如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
我把上面的答案用在Scala程序中
import java.lang.Math.{atan2, cos, sin, sqrt}
def latLonDistance(lat1: Double, lon1: Double)(lat2: Double, lon2: Double): Double = {
val earthRadiusKm = 6371
val dLat = (lat2 - lat1).toRadians
val dLon = (lon2 - lon1).toRadians
val latRad1 = lat1.toRadians
val latRad2 = lat2.toRadians
val a = sin(dLat / 2) * sin(dLat / 2) + sin(dLon / 2) * sin(dLon / 2) * cos(latRad1) * cos(latRad2)
val c = 2 * atan2(sqrt(a), sqrt(1 - a))
earthRadiusKm * c
}
我对函数进行了压缩,以便能够轻松地生成具有两个固定位置之一的函数,并且只需要一对lat/lon来生成距离。
其他回答
这是“Henry Vilinskiy”为MySQL和km改编的版本:
CREATE FUNCTION `CalculateDistanceInKm`(
fromLatitude float,
fromLongitude float,
toLatitude float,
toLongitude float
) RETURNS float
BEGIN
declare distance float;
select
6367 * ACOS(
round(
COS(RADIANS(90-fromLatitude)) *
COS(RADIANS(90-toLatitude)) +
SIN(RADIANS(90-fromLatitude)) *
SIN(RADIANS(90-toLatitude)) *
COS(RADIANS(fromLongitude-toLongitude))
,15)
)
into distance;
return round(distance,3);
END;
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
下面是答案中的Swift实现
func degreesToRadians(degrees: Double) -> Double {
return degrees * Double.pi / 180
}
func distanceInKmBetweenEarthCoordinates(lat1: Double, lon1: Double, lat2: Double, lon2: Double) -> Double {
let earthRadiusKm: Double = 6371
let dLat = degreesToRadians(degrees: lat2 - lat1)
let dLon = degreesToRadians(degrees: lon2 - lon1)
let lat1 = degreesToRadians(degrees: lat1)
let lat2 = degreesToRadians(degrees: lat2)
let a = sin(dLat/2) * sin(dLat/2) +
sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2)
let c = 2 * atan2(sqrt(a), 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;
}
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。