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    Re: Star-star distances for arc error
    From: Peter Hakel
    Date: 2009 Jun 19, 19:31 -0700
    Refraction shrinks all of the constellations under all observing conditions by lifting the stars towards the zenith. But for altitudes above 45 degrees, the "shrinking" is nearly uniform. That is, it is nearly proportional to the zenith distance (since the refraction is proportional to tan(z.d.)). And therefore all of the apparent distances between all of the stars are reduced by the same proportional amount. Thus to clear the distances between stars (when both stars are above 45 degrees altitude), all you have to do is multiply the measured observed distance by 1.00034. By my calculations, which somebody should check, the results never differ by more than 7 seconds of arc and usually less than 3 seconds of arc from the exact calculation which is pretty good for such a simple trick.

    ==========================================

    I did six test cases. The notation is:

    Ha1, Ha2: apparent altitudes (input in degrees)
    dm: measured distance (input in degrees)
    dc: cleared distance (output in degrees)
    diff = | 1.00034*dm - dc |  (output in arcseconds)

    For refraction I took the formula from the Nautical Almanac at the standard conditions of T = 10 C and P = 1010 mb.

    Results:

    Ha1 = 45
    Ha2 = 45
    dm = 90, most extreme case
    dc/dm = 1.0003704
    diff = 10.0

    Ha1 = 45
    Ha2 = 45
    dm = 40
    dc/dm = 1.0003033
    diff = 5.3

    Ha1 = 45
    Ha2 = 45
    dm = 1
    dc/dm = 1.0002909
    diff = 0.2

    Ha1 = 45
    Ha2 = 80
    dm = 35, stars on the same azimuth
    dc/dm = 1.0003810
    diff = 5.2

    Ha1 = 45
    Ha2 = 80
    dm = 55, stars on reciprocal azimuths
    dc/dm = 1.0003636
    diff = 4.7

    Ha1 = 50
    Ha2 = 60
    dm = 40
    gives dc/dm = 1.0003160
    diff = 5.0

    The diff's are within 7'' except for the first, most extreme case.  Neat trick!


    Peter Hakel



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