8 replies to this topic

[Patrick]

I've used the Haversine formula many times when calculating distances between two points on this big blue marble but it's normally over long distances. Well, I was playing with my Netduino and wired a simple homing beacon style GPS application that blinks the LED faster as you approach a target lat/lon.

private double calculateDistance(double lat1, double lon1, double lat2, double lon2)
{
double R = 20902230.97112861; //feet
double dLat = (lat2-lat1) * (exMath.PI / 180.0);
double dLon = (lon2-lon1) * (exMath.PI / 180.0); 
double a = exMath.Sin(dLat/2.0) * exMath.Sin(dLat/2.0) +
        exMath.Cos(lat1 * (exMath.PI / 180.0)) * exMath.Cos(lat2 * (exMath.PI / 180.0)) * 
        exMath.Sin(dLon/2) * exMath.Sin(dLon/2); 
double  c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1-a)); 
double d = R * c;

return d;
}

The problem arises when, *I think*, the distance gets small between the two points (say 1000 feet or so), the

exMath.Cos(lat1 * (exMath.PI / 180.0)) * exMath.Cos(lat2 * (exMath.PI / 180.0))

part goes to zero instead of being a really small number and then I get an incorrect result. Like the double in the .Net micro has less precision than it's desktop counterpart...

Any insight here would be greatly appreciated!

Ps.. I'm using this library for the math operations.

[Chris Seto]

Are you sure your GPS is giving you correct data? Garbage in, garbage out.

[Patrick]

Thanks Chris, but I'm positive. I tested with hard coded values that I verified against an online calculator.

[Chris Walker]

Can you give us sample data (along with the set of math conversions) for the desktop version and the .NET MF library you're using--and then we can take a look to see what we can do to help? Chris

[Patrick]

Thanks for the reply Chris...

This code on the desktop .Net Framework returns 45.016489891829075 Notice the hardcoded values... Accurate. Perfect!

static double calculateDistance(double lat1, double lon1, double lat2, double lon2)
        {
            double dist = 0.0;

            lat1 = 41.9243;
            lon1 = -88.1007;

            lat2 = 41.924399;
            lon2 = -88.100799;
            
            double rLat1 = lat1 * 0.017453293;
            double rLon1 = lon1 * 0.017453293;
            double rLat2 = lat2 * 0.017453293;
            double rLon2 = lon2 * 0.017453293;

            double R = 20902230.97112861;

            //Haversine
            double dLat = (rLat2 - rLat1);
            double dLon = (rLon2 - rLon1);

            double a = Math.Sin(dLat / 2.0) * Math.Sin(dLat / 2.0) +
                       Math.Cos(rLat1) * Math.Cos(rLat2) *
                       Math.Sin(dLon / 2.0) * Math.Sin(dLon / 2.0);

            double c = 2.0 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1.0 - a));

            dist = R * c;

            return dist;
        }

This code on the .Net MF returns 4082.6324955905925. Note that the exMath.Atan2 arguments needed to be flip flopped.

static double calculateDistance(double lat1, double lon1, double lat2, double lon2)
        {
            double dist = 0.0;

            lat1 = 41.9243;
            lon1 = -88.1007;

            lat2 = 41.924399;
            lon2 = -88.100799;

            double rLat1 = lat1 * 0.017453293;
            double rLon1 = lon1 * 0.017453293;
            double rLat2 = lat2 * 0.017453293;
            double rLon2 = lon2 * 0.017453293;

            double R = 20902230.97112861;

            //Haversine
            double dLat = (rLat2 - rLat1);
            double dLon = (rLon2 - rLon1);

            double a = exMath.Sin(dLat / 2.0) * exMath.Sin(dLat / 2.0) +
                       exMath.Cos(rLat1) * exMath.Cos(rLat2) *
                       exMath.Sin(dLon / 2.0) * exMath.Sin(dLon / 2.0);

            double c = 2.0 * exMath.Atan2(exMath.Sqrt(1.0 - a), exMath.Sqrt(a));

            dist = R * c;

            return dist;
        }

[Chris Walker]

Just to confirm...so you're getting ~45 on the desktop and ~4082 using the exMath library? I'll also put in an internal bug report and see if there's something we can do to boost the accuracy... Chris

[Patrick]

Just to confirm...so you're getting ~45 on the desktop and ~4082 using the exMath library?

That's correct.

[Chris Walker]

Great, thanks! We didn't write the 3rd-party exMath library but we'll take a look and see what we can do...

[Patrick]

Chris,

I tracked it down to an error in the exMath.Pow function when executing:

exMath.Sqrt(a)

I would get a much larger value on the .Net MF. I can't say where in that function it goes wrong however or why. However, I get around it by using the following code:

static double calculateDistance(double lat1, double lon1, double lat2, double lon2)
        {
            double dist = 0.0;

            lat1 = 41.9243;
            lon1 = -88.1007;

            lat2 = 41.924399;
            lon2 = -88.100799;

            double rLat1 = lat1 * 0.017453293;
            double rLon1 = lon1 * 0.017453293;
            double rLat2 = lat2 * 0.017453293;
            double rLon2 = lon2 * 0.017453293;

            double R = 20902230.97112861;

            //Haversine
            double dLat = (rLat2 - rLat1);
            double dLon = (rLon2 - rLon1);

            double a = exMath.Sin(dLat / 2.0) * exMath.Sin(dLat / 2.0) +
                       exMath.Cos(rLat1) * exMath.Cos(rLat2) *
                       exMath.Sin(dLon / 2.0) * exMath.Sin(dLon / 2.0);

            //using Math.Pow(a, 0.5) instead of exMath.Sqrt(a)
            double c = 2.0 * exMath.Atan2(System.Math.Pow((1.0 - a), 0.5), System.Math.Pow(a, 0.5));
  
            dist = R * c;

            return dist;
        }