Way back in Book V we determined the angular diameter of the moon as well as earth’s shadow at apogee. In the last post, we repeated the procedure for perigee. In the Almagest, Ptolemy doesn’t actually say what those calculations are for and instead, starts working out some figures for the sun. However, to try to keep things in a more reasonable flow (in my opinion), I’m going to skip to the end of this chapter and discuss why we care about the moon’s diameter and earth’s shadow.
In short, lunar eclipses can only happen near the lunar nodes. But, it doesn’t have to be exactly at a node. First off, the earth’s shadow has some width to it. In addition, the anomalies of the sun and moon play a role, which means the actual range the eclipse could occur in is surprisingly wide. So in this post, we’ll work on that.
Before jumping in, let’s make a handy reference guide for ourselves from the figures in the previous two posts:
| Apogee | Perigee | |
| Radius of Moon | $0;15,40º$ | $0;17,40º$ |
| Radius of Earth’s Shadow | $0;40,40º$ | $0;45,46º$1 |
Now, let’s consider the case where an eclipse just misses and the edge of the moon just touches the earth’s shadow. Let’s sketch that out2.
I’ve sketched this out with the moon just past its ascending node and just touching the earth’s shadow. When this happens, the line from the center of the earth’s shadow to the center of the moon forms a right angle at the moon’s circle. In that case, the distance between their two centers is the sum of the radii of the moon and earth’s shadow. At perigee, that means $0;17,40º + 0;45,56º = 1;03,36º$.
Ptolemy now wants to ask what the distance of the moon is in this case since, if the moon and the shadow were any closer together (i.e., more rightward) then there would be an eclipse. Knowing this tells us how far in ecliptic longitude the moon can be from the node and still have an eclipse occur.
This can be done by using the demi-degrees method on the right triangle above, which is estimating it as a planar triangle and probably not the best solution given we’re talking fairly large portions of the moon’s path here. However, Ptolemy doesn’t show any math, but comes up with $12;12º$.
Ptolemy doesn’t show any math on how he arrived at this value.
From this, Ptolemy concludes that anything less than $12;12º$ from a node in ecliptic longitude, to either side, would produce some sort of eclipse.
However, this is the position of the true moon. If you recall, what we’re really after in this Book is to be able to quickly determine based on the position of the mean moon if an eclipse might be possible. Ptolemy doesn’t explain exactly how he did this conversion and only hints that it is similar to the process we discussed in this post. There, we used the mean conjunction to estimate how long until true conjunction. To do so, we’d determined how far apart the moon and sun were based on their anomalies and then estimated their approximate speeds to figure out how long it would take the close that gap. Here, we’ll need to go the other way around.
First, let’s figure out what gap they need to close. At worst, they’d have to close the maximum amount of their anomaly. If we glance at our solar and lunar tables of anomaly, we can see that the sun’s maximum anomaly is $2;23º$ while the moon’s is $5;01º$.
The question is how long that would take and how much did their position along the ecliptic change in that time?
Again, Ptolemy doesn’t worry about the instantaneous motion and instead just substitutes the mean motion as the extra motion caused by the objects changing their anomaly is likely to be small over this period. Here’s a sketch that will hopefully make that make more sense.
Here I’ve drawn the system at two points in time3. The top line shows the mean conjunction, as shown by the mean sun and moon (the ones with a bar over their symbols) both being at the same point on the ecliptic. The true sun is at its greatest anomaly to the left (east), of $2;23º$. Meanwhile the moon is at its greatest anomaly to the right (west) of $5;01º$.
Then some time passes until it’s the true conjunction. The mean and true moon both move leftwards (eastwards), keeping their same spacing of $5;01º$ relative to one another. While they do that, the mean and true sun are also moving leftwards, but more slowly, also keeping their spacing of $2;23º$. This happens until the true sun and true moon are at the same point on the ecliptic. This is now true conjunction.
What we want to find is the distance between mean and true conjunction, which I’ve labeled as $x$4.
First, Ptolemy states that
in the time it takes the moon to traverse [the distance between the true moon and the true sun, $7;24º$, the sun will traverse an extra distance of about $\frac{1}{13}$ of that amount, i.e., $0;34º$.
Where did fraction of $\frac{1}{13}$ come from? This is the ratio of speeds of the sun and moon which Ptolemy called $1:13$ in this post5. So what Ptolemy is saying is that, initially, the moon only has to catch up by $7;24º$. However, it takes some time for it to do that, and since the sun is moving at $\frac{1}{13}$ of the moon’s speed, it must have moved an extra $0;34º$ in that time which the moon now needs to make up.
But in that time, the sun will move again and
while the moon is traversing that extra $0;34º$, the sun will again traverse an extra $\frac{1}{13}$ of that, or $0;03º$.
We could continue repeating this process6, but it quickly becomes diminishing returns, so Ptolemy cuts it off there stating
a $\frac{1}{13}^{th}$ of the latter ]$0;03º$ is negligible.
So all in all, the moon travels an extra $0;34º + 0;03º = 0;37º$ before the extra distance becomes negligible. Ptolemy notes that doing this is the same as taking $\frac{1}{12}$ of our original distance7 8.
So, the total distance between the mean conjunction and the true conjunction is $2;23º$, the sun’s anomaly, plus this additional motion of $0;37º$. Adding those together we find that the distance between the mean and true conjunction is $3;00º$.
Thus, looking at the extreme case of where an eclipse would just occur, we showed above that it can happen for a $12;12º$ away from a node for a true conjunction, but then the mean conjunction can be an addition $3º$ away from that. So when we’re quickly scanning through possible eclipse dates, we can immediately omit any where the mean conjunction is more than $15;12º$ from a node along the moon’s circle. Since the nodes are at $90º$ and $270º$ from the northern limit, that means we should be doing more thorough checking for eclipses if the distance from the northern limit of the moon is between $74;48º$ and $105;12º$ for the descending node and $254;48º$ to $285;12º$ for the ascending node.
That’s the end of the post as far as the Almagest is concerned. However, this post was very timely because I saw a real world application of this exact material recently. If you’re not familiar with the Clickspring YouTube channel, I highly recommend it. Their main project that I’ve been following for a few years now is the reproduction he’s working on for the Antikythera mechanism. However, there’s another project in which he rebuilds a similar geared device known as the Byzantine sundial9. One of the features of this device is that it can be used for eclipse prediction and this window shows up on one of the gears:
This is one of the wheels that goes under the base and on it, you can see two white pieces which represent the windows in which eclipses can occur! And in one of the Patreon videos, you can see what angle he used for this.
That’s a slight bit higher than the $30;24º$ window we just derived, but you certainly won’t miss any eclipses that way!
UPDATE 11/16/2021: I asked Chris why he used a $34º$ window via his Patreon and here was the response:
Hello mate, that came from a NASA/JPL discussion where they put the average eclipse season at 34 degrees (when considering all factors). The eclipse prediction feature is an additional capacity not present in the original device, hence the use of the modern figure – Cheers 🙂
UPDATE 1/5/2026: After some consideration, I realized that the diagram at the beginning of this post was incorrect. It was originally drawn as follows:
This had the right angle along the ecliptic. However, while the moon is just touching the earth’s shadow at that moment in time, you can see that, if we rewind the moon towards the node, it will actually enter earth’s shadow. Thus, this did not truly represent the instance in which the moon had a near miss of an eclipse. Rather, that happens when the right angle is at along the moon’s path as the corrected illustration above demonstrates.
This should have been clear to me as Ptolemy (towards the end of this chapter) states that we’re measuring the distance between the moon’s center and the center of the earth’s shadow
along the great circle drawn through the poles of the moon’s inclined orbit).
As I had originally drawn it, this was drawn along a great circle through the poles of the ecliptic.
I’m fairly certain my confusion came from consulting Neugebauer’s History of Ancient Mathematical Astronomy for help understanding this portion. There, Neugebauer draws the figure as I originally did, having the distance between these centers as the ecliptic latitude (See page $125$ and Fig $109$ in his work). Ironically, Pedersen largely skips this chapter in his Survey of the Almagest, only covering it in a highly conceptual manner, but manages to draw the diagram correctly.
Regardless, we can ask how serious of an error this is and the answer is that it is fairly small. Ptolemy doesn’t explain how he arrived at a value of $12;12º$ along the moon’s inclined circle, but if we use some trig on this right triangle, we would find that the hypotenuse (whether it be along the moon’s circle or the ecliptic) should be $12;09,44º$ and the adjacent side (which would be the other circle) would then be $12;06,57º$. Obviously, neither matches Ptolemy’s value, but they’re only a few minutes of arc different.
And because of this, the “correct” diagram is not critically important for the discussion coming up in Chapter 6. Indeed, Ptolemy’s own discussion in that chapter will treat these circles as if they’re of the same length.
This imprecise treatment is a great source of confusion for this Book as evidenced by the fact that I’m coming back to it over four years later.
- This value seems to come out of nowhere. After all, we previously determined this value to be $0;45,50º$. Ptolemy then called that approximately $2 \frac{3}{5}$ times the moon’s radius. The value he presents here is exactly $2 \frac{3}{5}$ times the radius of the moon.
- I originally drew this image incorrectly. See my 1/5/2026 update at the bottom of this post for information on this which I believes foreshadows some of the sources of confusion readers are likely to have with Book VI. Furthermore, if you’re making use of Neugebauer’s History of Ancient Mathematical Astronomy, my error was a result of an error there which I explain below as well.
- I just noticed I made a mistake and had the solar anomaly as $2;21º$ when it should be $2;23º$. I’ll try to fix that later, but right now I’m on the wrong computer as my laptop doesn’t love doing art.
- In what follows, I’ll be walking through Ptolemy’s method. In my original version of this post, I followed through Neugebauer’s as Ptolemy’s was extremely dense. However, there is a key piece of information that Neugebauer’s obscures. As that will pop up again in a later post, I decided to take the time to better understand Ptolemy’s method and explain it here. However, if you’re following along with Neugebauer and need some help understanding that, I’ve preserved my explanation of his work in a separate post.
- Even though that was a little low for the moon’s motion in relation.
- Does this feel like Zeno’s paradox with Achilles and the tortoise to anyone else?
- This isn’t just a happenstance that only works in this case. Essentially what we have here is an infinite series of $\frac{x}{13^n}$. When summed from $n=1$ to infinity, the series does in fact converge to $\frac{1}{12}x$.
- This is the part that’s going to come back in a few posts as this method of finding the distance between the mean and true conjunction is one we’ll continue to use!
- The detailed videos are only available to Patreon supports.