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    Re: Error in H.O. 229 Introduction/Explanation?
    From: Greg Rudzinski
    Date: 2013 Apr 18, 11:30 -0700

    Sean,

    The Pub 229 tables for some reason uses the north hemisphere azimuth rules for Z when Hc becomes a negative altitude so for your example LHA 240 Lat 45 S Dec 5 S Z = Zn.

    When using the NGA 229 calculator the Z rules don't change when the altitude becomes negative (see attached pic 175).

    When using my palm pilot sight reduction calculator I get Hc = 343* 6.4' azimuth 115.6*

    Greg Rudzinski

    Error in H.O. 229 Introduction/Explanation?
    From: Sean C
    Date: 2013 Apr 18, 10:11 -0700
    Gentlemen,

    Yesterday, I was working problem No. 5-3 from "100 Problems in Celestial Navigation" using H.O. 229.

    I completely botched it. :)

    I've never had a problem using the tables before. To be sure my error was not in my use (or misuse) of the tables, I re-read the introduction. I was using the calculator on my mobile phone and the formula for direct computation found in the Nautical Almanac to follow along (again, just to be sure). And then I read it again...and again. To my surprise, my calculations did not match what the introduction gives as the true azimuths of the hypothetical body in the example. I would like to ask the members here, Frank especially, to double check my work so I know I'm not crazy!

    The text from H.O. 229 (volume 3 in this example, but the error seems to persist throughout the 4 volumes I own), bottom of page viii, reads:

    "As an additional example, the following arguments are used for entering the tables:

    LHA: 240° (t 120°E)
    Latitude: 45° S (Same Name as Declination)
    Declination: 5° S
    However, inspection of the diagram on the plane of the celestial meridian in figures 2 and 4b reveals that the altitude is 16°53.6' below the celestial horizon; the tabular azimuth angle is the supplement of the actual azimuth angle of 64.4°. Further inspection of these figures reveals that with the LHA and and latitude (Same Name) remaining constant, the altitude of the body increases as the declination increases. Between values of declination of 26° and 27° the body crosses the celestial horizon. When the declination reaches 35°, the altitude is 6°39.6' above the celestial horizon; the tabular azimuth angle is the actual azimuth angle of 45.6°."

    Bolding is mine and indicates the erroneous text. By my calculation, with the initial declination value of 5° S, the actual azimuth should be equal to the tabular azimuth of 115.6°. Note that within the tables themselves, and on the excerpt pages vi and vii, the rules given at the bottom of the page and the tabular values are correct. It is only the text of the introduction that is in error.

    Likewise, in the second part, with a declination value of 35° S, by my calculation the azimuth should be the supplement of the tabular azimuth angle of 45.6°, which is 134.4°. In other words, I think the example has it backwards, resulting in completely erroneous azimuths.

    Again, I ask that NavList members independently verify my results. Thanks!

    -Sean C.

    P.S.

    I have no idea what my original error was. I've been checking and re-checking this error ever since I found it in the hopes that I don't embarass myself! :D
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