home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!wupost!zaphod.mps.ohio-state.edu!rpi!uwm.edu!psuvax1!rutgers!ub!acsu.buffalo.edu!kriman
- From: kriman@acsu.buffalo.edu (Alfred M. Kriman)
- Newsgroups: sci.math
- Subject: Re: Tiling sphere by triangles (Re: 3 space terahedron-packing)
- Summary: Two errata corrected.
- Message-ID: <BuHpJr.Frv@acsu.buffalo.edu>
- Date: 13 Sep 92 00:04:39 GMT
- References: <f#tng3h.spworley@netcom.com> <1992Sep11.182727.28044@nntp.uoregon.edu> <BuGKn5.Lnt@acsu.buffalo.edu>
- Sender: nntp@acsu.buffalo.edu
- Organization: UB
- Lines: 23
- Nntp-Posting-Host: lictor.acsu.buffalo.edu
-
- In article <BuGKn5.Lnt@acsu.buffalo.edu> kriman@acsu.buffalo.edu
- (I, Alfred M. Kriman) answered a query of goodman@bright.uoregon.edu
- (Albert Goodman) in article <1992Sep11.182727.28044@nntp.uoregon.edu>,
- whom I thank for pointing out an error in my posting.
-
- I was sloppy in my response. Two errors are noted below:
- >
- > It may provide insight into the problem to recognize that regular
- >spherical polygons of different sizes cannot be congruent. The reason
- I should have written something like "similar" ->^^^^^^^^^. The point
- is that scaling the lengths of a spherical polygon yields a spherical
- polygon which lies in the surface of a scaled (i.e., different-radius)
- sphere. This does not yield a new tiling for the original sphere, it
- just yields a similar tiling for a different sphere.
- >
- > More specifically, the sum of the exterior angles of a spherical
- >triangle (in radians) is pi plus the angular area of the triangle (in
- >steradians). ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
- Wrong. I should have written TWO times pi, MINUS the angular area ...
- For small triangles one recovers the Euclidean value 2*pi. For very large
- triangles that cover nearly the whole sphere, one obtains -2*pi, but the
- convention of "exterior" angle yields minus the intuitive values, so all
- is well.
-