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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
> Browse Files

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