在 PHP 中测量两个坐标之间的距离

Question

嗨，我需要计算具有纬度和经度的两点之间的距离。

我想避免对外部 API 的任何调用。

我尝试在 PHP 中实现半正矢公式：

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;

      }
    }

用一些具有公共距离的给定点进行测试，我没有得到可靠的结果。

不明白原来的公式或者我的实现是否有错误

Answer 1

不久前我写了一个半正弦公式的例子，并将其发布在我的网站上：

/**
 * 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，您得到的距离与您通过参数

$earthRadius

传入的单位相同。默认值为 6371000 米，因此结果也将以 [m] 为单位。要获得以英里为单位的结果，您可以例如通过 3959 英里作为

$earthRadius

，结果将以 [mi] 为单位。在我看来，如果没有特殊原因，坚持使用 SI 单位是一个好习惯。

编辑：

正如 TreyA 正确指出的那样，由于舍入误差，半正矢公式在对映点方面存在弱点（尽管它对于小距离来说是稳定的）。要绕过它们，您可以使用 Vincenty 公式。

/**
 * 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;
}

Answer 2

我找到了这个代码，它给了我可靠的结果。

function distance($lat1, $lon1, $lat2, $lon2, $unit) {

  $theta = $lon1 - $lon2;
  $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
  $dist = acos($dist);
  $dist = rad2deg($dist);
  $miles = $dist * 60 * 1.1515;
  $unit = strtoupper($unit);

  if ($unit == "K") {
      return ($miles * 1.609344);
  } else if ($unit == "N") {
      return ($miles * 0.8684);
  } else {
      return $miles;
  }
}

结果：

echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";

Answer 3

这只是 @martinstoeckli 和 @Janith Chinthana 答案的补充。对于那些好奇哪种算法最快的人，我编写了性能测试。最佳性能结果显示来自 codexworld.com 的优化功能：

/**
 * Optimized algorithm from http://www.codexworld.com
 *
 * @param float $latitudeFrom
 * @param float $longitudeFrom
 * @param float $latitudeTo
 * @param float $longitudeTo
 *
 * @return float [km]
 */
function codexworldGetDistanceOpt($latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo)
{
    $rad = M_PI / 180;
    //Calculate distance from latitude and longitude
    $theta = $longitudeFrom - $longitudeTo;
    $dist = sin($latitudeFrom * $rad) 
        * sin($latitudeTo * $rad) +  cos($latitudeFrom * $rad)
        * cos($latitudeTo * $rad) * cos($theta * $rad);

    return acos($dist) / $rad * 60 *  1.853;
}

这是测试结果：

Test name       Repeats         Result          Performance     
codexworld-opt  10000           0.084952 sec    +0.00%
codexworld      10000           0.104127 sec    -22.57%
custom          10000           0.107419 sec    -26.45%
custom2         10000           0.111576 sec    -31.34%
custom1         10000           0.136691 sec    -60.90%
vincenty        10000           0.165881 sec    -95.26%

Answer 4

这是一个很老的问题，但是对于那些对返回与 Google 地图相同结果的 PHP 代码感兴趣的人，可以使用以下代码：

/**
 * Computes the distance between two coordinates.
 *
 * Implementation based on reverse engineering of
 * <code>google.maps.geometry.spherical.computeDistanceBetween()</code>.
 *
 * @param float $lat1 Latitude from the first point.
 * @param float $lng1 Longitude from the first point.
 * @param float $lat2 Latitude from the second point.
 * @param float $lng2 Longitude from the second point.
 * @param float $radius (optional) Radius in meters.
 *
 * @return float Distance in meters.
 */
function computeDistance($lat1, $lng1, $lat2, $lng2, $radius = 6378137)
{
    static $x = M_PI / 180;
    $lat1 *= $x; $lng1 *= $x;
    $lat2 *= $x; $lng2 *= $x;
    $distance = 2 * asin(sqrt(pow(sin(($lat1 - $lat2) / 2), 2) + cos($lat1) * cos($lat2) * pow(sin(($lng1 - $lng2) / 2), 2)));

    return $distance * $radius;
}

我已经用各种坐标进行了测试，效果非常好。

我认为它也应该比某些替代方案更快。但没有测试过。

提示：Google 地图使用 6378137 作为地球半径。因此，将其与其他算法一起使用也可能有效。

Answer 5

这里是计算两个纬度和经度之间距离的简单而完美的代码。从这里找到以下代码 - http://www.codexworld.com/distance- Between-two-addresses-google-maps-api-php/

$latitudeFrom = '22.574864';
$longitudeFrom = '88.437915';

$latitudeTo = '22.568662';
$longitudeTo = '88.431918';

//Calculate distance from latitude and longitude
$theta = $longitudeFrom - $longitudeTo;
$dist = sin(deg2rad($latitudeFrom)) * sin(deg2rad($latitudeTo)) +  cos(deg2rad($latitudeFrom)) * cos(deg2rad($latitudeTo)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;

$distance = ($miles * 1.609344).' km';

Answer 6

对于那些喜欢更短更快的人（不调用 deg2rad()）。

function circle_distance($lat1, $lon1, $lat2, $lon2) {
  $rad = M_PI / 180;
  return acos(sin($lat2*$rad) * sin($lat1*$rad) + cos($lat2*$rad) * cos($lat1*$rad) * cos($lon2*$rad - $lon1*$rad)) * 6371;// Kilometers
}

Answer 7

尝试这个函数来计算纬度和经度点之间的距离

function calculateDistanceBetweenTwoPoints($latitudeOne='', $longitudeOne='', $latitudeTwo='', $longitudeTwo='',$distanceUnit ='',$round=false,$decimalPoints='')
    {
        if (empty($decimalPoints)) 
        {
            $decimalPoints = '3';
        }
        if (empty($distanceUnit)) {
            $distanceUnit = 'KM';
        }
        $distanceUnit = strtolower($distanceUnit);
        $pointDifference = $longitudeOne - $longitudeTwo;
        $toSin = (sin(deg2rad($latitudeOne)) * sin(deg2rad($latitudeTwo))) + (cos(deg2rad($latitudeOne)) * cos(deg2rad($latitudeTwo)) * cos(deg2rad($pointDifference)));
        $toAcos = acos($toSin);
        $toRad2Deg = rad2deg($toAcos);

        $toMiles  =  $toRad2Deg * 60 * 1.1515;
        $toKilometers = $toMiles * 1.609344;
        $toNauticalMiles = $toMiles * 0.8684;
        $toMeters = $toKilometers * 1000;
        $toFeets = $toMiles * 5280;
        $toYards = $toFeets / 3;


              switch (strtoupper($distanceUnit)) 
              {
                  case 'ML'://miles
                         $toMiles  = ($round == true ? round($toMiles) : round($toMiles, $decimalPoints));
                         return $toMiles;
                      break;
                  case 'KM'://Kilometers
                        $toKilometers  = ($round == true ? round($toKilometers) : round($toKilometers, $decimalPoints));
                        return $toKilometers;
                      break;
                  case 'MT'://Meters
                        $toMeters  = ($round == true ? round($toMeters) : round($toMeters, $decimalPoints));
                        return $toMeters;
                      break;
                  case 'FT'://feets
                        $toFeets  = ($round == true ? round($toFeets) : round($toFeets, $decimalPoints));
                        return $toFeets;
                      break;
                  case 'YD'://yards
                        $toYards  = ($round == true ? round($toYards) : round($toYards, $decimalPoints));
                        return $toYards;
                      break;
                  case 'NM'://Nautical miles
                        $toNauticalMiles  = ($round == true ? round($toNauticalMiles) : round($toNauticalMiles, $decimalPoints));
                        return $toNauticalMiles;
                      break;
              }


    }

然后使用函数作为

echo calculateDistanceBetweenTwoPoints('11.657740','77.766270','11.074820','77.002160','ML',true,5);

希望有帮助

Answer 8

尝试这个会产生很棒的结果

function getDistance($point1_lat, $point1_long, $point2_lat, $point2_long, $unit = 'km', $decimals = 2) {
        // Calculate the distance in degrees
        $degrees = rad2deg(acos((sin(deg2rad($point1_lat))*sin(deg2rad($point2_lat))) + (cos(deg2rad($point1_lat))*cos(deg2rad($point2_lat))*cos(deg2rad($point1_long-$point2_long)))));

        // Convert the distance in degrees to the chosen unit (kilometres, miles or nautical miles)
        switch($unit) {
            case 'km':
                $distance = $degrees * 111.13384; // 1 degree = 111.13384 km, based on the average diameter of the Earth (12,735 km)
                break;
            case 'mi':
                $distance = $degrees * 69.05482; // 1 degree = 69.05482 miles, based on the average diameter of the Earth (7,913.1 miles)
                break;
            case 'nmi':
                $distance =  $degrees * 59.97662; // 1 degree = 59.97662 nautic miles, based on the average diameter of the Earth (6,876.3 nautical miles)
        }
        return round($distance, $decimals);
    }

Answer 9

对于精确值，这样做：

public function DistAB()
{
      $delta_lat = $this->lat_b - $this->lat_a ;
      $delta_lon = $this->lon_b - $this->lon_a ;

      $a = pow(sin($delta_lat/2), 2);
      $a += cos(deg2rad($this->lat_a9)) * cos(deg2rad($this->lat_b9)) * pow(sin(deg2rad($delta_lon/29)), 2);
      $c = 2 * atan2(sqrt($a), sqrt(1-$a));

      $distance = 2 * $earth_radius * $c;
      $distance = round($distance, 4);

      $this->measure = $distance;
}

嗯，我认为应该这样做......

编辑：

对于公式和至少 JS 实现，请尝试：http://www.movable-type.co.uk/scripts/latlong.html

敢不敢...我忘了对圆函数中的所有值进行 deg2rad...

Answer 10

由于此处所写的大圆距离理论，乘数在每个坐标处都会发生变化：

http://en.wikipedia.org/wiki/Great-circle_distance

您可以使用此处描述的公式计算最接近的值：

http://en.wikipedia.org/wiki/Great-circle_distance#Worked_example

关键是将每个度分秒值转换为所有度值：

N 36°7.2', W 86°40.2'  N = (+) , W = (-), S = (-), E = (+) 
referencing the Greenwich meridian and Equator parallel

(phi)     36.12° = 36° + 7.2'/60' 

(lambda)  -86.67° = 86° + 40.2'/60'

Answer 11

最简单的方法之一是：

$my_latitude = "";
$my_longitude = "";
$her_latitude = "";
$her_longitude = "";

$distance = round((((acos(sin(($my_latitude*pi()/180)) * sin(($her_latitude*pi()/180))+cos(($my_latitude*pi()/180)) * cos(($her_latitude*pi()/180)) * cos((($my_longitude- $her_longitude)*pi()/180))))*180/pi())*60*1.1515*1.609344), 2);
echo $distance;

四舍五入至小数点后两位。

Answer 12

您可以尝试该库geo-math-php

composer require rkondratuk/geo-math-php:^1

示例：

<?php

use PhpGeoMath\Model\Polar3dPoint;

$polarPoint1 = new Polar3dPoint(
    40.758742779050706, -73.97855507715238, Polar3dPoint::EARTH_RADIUS_IN_METERS
);

$polarPoint2 = new Polar3dPoint(
    40.74843388072615, -73.98566565776102, Polar3dPoint::EARTH_RADIUS_IN_METERS
);

$geoDistance = $polarPoint2->calcGeoDistanceToPoint($polarPoint1);

Answer 13

您好，使用两个不同的纬度和经度获取距离和时间的代码

$url ="https://maps.googleapis.com/maps/api/distancematrix/json?units=imperial&origins=16.538048,80.613266&destinations=23.0225,72.5714";



    $ch = curl_init();
    // Disable SSL verification

    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
    // Will return the response, if false it print the response
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    // Set the url
    curl_setopt($ch, CURLOPT_URL,$url);
    // Execute
    $result=curl_exec($ch);
    // Closing
    curl_close($ch);

    $result_array=json_decode($result);
print_r($result_array);

您可以检查下面链接的示例使用 php 中的纬度和经度获取两个不同位置之间的时间