It is well known that a convenient mapping of a spherical surface to a flat surface involves projecting from the center onto the circumscribed cube, then unfolding the cube. As with any mapping of a spherical surface to a flat one, there is some amount of distortion. Of course, one could simply project onto an icosohedron instead and reduce the distortion, but using a cube is particularly convenient because one can use square arrays of pixels to represent the faces. (Of course there are hundreds of years of other kinds of projections.) So my question is, keeping these conveniently square faces, how much less distortion can you get by altering the mapping onto a face? For my purposes, what I care about is how distorted the square pixels become when mapped back to the sphere, because this determines the resolution of the flat representation. Is this an already solved problem? My thoughts follow... Using x,y,z coordinates on a unit sphere (x^2+y^2+z^2=1), and centering the cube's top face at <0,0,1>, the unadulterated projection to said top face is <x,y,z> -> <x/z,y/z> (where |x| and |y| are <= z). For this mapping, the linear density (Cartesian distance between two closely spaced points on the face divided by angular distance between the corresponding points on the sphere) varies between 1 and 3 over the face (although only by a factor of at most sqrt(3) as you vary direction at any given point); while the areal [is that a word?] density varies between 1 and 3sqrt(3) over the face. If we adulterate the projection so <x,y,z> -> <f(x/z),f(y/z)>, where f is a monotonic function, which may as well be antisymmetric about 0, and with f(0) = 0 and f(1) = 1, we can do better (and is that the only way to do better while keeping the square a square?). If we define g(s) = df(s)/ds, the areal density at <x,y,z> is g(x/z)g(y/z)/|z^3|, while the linear density is sqrt((g^2(x/z)((1-y^2)c+xys)^2 + g^2(y/z)((1-x^2)s+xyc)^2)/(1-(xs-yc)^2))/z^2 where c and s are the cos and sin of the direction (I'm pretty sure I computed this correctly). One obvious idea is to make the linear density constant along the face diagonals, giving f(s) = atan(sqrt(2)s)/atan(sqrt(2)), or constant along the horizontal and vertical lines through the center, giving f(s) = atan(s)/atan(1), or, more generally, f(s) = atan(ks)/atan(k) for some k. (The limit as k->0 gives the unadulterated projection.) For k = 1, the linear density ratio (max/min over the face) is sqrt(3), and the areal density ratio (max/min over the face) is sqrt(2). Perhaps a better idea is to make the areal density constant along the face diagonals, giving f(s) = integral((1+2s^2)^-(3/4),0,s)/integral((1+2s^2)^-(3/4),0,1) which I believe gives the least possible areal density ratio ((2/sqrt(3))^(3/2), a bit less than 5/4), although its linear density ratio (sqrt(2sqrt(3)), a bit less than 15/8) is not the least possible ratio between max and min linear densities (which would be sqrt(3), found along vs across the diagonal at the corners), but once you determine g(0) and g(1) the linear and areal densities at <0,0>,<1,0>, and <1,1> are all determined. BTW, this f(s) is suprisingly close to atan(ks)/atan(k) with k = 1.1488. --ms