Rusty Mathmatics

I have discovered, once again, that my mathematics abilities aren’t adequate. I mean I can do simple calculations and try to stay ahead of the cashiers in the supermarket by adding in my head what they add up with the cash register, and I have on occasion found myself adding arbitrary telephone numbers (don’t ask, I don’t know either), but I utterly fail at even slightly more complex tasks.

Case in point. I wanted to find out what radius and speed a Ringworld would need if I want to simulate a 24 hour day and want the “centrifugal” force to amount to an acceleration of 9.78 meters per second – in other words, simulates Earth gravity on the inner surface of the ring.

I consulted the (German) Wikipedia on the issue, but I had a hard time understanding the formula provided. Okay, didn’t help that it was quite late at night and I was tired, but still, makes me feel stupid. I did find an example which cleared things up quite considerably.

I wonder though, if maybe I ought to take some math courses. I guess I’ll never get around to it anyway – you know how it is with those things “I really should do some day”.

Oh, the answer to my problem, by the way, is 1.85 million kilometers and 134.5 km/s. Works out to an even 24 hours and exactly 9.78 m/s centrifugal force.

6 thoughts on “Rusty Mathmatics”

1. Martin says:

I don’t even want to talk about whether it’s 9.78 meters per second squared (!) or 9.81 — but you seem to have confused radius and circumference.

I get a radius of roughly 11.7 million kilometers (which roughly equals the circumference of a circle with 1.85 million kilometers radius). The result is a ring of 73.2 million kilometers circumference. The speed would thus be 848 km/s.

SCNR

2. Nils says:

You are right on the 9.78 m/s^2 of course (I picked the gravity at the equator, which is what Wilkipedia lists). But are you sure about the rest?

If I want to calculate the centrifugal force, the formula should be F = [(v^2 * m) / r].

Mass is 1 (due to gravity being expressed for 1 kg), so it doesn’t matter for us.

==> F = v^2 / r;

==> F*r = v^2;

==> sqrt(F*r) = v;

I then just plugged in different radi to see what happens.

For a radius of 1850000 km I get v = 134510.22 m/s.

The circumfence of that ring is pi*2r or 11623892818.28 m, divided by the speed makes 86416,43 seconds for one rotation, or 24h and 16.43 seconds – close enough to my goal of 24h for my purposes. I’d probably slow the ring down a tad so I have exact 24 hours, which helps with the clocks, and the somewhat lighter pseudo-gravity shouldn’t make a difference.

Where’s my mistake? I honestly don’t see it, but that wouldn’t be so surprising considering the topic.

3. Martin says:

Your formula ist alright. When you start throwing out the mass though it gets odd… When punching in numbers, what force did you use?

Here is my approach, based on your formula:

(1) F=m(v^2)/r
(2) F=ma
(3|1,2) a=(v^2)/r
(4) w=v/r (angular velocity, omega, here i use w)
(5|3,4) a=(w^2)r
(5) r=a/w^2

now a is 9.81 m/s^2 and w is 2pi/86400s

thus

r=9.81m(86400^2)/4(pi^2)

and just now I realize I forgot to sqaure 2pi in my original calc.

Good we talked, tho

4. Nils says:

I punched in the radius and the spreadsheet spat out the time of one rotation. I am not “throwing out” mass, mass is simply 1 🙂 I guess my calculation wasn’t too elegant but it seems to have worked. 😀 Thanks for doublechecking.

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