Calculate SW & NE Lat/Longs of Square X miles from a Centre Lat/Long

Question

I am working on getting data out of the Noaa API.

When talking to the Noaa API you can get a list of weather stations within a square. They call them "extents" and they are a 2 sets of lat/longs. The bottom left lat/long and the top right lat/long

As detailed here:

http://www.ncdc.noaa.gov/cdo-web/webservices/v2#stations

I have been given a list of lat/longs for a number of cities in the US and I am trying to work out the best way of putting a box around.

So, my first guess would be to take the central lat/long, work out the lat/long 75 miles to the West, then work out the lat/long of the point 75 miles South of that point.

Ideally I would like to have this as a c# function.

Has anyone got any ideas on the best way to code this please?

Thanks

Answer 1

Yippee! - found the solution...

First a simple class:

public class LatLonAlt
    {
        public double Latitude { get; set; }
        public double Longitude { get; set; }
        public double Altitude { get; set; }
    }

Then a function to calculate a new position:

public static HelpersModel.LatLonAlt CalculateDerivedPosition(HelpersModel.LatLonAlt source, double range, double bearing)
    {
        double latA = Convert.ToDouble(source.Latitude) * (Math.PI / 180);
        double lonA = Convert.ToDouble(source.Longitude) * (Math.PI / 180);
        double angularDistance = range / 6371;
        double trueCourse = bearing * (Math.PI / 180);

        double lat = Math.Asin(
            Math.Sin(latA) * Math.Cos(angularDistance) +
            Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));

        double dlon = Math.Atan2(
            Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
            Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));

        double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;

        HelpersModel.LatLonAlt results = new HelpersModel.LatLonAlt();
        results.Latitude = lat * (180 / Math.PI);
        results.Longitude = lon * (180 / Math.PI);
        results.Altitude = source.Altitude;

        return results;
    }

Then, and I know I can do this better.. but it works for now...

2 functions that work out the bottmleft and topright extent:

public static HelpersModel.LatLonAlt FindBottomLeftExtent(HelpersModel.LatLonAlt startpoint)
    {
        // first move left
        HelpersModel.LatLonAlt movedleft = CalculateDerivedPosition(startpoint, 72.42, 270);
        // move down
        HelpersModel.LatLonAlt moveddown = CalculateDerivedPosition(movedleft, 72.42, 180);

        return moveddown;
    }
    public static HelpersModel.LatLonAlt FindTopRightExtent(HelpersModel.LatLonAlt startpoint)
    {
        // first move right
        HelpersModel.LatLonAlt movedright = CalculateDerivedPosition(startpoint, 72.42, 90);
        // move up
        HelpersModel.LatLonAlt movedup = CalculateDerivedPosition(movedright, 72.42, 0);

        return movedup;
    }

HTH!

Trev