一、关于“面包屑”方法

地球半径在不同的纬度上是不同的。在Haversine算法中必须考虑到这一点。 考虑轴承的变化，它将直线变成拱门(更长的) 考虑到速度变化将把拱门变成螺旋(比拱门更长或更短) 高度变化将使平面螺旋变成3D螺旋(再次变长)。这对丘陵地区非常重要。

下面是考虑#1和#2的C语言函数:

double   calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
       double rLat2, double rLon2, double rHeading2){
  double rDLatRad = 0.0;
  double rDLonRad = 0.0;
  double rLat1Rad = 0.0;
  double rLat2Rad = 0.0;
  double a = 0.0;
  double c = 0.0;
  double rResult = 0.0;
  double rEarthRadius = 0.0;
  double rDHeading = 0.0;
  double rDHeadingRad = 0.0;

  if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
              || (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
              || (rLon2 > 180.0)) {
        return -1;
  };

  rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
  rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
  rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
  rLat2Rad = rLat2 * DEGREE_TO_RADIANS;

  a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
              rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);

  if (a == 0.0) {
        return 0.0;
  }

  c = 2 * atan2(sqrt(a), sqrt(1 - a));
  rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
              / 2.0));
  rResult = rEarthRadius * c;

  // Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns

  if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
              && (rHeading2 < 360.0)) {
        rDHeading = fabs(rHeading1 - rHeading2);
        if (rDHeading > 180.0) {
              rDHeading -= 180.0;
        }
        rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
        if (rDHeading > 5.0) {
              rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
        } else {
              rResult = rResult / cos(rDHeadingRad);
        }
  }
  return rResult;
}

2有一种更简单的方法，效果很好。

按平均速度。

Trip_distance = Trip_average_speed * Trip_time

由于GPS速度是由多普勒效应检测的，与[Lon,Lat]没有直接关系，如果不是主要的距离计算方法，至少可以考虑作为次要的(备份或校正)。

PHP版本:

(删除所有deg2rad()如果您的坐标已经是弧度。)

$R = 6371; // km
$dLat = deg2rad($lat2-$lat1);
$dLon = deg2rad($lon2-$lon1);
$lat1 = deg2rad($lat1);
$lat2 = deg2rad($lat2);

$a = sin($dLat/2) * sin($dLat/2) +
     sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2); 

$c = 2 * atan2(sqrt($a), sqrt(1-$a)); 
$d = $R * $c;

如果你使用的是。net，不要重新启动轮子。看到System.Device.Location。在另一个答案的评论中赞扬fnx。

using System.Device.Location;

double lat1 = 45.421527862548828D;
double long1 = -75.697189331054688D;
double lat2 = 53.64135D;
double long2 = -113.59273D;

GeoCoordinate geo1 = new GeoCoordinate(lat1, long1);
GeoCoordinate geo2 = new GeoCoordinate(lat2, long2);

double distance = geo1.GetDistanceTo(geo2);

我把上面的答案用在Scala程序中

import java.lang.Math.{atan2, cos, sin, sqrt}

def latLonDistance(lat1: Double, lon1: Double)(lat2: Double, lon2: Double): Double = {
    val earthRadiusKm = 6371
    val dLat = (lat2 - lat1).toRadians
    val dLon = (lon2 - lon1).toRadians
    val latRad1 = lat1.toRadians
    val latRad2 = lat2.toRadians

    val a = sin(dLat / 2) * sin(dLat / 2) + sin(dLon / 2) * sin(dLon / 2) * cos(latRad1) * cos(latRad2)
    val c = 2 * atan2(sqrt(a), sqrt(1 - a))
    earthRadiusKm * c
}

我对函数进行了压缩，以便能够轻松地生成具有两个固定位置之一的函数，并且只需要一对lat/lon来生成距离。

c#版本的Haversine

double _eQuatorialEarthRadius = 6378.1370D;
double _d2r = (Math.PI / 180D);

private int HaversineInM(double lat1, double long1, double lat2, double long2)
{
    return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
}

private double HaversineInKM(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * _d2r;
    double dlat = (lat2 - lat1) * _d2r;
    double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
    double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
    double d = _eQuatorialEarthRadius * c;

    return d;
}

这里有一个。net小提琴，所以你可以用你自己的Lat/ long测试它。

在Python中，你可以使用geopy库使用WGS84椭球来计算测地线距离:

from geopy.distance import geodesic
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(geodesic(newport_ri, cleveland_oh).km)