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Deducing Craster's Parabolic Projection

Suppose a pointed-polar equal-area pseudocylindrical projection in the equatorial aspect whose boundary meridians are parabolic arcs with axes on the Equator, which is twice as long as the central meridian.

Craster parabolic projection

Schematic development of Craster's parabolic projection

Since the projection is pseudocylindrical,

dy/dlambda = 0

in other words,


Due to symmetry around the Equator and central meridian, with no loss of generality let us consider only the northeastern quadrant where 0<=phi<=pi/2, 0<=lambda<=pi.

For H>0, the following constraints hold:

0<=x'<=2H; 0<=y<=H; xb=2H-x'

For xb=0, y=H therefore k=2/H and the boundary meridian is defined by:


As seen in the development of the sinusoidal projection, half the area on Earth between the Equator and the parallel phi is pi R^2 sin(phi). This time, instead of “discovering” that a projection is equal-area, we will use that property as a constraint to calculate the ordinate y. Half the area on the map between the Equator and any given ordinate Y is

int 0 to Y xbdy = int 0 to Y 2H-2y^2/Hdy = 2HY-2Y^3/3H

For phi=pi/2, Y=H and

S=pi R^2=2H^2-2H^2/3=4H^2/3

Therefore H=sqrt(3pi)R/2 and

xb=sqrt(3pi)R - 4y^2/(sqrt(3pi)R)


S(y)=-4y^3/3sqrt(3pi)R + sqrt(3pi)Ry

Solving the cubic equation S(y)-pi R^2 sin phi=0,

a=-4/(3sqrt(3pi)R),b=0,c=sqrt(3pi)R,d=-piR^2sin phi, delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2=16piR^2 cos^2phi

and because delta>=0 there are three real roots; however, since

-27a^2\Delta = -256\cos^2\varphi \le 0

the root terms involve complex expressions. With the substitutions for the depressed cubic

p=(3ac-b^2)/3a^2 = -9piR^2/4, q=(2b^3-9abc+27a^2d)/27a^3=R^3sin phi (3pi)^(3/2)/4
Graph of S(y)

Graph of S(y).

and using Viète's method, the roots are, for k=0,1,2:

tk=2sqrt(-p/3)cos((arccos sqrt(-3p)3q/2p)/3-2kpi/3)


4p^3+27q^2=729pi^3R^6(sin^2 phi-1)/16<=0

all three roots are real and t2<=t1<=t0.

An inspection of the graph of S(y) shows that the desired minimum positive root is the intermediary one, t1.

t1=2sqrt(-p/3)cos(arccos(3q sqrt(-3/p)/2p)/3-2pi/3)=sqrt(3)|R|cos(arccos(-sin phi|R|/R)/3-2pi/3)=sqrt(3pi)R cos(arccos(-sin phi)/3-2pi/3)

For -pi/2<=alpha<=pi/2,arccos(-sin alpha)=alpha+pi/2

t1=\sqrt(3pi)R cos((phi+pi/2)/3-2pi/3)=sqrt(3pi)R cos (phi/3-pi/2)=sqrt(3pi)R sin(phi/3)=y
x=xb lambda/pi = sqrt(3/pi)R lambda(1-4sin^2phi/3)

Or, since sin^2alpha=(1-cos2alpha)/2,

x=sqrt(3/pi)Rlambda(2cos(2phi/3)-1), y=sqrt(3pi)Rsinphi/3

Those are the forward equations of the parabolic equal-area projection, the best known of all projections presented by J.E.E.Craster.

Craster parabolic projection

J.E.E.Craster's parabolic projection

Angular deformation patterns for Craster parabolic and sinusoidal projections Angular deformation scale

Compared deformation patterns of Craster's parabolic (top left and bottom right quadrants) and the sinusoidal projections

The overall shapes of the sinusoidal and parabolic projections are easily confused, but there are differences:

HomeSite MapSinusoidalHow Projections are CreatedMollweide — August  5, 2018
Copyright © 2013 Carlos A. Furuti