如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
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
}
其他回答
我需要在PowerShell中实现这个,希望它可以帮助其他人。 关于这种方法的一些注意事项
Don't split any of the lines or the calculation will be wrong To calculate in KM remove the * 1000 in the calculation of $distance Change $earthsRadius = 3963.19059 and remove * 1000 in the calculation of $distance the to calulate the distance in miles I'm using Haversine, as other posts have pointed out Vincenty's formulae is much more accurate Function MetresDistanceBetweenTwoGPSCoordinates($latitude1, $longitude1, $latitude2, $longitude2) { $Rad = ([math]::PI / 180); $earthsRadius = 6378.1370 # Earth's Radius in KM $dLat = ($latitude2 - $latitude1) * $Rad $dLon = ($longitude2 - $longitude1) * $Rad $latitude1 = $latitude1 * $Rad $latitude2 = $latitude2 * $Rad $a = [math]::Sin($dLat / 2) * [math]::Sin($dLat / 2) + [math]::Sin($dLon / 2) * [math]::Sin($dLon / 2) * [math]::Cos($latitude1) * [math]::Cos($latitude2) $c = 2 * [math]::ATan2([math]::Sqrt($a), [math]::Sqrt(1-$a)) $distance = [math]::Round($earthsRadius * $c * 1000, 0) #Multiple by 1000 to get metres Return $distance }
在Python中,你可以使用geopy库使用WGS84椭球来计算测地线距离:
from geopy.distance import geodesic
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(geodesic(newport_ri, cleveland_oh).km)
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
我猜你想让它沿着地球的曲率运动。你的两点和地心在一个平面上。地球的中心是这个平面上的圆心,这两个点(大致)在这个圆的周长上。由此你可以通过求一点到另一点的角度来计算距离。
如果点的高度不一样,或者如果你需要考虑地球不是一个完美的球体,这就有点困难了。