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Deducing the Sanson-Flamsteed (sinusoidal) Projection

Imagine the equatorial aspect of a projection with some similarities to the Kavrayskiy VII but more strict properties:

Dimensions on a sphere
Spherical dimensions relevant to the geometry of the sinusoidal projection.

Consider a point P on Earth. Its distance to the Equator is m = phi R (this simple expression is one of the advantages of expressing coordinate angles as radians), which is the mapped point's ordinate. The parallel passing through P has radius r=R cos phi and circumference p=2pi r. Due to the pseudocylindrical property, the abscissa is proportional to half this value, with a factor lambda/pi. Thus,

x=R lambda cos phi; y=phi R
Sinusoidal map
Sinusoidal map in the equatorial aspect

The result is a classic projection, historically attributed to several authors, most notably Mercator, Sanson and Flamsteed. Today it is better known as the sinusoidal projection, after the shape of its meridians, which are of the form x=k sin(y+pi/2) with -1<=k<=1.

Despite its antiquity, the fact that the sinusoidal projection is equal-area seems to have been ignored or overlooked for a long time. This is surprising, because the sinusoidal's areal correspondence with the sphere may be intuitively determined by Cavalieri's principle, or by analogy to a stack of equidistant cylindrical maps. Let us present an informal proof based on calculus.

Corresponding infinitesimal area elements along parallels of sphere and sinusoidal map. Area element on a sphere
Area element on a sphere

Let us define an area element on the sphere, given by a thin ring at latitude phi. We know the ring's radius, R cos phi; its thickness, again because angles are expressed in radians, is R dphi, therefore its area is 2pi R^2 cos phi dphi. Let Ss be the area between the Equator and parallel Phi:

Ss=int(2pi R^2 cos phi dphi, 0, Phi)=2pi R^2 sin Phi

On the map, the corresponding area element is a horizontal strip with width determined by lambda=pi: 2R pi cos(phi). Its height is dy=Rdphi, and its area 2R^2pi cos phi dphi. The area Sm on the map between the Equator and Phi is

Sm=int(2 R^2 pi cos phi dphi, 0, Phi)=2pi R^2 sin Phi = Sm

The area of a "slice" between any two parallels can be calculated simply by changing the integration limits and remains identical on both Earth and the map. Besides, given a slice its area between any two meridians is identical on Earth and map because the scale is constant along all parallels. Therefore, the area is the same in any corresponding "cells" defined by two pairs of parallels and meridians. Finally, by composition, any corresponding regions on Earth and map have equal areas.



HomeSite MapKavrayskiy VIIHow Projections are CreatedCraster Parabolic  www.progonos.com/furuti    January  7, 2013
Copyright © 1996, 1997 Carlos A. Furuti