Ruben Nasser
New member
As I understand it, the force that causes the slide/barrel to accelerate backwards (this is, reaction -recoil- from the moving bullet) ceases after the bullet exits the barrel and the pressure is released (well before unlocking, in a properly timed pistol, the slide recoiling only about 0.1"). Does this means that slide velocity is highest at this point? From this point on the slide is working against the recoil spring, friction in the locking lugs, extracting and ejecting the case, resetting the disconector, owering the hammer, etc. Do you have any idea of the slide velocity in a 1911? (I contacted a tech representative from ParaOrd, and he says that Rob Leatham told him it is about 0.06 seconds in a stock 45 1911, that would translate into 1000 rpm, and sounds logical , I was thinking above 800 rpm, higher than a typical submachine gun).
Regarding the locking time, this is covered very confusingly (...at least for me) in the Kuhnhausen book. It seems he implies that the slide and barrel are somehow in a static position while the bullet is moving in the barrel, and that they only start moving backwards when the bullet exits. This is impossible, as equilibrium of momentum dictates that the slide/barrel must be moving backwards (albeit at a much slower velocity due to their larger mass) while the bullet is moving forward. Does the recoil spring contributes in this stage to the locking of the slide/barrel?
Do you have any idea of the cycle time in a 1911?
I did some ballpark calculations and the barrel time is in the order of 0.4 miliseconds or lower, wich seems about right.
I also did a mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other, getting an initial (free recoil) slide/barrel
velocity of about 28-30 fps, wich "feels" slow considering I even disregarded the influence (if any) of the recoil spring and other factors. On the other hand, this initial slide velocity is more that enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.2") in this time is only about 5.8 fps, or 20% of this initial slide velocity.
Please help me understand this!!!
[Edited by Ruben Nasser on 02-23-2001 at 09:20 PM]
Regarding the locking time, this is covered very confusingly (...at least for me) in the Kuhnhausen book. It seems he implies that the slide and barrel are somehow in a static position while the bullet is moving in the barrel, and that they only start moving backwards when the bullet exits. This is impossible, as equilibrium of momentum dictates that the slide/barrel must be moving backwards (albeit at a much slower velocity due to their larger mass) while the bullet is moving forward. Does the recoil spring contributes in this stage to the locking of the slide/barrel?
Do you have any idea of the cycle time in a 1911?
I did some ballpark calculations and the barrel time is in the order of 0.4 miliseconds or lower, wich seems about right.
I also did a mass x velocity equilibrium equation for the bullet and powder gasses on one side, and slide/barrel on the other, getting an initial (free recoil) slide/barrel
velocity of about 28-30 fps, wich "feels" slow considering I even disregarded the influence (if any) of the recoil spring and other factors. On the other hand, this initial slide velocity is more that enough to get the pistol to cycle in .06 sec, since the average slide velocity required to complete the cycle (about 4.2") in this time is only about 5.8 fps, or 20% of this initial slide velocity.
Please help me understand this!!!
[Edited by Ruben Nasser on 02-23-2001 at 09:20 PM]