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Thread: Questions: Self-loading Rifle Gas System

  1. #1 Questions: Self-loading Rifle Gas System 
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    Jan 2013
    Yet another imponderable which has troubled me for years! Perhaps someone here has had experience with, or possesses knowledge of, the dynamics behind gas operated rifle operation? I'm using the U.S. M-14 as an example, as I'm familiar with the relative sizes of the parts; that's the crux of the main question: Are theoretical principles applied to such design, or is a successful design the result of a lot of hit-and-miss?

    The diagram shows the parts concerned, disregard the text, I couldn't get rid of it. Here's how it works: The gas piston, about 1/2" in diameter, resides inside of the gas cylinder, held in place there by the gas cylinder plug, which screws into the end of the cylinder; the gas cylinder lock may be ignored here, not important to operation.

    The gas cylinder slips over the rifle's barrel, held in location by the gas cylinder lock, and a tiny hole drilled through the barrel wall into the bore through which the bullet passes, lines up with a hole in the gas cylinder, thereby allowing gas which is propelling the bullet down toward the barrel's muzzle to enter the gas cylinder, where it impinges on a surface of the piston accessed through one of the holes visible on the shiny part of the piston. Gas pressure drives the piston to the right, in this drawing, where the right end of the piston contacts the surface of the front end of an operating rod reaching all the way back (the gas piston/cylinder assembly is very near the muzzle end of the barrel) to the other end, the breech, of the barrel, the operating rod being there connected to the bolt, which confines the round of ammunition in the barrel's chamber, causes it to fire, the operating rod then forcefully drives the bolt back rearwards, against spring tension, the spent cartridge casing is ejected outwards and away, the spring then causing the operating rod to force the bolt back forwards into battery position, a fresh round of ammunition being stripped from the magazine below the bolt, and forced into the chamber ahead of it. This then leaves the rifle ready to be fired again.

    Now the big question: (and ya gotta try to picture this as it's happening): Gas piston is limited in motion to only about 1 inch of travel, while the operating rod and bolt travel perhaps 3-1/2 inches, all against high spring force. This means that the piston is in contact with the op rod for a distance of less than an inch, yet the rod is forcefully and very quickly driven rearward, leaving the piston stationary in it's little pocket within the gas cylinder. This also means, and this is what troubles me the most: the op rod is HIT with an extremely fast, high-force hammer-like blow, in order for it to continue on back and "do it's thing", agreed?

    Peak gas pressure is over 50,000 psi, which likely occurs very nearly as the bullet passes the gas bleed hole in the barrel, bear in mind now, that bullet is ready to leave the barrel at perhaps 2,800 feet per second. The gas bleed hole is located perhaps 6 inches, or less, back from the muzzle of the barrel. DO THE MATH: At 2,800 fps, HOW LONG is the time interval during which gas pressure must enter the gas cylinder, and drive the piston rearward against the op rod, before the bullet exits the barrel, allowing pressure to quickly drop as the spent gases leave the barrel?

    My "gut" feel suggests this CAN'T WORK! 'Course, I said that about television when I studied it in 1962! Does anyone here have a "handle" on just how these gas and mechanical dynamics are applied by the designers of such weaponry?

    Technical considerations such as these are what have always made me fascinated by firearms. I also enjoy developing the skill to use them accurately. jocular

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  3. #2  
    Join Date
    Apr 2007
    I can't help you with the calculations, but I think you may have a misconception about how fast the pressure will drop after the bullet exits the muzzle. The speed of the gas exiting the muzzle will be higher than the muzzle velocity of the bullet, but not as much as one might think. See "choked flow."

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  4. #3  
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    Quote Originally Posted by Harold14370 View Post
    I can't help you with the calculations, but I think you may have a misconception about how fast the pressure will drop after the bullet exits the muzzle. The speed of the gas exiting the muzzle will be higher than the muzzle velocity of the bullet, but not as much as one might think. See "choked flow." Flow of Gases pg.48.pdf
    Harold, thank you for that info. Here is a quick "tickle" of the mind. Piston diam. 1/2", Area 0.196 sq. in., thus given 50,000 psi impingement at however short a time interval, force on piston would be about 9800 lbs. Having had only one course ion Statics & Dynamics, a bit short-handed on this kind of expertise, nonetheless, some rule stated that in impact loading the MINIMUM force exerted is 2X the force present, thus as piston impacts the op-rod, almost 20,000 lbs force present, thus creating a most hammer-like blow to the op-rod, which responds in kind: it flies backwards like hell! A lot of inertial considerations also, as op-rod is stationary as piston strikes, and begins to move it. As this action occurs, the bullet leaves the barrel, maybe right around the time of piston/op-rod contact (just an imaginary guess).

    At any rate, the only military rifle of which I have knowledge which takes all this variation into consideration is the Belgian FN-FAL, which was also adopted by many military agencies: Israel, Argentina, and many more which I have forgotten, the FN design incorporated an adjustable gas port, to allow changes to be made to optimize cyclic control, as various parameters of ammunition varied. jocular
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