如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
Scala版本
def deg2rad(deg: Double) = deg * Math.PI / 180.0
def rad2deg(rad: Double) = rad / Math.PI * 180.0
def getDistanceMeters(lat1: Double, lon1: Double, lat2: Double, lon2: Double) = {
val theta = lon1 - lon2
val dist = Math.sin(deg2rad(lat1)) * Math.sin(deg2rad(lat2)) + Math.cos(deg2rad(lat1)) *
Math.cos(deg2rad(lat2)) * Math.cos(deg2rad(theta))
Math.abs(
Math.round(
rad2deg(Math.acos(dist)) * 60 * 1.1515 * 1.609344 * 1000)
)
}
其他回答
这取决于你需要它有多准确。如果你需要精确到毫米的精度,最好看看使用椭球的算法,而不是球体,比如Vincenty的算法。
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。
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。
打印稿版本
export const degreeToRadian = (degree: number) => {
return degree * Math.PI / 180;
}
export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => {
const earthRadiusInKm = 6371;
const dLat = degreeToRadian(lat2 - lat1);
const dLon = degreeToRadian(lon2 - lon1);
lat1 = degreeToRadian(lat1);
lat2 = degreeToRadian(lat2);
const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return earthRadiusInKm * c;
}