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Deducing the Kavrayskiy VII Projection

Consider Kavrayskiy (Kavraisky)'s VII, a map projection I will describe, in the equatorial aspect, as: pseudocylindrical, poles half as long as the Equator, equidistant parallels, meridian 120° contained in a circle centered on the map. What does that mean?

in the equatorial aspect only, all parallels are horizontal straight lines, while meridians are arbitrary curves
poles half as long as the equator
from the definition, the poles in pseudocylindrical projections are either points or straight lines ("polelines"). Compared with "pointed-polar" projections, "flat-polar" designs usually have lesser shape distortion in high latitudes. Most flat-polar maps have simple pole length/Equator length ratios, like 1/2 or 1/3
equidistant parallels
all but one equal-area pseudocylindrical projections have variable distance between parallels; in constrast, the non-equal-area Kavrayskiy VII has constant parallel spacing, which does not of course mean that the scale along its meridians is identical or even constant. Without further constraints, y=phi R making the central meridian alone a standard line
meridian 120° contained in a circle centered on the map
in every pseudocylindrical projection the scale along any parallel is constant. Therefore, all meridians have similar shapes, except the central one (considered 0° here) which is always straight. Shapes are affected by the longitude. In this case, if the meridian at 120° (i.e., 2pi/3) is an arc of circle, all others but the central one are elliptical arcs, flattened towards the central meridian, elongated towards the map's boundary
Mathematics of Kavrayskiy VII projection
Top right quadrant of Kavrayskiy VII map

Let us sketch the northeastern quadrant of the Kavrayskiy VII projection. Let W be half the poleline's length, and H its distance from the Equator; H=R pi/2 due to the equidistant property. For a given latitude lambda, let xc be the abscissa of the circular 120° meridian, and xb the abscissa of the 180° meridian, the fundamental boundary meridian.

Because true distances are linear along each pseudocylindrical parallel, both xb and the general abscissa x at any given latitude are proportional to xc:

xb=180xc/120=3xc/2; x=lambda xb/pi

The radius 4W/3 of the reference circle follows immediately. And from the Pithagorean theorem,

H2=(2W/3)2+(4W/3)2;9H2+4W2=16W2; H=W sqrt(4/3)

Half the length of the Equator is:

2W=H sqrt(3)=R pi sqrt(3)/2
Map in Kavrayskiy VII projection
Kavrayskiy VII map

Again from Pithagoras's theorem,

xc^2+y^2=16W^2/9=pi^2R^2/3; xc^2=pi^2R^2/3-phi^2R^2; xc=R sqrt(pi^2/3-phi^2)
x=3 lambda R sqrt(pi^2/3-phi^2)/2pi; y=R phi

Easy to understand and compute, Kavrayskiy's 7th projection has no special properties, preserving neither shapes nor areas. It simply has a good and balanced appearance, which for many purposes is often good enough.

HomeSite MapStereographic CylindricalHow Projections are CreatedSinusoidal    January  7, 2013
Copyright © 2012 Carlos A. Furuti