Comic Discussion > QUESTIONABLE CONTENT
Spinning space station design
celticgeek:
Here are My Slide Rules.
And the father of one of my friends in High School had a "double length" slide rule; a full twenty inches long!
DSL:
--- Quote from: Precipice on 20 Jan 2012, 13:34 ---
--- Quote from: DSL on 20 Jan 2012, 10:10 ---For some entertaining nonsense, the original "Star Fleet Technical Manual" from the 1970s contains schematics for a giant spinning space station with, among other things, docking facilities *around the rim* for Enterprise and her fleetmates. Quite apart from the increased gravity, I'd hate to be the one who had to navigate a ship into one of those docks. Mr. Sulu'd be earning his pay, he would ...
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Take a closer look at those schematics. It doesn't say anywhere that it spins, and it should be obvious that it doesn't. The floors don't follow the curve of the rim, they're flat. With a spin-induced gravity, if you stood near the end of a pie-shaped segment, it'd feel like the floor is tilted at a 30 degree angle. It must generate artificial gravity using the same technology as the Enterprise.
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Huh. I remember my copy (long gone) saying it spun. And remembering, even as a sixth-grader, that it made no sense.
Akima:
--- Quote from: Skewbrow on 20 Jan 2012, 01:33 ---But in a way the fact that you resorted to a frigging calculator to (effectively) estimate Pi over 30 (radians per second) squared up to the whopping accuracy of a single significant digit makes me sad. :cry: I know, that's the way it's done nowadays. This is not meant as a personal criticism, please don't take it as such. Just my old school thinking, where mental arithmetic is the default.
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Actually I didn't use a calculator (the noises were artistic license) :-P. I had a spreadsheet already set up to run the numbers for a range of radii and centripetal accelerations :lol:, but if I hadn't, I certainly would have used a calculator. I'm just not confident that I could calculate in my head the necessary radius for a space station given only that the rim would be travelling in uniform circular motion at one RPM, and the required centripetal acceleration at the rim was 1g (which I approximated as 9.81m/s). I take comfort from knowing enough physics and maths to do the calculation at all, and being curious enough to want to do it. I'm not nearly ancient enough to feel nostalgic about slide rules, but this was my first calculator (and the suànpán can easily handle hex as well as decimal owing to the peculiarities of traditional Chinese weights and measures):
--- Quote from: Skewbrow on 20 Jan 2012, 03:15 ---I realize that the docking spacecraft would be small, so 1RPM won't take much. May be this isn't a problem at all?
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I've always assumed that the incoming spacecraft would match its spin to that of the space-station. A 5m radius spacecraft (larger than anything in service today; Soyuz and Shénzhōu are both under 3m in diameter) rotating at one RPM would generate so little pseudo-gravity (about 0.06g) that its crew would probably not notice.
Is it cold in here?:
I'd consider using a magnetic bearing to avoid vacuum welding problems.
Skewbrow:
@Akima, a spreadsheet at hand is, of course, an acceptable excuse for not doing mental arithmetic. I just happened to be working with that single data point that you gave us first: omega=1RPM=Pi/30 rad/s is approximately (using the Hoosier approximation Pi=3) 0.1 rad/s, so to find a radius R giving the target acceleration of R*omega^2=1g=10m/s^2 we immediately see that R=1000 m. IOW for this particular set of inputs at this level of precision the "mental arithmetic" amounts to moving the decimal point around :-D
You're, of course, correct about the problem of matching spins being trivial. I knew that the crew wouldn't need to use barf bags, but for some reason it wasn't clear to me that matching spins is no more difficult than matching speeds. My lack of knowledge about steering in space showed.
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