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- From: djpegg@utkvx2.utk.edu (PEGG, DAVID J)
- Subject: Re: Three-sided coin
- Message-ID: <11NOV199218361868@utkvx2.utk.edu>
- News-Software: VAX/VMS VNEWS 1.41
- Sender: usenet@utkux1.utk.edu (USENET News System)
- Organization: University of Tennessee Computing Center
- References: <1992Nov10.032643.10467@galois.mit.edu> <1dp0m9INNkq6@agate.berkeley.edu> <1992Nov11.061630.22658@galois.mit.edu> <1drt9sINN7hu@darkstar.UCSC.EDU>
- Date: Wed, 11 Nov 1992 22:36:00 GMT
- Lines: 22
-
- 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.
- >
- >
- Don't forget that the coin presents its edge twice as often as a head
- or tail. So I'd say a first approximation for "floating thru honey"
- would be 2h=d. But then the center o' mass is different. It's taller
- and narrower when landing on an edge and thus must land in a smaller
- range of angles to settle onto its edge.
- Let's see, we need 120 degrees per face (60 per edge). That gives
- 1.73h=d where d is diameter. (I think this'll work even without the
- honey.)
-