Re: How to turn ecliptic longitude into solar declination?
My thanks go Werner for his detailed and helpful response to my question, and Fabio for his interesting comments on the astrolabe. I learned some new things today, and it was nice to see a diagram of the offset circles on the back of the astrolabe. Clever. Cheers, Steve --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
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 > 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 > > > > > > --- > https://lists.uni-koeln.de/mailman/listinfo/sundial > --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
I emphasize that saying that each third of an ecliptic-month is 10 degrees is not an approximation. An ecliptic-month is defined as exactly 1/3 of an astronomical- quarter…1/3 by ecliptic-longitude, not by time or days. An astronomical-quarter is the ecliptic interval between a solstice & an equinox…90 degrees along the ecliptic. On Fri, Oct 14, 2022 at 11:33 PM Michael Ossipoff wrote: > BTW, I like sundials that tell the ecliptic-months, Aries thru Pisces. > > …for which one would need the Solar declinations for the beginning of each > ecliptic-month, & preferably also for some fractions of each > ecliptic-month, such as 1/3 & 2/3. > > On Fri, Oct 14, 2022 at 10:16 PM Michael Ossipoff > wrote: > >> >> >> -- Forwarded message - >> From: Michael Ossipoff >> Date: Fri, Oct 14, 2022 at 10:16 PM >> Subject: Re: How to turn ecliptic longitude into solar declination? >> To: Steve Lelievre >> >> >> >> >> Or you could just use the ecliptic longitude, reckoned as usual from the >> Vernal Equinox…multiply its sine by the sine of the obliquely & take the >> inverse sine of the result. >> >> I’d suggested that other way because there are some spherical >> trigonometry formulas that require an argument between 0 & 90 degrees. >> >> …but that isn’t one of them. >> >>> >>> >>> On Fri, Oct 14, 2022 at 6:49 PM Michael Ossipoff >>> wrote: >>> Multiply the sine of ecliptic longitude (reckoned forward or backwards from the nearest equinox) by the sine of 23.438 or whatever the current obliquity’s exact value is). Take the inverse sine of the result. On Fri, Oct 14, 2022 at 4:57 PM Steve Lelievre < steve.lelievre.can...@gmail.com> wrote: > >>> Of course you’ll know when the declination is negative or positive, so >>> mark it accordingly. >>> >>> >>> >>> 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 > > > > > > --- > https://lists.uni-koeln.de/mailman/listinfo/sundial > > --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
BTW, I like sundials that tell the ecliptic-months, Aries thru Pisces. …for which one would need the Solar declinations for the beginning of each ecliptic-month, & preferably also for some fractions of each ecliptic-month, such as 1/3 & 2/3. On Fri, Oct 14, 2022 at 10:16 PM Michael Ossipoff wrote: > > > -- Forwarded message - > From: Michael Ossipoff > Date: Fri, Oct 14, 2022 at 10:16 PM > Subject: Re: How to turn ecliptic longitude into solar declination? > To: Steve Lelievre > > > > > Or you could just use the ecliptic longitude, reckoned as usual from the > Vernal Equinox…multiply its sine by the sine of the obliquely & take the > inverse sine of the result. > > I’d suggested that other way because there are some spherical trigonometry > formulas that require an argument between 0 & 90 degrees. > > …but that isn’t one of them. > >> >> >> On Fri, Oct 14, 2022 at 6:49 PM Michael Ossipoff >> wrote: >> >>> Multiply the sine of ecliptic longitude (reckoned forward or backwards >>> from the nearest equinox) by the sine of 23.438 or whatever the current >>> obliquity’s exact value is). >>> >>> Take the inverse sine of the result. >>> >>> On Fri, Oct 14, 2022 at 4:57 PM Steve Lelievre < >>> steve.lelievre.can...@gmail.com> wrote: >>> >> Of course you’ll know when the declination is negative or positive, so >> mark it accordingly. >> >> >> >> 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 --- https://lists.uni-koeln.de/mailman/listinfo/sundial --- https://lists.uni-koeln.de/mailman/listinfo/sundial
