# Physics of automatic weapons, mass of 5.56?

Suppose a man jumps out of a tall building. Somehow, he has braced two M-16 A2 automatic rifles under his arms, so that he can fire both continuously straight down.

According to this site the rifles can fire 90 rounds/minute each. They have a muzzle velocity of 3,100 feet/second. I don’t know the mass of the rounds, but they are 5.56 mm.

Suppose the man weighs 160 pounds.

How much will the rifles slow his fall? (And also, how much do 5.56 mm rounds weigh?)

Thanks.

I just looked up the weight for the bullet and 77 grains is a common version. That converts to 4.99 grams. Also, we need to know how tall the building is.

There would also be some thrust produced by the venting of the gases that drove the bullet.

If you want to be accurate, you also have to consider the hot gases that come out of the muzzle at high speed after the bullet has exited the barrel. The barrel can be considered a rocket, with an exhaust of bullets and hot gas.

Further, assume this man is Yosemite Sam.

A somewhat trivial point but 77 grains is extremely heavy for a .22 caliber bullet. caliber. 68-69 grains is more typical for a match hollowpoint and 55 of 62 grains is normal for military ammunition.

You are probably right about that but I picked the heaviest one because I was trying to maximize “thrust” although it didn’t really look at muzzle velocities because the OP gave it to us.

If you ignore gravity…
Man: 160 Lb = 72 575 grams, v[sub]terminal[/sub] = 100 mph = 44.7 meters per second.
E = 1/2 m v[sup]2[/sup] = 72,500,000 gm[sup]2[/sup]/sec[sup]2[/sup]

Bullet: 5 gram, 3100 feet/s = 945 meters per second.
E = 2,200,000 gm[sup]2[/sup]/sec[sup]2[/sup]
72,500,000/2,200,000 = 33
It’d take 33 bullets to stop the man. At 180 rounds per minute, that’d take about 11 secs of continuous fire. If all the combustion products went out the muzzle, it’d go a bit quicker, but IRL side venting would probably give you a wicked spin.

That 11 seconds needed to stop the guy from 100mph (sans gravity) corresponds to an acceleration of 4 meters per second[sup]2[/sup] (Final velocity = intial velocity - acceleration * time).
Sadly, that’s less than the 9.8 meters per second[sup]2[/sup] acceleration due to gravity, so the guy would still hit the ground hard. Now if he had 5 guns, things would be different.

or Dante from Devil May Cry who could also stop fallling by shooting his guns downward.

Naw, he just needs bigger guns. But don’t forget to add the mass of the guns into the equation.

…Maybe the back exhaust from a couple of shoulder fired missiles fired straight up would do the trick. Although that is technically recoiless…hmmm…Dunno.

So 4m/s[super]2[/super] is a pretty good number, eh? That isn’t too bad; he might be able to fall an extra story – so a 3-story fall might be like a 2-story fall, eh?

You’ve misundertsood the FM.

I believe you want to know about a person falling and holding down a trigger on his M16 or M4 machine gun. The cyclic rate of an M16/M4 is about 800 rounds per minute not 90. The capacity of the largest magazine is 100 rounds.
So the guy in the scenario has 200 rounds that he can fire at 1600 rounds per minute.

As far as math goes… you guys work that out. I just want to provide better information.

You need to look at inertia, not KE. If KE translated into recoil then no one could fire an M16 on auto. Anyone who has fired one knows they have extremely mild recoil.

If we look at inertia we’re being a bit more realistic. Let’s use a 62 grain bullet at 3,000fps. 62 grains is about 1/18,000 the mass of the 160lb human (I’m rounding just a bit here) 3,000fps is 2,045mph or 20.45 times the benchmark 100mph fall. If we want to stop a human going 100mph you’ll need to fire 880 rounds…all at once. That’s almost thirty military issue magazines.

If you want to take gravity into account, things get a little more complicated. Warning: semi-complicated math ahead.

This is essentially the rocket equation in a constant gravitational field. (Unfortunately, the equations on that page are broken, but I couldn’t find anything better on the web; I’m following my old intermediate mechanics book.) The end result is that the guy’s velocity as a function of time is given by

v(t) = -gt - u ln(1 - q t/m[sub]0[/sub])

where g = 9.8 m/sec[sup]2[/sup] is the acceleration due to gravity, u is the muzzle velocity of the bullets (in m/s), q is the rate at which mass is ejected (in kg/s), and m[sub]0[/sub] is the initial mass of the guy, his guns, and the bullets he has with him when he steps off the building.

Now suppose we want the guy to be able to step off the building, fire his guns straight downwards, and reach the ground with exactly zero velocity after some amount of time T. This means that the total mass of the bullets he’ll be firing is q T, and the mass of him and his guns is M = m[sub]0[/sub] - q T. If we want v(T) = 0, then we can jigger around the equation above to find that

v(T) = 0 = - g T + u ln (1 + q T / M)

or g T = u ln(1 + q T / M). If we plot the two curves y = g t and y = u ln(1 + q t / M), we find that they intersect at a non-zero value of T if and only if q u / M > g, or alternately, M < q u / g.

So what are the numbers? If the guns can really fire 800 rounds/minute, at 5 gms/round, and the guy has two guns, then q = 0.133 kg/s; u = 3100 ft / s = 940 m/s; and so our condition is that M < 12.8 kg, or 28.4 lbs. So unless the guy is really skinny, it won’t work (and don’t forget that this weight limit includes the mass of the rifles.)

I’m not sure how that equation was derived, but I think we need to make sure that two items are accounted for. First, the mass of the gunpowder (approximately 25 grains) can be added to the mass of the bullet, since this is also expelled downward. Second, the change in mass of the falling object isn’t just the mass of the bullets and gunpowder, but also of the brass cartridge casing which is not expelled downward but sideways at a relatively slow velocity (that’s probably not going to change the answer very much, but it is a valid tweak).

Finally, as a practical matter, the guy has to start worrying about ricochets as he gets closer to the ground. No one wants to fire full auto straight into a sidewalk at point blank range. He may want to stop falling (and firing) at say, 10 ft above the ground. Once he stops firing, he’ll start to fall again, but now it’s only a 10 ft fall.

It’s in chapter 3 (I think) of Classical Dynamics of Particles and Systems, by Marion & Thornton, if you want to look it up. Basically you assume that the rate of change of momentum of the entire system is equal to the gravitational force on it, then find a differential equation for the velocity as a function of time.

If the combustion products are ejected at the same speed as the bullet, then you can just add the mass of the gunpowder to the mass of the bullet and perform the calculations again. If not, then things get more complicated.

No, probably not, but I’ll look at it anyways if I have a free moment this afternoon. What’s the mass of a casing, roughly speaking?

6.4 grams, according to this site.

Sounds like the OP has been listening to too much Steppenwolf:

If you’re going to sweat casing weight, you also need to take wind resistance into account. This’ll make things much more complicated.