HomeSite MapMollweideHow Projections are CreatedEquidistant Cylindrical, Winkel I/IIMap Projections

Two Aspects for Two Arbitrary Azimuthal Projections

General Polar Azimuthal Projections

General plan of azimuthal projections General azimuthal projection, mapped
The general case of azimuthal projections is mainly defined by a function that transforms distances from the center of projection T.

The mathematical development of the azimuthal orthographic projection is purely geometric. Even though several azimuthal projections - like the two explained here - do not follow such a perspective process, all can be reduced to a general pattern.

On a sphere, two angles determine the distance and position of any point P relative to the center of projection T. For north polar aspects, mu equals the longitude lambda, while psi equals the colatitude Phi, or pi/2-phi.

On the map, due to the azimuthal property,

theta=mu

The fundamental characteristic of an azimuthal projection is a function which transforms distances from the center of the map, and therefore determines the spacing of parallels in the polar aspects:

rho=f(R psi)
General plan of azimuthal projections
The defining distance function for some azimuthal projections

And

x=rho cos theta, y=rho sin theta

The Azimuthal Equidistant Projection

For the azimuthal equidistant, an important projection for navigational applications, the distance rho from the center of the map to any other point is directly proportional to its true radial distance from the center of projection. In the north polar aspect:
rho=(pi/2-phi)R

The austral aspect is just as easy, with

rho=(pi/2+phi)R, theta=-lambda

Lambert's Azimuthal Equal-area Projection

In the only projection both azimuthal and equal-area, created by Lambert and suitable for world maps, the relative distance of points from the center of the map is progressively reduced in order to keep areal equivalence. Formulas follow from basic integral calculus; first we define an area element on both Earth and map.
Area element on the sphere

Let a thin spherical zone be the element, given the colatitude Phi:

ds=2piRsin Phi RdPi=2piR^2sin Phi dPhi

The corresponding element on a polar azimuthal map is a ring with area:

ds=2pi rho d rho
Area element on a polar azimuthal map

For any given colatitude Phi1, we want rho1 such that the spherical cap bounded by Phi1 and the disc bounded by rho1 on the map have identical areas. This is enough for areal preservation because the scales along circumferences on both the zone and ring, although different, are constant - again, due to the azimuthal property.

int from 0 to Phi1 2piR^2sin Phi dPhi=int from 0 to rho1 2pi rho d rho;-R^2cosPhi from 0 to Phi1 = rho^2/2 from 0 to rho1; -2R^2(cos Phi1-1)=rho1^2
rho1=Rsqrt(2)sqrt(1-cos Phi1)=2Rsqrt((1-cos Phi1)/2)=2R sin(Phi1/2)
rho=2Rsin((pi/2-phi)/2)

A similar sign change applies if the south polar aspect is intended.

Combined north polar azimuthal equidistant and equal-area projections
Combined north polar azimuthal equidistant (top) and equal-area (bottom) maps.
The combined map shows both projections virtually identical at latitudes north of 60°N (as predicted by the graph: Phi = 0.534). Beyond that, parallels get closer and closer together in Lambert's half, while remaining equally spaced in the azimuthal equidistant portion. The resulting areal difference is clearly visible in Antarctica.

General Equatorial Aspect for Azimuthal Maps

Calculating other aspects for azimuthal maps is possible by introducing coordinate transformations and rotations in space. However, the important equatorial aspect can be obtained in a more direct way, using two properties of triangles on a spherical surface.

The laws of sines and cosines for spherical triangles
Given the angles A, B, C on a spherical triangle's vertices, and the corresponding angles alpha, beta, gamma between edges connecting triangle vertices and the center of sphere O:   A spherical triangle
Law of sines sinA/alpha=sinB/beta=sinC/gamma
Law of cosines cos gamma=cos alpha cos beta+sin alpha sin beta sin C
A spherical triangle
A triangle on the sphere is defined by the center of projection T, the point to be projected P and an equatorial point with the same longitude.

On the equatorial aspect, the center of projection T lies on the intersection of the Equator and an arbitrary central meridian. It is one vertex of a spherical triangle, the second is projected point P, and the third one lies on the Equator at the same longitude as P. The respective angles at the center of the sphere are phi, lambda and alpha, and we want the angle at T, theta which corresponds to mu in the first diagram.

sin theta/sin phi = sin(pi/2)/sin alpha
cos alpha=cos phi cos lambda + sin phi sin lambda cos(pi/2); alpha=arccos(cos phi cos lambda)
cos phi=cos alpha cos lambda + sin alpha sin lambda cos theta
cos theta=(cos phi - cos alpha cos lambda)/sin alpha sin lambda=cos phi(1-cos2 lambda)/(sin alpha sin lambda)
sin theta=sin phi/sin alpha
cos theta=cos phi sin lambda/sin alpha

The Equatorial Azimuthal Equidistant

Just substituting the expressions for cos theta and sin theta, rho=r=alpha R:
alpha=arccos(cos phi cos lambda); x=alpha R cos phi sin lambda/sin alpha, y=alpha R sin phi/sin alpha

If lambda=phi=0, sin alpha=0 but x=y=0

The Equatorial Azimuthal Equal-Area

In the equations for Lambert's equal-area azimuthal projection, substitute alpha for Phi1:
x=rho cos theta = rho cos phi sin lambda/sin alpha; y=rho sin theta=rho sin phi / sin alpha

Now using two trigonometric identities,

sin2a=2sin a cos a;cos2a=cos^2a-sin^2a=2cos^2a-1;(cos2a+1)=cos^2a

the common factor can be expanded:

rho/sin alpha=2Rsin(alpha/2)/sin alpha=2Rsin(alpha/2)/2sin(alpha/2)cos(alpha/2)=R/cos(alpha/2)=R/sqrt((cos alpha+1)/2)=Rsqrt(2)/sqrt(cos phi cos lamda+1)

Finally,

k=R sqrt(2/(1+cos phi cos lambda));x=k cos phi sin lambda; y=k sin phi
Combined equatorial azimuthal equidistant and equal-area projections
Combined azimuthal equidistant (top) and equal-area (bottom) equatorial maps.
Again, both projections are very similar near the center of projection: the northern and southern portions of Africa join almost seamlessly.

HomeSite MapMollweideHow Projections are CreatedEquidistant Cylindrical, Winkel I/II  www.progonos.com/furuti    December  3, 2012
Copyright © 2001 Carlos A. Furuti