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    Re: Star - Star Observations
    From: Peter Hakel
    Date: 2010 Mar 9, 14:43 -0800
    Refraction increases the altitude of both stars, so they will appear closer to each other when compared to the true computed (nominal) distance.  If they are opposite in azimuth, I would get the observable distance by subtracting their refraction corrections from their nominal distance.

    Peter Hakel

    From: Brad Morris <bmorris@tactronics.com>
    To: NavList@fer3.com
    Sent: Tue, March 9, 2010 1:48:30 PM
    Subject: [NavList] Star - Star Observations

    [patrs deleted by PH]

    Here is the dilemma. When I get to larger angles, I need to go beyond my zenith. For example, I have been looking at Polaris vs Sirius. My latitude is about 40 degrees north. So Sirius is to my south, Polaris, naturally, is to my north. The nominal distance works out to about 106 degrees 20 odd minutes (forgive me, I don’t have the exact numbers in front of me). Because each object is on either side of my zenith, both objects will appear to be lower in the sky compared to the horizon, due to refraction. Yet because they oppose each other in azimuth, the observable distance between them should be larger by the sum of the refraction corrections, not reduced by the difference of the refraction corrections. That is, compute the true distance without refraction. Since each object is lowered by refraction, but in opposite directions, shouldn’t we add the refraction corrections to the nominal distance to obtain the observable distance?

    Best Regards

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