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    Re: Eqn. of time
    From: Bill B
    Date: 2005 Jan 12, 16:53 -0500

    The spreadsheet appears to be working correctly.
    A few of questions:
    Are the resulting times at Greenwich?
    To calculate 2006 it appears one could:
    1. Continue the current spreadsheet, with 1 Jan, 2006 DT as 366
    2. Use a new Sun mean longitude and anomaly for Jan 1 2006 (which may, or
    may not be 280.7506 + 0.9856481*366 and 357.7244+0.9856003*366
    I have been unable to determine by inspection what the Sun's mean longitude
    actually means or how it is calculated for say, Jan. 1, 2005.  Can you shed
    some light on that for me please?
    3.  Same questions for the Sun's mean anomaly.
    > Bill wrote five days ago:
    > "Would like one year in 12 hour increments along the x axis and plus/minus  17
    > minutes with 6 second resolution on the y axis.  Basically 730 x  340
    > resolution.  That may be outside the range of Excel to chart, but  could
    > easily be broken into chunks Excel could chew on and the resulting  charts
    > assembled graphically."
    > Frank replied:
    > 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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