On Oct 28, 2015, at 5:27 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
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.
No, that's not what I doubt at all. What I doubt is that the pleated strip (as depicted in Martin Gardner's "6th Book of Mathematical Diversions", p. 63) can be made into a Moebius band as described, isometrically into 3-space). —Dan
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!