Topological III is a large brass sculpture of this, in the lobby of the Harvard Science Center. http://mathtourist.blogspot.com/2011_12_01_archive.html, and scroll down to the second image. Andy On Fri, Oct 30, 2015 at 4:20 PM, Hilarie Orman <ho@alum.mit.edu> wrote:
I have similar questions about a twisted torus. If I have a pipe of triangular cross section and twist it 120 degrees around the vertical axis, then 360 degrees in the plane, and join the two ends, it has one exterior edge. But realizing this in a physical model is difficult because few materials are pliable enough to distribute the torsion uniformly.
I have a wire model of such a thing with 5 triangles. The interior is very thin. I would like to make a better model from flat pieces, with a more generous interior.
Hilarie
From: James Propp <jamespropp@gmail.com>
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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