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Posted 27 Dec 2007

Distance using Longitiude and latitude using c++

, 27 Dec 2007
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exactly what the title says
Title:       Distance using Longitude and latitude
Author:      Chhibs 
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


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

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 <span class="code-keyword"><cmath></span>
        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;


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


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Comments and Discussions

GeneralThanks Pin
Steve Gee27-Jun-09 12:24
memberSteve Gee27-Jun-09 12:24 
GeneralMathForum Pin
yarp28-Dec-07 20:05
memberyarp28-Dec-07 20:05 
GeneralYour title word "Longitiude" spell wrong. Pin
Jamesmeng28-Dec-07 14:11
memberJamesmeng28-Dec-07 14:11 
GeneralRe: Your title word "Longitiude" spell wrong. Pin
Mattamoo28-Dec-07 17:50
memberMattamoo28-Dec-07 17:50 
GeneralRe: Your title word "Longitiude" spell wrong. Pin
Mattamoo28-Dec-07 17:58
memberMattamoo28-Dec-07 17:58 

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