No snide remark intended,simply following the line plot.
Okay. Maybe I was a bit hasty. I've been conditioned for it.
First, a quick primer on recoil.
Newton 3 states that for every action, there must be an equal and opposite reaction. This is Newton's Law...not Newton's Theory. Push on an object and you immediately get pushed away from that object by the force vector that exists between you. If that force is sufficient to set both objects in motion, then both will move at the same instant. They don't have a choice.
Recoil is nothing more than the reaction side of an action/reaction event. A result of force forward/force backward. An action/reaction system requires three things. Two interacting objects and a compelling force between them.
Remove any one of those components, and you can't have an action/reaction event. Once the bullet has left, the action side of the system is gone, and the compelling force with it. If there is no action, there can be no reaction. If there is no force, there can be no acceleration, and hence no movement.
i.e. If the slide hasn't moved until after the bullet exits, the slide won't move.
Kuhnhausen's "Balanced Thrust/Force Vector" description was well written, nicely illustrated, and interesting...and utterly wrong.
Here is the description of the firing/recoil cycle. To simplify it, we'll assume a condition of zero headspace and proper barrel linkdown timing.
Bang! The bullet and slide start to move at the same instant. Due to the slide/barrel assembly's greater mass, the bullet is accelerating some 33 times faster.
The barrel is under a forward drag from the bullet's frictional contact. The rearward moving slide grabs it by the lugs and hauls it backward with it...against the resistance of that forward drag. Newton 3 again. Whatever resistance is imposed by friction in one direction, is imposed in the other direction in equal measure. Whatever force resists the barrel's rearward motion, resists the slide's rearward motion.
The bullet exits at nominally 1/10th inch of slide/barrel movement. At the instant that the bullet base clears the muzzle the link just starts to tug on the barrel. The barrel has not dropped yet. The lugs are still fully engaged vertically.
At .200 inch of slide travel, the upper barrel lugs are just clear of the slide's lugs, but the barrel is still moving rearward on its own momentum. The link is still pulling on it.
At .250 inch of travel, the barrel is linked down as far as the link will bring it, and its rearward momentum is stopped by the vertical impact surface. (VIS) If all is within spec, there is .003 inch of clearance between the bottom of the barrel and the top of the frame bed...and the barrel free-falls to the bed by gravity. If all is within spec, and the linkdown timing is right, there is .012-.015 inch of clearance between the top of the barrel and the underside of the slide's first lug.
NOW...you'll start to see the barrel protrude from the front of the slide.
The slide continues rearward via the momentum that it conserved during the acceleration phase...while the bullet was still present.
You can see the point of full linkdown/barrel impact on the VIS by simply pushing the muzzle straight back until it stops. (Cock the hammer first.)
Cycling it manually on the bench, you'll see a little barrel drop almost immediately because the lower lug is angled slightly, and gravity will pull it down as soon as the lug moves off the slidestop pin...but that won't happen during firing because the upper lugs are horizontally engaged in opposition under force. You can see how that works by using a tight-fitting plug in the barrel...such as an oversized cleaning patch on a rod...and pulling forward on the barrel while you have someone move the slide slowly. The barrel won't start to drop until the link has swung far enough...or about .100 inch...and will continue to drop until it contacts the VIS.
In the photo, the slide appears to have moved about .075 inch...so the barrel couldn't have tilted at the rear.