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From: rb@ccird1.UUCP (Rex Ballard)
Newsgroups: net.sci
Subject: Re: How will it fall?
Message-ID: <415@ccird1.UUCP>
Date: Sun, 11-May-86 05:13:24 EDT
Article-I.D.: ccird1.415
Posted: Sun May 11 05:13:24 1986
Date-Received: Wed, 14-May-86 05:39:01 EDT
References: <632@tekigm2.UUCP> <3461@hplabsb.UUCP>
Reply-To: rb@ccird1.UUCP (Rex Ballard)
Organization: CCI Rochester Development, Rochester NY
Lines: 30
Summary: Would the curve be noticed?

In article <3461@hplabsb.UUCP> bl@hplabsb.UUCP (Bruce T. Lowerre) writes:
>> Hello Out There,
>> 
>> Here's a little thought problem that might stir things up a bit.
>> 
>> Picture yourself on a space station similar to the one used in 2001
>> (i.e. a spinning ring or toroid). You are standing in the middle of
>> one of the decks near the outside edge of the ring and the spin of
>> the station is providing a "gravity" about equal to that found on
>> the surface of the Earth.
>> 
>> If you were to drop a ball (a simple release with no additional
>> forces applied), would it fall straight down (along a line through
>> the center of the ring and the point of release) or would it follow
>> another path (relative to the aforementioned line)?
>
>Due to angular momentum (or lack there of) it will fall in a curved
>path toward the opposite direction of the rotation of the toroid.

Theoretically, there should be the same curved path for a ball dropped
on Earth.  It is not noticable because the earths rotation is so slow,
and gravity so strong, that it would appear to fall straight.

As to whether you would see an observable curve, depends on the,
diameter of the toroid.  Since you have specified pseudogravitation
equal to earths,  a small toroid would have to spin very quickly, and
curvature would be quite pronounced.  If the toroid were sufficiently
large, the arc might only be a few seconds.

Does anybody have a formula for the relationship?