function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2,units) {
  var R = 6371; // Radius of the earth in km
  var dLat = deg2rad(lat2-lat1);  // deg2rad below
  var dLon = deg2rad(lon2-lon1); 
  var a = 
    Math.sin(dLat/2) * Math.sin(dLat/2) +
    Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * 
    Math.sin(dLon/2) * Math.sin(dLon/2)
    ; 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  var d = R * c; 
  var miles = d / 1.609344; 

if ( units == 'km' ) {  
return d; 
 } else {
return miles;
}}

查克的解决方案，也适用于英里。

//JAVA
    public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
    final int RADIUS_EARTH = 6371;

    double dLat = getRad(latitude2 - latitude1);
    double dLong = getRad(longitude2 - longitude1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return (RADIUS_EARTH * c) * 1000;
    }

    private Double getRad(Double x) {
    return x * Math.PI / 180;
    }

下面是VB的实现。NET，这个实现将根据您传递的Enum值以KM或Miles为单位给您结果。

Public Enum DistanceType
    Miles
    KiloMeters
End Enum

Public Structure Position
    Public Latitude As Double
    Public Longitude As Double
End Structure

Public Class Haversine

    Public Function Distance(Pos1 As Position,
                             Pos2 As Position,
                             DistType As DistanceType) As Double

        Dim R As Double = If((DistType = DistanceType.Miles), 3960, 6371)

        Dim dLat As Double = Me.toRadian(Pos2.Latitude - Pos1.Latitude)

        Dim dLon As Double = Me.toRadian(Pos2.Longitude - Pos1.Longitude)

        Dim a As Double = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Cos(Me.toRadian(Pos1.Latitude)) * Math.Cos(Me.toRadian(Pos2.Latitude)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2)

        Dim c As Double = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)))

        Dim result As Double = R * c

        Return result

    End Function

    Private Function toRadian(val As Double) As Double

        Return (Math.PI / 180) * val

    End Function

End Class

你也可以使用像geolib这样的模块:

安装方法:

$ npm install geolib

使用方法:

import { getDistance } from 'geolib'

const distance = getDistance(
    { latitude: 51.5103, longitude: 7.49347 },
    { latitude: "51° 31' N", longitude: "7° 28' E" }
)

console.log(distance)

文档: https://www.npmjs.com/package/geolib

计算距离——尤其是大距离——的主要挑战之一是解释地球的曲率。如果地球是平的，计算两点之间的距离就会像计算直线一样简单!哈弗辛公式包括一个常数(下面是R变量)，它表示地球的半径。根据你是用英里还是公里来测量，它分别等于3956英里或6367公里。 基本公式是:

Dlon = lon2 - lon1 dat = lat2 - lat1 = (sin (dlat / 2)) ^ 2 + cos (lat1) * cos (lat2) * (sin (dlon / 2)) ^ 2 C = 2 * atan2(√(a)，√(1-a)) distance = R * c(其中R为地球半径) R = 6367公里OR 3956英里

     lat1, lon1: The Latitude and Longitude of point 1 (in decimal degrees)
     lat2, lon2: The Latitude and Longitude of point 2 (in decimal degrees)
     unit: The unit of measurement in which to calculate the results where:
     'M' is statute miles (default)
     'K' is kilometers
     'N' is nautical miles

样本

function distance(lat1, lon1, lat2, lon2, unit) {
    try {
        var radlat1 = Math.PI * lat1 / 180
        var radlat2 = Math.PI * lat2 / 180
        var theta = lon1 - lon2
        var radtheta = Math.PI * theta / 180
        var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
        dist = Math.acos(dist)
        dist = dist * 180 / Math.PI
        dist = dist * 60 * 1.1515
        if (unit == "K") {
            dist = dist * 1.609344
        }
        if (unit == "N") {
            dist = dist * 0.8684
        }
        return dist
    } catch (err) {
        console.log(err);
    }
}

精确计算中长点之间距离所需的函数是复杂的，陷阱也很多。我不推荐哈弗辛或其他球形的解决方案，因为有很大的不准确性(地球不是一个完美的球体)。vincenty公式更好，但在某些情况下会抛出错误，即使编码正确。

与其自己编写函数，我建议使用geopy，它已经实现了非常精确的地理库来进行距离计算(论文来自作者)。

#pip install geopy
from geopy.distance import geodesic
NY = [40.71278,-74.00594]
Beijing = [39.90421,116.40739]
print("WGS84: ",geodesic(NY, Beijing).km) #WGS84 is Standard
print("Intl24: ",geodesic(NY, Beijing, ellipsoid='Intl 1924').km) #geopy includes different ellipsoids
print("Custom ellipsoid: ",geodesic(NY, Beijing, ellipsoid=(6377., 6356., 1 / 297.)).km) #custom ellipsoid

#supported ellipsoids:
#model             major (km)   minor (km)     flattening
#'WGS-84':        (6378.137,    6356.7523142,  1 / 298.257223563)
#'GRS-80':        (6378.137,    6356.7523141,  1 / 298.257222101)
#'Airy (1830)':   (6377.563396, 6356.256909,   1 / 299.3249646)
#'Intl 1924':     (6378.388,    6356.911946,   1 / 297.0)
#'Clarke (1880)': (6378.249145, 6356.51486955, 1 / 293.465)
#'GRS-67':        (6378.1600,   6356.774719,   1 / 298.25)

这个库的唯一缺点是它不支持向量化计算。 对于向量化计算，您可以使用新的gevectorslib。

#pip install geovectorslib
from geovectorslib import inverse
print(inverse(lats1,lons1,lats2,lons2)['s12'])

lat和lon是numpy数组。Geovectorslib是非常准确和非常快!我还没有找到改变椭球的方法。标准采用WGS84椭球，是大多数用途的最佳选择。