如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
private double deg2rad(double deg)
{
return (deg * Math.PI / 180.0);
}
private double rad2deg(double rad)
{
return (rad / Math.PI * 180.0);
}
private double GetDistance(double lat1, double lon1, double lat2, double lon2)
{
//code for Distance in Kilo Meter
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
return (dist);
}
private double GetDirection(double lat1, double lon1, double lat2, double lon2)
{
//code for Direction in Degrees
double dlat = deg2rad(lat1) - deg2rad(lat2);
double dlon = deg2rad(lon1) - deg2rad(lon2);
double y = Math.Sin(dlon) * Math.Cos(lat2);
double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
if (direct < 0)
direct = direct + 360;
return (direct);
}
private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
return (Speed);
}
private double GetDuration(DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
return (TimeDifferenceInSeconds);
}
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
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;
}
Unity版本c#
Haversine Algorithm。
public float Distance(float lat1, float lon1, float lat2, float lon2)
{
var earthRadiusKm = 6371;
var dLat = (lat2 - lat1) * Mathf.Rad2Deg;
var dLon = (lon2 - lon1) * Mathf.Rad2Deg;
var a = Mathf.Sin(dLat / 2) * Mathf.Sin(dLat / 2) +
Mathf.Sin(dLon / 2) * Mathf.Sin(dLon / 2) *
Mathf.Cos(lat1 * Mathf.Rad2Deg) * Mathf.Cos(lat2 * Mathf.Rad2Deg);
var c = 2 * Mathf.Atan2(Mathf.Sqrt(a), Mathf.Sqrt(1 - a));
return earthRadiusKm * 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)