Dear Steven,

The relation of solar declination delta(t)  to ecliptic longitude lambda(t) 
delta(t) = ArcSin[Sin[23.44]*Sin[lambda[t]]

You are interested in the relation of solar declination to time since the 
equinox.

Your formula delta(t) = 23.44*Sin(t), with t being the time (in degrees) since 
the spring equinox, is mathematically the 'first order Taylor expansion in  
obliquity phi’ of the precise expression for the solar declination.
The difference of your formula to the accurate expression is around 0.9 degrees 
maximum.

The 'first order Taylor expansion in eccentricity ecc’ is
delta(t) = ArcSin[Sin[phi]*Sin[t]] + ecc * Sin[phi] * Sin[2*t] / Sqrt[1 - 
Sin[phi]^2*Sin[t]^2] 
which is accurate to 0.015 degrees. (phi=23.44 degrees)

It is a much better approximation because for a Taylor approximation the 
argument should be much smaller than unity. 
phi=2*Pi/360*23.44 = 0.41 is not so small, but ecc=0.0167 is very small ...

You could still approximate the above expression to 
delta(t) = ArcSin[Sin[phi]*Sin[t]] + ecc * Sin[phi] * Sin[2*t] 
which is accurate to 0.03 degrees, and it does take into account the 
eccentricity of the orbit.

Other approximations can quickly become more complicated than using directly 
the correct formulas.

The numbers are still in my head because I recently  discussed this point in 
the NASS293 article on the analemma (Eq. 19, Eq. 20).

cheers
Werner



> On 15 Oct 2022, at 01:56, Steve Lelievre <steve.lelievre.can...@gmail.com> 
> wrote:
> 
> Hi,
> 
> For a little project I did today, I needed the day's solar declination for 
> the start, one third gone, and two-thirds gone, of each zodiacal month (i.e. 
> approximately the 1st, 11th and 21st days of the zodiacal months).
> 
> I treated each of the required dates as a multiple of 10 degrees of ecliptic 
> longitude, took the sine and multiplied it by 23.44 (for solstitial solar 
> declination). At first glance, the calculation seems to have produced results 
> that are adequate for my purposes, but I've got a suspicion that it's not 
> quite right (because Earth's orbit is an ellipse, velocity varies, etc.)
> 
> My questions: How good or bad was my approximation? Is there a better 
> approximation/empirical formula, short of doing a complex calculation?
> 
> Cheers,
> 
> Steve
> 
> 
> 
> 
> 
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