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--- Begin Message ---Thank you for reworking your equations for me. I have to confess that I'm weak at math and matrices make me a little queasy. The spherical trigonometry of Morrison's solution is territory I feel safer exploring, once I get my quadrants under control. > On Jul 20, 2024, at 5:44 PM, Steve Lelievre <steve.lelievre.can...@gmail.com> > wrote: > > John, > > Here's my second attempt. > > Steve > > >>>>>> > > > > Sun position vector S is  where 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> <mailto: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 >>>
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