Roderick, Your question about the equation of time with aspect to the tilt of the earth-axis.
Draw a circle with radius R. Draw a vertical and horizontal diameter. The vertical diameter intersects the circle in the poles, the horizontal diameter is the equator. On the equator draw points at distance R.sin30 and R.sin60 from the center. (0.5R and 0.866R) Do this at both sides of the center. Draw 4 arcs through the 2 poles and the just constructed points. Now you have a part of the equatorial coordinate system. You can also see this drawing as the whole hemisphere. Draw in this figure ( with other color if you like ) in the same way a second coordinate system, but with the 2 diameters rotated over 23.5 degrees to the left ( or right if you like it that way). This is the ecliptical coordinate system. You now have a drawing with 2 coordinate systems. The real sun runs along the ecliptic. The mean sun runs along the equator. Lets start at the center of the drawing. Then both the real sun and the mean sun are at the same position and the declination of the sun is 0 degrees. ( march 21) Assume the sun has ran 30 degrees along the ecliptic ( with uniform speed ). In the ecliptic coordinate system the sun is in point R.sin30 on the diameter. Draw an arc through this point and the 2 poles of the equatorial system. On the diameter of the equatorial system you see the sun has not reached the R.sin30 point. So in this period the real sun runs faster than the mean sun. Do the same when the sun has ran 60 degrees. When the sun has ran 90 degrees on the ecliptic, also in the equatorial system the sun has ran 90 degrees. So first the real sun is fast to the mean sun and then it slows down. This happens 4 times a year, but each periode of 90 degrees in reversed way. In formula : e = 23.5 sin d = sin e * sin l sin t = cos e * sin l / cos d e = obliquity d = suns declination l = ecliptical length of the sun t = time in equatorial system In table : l d t delta t 0 0 0 -- 30 11.50 27.90 27.9 60 20.20 57.80 29.9 90 23.50 90.00 32.2 You see, the first 30 degrees on the ecliptic is only 27.9 degrees on the equator, but the last 30 degrees is 32.2 degrees on the equator. The max. difference between l and t is about 2.4 à 2.5 degrees, that is about 9.6 à 10 minutes of time. The same problem occurs in a days period when comparing the change of time ( on the equator ) with the change of the azimuth ( on the horizon ). As the time changes uniform, the azimuth usely doesnt. The same coordinate systems can be used, but rotate one with an angle equal to 90 minus latitude. In the extremes at the poles both coordinate systems coincide and indeed, the azimuth changes uniform with the time. At latitude 0 the systems have an angle of 90 degrees and ( with the suns declination = 0 ) the azimuth stays east for the periode of half a day and then jumps to west for the second half of the day. For more experiments make a fine divided coordinate system and a second ( on a sheet for an overhead projector ) which can rotate on the first. Does this visualise your problem? Fer, Netherlands. [EMAIL PROTECTED]
