Now that Ptolemy has explained why the eccentric eccentre on which the center of the epicycle resides is necessary, Ptolemy informs us
we have still to demonstrate the position of the point on $\overline{AB}$ about which takes place the annual revolution of the epicycle in uniform motion towards the rear with respect to the signs, and the distance from $Z$ of the centre of that eccentre which performs its revolution in advance in the same period [as the previous].
In other words, it’s time to start figuring out all the parameters to calibrate Mercury’s model. In this post, we’ll explore the various eccentricities. Ptolemy has previously told us that they’re all equal, but now he will demonstrate this.
For this investigation we used two observation of greatest elongations, one as a morning-star and one as evening-star, but in both of which the mean position [of Mercury’s epicycle] was a quadrant from the apogee on the same side: that is the situation in which, approximately, the greatest equation of ecliptic anomaly occurs.
This statement is rather tricky and it’s a big part of what slowed me down on working through this section. In previous posts, we’ve been discussing how the movement closer to and further from the earth would magnify the anomaly. So this is what comes to mind when discussing the anomaly.
But that’s actually not what Ptolemy is discussing here. Rather, he’s specifically discussing the ecliptic anomaly. In short, he’s trying to determine the effect of Mercury’s center of rotation being offset at point $D$, instead of at $E$, in our original diagram.
We mostly explored the eccentric model’s fundamentals in Book III, but one of the things we never explicitly stated is that the anomaly from an eccentric is maximized near $90º$ from apogee as can be clearly seen by taking a look at the Table of the Solar Anomaly. This is why Ptolemy wants to explore it in this portion of the sky as opposed to where Mercury is closest. This happens when the center of Mercury’s epicycle is $\approx 10º$ into Cancer and Capricorn.
Now Ptolemy introduces two new observations:
[$1$] In the fourteenth year of Hadrian, Mesore [XII] $18$ in the Egyptian calendar [$130$ CE July $4$], in the evening, as we found in observations we got from Theon1. He says that [Mercury] was at its greatest distance from the sun, $3 \frac{5}{6}º$ behind [i.e., to the rear of] the star on the heart of Leo. Thus, according to our coordinates, its longitude was about $6 \frac{1}{3}º$ into Leo, while the longitude of the man sun at that moment was about $10 \frac{1}{12}º$ into Cancer. Thus, the greatest evening elongation was $26 \frac{1}{4}º$.
[$2$] In the second year of Antoninus, Mesore [XII] [$20$]$/21$2 in the Egyptian calendar [$139$ CE July $4/5$], at dawn, we observed its greatest distance by means of the astrolabe. Sighting it with respect to the bright star in the Hyades, we found its longitude as $20 \frac{1}{12}º$in Gemini. The mean sun was, again near $10 \frac{1}{3}º$ into Cancer. Thus, the greatest morning elongation was $20\frac{1}{4}º$.
Here, Ptolemy is giving two observations, both with the sun (and therefore the center of Mercury’s epicycle) at the point $90º$ from apogee (about its own center, $H$). In the two observations, Mercury is on opposite sides of the sun which is to say, opposite sides of the epicycle. This makes Mercury appear back in Gemini in one observation and forward in Leo in the other.
He then sets this up as a simple diagram:
In this, $\overline{AG}$ is the line of apsides with $A$ as the point towards apogee and $G$ as the point towards Aries. $Z$ is the point about which the eccentre carrying the epicycle’s center rotates. $H$ is the point that rotates the epicycle. $B$ is the the observer at the center of the ecliptic. $\Theta$ is the center of the epicycle with $\overline{\Theta K}$ and $\overline{\Theta L}$ as radii such that points $K$ and $L$ are tangent (i.e., at greatest elongation).
The first thing Ptolemy will determine is $\overline{BH}$.
As discussed above, we’ll let $\angle AH \Theta = 90º$.
Turning to the observations, the greatest elongations observed where $26 \frac{1}{4}º$ and $20\frac{1}{4}º$. So where are those angles in this diagram?
They’re not really there. So let me add a bit to the diagram to help illustrate:
This is the same diagram as the previous one, but here, I’ve added the Sun at $S$. As noted in the last post, the rotation of the center of Mercury’s epicycle is directly tied to that of the sun. However, that’s only from the point of view of the center of rotation of the eccentre – not the observer. In other words, it’s $\overline{HS}$ always has $\Theta$ on it – not $\overline{B \Theta}$.
Thus, $\angle SEL$ is different than $\angle SEK$ and are the observed greatest elongations.
But, what we can also note is that their sum,
$$\angle SBL + \angle SBK = \angle KB \Theta + \angle \Theta BL = \angle KBL$$
Thus, $\angle KBL = 20\frac{1}{4}º + 26 \frac{1}{4}º = 46;30º$.
Next, we can notice that $\triangle KB \Theta = \triangle \Theta BL$. This is because they both have a side that is tangent to the epicycle meaning they have a right angle at that point. They also each have a side that is a radii of the epicycle (and thus equal) and also share a side ($\overline{B \Theta}$). Thus, two of their three sides are the same in a right triangle, meaning they must be equal.
This means that $\angle KB \Theta = \angle \Theta BL$. In other words, each is half of $46;30º$ which is $23;15º$
Ptolemy now considers $\triangle KB \Theta$ and uses the demi-degrees method to solve it.
In this circle we now know $\angle KB \Theta = 23;15º$.
Thus, the angle from the center of the demi-degrees circle is twice that, or $46;30º$.
The corresponding chord would then be $\overline{\Theta K} = 47;22^p$ in the context in which the hypotenuse, $\overline{B \Theta} = 120^p$.
However, we determined earlier that the radius of the epicycle is $39;09^p$ in a context in which $\overline{AH} = 120^p$. Thus, we’ll want to convert to that context to keep everything consistent.
So we’ll convert $\overline{B \Theta}$ to that context3:
$$\frac{47;22^p}{39;09^p} = \frac{120^p}{\overline{B \Theta}}$$
Solving, I find $\overline{B \Theta} = 99;11^p$. Again, Ptolemy rounds to $99;09^p$. If that sounds familiar, it’s because we got exactly the same result in the last post.
Ptolemy next reminds us that, in this same context, we previously determined, $\overline{BZ} = 10;25^p$.
Again, the difference between the above greatest elongations, $6º$, comprises twice the equation of the ecliptic anomaly; and the latter is represented by $\angle B \Theta H$, as we proved previously4.
The first portion of this sentence states that the difference in the elongations is going to be twice the anomaly. The phrasing here seems to indicate that Ptolemy has already demonstrated this, but I cannot find where. However, it’s fairly easy to understand if we look at the previous diagram in which I added the sun.
If we imagine that $S$ were on $\overline{B \Theta}$ instead of $\overline{H \Theta}$, then $\angle KB \Theta$ would always be equal to $\angle KBS$. Similarly, $\angle \Theta BL = \angle SBL$. In other words, there would be no difference in elongation on either side of the sun and the maximum elongation would always be exactly equal to the anomaly.
But that’s not the case; $S$ lies on $\overline{H \Theta}$.
This has the effect of decreasing $\angle SBL$ when compared to $\angle \Theta BL$ in this diagram. However, because of the equivalences we stated above, by whatever amount $\angle SBL$ is decreased by, $\angle KBS$ must be increased by the same amount. This means that, if $S$ were $1º$ away from $\Theta$, the $\angle KB \Theta$ would increase by $1º$ while $\angle \Theta BL$ would decrease by $1º$. Thus, the difference in elongation on either side would be $2º$.
So taking that the other way around, the difference in the elongations tells us the anomaly. And as we’ve seen previously, the anomaly can also be represented by, what in this diagram is $\angle H \Theta B$.
Referring back to our elongations, the difference was $6º$, so the anomaly must be half of that, or $3º$.
Ptolemy now turns to solve $\triangle B \Theta H$. In this, the $\angle B \Theta H = 3º$ and we’ll create the context in which its hypotenuse, $\overline{B \Theta} = 120^p$.
In that context, $\overline{BH} = 6;17^p$.
However, in our previous context, we showed that $\overline{B \Theta} = 99;09^p$, so we’ll convert $\overline{BH}$ to that context:
$$\frac{\overline{BH}}{6;17^p} = \frac{99;09^p}{120;00^p}$$
$$\overline{BH} = 5;11;30^p$$
Ptolemy rounds to $5;12^p$ which I will, again, adopt for consistency.
What is interesting about this is that this implies that $\overline{BH}$ is half of $\overline{BZ}$. Previously, we’ve stated this was true, but now we have confirmation.
However, we’ve also stated that this distance should be equal to the distance between the true center and the rotational center of the third eccentre which controls the distance to the center of Mercury’s epicycle. Ptolemy now wants to verify this as well and so draws a new diagram:
Repeating what I’ve just said in the context of this new diagram, $\overline{ZH} = \overline{HB}$ and we want to check whether that’s equal to $\overline{ZM}$.
We’ll also be letting $\overline{ZA} = \overline{ZN}$, essentially saying $A$ is the point $N$ would be at when Mercury is at apogee.
Ptolemy now makes a very weird simplification. He states that $\angle MZH$ is a right angle, which is perfectly fine. However, this should also mean that $\angle ZH \Theta$ is as well, since the two eccentres rotate equally but opposite. However, Ptolemy instead treats $\angle \Theta ZH$ as a right angle.
Toomer notes that this is a simplification that is necessary to solve the problem at all.
Effectively, what it does is makes $\overline{NZ \Theta}$ a straight line, even though it really has a bend in it. And as long as $\overline{Z \Theta} \gg \overline{ZH}$, it’s not a horrible assumption5.
So if we treat this as a straight line, we know from this post that $\overline{AZ} = 109;34^p$. Furthermore, $\overline{Z \Theta} = \overline{B \Theta}$6 for which we’ve shown (both in that previous post and this one) that $\overline{B \Theta} = 99;09^p$.
Thus, if we pretend this is all a straight line, $\overline{NZ \Theta}$ is the sum of the two parts or $208;43^p$.
Next, recall that $\Theta$ must also be on the perimeter of the eccentre which rotates around $Z$, but has its true center at $M$. This means that $M$ is half of this value from each of these points or $104;22^p$ from either one.
Subtracting that from the length of $\overline{NZ}$ we then get $\overline{MZ} = 5;12^p$, which matches what we’ve gotten previously for $\overline{ZH}$ and $\overline{HB}$ thereby justifying Ptolemy’s insistence that each of these is equal.
Ptolemy now summarizes:
Thus, we have computed that
where the radius of the eccentre is $104;22^p$
each of the distances between the centers [$\overline{BH}$, $\overline{HZ}$, and $\overline{ZM}$] is $5;12^p$
and the radius of the epicycle is $39;09^p$.
He then converts this to a context in which the radius of the eccentres is $60^p$. In that case
each of the distances between the centers [$\overline{BH}$, $\overline{HZ}$, and $\overline{ZM}$] is $3;00^p$7
and the radius of the epicycle is $22;30^p$.
I’ll leave this post here for now, but in the next post, we’ll be looking at the apparent angular size of Mercury’s epicycle when at perigee.
- Toomer notes that the identity of Theon is uncertain due to how common the name was.
- Toomer notes that there is some disagreement on this date but doesn’t give much details. They can be found in Neugebauer who notes that all texts give the date as Mesore $24$. However, this is an incorrect reading of ka as kd, an easy mistake to make, that evidently occurred at some point very early in the lineage of manuscripts. The misreading is evident if the solar position is calculated for the two dates as the given position is only correct for the $21^{st}$ (ka).
- Recalling that $\overline{B \Theta}$ is the hypotenuse and thus has a length in the context from which we’re converting of $120^p$.
- Toomer takes this statement to refer to IX.6, but notes that Ptolemy didn’t actually prove this there. Rather, he just assumed that this angle would be the anomaly. However, I submit that Ptolemy was actually referring to III.5.
- We’ll be finding out it’s a ratio of $20:1$ here shortly.
- Because they’re both right triangles with $\overline{ZH} = \overline{HB}$ and the side $\overline{H \Theta}$ in common, the third sides must be equal as well.
- This is the ratio of $20:1$ I mentioned in a previous footnote.