Welcome to the NavList Message Boards.


A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

Compose Your Message

Add Images & Files
    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.

    Browse Files

    Drop Files


    What is NavList?

    Join NavList

    (please, no nicknames or handles)
    Do you want to receive all group messages by email?
    Yes No

    You can also join by posting. Your first on-topic post automatically makes you a member.

    Posting Code

    Enter the email address associated with your NavList messages. Your posting code will be emailed to you immediately.

    Email Settings

    Posting Code:

    Custom Index

    Start date: (yyyymm dd)
    End date: (yyyymm dd)

    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site
    Visit this site