如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
计算两个坐标之间的纬度和经度的距离,包括一个Javascript实现。
西部和南部的位置是负的。 记住,分和秒是60度,所以S31 30'是-31.50度。
别忘了把角度转换成弧度。许多语言都有这个功能。或者它是一个简单的计算:弧度=角度* PI / 180。
function degreesToRadians(degrees) {
return degrees * Math.PI / 180;
}
function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
var earthRadiusKm = 6371;
var dLat = degreesToRadians(lat2-lat1);
var dLon = degreesToRadians(lon2-lon1);
lat1 = degreesToRadians(lat1);
lat2 = degreesToRadians(lat2);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return earthRadiusKm * c;
}
下面是一些用法的例子:
distanceInKmBetweenEarthCoordinates(0,0,0,0) // Distance between same
// points should be 0
0
distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
// to Arlington
5918.185064088764
其他回答
我认为R中的一个算法版本仍然缺失:
gpsdistance<-function(lat1,lon1,lat2,lon2){
# internal function to change deg to rad
degreesToRadians<- function (degrees) {
return (degrees * pi / 180)
}
R<-6371e3 #radius of Earth in meters
phi1<-degreesToRadians(lat1) # latitude 1
phi2<-degreesToRadians(lat2) # latitude 2
lambda1<-degreesToRadians(lon1) # longitude 1
lambda2<-degreesToRadians(lon2) # longitude 2
delta_phi<-phi1-phi2 # latitude-distance
delta_lambda<-lambda1-lambda2 # longitude-distance
a<-sin(delta_phi/2)*sin(delta_phi/2)+
cos(phi1)*cos(phi2)*sin(delta_lambda/2)*
sin(delta_lambda/2)
cc<-2*atan2(sqrt(a),sqrt(1-a))
distance<- R * cc
return(distance) # in meters
}
对于java
public static double degreesToRadians(double degrees) {
return degrees * Math.PI / 180;
}
public static double distanceInKmBetweenEarthCoordinates(Location location1, Location location2) {
double earthRadiusKm = 6371;
double dLat = degreesToRadians(location2.getLatitude()-location1.getLatitude());
double dLon = degreesToRadians(location2.getLongitude()-location1.getLongitude());
double lat1 = degreesToRadians(location1.getLatitude());
double lat2 = degreesToRadians(location2.getLatitude());
double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.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;
}
对于任何寻找Delphi/Pascal版本的人:
function GreatCircleDistance(const Lat1, Long1, Lat2, Long2: Double): Double;
var
Lat1Rad, Long1Rad, Lat2Rad, Long2Rad: Double;
const
EARTH_RADIUS_KM = 6378;
begin
Lat1Rad := DegToRad(Lat1);
Long1Rad := DegToRad(Long1);
Lat2Rad := DegToRad(Lat2);
Long2Rad := DegToRad(Long2);
Result := EARTH_RADIUS_KM * ArcCos(Cos(Lat1Rad) * Cos(Lat2Rad) * Cos(Long1Rad - Long2Rad) + Sin(Lat1Rad) * Sin(Lat2Rad));
end;
我对这个代码没有任何功劳,我最初是在一个公共论坛上发现Gary William发布的。
基于Roman Makarov对这个线程的回复的Java版本的Haversine算法
public class HaversineAlgorithm {
static final double _eQuatorialEarthRadius = 6378.1370D;
static final double _d2r = (Math.PI / 180D);
public static int HaversineInM(double lat1, double long1, double lat2, double long2) {
return (int) (1000D * HaversineInKM(lat1, long1, lat2, long2));
}
public static 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;
}
}