In the last post, we explored how the eccentric and epicyclic models could produce the same result even if you didn’t have the same period of anomaly and ecliptic longitude. This was done by allowing the center of the eccentre to rotate around the Earth. In this post, we’ll explore how they can still produce the same result even
if [the members of] the ratios are unequal, and the eccentre is not the same size as the deferent…provided the ratios are similar
To prove this, Ptolemy draws the diagram for each model separately, starting with the epicycle model:
Here we have circle ABG as the deferent on which epicycle EZ travels on center G. The moon is at position Z.
B is a holdover from the diagram we made in the last post which we’ll keep as a reference point. I’ll explain more when we need it.
For the eccentric model, L is the center of the eccentre (circle HΘK) with the Earth at M. Here, the moon is at K.
We’ll say that $\frac{\overline{DG}}{\overline{GE}} = \frac{\overline{\Theta L}}{\overline{LM}}$, which is simply one mathematical statement of the condition we’re trying to prove in which “the ratios are similar”1.
Next, we’ll assume that,
in the same time as the epicycle has moved through $\angle{ADG}$, the moon has again moved through $\angle{EGZ}$, the eccentre through $\angle{HM \Theta}$, and the moon, again through $\angle{\Theta LK}$.
Since the moon traverses either of the circles on which it travels at the same rate, this gives us a relationship between these motions allowing us to state $\angle{EGZ} = \angle{\Theta LK}$2.
Ptolemy also states that this tells us that $\angle{ADG} = \angle{HM \Theta} + \angle{\Theta LK}$. This one is a little less obvious. To understand, I feel like it helps to actually consider what each of these represent. On the left side, we have $\angle{ADG}$ which is the primary motion of the the epicycle plus secondary motion which is the moon on the epicycle. However, because the epicycle moves the opposite direction, this is effectively a negative.
In the case of the other, $\angle{HM \Theta} + \angle{\Theta LK}$, this is again the primary motion of the deferent rotating plus the secondary motion of the moon on the deferent, this time additive since they’re in the same direction.
So with all that laid out, what we’re trying to prove is that
the moon will again appear to have traversed an equal [apparent] arc in the same time according to either hypothesis.
Stating that mathematically, we’re out to show that: $\angle{ADZ} = \angle{HMK}$.
Another thing we can note, is that the eccentric model is drawn at the second point in time, where point L, the center of the eccentre, is straight up in the drawing. At the first point in time (i.e., when the moon would have been at apogee in the eccentric model), point L would have been along $\overline{MH}$ which means $\overline{MH}$ represents the direction of apogee as well.
Now to get started. First, let’s return to what was up with the point B in the epicyclic diagram. If you refer to this image from the last post, you’ll see where it came from. In that image, $\angle{ADB}$ was the angle the center of the eccentre travelled which left $\angle{BHZ}$ as the angle the moon actually travelled on the eccentre. And the way we set things up, that angle was also equal to the angle the moon travelled on the epicycle in the epicyclic hypothesis ($\angle EGZ$) as well as the central angle ($\angle{EDB}$). So since all the angles are the same so too must their arcs be related.
Thus $arc \; BG$ is similar to $arc \; EZ$. However, in the eccentric diagram, the angle actually traversed on the eccentre by the moon is $\angle{\Theta LK}$ so we can add its arc, $arc \; \Theta K$ into that mix as well as similar to the others.
Next up, another part of the requirement Ptolemy gave for this was that the “ratios are similar”. Specifically this means $\frac{\overline{DG}}{\overline{GZ}} = \frac{\overline{KL}}{\overline{LM}}$.
And if the ratio of two sides is equal, so too must be the angle between them. Thus, $\angle{ZGD} = \angle{KLM}$ and this is sufficient to state that the two triangles, $\triangle{GDZ}$ and $\triangle{KLM}$, are similar.
With that established, we can then relate the other angles to state $\angle{GZD} = \angle{LMK}$.
We can also state that $\angle{BDZ} = \angle{GZD}$. It’s hard to see it from these images, but if you again refer to the image from the last post, it’s apparent why: $\overline{ZG} \parallel \overline{BD}$ with $\overline {ZD}$ as a diagonal between them, which makes $\angle{BDZ}$ and $\angle{GZD}$ alternate interior angles which are equal.
Thus, we can say that $\angle{GZD} = \angle{BDZ} = \angle{LMK}$.
Now let’s turn our attention to $\angle{ADB}$ again. This angle was the angle the center of the eccentre moved when we set it up in the last post. In this post, that same definition is in the eccentric diagram as $\angle{HM \Theta}$.
So these are equal which is to say: $\angle{ADB} = \angle{HM \Theta}$.
Now let’s grab two of those angles we just showed were equal and add one to each side:
$$\angle{ADB} + \angle{BDZ} = \angle{HM \Theta} + \angle{LMK}$$
This reduces to:
$$\angle{ADZ} = \angle{HMK}$$
which is what we set about to prove.
Thus, so long as the ratio of the radius of the deferent to the radius of the epicycle is equal to the ratio of the eccentre to the distance between the center of the eccentre and Earth are equal, these two models will produce identical results.
In the next post, we’ll start diving into some eclipses to demonstrate the first anomaly of the moon!
