does the bullet exit before the slide moves?

[Ownerus]

At the risk of belaboring this to death, I'll take one more stab at it.

.... The barrel and slide are 60 times the mass of the bullet. If we add the 2 pounds of spring force to the weight of the barrel & slide we get 3 pounds or 21,000 grains. Divide that by the weight of the bullet, 115 we get 182.608. The rear resistance of the barrel, slide and recoil spring is 182 times the weight of the bullet.....

I'll grant you your numbers. Close enough for discussion.

I don't think they ever use the weight of the powder charge verses the weight of the bullet. It would be he energy released by the burning of the powder verses the bullet weight.

It's not the weight of the powder versus the bullet, it's the powder PLUS the bullet. That's the actual mass being ejected. The small mass of the powder is, on average moving at half the velocity of the bullet (the gas at the breech face has essentially zero velocity, the gas at the base of the bullet is bullet velocity, the average of the two is 1/2 the bullet velocity)

As a general rule only 30% of the peak pressure remains when the bullet exits the barrel. Of a conservative 30,000 psi peak pressure for 9mm only 1000 psi would exhaust out the end of the barrel when the bullet leaves. This 1000 psi exhausting out the end of the barrel is the "recoil" that operates the slide of a semi-automatic handgun. Also that 1000 psi exhausting is the action Newton referred to and the reaction is the slide cycling, extracting the spent case then picking up and chambering a new round.

Again, to use your numbers, 30,000 psi peak pressure (I don't know the actual average pressure over the duration of the bullet travel through the barrel, maybe in the neighborhood of 10,000 psi? ) The force on your 9mm (.354" dia, .0984 sq inches)bullet at 30K psi is 2,952 lbs force. It's psi, pounds per square inch. The force is a function of the area it's acting on. Using my guesstimate of 10K psi average pressure, the force is still 984 lbs. As an average it's acting on the bullet for the 4-5" of travel thru the barrel depending on barrel length . That same force is pushing the opposite direction against the barrel/breech block assembly.

Now to address your exhausting gas theory. Essentially a rocket effect. Yes, I'll grant there is some of that but it is in no way the predominant force. If, as you say, there is 1000 psi as the bullet leaves the muzzle ( actually 30% of 30,000psi is 9000psi) it almost instantly decays to zero. Probably well before the bullet is even an inch from the muzzle. Even pretending that the pressure holds that long that's still a much smaller force (885 lbs at 9K psi) for a much shorter time.
I don't think the barrel and slide will ignore the 980+ pounds of force applied for the time the bullet takes to travel 4-5" and not move and then suddenly give a snappy response to a much smaller force applied for much less time as the bullet leaves the barrel.

What people have a hard time grasping is the fact that an extremely high force applied over a short distance can impart a lot of kinetic energy into an object (in this case the barrel and slide). A 1911 slide/barrel moves about .1" (1/8"=.125" for reference) before the barrel starts to drop out of locked engagement leaving plenty of time for the bullet to leave and pressure to drop.

Another example of high force/short time and short distance to impart enough energy to cycle an action is the short tappet gas system of the M1 Carbine. The tappet stroke is around .2" yet it imparts enough energy into the operating rod to cycle the rifle thru the whole unlock/extract/eject/feed/chamber cycle of a couple of inches.

Consider the M1 Garand. The gas port is only 1-1/2" from the muzzle (where the bullet is at almost full velocity, not leaving much time for the gas to act on the op rod piston). It gathers enough force to move a lot of machinery in that op rod, bolt etc. Neither of these have any opportunity for any "rocket effect" of exhaust gas to cycle the action.

Finally, take the Colt Ace .22 conversion for the 1911 .45. It cycles essentially a full weight steel slide with the pressure of a .22 rimfire cartridge. Not a lot of exhaust gas rocket action there. It does it by using a differential piston "floating chamber" (more square inches for the gas pressure to act on) to impart enough energy into the slide over around 1/8" of travel to carry it through the rest of the cycle.

As I said to begin with, YES, the slide moves at the same time as the bullet and for the same reason, just not nearly as fast or as far in the time it take the bullet to exit.

 

[EPS]

At the risk of belaboring this to death, I'll take one more stab at it.


I'll grant you your numbers. Close enough for discussion.



It's not the weight of the powder versus the bullet, it's the powder PLUS the bullet. That's the actual mass being ejected. The small mass of the powder is, on average moving at half the velocity of the bullet (the gas at the breech face has essentially zero velocity, the gas at the base of the bullet is bullet velocity, the average of the two is 1/2 the bullet velocity)


Again, to use your numbers, 30,000 psi peak pressure (I don't know the actual average pressure over the duration of the bullet travel through the barrel, maybe in the neighborhood of 10,000 psi? ) The force on your 9mm (.354" dia, .0984 sq inches)bullet at 30K psi is 2,952 lbs force. It's psi, pounds per square inch. The force is a function of the area it's acting on. Using my guesstimate of 10K psi average pressure, the force is still 984 lbs. As an average it's acting on the bullet for the 4-5" of travel thru the barrel depending on barrel length . That same force is pushing the opposite direction against the barrel/breech block assembly.

Now to address your exhausting gas theory. Essentially a rocket effect. Yes, I'll grant there is some of that but it is in no way the predominant force. If, as you say, there is 1000 psi as the bullet leaves the muzzle ( actually 30% of 30,000psi is 9000psi) it almost instantly decays to zero. Probably well before the bullet is even an inch from the muzzle. Even pretending that the pressure holds that long that's still a much smaller force (885 lbs at 9K psi) for a much shorter time.
I don't think the barrel and slide will ignore the 980+ pounds of force applied for the time the bullet takes to travel 4-5" and not move and then suddenly give a snappy response to a much smaller force applied for much less time as the bullet leaves the barrel.

What people have a hard time grasping is the fact that an extremely high force applied over a short distance can impart a lot of kinetic energy into an object (in this case the barrel and slide). A 1911 slide/barrel moves about .1" (1/8"=.125" for reference) before the barrel starts to drop out of locked engagement leaving plenty of time for the bullet to leave and pressure to drop.

Another example of high force/short time and short distance to impart enough energy to cycle an action is the short tappet gas system of the M1 Carbine. The tappet stroke is around .2" yet it imparts enough energy into the operating rod to cycle the rifle thru the whole unlock/extract/eject/feed/chamber cycle of a couple of inches.

Consider the M1 Garand. The gas port is only 1-1/2" from the muzzle (where the bullet is at almost full velocity, not leaving much time for the gas to act on the op rod piston). It gathers enough force to move a lot of machinery in that op rod, bolt etc. Neither of these have any opportunity for any "rocket effect" of exhaust gas to cycle the action.

Finally, take the Colt Ace .22 conversion for the 1911 .45. It cycles essentially a full weight steel slide with the pressure of a .22 rimfire cartridge. Not a lot of exhaust gas rocket action there. It does it by using a differential piston "floating chamber" (more square inches for the gas pressure to act on) to impart enough energy into the slide over around 1/8" of travel to carry it through the rest of the cycle.

As I said to begin with, YES, the slide moves at the same time as the bullet and for the same reason, just not nearly as fast or as far in the time it take the bullet to exit.

I'll take your word for it I'm not reading the whole thing lol

 

[Dapapadon]

Yep, you got me, I was lazy and didn't do the math. You didn't do the research, did you? I tried to explain the exhausting gasses to you and you coined the phrase "rocket effect". It is actually called secondary recoil. You don't think it matters. Go borrow a 50 BMG bolt gun and take a shot. Then remove the muzzle brake and take a second shot. Come back and tell us how that went. Tell us how those exhausting gasses don't matter.

Me, I'm done with this discussion.

To the OP, sorry this got off topic.

My opinion The Heretic got it right. The barrel and slide don't move till the bullet leaves the barrel.

 

[Ownerus]

An interesting photo I found in one of my books. An old x-ray of a 1911 being fired. The bullet is still in the barrel near the muzzle. I hope some poor duck wasn't holding the gun in his hand with an x-ray that intense! It's fuzzy and a little indistinct so it probably won't convince anyone who believes the slide magically waits until the bullet has left to move but look closely. The back of the slide appears to be a little behind the frame and if you look closely, you can see the bottom barrel lug has moved off of the slide stop pin. It is against the pin before firing. It's the part that keeps the barrel and slide from launching off the front of the frame. The barrel and slide are still locked together because the link is traveling through the part of its arc that's mostly horizontal. It will pull the back of the barrel down later in its travel.

Anyway, interesting photo.

 

[Jabba]

Dapapadon, you can test the theory pretty easily. Fire a blank cartridge with the same amount of gunpowder as a normal round. Without the a bullet to accelerate and create an opposing force, the barrel/slide won't have enough energy acting on it to cycle(I'm guessing). If it's all due to the gunpowder rushing out of the barrel, nothing should change.

 

[cookie]

The brass expands and is tight in the chamber. When the bullet leaves the barrel the pressure drops and allows the brass to [shrink back to closer to the prefired size] be extracted. It is a whole timing / pressure thing.