Curvahedra
Curvahedra
I'm now a backer of this new kickstarter by Edmund Harriss.
I got to see an earlier incarnation of this project a few years ago at the JMM. What I probably like most about it is that it can be used to easily create surfaces of positive, zero, or negative curvature.
https://www.kickstarter.com/projects/2100742888/curvahedra-a-new-way-to-make-beautiful-geometry
I'm now a backer of this new kickstarter by Edmund Harriss.
I got to see an earlier incarnation of this project a few years ago at the JMM. What I probably like most about it is that it can be used to easily create surfaces of positive, zero, or negative curvature.
https://www.kickstarter.com/projects/2100742888/curvahedra-a-new-way-to-make-beautiful-geometry
Do you know if Edmund Harriss tried to sell his design to a toy company (these days through an agent like this one: https://goo.gl/TxGcVE)?
ReplyDeleteI don't, but hopefully he can chime in to answer your question.
ReplyDeleteHey, can I ask a question related to this image of yours on WIkipedia?: https://commons.wikimedia.org/wiki/File:633_honeycomb_one_cell_horosphere.png
ReplyDeleteIt is seems to me that this would be a nice way to "fake" a finite hexagonal tiling of the sphere (which, as we know, is impossible). The main idea is to make sure the viewer doesn't see the cluster-point of infinitely many hexagons located at the North Pole. Simply remove the grey outer sphere, make the inner sphere opaque, and position the camera so that the camera is directly below the South Pole looking up.
In addition, we can reinforce the illusion by "rotating" the sphere, giving the illusion that we're showing the viewer the entire sphere. In order not to reveal the cluster-point, this could be done by taking advantage of how the horosphere is an embedding of the Euclidean plane into hyperbolic space. We would essentially just translate the tiling. Specifically: there's a map from the horosphere to the plane that preserves distances (assuming the horosphere is sitting in hyperbolic space), so map the horosphere to the plane, translate it, and map it back.
This would make a nice GIF and a nice applet.
Do you think you could make this (either the image or the GIF)? I think it would look cool, and it would be a cool way to fool people into thinking that a purely hexagonal polyhedron could exist.
commons.wikimedia.org - File:633 honeycomb one cell horosphere.png - Wikimedia Commons
Hi Naomi Weinberger, that's a cool idea, and definitely possible. I especially like your thought to do a parabolic transformation (limit-rotation) of the "apeirohedron". I have a few comments...
ReplyDeleteI've made an animation reminiscent of that, though it's different than what you are picturing in other ways.
https://youtu.be/wjeeMlHsYjE
I also thought I'd point you to a program that you can use to look at these. The image below is a snip from it. Even though it is a puzzle program, it's a cool way to zoom around hyperbolic space with {6,3}-apeirohedra in it. They look like polyhedra because of the dihedral angle they take on in hyperbolic space. You can even use the twisting of the puzzle to rotate the polyhedra (though it is a normal rotation, not a limit-rotation).
http://astr73.narod.ru/M3dHT633/M3dHT633.html
I added your thought to my project idea list, so I'll keep it in mind as something cool to do. (This is the list I look at when I find extra time and motivation to go be creative.)
https://lh3.googleusercontent.com/wUyVKzSSt0Xt4AjmYtXeuyZNsbrogzv6qiCjwG6LJNK_itx798bExXkwfRb0E8FmH8-rzOceeu7NUFlcoYvYLdbtCnhWSNHaIdmQ=s0
Naomi Weinberger, I took a first stab at what you asked. See it here:
ReplyDeleteplus.google.com - What polyhedron is this? Hints: * The dihedral angle is 120 degrees. * It's…
That pictures a limit-rotation like you suggested. I think I'll experiment with some different transformations as well.
Thank you! About only seeing three faces at a time — since the horosphere is the same as the plane, couldn't you have chosen a hexagonal tiling of the plane with smaller hexagons? Roice Nelson
ReplyDeleteNaomi Weinberger, interesting question. I think you are right!
ReplyDeleteI chose the smallest hexagons with the restriction that the polyhedron also works as a cell of a regular honeycomb (120 degree dihedral angle, so that 3 polyhedra fit around each edge), but I could pick a hexagonal tiling with a larger dihedral angle even though the polyhedron won't be able to tile the space. I'll give it a shot.
Naomi Weinberger, what do you think of this?
ReplyDeleteI could go as small as we'd like, but this one seems like a nice balance.
https://lh3.googleusercontent.com/A-we7c87Mw5FBmirt7cnAa_CRX9pD47xPCed_n29xfS2Xst52QdVHupDKnLKkPZH9ZuUHaLXHF6oOWayVgbuazxw1wNeTIGZXjNl=s0