Firing MG's and cannons

[Shortround6]

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".

You are over complicating things.

The bullet weight X velocity + propellant weight X velocity (usually a constant) resulting in a recoil impulse figure is used by Tony Williams in his Assault rifle book and in some recoil calculators like this one: ShootersCalculator.com | Recoil Calculator

Granted this is for a single shot.

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.

having a worn barrel changes things because the bullet will exit at a lower velocity (less momentum).

 

[GregP]

Hi Shortround,

Sorry for getting technical. A rocket burns for awhile and has a nozzle. A one-shot thruster has fuel that literally detonates and burns for a VERY short while, like a rifle cartridge going off. The main difference is duration and the type of nozzle used.

Either way, it doesn't seem to do much to a 9,500 pound airplane with six 50's. I could easily do it for a plane with 2/3 the mass and four 50's plus two 20's and things might get a bit more interesting. I'm sure you could, too.

 

[Shortround6]

Running your numbers for bullet weight, velocity and adding the propellant weight through the calculator in the listed website in previous post and using 9500lbs for the weight of the "gun" gives velocity of 0.04fps for the "gun" (airplane) and multiplying times 80 for the rate of fire give 3.2 fps or 2.18mph which is more than close enough considering you were using a 670 grain bullet and I added 240 grains worth of propellant ( calculator more than likely uses a constant for escaping gas velocity).
Upping the bullet weight to 714 grains and velocity 2900fps increases the "speed" (actually negative) to 2.72mph for 80 rounds fired.

Your method and mine both work and I would say they are close enough.

BTW cutting the propellant weight to 2 grains instead of 240 ( you have to put something in the box) and using your bullet weight and velocity gives 1.63mph so we are really close and I don't know what got rounded off where.

 

[Garyt]

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.

That I would agree with. Question is, how much of the discharge is vectored in other areas? Obviously to chamber a round, and the older a gun is and less "tight" it is, the more that will escape in a way not congruent with decreasing the planes velocity.

But you do not add the force of joules of the explosion AND the energy imparted to the bullet. I would guess the site you link to has some method of calculating the efficiency of the gun as well.

I think the difference is I am in my head calculating by hand as it were all the steps, while you are using an online calculator that has all the formulas figured into it if you input the base info. But it does not illustrate all the steps it goes through to arrive at a calculation.

 

[Garyt]

I found what I was looking for to explain gun efficiency. It's Wikepedia, but seems to be accurate.

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

It shows this firearm to be 32% efficient.

That was my point - it's not as simple a factoring joules from the explosion in it's TNT equivalent and factoring this all as energy.

Now either the gun is not a blowback type with the gasses being used to chamber a round, or the blowback effect is already calculated in the form of "Hot Gases". If this is indeed the case, less of these "Hot Gases" apply to the energy being used to stop said plane, as they are being vectored the wrong way.

Of course the barrel length increases efficiency, as the gases are kept on the proper vector longer, as well as being confined longer.

The other issue which is not illustrated by the above numbers which I touched on the the chambering of the round - how much of the gases, which constitute 34% of the explosions's energy, are vectored properly to reduce speed of the plane? And I am not sure as to the answer to this, again as with bullet velocity a longer barrel would help vector the escaping gases better.

 

[gumbyk]

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.

I can't argue with that math...

I suspect that most of the 'paused in flight' and 'loss of 20-30 mph' anecdotes are just that, stories (even if that is what it seemed like to the pilots at the time). How many pilots would have been looking at their airspeed indicator while firing at the enemy, or even in practice?

 

[GregP]

I think that (the breakdown) was said a earlier (at least the part about the bullet only getting a fraction of the energy imparted), but is certainly worth remembering. All the shooter cares about is how hard it will kick.

The aircraft designer must care about the kickback, getting rid of the spent shells and links, venting the gun gas somewhere not connected to the cockpit or cockpit ventilation system, and making sure the hot gases don;t vent against a flammable tank of liquid.

Maybe we are closing in on letting this one go until we can find some real-world data.

 

[Shortround6]

I just found the calculator a few minutes ago but I have had Tony Williams book for a while.

This where I believe you are getting confused "But you do not add the force of joules of the explosion AND the energy imparted to the bullet.

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.

FWI, many guns back in the 1890s and early 1900s saw a noticeable reduction in recoil when their cartridges switched from black powder to smokeless powder. You might call that efficiency but there was a lot less "stuff" going out the muzzle as the smokeless powder charge was often 1/2 down to 1/6 as much as the black powder charge had been. The Black powder gas may exit at a different velocity though. Source on this is a 1934 Winchester catalog with recoil figures for different loadings of some cartridges.

And as far as the powder gases going out the muzzle affecting things..........muzzle brakes would NOT work if there was not mass moving at speed acting on them.

 

[GrauGeist]

Maybe we are closing in on letting this one go until we can find some real-world data.

Better still...to those that can't grasp the physics involved, I suggest you get on a bicycle and while riding, shoulder your 12 gauge shotgun and fire a round dead ahead.

Let us know how that works out for you...

 

[GregP]

I can understand the "lost 20 - 30 mph while firing" stories.

Perhaps these guys were in cruise and left the throttle and mixture at cruise while they were switching on the guns and turning on the gunsight. If they were cruising at, say, a constant 250 mph and pulled up while firing, they WOULD probably lose some 20 - 30 mph until they DID something to change that. I'd think they would drop the nose, go to auto-rich and high rpm, and cobb the throttle, and then get back in the fight, but forget they had forgotten to change the mixture and rpm, and advance the throttle before doing it.

Might not have happened that way, but seems logical to me as a possible answer.

The early P-38 Lighting guys had to handle TWO mixtures, rpm's, and throttles, plus the guns and gunsight before they were ready, and a lot got shot down while they were doing these very actions after being bounced while in cruise and not seeing the enemy first. Why should the P-51 / P-47 guys be immune to the same results? Surely we had rookies flying those planes.

 

[tyrodtom]

I race cars, and I let other people race my car. For 4 years now there is always a Go-Pro camera inside the car viewing the driver and a lot of the action out front and to the sides.

Before installing those Go-Pro cameras we relied on videos taken of the race.

Viewing those videos , from both views often tell a different story from the driver, even when the driver was me.

I'm come to the conclusion that the more excited, or excitable, a person is, the less reliable his recall is of events.

 

[Garyt]

I'm come to the conclusion that the more excited, or excitable, a person is, the less reliable his recall is of events.

That probably is as important to the topic as any of our calculations. Perception to someone is reality.

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.

Actually Shortround, recoil indeed does "give a hoot".

It's a rather simple concept. A round going of is an explosion that generates energy. Recoil is based how how much of that force is vectored to push someone back, and also how much force is NOT wasted on other things, such as heat, barrel friction, etc.

A bomb does not really have any "recoil" in most situation. Why? Force is not being vectored in any particular area. This can change in a semi-enclosed room, which is in essence what a gun barrel is, a semi-enclosed room designed to vector force a specific direction.

Recoil indeed does "give a hoot" - but we are looking at it from two different ways. You are looking at your recoil calculator - I am going through the processes you recoil calculator does, although it fails to explain these processes to you. If you understand what I am saying, you will have a better idea how your recoil calculator comes up with these numbers.

 

[stona]

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.
Cheers
Steve

 

[Garyt]

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.

Not quite that simple. Let's apply Newtons third law. Sure. What is the force that is being applied in one direction that must be equaled? In other words, how do we calculate this?

 

[stona]

Newton's second law.

Steve

 

[Garyt]

Newtons second law:

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.

OK. How do you come up with the mass of the object times the acceleration - Look at the muzzle velocity and mass of projectile.

What you are not accounting for however, is the escape of hot gas that is vectored in the correct way. As per my post earlier, the energy imparted to the bullet is very similar to the energy imparted to hot gases (which are NOT accelerating the object). The true question here than is which of these are vectored "correctly". and which are vectored in other direction.

It seems you merely want to take the energy imparted upon the projectile- that would be great id guns were a 100% efficient machine in transfer of energy. But they are not. They are closer to 33%.

This is why Shortrounds online calculator also takes into account the grams of explosive in the firing charge.

 

[Shortround6]

You have a formula (or more correctly a breakdown) for efficiency. It tells you how well a particular gun/projectile/propellant combination does at converting the power in the propellant into kinetic energy in the projectile.

When figuring for recoil the "efficiency" means very little. The ONLY way efficiency enters into it all is that a more "efficient" powder/propellant will be present in smaller quantity and so add less mass to the total going out the end of the barrel.

You have your propellant ignited (by the way it does not explode, it burns, very rapidly but it burns, not explode) and force is applied to the base of the projectile and to the head of the cartridge and to the walls, equal force in all directions. The projectile starts moving first due to inertia (the projectile being much lighter than the gun) the pressure against the side walls (and chamber walls) can be ignored as aside from a little temporary stretch nothing much happens and as far as movement goes the force up cancels the force down and the same for left and right and so on. The force pushing against the case head on the other hand does do something. It wants to move the gun in the opposite direction the bullet is traveling. The pressure in the chamber/beginning of the barrel rises to a peak ( for high velocity small arms usually just a few inches of bullet travel) and then starts to drop as the volume of the combustion space increases the the further down the barrel the projectile travels. The propellant charge is still burning, at exactly what rate of gas generation and pressure is dependent on the exact propellant. The much heavier gun/human/gun mount/vehicle has also started to move although much slower due to weight and inertia. Once the projectile clears the end of the barrel pressure drops pretty much to zero and force acting on both the projectile and the gun/human/gun mount/vehicle stops and both continue in motion with the force already imparted.

Now I have rifles that can achieve the same projectile velocity using the same bullet but using 35 grains of propellant A or 45 grains of propellant B ( or a variety inbetween). This may show a rather large difference in efficiency, they will show differences in peak pressure and in flame temperature but the only difference in recoil (and too little to feel with the shoulder) is that 10 grain difference in propellant weight.

Can your theory explain the figures in the Winchester catalog?
cartridge....bullet.....MV fps ....ME ft/lb.....smokeless.....Blackpowder
38-55.........255......1320........990............6.0..............8.5
40-82.........260......1490........1285..........9.0.............12.0
44-40.........200......1300........750............4.0..............5.5
45-70.........405......1320.......1560...........12.0............16.0
50-110.......300.......1605.......1720..........11.5............19.5

Numbers under the smokeless and Blackpowder headings are the free recoil in ft. lbs.

Since the exterior ballistics are the same something is going on inside the gun. You may say that the smokeless powder is more "efficient" (and it is) but we have an increase of 33%-69% in recoil energy going from the light weight smokeless powder to the heavy blackpowder depending on cartridge with NO change in projectile energy.

Something is increasing the rearward movement of the gun and it is NOT heat or barrel friction or.......
It can be explained, and rather easily, as increased mass moving down and out the barrel at high speed.

 

[Garyt]

I guess one question, what is the "burn efficiency" of both powders?

If the black powder burns less efficiently, it would have more mass left in the barrel. This means more mass propelled, so in essence you are adding to the mass of the projectile but not speeding up the projectile at all, thereby increasing recoil for no addition to performance. It may also burn less efficiently in regards to not providing the same level of acceleration, but that should not make a difference on recoil, as both the projectile and the recoil are effected the same.

There is a lot to here that we are leaving out. Newtons laws, which applies to most of this discussion, work great in a vacuum.

But we have an atmosphere, gravity, and other issues which all come into play as well, so for any true calculations we should include these effects that apply as well. I think we often assume these effects are negligible, but sometimes they are not.

 

[stona]

The mass of gases leaving the barrels should be added to the mass of the projectiles leaving the barrels. In the equation it is the sum of all mass being projected from the aircraft that is relevant. 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. Any gases or anything else that remain in the aircraft, or weapon, are not relevant.
Cheers
Steve

 

[Garyt]

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.

I think it is a lot more than you think actually. Two points:

Barrel friction 2%
Projectile Motion 32%
Hot gases 34%
Barrel heat 30%
Unburned propellant 1%.

The .300 Hawk round delivers as much energy through the hot gases leaving the barrel as it does the round itself. A bit more actually than from the round. I would guess this is similar to most other firearms.

Secondly, I used Shortround's recoil calculator. Both times I used a 1000 grain bullet, 3000 feet per second in velocity, and a 200 pound "gun". The only change was in the propellant, I used 1 grain for one calculation and 500 grains for the other. Bear in mind there was no chance in the velocity or mass of the projectile.

The 500 grain example had a recoil energy of 47.97 foot pounds. The 1 grain example was 14.32 foot pounds of recoil energy.

ShootersCalculator.com | Recoil Calculator

I think a few important things to remember - Gun powder does not completely burn, so the mass of the gases also include the mass of the unburned powder. I think black powder burns with close to a 50% waste product, not sure about smokeless. The other issue is that energy is mass times acceleration squared. Even if the inert powder and gases are a smaller percentage of the weight of the projectile, they are being propelled much faster than the projectile, so would have much higher "energy per ounce" than the projectile would.