You are correct in noting that I did not include the minus sign in the equation for sunset time. It is obvious that the sign is negative from the simplification of the equation below for the special case when Ho = 0. I did not bother to include it this time as it is important for the user of the equation to place the solution in the appropriate quadrant.
The second equation quoted is similarly derived, but not from the usual solution of the navigation triangle for azimuth using the Law of Sines. This is the familiar equation Sin Z = Cos D x Sin t / Cos H. The solution using two sides (L & H) and the contained angle (Z) gives the equation: Sin D = Sin L x Sin H + Cos L x Cos H x Cos Z. The form is similar to the altitude equation quoted. In the special case for H = 0, this reduces in a similar fashion as before to Cos Z = Sin D / Cos L. QED. The third equation is from W. M. Smart's "Textbook on Spherical Astronomy". He leads the reader through the above derivations but leaves as an exercise for the reader, "If psi is the angle which the bodies path, at rising, makes with the horizon, prove that Cos psi = Sin L / Cos D." I have quoted the equation but, like Smart, leave the challenge to prove it open. Roger Bailey Walking Shadow Designs N 51 W 115 ps. For those who have not been introduced to the topic, welcome to the wonderful world of spherical trigonometry. It is like plane Euclidean geometry but with a few curves thrown in. In this world, the world we live on, parallel lines meet, the sides of triangles are angles and the sum of the angles in a triangle can be greater than 180 degrees. Otherwise everything is normal and you can solve all these useful equations on a pocket calculator. RTB At 09:31 AM 9/12/99 -0800, Luke Coletti wrote: >Hello Fer et al, > > Meeus mentions (as you point out as well), that this formula (1), which >is derived from the well known local horizontal altitude (h) formula (2) >and is a special case where h=0, computes the hour angle (Ho) >corresponding to the time of rise or set of the sun (celestial object). >I now see what you meant by the length of the half day arc. > >(1) cos(Ho) = - tan(lat) * tan(dec) > >(2) sin(h) = sin(lat)*sin(dec) + cos(lat)*cos(dec)*cos(H) > > >Luke > >> Some messages have been on this list about the formula to calculate >> sunrise and so on. >> The first one ( Cos t = Tan L x Tan D ) in fact calculates the length of >> half the day arc. >> But in that formula must be a negative sign, thus the formula is : >> >> Cos t = - Tan L x Tan D >> >> Best, Fer. >> >> -- >> Fer J. de Vries >> [EMAIL PROTECTED] >> http://www.iaehv.nl/users/ferdv/ >> lat. 51:30 N long. 5:30 E >
