如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
当前回答
下面是Haversine公式的java实现。
public final static double AVERAGE_RADIUS_OF_EARTH_KM = 6371;
public int calculateDistanceInKilometer(double userLat, double userLng,
double venueLat, double venueLng) {
double latDistance = Math.toRadians(userLat - venueLat);
double lngDistance = Math.toRadians(userLng - venueLng);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(userLat)) * Math.cos(Math.toRadians(venueLat))
* Math.sin(lngDistance / 2) * Math.sin(lngDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (int) (Math.round(AVERAGE_RADIUS_OF_EARTH_KM * c));
}
请注意,这里我们将答案四舍五入到最近的km。
其他回答
下面是postgres SQL中的一个示例(以公里为单位,为英里版本,将1.609344替换为0.8684版本)
CREATE OR REPLACE FUNCTION public.geodistance(alat float, alng float, blat
float, blng float)
RETURNS float AS
$BODY$
DECLARE
v_distance float;
BEGIN
v_distance = asin( sqrt(
sin(radians(blat-alat)/2)^2
+ (
(sin(radians(blng-alng)/2)^2) *
cos(radians(alat)) *
cos(radians(blat))
)
)
) * cast('7926.3352' as float) * cast('1.609344' as float) ;
RETURN v_distance;
END
$BODY$
language plpgsql VOLATILE SECURITY DEFINER;
alter function geodistance(alat float, alng float, blat float, blng float)
owner to postgres;
这是一个简单的javascript函数,从这个链接可能是有用的。不知何故相关,但我们使用谷歌地球javascript插件而不是地图
function getApproximateDistanceUnits(point1, point2) {
var xs = 0;
var ys = 0;
xs = point2.getX() - point1.getX();
xs = xs * xs;
ys = point2.getY() - point1.getY();
ys = ys * ys;
return Math.sqrt(xs + ys);
}
单位不是距离,而是相对于坐标的比率。还有其他相关的计算,你可以在这里代替getApproximateDistanceUnits函数链接
然后我使用这个函数来查看经纬度是否在半径内
function isMapPlacemarkInRadius(point1, point2, radi) {
if (point1 && point2) {
return getApproximateDistanceUnits(point1, point2) <= radi;
} else {
return 0;
}
}
点可以定义为
$$.getPoint = function(lati, longi) {
var location = {
x: 0,
y: 0,
getX: function() { return location.x; },
getY: function() { return location.y; }
};
location.x = lati;
location.y = longi;
return location;
};
然后你可以做你的事情,看看一个点是否在一个半径范围内,比如:
//put it on the map if within the range of a specified radi assuming 100,000,000 units
var iconpoint = Map.getPoint(pp.latitude, pp.longitude);
var centerpoint = Map.getPoint(Settings.CenterLatitude, Settings.CenterLongitude);
//approx ~200 units to show only half of the globe from the default center radius
if (isMapPlacemarkInRadius(centerpoint, iconpoint, 120)) {
addPlacemark(pp.latitude, pp.longitude, pp.name);
}
else {
otherSidePlacemarks.push({
latitude: pp.latitude,
longitude: pp.longitude,
name: pp.name
});
}
下面是VB的实现。NET,这个实现将根据您传递的Enum值以KM或Miles为单位给您结果。
Public Enum DistanceType
Miles
KiloMeters
End Enum
Public Structure Position
Public Latitude As Double
Public Longitude As Double
End Structure
Public Class Haversine
Public Function Distance(Pos1 As Position,
Pos2 As Position,
DistType As DistanceType) As Double
Dim R As Double = If((DistType = DistanceType.Miles), 3960, 6371)
Dim dLat As Double = Me.toRadian(Pos2.Latitude - Pos1.Latitude)
Dim dLon As Double = Me.toRadian(Pos2.Longitude - Pos1.Longitude)
Dim a As Double = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Cos(Me.toRadian(Pos1.Latitude)) * Math.Cos(Me.toRadian(Pos2.Latitude)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2)
Dim c As Double = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)))
Dim result As Double = R * c
Return result
End Function
Private Function toRadian(val As Double) As Double
Return (Math.PI / 180) * val
End Function
End Class
如果你正在使用python; PIP安装地质
from geopy.distance import geodesic
origin = (30.172705, 31.526725) # (latitude, longitude) don't confuse
destination = (30.288281, 31.732326)
print(geodesic(origin, destination).meters) # 23576.805481751613
print(geodesic(origin, destination).kilometers) # 23.576805481751613
print(geodesic(origin, destination).miles) # 14.64994773134371
你可以用Haversine公式计算它,它是:
a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
下面给出了一个计算两点之间距离的例子
假设我要计算从新德里到伦敦的距离,那么我该如何使用这个公式:
New delhi co-ordinates= 28.7041° N, 77.1025° E
London co-ordinates= 51.5074° N, 0.1278° W
var R = 6371e3; // metres
var φ1 = 28.7041.toRadians();
var φ2 = 51.5074.toRadians();
var Δφ = (51.5074-28.7041).toRadians();
var Δλ = (0.1278-77.1025).toRadians();
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // metres
d = d/1000; // km