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    Re: Take two: Latitude and longitude around noon
    From: Brad Morris
    Date: 2017 Apr 8, 17:20 -0400
    The input to the USNO was 
    E 81°33'46", S66°33'46"

    A bunch of you wrote about the ambiguity in which limb was observed.  There was no ambiguity, the data was presented in the text.  I stated "
    As the Antarctic Winter nears its end, I managed to obtain the apparent sun altitudes, as found [below, see PS].  The apparent position corresponds to the observed altitude."

    From the USNO site, we have:
    A calculated apparent position corresponds most closely to the observed position of an object on the celestial sphere. The aberration of light (due to the velocity of the observer) and the relativistic bending of light (due to the Sun's gravitational field) are taken into account. For solar system objects, light propagation time is also included.

    From Dutton, we know:
    The "observed position", labeled Ho, is obtained from the sextant reading with all corrections applied.

    And so you had Ho for every reading.  The height of eye, the refraction and index correction were all red herrings.  Every professor tells you to read the question carefully.  Extra data lead to confusion! 
    Herman Dekker came very close to stating the correct conclusion.  The data provided was the center of the sun, neither limb is correct.

    Frank wrote:
    Incidentally, as one approaches the pole, a noon sun curve like this becomes flatter, and the ability to determine the axis of symmetry of the observations becomes more difficult. 

    I thought about getting much closer to the pole, as it's obvious where this breaks down. I decided to put the position right on the Antarctic Circle, so as to bring that point to the foreground.  I kept the data without noise however, for an entirely different reason.

    If we know the latitude, then the semi-parabolic arc, described by the earth rotation and sun declination change, is perfectly defined.  Leaving the data pure (unrandomized) provided nearly every NavList estimation to be relatively spot on. A demonstration of how good it could be with a pure arc.  Some analysis provided an answer to within 1'.  My assertion then, is that we should create that semi-parabolic arc via declination, equation and latitude, and then fit that pure semi-parabolic arc to the actual data using a least squares fit.  

    We should not derive the semi-parabolic arc from the randomized data.  When we did that with Frank's Rhode Island observation, the longitude spread was enormous, but the latitude spread was small.  If we had used a pure, generated semi-parabolic arc, and then fit it to the data, I think the results would have had a narrower spread.

    Five answers were provided.  The precise distance in nautical miles was calculated.
    Greg 8.566nm
    Herman 7.795 nm
    Peter 8.553 nm
    Robin 9.226 nm
    Frank --- did not provide a latitude ---

     Greg had the longitude to within 0.1' (!!!), but Herman came the closest overall.  Perhaps another round is in order, Herman, would you pick your position?


    On Apr 6, 2017 8:59 PM, "Brad Morris" <NoReply_Morris@fer3.com> wrote:
    I am monitoring this.
    You do have all the information.
    There is no ambiguity.
    Please read the text carefully.

    I would prefer not to disclose the correct answer just yet


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