3 ms·
Ah, of course, thanks.
by gibybo 13y ago
Ah, of course, thanks.
- andrewcooke 13y agoyou can use longitude and z (height "up the axis") (it's uniform because a ring of thickness dz at height z is of area 2pi * z * cos(lat) * dz/cos(lat) (so the latitude cancels nicely - it's a trick used for equal area map projections).