Message Boards Message Boards

GROUPS:

[GIF] Fall Out (Rotating circle on the projective plane)

Posted 11 months ago
1578 Views
|
1 Reply
|
5 Total Likes
|

Rotating circle on the projective plane

Fall Out

This was a result of various experiments mapping from the sphere to the plane.

In this case, I'm taking a circle of disks on the sphere, then mapping to the plane by the following map: each point on the sphere (except those on the equator) is sent to the point on the $z=1$ plane lying on the same line through the origin (this map arises as a way of identifying [most of] the projective plane with an actual plane). Then, apply the inversion in the unit circle $z \mapsto \frac{z}{|z|^2}$.

The circle on the sphere is the orbit of the point $p = (0,1/2,\sqrt{3}/2)$ under rotations around $(\cos s, 0 \sin s)$. Here $s$ is treated as the time parameter and varies from $0$ to $\pi$.

Here's the code:

inversion[p_] := p/Norm[p]^2;

With[{n = 141, d = .01, p = {0, 1/2, Sqrt[3]/2}, 
  b = NullSpace[{N[{0, 1/2, Sqrt[3]/2}]}], 
  cols = {Black, GrayLevel[.95]}},
 Manipulate[
  Graphics[
   {PointSize[.01], cols[[1]],
    Polygon /@
     Table[inversion[#1[[1 ;; 2]]/#1[[3]]] 
       &[RotationMatrix[t, {Cos[s], 0, Sin[s]}].(Cos[d] p + Sin[d] (Cos[θ] b[[1]] + Sin[θ] b[[2]]))],
      {t, 0., 2 π, 2 π/n}, {θ, 0., 2 π - 2 π/20, 2 π/20}]},
   PlotRange -> 4, ImageSize -> 540, Background -> cols[[-1]]],
  {s, 0., π}]]

enter image description here - Congratulations! This post is now a Staff Pick as distinguished by a badge on your profile! Thank you, keep it coming!

Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract