Handgun physics question


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Hagios
March 3, 2011, 04:55 PM
Ok, I'm confused. My not-so-uber knowledge of physics suggests that recoil should be much more violent than it really is.

Let's suppose you are shoot a 45 ACP round. That means the bullet weighs about 15 grams and goes about 288 meters per second.

Kinetic energy is 1/2 (mass)(velocity)^2.

So the total kinetic energy is 1/2 (.015kg)(288)^2 = 622 Joules

For every action there is an equal and opposite reaction. Let's assume that *all* of that energy goes into recoil (is this where I'm going wrong). In the real world some of that will go into the slide and perhaps elsewhere. And let's suppose that the shooter weighs 80 kilos (a normal sized guy).

Then we use the formula for kinetic energy and solve for velocity.

1/2 (80)v^2 = 622

velocity = 3.94 mph

That doesn't make any sense. The recoil of a gun makes your hand go back at about that speed, not your whole body. If you were shooting on ice skates you'd get launched backwards like how the guy who gets hit by a shotgun in the movies gets launched.

What am I doing wrong? Or do handguns really generate that much recoil, and it's just that you are braced and gets lost in the friction on the ground?

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Guiding101
March 3, 2011, 05:04 PM
your math is correct....however it does not account for your bodies reaction to forces impossed upon it. Your joints work as a cushion, a do your muscles, tendons, and skeletal structure. When punched in the stomach.....which by a typical adult male will result in around 300-400 psi of energy, your body simply gives and absorbs the force and distributes it to surrounding structure.
Also the weight of the gun, spring and if in an auto, the slide absorb some of the impact.......and yes if you were to firmly mount a gun to a sled on ice and fire, the sled would move backwards......the degree of wich is dependent on the recoil of any said firearm.

Manco
March 3, 2011, 05:10 PM
Try this again using momentum (mass*velocity) rather than kinetic energy.

velocity = 3.94 mph

And meters/second rather than mph. ;)

J-Bar
March 3, 2011, 05:17 PM
Momentum is conserved, but kinetic energy is not. The firearm's momentum (mass x velocity) equals the mass x velocity of the bullet and gases/powder moving forward. Kinetic energy is a function of velocity squared, and the bullet has a lot more of that than the recoiling gun.

J-Bar
March 3, 2011, 05:18 PM
Manco beat me to the draw!

glockcompact
March 3, 2011, 06:42 PM
First time poster here. I'm poking my nose in and i'm always interested in physics. Shouldn't this also have "time" involved in the equation. The slide functions which adds time. This is one reason why a givin revolver weighing the same and having the same muzzle energy as a semi would have more recoil. Not including ergonomics. To me a slide effectively disperses the force over a longer amount of time which gives less felt recoil. If it doubles the amount of time it would be considerable. Am I wrong with this?

Caliper_RWVA
March 3, 2011, 09:51 PM
Semi-autos spread recoil over a longer time, which does reduce *perceived* recoil, but the same recoil impulse is present in a revolver and a semi-auto of the same cartridge.

Think of it this way - a rubber recoil pad reduces perceived recoil by spreading the recoil over a longer time period, but the same recoil is present.

For the OP: look up elastic collision and inelastic collision on Wikipedia. The same principles apply in reverse when firing the gun.

hso
March 3, 2011, 10:05 PM
The change in momentum over time due to an applied force is known as impulse. Factor in that the springs and moving slide of a semi auto are absorbing and redirecting some of the force. Consider that the 80kilo shooter isn't on a frictionless surface and mechanically redirects more of the recoil into the ground along with all that and you can see that qualitatively you don't have to worry about being flung back.

ArthurDent
March 4, 2011, 03:52 AM
Yes to the replies, above. Let me add some more. I've been trying to read-up on the physics. Here's some of what I've understood so far:

Momentum is conserved. Pick an inertial reference frame. The sum of all (mass * velocity) adds up the same before and after the shot. The part that you may not have considered is the mass of the burned powder! One figure I've seen says that about 4000 ft/sec is a good guess for rifles, regardless of the muzzle velocity of the round. However, this figure has to vary with barrel length, so handguns should have a lower number. Of course this depends on how fast the powder burns...

Total energy is conserved, but mechanical energy is NOT conserved. Again, from various sources, I've read that a firearm is only about 1/3 efficient. Figure out how much burning energy is contained in the powder, and you've only got about 1/3 of that left to propel the projectile. The rest goes into heat and muzzle blast and sound and...

And, as stated above, perceived recoil is not necessarily related to either of these numbers, because of bunches of factors, including the shape of the grips, the time distribution of the kick, etc.

As nearly as I can tell, there is no end to how far you can pursue knowledge about firearms. Every answer creates 3 new questions. Welcome to the confusion! :)

Laserslug
March 4, 2011, 05:01 AM
The conservation of energy includes kinetic energy plus work plus heat. That makes the calculation too difficult for tonight because of the energy lost heating up the barrel and the flames blowing out into the air. Momentum is easier to calculate. The conservation of momentum says the mv of the bullet and gasses is equal and opposite the mv of the gun if it is held very loosely.

bullet momentum:
mv = (.015kg)(288m/s) = 4.32 kg m/s
(English units of momentum is 31 pound seconds)

gasses momentum: gunpowder weighs more than the gasses. 15.4 grains per gram. Use .5 grams of gunpowder.
mv = *(.0005kg)(288m/s) = .144 kg m/s
total .144 + 4.320 is less than 4.464 kg m/s for bullet plus gunpowder gasses.

gun momentum = 4.464 kg m/s
gun mass 1 kg
gun velocity = momentum/mass = 4.464 m/s = 14.6 ft/second
That is 10 miles per hour gun recoil speed. Enjoy!

J-Bar
March 4, 2011, 07:29 AM
One picky point about "gunpowder weighs more than the gasses"...no.

The mass of the gasses pushing the bullet down the barrel is the same as the mass of the powder before ignition, assuming complete burning. The material has changed from solid to gas, but the number of atoms in there is the same...it is a completely closed system.

Carl N. Brown
March 4, 2011, 08:10 AM
On gun powder, you must subtract the mass not expelled , but left in the barrel as unburned or as solid residue, which is considerable in black powder, Wolf or Walmart ammo. :)

Laserslug
March 4, 2011, 08:56 AM
Energy efficiency estimate of 26% is calculated next...

The gunpowder burns in oxygen to make gasses, but it leaves some dirty stuff in the barrel. So if .5 grams of gunpowder burns, oxygen adds weight, but residue is lost in the barrel. I guess that the expelled gasses weigh less than .5 grams. I could be wrong. Maybe the added oxygen results in .9 grams of gasses. Good point, J-Bar.

Estimate the energy efficiency ratio of bullet kinetic energy divided by chemical energy of gunpowder:

bullet kinetic energy is 1/2 (.015kg)(288)^2 = 622 Joules
Gun weighs 35 ounces on Earth. That is 1 kg anywhere else. From the momentum calculation, the gun recoils at 10 mph. or 4.464 meters per second.
gun kinetic energy is 1/2 (1 kg)(4.464m/s)^2 = 10 Joules

gunpowder energy 470,000 joules per mole
potassium nitrate is KNO3, similar to black gunpowder
K 39 g/mole
N 14
O3 16*3 = 48 grams per mole
(A mole is 602210000000000000000000 molecules)
KNO3 is 101 grams/mole
gunpowder mass is assumed to be .5 grams
.5g/(101g/mole) = .005 mole
energy is .005 mole* 470,000 J/mole = 2350 J

gun efficiency 622 J / 2350 J = 26%

txhoghunter
March 4, 2011, 08:59 AM
Others have probably said it, but Energy is also given off as heat from friction.

And it is momentum that is conserved between the shooter and the gun as opposed to KE

benEzra
March 4, 2011, 09:14 AM
The gunpowder burns in oxygen to make gasses
That only occurs when the hot gases produced by gunpowder deflagration meet the external atmosphere, i.e. when they exit the muzzle.

Modern powders are primarily nitrocellulose, plus nitroglycerin in double-base powders, as I recall. The combustion of nitrocellulose is approximately as follows (approximate because it's a polymer):

C24H30(NO2)10O20 → 12C02+ 12C0 +4H20 + 11H2 + 5N2
nitrocellulose
(approximation)

Note that atmospheric oxygen is not involved at all; all of the oxygen involved comes from the nitrocellulose molecule itself. The same is true of the nitroglycerin in double-base powders. It doesn't matter if the cartridge case started with traces of air, inert fill gas, or hard vacuum.

Notice that the combustion products include quite a bit of hydrogen and carbon monoxide, and they are quite hot. When 1000-degree hydrogen and CO slam into fresh atmospheric oxygen at two or three thousand mph, they burn almost instantly, creating the muzzle flash as they flash-burn into H2O and CO2. But the products doesn't contribute to the gun's recoil momentum, because practically all of this combustion occurs outside the barrel.

As to the OP, your mistake is using kinetic energy instead of momentum, as others have pointed out. The bullet carries far more kinetic energy than the recoiling gun does; guns are essentially directed-energy weapons, after all. Only the momentum is equal in the forward and reverse directions.

CraigC
March 4, 2011, 11:17 AM
Do we not also have to take into account that the bullet is accelerating down the bore? That it does not go from zero to muzzle velocity instantly?

DPris
March 4, 2011, 11:22 AM
You might.
I just shoot the things. :D
Denis

J-Bar
March 4, 2011, 11:27 AM
The momentum of the recoiling gun increases as the momentum of the bullet and gasses and any unburned powder increases. That's what "conservation" means; the opposing momenta balance each other out during the whole process. Recoil begins the instant the bullet starts to move.

CraigC
March 4, 2011, 11:59 AM
I just shoot the things.
Me too!

papa_bear
March 4, 2011, 12:11 PM
my pocket protector doubles as a holster :neener:

1911Tuner
March 4, 2011, 10:06 PM
Do we not also have to take into account that the bullet is accelerating down the bore? That it does not go from zero to muzzle velocity instantly?

Absolutely. Depending on the powder burn rate and the barrel length, the bullet may even be moving faster before it leaves the muzzle than it does at exit.

Peak pressure...and therefore peak force...occurs early on in the process. Even using slow rifle powders, it's usually within 2-4 inches of bullet travel. With fast pistol powders, peak pressure/force can occur before the bullet has completely passed from the chamber leade into the rifling. With really fast numbers like Bullseye...even before the bullet base has cleared the case mouth.

Recoil is acceleration of the gun, and can only occur while the bullet is still present, and force is on the system. Once the bullet has exited, the force is gone, and the gun can't accelerate any more, other than from the residual gas and particulate ejecta that's behind the bullet. Even if we assume that a quick handgun powder blows gas at 2,000 fps...which I doubt...5 grains of mass at 2,000 fps doesn't create a lot of recoil impulse...roughly 1/4th that of a standard velocity .22 Short.

Actual recoil from the ballistic event is over by the time our brains can even process the fact that it's occurred. What we perceive as recoil is mostly momentum after the event...and in autopistols...well after the event. With no solid connection beween the gun and the gun mount, most of the muzzle flip comes from the slide impacting the frame. By the time that happens, the bullet is about 20 yards downrange, assuming a 230 grain bullet at 800 fps.

gbran
March 4, 2011, 10:30 PM
I'm not sure mathmatically how to factor in the type and weight of the platform and the effect it has on recoil. We all know the difference in recoil between 13 ounce and 50+ ounce .357 mag revolvers.

smokey262
March 5, 2011, 12:24 AM
You guys are good. If I would have had you as math teachers and used guns in the examples all those years ago maybe I would have retained more.

DoubleTapDrew
March 5, 2011, 12:36 AM
Hmm this all seems more complex than the typical "if the bullet will knock down the shootee it would knock down the shooter, equal and opposite reaction" response

jim243
March 5, 2011, 01:41 AM
Let me see, a 45 ACP bullet weighs 230 grains, and you weigh HOW MUCH??

Jim

ArthurDent
March 5, 2011, 02:33 AM
Just buy this book, and then spend many fun hours pondering:

Understanding Firearm Ballistics, by Robert Rinker (http://www.amazon.com/Understanding-Firearm-Ballistics-Robert-Rinker/dp/0964559854/)

:D

(Note: if you have a real math brain this may be painful to read, as the author works very hard to present things in a non-mathematical, high-school-or-below level. However, the math is there, and information seems to be quite accurate. There are many tables addressing this specific question.)

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