At 10:53 AM 12/27/99 -0200, Fernando Cabral wrote: > > >Given the longitude, latitude and altitude of two points, plus >the day of the year, is there an easy and straightforward >way to find which one will see the sun first? > >Let's say we have place "X" at 8 S 34 W altitude 200m and >place Y at 22S 42 W, altitude 1000m. > >It is winter Solstice in the North Hemisphere. > >Supposing there is a sundial at each place, which one >will mark the hour first? > >- fernando > >Hi Fernando,
Welcome back from your desert trek to ask my favourite question, the time of sunrise. The mathematical solution involves solving the Sunrise Equation "Cos t = - Tan D x Tan L. This formula is mathematically exact for the time (t) when the zenith distance of the center of the sun is 90 degrees. To get to the time of sunrise, there are corrections for refraction, dip and semidiameter. More on this later. For your examples, the solution for the time angle t from local noon is straight forward. This gives me a chance to use the new scientific calculator I got for Christmas that does trig and decimal degree to DMS conversions. 1. L=8, D=-23.5, Cos t =-Tan 8 x Tan -23.5 so t = 86.496 degrees or 5:45:59 2. L=22, D=-23.5, Cos t=-Tan22 x Tan -23.5 so t = 79.882 degrees or 5:19:32 These times are local solar times. To put them on a consistent basis, convert to UT by adding the longitude correction (LC). Longitude as an angle in degrees /15 = longitude as time is hours. 1. Long = 34 W so LC = 2:16 and sunrise UT is 5:45:59 + 2:16 = 8:01:59. 2. Long = 42 W so LC = 2:48 and sunrise UT is 5:19:32 + 2:48 = 8:07:32. This calculation does not consider the corrections required for real sunrise, the instant of the first ray of sunlight above the horizon. For this you have to consider =semidiameter ~16', refraction ~32' and dip due to the height of the eye above the horizon (5 ft ~ 2'). Altitude can come into the correction a couple of ways depending on the horizon you are using. Generally it is the height of eye over a common horizon like sea level. In this case the correction is .97 x square root of height in feet. Altitude also affects the refraction correction as does barometric pressure and temperature. Most published calculations are based on a total correction of the altitude of -50' or a zenith distance of 90:50. The longer solution of the navigation triangle must now be used. Sin Altitude = Sin L x D + Cos L x D x Cos t. Rearranged this gives Cos t = (Sin (-50')- Sin L x Sin D)/ Cos L X Cos D Solving this using my new $7 calculator, the times are: for 8 S, 34 W, 5:42:18 local solar (7:58:18 UT) and for 22 S, 42 W, 5:15:32 local solar (8:03:32). As expected, these times are a few minutes earlier. For an explanation of the spherical trigonometry, see my previous posting at <http://dialist.webjump.com> one of the sites David Bell set up for storing attachments. Happy Solstice, Roger Bailey N 51 W 115
