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    Re: Bennett's '...Celestial Navigator' --An improved Zn calculation
    From: George Huxtable
    Date: 2005 Nov 26, 00:30 -0000

    D. Walden wrote-
    > With some trepidation, I raise  again the question of using '...Celestial
    > Navigator' to obtain  Azimuth.
    Wow, that WAS a long time ago! It was a matter that seemed to excite great
    passion in the breasts of Bennett's adherents; or in one such breast, at
    D.Walden doesn't quote my original posting in detail, or give its date, and
    I no longer have easy access to it, so I will have to go a bit by memory.
    For new readers: it concerned the azimuth tables in "The complete on-board
    celestial navigator" 2003-2007 edition, by George G Bennett, about which it
    was claimed- "No interpolation is required, and it is one of the simplest
    techniques for finding azimuth with an accuracy of one or two degrees."
    I decided to put that claim to the test, and put forward what I thought to
    be a worst-case set of inputs for which the azimuth tables gave a result
    that was all of 15 degrees different from the true azimuth. These values are
    what Walden refers to when he writes-
    >Using the infamous Huxtable example:
    >  Dec=55-30
    >  LHA=54-30
    >  Alt=61-30
    To keep the record straight, I don't recognise those numbers as exactly the
    same worst-case as I suggested. Just to ensure that the worst-case choice
    was made between adjacent rows and columns, when finding the nearest value
    without interpolation, I think I actually proposed-
    >  Dec=55-29
    >  LHA=54-31
    >  Alt=61-31
    When those values are plugged in to the Bennett azimuth table as nearest
    whole-number degrees, we get dec = 55, lat = 55, which gives a resultant
    value for X of 469. Combining that with an alt = 62, the azimuth table gave
    a resulting value of 270 degrees, or due West. (In my initial mailing, I had
    in error stated that to be 90 degrees, but that's by the way). This result
    of 270 is nearly 15 degrees out from the azimuth you get if you make a
    calculation with spherical trig., using-
    Az = arcsin ( - sin LHA * cos dec / cos alt),
    which gives Az as 255deg 21' or 284deg 39'
    That said, I'm quite happy for Walden to round off those inputs to the
    nearest half-degree, as he did.
    Walden continued-
    >  George H. didn't give the corresponding Lat, but it can be found to be:
    >  Lat=60-18
    Let's pause again here. How, I wonder, has he arrived at that value, which I
    do not dispute. He doesn't explain.
    Actually, there are two possible such latitudes. There's a position circle,
    centred on the geographical position of the body given by dec, with radius
    (90 - alt) degrees, and any meridian that crosses it, defined by LHA, must
    cross it twice. In this case, one such latitude is at 60 deg 18', as Walden
    states, and the other is at 76deg 11'.
    As for me, if I needed lat given those quantities, I would work it out using
    lat = arctan(tan dec / cos LHA)
    + arccos ( sin ( arctan ( tan dec / cos LHA) )*sin alt /sin dec)
    Because arc cos has two possible values, this gives two solutions, as above.
    There are other ways of expressing it, which may be simpler.
    How did Walden obtain that value for lat, which he needed to use in his
    proposed different way of using the Bennett table, "backwards"?.
    Walden then abandons the Bennett azimuth table, the thing that causes all
    the trouble. Instead, he takes the procedure and tables that Bennett
    to calculate the altitude, and uses them to obtain the azimuth instead, by
    twisting the astronomical triangle round. In those tables, quantities can be
    entered to the nearest minute, unlike Bennett's azimuth tables, in which
    they are only to the nearest degree. So it's no surprise that he ends up
    with a much better result.
    .>  First going through the altitude  calculation using the Bennett work
    form, on page 168, to generate  altitude from given Dec, LHA, and Lat.
    >  line 13 Local Hour Angle          54-30 -> 8841
    >  line 14 DR Latitude             N 60-18 -> 3974
    >  line 15 Declination             N 55-30 -> 3217
    >                                             ______
    >  line 16  (theta=28-04)                  SUM 16032 -> RES  11760
    >  line 17 Latitude ~ Declination
    >           (ABS(Lat-Dec))              4-48  ------------>    351
    >                                                          ______
    >  line 18 Computed Altitude         61-30.5 <------  ALT  12111
    >  Now for the new method to calculate  Zn.  In a sentence, use Bennett's
    > table 'backwards' substituting  Alt for Dec, and Dec for Alt.  The final Z
    > will be the LHA  value.  Continuing with the infamous example from above:
    >  remember, substitute Dec for Alt
    >  line 18 Computed Altitude         55-30  --------> ALT  17587
    >  now, Lat~Dec becomes  Lat~Alt                            ______
    >  line 17 Latitude ~  Declination     1-12   -------->         22
    >  now, calculate what RES must be for sum to equal ALT, (17587-22)
    >  line 16   (theta=34-29)                 SUM 13763 <- RES  17565
    >                                             ______
    >  now, we have the sum of three, we know two, so we can solve for the
    > third.
    >  remember, substitute Alt for Dec
    >  line 15 Declination              61-30.5 -> 4187
    >  note, line 14 is the same as above
    >  line 14 DR  Lat                  N 60-18   -> 3974
    >  for SUM to be correct, line 13 must be 13763-4187-3974
    >  note use top of column LHA value as Z
    >  line 13 Local Hour Angle           75-07 <- 5602
    >  now, we apply our one rule, if LHA(the real LHA)<180, Zn=360-Z, else Zn=Z
    >  So, Zn=360-(75-07)=284-53  Exactly the ATAN2 formula result!
    >  Note, there is a typo, which I  don't recall seeing mentioned before, in
    > Bennett's response to  Huxtable: "If, however, the Tables are interpolated
    > (X=460) the azimuth  is found to be 255 or 285 (not 075 or 105) which
    > compares favourably  with the results from direct calculation of 255.3 and
    > 254.8."  The  last number should be 284.8, as above.
    I agree, that must be a typo. Bennett's argument, that if you interpolate
    you get a much better answer, though true, was beside the point, because his
    book specifically stated that interpolation wasn't needed. He accepted that
    at some angles large errors could result, and promised to put a note to that
    effect in future printings of his book. That was what I was after. I presume
    that he has done so.
    >  Pretty slick, eh?
    I can only agree. At first sight, pretty slick! But it needs a bit more than
    that single example to fully convince us if the virtues of Walden's
    The next sentence is a bit ominous...
    >  (Some adjustments of signs for special cases are left to the reader as an
    > exercise.)
     D. Walden should be congratulated on a useful piece of lateral thinking.

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