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The answers I seem to be getting here are that this is not possible, but I have come across a formulae for subdividing an icosahedron (which I dont have to hand, but goes something like this)

$V = 10 u^2+2$

$E = 30 u^2$

$F = 20 u^2$

Where $u=\text{ frequency of subdivision}$

(I may have it the wrong way around but you get the picture) this allows us to calculate the amount of vertices, edges and faces for each subdivision.

Using this formula (or the correct one anyway) you can incrediment the frequency untill the number of faces reach 1000. It would appear in theory this will allow me to imitate a sphere with 1000 congruent polygon's (faces) by constructing a subdivided icosahedron because using $V-E+F=2$ I get $502-1500+1000=2$

Can anyone confirm if the faces will wrap around to make a full icosahedron without any missing faces?

Know the answer? reply and you will most likely get the vote.


Complete icosahedron with $20,40,80,120,160,200,240,280,320...1000$ faces, is it true?

Hi, Does anyone know if it possible to represent a sphere with $1000$ equal surfaces?

When I say equal surfaces, I am talking in terms of area. However it would also be preferable if the shape of these surfaces match.

For example, I am looking at subdivided icosahedrons, all surfaces are equal triangles in terms of shape and area, but to produce a nice sphere it would seem the number of surfaces would need to be $20, 80, 320$ etc.

Also I have come across zonohedrons - Does this allow any amount of surface? I can see that the surfaces are different shapes (but i should hope they have the same area)

I know this may sound a bit confusing, but any feedback will be much appreciated.

Thanks, Chris

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By "equal surfaces", you meant "congruent polygons" I suppose. In Mathematica at least, there is the Geodesate[] command, but the polygons are not guaranteed to be congruent. – J. M. Nov 23 '10 at 11:31
Congruent would be the correct terminology. My problem is with the amount of congruent surfaces that make up a spherical object ^.^ – Chris Nov 23 '10 at 11:54
Depends on what you mean by "represent." – Qiaochu Yuan Nov 23 '10 at 12:09
Ok let me rephrase the question. Is it possible to take 1000 congruent polygons and position them side by side in 3d space to look like a sphere, with no spaces inbetween the polygons. I want to make a sphere "look-a-like" with specifically 1000 faces. I am beginning to wonder if this is even mathmatically possible... any clarification? – Chris Nov 23 '10 at 12:17
It also depends on what you mean by "surface." Could you tell us what you want this for? – Qiaochu Yuan Nov 23 '10 at 12:25
up vote 7 down vote accepted

If you want the faces congruent, it will not work. If you try to use congruent polygons you cannot satisfy Euler's formula. This says if you use pentagons and hexagons there must be 12 pentagons. The truncated icosahedron, which has 12 pentagonal and 20 hexagonal faces is quite close to a sphere and there are only two kinds of faces. I thnk you could subdivide the hexagons into 7 and pentagons into one pentagon and five hexagons to get 200 hexagons and 12 pentagons, but I don't think the hexagons are all the same shape any more. Another subdivision has 1460 hexagons and 12 pentagons. I think there are more in between, but a quick web search didn't turn up any good references.

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I would vote on this if I could. Thankyou for getting straight to the point and providing a sufficient answer. – Chris Nov 23 '10 at 14:12
"truncated icosahedron" - i.e., the soccer ball – J. M. Nov 23 '10 at 14:13

Draw 1000 meridians enclosing an angle of 0.36 degrees between two successive ones. Or, if you prefer 1000 congruent triangles: Draw an equator and 500 meridians enclosing an angle of 0.72 degrees between two successive ones.

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+1 I don't think it is in the spirit of what was wanted, but it does solve the problem with matching shapes. – Ross Millikan Nov 23 '10 at 14:52

I've been told there are are only 3 ways to subdivide the starting triangles to make a geodesic dome:

  • class 1 (alternative method): Starting with one triangle, divide each edge into u segments, dividing the entire triangle into u^2 smaller triangles, using the original edges and new edges more or less parallel to the 3 original edges. This gives

Starting with a 20 sided icosahedron, the class 1 subdivision gives 20*u^2 triangles that cover the entire sphere: 20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, etc. triangular faces.

  • class 2 (triacon method): Starting with one triangle, pick some even integer u, divide each edge into u segments, and divide the entire triangle into 3*(u/2)^2 smaller triangles with new edges at right angles to the 3 original edges (erasing the original edges; the non-equilateral pieces along the edges count as as "half a triangle").

Starting with a 20 sided icosahedron, the class 2 subdivision gives 20*3*(u/2)^2 triangles that cover the entire sphere: 60, 240, 540, 960, 1500, 2160, etc. triangular faces.

  • class 3: ... alas, I don't seem to have the formula at hand.

So you won't get a geodesic sphere with exactly 1000 faces, but you can get one with around 1000 faces, with a frequency of 8 or 9.

While sub-dividing appears to create congruent equilateral triangles, stopping there gives small flat coplanar triangles -- the next design steps to "inflate" those vertices so they touch the surface of a sphere result in lots of triangles that are approximately equilateral, but not exactly -- even worse, they are not exactly congruent to each other, either. A 3v Icosahedron Dome uses 2 kinds of faces and 3 kinds of strut lengths.

It is possible to tweak the edge lengths to make thousands of triangles have exactly equal areas, even though they have slightly different shapes.

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