How do I calculate the Azimuth (angle to north) between two WGS84 coordinates in a single T-SQL query?

Question

I found the solution for this question in C#, but I can't translate it to a single query T-SQL, since my C# implementation requires branching (if then else).

I also found the following C# solution, which could be translated to a single query T-SQL but it doesn't produce the correct results

public static double GetAzimuth(WGSCoord c1, WGSCoord c2) { 
     var lat1 = DegToRad(c1.Latitude); 
     var lon1 = DegToRad(c1.Longitude); 
     var lat2 = DegToRad(c2.Latitude); 
     var lon2 = DegToRad(c2.Longitude);

     return RadToDeg(Math.Asin(Math.Sin(lon1 – lon2) * Math.Cos(lat2) / Math.Sin(Math.Acos(Math.Sin(lat2) * Math.Sin(lat1) + Math.Cos(lat1) * Math.Cos(lat2) * Math.Cos(lon2 – lon1))))); 
}

Code from Tamir Khason – Just code

Could someone correct the code above or provide an alternate solution?

Answer 1

Have you considered creating an assembly with a SP in C# for sql server? Thats the route I'd probably go.

Answer 2

In T-SQL, you could use the CASE expression

e.g.

SELECT ...
CASE 
    WHEN latD = 0 AND longD < 0 THEN ....
    WHEN latD < 0 AND longD = 0 THEN ....

etc.

Answer 3

Replace ifs with CASE expressions:

   if (latitudinalDifference == 0)
            {
                if (longitudinalDifference != 0)
                {
                    azimuth = Math.PI / 2d;
                }
            }

replace with:

SELECT CASE WHEN @latitudinalDifference = 0 AND @longitudinalDifference <> 0 THEN ...
 ELSE ... END AS azimuth

replace consecutive ifs with nested selects:

if(some condition)
{
  i=1; 
}
else
{
 i=2;
}
if(some other condition)
{
  i++; 
}

replace with

SELECT i + CASE WHEN (some other condition) THEN 1 ELSE 0 END
FROM(
SELECT CASE WHEN (some condition) THEN 1 ELSE 2 END AS i
) AS t

Answer 4

There's quite a lot of the necessary spherical trigonometry in the answer to SO 389211. Copying and modifying what I wrote there:

Consider a sperical triangle with angles A, B, C at the vertices and sides a, b, c opposite those vertices (that is, side a is from B to C, etc.). Applying this to the problem, we can call the two points given B and C, and we create a right spherical triangle with a right angle at A.

Consider this diagram:

                  + C
                 /|
                / |
            a  /  | b
           |  /   |
           |X/    |
           |/     |
         B +------+ A
              c

You are given two points B and C, and you want to determine the angle X = 90º - B. The side c is equal to the difference in longitude, Δλ; the side b is equal to the difference in latitude, Δφ; the angle A is 90º, so sin A = 1 and cos A = 0. To determine X, we want the value of B given b, c and A.

Looking at the problem from first principles, we need the two main spherical trigonometry equations:

1. The Sine Formula:

   sin A   sin B   sin C
   ----- = ----- = -----
   sin a   sin b   sin c
   

2. The Cosine Formula:

   cos a = cos b . cos c + sin b . sin c . cos A
   

Therefore, I believe an equation for a is:

cos a = cos Δλ . cos Δφ + sin Δλ . sin Δφ . cos 90º

a = arccos (cos Δλ . cos Δφ)

Given a, b and A, we can use the Sine Formula to determine B:

sin a   sin b
----- = ----
sin A   sin B

Or

        sin b . sin A
sin B = -------------
            sin a

Or, since A = 90º, sin A = 1, and sin B = sin (90º - X) = cos X:

        sin b
cos X = -----
        sin a

I rather suspect that if I bent my mind to it (or you bent your mind to it), you could come up with an answer that didn't involve using arccos followed by sin.