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    Re: Another "emergency navigation" sight reduction method
    From: Gary LaPook
    Date: 2015 Jul 9, 00:44 +0000

    An easy way to check the computation on a Bygrave is to do the same computation on a calculator since this allows you to check the intermediate steps.

    Just use the standard Bygrave formulas in the three step process following along on the form I have posted.

    First calculate co-latitude and save it in a memory in the calculator. If you are using a value for hour angle that is not a whole number of degrees you might want to make the conversion to decimal degrees and save it in a memory since it will we used twice. If you are using whole degrees then this step is not necessary.

    Then you calculate "W" using the formula:

    tan W = tan D / cos H

    and sum it to the memory where you have saved co-latitude which is then X and then make any adjustment necessary to convert X to Y. (If you are just making trials you can avoid this step by your choice of the trial values.) There is no reason to store W itself since it is not used again. You can then convert W to degree and minute format to compare with the Bygrave derived value.

    Then you compute azimuth angle using the formula:

    tan Az = (cos W / cos Y ) x tan H.

    If you want you can also convert Az to degree and minute format to compare with the Bygrave.

    The last step is to calculate altitude with the formula:

    tan Hc = cos Az x tan Y.

    Then convert to degree and minute format to compare with the Bygrave result. 

    (When entering values in the format of degrees minutes seconds, change decimal minutes to seconds, 6 seconds per tenth of a minute, in your head  before punching in the assumed latitude, declination  and hour angle if necessary.)

    Using whole degrees for declination, assumed latitude and hour angle, using a TI-30 with only 3 memory locations the key strokes are:


                 (co-latitude = 90 - Assimed latitude)

    Assumed Lat
    STO 1  (co-latitude stored in memory 1)

    (tan W = tan D / cos H)


    tan    (computed W)
    SUM 1  (X now stored in memory 1)(change X to Y if necessary)


    (tan Az = (cos W / cos Y ) x tan H)

    cos    (of W from prior step)   

    RCL 1  (recalls Y from memory 1)
    tan (computed Azimuth angle)


    (tan Hc = cos Az x tan Y)

    cos    (of Az from prior step)   

    RCL 1  (recalls Y from memory 1)
    tan  (computed altitude, Hc)

    D.D - DMS (changes Hc in decimal degrees to degrees, minutes and seconds)


    From: Bob Goethe <NoReply_Goethe@fer3.com>
    To: garylapook@pacbell.net
    Sent: Wednesday, July 8, 2015 3:26 PM
    Subject: [NavList] Re: Another "emergency navigation" sight reduction method

    This PDF was just the sort of thing I was looking for, with a concise statement of the formulae for Bygrave.  That said, my first attempt at using the equations came up with nothing like what I got from Pub. 249.  I would appreciate you (or perhaps others) helping me sort out where I am making my errors.
    First, I recast the equations slightly to make them fit the way my mind works.
    tan(x) = tan(d) ÷ cos(LHA)  I recast as
            1. tan(d) ÷ cos(t) = tan(x)
    ...where d = declination of the GP, and
                t = meridian angle
    I inserted a line for myself of
             2.  lat ~ X = Y
    ...where I am looking for the difference between the latitude of the AP (i.e. "lat") and the latitude of point X.  This difference I designate as Y.
          3.  (tan(t) * cos(X)) ÷ cos(Y) = tan(Az);   
    I inserted another line:
          4.  180 - Az = Z
    ...where Z is the azimuth angle to the GP (e.g. N 100° W).  From that, I can derive Zn, the azimuth (e.g. 260°)
    Finally, where Ronald uses "a" in his final equation, I designate that as Hc.  Hence:
          5.  cos(Az) * tan(Y) = tan(Hc);   
    I *think* I have wrapped my head around the equations...except that I come up with answers that don't match with Pub. 249.  So either I have misunderstood the equations, or I am not using my calculator properly.
    Here is the scenario I used.
    GP of the sun:   GHA = 48° 03.6'            Declination = N 11° 44.6'
    AP = 53° N    114° 03.6' W
    Hence, the meridian angle = t = 66°
    1.  tan(11.743°) ÷ cos(66°) = tan(X)
          0.208 ÷ 0.407 = 0.511
          X = arctan(0.511) = 27.1°
    2.  53° ~ 27.1° = Y
         Y = 25.9°
    3. (tan(66°) * cos(27.1°)) ÷ cos(25.9°) = tan(Az);   
        (2.25 * 0.890) ÷ 0.899 = tan(Az)
        2.00 ÷ .899 = 2.22
        Az = arctan(2.22) = 65.8°
    4.  180 - 65.8 = Z
          Z = N 114° E
    5. cos(65.8°) * tan(25.9°) = tan(Hc);   
         0.410 * 0.485 = 0.199
        arctan(0.199) = 11.2°
    Now, when I do this with Pub. 249, I get an azimuth angle "Z" of N 103° E and an Hc of 23° 44'.
    Clearly, I am doing some things very wrong.  Do you observe what they might be?
    Thank you for giving thought to this.

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