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 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 Perga2.
Both lemmas are pretty dense, so I’ll try to break them down.
[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,
In this portion, Ptolemy is simply stating that this will apply to the basic epicyclic model – a simple epicycle with the deferent’s center on the Earth, with both the deferent and epicycle rotating counter-clockwise.
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
Let’s draw that out:
Here, I’ve drawn an epicycle, centered on $E$ with the observer at $Z$. This makes $A$ the apogee and $G$ the perigee. We’ll draw the planet in at $H$ which makes the line of sight to the planet $\overline{ZH}$, which we’ll extend to the opposite side of the epicycle at $B$. We’ll also drop a perpendicular from $E$ such that it falls on this line perpendicularly at $P$. This means that it bisects $\overline{BH}$ making $\overline{BP} = \overline{PH}$.
What Ptolemy is telling is is that the ratio of
$$\frac{\overline{BP}}{\overline{HZ}} \; or \; \frac{\overline{PH}}{\overline{HZ}}$$
is going to have some special meaning. He tells us that, if it
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.
“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.
So, reviewing this last section: if the above ratio is equal to the ratio of the speed of the planet about the epicycle to the mean speed, then that point, $H$ is where the planet will be when it changes from prograde to retrograde motion.
He’s not proving this just yet. For now, he’s just laying out the theorem as presented by Apollonius.
However, there’s a second one.
[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,
While Ptolemy refers to this as the “eccentric hypothesis” it’s not the simple eccentric we had for the sun in which the center of the eccentre is fixed. In this version, its center rotates counter-clockwise about the earth and the eccentre itself rotates clockwise. This isn’t something we’ve really encountered in the Almagest3, so it’s unclear why this is being presented at this point. Perhaps it was part of Apollonius’ theorem and Ptolemy is simply presenting it for completeness?
Pedersen writes in A Survey of the Almagest:
There remains the question of why [Ptolemy] bothers to introduce an eccentric model of this kind. It is very dissimilar to the model he actually chose for describing the motion of the superior planet, with a fixed eccentric deferent to account for the first anomaly, and an epicycle to account for the second. Perhaps the most plausible reason is that the following proof of Apollonius’ theorem in terms of the eccentric model seems to be a piece of original work, which Ptolemy found natural to present inside his general exposition of planetary theory, although it is without any immediate use.
As an additional note, Ptolemy states that this version of the eccentric model only works for the outer planets, but it is just as applicable for Mercury and Venus as well.
Returning to Ptolemy:
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
Again, let’s draw a picture to help:
Here, we have circle $AHGD$ as the eccentre, centered on $E$, with the planet still at $H$. As instructed, we’ll draw a line through the observer at $\Theta$, extending it until it hits the other side of the eccentre at $D$.
We then want half of this line, which we again find this by dropping a perpendicular from $E$ at $P$.
Then, the ratio we want is half of the line (i.e., $\overline{HP}$ or $\overline{PD}$) to the shorter of the two segments when split by the observer at $\Theta$. This means $\overline{H \Theta}$.
So, the ratio this time is:
$$\frac{\overline{HP}}{\overline{H \Theta}} \; or \; \frac{\overline{PD}}{\overline{H \Theta}}.$$
Now that we know that, what ratio of speeds is this equal to?
is equal to the ratio of the speed of the eccentre to the speed of the planet,
Again, Ptolemy doesn’t clearly define the “speed of the eccentre” or the “speed of the planet” here, but we’ll come to an example a bit later which will make this clearer.
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.
So that’s the two theorems which Ptolemy has, thus far, presented without proof. But before Ptolemy presents his proof, he needs to present a few preliminary proofs to assist with this.
Ptolemy has a statement that this proof will
contain both hypotheses combined in a common [figure], to demonstrate their agreement and similarity in these ratios of theirs too.
Proof 1
We’ll begin with the following diagram.
In this, Ptolemy tells us he wants to prove that
$$\frac{\overline{AZ}}{\overline{ZG}} = \frac{\overline{AK}}{\overline{KG}}.$$
In this image, we have the epicycle as $ABGD,$ centered on $E.$ We’ll draw in the observer on Earth at $Z$4.
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.
With that out of the way, 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 $\overline{LM} \parallel \overline{AD}$ $\angle DGH$ and $\angle DGM$ are right angles as well.
Additionally, $\angle GDH = \angle GD \Theta$5.
Now, let’s focus on $\triangle GLD$ and $\triangle GMD.$ These two triangles are congruent. So let’s divide $\overline{AD}$ by one of the equal portions of each of these triangles:
$$\frac{\overline{AD}}{\overline{GL}} = \frac{\overline{AD}}{\overline{GM}}.$$
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:
$$\frac{\overline{AD}}{\overline{GM}} = \frac{\overline{AZ}}{\overline{GZ}}.$$
That’s gives us something we can substitute for the right hand side of the above equation.
Similarly, $\triangle ADK$ is similar to $\triangle GLK$ allowing us to write:
$$\frac{\overline{AD}}{\overline{GL}} = \frac{\overline{AK}}{\overline{KG}}.$$
That gives us something we can substitute for the left hand side of the above equation.
Now, substituting both sides, we get:
$$\frac{\overline{AZ}}{\overline{ZG}} = \frac{\overline{AK}}{\overline{KG}}.$$
Closing out this proof, Ptolemy 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 is where Ptolemy has demonstrated the equivalence between the two models. That the two models return similar results should be unsurprising since we’ve already shown some equivalences back in III.3.
But what we now have is 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.
Proof 2
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 although the lines in question will remain the same:
Here, 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
$$\frac{\overline{DZ}}{\overline{Z \Theta}} = \frac{\overline{BK}}{\overline{K \Theta}}.$$
That concludes what Ptolemy wanted to prove in this section.
Componendo & Dividendo
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, allowing you to be 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 it6.
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:
$$\frac{\overline{OZ}}{\overline{Z \Theta}} = \frac{\overline{P \Theta}}{\overline{K \Theta}}.$$
I’ll pause here for a moment and lets examine the right hand side of this equation. Although we’ve relabeled things $\frac{\overline{P \Theta}}{\overline{K \Theta}}$ is exactly the ratio from the second theorem above for the eccentric model: Half of the line drawn through the planet and the observer, divided by the line from the observer to the planet.
Moving on, this is still an equivalence of ratios, so he 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:
$$\frac{\overline{O \Theta}}{\overline{Z \Theta}} = \frac{\overline{PK}}{\overline{K \Theta}}.$$
This gives us a third handy set of ratios.
Let’s pause and take a look at the left hand side of this equation. Although we’ve changed what our lettering is, and which side of the diagram we’re looking at, this ratio is half of the line inside the epicycle to the portion of the line outside of it. In other words, it’s the ratio from Apollonius’ first theorem regarding the epicyclic model. The right hand side simply equates this to the eccentric model.
Relationship to Speeds
Ptolemy then has a section relating these ratios to speeds of the various components of the eccentric and epicyclic models. This bit is very confusing, and I’ll admit, I haven’t made complete sense of it yet. Hence why I’m breaking this out. I’ll go ahead and quote Ptolemy here, but will be light on discussion in hopes that I can return to this later with more insight7.
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.
This section makes some amount of sense to me. We’ve just established that $\frac{\overline{O \Theta}}{\overline{Z \Theta}} = \frac{\overline{PK}}{\overline{K \Theta}}.$ But what we haven’t really established is that:
$$\frac{\overline{O \Theta}}{\overline{Z \Theta}} = \frac{speed \; of \; the \; epicycle}{\overline{speed \; of \; the \; planet}}.$$
As far as I see it, that statement is what Apollonius’ theorem is all about which we haven’t proven yet. So my feeling is that Ptolemy is kind of skipping ahead, taking for granted that we accept Apollonius’ theorem, even though it remains to be proven.
Ptolemy continues on:
The reason that in this case [i.e. in the eccentric hypothesis] we do not use this ratio obtained by 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 epicycles 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.
As best I can tell, this statement is pointing out that the requirement for the second part of Apollonius’ theorem (regarding the eccentric model) made use of the first “dividendo” (in which we got $\frac{\overline{P \Theta}}{\overline{K \Theta}}$) instead of the final result, after the second dividendo (in which we got $\frac{\overline{PK}}{\overline{K \Theta}}$ for the right hand side).
I’m still confused as to how this relates to the speeds.
But assuming the speeds, Ptolemy then provides an example:
Thus, e.g. for Mars,
$$\frac{speed \; of \; epicycle}{speed \; of \; planet} \approx \frac{42}{37}$$
(for that, approximately, is the ratio which, as we demonstrated, holds between the [mean] motions in longitude and anomaly).
Although I’m still uncertain how Ptolemy is relating these speeds to the components in the diagram, it is clear where these numbers came from. They’re from our discussion on the planetary periods from IX.3. There we said that there were $37$ returns in anomaly per $42$ revolutions of the planet.
Ptolemy again tells us that this is equal to $\frac{\overline{O \Theta}}{\overline{\Theta Z}}$ from Apollonius first portion regarding the epicyclic model.
However, for the epicyclic model, he reminds us that:
But
$$\frac{speed \; of \; eccentre}{\overline{speed \; of \; planet}} \approx [42 + 37 =] \frac{79}{37}.$$
This, Ptolemy tells us, corresponds to
the ratio obtained componendo, $\frac{\overline{P \Theta}}{\overline{\Theta K}}$
since we found that the divided ratio, $\frac{\overline{PK}}{\overline{K \Theta}}$, is equal to $\frac{\overline{O \Theta}}{\overline{\Theta Z}} $(i.e. $\frac{42}{37}$).
That concludes the discussion relating these ratios to the speeds. However,
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.
In other words, after all this, we still need to prove Apollonius’ theorem. So, we’re ready to do that now, right?
Well, no.
Ptolemy states that (following Apollonius), we’ll need an additional preliminary lemma.
Additional 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:
$$\frac{\overline{GD}}{\overline{BD}} > \frac{\angle ABG}{\angle BGA}.$$
To prove this, 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:
$$[1] \; \triangle AEZ > sector \; AEH$$
since the sector is clearly contained within the triangle
and
$$[2] \; \triangle AEG < sector \; AEG$$
since the triangle is clearly contained within the sector, even when $\overline{AE} = \overline{AG}.$
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:
$$\frac{sector \; AEH}{sector \; AEG} = \frac{\angle EAZ}{\angle EAG}.$$
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.
- Neugebauer in HAMA informs us that Ptolemy’s treatment of these theorems is the only information that survives about them. Furthermore, “it remains unknown how far Apollonius tried to coordinate his cinematic theory with empirical data.”
- Pedersen notes that it did crop up briefly in IV.6 and assumes that the passages there are also referencing Apollonius.
- Note that, at this time, Ptolemy is not referencing the eccentric model at all. We will and, as you’re probably expecting, the observer will be at $K$ in that model.
- 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.
- I’ve tried referring to Neugebauer, but he skips pretty much this entire section and instead replaces it with a very short modern proof. Pedersen’s treatment is a more in depth, but seemingly displaces the discussion until later, after Apollonius’ theorem is proven. I’m hoping that by following suit, it will make more sense.