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authorKristian Evers <kristianevers@gmail.com>2017-08-14 09:48:50 +0200
committerGitHub <noreply@github.com>2017-08-14 09:48:50 +0200
commitd00501750b210a73f9fb107ac97a683d4e3d8e7a (patch)
treee6b71867033a48f53927a4e503d33bd51c3f3216 /docs/source
parent78fcfdefa152f1f1e12c85799bf454ca3d45f247 (diff)
parent6492c4e2b421a5a7ef34be6a77cc6e9ecf930b67 (diff)
downloadPROJ-d00501750b210a73f9fb107ac97a683d4e3d8e7a.tar.gz
PROJ-d00501750b210a73f9fb107ac97a683d4e3d8e7a.zip
Merge pull request #556 from lukecampbell/master
Uses the mathjax formatting for geodesic equations
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diff --git a/docs/source/geodesic.rst b/docs/source/geodesic.rst
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@@ -66,16 +66,18 @@ the earth. This from Mikael Rittri on the Proj mailing list:
For the fixed Earth radius, I would choose the average of the:
- c = radius of curvature at the poles,
- b^2^ / a = radius of curvature in a meridian plane at the equator,
+ :math:`c` = radius of curvature at the poles,
+
+ :math:`\frac{b^2}{a}` = radius of curvature in a meridian plane at the equator,
since these are the extreme values for the local radius of curvature of the
earth ellipsoid.
If your coordinates are given in WGS84, then
- c = 6 399 593.626 m,
- b^2^ / a = 6 335 439.327 m,
+ :math:`c` = 6 399 593.626 m,
+
+ :math:`\frac{b^2}{a}` = 6 335 439.327 m,
(see http://home.online.no/~sigurdhu/WGS84_Eng.html) so their average is 6,367,516.477 m.
The maximal error for distance calculation should then be less than 0.51 percent.