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    Re: shortest twilight problem...
    From: Robin Stuart
    Date: 2010 Jul 1, 11:41 -0700

    Knowing that there is light at the end of the tunnel is a great incentive to try to push the algebra through to the end. Adopting Frank Reed's terminology :-), the "modern" formulation can be beaten into the formula given by Joel Silverberg as follows:

    Starting from

    0 = sin(d)*sin(L) + cos(d)*cos(L)*cos(LHA1)
    cos(90+18) = sin(d)*sin(L) + cos(d)*cos(L)*cos(LHA2)

    where
    LHA1 is the Sun's LHA at sunrise/sunset
    LHA2 is the Sun's LHA at the start/end of twilight
    d is the Sun's declination
    L is the observer's latitude

    this can be rewritten as

    LHA1 = acos( -tan(d)*tan(L) )
    LHA2 = acos( ( cos(90+18) - sin(d)*sin(L) ) / ( cos(d)*cos(L) ) )

    Then minimize LHA1 – LHA2 by differentiating with respect to declination. In practice this is done by considering the condition

    (LHA1')^2 = (LHA2')^2

    where the prime (') denotes differentiation and the squaring (^2) is done to remove square-roots arising from the derivative of the inverse cosine function (acos). The resulting expression can eventually be reduced, using standard straightforward methods, to

    2*sin(d)*sin(L) – ( sin^2(d) + sin^2(L) ) * cos(90+18) = 0

    and then applying the identity

    cos(90+h) = -sin(h) = -2*t / ( 1 + t^2 ) where t = tan(h/2)

    yields the formula

    sin(d) = - tan( h/2 ) * sin(L) for h = 18 degrees in this case.

    The minus sign comes about naturally and is a "modern" way of expressing Joel's caveat "that the solar declination on that day will be negative if the observer is in the Northern hemisphere and positive if he/she is in the Southern hemisphere".

    This is a "modern" rather brute force and mechanical approach to the problem and I admit using the "modern" algebraic manipulation program Mathematica to ease the otherwise tedious manual algebra in intermediate steps. It is far less elegant than the derivations presented in the texts that Frank has kindly provided. In practice, however, (although I can't speak for Laplace!) results are often derived by brute force first and then the elegant shortcut comes later at which point it is proclaimed to be obvious,

    Regards,
    Robin Stuart

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