If you want a stable method of computing geodesic distances, I recommend Richie Carmichael's wrapper for ESRI's Projection Engine.
Update: I just tried Richie's code with ArcGIS 10.0 on Vista64 and get an exception after calling LoadLibrary. I'll look into that more later.
For now though, here is some code in response to questions in the comments of another answer.
The code compares IProximityOperator for points with and without spatial references. Then it shows how to use an azimuthal equidistant projection (with first point being the point of tangency) to find the great circle distance.
private void Test()
{
IPoint p1 = new PointClass();
p1.PutCoords(-98.0, 28.0);
IPoint p2 = new PointClass();
p2.PutCoords(-78.0, 28.0);
Debug.Print("Euclidian Distance {0}", EuclidianDistance(p1, p2));
Debug.Print("Distance with no spatialref {0}", GetDistance(p1, p2));
ISpatialReferenceFactory srf = new SpatialReferenceEnvironmentClass();
IGeographicCoordinateSystem gcs =
srf.CreateGeographicCoordinateSystem((int)esriSRGeoCSType.esriSRGeoCS_WGS1984);
p1.SpatialReference = gcs;
p2.SpatialReference = gcs;
Debug.Print("Distance with spatialref {0}", GetDistance(p1, p2));
Debug.Print("Great Circle Distance {0}", GreatCircleDist(p1, p2));
}
private double GetDistance(IPoint p1, IPoint p2)
{
return ((IProximityOperator)p1).ReturnDistance(p2);
}
private double EuclidianDistance(IPoint p1, IPoint p2)
{
return Math.Sqrt(Math.Pow((p2.X - p1.X),2.0) + Math.Pow((p2.Y - p1.Y), 2.0));
}
private double GreatCircleDist(IPoint p1, IPoint p2)
{
ISpatialReferenceFactory srf = new SpatialReferenceEnvironmentClass();
IProjectedCoordinateSystem pcs =
srf.CreateProjectedCoordinateSystem((int)esriSRProjCSType.esriSRProjCS_WGS1984N_PoleAziEqui);
pcs.set_CentralMeridian(true, p1.X);
((IProjectedCoordinateSystem2)pcs).LatitudeOfOrigin = p1.Y;
p1.SpatialReference = pcs.GeographicCoordinateSystem;
p1.Project(pcs);
p2.SpatialReference = pcs.GeographicCoordinateSystem;
p2.Project(pcs);
return EuclidianDistance(p1, p2);
}
Here's the output:
Euclidian Distance 20
Distance with no spatialref 20
Distance with spatialref 20
Great Circle Distance 1965015.61318737
I think it would be interesting to test this against the projection engine dll (pe.dll). Will post results if I ever get Richie's code to work.
Update:
Once I changed Richies code to compile for x86, I got it to run. Interesting ... the great circle distance it give me is 1960273.80162999 - a significant difference from that returned from the azimuthal equidistant method above.
Best Answer
The answer is "None".
It is not possible to create a projection of the world which preserve the scale in all direction from any point of the world.
The cylindrical equidistant projection, you only have true distance along the meridians and along the equator. So if your points don't have the same longitude, you get it wrong.
With an azimuthal equidistant projection, you would need to center it on every point from layer in order to have true distance.
So, my recommendation is to use Python in order to get your "great circle" distance. You can see this post based on the geopy package. First you compute the geographic coordinates of your points (calculate geometry), then you measure your distances in a loop.