CODEWITHSUNDEEP
c#geolocationgeometry
Paul Stanley
Paul Stanley

Reputation: 2161

Map Coordinates C#

I have a set of latitude Longitude points. If I wanted to test if a new point was within x metres of any of the existing points would this be possible? Would it be best if I use this way?

ForEach(Coordinate coord in Coordinates)
{
     var distance =  GeoCoordinate.GetDistance(lat,lon);
     if(distance <= x)
     {
         addToQualifyingList(coord);
     }
}

and compare the new coordinate with every point in the set and check to see it is within x metres?

Upvotes: 1

Views: 2273

Answers (2)

fubo
fubo

Reputation: 45947

Here is a method to calculate the distance between 2 points (lat1, lon1 and lat2, lon2)

public enum DistanceUnit { miles, kilometers, nauticalmiles }
public double GetDistance( double lat1, double lon1 , double lat2 , double lon2, DistanceUnit unit)
{
    Func<double, double> deg2rad = deg => (deg * Math.PI / 180.0);
    Func<double, double> rad2deg = rad => (rad / Math.PI * 180.0);
    double theta = lon1 - lon2;
    double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
    dist = Math.Acos(dist);
    dist = rad2deg(dist);
    dist = dist * 60 * 1.1515;

    if (unit == DistanceUnit.kilometers)
    {
        dist = dist * 1.609344;
    }
    else if (unit == DistanceUnit.nauticalmiles)
    {
        dist = dist * 0.8684;
    }
    return (dist);
}

To determine all Coordinates with distance below 1 kilometer:

List<Coordinate> result = Coordinates.Where(x => GeoCoordinate.GetDistance(lat,lon, x.lan, x.lon, DistanceUnit.kilometers) < 1).ToList();

Upvotes: 2

Git
Git

Reputation: 214

Latitude and Longitude are both needed to know the position on earth or any other sphere shaped object. These are given in degree from their respective zero line / point.

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

Has a solution in Java

>    var R = 6371e3; // metres
>     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;

Apparently, this is this is one of the best solutions for small distances.

You should be able to just copy and paste it to C#, youll jsut have to change the variable names.

Upvotes: 0

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