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Hello folks,

This is my first time posting here, so forgive me if this is not the correct spot to pose this question.

I am very interested in the specifics about blowback gun physics. Not with bullets mind you, but the pressure they exert on guns and how that pressure is distributed to the slide, casing and .

Here are my questions:

-How is the pressure inside the chamber measured? Pounds of force?
-About how much pressure is exerted onto the slide?
-Are there specs availiable to people like me who want to know how muc pressure is required to cycle say, a Glock 19, Sig P226, or 1911?

I have done some google searching tonight, but I can't find any of this information.

Any help would be appreciated.

-R
 

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A warm welcome.

I'll put this in simple terms and Uncle Nick will come along with the full scientific lingo; pick the version that leaves you scratching your head less.

Chamber pressure is measured in pounds per square inch - or that's the idea. In earlier days they used a measuring method that didn't exactly measure psi, but they called it psi. It really doesn't mater if it's called beezers per square gidgin as long as it's applied uniformly. Guns work at some amazingly high pressures, with shotguns being lowest, then rimfires, then handguns and finally rifles at the highest pressures.

Blowback pistols are simple devices because they don't lock the barrel closed mechanically. It's actually the operating pressure of the round firing that expands the brass case, and the tightly expanded case is what holds the action closed - nothing else. The action stays closed until the bullet exits the muzzle and the pressure starts to drop. The first thing that happens is that the case loses its grip. But there's still a lot of pressure remaining in the barrel, even though the bullet is gone, so it is that residual pressure that starts moving the slide back, using the case as a piston of sorts. There's more than enough pressure there to compress the recoil spring, cock the hammer and move the slide all the way back to its stop. After the slide stops, the recoil spring takes over and shoves it forward again, chambering the next round. Simple, yet quite an intricate balance of forces, masses and geometries.

I can't give you exact numbers, but most blowback pistols are limited to about 15,000 psi compared to pistols with mechanically locked breeches that operate in the 35,000 psi range. Quite the difference. Just a guess, but after bullet exit, I'd say that the pressure that actually operates the slide is perhaps 5,000 psi. It drops very rapidly from 15,000 to zero after bullet exit, but there's enough there for long enough to work the action, obviously.

How's that?
 

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The pressure needed varies with the cartridge. the needed pressure depends upon the inertia(weight) of the slide/breech block/etc AND the spring loading on the breech.

If you consider a simple design like a 22 auto, if it is in 22 it has a given spring. but the same gun in 22 mag will usually have a heavier spring. So there is no answer to your question, one would need to know the specific gun and cartridge. The firearm is designed for the presssure presented by the given cartridge. So the pressure needed for a given cartridge is the SAMMI spec pressure.
Going below the standard cartridge pressure will eventually cause the firearm to fail to fully cycle; going over the standard pressure will cause energy to be wrongly applied to the internals (ruin your gun).
 

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By now, there's probably a proprietary computer program owned by the firearms companies or the government that calculates all this. In the past, IIRC, it starts off with best calculations on paper then trial and error using a prototype firearm refining the design to eliminate the glitches.

IMO, the genius who made it all happen before computers was J.M. Browning who instinctively knew what he had to do to make things work.

There's way too many things to take into account for me to even try to answer your questions and not enough computer space here if I could. I'm just happy to know if I buy something at my local gun shop, it's going to work.:)
 

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Really pretty simple calulation. Take the pressure of the load and divide it by the case head area. that is the force on the cartridge head. Also need to know the dwell time of the bullet in the barrel (depends upon barrel length), and the weight of the breech block. the distance that the action has to move to cycle properly, and then get the spring to absorb that energy.
 

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Simple in principle, but a pain in practice because the pressure changes constantly throughout the launch of the bullet and expulsion of propellant gas.

What matters to the slide's blowback function is that it pick up enough momentum to carry it through cocking the hammer (where there is one) and compressing the recoil springs and arriving fully in counter-battery with enough momentum to spare to compensate for build up of dirt, cold temperature thickening of lubricant, etcetera, so the gun operates reliably.

Momentum is mass times velocity, so in order to have that momentum the slide has to have been accelerated. Acceleration results from the application of some amount of force for some period of time. Force×time is called impulse. It has units of Newton-seconds, but those convert to units of kg-m/s or lb-ft/sec, which happen to be the same units as momentum has. The impulse that introduces momentum is thus exactly equal to the difference in momentum between the start and end of the impulse.

The mathematical way to quantify the impulse applied the slide is to integrate the force applied rearward with respect to time from start to finish. That's calculus 101 applied to physics 101. It would be simple, within those disciplines, except that to proceed you first need a function that describes the force as a time dependent variable. That will equal breech pressure as a time dependent variable after multiplying by the cartridge head area. But the pressure function varies with every powder and bullet combination you choose. Indeed, a blowback operated .22 LR or .25 Auto often has a fairly light slide, and with a stiff enough recoil spring, that is good enough. But when you get to a 9 mm or .45 Auto blowback operated gun, like the Sten gun and M3 Grease gun, respectively, you've got to have either very stiff springs or a very heavy bolt mass to prevent the force from pushing the bolt back a significant distance before the bullet can clear the muzzle and pressure has dropped. Otherwise you would burst the cases.

There are additional complexities in how you allow for friction of the brass in the chamber, the force needed for brass stretching, where it occurs, and so on. For that reason, you have to decide whether you want to know the total force pushing the inside of the case or just the portion left over after friction; the force actually applied to the slide.

You can arrive at rough estimates of those. For overall momentum applied to the gun, you can take advantage of Newton's third law of motion that says equal and opposite forces act on the gun and expelled mass, which means you have an equal and opposite impulse, which means you wind up with equal and opposite momentum. So, you take the momentum of the bullet, less an allowance of perhaps 3% or so of velocity being gained from muzzle blast after leaving the barrel, add momentum of the portion of the gas mass accelerated inside the bore, and finally the momentum imparted to the gas mass during its very rapid acceleration as muzzle blast (rocket effect). Your gun's momentum will be equal and opposite to all that.

SAAMI has a free publication with recommended values for estimating the above, here. SAMMI is interested in perceived recoil, so they calculate estimated free recoil energy, but they give you the needed elements of combined velocity and mass to arrive at momentum instead. Just multiply the mass times the velocity (the top two formulas on the third page). You now have a good estimate of the total rearward momentum. The portion that goes to bolt will be what is left over after subtracting the portion spent starting back against the initial force of the recoil springs and the friction of the case drag in the chamber under pressure.

How much all that adds up to can be estimated pretty well for a particular gun, too. You just need to carefully empty the gun, then see how much force it takes to cock the gun by using a dowel in through the muzzle to push down on the breech while the butt is against a scale. You probably want to take that force every quarter inch or so and be sure to include the cocking of the firing mechanism. From that, you can plot force over distance, arriving at the average force over each quarter inch. You then take the mass of the bolt assembly and half the mass of the recoil spring and apply those forces to them to see how fast each quarter inch would accelerate that mass if it was a push rather than a resistance. At the end you'll have the minimum slide momentum needed to cock the gun and reach full counter-battery, except for the extraction friction of the case. This will also will not take the inertia of the mass of the hammer and mainspring into account, but you can figure some of that out, too, once you have approximate velocities and if you really want more detail. Friction is omitted, too, but if you subtract the resultant momentum from this approximation from the total momentum you found for the recoiling gun, then you'll have a range of difference within which all the missing elements fall.

The above is not analytically perfect, but will get you in the ballpark.
 

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I just didn't want to make you look like your powers of prognostication had foiled you.

One thing I meant to include, but got distracted before I did, is why impulse is an important property. Suppose you fire a load at 1000 fps. Then you load down to 500 fps. The average force on the bullet base is cut in half, but the time it takes for the bullet to get out of the barrel is doubled. Since the impulse is force × time, the impulse is the same. You would figure that means the gun should operate the same, but the problem is the recoil spring force has to be overcome by the force before it starts giving momentum to the slide. When the force is cut in half, that same spring force is now twice the percentage of the average force it was before, and that can spoil operation. But over a range where that change in percentage is still within the excess impulse available, the gun will keep operating.
 

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Thanks for upholding my vaticinatory reputation!

And so once again we discover that the answer to this - as to so very many other shooting questions - is: It depends.
 

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Yup - that's an answer that applies to 99% of the questions asked on the board, Rocky! :p
 

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. . . and Newton worked out all the necessary physics before 1700, when there were no cartridge firearms at all. ;)
 

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This all happens over a period of time rather than an instant of time. Newton's laws are instantaneous and require that the little bits all be added up, a process called integration. So Newton invented calculus to do this. Recoil pads, springs, limp wrists and low chamber pressure etc, slowdown the application of the impulse and spread the reaction over time so that its instantaneous values are reduced. This is much easier to just figure out empirically which is why Browning did not need all this crap.
 

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The OP asked specifically about the physics of what was going on because he's interested in it. Doesn't mean you need to be. Dumping on his interest as useless crap isn't contributing. I feel that way about collecting Hummel figurines, but if someone asks about them, my feelings don't address their question or interest.

Also sounds like you're making a gratuitous assertion about Browning's sophistication. Though he had only a grade school level of formal education, workbooks left over from his day of the one-room school houses show kids were better at geometry and trigonometry by the time they cleared 8th grade than modern high school graduates are. My point is, it's easy to underestimate what they knew based on when they stopped going to school. Modern grade levels are not really comparable.
 

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Harry S to find force you multiply pressure (force/unit area) times area. Uncle Nick of course Browning did not have the tools to do all this crap. If I had the ordinance lab I supervised after I finished engineering school and I wanted to know how much powder it took to operate my pistol I would get some primed brass and work up a load. By the way for those who have missed it I have expressed doubts about our ability to measure pressure in gun barrels.Uncle Nick and Rocky don't agree with me.
 

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The Troll Whisperer (Moderator)
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That's what makes horse races, olafhafdt!

Differences of opinion, when conducted in the gentlemens way this discussion has gone, is what makes for informed opinions of the masses. I've learned a great deal by discussions such as this on the board over the years. Always like to hear all sides of the argument.
 

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Olaf,

I think we may be talking past each other. I don't normally make calculations to develop a load, either (though I do use QuickLOAD a lot, which is based on physics and which often predicts velocities with good accuracy). But it's a moot point, as the OP didn't ask anything about load development or even the precision limits of making operating calculations. All he said was he was interested in the physics of blowback operation, then asked for three details which I read as:

  1. What units of measure are used for pressure in firearms?
  2. About how much pressure acts on the slide?
  3. Are pressure specs available?
I believe those have all been addressed.

I took your comment re Browning to mean you knew for a fact he had no engineering sophistication; i.e., he would be unable to calculate compression force from a spring constant or calculate a leverage or the force of steam pressure on a piston, or any other practical physics computation. My expectation is he likely could do those kinds of things. Unfortunately, I don't have access to his personal notebooks or work diaries, so I have no way to know to be able to state it as a certainty one way or the other.

Newtonian physics is mathematical descriptions of mechanics. That's why it is included under the more general heading of Classical Mechanics. As with any practical calculations of events in the physical world, the accuracy of the result is dependent on being able to input accurate arguments, so the computer adage, "garbage in, garbage out", applies.
 

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In my opinion most of what we think we know is very incomplete. The things that we know that work are those things we have found out by trial and error. We really have no idea what the instantaneous ballistic values are. What we have is a bunch of ideas and theories. Everything between the instant the firing pin drops untill the bullet comes to rest is conjecture. I think we are produceing a lot of numbers that have much less revelance than we give them credit for.
 
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