如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
下面是Kotlin的一个变种:
import kotlin.math.*
class HaversineAlgorithm {
companion object {
private const val MEAN_EARTH_RADIUS = 6371.008
private const val D2R = Math.PI / 180.0
}
private fun haversineInKm(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double {
val lonDiff = (lon2 - lon1) * D2R
val latDiff = (lat2 - lat1) * D2R
val latSin = sin(latDiff / 2.0)
val lonSin = sin(lonDiff / 2.0)
val a = latSin * latSin + (cos(lat1 * D2R) * cos(lat2 * D2R) * lonSin * lonSin)
val c = 2.0 * atan2(sqrt(a), sqrt(1.0 - a))
return MEAN_EARTH_RADIUS * c
}
}
其他回答
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;
}
PHP版本:
(删除所有deg2rad()如果您的坐标已经是弧度。)
$R = 6371; // km
$dLat = deg2rad($lat2-$lat1);
$dLon = deg2rad($lon2-$lon1);
$lat1 = deg2rad($lat1);
$lat2 = deg2rad($lat2);
$a = sin($dLat/2) * sin($dLat/2) +
sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$d = $R * $c;
下面是c#语言(用纬度和弧度表示):
double CalculateGreatCircleDistance(double lat1, double long1, double lat2, double long2, double radius)
{
return radius * Math.Acos(
Math.Sin(lat1) * Math.Sin(lat2)
+ Math.Cos(lat1) * Math.Cos(lat2) * Math.Cos(long2 - long1));
}
如果你的纬度和长度是用角度表示的,那么除以180/PI就可以转换成弧度。
飞镖版本
Haversine Algorithm。
import 'dart:math';
class GeoUtils {
static double _degreesToRadians(degrees) {
return degrees * pi / 180;
}
static double 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 = sin(dLat/2) * sin(dLat/2) +
sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
var c = 2 * atan2(sqrt(a), sqrt(1-a));
return earthRadiusKm * c;
}
}