Now that we’ve squared away the most accurate motion in latitude Ptolemy can muster, it’s time to use it to reverse calculate the lunar latitude at the beginning of the epoch, but to do so, we’ll need to calculate the precise position in latitude at some point in time. To do so, Ptolemy again turns to eclipses, with all the same considerations as the last post, except we want ones at opposite nodes instead of near the same one.
For his eclipses Ptolemy starts with the same one from when we looked at the epoch for anomaly. Specifically the one that occured
in the second year of Mardokempad, Thoth [I] 18/19 in the Egyptian calendar [-719 Mar. 8/9], at midnight in Babylon and $\frac{5}{6}$ of a equinoctial hour before midnight in Alexandria… [in which] the moon was obscured 3 digits from the south.
To that, we’ll add one from
the twentieth year that Darius who succeeded Kambyses, Epiphi [XI] 28/29 in the Egyptian calendar [-501 Nov. 19/20], when $6 \frac{1}{3}$ equinoctial hours of night had passed, …[and] the moon was, again, obscured from the south $\frac{1}{4}$ of its diameter. The middle of the eclipse was $\frac{2}{5}$ of an equinoctial hour before midnight in Babylon (for the length of half the night was about $6 \frac{3}{4}$ equinoctial hours on that date) and [hence] $1 \frac{1}{4}$ equinoctial hours before midnight in Alexandria.
As we did in the last post, we can check where the moon was on the epicycle at the time, and if we did so we would determine
these eclipses occurred when the moon was near its greatest distance, but the first was near the ascending node, while the second was near the descending node.
Let’s take a look at how this looks from a top down view of the deferent:
In this view, points $A$ and $G$ are the two nodes wherein $A$ is the ascending node, and $G$ is the descending node. $B$ is the point where the latitude reaches its greatest latitude above the ecliptic, which Ptolemy calls the “northernmost point”. We’ll take the first of the two above eclipses to be at point $D$ with the second at $E$ and since these should be roughly equal distances from the nodes, this means $arc \; AD = arc \; EG$.
Now if we consider the time from the beginning of the epoch to the first eclipse, Ptolemy gives it as $27$ Egyptian years, $17$ days, $11 \frac{1}{6}$ hours. If we walked through the mean motion table we would determine that the moon’s distance from apogee1 was $12;24º$ and the mean position was greater than the true by $0;59º$. So now, we can define point Z as the mean position of the moon and thus, $arc \; DZ = 0;59º$.
Doing the same for the second eclipse, Ptolemy gives a period of $245$ Egyptian years $327$ days, and $10 \frac{1}{4}$ hours which puts the moon $2;44º$ from apogee and the mean position greater than the true by $0;13º$ which means $arc \; EH = 0;13º$.
We can similarly take the time between eclipses which Ptolemy gives as $218$ years, $309$ days, and $23 \frac{1}{12}$ hours over which time the moon moves2 $160;4º$. This is $arc \; ZBH$.
That’s all our givens. So now we can start playing with these numbers to determine other pieces:
$$arc \; DE = arc \; DZ + arc \; ZBH – arc \; EH = 160;50º$$
If we then subtract this from $arc \; ABG$ which is $180º$, this would give us the the leftover bits which is to say $arc \; AD + arc \; EG = 19;10º$.
But since we said above that these two arcs are equal, the length of either one is $arc \; AD = arc \; EG = 9;35º$.
In other words, the true position of the moon at the first eclipse was that distance after after the ascending node, while at the second eclipse, it was that distance before the descending node.
From these, we can easily determine the position of the mean moon for each from their respective nodes.
$$arc \; AZ = arc \; AD + arc \; DZ = 10;34º$$
$$arc \ HG = arc \; EG – arc \; EH = 9;22º$$
Recalling that the moon travels counter-clockwise around the deferent, if we start from B3, is $280;34º$ from point $Z$ which was the position of the mean moon. Similarly, we can determine the position of the mean moon for the second eclipse to be $80;38º$ after the northernmost point.
Next Ptolemy calculates the motion of the mean motion in latitude from the beginning of the epoch to the first eclipse and derives a motion of $286;19º$. If we subtract this from the position of the mean moon at the first eclipse we just derived, we determine the mean position in latitude (from the northern imit) of $354;15º$.
In terms of following along with the Almagest, I’ll stop here for the rest of this post for two reasons. The first is that the short remainder of this chapters moves on to prefracing the next chapter so I’ll cover it in the next post, and second, we should finally take a moment to stop and reflect on what some things mean.
In particular, we’ve finally started talking about the mean motion in latitude. But what does that mean? It’s something that first came up in Chapter 2 of this book, but I’ve shied away from what it means in terms of the model until now. And I feel like that needs to be remedied. When we’ve talked about the previous motions they’ve been motions around a particular circle; the motion in longitude is the motion along the deferent projected onto the ecliptic and the anomalistic motion is the motion around the epicycle.
The same will be true here, but before explaining, let’s recall one other thing about those two other mean motions: When we calculated the position for each at epoch, it was always in relation to something. For longitude, those initial epoch positions were ecliptic longitude. For anomaly, it was with respect to apogee.
So in this case, it’s again motion along the ecliptic, but this time with respect to the northernmost point which, along with the nodes, rotates clockwise slowly. Let’s use some pictures to help us understand:
Here, let’s start with the case where the moon ($M$), the vernal equinox, and the northernmost point ($N$) are all at the same point at some moment in time. Now, we’ll let a day pass. I’ll exaggerate the angles here to illustrate the point, but where what it would look like:
Here, the moon has moved around the deferent counter-clockwise from its starting point. But the question is, by how much? If we measure with respect to the vernal equinox, it’s moved $13;10,34º$. But since $N$ has rotated clockwise away from $M$, if we measure with respect to $N$, then it’s moved further. Specifically, by the amount the northernmost point moved clockwise from the vernal equinox. This is about $0;3,11º$ per day. So the distance $M$ would have moved is $13;13,45º$. Thus, both motions are along the deferent, but the two different reference points mean you get different amounts of motion. Ecliptic longitude is measured with respect to the vernal equinox, whereas the latitude is a function of the distance from the northernmost point4.
But note I said “a function of.” The moon isn’t climbing in latitude, getting further and further from the ecliptic until it swings back around the other side. So much like we’ve seen for the anomaly and the sun, we’ll want to produce a table that we can look up the latitude based on the distance from the northernmost point. And truthfully, we’ll also need to determine where that northernmost point was at the beginning of epoch as well.
Which is what you’d expect Ptolemy to do next. But he just…. doesn’t.
I suspect the reason is there’s going to be some important effects from the mysterious “second anomaly” Ptolemy keeps hinting at that will be a major focus in the next book. As such, there’s probably no reason to do so now as anything we did would get replaced. So while we’ve done some good preliminary work, we’ll have to press pause on this topic for now.
- Keep in mind that the picture as drawn here is the deferent. The epicycle is not drawn in. So point $B$ is not the apogee.
- After subtracting out complete revolutions.
- Which is $90º$ before $G$, and therefore $270º$ before $A$.
- In modern astronomy, we determine it with respect to the ascending node instead, but same idea.