-Z- (z@gundam.com)
Tue, 7 Nov 2000 22:28:39 -0800

> -----Original Message-----
> From: owner-gundam@1u.aeug.org [mailto:owner-gundam@1u.aeug.org]On
> Behalf Of Vince Leon
> Sent: Tuesday, November 07, 2000 16:13
> To: 'gundam@aeug.org'
> Subject: RE: [gundam] Gravity within a colony (was: Funnels and flying
> MS)
>
> So in calculated a colonies' rotation with a radius of 500 meters I would
> do it as so:
> Calculate circumference (2 * Pi * 500 meters) = 2 * 3.14 * 500m = 3,149.59
> meters
>
> then velocity if colony surface^2 (gravity * radius) = 9.8m/s * 500m =
> 4,900m^2/s
>
> then take the sq root of the velociy = 4,900m^2/s = 70 m/s
>
> so the time for one revolution would be (circumference / velocity) =
> 3,149.6m / 70m/s
>
> this comes to 44.88 seconds for one revolution. Is this incorrect?

It sounds about right to me. A radius of 895 meters (diameter of 1.79 km)
rotated at 1 RPM yields exactly 1 G at the rim, so a radius of 500 meters
(diameter of 1 km) should rotate at about 1.34 RPM to achieve the same effect
and 1.34 RPM = 1 R every 44.8 minutes.

-Z-

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