Fred has answered a question I should have asked, but didn't! In a similar vein we can ask, What's the smallest possible number of faces a Mobius strip can have? You are permitted to use a rectangle of any aspect ratio you like. Jim Propp On Thursday, October 29, 2015, Fred Lunnon <fred.lunnon@gmail.com> wrote:
From the context, evidently "any" meant "every" --- so my kaleidocycle hardly constituted a great leap forward concerning either Moebius band problem.
It did however have the merit of being isometric, which is a start. I haven't caught up with the background reading at this stage, so may still be blundering around --- but nobody seems to have pointed out yet that the naïve embedding, in which the long edges of the strip spiral around a torus while the central axis describes a circle, does not quite "work" in the first place.
[ Though smooth it is clearly not isometric: to a first approximation anyway, I find the ratio of their embedded lengths equals sqrt( 1 + (x pi/4)^2 ) ~ 1 + (1/2)(x pi/4)^2 where x denotes width/length. ]
A pleated wide strip amounts in effect to a solid oblong with depth << width << length , or say y << x << 1 , increasing the discrepancy further. A smooth isometric embedding might perhaps overcome this by somehow shadowing a kaleidocycle; but if this involves the number of cells k -> oo , the resulting surface may turn out rather complicated --- compare the Nash embedding https://www.youtube.com/watch?v=RYH_KXhF1SY
Fred Lunnon
On 10/29/15, Fred Lunnon <fred.lunnon@gmail.com <javascript:;>> wrote:
Apologies --- in fact, it was actually Jim Propp who asked
<< Regarding Mobius bands, Dan's remark ... makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band? >>
I am assuming that the strip ends are permitted to be glued together. But I am unsure whether "any" means "some" or "all" in this context ...
WFL
On 10/29/15, Dan Asimov <dasimov@earthlink.net <javascript:;>> wrote:
On Oct 28, 2015, at 5:27 PM, Fred Lunnon <fred.lunnon@gmail.com
<javascript:;>> 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!
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