A simple geodesic dome and Fibonacci sphere mesh generator

This repository has two scripts, geodome.py for geodesic dome and fibonacci.py for Fibonacci sphere. Each script generates a triangle mesh aligned with the unit sphere, saves it as PLY, and plots it in 3D by matplotlib.

geodome.py for Geodesic dome

This is a tiny (100 lines or so) script to calculate the geometry of geodesic domes by subdividing an initial icosahedron recursively. The calculated geodesic domes satisfy:

• The first vertex is always at (0, 0, 1), and the 12th vertex is always at (0, 0, -1).
• The 25th and 32nd vertices are always at (0, 1, 0) and (0, -1, 0) respectively (available if tessallation lv >= 1).
• The 102th and 125th vertices are always at (-1, 0, 0) and (1, 0, 0) respectively (available if tessallation lv >= 2).
• The vertices are on the unit shpere of radius 1; meaning that the vertex position is identical to the vertex normal.
• The face vertices are stored in CCW order.

fibonacci.py for Fibonacci sphere

This is a numpy version of Evenly distributing n points on a sphere. The script computes triangle edges using scipy.spatial.ConvexHull.

• The first vertex is always at (0, 0, 1), and the last vertex is always at (0, 0, -1).
• The vertices are on the unit shpere of radius 1; meaning that the vertex position is identical to the vertex normal.
• The face vertices are stored in CCW order.

Usage

Geodesic Dome

# Icosahedron
g = GeodesicDome()

# Subdivide N times
g.tessellate(3)

# Check the numbers of vertices / faces
print(f'num of vertices = {len(g.v)}, num of faces = {len(g.f)}')

# Save as PLY
g.save_as_ply('a.ply')

# 3D plot
g.plot3D()

Fibonacci sphere

# Fibonacci sphere of 1000 points
f = FibonacciSphere(1000)

# Check the number of vertices / faces
print(f'num of vertices = {len(f.v)}, num of faces = {len(f.f)}')

# Save as PLY
f.save_as_ply('a.ply')

# 3D plot
f.plot3D()

Algorithm

Geodesic Dome

Level 0 == Icosahedron

As explained in Wikipedia, a unit icosahedron can be defined as a circular permutations of (0, +/-1, +/-φ), where φ = (1 + sqrt(5)) / 2. This definition is simple, but not easy to use for some scenarios since none of the vertices are on an axis.

Instead, as explained in https://mae.ufl.edu/~uhk/PLATONIC-SOLIDS.pdf, we can define the vertices in cylindrical coordinates (r, θ, z) as follows.

1. Suppose the top and the bottom vertices of the icosahedron are on the Z axis. The rest of the vertices form two rings of pentagons orthogonal to the Z axis.
2. The distance b from a vertex on the pentagons to the Z axis is b = 1 / (2 sin(π/5)).
3. The distance a from the middle of one of the pentagon edges to the polar axis is a = 1 / (2 tan(π/5)).
4. The height c of the pentagonal pyramid (the distance from the pentagon to the top) is c = sqrt(3/4 - a^2).
5. The distance d between the pentagons is d = sqrt(3/4 - (b-a)^2).
6. The top/bottom vertices are (0, 0, +/- (c + d/2)).
7. The vertices of the upper pentagon are (b, 2nπ/5, d/2), where n=0, ..., 4.
8. The vertices of the lower pentagon are (b, (2n+1)π/5, -d/2), where n=0, ..., 4.

Finally we can convert the cylindrical coordinates to Cartesian by (x, y, z) = (r cosθ, r sinΘ, z) as usual.

Level n == Tessellation of Level (n-1)

The tessellation is implemented by subdividing each triangle into four subtriangles.

• For each triangle,
  1. insert new vertices at the midpoints of its edges,
  2. move (push) the new vertices in the radial direction so as to be on the unit sphere,
  3. replace the triangle with the four subtriangles as follows.

  O-----------O        O-----x-----O
   \         /          \   / \   /
    \       /            \ /   \ /
     \     /     ==>      x-----x
      \   /                \   /
       \ /                  \ /
        O                    O
  

Numbers of Vertices / Faces / Edges

• Level 0 == icosahedron has 12 vertices, 20 faces, and 30 edges (and satisfies Euler formula V + F - E = 2, of course).
• Level n has 10 * 4^n + 2 vertices, 20 * 4^n faces, and 20 * 4^n * 3 / 2 edges.
  • Level 1: 42 vertices, 80 faces
  • Level 2: 162 vertices, 320 faces
  • Level 3: 642 vertices, 1280 faces

Fibonacci sphere

Please check the following references.

• Evenly distributing n points on a sphere
• Álvaro González, "Measurement of areas on a sphere using Fibonacci and latitude-longitude lattices"