Yet the only evidence to the contrary you provide is some formula you came up with for a pistol which isn't relative to an AR. You don't actually reply to the points given and you refuse to acknowledge the other party said they didn't want to get into the debate and trash the thread in the first place. Instead you're trying to bait under the guise of honest intellectual discussion when you and I both know you're waiting to pounce with the same rhetoric that gets brought into the same debate every time this topic gets brought up. Stick to engineering Clark because your bullshitting skills are preschool at best.
Seriously, I have never heard of this debate before, and I have been posting every day on gun forums for 20 years, since before there were www gun forums. I was the #1 poster on rec.guns on the old usenet part of the internet, with over 100,000 posts.
If I pull one of my ARs, I see the gas is tapped 10 inches from the muzzle and 13 inches from the breech.
I know that 25 gr 4895 55 gr Vmax is a typical load.
Quickload shows that load with a 23" barrel having pressure fade from 20kpsi to 10kpsi in the 0.3 ms it takes the bullet to get from the gas port to the muzzle.
The bolt assembly weighs 11 ounces.
The piston [bolt carrier key] has an inside diameter of 0.18" = 0.025 sq in = 1.6 EE -5 Meters squared
30kpsi = 2 EE 8 Newtons/Meter squared
Force = 2 EE 8 x 1.6 EE- 5 = 3.2 EE3 Newtons
11 ounces = .31 kg
Acceleration = F/M = 3.2EE3 Newtons/ .31 kg = 1.03 EE4 meters/ second second
time = 0.3 ms
Velocity = acceleration times time = 1.03 EE4 m/ss x 3 EE-4 s = 3.1m/s = 121 inches per second max speed when the race is over.
The bolt has averaged half that speed = 60 inches /second and has 4 inches to go to lock back.
While the bullet's average speed is 34,800 inches/sec and has 10 inches to go.
In .0003 seconds the bolt has travelled 0.018" and is 0.4% done
In .0003 seconds the bullet has traveled 10" and is 100% done and gone