Formulas for 4 linear Power-Function maps

[RogerOwens]

First, the general formula for these 4 linear Power-Function maps:

The origin of the co-ordinate system is the map's center. Longitudes left (west) of the central meridian are negative.

lat is latitude, in degrees.
lon is longitude, in degrees.

The distance from the origin to the north pole is one unit.

y = lat/90

x = f(lon/180)(2 - 2(abs(y))^p)

= 2f(lon/180)(1-(abs(y))^p)

f is the x-adjustment factor that moves the central-meridian conformal-point up to lat 45. It also somewhat reduces the high-latitude shear distortion.

I propose four maps:

p = 3 and f = 1
P = 3 and f = .808122036

p = 8.32 and f = 1
p = 8.32 and f = .709326391

The p = 3 map is intended to have linearity, with better areas than it would have if it were quartic, but with less shear-distortion than the Parabolic--maybe not enough to bother people much.

The p = 8.32 map has the same space-efficiency as Eckert III and Eckert IV, but has a point-pole.

The shear-distorition at high latitude might be excessive. The outer meridian, when it reaches the north pole, has a slope of only 1 in 16, only about 4 degrees from the horizontal. Other than that, I haven't calculated shear, and I haven't drawn the map.

So I don't know whether or not it has excessive high-latitude shear.

If it high-latitude shear makes a point-pole incompatible with high space-efficiency, then I must retract my statement that pseudocylindricals don't need a flat pole, and that giving them one amounts to "cartographic dishonesty".

If so, then, high-space efficiency pseudocylindricals need a flat pole, to get some of the advantages that a cylindrical map gains thereby.

Michael Ossipoff