Doing Some Math

I'm not necessarily very good at math, but sometimes I get drawn into strange mathematical vacations at work. I remember once years ago I was trying to calculate roughly how many beans it would take to construct an actual hill of beans. The answer depended strongly on the angle of repose of said beans, and since I had no way of determining their angle of repose by empirical means, I concluded that my calculations were as unscientific as string theory. Don't like the answer you got? Well, just perturb some variables and try it again!

My boss came by and caught me in the middle of this furious pinto-bean math-fest and asked me what I was doing. I replied (and the trick, really, is to reply with gusto): "I'm doing some math, Jefe!" He assumed I was doing important math and proceeded to stammer "Oh, well, in that case, um, carry on, I'll leave you alone..."

So I did some math today about the Navy rail run described below. And wouldn't you know, I believe I found something wrong with the article. I happen to know, courtesy of http://www.fas.org/, that a 5-inch/54 fires a 32 kilogram projectile at 808 meters per second. By my math, using the infamous ke = 1/2 mv2 bruiser, I come up with a bit over 10 MJ. In fact, reverse-calculating the muzzle velocity based on a 10 MJ of energy and a 32 kilogram projectile yields a muzzle velocity of 790 meters per second - close to 808, but a hair below. So I'm not sure why the article insists that these guns produce 9 MJ when a bit of elementary physics suggests that they produce over 10 MJ.

Now, let's assume that our 10 MJ rail gun is 100% efficient. That is, it produces no heat, has no friction and no electrical resistance. I happen to know that there are 3,600 Joule in a kilowatt-hour of electricity, so a 10 MJ gun would consume 2.8 kilowatt-hours of electricity per shot (Editor's Note: there are in fact 3600 Joules in a watt-hour of electricity. A kilowatt-hour is 3.6 million Joules. The math is right; I simply cited the wrong number.) Electricity was going for about ten cents per kilowatt-hour in 2001, so per shot, the gun would cost 28 cents. That's pretty economical, considering you can't even get a Diet Coke for 28 cents.

However, we know the gun can't be perfectly efficient, if only because the rails aren't superconductors. Let's say it's 10% efficient, once you add up all the various sources of inefficiency. That means it uses 28 kilowatt-hours of electricity per shot, at a cost of $2.80. That's still pretty reasonable - I mean, I'd pay $2.80 to pull the trigger on a ten million Joule rail gun! Darn tootin!

But that 28 kilowatt-hour number is starting to get sizeable. That's a lot of power.

Let's go back to our 100% efficient rail gun for a minute and assume that we want to fire it every ten seconds. That is, every ten seconds we have to cram 2.8 kilowatt-hours of power down its throat. Conversion charts seem to tell me that that's roughly the equivalent of 135 horsepower. Double that because no generator is 100% efficient and it would seem to require the services of a 270-horsepower generator to keep a 100% efficient 10 MJ rail gun in power if it was going to fire six shots a minute.

But multiply by ten because the rail gun is really only 10% efficient, and suddenly you need a 2700-horsepower generator to run thing. That's getting into a size and weight bracket rarely seen in home generators. It's not, to put it mildly, the sort of thing you're going to be able to power up with your Homelite or Honda generator. In fact, I'm not sure but I think we're talking about engines the size of those found in very large mining trucks like the Caterpillar 797.

Well, maybe a rail gun isn't such a hot idea after all. At least not until room-temperature superconductivity becomes real.

And thus, I stop doing some math.