Timeline for What kinds of line segments/edges require high accuracy in a true surface-of-the-ellipsoid representation?
Current License: CC BY-SA 3.0
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| Aug 19, 2014 at 23:00 | history | edited | cffk | CC BY-SA 3.0 |
Add information on the closest approach problem
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| Jul 21, 2011 at 3:35 | comment | added | cffk | @Dan, An ellipsoidal geodesic does not lie in a plane. (If it did, it would have to be a plane curve; and yet we know that, in general, on each circuit of the globe the geodesic falls short by an amount O(f).) The fault in your reasoning is (b) -- the connection between the aux. sphere and the ellipsoid is not linear. The latitude transformation is equivalent to a stretch in the z direction and so is linear. However the longitudes are related by a elliptic integral and this prevents there being any simple linear relationship. | |
| Jul 21, 2011 at 0:29 | comment | added | Dan S. | Here goes, as concisely as I could make it. My reasoning is expressed in 3D Cartesian, not angular coords: (a) On a sphere, all points in a great circle are coplanar. (b) The transformation to the auxiliary sphere is linear and invertible. (Mistaken thinking?) (c) All points in an elliptical geodesic transform to points along a great circle on the aux. sphere. (d) All points on an elliptical geodesic are coplanar as well, due to (b). Finally, (e): Due to coplanarity, two candidate geodesic intersection points on the ellipsoid can be found by plane intersection. | |
| Jul 21, 2011 at 0:23 | comment | added | Dan S. | I'm quite sure I'm off track, but not sure where; I'd love it if you could help debug my reasoning. (Next comment.) Otherwise: thanks a heap for the helpful code + commentary + link; it's tremendously useful. | |
| Jul 20, 2011 at 20:53 | history | answered | cffk | CC BY-SA 3.0 |