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
我猜你想让它沿着地球的曲率运动。你的两点和地心在一个平面上。地球的中心是这个平面上的圆心,这两个点(大致)在这个圆的周长上。由此你可以通过求一点到另一点的角度来计算距离。
如果点的高度不一样,或者如果你需要考虑地球不是一个完美的球体,这就有点困难了。
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;
}
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
}
}
c#版本的Haversine
double _eQuatorialEarthRadius = 6378.1370D;
double _d2r = (Math.PI / 180D);
private int HaversineInM(double lat1, double long1, double lat2, double long2)
{
return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
}
private 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;
}
这里有一个。net小提琴,所以你可以用你自己的Lat/ long测试它。
在Python中,你可以使用geopy库使用WGS84椭球来计算测地线距离:
from geopy.distance import geodesic
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(geodesic(newport_ri, cleveland_oh).km)