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
