在PHP中测量两个坐标之间的距离

问题描述：

我需要计算纬度和经度的两点之间的距离.

Hi I have the need to calculate the distance between two points having the lat and long.

我希望避免对外部API的任何调用.

I would like to avoid any call to external API.

我试图在PHP中实现Haversine公式:

I tried to implement the Haversine Formula in PHP:

这是代码:

class CoordDistance
 {
    public $lat_a = 0;
    public $lon_a = 0;
    public $lat_b = 0;
    public $lon_b = 0;

    public $measure_unit = 'kilometers';

    public $measure_state = false;

    public $measure = 0;

    public $error = '';



    public function DistAB()

      {
          $delta_lat = $this->lat_b - $this->lat_a ;
          $delta_lon = $this->lon_b - $this->lon_a ;

          $earth_radius = 6372.795477598;

          $alpha    = $delta_lat/2;
          $beta     = $delta_lon/2;
          $a        = sin(deg2rad($alpha)) * sin(deg2rad($alpha)) + cos(deg2rad($this->lat_a)) * cos(deg2rad($this->lat_b)) * sin(deg2rad($beta)) * sin(deg2rad($beta)) ;
          $c        = asin(min(1, sqrt($a)));
          $distance = 2*$earth_radius * $c;
          $distance = round($distance, 4);

          $this->measure = $distance;

      }
    }

用一些距离公共的给定点进行测试，我没有得到可靠的结果.

Testing it with some given points which have public distances I don't get a reliable result.

我不知道原始公式或实现中是否存在错误

I don't understand if there is an error in the original formula or in my implementation

答

不久前，我写了一个hasversine公式的示例，并将其发布在我的网站上:

Not long ago I wrote an example of the haversine formula, and published it on my website:

/**
 * Calculates the great-circle distance between two points, with
 * the Haversine formula.
 * @param float $latitudeFrom Latitude of start point in [deg decimal]
 * @param float $longitudeFrom Longitude of start point in [deg decimal]
 * @param float $latitudeTo Latitude of target point in [deg decimal]
 * @param float $longitudeTo Longitude of target point in [deg decimal]
 * @param float $earthRadius Mean earth radius in [m]
 * @return float Distance between points in [m] (same as earthRadius)
 */
function haversineGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $latDelta = $latTo - $latFrom;
  $lonDelta = $lonTo - $lonFrom;

  $angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
    cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
  return $angle * $earthRadius;
}

➽请注意，返回的距离与使用参数$earthRadius传递时返回的单位相同.默认值为6371000米，因此结果也将以[m]为单位.为了获得以英里为单位的结果，您可以例如将3959英里作为$earthRadius传递，结果将以[mi]为单位.我认为，如果没有特别的理由，坚持SI单位是个好习惯.

➽ Note that you get the distance back in the same unit as you pass in with the parameter $earthRadius. The default value is 6371000 meters so the result will be in [m] too. To get the result in miles, you could e.g. pass 3959 miles as $earthRadius and the result would be in [mi]. In my opinion it is a good habit to stick with the SI units, if there is no particular reason to do otherwise.

正如TreyA正确指出的那样，由于舍入误差，Haversine公式具有对映点的弱点 (尽管可以在短距离内保持稳定).要绕过它们，您可以改用 Vincenty公式.

As TreyA correctly pointed out, the Haversine formula has weaknesses with antipodal points because of rounding errors (though it is stable for small distances). To get around them, you could use the Vincenty formula instead.

/**
 * Calculates the great-circle distance between two points, with
 * the Vincenty formula.
 * @param float $latitudeFrom Latitude of start point in [deg decimal]
 * @param float $longitudeFrom Longitude of start point in [deg decimal]
 * @param float $latitudeTo Latitude of target point in [deg decimal]
 * @param float $longitudeTo Longitude of target point in [deg decimal]
 * @param float $earthRadius Mean earth radius in [m]
 * @return float Distance between points in [m] (same as earthRadius)
 */
public static function vincentyGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $lonDelta = $lonTo - $lonFrom;
  $a = pow(cos($latTo) * sin($lonDelta), 2) +
    pow(cos($latFrom) * sin($latTo) - sin($latFrom) * cos($latTo) * cos($lonDelta), 2);
  $b = sin($latFrom) * sin($latTo) + cos($latFrom) * cos($latTo) * cos($lonDelta);

  $angle = atan2(sqrt($a), $b);
  return $angle * $earthRadius;
}

在PHP中测量两个坐标之间的距离

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