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 }

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;

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
    }
}

计算两个坐标之间的纬度和经度的距离，包括一个Javascript实现。

西部和南部的位置是负的。 记住，分和秒是60度，所以S31 30'是-31.50度。

别忘了把角度转换成弧度。许多语言都有这个功能。或者它是一个简单的计算:弧度=角度* PI / 180。

function degreesToRadians(degrees) {
  return degrees * Math.PI / 180;
}

function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
  var earthRadiusKm = 6371;

  var dLat = degreesToRadians(lat2-lat1);
  var dLon = degreesToRadians(lon2-lon1);

  lat1 = degreesToRadians(lat1);
  lat2 = degreesToRadians(lat2);

  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
          Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2); 
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
  return earthRadiusKm * c;
}

下面是一些用法的例子:

distanceInKmBetweenEarthCoordinates(0,0,0,0)  // Distance between same 
                                              // points should be 0
0

distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
                                                          // to Arlington
5918.185064088764

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

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;
}