如何计算两个GPS坐标之间的距离(使用经纬度)?


当前回答

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。

其他回答

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

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;

飞镖版本

Haversine Algorithm。

import 'dart:math';

class GeoUtils {

  static double _degreesToRadians(degrees) {
    return degrees * pi / 180;
  }

  static double 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 = sin(dLat/2) * sin(dLat/2) +
        sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2);
    var c = 2 * atan2(sqrt(a), sqrt(1-a));
    return earthRadiusKm * c;
  }
}
    private double deg2rad(double deg)
    {
        return (deg * Math.PI / 180.0);
    }

    private double rad2deg(double rad)
    {
        return (rad / Math.PI * 180.0);
    }

    private double GetDistance(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Distance in Kilo Meter
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
        return (dist);
    }

    private double GetDirection(double lat1, double lon1, double lat2, double lon2)
    {
        //code for Direction in Degrees
        double dlat = deg2rad(lat1) - deg2rad(lat2);
        double dlon = deg2rad(lon1) - deg2rad(lon2);
        double y = Math.Sin(dlon) * Math.Cos(lat2);
        double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
        double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
        if (direct < 0)
            direct = direct + 360;
        return (direct);
    }

    private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
        double theta = lon1 - lon2;
        double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
        dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
        double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
        return (Speed);
    }

    private double GetDuration(DateTime CurTime, DateTime PrevTime)
    {
        //code for speed in Kilo Meter/Hour
        TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
        double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
        return (TimeDifferenceInSeconds);
    }

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。