Firing MG's and cannons

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OK, I'm convinced that momentum rather than energy is the correct approach.

As has been pointed out, the momentum added the projectile is balanced by momentum removed from the airframe. Each projectile will remove the same amount of momentum from the airframe, and as the relative momentums are different by several orders of magnitude we can probably forgo complex differential equations and treat the repeated discrete firing events as a continuous linear force.

Force is equal to the rate of change of momentum with time. So the equivalent force (in Newtons) due to recoil is the mass of the projectile (in kg) times the muzzle velocity (m/s) times the rate of fire (rps) times the number of guns.

So taking a Tempest V with 4 HS.404 cannon (145 g projectile, 880 m/s, 700 rpm), the total recoil force is about 6 kN. Or about 600 kg under 1G. For comparison, the total drag on the airframe can be estimated as the engine power divided by the speed: 1600 kW / 200 m/s gives airframe drag as about 8 kN.

This is very rough stuff, but I think it should at least be in the right ballpark. And it certainly squares with the anecdotal evidence.
 
20 or 30 mph is a awfully imprecise figure.

There's probably thousands of pilot reports out there of encounters where they fired at other aircraft.

Are there after action reports where they indicated firing slowed them do so much they had to quit firing to keep up ??

If it actually slowed them that much, that would have had to occur occasionally .
 
This is getting interesting.

This is may be a closed system, but it not a conservative system. When you shoot a round, the very small bullet gains speed (momentum and energy), but the vast majority of the shell is lost mass since it gets ejected overboard along with the belt link at the current speed of the aircraft. Most of the weight of the powder is lost when it is converted into hot gas to propel the bullet, and the spring in the breech absorbs a lot of energy and distributes the impulse over the time to fully retract ... never mind the spring-powered rebound.

The potential energy of the unignited powder enters into the conservation, too.

It would be a LOT easier if the shell was in a rifle and there was no spring recoil absorption.

For those of you who like such exercises, the typical 50-Cal round is .717 kg with the bullet being .0434 kg (43.4 g), the powder being an average of 14.5 g, and the shell itself being .6594 g. A typical 50-Cal belt link comes in around .013608 kg / link or so.

When I do a real first-order calculation using conservation of energy and a 1 second burst with 6 Browning M2 50's with a cyclic rate of fire of 800 rpm, I get a theoretical speed loss of 6.956 mph for a 1 second burst from a starting speed of 300 mph. All my calculations were done in metric units and converted back.

Conservation of momentum is great for a closed system consisting of collisions, both elastic and inelastic, but a machine gun round is not a collision ... at least until the bullet hits something ... and outside forces are acting on it (gunpowder, rifle barrel friction, and heat generation and dissipation, not to mention gravity).

This isn't an easy, first-order calculation, and I have yet to find an authoritative source for the real life speed loss caused. If you hear 50 fighter pilots from WWII speak, maybe 2 will mention this ... IF asked. Neither agree on the speed so lost.

As a former test manager of a place that produced propellants, I can say a ballistic pendulum gives a very good energy analysis of a bullet hitting a block inelastically, and can also give a great energy analysis of propellant fired. But we had no spring recoil absorption and we weren't ejecting mass when we fired anything. We were analyzing the energy capacity of propellant and defining the energy imparted when a bullet impacted on a target. Most of our calculations were energy based and followed standard DOD formulas and procedures.
 
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Actually, this can be solved, but as you so correctly implied, Graugeist, this isn't a math forum. There are several solution of 1st, 2nd, and 3rd order approximations ... but we don't have any real data to compare them with. That is, known rates of fire, known weights, known speeds before and after firing and the length of the burst. All we have is some general heresay.

The quote from Zemke above is nothing, He was just trying to not be hogtied by flying close formation.

It WOULD be interesting to check the speed loss with real data and then verify the calculations.
 
IF it was that significant wouldn't there at least be some mention of it in the flight manual ?

If it could drop your speed 20 to 30 mph, ( strange that he said mph instead of knots) that would been something you'd need to know in case you got into a tight turning fight close to the ground.
 
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Well, if MG's or cannons really effected speed that much, they should have been reversible, so you can get that added velocity when needed

I must add, some of the calculations have shown guns being fired for a minute when calculating (this may have been from the link of prior conversation on the subject). I don't enough ammo is carried even for a full minute. We need to look into a few seconds here or there.

Also, the propulsion lost is based upon the velocity of the projectile only - not the entire round, as it is not propelled forward.
 
It may be best to think of the guns as little rocket motors.
Bullet mass X velocity + Propellant mass X velocity X firing impulses per second (rate of fire) X number (and type) guns (motors) = total thrust forward (and acting rearward) per second.

Thrust may not be best word.

In most cases this force will be acting on the weight of the airplane.

Figure out your speed loss for one second and then multiply as necessary for length of burst. I seriously doubt that Hurricanes with 40mm cannon used an entire "drum" of ammo in one burst. Cycle rate was around 100 rpm (1.66rps) or 9 seconds and a plane doing even 240mph will cover 0.6 miles in 9 seconds. Any pilot opening fire at over 1000 yds at a target the size of a tank needs to go back to gunnery school.
 
Shells aren't rocket motors, they don't have a nozzle. If it had a nozzle, the force would be multiplied many times over. What you get is an impulse of short duration, confined to kickback by the shell chamber. Much of the force is absorbed by the recoil spring and is transmitted to the gun mount during recoil and as the bolt returns to the closed position. The amount of kickback is predictable and the math models can be tweaked to predict it ... IF we have real world data to model.

Otherwise we have an unverified model. Having asked this question of several former WWII pilots, I can say they didn't give answers that showed any consistency at all, and that's why I haven't posted it to date. Some said the speed loss was not noticeable and some said it was significant. Typical agreement in the fog of combat. In all, I have asked some 5 - 6 pilots and the answers are all over the place.

The force will be acting on the gun mount that is probably riveted to the wing spar or a fuselage former and bulkhead, with attendant stiffness. You'd have to know the force to properly design the gun mounts with properly-sized rivets, but I doubt those analyses are available for public dissemination at this late date if they haven't surfaced already. I seriously doubt we'll find well-documented data, but hope we do.

Seems like an open-ended discussion unless we can bet some measured data for a known model at a known weight firing known rounds. Not all 50-cal round have the same amount of powder, same weight, or even same shape.
 
Bullet mass X velocity + Propellant mass X velocity
Click to expand...

We can't really use both the bullet and propellant. The propellant is used to power the bullet. The only way that it could be determined that the propellant could be factored in would be to determine the gun "efficiency", and determine what is still "left over". And then one would have to see where that force is vectored - most of it the same direction as the bullet, but there will be some vectored in other directions as well. Loading the next round is one area where this energy not going forward goes.
 
The recoil of modern rifle and MG round is rather well known in armament circles. I used tobe in the armament circle but didn't bring too many references with me when I left, so I don't have those tables.

The mass of the aircraft is digitally decreasing ... that is, as each round is fired, a certain mass is lost with the bullet being fired and the shell and link dropping away from the aircraft. The modeling of this is not that hard.

What we lack are hard data to tweak the model(s).
 
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"I've read that crews of B-25's w/75 mm guns reported that the plane actually paused in flight when the gun was fired!"

in reverse order.

I've read that crews of B-25's w/75 mm guns reported that the plane actually paused in flight when the gun was fired!
Click to expand...

Lets think about that one.
Plane "paused"???? Air speed reduced to zero and then accelerated back up to flying speed?

Try driving 60mph and hitting the brake hard for a fraction of a second and then releasing. You may be thrown against the straps (seat belt) but the car will still be traveling at a good rate of speed. the car didn't "pause" even if all the junk piled on the back seat is no on the floor.

The mass of the aircraft is digitally decreasing ... that is, as each round is fired, a certain mass is lost with the bullet being fired and the shell and link dropping away from the aircraft. The modeling of this is not that hard.
Click to expand...

It may not be hard but it is totally unnecessary. The fact that the plane shot off even 100lbs of ammo (and .50 cal ammo weighs about 30lbs per hundred for bullets, propellant, cases and links) is a change of 1.25% of the weight of an 8000lb airplane. A change of 1.25% in aircraft weight is not going to change the answer much.

We can't really use both the bullet and propellant. The propellant is used to power the bullet.
Click to expand...

Actually we do need to use the propellant. It is mass being ejected from the gun barrel and fits into the 'equal and opposite reaction' it is also exiting the barrel at high speed ( 1,200 meters per second is often used as a constant for guns in this velocity range), higher than the bullet/projectile. Please note that this is not energy. It is the weight of the solid propellant (turned into a gas) and exiting the gun though the muzzle. There may be a very small residual amount that leaks out through the breach between one round being extracted and other round be chambered. We are not concerned with the the efficiency of the propellant, it's energy content, it's flame temperature or anything except it's mass and exit velocity.

Shells aren't rocket motors, they don't have a nozzle.
Click to expand...

I believe I said gunsand not "Shells".

You have, for example, a single .50 cal machine guns and it is ejecting a 46 gram projectile at 870 meters a second and about 15.6 grams of propellant at about 1200 m/s 13-14 times a second.

It's effect on the aircraft's flight will be somewhat like a little retro-rocket.

The recoil springs in the gun don't do a whole lot to reduce the recoil. The gun starts recoiling as the bullet and gas go down the barrel, on the Browning the the barrel is moving backwards a short distance. The breechblock/bolt is unlocked during this movement and as the barrel stops the breechblock/bolt continues rearward to extract the shell. At the point the shell casing begins to be exposed the bullet (and a fair amount of the gas) have already left the barrel and gas pressure is dropping fast. Everything that happens from this point on is done by the impulse already imparted to the parts in question. The 29kg of the M2 .50 cal gun are already in motion, they started when the barrel and breechblock were locked together. The gun does NOT begin to recoil when the breechblock hits the buffer after traveling it's full distance. Hitting the buffer may give a little 'bump' to the recoil at best. The springs in the gun only store enough energy to cycle the action, return barrel to forward position and send the breechblock back forward.
In the case we are looking at worst case is that the gun/s is/are bolted to plane with no movement/absorption.
If we can figure out the worst case and it does not come up to the 20-30mph speed loss of "legend" then we can argue over how much less of a speed reduction the springs and mounts might cause.
 
Hi Shortround,

Guns are nothing like rockets. They act much more like small thrusters, and a thruster is not a rocket ... at least not in the business. To a layman, perhaps. The propellants are nothing remotely alike and the duration is off by tens or even hundreds of orders of magnitude.

The recoil tends to be drawn out way past the actual impulse since the bolt is retracting against a spring in classic spring action and, once the shell is ejected, the exact same force is moving the bolt forward to close the chamber. As I clearly stated above, these factors are well known in armament circles. You can make as sloppy or as concise a model as you like, but without real-world experimental data to compare your model with, you are spitting into the wind and cannot be proven correct by any means. Indeed you can't be proven wrong, either.

If you have no real world data, you do the best you can, construct the device and go exercise it, with accurate instrumentation in place to record the results. If you were close, you did a good job. If not you adjust your model to fit the real world and try again. That's how they invented orbital space flight. There were many failures before they achieved a controlled orbit.

C'mon Shortround, you aren't a neophyte and this ins't the first time around the block for either of us. You can't successfully argue a correct solution without something to compare it with that supports your theory. Even if it DOES support your theory, it doesn't prove it if someone else comes up with real-world data that show your theory to be wrong.

It's the scientific method. Nothing new to either of us.

I'd love to exchange Excel spreadsheets with the proposed solutions with you, but we still need real WWII data to home in on it since it is unlikely that I can find data from a P-51 Mustang taken at 9,500 pounds shooting exactly a 1 second burst.

But ... if we CAN find some data, we could both show our models when such data emerges into the sunlight from wherever it is currently hidden.
 
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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".
 
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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 believe I said it would help to think that the guns were like rockets not that they were actual rockets. Later I said " It's effect on the aircraft's flight will be somewhat like a little retro-rocket"

Now from wiki: "A retrorocket (short for retrograde rocket) is a rocket engine providing thrust opposing the motion of a vehicle, thereby causing it to decelerate. They have mostly been used in spacecraft, with more limited use in short-runway aircraft landing."

"Rockets were fitted to the nose of some models of the DFS 230, a World War II, German Military glider"

And I believe some use was made of retro-rocket on a model of airborne depth charge for a more vertical drop.

Thruster may be a more modern term.

I may not be thinking out the problem but I believe it is this type of effect that we are looking at. Can the firing of the guns on an aircraft cause it to decelerate? as in can the firing of the P-47s guns cause it to drop 20-30mph in speed and if so how long would the firing time need be?

Now to me "worst case" is assuming that ALL of the recoil force (or reverse thrust) goes into slowing the plane down. No allowing for reduced force due to springs or mounts. If the calculations say the plane would only be slowed down 8mph with a 3 second burst then we can say 'yes, the plane would be slowed down' but we can say the 20-30mph figure is either way off or the guns need to be fired for 8-10 seconds. If the calculations say the P-47 would be slowed by 21mph or something close then it is possible and we need to go further into the situation. Do you get full reverse thrust/recoil or is it reduced for one or more reasons.

Right now I don't really care if the 3 second burst slows the P-47 down by 8.1 or 8.7 mph . That kind of precision can later if anybody really cares and if we can get better data.
 
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