Dear John Carmichael and all, My old trig teacher said "Don't speculate, calculate!" so here goes my response to your questions on sunset phenomenon..
On this mailing list last year, we have had fun with the Sunset Equation which shows that the local time of sunset (or sunrise) is a simple trig function of latitude and declination. Cos (t) = - Tan (Dec) * Tan (Lat) This gives the local time, as an hour angle from noon for the theoretical sunset, when the altitude of the sun is zero. Apparent sunset, when the last crack of sun disappears is a bit later. The corrections in the altitude angle are about 50', 16' for semi diameter and 34' for average refraction. For this, no altitude or dip corrections have been made for the real horizon. The above equation shows when sunset occurs. There are two related equations which can be used to answer your questions on where and how sunset occurs. The azimuth of the sun at the theoretical sunset is given by: Cos (Az) = Sin (Dec) / Cos (Lat) The angle (Psi) that the sun's path makes as it crosses the theoretical horizon is given by: Cos (Psi) = Sin (Lat) / Cos (Dec) These equations are all simplifications for the special case when the altitude is zero. They can be used answer your questions on the apparent sunset by setting up a simple right angle triangle and solving for the required corrections. A sketch is required here but text e-mail does not do sketches. Visualize a simple right angle triangle. The vertical side is the correction in altitude, the 50' discussed above. The angle with the horizontal is Psi. The horizontal side, the correction in azimuth, from simple trig is 50'/Cos(Psi). The hypotenuse is the correction in time along the path of the sun. This is simply 50' / Sin(Psi). Convert the angle to time the usual way 4 minutes per degree (as 360 degrees = 24 hours). So what does all this mean. The equations show that sunset phenomenon corrections depend on latitude and declination. In the tropics, the sun sinks fast, plonk and it is dark! At high latitudes, the twilight goes on forever, allowing a full golf game after dinner on summer evenings. We are paying for that now with less than 8 hours of daylight. The theoretical phenomenon, based on the mathematical concept of altitude = 0 is easily corrected to find the delay in time and azimuth shift for the apparent phenomenon, the last crack o sun corrected for semi diameter and average refraction. The corrections are reasonable approximations. Do not extrapolate to far as Psi is correct only for the instant the sun crosses the horizon. For your case, assuming Phoenix at Lat 33:30 N, the time correction (50'/60'*4/Sin(Psi)) is greater than the 3 min 20 sec that 50' gives at the equator: Dec = 0, time = 3:59; Dec = 10, time = 4:01; Dec = 20, time = 4:07. Similarly the azimuth correction (50'/60//Cos(psi)) gives for Lat = 33:30: Dec = 0, Az = 1:30.6; Dec = 10, Az = 1:19.2; Dec = 20, Az = 1:15.1. This is the azimuth correction for the apparent sunset on top of the sunset azimuth calculated for the theoretical sunset by Cos(Az) = Sin(Dec) / Cos(Lat). For latitude = 52 degrees, the time correction calculation gives: Dec = 0, time = 5:25, Dec = 10, time = 5:33, Dec = 20, time = 6:07. These are similar to the values that Patrick Powers provided. The reference for the equations is "Textbook on Spherical Astronomy' by W. M. Smart. Although this is probably more than you were looking for, I hope it is useful. Your questions were useful to me in provoking this study of the sunset phenomenon. Now we have three equations for the sunset tee shirt. Two versions of the original concept with just the Sunset Equation "Cos T = - Tan (Dec) * Tan(Lat)" were displayed at the NASS meeting in Seattle last September. We have an improvement for next year which shows not just when but where and how sunrise and sunset occur ;-) Roger Bailey, Walking Shadow Designs, 51 N, 115 W At 09:07 AM 1/22/99 -0700, Philip P. Pappas, II wrote: > Hello All: > >I've been enjoying the discussion on the delay of sunset due to refraction. >Somewhere I heard that at mid latitudes this is about three minutes. Can >anyone confirm this? > >Here is the reason I ask. On the outer edge of my horiz. sundials I engrave >little sunrises and sunsets which mark the time and direction of sunrise and >sunset on the equinoxes and solstices. Not knowing how to calculate the >placement of the engravings with math and formulas, I used a simpler (for >me) technique. > >Before starting I already knew that the sun should set due west on the >equinox. To find out where the sun rises and sets on the solstices I first >obtained U S Naval Observatory printouts of the times of sunrises and sets >and then plotted these times onto the sundial face. This automatically >gives the directions. ( I use the average times over several years as the >time of rise and sets vary by a couple of minutes each year). > >A problem arose, however, when I plotted the equinoxes. The direction of the >rises and sets was a several degrees SOUTH of due east and due west! > >Could this be that in theory the sun sets due west on the equinox (as it >would if the earth had no atmosphere) but in reality is sets slightly south >of due west due to refraction, and would this explain the time delay? This >has really bothered me, because I want my dials to be correct.(my dials are >corrected for longitude and I corected the observatory times with the EOT) > >Please comment on this somebody! > >John Carmichael > >thanks > >
