Area and volume of a geodesic sphere

[Dick Fischbeck]

I'm looking for criticisms. 

Frequency = root [(n-2)/10]

This is the inverse of n= 10ff+2, which is well known and appears in the linked page from Kirby.

https://www.flickr.com/photos/kirbyurner/4884381012/

Triangles = 2(n-2)

This is well known as well and appears as the relationship between vertexes and faces of a geodesic sphere in that linked page.

If we multiply these two equations, we get:

Tetrahedrons = root [(2/5)((n-2)^3)]

If all the triangles on an omni-triangulated geodesic sphere are equal in size, then the number of triangles equals the area of that sphere.

What is not well known, afaik, is that we can determine the volume of that sphere (in equal size tetrahedrons) by only knowing the number of vertexes of that sphere. Volume is a function the number of vertexes.

Dick

[Gerry in Quebec]

Hi Dick,

Perhaps I don't fully understand your assumptions, but let's start with your statement, "If all the triangles on an omni-triangulated geodesic sphere are equal in size, then the number of triangles equals the area of that sphere."

We can test this by looking at a simple example of a geodesic sphere, a regular icosahedron. All 20 faces are equilateral triangles. If the polyhedron radius is 1 unit in length, then the surface area is 9.57 square units, not 20 sq. units.

I'll take a look at the other items in your post later.

Cheers,

- Gerry in Québec

[Dick Fischbeck]

Hi Gerry

I have a stipulation that I didn't mention in that all triangles can be scalene and all of them are equal in area. I'm not using square areas.

On Mon, May 24, 2021 at 4:43 PM Gerry in Quebec <toomey...@gmail.com> wrote:rea

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[Gerry in Quebec]

Hi Dick,

Okay, I think I understand you now. You're using the volume of the geodesic sphere's basic building block, a tetrahedron, as a unit of volume measurement, where all the building blocks are identical. Here's something to consider, though. Knowing the number of vertices on the sphere's surface, as well as the number of identical faces, and the number of edges does not let you calculate the volume, except in a limited number of cases. You also need chord factors.

 The image above is an example: two convex polyhedra that can be considered geodesic spheres. They have the same number of faces (24), edges (36) and vertices (14), and in each instance the constituent isosceles triangle faces (and related tetrahedra) are identical. But the single face type of one polyhedron differs from that of the other polyhedron. As a result, these two polyhedra have different surface areas and volumes.

- Gerry

[Dick Fischbeck]

I'll just add for now a picture that I like a lot.

dateposted-public

The idea is to multiply the frequency of a geodesic ball by the number of triangles. Frequency and the number of triangles are both a function of the number of vertexes.

f vertexes

The number of triangles (which can be of the same area and be the base of a tetrahedron) times the frequency (which is the height of the skinny tetrahedrons) has to equal the total volume (in tetrahedral units). So volume can be a function of the number of vertexes. That's all I'm saying.

To view this discussion on the web visit https://groups.google.com/d/msgid/geodesichelp/199c173b-1bc5-4d2b-8e37-84cb548e0c80n%40googlegroups.com.

[Levente Likhanecz]

hi Dick,
nothig to contribute to your math, but i love that pic on the moon.
that way (the triangular trumpets) i reconstructed many domes in
sketchup, knowing only chord factors, and lazy to calculate much.
especially taffgoch's rotegrities by chords or some heavily iterated
"non geometrical" solutions.

best regards, lev