Check out this new web toy I made. Expl

Check out this new web toy I made. Explore the tilings of the hyperbolic plan...
Tim HuttonTim Hutton - 2014-09-13 00:22:48+0000 - Updated: 2014-09-13 20:42:36+0000
Check out this new web toy I made. Explore the tilings of the hyperbolic plane by moving your mouse around to smoothly change the curvature of space. Take the curvature positive to discover the polyhedra.


Dmitry Shintyakov - 2014-09-13 18:21:27+0000 - Updated: 2014-09-14 13:15:08+0000
Cool. I had experimented with cellular automata on hyperbolic tilings (on the some {5;4} tiling, in fact), and wrote a simple simulator for them, it's in Java:
And I have even found some spaceships:

Unfortunately, {5;4} tiling appeared boring: rules with spaceships are instable. I tried to write a simulator for the {4;5} tilings, but failed to develop a method of cell addressing.
Roice Nelson - 2014-09-13 19:40:39+0000
Very nice +Tim Hutton.  I like the addative coloring of the overlapping edges and polygons.  I like how you can move among all the geometries too.  Well done!
Owen Maresh - 2014-09-14 02:13:00+0000 - Updated: 2014-09-14 02:14:26+0000
You need to find someone who studies McKay correspondence and have them take a look at this. ( has relevant information, tagging +John Baez)
Owen Maresh - 2014-09-14 04:13:21+0000
Also, I think it might be prudent to make a table of curvatures which correspond to the table in the above paper. It's sort of funny that the dodecahedron has curvature 0.61 
Christopher Hanusa - 2014-09-14 12:32:30+0000
Refurio Anachro - 2014-09-14 14:50:41+0000
Cool. Maybe explain the mouse controls also on the index.html. Huh, did i just find an icosahedron among pentagons? Hah, not quite :)
Tim Hutton - 2014-09-14 18:08:31+0000 - Updated: 2014-09-14 19:48:08+0000
Thanks everyone. I'm still learning all this stuff so let me know if I've got anything wrong. +Owen Maresh, I think the curvature makes sense. Its reciprocal is the radius of the sphere, so 0.61 is what you'd expect for a dodecahedron in our model, where each vertex is distance 1 from the face center.
Tim Hutton - 2014-09-14 19:46:24+0000 - Updated: 2014-09-15 09:36:47+0000
+Refurio Anachro Yes, I noticed that! It's at curvature 0.894 in the {5,q} family. After some exploration I think it must be the great dodecahedron: (Better rendering would help here.) Now I'm looking for other things! I think the great icosahedron is there too, at 0.983 in the {3,q} family.
Refurio Anachro - 2014-09-14 20:07:00+0000
I bet you're right +Tim Hutton.

> I think the great icosahedron is there too, at 0.98 in the {3,q} family.

Of course it is! I would have missed that one, thanks!
Tim Hutton - 2014-09-14 20:14:53+0000 - Updated: 2014-09-14 20:15:18+0000
I've just updated the app with constant 3D rotation, to help with understanding what we're seeing. But it looks now like I need proper 3D rendering to show things clearly.
Tim Hutton - 2014-09-14 20:30:37+0000
Question: what is the weird spikiness at extreme negative curvatures? Is this an artefact of our circle inversion somehow or a genuine phenomenon?
Owen Maresh - 2014-09-14 20:53:54+0000 - Updated: 2014-09-15 00:45:12+0000
very roughly:

-1.115 Schlegel diagram of tetrahedron;
-1.054 Schlegel diagram of icosahedron
0.61 icosahedron,
0.815 octahedron
0.944 tetrahedron
0.983 first stellation of the dodecahedron
Tim Hutton - 2014-09-15 12:20:55+0000 - Updated: 2014-09-15 12:22:01+0000
+Owen Maresh I think it must be the great icosahedron at 0.983 in the {3,q} family, not the first stellation of the dodecahedron, because the base shape is a triangle not a pentagram. But now I realise a cool way to extend the app to include the stellated dodecahedron: if we had a slider for each dimension. The first would turn a line segment into various polygons, including star polygons because of the way the reflection works. Then the second slider would reflect that polygon in its edges, making all the tilings and polyhedra we see in the current version but also all the polyhedra made from star polygons. Then a third slider could turn those into space-filling tessellations and 4D polyhedra like the hypercube. We'd probably stop at three sliders to avoid hurting my brain even more...
Tim Hutton - 2014-09-17 15:42:23+0000 - Updated: 2014-09-30 09:57:10+0000
I've now made a version with 3 sliders:
It allows you to share the link directly to shapes you find. e.g. {5,3,4} is here: Here's {4,3,4}:
The curvatures are now based off the edge length of 1. Previously it was based on the polygon circumcircle radius of 1, which wasn't really sensible in this new multi-dimensional setting.
Share the shapes you find!
Christopher Hanusa - 2014-09-17 15:46:52+0000
Tim Hutton - 2014-09-17 15:54:43+0000 - Updated: 2014-09-17 16:26:44+0000
Oh heck, I've got a bug in the curvatures. No more sharing links until it's fixed please! Sorry! Hold the line.
Tim Hutton - 2014-09-17 16:04:13+0000 - Updated: 2014-09-17 16:08:42+0000
Bug fixed. I've edited my links above. Sorry about that.
+Christopher Hanusa your amusing hair one was something like this I think:
Tim Hutton - 2014-09-17 16:18:47+0000 - Updated: 2014-09-30 09:57:43+0000
Tim Hutton - 2014-09-17 16:22:21+0000 - Updated: 2014-09-30 09:58:45+0000
So now we can do star polygons and star polyhedra. Here's the pentagram:
which you can then bend into the first stellation of the dodecahedron here:
Tim Hutton - 2014-09-17 16:25:38+0000 - Updated: 2014-09-30 09:59:17+0000
Christopher Hanusa - 2014-09-17 17:08:57+0000
Tim Hutton - 2014-09-17 18:09:42+0000 - Updated: 2014-09-17 19:23:31+0000
+Christopher Hanusa: The one you mentioned here: is the tiling of space that would happen if regular tetrahedra worked. They don't, quite. Instead they leave little gaps. Famously, Aristotle thought they did fill space.
Christopher Hanusa - 2014-09-17 18:19:25+0000
Thanks +Tim Hutton for that elucidation.
Tim Hutton - 2014-09-18 08:43:09+0000 - Updated: 2014-09-18 10:12:40+0000
The third slider now allows positive curvature too, using stereographic projection from the 4th dimension.
Tim Hutton - 2014-09-18 08:55:00+0000
I'm making a list of useful coordinates here:
Anyone is welcome to add more. Just make a github account and edit directly or send them to me.
Tim Hutton - 2014-09-18 10:46:33+0000 - Updated: 2014-09-30 10:00:34+0000
Following on from the Aristotle story, if you bend space with the third slider only a tiny bit you can get the tetrahedra to close those gaps and fill space. You end up with {3,3,5}: (although we only show part of the mesh)
It has positive curvature, so it closes back on itself and actually makes a polyhedron in 4D made of 600 tetrahedra joined face-to-face:
Shamanic Harmonics - 2015-10-26 16:21:17+0000
Is there a way to export the tiling patterns? I'd like to use those in combination with some pattern generating software like Taprats
Tim Hutton - 2015-10-26 20:19:55+0000
+Shamanic Harmonics There's no export feature at the moment. What format should it output as?
J Davis - 2015-10-26 23:31:11+0000
First off Tim, let me congratulate you on an amazing piece of work. It is really amazing the shapes and tiling patterns that can be created just by moving he mouse on the simpler version! (I love both versions :)

It would be great to be able to freeze the geometry and export -
 If a all possible, Encapsulated Post Script so it can be vector and used in Illustrator.

Since it is also wrapping 3D maybe an export to .stl or .obj also?

THANKS again for this!

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