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George Huxtable suggests using Chuck's Venus and Moon altitudes
to locate his position, and shows how we could get an AP to
start with by drawing circles on a globe:

> For a first shot (as we have no idea yet where Chuck was observing
from)
> the geographical positions of Moon and Venus at that time could be
marked
> on a globe. Chuck's measured altitudes for these bodies should be
corrected
> in the ordinary way (ie corrections as for altitudes, not corrections
as
> for lunar distances), and then the corrected altitudes converted to
zenith
> distance by taking the 90-degree complement. Now draw circles of that
> radius about the two GPs on the globe. These intersect in two places,
so
> you have to guess which one is most likely. This paragraph could be
> bypassed if we had a rough idea of whereabouts Chuck lived, in
advance.

Instead of using this procedure, we could use the method of
"double altitudes" to find the latitude directly.  It involves
three spherical triangles, and repeated use of the formula
    cos a = cos b * cos c + cos A * sin b * sin c
where lower-case letters represent the length of a side, and
upper-case letters represent the angle opposite that side.
This is the same formula used in sight reduction, except that
altitude, latitude, and declination are each the complement of
a side in the navigational triangle, so sines and cosines of
the sides get interchanged.  The formula as I've presented
it directly gives the third side if we know two sides and
their included angle; a little algebra rearranges it to
give an angle if we know three sides:
    cos A = (cos a - cos b * cos c) / (sin b * sin c)

The first spherical triangle has vertices of the pole and the
two bodies (Venus and the Moon, in this case).  We know the
two declinations, which gives us two of the sides, and normally
know the difference in GHA, so we can use the formula to compute
the distance between the two bodies.  In our example we already
know the distance between the bodies, too, so we have all three
sides of this triangle.  We need to use the formula once more
to compute the azimuth angle from one body (say Venus) to the
other -- we'll use this for solving the third triangle later on.

The second spherical triangle has vertices of the two bodies
and the observer's zenith.  Two of its sides are the zenith
distances of the bodies (complements of the measured altitudes,
after correction for IE, dip, refraction, semi-diameter, and
parallax).  The third side is the distance between the bodies,
either computed from the first triangle, or using the cleared
distance from our lunar work.  We need to use the formula to
compute the angle of this triangle at the same body as we did
above.  So in our example, this is the angle Zenith-Venus-Moon.

The third spherical triangle has vertices of the pole, the
observer's zenith, and the chosen body (Venus).  Its sides are
the body's co-declination, the measured zenith distance
(corrected), and the observer's unknown latitude.  Further,
we know something about the angle of this triangle at the chosen
body.  Because we already know the angles Pole-Venus-Moon and
Zenith-Venus-Moon, a simple addition or subtraction will give
the angle Pole-Venus-Zenith.  (A diagram is probably needed
here, to decide whether to add or subtract.)  From this
angle and the other two sides, the formula gives us the third
side - our co-latitude.

This method of finding latitude by two observations can be
used without a chronometer, as long as we can get the declinations
of the bodies and their difference in GHA.  For two fixed stars,
you hardly even need a calendar for that; for the sun it helps
to know the approximate hour and the date.  If the two bodies
(or even the same one) are observed at different times from
the same location, we need a watch to determine the difference
in GHA, but it need not be set to any particular time, and need
not have particularly good long-term accuracy.

Once we have latitude, we can use either (or both) of the
observations as a "time sight".  This uses the same formula to
give us the LHA of the observed body, given latitude,
declination, and altitude.  Since in our case we know the
time, and thus GHA, our longitude follows directly.

Perhaps this is only of academic interest -- certainly the
method that finds an approximate position first lets us
use a more familiar method (and tables).  But I was
fascinated when I first read about it, with pictures, here:
http://www.northwestjournal.ca/dtnav.html and go to Article VI.

   

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