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.
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 equatorial ring with the celestial equator, and ensure that, when the star 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 takes the mean motion tables and calculates the motion beyond full revolutions that the planets should have been expected to have moved about their deferent as this will be how much the center of the epicycle moved about the center of motion 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 $67;50º$ which agrees exactly with Ptolemy. For the second I get $93;44º$, 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, we’ll
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 univorm 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.
Here’s this diagram as we’ve laid it out so far. It may look a bit confusing, but essentially all we’ve done is drawn circles about each of our three centers; the center of uniform speed, the center of fixed 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 $D$), we’ve drawn Mars at these three points on that circle.
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 (that of motion and of the observer) to the positions of Mars. This adds the points $K$, $L$, and $M$ which are the projections of Mars at the times of the three observations onto the circle centered on the observer. It also adds $E$, $Z$, and $H$ which are the projections for Mars onto the circle that is the center of 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, and then removed them 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 $N$.
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. However, as it is, they subtend arcs $AB$ and $BG$ of the middle eccentre, which are not given; and, if 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. 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.
Here’s the additional points that Ptolemy suggests adding, essentially, taking the positions that Mars would have been at thanks to its motion about the equant circle and projecting those onto the circle of the ecliptic.
In short, we don’t really have the right arcs. We have the ones on the circle representing the center of motion/equant. We want the ones on the circle representing the ecliptic, centered on the observer which would be points, $K$, $T$, and $Y$. Unfortunately, what we have on that circle is points $S$, $L$, and $M$.
So what are we to do?
[E]ven without the previous precise determination of the eccentricity and apogee, the arcs are given approximately since the difference arcs are not so lare. Therefore, we shall first carry out the calculation as if the arcs $ST$ and $TY$ did not differ significantly from the arcs $KL$ and $LM$.
What Ptolemy is saying here is that we’ll fudge it for now. Looking at the diagram, we can see that $arc \; ST$ is larger than $arc \; KL$, but not overwhelmingly so. Similarly, we see that $arc \; TY$ is only slightly larger than $LM$. So we’ll do a bit of mathematical cheating here and simply pretend that the arcs observed on the ecliptic are equal to the ones with which we’re concerned.
This will be a first pass to get us a rough estimation of the eccentricity and position of apogee, and then we’ll use our first guess to refine the calculation through an iterative process.
To help us with this, Ptolemy redraws his diagram to focus on the most important pieces for now.
In this figure, the circle $ABGE$, is the “eccentre of mean motion”, i.e., the circle about which Mars moves with uniform speed or the equant circle1.
Point $D$ will be the observer at the center of the ecliptic.
As with before, points $A$, $B$, and $G$ are Mars at the three observations. These will be joined to form the line of sight with the observer as $\overline{DA}$, $\overline{DB}$, and $\overline{DG}$.
Line $\overline{DG}$ is also extended until it meets the side of the circle opposite $G$ at $E$2
We’ll also draw $\overline{AB}$ and $\overline{AE}$ and drop perpendiculars from $E$ onto $\overline{AD}$ at $Z$, from $A$ onto $\overline{BE}$ at $\Theta$, and from $E$ onto $\overline{BD}$ at $H$.
Previously, we said that the motion Mars moved (from the point of view of the observer) about the ecliptic was $93;44º$. That means that $\angle BDG = 93;44º$. However, this does not mean $arc \; BG$ is since $\angle BDG$ is not at the center of this circle. Additionally, we know $arc \; BG = 95;28º$ as this was the mean motion about this circle.
Thus, its supplement, $\angle EDH = 86;16º$.
We can now imagine a demi-degrees circle about $\triangle EDH$. In it, $arc \; EH = 172;32º$ as its twice the angle it subtends.
Therefore, in this context, where the hypotenuse, $\overline{ED} = 120^p$, $\overline{EH} = 119;45^p$.
We’ll hold onto that for a bit, and now focus on $arc \; BG$ which, as we just mentioned, is $95;28º$. Therefore, we can state that the angle subtending this on the far side of the circle, $\angle BEG = 47;44º$. This means we now know the size of two angles in $\triangle BED$. Thus, the remaining angle, $\angle DBE = 46;00º$.
Knowing that, we can now focus on $\triangle BEH$. In it, $arc \; EH = 92;00º$ since it is opposite the angle we just found.
Therefore, the chord, $\overline{EH} = 86;19^p$ in this context where the hypotenuse, $\overline{BE} = 120^p$.
But we just found this same segment in the previous context which will allow us to convert $\overline{BE}$ to that same previous context:
$$\frac{119;45^p}{86;19^p} = \frac{\overline{BE}}{120;00^p}$$
$$\overline{BE} = 166;29^p$$
in the context where $\overline{ED} = 120^p$.
We’ll now turn to $\angle ADG$. This angle is the motion of Mars (from the point of view of the observer) from the first observation to the last. In other words3,
$$\angle ADG = 93;44º + 67;50º = 161;34º.$$
Next, we’re after $\angle ADE$. This angle is:
$$180º – \angle ADG = 180º – 161;34º = 18;26º.$$
Now we can focus on $\triangle ADE$, imagining a demi-degrees circle about it.
In that, $arc \; EZ = 36;52º$ as it’s twice $\angle ADE$. Thus, the corresponding chord, $\overline{EZ} = 37;57^p$, again in the context where $\overline{ED} = 120^p$.
Now let’s actually consider $arc \; AG$ on the circle we actually have. As a reminder, this circle is the circle about the equant or the circle about which the mean motion takes place. Thus, $arc \; AG$ is the sum of the mean motion over this interval or:
$$81;44º + 95;28º = 177;12º.$$
This can be related to $\angle AEG$ which would have half that measure or $88;36º$.
That means we now know two of the angles in $\triangle ADE$ ($\angle AED$ and $\angle ADE$). Thus, we can determine the remaining angle:
$$\angle DAE = 180º – 88;36º – 18;26º = 72;58º$$
And knowing that, we can turn to $\triangle AEZ$, again imagining a demi-degrees circle about it.
In it, the arc opposite the angle just described $arc \; ZE$ is twice the angle or $145;56º$, the corresponding chord of which, $\overline{EZ} = 114;44^p$, in this context where the hypotenuse $\overline{AE} = 120^p$.
We’ve now established $\overline{EZ}$ in two contexts, so we can use that previous one to convert everything to that context:
$$\frac{37;57^p}{114;44^p} = \frac{\overline{AE}}{120^p}$$
$$\overline{AE} = 39;42^p.$$
As a reminder, the context we just converted to was the one in which $\overline{DE} = 120^p$.
Putting that aside from now, we can also recall that $\angle AEB = 81;44º$ as this was the apparent change in Mars’ position between the first observation to the second from the point of view of the observer at $D$.
We’ll use that as we imagine another demi-degrees circle about triangle $AE \Theta$ where the hypotenuse $\overline{AE} = 120^p$.
In it, $arc \; A \Theta = 81;44º$ as well and the corresponding chord, $\overline{A \Theta} = 78;31^p$.
I want to pause here for a moment to consider the context we’re talking about right now. The angle we just referred to to start off this calculation was $\angle AEB$. This is important because that angle was determined because we knew the change in position along the ecliptic from the point of view of the observer.
The circle we’re talking about isn’t the circle of the ecliptic but because, in our previous diagram, Ptolemy stipulated that all of the circles were the same diameter, that means that the pieces we just determined are in the context in which the diameter of the circles, this one included, are $120^p$. This will become important shortly on as we’re ultimately going to convert everything to this context.
Continuing on, we can then use the Pythagorean theorem on the remaining side to determine $\overline{E \Theta} = 90;45^p$.
Again, we’ll convert this to the context in which $\overline{DE} = 120^p$ using $\overline{AE}$ as the common side in these contexts4:
$$\frac{39;42^p}{120;00^p} = \frac{\overline{A \Theta}}{78;31^p}$$
$$\overline{A \Theta} = 25;58^p.$$
Ptolemy uses the demi-degrees conversion to also convert $\overline{E \Theta}$ but we can also convert using the Pythagorean theorem which gives me $\overline{E \Theta} = 30;02^p$ in that context.
Now, recall that we previously showed that the entire length of $\overline{BE} = 166;29^p$. We can subtract $\overline{\Theta E}$ which we just determined off of this to determine $\overline{B \Theta}$:
$$\overline{B \Theta} = 166;29^p – 30;02^p = 136;27^p.$$
Turning now to $\triangle AB \Theta$, we then know two sides of it, so can use the Pythagorean theorem to determine $\overline{AB} = 138;53^p$5 in the context in which $\overline{ED} = 120^p$.
Now we’re finally ready to convert into the context in which the diameter of this circle is $120^p$. To do so, we’ll make use of $\overline{AB}$.
In that case:
$$\overline{78;31^p}{138;53^p} = \frac{\overline{ED}}{120^p}$$
$$\overline{ED} = 67;50^p.$$
And for $\overline{AE}$:
$$\overline{78;31^p}{138;53^p} = \frac{\overline{AE}}{39;42^p}$$
I’m forced to pause here because Ptolemy makes a few mistakes in a row.
First, in solving for $\overline{AE}$ I find it to be $22;27^p$ which Toomer also reports is correct. Yet, somehow, Ptolemy comes up with $22;44^p$.
Next, Ptolemy converts the chord to the corresponding arc, $arc \; AE$.
When I do so with Ptolemy’s value, I find it to be $21;50º$. Ptolemy manages to come up with $21;41º$. If I use the correct value for the chord, I find the arc to be $21;34º$, which Toomer again affirms is correct.
Thus, it’s hard to explain where Ptolemy came up with these values. Toomer has a few notes on what values are found in various manuscripts but concludes that it is most likely that the error is Ptolemy’s and not simply an error in transcription. Please refer to his footnote $40$ for details.
As far as what to do with this error, I’ll adopt Ptolemy’s value of $arc \; AE = 21;41º$ for the arc as we’ll now add it to $arc \; AG$ to get $arc \; EABG$:
$$177;12º + 21;41º = 198;53º.$$
This means that the remaining $arc \; GE = 161;07º$.
The corresponding chord, $\overline{GE} = 118;22^p$, again in the context where the diameter of this circle is $120^p$.
So what of it?
Now, if $\overline{GE}$ had been found equal to the diameter of the eccentre, it is obvious that the centre would lie on $\overline{GE}$, and the ratio of the eccentricity would immediately be apparent. But, since it is not equal [to the diameter], but makes [arc] $EABG$ greater than a semi-circle, it is clear that the centre of the eccentre will fall within the latter.
What Ptolemy is saying here is that if $\overline{GE}$ would have equaled $120^p$, then it would have meant it passed through the center of the circle. But it doesn’t, so point $D$ can’t be the center of the circle but is, instead, some distance away, in the part of the circle with $arc \; EABG$ on its edge.
So we’ll still have to do a bit more work to find out how far off of the center $D$ ended up being. But as this point has gotten quite long, we’ll explore that in the next post!
- Centered on $\Theta$ in our last diagram.
- At this point in the description, Ptolemy becomes rather agnostic about which points are used which way. Instead of extending $\overline{GD}$, we could have extended any of the lines from the observations in the same manner. We would then have to adjust the rest of the lines we’ve drawn. For the sake of clarity, I’m describing the diagram as drawn instead of giving the more general description as Ptolemy does.
- Ptolemy seems to incorrectly imply that these are equal to the arcs in this circle, but they are not.
- I come up with $\overline{A \Theta} = 25;59,33^p$ which should round to 26;00^p, but take Ptolemy’s value here for consistency.
- I get $\overline{AB} = 138;54^p$, but again adopt Ptolemy’s value for consistency.