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    Re: Eqn. of time
    From: Bill B
    Date: 2005 Jan 10, 23:47 -0500

    Thank you Frank, after a few passes it worked like a charm.
    
    A lot of numbers to chew on, even with a computer.  Was this all done by
    hand with a crow-quill pen and candles or lanterns light days gone by?  Wow.
    
    Bill
    
    > From the sound of it, the best approach would be to calculate EqT at each
    > date and do whatever graphing you find useful. Here's what you do:
    >
    > 1) calculate the number of days elapsed since Jan. 1, 2005 0h UT. Typical
    > date functions in spreadsheets do this with a simple formula like
    > "=date(yy,mm,dd)-date(2005,1,1)". Let's call that DT.
    > 2) calculate the Sun's mean longitude "=280.7506+0.9856481*DT". Call that  L.
    > 3) calculate the Sun's mean anomaly "=357.7244+0.9856003*DT". Call that  M.
    > 4) if required by your spreadsheet software, divide these two angles by the
    > number of degrees in a unit angle (one radian). That is, divide M and  L by
    > 57.29578. This operation can be folded into the above steps.
    > 5) calculate the EqT in seconds of time using
    > "=591.7*sin(2*L)-459.5*sin(M)+39.5*sin(M)*cos(2*L)-12.7*sin(4*L)-4.8*sin(2*M).
    > 6) verify that everything is working right. For 0h UT on Jan. 10, 2005, you
    > should get -444.14 seconds.
    >
    > You could set this up in a spreadsheet with DT as a simple running count
    > from 0 to 364 in the A column and L, M, etc. and EqT in succeeding columns.
    > Graphing the results would then be simple.
    >
    >
    > Some details on this calculation can be found in Smart's "Spherical
    > Astronomy". It's been quoted frequently in other "calculation  cookbooks". It
    > gives
    > values of EqT to within about two seconds  for a couple of decades around the
    > present date (including leapyears). If you  ever decide to go beyond that,
    > you'll need to adjust the coefficients in step 5  and calculate L and M more
    > carefully but this should serve your practical needs,  I think.
    >
    > -FER
    > 42.0N  87.7W, or 41.4N  72.1W.
    > www.HistoricalAtlas.com/lunars
    >
    
    
    

       
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