The EPSG code for 3857 has this in its "Remarks" section (https://epsg.org/crs_3857/WGS-84-Pseudo-Mercator.html) :

Not a recognised geodetic system. Uses spherical development of ellipsoidal coordinates. Relative to WGS 84 / World Mercator (CRS code 3395) gives errors of 0.7 percent in scale and differences in northing of up to 43km in the map (21km on the ground).

Does "Spherical development of ellipsoidal coordinates" mean that it takes WGS84 coordinates, which are geodetic lat-long on the ellipsoid as defined in EPSG:4326, and ignores the ellipsoid, pretends they are lat-long on a perfect sphere, and then does the Mercator transform?

And this is in contrast to EPSG:3395, which treats EPSG:4326 coordinates correctly as ellipsoidal coordinates, and does the Mercator transform with them?

I can't find any clear explanation of what that phrase means, and I want to make sure I'm interpreting it correctly.

  • 1
    This article tries to explain the thing alastaira.wordpress.com/2011/01/23/…. It is 12 years old and references to certain software and libraries may be outdated.
    – user30184
    Commented Jul 18, 2023 at 10:31
  • I think the answer is "its complicated", and is summarised by "The problem is that there is no standardised way to represent this “dual”-ellipsoid system, in which a different ellipsoid is used to interpret the geographic coordinates than is used to project them."
    – Spacedman
    Commented Jul 18, 2023 at 11:34

1 Answer 1


You are correct.

I heard the argument that considering the WGS 84 coordinates as if they were measured on a sphere helped to reduce the computing time, which is usefull for Web Mapping, but I didn't get a confirmation that this is effectively faster.

Note that the radius of this sphere is the radius of the WGS84 ellipsoid at the equator, which further increase the differences at the pole compared with "average" Earth radius in other sphere-based CRS.

In any case, many institutions discourage the use of this projection for GIS applications because of the distortions at fine scale.

  • Yes, it's faster because the spherical Mercator algorithm is simpler than the ellipsoidal version. Probably not too much as even the ellipsoidal equations are straightforward. x equation is the same; the y equation has around 6-7 more operations.
    – mkennedy
    Commented Jul 20, 2023 at 2:39

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