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    Re: Azimuth Formula Questions
    From: Bill B
    Date: 2005 Oct 31, 21:27 -0500

    George wrote:
    > Let's say the observer was just a few feet North of the Equator, with
    > latitude = 0.001 degrees.
    > Then apply that formula quoted above, with L = +0.001 deg, d = +1 deg, LHA =
    > +1deg
    > That gives me Z = 45.024 degrees. With observer in the North hemisphere
    > (just) and LHA less than 180, applying the relevant quoted rule of Zn =
    > 360 - Z gives Zn = 314.975 degrees,. Just as you would expect, its bearing
    > is nearly 45 nearly Northwest from the observer. Seems entirely reasonable,
    > doesn't it?
    > But now let the observer travel a few feet South, across the equator, to a
    > lat of -.001 degrees South.
    > Plugging in that new value of lat changes Z to 44.967, almost exactly the
    > same as before (jst as we would expect, for such a small movement of the
    > observer). But now, the observer's Lat is in the South hemisphere, so we
    > have to apply a different rule to put azimuth in the right quadrant. With
    > LHA less than 180, we now must apply Zn = 180 + z, , or 224 967 degrees, so
    > its calculated bearing is now SW of the observer. This is obviously quite
    > wrong, if we sketch the picture out. So either I have made a silly error, or
    > HO229 has, or there's a further error in the transcription. Which is it?
    Neither a silly or transcription error.
    Located the 229 tables online thanks to an old post from Frank.
    Looking at LHA 1, Dec 1, Lat 0, it gives "Same Name" Z of 45d, and "Contrary
    Name" Z of 135.  Clearly the Contrary value is the supplement of the Same
    (calculated) value.  Apply their rules and Zn = 315d in both cases.  That
    same supplementary relationship exists for Z when you cross over from Same
    Name to Contrary Name on the right hand page tables.
    I imagine that this may be spelled out or diagramed in the introduction if
    one knows what to look for, but is above my ability to ferret out.
    Having solved that mystery, the scenario you proposed is pure hell for
    tabular solutions.  A different formula for calculating Z, a special set of
    interpolations for body near the zenith, and yet another 6+ step process for
    interpolating Z for the .001d/3.6" angle. Probably a double-second
    difference calculation to boot?  Also a resolution problem for my little
    All for a observation that is a poor choice to begin with.
    Thanks to you and many others on the list, I have exceeded my initial goal
    of 229 reductions, and can now knock off a reduction on a $10, 3-memory
    TI-30XA  using only the daily pages in less than two minutes.  But still so
    much more to learn.
    Hope you will give 229 a look.  Not ideal in strange circumstances, but does
    work well for recommended observations.  Coupled with an almanac, no
    electricity, calculators or computers required (other than generating the
    volumes ;-)  Pencil and paper only.  A great backup to GPS backup. 
    It also has other tricks to play with the tables, such as star location or
    identification, great-circle route calculation, and a table of offsets to
    correct the LOP from a long intercept to more closely approximate a COP.
    Not as painless as a calculator, but not rocket science either.  Heck, even
    I can do it!

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