[SOLVED] Inverted Distance Calculator

[raindropz]

Hello all.

I'm stucked with something, maybe someone can help me.  ???

 

-I've got a function that calculates the distance between two points of the world (given the latitude/longitude coordinates).

-I have a mysql table with the latitudes/longitudes of different cities.

-What I need to do is the invertion of the script. I mean: given an initial point (longitude/latitude) and a radius distance, then select each cities in this radious (Example: Select All cities inside a distance of 50 miles from CityX).

-Could PLEASE somebody help me with the MySql Query that I need to perform? I'm truly lost about this.  :'(

 

Well, this is what I got:

 

1. This is the table structure:

 

city            varchar(90)

latitude      decimal(10,7)

longitude    decimal(10,7)

 

2. This is the function I have:

 

[color=blue]<?php

/*::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
/*::                                                                         :*/
/*::  this routine calculates the distance between two points (given the     :*/
/*::  latitude/longitude of those points). it is being used to calculate     :*/
/*::  the distance between two zip codes or postal codes using our           :*/
/*::  zipcodeworld(tm) and postalcodeworld(tm) products.                     :*/
/*::                                                                         :*/
/*::  definitions:                                                           :*/
/*::    south latitudes are negative, east longitudes are positive           :*/
/*::                                                                         :*/
/*::  passed to function:                                                    :*/
/*::    lat1, lon1 = latitude and longitude of point 1 (in decimal degrees)  :*/
/*::    lat2, lon2 = latitude and longitude of point 2 (in decimal degrees)  :*/
/*::    unit = the unit you desire for results                               :*/
/*::           where: 'm' is statute miles                                   :*/
/*::                  'k' is kilometers (default)                            :*/
/*::                  'n' is nautical miles                                  :*/
/*::  united states zip code/ canadian postal code databases with latitude & :*/
/*::  longitude are available at http://www.zipcodeworld.com                 :*/
/*::                                                                         :*/
/*::  For enquiries, please contact sales@zipcodeworld.com                   :*/
/*::                                                                         :*/
/*::  official web site: http://www.zipcodeworld.com                         :*/
/*::                                                                         :*/
/*::  hexa software development center © all rights reserved 2004            :*/
/*::                                                                         :*/
/*::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::*/
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;
      }
}

echo distance(32.9697, -96.80322, 29.46786, -98.53506, "m") . " miles
";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "k") . " kilometers
";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "n") . " nautical miles
";[/color]

 

Thanks for all! 

 

-Raindropz 

[toplay]

You didn't post it in the wrong forum.

 

I moved it here.

 

If you're asking for a MySQL query help, then it belongs here!

 

 

[raindropz]

Oh... I see... Well it's good to know that!, I was just starting to think that I was crazy lol

[Barand]

Something like

 

SELECT * FROM cities
WHERE
    60 * 1.1515 * DEGREES(ACOS(SIN(RADIANS($cityXlat))
    * SIN(RADIANS(latitude)) +  COS(RADIANS($cityXlat))
    * COS(RADIANS(latitude)) * COS(RADIANS($cityXlong-longitude)))) <= 50

[raindropz]

It works 

Thanks for the support!!

 

-raindropz