geometry - Partition the Unit Sphere -

i'd make lookup table unit vectors. each unit vector map bin in table, , bin contain information small set of vectors similar directions.

i represent vector using ($\theta$, $\phi$, 1), , chop angle ranges bins make 2d lookup table (so first bin theta in range of 0 $2*\pi / n$, n number of bins want theta direction). trouble think bins going represent larger areas on surface of unit sphere others, , regions of same size.

am wrong in thinking evenly dividing angle range make bins larger others? if not, know better way of making lookup table?

i found papers , presentations this one, i'm not going lie, don't understand (i've heard of lebesgue measure, i'll damned if know means), , doesn't particularly promising anyway.

if split longitude n equal size segments, then, equal area domains on unit sphere, have have "uneven" segments along latitude dimension. area of spherical segment between 2 lines of constant latitude (parallels) depends on "height", i.e., length of projection of segment vertical axis. means if split vertical axis equal length parts, splitting sphere in equal area domains.

the bottom line is: following n*m domains have equal areas:

• 2*k*pi/n < longitute < 2*(k+1)pi/n, k=0...n-1
• -1 + 2*j/m < sin(latitude) < -1 + 2*(j+1)/m, j=0...m-1