How does one parameterize the surface formed by a *real paper* Möbius strip?
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Here is a picture of a Möbius strip, made out of some thick green paper:
I want to know either an explicit parametrization, or a description of a process to find the shape formed by this strip, as it appears in the picture. Now before you jump up and declare, "That's easy! It's just $left(left[1+u cos frac theta2right]cos theta,left[1+ucosfrac theta2right]sin theta,usin frac theta2right)$ for $uinleft[-frac12!,frac12right]$, $thetain[0,2pi)$!" Understand that this parametrization misses some features of the picture; specifically, if you draw a line down the center of the strip, you get a circle, but the one in the picture is a kidney-bean shape, and non-planar. What equations would I need to solve to get a "minimum-energy" curve of a piece of planar paper which is being topologically constrained like this? Is it even true that the surface has zero curvature? (When I "reasonably" bend a piece of paper into a smooth shape, will it have zero curvature across the entire surface, or does some of paper's resistance correspond to my imparting non-zero curvature to the surface?)
This question is thus primarily concerned with the equilibrium shapes formed by paper and paper-like objects (analogous to minimal surface theory in relation to soap-bubble models). Anyone know references for this topic?
reference-request ordinary-differential-equations physics differential-topology minimal-surfaces
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
$endgroup$
|
show 14 more comments
$begingroup$
Here is a picture of a Möbius strip, made out of some thick green paper:
I want to know either an explicit parametrization, or a description of a process to find the shape formed by this strip, as it appears in the picture. Now before you jump up and declare, "That's easy! It's just $left(left[1+u cos frac theta2right]cos theta,left[1+ucosfrac theta2right]sin theta,usin frac theta2right)$ for $uinleft[-frac12!,frac12right]$, $thetain[0,2pi)$!" Understand that this parametrization misses some features of the picture; specifically, if you draw a line down the center of the strip, you get a circle, but the one in the picture is a kidney-bean shape, and non-planar. What equations would I need to solve to get a "minimum-energy" curve of a piece of planar paper which is being topologically constrained like this? Is it even true that the surface has zero curvature? (When I "reasonably" bend a piece of paper into a smooth shape, will it have zero curvature across the entire surface, or does some of paper's resistance correspond to my imparting non-zero curvature to the surface?)
This question is thus primarily concerned with the equilibrium shapes formed by paper and paper-like objects (analogous to minimal surface theory in relation to soap-bubble models). Anyone know references for this topic?
reference-request ordinary-differential-equations physics differential-topology minimal-surfaces
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
$endgroup$
4
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I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
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– MJD
Dec 17 '12 at 17:54
3
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+1 for the sheer aesthetic pleasure given by your picture.
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– Georges Elencwajg
Dec 17 '12 at 17:57
2
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So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
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– Thomas Andrews
Dec 17 '12 at 18:12
3
$begingroup$
This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
$endgroup$
– Jyrki Lahtonen
Dec 17 '12 at 19:23
3
$begingroup$
@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
$endgroup$
– MJD
Dec 17 '12 at 20:09
|
show 14 more comments
$begingroup$
Here is a picture of a Möbius strip, made out of some thick green paper:
I want to know either an explicit parametrization, or a description of a process to find the shape formed by this strip, as it appears in the picture. Now before you jump up and declare, "That's easy! It's just $left(left[1+u cos frac theta2right]cos theta,left[1+ucosfrac theta2right]sin theta,usin frac theta2right)$ for $uinleft[-frac12!,frac12right]$, $thetain[0,2pi)$!" Understand that this parametrization misses some features of the picture; specifically, if you draw a line down the center of the strip, you get a circle, but the one in the picture is a kidney-bean shape, and non-planar. What equations would I need to solve to get a "minimum-energy" curve of a piece of planar paper which is being topologically constrained like this? Is it even true that the surface has zero curvature? (When I "reasonably" bend a piece of paper into a smooth shape, will it have zero curvature across the entire surface, or does some of paper's resistance correspond to my imparting non-zero curvature to the surface?)
This question is thus primarily concerned with the equilibrium shapes formed by paper and paper-like objects (analogous to minimal surface theory in relation to soap-bubble models). Anyone know references for this topic?
reference-request ordinary-differential-equations physics differential-topology minimal-surfaces
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
$endgroup$
Here is a picture of a Möbius strip, made out of some thick green paper:
I want to know either an explicit parametrization, or a description of a process to find the shape formed by this strip, as it appears in the picture. Now before you jump up and declare, "That's easy! It's just $left(left[1+u cos frac theta2right]cos theta,left[1+ucosfrac theta2right]sin theta,usin frac theta2right)$ for $uinleft[-frac12!,frac12right]$, $thetain[0,2pi)$!" Understand that this parametrization misses some features of the picture; specifically, if you draw a line down the center of the strip, you get a circle, but the one in the picture is a kidney-bean shape, and non-planar. What equations would I need to solve to get a "minimum-energy" curve of a piece of planar paper which is being topologically constrained like this? Is it even true that the surface has zero curvature? (When I "reasonably" bend a piece of paper into a smooth shape, will it have zero curvature across the entire surface, or does some of paper's resistance correspond to my imparting non-zero curvature to the surface?)
This question is thus primarily concerned with the equilibrium shapes formed by paper and paper-like objects (analogous to minimal surface theory in relation to soap-bubble models). Anyone know references for this topic?
reference-request ordinary-differential-equations physics differential-topology minimal-surfaces
reference-request ordinary-differential-equations physics differential-topology minimal-surfaces
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
edited Dec 23 '18 at 22:44
Mario Carneiro
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
asked Dec 17 '12 at 17:49
Mario CarneiroMario Carneiro
18.5k34090
18.5k34090
4
$begingroup$
I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
$endgroup$
– MJD
Dec 17 '12 at 17:54
3
$begingroup$
+1 for the sheer aesthetic pleasure given by your picture.
$endgroup$
– Georges Elencwajg
Dec 17 '12 at 17:57
2
$begingroup$
So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
$endgroup$
– Thomas Andrews
Dec 17 '12 at 18:12
3
$begingroup$
This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
$endgroup$
– Jyrki Lahtonen
Dec 17 '12 at 19:23
3
$begingroup$
@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
$endgroup$
– MJD
Dec 17 '12 at 20:09
|
show 14 more comments
4
$begingroup$
I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
$endgroup$
– MJD
Dec 17 '12 at 17:54
3
$begingroup$
+1 for the sheer aesthetic pleasure given by your picture.
$endgroup$
– Georges Elencwajg
Dec 17 '12 at 17:57
2
$begingroup$
So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
$endgroup$
– Thomas Andrews
Dec 17 '12 at 18:12
3
$begingroup$
This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
$endgroup$
– Jyrki Lahtonen
Dec 17 '12 at 19:23
3
$begingroup$
@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
$endgroup$
– MJD
Dec 17 '12 at 20:09
4
4
$begingroup$
I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
$endgroup$
– MJD
Dec 17 '12 at 17:54
$begingroup$
I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
$endgroup$
– MJD
Dec 17 '12 at 17:54
3
3
$begingroup$
+1 for the sheer aesthetic pleasure given by your picture.
$endgroup$
– Georges Elencwajg
Dec 17 '12 at 17:57
$begingroup$
+1 for the sheer aesthetic pleasure given by your picture.
$endgroup$
– Georges Elencwajg
Dec 17 '12 at 17:57
2
2
$begingroup$
So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
$endgroup$
– Thomas Andrews
Dec 17 '12 at 18:12
$begingroup$
So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
$endgroup$
– Thomas Andrews
Dec 17 '12 at 18:12
3
3
$begingroup$
This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
$endgroup$
– Jyrki Lahtonen
Dec 17 '12 at 19:23
$begingroup$
This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
$endgroup$
– Jyrki Lahtonen
Dec 17 '12 at 19:23
3
3
$begingroup$
@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
$endgroup$
– MJD
Dec 17 '12 at 20:09
$begingroup$
@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
$endgroup$
– MJD
Dec 17 '12 at 20:09
|
show 14 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.
Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.
You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf
edited May 26 '13 at 8:35
azimut
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
$endgroup$
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
add a comment |
$begingroup$
MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative.
However, your question does address a very common misconception. In a sense, the same one addressed by the original question.
It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist and joining the ends is described by the parametrization you give.
There are some very real differences between the two that help explain why this isn't so.
FIRST
The parametrization describes a Mobius strip whose center line is a circle.The paper model is like the recycling symbol and has a triangular center line.
SECOND
The parametrization describes the path of a straight line centered on a circle; the straight line travels around the circle and completes a single 180 degree rotation (half twist) by the time it returns to its starting point. At no time is the straight line ever bent in 3D space.
If you draw straight lines crosswise on a strip of paper, then make a model of a Mobius strip, you will discover that in the three corners those lines are bent in 3D space. If you take the time to look seriously at those corners you will discover that each one incorporates a half twist. So a paper Mobius strip actually has three half twists.
THIRD
If you cut the paper model in half crosswise, it will revert to being a flat strip of paper. The Mobius strip being modeled is therefore developable.
If you cut a model of the parametrization in half, it will not lie flat. It is not developable.
To answer your question, the reason I say no one has created a parametrization of a developable Mobius strip, is because as far as I know no one has. As I pointed out, E.L.Starostin and G.H.M. van der Heijden have been able to mathematically model a developable Mobius strip; but their model is not a simple parametrization.
I suggest you go back to your university and point out that they missed the boat on this one. It's as if they told you that topologically a cube, a sphere, and a tetrahedron were the same, then said the formula for a cube describes them all.
By the way, these are not the only two geometrical forms a Mobius strip can take. There are more out there still waiting to be modeled mathematically.
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
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$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
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– Mario Carneiro
Aug 30 '14 at 19:16
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Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
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– Wayne Kollinger
Aug 30 '14 at 21:46
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Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
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– Mario Carneiro
Aug 30 '14 at 21:57
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I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
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– MJD
Aug 30 '14 at 23:27
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Wayne, I encourage you to click thecontact uslink at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?
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– davidlowryduda♦
Aug 31 '14 at 3:10
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.
Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.
You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf
edited May 26 '13 at 8:35
azimut
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
$endgroup$
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
add a comment |
$begingroup$
The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.
Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.
You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf
edited May 26 '13 at 8:35
azimut
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
$endgroup$
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
add a comment |
$begingroup$
The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.
Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.
You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf
edited May 26 '13 at 8:35
azimut
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
$endgroup$
The Möbius strip you show is a developable surface. No one, as far as I know, has been able to create a parametrization of it.
Since 1858, when the Möbius strip was discovered, mathematicians have been looking for a way to model it. The problem was finally solved in 2007 by E.L. Starostin and G.H.M. van der Heijden.
You might want to read their paper "The equilibrium shape of an elastic developable Möbius strip" by going to this site - http://www.ucl.ac.uk/~ucesgvd/pamm.pdf
edited May 26 '13 at 8:35
azimut
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
edited May 26 '13 at 8:35
azimut
16.4k1051100
edited May 26 '13 at 8:35
azimut
16.4k1051100
edited May 26 '13 at 8:35
azimut
16.4k1051100
16.4k1051100
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
answered Dec 18 '12 at 3:17
Wayne KollingerWayne Kollinger
31632
31632
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
add a comment |
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
1
1
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
$begingroup$
Wow! I can't believe someone's really looked into this in that much detail. +2! Also, it seems like they did a good job of a parametrization of the surface, since they have plots.
$endgroup$
– Mario Carneiro
Dec 18 '12 at 4:11
2
2
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
$begingroup$
The standard parameterization I saw in university was something like the one given by Wikipedia, or by the original question in this post: $$x(u,v)= left(1+frac{v}{2} cos frac{u}{2}right)cos u\ y(u,v)= left(1+frac{v}{2} cosfrac{u}{2}right)sin u\ z(u,v)= frac{v}{2}sin frac{u}{2}.$$ It seems clear that this parametrization has everywhere-zero curvature. Why do you say “No one… has been able to create a parametrization of it.”?
$endgroup$
– MJD
Mar 31 '14 at 15:01
1
1
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
$begingroup$
Pinging @MJD because Wayne's second answer appears to be targeted at you. As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2pi$, but the edge lines each have length $2pi(1+v_0^2/32+O(v_0^4))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:38
add a comment |
$begingroup$
MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative.
However, your question does address a very common misconception. In a sense, the same one addressed by the original question.
It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist and joining the ends is described by the parametrization you give.
There are some very real differences between the two that help explain why this isn't so.
FIRST
The parametrization describes a Mobius strip whose center line is a circle.The paper model is like the recycling symbol and has a triangular center line.
SECOND
The parametrization describes the path of a straight line centered on a circle; the straight line travels around the circle and completes a single 180 degree rotation (half twist) by the time it returns to its starting point. At no time is the straight line ever bent in 3D space.
If you draw straight lines crosswise on a strip of paper, then make a model of a Mobius strip, you will discover that in the three corners those lines are bent in 3D space. If you take the time to look seriously at those corners you will discover that each one incorporates a half twist. So a paper Mobius strip actually has three half twists.
THIRD
If you cut the paper model in half crosswise, it will revert to being a flat strip of paper. The Mobius strip being modeled is therefore developable.
If you cut a model of the parametrization in half, it will not lie flat. It is not developable.
To answer your question, the reason I say no one has created a parametrization of a developable Mobius strip, is because as far as I know no one has. As I pointed out, E.L.Starostin and G.H.M. van der Heijden have been able to mathematically model a developable Mobius strip; but their model is not a simple parametrization.
I suggest you go back to your university and point out that they missed the boat on this one. It's as if they told you that topologically a cube, a sphere, and a tetrahedron were the same, then said the formula for a cube describes them all.
By the way, these are not the only two geometrical forms a Mobius strip can take. There are more out there still waiting to be modeled mathematically.
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
$endgroup$
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
Wayne, I encourage you to click thecontact uslink at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?
$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
add a comment |
$begingroup$
MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative.
However, your question does address a very common misconception. In a sense, the same one addressed by the original question.
It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist and joining the ends is described by the parametrization you give.
There are some very real differences between the two that help explain why this isn't so.
FIRST
The parametrization describes a Mobius strip whose center line is a circle.The paper model is like the recycling symbol and has a triangular center line.
SECOND
The parametrization describes the path of a straight line centered on a circle; the straight line travels around the circle and completes a single 180 degree rotation (half twist) by the time it returns to its starting point. At no time is the straight line ever bent in 3D space.
If you draw straight lines crosswise on a strip of paper, then make a model of a Mobius strip, you will discover that in the three corners those lines are bent in 3D space. If you take the time to look seriously at those corners you will discover that each one incorporates a half twist. So a paper Mobius strip actually has three half twists.
THIRD
If you cut the paper model in half crosswise, it will revert to being a flat strip of paper. The Mobius strip being modeled is therefore developable.
If you cut a model of the parametrization in half, it will not lie flat. It is not developable.
To answer your question, the reason I say no one has created a parametrization of a developable Mobius strip, is because as far as I know no one has. As I pointed out, E.L.Starostin and G.H.M. van der Heijden have been able to mathematically model a developable Mobius strip; but their model is not a simple parametrization.
I suggest you go back to your university and point out that they missed the boat on this one. It's as if they told you that topologically a cube, a sphere, and a tetrahedron were the same, then said the formula for a cube describes them all.
By the way, these are not the only two geometrical forms a Mobius strip can take. There are more out there still waiting to be modeled mathematically.
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
$endgroup$
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
Wayne, I encourage you to click thecontact uslink at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?
$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
add a comment |
$begingroup$
MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative.
However, your question does address a very common misconception. In a sense, the same one addressed by the original question.
It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist and joining the ends is described by the parametrization you give.
There are some very real differences between the two that help explain why this isn't so.
FIRST
The parametrization describes a Mobius strip whose center line is a circle.The paper model is like the recycling symbol and has a triangular center line.
SECOND
The parametrization describes the path of a straight line centered on a circle; the straight line travels around the circle and completes a single 180 degree rotation (half twist) by the time it returns to its starting point. At no time is the straight line ever bent in 3D space.
If you draw straight lines crosswise on a strip of paper, then make a model of a Mobius strip, you will discover that in the three corners those lines are bent in 3D space. If you take the time to look seriously at those corners you will discover that each one incorporates a half twist. So a paper Mobius strip actually has three half twists.
THIRD
If you cut the paper model in half crosswise, it will revert to being a flat strip of paper. The Mobius strip being modeled is therefore developable.
If you cut a model of the parametrization in half, it will not lie flat. It is not developable.
To answer your question, the reason I say no one has created a parametrization of a developable Mobius strip, is because as far as I know no one has. As I pointed out, E.L.Starostin and G.H.M. van der Heijden have been able to mathematically model a developable Mobius strip; but their model is not a simple parametrization.
I suggest you go back to your university and point out that they missed the boat on this one. It's as if they told you that topologically a cube, a sphere, and a tetrahedron were the same, then said the formula for a cube describes them all.
By the way, these are not the only two geometrical forms a Mobius strip can take. There are more out there still waiting to be modeled mathematically.
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
$endgroup$
MJD, since you refer back to the original question, and given your previous comments, I assume you are trying to be provocative.
However, your question does address a very common misconception. In a sense, the same one addressed by the original question.
It is common to think that the Mobius strip that can be modeled by giving a strip of paper a half-twist and joining the ends is described by the parametrization you give.
There are some very real differences between the two that help explain why this isn't so.
FIRST
The parametrization describes a Mobius strip whose center line is a circle.The paper model is like the recycling symbol and has a triangular center line.
SECOND
The parametrization describes the path of a straight line centered on a circle; the straight line travels around the circle and completes a single 180 degree rotation (half twist) by the time it returns to its starting point. At no time is the straight line ever bent in 3D space.
If you draw straight lines crosswise on a strip of paper, then make a model of a Mobius strip, you will discover that in the three corners those lines are bent in 3D space. If you take the time to look seriously at those corners you will discover that each one incorporates a half twist. So a paper Mobius strip actually has three half twists.
THIRD
If you cut the paper model in half crosswise, it will revert to being a flat strip of paper. The Mobius strip being modeled is therefore developable.
If you cut a model of the parametrization in half, it will not lie flat. It is not developable.
To answer your question, the reason I say no one has created a parametrization of a developable Mobius strip, is because as far as I know no one has. As I pointed out, E.L.Starostin and G.H.M. van der Heijden have been able to mathematically model a developable Mobius strip; but their model is not a simple parametrization.
I suggest you go back to your university and point out that they missed the boat on this one. It's as if they told you that topologically a cube, a sphere, and a tetrahedron were the same, then said the formula for a cube describes them all.
By the way, these are not the only two geometrical forms a Mobius strip can take. There are more out there still waiting to be modeled mathematically.
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
answered Aug 30 '14 at 15:10
Wayne KollingerWayne Kollinger
111
111
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
Wayne, I encourage you to click thecontact uslink at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?
$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
add a comment |
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
Wayne, I encourage you to click thecontact uslink at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?
$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
You didn't make it obvious in this post, but the triangular "recycle sign" Mobius strip is only developable away from the three folds, so it is not a true solution. The true solution must be bent in 3D space. As for "three half twists", that is a choice as well. The photo in the OP contains one half twist, and (as long as the width is sufficiently small compared to the loop length) there are solutions for any odd number of half-twists (well, even for an even number of half twists, but then it's not a Mobius strip).
$endgroup$
– Mario Carneiro
Aug 30 '14 at 19:16
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Mario, thank you for pointing out to MJD that reason the "parametrized" Mobius strip won't lie flat when cut is the ration of the length of the side to the center line. I fell asleep at the switch on that one.
$endgroup$
– Wayne Kollinger
Aug 30 '14 at 21:46
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
Actually, I'd love to see how to make that argument rigorous, especially since there is no assumption that the final shape of the flattened surface is a rectangle, although it's not hard to see that the short ends are straight lines and the two sides are curves of some sort which are separated by no more than the strip width.
$endgroup$
– Mario Carneiro
Aug 30 '14 at 21:57
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
I wasn't trying to be provocative. By March 2014 I had forgotten what I had said back in December of 2012.
$endgroup$
– MJD
Aug 30 '14 at 23:27
$begingroup$
Wayne, I encourage you to click the
contact us link at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
$begingroup$
Wayne, I encourage you to click the
contact us link at the bottom to get your two accounts merged. This can help the organization of these answers, and why not?$endgroup$
– davidlowryduda♦
Aug 31 '14 at 3:10
add a comment |
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$begingroup$
I upvoted for the "before you jump up and declare…" part because that's exactly what I was on the point of doing.
$endgroup$
– MJD
Dec 17 '12 at 17:54
3
$begingroup$
+1 for the sheer aesthetic pleasure given by your picture.
$endgroup$
– Georges Elencwajg
Dec 17 '12 at 17:57
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So you don't want the exact curve represented here? Because you go out of your way to say you want the exact curve here. The set of constraints on your problem are a physics problem. The math problem inherent here is, given a set of constraints, what would the equation be? As such, you haven't given us a math problem, but a physics problem.
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– Thomas Andrews
Dec 17 '12 at 18:12
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This is an intriguing problem, but I agree with Thomas Andrews that we need input from physicists first. What integral gives us the total tension? (that we are supposed to minimize, I guess) Furthermore, an idealistic paper with an idealistic `join/glue' would still have a group of symmetries amounting to adding a constant to your parameter $theta$ (modulo $2pi$). But the solution manifestly does not have that symmetry! So some kind of spontaneous breaking of symmetry would have to happen, or may be the seam affects the equation after all?
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– Jyrki Lahtonen
Dec 17 '12 at 19:23
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@JyrkiLahtonen I am a physics ignoramus, but I imagine that there is an unstable symmetrical solution, which, slightly perturbed, becomes a configuration that collapses into a stable asymmetrical solution that is a member of a symmetrical family of configurations; exactly which one depends on the perturbation. I suppose the situation is analogous to that of a perfectly symmetrical pencil balanced on end. Which way does it fall? The pencil doesn't fall until it is perturbed, and it is the perturbation which breaks the symmetry.
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– MJD
Dec 17 '12 at 20:09