How to measure distance using Haversine formula with MySQL? - mysql

Question

I get Latitude and Longitudes from Google Maps Reverse-Geocoding API and then I need something like this:
mysql_query("SELECT users.*, ".mysql_distance_column($lat,$lng)." FROM users ORDER BY DISTANCE";
function mysql_distance_column($lat=40 , $lng=-73) {
$defaultLatitudeColumn = 'user_lat';
$defaultLongitudeColumn='user_lng';
$defaultColumnName='user_distance';
return "((
(3956 * 2 * ASIN(SQRT( POWER(SIN(({$lat} - abs({$defaultLatitudeColumn}))
* pi()/180 / 2), 2) + COS({$lat} * pi()/180 )
* COS(abs({$defaultLatitudeColumn}) * pi()/180)
* POWER(SIN(({$lng} - {$defaultLongitudeColumn}) * pi()/180 / 2), 2) ))
)) ) as {$defaultColumnName} ";
}
UPDATE
I cant ge this to work
delimiter //
CREATE FUNCTION `GeoDistMiles`( lat1 FLOAT (10,6), lon1 FLOAT (10,6), lat2 FLOAT (10,6), lon2 FLOAT (10,6) )
RETURNS FLOAT
DETERMINISTIC
NO SQL
BEGIN
DECLARE pi, q1, q2, q3 FLOAT (10,6);
DECLARE rads FLOAT (10,6) DEFAULT 0;
SET pi = PI();
SET lat1 = lat1 * pi / 180;
SET lon1 = lon1 * pi / 180;
SET lat2 = lat2 * pi / 180;
SET lon2 = lon2 * pi / 180;
SET q1 = COS(lon1-lon2);
SET q2 = COS(lat1-lat2);
SET q3 = COS(lat1+lat2);
SET rads = ACOS( 0.5*((1.0+q1)*q2 - (1.0-q1)*q3) );
RETURN 3963.346 * rads;
END

Answer 1

Here is the formula I use. Remember that the Earth is not a perfect sphere, so the results will never be perfect.
CREATE DEFINER=`root`#`localhost` FUNCTION `GeoDistMiles`( lat1 FLOAT, lon1 FLOAT, lat2 FLOAT, lon2 FLOAT ) RETURNS float
BEGIN
DECLARE pi, q1, q2, q3 FLOAT;
DECLARE rads FLOAT DEFAULT 0;
SET pi = PI();
SET lat1 = lat1 * pi / 180;
SET lon1 = lon1 * pi / 180;
SET lat2 = lat2 * pi / 180;
SET lon2 = lon2 * pi / 180;
SET q1 = COS(lon1-lon2);
SET q2 = COS(lat1-lat2);
SET q3 = COS(lat1+lat2);
SET rads = ACOS( 0.5*((1.0+q1)*q2 - (1.0-q1)*q3) );
RETURN 3963.346 * rads;
END

Answer 2

I assume that you are trying to use http://en.wikipedia.org/wiki/Haversine_formula.
This is lightly tested, but I think that your formula should be:
(ROUND((3956 * 2 * ASIN(SQRT(POWER(SIN(({$lat} - {$defaultLatitudeColumn}) * pi() / 180 / 2), 2) + COS({$lat} * pi()/180 ) * COS({$defaultLatitudeColumn} * pi()/180) *POWER(SIN(({$lng} - {$defaultLongitudeColumn}) * pi()/180 / 2), 2) )) )*{$magicNumber}) )/{$magicNumber}
(I removed the abs calls.)

Answer 3

Hi I have a simple procedure you can use it for your work.
The procedure is very simple and calculate the distance between two cities.you can modify it in your way.
drop procedure if exists select_lattitude_longitude;
delimiter //
create procedure select_lattitude_longitude(In CityName1 varchar(20) , In CityName2 varchar(20))
begin
declare origin_lat float(10,2);
declare origin_long float(10,2);
declare dest_lat float(10,2);
declare dest_long float(10,2);
if CityName1 Not In (select Name from City_lat_lon) OR CityName2 Not In (select Name from City_lat_lon) then
select 'The Name Not Exist or Not Valid Please Check the Names given by you' as Message;
else
select lattitude into origin_lat from City_lat_lon where Name=CityName1;
select longitude into origin_long from City_lat_lon where Name=CityName1;
select lattitude into dest_lat from City_lat_lon where Name=CityName2;
select longitude into dest_long from City_lat_lon where Name=CityName2;
select origin_lat as CityName1_lattitude,
origin_long as CityName1_longitude,
dest_lat as CityName2_lattitude,
dest_long as CityName2_longitude;
SELECT 3956 * 2 * ASIN(SQRT( POWER(SIN((origin_lat - dest_lat) * pi()/180 / 2), 2) + COS(origin_lat * pi()/180) * COS(dest_lat * pi()/180) * POWER(SIN((origin_long-dest_long) * pi()/180 / 2), 2) )) * 1.609344 as Distance_In_Kms ;
end if;
end ;
//
delimiter ;

Answer 4

SELECT ACOS(COS(RADIANS(lat)) *
COS(RADIANS(lon)) * COS(RADIANS(34.7405350)) * COS(RADIANS(-92.3245120)) +
COS(RADIANS(lat)) * SIN(RADIANS(lon)) * COS(RADIANS(34.7405350)) *
SIN(RADIANS(-92.3245120)) + SIN(RADIANS(lat)) * SIN(RADIANS(34.7405350))) *
3963.1 AS Distance
FROM Stores
WHERE 1
HAVING Distance <= 50
Here's how I use it in PHP:
// Find rows in Stores within 50 miles of $lat,$lon
$lat = '34.7405350';
$lon = '-92.3245120';
$sql = "SELECT Stores.*, ACOS(COS(RADIANS(lat)) *
COS(RADIANS(lon)) * COS(RADIANS($lat)) * COS(RADIANS($lon)) +
COS(RADIANS(lat)) * SIN(RADIANS(lon)) * COS(RADIANS($lat)) *
SIN(RADIANS($lon)) + SIN(RADIANS(lat)) * SIN(RADIANS($lat))) *
3963.1 AS Distance
FROM Stores
WHERE 1
HAVING Distance <= 50";

Answer 5

3956 * 2 * ASIN(SQRT( POWER(SIN(($latitude -( cp.latitude)) * pi()/180 / 2),2) + COS($latitude * pi()/180 ) * COS( abs( cp.latitude) * pi()/180) * POWER(SIN(($longitude - cp.longitude) * pi()/180 / 2), 2) )) as distance
Returns miles or KM?