Continuing on with our corrections for the equant, we’ll work on the third opposition for Jupiter.
As with before, the only thing that’s really changed in this diagram is that we’ve now placed point $G$ as our third opposition.
We’ll first recall that we determined $arc \; NX = 32;51º$ as a first approximation. Thus, the central angle, $\angle NZX$ is as well.
We’ll then enter a demi-degrees context about $\triangle DZH$ in which the hypotenuse, $\overline{ZD} = 120^p$. In that circle, $arc \; DH = 65;42º$ allowing us to find the corresponding chord $\overline{DH} = 65;06^p$.
Similarly, we can find that supplementary $arc \; ZH = 114;18º$, so the corresponding chord, $\overline{ZH} = 100;49^p$.
We then convert these to the context in which the diameter of the eccentres is $120^p$:
$$\frac{2;42^p}{120^p} = \frac{\overline{DH}}{65;06^p}$$
$$\overline{DH} = 1;28^p$$
and
$$\frac{2;42^p}{120^p} = \frac{\overline{ZH}}{100;49^p}$$
$$\overline{ZH} = 2;16^p.$$
We’ll now look at $\triangle GDH$. In it, we know $\overline{GD} = 60^p$ since it’s a radius, and we just found $\overline{DH}$, so we can use the Pythagorean theorem to find $\overline{GH} = 59;59^p$.
Again, $\overline{\Theta H} = \overline{HZ}$ and $\overline{E \Theta} = 2 \cdot \overline{DH}$.
Therefore, we can subtract $\overline{H \Theta}$ from $\overline{GH}$ to determine $\overline{G \Theta} = 57;43^p$.
We can then look at $\triangle E \Theta G$. In it, we just determined $\overline{G \Theta}$ and we know that $\overline{E \Theta} = 2;56^p$, so we can use the Pythagorean theorem to determine $\overline{EG} = 57;47^p$.
We’ll then convert into a demi-degrees context about that triangle in which the hypotenuse, $\overline{EG} = 120^p$:
$$\frac{120^p}{57;47^p} = \frac{\overline{E \Theta}}{2;56^p}$$
$$\overline{E \Theta} = 6;06^p$$
although Ptolemy rounds down to $6;05^p$.
We can then find the corresponding arc $arc \; E \Theta = 5;49º$ although Ptolemy comes up with $5;48º$.
Thus, the angle this arc subtends on the opposite side of the demi-degrees circle $\angle EG \Theta = 2;54º$.
We’ll now look at $\overline{GX}$ which is a radius, so has a measure of $60^p$. We can then subtract off $\overline{Z \Theta}$ to find that $\overline{\Theta X} = 55;28^p.$
That gives us two of the sides in $\triangle E \Theta X$, so we can find the remaining side, $\overline{EX} = 55;33^p.$
We’ll then convert into a demi-degrees context about that triangle:
$$\frac{120^p}{55;33^p} = \frac{\overline{E \Theta}}{2;56^p}$$
$$\overline{E \Theta} = 6;20^p.$$
We can then look up the corresponding arc for which I find, $arc \; E \Theta = 6;03º$ although Ptolemy finds $6;02º$. Thus, the angle which this arc subtends on the demi-degrees circle, $\angle EX \Theta = 3;01º$.
We can then subtract:
$$\angle GEX = \angle EX \Theta – \angle EG \Theta$$
$$\angle GEX = 3;01º – 2;54º = 0;07º.$$
Hence, since the planet at the third opposition, when viewed along $\overline{EG}$, had a longitude of $14;23º$ into Aries, it is clear that, if it had been on $\overline{EX}$, it would have had a longitude of $14;30º$ into Aries.
So that takes care of our three corrections.
In the next post, we’ll take these into consideration and recalculate the eccentricity and line of apsides.
