RKG said:
Recoil of the rifle, on the other hand, almost entirely occurs after the bullet has exited the muzzle. There is an engineering explanation for this, but there are also a host of high speed video stills (and, before that, stroboscopic photographs) illustrating that the muzzle does not begin to move in recoil until the slug has exited.
Actually, none of that is correct. If recoil had no effect on muzzle position before the bullet cleared the muzzle, different load levels would not cause the elevation in point of impact (POI) to change by more than the difference in drop due to difference in velocity, which is considerably smaller than the change an Auddette ladder reveals. The change during the millisecond it takes the bullet to leave the muzzle is not much, and you usually have to put a stationary line on the slow motion movie to distinguish it, but it happens. If it didn't, barrel tuners would have no effect. (I note some slow motion images are from guns in machine rests, so these may exhibit no visible movement.)
Recoil has two stages. In a rifle, the first stage occurs while the mass of the bullet and about half the mass of the powder charge get pushed down the bore. The second stage is variously called "rocket effect" or "after-effect". It occurs when the bullet clears the muzzle, uncorking the pressure behind it, which then accelerates the mass of the gas inside forward at even higher velocity, causing additional push back. In some overbore cartridges, this can account for half the total recoil energy given to the gun. If you ever wondered why a muzzle brake that vents gas equally in all directions, pushing up as much as it pushes down, can still tame part of the recoil, this is the reason: it vents the gas sideways instead of forward, leaving little high pressure gas available to produce rearward rocket effect. It does nothing about the portion of the recoil due to the bullet mass going forward, so there is still recoil.
What drives the gun back is Newton's third law of motion. It is most commonly shortened to state: for every action there is an equal and opposite reaction. What this means is simply that the momentum of all the stuff pushed out of the muzzle, bullet, gas, unburned powder, etc.; aka, the
ejecta) will be equal and opposite in direction to the momentum imparted to the gun and, through it, to the shooter, and through the shooter, to the earth. You cannot move the bullet down the bore without picking up momentum, due to the equal and opposite force applied by pressure to the bullet and to the breech of the gun. It's inescapable.
Fortunately for the shooter, while momentum is equal and opposite, energy is not. This is what keeps the gun from damaging the shooter as much as it does the target. Yes, yes, the bullet is pointy and contacts less area, but when you look at the vast permanent cavity of damage caused by a high power rifle bullet that expanded or tumbled, you realize the diameter of that damage is often even bigger than the butt of a rifle, yet the shooter didn't experience such damage.
SAAMI has a free document describing
how to calculate gun recoil energy (it is mainly what you perceive as recoil).
As to the velocity of the recoiling gun, that's very easy. Because momentum is equal and opposite and created simultaneously, if the rifle were free floating and if you have a perfect muzzle brake to take all the rocket effect out, the rearward speed of the rifle would just be the speed of the bullet divided by how many times heavier gun is than the weight of the bullet plus about half the powder charge. If the gun weighs 8 lbs and fires a 150 grain bullet with a 40 grain charge of powder to 2800 fps, then 8 lbs is 56,000 grains, and 56,000 grains divided by 150+20 grains is 329.4, and if you divide 2800 fps by 329.4, you get 8.5 ft/s. Add in rocket effect for a .308 W and you might be in the range of 11 fps. During the acceleration of the bullet the average velocity of the gun will be half that, or 5.5 fps, so in the millisecond it takes the bullet to clear the muzzle, the gun will have moved 0.0055 feet, or 0.066 inches or a little over a 1/16th of an inch. And this assumes you are making no contact with it, not loading it with your own body mass (which adds to that 8 lbs). It's not much.
To the OP's question: The effect of this recoiling movement does not change the pressure on the bullet or on the gun breech, so the gun's velocity is not subtracted directly from the bullet's, as it would be if the gun were moving backward at the start of the shot. That would be subtracted because the backward velocity would already be in the bullet. But in the recoiling gun, the gun doesn't move until the bullet does. So the bullet never has rearward velocity to subtract. At that point you are just looking at two independently moving bodies with the same pressure acting on both. Instead, the recoil velocity effect will be that of leaving the bullet with 0.066 fewer inches of barrel length by the time it reaches the position of the muzzle. For the 150 grain .308 W I described, it amounts to about 1.5 fps velocity loss.
I think most of the difference in velocity claimed for different shooting positions has to do with a combination of powder position in the cartridge being changed and with firing angle over the chronograph being changed and introducing a measuring error. The theoretical difference for the free recoiling gun is just too small, and one pressed against a body will lose even less.
As to energy, that is equal to the work done on the bullet. It is the distance of travel in the gun times the average (not the peak) force in pounds applied to the base of the bullet by pressure. It even has the same units as energy: ft-lbs. So if the bullet has 2611 ft-lbs at the muzzle and the barrel was 23.934 inches long (after subtracting 0.066 inches), you had and average force of 1305.5 lbs at its base during its whole time in the barrel. That ignores the equivalent of half of the powder mass traveling with it. That would have gained about 12% of the energy of the bullet, so the total force applied to both was actually about 1480 lbs, and that's what the rifle will see. Applied to the recoil distance of the gun, you have 0.0055 ft time 1480 lbs force, or about 8.14 ft-lbs of energy in the gun. That's ideal. In closer approximation, QuickLOAD identifies both unburned powder mass and half the burned mass moving forward and comes out with more like 9.24 ft-lbs. The rocket effect raises that to almost 15 ft-lbs. But the bullet has 2611 ft-lbs. No comparison, and the shooter feels that much less.