Fastest Way to Find Distance Between Two Lat/Long Points

I currently have just under a million locations in a mysql database all with longitude and latitude information.

I am trying to find the distance between one point and many other points via a query. It’s not as fast as I want it to be especially with 100+ hits a second.

Is there a faster query or possibly a faster system other than mysql for this? I’m using this query:

   ( 3959 * acos( cos( radians(42.290763) ) * cos( radians( ) ) 
   * cos( radians(locations.lng) - radians(-71.35368)) + sin(radians(42.290763)) 
   * sin( radians( AS distance 
FROM locations 
WHERE active = 1 
HAVING distance < 10 
ORDER BY distance;

Note: The provided distance is in Miles. If you need Kilometers, use 6371 instead of 3959.


Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

  • Create your points using Point values of Geometry data types in MyISAM table. As of Mysql 5.7.5, InnoDB tables now also support SPATIAL indices.
  • Create a SPATIAL index on these points
  • Use MBRContains() to find the values:
      SELECT  *
      FROM    table
      WHERE   MBRContains(LineFromText(CONCAT(
              , @lon + 10 / ( 111.1 / cos(RADIANS(@lat)))
              , ' '
              , @lat + 10 / 111.1
              , ','
              , @lon - 10 / ( 111.1 / cos(RADIANS(@lat)))
              , ' '
              , @lat - 10 / 111.1 
              , ')' )

, or, in MySQL 5.1 and above:

    SELECT  *
    FROM    table
    WHERE   MBRContains
                            Point (
                                    @lon + 10 / ( 111.1 / COS(RADIANS(@lat))),
                                    @lat + 10 / 111.1
                            Point (
                                    @lon - 10 / ( 111.1 / COS(RADIANS(@lat))),
                                    @lat - 10 / 111.1

This will select all points approximately within the box (@lat +/- 10 km, @lon +/- 10km).

This actually is not a box, but a spherical rectangle: latitude and longitude bound segment of the sphere. This may differ from a plain rectangle on the Franz Joseph Land, but quite close to it on most inhabited places.

  • Apply additional filtering to select everything inside the circle (not the square)
  • Possibly apply additional fine filtering to account for the big circle distance (for large distances)

Method 2

Not a MySql specific answer, but it’ll improve the performance of your sql statement.

What you’re effectively doing is calculating the distance to every point in the table, to see if it’s within 10 units of a given point.

What you can do before you run this sql, is create four points that draw a box 20 units on a side, with your point in the center i.e.. (x1,y1 ) . . . (x4, y4), where (x1,y1) is (givenlong + 10 units, givenLat + 10units) . . . (givenLong – 10units, givenLat -10 units).
Actually, you only need two points, top left and bottom right call them (X1, Y1) and (X2, Y2)

Now your SQL statement use these points to exclude rows that definitely are more than 10u from your given point, it can use indexes on the latitudes & longitudes, so will be orders of magnitude faster than what you currently have.


select . . . 
where between X1 and X2 
and   locations.Long between y1 and y2;

The box approach can return false positives (you can pick up points in the corners of the box that are > 10u from the given point), so you still need to calculate the distance of each point. However this again will be much faster because you have drastically limited the number of points to test to the points within the box.

I call this technique “Thinking inside the box” 🙂

EDIT: Can this be put into one SQL statement?

I have no idea what mySql or Php is capable of, sorry.
I don’t know where the best place is to build the four points, or how they could be passed to a mySql query in Php. However, once you have the four points, there’s nothing stopping you combining your own SQL statement with mine.

select name, 
       ( 3959 * acos( cos( radians(42.290763) ) 
              * cos( radians( ) ) 
              * cos( radians( locations.lng ) - radians(-71.35368) ) 
              + sin( radians(42.290763) ) 
              * sin( radians( ) ) ) ) AS distance 
from locations 
where active = 1 
and between X1 and X2 
and locations.Long between y1 and y2
having distance < 10 ORDER BY distance;

I know with MS SQL I can build a SQL statement that declares four floats (X1, Y1, X2, Y2) and calculates them before the “main” select statement, like I said, I’ve no idea if this can be done with MySql. However I’d still be inclined to build the four points in C# and pass them as parameters to the SQL query.

Sorry I can’t be more help, if anyone can answer the MySQL & Php specific portions of this, feel free to edit this answer to do so.

Method 3

I needed to solve similar problem (filtering rows by distance from single point) and by combining original question with answers and comments, I came up with solution which perfectly works for me on both MySQL 5.6 and 5.7.

    (6371 * ACOS(COS(RADIANS(56.946285)) * COS(RADIANS(Y(coordinates))) 
    * COS(RADIANS(X(coordinates)) - RADIANS(24.105078)) + SIN(RADIANS(56.946285))
    * SIN(RADIANS(Y(coordinates))))) AS distance
FROM places
        Point (
            24.105078 + 15 / (111.320 * COS(RADIANS(56.946285))),
            56.946285 + 15 / 111.133
        Point (
            24.105078 - 15 / (111.320 * COS(RADIANS(56.946285))),
            56.946285 - 15 / 111.133
HAVING distance < 15
ORDER By distance

coordinates is field with type POINT and has SPATIAL index
6371 is for calculating distance in kilometres
56.946285 is latitude for central point
24.105078 is longitude for central point
15 is maximum distance in kilometers

In my tests, MySQL uses SPATIAL index on coordinates field to quickly select all rows which are within rectangle and then calculates actual distance for all filtered places to exclude places from rectangles corners and leave only places inside circle.

This is visualisation of my result:


Gray stars visualise all points on map, yellow stars are ones returned by MySQL query. Gray stars inside corners of rectangle (but outside circle) were selected by MBRContains() and then deselected by HAVING clause.

Method 4

The following MySQL function was posted on this blog post. I haven’t tested it much, but from what I gathered from the post, if your latitude and longitude fields are indexed, this may work well for you:


DROP FUNCTION IF EXISTS `get_distance_in_miles_between_geo_locations` $$
CREATE FUNCTION get_distance_in_miles_between_geo_locations(
  geo1_latitude decimal(10,6), geo1_longitude decimal(10,6), 
  geo2_latitude decimal(10,6), geo2_longitude decimal(10,6)) 
returns decimal(10,3) DETERMINISTIC
  return ((ACOS(SIN(geo1_latitude * PI() / 180) * SIN(geo2_latitude * PI() / 180) 
    + COS(geo1_latitude * PI() / 180) * COS(geo2_latitude * PI() / 180) 
    * COS((geo1_longitude - geo2_longitude) * PI() / 180)) * 180 / PI()) 
    * 60 * 1.1515);
END $$


Sample usage:

Assuming a table called places with fields latitude & longitude:

SELECT get_distance_in_miles_between_geo_locations(-34.017330, 22.809500,
latitude, longitude) AS distance_from_input FROM places;

Method 5

if you are using MySQL 5.7.*, then you can use st_distance_sphere(POINT, POINT).

Select st_distance_sphere(POINT(-2.997065, 53.404146 ), POINT(58.615349, 23.56676 ))/1000  as distcance

Method 6

SELECT * FROM (SELECT *,(((acos(sin((43.6980168*pi()/180)) * 
sin((latitude*pi()/180))+cos((43.6980168*pi()/180)) * 
cos((latitude*pi()/180)) * cos(((7.266903899999988- longitude)* 
pi()/180))))*180/pi())*60*1.1515 ) as distance 
FROM wp_users WHERE 1 GROUP BY ID limit 0,10) as X 

This is the distance calculation query between to points in MySQL, I have used it in a long database, it it working perfect! Note: do the changes (database name, table name, column etc) as per your requirements.

Method 7

set @latitude=53.754842;
set @longitude=-2.708077;
set @radius=20;

set @lng_min = @longitude - @radius/abs(cos(radians(@latitude))*69);
set @lng_max = @longitude + @radius/abs(cos(radians(@latitude))*69);
set @lat_min = @latitude - (@radius/69);
set @lat_max = @latitude + (@radius/69);

SELECT * FROM postcode
WHERE (longitude BETWEEN @lng_min AND @lng_max)
AND (latitude BETWEEN @lat_min and @lat_max);


Method 8

   (((acos(sin(('$latitude'*pi()/180)) * sin((`lat`*pi()/180))+cos(('$latitude'*pi()/180)) 
    * cos((`lat`*pi()/180)) * cos((('$longitude'- `lng`)*pi()/180))))*180/pi())*60*1.1515) 
    AS distance
    from table having distance<22;

Method 9

A MySQL function which returns the number of metres between the two coordinates:

RETURN ACOS( SIN(lat1*PI()/180)*SIN(lat2*PI()/180) + COS(lat1*PI()/180)*COS(lat2*PI()/180)*COS(lon2*PI()/180-lon1*PI()/180) ) * 6371000

To return the value in a different format, replace the 6371000 in the function with the radius of Earth in your choice of unit. For example, kilometres would be 6371 and miles would be 3959.

To use the function, just call it as you would any other function in MySQL. For example, if you had a table city, you could find the distance between every city to every other city:

    ROUND(DISTANCE_BETWEEN(`city1`.`latitude`, `city1`.`longitude`, `city2`.`latitude`, `city2`.`longitude`)) AS `distance`
    `city` AS `city1`
    `city` AS `city2`

Method 10

The full code with details about how to install as MySQL plugin are here:

I posted this last year as comment. Since kindly @TylerCollier suggested me to post as answer, here it is.

Another way is to write a custom UDF function that returns the haversine distance from two points. This function can take in input:

lat1 (real), lng1 (real), lat2 (real), lng2 (real), type (string - optinal - 'km', 'ft', 'mi')

So we can write something like this:

SELECT id, name FROM MY_PLACES WHERE haversine_distance(lat1, lng1, lat2, lng2) < 40;

to fetch all records with a distance less then 40 kilometers. Or:

SELECT id, name FROM MY_PLACES WHERE haversine_distance(lat1, lng1, lat2, lng2, 'ft') < 25;

to fetch all records with a distance less then 25 feet.

The core function is:

haversine_distance( UDF_INIT* initid, UDF_ARGS* args, char* is_null, char *error ) {
    double result = *(double*) initid->ptr;
    /*Earth Radius in Kilometers.*/ 
    double R = 6372.797560856;
    double DEG_TO_RAD = M_PI/180.0;
    double RAD_TO_DEG = 180.0/M_PI;
    double lat1 = *(double*) args->args[0];
    double lon1 = *(double*) args->args[1];
    double lat2 = *(double*) args->args[2];
    double lon2 = *(double*) args->args[3];
    double dlon = (lon2 - lon1) * DEG_TO_RAD;
    double dlat = (lat2 - lat1) * DEG_TO_RAD;
    double a = pow(sin(dlat * 0.5),2) + 
        cos(lat1*DEG_TO_RAD) * cos(lat2*DEG_TO_RAD) * pow(sin(dlon * 0.5),2);
    double c = 2.0 * atan2(sqrt(a), sqrt(1-a));
    result = ( R * c );
     * If we have a 5th distance type argument...
    if (args->arg_count == 5) {
        if (strcmp(args->args[4], "ft") == 0) result *= 3280.8399;
        if (strcmp(args->args[4], "mi") == 0) result *= 0.621371192;

    return result;

Method 11

A fast, simple and accurate (for smaller distances) approximation can be done with a spherical projection. At least in my routing algorithm I get a 20% boost compared to the correct calculation. In Java code it looks like:

public double approxDistKm(double fromLat, double fromLon, double toLat, double toLon) {
    double dLat = Math.toRadians(toLat - fromLat);
    double dLon = Math.toRadians(toLon - fromLon);
    double tmp = Math.cos(Math.toRadians((fromLat + toLat) / 2)) * dLon;
    double d = dLat * dLat + tmp * tmp;
    return R * Math.sqrt(d);

Not sure about MySQL (sorry!).

Be sure you know about the limitation (the third param of assertEquals means the accuracy in kilometers):

    float lat = 24.235f;
    float lon = 47.234f;
    CalcDistance dist = new CalcDistance();
    double res = 15.051;
    assertEquals(res, dist.calcDistKm(lat, lon, lat - 0.1, lon + 0.1), 1e-3);
    assertEquals(res, dist.approxDistKm(lat, lon, lat - 0.1, lon + 0.1), 1e-3);

    res = 150.748;
    assertEquals(res, dist.calcDistKm(lat, lon, lat - 1, lon + 1), 1e-3);
    assertEquals(res, dist.approxDistKm(lat, lon, lat - 1, lon + 1), 1e-2);

    res = 1527.919;
    assertEquals(res, dist.calcDistKm(lat, lon, lat - 10, lon + 10), 1e-3);
    assertEquals(res, dist.approxDistKm(lat, lon, lat - 10, lon + 10), 10);

Method 12

Here is a very detailed description of Geo Distance Search with MySQL a solution based on implementation of Haversine Formula to mysql. The complete solution description with theory, implementation and further performance optimization. Although the spatial optimization part didn’t work correct in my case.

Method 13

Have a read of Geo Distance Search with MySQL, a solution
based on implementation of Haversine Formula to MySQL. This is a complete solution
description with theory, implementation and further performance optimization.
Although the spatial optimization part didn’t work correctly in my case.

I noticed two mistakes in this:

  1. the use of abs in the select statement on p8. I just omitted abs and it worked.
  2. the spatial search distance function on p27 does not convert to radians or multiply longitude by cos(latitude), unless his spatial data is loaded with this in consideration (cannot tell from context of article), but his example on p26 indicates that his spatial data POINT is not loaded with radians or degrees.

Method 14

$objectQuery = "SELECT table_master.*, ((acos(sin((" . $latitude . "*pi()/180)) * sin((`latitude`*pi()/180))+cos((" . $latitude . "*pi()/180)) * cos((`latitude`*pi()/180)) * cos(((" . $longitude . "- `longtude`)* pi()/180))))*180/pi())*60*1.1515  as distance FROM `table_post_broadcasts` JOIN table_master ON table_post_broadcasts.master_id = WHERE table_master.type_of_post ='type' HAVING distance <='" . $Radius . "' ORDER BY distance asc";

Method 15

Using mysql

SET @orig_lon = 1.027125;
SET @dest_lon = 1.027125;

SET @orig_lat = 2.398441;
SET @dest_lat = 2.398441;

SET @kmormiles = 6371;-- for distance in miles set to : 3956

SELECT @kmormiles * ACOS(LEAST(COS(RADIANS(@orig_lat)) * 
 COS(RADIANS(@dest_lat)) * COS(RADIANS(@orig_lon - @dest_lon)) + 
 SIN(RADIANS(@orig_lat)) * SIN(RADIANS(@dest_lat)),1.0)) as distance;




NOTE: LEAST is used to avoid null values as a comment suggested on

Method 16

I really liked @Māris Kiseļovs solution, but I like many others may have the Lat and lng’s POINTS reversed from his example. In generalising it I though I would share it. In my case I need to find all the start_points that are within a certain radius of an end_point.

I hope this helps someone.

SELECT @LAT := ST_X(end_point), @LNG := ST_Y(end_point) FROM routes  WHERE route_ID = 280;
  (6371e3 * ACOS(COS(RADIANS(@LAT)) * COS(RADIANS(ST_X(start_point))) 
  * COS(RADIANS(ST_Y(start_point)) - RADIANS(@LNG)) + SIN(RADIANS(@LAT))
  * SIN(RADIANS(ST_X(start_point))))) AS distance 
FROM routes
    Point (
            @LNG + 15 / (111.320 * COS(RADIANS(@LAT))),
            @LAT + 15 / 111.133
    Point (
    @LNG - 15 / (111.320 * COS(RADIANS(@LAT))),
        @LAT - 15 / 111.133
HAVING distance < 100
ORDER By distance;

All methods was sourced from or, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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