如何计算由经纬度指定的两点之间的距离?

为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。


当前回答

这个链接可能对你有帮助,因为它详细介绍了使用哈弗辛公式来计算距离。

摘录:

这个脚本计算两点之间的大圆距离 也就是说,在地球表面上的最短距离-使用 “半正矢”公式。

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2) {
  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; // Distance in km
  return d;
}

function deg2rad(deg) {
  return deg * (Math.PI/180)
}

其他回答

下面是移植到Java的已接受的答案实现,以备任何人需要。

package com.project529.garage.util;


/**
 * Mean radius.
 */
private static double EARTH_RADIUS = 6371;

/**
 * Returns the distance between two sets of latitudes and longitudes in meters.
 * <p/>
 * Based from the following JavaScript SO answer:
 * http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
 * which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
 */
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
    double dLat = toRadians(lat2 - lat1);
    double dLon = toRadians(lon2 - lon1);

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

    return d;
}

public double toRadians(double degrees) {
    return degrees * (Math.PI / 180);
}

正如指出的那样,精确的计算应该考虑到地球不是一个完美的球体。以下是这里提供的各种算法的一些比较:

geoDistance(50,5,58,3)
Haversine: 899 km
Maymenn: 833 km
Keerthana: 897 km
google.maps.geometry.spherical.computeDistanceBetween(): 900 km

geoDistance(50,5,-58,-3)
Haversine: 12030 km
Maymenn: 11135 km
Keerthana: 10310 km
google.maps.geometry.spherical.computeDistanceBetween(): 12044 km

geoDistance(.05,.005,.058,.003)
Haversine: 0.9169 km
Maymenn: 0.851723 km
Keerthana: 0.917964 km
google.maps.geometry.spherical.computeDistanceBetween(): 0.917964 km

geoDistance(.05,80,.058,80.3)
Haversine: 33.37 km
Maymenn: 33.34 km
Keerthana: 33.40767 km
google.maps.geometry.spherical.computeDistanceBetween(): 33.40770 km

在小范围内,Keerthana的算法似乎与谷歌Maps的算法一致。谷歌Maps似乎没有遵循任何简单的算法,这表明它可能是这里最准确的方法。

不管怎样,这里是Keerthana算法的Javascript实现:

function geoDistance(lat1, lng1, lat2, lng2){
    const a = 6378.137; // equitorial radius in km
    const b = 6356.752; // polar radius in km

    var sq = x => (x*x);
    var sqr = x => Math.sqrt(x);
    var cos = x => Math.cos(x);
    var sin = x => Math.sin(x);
    var radius = lat => sqr((sq(a*a*cos(lat))+sq(b*b*sin(lat)))/(sq(a*cos(lat))+sq(b*sin(lat))));

    lat1 = lat1 * Math.PI / 180;
    lng1 = lng1 * Math.PI / 180;
    lat2 = lat2 * Math.PI / 180;
    lng2 = lng2 * Math.PI / 180;

    var R1 = radius(lat1);
    var x1 = R1*cos(lat1)*cos(lng1);
    var y1 = R1*cos(lat1)*sin(lng1);
    var z1 = R1*sin(lat1);

    var R2 = radius(lat2);
    var x2 = R2*cos(lat2)*cos(lng2);
    var y2 = R2*cos(lat2)*sin(lng2);
    var z2 = R2*sin(lat2);

    return sqr(sq(x1-x2)+sq(y1-y2)+sq(z1-z2));
}

下面是Erlang实现

lat_lng({Lat1, Lon1}=_Point1, {Lat2, Lon2}=_Point2) ->
  P = math:pi() / 180,
  R = 6371, % Radius of Earth in KM
  A = 0.5 - math:cos((Lat2 - Lat1) * P) / 2 +
    math:cos(Lat1 * P) * math:cos(Lat2 * P) * (1 - math:cos((Lon2 - Lon1) * P))/2,
  R * 2 * math:asin(math:sqrt(A)).
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。

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);
    }
}