Having worked out a preliminary eccentricity and position for the line of apsides, we now turn to finding a better approximation iteratively. Fortunately, Ptolemy only finds one iteration necessary for Jupiter.
To get started, we’ll produce a new diagram:
In this diagram we have
the eccentre carrying the epicycle center be $LM$ on center $D$, and the eccentre of the planets mean motion be $NX$ on center $Z$, equal [in diameter] to $LM$.
We’ll then draw the line of apsides through these centers, creating $\overline{NM}$ with $L$ at the intersection with the circle of mean motion and $E$ as the observer on Earth.
We’ll place the center of the epicycle the first opposition be $A$, connecting that to points $D$ and $E$, as well as extending $\overline{ZA}$ in both directions, producing point $X$. In the other direction, it is extended to $\Theta$ where it meets a perpendicular dropped from $E$. Similarly, a perpendicular is dropped onto $\overline{X \Theta}$ from $D$ producing point $H$.
In the last post, we concluded that $\angle NZX = 79;30º$ as this was the motion along the circle of mean motion between apogee and the first opposition.
Thus, its vertical angle, $\angle DZH$ is as well.
So, we’ll create a demi-degrees context about $\triangle DZH$ in which the hypotenuse, $\overline{DZ} = 120^p$.
We can then state that the arc opposite $\angle DZH$, $arc \; DH = 159;00º$. Similarly, the supplement, $arc \; ZH = 21º$.
This allows us to find the corresponding chords. Doing so, I find, $\overline{DH} = 117;59^p$ and $\overline{ZH} = 21;52^p$.
Next, let’s refer back to the eccentricity. In the last post, we determined that the distance between the observer on Earth and the center of mean motion was $5;23^p$. In this diagram, that’s $\overline{EZ}$. So half that is $\overline{DZ} = 2;42^p$.
This gives us $\overline{DZ}$ in the context in which the diameter of the eccentre is $120^p$, which we also know in the demi-degrees context we were just working with, so we can use it to convert the other sides:
$$\frac{2;42^p}{120^p} = \frac{\overline{DH}}{117;59^p}$$
$$\overline{DH} = 2;39^p$$
and
$$\frac{2;42^p}{120^p} = \frac{\overline{ZH}}{21;52^p}$$
$$\overline{ZH} = 0;29^p$$
which Ptolemy rounds to $0;30^p$.
Next, let’s focus on $\triangle DAH$. In it, we know that $\overline{DA} = 60^p$ since it’s a radius and we just found $\overline{DH}$. Thus we can use the Pythagorean theorem to find:
$$\overline{HA} = \sqrt{60^2 – 2;39^2}$$
$$\overline{HA} = 59;56^p.$$
As when we did this for Mars, $\overline{ZH} = \overline{H \Theta}$ and $\overline{E \Theta} = 2 \cdot \overline{DH}$.
This allows us to determine
$\overline{A \Theta} = 59;56^p + 0;29^p = 60;25^p.$
However, Ptolemy is evidently calculating with higher precision but still only showing the first sexagesimal place as he comes up with $60;26^p$.
Additionally, $\overline{E \Theta} = 5;18^p$.
That’s two sides of $\triangle AE \Theta$, so we can use the Pythagorean theorem to determine the remaining side, $\overline{AE}$:
$$\overline{AE} = \sqrt{60;25^2 + 5;18^2}$$
$$\overline{AE} = 60;39^p.$$
And again Ptolemy has been doing some rounding as he comes up with $60;40^p$
Now, let’s enter a demi-degrees context about this triangle. In it, the hypotenuse $\overline{AE} = 120^p$ which we can use as our conversion piece:
$$\frac{120^p}{60;40^p} = \frac{\overline{E \Theta}}{5;18^p}$$
$$\overline{E \Theta} = 10;29^p.$$
We can then find the corresponding arc, $arc \; E \Theta = 10;01º.$
This means that the angle opposite in the circle, $\angle EA \Theta = 5;00,30º.$ Ptolemy has rounded this to $5;01º.$
We’ll set that aside and look at $\overline{X \Theta}$. This is $\overline{XZ}$ (which is $60^p$ since it’s the radius) plus $\overline{Z \Theta}$ which is $1^p$. Thus, the total length of $\overline{X \Theta} = 61;00^p$.
That gives us two sides of $\triangle E \Theta X$, so we can again use the Pythagorean theorem to find the remaining side:
$$\overline{EX} = \sqrt{61;00^2 + 5;18^2}$$
$$\overline{EX} = 61;14^p.$$
We’ll now create a demi-degrees context about $\triangle EX \Theta$ in which, the hypotenuse, $\overline{EX} = 120^p$. We know that piece in two contexts, so we can use it to convert $\overline{E \Theta}$
$$\frac{120^p}{61;14^p} = \frac{\overline{E \Theta}}{5;18^p}$$
$$\overline{E \Theta} = 10;23^p.$$
We can then look up the corresponding chord. I find $arc \; E \Theta = 9;56º$ but Ptolemy finds it to be $9;55º$. Taking his value, the of the angle opposite it in this circle, $\angle EX \Theta = 4;58º$.
Thus, we can now find $\angle AEX$ which is the difference between $\angle EA \Theta – \angle EX \Theta$:
$$\angle AEX = 5;01º – 4;58º = 3;00º.$$
But at the first opposition, the planet, viewed along $\overline{EA}$, had an apparent longitude of $23;11º$ into Scorpio. Thus, it is clear that, if the epicycle center were carried, not on eccentre [of mean distance] $LM$, but on [eccentre of mean speed] $NX$, it would have been at point $X$ on that eccentre, and the planet would have appear along $\overline{EX}$, differing by $0;03º$ [from the actual position], and thus, would have had a longitude of $23;14º$ into Scorpio.
So there’s our first correction. In the next post, we’ll do the same for the second opposition.
