利用 Haversine 公式计算球体上两点间的距离

Haversine 公式利用两点的纬度和经度，计算出球体表面两点间的最短距离。它在导航领域有着非常重要的作用。从三角函数的角度来看，haversine 可以表示为：

haversine(\theta)=sin^2\Big(\frac{\theta}{2}\Big)

我们通过以下公式计算中心角（即 d/r）的 haversine 值：

\largehaversine\Big(\frac{d}{r}\Big)=haversine(\Phi2-\Phi1)+ cos(\Phi1)cos(\Phi2)haversine(\lambda2-\lambda1)

其中，r 是地球半径（6371 公里），d 是两点之间的距离，\phi1 和 \phi2 是两点的纬度，\lambda1 和 \lambda2 则分别是两点的经度。

我们可以通过应用反 haversine 函数或使用反正弦函数来求解 d，公式如下：

d = r hav^{-1}(h) = 2r sin^{-1}(\sqrt{h})

或者

d = 2r sin^{-1}\bigg(\sqrt{sin^2\Big(\frac{\Phi2-\Phi1}{2}\Big)+cos(\Phi1)cos(\Phi2)sin^2\Big(\frac{\lambda2-\lambda1}{2}\Big)}\ \bigg)

以伦敦的大本钟（51.5007° N, 0.1246° W）和纽约的自由女神像（40.6892° N, 74.0445° W）为例，它们之间的距离为 5574.8 公里。请注意，这并不是完全精确的测量值，因为该公式假设地球是一个完美的球体，但实际上地球是一个扁球体。

下面是上述公式的具体实现代码：

C++

// C++ program for the haversine formula
// C++ program for the
// haversine formula
#include 
#include 
using namespace std;

static double haversine(double lat1, double lon1,
                        double lat2, double lon2)
    {
        // distance between latitudes
        // and longitudes
        double dLat = (lat2 - lat1) *
                      M_PI / 180.0;
        double dLon = (lon2 - lon1) * 
                      M_PI / 180.0;

        // convert to radians
        lat1 = (lat1) * M_PI / 180.0;
        lat2 = (lat2) * M_PI / 180.0;

        // apply formulae
        double a = pow(sin(dLat / 2), 2) + 
                   pow(sin(dLon / 2), 2) * 
                   cos(lat1) * cos(lat2);
        double rad = 6371;
        double c = 2 * asin(sqrt(a));
        return rad * c;
    }

// Driver code
int main()
{
    double lat1 = 51.5007;
    double lon1 = 0.1246;
    double lat2 = 40.6892;
    double lon2 = 74.0445;
    
    cout << haversine(lat1, lon1,
                      lat2, lon2) << " K.M.";
    return 0;
}

// This code is contributed
// by Mahadev.

Java

// Java program for the haversine formula
public class Haversine {

    static double haversine(double lat1, double lon1,
                            double lat2, double lon2)
    {
        // distance between latitudes and longitudes
        double dLat = Math.toRadians(lat2 - lat1);
        double dLon = Math.toRadians(lon2 - lon1);

        // convert to radians
        lat1 = Math.toRadians(lat1);
        lat2 = Math.toRadians(lat2);

        // apply formulae
        double a = Math.pow(Math.sin(dLat / 2), 2) + 
                   Math.pow(Math.sin(dLon / 2), 2) * 
                   Math.cos(lat1) * 
                   Math.cos(lat2);
        double rad = 6371;
        double c = 2 * Math.asin(Math.sqrt(a));
        return rad * c;
    }

    // Driver Code
    public static void main(String[] args)
    {
        double lat1 = 51.5007;
        double lon1 = 0.1246;
        double lat2 = 40.6892;
        double lon2 = 74.0445;
        System.out.println(haversine(lat1, lon1, lat2, lon2) + " K.M.");
    }
}

Python 3

# Python 3 program for the 
# haversine formula
import math

# Python 3 program for the
# haversine formula
def haversine(lat1, lon1, lat2, lon2):
    
    # distance between latitudes
    # and longitudes
    dLat = (lat2 - lat1) * math.pi / 180.0
    dLon = (lon2 - lon1) * math.pi / 180.0

    # convert to radians
    lat1 = (lat1) * math.pi / 180.0
    lat2 = (lat2) * math.pi / 180.0

    # apply formulae
    a = (pow(math.sin(dLat / 2), 2) + 
         pow(math.sin(dLon / 2), 2) * 
             math.cos(lat1) * math.cos(lat2));
    rad = 6371
    c = 2 * math.asin(math.sqrt(a))
    return rad * c

# Driver code
if __name__ == "__main__":
    lat1 = 51.5007
    lon1 = 0.1246
    lat2 = 40.6892
    lon2 = 74.0445
    
    print(haversine(lat1, lon1,lat2, lon2), "K.M.")

# This code is contributed 
# by ChitraNayal

C#

// C# program for the haversine formula
using System;
class GFG
{

static double haversine(double lat1, double lon1,
                        double lat2, double lon2)
{
    // distance between latitudes and longitudes
    double dLat = (Math.PI / 180) * (lat2 - lat1);
    double dLon = (Math.PI / 180) * (lon2 - lon1);

    // convert to radians
    lat1 = (Math.PI / 180) * (lat1);
    lat2 = (Math.PI / 180) * (lat2);

    // apply formulae
    double a = Math.Pow(Math.Sin(dLat / 2), 2) + 
               Math.Pow(Math.Sin(dLon / 2), 2) *