I downloaded from somewhere-or-other a doctoral thesis (2006) by a Paul Leopardi at UNSW which gives a method for dividing any N-sphere into sectors of equal area (and reasonable compactness) by lines of latitude and longitude. I haven’t finished looking through it but I suspect that its approach can be adapted to suggest a generalization of the spiral method to N dimensions.

for 0 ≤ j < N :
x[j] = (width/N) * (0.5 + (j*P)%N)
y[j] = (height/N) * (0.5 + (j*Q)%N)

function p = spherepointsgolden(n)
%SPHEREPOINTSGOLDEN Place points evently on a unit sphere
% Generate a set of evenly distributed points on a unit sphere
% using the method of the Golden Section Spiral method.
%
% Output is a 3xN matrix of points for X,Y,Z.
%
% Based on python code from Patric Boucher on http://www.xsi-blog.com

inc = pi * (3-sqrt(5));
off = 2/n;

k = 0:n-1;
y = k * off – 1 + (off/2);
r = sqrt(1 – (y.^2));
phi = k * inc;

p = [cos(phi).*r; y ;sin(phi).*r]‘;
end

I’m looking to find a way to give me a set of the most unique vectors. This kind of thing would do it i think. the most seperated points on the 10D equivalent of a sphere should be the most unique set of vectors with 10 elements right?

public function pointsOnSphere ( num:uint, radius:Number=1.0 ) :Array
{
var pts:Array, inc:Number, off:Number, i:uint, y:Number, r:Number, phi:Number;

num = uint( num );
pts = new Array( num );
inc = Math.PI * ( 3 – Math.sqrt( 5 ) );
off = 2 / num;
i = num;
while ( i– ) {

y = i*off – 1 + ( off / 2 );
r = Math.sqrt( 1 – y*y );
phi = i*inc;
pts[ i ] = { x:Math.cos( phi )*r*radius, y:y*radius, z:Math.sin( phi )*r*radius };
}

return pts;
}