So far in this chapter, we’ve been looking at the two different hypotheses to explain the non-constant angular motion of objects in the sky. Ptolemy claimed that these were equivalent under certain circumstances and, in the last post, we showed how they do indeed produce the same results in the specific case of the object travelling 90º in apparent motion from apogee1 and that it always takes longer for the object to go from slowest motion to mean, than it does mean to fastest.
But that’s not really a full demonstration that they’re functionally the same. So in this post, we’ll show that their apparent angular position from the mean (known as the equation of the anomaly) is always the same, so long as there’s a few things that are consistent between models.
Let’s start with the diagram combining the two models as that will help explain the requirements.
In this setup, the eccentric model is shown as circle EZH (which Ptolemy refers to as the “eccentre”) is the circle with center Θ, from which the observer would be offset at D. Meanwhile, the epicyclic model is shown by circle ABG (which Ptolemy also calls the “concentre”) is the deferent at center D, which carries epicycle KZ.
The requirements for things to be similar is that2:
1. The ratio of the distance between the true center of the eccentre and the observer to that of the radius of the eccentre be the same as the the ratio of the radius of the epicycle to the radius of the deferent. Stated mathematically:
$$\frac{\overline{D \Theta}}{\overline{\Theta E}} = \frac{\overline{BZ}}{\overline{DB}}$$
2. The deferent should be rotating rearwards with respect to the heavens (counter-clockwise) while the epicycle rotates forwards (clockwise) at the same angular speed.
If those arguments are fulfilled, then the results produced by the two hypotheses should be the same.
To illustrate this, we’ll start with a simple case for the first requirement and require that $\overline{BZ} = \overline{D \Theta}$ as well as the radii of both the eccentre and the deferent. As we did in the last post, we’ll consider the object having started at point E (which is where point K would have been when the epicycle was there), and let time pass until the object is at point Z. Fulfilling requirement two, as it moves counter-clockwise on the deferent, it will be moving the same angular distance clockwise on the epicycle which tells us that $\angle{E \Theta Z} = \angle{ADB} = \angle{KBZ}$.
In this setup, we can look at the angle between the mean and the actual, again, known as the “equation of the anomaly”. For the eccentric model, we demonstrated how this was $\angle{DZ \Theta}$ in the last post3. In epicyclic model we showed, the equation of anomaly to be $\angle{BDZ}$. These two angles are quite clearly equal to one another because they’re alternate interior angles which means that the effects for both models are identical.
And this doesn’t just hold true for this point. Because the rotation of the epicycle is always the exact opposite of the motion around the deferent, $\overline{BZ}$ always maintains its orientation which is to say it stays parallel to $\overline{D \Theta}$. In addition, $\overline{DB}$ and $\overline{\Theta Z}$ always remain parallel as well, which means that, no matter where around the orbit the object goes, we always have parallelogram BDΘZ and the two angles mentioned above will always be equal as well.
Simply looking at the diagram, this makes perfect sense. If you mentally remove circle EZH, and just picture the point of the object as it would go around its orbit, it simply traces a path that’s distance $\overline{BZ}$ above the circle the epicycle rides on. And because $\overline{BZ} = \overline{D \Theta}$, it happens to be the eccentre.
But what happens if $\overline{BZ}$ is not equal to $\overline{D \Theta}$ because the eccentre is larger or smaller than the deferent?
According to Ptolemy:
provided their ratios are similar [as described by the first rule above], the same phenomena will result.
To get us started on the proof for this, let’s begin with just the epicyclic model drawn out:
Since this is essentially the same setup as we’ve had for the epicyclic model in the past few setups, I won’t bother explaining things again, but we can again note that $\angle{ADE} = \angle{ZBE}$ again due to the angular rotation of the epicycle being equal in magnitude and opposite in direction to the rotation of the epicycle on the deferent. Also, $\angle{BDZ}$ is the equation of the anomaly.
What we’ll try to prove here is that
the body will also appear to lie on the same line, $\overline{DZ}$, according to the eccentric hypothesis, whether the eccentre is greater or smaller than the concentre [deferent], ABG, provided only that one assumes that the ratios [as described in the first rule] are similar and that the periods of rotation are the same.
As Ptolemy states, the eccentre could be either larger or smaller than the deferent. We could prove each of these in turn, but Ptolemy goes full out and tackles both situations at once. So we’ll add both a larger eccentre (circle HΘ on center K) and a smaller one (circle LM on center N).
In addition, we’ve also added $\overline{NM}$ and $\overline{K \Theta}$ which are lines from the center of their respective eccentres to the apparent position of the body on their eccentre and are useful for noting the angle from the apogee.
One of the key other features here is the bit Ptolemy stated above about “the ratios are similar”. Specifically what that means is that4
$$\frac{\overline{DB}}{\overline{BZ}} = \frac{\overline{\Theta K}}{\overline{KD}} = \frac{\overline{MN}}{\overline{ND}}$$
Since this is a condition of the hypothesis to be proved, we can treat it as a given for this proof. We can also state that $\angle{BZD} = \angle{MDN}$ since $\overline{BZ} \parallel \overline{DA}$5.
This means that the three triangles, $\triangle{ZDB}$, $\triangle{D \Theta K}$, and $\triangle{DMN}$ are all similar because the ratio of their sides is the same as in the angle between those two sides6. Thus, the corresponding angles of all three triangles are equal allowing us to state $\angle{BDZ} = angle{D \Theta K} = angle{DMN}$.
Here I’ve taken away some of the noise and colored the congruent angles to make things easier to see.
In addition to the angles in the triangles we just described, I’ve also marked $\angle{BDG}$ as congruent to all the other red angles because it’s an alternate interior angle to $\angle{ZBD}$. And notice those three congruent red angles along $\overline{AG}$. The first thing that it shows is that, $\overline{DB} \parallel \overline{NM} \parallel \overline{K \Theta}$ since they’re all the same angle from the same line.
Second, consider each of their supplementary angles: $\angle{AK \Theta}$, $\angle{ANM}$, and $\angle{ADB}$. They too would be the same and each of these angles is the equation of the anomaly and equal, thus demonstrating Ptolemy’s proposition that, so long as the ratios described above are equal7, the two models return the same result even if the eccentric is smaller than the deferent and the distance between its center and the observer larger or smaller than the radius of the epicycle.
This ends the portion of the chapter where Ptolemy looks at the similarities in the equation of the anomaly. There’s still a bit more about similarities in apparent distances from apogee and perigee, but we’ll explore that in the next post.
- So long as, in the epicyclic model, the motion of the epicycle is rearwards and the motion of the object on the epicycle is forwards.
- These are my phrasings of the requirements. Ptolemy does not state them so concisely nor number them.
- Although the points were labeled differently.
- This is just the reciprocals of the first requirement to match how Ptolemy stated it.
- If that’s still hard to follow here’s a picture without all the other chaos going on and the angles explicitly marked.
It’s another case of alternate interior angles being congruent. - Side-Angle-Side similarity.
- And the bit we stated about the epicycle moving forward as the deferent moves rearwards.
