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    Re: A celestial navigation problem
    From: Gary LaPook
    Date: 2011 Nov 17, 11:08 -0800
    That explains it since my position came out slightly west of the island. It should have alerted me to go back and check my work.

    BTW, I bring up from time to time the question of how precisely you need to know your position. The first approximation, 11.5° S, 150° W , coupled with standing on dry land is all the information you need to determine you are on Flint Island since no other island is near those coordinates.


    --- On Thu, 11/17/11, slk1000@aol.com <slk1000@aol.com> wrote:

    From: slk1000@aol.com <slk1000@aol.com>
    Subject: [NavList] Re: A celestial navigation problem
    To: NavList@fer3.com
    Cc: slk1000@aol.com
    Date: Thursday, November 17, 2011, 6:14 AM

    A pat on the back for Gary!  Nice job throughout, including the intermediate answers, except for one oversight!  Now I am confident that my solution is correct and the students should be able to handle it.

    The students have the dip correction table, sun altitude correction table, daily pages, etc., provided with the Student Manual.  I have provided them all with old Nautical Almanacs for the Increments and Corrections.  They have everything they need, but I'm sure they appreciate your concern for them

    The only difference I see between my solution and Gary's is that watch error was overlooked, so the time of LAN should have been 22-07-41.

     Also, FYI, the Student Manual uses 23.4º instead of 23.5º at the solstices.  (I always say "about twenty-three and a half degrees".)

    Thanks for taking the time to look at this.


    -----Original Message-----
    From: Gary LaPook <garylapook@pacbell.net>
    To: NavList <NavList@fer3.com>
    Sent: Thu, Nov 17, 2011 4:20 am
    Subject: [NavList] Re: A celestial navigation problem

    You didn't include the sextant correction table, do you expect the students to figure them in their heads? Dip approx square root of 6, somewhere between 2 and 3 minutes, call it 2.5 (2.4 from table.)

    Refraction= zero above 63° (to the precision on 1') or 0.2' from table.
    Or do you use the S.D. of +16.3 minus the 0.2' refraction, total +16.1 or from the sun table of 16.0'.

    Increment for GHA for time of noon average of the times for both of the 77° 44.1' altitudes, 22:07:46 Z. Do they do the increment in their heads, 15' X 7 minutes plus 46/4' per second = 1° 56.5', same as in increments table.

    Approx longitude 150° W based on 2200Z time of noon.

    Sun moves from left to right so must be south.

    Approx declination 23.5° S,  around winter solstice. Approx Ho = 78° so ZD approx 12°  so approx lat is 11.5° S.

    Longitude at noon, 22:07:46  = GHA =149°53.4' plus 1°56.5' so longitude is 151°49.9' W.

                       Hs = 77° 48.7'
    I.C.                +          1.3'
    Dip                 -           2.4'
    S.D.               +        16.3'
    Ref                 -           0.2'
    Ho =                   78°  03.5'

    ZD =                    89° 60.0'
                         -     78° 03.5'
                               11° 56.3'

    Lat                       23° 22.6'
                         -      11° 56.3'
                                11° 26.3' S.

    So you are on Flint Island.

    It is easy to memorize the refraction corrections to  a precision of 1';

    5' above 10°
    4           12
    3           16
    2           21
    1           33
    0           63

    It's a good test, student must be able to think through the problem.


    --- On Wed, 11/16/11, slk1000@aol.com <slk1000@aol.com> wrote:

    From: slk1000@aol.com <slk1000@aol.com>
    Subject: [NavList] A celestial navigation problem
    To: NavList@fer3.com
    Cc: slk1000@aol.com
    Date: Wednesday, November 16, 2011, 6:44 PM

    I have been teaching the Power Squadrons Navigation and Junior Navigation courses for quite a while now, but I started teaching the "new" Junior Navigation course for the first time in September.  The course is quite a bit watered down compared to the "old" course, with the sun being the only body of interest.  In the past I created or borrowed what I call SuperNavigator problems for the students, generally a bit beyond the scope of the material, but requiring the students to do some thinking rather than just plugging in numbers.

    Due to the limited amount of material in the new course, I think I was able to incorporate all aspects of celestial navigation included in the course into one SuperNavigator problem, which I have attached, along with the one page of the 1972 Nautical Almanac required.  I have not included my solution, for fear of making a fool of myself
    :-).  I would appreciate it if you would take a look at it and tell me if you think it makes sense and is reasonable.  I think the intermediate answers require the students to prove they have a real understanding of the material more than the final answer does, but I would like to know what you think.

    Thanks for your support.

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