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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 
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 <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;


License

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

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About the Author

chhibs

United States United States
No Biography provided

Comments and Discussions

 
GeneralThanks PinmemberSteve Gee27-Jun-09 12:24 
GeneralMathForum Pinmemberyarp28-Dec-07 20:05 
GeneralYour title word "Longitiude" spell wrong. PinmemberJamesmeng28-Dec-07 14:11 
GeneralRe: Your title word "Longitiude" spell wrong. PinmemberMattamoo28-Dec-07 17:50 
GeneralRe: Your title word "Longitiude" spell wrong. PinmemberMattamoo28-Dec-07 17:58 

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