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    Re: Delta T long term formula
    From: Paul Hirose
    Date: 2018 Feb 14, 11:17 -0800

    On 2018-02-05 7:38, Antoine Couëtte wrote:
    > This is absolutely EXCELLENT ! , including your own "best fit" outside the [-720 - 2016] time frame.
    
    Thank you, but I didn't do anything except integrate the lod (length of
    day) formula from the paper by Stephenson et al:
    
    lod (ms) = 1.78 * t - 4 * sin(2 * π * t / 15)
    
    where t is centuries after 1825.0 and the sine argument is radians. For
    example, lod at 2018.0 is .54 ms per the formula. More precisely, that's
    the excess length of day: the amount the UT1 day exceeds 86 400 SI seconds.
    
    To predict ∆T it's more convenient if t is years after 1825 and the
    result is excess length of year (seconds). Then the formula is:
    
    loy (s) = .006501 * t - 1.461 * sin(.004189 * t)
    
    Its integral is
    
    .00325 * t^2 + 349 * cos(.00419 * t)
    
    To that I add -293.600 to make a smooth connection to the last ∆T
    polynomial when it ends at 2016.0. (However, there's an abrupt change in
    the first derivative.) So, for an epoch after 2016.0:
    
    ∆T = -293.600 + .00325 * t^2 + 349 * cos(.00419 * t)
    
    The t^2 coefficient is identical to the parabolic formula (∆T = -320 +
    32.5 * t^2) from the paper, if you remember t in the latter is centuries
    instead of years.
    
    In other words, the above formulas for lod, loy, and ∆T all imply the
    same constant deceleration in Earth's rotation, plus the 1500 year
    sinusoidal lod oscillation detected by Stephenson et al. The exception
    is the parabolic formula, which omits the oscillation.
    
    Even if you include the 1500 year oscillation, the result is only a long
    term average. For instance, at 1975 the loy formula predicts .32 s
    excess length of year, which implies a leap second every 3 years. In
    reality there was about one per year the 1970s.
    

       
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