Determine longitudes and latitudes within a range

Question

I have locations in my database. A location has the attributes latitude and longitude (taken from google maps, example: 48.809591). Is there any query that could help me retrieve the locations within a range of another location?

Example: I have the location A with latitude = 48.809591, and longitude = 2.124009 and want to retrieve all location objects in my database that are within 5 miles of location A

My first thought was to retrieve the locations in a square where location.latitude < A.latitude + 5 miles and location.latitude > A.latitude - 5 miles and location.longitude < A.longitude + 5 miles and location.longitude > A.longitude - 5 miles, and then remove the irrelevant locations from the returned array with the help of something like http://www.movable-type.co.uk/scripts/latlong.html

Any ideas?

Answer 1

Just in case you're using MySQL as your DBMS¹, you may be interested in checking out the following presentation:

• Geo/Spatial Search with MySQL² by Alexander Rubin

The author describes how you can use the Haversine Formula in MySQL to order spatial data by proximity and limit the results to a defined radius. More importantly, he also describes how to avoid a full table scan for such queries, using traditional indexes on the latitude and longitude columns.

---

¹ Even if you aren't, this is still interesting and applicable.
² There is also a pdf version of the presentation.

Answer 2

The calculation you want, i think, is called the great circle distance:

http://en.wikipedia.org/wiki/Great-circle_distance

Answer 3

You would need a distance function.

For Sql Server it would look something like this (note that distance is in kilometers)

    CREATE FUNCTION distance
    (
      @startLatitude float, 
      @startLongitude float, 
      @endLatitude float,
      @endLongitude float
    )
    RETURNS float
    AS
    BEGIN

      DECLARE @distance float;

      set @distance = 
        6371 * 2 * atn2(sqrt(power(sin(pi() / 180 * (@endLatitude - @startLatitude) / 2), 2) +
        power(cos(@startLatitude * pi() / 180), 2) *
        power(sin(pi() / 180 * (@endLongitude - @startLongitude) / 2), 2)),
        sqrt(1 - power(sin(pi() / 180 * (@endLatitude - @startLatitude) / 2), 2) +
        power(cos(@startLatitude * pi() / 180), 2) *
        power(sin(pi() / 180 * (@endLongitude - @startLongitude) / 2), 2)));
      RETURN @distance
    END