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
我已经创建了这个小Javascript LatLng对象，可能对某人有用。
var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2);
代码:
/**
* latLng point
* @param {Number} lat
* @param {Number} lng
* @returns {LatLng}
* @constructor
*/
function LatLng(lat,lng) {
this.lat = parseFloat(lat);
this.lng = parseFloat(lng);
this.__cache = {};
}
LatLng.prototype = {
toString: function() {
return [this.lat, this.lng].join(",");
},
/**
* calculate distance in km to another latLng, with caching
* @param {LatLng} latLng
* @returns {Number} distance in km
*/
distanceTo: function(latLng) {
var cacheKey = latLng.toString();
if(cacheKey in this.__cache) {
return this.__cache[cacheKey];
}
// the fastest way to calculate the distance, according to this jsperf test;
// http://jsperf.com/haversine-salvador/8
// http://stackoverflow.com/questions/27928
var deg2rad = 0.017453292519943295; // === Math.PI / 180
var lat1 = this.lat * deg2rad;
var lng1 = this.lng * deg2rad;
var lat2 = latLng.lat * deg2rad;
var lng2 = latLng.lng * deg2rad;
var a = (
(1 - Math.cos(lat2 - lat1)) +
(1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
) / 2;
var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)
// cache the distance
this.__cache[cacheKey] = distance;
return distance;
}
};
其他回答
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));
}
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)
}
我已经创建了这个小Javascript LatLng对象，可能对某人有用。
var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2);
代码:
/**
* latLng point
* @param {Number} lat
* @param {Number} lng
* @returns {LatLng}
* @constructor
*/
function LatLng(lat,lng) {
this.lat = parseFloat(lat);
this.lng = parseFloat(lng);
this.__cache = {};
}
LatLng.prototype = {
toString: function() {
return [this.lat, this.lng].join(",");
},
/**
* calculate distance in km to another latLng, with caching
* @param {LatLng} latLng
* @returns {Number} distance in km
*/
distanceTo: function(latLng) {
var cacheKey = latLng.toString();
if(cacheKey in this.__cache) {
return this.__cache[cacheKey];
}
// the fastest way to calculate the distance, according to this jsperf test;
// http://jsperf.com/haversine-salvador/8
// http://stackoverflow.com/questions/27928
var deg2rad = 0.017453292519943295; // === Math.PI / 180
var lat1 = this.lat * deg2rad;
var lng1 = this.lng * deg2rad;
var lat2 = latLng.lat * deg2rad;
var lng2 = latLng.lng * deg2rad;
var a = (
(1 - Math.cos(lat2 - lat1)) +
(1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
) / 2;
var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)
// cache the distance
this.__cache[cacheKey] = distance;
return distance;
}
};
下面是另一个转换为Ruby代码的代码:
include Math
#Note: from/to = [lat, long]
def get_distance_in_km(from, to)
radians = lambda { |deg| deg * Math.PI / 180 }
radius = 6371 # Radius of the earth in kilometer
dLat = radians[to[0]-from[0]]
dLon = radians[to[1]-from[1]]
cosines_product = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(radians[from[0]]) * Math.cos(radians[to[1]]) * Math.sin(dLon/2) * Math.sin(dLon/2)
c = 2 * Math.atan2(Math.sqrt(cosines_product), Math.sqrt(1-cosines_product))
return radius * c # Distance in kilometer
end
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
