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Dear Herbert (and Dave),

Thank you for the excellent "counter example" (below) and the idea of thinking
about the three lunar distances in pairs. Mea Culpa! If there is only parallax
of the moon at work and the three stars are fixed it is clear that the third lunar
distance equation contribues nothing to the problem. We are talking about
measured distances from fixed points on a two-dim surface, and two of these
determine an intersection (the moon's apparent position) uniquely without
any need of the third. And this is independent of GMT.

So as you have said all along the only hope that the three lunar distance
equations (in three unknowns) are independent and solvable is to include
refraction in the problem. Even though refraction is a small correction
(usually) with respect to parallax for the moon, it also perturbs the positions
of the three stars in a GMT dependent fashion. Presumably this means
that adding the third lunar distance can lead to a unique solution for
lat, long, and GMT. But I certainly can't prove it...

Cheers,
George

>On 2011-08-21, Herbert P wrote:
>
>Amy, Bob and Jim have together 15 candies. Amy and Bob together have 9.
>Can you tell me how many candies each child has? Oops, you need a third
>equation. Jim has 6. Can you tell me now?
>
>Herbert

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