4 ms·
After a bit of further thought, I realized that the triangular sectors aren't equilateral after just one iteration. The angles at the equator are 90°, and the p
by sesqu 16y ago
After a bit of further thought, I realized that the triangular sectors aren't equilateral after just one iteration. The angles at the equator are 90°, and the polar angles are 180°/n - works when n=4. But then the four nested ones will be 3x(90° 60° 60°) and one equilateral, and so no longer identical (and probably not of equal areas).
For area-equality, shape-inequality, a recursive three-division becomes also worthy of consideration.
- eru 16y agoMight work. Here's what I thought of: Use a Cartesian coordinate system with the centre of the sphere as its origin and the radius of the sphere normalized to 1. Each point on the square has coordinates x_1,x_2,x_3. The new coordinates will live in [0,1]^3 (i.e. three real numbers between 0 and 1). To convert, do the following: Generate a random number r from a N(0,1) normal distribution. For each coordinate determine the quantile of q_i := rx_i in the N(0,1) normal distribution. (This results in 0 <= q_i <= 1.) Take the triple (q_1,q_2,q_3) and express it as a binary number. Bonus points for using octal numbers like this: Q(j) = q_1(j) + 2q_2(j) + 4*q_3(j) where Q(j) is the j-th octal digit of the number Q after the comma and q_i(j) is the j-th digital digit of the number q_i after comma. (This octal scheme has the property that you can cut off at any time for getting a less accurate description of the location.) This coordinate scheme is redundant. Also a uniform distribution of random points on the surface of the sphere, will lead to a uniform distribution of coordinates in this scheme.