In the last Almagest post, Ptolemy explained that for the models of the superior planets, we’ll be needing to use oppositions. So with that explained, we’ll begin by working to develop the model for Mars by looking at its eccentricity and apogee. However, this isn’t going to be nearly as straightforward as it has been previously. As we’ll see, we’re missing some of the angles and arcs that would be needed for this. Thus, Ptolemy is going to use an iterative method to close in on them.
As with many previous posts, we’ll start with a series of observations. He tells us that these observations were taken using the “astrolabe”. However, we need to recall that what Ptolemy called the astrolabe, we would think of as an armillary sphere. Thus, the methodology here was to align the ecliptic ring with the ecliptic, and ensure that, when the planet was sighted, its position was $180º$ from the calculated position for the mean sun.
So here are the observations:
$1$. The first in the fifteenth year of Hadrian, Tybio [V] $26/27$ in the Egyptian calendar, [December $14/15$, $130$ CE], $1$ equinoctial hour after midnight, at about $21º$ into Gemini.
$2$. The second in the nineteenth year of Hadrian, Pharmouthi [VIII] $6/7$ in the Egyptian calendar [February $21/22$, $135$ CE], $3$ hours before midnight, at about $28;50º$ into Leo.
$3$. The third in the second year of Antoninus, Epiphi [XI] $12/13$ in the Egyptian calendar [May $27/28$, $139$ CE] $2$ equinoctial hours before midnight, at about $2;34º$ into Sagittarius.
These then get paired up and broken into intervals:
From opposition [$1$] to [$2$], [the period was] $4$ Egyptian years, $69$ days, and $20$ equinoctial hours.
From [$2$] to [$3$], $4$ years, $96$ days, and $1$ equinoctial hour.
Ptolemy then looks to the mean motion tables and calculates the motion beyond full revolutions that the center of Mars’ epicycle moved about the equant in that time.
For the first interval, I come up with $81;43,27º$ which is in decent agreement with Ptolemy’s $81;44º$. For the second interval I come up with $95;27,31º$ which rounds exactly to Ptolemy’s $95;28º$.
Ptolemy also looks at the arcs from the actual observations, (i.e., the apparent distance moved as observed by someone on Earth) which is just the difference of ecliptic longitude between each of the pairs of observations.
The increase in the first interval I find to be $67;50º$1 which agrees exactly with Ptolemy. For the second I get $93;44º$2, again, agreeing with Ptolemy.
Now we’ll produce the diagram we’ll be using to work things out. I’ll go ahead and do this in several steps as this diagram is going to be rather busy.
First,
let there be drawn, in the plane of the ecliptic, three equal circles: let the circle carrying the epicycle centre of Mars be $ABG$ on centre $D$, the eccentre of uniform motion $EZH$ on centre $\Theta$, and the circle concentric with the ecliptic [be] $KLM$ on centre $N$, and let the diameter through all [three] centres be $XOPR$. Let $A$ be the point at which the epicycle centre was at the first opposition, $B$ the point where it was at the second opposition, and $G$ the point where it was at the third opposition.
I’m going to break this down a bit more slowly than Ptolemy does, so I’ll be skipping a few points for now.
First, let’s start with the three circles Ptolemy describes.
1. The topmost one with $X$ and $P$ on its circumference is the equant circle – the circle of mean motion, and is centered on $\Theta$.
2. The second one, which includes points $A$, $B$, and $G$ on its circumference, is the eccentric of mean distance. This one is centered on $D$ and the center of Mars’ epicycle is always on this circle.
3. The bottommost circle, with $O$ and $R$ on the circumference is the ecliptic, which has the observer, on Earth, at its center, $N$.
Here’s this diagram as I’ve described it so far. It may look a bit confusing, but essentially all we’ve done is drawn three circles about each of our three centers; the center of uniform speed (equant), the center of mean distance, and the center of the ecliptic (i.e., the observer).
Since the center of Mars’ epicycle rides on the second of these (i.e., about $D$), I’ve drawn Mars at these three points on that circle, which are $A$, $B$, and $G$.
I’ve cheated a bit and looked ahead to know where to place them on that circle, so let’s explore at least part of that now.
Join, $\overline{\Theta AE}$, $\overline{\Theta BZ}$, $\overline{\Theta HG}$, $\overline{NKA}$, $\overline{NLB}$, and $\overline{NGM}$. Then $arc \; EZ$ of the eccentric [equant] is $81;44º$, the amount of the first interval of mean motion, and the $arc \; ZH$ is $95;28º$, the amount of the second interval.
In this step, we’ve extended lines from two of our centers (the equant and of the observer) to the positions of Mars for each of the observations. This adds the points $K$, $L$, and $M$ which are the projections of Mars at the times of the three observations onto the ecliptic from the point of view of the observer. It also adds $E$, $Z$, and $H$ which are the projections for Mars onto the equant circle from the point of view of the center of mean motion.
The last portion of Ptolemy’s instructions is also how I knew where to place $A$, $B$, and $G$ relative to one another because I’d actually drawn in $E$, $Z$, and $H$ with the spacings given by the motion in the mean motion tables3, and then removed them from the previous diagram temporarily so as not to complicate the last step.
That still doesn’t tell us how I knew where to position them relative to the line of apsides, but we’ll come to that in time.
We also know is that $arc \; KL$ is $67;50º$ and $arc \; LM$ is $93;44º$ as these we there apparent changes from the point of the view of the observer at $N$ as calculated above.
If this setup is feeling familiar, it should, because it’s quite similar to the one we set up in IV.6 when working on the moon. However, we can’t quite use the same tricks here because the center of motion/equant being removed from the center of motion changes things slightly. Or, as Ptolemy puts it,
[I]f arcs $EZ$ and $ZH$ of the eccentric [equant] were subtended by arcs $KL$ and $LM$ of the ecliptic, that would be all we would need in order to demonstrate the eccentricity.
I’ll break from Ptolemy here to explain the problem a bit better.
What Ptolemy is really doing is looking at angles. When he talks about $arc \; EZ$ and $arc \; ZH$ (both on the equant circle), he’s similarly referring to the central angles they subtend: $\angle E \Theta Z$ and $\angle Z \Theta H$ respectively. These are the angles Mars would have appeared to move between observations if viewed from the equant, $\Theta$.
However, we’re not sitting at the equant. Rather, we’re sitting at $N$ and we observe the change between oppositions along the ecliptic circle, so $\angle KNL$ for the first interval and $\angle LNM$ for the second interval.
Ptolemy is comparing these, noting that, for the first interval, the angular change when viewed from the equant ($\angle E \Theta Z$) is greater than it if it were viewed from Earth ($\angle KNL$).
Similarly, for the second interval, the angular change when viewed from the equant ($\angle Z \Theta H$) is less than if it were viewed from Earth ($\angle LNM$).
This is problematic because we’re going to be determining the line of apsides and eccentricity assuming we know what the changes look like from $\Theta$ when in fact, the observations are from the point of view of us here on Earth at $N$.
However, as it is, they subtend arcs $AB$ and $BG$ of the middle eccentre, which are not given;
What Ptolemy is pointing out here is that $\angle KNL = \angle ANB$ and $\angle LNM = \angle BNG$. However, the vertexes of the angles I just gave are all from the observer at $N$. Since $N$ isn’t the center of the middle circle ($D$ is), this means that the arcs he just described won’t have the same measure.
So, as Ptolemy states, we don’t know $arc \; AB$ or $arc \; BG$.
[I]f we join $\overline{NSE}$, $\overline{NTZ}$, and $\overline{NHY}$, we again find that arcs $EZ$ and $ZH$ of the eccentric [ecliptic] are subtended by arcs $ST$ and $TY$ of the ecliptic, which are, obviously, not given either.
Here’s the additional points that Ptolemy is adding, essentially, taking the projected positions of Mars onto the equant circle as viewed from the equant (points $E$, $Z$, and $H$), and connecting those to the observer at $N$, noting where they intersect the ecliptic, at points $S$, $T$, and $Y$ respectively.
Ptolemy continues:
Hence, the difference arcs, $KS$, $LT$, and $MY$, must first be given, in order to carry out a rigorous demonstration of the ratio of the eccentricity starting from the corresponding arcs, $EZ$, $ZH$, $ST$, and $TY$. But the latter [arcs $ST$ and $TY$] cannot be precisely determined until we have found the ratio of the eccentricity and [the position of] the apogee.
[E]ven without the previous precise determination of the eccentricity and apogee, the arcs are given approximately since the difference arcs are not so large. Therefore, we shall first carry out the calculation as if the arcs $ST$ and $TY$ did not differ significantly from the arcs $KL$ and $LM$.
Here, Ptolemy is starting to preview what we’re going to be doing in the rest of this very long chapter.
Going beyond what Ptolemy tells us here, our method will be as follows:
1. We lack the correct angle/arc to be able to correctly calculate the eccentricity and line of apsides. So instead, we’ll use one that’s pretty close which we do know. Specifically, we need to know $\angle ZNH$, but what we actually know is $\angle LNM$ which is close as it differs by angles $\angle LNT$ and $\angle YNM$.
2. We’ll use this first approximation to determine these angles to get a better grasp of what we should have been using in (1). However, this will still be wrong because the calculation of those angles was still based on an incorrect angle in the first place. But it will give us a better approximation for (1).
3. Having determined better values for this, we’ll repeat (1), and then repeat (2) in order to calculate even better values for (1) in an iterative process, repeating until Ptolemy is satisfied with the results4.
We’ve covered a lot of ground already, so I’m going to break here and we’ll start on the first iteration in the next post.
NOTE: This post was originally part of the following post, but after making several changes based on insight gained later in the chapter, I’ve decided to break it into its own post.
- We’ll be calling this $arc \; KL$ shortly.
- Which we’ll be calling $arc \; LM$.
- I.e., $arc \; EZ$ and $arc \; ZG$ were $81;44º$ and $95;28º$ respectively.
- The good news is that there will only be three iterations for Mars, and only two for Jupiter and Saturn. However, past the first iteration, Ptolemy doesn’t show any work, which means the number of posts coming out of this chapter is going to be disproportionately large since I do want to walk through the calculations.