One of the biggest keys to understanding the model of Mercury1, is that the eccentre which drives the main motion2, is tied to the motion of the sun. Again, we can see this by comparing the values in the mean motion table for this rotation to those of the sun and seeing they’re identical.
However, we’ll also need to contend with the motion of the second eccentre which controls Mercury’s distance. We’ve talked a fair bit about this, but now it’s time to start seeing how it impacts the basic motion and anomaly. In short, it will offer a “correction” to these. So let’s get started.
Ptolemy begins with
two reliable observations; one from among those recorded in our time, and the other from the ancient observations.
[Firstly], we observed the planet Mercury in the second year of Antoninus (which was the $886^{th}$ year from Nabonassar), Epiphi [XI] $2/3$ in the Egyptian calendar [$139$ CE, May $17/18$], by means of the astrolabe instrument. It had not yet reached its greatest elongation as evening-star. When sighted with respect to the star on the heart of Leo it was observed at a longitude of Leo, $17 \frac{1}{2}º$; and at that moment, it was also $1 \frac{1}{6}º$ to the rear of the moon’s centre. The time at Alexandria was $4 \frac{1}{2}$ equinoctial hours before midnight of [Epiphi $2/$]$3$, since, according to the astrolabe, the $12^{th}$ degree of Virgo was culminating, while the sun was in about the $23^{rd}$ degree of Taurus. Now, at that moment, the positions according to the hypotheses we have demonstrated [previously] were as follows:
mean longitude of the Sun: Taurus $22;34º$
mean longitude of the Moon: Gemini $12;14º$
anomaly of the moon from the apogee of the epicycle: $281;20º$
hence, by computation, true position of the moon’s centre: Gemini $17;10º$
apparent position of the moon’s centre: Gemini $16;20º$Thus, from this [computation] too, we find that Mercury’s longitude was $17 \frac{1}{2}º$ into Gemini (since it was $1 \frac{1}{6}º$ to the rear of the moon’s centre).
The first goal, which I will limit this post to, is to determine the position of Mercury about the epicycle at the time of this observation.
To do so, Ptolemy produces the following diagram:
In this diagram, we again let $\overline{AE}$ be the line of apsides with $A$ as apogee, $B$ as the point about which the epicycle that rotates clockwise and controls the distance to Mercury’s epicycle rotates, $G$ as the one about which Mercury’s epicycle performs its uniform motion, and $D$ as the observer on Earth.
Next, we have the epicycle, centered on $Z$ with points $\Theta$, $K$, and $L$ on its circumference where $K$ is an extension of $\overline{DZ}$3, $\Theta$ is an extension of $\overline{GZ}$, and Mercury is at $L$. The center of that second eccentre is at point $H$.
Now, we’ll start forming a bunch of right triangles.
First, we’ll form $\overline{ZL}$ and drop a perpendicular from $Z$ onto $\overline{DL}$ at $X$.
Next, we’ll extend a perpendicular from $D$ onto $\overline{G \Theta}$ at $N$.
We’ll also extend $\overline G \Theta$ to the left, until it meets a perpendicular dropped from $H$ at $M$. NOTE: $M$ is NOT on $\overline{HB}$. It just looks that way as it’s quite close, but it’s slightly to the right of it.
Lastly, we’ll form $\overline{HG}$.
Here’s how that looks:
Now, at this time, the position of the mean sun (i.e., if it had been viewed from point $G$) was, as stated above, at Taurus $22;34º$. Recall that point $E$ is $10º$ into Aries, which means that $\angle EGZ$ is $42;34º$4 as is $\angle GBH$.
We’ll first concentrate on $\triangle GBH$. Within this triangle, we just stated that we know $\angle GBH$. However, we also know $\overline{BH} = \overline{GB}$ by hypothesis. That makes this an isosceles triangle in which $\angle BGH = \angle BHG$.
We can then determine the measure of these two angles since their sum must be $137;26º$ ($180º – 42;34º$). Since these two angles are equal, each one must then be half of this, or $68;43º$.
Now we want to solve the sides of this triangle. But there’s a problem: This isn’t a right triangle which means we can’t use this method.
Ptolemy doesn’t really give us much in the way of explanation on how he goes about solving this. Toomer takes this as an indication that Ptolemy knows an “equivalent of the sine theorem5 [for] a triangle which is not right-angled.” However, in my mind, it is also entirely possible that Ptolemy bisected the triangle by dropping a perpendicular from $B$ onto $\overline{HG}$ to form right triangles, solved it, and simply didn’t show those steps.
Regardless, Ptolemy determines that $\overline{GH} = 81;10^p$ and $\overline{BG} = 111;49^p$ in this demi-degrees context. Note that we can’t further specify the context by noting which hypotenuse we’ve taken to be $120^p$ since there wasn’t one in this case. But it doesn’t particularly matter for us to remember this for long, because Ptolemy immediately converts into the larger context we’ve been working in in which $\overline{BG} = 3^p$.
Converting:
$$\frac{3^p}{111;49^p} = \frac{\overline{GH}}{81;10^p},$$
$$\overline{GH} = 2;11^p.$$
Next, we’ll note that, $\angle BGM = 42;34º$. We know this because it’s a vertical angle with $\angle DGN$ which represents the angle that the center of the epicycle has rotated past the line of apsides.
We can then subtract:
$$\angle MGH = \angle BGH – \angle BGM = 68;43º – 42;34º = 26;09º.$$
Next, we’ll focus on $\triangle MHG$. Within it, we just determined $\angle HGM$. We’ll now draw a demi-degrees circle about it.
This means that $arc \; HM = 52;18º$ and looking up the corresponding chord, we determine that $\overline{HM} = 52;53^p$ in the context in which the hypotenuse, $\overline{HG} = 120^p$.
We can also determine, in this circle, that $arc \; GM$ (which is the supplement to $arc \; GM = 127;42º$ and the corresponding chord, $\overline{GM} = 107;43^p$.
Now, we’ll convert to the overall context we’re working towards where each of the eccentricities have a length of $3^p$. In that context, we previously determined that $\overline{GH} = 2;11^p$. This allows us to convert:
$$\frac{\overline{HM}}{52;53^p} = \frac{2;11^p}{120^p},$$
$$\overline{HM} = 0;58^p.$$
We can then determine $\overline{GM} = 1;58^p$, either by doing another context change, or more simply by using the Pythagorean theorem.
As a quick reminder, in this context, the radius of the eccentre $\overline{HZ} = 60^p$.
Ptolemy now does a bit of fuzzy math. He notes that,
$\overline{MZ}$, being a negligible amount less than $\overline{HZ}$, the hypotenuse [of $\triangle HMZ$], is [therefore] the same, $60^p$.
I don’t find any commentary on how true this is in either Toomer or Neugebauer6 and I can’t find a good way to do any sort of proof on it. So, I’ll just take Ptolemy at his word for now.
Taking this to be the case, we can then state that $\overline{MZ} = 60^p$ and can therefore subtract out $\overline{GM}$ to find:
$$\overline{GZ} = 60;00^p – 1;58^p = 58;02^p.$$
Now we’ll take a look at $\triangle DGN$ and create a demi-degrees circle about it. In this, we know that $\angle DGN = 42;34º$. Thus $arc \; DN = 85;08º$ and thus, the corresponding chord $\overline{DN} = 81;10^p$. This is for the context in which the hypotenuse, $\overline{GD} = 120^p$.
Similarly, we can determine that $\overline{GN} = 88;23^p$, again, either by the demi-degrees method or the Pythagorean theorem.
And again, we’ll translate this to the context in which the eccentricities (which includes $\overline{GD}$ in this triangle) are $3^p$:
$$\frac{\overline{GN}}{88;23^p} = {3^p}{120^p},$$
$$\overline{GN} = 2;13^p.$$
We can then find $\overline{DN} = 2;02^p$ using either another demi-degrees conversion or the Pythagorean theorem.
Now, we can subtract:
$$\overline{NZ} = \overline{GZ} – \overline{GN} = 58;02^p – 2;13^p = 55;49^p$$
If we now look at $\triangle DZN$, we now know the two sides and can therefore find the hypotenuse, $\overline{DZ}$ using the Pythagorean theorem.
Doing so I find it to be, $55;51^p$.
Next, we’ll take a look at $\triangle DZN$ and make a demi-degrees circle about it. In that context, $\overline{DZ} = 120^p$ since it’s the hypotenuse. We can then use that to determine $\overline{DN}$ in this context:
$$\frac{\overline{DN}}{2;02^p} = \frac{120^p}{55;51^p},$$
$$\overline{DN} = 4;22^p.$$
This can then be used to determine the corresponding arc, $arc \; DN$ by using the chords table. Doing so, we find that $arc \; DN = 4;11º$. Thus, the angle it subtends, $\angle DZN = 2;05º$.
Next, let me write out three statements:
$$\angle EDZ + \angle NDZ + \angle GDN = 180º.$$
$$\angle GDN + \angle DGN = 90º.$$
$$\angle DZN + \angle NDZ = 90º.$$
I’ll then solve the second of these for $\angle GDN$ and the third for $\angle NDZ$ and substitute them into the first.
$$\angle EDZ + 90º – \angle DZN + 90º – \angle DGN = 180º.$$
Simplifying:
$$\angle EDZ = \angle DZN + \angle DGN.$$
We just calculated $\angle DZN$ and stated $\angle DGN$ is the angle that the center of the epicycle has rotated past the line of apsides. So we know both of these and can state that7
$$\angle EDZ = 2;05º + 42;34º = 44;39º.$$
Now, let’s recall the initial observation from the beginning of the post. There, Mercury was observed position was $17 \frac{1}{2}º$ into Gemini. That’s $67;30º$ past $E$ ($10º$ into Aries). In other words, $\angle EDL = 67;30º$.
We can then subtract $\angle EDZ$ from this to determine $\angle XDZ = 22;51º$.
Considering a demi-degrees circle about $\triangle DZX$, we can then use this to determine $\overline{ZX} = 46;35^p$ in this context where the hypotenuse, $\overline{DX} = 120^p$.
Again, we’ll convert to the same context we have previously8.
$$\frac{\overline{ZX}}{46;35^p} = \frac{55;51^p}{120^p},$$
$$\overline{ZX} = 21;41^p.$$
Next, we’ll focus on $\triangle ZLX$, drawing a demi-degrees circle about it.
In that demi-degrees context, $\overline{ZL}$, the hypotenuse is $120^p$. We’ll also convert $\overline{ZX}$:
$$\frac{\overline{ZX}}{21;41^p} = \frac{120^p}{22;30^p},$$
$$\overline{ZX} = 115;39^p.$$
The corresponding arc, $arc \; ZX$ can then be determined using the table of chords to be $149;02º$. Therefore, the corresponding angle, $\angle ZLX = 74;31º$.
Next, we can state9:
$$\angle LZK = \angle XDZ + \angle ZLX.$$
We now know the right hand side of this equation so, plugging in:
$$\angle LZK = 22;51º + 74;31º = 97;22º.$$
This angle is the distance of Mercury about its epicycle from the apogee of the epicycle, which is what we were trying to find, for now.
- And that of Venus.
- That which was centered on $G$ in the last diagram we looked at which I’m putting below:
- Ptolemy does not explicitly state it, but it will be apparent late in this derivation as we’ll be discussing its vertical angle, $\angle GZD.
- $20º$ remaining in Aries and the $22;34º$ into Taurus.
- I.e., the law of sines
- Although Neugebauer doesn’t follow Ptolemy very closely in this section, and I haven’t read carefully enough to ensure that I’m not missing something.
- As a quick aside, Ptolemy does this portion staying in his demi-degrees context. In these contexts, he expresses angles as double their normal value but uses the double degree symbol (ºº) to denote that he is in such a context.
- The one in which the eccentricities all have a value of $3^p$.
- The proof for this is identical as the one above in which I wrote three equations and substituted the second and third into the first.
