Aces High Bulletin Board
General Forums => Aces High General Discussion => Topic started by: Nath-BDP on August 12, 2000, 09:56:00 PM
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I remember seeing a history channel episode a few years back and they had a F86 pilot talking about how firing 6 x .50 cal would slow the plane down 40-50 mph.
Gun recoil effecting aircraft speed is present in AH but it only effects speed minimally, slows aircraft down maybe 5 or 10 mph in my experience.
This should be modeled accuratly even though it will help runners.
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Here's the formula to figure recoil. Yes, HT has this exact calculation and uses it to figure the recoil of the guns here.
Here's how to figure recoil for ANY weapon. If you know where to look, you can also find the amound of kenetic energy of a given round.
[Excerpt from Shooting Times Magazine, September 1991 Article by Lane Pearce "A handful of Fire and Fury!"]
What is recoil energy? Or understanding why guns kick
"There's a big difference between the energy and the momentum of a moving object, as you can see in the following equations:
[a] Kinetic energy = 0.5 x mass x (velocity2)
Momentum = mass x velocity
Because the velocity term in the first equation is squared, velocity influences the bullet's energy much more than momentum.
Further searching led me to Hatcher's Notebook, by Major General Julian Hatcher, a compendium of useful information on firearms-related topics. Two chapters of the book cover Hatcher's observations and theory associated with recoil. He described this complicated phenomena using classical physics and a good measure of empirical calculations based on extensive testing.
Studying Hatcher, I confirmed that for practical purposes, the recoil or "kick" of a firearm can be approximated using the simple equations shown above. For example, let's predict the energy of a S&W Model 29 .44 Magnum revolver firing a 240-grain bullet, propelled by 24 grains of powder, at 1,400 fps [feet per second].
To do this you first convert the weight of the bullet and powder in grains to their equivalent value in terms of mass. In the language of physics, the weight is expressed in pounds-force [lb] and mass is expressed in pounds mass [lbm]. To convert pounds-force to pounds-mass, you devide by the gravitational constant, 32.174. (This conversion is necessary to make the use of physics as difficult as possible, much like Mr. Newton must have intended!)
To find the energy of the bullet, devide the weight of the bullet in grains by 7,000 to convert to pounds, then devide that figure by 32.174. The result of these calculations is 0.001066 [lbm], which is the mass of the bullet. That's a tiny number, but when you multiply by the square of the velocity [1,400 X 1,400] and devide that number in half (look back at equation [a] above) you get 1044. This is the muzzle energy in foot-pounds [ft-lbs] of our sample .44 Magnum bullet.
To calculate the recoil energy of the gun, remember, you must first determine the momentum of the bullet and powder charge. Take the tiny number for the mass of the bullet shown above, and add it to the mass of the powder [24 grains/7000/32.174 = 0.0001065 lbm] and multiply the result by 1,400 (see equation ) and you get 1.641. This number is the momentum, in pounds-mass x fps [lbm-fps] of the bullet and powder gasses exiting the barrel. This value can be used to calculate the recoil velocity and the revolver and, in turn, its recoil energy.
Because the momentum of the gun is equal to the momentum of the bullet and powder gasses, we can assume that the momentum of the gun equals 1.641 lbm-fps. Then, using equation again, we can algebraically calculate the gun's recoil velocity.
The weight of the gun, in pounds, must be devided by the gravitational constant to get its mass. The three pound S&W Model 29 masses 0.093243 lbm. Dividing the momentum of the bullet and powder gasses by the mass of the gun produces a recoil velocity of 17.6 fps. Put the recoil velocity and the revolver's mass into the kinetic energy equation and you get 14.44 ft-lbs [of recoil energy].
The exact mathematical equation is:
V=[w + (k x c)] x v
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W
Where:
V= the gun's recoil velocity
w= the bullet weight, in pounds [take the weight in grains and divide by 7,000]
c= the weight of the powder charge, in pounds [again divide the weight in grains by 7,000]
v= the bullet's muzzle velocity
W= the guns weight in pounds [take the weight in ounces and divide by 16]
And finally:
k= an empirically determined constant that approximates the effect of the gasses escaping from the muzze of the firearm just after the bullet exits the barrel. To get this value exactly, you'd have to know the pressure distribution over time in the gun's chamber and barrel, the exit velocity of the gasses, and the effects of gas loss at the barrel/cylinder gap of the specific test revolver and a host of other variables. Again, however, we can approximate this number for any practical application of the equation. Hatcher reported that the value for "k" ranges between 1 and 2 depending on the gun/cartridge interior ballistics. For shotguns and low pressure handguns, 1.25 is a good approximation. For magnum handgun loads, and low-intensity rifle cartridges, e.g., .30-40 Krag, .30-30 WCF, etc... you would use 1.5. For .30-06 and other high-(>45,000 psi) pressure firearms, 1.75 may be used fo predict the recoil velocity and energy."
Flakbait
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Nath I did a test of speed loss due to recoil a while back: http://bbs.hitechcreations.com/smf/Smileys/default/Forum1/HTML/002844.html (http://bbs.hitechcreations.com/smf/Smileys/default/Forum1/HTML/002844.html)
I only looked at one plane, but it came out very close to the correct recoil.
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Yea but how long did it take you to reach the speed at which you stopped deaccelerating and how many rounds did it take? IMHO you should deaccelerate rapidly, no?
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I didn't check the time to decelerate. For me to calculate that would require running a simulation of my own, and I'd have to redo the test and record speed vs. time data. I'm too lazy for that. (http://bbs.hitechcreations.com/smf/Smileys/default/smile.gif)
Keep in mind that the forces of thrust and drag still act on the plane when the guns are firing.
As the plane slows down, the thrust becomes greater and the drag becomes smaller. Recoil does not change though.
So the decceleration caused by the recoil force will become smaller as the plane goes slower. At very low speeds (say 200 mph) the plane will actually accelerate when you are firing. It's only when you are at top speed that the plane will deccelerate quickly.
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OK I'm really bored. (http://bbs.hitechcreations.com/smf/Smileys/default/smile.gif)
I took all my theoretical predictions from the other thread and got a weight for a P-51D with maximum internal fuel from America's Hundred Thousand (10,208 lb). Then I made a simulation (using Simulink) of what would happen if 6 x .50 cal opened fire when the plane was at top speed (355 mph MIL power) at sea level.
(http://www.raf303.org/funked/p51recoil.jpg)
Now go forth and do flight testing. I did a quick check in AH and I got 20 seconds to deccelerate to 325 mph. Seems to match my sim pretty well.
[This message has been edited by funked (edited 08-13-2000).]
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Heh funked.
You rule.
But how'd you do that sumulation?
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StSanta
JG54 "Grünherz"
"If you died a stones throw from your wingie; you did no wrong". - Hangtime
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I can toss in this little anecdotal piece for you.
My dad flew a B-25H over to New Guinea. It went to another squadron, not his, so he didn't fly it in combat. However, the other pilot (his best friend, who sadly did not return) on the flight over went to that squadron with that aircraft.
Charley Knight told my dad that every time you tripped the trigger on the 75mm in the nose it was an almost instant loss of 10 MPH. BOOM/-10MPH.
FWIW. (http://bbs.hitechcreations.com/smf/Smileys/default/wink.gif)
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Isn't physics cool, use same physics get the same results.
HiTech
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Originally posted by hitech:
Isn't physics cool, use same physics get the same results.
HiTech
It's not just cool, it's the law. And it's not subject to appeal.
(http://bbs.hitechcreations.com/smf/Smileys/default/smile.gif)
- Jig
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Santa I used a simulation package called Simulink. It is part of a mathematics package called Matlab. Simulink lets you graphically build block diagrams for simulations and sends the results to the Matlab workspace. I also use an add-on package that lets me interface with equipment and sensors in real time. So I can simulate controlling a robot or hydraulic power system, design a good controller, and then hit a switch and operate the real system with the controller I built in the simulation.
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funked
I have a Student's Edition of Mathlab, but quite frankly haven't used it too much (other than for solving equations (http://bbs.hitechcreations.com/smf/Smileys/default/biggrin.gif))
You an engineer by any chance?
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StSanta
JG54 "Grünherz"
"If you died a stones throw from your wingie; you did no wrong". - Hangtime
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Santa yeah I'm an enginerd. The student edition doesn't have Simulink AFAIK. I think there are still ways to run simulations in Matlab without Simulink but you can only do linear systems (which airplanes aren't). (http://bbs.hitechcreations.com/smf/Smileys/default/biggrin.gif)
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Reading about the "Flying Tigers" apparenlty in the P-40E you could fire the guns on one side of the plane (I guess don't charge them, individual charging handles?)
Anyhow one of the Tigers was bombing the Salween River Bridge or Japanes units near there and was having difficulty pulling out of the dive. He fire the guns on one side of the P-40 and that yawed the plane enough to allow him to pull out of the dive.
Also In Erik Shilling's Book "Destiny, A Flying Tigers Rendevous with Fate." He reccounts test firing the guns on an E after lift off in India. Only the guns on 1 side were loaded, he was low firing into the indian ocean and very nearly went into a flat spin at low altitude over water. (Just quickly test firing the guns.)
The A-10 can stall firing it's vulcan cannon also.
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(http://www.tir.com/~lkbrown1/dtahcard.gif)
"Downtown" Lincoln Brown.
lkbrown1@tir.com
http://www.tir.com/~lkbrown1 (http://www.tir.com/~lkbrown1)
Wrecking Crews "Drag and Die Guy"
Hals und beinbruch!
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I'd believe that, except the A-10 doesn't have a Vulcan.
If you've ever seen an AH-1 with a Vulcan, you'll notice 3 barrels are removed for that same reason - recoil can exceed thrust!
[This message has been edited by funked (edited 08-14-2000).]
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Also I remember a story of a Spitfire pilot thats right Hispano jammed and when he fired the left one his aircraft yawed to the left excessivly, this isn't modeled.
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Well there is definitely some rocking from the recoil. You can see this if you look at the plane on external view when firing from the ground or the air. Recoil from a single Hispano would be about 250 lbs.
HT can you tell us if the recoil forces are input at the locations of the guns on the airframe, or are they lumped together and subtracted from the thrust?
[This message has been edited by funked (edited 08-14-2000).]
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Of all the statements being made in this thread, the silliest is the idea that the A-10 can stall when firing the gun...unless the person was talking about the engine disruptions early on due to gun gas ingestion.
Good grief, people, this is only a game!
Andy
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Game??? What exactly are you talking about?!? (http://bbs.hitechcreations.com/smf/Smileys/default/wink.gif)
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HT can you tell us if the recoil forces are input at the locations of the guns on the airframe, or are they lumped together and subtracted from the thrust?
Yes I can tell you.
HiTech
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<SPITS COKE ALL OVER THE MONITOR>
LOL!!!!!