In our last post, I noted that Ptolemy had identified the line of apsides, but was very careful not to state which point was the apogee and which was the perigee. He’s now ready to start looking into that:
In accordance with the above, we investigated the size of the greatest elongations which occur when the mean longitude of the sun is exactly in the apogee, and again, when it is diametrically opposite that point.
He then tells us he will only be able to conduct this investigation for his present time as the historic records did not provide the necessary observations:
We cannot derive this from the ancient observations, but we can do so from our own observations made with the astrolabe.
He then praises the usefulness of this tool stating,
For it is in this situation that one can best appreciate the usefulness of this way of making observations, since, even if those stars with previously determined positions which are visible are not near the planet being observed (which is generally the case with Mercury, since, for the majority of the fixed stars, it is rare that they are visible when they are [only] as far from the sun as Mercury is), one can still determine positions of the planet in question accurately in latitude and longitude, by sighting stars which are at a considerable distance.
In short, Ptolemy is addressing one of the issues I noted in the last post. Specifically, when viewing Mercury as an evening-star, it must be done just before sunset at which point it will be one of the most visible objects in the sky (at greatest elongation). However, if one wishes to check this distance at that same moment by measuring Mercury against some star, then there’s not many options of other stars which would be visible. Thus, you would need to measure over a long distance which Ptolemy tells us the armillary sphere is uniquely suited for.
He then presents his observations:
[Firstly] then, in the nineteenth year of Hadrian, Athyr [III] $14/15$ in the Egyptian calendar [$134$ CE Oct. $2/3$], at dawn, Mercury, which was around its greatest elongation, was sighted with respect to the star on the heart of Leo1, and was seen to have a longitude of $20 \frac{1}{5}º$ into Virgo. The mean sun was at about Libra $9 \frac{1}{4}º$, so the greatest elongation was $19 \frac{1}{20}º$.
[Secondly], in that same year, Pachon [IX] $19$ [$135$ CE Apr. 5], in the evening, [Mercury], which was again around its greatest elongation, was sighted with respect to the bright star in the Hyades, and was seen to have a longitude of Taurus $4 \frac{1}{3}º$. The mean sun had a longitude of Aries $11 \frac{1}{12}º$. Hence, in this case, one calculates the greatest elongation as $23 \frac{1}{4}º$, and it is immediately obvious that the apogee of the eccentre is in Libra and not in Aries.
And there we have it. The apogee is placed in Libra while the perigee is (presumably) in Aries.
How did Ptolemy arrive at this conclusion?
Recall that the greatest elongation is created when the planet is located at the point on its epicycle tangent to the observer. Next, recall circle on which the center of the epicycle rides is off-center from the observer which will naturally bring it closer and further at times.
When it’s closer, that is perigee and it will have a larger greatest elongation at that time since bringing the epicycle closer enlarges the anomaly created by the epicycle. Conversely, when it is furthest away (i.e., at apogee), it will reduce the greatest possible elongation in that portion of the sky since the epicycle is pushed further away.
And that’s precisely what we see here. In Libra, the greatest elongation is $19 \frac{1}{20}º$ while in Aries it is $23 \frac{1}{4}º$. So apogee is in Libra and, presumably, perigee is in Aries.
Except that we’ll soon find out it’s not.
Ptolemy now introduces a diagram:
Here, $B$ is the observer at the center of the ecliptic. We then depict the position of the epicycle at each of the locations on either side of the line of apsides – $G$ being the point $10º$ into Aries and at $A$ into Libra.
Points $D$ and $E$ represent the planet on the epicycle at its greatest elongation with $\overline{AD} = \overline{GE}$ since they’re both the same epicycle radius. . Lastly, $\overline{BE}$ and $\overline{BD}$ are the line of sights for the observer to the planet at greatest elongation (i.e., tangent to the epicycles).
We also know the angles $\angle GBE$ and $\angle ABD$ from the observations Ptolemy just gave. They are $23;15º$ and $19;03º$ respectively.
Now, let’s consider $\triangle ABD$. As usual, to solve this right triangle, Ptolemy employs the demi-degrees method.
It’s been awhile since we’ve done this, so as a refresher, this involves inscribing this triangle in a circle. This necessarily places the hypotenuse ($\overline{AB}$) as the diameter. Then, $\angle ABD$ is an angle on the perimeter across from the chord we’re interested in, $\overline{AD}$. The table of chords works using the central angle in the circle, which will be twice $\angle ABD$ or $38;06º$. If we use that, we can determine the length of $\overline{AD} = 39;09^p$ in the context of this circle where the diameter, $\overline{AB} = 120^p$.
Doing the same for $\triangle{BGE}$, we find that $\overline{GE} = 47;22^p$ in the context of its demi-degrees circle in which $\overline{BG} = 120^p$.
However, that’s two different contexts we’ll need to square. Which we can do because we know that $\overline{AD}$ should equal $\overline{GE}$ since they’re both radii of the epicycles.
Thus, Ptolemy determines what the length of $\overline{BG}$ would be in a context in which $\overline{GE} = 39;09^p$, bringing it into the same context as our first demi-degrees circle.
I’ll set this up as follows:
$$\frac{47;22^p}{39;09^p} = \frac{120^p}{\overline{BG}}$$
Solving this, I find $\overline{BG} = 99;11^p$. Ptolemy comes up with $99;09^p$ which is pretty close and I’ll go with so we stay in line with his work.
We can then add this to $\overline{AB}$ to determine the full length of $\overline{AG} = 219;09^p$.
Now let’s focus on $Z$. This point, Ptolemy now tells us, is drawn such that it bisects this line indicating $\overline{AZ} = 109;34^p$.
And if we then subtract $\overline{BG}$ from this, we determine $\overline{ZB} = 10;25^p$.
But what is $Z$ in the context of our previous diagram?
Ptolemy proposes that it is
the centre of the eccentre on which the centre of the epicycle is always located [which is] the centre of that [eccentre which] moves about point $Z$. For those are the only conditions under which the centre of the epicycle could be equidistant from $Z$ at both the above diametrically opposite situations, as demonstrated.
This statement is a very confusing because it’s not clear if Ptolemy is referring to the geometric center or the center of rotation2.
So I’ll pause for a moment and dive into the model a bit more. I’ll do so by bringing back the model we’d produced for Mercury a few posts ago:
You may recall, that when we set up this model, one of the things we noted is that the circle rotating about $Z$ (in this diagram) rotates clockwise at the same speed as the circle that rotates around $D$, which rotates counter-clockwise.
As a reminder, the former is the one that carries the center of the epicycle, but the epicycle is actually rotating around point $D$, but not at a fixed radius, $\overline{DK}$. What this means is that the circle rotating about $Z$ really just changes the distance of $K$ while the circle around $D$ controls the angular position of $K$ about $D$.
So what does this do to the shape of Mercury’s orbit?
Here, I’ll steal a diagram from Pedersen’s Survey of the Almagest3
Here, Pedersen has added a few things. Firstly, $e$ as the length of $\overline{ED} = \overline{DZ} = \overline{ZH}$. This is commonly known as the eccentricity.
Also, $R$, which is the radius of the eccentre that carries the center of the epicycle.
As you can see, as the ratio of $\frac{e}{R}$ increases4, we can see that the shape changes, pinching in. That will become important later.
But for now, what Ptolemy is asking is, which point, $Z$, $D$, or $E$, is the half way point along the line from the top of the shape to the bottom?
It turns out it’s $Z$. This is what Ptolemy was trying to state when he said,
the centre of the epicycle could be equidistant from $Z$ at both the above diametrically opposite situations.
Except there’s a wrinkle. One which you may be anticipating if you’ve been paying close attention to the observations Ptolemy gave previously, or gave a good bit of thought to the above diagram of the shape of the orbit of Mercury’s epicycle.
But if $Z$ were the actual centre of the eccentre on which the epicycle centre is always located5, that eccentre would be stationary, and the situation in Aries would be the closest to the earth of all situations [i.e., Aries would be the perigee], since $\overline{BG}$ is the shortest of [all] lines drawn from $B$ to the circle described on centre $Z$6.
The way I read this, Ptolemy is asking the hypothetical of what would happen if the center of the eccentre that carried the center of the epicycle7 wasn’t offset (i.e., it was at $Z$ instead of offset at $H$ as he’s defined it)?
In that case, the perigee would be located in Aries.
But, surprise! It’s not.
However, we find that the situation in Aries is not the closest to earth of all, but the situations in Gemini and Aquarius are even closer to the earth than [Aries], and approximately equal to each other.
Recall that the greatest elongation found in Aries was $23 \frac{1}{4}º$. However, if you look back at the previous post where he determined the line of apsides, there were two observations in there with even greater elongations indicating the center of the eccentre was somehow brought even closer than in Aries. This was observation [$4$] in Aquarius and [$6$] in Gemini both of which had elongations of $26 \frac{1}{2}º$.
This is why Ptolemy needed to set the center of the eccentre that carries the epicycle off center at $H$ instead of $Z$, it brings Mercury closer to the earth at two points instead of one. And, it does so in a manner such that the greatest elongations appear in Gemini and Aquarius.
I’m going to break with Ptolemy’s narrative temporarily because there’s something extremely important to understand that Ptolemy just isn’t saying clearly. It’s that the center of Mercury’s epicycle is always in line with the sun. You can see a hint of this when you look at the mean motion table for Mercury, specifically the mean motion in longitude. If you compare that against the solar mean motion table, you’ll see they’re identical.
Why is this important?
It’s because it means that the elongation we’ve been discussing this entire post has two factors: Mercury’s position about the epicycle (i.e., its anomaly) and how close the epicycle is brought to the earth (which magnifies the anomaly). What it doesn’t include is the distance between the sun and the center of Mercury’s epicycle – because there is no difference. We never have to worry about the elongation being the anomaly plus the distance between the sun and the center of Mercury’s epicycle.
At this point, it’s hard to say exactly where the closest point is from the above diagram, and certainly, it will depend on the ratio of $\frac{e}{R}$. But because Ptolemy finds the greatest elongations in Gemini and Aquarius, we know the closest points have to be $26 \frac{1}{2}º$ from those points. This would place them in Cancer and Capricorn8.
Ultimately, this is what Ptolemy means when he says that “Mercury comes closest to earth twice in one revolution” – There are two points where Mercury gets pulled in making a double perigee.
But what of the “too” in the chapter title?
That refers to the moon which, due to its moving eccentre also exhibits this behavior.
Bonus Content: The material here left me feeling like something was missing. Specifically, Ptolemy determines that the closest point of Mercury’s epicycle is $120º$ from apogee using observations, but does this model really produce that?
In this post, I’ve explored that distance as a function of the angle from apogee using modern math and created a Google sheet that allows users to play with the parameters.
TL:DR – Ptolemy’s very close. I come up with a minimum distance at $120.5º$ from apogee.
- α Leo aka. Regulus.
- In this same chapter, Ptolemy will differentiate between the two calling the true center (i.e., $H$) the “actual center”. So the fact that he’s not specific here supports my conclusion that he’s referring to the center of rotation, $Z$.
- I’ve relabeled this diagram to be consistent with the labeling of the points from Toomer, which I have followed.
- Pedersen is evidently holding $e$ constant in this and allowing $R$ to decrease to increase the ratio.
- I.e., if $Z$ were both the rotational and geometrical center of this eccentre thereby removing $H$.
- Toomer cites Euclid’s Elements III.7 as a proof for this.
- That phrase is already getting really annoying to use over and over. Other authors simply refer to this as the deferent which I think is a far better choice. But as I’m trying to stay with Ptolemy, I’ll keep using his phrasing. *sigh*
- I’m not certain why Ptolemy concludes that the closest points are back towards apogee, in Libra instead of towards Aries. Perhaps he understood the shape Mercury’s orbit was making as illustrated here better than he lets on as it precludes that possibility, but if so, he makes no mention of it and leaves it for the reader to puzzle over.