How can I visualize the standard topological "rubber-sheet" construction of a torus, that is, morphing a square into a torus?
How can I start or are there any examples in the Mathematica documentation respectively?
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$\begingroup$ Reminded me of mathematica.stackexchange.com/questions/7545/… $\endgroup$– Dr. belisariusFeb 18, 2014 at 13:42
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$\begingroup$ "Folding" in the origami sense will only be able to approximate a torus by faces with zero Gaussian curvature... $\endgroup$– Yves KlettFeb 18, 2014 at 14:27
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$\begingroup$ There is a demonstration of actual folding (though into a paper airplane rather than a torus): demonstrations.wolfram.com/OrigamiPaperPlanes $\endgroup$– bill sFeb 18, 2014 at 15:18
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$\begingroup$ @holistic since you've not claryfied the question I gave myself a freedom to reformulate it a little bit so it fits SE format and is clear now. If you dissagree feel free to revert it. $\endgroup$– Kuba ♦Feb 20, 2014 at 8:22
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2 Answers
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Edit I had some time so I've added full surface torus. Old code in edit history.
DynamicModule[{x = 2., l = 100., x2 = 2., l2 = 100., grid, fast, slow},
Grid[{{
Graphics3D[{
Dynamic[Map[{Blue, Polygon[#[[{1, 2, 4, 3}]]]} &,
Join @@@ (Join @@ Partition[#, {2, 2}, 1])
]&[
ControlActive[fast[l, l2], slow[l, l2]]]
]
}, PlotRange -> {{-7, 7}, {-7, 7}, {-1, 2}}, ImageSize -> 600,
Axes -> True, BaseStyle -> 18]
,
Column[{
Slider[Dynamic[x, (l = 10.^#; x = #) &], {.0001, 2.}],
Slider[Dynamic[x2, (l2 = 10.^#; x2 = #) &], {.0001, 2.}] }]
}}]
,
Initialization :> (
grid[l_, l2_, n_, m_] := Outer[Compose,
Array[RotationTransform[# Pi/l2, {0, 0, 1.}, {0, -l2, 0}] &, n, {-1, 1}],
Array[RotationTransform[# Pi/l, {1., 0, 0}, {0, 2, l}][{0, 2, 0}] &, m, {-1, 1}],
1];
fast[l_, l2_] = grid[l, l2, 10, 10];
slow[l_, l2_] = grid[l, l2, 50, 25];
)]
For < V.9 please switch Array to Table. This syntax for Array was introduced, silently, in V.9. linespace equivalent in MMA
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1$\begingroup$ @RunnyKine Thanks, I've always wanted to do this :P $\endgroup$– Kuba ♦Feb 20, 2014 at 8:18
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$\begingroup$ @Kuba: Thank you. That's what I imagined :) $\endgroup$– holisticFeb 20, 2014 at 9:24
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$\begingroup$ @Kuba yeah nevermind, I found out myself, sorry. Nice stuff! really well done. $\endgroup$– EllieNov 6, 2014 at 13:45
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Folding or mapping?
Mapping is simple. Say you have a rectangular piece of paper of width $a$ and height $b$, so that any point $P$ on the paper has the coordinates $(x,y)$ with $0\leq x\leq a$ and $0\leq y\leq b$ then you can simply use the same coordinates on the torus and identify $x=0$ with $x=a$ (and $y=0$ with $y=b$), respectively.
If you are thinking of the surface of a torus in 3D and you want to explicitly map a point on the paper onto the torus, you first need to parametrize the surface of the torus accordingly. If your torus is centered on $O=(0,0,0)$ with the "equatorial plane" in the $x-y$-plane, one possible parametrisation of the surface is given by $( R \cos\Phi + r \cos\Phi\cos\phi, R\sin\Phi - r \sin\Phi\cos\phi, r \sin\phi)$, where $R$ is the radius of the "centerline" and $r$ the radius of "tube".
Here, you can simply map $\Phi \leftrightarrow \frac{2\pi x}{a}$ and $\phi \leftrightarrow \frac{2\pi y}{b}$.
Now, folding (as in: building a paper model) is another story...
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$\begingroup$ I guess I meant both :). I wanted to map any point on the square to a point on the torus because I thought this is necessary to know if I want to fold a square to a torus with mathematica. Just like it is done with a piece of paper by rolling it to a zylinder and then connecting the ends of the zylinder. $\endgroup$– holisticFeb 18, 2014 at 22:34
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$\begingroup$ Rolling up the paper to make a zylinder is easy. Connecting the ends of the zylinder together is a problem, at least with any kind of paper I know of. Maybe someday such a cooperative material will be available at Amazon. $\endgroup$ Jul 9, 2017 at 22:25