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    Re: Longitude from Local Apparent Noon Altitudes
    From: Bruce J. Pennino
    Date: 2012 Aug 19, 20:50 -0400

    If you look at an aerial of Provincetown ,MA,  Snail Road leaves Rt 6 and goes toward bay. Snail Road meets 6A.  I was just 300-400 ft  east  of there standing on a sun bathing deck adjacent to the water.Actually, the deck overhangs the beach/water at high tide. This location is directly north of the point you determined.

     From "eyeballing data" and  folding graph, I guesstimated UT of 16h43m56sec.

    Based on an online calculator,  the sun deck location is N 42 04'   W 070 10'. Your analysis...very well done!

    My "analysis" gave me N 42 06 W 70 00.......learning!

    Thanks all


    On Sun, Aug 19, 2012 at 3:27 PM, =?utf-8?Q?Antoine Cou=C3=ABtte?= wrote:

     Hello Bruce,

    Provided I understood everything correctly (I figured out that the first UT was 16h22m02.2s and so on ....) and provided I correctly entered all ... 325 numbers on my Calculator - while knowing that just one single typo can ruin all the results - I found the following results (7th order regression) :
    UT of Transit (i.e. body full south of Observer's Position) 16h44m12,8 , hence a Longitude of W070°04'3 ,
    UT of Culmination occurred 8,23s earlier, i.e. at UT = 16h44m04s6 . The 8.23s difference between both comes from the Sun Declination which is decreasing. In this case, a first order value for the time difference in seconds of time is (48/Pi)* Declination change in arc minutes per hour.
    I assummed that the Observer is steady on Earth.
    Observed latitude N42°02'3 S
    Standard Deviation of the Observations : 0.9 NM (defines quality of Observations)
    Longitude Uncertainty / Latitude uncertainty (both in arc minutes) <= 3.9
    In other words, the Latitude should be accurate to +/- 2.7 NM at 3 Sigma, and the Longitude should be accurate to +/- 10'5 at 3 Sigma, as long as I have correctly entered the 64 different values for these 32 Observations.
    Best Regards to all
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