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Deducing Mollweide's Projection

Although equal-area, mathematically very simple, and preserving parallel spacing, the previous sinusoidal/Sanson-Flamsteed projection is not completely satisfactory at high latitudes, due to excessive shearing and crowded meridians. Craster's parabolic projection has meridians a bit more rounded, but the poles are still sharp. A slightly more complicated analysis leads to Mollweide's projection.

Plan of Mollweide projection
Plan of Mollweide's projection

Suppose the equatorial aspect of an equal-area projection with the following properties:

Since the projection is pseudocylindrical with predetermined meridian shapes, let us repeat the approach for determining the equations of the parabolic design: for any parallel, find an ordinate that equates corresponding areas on map and Earth.

Consider an ellipse centered on the origin, with major axis a on the x-axis:

x^2/a^2+y^2/b^2=1
x^2=a^2(1-y^2/b^2)

For -b<=y<=b, x=a sqrt(b^2-y^2)/b.
The area between the x-axis and the parallel mapped into y=Y is

Sm=2 integral from 0 to Y (xdy) = 2a integral from 0 to Y (sqrt(b^2-y^2)dy)/b

For 0<=y<=b, let y=b sin theta: 0<=theta<=pi/2, dy=b cos theta dtheta

int sqrt(b^2-y^2)dy=int sqrt(b^2(1-sin^2theta))dy=int b cos theta b cos theta dtheta = b^2 int cos^2 theta dtheta

Since

cos^2 alpha=(1+cos 2alpha)/2; int cos n alpha dalpha=(1/n)sin n alpha + C
b^2 int cos^2 theta dtheta=(b^2/2)(int dtheta + int cos 2theta dtheta)=b^2(theta+sin2theta/2)/2+C

and

Sm=2ab/2 (2theta+sin2theta)/2+C = ab(2theta+sin 2theta)/2

for some 0<=theta<=pi/2.

Area bounded by the Equator and another parallel
The area bounded by the Equator and another parallel

Because a=2b, the area of the full ellipse is ab pi = a^2 pi/2.
The area of a spherical Earth is 4piR^2, therefore

a=R sqrt 8
y=(R sqrt 8 sin theta)/2
Sm=2R^2(2theta+sin 2theta)

From the development of the sinusoidal projection, we know that the region on a sphere bounded by the Equator and a parallel phi is a spherical zone with area

Ss=2piR^2sin phi

Making Sm=Ss,

2theta+sin 2theta = pi sin phi
Mollweide projection
Mollweide's projection

Unfortunately, unlike for Craster's, there is not a closed algebraic solution that directly converts phi (via theta) to y. We must resort to numerical root solving, which essentially comprises repeatedly "guessing" approximate values for theta and evaluating differences until a desired precision is achieved. This task is ideally suited to electronic computers; previously, human "computers" (the original meaning of the word) composed interpolation tables by laboriously calculating values for selected latitudes. Nevertheless, iterative numerical algorithms like the secant and Newton-Raphson methods converge relatively quickly if the initial guess is about phi itself, except near - but not at - the poles.

Finally, from the ellipse equation, the Eastern boundary meridian xb is given by

xb=2sqrt(2R^2-y)=2sqrt(2R^2-2R^2sin^2theta)=2sqrt2sqrt(1-sin^2theta)

Like in all pseudocylindrical designs, x=lambda xb / pi, therefore the equations for Mollweide's projection are:

x=2sqrt(2)R(lambda/pi)cos theta; y=sqrt(2)Rsin theta

HomeSite MapCraster ParabolicHow Projections are CreatedAzimuthal Equal-Area/Equidistant  www.progonos.com/furuti    January 28, 2013
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