seeing...but it's there. The blow-by gasses escape ahead of the bullet...before obturation and bore seal. The slide moves for a very short distance...maybe .a 32nd inch...hesitates...then suddenly accelerates rearward as the full force of recoil bears on it. Much easier to see is the gun pitching slightly forward just before the slide starts to move backward in the violent recoil phase. Gasses...slide movement...gun forward movement...slide stops...slide recoils...bullet exits. Just like that.

It's right there...at the point that the bullet encounters the land diameter and stops...that the slide is delayed. The bullet overcomes the resistance, and screams toward the muzzle under the full force of the pressure peak...while the slide sluggishly fights to regain its lost momentum. And it's right...THERE...that Kunhausen's balanced thrust vector exists.
Balanced. Suspended briefly in a state of equilibrium. So...By the time that the bullet has fully entered the rifling, and the slide has regained most of its lost velocity and momentum...10% of the bullet's trip is complete.

All this is very brief...taking maybe a millisecond from start to finish for the cycle of static friction to kinetic friction...back to static, and finally...back to kinetic. By the time the bullet has reached roughly the mid-point of the barrel...the momentum cancels out any resistance offered to the slide by the bullet's passage...even though opposing forces still bear on the lugs...and the bullet finally exits. (Now, your momentum/resistance equations will balance out.)

Adding up the effects of the delaying forces, we also have to understand that they don't all bring their full effects to bear at the same instant. More like handing off the baton in a relay race. First, of course, is the inertial mass of the slide and barrel. Then the bullet slamming into the rifling...the big one...Then comes the hammer and mainspring. Finally...after the momentum has reached the point that its effects counteract the resistance...the recoil spring comes into its own and begins to bear hard on the slide after the bullet is long gone.

Everything that has the opportunity to slow the slide WILL slow the slide. Nothing can be discounted, no matter how small or insignifigant. Not even a fly that happens to get smacked by the back of the slide as it moves. If it CAN affect it. It will.

A final clue here is this...and it requires thought to fully understand what it's saying:

A balanced force system is a necessary condition for equilibrium, but it is not a sufficient condition to ENSURE equilibrium.

That's a riddle of sorts...