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    Re: Sumner's Line (Navigation question)
    From: Bill B
    Date: 2006 Feb 11, 15:43 -0500

    You wrote:
    "So what's the math that converts a measured altitude into an  LHA? This is
    an easy spherical triangle problem. The Sun's local hour angle is  the angle
    measured at the elevated celestial pole between the observer's  meridian and
    the arc from the pole to the Sun (draw it!). To calculate that  angle, we
    can use the three sides of the spherical triangle made by the Sun,  Zenith,
    and the Elevated Celestial Pole. I won't spoil the fun. You try it out  for
    yourself. Draw that triangle and solve for LHA using the cosine formula."
    I believe I understand you, but want to check my reasoning.  My derivation
    of the Hc formula is:
    cos LHA = (sine Hc - (sine lat sin dec)) / (cos lat cos dec)
    When I use my derivation of the Hc formula to solve for LHA, I do not get
    LHA in the case of a body east of the meridian.  In your example, an angle
    of 45d (0 Eqn of Time) would yield a local time of 3 PM or 9 AM.  Clearly
    the 9 AM LHA would be 315 if we measure from the observer west to the body.
    Of course my calculator does not know whether the body is pre or post
    meridian passage, and cos 45d = cos 315d.
    At least the way I am doing it for now, I believe I am getting angle t, the
    angle formed from the observer to the elevated pole to the the body, which
    is not the LHA when the body is east of the meridian.
    You also wrote:
    "...then the Sun's  altitude, which can be measured accurately with a
    sextant, automatically yields the Sun's local hour angle. This hour angle is
    exactly the same thing as the observer's local apparent time."
    Perhaps a matter of semantics, but in the case of a body east of the
    meridian, to get local time I think I should subtract the arc-to-time
    conversion from 12 to get local time.
    Have I missed something?

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