Changes between Version 1 and Version 2 of Ticket #12427, comment 27


Ignore:
Timestamp:
2016-01-26T05:49:22+01:00 (10 years ago)
Author:
cmuelle8

Legend:

Unmodified
Added
Removed
Modified

  • Ticket #12427, comment 27

    v1 v2  
    1111 If the arcs obtained by intersecting the last two planes are unequal, I suggest these to be the true shortest arcs.
    1212
    13  If the normals ''do'' intersect, and {{{IP}}} is ''neither in'' equatorial plane ''nor on'' polar axis, all three arcs will be equal ''and'' be the true shortest arc between the points. The same applies if {{{IP}}} lies ''uncentered'' in equatorial plane, but at least one of the normals lies not. (The great ellipse through the points coincides with the rotated ellipse defining the ellipsoid in these cases.)
     13 If the normals ''do'' intersect, and {{{IP}}} is ''neither in'' equatorial plane ''nor on'' polar axis, all three arcs will be equal ''and'' be the true shortest arc between the points. The same applies if {{{IP}}} lies ''uncentered'' in equatorial plane, but at least one of the normals lies not.
     14 (The great ellipse through the points coincides with the rotated ellipse defining the ellipsoid in these cases.)
    1415
    1516 If both normals lie in equatorial plane, {{{IP}}} coincides with the center of the ellipsoid. Other two points, displaced from the center along the polar axis in proportion to the eccentricity need to define planes ''P2'', ''P3'' in this case. The surface normals are not helpful to find the true shortest arcs left and right to equator in this case. Similar applies if {{{IP}}} lies ''on'' polar axis.