...how it works with out pressure amounts ? I get that bullet weights and slide weights all mean something but doesn't how much pressure is created matter more ? Isn't everything the gun needs to do to operate dependent on the amount of pressure created .
That's the beauty of conservation of momentum. Once you know the momentum of what comes out of the barrel, you also know the recoil momentum.
Barrel friction, pressure, force, all of that wraps into one single number--the muzzle momentum.
NOT because those factors (pressure, force, barrel friction) are unimportant, but because they are already taken into account when one considers muzzle momentum.
Or does it matter if there's enough pressure to blow up the gun or create a squib , the slide will move the same amount at the same timing regardless of the pressures created ?
The momentum of the recoiling mass (in this case the slide, barrel, bushing, recoil spring plug, and half the recoil spring) will be equal to the momentum of what comes out of the muzzle--gases and bullet.
Once you know the momentum of the recoiling mass, you can calculate its velocity by dividing the momentum by the combined weight of the recoiling mass. Then with the velocity you can figure how much the slide moves in a given amount of time.
or how far down the barrel the bullet is when said pressures are high enough to start causing the slide and other parts to move ?
The slide/barrel begins to move when the bullet begins to move. If the bullet has moved part of the way down the barrel then the slide has already begun moving.
Because the semiauto has all these allowable ( micro movements ) the over all movement ( muzzle rise ) is less then if it was a fixed single shot firearm because these( micro movements ) are nullifying/canceling out some of the momentum ???
Very close.
Muzzle rise happens when recoil momentum drives a recoiling mass AND the resistance to the recoiling mass is below, not directly behind the muzzle. That unbalanced resistance results in the muzzle rising as the gun torques around the center of resistance.
In a revolver, the WHOLE gun is the recoiling mass and the resistance to the recoiling mass is well below the bore. So the instant that the barrel starts recoiling, the muzzle starts rising.
In a locked breech Browning type recoil operated pistol, the resistance to the recoiling mass--UNTIL THE BARREL UNLOCKS FROM THE SLIDE--is very small. So the slide and barrel move almost straight back until the barrel unlocks from the slide. There's very little muzzle rise during this phase.
The other half of the equation is that, BY DESIGN, a locked breech Browning type recoil operated pistol does not unlock the slide from the barrel until the bullet is out of the bore.
Combine those two facts and you end up with the result that there is very little muzzle rise until the bullet leaves the bore.
That math is WAY over my head...
No, it's not. Here's what's going on.
First of all, the math calculates the momentum of the bullet (bullet weight times muzzle velocity). Momentum is mass times velocity.
Next the math calculates (approximates) the momentum of the gases that escape the muzzle. That's powder weight times 1.5 the muzzle velocity of the bullet. That's a reasonable approximation for the gas momentum in a pistol.
Then the two momentums are summed into one total momentum figure of M.
NOW we know the total momentum of what comes out of the bore. But we really want to know the slide velocity.
Conservation of momentum tells us that the momentum of the recoiling mass is equal to the total muzzle momentum.
So the total muzzle momentum figure (M) is ALSO the momentum figure for the recoiling mass.
Since momentum is mass times velocity, if we divide the momentum figure by the weight of the recoiling mass, we can calculate its velocity.
(Mass x velocity)/Mass = velocity
So next we figure up the total weight of the things that make up the recoiling mass. The slide, barrel, half the recoil spring weight, the bushing, and the recoil spring plug.
Now we have the weight, we can divide the momentum figure by that weight to get the velocity of the recoiling mass at the instant that the bullet exits the muzzle.
The physicists will note that I'm being sloppy by pretending that mass and weight are identical. While that is not actually true, in this case, because we're going to cancel out the weight and just be left with velocity, we don't need to worry about the correction factors since they will cancel out too.
.This momentum and movement ratio would seem to indicate a fixed object/firearm . Meaning a bolt action or single shot firearm or revolver . That ratio can't be accurate if there is a slide being resisted by a spring which is being held by a not so stable hand , correct ? There would seem to be lots of give/extra allowable movement in the equation when using a semi auto ? You'd need to know how much the spring is resisting the movement as well as how much the hand hold actually freely lets the frame move . Maybe not freely but your hand is no vise so the frame is going to move hundreds if not tenths of an inch with relatively no resistance based on the soft tissue of the hand . 
