Efficient and speedy logic to find distance between multiple latitude and longitude in java?

Question

I have a product table which I maintain product details and location in latitude and longitude of each product. Normally I have a simple search based on name, price etc.

Now I going to implement a search nearby location. For that I found some logic for finding distance between two latitude and longitude.

Sample links to find distance
http://stackoverflow.com/questions/27928/how-do-i-calculate-distance-between-two-latitude-longitude-points?rq=1
http://stackoverflow.com/questions/3694380/calculating-distance-between-two-points-using-latitude-longitude-what-am-i-doi

But my scenario is I have more than 10,000 products, each search I wants to find distance between my location with 10,000 product is much more complicated.

Is there any simple way to find?

Note : My location is Dynamic.

Answer 1

Basically its not a Java problem what you need an algo which can provide you the distance between two points represented by the Longitude and Latitude.

The logic would be as follows-
1. Find your longitude and latitude (I hope you already have this) say point A.
2. Find the distance between you and all the products i.e. between A and all the products, say d.
3. Sort all the products based on D
4. You are good to go.

Formula for the distance

a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c

where φ is latitude, λ is longitude, R is earth’s radius (mean radius = 6,371km);
note that angles need to be in radians to pass to trig functions!

JavaScript Code
var R = 6371; // km
var φ1 = lat1.toRadians();
var φ2 = lat2.toRadians();
var Δφ = (lat2-lat1).toRadians();
var Δλ = (lon2-lon1).toRadians();

var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) + Math.cos(φ1) * Math.cos(φ2) * Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;

Credit: http://www.movable-type.co.uk/scripts/latlong.html