An Earthy Equation

[FlapsOut]

Anyone studying Physical Geometrics knows that the Earth, rather than being a bedrock hard globe in the heavens, is more like a water-balloon...constantly expanding and contracting as it rotates.

I would like to know that; If I were to calculate a 'Curve-of-the-Earth' distance calculation from, for instance:

'KIAH' to '13TE'? Would anyone have an argument against this formulae? And, if so? What argument?

In C#, BTW...

double distance = CalculateDistanceByTwoLatLngs(30.6560320034623, -96.5383015573025, 29.959241723995, -95.3403444794987);

private double CalculateDistanceByTwoLatLngs(double Latitude1, double Longitude1, double Latitude2, double Longitude2)
{
double Lat1, Lng1, Lat2, Lng2;

Lat1 = (Latitude1 * 3.141592657) / 180;
Lng1 = (Longitude1 * 3.141592657) / 180;
Lat2 = (Latitude2 * 3.141592657) / 180;
Lng2 = (Longitude2 * 3.141592657) / 180;

double distance = Math.Acos(Math.Sin(Lat1) * Math.Sin(Lat2) + Math.Cos(Lat1) * Math.Cos(Lat2) * Math.Cos(Lng2 - Lng1));

if (distance < 0)
distance += Math.PI;

distance = 3959.0 * distance; // 3959.0 being the 'Average Radius of the Earth in Miles'

return (distance);
}

Then? distance = 86.167549658405591 Miles...er, 86.17 miles.

Flaps ;)

P.S. Please do not answer with, 'Well, FSNav says...' or 'Well, FS Commander says...' :o

[arno]

Hi,

The only thing I see right away is that you use an average radius of the earth, while the SDK clearly lists the two radii that FS uses for its earth model. So I think that would make it not 100% accurate.

Here are the figures from the SDK:

Equatorial diameter=12756.27 km
Equatorial circumference=40075.0 km
Polar diameter=12734.62 km
Polar circumference=40007.0 km

But I never needed a formula to calculate the distance between two lat/longs over a long distance (only when using meters from a reference point in flat earth), so I don't know if the actual formula you use is correct.

[rhysa]

If you are going to use a spherical model then why not just use the Haversine method?

Then there is the Vincenty method, which I found here (http://www.movable-type.co.uk/scripts/latlong-vincenty.html) that uses an ellipsoidal model if you want to be very accurate. You can use the WGS84 numbers in this method and so you should get a good match with FSX.

FSX uses the WGS-84 standard, it is mentioned in the terrain SDK I think, but also you will see that arno's numbers match up with the WGS84 standard definition (http://home.online.no/~sigurdhu/WGS84_Eng.html).

I've used Haversine mainly because we only do short distance calculations where the error will be very small, and the results we see in FSX back that up.

[FlapsOut]

Thanks Guys,

Two great answers!

I'm going to add that info and those formulas to my handy-dandy Flight Sim Library! :)

Flaps