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Distance using Longitiude and latitude using c++
exactly what the title says
Title: Distance using Longitude and latitude Author: Chhibs Email: annum0@gmail.com Member ID: 12345 Language: C++ Platform: Windows, Linux Technology: WiMAX,WiFi, WDM Level: Intermediate, Advanced Description: A quick code dump for haversine formula Section Language c++ SubSection Howto
Introduction
I have seen code for calculating distance using the haversine formula using C# etc on the site, but nothing using c++, so here is the code that just does that
Background (optional)
I saw couple fo articles doing thie distance calculation using the haversine formula but using .Net instead, since I code in C++(no .Net until really needed), I ported to c++ and below is the code
From math forum
http://mathforum.org/library/drmath/view/51879.html
Presuming a spherical Earth with radius R (see below), and that the locations of the two points in spherical coordinates (longitude and latitude) are lon1,lat1 and lon2,lat2, then the Haversine Formula (from R. W. Sinnott, "Virtues of the Haversine," Sky and Telescope, vol. 68, no. 2, 1984, p. 159): dlon = lon2 - lon1 dlat = lat2 - lat1 a = (sin(dlat/2))^2 + cos(lat1) * cos(lat2) * (sin(dlon/2))^2 c = 2 * atan2(sqrt(a), sqrt(1-a)) d = R * c will give mathematically and computationally exact results. The intermediate result c is the great circle distance in radians. The great circle distance d will be in the same units as R.
Using the code
I am basically just posting a snippet of code that I use in a class which does the distance calculation for me// Sample format for latitide and longitudes // double lat1=45.54243333333333,lat2=45.53722222,long1=-122.96045277777778,long2=-122.9630556; // Below is the main code #include <cmath> double PI = 4.0*atan(1.0); //main code inside the class double dlat1=lat1*(PI/180); double dlong1=long1*(PI/180); double dlat2=lat2*(PI/180); double dlong2=long2*(PI/180); double dLong=dlong1-dlong2; double dLat=dlat1-dlat2; double aHarv= pow(sin(dLat/2.0),2.0)+cos(dlat1)*cos(dlat2)*pow(sin(dLong/2),2); double cHarv=2*atan2(sqrt(aHarv),sqrt(1.0-aHarv)); //earth's radius from wikipedia varies between 6,356.750 km — 6,378.135 km (˜3,949.901 — 3,963.189 miles) //The IUGG value for the equatorial radius of the Earth is 6378.137 km (3963.19 mile) const double earth=3963.19;//I am doing miles, just change this to radius in kilometers to get distances in km double distance=earth*cHarv;