在php中从两个纬度和经度计算最近的经度和经度

Question

假设我有lat1 = 21.154533＆amp; long1 = 79.045760 精度为x米，另一个lat2 = 21.153184＆amp; long2 = 79.045882。所以可以 我找到了与lat1和amp;相关的最近位置（lat，long）。 long1借助lat2＆amp; long2方向？

Answer 1

最后，我找到了自己问题的答案。

  

  
• 使用Lat1，Long1＆amp;计算轴承/角度。 Lat2，Long2
  

    function bearing($lat1, $long1, $lat2, $long2){
    $bearingradians = atan2(asin($long1-$long2)*cos($lat2),
    cos($lat1)*sin($lat2) - sin($lat1)*cos($lat2)*cos($long1-$long2)); 
    $bearingdegrees = abs(rad2deg($bearingradians));
    return $bearingdegrees;
    }

  

  
• 计算Lat1，Long1和＆amp;之间的距离。 LAT2，Long2
  

function distance($lat1, $lon1, $lat2, $lon2) {
      $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);
      $distanceMeters = $dist * 60 * 1000;
     return $distanceMeters; 
    }

  

  
• 然后在Lat1，Long2附近生成新的Lat Long。
  

$earthRadius = 6371000;
$lat1 = deg2rad(21.156845);
$lon1 = deg2rad(79.053331);
$lat2 = deg2rad(21.156805);
$lon2 = deg2rad(79.057634);

//$bearing = $this->bearing($lat1,$lon1,$lat2,$lon2); //you can get value from above function
//$distance = $this->distance($lat1,$lon1,$lat2,$lon2);//you can get value from above function
$bearing = deg2rad(90);
$distance =100;
$Newlat = asin(sin($lat1) * cos($distance / $earthRadius) + cos($lat1) * sin($distance / $earthRadius) * cos($bearing));
$Newlon = $lon1 + atan2(sin($bearing) * sin($distance / $earthRadius) * cos($lat1), cos($distance / $earthRadius) - sin($lat1) * sin($Newlat));
echo 'distance: ' .$distance;
echo 'bearing: ' .$bearing;
echo 'LAT: ' . rad2deg($Newlat);
echo 'LNG: ' . rad2deg($Newlon);

  

结论是我们得到不正确的lat2，long2值。但我们可以   精确计算lat，长度接近lat1和lat2，其精确角度。我们在整合信标技术时发现了这一挑战。