ATM Sun's altaz.

["James W. Burrows" <burrjaw@halcyon.com>]

On Thu, 15 May 1997, Kip Larsen wrote:

> I've also got a question that may be slightly off topic but I thought one 
> of you atm'ers out there might have the answer to.  I have been looking 
> for an astronomical formulae to determine the altitude and azimuth of 
> the SUN given a particular latitude, longitude, date and time of day.  
> Anyone know this formulae or know where I might look to try to find it?

OK, here goes, you saw it first on the ATM newslist:

Input
   M      month     (1..12)
   DAY              (1..31)
   Y      year      (1900..2099)
   U      UTC       (day (=(h+(min+s/60)/60)/24)
   LAT    latitude  (deg)
   LONG   longitude (cycle (=deg/360), W-)
   
J, days from J2000.0
   M = M + 1
   if M < 4 then
      Y = Y - 1
      M = M + 12
   end if
   J = int(365.25*Y) + int( 30.6*M) + DAY + U - 730563.5
      int() means chop or truncate to integer
   
S, local sidereal time (day)
   S = .779057 + .002737909*J + J + LONG
      The multiplication here needs all 7 digits in the factor.  Only
      need the fractional part of S.

Sun's position
   L = 280.466 + .9856474*J   mean longitude (deg)
   g = 357.528 + .9856003*J   mean anomaly (deg)
   l = L + 1.915*sin(g) + .020*sin(2*g) ecliptic longitude (deg)
   e = 23.440 - .0000004*J    obliquity of ecliptic (deg)
   RA = arctan(cos(e)*tan(l)) right ascension (change to cycles)
   DEC = arcsin( sin(e)*sin(l))    declination (deg)

Alt & azm
   H = S - RA  local hour angle (change to deg)
   alt = arcsin(sin(LAT)*sin(DEC) + cos(LAT)*cos(DEC)*cos(H))
   azm = arctan( x=> tan(DEC)*cos(LAT) - sin(LAT)*cos(H), y=> -sin(H))
      This arctan means a change to polar coordinates from rectangular
      coordinates x, y.

Hope I got all the "simplifying" numbers right.

     -- Jim Burrows:     (206)244-2933, fax -0294
     -- E-mail:     burrjaw@halcyon.com
     -- Seattle:    WGS84 lat 0.828547, long -2.135692 (rad)