Solunars by Aldebaran Sidereal Academy

Hello,

Can someone please tell me what the "starting point" for the mean north node of the Moon is? I have tried to look this up online, unsuccessfully.
I understand that this node moves retrograde at a constant (mean) velocity over its ~18.6 year cycle. What I can not figure out is how was the starting point determined? I ask this because if you compare the Mean north node to the True north node at times when the Moon is on the Ecliptic, they are not in the same position and often are over 1 degree apart.

So, since they are close by in position, what is the reasoning for the determined position of the Mean north node? It seems arbitrary, but I would like to know the reasoning/geometry behind it.

Thank you.

Good-morning, BlueKnight.

I'm not really clear on the question: I'll try to answer several points.'

First, we strongly recommend that if you're going to use Moon's node, you use the True Node. The mean (average) position of the True Node without all the calculations being applied. Using it is like using the mean (average) Sun by advancing Sun a little more than 0°59'/day regardless of its real speed. As you noted, the mean node can be quite a bit in error.

That being said, whichever you use - it is a point on the zodiac (two opposite points. Moon's nodes are the two opposing points on the zodiac where the Moon's orbit crosses the ecliptic (zodiac). This doesn't have a starting point any more than anything else in a horoscope has a starting point; e.g., where would you say "Mercury's starting point" is? It's the same sort of question. Like Mercury or Sun or Moon or whatever, Moon's nodes rotate the zodiac; except, like the vernal point, they move retrograde (the Mean Node is always retrograde, the true node loops between retrograde and occasionally direct).

BlueKnight22 wrote: Mon Nov 22, 2021 8:38 am So, since they are close by in position, what is the reasoning for the determined position of the Mean north node? It seems arbitrary, but I would like to know the reasoning/geometry behind it.

This would be easy if you understood spherical geometry as used in positional astronomy. I more or less gave the answer above, though: When one calculate the nodal positions, the mean position is one step along the way to calculating the true position. Mean means "average." These aren't two different things: It would help if you add the words "position of" in their names: The mean (average) position of Moon's north node is one stop along the way to calculating the true position of Moon's north node.

Do you have a web page that has the equations for the calculation of the Mean node?

Not handy. I used to do it by formula on a calculator, but haven't needed to do that in a long time (decades). In general, you advance it an entire circle in 18.599525 years, or 19.35533299909541° per sidereal year.

I understand why the mean rate is what it is but I don't understand how the position is oriented. For example, if I were to make my own version of a Moon's Mean Nodes ephemeris and I was going to orient it to the next time the Moon is on the ecliptic (December 3rd 2021) and then print out the ephemeris forwards and backwards in time using the fixed mean rate, I would have an ephemeris that was close to but not the same as the standard ephemeris for the True and Mean nodes.

I am trying to understand if my hypothetical ephemeris is any less valid than the traditionally used Mean node ephemeris?

BlueKnight22 wrote: Mon Nov 22, 2021 11:17 am I understand why the mean rate is what it is but I don't understand how the position is oriented. For example, if I were to make my own version of a Moon's Mean Nodes ephemeris and I was going to orient it to the next time the Moon is on the ecliptic (December 3rd 2021) and then print out the ephemeris forwards and backwards in time using the fixed mean rate, I would have an ephemeris that was close to but not the same as the standard ephemeris for the True and Mean nodes.

It would be roughly similar, but not matching either (except for the moment of the ecliptic cross).

I don't understand what you mean by the word "oriented" above.

I am trying to understand if my hypothetical ephemeris is any less valid than the traditionally used Mean node ephemeris?

Yes, it is, because it doesn't show either node's actual position during that time.

OK, thanks. If I do find info on how to calculate the mean node, I will post it here.