Changes between Initial Version and Version 1 of Ticket #12427, comment 33


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Timestamp:
2016-01-27T06:32:04+01:00 (10 years ago)
Author:
cmuelle8

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  • Ticket #12427, comment 33

    initial v1  
    66
    77Imo you should think so: Given two non-antipodal, disjunct points on equator, let A be the area between the two possible geodesics, such that the shortest arc on equator is contained in A and halving A. Since the ellipsoid surface is continuous, A ''must'' shrink to zero first if both points are displaced gradually into different "hemiellipsoids" (defined by equator), for the geodesic to eventually become unique.
     8
     9EDIT: You can also find an access to it like this:
     10 Let A, B be non-antipodal, disjunct points on equator.
     11 Let N, S be disjunct intersection points of polar axis and ellipsoid.
     12
     13 Then the geodesic on the northern hemiellipsoid between A, B defines an ellipsoidal triangle, together with the meridian arcs from N to A and N to B.
     14
     15 Another, non-intersecting ellipsoidal triangle with disjunct sides is defined by the geodesic on the southern hemiellipsoid between A, B, together with meridian arcs from S to A and S to B.
     16
     17 Let 2G be the bi-gon between the two geodesics.
     18
     19We have essentially tri-sected the area between the two meridians defined by N-A-S and N-B-S. What will happen to 2G, if A and B are moved along their meridians, in opposite direction?