Roger,
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
> >
