Our next task is to demonstrate the type and size of the moon’s anomaly.
In chapter 2 of this book, we spent quite a bit of time talking about the moon’s anomaly, describing a method by which Hipparchus could have used periods of eclipses to determine the anomaly’s period. While we never actually completed the method, Ptolemy still gave us the period Hipparchus supposedly derived. Now we’re going to put that to use to start building our first lunar model.
I say first because we’re going to treat the moon’s anomaly
as if it were single and invariant.
In other words the moon actually has a second anomaly necessary to properly model it which astronomers prior to Ptolemy were apparently unaware of. That anomaly is
linked to its distance from the sun; this [second anomaly] reaches a maximum round about both [waxing and waning] half-moons, and goes through its period of return twice a month, [being zero] precisely at conjunction and opposition.
This second anomaly is one of the main focuses of Book V. We won’t deal with it right now because
it is impossible to determine the second [anomaly] apart from the first, which is always combined with it, whereas the first can be found apart from the second, since it is determined from lunar eclipses, at which there is no perceptible effect of the anomaly connected with [the distance from] the sun.
In other words, this second anomaly went undiscovered because it is zero at conjunction or opposition, and since lunar eclipses always happen at opposition, it played no role. Thus, we can worry about this first anomaly all by itself for now.
So neglecting that second anomaly for now, how will we progress with the first model?
We [like Hipparchus], using three lunar eclipses, shall derive the maximum difference from mean motion and the epoch of the [moon’s position] at the apogee, on the assumption that only this [first] anomaly is taken into account and that it is produced by the epicyclic hypothesis.
So as previously hinted at, we’ll be using the epicycle model to explain the first anomaly because
we shall find the [the epicycle hypothesis] more suitable to represent the second anomaly…when we come to combine both anomalies. However, the same phenomena will in all cases result from both the hypotheses we have described, whether, as in the situation describe for the sun, the period of return in anomaly and the period of return in the ecliptic [i.e., in longitude] are both equal, or whether, as in the case of the moon, they are unequal, provided only that the ratios [of epicycle to deferent and eccentricity to eccentre] are taken as identical.
In short, in Book III, we proved that the two models were equivalent for the sun. But that was when the anomalistic period was exactly equal to the return in ecliptic longitude (1 year).
Since the moon completes its return with respect to the ecliptic sooner than its return with respect to this anomaly, it is clear that, in the epicyclic hypothesis, over a given period of time, the epicycle will always traverse a greater arc of the circle concentric to the ecliptic than the arc of the epicycle traversed by the moon in the same time.
In other words, when the epicycle completes a full circle on the deferent, the moon on the epicycle still has further to go. So what do we need to tweak in order for these two models to be the same when the return in ecliptic longitude doesn’t line up with the return in anomaly?
In the eccentric hypothesis, the arc traversed by the moon on the eccentre will be similar to the arc traversed by it on the epicycle [in the epicyclic hypothesis], while the eccentre will move about the center of the ecliptic in the same direction as the moon by an amount equal to the increment of the motion in longitude over the motion in anomaly [in the same time] (this corresponds to the increment of the arc of the deferent over the arc of the epicycle [in the epicyclic hypothesis]). In this way we can preserve the equality of the periods of both motions [i.e., in longitude and anomaly], as well as the equality of the ratios, in both hypotheses.
Yikes. That was almost all one sentence. Instead of trying to parse it right now, let’s start looking at the diagram:
To begin, I’ve started with just a very simple comparison between the epicyclic model and the eccentric model. Here, H is the center of the eccentre which I’ve colored blue to help distinguish it. E is apogee which is where we’ll start the moon. D is the center of the deferent on which the ecliptic, centered on A, rides.
Now we’ll let time pass and draw things out again.
This is pretty busy so let’s take a moment to digest.
I’ve left the initial position of the epicycle in for reference, but now the epicycle has moved to point G on the deferent and the moon has moved clockwise on the epicycle to point Z.
Let’s pause to notice that the epicycle has moved 90º, but the moon around the epicycle has moved some angle less than that since we’re requiring that the anomaly (driven by the rotation on the epicycle) take longer than the time the epicycle takes to traverse the deferent.
However, if you look back at the first picture, you should be able to tell that, at point Z, there was no way it could have remained on the eccentre if the center remained at H, just as E is no longer on the eccentre.
So Ptolemy’s solution is to have the eccentre rotate around the Earth as well. In doing so, the point that was previously E, gets swept around to point Θ, while the moon travels on the eccentre to point Z, sweeping out $arc \; \Theta Z$ in the same period of time as the moon would travel from E to Z on the epicycle. This means that their respective angles, $\angle{\Theta H Z}$ and $\angle{EGZ}$ are equal as well.
We’ll also have $\overline{GZ} \parallel \overline{DH}$. Line $\overline{HZ}$ is drawn in so it is parallel to $\overline{DG}$ thus forming parallelogram $DHZG$.
Point B is included as reference.
Now what we’ll be trying to prove is that $\frac{\overline{ZH}}{\overline{HD}} = \frac{\overline{DG}}{\overline{GZ}}$. In addition, while we’ve provisionally stated it above, we’ll formalize that $arc \; Z \Theta = arc \; EZ$.
The first of these comes immediately from the fact that we have the aforementioned parallelogram. The short sides ($\overline{GZ}$ and $\overline{DH}$) are equal as are their long sides ($\overline{HZ}$ and $\overline{DG}$). Thus taking the ratio of long to short sides must also be equal1.
Next, up proving $arc \; Z \Theta = arc \; EZ$.
In the amount of time we’ve sketched out, recall that the moon has moved from Θ to Z along the eccentre, which swept out $\angle{\Theta H Z}$ with respect to the center of the eccentre. However, since $\overline{HZ}$ is an extension of $\overline {\Theta H}$, and $\overline{ZH} \parallel \overline{GZ}$ this means $\angle{\Theta H Z} = \angle{\Theta D E}$.
In addition, we have that $\angle{\Theta D E} = \angle{EGZ}$, again due to the parallelogram.
Therefore the moon has reached point Z in the same time according to either hypothesis, since the moon itself has traversed $arc \; EZ$ on the epicycle and $arc \; \Theta Z$ on the eccentre, which we have shown to be similar, while the epicycle centre has moved through $arc \; AG$ and the eccentre through $arc \; AB$, which is the increment of $arc \; AG$ over $arc \; EZ$.
So this proves the general principle that, by allowing the center of the eccentre to rotate as well, we can create the situation in which the moon would not progress with an equal return in anomaly and return in ecliptic longitude.
In the next post, we’ll explore a bit further on this topic and show what else we can change and still have this hold true.
- Toomer points to I.33 in Euclid’s Elements which states that “Straight lines which join the ends of equal and parallel straight lines in the same directions are themselves equal and parallel.”