Re: How to turn ecliptic longitude into solar declination?
Okay, that’s good to hear. …& thanks clearing it up. On Sun, Oct 16, 2022 at 3:54 PM Steve Lelievre < steve.lelievre.can...@gmail.com> wrote: > Michael, > > On 2022-10-16 1:40 p.m., Michael Ossipoff wrote: > > Thank you for mentioning that I answered Steve's question. > > ...something not acknowledged by Steve for some reason. > > > Please be assured that no slight was intended. Thank you for taking the > time to reply to my question. > > I did not acknowledge your response because I had not seen it. My email > software treated your messages as spam so I didn't see them until > Frank's message prompted me to check the junk folder. Just as soon as I > figure out the applicable setting, I'll change it. > > Steve > > > --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
Michael, On 2022-10-16 1:40 p.m., Michael Ossipoff wrote: Thank you for mentioning that I answered Steve's question. ...something not acknowledged by Steve for some reason. Please be assured that no slight was intended. Thank you for taking the time to reply to my question. I did not acknowledge your response because I had not seen it. My email software treated your messages as spam so I didn't see them until Frank's message prompted me to check the junk folder. Just as soon as I figure out the applicable setting, I'll change it. Steve --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
Frank-- Thank you for mentioning that I answered Steve's question. ...something not acknowledged by Steve for some reason. I didn't notice that when I first read your post. Thanks for setting the record straight ! So, to the list I just want to clarify that, when Steve asked how to determine declination from ecliptic-longitude, I was the first to answer his question, when I gave the following instruction: "Multiply the sine of the ecliptic-longitude by the sine of the obliquity, & then take the inverse-sine of the result." October 16th --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
[quote] At the moment we are in Vintagarious, the first month, and you will see that each day has the symbol for Aries. [/quote] Then you have an error, because Vendemiaire doesn't roughly approximate Aries. Vendemiaire roughly approximates Libra. As for the nature of the French Republican Calendar's rough approximation of the ecliptic-months, due to its piling up its excess 5 or 6 says all at the end of the year, IL amply covered that in earlier posts. The Indian National Calendar does a much better job, when it gives 31 days to Taurus thru Virgo. The Indian National Calendar isn't a fixed calendar. No blank days & no periodic-error-increase due to a leapweek. --- https://lists.uni-koeln.de/mailman/listinfo/sundial
Re: How to turn ecliptic longitude into solar declination?
Dear Steve, Michael, Werner and Fabio have provided some excellent responses to your question. If you are ONLY interested in relating three ANGLES - solar longitude, solar declination and the obliquity - then this relationship is indeed all you need: sin(lambda).sin(obliquity) = sin(declination) Importantly, the cusps (starts) of the 12 Zodiacal months are DEFINED as being at 30-degree intervals round the ecliptic. If you wish to use one-third months, then using 10=degree intervals is the right thing to do. [ASIDE: I deliberately overlook numerous little details such as 12x30 not being quite 360 degrees because the sun doesn't QUITE get back to where it started, but you can usually ignore that unless you want to discuss fine details with Werner :-)] As every diallist knows, equal angles do not generally translate into equal intervals of time. The sun doesn't travel round the ecliptic at a uniform speed. Fabio's ring illustrates this beautifully BUT Fabio's French Republican Calendar also tells you about this, not just on the inside front cover but also... By looking at the information provided for each day, you can do some simple investigation as to how the lengths of the Zodiacal months vary... At the moment we are in Vintagarious, the first month, and you will see that each day has the symbol for Aries. Next month is Fogarious and you will see that each day has the symbol for Scorpio. The following month is Frostaious and you will see that each day has the symbol for Sagittarius. This seems too good to be true. Seemingly the 30-day months are precisely in sync with the Zodiacal months. Sadly this isn't quite the case because... The following month is Snowous and only the first 29 days shows the symbol for Capricorn. We are at our closest to the sun here and actually cover more than one degree a day. You will see Day 30 shows the symbol for Aquarius. The Calendar is running one day slow! Here is a table of the crude end-of-month errors in days slow (-) and days fast (+): Vintagearious 0 Fogarious 0 Frostarious 0 Snowous-1 Rainous-1 Windous-1 Buddal -1 Floweral0 Meadowal +2 Reapidor +3 Heatidory +4 Fruitidor +5 I find it astonishing that it is only in the last four months that we end the month more than a day fast or slow. Of course we end the 12-month period five days early which is why the calendar ends with five (or six) Complementary Days. This is a nice simple way of showing that the angular rate of change in solar longitude doesn't vary very much! I find that the more I live with this Calendar the more it grows on me :-) You will find this too! Very best wishes Frank --- https://lists.uni-koeln.de/mailman/listinfo/sundial
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
Fwd: How to turn ecliptic longitude into solar declination?
-- 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?
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: > 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
How to turn ecliptic longitude into solar declination?
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
