How to calculate azimuth (angle to the north) between two WGS84 coordinates in one T-SQL query?

Question

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

I also found the following C # solution that could be translated into a single T-SQL query, but it does not give 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 Hasson - Code Only

Can someone fix the code above or provide an alternative solution?

Answer 1

Replace ifs with CASE expressions:

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

replaced by:

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

replace consecutive ifs with nested selections:

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

replace

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 2

SP # SQL-? , , , .

Answer 3

SO 389211 . , :

A, B, C a, b, c (.. a B C ..). , , B C, A.

:

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

B C, X = 90º - B. c , Δλ; b , Δφ; A 90º, sin A = 1 cos A = 0. X, , B b, c A.

, :

• :

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

• :

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

, a:

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

a = arccos (cos Δλ . cos Δφ)

a, b A, B:

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

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

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

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

, , ( ), , arccos, .

Answer 4

In T-SQL you can use a CASE expression

eg.

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

and etc.