Having gotten our preliminary estimate for the eccentricity and position of apogee
let us investigate the differences which can be derived from them in the ecliptic arcs which we week to determine at each of the oppositions [in turn].
In this post, we’ll cover Ptolemy’s method for the first opposition.
To get us started, we’ll return to the first diagram we produced for this chapter, but only concentrate on the first observation at point $A$. However, we’ll add a few more lines in:
Here things remain largely the same, with $\Theta$ as the center of uniform motion (the equant), $D$ as the center of uniform distance, and $N$ the observer.
At the time of the first observation, Mars was observed at $K$ which means the center of its epicycle was at $A$.
We’ll also consider the projection of $A$ from the equant onto the equant circle to form $\overline{\Theta E}$ which goes through $A$. We won’t need the intersection of the ecliptic this time.
Next, that line will be extended the opposite direction and perpendiculars dropped on it from $N$ and $D$ at $Q$ and $F$ respectively.
We’ll also extend a line from $D$ to $A$, producing $\overline{DA}$, which doesn’t get extended in either direction, and we won’t draw in a point where it intersects the ecliptic.
We’ll extend a line from $N$ to $E$, forming $\overline{NE}$ again, marking the point it intersects the ecliptic as $S$.
Lastly, we’ll mark the line of apsides as $\overline{XN}$ where $X$ is where the line intersects the equant circle near apogee. We’ll also mark $O$ which is the point where the line intersects the ecliptic near apogee.
That’s everything we need to understand the diagram, but before we jump into the math, I encourage a quick look back at the last post to recall what we were working with there. Specifically, the circle we focused on was the equant circle.
In this diagram, that’s the one centered on $\Theta$ with $X$ on its perimeter. This is important because it means anything we solved for that circle in that post, we can now use.
Which is what we’ll do immediately as we found that the arc between perigee at $X$ and $E$ was $36;31º$ as is the angle it subtends, $\angle E \Theta X$ as well as its vertical angle, $\angle D \Theta F$.
Now we’ll create a demi-degrees circle about $\triangle D \Theta F$ which contains this angle and will have its hypotenuse, $\overline{\Theta D} = 120^p$ in this demi-degrees context.
The arc opposite that angle, $arc \; DF = 73;02º$ and its corresponding chord, $\overline{DF} = 71;24^p$ although Ptolemy rounds up to $71;25^p$, which I’ll use for consistency.
We can then determine $\overline{F \Theta} = 96;27^p$, either by another application of the demi-degrees circle, or by the Pythagorean theorem.
We can now convert this back to the context in which the diameter of the eccentres is $120^p$ as we found the distance between the center of mean motion and the observer was $13;07^p$ in the last post. As I noted there, because the center of distance falls half way between them, the eccentricity would be half this which is $6;33,30^p$ although Ptolemy failed to call that out then. However, we can now use that fact to state that $\overline{\Theta D} = \overline{DN} = 6;33,30^p$ and use the former of these to convert contexts:
$$\frac{6;33,30^p}{120^p} = \frac{\overline{DF}}{71;25^p}$$
$$\overline{DF} = 3;54^p.$$
And again, we can either convert contexts or use the Pythagorean theorem to determine $\overline{F \Theta} = 5;16^p$.
Now, let’s turn our attention to $\triangle DFA$, but not considering it in a demi-degrees context. In this triangle, we just determined $\overline{F \Theta}$ and we know that $\overline{DA} = 60^p$ as it’s a radius. Therefore, we can use the Pythagorean theorem to determine, $\overline{FA} = 59;52^p$. Again, this is in the context in which the diameters of the eccentres is $120^p$.
Next, let’s look at $\overline{QF}$ and $\overline{F \Theta}$. These two lines are the same length as $\triangle \Theta QN$ and $\triangle \Theta FD$ are effectively similar triangles in which the latter is half the size of the former. Thus, we know that $\overline{QF} = 5;16^p$ as well.
This can be added to $\overline{FA}$ to determine $\overline{QA} = 65;08^p$.
Similarly, $\overline{QN} = 2 \cdot \overline{DF}$ which allows us to state that $\overline{QN} = 7;48^p$.
That gives us two sides of the right triangle, $\triangle NAQ$. So we can use the Pythagorean theorem to determine the hypotenuse, $\overline{NA} = 65;36^p$.
Now we’ll want to solve the angles in this triangle, so we’ll create a demi-degrees context about it in which the hypotenuse, $\overline{NA} = 120^p$. Converting $\overline{QN}$ into that context:
$$\frac{120^p}{65;36^p} = \frac{\overline{QN}}{7;48}$$
$$\overline{QN} = 14;16^p.$$
We can then look up the corresponding angle. I find it to be $13;39º$, but Ptolemy finds $13;40º$. Again, this is the central angle, so the actual angle opposite it on the demi-degrees circle would be half that, which is to say $\angle NAQ = 6;50º$.
Next, we’ll look at $\overline{QE}$. This is composed of $\overline{Q \Theta} + \overline{\Theta E}$. The former of these is $2 \cdot \overline{F \Theta} = 10;32^p$ and the latter is $60^p$ since it’s a radius. Putting those together we find that $\overline{QE} = 70;32^p$.
If we look at $\triangle QNE$ we now know two sides: $\overline{QE}$ ($70;32^p$) and $\overline{QN}$ ($7;48^p$), which we can plug into the Pythagorean theorem to find the hypotenuse, $\overline{NE} = 70;58^p$ which Ptolemy rounds off to $71^p$.
We’ll now consider a demi-degrees circle about this triangle. In it, the hypotenuse, $\overline{NE} = 120^p$. Converting $\overline{QN}$ into this context as well:
$$\frac{120^p}{71^p} = \frac{\overline{QN}}{7;48^p}$$
$$\overline{QN} = 13;11^p.$$
Ptolemy finds it to be $13;10^p$.
Looking up the corresponding angle in the chord table, I find it to be $12;36º$ which agrees with Ptolemy. But again, this would be the central angle. For the angle on the edge of the circle, it would be half of this which is to say $\angle NEQ = 6;18º$.
Next, let’s take a look at $\angle ANE$. This is
$$\angle ANE = \angle NAQ – \angle NEQ.$$
As a quick proof of this, let’s look at $\triangle AEN$. We can write:
$$\angle AEN + \angle ANE+ \angle NAE = 180º.$$
However, $\angle NAE = 180º – \angle NAQ$ since it’s the supplement. Thus, we can substitute:
$$\angle AEN + \angle ANE + (180º – \angle NAQ) = 180º.$$
We can cancel the $180º$ from each side and rearrange a bit to determine
$$\angle ANE = \angle NAQ – \angle AEN$$
and $\angle AEN = \angle NEQ$, so we get
$$\angle ANE = \angle NAQ – \angle NEQ.$$
Plugging in:
$$6;50º – 6;18º = 0;32º.$$
That, then, is the amount of $arc \; KS$ of the ecliptic.
Ptolemy doesn’t tell us what we’re going to do with this just yet. Instead, we’ll repeat this procedure for the other two observations, which I’ll again split that into separate posts.
However, if we think back to the first post where we started trying to calculate the line of apsides and the eccentricity, the initial issue was faced was that we used an angle from the point of view of the observer in place of one that should have been from the equant.
Again, let’s take a quick look at the differences between those angles:
As a reminder, what Ptolemy wanted in that first post was $\angle Z \Theta H$ which we didn’t have. Instead, we used $\angle LNM$.
But the differences between these two angles are pretty small. Indeed, they’re really differentiated by $\angle LNT$ on one side and $\angle YNM$ on the other. Both of these little angles are on the ecliptic circle.
So, if we could find those, we could correct our initial angle.
What we just did in this post was find $\angle KNS$, also on the ecliptic circle which basically shows us that we can find those other two angles.
Except for one small hiccup. Our determination of all these little angles was calculated based on the wrong angle. So the corrections won’t be perfect either. However, they’ll get us closer.
And herein lies the iterative process I’ve kept mentioning. Which we’ll come to in due time…
