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    Re: Raw data for bubble
    From: Bill B
    Date: 2007 Mar 22, 17:43 -0400

    > From: "Bill Noyce" 
    > I think the original question was, how to predict the time at which
    > the sun has an azimuth of 90 or 270 degrees, given a known or assumed
    > latitude.  In this case, you have the co-latitude, the co-declination
    > of the sun, and the azimuth (which is a right angle) as three known
    > parts of the triangle.  (The sun's declination comes from a calendar,
    > plus perhaps an estimate of longitude to the nearest 15 degrees.  You
    > would have to be more careful if using the moon.)
    > The formulas will come out something simple like
    > cos LHA = tan d / tan L  and  sin h = sin d / sin L
    Interesting.  For Peter to calculate the exact moment the body is at 90/270d
    he must know his location exactly.
    While you state co-latitude is a known, it cannot be known exactly until the
    exact time of transit is known.  However delination is a relatively slowly
    changing variable, so if we can estimate the time of 90/270 to a half hour
    or so we can derive a good estimate of LHA from AP Lat (LHA = tan d /tan L).
    With LHA and AP lon we can estimate GHA, from which we can derive estimated
    time of transit within a few minutes or better.
    With time of transit, we get a closer approximation of declination and GHA.
    We could use the direct computation of azimuth from the formula George
    posted a while back as a sanity check.
    "Tan Z = sin LHA / (cos LHA  sin lat - cos lat  tan dec)
    and the rules for putting Z into the right quadrant, 0 to 360, clockwise
    from North, are:
    --If tan Z was negative, add 180 deg to Z.
    --If hour-angle was less than 180 deg, add another 180 deg to Z"
    If the computed azimuth is within a degree or so of 90/270d, we are pretty
    much good to go as we have seen earlier. One could be off the actual time
    for a given AP by several minutes or more with little damage done unless AP
    latitude was horribly off.
    If off more than a degree or so, make an estimate of the change in time
    needed to move the azimuth to the desire angle and try again.
    > I think the intent is to observe until the predicted h occurs, and
    > record the resulting GMT.  From that and the matching LHA you compute
    > GHA and Longitude.  Right?  And you use the graphical method to
    > interpolate to hit the desired h exactly.
    You would have to ask Peter about that.  My impression is once he had the
    time of transit, he computed Hc.  Any difference between Hc or Ho was
    plotted east/west from AP for longitude.
    > From the modern perspective, a sight at this special time gives a LOP
    > like any other; its main advantage is that the LOP runs exactly north
    > and south if you're at the assumed latitude, so any error in the
    > latitude has only a tiny effect on the resulting longitude.
    Bill B
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