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    Re: NASR question on large F values
    From: David Iwancio
    Date: 2021 Jun 3, 02:30 -0700

    John:

    First off, I made a mistake earlier in describing the formula for B (I think I was confusing it with A).  The real math is:

    tan(B) = cos(LHA) / tan(lat)

    With that out of the way, you were talking earlier about finding an F value of around -91°.  Considering the sun, the greatest dec value is going to be about 20°

    F = B + dec

    To get F to be negative we once again consider a contrary dec.

    -91° = B - 20°

    So our target B is -71°.

    With B between 0° and -90°, tan(B) must also be negative.  Since tan(lat) will always be positive (with lat always between 0° and +90°), the sign of B is dictated by cos(LHA).  This is where the rule for the sign of B being negative for LHA's between 90° and 270° comes from.

    To increase the chances of a body being at least near the horizon, an LHA at or near 91° (or 269°) is probably the best bet.  This is near the bottom of the columns in the tables 

    Glancing through the tables, for latitude 1°, B = -71°33' at LHA = 93°.  Setting dec = -20° and rounding to the nearest degree, this yields F = -92° and A = 87°.  H comes out to -3°, which puts the sun well below the horizon.

    On the equator, if we arbitrarily declare dec to be negative, with LHA = 91° the table gives an F value of -110° and altitude of -56'.  This is a possibility if your bridge wing is high enough above the waterline. 

       
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