LAST EDITED ON Nov-20-12 AT 10:10 PM (EST)
 
>The main gun of the Warthog (and its Russian counterpart)
>showed up on XKCD rather unexpectedly...

By a curious coincidence, I did a similar, but admittedly rather less wacky, bit of random math a little while ago. (The two or three people left who follow my LJ will have seen an earlier version of this before; some of the historical data is slightly refined in this one, but the math is the same.) I was watching a documentary about the development of the machine gun, and there was a segment in which the Consulted Authorities talked about the discovery that a skillful machine gunner could use the weapon to deny an area. This is done by making it too dangerous for enemy troops to enter, even if the gunner is not technically shooting at them specifically.

I was contemplating that in the abstract, and it occurred to me to wonder: how far apart are the individual bullets in the stream of fire coming out of a machine gun? I mean, obviously this isn't germane to the issue of area denial, because it's not like anyone's fast enough to walk between them, but still, I was curious, so I decided to sit down and see if I could work it out.

Naturally, this is going to vary somewhat in practice, due to variations in the weapons, ammunition, conditions, and so forth, but if it's viewed as an "ideal world" basic physics thought experiment, it works out like this. The two easiest pieces of information to find about any machine gun are its rate of fire and the muzzle velocity of the round it fires. I took as my reference sample the Maxim gun, which was extensively used during World War I (when the machine gun is widely seen to have come into its own as a battlefield weapon).

First we need to take the rate of fire and work out how much time an individual round has spent traveling before the next one is fired. A WWI-vintage Maxim, such as the one built for the British army by Vickers Ltd., had an "official" rate of fire of 450-600 rounds per minute - call it 500 RPM in practice. (All figures shown below are rounded to three significant digits, but weren't rounded in my calculator before use in the next step.)

500 rd/min * 1 min/60 sec = 8.33 rd/sec; the reciprocal of this is 1 sec / 8.33 rd = 0.120 sec/rd, that is, each round travels for 0.120 seconds before the next one is fired.

With this and the muzzle velocity (discounting the effects of friction and with it aerodynamic drag), we can find the distance each round travels before the next emerges, or the distance between them in the bullet stream. The Vickers gun was chambered for the standard .303 British rifle cartridge, and the Mk VII .303 British round in common use during WWI had a rated muzzle velocity of 2,440 feet per second.

2,440 ft/sec * 0.120 sec/rd = 288 ft/rd

That is, if you were able to discount the effects of gravity and friction while firing a Vickers gun, and stop time while the gun was firing at its cyclic rate, you would find that each individual bullet in the stream of fire was 288 feet from both the one ahead of it and the one behind.

Or, to put it another way (as a friend of mine pointed out), if you're firing at a target less than about 100 yards away, you'll only have one bullet in the air between you and the target at a time. At a Vickers gun's maximum effective range - about a thousand yards - there would be 10, with another one leaving just before the first arrives.

I freely concede that this information is of no use to anyone, but I had a bit of nerdy fun working it out.

--G.
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Benjamin D. Hutchins, Co-Founder, Editor-in-Chief, & Forum Mod
Eyrie Productions, Unlimited http://www.eyrie-productions.com/
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