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    Re: Lunar Distance Puzzle
    From: UNK
    Date: 2011 Aug 17, 04:31 +0000

    Dave,
    
    Frank is right when he says that in a universe without refraction, the 
    third distance does not add any information that isn't already contained 
    in the first two. Therefore you cannot leave refraction out of the game.
    
    With 0 refraction you only get a position line that looks like the 
    eclipse diagrams in the N.A. It's the exact same problem: There you have 
    a slowly moving line in space that connects the centers of the eclipsed 
    and eclipsing bodies. In your puzzle you have a slowly moving line 
    connecting the moon and that point on the celestial sphere that is 
    uniquely defined by its distances from two (!) stars. When you intersect 
    that line in space with the rotating earth, you get said position line.
    
    To obtain a unique position you need more information, which I 
    understood to be the refraction hidden in the three distorted distances. 
    Refraction tells us something about altitude, hence the rotation angle 
    of the earth, i.e. time. So it would appear that we may deduce the 
    rotation angle of the earth from the given data and find the unique 
    solution by intersecting the magic line in space with the surface of the 
    earth at a given time. This is how I understood your thought experiment.
    
    In reality, however, the refraction model that we use sets a limit to 
    how the measured refractions can be exploited. The function for 
    refraction cannot be inverted to the degree of accuracy that is required 
    here. Certainly the simple formula in the N.A. is not designed to do 
    this job. That's why I responded by saying that your problem cannot be 
    solved. My comment has been misconstrued as a reference to a practical 
    limitation and incorrectly quoted as "the method is sensitive to 
    refraction", as if refraction were detrimental to finding the exact 
    solution. That's not what I said. On the contrary, refraction is 
    essential to the solution of your puzzle. It depends on it, and that's 
    why it's not feasible.
    
    It is only now that I am really puzzled because not only do you not seem 
    to think that you need refraction, you even found a solution in a 0 
    refraction model. Would you please show the details?
    
    Herbert
    
    P.S.
    
    When trying to verify your topocentric distances with MICA, I get 
    discrepancies of up to 1.5" between yours and mine. It could be the 
    underlying ephemeris, could be delta T. Normally, I would not care, but 
    in this case it seems important. What is your geoc. and topoc. moon 
    position at  +3h 27m 0.4s?
    
    
    
    
    On 2011-08-16 01:31, Dave Walden wrote:
    >
    > The Lunar distances for a "no refraction" case:
    >
    >
    > +28� 0' 58.3"
    > +37� 5' 33.8"
    > +61� 36' 0.8"
    >
    > My solution
    >
    > +3h 27m 0.4s
    >
    > +38� 55' 39.2"N
    > +77� 12' 45.3"W
    >
    > Still a pretty good solution. It's the parallax, not the refraction.
    >
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