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    Re: The Great Circle Challenge
    From: Gary LaPook
    Date: 2014 Dec 30, 00:31 -0800
    Doing it with inspection tables.

    HO 214

    Starting position (AP.)  Lat.  41 30.0' S         Lon. 174 30' E
    Destination lat (decl)                 21 57.0' N    Lon   1 59 40' W   Dlo 25 50' (LHA)

    Entering the table with the closest tabulated values Lat 41   Dec 22 contrary LHA 26

    Azimuth angle of 153.9 and tab Hc   22 31.8   delta dec + .93 ,    delta t +.34, signs determined by
    inspection, comparing the tabulated value with the next value for declination and for the next
    value of LHA. Since we are using the declination of 22 since it is closest, the difference of
    declination is only -03', and since the declination is between 22 and 21 we compare with decl 21
    to get the sign of the delta dec correction factor. Since we are using an LHA of 26 the difference
    of LHA (t) is -10' and the LHA is between 26 and 25 we compare LHA .25 to get the sign of the
    delta t correction factor.  These delta values are the equivalent to the delta dec value in HO 229
    except HO 214 uses decimals of a degree and HO 229 uses minutes. 

    So, looking at the multiplication table in HO 214 we see that the delta dec +.93 times the 3
    minutes of declination difference produces a correction of +2.8' and the difference of 10' of LHA
    produces a correction of +3.4.

    SO:

     Tab Hc     22 31.8
    Delta d     +      2.8
    Delta t       +     3.4
                   --------------
                      22 38.0'

    Now we look at the Delta latitude correction table, entering with the azimuth angle and the
    difference in latitude. And extract the correction for the DR latitude of -27.0'.

                       22.38.0'
                    -        27.0'
                      -----------

                        22 11.0 for the Hc which we subtract from 90 degrees (89 60.0')
    22 11.0'
                                                                                                              ---------------  
                                                                                                                 67 49.0

    67 X 60 = 4020.0
                   +    49.0
                     
                       4069.0 NM

    You can do the same calculation with HO 229 using the printed delta dec correction and figuring
    our your own delta t correction by subtracting the Hc for dec 22, LHA 26 Lat 41,    22 31.8, delta
    dec +.56.
    From that for the next LHA value of 25,    22 51.4 - 22 31.8 producing the correction factor of +19.6 which is
    the equivalent of the + .34 delta t number in HO 214 
    Using the multiplication tables in HO 229

                          Tab Hc                      22 31.8
                           Declination corr            +2.8
                           LHA corr                      + 3.3
                                                           
                                                            22 37.9'
                                                          
    Applying the same DR latitude corr     -27.0
                                                            
                                                               -----------
                                                               22 10.9' almost identical as the HO 214 Hc.

    gl


    From: Gary LaPook <NoReply_LaPook@fer3.com>
    To: garylapook---.net
    Sent: Monday, December 29, 2014 10:41 PM
    Subject: [NavList] Re: The Great Circle Challenge

    Here is a link to those tables:


    gl


    From: Gary LaPook <NoReply_LaPook@fer3.com>
    To: garylapook---.net
    Sent: Monday, December 29, 2014 12:06 PM
    Subject: [NavList] Re: The Great Circle Challenge

    I have posted a long time ago the interpolation tables from HO 214 that alows the use of a DR position. Search for them on navlist and you can use them width HO 229 to figure the GC route.
    Gl


     
     




       
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