Francois, I saw your equation of time question on the remailer list. The equation that I use is from a book called "Practical Astronomy with your Calculator" and is more involved. I think that it is similar to yours but not sure. Here is what I use I hope there are no transcription errors. For July 27th 1980 at noon eot = -6 min 25sec (Remember noon is July 27.5) I have written it more in "computereze" than mathamatical. There is a more precise method but I think this is close enough for sundials.
++ron 1. Calculate the right ascension of the Sun in decimal hours: D = Number of days since January 0.0 = (209.5) Add/subtract number of days since/till 1990 D = 209.5 - 3653 = -3443.5 N = 360/365.242191 * D = -3394.0767 Subtract or add multiples of 360 until N is in the range 0-360. N = 205.92332 ' The following Three variables can be calculated for other ' epochs EL = 279.403303 ecliptic longitude at epoch 1990.0 PL = 282.768422 ecliptic longitude of perigee OE = 23.441884 ecliptic obliquity E = .016713 eccentricity of the orbit M = N + EL - PL = 202.5582 If M is negative add 360 EC = 360/PI * E * sin(M) = -0.7347003 L = N + EC + EL = 484.59192 Sun's geocentric ecliptic longitude If L more than 360, subtract 360 L = 124.59192 RA = atan( sin(L)*cos(OE)/cos(L)) = -53.068296 Remove abiguity of atan RA = RA + 180 = 126.9317 H = RA/15 = 8.462113 2. Take H as GST and convert to Universal Time. JD = Julian Date of 0hour on this calender date. = 2444447.5 S = JD - 2451545.0 = -7097.5 T = S/36525.0 = -0.194319 T0 = 6.697374558 + (2400.051336 * T) + (0.000025862 * T^2) T0 = -459.6781 Reduce T0 to the range of 0-24 by adding or subtracting multiples of 24 T0 = 20.321904 UT = H - T0 = -11.859791 Reduce UT to the range of 0-24 by adding or subtracting multiples of 24 UT = 12.140209 UT = UT * 0.9972695663 = 12.10706 3. Calculate EOT in decimal hours EOT = 12 - UT = -0 .10706 hours EOT = -6m 25.4s
