From: Henry Spencer <henry@zoo.toronto.edu> Newsgroups: sci.space.tech Subject: Re: space colony construction (was Re: Terraforming...) Date: Sun, 10 May 1998 23:46:01 GMT In article <rmn2cshhtq.fsf@fluent.com>, John Stoffel <jfs@fluent.com> wrote: >Does anyone know how to compute the instability of a revolving object? >I know small radius vs a long axle length is unstable (i.e. bottle), >and large radius, small axle (bicycle tire) is more stable. Where >does the trade off happen? When the axle and radius are equal? Or is >it a function of the average mass (moment?) from the center of >rotation? It's a matter of the moment of inertia (integral of radius^2 * mass). In the short term, spin is stable only around the axis of either highest or lowest moment of inertia; in the long term, for most practical purposes, it is stable only around the axis of highest moment of inertia. (The spacecraft has both angular momentum and kinetic energy of spin. In an isolated spinning spacecraft, angular momentum is conserved but energy can be dissipated, e.g. as heat in wiggling flexible parts. Since momentum is proportional to velocity and energy to velocity squared, you get minimum energy for a given amount of angular momentum with the mass farthest out from the center, because that gives the slowest spin. So any energy-dissipation mechanism -- fluid sloshing in tanks, flexible antennas flexing, etc. -- strongly biases the spacecraft toward ending up spinning around the axis of highest moment of inertia, giving the slowest possible spin.) -- Being the last man on the Moon | Henry Spencer is a very dubious honor. -- Gene Cernan | henry@zoo.toronto.edu

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