Why no makarov in 9mm?

"Strong springs dont do anything to counteract pressure."

That is absolutely incorrect.

A stronger spring serves the exact same purpose as a heavier breech block -- preventing the action from opening until the bullet has left the barrel and pressure begins to drop.

It is entirely possible to design a submachine gun with an extremely heavy spring and a very light bolt that will work just as well as one with a very heavy bolt and a light spring.

That is, however, impractical from the point that the stronger the spring, the harder it is cock the gun, so in practice spring strength and bolt weight is balanced.

A perfect example of a blowback gun that uses an extremely heavy spring the Astra 400 handgun.

The slide quite heavy, but the springs in an Astra are VERY heavy, making the gun far more difficult to cock than virtually any other handgun out there.

The Astra 600, a variant of the 400, was chambered specifically for the 9mm Luger, and was also blowback, and is also a beast to cock.

Spring rates for both handguns are apparently in the 40 to 50 pound range, FAR heavier than any other handgun available today.

Remember, even though cartridges operate at thousands of pounds per square inch internal pressure, they don't transfer that pressure mechanically in a 1:1 manner. Cartridge grip on the chamber walls, static inertia inherent in the design, spring rate, etc., all play a factor in this.

But to say that springs do nothing other than return the bolt to battery is absolutely and 100% incorrect.
 
"Like you stated, it is your own theory!"

Yes, it is my theory.

But it is a theory that is backed up FAR better by existing data and historical fact than the theory that "Well, they did X so that we couldn't use it in our Y."

This has been said about so many things, not just small arms ammunition, that it's risen (dropped?) to the level of internet meme/urban legend than anything else.

It simply has no solid basis in fact.
 
And once again, what does this non topic bloviating have to do with 9x19 Makarov pistols.?????:confused: If I do something like that you would send me PM's and warnings!:eek:
 
We now have a confliction. I don't mind being wrong. I would rather be wrong on the issue of spring weight vs. bolt weight because your argument makes more sense to me. But some of the information I've read on this subject seems to support the idea that the compressive nature of a spring inherently allows for quicker reaction to force applied as opposed to a solid mass.
 
"But some of the information I've read on this subject..."

If it's available, share the source of what you've read.

"...seems to support the idea that the compressive nature of a spring inherently allows for quicker reaction to force applied as opposed to a solid mass."

That's where the strength of the spring, or the spring's "rate," as it is known, comes into play, but what it comes down to is that the intertia inherent in a solid mass can be duplicated with a spring of the proper rate.

The question is, then, does that make that spring suitable for use in the gun?

Not necessarily. As I've indicated there are other factors that have to be woven into the design, as well.

Designing a blowback operated firearm, especially a submachine gun, is an intricate balancing match.
 
What does that have to do with Makarov never being chamber in 9x19?

Follow the conversation. The Walther video talks about the new innovative methods used on the old PPK (and thereby Makarov) design to use 9mm in a very Makarov like design.

Oh. By the way, it's an 8 round single stack. Looks to be a very nice gun. While it can't be considered a Makarov, it should be about as close as you can get in 9mm.
 
If it's available, share the source of what you've read.

http://www.orions-hammer.com/blowback/

That last point bears repeating. From the previously cited Chinn Vol 4, page 15 (underline added by me):
"NOTE: There is one point which requires special clarification at this time. In many descriptions of blowback actions, it is strongly implied that the driving spring contributes a substantial portion of the resistance which limits acceleration imparted to the bolt by the powder gases. Actually, this is not so. Although it is true that the driving spring absorbs the kinetic energy of the recoiling bolt and thus limits the total distance it moves, the resistance of the spring does not have any real effect in the early phase of the cycle of operation. The bolt acceleration occurs mainly while the powder gas pressures are high and are exerting a force of many thousands of pounds on the bolt. The driving spring, in order to permit the bolt to open enough to allow feeding, must offer a relatively low resistance. Although this resistance is sufficient to absorb the bolt energy over the comparatively great distance through which the bolt moves in recoil, it is not great enough to offer significant opposition to the powder gas pressure until the chamber pressure has dropped to a relatively low level well after the projectile has left the muzzle."
The myth that "a stronger recoil spring will prevent case head separations" persists on the internet to this day. This is a myth.
 
The way I see it, you're both kind of right. What keeps a blow back closed? it is the inertia of the total system. Bolt mass, AND spring tension.

The easy way (and the only really practical way) is to have the bolt mass do most, or even all of the work. But it could work, mechanically, the other way around.

To work by using the spring as the main supplier of the force, it would have to exert a huge amount of force on the bolt when the bolt was closed, and would be too much to cock by hand. We don't build them like that, because it would be impractical. A gun that requires the equivalent of a car jack to open the action simply would not be accepted if any other possibility was available.

The way we build blowbacks the spring may not add anything significant to keeping the action closed during peak pressure, but it does add something. It may not do any "real" work, but it is there.
 
Isn't the Remington R51 a blow up design?

Sorry, couldn't help myself.:D:D:D
It's actually a locking breech block action, not straight blow back. There is more going on than just the slide being pushed back.

However, the notable feature is the use of a locking breech block within the slide utilizing the "hesitation-locked" action originally developed by John Pedersen.[8][3] When the pistol is in battery, the breech block rests slightly forward of the locking shoulder in the frame. When the cartridge is fired, the bolt and slide move together a short distance rearward powered by the energy of the cartridge as in a standard blowback system.[7] When the breech block contacts the locking shoulder, it stops, locking the breech.[7] The slide continues rearward with the momentum it acquired in the initial phase.[7] This allows chamber pressure to drop to safe levels while the breech is locked and the cartridge slightly extracted. Once the bullet leaves the barrel and pressure drops, the rearward motion of the slide lifts the breech block from its locking recess through a cam arrangement, continuing the operating cycle.
http://m.youtube.com/watch?v=EOxeLEQZNxo

And despite my silly quote change, I still want one!
 
Chin's work is certainly not to be dismissed or ignored, but there's an important consideration that needs to be remembered...

His work addressed machine guns firing rifle- to cannon-type cartridges generating far greater pressures than the handgun cartridges we're talking about. I don't believe that any of his work addressed submachine guns firing handgun class cartridges.

I'd say that what is true of a blowback gun firing a rifle-class, bottlenecked, tapered cartridge is likely not applicable, or at least not as applicable, to a gun firing a handgun class cartridge.

There have been virtually no truly successful blowback firearms that fire rifle-class cartridges. The demands are simply too great.

At least that's my theory.
 
I don't believe that any of his work addressed submachine guns firing handgun class cartridges.

Actually, Mike, some of it did, but it was far from his main focus. Its been a long time since I've read "The Machine Gun", but I do remember that he has (along with a number of others) the formula for calculating the mass needed for a blowback bolt, (and I think spring tension, too but can't remember clearly).

The math is over my head, (I basically stop at quadratic equations), but I can recognize the principles. Given projectile, velocity and pressure, (and maybe some other values to plug in), it would apply to a blowback design, either SMG or handgun.

Also one has to look at the context, I don't disagree with Chin's quoted statement , but I believe he was speaking in the context of existing designs. Where, he is entirely correct.

In the hypothetical design where the spring tension provides the inertia, rather than the bolt mass, Chin's statement would not apply.
 
Chinn is correct, even for handguns. A spring heavy enough to hold the bolt shut against initial blowback pressure would be too heavy to allow the bolt to move at all. With a spring of reasonable weight (and mass) the bolt will always move some before the spring can reassert itself, and that can be enough to allow the case to be unsupported and burst. If the spring is heavy enough to prevent that movement, it will be very heavy (Astra 400) and probably too heavy for the shooter to operate the slide without some means of disconncecting the spring (.35 S&W). (The Astras, of course, have a heavy slide, plus a heavy hammer spring that acts to delay the blowback, so the recoil spring is not acting alone to keep the bolt closed.)

Jim
 
I did these calculations several years ago for a discussion on another forum. Came up with a recoil force of about 850 pounds reacting on the slide of a .45 ACP. As others have said, you can see where the 16 pounds of force or so from the recoil spring is pretty irrelevant.

If anyone is really interested, you can use Newton’s stuff from 400 years ago to understand how your gun works. We used it to get us to the moon back in the 60’s, we use it on the rockets and guns we’re designing today. If you don’t care to know, don’t worry about it, keep believing whatever makes you happy! No big deal.

For example, how long does your bullet remain in your gun barrel, and what kind of acceleration does the bullet experience?

What kind of acceleration does your slide experience, and how fast is it moving when the bullet exits? How far has the slide moved rearward when the bullet exits?

This kind of stuff is REAL important if you’re designing things like rocket mechanisms or artillery fuses or proximity detonators.

Not real easy to effectively transmit mathematical concepts in a forum format, but we can try. I’m going to ignore the contribution of the unburned powder and gases exiting the barrel with the bullet.

We’ll call the starting position of the bullet the zero of our coordinate system, so X(initial) = 0. We’ll abbreviate it as Xi cause I’m lazy. That means Xi = 0.

V(initial) is the starting velocity. We’ll call that Vi. For a bullet in a gun, it’s initial velocity is 0. So Vi = 0.

We’ll call the last position we consider, the end of the barrel, X(final). I’ll abbreviate that as Xf.

T = time in seconds.
T**2 means “T raised to the second power” or "T squared" or “T x T”.
X = position in inches.
V = velocity in feet per second.
A = acceleration in feet per second squared (fps**2).

The basic motion equation in terms of time is:

Xf = Xi + V x T + (A x T**2) / 2

If you want more on this basic motion equation, you can go here:

http://en.wikipedia.org/wiki/Equations_of_motion

Introductory calculus teaches us that the first derivative of position is velocity, and the first derivative of velocity and the second derivative of position is acceleration (I’m not going into calculus for you).

So Velocity = Vi + Acceleration x Time, or

Vf = Vi + A x T

We’ll look at a typical .45 caliber 1911. First thing we have to do is get everything into common units.

Bullet weight = 230 grains = 230/7000 pounds = .033 pounds = .001025 slugs (slug is mass unit in english system).
Barrel length = 5 inches = 5/12 feet = .417 feet
Bullet velocity when it exits the barrel at 5 inches = 830 feet per second.

Now we just use the basic motion equation:

Xf = Xi + Vi x T + (A x T**2) / 2

Put in our 1911 values that we know:

.417 = 0 + 0 x T + (A x T**2) / 2 or

.417 = (A x T**2) / 2 (NOTE – This will be EQ1)

Notice that we have 2 unknowns (A and T) but so far only one equation. From basic algebra we know that we need as many equations as unknowns if we hope to solve them.

This is where we use the rest of the stuff we know and the velocity equation:

Vf = Vi + (A x T)

We know Vf (Velocity final) is 830 fps. We know Vi (Velocity initial) is 0, so

830 = 0 + (A x T)

Since we loved algebra, we solve for A in the above equation and find

A = 830 / T (NOTE – This will be EQ2)

Now we can cleverly plug the above relationship back into our basic motion equation EQ1, replacing A with 830 / T.

.417 = (830/T) x (T**2) / 2

Simplifying this (basic algebra again) gives

.417 = 415 x T

Now we solve for T:

T = .417/415 = .0010048 seconds

In other words, T is right at 1/1000 of a second.

Since we now know T, we can easily solve for A from EQ2:

A = 830/ T = 830 / .001 = 830,000 feet per second squared.

Acceleration due to gravity = 32.2 fps**2, so our bullet is experiencing:

830,000 / 32.2 = 25776.4 G’s as it goes down the barrel.

Kind of important to know if you want to hang some electronics on your bullet. They better be able to handle more than 25,000 G’s.

How much force is the locked together barrel/slide seeing, and how fast is it accelerating?

We know that the force from the bullet equals its mass times acceleration, and we know that for every force there is an equal and opposite reaction (that Newton stuff again).

So the force delivered to the locked together barrel/slide from the bullet is:

F = M x A

F = mass of bullet x acceleration of bullet

F = .001025 x 830,000 = 851 pounds

How fast will 851 pounds accelerate the barrel/slide?

F = M x A = 851

Algebra again:

A = F / M = 851 / M

I’m going to say 1.5 pounds for the slide and .5 pounds for the barrel, mass would be 2/32.2 = .062 slugs.

So the acceleration of the slide is:

A = 851 / .062 = 13695 feet per second squared.

13,695 / 32.2 = 425.31 G's

Now we know the acceleration, so how fast is the slide moving when the bullet exits the barrel?

We know the bullet is in the barrel for .001 seconds, and once it leaves the barrel there is no more reaction with the gun. So the slide/barrel acceleration is applied for .001 seconds.

V = A x T = 13695 x .001 = 13.695 feet per second.

The slide is moving rearward at 13.7 feet per second when the bullet leaves the barrel.

How far has the slide moved when the bullet leaves the barrel?

Again we use the basic motion equation. Now that we know the acceleration and time, we can find the distance the slide moves:

Xf = Xi + Vi x T + (A x T**2) / 2

Xf = 0 + 0 x .001 + (13695 x .001**2) / 2

Xf = .0068475 feet = .082 inches, or just over 80 thousands of an inch.

You’ll find that with your 1911 slide .082 inches retracted, the barrel and slide will still be locked together. John Browning did all the stuff I did above over 100 years ago (once smokeless powder was invented so everything didn't get gummed up) when he started designing machine guns and semi-automatic pistols.

Recap:

Bullet velocity at barrel exit: 830 fps
Time bullet is in barrel: .001 second
Bullet acceleration while in barrel: 830,000 fps**2, or just over 25,000 G’s.
Slide acceleration while bullet is in barrel: 13,695 fps**2 or just over 425 G's
Slide velocity when bullet exits: 13.7 fps
Distance barrel/slide moves while bullet is in it: .082 inches

Note: The slide is accelerated ONLY while the bullet is in the barrel, so it’s only accelerated for .001 seconds. You can see where the recoil spring force (16 or 18 pounds or so) is pretty irrelevant while the bullet is accelerating.

All of the stuff above is freshman year engineering Calculus 1, Physics, and Statics. No way anyone is going to explain it on an internet forum to anyone who does not have the fundamental background to understand it. I can see where it aggravates some people who understand trying to explain it!

Since we know the slide velocity, slide mass, barrel mass, recoil spring rate, and location where the slide unlocks, we could go on to calculate the slide velocity when it unlocks, the barrel velocity when it hits it's hard stop, the slide velocity when it hits it's hard stop in recoil, the time it takes to reach full recoil, and how long it will take to return to battery. We ignored the contribution of the recoil spring, hammer spring, unburnt powder, and gases exiting the barrel. Feel free to add them into the calculations if you feel the need.
 
I did these calculations several years ago for a discussion on another forum. Came up with a recoil force of about 850 pounds reacting on the slide of a .45 ACP. As others have said, you can see where the 16 pounds of force or so from the recoil spring is pretty irrelevant.
Nicely done. However, the calculation is for a locked breech design which operates differently from the blowback operated design under discussion.

Given that the recoil springs in blowback designs tend to be much stiffer than roughly comparable locked breech designs, it seems like they may play a more important role in delaying the slide than in a locked breech gun.

(Meaning, of course, I'd love to see the math worked out for a blowback... :D)
 
45_Auto's calculations remind me of my physics classes so many years ago. The only formula I can still remember is "F = Your Grade".
 
Back
Top