In the previous two posts, we’ve been looking at how the apparent position of the center of a superior planet’s epicycle shifts position along the various circles. Specifically, we’ve been looking at how the angle between oppositions differs between the eccentre of mean distance and the equant circle.
In this post, we’ll do the same for the third opposition Ptolemy gave us. Unsurprisingly, the process will be largely the same as in the past two posts.
First, we’ll begin with a diagram set up for this observation:
The majority of the points and lines in this diagram are simply repeated from the diagram in this post. But, as a refresher, we have $\Theta$ being the equant, $D$ as the center of mean distance, and $N$ as the observer on Earth.
The center of the planet’s epicycle will be at $G$. We’ll extend a line from the observer through that to point $M$ which lies on the circle of the zodiac.
We’ll also draw in $\overline{\Theta G}$ marking on it, point $H$ where it intersects with the equant circle.
We’ll next draw in $\overline{NY}$ such that it goes through $H$ and has $Y$ on the circle of the zodiac.
Points $P$ and $R$ are drawn at the intersections of the equant circle and circle of the zodiac where they meet the line of apsides, respectively.
Lastly, we’ll extend lines from $N$ and $D$ such that they fall at right angles to $Q$ and $F$ respectively.
To begin, we have previously determined that $arc \; PH = 39;19º$1. That’s the same as $\angle P \Theta H$ since that arc lies on the equant circle.
We’ll first create a demi-degrees circle about $\triangle D \Theta F$.
In it, the hypotenuse, $\overline{\Theta D}$ will have a measure of $120^p$. Then, we have $arc \; DF = 78;38º$. We can then find the corresponding chord, $\overline{DF} = 76;02^p$.
We can then use the Pythagorean theorem to determine the remaining side, $\overline{\Theta F} = 92;50^p$.
We can then convert back to our context in which the eccentricity (i.e., $\overline{D \Theta} = \overline{DN} = 6;33,30^p$:
$$\frac{6;33,30^p}{120^p} = \frac{\overline{DF}}{76;02^p}$$
$$\overline{DF} = 4;09^p$$
$$\frac{6;33,30^p}{120^p} = \frac{\overline{\Theta F}}{92;50^p}$$
$$\overline{\Theta F} = 5;04^p.$$
Next, let’s look at $\triangle GDF$. In it, $\overline{GD} = 60^p$ as it’s the radius of the eccentre of mean distance and we also just determined $\overline{DF}$. Therefore, we can use the Pythagorean theorem to solve for the remaining side, $\overline{GF}$. Doing so, I find it to be $59;51^p$ which agrees with Ptolemy.
As we’ve seen in the past two posts, we can also state that $\overline{QF} = \overline{\Theta F}$.
That means we can determine the length of $\overline{GQ}$ by subtracting $5;04^p$ from $\overline{GF}$. Doing so, I find $\overline{GQ} = 54;47^p$.
We’ll now look at $\triangle NGQ$. We just found one side of this triangle. But we also know that $\overline{NQ} = 2 \cdot \overline{FD} = 8;18$. So we can again use the Pythagorean theorem to determine the hypotenuse, $\overline{NG} = 55;25^p$.
Next, Ptolemy wants to explore the angles in this triangle, so we’ll create a demi-degrees circle about that triangle. In it, the hypotenuse $\overline{NG} = 120^p$ which we can use to convert the sides.
$$\frac{120^p}{55;25^p} = \frac{\overline{NQ}}{8;18^p}$$
$$\overline{NQ} = 17;58^p$$
which Ptolemy rounds to $17;59^p$ and I will adopt for consistency.
We can then look up the corresponding arc, $arc \; NQ = 17;14º$.
Therefore, the angle this arc subtends, $\angle NGQ = 8;37º$.
We’ll set that aside for now and next look at $\overline{\Theta H}$. This is a radius of the equant, so has a measure of $60^p$. We subtract $\overline{Q \Theta}$, which is $2 \cdot \overline{F \Theta} = 10;08$, to determine $\overline{HQ} = 49;52^p$.
Now, if we look at $\triangle NHQ$, we again know two sides – $\overline{HQ}$ and $\overline{QN}$ (which was $2 \cdot \overline{DF} = 8;18^p$. Thus, we can determine the hypotenuse, $\overline{NH} = 50;33^p$.
As with the previous triangle, we’ll now explore the angles in this by creating a demi-degrees circle about it. In that context, the hypotenuse, $\overline{NH} = 120^p$ which we can use to convert the other sides:
$$\frac{120^p}{50;33^p} = \frac{\overline{QN}}{8;18^p}$$
$$\overline{QN} = 19;42^p.$$
We can then look up the corresponding arc. I find $arc \; QN = 18;54^p$. The corresponding angle subtended by this arc, $\angle NHQ = 9;27º$.
Ptolemy then notes that
$$\angle GNH = \angle NHQ – \angle NGQ = 9;27º – 8;37º = 0;50º.$$
The vertex of this angle is the center of the ecliptic, so $arc \; MY$ has the same measure.
So now let’s zoom out again and return to our larger diagram showing all three oppositions:
As with the last post, what we’re really trying to find here is $\angle{ZNH}$.
In this post, we determined that $arc \; LM = 93;44º$. Therefore the angle it subtends at the center of its circle2, $\angle LNM$ also has this measure.
In the above diagram, we can see that $\angle ZNH$ is less than $\angle LNM$ by $\angle BNZ$ and $\angle YNM$. Thus, we can subtract:
$$\angle ZNH = \angle LNM – \angle BNZ – \angle YNM = 93;44º – 0;33º – 0;50º$$
$$\angle ZNH = 92;21º.$$
To recap: So far in this chapter, we’ve determined the arcs along the ecliptic by using the apparent change along the ecliptic between oppositions. However, what we were really after was the change in apparent position along the equant circle. Thus, in this post as well as the last two, we’ve been working on changing the former to get to the latter.
Ptolemy still hasn’t told us why, but there’s still five pages left in this chapter, so I expect we’ll be finding out soon!
