Rogier (and David too :) This is quite exciting. Based on this, I think I can formulate something. Thank you for making me understand.
have a great day !! ~~ ----- Original Message ----- From: "Rogier Wolff" To: "Fire Girl" Subject: Re: [OSM-dev] Question about Calculating Radius from GPS cord Date: Fri, 8 Aug 2008 21:41:52 +0200 On Fri, Aug 08, 2008 at 11:02:15AM -0500, Fire Girl wrote: > I think I am following you! :) > > Here is the math I have so far.... > > Calculation: 5 miles I'm assuming radius, not diameter. So we'll get a ~10mile square. > of: 45.5 latitude > and: 25.5 longitude Note that Milo thinks this is inaccurate. He's right. But these "quick" calculations are very efficient. If for example, you want to select all OSM objects within 5 miles distance from a certain coordinate, you damn well better do THIS math first, and only then use proper geo-valid formulas to select the real ones. The reason is that if you put all objects in the database through the "calculate-distance-to-reference-point" formula, this will take ages, but selecting on a 10 mile min longitude / max longitude and min latitude max latitude first will be fast: There should be an index on those elements of the database. Now for some applications, the "quick-n-dirty" method is already good enough. So.... here we go. kmpermile = 1.609; circumferencekm = 40000; circumferencemile = circumferencekm / kmpermile; radius = 5; // Miles! latitude = 45.5; // North longitude = 25.5; // east! delta = radius / (circumferencemile / 360); latmin = latitude - delta; latmax = latitude + delta; longmin = longitude - delta / cos (latitude); longmax = longitude + delta / cos (latitude); which according to my calculator comes to: latmin = 45.42759500000000000000 latmax = 45.57240500000000000000 longmin = 25.39669848650172993913 longmax = 25.60330151349827006087 Apparently the cosine changes by about 1 in 776 on this 5 mile stretch of earth. So this square is only about 9.987 miles wide at the north end, and 10.013 miles wide a the south end. Similar differences can be expected to crop up because the earth is not a perfect sphere. Roger. > > ------------------------------- > > 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 > > _______________________________________________ > dev mailing list > [email protected] > http://lists.openstreetmap.org/listinfo/dev -- ** [EMAIL PROTECTED] ** http://www.BitWizard.nl/ ** +31-15-2600998 ** ** Delftechpark 26 2628 XH Delft, The Netherlands. 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