Geometry help needed

[Dominatus]

Hi, working on a independent project, and I need some help with the geometry... A structure that resembles half a sphere, composed of hexagon plates separated by a frame. Basically I need to know the easiest way to solve for SA of the frame and plates, respectively. I also need to figure out how many plates I need exactly... Any advice?

[K^2]

Do all faces have to be hexagons? That won't be very neat. A full sphere has a non-zero Euler Characteristic, so you can't build it out of hexagons. Half a sphere is possible, but it wouldn't be very even or pretty.

Simplest solution for a sphere is a truncated icosahedron. Basically, a soccer ball shape. Half of that should work for you pretty well, but you'll need a few pentagonal faces.

In practice, when large hemispherical domes are built, all faces are triangular. That gives the most stable construction. Each vertex of the frame is connected to 5 or 6 other vertices. In a full sphere, you'll have exactly 12 vertices with 5 neighbors. Rest will have 6. So for a hemisphere, you'd build the most stable structure with 6 such 5-neighbored vertices.

Surface area will depend on the choice of structure, of course. Truncated icosahedron is relatively easy to compute, if that's what you are leaning towards.

[Dominatus]

Okay, thanks. That should really help out.

[Idobox]

K^2, you have 12 vertices with 5 neighbours if you start from a dodecahedron, and turn each pentagonal face into 5 triangles. If you use a different scheme, you will different numbers.

For example, an icosahedron has 20 vertices, all connected to 5 neighbours. If you turn each triangle into 4 triangles (think triforce), you keep the 20 vertices with 5 edges, and add 30 new ones with 6 edges.

Dominatus,

What you're looking for is called a geodesic dome. If you don't want to solve the spatial geometry yourself, you can try to look for DIY projects, or alternatively use a 3D program, create a sphere with low polygon count, and extract the dimensions of each face. Be aware that the faces are not all the same, the triangles have different angles when you get closer to one of the 12 or 20 original vertices.

If you want hexagons instead of triangles, you have to take the dual, ie take the center of each triangle and connect them. Of course, you will have a few (12 or 20) pentagons. Actually, if you start from a simpler solid, like a cube or tetrahedron, you would get less irregular faces (8 or 4) but they would be square or triangle.

[ZetaX]

K^2, you have 12 vertices with 5 neighbours if you start from a dodecahedron, and turn each pentagonal face into 5 triangles. If you use a different scheme, you will different numbers.

For example, an icosahedron has 20 vertices, all connected to 5 neighbours. If you turn each triangle into 4 triangles (think triforce), you keep the 20 vertices with 5 edges, and add 30 new ones with 6 edges.

An icosahedron has only 12 vertices. The statement that any construct of triangles with 5 or 6 neighbours each has exactly 12 with 5 neighbours is correct, and you could write down a more general statement concerning triangular constructs with any amount of neighbours and any final shape. There are two ways to imagine what is going on:

- by the combinatorial/topological properties of triangular (i.e. "simplicial") complexes; this is where Euler's formul and the Euler characteristic come in

- by the curvature, which has a fixed total value for a sphere (or any other "object"), and every vertex adds some value to it, namely the 2 pi minus the sum of the angles of the polygons bordering it.

[K^2]

A useful thing to remember is that icosahedron is a dual of the dodecahedron. If the later has 12 faces and 20 vertices, the former has 20 faces and 12 vertices. Makes it easy to remember. In general, every platonic solid is a dual of a platonic solid, with tetrahedron being its own dual.

The statement that any construct of triangles with 5 or 6 neighbours each has exactly 12 with 5 neighbours is correct

It's probably worth clarifying to Idobox or anyone else not familiar with theory, that this is true for anything that's topologically a sphere. If we are still talking about a triangular mesh (or pure simplical 2-complex as a topologist might call it) then roughly speaking, it is equivalent to a sphere if you can drag all of its vertices and re-position them, without overlapping, to make something resembling a sphere. So a 3D model of a sofa is probably going to be equivalent to a sphere. While a common example of a surface equivalent to a torus is a surface of a coffee cup. A torus has Euler Characteristic of 0, so it can be built out of all vertices having exactly 6 neighbors. But more realistically, since you'll encounter positive and negative curvature, you'll be looking at vertices with 5 and 7 neighbors, but there has to be the same amount of both.

Actually, if anyone has interest in geometry or 3D modeling, I would recommend reading up on Euler Characteristic. Wikipedia article is pretty easy to follow, and it can be useful in many situations. I've made great use of the concept in some simple simulations of deforming objects. If an Euler Characteristic has changed, it likely signals that the body you are simulating has fragmented. Made for some good optimization.

@ZetaX: For some reason, I never thought about it in terms of conserved total curvature. That's a good way to look at it. Thanks.

[Dominatus]

Okay, these ideas are fascinating to me, a cat-watching-the-fishbowl sort of way. I won't pretend to understand most of it, but will try and educate myself some time.

It just occurred to me that I may not have been clear enough regarding some of the dimensions.. This was to be composed of unknown number of hexagon plates, say 48 cm across, set flush with a frame separating the parallels of adjoining hexagons by 12 cm. The total height of the dome gives it a radius of 10m. Hexagons at the base are divided at whatever point is required to make them flush with the surface.

So if that information helps or changes anything, good. I do enjoy reading these, even if I barely comprehend them. Keep it coming.

[K^2]

Do they have to be regular hexagons? Do they have to be all of the same size?

[Idobox]

oh, sorry, screwed up the vertices and faces on the icosahdron.

I've done a little of topology, and know the difference between simply and doubly connected surfaces for example, and that kind of stuff, but I haven't ever touched Euler characteristic or curvature questions, I'll check that, it sounds fascinating.

The 12 point thing, I assume, is only true for spheres in 3D. For a half-sphere, it is totally possible to have only one vertex with 5 edges.

Dominatus, you can't use regular hexagons (6 times the same angle), or you'll end up with a flat surface. If you want on order of magnitude approximation, you can calculate the surface of a half-sphere 2*pi*r², and divide by the surface of your hexagons (don't know the formula, but you can find it on google).

Unless you have a good reason to use hexagons, I advise using triangles. Making a flat triangle is easy, as long as the edges are straight. With other surfaces, it becomes much more difficult, and require higher precision.

[PakledHostage]

It just occurred to me that I may not have been clear enough regarding some of the dimensions..

Can you provide a sketch of what you're trying to do? A picture, as they say, is worth a thousand words... Here's a photo of a dome that may be close to what you (and many of the other people posting in this thread) are talking about. Is finding the surface area of the hexagonal faces of the dome what you're trying to solve?

[K^2]

The 12 point thing, I assume, is only true for spheres in 3D.

You can generalize rule to an arbitrary number of dimensions, but we are talking about a closed surface, yes. Euler Characteristic for a hemisphere is same as that of a disk, 1. Which means you can build a hemisphere out of triangles with every interior vertex having 6 neighbors, so long as on the boundary, for example, most vertices have 4 neighbors, with exactly 6 vertices having 3 neighbors.

In fact, lets generalize this a little bit. Consider an interior vertex with n neighbors. It accounts for n/2 edges and n/3 faces. So its contribution to Euler Characteristic is 1 - n/2 + n/3 = 1 - n/6. So for n = 6, this is exactly 0. For n = 5, this is +1/6. So in order to get +2 of the sphere, where all vertices are going to be interior vertices, you need to have 12 vertices with 5 neighbors, each adding 1/6 to Euler Characteristic giving you total of +2. If you have a vertex with 7 neighbors, it contributes -1/6, so you must balance it with another 5-neighbor vertex.

Example: Consider an octahedron. It is built out of 8 triangles having 6 vertices. At each vertex you have 4 triangles, so each vertex has 4 neighbors and. Therefore, each one contributes 1 - 4/6 = +1/3 to Euler Characteristic. That gives you a total of +2, which is same as for a sphere. And indeed, all regular polyhedra are topologically equivalent to spheres.

On the boundary, however, each vertex is going to have 1 more edge than face. So if it has n neighbors, it contributes 1 - n/2 + (n-1)/3 = 2/3 - n/6 to Euler Characteristic. So now, you get 0 contribution with n = 4. While n = 3 neighbored vertices are going to add +1/6. So if all of the interior vertices have 6 neighbors, for a total Euler Characteristic of 0, then on the edge you need to accumulate +1 of the disk, which can be done with 6 vertices of 3 neighbors each.

Example: Consider a regular pentagon broken up into 5 triangles. The interior vertex has five neighbors, so it contributes +1/6 to the total. There are 5 vertices on the boundary having 3 neighbors each, providing +1/6 each for a total of +5/6. The total for this object is +1, which is equivalent to that of the disk.