How to find distance from the latitude and longitude of two locations?

Question

I have a set of latitudes and longitudes of locations.

• How to find distance from one location in the set to another?
• Is there a f

Answer 1

Apply the Haversine formula to find the distance. See the C# code below to find the distance between 2 coordinates. Better still if you want to say find a list of stores within a certain radius, you could apply a WHERE clause in SQL or a LINQ filter in C# to it.

The formula here is in kilometres, you will have to change the relevant numbers and it will work for miles.

E.g: Convert 6371.392896 to miles.

    DECLARE @radiusInKm AS FLOAT
    DECLARE @lat2Compare AS FLOAT
    DECLARE @long2Compare AS FLOAT
    SET @radiusInKm = 5.000
    SET @lat2Compare = insert_your_lat_to_compare_here
    SET @long2Compare = insert_you_long_to_compare_here

    SELECT * FROM insert_your_table_here WITH(NOLOCK)
    WHERE (6371.392896*2*ATN2(SQRT((sin((radians(GeoLatitude - @lat2Compare)) / 2) * sin((radians(GeoLatitude - @lat2Compare)) / 2)) + (cos(radians(GeoLatitude)) * cos(radians(@lat2Compare)) * sin(radians(GeoLongitude - @long2Compare)/2) * sin(radians(GeoLongitude - @long2Compare)/2)))
    , SQRT(1-((sin((radians(GeoLatitude - @lat2Compare)) / 2) * sin((radians(GeoLatitude - @lat2Compare)) / 2)) + (cos(radians(GeoLatitude)) * cos(radians(@lat2Compare)) * sin(radians(GeoLongitude - @long2Compare)/2) * sin(radians(GeoLongitude - @long2Compare)/2)))
    ))) <= @radiusInKm

If you would like to perform the Haversine formula in C#,

    double resultDistance = 0.0;
    double avgRadiusOfEarth = 6371.392896; //Radius of the earth differ, I'm taking the average.

    //Haversine formula
    //distance = R * 2 * aTan2 ( square root of A, square root of 1 - A )
    //                   where A = sinus squared (difference in latitude / 2) + (cosine of latitude 1 * cosine of latitude 2 * sinus squared (difference in longitude / 2))
    //                   and R = the circumference of the earth

    double differenceInLat = DegreeToRadian(currentLatitude - latitudeToCompare);
    double differenceInLong = DegreeToRadian(currentLongitude - longtitudeToCompare);
    double aInnerFormula = Math.Cos(DegreeToRadian(currentLatitude)) * Math.Cos(DegreeToRadian(latitudeToCompare)) * Math.Sin(differenceInLong / 2) * Math.Sin(differenceInLong / 2);
    double aFormula = (Math.Sin((differenceInLat) / 2) * Math.Sin((differenceInLat) / 2)) + (aInnerFormula);
    resultDistance = avgRadiusOfEarth * 2 * Math.Atan2(Math.Sqrt(aFormula), Math.Sqrt(1 - aFormula));

DegreesToRadian is a function I custom created, its is a simple 1 liner of"Math.PI * angle / 180.0

My blog entry - SQL Haversine

Answer 2

Use the Great Circle Distance Formula.

Answer 3

Are you looking for

Haversine formula

The haversine formula is an equation important in navigation, giving great-circle distances between two points on a sphere from their longitudes and latitudes. It is a special case of a more general formula in spherical trigonometry, the law of haversines, relating the sides and angles of spherical "triangles".

Answer 4

here is a fiddle with finding locations / near locations to long/lat by given IP:

http://jsfiddle.net/bassta/zrgd9qc3/2/

And here is the function I use to calculate the distance in straight line:

function distance(lat1, lng1, lat2, lng2) {
        var radlat1 = Math.PI * lat1 / 180;
        var radlat2 = Math.PI * lat2 / 180;
        var radlon1 = Math.PI * lng1 / 180;
        var radlon2 = Math.PI * lng2 / 180;
        var theta = lng1 - lng2;
        var radtheta = Math.PI * theta / 180;
        var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
        dist = Math.acos(dist);
        dist = dist * 180 / Math.PI;
        dist = dist * 60 * 1.1515;

        //Get in in kilometers
        dist = dist * 1.609344;

        return dist;
    }

It returns the distance in Kilometers

Answer 5

The Haversine formula assumes a spherical earth. However, the shape of the earh is more complex. An oblate spheroid model will give better results.

If such accuracy is needed, you should better use Vincenty inverse formula. See http://en.wikipedia.org/wiki/Vincenty's_formulae for details. Using it, you can get a 0.5mm accuracy for the spheroid model.

There is no perfect formula, since the real shape of the earth is too complex to be expressed by a formula. Moreover, the shape of earth changes due to climate events (see http://www.nasa.gov/centers/goddard/earthandsun/earthshape.html), and also changes over time due to the rotation of the earth.

You should also note that the method above does not take altitudes into account, and assumes a sea-level oblate spheroid.

Edit 10-Jul-2010: I found out that there are rare situations for which Vincenty inverse formula does not converge to the declared accuracy. A better idea is to use GeographicLib (see http://sourceforge.net/projects/geographiclib/) which is also more accurate.

Answer 6

Here's one: http://www.movable-type.co.uk/scripts/latlong.html

Using Haversine formula:

R = earth’s radius (mean radius = 6,371km)
Δlat = lat2− lat1
Δlong = long2− long1
a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)
c = 2.atan2(√a, √(1−a))
d = R.c