Why can't points be plotted on a sphere equally such that a sphere could be divided where all the legs are equal say many triangles? The rate of change is the same in all directions so why are some legs longer in a geodesic dome? This is counter-intuitive for me.
2026-03-25 11:20:17.1774437617
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Rate of change when moving on a sphere is the same in all directions. So why different leg lengths of triangles required for a geodesic dome?
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If you are looking for much discussion on how to "optimize" the number of struts for geodesic dome tesselation, have a look at Google's group "Geodesic Help Group".
Here is a fun experiment to do to understand how equilateral triangles can go together to make regular polyhedra. You will need construction paper and tape.
Cut equally sized equilateral triangles. You will need $32$ of them. To avoid writing "equilateral" every time, when I say triangle I mean equilateral triangle.
We are going to construct polyhedra by taping our triangles together with a fixed number of them meeting at a vertex.
If we insist that only one triangle can be at a vertex then the only shape we can construct is the triangle itself. Try $2$ triangles at a vertex and it is just as uninteresting.
Tape $3$ triangles together so they meet at a vertex. The three $60^{\circ}$ angles that meet at the vertex will add up to less than $360^{\circ}$ so when you tape the next triangle you will get a convex shape. In fact, you will only be able to tape one more triangle to get a tetrahedron.
Now do the same thing with $4$ triangles at a vertex. You will get an octahedron.
Five triangles at a vertex will give you an icosahedron.
What happens when we try $6$ triangles at a vertex? The $6$ interior angles add up to $360^{\circ}$ so adding more triangles tiles the plane; it does not give you a convex shape.
Therefore, the tetrahedron, the octahedron and the icosahedron are the only regular polyhedra that can be formed with equilateral triangles and none of these approximate the sphere that well.
As a bonus, repeat the same experiment with squares and pentagons to get the cube and the dodecahedron.
Since regular hexagons already tile the plane, no Platonic solid can be formed with a regular $n$-gon where $n\ge 5$.