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On 10/4/2023 8:56 PM, NavList Community wrote:
> As I am not using any "modern" software - Stellarium and the like - I am
> left with just crunching numbers + paper sheet and pencil and I remain a
> bit frustrated at not having found one sufficiently reliable method -
> using correct and valid linear functions of time*/if any .../*- to speed
> up the process of solving this problem.

I don't know how to avoid trial and error, but I can do the last step
with a linear solution.

+28.000° north lat
  -90.000° east lon
0 height (m)

UTC, Sun az, Sun alt, Sun SD, Moon az, Moon alt, MoonSD. Refraction and
the value of UT1-UTC are ignored. The latter has no practically no
effect since the relative positions of the bodies are important, not
azimuths and altitudes.

16:30:00 150.325° +49.385° 0.267° 150.539° +49.569° 0.254°
17:00:00 161.259° +52.102° 0.267° 161.416° +52.117° 0.254°
17:30:00 173.325° +53.557° 0.267° 173.330° +53.429° 0.254°
18:00:00 185.844° +53.597° 0.267° 185.614° +53.368° 0.254°
18:30:00 197.956° +52.219° 0.267° 197.453° +51.942° 0.254°

Calculate separation (center to center) and Sun to Moon position angle.
PA is zero toward the zenith and increases counterclockwise. In this
case PA continuously decreases, so the Moon moves CW with respect to the
Sun.

16:30:00 0.230° 322.815°
17:00:00 0.097° 278.873°
17:30:00 0.128° 181.431°
18:00:00 0.267° 149.033°
18:30:00 0.415° 131.635°

The semidiameters and separation angle are the sides of a spherical
triangle. Its vertices are the Sun and Moon centers and the point where
their limbs intersect. (Actually there are two intersections and two
triangles. But the triangles are mirror images so one is sufficient.)

At each time in the above table, solve for angle B: the angle at the Sun
center. This is half the angle subtended by the "cookie bite" in the
Sun's limb. Since no side is more than about 1/2 degree, a plane
triangle calculation is accurate enough.

Also compute Moon PA minus B at each time. This is the PA at the CW edge
of the bite.

16:30:00 60.799° 262.016°
17:00:00 71.586° 207.287°
17:30:00 70.094° 111.338°
18:00:00 56.836° 92.197°
18:30:00 36.137° 95.498°

The CW edge of the bite moves approximately from 3 o'clock to 9 o'clock
on the Sun limb. It is at PA 180 between 17:00 and 17:30. By trial and
error identify the correct minute:

17:06:00 71.830° 186.991°
17:07:00 71.830° 183.142°
17:08:00 71.827° 179.189°
17:09:00 71.823° 175.155°

The size of the bite is almost constant, but its edge moves about 4° per
minute in PA.

At this point we can do a linear inverse interpolation. The PA second
differences indicate linear interpolation should be accurate to 1 second
of time. (Linear interpolation max error is 1/8 the mean 2nd
difference.) I get 17:07:48 as the time when PA = 180.

Of course if you have automated the table computation, as I have, it's
easier to simply reduce the interval. But Kermit wanted a linear
solution. Sorry, I couldn't find one that did not require some trial and
error. But the 17:07:48 calculated by linear interpolation was accurate:

17:07:46 71.827° 180.120°
17:07:47 71.827° 180.053°
17:07:48 71.827° 179.987°
17:07:49 71.827° 179.921°

By a similar process, calculate when the other edge of the bite is at PA
180. The difference is that you add B to Moon PA instead of subtracting.
That gives the PA at the CCW edge of the bite.

18:20:39 43.591° 180.024°
18:20:40 43.578° 180.002°
18:20:41 43.566° 179.981°
18:20:42 43.554° 179.960°

Summary: Sun LL is defective from 17:07:48 to 18:20:40 UTC.

I ignored refraction. Thought experiment: on Earth with no atmosphere,
suppose one edge of the cookie bite is on the same vertical circle as
the Sun center. Now introduce the atmosphere. Both points increase
altitude (not equally) due to refraction, but they remain on the same
vertical circle. So refraction does not affect the time when the limb is
defective.

P.S. A re-computation with a spherical triangle solution (not the plane
triangle approximation) gives identical position angles within 0.001°.

--
Paul Hirose
sofajpl.com

   

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