如果你需要更准确的数据，可以看看这个。

Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a) They are based on the assumption that the figure of the Earth is an oblate spheroid, and hence are more accurate than methods such as great-circle distance which assume a spherical Earth. The first (direct) method computes the location of a point which is a given distance and azimuth (direction) from another point. The second (inverse) method computes the geographical distance and azimuth between two given points. They have been widely used in geodesy because they are accurate to within 0.5 mm (0.020″) on the Earth ellipsoid.

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。

下面是答案中的Swift实现

func degreesToRadians(degrees: Double) -> Double {
    return degrees * Double.pi / 180
}

func distanceInKmBetweenEarthCoordinates(lat1: Double, lon1: Double, lat2: Double, lon2: Double) -> Double {

    let earthRadiusKm: Double = 6371

    let dLat = degreesToRadians(degrees: lat2 - lat1)
    let dLon = degreesToRadians(degrees: lon2 - lon1)

    let lat1 = degreesToRadians(degrees: lat1)
    let lat2 = degreesToRadians(degrees: lat2)

    let a = sin(dLat/2) * sin(dLat/2) +
    sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2)
    let c = 2 * atan2(sqrt(a), sqrt(1 - a))
    return earthRadiusKm * c
}

寻找带谷歌的哈弗辛;以下是我的解决方案:

#include <math.h>
#include "haversine.h"

#define d2r (M_PI / 180.0)

//calculate haversine distance for linear distance
double haversine_km(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 6367 * c;

    return d;
}

double haversine_mi(double lat1, double long1, double lat2, double long2)
{
    double dlong = (long2 - long1) * d2r;
    double dlat = (lat2 - lat1) * d2r;
    double a = pow(sin(dlat/2.0), 2) + cos(lat1*d2r) * cos(lat2*d2r) * pow(sin(dlong/2.0), 2);
    double c = 2 * atan2(sqrt(a), sqrt(1-a));
    double d = 3956 * c; 

    return d;
}

    private double deg2rad(double deg)
    {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad)
    {
        return (rad / Math.PI * 180.0);
    }

    private double GetDistance(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Distance in Kilo Meter
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
        return (dist);
    }

    private double GetDirection(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Direction in Degrees
        double dlat = deg2rad(lat1) - deg2rad(lat2);
        double dlon = deg2rad(lon1) - deg2rad(lon2);
        double y = Math.Sin(dlon) * Math.Cos(lat2);
        double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
        double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
        if (direct < 0)
            direct = direct + 360;
        return (direct);
    }

    private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
        double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
        return (Speed);
    }

    private double GetDuration(DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
        return (TimeDifferenceInSeconds);
    }

打印稿版本

export const degreeToRadian = (degree: number) => {
  return degree * Math.PI / 180;
}

export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => {
    const earthRadiusInKm = 6371;

    const dLat = degreeToRadian(lat2 - lat1);
    const dLon = degreeToRadian(lon2 - lon1);

    lat1 = degreeToRadian(lat1);
    lat2 = degreeToRadian(lat2);

    const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
    const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));

    return earthRadiusInKm * c;
}