如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
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
其他回答
Unity版本c#
Haversine Algorithm。
public float Distance(float lat1, float lon1, float lat2, float lon2)
{
var earthRadiusKm = 6371;
var dLat = (lat2 - lat1) * Mathf.Rad2Deg;
var dLon = (lon2 - lon1) * Mathf.Rad2Deg;
var a = Mathf.Sin(dLat / 2) * Mathf.Sin(dLat / 2) +
Mathf.Sin(dLon / 2) * Mathf.Sin(dLon / 2) *
Mathf.Cos(lat1 * Mathf.Rad2Deg) * Mathf.Cos(lat2 * Mathf.Rad2Deg);
var c = 2 * Mathf.Atan2(Mathf.Sqrt(a), Mathf.Sqrt(1 - a));
return earthRadiusKm * c;
}
下面是Kotlin的一个变种:
import kotlin.math.*
class HaversineAlgorithm {
companion object {
private const val MEAN_EARTH_RADIUS = 6371.008
private const val D2R = Math.PI / 180.0
}
private fun haversineInKm(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double {
val lonDiff = (lon2 - lon1) * D2R
val latDiff = (lat2 - lat1) * D2R
val latSin = sin(latDiff / 2.0)
val lonSin = sin(lonDiff / 2.0)
val a = latSin * latSin + (cos(lat1 * D2R) * cos(lat2 * D2R) * lonSin * lonSin)
val c = 2.0 * atan2(sqrt(a), sqrt(1.0 - a))
return MEAN_EARTH_RADIUS * c
}
}
PHP版本:
(删除所有deg2rad()如果您的坐标已经是弧度。)
$R = 6371; // km
$dLat = deg2rad($lat2-$lat1);
$dLon = deg2rad($lon2-$lon1);
$lat1 = deg2rad($lat1);
$lat2 = deg2rad($lat2);
$a = sin($dLat/2) * sin($dLat/2) +
sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$d = $R * $c;
如果你需要更准确的数据,可以看看这个。
Vincenty's formulae are two related iterative methods used in geodesy to calculate the distance between two points on the surface of a spheroid, developed by Thaddeus Vincenty (1975a) They are based on the assumption that the figure of the Earth is an oblate spheroid, and hence are more accurate than methods such as great-circle distance which assume a spherical Earth. The first (direct) method computes the location of a point which is a given distance and azimuth (direction) from another point. The second (inverse) method computes the geographical distance and azimuth between two given points. They have been widely used in geodesy because they are accurate to within 0.5 mm (0.020″) on the Earth ellipsoid.
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);