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Comments by "" (@neutronalchemist3241) on "How Does it Work: Blowback Action" video.
@motohead077 Let's put it like this. Recoil accelerates the slide back for the first 1mm ot its run. Then the pressure drops and don't push it back any more. The recoil spring then decelerate the slide for the residual 4cm (that's 40 times the distance, and even more of the time) of its run. And the slide still slams into the receiver. So the K of the recoil spring is less (much less) than 1/40 of the force of the recoil. That's why it's not taken into consideration.
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Unfortunately it's the thing that will facilitate the case rupture (since the front of the case is clinged to the chamber and the back is pushed out of it, the case is stretched and breaks). Infact, the fluted chamber was invented to aid extraction and allow the cartridge to start moving when the pressure is still relatively high. However a mechanism, the "ring delayed blowback" had been invented to instead use this pressure to delay the opening. However is debated if it really worked or not.
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Actually, to be sold on the market, any firearm must have shown to be capable to safely shoot proof cartridges that produces 25% more pressure than the maximum pressure limit for the same cartridge in its commercial version. Blowback weapons are as safe as breechlock weapon in firing +P ammos. The limiting factor, for both, is simply the resistance of the frame/slide assembly to the harder beating created by the more powerful (10% more in case of the +P) ammos. In reality, for pistol cartridges, the weight of the bolt/slide in blowback weapons is almost always much higher that the one needed to simply avoid case separation, even for proofloads. An heavier slide/bolt is used to tame the perceived recoil and, in SMGs, to reduce the fire rate.
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Actually the bolt's weight in a SMG is usually much more than what's simply needed to keep it closed for enough time to safely shoot the cartridge. The weight is used to slow down the fire rate, that otherwise would easily exceed the 1000RPM (that infact were exceeded in the first SMG, the Villar Perosa of WWI).
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The real reason is Pascal's principle. Any square mm of the case is subject to the same pressure. But the bolt is pushed back by the base of the cartridge. If the base of the cartridge has 4 times the area of the base of the bullet, the force pushing back the bolt is 4 times the one that's propelling the bullet. So, if a .223 has roughly the same momentum as a 9mm, but the base of the cartridge has four timesthe area of the base of the bullet, it needs a 4 times heavier bolt.
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No. Let's put like that. In a 9mm firearm, the pressure in the chamber propels the slide backward for the first 1mm of it's run. Then the bullet leaves the barrel, the pressure drops, and only inertia keeps on pushing the slide, while the recoil spring slow it dow, for the residual 40mm or run. And yet the slide slams into the receiver. That means that a standard recoil spring resistance is less (much less) than 1/40 of the initial recoil impulse. That's why, in calculating the necessary mass of the bolt in blowback firearms, the resistance of the spring is not even taken into account. A human wouldn't be able to rack a slide/bolt with a recoil spring so rigid to really delay the opening of the bolt.
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