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    Re: Ex-Meridian Turkey Shoot Challenge Solution
    From: Jaap vd Heide
    Date: 2011 Nov 25, 04:29 -0800

    Dear Greg,

    Let me start by thanking you for triggering me to submerge myself in ex-meridian positioning! :)

    I have taken an approach that differs slightly, as I was asking myself whether I would be able to solve this without a given value 1.9 for factor a.

    What I came to is this:
    Imagine the challenge to be two ex-meridian sun observations *roughly to the South*, one before meridian passage, one after, on nov 24, 2011.
    Observation XMb (before MP)
    UT 19:30:00 (which will later be reffered to as Txmb)
    Ho 34°47.2' (which will later be referred to as Hob)

    Observation XMa (before MP)
    UT 20:00:00 (which will later be reffered to as Txma)
    Ho 34°53.3' (which will later be reffered to as Hoa)

    Hint: use ex-meridian height correction from "Bowditch". (I used the NGA 2002 download edition [http://msi.nga.mil/MSISiteContent/StaticFiles/NAV_PUBS/APN/pub9.zip] where this can be found in tables 24/25 and the explanation to these tables)

    Note: delta Ho or dHo = 6.1'

    For solar data I used Hemming Umland's "long term almanac for sun, moon and Polaris 1.09" - on Android, as I did most during my commute on the train - [http://www.celnav.de/longterm.htm] with dT 66.584 s [http://maia.usno.navy.mil/ser7/ser7.dat])

    From the almanac:
    Xmb -> Sun GHA 115°49'55"; dec S 20°35'33"
    Xma -> Sun GHA 123°19'49"; dec S 20°35'48" [note: delta dec +15" or +0.25']

    1st estimate of L
    Hob = 34°47.2' => co-alt = 55°12.8' => L ~ 34.6°
    Hoa = 34°53.3' => co-alt = 55°6.7' => L ~ 34.5°

    As the position will be on the intersection of circles originating in de GP of the sun at the time of the two observations, the position will be shifted very slightly towards the equator with respect to the two 1ste estimates. I therefor used L ~ 34.5° in the equation from Bowdith and checked the sensitivity by also calculating a from L estimates of 33°, 34° and 35°. In the end I continued with some confidence using a = 1.85

    ( Bowditch formula: a = 1.9635 • cos L • cos d • csc (L~d) => a(34.5) = 1.847 )
    The correction according to Bowditch (C = a•t^2/60) thus would become C = 1.85•t^2/60 => dHo = 1.85•t^2/60

    Now let there be a time-span x from Txmb to LAN and a time-span y from LAN to Txma.
    As Hob < Hoa and the max value of Ho would be at LAN, y < x (LAN will be closer to Txma)

    With the above we get

    (1) x+y = (Txma-Txmb)

    (2) dHo = a/60•(x^2-y^2) (without reasoning why in the case at hand y < x use dHo = |a/60•(x^2-y^2)|

    From (1) -> (3) y^2 = (Txma-Txmb)^2-2x(Txma-Txmb)+x^2

    Substituting (3) in (2) -> (4) dHo = a/60•(x^2-(Txma-Txmb)^2+2x(Txma-Txmb)-x^2)

    (4) can be re-writen as
    (5) x = (Txma-Txmb)/2+30•dHo/(a•(Txma-Txmb))

    Filling in (Txma-Txmb)=30, dHo=6.1 and a=1.85 renders x = 18 min 18 s

    As LAN = Txmb + x => LAN occurs at 19:48:18

    From the almanac => GHA(sun) = 120°24'21"; dec(sun) = S 20°35'42" (dec(sun) = S 20°35.7')
    Lon at LAN = GHA(sun) => longitude sought is W 120°24'21" or W 120°24.4'

    Bowditch's C at LAN would be a•y^2/60 = a•((Txma-Txmb)-x)^2/60 => 4'13" or 4.2'
    => Ho(LAN) would have been Hoa+4.2' = 34°57.5' => co-alt 55°5.5'
    L then would be 55°5.5' - 20°35.7' = 34°26.8'

    So concluding, I would have expected the turkey shoot @ 34°26.8'N, 120°24.4'W

    Regards,

    Jaap
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