Best way to tile a sphere ?
June 2, 2008 7:07 AM   Subscribe

How can I tile a sphere using the minimum number of differently shaped units? I have a 3 meter diameter sphere and I need to tile it using units that are each about 400 square cm (ie like a 20 x 20cm square). The units must be flat (ie, not spherical triangles).

I'm currently thinking of using a geodesic tiling, but a (for instance) a 5V geodesic tiling results in (i think) 6 different tile shapes, and also has the problem that the density of tiles isn't constant over the surface - there are obvious visual artefacts where the tiles bunch up. I need to try to minimise the number of different types of unit, and to optimise the appearance so that it's as even as possible. Are there any better ways of tiling a sphere?
posted by silence to Computers & Internet (14 answers total) 6 users marked this as a favorite
 
If I'm understanding this problem correctly, you should be able to do this with one tile shape. Imagine splitting the sphere across the hemisphere. You can now make each side have any amount of triangles, with base upon the hemisphere and tip on the pole of the circle.
posted by LSK at 7:15 AM on June 2, 2008


I dont understand what you mean by "the units must be flat." You can't tile a sphere with perfectly flat shapes of any type. They are different geometries with different Gaussian curvatures.
posted by vacapinta at 7:19 AM on June 2, 2008


Response by poster: I'm obviously not expressing the problem very well - since I'm certainly no expert in geometry I'm probably not using the right terms. I'm only attempting to approximate the surface of the sphere, since I understand that it's obviously impossible to make a properly curved surface with flat surfaces. However, I would like to do a bit better than the kind of diamon shape that would be produced by a circle of triangles meeting at a point - imagine a geodesic sphere like this one. Each of the individual faces is flat, but they tile the sphere without any gaps between the edges - that's basically what i need.
posted by silence at 7:28 AM on June 2, 2008


Well, you could try a dymaxion map if I'm understanding you correctly, but any polyhedral projection like that will obviously have oddities at the vertices.
posted by vacapinta at 7:34 AM on June 2, 2008


20 hexagons and 12 pentagons is a common method of doing it.
posted by TedW at 7:37 AM on June 2, 2008


Poly is a free program that lets you play around with all sorts of polyhedra, including geodesic spheres. You can print out flattened versions and scale them up. Unfortunately, it doesn't seem to offer any help in determining the proper scale.
posted by hydrophonic at 7:39 AM on June 2, 2008


What is it about the geodesic sphere that you don't like?
posted by signal at 7:41 AM on June 2, 2008


Response by poster: thanks for the link to the poly software, that might help.

The problem with constructing it like a football is that the individual faces would be far too large - i need them to be around 400cm2.

I should probably explain what this is about - the intention is to build a large spherical display screen out of flat LED modules. If the modules are too big they will be too complex and expensive to manufacture, similarly the more different sizes and shapes of module there are the more expensive it becomes.
posted by silence at 7:45 AM on June 2, 2008


Response by poster: There are two problems with the geodesic. Firstly, in order to subdivide the sphere adequately I would need to use 6 different shapes of triangle (a 5V geodesic), and secondly the tiling tends to pinch visually which leads to unpleasant visual artefacts. If there's no better way to do it then I can probably live with this, but since my knowledge of geometry is extremely limited I'm thinking that there might be a better solution out there.
posted by silence at 7:50 AM on June 2, 2008


Best answer: Browsing the lists of polyhedra on wikipedia turns up some possibilities. You could start with a deltoidal hexecontahedron, and then make each face out of 6 or 12 triangles. I think you'd have only 2 tile shapes, and for bonus points you could make each face from the original polyhedron slightly convex.
posted by sfenders at 8:25 AM on June 2, 2008


Why not use projectors, either internal or external? It would seem the logistics of overlapping projected images to be seamless would be much less cumbersome (not to mention cheaper) than manufacturing non-rectangular LED modules.

Or there's POV (persistence of vision) display technoology, as here: YouTube link.
But 3 meters might be outside the range of the practicable limits.
posted by bartleby at 8:33 AM on June 2, 2008


Response by poster: sfenders - brilliant, thanks for the link. That's exactly the kind of thing I was looking for.

bartleby -
I've got quite a lot of experience with blending multiple projectors etc - and I'd love to do it like that. The problem is that it's just not possible to get the display bright enough or high enough contrast with projectors (it's got to work in daylight). I looked at POV systems too - but I don't like the flickeryness or the idea of having moving parts - plus I think scaling up to 3m diameter would be a real issue.
posted by silence at 8:49 AM on June 2, 2008


Best answer: Ok, so you are looking for a polyhedron with lots of faces that are all the same shape, where the polyhedron is "roughly spherical"? If you can make tiles of any shape (but they all need to be the same shape), you might find one of the Catalan Solids to be useful: they are the duals of the Archimedean solids (which have every vertex the same) and so are monohedral polyhedra---that is, every face is the same.

In terms of things that look spherical, the DisdyakisTriacontahedron and the PentagonalHexecontahedron look pretty good (the second one might be a little easier to make, or maybe not). But I haven't checked your desired tile size against the given constraints very well.

(Ok, I did a little bit of poking at it, and the surface area of a sphere that is three meters in diameter should be, in centimeters, 4*Pi*(150)^2 = 282743 cm^2. Dividing that by 400, you get approximately 706 tiles. But that seems like too many tiles for what you're describing, so I'm thinking probably I've got a stupid arithmetic error).

If you really are looking for about 700 tiles with not very many face shapes, it's possible that zonohedra may be helpful; what's not clear on a cursory glance is how many classes of rhombi are being used. More info is given here which may give you some information on how to construct the zonohedron you want: you would, I guess, want one with n(n-1) about 700, so n would be about 25.
posted by leahwrenn at 9:08 AM on June 2, 2008


I'd be interested in knowing what you finally end up with---this is an interesting problem.
posted by leahwrenn at 9:44 AM on June 3, 2008


« Older The Ghost in my Wife's Samsung Mobile   |   Nepotism at my Job is Driving me crazy. How do I... Newer »
This thread is closed to new comments.