<p>Attempting to answer my own question, here's what seems to be happening:</p>
<p>Since a Saros is slightly shorter than 18.6 years (the lunar nodes period of rotation), after 1 Saros cycle the nodes will be short a few degrees of a full 360° rotation relative to the stars :</p>
<pre><code>360° - 360° * 6585.33 / 6793 = 11° short
</code></pre>
<p>Since 1 Saros is not an integer number of years, the Sun won't be in the same position either relative to the stars :</p>
<pre><code>with 1 sidereal year = 365.256 days
1 Saros = 6585.33 days = 18 * 365.256 + 10.722 days
10.722 days corresponds to about:
360° * 10.722 / 365.256 = 10.57° rotation
</code></pre>
<p>So 1 Saros later the Sun will be 10.57° further away on the ecliptic.
As seen from Earth the Sun and nodes move in opposite directions however, so this almost cancels out, and the Sun-Nodes alignment is almost exactly preserved :</p>
<pre><code>11° - 10.57° = 0.43° alignment difference
</code></pre>
<p>The Sun must be within about 17° of the nodes for an eclipse to be possible, so if it was close to the nodes it will take many Saros cycles for the alignment to be broken.</p>

CC BY-SA 4.0
2024-11-14T11:56:36.473Z
2024-11-14T11:56:36.473Z
lemonsqueeze
<p>I'm trying to understand how the Earth, Moon, Sun and stars line up after one Saros cycle is over.</p>
<p>According to this <a href="https://eclipsewise.com/solar/SEhelp/SEperiodicity.html" rel="nofollow noreferrer">Article</a> a Saros cycle is a period of 223 synodic months after which the Sun, Earth, and Moon return to approximately the same geometry. &quot;Any two eclipses separated by one Saros cycle share similar characteristics : they occur at the same node with the Moon at nearly the same distance from Earth and at the same time of year&quot;.</p>
<p>223 synodic months is 6585.32 days, or about 18 years, 11 days and 8h.
So ignoring the 8h shift, the earth, moon and sun will be in the same relative geometry, and almost same geometry relative to the stars (same time of year + ~11 days).</p>
<p>The rate of lunar <a href="https://en.wikipedia.org/wiki/Lunar_precession" rel="nofollow noreferrer">Nodal Precession</a> however is about 6793 days, or 18.6 years.
18.6 years don't fit into a Saros nicely. What i don't understand is how the lunar nodes can be back to where they were after one Saros cycle is over if their rotation period is greater than one Saros.</p>
<p>To clarify, i'm aware of the structure of a Saros cycle and what makes it work for eclipse predictions. What is asked here is how to resolve the apparent contradiction with the Nodal Precession cycle, whose period seems to imply that one Saros after an eclipse there won't be another eclipse because the nodes will be out of alignment.</p>

CC BY-SA 4.0
2024-11-07T10:03:06.123Z
From https://astronomy.stackexchange.com/a/25756/16685 "The Saros cycle of 18.03 years is 223 synodic months, 242 draconic months, or 239 anomalistic months (27.55 days between lunar perigees or apogees). This multiple coincidence of phase, node, anomaly, and time of year makes the saros especially useful for prediction".
PM 2Ring
CC BY-SA 4.0
2024-11-07T10:09:33.913Z
Also see https://math.stackexchange.com/q/4412720/207316
PM 2Ring
CC BY-SA 4.0
2024-11-07T10:22:26.750Z
The linked question isn't an exact duplicate, but I *think* it covers your question. Yes, a Saros cycle is ~6 months short of a nodal precession cycle, but because it's (almost) an integer number of draconic months, the Moon will be at the node, and similarly, the Moon's position relative to its perigee is the same because it's (almost) an integer number of anomalistic months, so the angular size of the Moon is also the same.
PM 2Ring
CC BY-SA 4.0
2024-11-07T13:10:49.440Z
This question is similar to: [Spacing (2422 days) between solar eclipses - unrelated saros numbers](https://astronomy.stackexchange.com/questions/25749/spacing-2422-days-between-solar-eclipses-unrelated-saros-numbers). If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.
Starship
CC BY-SA 4.0
2024-11-08T13:50:18.310Z
A Saros cycle corresponds to an integer number of returns of the moon to the same node. A nodal cycle corresponds to the nodes return to the same sidereal position. In order for the two to be equivalent the Saros would have to also correspond to an integer number of sidereal months. While almost true: 223 months $\approx$ 242.03 months, it is not close enough.
Ryan Cole
CC BY-SA 4.0
2024-11-09T19:36:09.543Z
Thanks for the comments, added clarifications to the question. As far as i can tell none of the Saros related questions already posted address this topic.
lemonsqueeze
CC BY-SA 4.0
2024-11-12T00:10:47.260Z
*"one Saros after an eclipse there won't be another eclipse because the nodes will be out of alignment."* The Moon returns to a node each draconic month, which has a mean length of   27.212220815 days (J2000). The true month lengths vary because the Moon's motion is complex. I have details (primarily in reference to the synodic month) in https://astronomy.stackexchange.com/a/55112/16685
PM 2Ring
<the-moon><precession><eclipse>
Eclipse Saros Cycle vs Nodal Precession Cycle
lemonsqueeze
# Eclipse Saros Cycle vs Nodal Precession Cycle

(Asked by: username_0 on 2024-11-07)

I'm trying to understand how the Earth, Moon, Sun and stars line up after one Saros cycle is over.

According to this [Article](https://eclipsewise.com/solar/SEhelp/SEperiodicity.html) a Saros cycle is a period of 223 synodic months after which the Sun, Earth, and Moon return to approximately the same geometry. "Any two eclipses separated by one Saros cycle share similar characteristics : they occur at the same node with the Moon at nearly the same distance from Earth and at the same time of year".

223 synodic months is 6585.32 days, or about 18 years, 11 days and 8h.
So ignoring the 8h shift, the earth, moon and sun will be in the same relative geometry, and almost same geometry relative to the stars (same time of year + ~11 days).

The rate of lunar [Nodal Precession](https://en.wikipedia.org/wiki/Lunar_precession) however is about 6793 days, or 18.6 years.
18.6 years don't fit into a Saros nicely. What i don't understand is how the lunar nodes can be back to where they were after one Saros cycle is over if their rotation period is greater than one Saros.

To clarify, i'm aware of the structure of a Saros cycle and what makes it work for eclipse predictions. What is asked here is how to resolve the apparent contradiction with the Nodal Precession cycle, whose period seems to imply that one Saros after an eclipse there won't be another eclipse because the nodes will be out of alignment.

**Tags:** the-moon, precession, eclipse
> **Comments:**
> **username_1**: From https://astronomy.stackexchange.com/a/25756/16685 "The Saros cycle of 18.03 years is 223 synodic months, 242 draconic months, or 239 anomalistic months (27.55 days between lunar perigees or apogees). This multiple coincidence of phase, node, anomaly, and time of year makes the saros especially useful for prediction".  
> **username_1**: Also see https://math.stackexchange.com/q/4412720/207316  
> **username_1**: The linked question isn't an exact duplicate, but I *think* it covers your question. Yes, a Saros cycle is ~6 months short of a nodal precession cycle, but because it's (almost) an integer number of draconic months, the Moon will be at the node, and similarly, the Moon's position relative to its perigee is the same because it's (almost) an integer number of anomalistic months, so the angular size of the Moon is also the same.  
> **username_2**: This question is similar to: [Spacing (2422 days) between solar eclipses - unrelated saros numbers](https://astronomy.stackexchange.com/questions/25749/spacing-2422-days-between-solar-eclipses-unrelated-saros-numbers). If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem.  
> **username_3**: A Saros cycle corresponds to an integer number of returns of the moon to the same node. A nodal cycle corresponds to the nodes return to the same sidereal position. In order for the two to be equivalent the Saros would have to also correspond to an integer number of sidereal months. While almost true: 223 months $\approx$ 242.03 months, it is not close enough.  
> **username_0**: Thanks for the comments, added clarifications to the question. As far as i can tell none of the Saros related questions already posted address this topic.  
> **username_1**: *"one Saros after an eclipse there won't be another eclipse because the nodes will be out of alignment."* The Moon returns to a node each draconic month, which has a mean length of   27.212220815 days (J2000). The true month lengths vary because the Moon's motion is complex. I have details (primarily in reference to the synodic month) in https://astronomy.stackexchange.com/a/55112/16685  

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## 1 answers

### Answer by username_0 on 2024-11-14. Score: 2

Attempting to answer my own question, here's what seems to be happening:
Since a Saros is slightly shorter than 18.6 years (the lunar nodes period of rotation), after 1 Saros cycle the nodes will be short a few degrees of a full 360° rotation relative to the stars :
360° - 360° * 6585.33 / 6793 = 11° short

Since 1 Saros is not an integer number of years, the Sun won't be in the same position either relative to the stars :
with 1 sidereal year = 365.256 days
1 Saros = 6585.33 days = 18 * 365.256 + 10.722 days
10.722 days corresponds to about:
360° * 10.722 / 365.256 = 10.57° rotation

So 1 Saros later the Sun will be 10.57° further away on the ecliptic.
As seen from Earth the Sun and nodes move in opposite directions however, so this almost cancels out, and the Sun-Nodes alignment is almost exactly preserved :
11° - 10.57° = 0.43° alignment difference

The Sun must be within about 17° of the nodes for an eclipse to be possible, so if it was close to the nodes it will take many Saros cycles for the alignment to be broken.


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