From: aephraim@physics.Berkeley.EDU (Aephraim M. Steinberg)
Newsgroups: sci.math,sci.physics
Subject: Re: Three-sided coin
Date: 12 Nov 1992 03:30:24 GMT
Organization: University of California, Berkeley
NNTP-Posting-Host: physics.berkeley.edu

In article <1drt9sINN7hu@darkstar.UCSC.EDU>, ask@ucscb.UCSC.EDU (Andrew Stanford Klingler) writes...
> 
>Coins "flip nicely" because they're nearly two-dimensional symmetric objects.
>If I remember my mechanics correctly, the only stable rotations are about the
>inertial axes with the highest and lowest moments.  In this case that means
>the "vertical" axis and any "horizontal" axis.  I don't recall how to
>compute the angular trajectory of an object with an arbitrary initial
>orientation and spin, but the general answer will not be proportional times
>for proportional solid angles.  If the coin is "nicely flipped" it should
>be proportional times for proportional pure angles, giving a "floating
>through honey" solution h=d.  But as you say, this is far from relevant
>to a real throw.

The terms "vertical" and "horizontal" here are misleading.  What you say
is correct, but there is no guarantee that in a "nice flip" (by which I
assume you mean one in which the coin rotates about the diameter and
not the cylinder axis) the rotation axis will be oriented horizontally.
In fact I think that, albeit photogenic, such coin tosses are probably
relatively rare in practice (though certainly easy with practice :-).
I believe that, even if we assume all rotation to be about one of these
"special" axes (except for the precession due to gravity, of course),
the random distribution of where in real space that axis points will give
back my initial solid angle response.

I'm also not convinced that the mellifluous approximation is as bad as
it has been given credit.  The essential point is that the coin not
bounce around so violently upon landing that it invariably ends up in
one of the lower-energy states (heads or tails, in the case of radius
being 1.4 times thickness).  All of the landing states are insensitive
to _small_ perturbations, since what the solid angle approach means is
that it lands on whatever is directly below the center of mass.  As the
edge hits the table, the coin begins rotating about the horizontal (true
horizontal, now), and if it is damped sufficiently, it will never fall
over nor climb up onto an edge.  If, on the other hand, it bounces back
off the table, the rotation will continue freely, and at that point the
probability of how it will next hit is presumably determined by polar,
not solid, angles.  Of course, related to the fact that heads and tails
are lower energy states, it is not guarranteed that once it lands heads
and bounces, it bounces high enough to rotate around to its edge-- whereas
if it bounces off its edge, it is generally free to rotate around to
one face or the other.  In the limit of many bounces, this becomes like
optical pumping, and makes the higher-energy state very unlikely,
possibly bringing us back to John Baez's original solution.  If there
are no (or few) bounces, the system never reaches thermal equilibrium,
and I think the honey approach is approximately correct.

So much time spent discussing items I earn so few of :-).

-- 
Aephraim M. Steinberg                    | "WHY must I treat the measuring
UCB Physics                              | device classically??  What will
aephraim@physics.berkeley.edu            | happen to me if I don't??"
                                         |       -- Eugene Wigner