如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
如果你使用的是。net,不要重新启动轮子。看到System.Device.Location。在另一个答案的评论中赞扬fnx。
using System.Device.Location;
double lat1 = 45.421527862548828D;
double long1 = -75.697189331054688D;
double lat2 = 53.64135D;
double long2 = -113.59273D;
GeoCoordinate geo1 = new GeoCoordinate(lat1, long1);
GeoCoordinate geo2 = new GeoCoordinate(lat2, long2);
double distance = geo1.GetDistanceTo(geo2);
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;
}
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
这是我在Elixir中的实现
defmodule Geo do
@earth_radius_km 6371
@earth_radius_sm 3958.748
@earth_radius_nm 3440.065
@feet_per_sm 5280
@d2r :math.pi / 180
def deg_to_rad(deg), do: deg * @d2r
def great_circle_distance(p1, p2, :km), do: haversine(p1, p2) * @earth_radius_km
def great_circle_distance(p1, p2, :sm), do: haversine(p1, p2) * @earth_radius_sm
def great_circle_distance(p1, p2, :nm), do: haversine(p1, p2) * @earth_radius_nm
def great_circle_distance(p1, p2, :m), do: great_circle_distance(p1, p2, :km) * 1000
def great_circle_distance(p1, p2, :ft), do: great_circle_distance(p1, p2, :sm) * @feet_per_sm
@doc """
Calculate the [Haversine](https://en.wikipedia.org/wiki/Haversine_formula)
distance between two coordinates. Result is in radians. This result can be
multiplied by the sphere's radius in any unit to get the distance in that unit.
For example, multiple the result of this function by the Earth's radius in
kilometres and you get the distance between the two given points in kilometres.
"""
def haversine({lat1, lon1}, {lat2, lon2}) do
dlat = deg_to_rad(lat2 - lat1)
dlon = deg_to_rad(lon2 - lon1)
radlat1 = deg_to_rad(lat1)
radlat2 = deg_to_rad(lat2)
a = :math.pow(:math.sin(dlat / 2), 2) +
:math.pow(:math.sin(dlon / 2), 2) *
:math.cos(radlat1) * :math.cos(radlat2)
2 * :math.atan2(:math.sqrt(a), :math.sqrt(1 - a))
end
end
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。
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。