如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
在SQL Server 2008中使用地理类型非常容易做到这一点。
SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm
4326是WGS84椭球地球模型的SRID
其他回答
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;
}
下面是我在Python中使用的Haversine函数:
from math import pi,sqrt,sin,cos,atan2
def haversine(pos1, pos2):
lat1 = float(pos1['lat'])
long1 = float(pos1['long'])
lat2 = float(pos2['lat'])
long2 = float(pos2['long'])
degree_to_rad = float(pi / 180.0)
d_lat = (lat2 - lat1) * degree_to_rad
d_long = (long2 - long1) * degree_to_rad
a = pow(sin(d_lat / 2), 2) + cos(lat1 * degree_to_rad) * cos(lat2 * degree_to_rad) * pow(sin(d_long / 2), 2)
c = 2 * atan2(sqrt(a), sqrt(1 - a))
km = 6367 * c
mi = 3956 * c
return {"km":km, "miles":mi}
在SQL Server 2008中使用地理类型非常容易做到这一点。
SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm
4326是WGS84椭球地球模型的SRID
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
}
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;
}
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