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> I hereby display the solutions to this "one-body fix" problem, if only as a 
curiosity. �I expect that these have been worked out before but I haven't 
found them anywhere so far (references would be much appreciated). �The known 
quantities are the GP (Dec, GHA), observed altitude (Ho), and azimuth to GP 
(Zn). �From the navigation triangle we can calculate the LHA and then the 
Latitude:
>
> sin LHA = sin Zn * cos Ho / cos Dec
>
> sin Latitude = ( sin Ho * sin Dec �- �cos Ho * cos Dec * cos LHA * cos Zn ) 
�/ �( 1 �- �cos Ho * cos Dec * sin LHA * sin Zn )
>
> Peter Hakel


I don't know of a reference for Peter's second equation above.  But, I
easily calculated a different one from the standard altitude and
azimuth equations:

Sin L = (cosA cosH sinH - sind cosd cosLHA)/(sind cosA cosH - cosd
sinH cosLHA).

Since both this and Peter's check out for the special case at LAN,
they're probably both OK.
--JK


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