Fwd: How to turn ecliptic longitude into solar declination?

2022-10-14 Thread Michael Ossipoff
-- 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
>>>
>>>
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Re: How to turn ecliptic longitude into solar declination?

2022-10-14 Thread Michael Ossipoff
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?

2022-10-14 Thread Steve Lelievre

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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