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    Re: Astronomical Refraction: Computational Method for All Zenith Angles
    From: Marcel Tschudin
    Date: 2005 Aug 17, 15:37 +0300

    Frank wrote
    > No source code, but there's an excellent short article by  Auer and
    > Standish
    > on this topic including sample calculations for various  temperatures and
    > pressures and even for negative altitudes as seen from 15,000  meters
    > observer
    > height (just the thing that got you started on this!) in the  Astronomical
    > Journal for May, 2000 [AJ,119:2472-2474,2000 May]. This is the
    > calculational
    > approach recommended in the 'Explanatory Supplement to the  Astronomical
    > Almanac',
    > so it doesn't get much better. Checking against the  tables in the
    > Nautical
    > Almanac, the refraction values generated by this  algorithm are an exact
    > match
    > except for differences of a couple of tenths of a  minute of arc very
    > close to
    > the horizon.
    >
    > You can get the article in .pdf  format from ADS  here:
    > http://adsabs.harvard.edu/advanced_fulltext_service.html .
    >
    > But  to save searching and for those list members who want the article but
    > don't get  along well with pdf files, I've made a link to .gif copies of
    > the
    > article's  three pages here:
    > http://www.HistoricalAtlas.com/lunars/ref.html  .
    
    Thank you, Frank, for indicating this paper. The content of it is at least
    partially not clear and raises further questions:
    
    Looking at table 2 with the results showing the two parameters hw and ho.
    According to the text they both indicate the height of the observer. But
    what does then mean e.g. the column with hw=0m and ho=2000m? A possible
    interpretation could be that the calculated refraction values are for the
    height of 2000m but normalised to sea level, this would also explain the
    values mentioned for Tw and Pw, which are standard sea level conditions. A
    comparison of the calculated refraction for a height of 2000m and an
    altitude of 0deg with the data of table 6 looks as follows:
    
    Table 2 of the paper:
    Refraction at 2000m = 1780.59 sec. (= 29.677 min.)
    
    Table 6 of the Pub. 249 with
    refraction values normalised to standard sea level conditions, from
    polynomial fits to the data:
    Refraction (height=1524m with T=15 deg C and P=1013.25hPa)  =  34.09 min.
    Refraction (height=3048m with T=15 deg C and P=1013.25hPa)  =  34.35 min.
    
    Table 6 of the Pub. 249 with
    refraction values at indicated height from polynomial fits to the data:
    Refraction (height=1524m with T=5.09 deg C and P=842.63hPa)  =  29.37 min.
    Refraction height=3048m with T=-4.81 deg C and P=696.24hPa)  =  25.21 min.
    
    This indicates that the refraction values shown are rather for the standard
    conditions at the indicated height and not for the standard conditions at
    sea level as mentioned in this table 2.
    
    An other point unclear to me is the input value for the refractive index.
    Table 1 mentions for alpha=0.0029241; this value corresponds to (n-1) at
    standard sea level conditions, therefore n=1.0029241. For air one would
    expect a value from some where around 1.0003. I assume that a zero got lost
    somewhere and it was mend n=1.0002941. Unfortunately the paper does not
    mention why this value was chosen or to what it corresponds to. According to
    the curve shown on the page from Andrew Young
    http://mintaka.sdsu.edu/GF/explain/optics/disp.html it would correspond to a
    wavelength at the red end of the visible spectrum and according to the
    values shown on page 17 of the script from Stanford
    http://www.stanford.edu/group/efmh/FAMbook/Chap10.pdf it would correspond to
    a value at the blue end.
    
    Marcel
    
    
    

       
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