Wikipedia:Reference desk/Archives/Mathematics/2018 January 8
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January 8
[edit]Distances on the Earth ellipsoid calculated by Googe Earth?
[edit]Hello you cunning people everywhere. I wonder how Google Earth can give instantly he distance between 2 points. I checked that they are not calculated on a sphere but on an ellipsoid. Ex : Google Earth meridian perimeter = 40 008 km (equal to the wikipedia one), equatorial perimeter = 40 066 km (wikipedia says 40 075).
I read here Ellipse#Circumference integrals formulas and here Elliptic_integral#Complete_elliptic_integral_of_the_second_kind that it's not easy to get a correct value of these distances. However, the distances are displayed INSTANTLY. You don't need to press the ENTER key to get the distance; you just move the mouse and the distance is there. How can it be done instantly! I thank you for your thingkings. Happy new year.--Jojodesbatignoles (talk) 12:39, 8 January 2018 (UTC)
- A relevant article: Geographical distance#Ellipsoidal-surface formulae.
- Hopefully someone else here will have a more specific answer about algorithms actually used, but if not, you may wish to ask over on the computing reference desk. -- ToE 14:17, 8 January 2018 (UTC)
- (edit conflict × 2) ToE beat me to it, but also see Vincenty's formula. Computation of elliptic integrals can also be done to high precision very quickly via the arithmetic-geometric mean. –Deacon Vorbis (carbon • videos) 14:32, 8 January 2018 (UTC)
- There are a few things to take into account here. First, you only need distances to within a few feet; it does no good to get much more precise than GPS. Second, the earth isn't an exact oblate spheroid so the ellipsoid distance is only an approximation anyway. Third, modern processors are very fast, so you can do a lot of computation in less time that it takes the internet to get the answer to you. Fourth, presumably most of the distances they need are precomputed; the length of a stretch of road isn't going to change (modulo a major earthquake) so you can easily look up the value rather than compute it each time. Anyway, is it documented that Google uses ellipsoidal distance? They could just be using spherical distance and applying a first order correction based on latitude. --RDBury (talk) 17:49, 8 January 2018 (UTC)
- They have a 3D model of the Earth surface and then can calculate any distance just by numerically integrating along the path on this surface using an appropriate quadrature rule. Ruslik_Zero 20:15, 8 January 2018 (UTC)
- Finding the minimum distance requires the calculus of variations though.--Jasper Deng (talk) 10:11, 10 January 2018 (UTC)
- They have a 3D model of the Earth surface and then can calculate any distance just by numerically integrating along the path on this surface using an appropriate quadrature rule. Ruslik_Zero 20:15, 8 January 2018 (UTC)
- There are a few things to take into account here. First, you only need distances to within a few feet; it does no good to get much more precise than GPS. Second, the earth isn't an exact oblate spheroid so the ellipsoid distance is only an approximation anyway. Third, modern processors are very fast, so you can do a lot of computation in less time that it takes the internet to get the answer to you. Fourth, presumably most of the distances they need are precomputed; the length of a stretch of road isn't going to change (modulo a major earthquake) so you can easily look up the value rather than compute it each time. Anyway, is it documented that Google uses ellipsoidal distance? They could just be using spherical distance and applying a first order correction based on latitude. --RDBury (talk) 17:49, 8 January 2018 (UTC)
- Jojo, how did you come up with 40 066 km? I just ran Google Earth Pro (the desktop application), created a path 0° 0° -> 0° 90° -> 0° 180° -> 0° -90° -> 0° 0° (or as close as I could get by clicking), and selected the Measurements tab which gave "Length: 40,075 kilometers". This corresponds to the 6 378 137.0 m Semi-major axis from WGS 84. (6378137.0 * 2π = 40 075 016.7 m) Your 40 066 km, besides being the subject of your question on Wikipédia:Oracle (the French equivalent of our Reference Desks), is given as the equatorial circumference in a number of Indian sources, such as this ISCE school atlas and this Civil Services Examination study book. 40 066 km corresponds to a 6376.7 km semi-major axis. I wonder which datum uses that. -- ToE 20:53, 8 January 2018 (UTC)
- It could be a Google Earth bug, it draws some weird paths when you try to connect two points that are nearly 180° apart. For example I just tried to repeat your experiment in a different way, I connected on equator 0°1'E and 179°59'E and it routed me over the western hemisphere. Though, when I got it to work properly, I got 40075.02 km. 93.139.63.123 (talk) 05:12, 10 January 2018 (UTC)
- Doesn't sound like a bug. Per Geodesics on an ellipsoid, the shortest path between two points on the equator does not necessarily run along the equator. So had your 40 066 km come from measuring a path which skirted the equator? Funny if it was just a coincidence. I wonder where those Indian publishers came up with it. -- ToE 06:17, 10 January 2018 (UTC)
- ...for example. the shortest path from 0°N,0°E to 0°N,180°E goes through the pole. --CiaPan (talk) 10:37, 10 January 2018 (UTC)
- Doesn't sound like a bug. Per Geodesics on an ellipsoid, the shortest path between two points on the equator does not necessarily run along the equator. So had your 40 066 km come from measuring a path which skirted the equator? Funny if it was just a coincidence. I wonder where those Indian publishers came up with it. -- ToE 06:17, 10 January 2018 (UTC)
- It could be a Google Earth bug, it draws some weird paths when you try to connect two points that are nearly 180° apart. For example I just tried to repeat your experiment in a different way, I connected on equator 0°1'E and 179°59'E and it routed me over the western hemisphere. Though, when I got it to work properly, I got 40075.02 km. 93.139.63.123 (talk) 05:12, 10 January 2018 (UTC)