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Frank's reply brings up a question I've wondered about for a very long
time, ever since I first learned celestial:

If I've got two Ho's (and obviously know what bodies were observed, what
time the observations were taken, etc, etc), is there a direct solution
to obtaining a position that does NOT require a DR, calculating two Hcs
and Zns, and crossing the resulting LOPs?

As someone who works in the computer field and has at least a
rudimentary knowledge of numerical analysis, I can easily see that this
could be set up as an iterative solution (ie, guess at a position,
calculate what the observed body altitudes would be, compare to actuals,
use differences to get a direction to move the assumed L/Lo, repeat...).
  This is, in fact, not much different from the way your friendly GPS
calculates its position.

But there's a part of me that says some expert in spherical trig came up
with a way to cross two spherical triangles centuries if not millennia
ago...

Lu Abel

Frank Reed wrote:
> Greg:
> "But as a practical matter (assuming  that we've got a current almanac and
> can get a reasonably accurate time  reading), even if we had no idea at all
> about a DR position couldn't we at least  narrow down ..."
>
> As long as you don't mind waiting around for the stars  to come out (and
> given your assumptions of an almanac and a good clock set to  GMT or some known
> zone time), you can get your position "by eye" within two or  three degrees
> easily and up to ten times better with careful land-based  observations simply by
> looking at the zenith --no sextant required. If I look  straight up and see
> Vega, for example, within a degree of the zenith at 0600 GMT  on July 1
> (Greenwich Date!), then I can write down my position immediately. The  declination of
> the zenith is the observer's latitude. So for the case of Vega in  the zenith,
> my latitude must be close to 39 degrees North. The right ascension  of the
> zenith is the observer's sidereal time. Of course navigators don't use RA  and
> SidT anymore, so for modern almanac data, you use the equivalent rule: the  GHA
> of the zenith is the observer's longitude (west). For Vega in the zenith,
> the SHA is about 81 degrees and the GHA of Aries is about 9 degrees at 0600GMT
> on this date, so my longitude is 90 west.

   

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