Calculations for recoil and KE

[flakbait]

I've postd this several times, either here or on the WB board. When you figure the recoil for a given weapon, you must know the kenetic energy value of the projectile. That KE calculation is included below, along with how to figure the recoil for just about ANY gun.
Just follow the steps, try to keep up, and have your algebra degree at the ready


[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
   ---------------------
            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."

If you need this information again, tell me and I'll repost it.

Flakbait

[-towd_]

FEAR MY BANG STICK !!!

[bloom25]

I'd say you are probably right about the momentum vs. KE statement you made.  Notice that momentum is the derivative of KE.  I.e. the momentum is the rate of change with respect to time of KE.  (BTW: Pounds mass is not the proper Imperial system term for mass.  Although the calculation is correct, the name for mass is the "slug"    Funny but true! )

One question for you though flakbait, do you think something is wrong with the game, or are you just posting this to educate us?  

 

------------------
bloom25
THUNDERBIRDS

[miko2d]

 First, the calculations are only complicated by fancy conversions if you are not using the  Metric System. Being an american citizen I am proudly using the archaic measures, but I am not blaming anyone, least of all Sir Newton for that.

 Second, why do we care about energy when it comes to recoil? I think that momentum is all we need to consider.

1. The plane and the bullets start at rest relatively to each other.
2. The plane and the bullets get pushed away from each other. They obtain the same momentum in different directions. Since the momentum is mass multiplied by velocity, the ratio of velocities is inversely proportiomal to the ratio of the masses.

M_plane x V_plane = M_bullets x V_bullets
V_plane = M_bullets x V_bullets / M_plane

Example:
 In a second the guns throw 1 pound of led at the velocity of 1600 feet/sec. The momentum is 1600 pounds feet/sec
 The plane receives the same momentum in the opposite direction.
 If it weights 10,000 pounds, its velocity will be 1600 / 10,000 = 0.16 feet/sec or 0.109 mils/hour.

 I disregard the gasses escaping the barrel and reduction in the weight with the ammo being expended. Also, as teh plane slows down the propeller becomes more efficient. It should not matter much if the shooting time is one or two seconds.

 So we could start a plane, run it at autopilot untill the speed stabilizes, shoot for two seconds, measure how much the speed dropped, calculate the same drop using the real weight of the plane, bullets and their velocity. Then we will see if the recoil modelling is correct.

 I have a feeling that it is weaker then it should be. I once was stupid enough to shoot a 50 cal (actually 12.5mm) tank-mounted MG holding it in my hands and you would not believe how far it threw me    
 If I am correct, then a plane shooting somebody's 6 with long bursts does not loose enough speed. to affect it's pursuit. That coupled with no ill effect on the barrel promotes spray and pray.

 Anyone cares to verify that with the real numbers?

miko--

[This message has been edited by miko2d (edited 04-18-2000).]

[flakbait]

I posted that information as a courtesy. Several people believe that there's not enough recoil present in the game, so I figured I'd post HOW to figure it out. There's also a LOT of talk regarding the kenetic energy a given projectile has. That's included in the recoil calculation, which is the second reason I posted it. Using those steps, you can not only tell which gun recoils with a given amount of force, but you also know the KE of the round. This KE figure is given as ft-lbs [foot-pounds]. Which, for the uneducated among us, means one pound of weight moved with one foot of distance.

HT has used that very calculation I posted to figure the recoil of the weapons in AH. Evidently he was using the momentum calculation twice, since he didn't have the energy equation. As he put it "I didn't know the energy calculation so I never got the recoil right. I'd fly along and fire the guns, and I'd suddenly be flying backwards at near 300 mph!". Scary ain't it?


Flakbait

[funked]

Miko you are correct that momentum is the simplest way to get an answer.  Energy is not very good, because the explosion of the gun powder adds energy to the system, really screwing up any energy balance one might attempt!

However You can't just subtract from the plane's velocity like that.  There are these things called Thrust and Drag.    

The engine thrust will remain roughly constant as you slow down (slight increase).  However the drag will decrease.  So for sustained (steady-state) firing, the new speed will be the speed where Thrust = Drag + Recoil.

The speed reduction found this way is far less than what you would find with your equation.

Also the momentum of the gases is NOT negligible.

[This message has been edited by funked (edited 04-19-2000).]

[funked]

I went and did the calcs.  If anything, Aces High has too much recoil.  
 http://bbs.hitechcreations.com/smf/Smileys/default/Forum1/HTML/002844.html