1
$\begingroup$

Suppose I want to triangulate a sphere, but the maximum edge length of a triangle is to be at most L.

What's known about how the number N of triangles needed increases as L decreases? Any research articles/blog posts/algorithmsetc I should look up to get answers to this question?

$\endgroup$
1
  • 2
    $\begingroup$ For small $L$ a subdivided Platonic solid (a.k.a. a geodesic polyhedron) should give a nice upper bound on $N$. $\endgroup$
    – user856
    Feb 19, 2020 at 6:43

1 Answer 1

1
$\begingroup$

As @Rahul commented, a geodesic polyhedron will give you a good idea of what a triangulation will look like. To estimate the number of triangles note first that the triangles are approximately equilateral and second that a triangle on the sphere has striclty bigger area that a triangle with the same side lengths in the flat plane, where the ratio of the areas goes to one as the side lengths go to zero. An equilateral triangle in the plane with side length $L$ has area $\frac{\sqrt{3}}{4}L^2$, the sphere (with unit radius) has area $4\pi$. Hence you need approximately $\frac{16\pi}{\sqrt{3}L^2}$ triangles to cover the sphere.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .