You can join four isosceles right triangles along hinges (I use tape) to form a square, and then join four such square panels (again with tape) to form a long rectangle, and lastly join the first and last panels together with tape to form a closed band. If you color the two sides of the band with different colors, it's a fun puzzle to turn the band inside out, exchanging the colors. I learned about this in the 80s, and amused myself by learning how to do the moves behind my back. Does anyone know the name and/or provenance of this puzzle? Jim
Don't know the answer, but it brings up the question: Given a cylinder X = S^1 x [0,h] in R^3 (where S^1 is a unit circle in R^2): What is the largest h (or sup) such that X can be everted? (That is, continuously deformed through smooth surfaces such that the inside and outside change places.) Any guesses, at least? * * * The related question: What is the smallest or inf of h such that [0,1] x [0,h] can be smoothly mapped onto a Moebius band M in R^3 via f: [0,1] x [0,h] —> M with f(t,0) ~ f(1-t,h) for all t in [0,1] ? appears to have the answer inf{h} = sqrt(3). (If differentiability is assumed to be only C^1 then the answer is inf{h} = 0, though this is very hard to imagine.) —Dan
On Oct 24, 2015, at 2:47 PM, James Propp <jamespropp@gmail.com> wrote:
You can join four isosceles right triangles along hinges (I use tape) to form a square, and then join four such square panels (again with tape) to form a long rectangle, and lastly join the first and last panels together with tape to form a closed band. If you color the two sides of the band with different colors, it's a fun puzzle to turn the band inside out, exchanging the colors. I learned about this in the 80s, and amused myself by learning how to do the moves behind my back.
Does anyone know the name and/or provenance of this puzzle?=
I spaced out when I originally wrote the (now-fixed) following message, having omitted mention that certain maps are to be *isometries*. Also, to avoid the Nash-Kuiper embedding theorem I've now asked that all maps be smooth (C^oo). The amended version is as follows. —Dan ---------------------------------- Given a cylinder X = S^1 x [0,h] = {{cos(theta), sin(theta), z) | 0 <= theta < 2pi, 0 <= z <= h} in R^3 (where S^1 is a unit circle in R^2): What is the largest h (or sup) such that X can be everted? (That is, smoothly deformed through smooth surfaces, each ISOMETRIC to the original, such that the inside and outside change places.) ((( Specifically: A C^oo isotopy (homotopy through embeddings) H: X x [0,1] —> R^3 such that for all p in the cylinder X (= S^1 x [0,h]) we have H(p,0) = p and H(p,1) = T(p) where for all (x, y, z) in R^3 we define T(x,y,z) := (x, y, h-z). ))) Any guesses, at least? * * * A related question: What is the smallest or inf of h such that [0,1] x [0,h] can be smoothly mapped ISOMETRICALLY onto a Moebius band M in R^3 via f: [0,1] x [0,h] —> M with f(t,0) = f(1-t,h) for all t in [0,1] ? appears to have the answer inf{h} = sqrt(3). (If differentiability is assumed to be only C^1, then the answer is inf{h} = 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
Regarding Mobius bands, Dan's remark (If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band? Jim Propp
Martin Gardner offered an argument (due to ?) that isn't quite valid: Take a 100 x 1 rectangle of paper and pleat it 1001 times in the 100 direction (i.e., make 1000 folds back and forth, 100/1001 apart from each other). You get a very thin rectangle that is 100/1001 x 1. Now bend it around with a half-twist and glue the pleated edges together to get a Moebius band. (Because there were an even number of folds, they match up.) This may "work" with paper, but not isometrically with a genuine flat 2-manifold in R^3. —Dan
On Oct 25, 2015, at 1:55 PM, James Propp <jamespropp@gmail.com> wrote:
Regarding Mobius bands, Dan's remark
(If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band?
I am not sure exactly what question is being debated. At least under some rules, many aspect ratios of Moebius strips are foldable. The tri-hexa-flexagon, for instance, is a Moebius strip. On Sun, Oct 25, 2015 at 6:59 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Martin Gardner offered an argument (due to ?) that isn't quite valid:
Take a 100 x 1 rectangle of paper and pleat it 1001 times in the 100 direction (i.e., make 1000 folds back and forth, 100/1001 apart from each other).
You get a very thin rectangle that is 100/1001 x 1. Now bend it around with a half-twist and glue the pleated edges together to get a Moebius band. (Because there were an even number of folds, they match up.)
This may "work" with paper, but not isometrically with a genuine flat 2-manifold in R^3.
—Dan
On Oct 25, 2015, at 1:55 PM, James Propp <jamespropp@gmail.com> wrote:
Regarding Mobius bands, Dan's remark
(If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I agree with Dan that this won't work without some additional pleats that may have to be chosen with some care. Or maybe there's no way to do it. I'd like to ask Erik Demaine what he knows about this. Dan, may I forward your email to him? Also, can anyone provide the Gardner reference Dan has in mind? Thanks, Jim Propp On Sunday, October 25, 2015, Dan Asimov <dasimov@earthlink.net> wrote:
Martin Gardner offered an argument (due to ?) that isn't quite valid:
Take a 100 x 1 rectangle of paper and pleat it 1001 times in the 100 direction (i.e., make 1000 folds back and forth, 100/1001 apart from each other).
You get a very thin rectangle that is 100/1001 x 1. Now bend it around with a half-twist and glue the pleated edges together to get a Moebius band. (Because there were an even number of folds, they match up.)
This may "work" with paper, but not isometrically with a genuine flat 2-manifold in R^3.
—Dan
On Oct 25, 2015, at 1:55 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Regarding Mobius bands, Dan's remark
(If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Oct 26, 2015, at 7:26 PM, James Propp <jamespropp@gmail.com> wrote:
I agree with Dan that this won't work without some additional pleats that may have to be chosen with some care. Or maybe there's no way to do it.
I'd like to ask Erik Demaine what he knows about this. Dan, may I forward your email to him?
No.
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. —Dan
Thanks,
Jim Propp
On Sunday, October 25, 2015, Dan Asimov <dasimov@earthlink.net> wrote:
Martin Gardner offered an argument (due to ?) that isn't quite valid:
Take a 100 x 1 rectangle of paper and pleat it 1001 times in the 100 direction (i.e., make 1000 folds back and forth, 100/1001 apart from each other).
You get a very thin rectangle that is 100/1001 x 1. Now bend it around with a half-twist and glue the pleated edges together to get a Moebius band. (Because there were an even number of folds, they match up.)
This may "work" with paper, but not isometrically with a genuine flat 2-manifold in R^3.
—Dan
On Oct 25, 2015, at 1:55 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Regarding Mobius bands, Dan's remark
(If differentiability is assumed to be only C^1, then the answer is inf{h}
= 0, because of the Nash-Kuiper isometric embedding theorem. It's very hard to imagine just how this works.)
makes me wonder what the situation is for PL embeddings. Can a clever origamist turn ANY rectangle into a Mobius band?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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, Jim Propp
<< and I don't know how to get my Mac to search for two-word phrases >> Put them inside quotation marks? WFL On 10/27/15, 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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Tried using quotes. Doesn't work in SmartFolder. And Spotlight under Yosemite searches the whole bloomin' web, not just my laptop. Jim Propp On Tuesday, October 27, 2015, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< and I don't know how to get my Mac to search for two-word phrases >>
Put them inside quotation marks? WFL
On 10/27/15, James Propp <jamespropp@gmail.com <javascript:;>> 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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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,
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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!
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> wrote:
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!
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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> 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> wrote:
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!
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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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It can certainly be done in six; I have a sketch of an example inspired by the tri-hexa-flexagon. On Fri, Oct 30, 2015 at 11:14 AM, James Propp <jamespropp@gmail.com> wrote:
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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I must cautiously backpedal. I haven't proven that my example can be realized with planar quadrilaterals, though I would be very surprised if it couldn't. On Fri, Oct 30, 2015 at 11:40 AM, Allan Wechsler <acwacw@gmail.com> wrote:
It can certainly be done in six; I have a sketch of an example inspired by the tri-hexa-flexagon.
On Fri, Oct 30, 2015 at 11:14 AM, James Propp <jamespropp@gmail.com> wrote:
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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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!
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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-- Andy.Latto@pobox.com
I suppose this question is about a Moebius strip embedded in R^3 as a polyhedron (a union of planar triangles that can intersect pairwise in only an edge, a vertex, or the empty set. Yes? —Dan P.S. Please note correct spelling of Moebius (unless you can use an umlaut).
On Oct 30, 2015, at 8:14 AM, James Propp <jamespropp@gmail.com> wrote:
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.
Yes, and moreover, I want the polyhedron to be flattenable to an ordinary rectangle. (Though I suppose that if one can reduce the number of faces by dropping this constraint, that'd be worth knowing.) Jim On Fri, Oct 30, 2015 at 5:51 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I suppose this question is about a Moebius strip embedded in R^3 as a polyhedron (a union of planar triangles that can intersect pairwise in only an edge, a vertex, or the empty set. Yes?
—Dan
P.S. Please note correct spelling of Moebius (unless you can use an umlaut).
On Oct 30, 2015, at 8:14 AM, James Propp <jamespropp@gmail.com> wrote:
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.
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What's the difference between this and a polyhedral torus? [ Eg. 20-face "polytore" from math-fun Sep. '09 ] Or is this asking for one-sided (immersed) polyhedra? WFL On 10/30/15, James Propp <jamespropp@gmail.com> wrote:
Yes, and moreover, I want the polyhedron to be flattenable to an ordinary rectangle. (Though I suppose that if one can reduce the number of faces by dropping this constraint, that'd be worth knowing.)
Jim
On Fri, Oct 30, 2015 at 5:51 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I suppose this question is about a Moebius strip embedded in R^3 as a polyhedron (a union of planar triangles that can intersect pairwise in only an edge, a vertex, or the empty set. Yes?
—Dan
P.S. Please note correct spelling of Moebius (unless you can use an umlaut).
On Oct 30, 2015, at 8:14 AM, James Propp <jamespropp@gmail.com> wrote:
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.
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Would anyone care to present all the conditions in a math question all in one place at one time? —Dan
On Oct 30, 2015, at 3:58 PM, James Propp <jamespropp@gmail.com> wrote:
Yes, and moreover, I want the polyhedron to be flattenable to an ordinary rectangle. (Though I suppose that if one can reduce the number of faces by dropping this constraint, that'd be worth knowing.)
Jim
On Fri, Oct 30, 2015 at 5:51 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I suppose this question is about a Moebius strip embedded in R^3 as a polyhedron (a union of planar triangles that can intersect pairwise in only an edge, a vertex, or the empty set. Yes?
—Dan
P.S. Please note correct spelling of Moebius (unless you can use an umlaut).
On Oct 30, 2015, at 8:14 AM, James Propp <jamespropp@gmail.com> wrote:
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.
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Reconciled to knowing diddly-squat about any of these questions, I resorted to (recommended) https://en.wikipedia.org/wiki/Moebius_band Though I fail to understand properly the little it says concerning Dan's original enquiry about embedding a corrugated wide strip. Another matter I haven't so far found discussed is the actual shape assumed by a (stiffish) flat strip when deformed into a Möbius band (isometrically, and assuming a perfect join). Presumably this can in principle be expressed as a variational problem minimising the total stress. Is there an explicit parametric solution? An implicit algebraic equation? Has there been any approximate numerical attack? Is the solution smooth, or analytic? And totally off-topic again, I can't resist quoting a link from the Wikipedia article to "A Planetary Möbius Gear System" http://mechproto.olin.edu/final_projects/average_jo.html --- is Oskar listening? Fred Lunnon On 10/31/15, Dan Asimov <dasimov@earthlink.net> wrote:
Would anyone care to present all the conditions in a math question all in one place at one time?
—Dan
Another question in this vein has to do with finding the right way to measure distortion, and to prove that the only way to evert a cylinder (or Mobius band, or Mobius-band-with-three-half-twists) is to introduce at least a certain amount of distortion en route. One obvious way to measure distortion of an embedded PL manifold-with-boundary is to measure the flexing happening along all the edges; the sum will be a dimensionless (angular) quantity. Or maybe we should multiply each flex-angle by the length of the edge being flexed, a la Dehn's invariant. Possibly we should draw inspiration from existing concepts in differentiable topology. Are there ways of measuring distortion of embedded or immersed manifolds, and results that (for instance) decree that if you want to evert a sphere, you need to distort it by at least a certain amount? Or maybe that's totally wrong-headed. After all, when we evert a band with some number of twists, we're not changing the local metric at all. Creases are not like cone-points. Jim Propp On Saturday, October 31, 2015, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Reconciled to knowing diddly-squat about any of these questions, I resorted to (recommended) https://en.wikipedia.org/wiki/Moebius_band Though I fail to understand properly the little it says concerning Dan's original enquiry about embedding a corrugated wide strip.
Another matter I haven't so far found discussed is the actual shape assumed by a (stiffish) flat strip when deformed into a Möbius band (isometrically, and assuming a perfect join). Presumably this can in principle be expressed as a variational problem minimising the total stress. Is there an explicit parametric solution? An implicit algebraic equation? Has there been any approximate numerical attack? Is the solution smooth, or analytic?
And totally off-topic again, I can't resist quoting a link from the Wikipedia article to "A Planetary Möbius Gear System" http://mechproto.olin.edu/final_projects/average_jo.html --- is Oskar listening?
Fred Lunnon
On 10/31/15, Dan Asimov <dasimov@earthlink.net <javascript:;>> wrote:
Would anyone care to present all the conditions in a math question all in one place at one time?
—Dan
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It's called a flexatube. Martin Gardner describged it in his 1961 book, "The Second Scientific American Book of Mathematical Puzzles & Diversions: A New Selection". Here's a picture and description: http://mathworld.wolfram.com/Flexatube.html I have often made these out of paper, putting red dots in the centers of the panels on one side and blue dots on the other. Tom
On 2015-10-24 15:49, Tom Karzes wrote:
It's called a flexatube. Martin Gardner describged it in his 1961 book, "The Second Scientific American Book of Mathematical Puzzles & Diversions: A New Selection". Here's a picture and description:
http://mathworld.wolfram.com/Flexatube.html
I have often made these out of paper, putting red dots in the centers of the panels on one side and blue dots on the other.
Tom
Vaguely related: http://www.jaapsch.net/puzzles/magic.htm It seems to be making a comeback. --rwg
I just finally looked at the Gardner column that Tom mentioned (and that Allan Wechsler reminded me about), and I see that this object is due to Arthur Stone, even though the MathWorld page doesn't mention Stone. Jim Propp On Sat, Oct 24, 2015 at 6:49 PM, Tom Karzes <karzes@sonic.net> wrote:
It's called a flexatube. Martin Gardner describged it in his 1961 book, "The Second Scientific American Book of Mathematical Puzzles & Diversions: A New Selection". Here's a picture and description:
http://mathworld.wolfram.com/Flexatube.html
I have often made these out of paper, putting red dots in the centers of the panels on one side and blue dots on the other.
Tom
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participants (9)
-
Allan Wechsler -
Andy Latto -
Dan Asimov -
Dan Asimov -
Fred Lunnon -
Hilarie Orman -
James Propp -
rwg -
Tom Karzes