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Posted 27 Dec 2007

# Distance using Longitiude and latitude using c++

, 27 Dec 2007
 Rate this:
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
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;```

## About the Author

 United States
No Biography provided

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## Comments and Discussions

 First Prev Next
 Thanks Steve Gee27-Jun-09 12:24 Steve Gee 27-Jun-09 12:24
 MathForum yarp28-Dec-07 20:05 yarp 28-Dec-07 20:05
 Your title word "Longitiude" spell wrong. Jamesmeng28-Dec-07 14:11 Jamesmeng 28-Dec-07 14:11
 Re: Your title word "Longitiude" spell wrong. Mattamoo28-Dec-07 17:50 Mattamoo 28-Dec-07 17:50
 Re: Your title word "Longitiude" spell wrong. Mattamoo28-Dec-07 17:58 Mattamoo 28-Dec-07 17:58
 Last Visit: 31-Dec-99 18:00     Last Update: 24-Jul-17 18:20 Refresh 1

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