Ad: This forum contains affiliate links to products on Amazon and eBay. More information in Terms and rules
I'm not sure where you are coming up with your calculation for this, but depending upon the formula used, we can't "double dip" in both the projectile and the explosion.
My point is this - we can easily calculate in Kilojoules the energy produces by the explosion, using a TNT equivalent. Now, we cannot use all of that. There is a certain amount of that force that is applied to the bullet to accelerate it. It's pretty easy to reverse calculate how much was applied to the projectile. Of the remaining, calculating how much is used to chamber the next round is tougher. And as for gun efficiency, that I am not sure about as to how much will be lost. Maybe take the energy lost in muzzle velocity from old gun/new gun testing, but that's only a part of it. Not sure how much inefficiency is "normal".
All that really matters is that the amount of "stuff" (bullet, powder gas, sabot, wads, etc) going out the muzzle at what ever various velocities match in momentum the amount of "stuff" (gun, sight/scope, extra ammo, etc) going the other way.
Firearm energy efficiency[edit]
From a thermodynamic point of view, a firearm is a special type of piston engine, or in general heat engine where the bullet has a function of a piston. The energy conversion efficiency of a firearm strongly depends on its construction, especially on its caliber and barrel length. However, for illustration, here is the energy balance of a typical small firearm for .300 Hawk ammunition:[1]
Barrel friction 2%
Projectile Motion 32%
Hot gases 34%
Barrel heat 30%
Unburned propellant 1%.
which is comparable with a typical piston engine
OK, I'll bite and post my first real try at it after some thought. Please, not poking holes at this unless you post your own alternate solution along with the pokes.
For a first-order approximation for a P-51 Mustang based on momentum, let's start with a classic example.
We fire a 240 grain bullet at 1,180 feet per second into a 170 pounds block of wood. What velocity will the target have when the bullet absorbed into the block. This is easy and classic. 240 grains = 0.0156 kg, 1,180 fps = 359.646 m/s, and 170 lbs = 77.1107 kg.
M v (bullet) = mv (target), so (M v bullet)/ m target = v target, and the target moves at (.0156 * 359.646) / (77.1107 + .0156) = 0.072564 m/s or 0.1622 mph.
---------------------------------
For a first-order approximation of the P-51, let's assume the target is a block of wood 5 cm in front of the muzzle and that all the bullets are absorbed into the block. The reason I assume that for this example is that the momentum of the block will be the same as the momentum of the P-51 if they are the same mass and it is close enough to use the muzzle velocity before it degrades and slows down. I am, of course, neglecting the mass lost as the shells and links are ejected. The error will be quite small.
The bullet has a mass of 0.0434 kg and it fires at a velocity of 800 m/s. The cyclic firing rate is 800 rounds per minute or 13.3333 rounds per second per gun. We have six guns so we will fire 80 rounds. The weight of the P-51 firing the bullets starts out at 9,500 pounds or 4,309.12755 kg, so our target will be the same mass. Therefore the velocity imparted to the target will be
[(0.0434 * 80) * 4,309.12755] / [(4,309.12755 + (0.0434 * 80)] = 0.708473 m/s or 1.584951 mph. Note I added the mass of the absorbed bullets to the mass of the target, but not doing so would probably affect the answer out in the 4th decimal place or so.
Therefore, as a first order approximation I'll say the P-51 will lose 1.58 mph for each 1 second burst it shoots.
If we had some real-world data, we could verify or debunk this first-order approximation, but it isn't a major speed loss and wouldn't be all that dangerous as fighters generally shot 1 - 3 second bursts to keep the barrels cool and to avoid spraying their loads into the air as the target moved around in front of them.
Maybe we are closing in on letting this one go until we can find some real-world data.
I'm come to the conclusion that the more excited, or excitable, a person is, the less reliable his recall is of events.
Recoil does't give a hoot about the "joules of the explosion" OR "the energy imparted to the bullet."
Forget about both concepts when it comes to recoil. forget about "the efficiency of the gun" as well. It doesn't matter if the gun is efficient or inefficient. All that matters is that balance in momentum.
You are all over thinking this. You simply apply Newton's third law. Everything else (explosive forces, re-cocking weapons etc) is simply a shoal of red herrings.
Second law: F = ma. The vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object.
I would contend (but I don't know much about firearms) that this mass of gas is a very small component of the total mass being ejected.
Barrel friction 2%
Projectile Motion 32%
Hot gases 34%
Barrel heat 30%
Unburned propellant 1%.