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
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
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
[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
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) =
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
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)
erably 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
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 lo
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>
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