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Re: Lewis and Clark lunars: more 1803 Almanac data
From: Paul Hirose
Date: 2004 Apr 18, 23:50 -0700

```George Huxtable wrote:
>
> So, if you take a starmap, and mark on it the Moon at dec 18.617, RA 8h 44m
> 41s. Then draw two lines through that point: one at an angle of (70 - 48)
> deg, or 22deg (measured clockwise from North, which is straight-up on the
> map); the other at (70 + 48) deg, or 118 deg clockwise. Extend these lines
> as great circles to an arc-length of 60deg 56.2' from the Moon. Then the
> mystery star ought to lie somewhere near the end of one of those arcs.

This requires solving a spherical triangle, two sides and their
included angle being known. It's the same kind of problem we face when
reducing a sextant shot. In this case, though, the included angle is
not at the pole, it's at the Moon. So imagine the Moon is the
triangle's "pole", and the north pole is the "observer".

The "LHA" is measured clockwise (viewed from outside the celestial
sphere) from the "observer" (pole) to the unknown star. It's 338
degrees (George's position angle of 22 degrees, reversed to correspond
to the way LHA is measured).

"Assumed latitude" is the Moon's declination, N 19 degrees. (I assume
it's north since George did not put a sign on the value.)

"Declination" is 90 minus the length of arc to the unknown star, 90 -
61 = N 29 degrees.

Looking up the solution in HO 229, Hc is actually the declination of
the unknown star, 68 degrees. And the tabulated Z, 59 degrees, is
added (since the star is west of the Moon) to the Moon's SHA to obtain
star SHA, 288 degrees.

For a check, rework the "sight reduction". This time put the parts in
their normal places, except use the Moon's position as the "observer".
LHA = 59. Assumed lat. = N 19. Dec. = N 68.

The HO 229 tabulated values rounded to the nearest degree are Hc = 29
and Z = 22. The arc length is 90 - 29 = 61 degrees. It looks good.

Right ascension of the star would be 4 h 50 m by my calculation.

By entering HO 229 with different LHA values (again, this represents
azimuth to the star), the declination and assumed latitude remaining
constant, you could get enough points to sketch a line of equal lunar
distance in a star atlas. How much error would be introduced by simply
plotting in a modern atlas, disregarding precession since 1803? I
don't know.

```
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