I would like to vote with my GIF on this attachment issue. Sometimes a picture is worth a thousand words. The following discussion would be incomprehensible to most readers without the drawing. For this reason I have bent the rules and attached a sketch of the "Navigational Triangle" to explain the solution of the "Psi Challenge". The clue to the challenge is that Psi, the angle of the sun's path as it crosses the horizon, is the third angle in the Navigational Triangle. Solving using the cosine rule, the solution of a triangle using two sides and the included angle gives the expression Sin L = Sin D x Sin H + Cos D x Cos H x Cos Psi. Plugging in for H = 0, Sin H = 0 and Cos H =1 reduces the equation for psi at sunset to Cos Psi = Sin L / Cos D. QED.
Thanks Luke for your answer. Yours was the only response to the challenge. You get good marks for the substitutions but I did not follow the original premise and subsequent logic. Have a look at the attached GIF showing the Navigational Triangle. All readers suffering from math phobia should exit here and delete this posting. To all others, welcome to the world of spherical trigonometry. The Navigation Triangle is very useful in solving the usual problems in sundial design and navigation. In one case, the question is "What is the altitude and azimuth of the sun at this time in this location." In the other it is "Where am I, given the altitude and azimuth of the sun at this time." The drawing shows the celestial sphere. Your position is in the center. Your zenith is straight up. >From your zenith to the equator line Q is your Latitude L. From your zenith to the pole P is your Co-Latitude or (90-L). This is the first side of the triangle.The horizon is the horizontal line through the center. The sun moves along the declination line D, parallel to the horizon but displaced by the declination. Any great circle from the pole to this declination line is the Co-Declination (90-D). This is the second side of the triangle. The third side is the great circle distance from your zenith position to the position of the sun on the declination line. This is the Co-Altitude (90-H). The Altitude is the remaining distance along this line down to the horizon line. The angles in the triangle are the following. The time or hour angle t is the angle at the pole from your meridian to the hour angle of the sun(RA or SHA). This is where time and longitude come in for navigation but for our purposes, t=0 at local solar noon and all time angles are relative to this. The angle at your position from your meridian to the sun is the Azimuth Z. The third angle of the triangle at the position of the sun is Psi. See how complicated the words are and how simple the drawing is. We need to be able to post small attachments. If this were regular plane geometry, the solution of the triangle, knowing two sides a and b and the included angle C is: c^2 = a^2 + b^2 - 2ab Cos C. The equivalent expression for this Cosine Rule in spherical trigonometry is: Cos C = Cos A x Cos B + Sin A x Sin B x Cos c. In the case of the Navigation Triangle we are dealing with the corresponding angles (90-angle) so the sines become cosines and vise versa. Knowing this we can write all the relationships in this form just by looking at the triangle. To solve for time t: Sin H = Sin L Sin D + Cos L Cos D Cos t To solve for where Z: Sin D = Sin H Sin L + Cos H Cos L Cos Z To solve for how Psi: Sin L = Sin D Sin H + Cos D Cos H Cos Psi For the sunset phenomena when the sun is on the horizon, H =0, Sin H = 0 and Cos H = 1. Substituting this in the above equations results in the following. Time of sunset t: Cos t = - Tan L Tan D Azimuth at sunset Z: Cos Z = Sin D / Cos L Sunset path angle Psi: Cos Psi=Sin L / Cos D These simple elegant equations define sunset phenomenon: when, where and how the sun sets as related to latitude and the declination of the sun. I recognize that this discussion has been incomprehensible to most even with the attached drawing but it can be fun to work it through. Cheers, Roger Bailey Walking Shadow Designs N 51 W 115 > > I believe the previous two provide the solution to the third, i.e., in >a spherical right triangle (special case were the Sun's altitude = 0), >let the hour angle (HA) be side_c, the azimuth (AZI) be side_b and the >subtended angle be psi. Then cos(psi) = (side_c/side_b) = >cos(HA)/cos(AZI) and reduces to the third with the cancellation of like >terms and the cos() from acos(). Perhaps the tee-shirt should include a >graphic of a spherical right triangle as well, much like the e and b >field pictorials that adorn the tee-shirts with Maxfield's EM field >equations on them. > >cos(psi) = cos(HA) / cos(AZI) >cos(HA) = sin(lat)/cos(lat) * sin(dec)/cos(dec) >cos(AZI) = sin(dec) / cos(lat) > >-Luke > >Roger Bailey wrote: >> >> 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 >> > > Attachment converted: MAC Hard Disk:Slide2.GIF (GIFf/JVWR) (0001B399)
