PHP Select random lon / lat in a specific radius

Let's say I have this lon / lat: 33.33333,22.22222

How can I randomly select another lon / lat within X miles / km?

Thank,

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4 answers

You can use this message to help you:

http://blog.fedecarg.com/2009/02/08/geo-proximity-search-the-haversine-equation/

So, in your example, you just pick a random number from 1 to 10 miles, where 10 is your "within a certain radius."

$longitude = (float) 33.33333;
$latitude = (float) 22.22222;
$radius = rand(1,10); // in miles

$lng_min = $longitude - $radius / abs(cos(deg2rad($latitude)) * 69);
$lng_max = $longitude + $radius / abs(cos(deg2rad($latitude)) * 69);
$lat_min = $latitude - ($radius / 69);
$lat_max = $latitude + ($radius / 69);

echo 'lng (min/max): ' . $lng_min . '/' . $lng_max . PHP_EOL;
echo 'lat (min/max): ' . $lat_min . '/' . $lat_max;

Update:

As Tomalak said in the comments below, this works under the assumption that the earth is a sphere and not an uneven geoid. Because of this, you get approximations rather than potentially (close) exact results.

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@MikeLewis , , .

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$distance (, $radius ) $distance . , . : .

, , (90-($distance/(pi*3959)*180). : .

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.. , , . , z (0,0,1) , y (0,1,0) , 0. , .

/**
 * Given a $centre (latitude, longitude) co-ordinates and a
 * distance $radius (miles), returns a random point (latitude,longtitude)
 * which is within $radius miles of $centre.
 *
 * @param  array $centre Numeric array of floats. First element is 
 *                       latitude, second is longitude.
 * @param  float $radius The radius (in miles).
 * @return array         Numeric array of floats (lat/lng). First 
 *                       element is latitude, second is longitude.
 */
 function generate_random_point( $centre, $radius ){

      $radius_earth = 3959; //miles

      //Pick random distance within $distance;
      $distance = lcg_value()*$radius;

      //Convert degrees to radians.
      $centre_rads = array_map( 'deg2rad', $centre );

      //First suppose our point is the north pole.
      //Find a random point $distance miles away
      $lat_rads = (pi()/2) -  $distance/$radius_earth;
      $lng_rads = lcg_value()*2*pi();


      //($lat_rads,$lng_rads) is a point on the circle which is
      //$distance miles from the north pole. Convert to Cartesian
      $x1 = cos( $lat_rads ) * sin( $lng_rads );
      $y1 = cos( $lat_rads ) * cos( $lng_rads );
      $z1 = sin( $lat_rads );


      //Rotate that sphere so that the north pole is now at $centre.

      //Rotate in x axis by $rot = (pi()/2) - $centre_rads[0];
      $rot = (pi()/2) - $centre_rads[0];
      $x2 = $x1;
      $y2 = $y1 * cos( $rot ) + $z1 * sin( $rot );
      $z2 = -$y1 * sin( $rot ) + $z1 * cos( $rot );

      //Rotate in z axis by $rot = $centre_rads[1]
      $rot = $centre_rads[1];
      $x3 = $x2 * cos( $rot ) + $y2 * sin( $rot );
      $y3 = -$x2 * sin( $rot ) + $y2 * cos( $rot );
      $z3 = $z2;


      //Finally convert this point to polar co-ords
      $lng_rads = atan2( $x3, $y3 );
      $lat_rads = asin( $z3 );

      return array_map( 'rad2deg', array( $lat_rads, $lng_rads ) );
 }
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x1, 0 x. x2, 0 x. (1/2) x1 + , (1/2) x2 + .

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Matlab .

function [lat, lon] = geosample(lat0, lon0, r0, n)
% [lat, lon] = geosample(lat0, lon0, r0, n)
%
% Return n points on the WGS84 ellipsoid within a distance r0 of
% (lat0,lon0) and uniformly distributed on the surface.  The returned
% lat and lon are n x 1 vectors.
%
% Requires Matlab package
%  http://www.mathworks.com/matlabcentral/fileexchange/39108

  todo = true(n,1); lat = zeros(n,1); lon = lat;

  while any(todo)
    n1 = sum(todo);
    r = r0 * max(rand(n1,2), [], 2);  % r = r0*sqrt(U) using cheap sqrt
    azi = 180 * (2 * rand(n1,1) - 1); % sample azi uniformly
    [lat(todo), lon(todo), ~, ~, m, ~, ~, sig] = ...
        geodreckon(lat0, lon0, r, azi);
    % Only count points with sig <= 180 (otherwise it not a shortest
    % path).  Also because of the curvature of the ellipsoid, large r
    % are sampled too frequently, by a factor r/m.  This following
    % accounts for this...
    todo(todo) = ~(sig <= 180 & r .* rand(n1,1) <= m);
  end
end

This code sample is evenly inside the circle on the azimuthal equidistant projection centered at lat0, lon0. Radial, respectively. azimuthal, the scale for this projection is 1, respectively. g / m Therefore, the range of distortion is r / m, and this is due to the adoption of such points with a probability of t / g.

This code also describes a situation where r0 is about half the circumference of the earth and avoids double sampling, almost antipodal points.

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