Hi Roger,
My answered is premised on the belief that the HA arc and AZI arc are
components of great circles with the intersection between them forming
the angle you are interested in, i.e., (psi). If true, then side_a of
the spherical right triangle is by definition psi. Hope this helps.
Best,
Luke
I wrote:
>
> 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
