Let z1 be the co-latitude of some parallel which is to be correctly represented, then 2h sin 1⁄2z1 / δh = sin z1, and h = cos2 1⁄2z1; putting this value of h in equation (ii.) the radius of any parallel
= ρ = 2 sin 1⁄2z sec 1⁄2z1
(iii.)
This is Lambert’s conical equal-area projection with one standard parallel, the pole being the centre of the parallels.
If we put z1=θ, then h = 1, and the meridians are inclined at their true angles, also the scale at the pole becomes correct, and equation (iii.) becomes
ρ = 2 sin 1⁄2z;
(iv.)
this is the zenithal equal-area projection.
Reverting to the general expression for equal-area conical projections
ρ = √{2 (C − cos z) / h},