How do I calculate the Azimuth (angle to north) between two WGS84 coordinates

Question

I have got two WGS84 coordinates, latitude and longitude in degrees. These points are rather close together, e.g. only one metre apart.

Is there an easy way to calcu

Answer 1

Has anyone tested this? It does not return the correct answers

This function assumes that you're on a flat surface. With the small distances that you've mentioned this is all fine, but if you're going to be calculating the heading between cities all over the world you might want to look into something that takes the shape of the earth in count;

Your flat earth has little to do with this. The error, as you call it, is because you are calculating an initial azimuth from a point. Unless you are heading straight to a pole, you relationship to the pole will change with distance. Regardless, the above program does not return correct results.

Answer 2

This would work only for small differences. Otherwise you can't just "latitudinalDifference / longitudinalDifference".

Answer 3

Here is the C# solution. Tested for 0, 45, 90, 135, 180, 225, 270 and 315 angles.

Edit I replaced my previous ugly solution, by the C# translation of Wouter's solution:

public double GetAzimuth(LatLng destination)
{
    var longitudinalDifference = destination.Lng - this.Lng;
    var latitudinalDifference = destination.Lat - this.Lat;
    var azimuth = (Math.PI * .5d) - Math.Atan(latitudinalDifference / longitudinalDifference);
    if (longitudinalDifference > 0) return azimuth;
    else if (longitudinalDifference < 0) return azimuth + Math.PI;
    else if (latitudinalDifference < 0) return Math.PI;
    return 0d;
}

public double GetDegreesAzimuth(LatLng destination)
{
    return RadiansToDegreesConversionFactor * GetAzimuth(destination);
}

Answer 4

I found this link

http://williams.best.vwh.net/avform.htm

given in the answer to

Lat/Lon + Distance + Heading --> Lat/Lon

This looks promising, especially the flat earth approximation given near the end.

Answer 5

I would recommend implementing a correction factor based on the longitude. I implemented a simular routine once to return all geocoded records within x miles of a specific spot and ran into simular issues. Unfortunately I don't have the code anymore, and can't seem to recall how I got to the correction number but you are on the right track.

Answer 6

The formulas that you refer to in the text are to calculate the great circle distance between 2 points. Here's how I calculate the angle between points:

uses Math, ...;
...

const
  cNO_ANGLE=-999;

...

function getAngleBetweenPoints(X1,Y1,X2,Y2:double):double;
var
  dx,dy:double;
begin
  dx := X2 - X1;
  dy := Y2 - Y1;

  if (dx > 0) then  result := (Pi*0.5) - ArcTan(dy/dx)   else
  if (dx < 0) then  result := (Pi*1.5) - ArcTan(dy/dx)   else
  if (dy > 0) then  result := 0                          else
  if (dy < 0) then  result := Pi                         else
                    result := cNO_ANGLE; // the 2 points are equal

  result := RadToDeg(result);
end;

• Remember to handle the situation where 2 points are equal (check if the result equals cNO_ANGLE, or modify the function to throw an exception);

• This function assumes that you're on a flat surface. With the small distances that you've mentioned this is all fine, but if you're going to be calculating the heading between cities all over the world you might want to look into something that takes the shape of the earth in count;

• It's best to provide this function with coordinates that are already mapped to a flat surface. You could feed WGS84 Latitude directly into Y (and lon into X) to get a rough approximation though.