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Physics problem

Any mathematicians or physicists out there? I'm working on a story that takes place on a torus-shaped space station. Assuming the torus is one mile in radius (1584 meters) and revolving around its center, I'm trying to calculate what the rotational period of the station would have to be in order for a person standing inside the rim to feel centripetal acceleration equivalent to 1 gee—9.8 m/s2, that is.

It's been a long time since college physics for me, but my calculations using the formula

ac = ω2r

where ac is centripetal acceleration, r is the radius and the angular velocity ω is

ω = Δθ

and solving for time Δt when the arc Δθ is 2π radians, seems to show that the rotational period would have to be about 79.88 seconds, or one rotation every one minute and twenty seconds. (Basically, I plugged the lower formula into the upper one and solved for Δt.)

Can anyone verify this result?


( 12 comments — Leave a comment )
Aug. 3rd, 2003 10:32 pm (UTC)
That looks about right. I checked it with my req program, which comes with convenient physical constants:

> 2*pi/sqrt(g_earth / 1 mile)

(from omega * T = 2 * pi, and omega^2 * R = g_earth, solve for T. I'm sure there are much better math programs, I'm just familiar with this one.)
Aug. 4th, 2003 09:26 am (UTC)
That's pretty damn slick. Thanks!
Aug. 13th, 2003 12:05 pm (UTC)
Thank you! I just ran into something even slicker: try searching google for

2*pi/sqrt(10 meters/second^2 / 1 mile)

and see what you get. :) (I tried it with "acceleration of gravity" in place of "10 meters/second^2" but it wasn't quite that smart.)
Aug. 13th, 2003 12:55 pm (UTC)
That's far too cool for school. Wow.

Playing around, I noticed that you can use G, the gravitational constant, in expressions. In fact, you can use it (as shown in one of Google's own examples) to calculate gravitational acceleration at Earth's surface:

G*mass of earth/radius of earth^2

I had spiffed up my own centripetal acceleration calculator:


But the Google calculator blows me away.
Aug. 13th, 2003 01:22 pm (UTC)
Turns out they have that quantity built-in as little g -- I wish that was documented.

So I got to wondering a few things -- could it replace your centripetal calculator? "2*pi/sqrt(g / d) for d in 1 mile, 2 miles, 3 miles" didn't work, alas. At least not yet. It does do some simple error correction on your input, filling in missing parentheses, but anything more ungrammatical just reduces to an ordinary search.
Aug. 4th, 2003 12:43 am (UTC)
That's what I get too. Had to sort it out from square one, it's
been a while, but I get that same answer. Seems sort of fast
but I don't see an error, and I got the same answer from a
completely different path.

Aug. 4th, 2003 08:30 am (UTC)
Re: Yeah..
Yeah, I thought intuitively that it would be able to turn much more slowly. I'm going to play with bigger values for the radius and see what happens.
Aug. 4th, 2003 01:04 am (UTC)
other effects..
You know, of course, about all the other effects of such a
situation? In the right circles this is rather old hat, but
it's not obvious to people outside those circles.

It makes a huge difference to projectiles, for instance,
whether they go spinward, anti-spinward, or perpendicular
to the radius.

On your station, the person standing on the rim may reasonably
be viewed as standing on an airport people-mover going at
124 meters/second. If I shoot a gun at someone 10 meters
away, the bullet takes 3 milliseconds to get there, during which time the floor has moved .37 meters, which the bullet,
flying free, has not moved.

Another way to view it is that the bullet is going in the
straight line dictated by the force of the power, AND the
momentum it had whilst spinning around in the station while
inside the gun. Me and my target are, however, still going
around in a big circle.
Aug. 4th, 2003 08:14 am (UTC)
Re: other effects..
Thanks for the info. Right now I don't recall any projectiles in the story—I'm getting down and dirty in revisions—but this information will be useful to keep in mind as I picture each scene.
Aug. 4th, 2003 12:02 pm (UTC)
Re: other effects..
Minor nit: the bullet shares the initial 124 m/s (as you say in the paragraph below that). You get the Coriolis effect from the gradient in velocities, not the velocity itself -- I don't think there's any direction you could fire it that would give you such a huge displacement of the bullet. A couple millimeters would be more in line. (a = 2 * omega cross v)

I agree with everything else you said.
Aug. 4th, 2003 07:48 am (UTC)
Query on method: Is it essential for the story that you know this? Or is it just for your comfort, so you don't mess up any description? I mean, when I design my fantasy worlds, and I draw the maps, I get really paranoid about landmass to water ratio, because once I drew a world and it didn't have enough water to support the land mass involved, and my boyfriend pointed this out, and I felt like the story just couldn't work until I got my world ratios sorted out.

Probably why I don't write more science fiction--I need more science to feel happy writing it.
Aug. 4th, 2003 08:23 am (UTC)
I probably wouldn't have started thinking about this except that the main character in the story tends to do physics problems in his head at times of stress, to take his mind off other things. In the first scene, he's trying to figure out how fast he would have to walk to maintain a fixed position as the station revolves around him. As it turns out, as those rotational velocities, he wouldn't be able to walk or run anywhere near fast enough, which he would no doubt already know so he wouldn't seriously consider it as a physical possibility.

I don't generally work things like this out unless they're vital to the plot or character, although judging from how badly I intuited this one in the first draft, I should probably do more working out like this in the future.
( 12 comments — Leave a comment )


William Shunn

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