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    Re: NASR question on large F values
    From: David Iwancio
    Date: 2021 May 21, 09:59 -0700

    John:

    By definition, a body with a negative F is below your astronomical horizon, so the more obvious example would be if you're using the tables to determine a great circle route to a destination closer to your antipode than to you (i.e. further away from you than 10,800 nmi).. 

    As a recap, F = dec + B.  The magnitudes of both dec and B range from 0° to 90°.  If you're specifically looking for F < -90°, then both dec and B must be negative.  Declination is treated as negative if its name is contrary to your latitude.

    sin(B) = cos(lat) * cos(LHA).  Since lat is always treated as positive, B will only be negative if cos(LHA) is negative (hence the rule about 90° < B < 270°).  B's maximum negative magnitude occurs at cos(LHA) = -1, when LHA is 180°.  We therefore need to be looking at a body relatively near lower transit.  For such a body to be above your (visible) horizon it needs to have a relatively high declination, where LHA has less influence on altitude.

    The maximum contrary declination you can see over your horizon is the complement of your latitude.  To see a body with a large contrary declination, you need a small latitude.

    To sum up, you need:

    • A contrary dec
    • A high dec
    • A small lat

    An example would be taking Polaris sights from a few miles south of the equator, probably flying at altitudes where the air is thin enough to see such a faint star so low.  And your Polaris tables fell behind some shelves.

    Or you're looking for a great circle route connecting Spain and New Zealand.

       
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