Changes between Version 3 and Version 4 of Ticket #12427, comment 36


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Timestamp:
2016-01-28T10:46:19+01:00 (10 years ago)
Author:
cmuelle8

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  • Ticket #12427, comment 36

    v3 v4  
    22> You are right, I just considered displacement to the same side.
    33
    4 Btw, if {{{abs(lon(a)-lon(b)) <= (1 − f)*pi && lat(a)==0 && lat(b)==0}}}
     4Btw, if {{{abs(lon(a)-lon(b)) <= (1−f)*pi && lat(a)==0 && lat(b)==0}}}
    55then equator ''is'' a shortest geodesic. I believe [https://en.m.wikipedia.org/wiki/Geodesics_on_an_ellipsoid#Solution_of_the_inverse_problem the wiki article] to be true in that respect.
    66 If a and b are moved ever closer together on equator, there must be a time when the shortest geodesic north, the one south and equator ''all'' have ''minimum curvature'' (wrt to the bi-gon surface), and hence the same length. Beyond this step the area grows in size and hence is not of interest anymore. It's like climbing a hill, once you reach a certain height, it's shorter crossing the top to descend to the same height you would otherwise reach on equidistant paths around it.