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From: desj@brahms.BERKELEY.EDU (David desJardins)
Newsgroups: net.sci
Subject: Re: How will it fall?
Message-ID: <13789@ucbvax.BERKELEY.EDU>
Date: Wed, 14-May-86 05:48:37 EDT
Article-I.D.: ucbvax.13789
Posted: Wed May 14 05:48:37 1986
Date-Received: Fri, 16-May-86 06:46:59 EDT
References: <632@tekigm2.UUCP> <3461@hplabsb.UUCP> <415@ccird1.UUCP>
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Reply-To: desj@brahms.UUCP (David desJardins)
Organization: University of California, Berkeley
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In article <415@ccird1.UUCP> rb@ccird1.UUCP (Rex Ballard) writes:
>>> Picture yourself on a space station similar to the one used in 2001
>>> (i.e. a spinning ring or toroid) .... 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 ... would it fall straight down ... or
>>> would it follow another path?
>>
>>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.
>
>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?

   The exact equation of motion in the coordinate frame of the observer
moving with the rotation (+x = spinward, +y = down, observer at (0,0),
R = radius of rotation, w = rate of rotation = sqrt (g/R)) is:

   (x, y) = (wRt cos wt - R sin wt,  wRt sin wt + R (cos wt - 1)).

To third order in time, and substituting sqrt (g/R) for w, this is:

   (x, y) ~= (-1/2 g sqrt (g/R) t^3,  1/2 g t^2).

The y term at least is not surprising!  Now, eliminating t and solving
for x in terms of y,

   x ~= - sqrt(2) y^3/2 R^-1/2.

So, if y = 1 meter, we have the following results:

   R          x   (distance of impact point from "straight down")
------      -----
  10 m      .44 m
 100 m      .14 m
1000 m      .04 m

   -- David desJardins