18
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ Reminded me of mathematica.stackexchange.com/questions/7545/… $\endgroup$ Feb 18, 2014 at 13:42
  • $\begingroup$ "Folding" in the origami sense will only be able to approximate a torus by faces with zero Gaussian curvature... $\endgroup$
    – Yves Klett
    Feb 18, 2014 at 14:27
  • $\begingroup$ There is a demonstration of actual folding (though into a paper airplane rather than a torus): demonstrations.wolfram.com/OrigamiPaperPlanes $\endgroup$
    – bill s
    Feb 18, 2014 at 15:18
  • $\begingroup$ @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

2 Answers 2

41
$\begingroup$

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];
   )]

enter image description here

For < V.9 please switch Array to Table. This syntax for Array was introduced, silently, in V.9. linespace equivalent in MMA

$\endgroup$
7
  • 2
    $\begingroup$ This is nice +1. $\endgroup$
    – RunnyKine
    Feb 20, 2014 at 0:28
  • 1
    $\begingroup$ @RunnyKine Thanks, I've always wanted to do this :P $\endgroup$
    – Kuba
    Feb 20, 2014 at 8:18
  • $\begingroup$ @Kuba: Thank you. That's what I imagined :) $\endgroup$
    – holistic
    Feb 20, 2014 at 9:24
  • $\begingroup$ Nice animation ++1 $\endgroup$
    – Junho Lee
    Nov 4, 2014 at 8:37
  • $\begingroup$ @Kuba yeah nevermind, I found out myself, sorry. Nice stuff! really well done. $\endgroup$
    – Ellie
    Nov 6, 2014 at 13:45
6
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – holistic
    Feb 18, 2014 at 22:34
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.