John,
Here's my second attempt.
Steve
>>>>>>
Sun position vector S iswhere /ω/ is hour angle, /φ/is latitude, and /δ/
is solar declination.//
Azimuth (from south) obeys so //
Dividing right hand side by /–cos δ /and then rearranging
Setting /a/= , /b/ = /1/, /c/ = gives
/a cos ω – b sin ω = c/
Use trig identity
/a cos ω – b sin ω ≡ R cos(ω + x) /where/R = ///and////
which converts our case to
/R cos(ω + x) = c/
which rearranges as
/ω = ///
or
/ω//= ///
//
To force correct quadrant
if /γ/> 90, use /ω//=/360 − Term 1 – Term 2
if /γ/< −90, use /ω//= /−Term 1 – Term 2
if /γ/= 0, use /ω/= 0
These rules are for latitudes north of Tropic of Cancer. I'll leave you
to figure out quadrant adjustments for other cases if you need them.
On 2024-07-20 9:13 a.m., John Goodman wrote:
No rush; thank you. I appreciate you applying your brain to my concerns!
On Jul 20, 2024, at 12:11 PM, Steve Lelievre
<steve.lelievre.can...@gmail.com> wrote:
Oh, no! I think I may have my initial term for tan gamma inverted, so
my result would be an angle relative to EW.
I have to go out for the day now, but will check my calculation when
I get home.
Steve
---------------------------------------------------
https://lists.uni-koeln.de/mailman/listinfo/sundial
