In Books X and XI, we’ve begun developing a theory for the motion of the planets and have that pretty much complete for all of the planets in longitude; we haven’t begun addressing their latitude. But, before getting to that, Ptolemy wants to explore retrograde motion.
While the models already account for this, Ptolemy is wanting to be able to predict when it will happen1. That will be our focus for the Book XII and we’ll get to latitudes in Book XIII.
Ptolemy tells us that he wants to “examine the greatest and least retrogradations associated with each of the five planets, and to show that the sizes of these, [as computed] from the above hypotheses, are in as close agreement as possible with those found from observations.”
While retrogradations were astrologically significant, exploring this and checking it against observation is also a strong test of the model itself, which is likely why Ptolemy is interested in it.
But before we can get into it, we’ll need to explore some preliminary lemmas. Specifically, Ptolemy wants to introduce us to a set of hypotheses from Apollonius of Perga. He tells us that:
[1] If [the synodic anomaly] is represented by the epicyclic hypothesis, in which the epicycle performs the [mean] motion in longitude on the circle concentric with the ecliptic towards the rear [i.e., in the order] of the signs, and the planet performs the motion in anomaly on the epicycle [uniformly] with respect to its own centre, towards the rear along the arc near apogee, and if a line is drawn from our point of view intersecting the epicycle in such a way that the ratio of half that segment of the line intercepted within the epicycle to that segment intercepted between the observer and the point where the line intersects the epicycle nearer its perigee is equal to the ratio of the speed of the epicycle to the speed of the planet, then the point on the arc of the epicycle nearer the perigee determined by the line so drawn is the boundary between forward motion and retrogradation, so that when the planet reaches that point it creates the appearance of station.
[2] If the anomaly related to the sun in represented by the eccentric hypothesis (which is a viable hypothesis only for the three [outer] planets which can reach any elongation from the sun), in which the centre of the eccentre moves [uniformly] about the centre of the ecliptic with the speed of the [mean] sun towards the rear [i.e., in the order] of the signs, while the planet moves on the eccentre in advance [i.e., in the reverse order] of the signs with a [uniform] speed with respect to the centre of the eccentre and equal to the [mean] motion in anomaly, and if a line is drawn in the eccentre through the centre of the ecliptic (i.e., the observer) in such a way that the ratio of half the whole line to the smaller of the two segments of the line formed by [the position of] the observer is equal to the ratio of the speed of the eccentre to the speed of the planet, then when the planet arrives at the point in which the above line cuts the arc of the eccentre near the perigee, it will produce the appearance of station.
“Station” is a term we haven’t encountered before. You can probably guess what it means from context – it’s simply the points where the planet’s motion appears to hold still as it changes between prograde and retrograde motion.
These two hypotheses are essentially telling us where about the orbits these standstills will be for the epicyclic models and eccentric hypotheses.
Ptolemy then tells us that he intends to explain both of these hypotheses with a
proof which contains both hypotheses combined in a common [figure], to demonstrate their agreement and similarity in these ratio of theirs too.
In other words, he’s going to prove them both at once and show that they’re similar in nature.
To do so, he begins with the following diagram:
In this image, we have the epicycle as $ABGD,$ centered on $E.$ We’ll draw in the observer on Earth at $Z.$
Then, we’ll take two arcs on either side of $G$ which are $arc \; HG$ and $arc \; G \Theta$ which are equal in length. We’ll extend lines through $H$ and $\Theta$ from $Z$ until the line hits the other side of the epicycle at $B$ and $D$ respectively.
We’ll draw $\overline{AD},$ $\overline{DG},$ $\overline{DH},$ and $\overline{B \Theta}.$ The point of intersection between $\overline{DH}$ and $\overline{B \Theta}$ we’ll label as $K$ and it falls on $\overline{AZ}.$
Lastly, we’ll draw a line parallel to $\overline{AD}$ upon which $G$ falls. We’ll cut this line off where it intersects $\overline{DH}$ and $\overline{DZ}$ at $L$ and $M$ respectively.
Ptolemy’s goal here will be to prove that
$$\frac{\overline{AZ}}{\overline{ZG}} = \frac{\overline{AK}}{\overline{KG}}.$$
We’ll begin by examining $\triangle ADG.$ This triangle has the diameter of the epicycle as one of its sides, which means that the angle on the circumference, $\angle ADG$ is a right angle. And since $\overline{HM} \parallel \overline{AD}$ $\angle DGH$ and $\angle DGM$ are right angles as well.
Additionally, $\angle GDH = \angle GD \Theta$2.
Now, let’s focus on $\triangle GLD$ and $\triangle GMD.$ These two triangles are congruent.
Therefore,
$$\frac{\overline{AD}}{\overline{GL}} = \frac{\overline{AD}}{\overline{GM}}.$$
Next, let’s look at $\triangle ADZ$ and $\triangle GMZ.$ These two triangles are similar triangles.
Therefore, we can write,
$$\frac{\overline{AD}}{\overline{AZ}} = \frac{\overline{GM}}{\overline{GZ}}.$$
Doing a bit of rearranging:
$$\frac{\overline{AD}}{\overline{GM}} = \frac{\overline{AZ}}{\overline{GZ}}.$$
Similarly, $\triangle ADK$ is similar to $\triangle GMZ$ allowing us to write:
$$\frac{\overline{AD}}{\overline{GL}} = \frac{\overline{AK}}{\overline{KG}}.$$
Thus, doing a bit of substitution:
$$\frac{\overline{AZ}}{\overline{ZG}} = \frac{\overline{AK}}{\overline{KG}}.$$
Ptolemy pauses there and says the following
So, if we imagine epicycle $ABGD$ to be the actual eccentre in the eccentric hypothesis, the point $K$ will be the centre of the ecliptic, and diameter $\overline{AG}$ will be divided by it in the same ratio as [the corresponding amounts] in the epicyclic hypothesis.
This statement takes a bit to parse, but I don’t really have much to say that can make it more clear. What Ptolemy has done is just given two ratios. The one of the left is a ratio of two lines within the epicyclic model. Ptolemy has demonstrated that these are equivalent to another ratio of lines in the eccentric hypothesis. In short, there are equivalences between the eccentric and epicyclic hypotheses.
This should be unsurprising since we’ve already shown some equivalences back in III.3.
But this doesn’t really seem to do much to prove what we set out to demonstrate: The positions at which the planet’s ecliptic motion seems to hold still. That must lie ahead.
Next, Ptolemy wants to show that
$$\frac{\overline{DZ}}{\overline{Z \Theta}} = \frac{\overline{BK}}{\overline{K \Theta}}.$$
To do so, we’ll need to make some modifications to our diagram:
In this, we’ve removed several lines and added a few. Most of the ones we’ve added are all perpendiculars. We’ve dropped them from $E$ onto $\overline{DZ}$ at $O,$ from $E$ onto $\overline{B \Theta}$ at $P,$ and from $\Theta$ onto $\overline{AZ}$ at $X.$ We’ve also added $\overline{BD}$ which will obviously be parallel to $\overline{X \Theta}$ and perpendicular to $\overline{AZ}.$
In this new setup, $\overline{BN} = \overline{ND}.$
This allows us to write
$$\frac{\overline{BN}}{\overline{X \Theta}} = \frac{\overline{ND}}{\overline{X \Theta}}.$$
Now, looking at $\triangle ZND$ and $\triangle ZX \Theta,$ these are similar triangles allowing us to write
$$\frac{\overline{ND}}{\overline{X \Theta}} = \frac{\overline{DZ}}{\overline{Z \Theta}}.$$
Similarly, $\triangle BNK$ is similar to $\triangle \Theta XK$ which allows us to state
$$\frac{\overline{BN}}{\overline{X \Theta}} = \frac{\overline{BK}}{\overline{K \Theta}}.$$
Thus, with a bit of substitution we find
$$\frac{\overline{DZ}}{\overline{Z \Theta}} = \frac{\overline{BK}}{\overline{K \Theta}}.$$
Again, this gives us a ratio, but it’s still not clear where we’re going with this.
Ptolemy now uses a theorem we don’t see much today known as componendo and dividendo. These are essentially a set of theorems about when you have to ratios equal to one another being able to write a few other equalities.
For the componendo portion, this allows us to write:
$$\frac{(\overline{DZ} + \overline{Z \Theta})}{\overline{Z \Theta}} = \frac{(\overline{BK} + \overline{K \Theta})}{\overline{K \Theta}}.$$
But note that
$$\overline{BK} + \overline{K \Theta} = \overline{B \Theta}.$$
Thus, we can simplify a bit:
$$\frac{(\overline{DZ} + \overline{Z \Theta})}{\overline{Z \Theta}} = \frac{\overline{B \Theta}}{\overline{K \Theta}}.$$
The dividendo portion here is not the standard one, which is about being able to express the ratios in a subtractive form. So instead of just invoking a theorem here, let’s walk through it3
You may have noticed above that we didn’t use the perpendiculars we dropped from $E.$ This is where we’ll make use of them.
First, let’s look at $\overline{DZ}.$ On it, we have point $O$ which is half way between $D$ and $\Theta$. Because of this, we can write
$$2 \cdot \overline{OZ} = \overline{DZ} + \overline{Z \Theta}.$$
This isn’t immediately obvious, so let’s do a quick proof:
$$\overline{OZ} = \overline{DZ} – \overline{DO}$$
and
$$\overline{OZ} = \overline{\Theta Z} + \overline{O \Theta}.$$
If we add these together we get:
$$2 \cdot \overline{OZ} = \overline{DZ} – \overline{DO} + \overline{\Theta Z} + \overline{O \Theta}.$$
However, because $O$ is half way between $D$ and $\Theta$, $\overline{DO} = \overline{O \Theta}.$
As such, we can substitute into the equation above, allowing the $- \overline{DO}$ term to cancel the $\overline{O \Theta}$ term which leaves us with
$$2 \cdot \overline{OZ} = \overline{DZ} + \overline{Z \Theta}.$$
Next, let’s look at $\overline{B \Theta}.$ We dropped point $P$ on this such that $\overline{EP}$ was perpendicular to $\overline{B \Theta}.$ This is a perpendicular to a chord, which means that it bisects $\overline{B \Theta}$ allowing us to state
$$\overline{B \Theta} = 2 \cdot \overline{P \Theta}.$$
Now, taking the result from the componendo portion, we can do a bit of substitution into the numerators on each side allowing us to write:
$$\frac{2 \cdot \overline{OZ}}{\overline{Z \Theta}} = \frac{2 \cdot \overline{P \Theta}}{\overline{K \Theta}}.$$
This can then be simplified slightly by dividing both sides by $2$ and giving us:
$$\frac{\overline{OZ}}{\overline{Z \Theta}} = \frac{\overline{P \Theta}}{\overline{K \Theta}}.$$
This is still an equivalence of ratios, so we can now invoke the usual dividendo theorem to write:
$$\frac{\overline{OZ} – \overline{Z \Theta}}{\overline{Z \Theta}} = \frac{\overline{P \Theta} – \overline{K \Theta}}{\overline{K \Theta}}.$$
And we can do a bit of simplification to the numerators:
$$\frac{\overline{O \Theta}}{\overline{Z \Theta}} = \frac{\overline{PK}}{\overline{K \Theta}}.$$
Ptolemy then explains what this is meant to prove:
Therefore, if, in the epicyclic hypothesis, $\overline{DZ}$ is drawn in such a way that the ratio of $\overline{O \Theta}$ to $\overline{Z \Theta}$ equals the ratio of the speed of the epicycle to the speed of the planet, in the eccentric hypothesis $\frac{\overline{PK}}{\overline{K \Theta}}$ will have that same ratio.
I’ll skip forward momentarily to where Ptolemy provides an example of what he means:
[F]or Mars:
$$\frac{speed \; of \; epicycle}{speed \; of \; planet} \approx \frac{42}{37}.$$
It should be noted that Ptolemy’s language here is very vague; “speed of the epicycle” is referring to the mean speed of the center of the epicycle about the eccentre. Similarly, “speed of the planet” refers to the speed of the planet about the epicycle.
With that cleared up, let’s pause and examine where his numbers come from. Back in IX.3, we look at the returns of Mars. There Ptolemy told us that “$37$ returns in anomaly correspond to $79$ solar years… and to $42$ revolutions of the planet.”
Thus, the speed of the epicycle’s center is
$$\frac{42}{79} \; \frac{returns}{year}$$
and for the planet about the epicycle
$$\frac{37}{79} \; \frac{returns}{year}.$$
If we take the ratio of these, the $79$ will cancel out leaving us with the ratio Ptolemy gave us.
What Ptolemy is telling us here is that this ratio of the periods is related to ratios of segments in each of the models: the ratio of $\frac{O \Theta}{\Theta Z}$ in the epicyclic model and $\frac{\overline{PK}}{\overline{K \Theta}}$ in the eccentric.
Skipping back a bit, Ptolemy also tells us that:
The reason that in this case [i.e., the eccentric hypothesis] we do not use this ratio obtained [via] dividendo (namely $\frac{\overline{PK}}{\overline{K \Theta}}$) to get the stations, but rather the undivided ratio (namely $\frac{\overline{P \Theta}}{\overline{K \Theta}}$), is that the epicycle’s speed is in the same ratio to the planet’s as the [mean] motion in longitude (alone) to the [mean] motion in anomaly, whereas the ratio of the eccentre’s speed to the planet’s is the same as that of the sun’s mean motion (i.e., the sum of the planet’s [mean] motions in longitude and anomaly) to the motion in anomaly.
This statement is pretty dense, and took me awhile to puzzle out (probably because all of these ratios start looking the same after awhile). But, as best I can tell, what Ptolemy is saying is that, for the eccentric model, we don’t have a speed around the epicycle to plug into this ratio.
Instead, what we do have is the eccentre’s speed, which Ptolemy tells us is equal to the sum of the planet’s mean motions in longitude and anomaly. This means that we’ll use $42+37 = 79$ for the numerator while still using the same denominator to get a ratio of $\frac{79}{37}$ and that this is instead equal to the ratio given before we did the dividendo, which was $\frac{P \Theta}{\Theta K}$.
So, this is just a slight twist on the way we could play with these ratios.
That being said it’s not clear at all why Ptolemy is particularly concerned with the eccentric model given that the planets do not use such a model. Pedersen speculates:
Perhaps the most plausible reason is that the…proof of Apollonius’ theorem in terms of the eccentric model seems to be a piece of original work, which Ptolemy found it natural to present inside his general exposition of planetary theory, although it is without any immediate use.
In other words, he speculates that what we are working through is an original proof by Ptolemy and he just stuck it in for the sake of completeness.
Regardless of why, we still aren’t at an actual theory of standstills. And Ptolemy is aware of this, stating the above was simply “preliminary theorems” and that
It remains to prove that, when one takes lines [corresponding to $\overline{ZD}$ and $\overline{B \Theta}$] divided in the ratio described, then in both hypotheses, $H$ and $\Theta$ represent the points in which station appears to take place, and [thus] $arc \; HG \Theta$ must be retrograde, and the remainder [of the circle] possessed forward motion.
So, we’re ready to do the actual proof now, right?
Well, no.
Ptolemy states that (following Apollonius), we’ll need an additional preliminary lemma.
First, we’ll begin with $\triangle ABG$. In this triangle, we need $\overline{BG} > \overline{AG}$.
Furthermore, we’ll create a point $D$ on $\overline{BG}$ such that $\overline{GD} \geq \overline{AG}$.
If the above are true, then Ptolemy tells us that:
$$\frac{\overline{GD}}{\overline{BD}} > \frac{\angle ABG}{\angle BGA}.$$
We’re about to go through a long proof of this but, before we do, we can at least do a quick sanity check: If we imagine stretching the triangle by moving $G$ further and further to the left, then $\angle AGB$ will decrease. Thus, the length of $\overline{GB}$ is inversely proportional to $\angle AGB$ – which we see expressed in the formula presented. The same is true for point $B$ and its angle: $\angle ABG$.
Now, let’s get started on the proof.
First, we’ll create a parallelogram: $ADGE$ and extend $\overline{BA}$ and $\overline{GE}$ such that they intersect at point $Z$.
Then, since
$$\overline{AE} [= \overline{GD}] \geq \overline{AG}$$
we’ll draw an arc of a circle with radius $\overline{AE}$ with one end of that arc on $\overline{BZ}$ at $H$. For the other end of the arc will fall either at $G$, if $\overline{AE} = \overline{AG}$, or somewhere “beyond” it (i.e., if we extended $\overline{AG}$ , it would intersect this line somewhere further from $A$ than $G$).
So here’s what we should have:
Note that, in this version, I’ve drawn things such that $\overline{AE} = \overline{AG}$ which is what Ptolemy instructs us to do, even if it’s not necessary for what he’s wanting to prove.
With this setup complete, we can make two observations:
$$[1] \; \triangle AEZ > sector \; AEH$$
and
$$[2] \; \triangle AEG < sector \; AEG.$$
But, as we can see, $\triangle AEZ > sector \; AEH$ by much more than $\triangle AEG < sector \; AEG$.
The result is that, if we take the ratio we get:
$$\frac{\triangle AEZ}{\triangle AEG} > \frac{sector \; AEH}{sector \; AEG}.$$
We can also state that:
$$\frac{sector \; AEH}{sector \; AEG} = \frac{\angle AEH}{\angle AEG}.$$
Again, no proof is provided, but all this is saying is that the size of a sector is directly related to the central angle of that sector.
Ptolemy then tells us that:
$$\frac{\triangle AEZ}{\triangle AEG} = \frac{\overline{ZE}}{\overline{EG}}.$$
Toomer tells us that this is justified by Proposition $1$, from Book VI of Euclid’s Elements which states that “Triangles… which are under the same height are to one another as their bases.”
In other words, because these two triangles share $\overline{EA}$ (which, if we rotate things, can be the height of both), then the ratio of the areas is equal to the ratios of the bases.
We can now do a bit of substitution.
We’ll be keeping the right hand side of this equation and substituting the left hand side to state that:
$$\frac{sector \; AEH}{sector \; AEG} < \frac{\overline{ZE}}{\overline{EG}}.$$
Flipping sides of the equation so that our final result will line up with Ptolemy:
$$\frac{\overline{ZE}}{\overline{EG}} > \frac{sector \; AEH}{sector \; AEG}$$
And then we can substitute again:
$$ \frac{\overline{ZE}}{\overline{EG}} > \frac{\angle ZAE}{\angle EAG}.$$
Next, Ptolemy tells us that:
$$\frac{\overline{ZE}}{\overline{EG}} = \frac{\overline{GD}}{\overline{DB}}$$
As with before, there’s no proof, but this is just applications of similar triangles with the triangles in question being $\triangle ZEA$ and $\triangle ZGB$, and also noting that $\overline{EA}$ from the first triangle has been substituted for $\overline{GD}$.
Also by similar triangles:
$$\angle ZAE = \angle ABG.$$
Furthermore,
$$\angle EAG = \angle BGA$$
because there are alternate interior angles.
Jumping back to where we left off with our substitutions before we can substitute again:
$$\frac{\overline{GD}}{\overline{DB}} > \frac{\angle ABG}{\angle BGA}.$$
Which is what we set out to prove for this lemma.
And it is obvious that, if $\overline{GD} (=\overline{AE})$ is [taken to be], not equal to $\overline{AG}$, but greater, [then] the difference in the ratios will be even greater.
This has been a pretty good bit of work, so we’ll leave off here and return to the whole question of how this has anything to do with the stations.
Pausing for several months to work on getting my YouTube channel started certainly put a dent in things. But, as I’m wanting to do some videos on the Almagest, I suppose I should get back into the swing of things!
- Much like he wanted to know when eclipses will occur.
- This follows from Euclid III.27.
- Toomer explains that this is a case in which, if we have a two ratios equal to each other in the form $\frac{a}{b} = \frac{c}{d},$ then we can write
$$\frac{\frac{a}{b}}{n} \frac{\frac{c}{n}}{d}.$$
This is the only time in the Almagest this theorem is invoked.