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I understand WGS84 models the equator as a circle with radius=GeodesyData["WGS84","SemimajorAxis"]. So I expected the two computations below to be the same. Where is my mistake?

Pi*GeodesyData["WGS84","SemimajorAxis"]
(* 2.00375 * 10^7 m *)

UnitConvert@GeoDistance[GeoPosition@{0,0},GeoPosition@{0,180}]
(* 2.00039 * 10^7 m *)
Ted Ersek
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1 Answers1

18

The shortest (geodesic) distance is not around the equator. Try

UnitConvert@ GeoDistance[
 {GeoPosition@{0, 0}, GeoPosition@{0, 90}, GeoPosition@{0, 180}}]
(*  Quantity[2.00375*10^7, "Meters"]  *)

and

UnitConvert@ GeoDistance[
 {GeoPosition@{0, 0}, GeoPosition@{90, 0}, GeoPosition@{0, 180}}]
(*  Quantity[2.00039*10^7, "Meters"]  *)
Michael E2
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  • Right, the shortest path crosses the North or South pole. We get half the perimeter around the equator using UnitConvert@ GeoDistance[{GeoPosition@{0, 0}, GeoPosition@{0, 90}, GeoPosition@{0, 180}}] (* 2.00375 * 10^7 m *) – Ted Ersek Dec 13 '21 at 20:56