Changes between Version 6 and Version 7 of Ticket #12427, comment 36


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Timestamp:
2016-01-28T11:45:45+01:00 (10 years ago)
Author:
cmuelle8

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  • Ticket #12427, comment 36

    v6 v7  
    44Btw, 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.
    6  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'' (when the bi-gon surface collapses to an arc), and hence the same length. Beyond this step the area would grow 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.
     6 If a and b are moved ever closer together on equator, there must be a time when the geodesic north, the one south and equator ''all'' have ''minimum curvature'' (when the bi-gon surface collapses to an arc), and hence the same length. Beyond this step the area would grow 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.
    77
    88> I don't get what you mean by maximum curvature. The curvature is different at each point along the curve. You could take some kind of average along the line, but then it is far from obvious why this would give great ellipses.