I think I am following you! :) Here is the math I have so far....
Calculation: 5 miles of: 45.5 latitude and: 25.5 longitude ------------------------------- so, for the latitude, (i.e. add and subtract 2.5/69ths of a degree of latitude) With 5 miles assumed, and using that forumla, the distance from 45.5 would be: 45.505797 (on the plus side) 45.494203 (on the minus side) Right? :) but on the Longitude side, (i.e. 2.5/f) .... f = 69 * sin(latitude) .... i am not so clear what Latitude number I should be using there and what that might look like. How do I look at least so far in my calculations? :) ----- Original Message ----- From: "David Earl" To: "Fire Girl" Subject: Re: [OSM-dev] Question about Calculating Radius from GPS cord Date: Fri, 08 Aug 2008 16:02:10 +0100 On 08/08/2008 14:30, Fire Girl wrote: > I am working with OSM data, and would like to be able to spec out > 5 mile bounding boxes from certain GPS points. > > After research into this problem, I am to understand that each > degree of latitude is approximately 69 miles (111 kilometers) > apart with a slight variance (68.703 - 69.407 miles) between the > equator and the poles, and that each degree of longitude is > widest at the equator @ 69.172 miles (111.321 kilometers) and > gradually shrinks to zero at the poles. : ) :) > > So what does this mean? If I want to take a input point, like lets say, > > 167.9 lat > -29.1 lon > > or > > -63.1 > 18.1 > > Can someone say with authority, what the 'calculus' would be to > definitivly construct a NSWE bounding box with a 5 mile radius > around those points?.... that would be basically close enough or > accurate? :) If you want accuracy, then you are asking the wrong question, because the "bounding box" will be a curved section of a surface of the earth (or, rather, the approximation to it defined by the ellipsoid for the datum you're using for the lat/lon coordinates - unless you want to start taking altitude and terrain into account), not a flat straight-edged box. So if you want to talk in terms of flat bounding boxes, you have to start asking "in what projection?" etc. If all you want is the lat/lon of the corners an approximate box parallel to lines of latitude and longitude approx 5 miles across, then you could indeed just add and subtract 2.5/69ths of a degree of latitude and 2.5/f of a degree of longitude, where f is the approx length of one degree at that latitude, i.e. f = 69 * sin(latitude) miles (with the reasonable approximation that the earth is a sphere, so the trigonometry is trivial). David -- Be Yourself @ mail.com! Choose From 200+ Email Addresses Get a Free Account at www.mail.com
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