from: http://www.movable-type.co.uk/scripts/LatLong.html -- Since the earth is not quite a sphere, there are small errors in using spherical geometry; the earth is actually roughly ellipsoidal (or more precisely, oblate spheroidal) with a radius varying between about 6,378km (equatorial) and 6,357km (polar), and local radius of curvature varying from 6,336km (equatorial meridian) to 6,399km (polar). This means that errors from assuming spherical geometry might be up to 0.55% crossing the equator, though generally below 0.3%, depending on latitude and direction of travel. An accuracy of better than 3m in 1km is good enough for me, but if you want greater accuracy, you could refine the result by using the local radius of curvature, as explained in the US Census Bureau GIS FAQ. -- Alternatively, you could use the Vincenty formulation: http://www.movable-type.co.uk/scripts/LatLongVincenty.html > Date: Mon, 10 Jul 2006 20:51:34 +0100 > From: "Armin M. Warda" <armin.warda at googlemail.com> > Subject: Re: Fwd: [maemo-users] how to determine track's length from > GPX-file > To: maemo-users at maemo.org, > Message-ID: > <11152561094663348731.1957747793.Armin.Warda at GoogleMail.Com> > Content-Type: text/plain; charset=UTF-8 > > Nils wrote: > > I do not want to start nitpicking here but that solution is rather inaccurate. > > It might serve as a first rough estimation > > but it neglects the fact that the earth is not a perfect sphere. > > How (in-)accurate is it? Possible to quantify? > Up to 5% error would be tolerable for me, > because I only use it for cycling. > > regards, Armin. > > -- Andrew Turner ajturner at highearthorbit.com 42.4266N x 83.4931W http://highearthorbit.com Northville, Michigan, USA
