# How can I fire a pistol without breaking my hand?

If I fire a bullet from a pistol, and it has enough force to break your ribs, why doesn’t the recoil break my hand? Newton’s Third Law, and all that.

(a) the recoil force is spread out over your whole hand, not focused into a bullet-diameter area
(b) the weight of the gun absorbs some of the force
© depending on the gun, the recoil may also be used to eject and feed the next cartridge
(d) the barrel is above your forearm generally, so it tends to rotate upwards, dispersing a bit more energy
(e) someone who know more about guns will be along shortly

I had a real nice, witty, answer that explained all of these. Then the SDMB ate it as it so often does, and now I’m just too ticked to go into it again. I’ll stick with your summary.

–Patch

Mostly, just conservation of momentum.

m[sub]bullet[/sub]v[sub]bullet[/sub] = m[sub]gun[/sub]v[sub]gun[/sub].

While v[sub]bullet[/sub] is high, m[sub]gun[/sub] is maybe 2 orders of magnitude higher than m[sub]bullet[/sub]. I’m sure someone who knows guns can actually look up real numbers that will give you v[sub]gun[/sub].

Same reason a baseball bat hitting a ball doesn’t break your hand, even though a solidly hit baseball can crack your skull open. The action <> reaction energy on the delivering (bat) end is spread out across a much more massive object.

Think about ahow knife works where the cutting/cleaving action occurs becasue the applied force is directed into a very small area. Also in a gun, a chunk of energy from the powder explosion is vented out of the gun, or used to do other work like ejecting cartridges or advancing and re-setting the gun mechanism in some fashion.

True. But even if the gun’s recoil were concentrated to the cross-sectional area of the bullet, the recoil wouldn’t be too bad.

I understand what you’re getting at here. But strictly speaking, the weight of the gun doesn’t “absorb” anything. Instead, the mass of the gun (and the mass of the person holding it) does play a part in the conservation of momentum equation. This means that, because the gun has quite a bit more mass when compared to the bullet, the velocity imparted on the gun will be low.

I’m not sure you’d like the gun to recoil at with that force while pressing against a point only the size of the point of the buwette. That has to be about a hundred times bigger than the point of the bullet, thereby decreasing the lb/sq. in. by about the same hundred times. That’s quite a difference.

I’m going to have to go ahead and disagree with you there. The size of the backstrap of the gun is an important consideration.

Another thing, which I don’t think anyone has touched on, is that when you shoot the gun will roll a bit in your hand, so that the impulse force of the recoil is reduced. Sort of like the difference between jumping down three stairs and landing with locked knees vs. rolling with the landing a bit.

A Baretta 9mm weighs 1.16kg (1160g), and the round weighs about 8g. Muzzle velocity is 365 m/sec. Let’s make believe ALL of the recoil is transferred to the shooter, and not used by the slide and spring to cycle the weapon. Using SmackFu’s equation,

8g x 365m/sec = 1160g x v[sub]gun[/sub]

v[sub]gun[/sub] = 2.5m/sec, or about 5.6 mph. Not that fast.

Actually, it’s not quite as simple as just calculating the velocity of the gun. 1 kg at 10 m/s is going to hurt you just as much as 10g at 1000 m/s.

The real question is, where does the energy go? If you get hit by a bullet, most of the energy goes into a shockwave which does nasty things to your ribs or your internal organs, depending on where it hits.

If you’re holding a gun properly, most of the energy goes into moving you backwards. Firing a heavy handgun this is very noticable. Firing a light handgun your arm will be forced upward, but otherwise you’ll hardly notice.

On the other hand, if you hold a handgun really loosely, you can break your hand with the recoil.

If you care to experiment, hold a handgun lightly against your chest and fire. It will hurt. It probably won’t break ribs, because there’s more surface area to absorb the shock (ie, slow the energy transfer).

It could be useful to consider energy as well as momentum.

Let mV be the bullet’s momentum (small mass, large velocity). Mv represents is the gun/hand. If momentum is conserved, then mV = Mv.

Let’s say the weight of the gun plus your hand is around 3 lbs, and the bullet weighs half an ounce. That a ratio of about 100:1.

If M = 100 * m, it follows that V = 100 * v.

But energy is proportional to mass times velocity squared. So the ratio of the bullet’s energy to the gun/hand’s energy is given by:
mVV/Mvv

Substituting, we get
mvv100100/m100v*v
which is easily reduced to 100
In other words, the moving bullet carries 100 times as much energy as was transferred to your hand, arm and body when you pulled the trigger. It will lose some of that as it plows through the air, but there’s more than enough left to do a lot more damage to the target than was done to you.

Hang on… doesn’t that break Newton’s 3rd law? “For every action, there is an equal and opposite reaction.”

Seems to me that according to ol’ Isaac’s law, the bullet leaves the barrel with EXACTLY the same amount of energy that went the opposite direction as recoil.

The main thing is that bullets are say… 124 grains for a 9mm, which is 0.28 ounces. My Ruger P95 is 28 ounces, so the mass is 100 times more than the bullet.

I’m not quite up on my physics calculations, but that leads me to believe that the pistol will only go back something like 1/10 as fast as the bullet leaves the barrel. Add the considerably larger surface area to spread this force over, and you have something that’s relatively non-violent compared to the bullet going 10 times faster, but weighing 100 times less, and impacting an area 9mm in diameter.

No, Newton’s third law refers to force. It covers momentum too, because the equal and opposite forces obviously act for the same time and momentum = force * time. This is the justification for the momentum conservation part of my argument.

Conservation of energy is a valid concept as well, but there’s no “equal and opposite” at work there. The chemical energy released by burning the powder in the cartridge was converted into the kinetic energy of the bullet’s motion. Some of this went to heat the barrel and some was transferred to motion of the air through which the bullet flies. Eventually, the remainder will be dissipated as the bullet strikes a target. Taking everything into account, there’s no net gain or loss of energy - just conversion.

No, that’s incorrect. The energy of the bullet is much greater than the energy the gun receives in recoil.

Momentum of gun = m[sub]gun[/sub] v[sub]gun[/sub]
Momentum of bullet = m[sub]bullet[/sub] v[sub]bullet[/sub]
Kinetic energy of gun = E[sub]gun[/sub] = 0.5 m[sub]gun[/sub] v[sub]gun[/sub][sup]2[/sup]
Kinetic energy of bullet = E[sub]bullet[/sub] = 0.5 m[sub]bullet[/sub] v[sub]bullet[/sub][sup]2[/sup]

As a starting point, you must assume the momentum of the gun is the same as the bullet’s momentum. Thus

m[sub]gun[/sub] v[sub]gun[/sub] = m[sub]bullet[/sub] v[sub]bullet[/sub]

solving for m[sub]gun[/sub]:

m[sub]gun[/sub] = m[sub]bullet[/sub] v[sub]bullet[/sub] / v[sub]gun[/sub]

Plugging m[sub]gun[/sub] into energy of gun equation:

E[sub]gun[/sub] = 0.5 m[sub]bullet[/sub] v[sub]bullet[/sub] v[sub]gun[/sub]

From the energy of bullet equation we know that:

m[sub]bullet[/sub] = 2 E[sub]bullet[/sub] / v[sub]bullet[/sub][sup]2[/sup]

Plugging this into the (new) Energy of gun equation gives us:

E[sub]gun[/sub] / E[sub]bullet[/sub] = v[sub]gun[/sub] / v[sub]bullet[/sub]

From the original conservation of momentum equation we know that

v[sub]gun[/sub] / v[sub]bullet[/sub] = m[sub]bullet[/sub] / m[sub]gun[/sub]

Thus

E[sub]gun[/sub] / E[sub]bullet[/sub] = m[sub]bullet[/sub] / m[sub]gun[/sub]

Let’s plug in some numbers.

Let’s say I’m shooting an SA58 FAL. The caliber is NATO 7.62 x 51 (.308 WIN). The rifle weighs approximately 9.5 lbs with full magazine. The bullet weight is 147 grains. There are 7000 grains to a pound, thus the bullet weighs 0.021 lbs. (The weight ratio is equivalent to the mass ratio.)

The gun is 452.4 times more massive than the bullet. The kinetic energy the gun receives is:

E[sub]gun[/sub] = E[sub]bullet[/sub] m[sub]bullet[/sub] / m[sub]gun[/sub]
E[sub]gun[/sub] = E[sub]bullet[/sub] / 452.4
E[sub]gun[/sub] = 0.22% E[sub]bullet[/sub]

In the above example, the kinetic energy the gun receives is less than one quarter of one percent of the bullet’s kinetic energy.

Per Fackler & associates, this is not so. While there is a shockwave, it isn’t the major wounding mechanism. It isn’t even the most important secondary wounding mechanism.

The single most important wounding mechanism is the crushed & torn perminant wound cavity, created as the bullet plows through the body. The second most important wounding mechanism is the perminant wound cavities created by bullet fragments and secondary projectiles (bone splinters) as they scatter through the body. The third wounding mechanism in rank of importances is sepsis. Lastly, and finally, comes the temporary wound cavity (shock wave), which in most cases does no dispruption of tissue at all, merely causing the rather elastic tissues of the body to stretch aside briefly. In some cases, where the temporary wound cavity includes more solid organ tissue, such as a kidney or the liver, the temporary cavity might stretch the organ beyond its elastic limit, rupturing the organ, but that’s a low-frequency event.

And yes, I know Cecil did a column on this. He also got it flat wrong, sad to say. Even the Perfect Master must miss on occasion, if only to provide the exception that proves the rule.

Not quite. A locomotive might weigh 200 tonnes, but it if bumps into you at 1 cm/second, you suffer no damage. By your math, you should also be able to handle a 10kg object hitting you at 720 km/hour just as easily.

I dunno, being nudged by a locomotive sounds a lot easier than trying to catch an artillery shell.

The amount of force a movie object delivers is geometric to its velocity, using the formula: F = mv[sup]2[/sup]

The 1kg object at 10 m/s:

F = 1 x 10[sup]2[/sup] = 100 Newtons

The 10g object at 1000 m/s:

F = 0.01 x 1000[sup]2[/sup] = 10,000 Newtons

You forgot the squared velocity part.

Tranquilis: You are 100% correct. I wish I had a nickel every time someone at the firing range starting babbling on about “hydrostatic shock” and “energy dumping.” Both are bullocks, for the most part.

Um, if you are flat against a wall and a locomotive hits you at 1cm/s, I’m pretty sure you will suffer damage.

Crafter_Man:
Well, “energy dumping” might be considered valid, if you take it from this point: Kinetic energy converted to work in the body per unit distance of penetration.

In other words: The amount of energy used to tear-up the internals of the victim, and how much of that shredding is done in each inch of wound depth.

There are better, and less deceptive ways of discussing this behavior, though, than “energy dumping.” I personally like the phrase “wounding capability.” That encompasses the effects of a bullet without making silly non-scientific claims.

Dr. Evil: Ready the unnecessarily slow-moving train mechanism!