Comments on: Finding distances using latitude and longitude http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/ The blog of John D. Cook Sat, 11 Feb 2012 22:42:11 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Journey away from the center of the Earth — The Endeavour http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/comment-page-1/#comment-103672 Journey away from the center of the Earth — The Endeavour Fri, 16 Sep 2011 12:26:12 +0000 http://www.johndcook.com/blog/?p=391#comment-103672 [...] is the shape of the Earth? Finding distances from coordinates Mercator projection Inverse Mercator projection ? [...] [...] is the shape of the Earth? Finding distances from coordinates Mercator projection Inverse Mercator projection ? [...]

]]> By: Spherical trigonometry — The Endeavour http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/comment-page-1/#comment-98079 Spherical trigonometry — The Endeavour Fri, 12 Aug 2011 02:52:12 +0000 http://www.johndcook.com/blog/?p=391#comment-98079 [...] is the shape of the earth? Finding distances using longitude and latitude ? [...] [...] is the shape of the earth? Finding distances using longitude and latitude ? [...]

]]> By: Walt http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/comment-page-1/#comment-49939 Walt Wed, 27 Oct 2010 23:55:30 +0000 http://www.johndcook.com/blog/?p=391#comment-49939 Sam, you can easily calculate that using a spherical triangle with the North pole, your starting point, and the ending point. Then you have two sides and the included angle, from which you can calculate the rest. The short distance approximation is always flat-plane, since the relative curvature depends on how far you travel. The only corrections needed to a spherical model are the deviations of the Earth from a sphere. Namely, changing radius with latitude, and the deviation if your starting and ending points are not at the same elevation. Sam, you can easily calculate that using a spherical triangle with the North pole, your starting point, and the ending point. Then you have two sides and the included angle, from which you can calculate the rest.

The short distance approximation is always flat-plane, since the relative curvature depends on how far you travel.

The only corrections needed to a spherical model are the deviations of the Earth from a sphere. Namely, changing radius with latitude, and the deviation if your starting and ending points are not at the same elevation.

]]> By: sam http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/comment-page-1/#comment-49921 sam Wed, 27 Oct 2010 20:21:57 +0000 http://www.johndcook.com/blog/?p=391#comment-49921 So what is the best way to do the reverse? Given a starting lat/lon, compass heading and distance, what is the lat/lon where you end? What would the best short distance approximation be and what would its limits be? It would seem that a more complex model than spherical would have to be used going this direction to be useful. So what is the best way to do the reverse? Given a starting lat/lon, compass heading and distance, what is the lat/lon where you end? What would the best short distance approximation be and what would its limits be? It would seem that a more complex model than spherical would have to be used going this direction to be useful.

]]> By: Walt Roscello http://www.johndcook.com/blog/2008/09/11/finding-distances-using-latitude-and-longitude/comment-page-1/#comment-31816 Walt Roscello Fri, 29 Jan 2010 17:30:23 +0000 http://www.johndcook.com/blog/?p=391#comment-31816 A useful calculation. Note that for distances below about 10 miles, you should use an expansion of arccos around 1, since the arccos loses precison. At that distance, though, you might as well just use pythagoras and delta_long * cos(avg_lat). A useful calculation. Note that for distances below about 10 miles, you should use an expansion of arccos around 1, since the arccos loses precison. At that distance, though, you might as well just use pythagoras and delta_long * cos(avg_lat).

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