Regarding the construction of a Moebius band by a triangulated strip --- Dan seemed to doubt that this was possible, but kept the reason close to his chest. On the other hand, it appears almost obvious to me that any sufficiently long strip can do the job, since with one end fixed the other end has 3 translational and 3 rotational degrees of freedom. Therefore provided self-intersection can be avoided, it may adopt any congruence within some region restricted only by strip length --- including meeting the first end after a half-twist. A formal robotics-style proof should be possible, investigating the rank of some appropriate set of 4x4 matrices; I'm not sure that this procedure would provide any worthwhile insight however! Consider instead a classical "kaleidocycle" (rotating ring) of 10 regular tetrahedra: one half of its exterior comprises a continuous strip of rhombi, each divided into two equilateral triangles by a single diagonal, forming a (5/2)-twist Moebius band. Now generalise to k (initially flat) tetrahedral cells with half-square faces, but one edge then shortened so that the twist angle equals pi/k . The strip of unadjusted faces yields the required Moebius band. Clearly for sufficiently large k , collisions are avoided by any nearly circular configuration. The third set of parallel creases is unused. And --- the band rotates like a smoke-ring! Fred Lunnon On 10/28/15, Dan Asimov <asimov@msri.org> wrote:
Actually turns out there's a paper:
"Inverting a cylinder through isometric immersions and isometric embeddings", "B. Halpern and C. Weaver, Trans AMS, vol 230, 1977.
that rigorously (?) shows both the theorem stated in the Tabachnikov-Fuchs book about the Moebius band, and also establishes bounds for when the open cylinder can be turned inside out.
But none of these consider the case of a triangulated surface whose edges are hinged and whose faces are required to remain flat and rigid.
—Dan
On Oct 27, 2015, at 6:28 PM, James Propp <jamespropp@gmail.com> wrote:
And it's even freely available on the web!:
http://www.math.psu.edu/tabachni/Books/taba.pdf
Chapter 14 deals with the smooth version of the problem, summarily dismissing the non-smooth version by invoking the construction that Dan and I are skeptical about.
Thanks, Dan. I'll see what Erik thinks about the construction of Figure 14.2 in the book
Jim Propp
On Tuesday, October 27, 2015, Dan Asimov <dasimov@earthlink.net> wrote:
A more useful reference is the book "Thirty Lectures on Classic Mathematics" by Dmitry Fuchs and Serge Tabachnikov, whose Chapter 14 is all about this problem.
—Dan
On Oct 27, 2015, at 8:03 AM, James Propp <jamespropp@gmail.com> wrote:
Dan Asimov wrote:
Dan, may I forward your email to him?
No.
Understood.
Also, can anyone provide the Gardner reference Dan has in mind?
6th Book of Mathematical Diversions, p. 63, with illustation.
Gardner says there is a proof in Stephen Barr's Experiments in Topology (1964), but Barr attributes the method to Gardner and provides no proof.
I'm trying to locate this in my CD of all Gardner's columns, but unfortunately when MAA re-released the books, they gave them different titles (and possibly traded material between books). Can anyone locate this column on the CD, or provide a word from the column that uniquely specifies it? (Lots of columns contain the words "Stephen", "Barr", and "Topology", and I don't know how to get my Mac to search for two-word phrases.)
Thanks,
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