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--- Begin Message ---Thank you Alfonso. Your reply is right on time. I've been surprised by how many different ways there are to solve this problem! I'm drawn back to Morrison's approach. His figures of spherical triangles have a foundation in observable orientations. I'm working on programming Morrison's equations and, once I account for the quadrants with the proper additions, subtractions, and inversions, I think I may have a solution that works for me. Thanks to all the mathematically astute members of this list who've given me their detailed explanations. John > On Jul 22, 2024, at 5:13 AM, Alfonso Pastor-Moreno <galvag...@gmail.com> > wrote: > > To John Goodman. > Hello John. > Sorry for the delay in my answer. > I think that the solution to clarify your question is in the Morrison's book. > Here you have the solution that Morrison gave to you in 2015, interleaved > with the explanation that Morrison gives in his book. > And if you continue reading in the page 264 is the complementary explanation, > that Morrison gives. > Here are the notes I wrote, paraphrasing Morrison's text. > For the figures of the two spherical triangles to which I make reference in > my text, I recommend to you to look to the figures of the Morrison's book of > the pages 264 and 265, because it is not possible to send here the figures to > you due to the rules of the SUNDIAL MAILING LIST, that allow only to send > text. > I hope this will help. > Have my best regards. > Alfonso Pastor-Moreno > galvag...@gmail.com > > From Morrison:Pages 262 and following > > It is not possible to calculate the Sun's Altitude directly, but Gunter > solved the problem in a clever way (see belowfor an outline of the proof). > > 1.- For a given Azimuth, calculate the Sun's Altitude when the Sun's > declination δ is zero; > 1. Calculate the Sun's Altitude (h0) when the Sun's declination (δ) is > zero:(i.e. the Sun is on the equator) from: tan ho = cos A / tan φ > > 2.- Then calculate an auxiliary angle, x from: sin x = (cos ho sin δ) / sin φ > 2. Calculate an auxiliary angle (x) from: sin x = (cos h0 sin δ) / sin φ > 3.- Calculate the Sun's altitude, h, for the azimuth and declination: > If A < 90°, h = x + ho > If A > 90°, h = x - ho > 3. Calculate sun's altitude, h, for the azimuth and declination from: > > If A < 90, h = x + h0 > If A > 90, h = x - h0 > Once again, a significant number of calculations are required but this is not > difficult with a spreadsheet program or a simple computer program. > > I assume that the azimuth at noon is 180 degrees which would make 90 degrees > due east. > Why would there be mathematical symmetry around a 90° azimuth? > > I would expect symmetry around noon when A is 180° degree > Can anyone explain the logic here for me? > Thank you all very much, John > > Here is what Hervé Guillemet wrote: Hi John, > If I understand your point, I think that the confusion comes from the fact > that for the gnomonists, the origin of azimuth is South , and not North like > it is for navigators (sail men , pilots ). In such case East is -90°, West is > +90°, North is +/- 180°. > > With this origin ( South = 0°), there is a symmetry in the trigonometric > circle : > sin (π - A) = sin A. As an example: > sin 120° = sin 60° = 0,866... > For the same sine value, you have 2 angles. > This is why you need to determine whether > Az is > 90° or < 90° > > I hope that it answers your point. Best regards > > (Now back to Morrison:) > > The problem to be solved is to calculate h given δ, φ and A. > There is a standard equation from spherical trigonometry that at first glance > appears to solve the problem: sin δ = (sin φ sin h) - (cos φ cos h cos A) > However, this equation cannot be solved directly for h, although it can be > solved by successive approximations. > Gunter needed a way to solve directly for h, and he solved the problem by > breaking it into two parts that can be solved. > > 1.- First. Gunter solved the Sun's altitude when it's declination is zero. > (i.e. when in the Equator) > 2.- Second. Gunter found the additional altitude of the Sun when the > declination is not zero, and the two components are added. > > > > In the first part (Triangle of the page 264 ) it is found the Sun's altitude > when the Sun is on the equator and when it is above the horizon, and with an > azimuth less than 90°. > > The construction is a right spherical triangle than can be solved using > Napier's rules as: > sin (90 - A) = tan ho tan φ or cos A = tan ho tan φ > > In the second part (Triangle of the page 365), Gunter finds the additional > altitude of the Sun when the declination is non-zero. The construction is for > a positive declination. The case for a negative declination is similar. > > Call x the Sun's altitude above its altitude when the declination is zero. > The construction uses two spherical triangles, both of which have the zenith > as the apex, one side is the meridian and the other side is the Sun's hour > circle. > > The interior angle at the apex is the Sun's azimuth. The base of the larger > triangle is the equator, and the sides are φ on the meridian and (90-ho) on > the Sun's hour circle. The base of the smaller triangle is the declination > path of the Sun. > > The angular distance from the equator to the declination path on the meridian > is the Sun's declination. The distance from the equator to the Sun on the > Sun's hour circle is the unknown, x. > > The two triangles are similar since the declination path and the equator are > parallel and the triangles share a common apex. Therefore, x : (90-ho) :: δ : > φ, and: > > sin x / sin (90 - ho) = sin x / cos ho = sin δ / sin φ > or sin x = cos ho sin δ / sin φ > > > > > --------------------------------------------------- > https://lists.uni-koeln.de/mailman/listinfo/sundial >
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