It seems like a simple question. Even stupid. And it probably is.

So I'm sketching out a space station for a science-fiction project, and after doing more simple math than I am accustomed to in order to come up with an approximate volume of over three-billion cubic meters, I now realize that the damn thing isn't big enough.

Certain components can stay the same size, for all I care. But since I'm already cheating the math on wedge volumes, is there any shortcut for recalculating the volume of a cylindrical component if I only double the radius? Or if I double both radius and height? Like multiplying the cylindrical volume by eight°? Or sixty-four if I change both r and h?

Seriously, my head freakin' hurts right now. I always hated math. I'm waiting until later today to do a new sketch and calculations.
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Notes:

° eight — You understand, of course, I'm not dealing this hand with a full deck. Eight seems like a reasonable number because it's two cubed. I have no other reason to presume it will work, and I'm unable to think it through at the moment. The more I think about it, the less frightened I am of doing the calculations all over again, but when I came up with a number like 2,454,369,260 cubic meters, it seemed like something worth asking. In fact, the more words I type, the more it seems like I should just shut up and redo the calculations. Even so, I'm still curious if there's a shortcut. Laugh if you want. It won't bother me.

V=pi * radius^2 * height

http://www.science.co.il/Formula.asp

Thank ye, Draqon, but, frankly, that part I was able to figure out myself. I had more fun with the cones, and only cheated on wedges because I couldn't figure out the diagram on the calculator site I dug up. But I need to double r at least, and maybe h as well. I thought it would be nice if I could just take the number I came up with (2.454 billion) and multiply it by x. There's a long story about me and math involved, but I'll skip it for now. Suffice to say that I will, someday, have my revenge on a man named Spanier. And also a man named Gavin, but that's a different ... okay, a different chapter of the story.

Well, if you double the radius only, then the volume increases by a factor of four. If you double the height, the volume only increases by a factor of two. So, as you'd expect (and as you pointed out), the volume increases by a factor of eight if you double the height AND radius.

Am I reading your question correctly?

Consider this a...um...fishing lesson :)

r \rightarrow 2r

V = \pi r^2 h \rightarrow \pi (2r)^2 h = 4\pi r^2 h= 4V

It seems to me that a torus-like shape is best for a space station. Perhaps a torus with a square or rectangular cross section rather than a circular cross section. Artificial gravity via rotation. Some provision for sealing off a section in the event of a meterite punching a hole in the station. With one section sealed off, you could still get to every other section while the hole was being repaired.

I was thinking the same thing as dinosaur.

The volume of a torus with a circular cross-section of revolution is simply the area of the circle multiplied by the distance of one revolution travelled by its centroid.

Also just use factors, as Ben said.

And if you don't like working with billions, just use cubic km. In the end one cubic km equals a billion cubic meters.

I admit the topic had slipped my mind in the last couple weeks; you know, 'tis the season, &c.

I would like to thank you all for your assistance and insight. And yes, Ben, you were reading the question correctly indeed. In fact, when I finally went back and tested the calculation and, for the record, having sliced off some decimal fractions here and there, I was amazed to find that the new figure was, after all that, only five cubic meters different from multiplying the former by eight.

• r=1.25 km, h=0.5 km: 2,454,369,260
• times eight: 19,634,954,080
• r=2.5 km, h=1.0 km: 19,634,954,085

And to me, that's really cool because, like I said, I don't do a whole lot of math.

Right now the thing looks like a big-assed hubcap floating above the Earth.

As a note to Dinosaur and Facial: What's funny is that as I was thinking about everything, I came to the conclusion that it would be trippy if I made the thing a massive sphere and then tried to apply artificial gravity as if the thing were a planet, so that your feet, as such, would be pointed toward the center of the thing. And then I remembered George Lucas. One of my favorite things about the new Battlestar Galactica is that they use ballistic weapons. We might recall that Mr. Lucas sued the original BG for being too similar in its use of laser bolts. When Galactica opened up with a massive firing solution as it met the Cylon base stars at the end of the three-hour pilot miniseries, I fell out of my chair, jaw open, and could only manage to say, "Take that, George, and f@ck off!" Quite obviously, a spherical station would get me sued in the long run (don't think I don't have cinematic ambitions attached to this one), and for the same reason I was concerned about the torus. (Further review (http://www.stardestroyer.net/TPMRevelations/TradeFed.jpg) assuages that concern somewhat but that would be one big-assed torus.)

But the discussion fascinates me; I'll open a SciFi topic on space station design.

It shouldn't be that surprising that it's off by so little, since essentially what you're doing is messing around with commutative properties.

But Ben shows the math quite good.

Oh, I'm just a whiny tw@t about math, Roman. To be honest, I wouldn't have said anything if it was off by like a hundred or something. But five just struck me oddly.

At any rate, the SciFi topic can be found here:

http://www.sciforums.com/showthread.php?t=75644

To uh, be perfectly honest, I learned my factors of eight from D&D. Enlarge Person doubles your size, but octuples your weight.

Well, nobody can say D&D doesn't teach you anything. I still want to see the book on craps played with 100-sided dice, though.

These are issues of "numeracy," which is the numerical analog of literacy.

We know that the area of any two-dimensional object is a quadratic function of its linear dimensions. We know this by extrapolating intuitively from a square of side x: A=x^2. If we double the length of the side, the area becomes (2x)^2 = 4(x^2). This makes intuitive sense because we have multiplied the linear dimensions by two and by golly that multiplies the area by two squared. Try the same exercise with a factor of three; try it with a rectangle of your chosen dimensions; try it with a circle. Try it with any two-dimensional figure that you can easily calculate the area of. You'll find if you increase both of its dimensions proportionally, so that it changes size but not shape, that you increase its area by the square of the multiplier.

The same principle holds with three-dimensional objects. If you increase their dimensions proportionally, so they change only size but not shape, you increase their volume by the cube of the multiplier. So if you double the width of a cube or the radius of a sphere, you multiply its volume by eight.

This is a principle that comes into play in biology. Suppose you double the size of an animal proportionally so that it looks identical to a smaller version; you couldn't tell them apart without a reference object or seeing them together. The volume of the animal increases by a factor of eight, and so, presumably, does its mass. But what happens to its muscles? It's their cross section that determines their strength, and the cross section only increases by a factor of four. Each muscle cell is supporting twice the weight of the muscle cells in the smaller animal, and this results in a weak animal. This is why elephants have such disproportionately massive legs, or why giraffes have such disproportionately delicate bodies: the relationship between body weight and muscle strength has been adjusted in two different ways.

And this is why insects don't get much larger than those extinct two-foot-long cockroaches. With a vertebrate, we can grow thicker bones to support our more massive tissues; we even do tricks like growing rib cages to support our complex inner organs. The two-dimensional cross-section of the bone increases disproportionately, to keep up with the more rapid increase in mass of the three-dimensional body. But insects have exoskeletons, and there's a limit to how much internal tissue can be supported by a surrounding cylinder before it simply collapses into a mass of goo. Notice that their close relatives the myriapods (centipedes and millipedes) solve this problem in a different way: they grow more legs, so each leg only has to support a small portion of the total weight, and those small muscles can easily do their job by being attached internally to a cylinder of exoskeleton. Their bodies also have multiple segments, so each segment is of a reasonable size for the exoskeleton to support its share of the internal organs.

This also explains why marine mammals can grow so much larger than terrestrial mammals. Their body weight is distributed over their entire lower skin surface, rather than being supported by legs. DNA analysis has discovered that the ancestors of whales were a type of hippopotamus, an animal that has one of the strongest, most massive sets of hind legs. It evolved into an animal that has not only lost those legs completely, but also the massive pelvis that anchored them!