如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
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。
其他回答
一个T-SQL函数,我用来根据中心的距离选择记录
Create Function [dbo].[DistanceInMiles]
( @fromLatitude float ,
@fromLongitude float ,
@toLatitude float,
@toLongitude float
)
returns float
AS
BEGIN
declare @distance float
select @distance = cast((3963 * ACOS(round(COS(RADIANS(90-@fromLatitude))*COS(RADIANS(90-@toLatitude))+
SIN(RADIANS(90-@fromLatitude))*SIN(RADIANS(90-@toLatitude))*COS(RADIANS(@fromLongitude-@toLongitude)),15))
)as float)
return round(@distance,1)
END
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
}
我需要在PowerShell中实现这个,希望它可以帮助其他人。 关于这种方法的一些注意事项
Don't split any of the lines or the calculation will be wrong To calculate in KM remove the * 1000 in the calculation of $distance Change $earthsRadius = 3963.19059 and remove * 1000 in the calculation of $distance the to calulate the distance in miles I'm using Haversine, as other posts have pointed out Vincenty's formulae is much more accurate Function MetresDistanceBetweenTwoGPSCoordinates($latitude1, $longitude1, $latitude2, $longitude2) { $Rad = ([math]::PI / 180); $earthsRadius = 6378.1370 # Earth's Radius in KM $dLat = ($latitude2 - $latitude1) * $Rad $dLon = ($longitude2 - $longitude1) * $Rad $latitude1 = $latitude1 * $Rad $latitude2 = $latitude2 * $Rad $a = [math]::Sin($dLat / 2) * [math]::Sin($dLat / 2) + [math]::Sin($dLon / 2) * [math]::Sin($dLon / 2) * [math]::Cos($latitude1) * [math]::Cos($latitude2) $c = 2 * [math]::ATan2([math]::Sqrt($a), [math]::Sqrt(1-$a)) $distance = [math]::Round($earthsRadius * $c * 1000, 0) #Multiple by 1000 to get metres Return $distance }
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。
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。