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Re: A triangle on the equator
From: Paul Hirose
Date: 2025 Oct 12, 20:32 -0700
From: Paul Hirose
Date: 2025 Oct 12, 20:32 -0700
> *From: *Bill Lionheart > *Date: *2025 Oct 7, 12:57 -0700 > > small right angle triangles will always fall a bit short of pythagoras > in this way , a^2 > b^2 +c^2 and the angle sum will always be a bit more > than 180 degrees. Geodetic surveyors called the angle discrepancy the "spherical excess." For instance, see U.S. Coast & Geodetic Survey Special Publication No. 8, "Formulas and Tables for the Computation of Geodetic Positions," 7th edition: https://babel.hathitrust.org/cgi/pt?id=uc1.31822009761388&seq=10 If the triangle is on the equator (φ = 0) and included angle C is 90°, the numerator simplifies to the product of the sides. Suppose they are both 10 000 meters. For Earth equatorial radius use 6 378 000 m and for squared eccentricity 0.006 694. Then I get about a quarter arcsecond spherical excess. Keep in mind that's how much the sum of all three angles exceeds 180. As for the length of an angular unit on a meridian, the appendix of that document (p. 93) gives an expression for the radius of curvature at any latitude. First, we need the ellipsoid parameters. For the GRS80, semimajor axis (a) is 6 378 137 m and eccentricity (e^2) is about 0.006 694 380 (the actual value is a transcendental number). Then, R = [a(1 - e^2)] / [1 - e^2 sin^2 φ]^1.5 At the equator (φ = 0), R = 6335 km At φ = 90°, R = 6400 km At the equator, one minute of latitude is 6335 km sin 1′ = 1.842 km. At the pole it's 6400 km sin 1′ = 1.862 km. -- Paul Hirose sofajpl.com
