Now that we’ve gotten an understanding of the eccentricity and line of apsides for Jupiter, Ptolemy wants to demonstrate that they are correct. He plans to demonstrate this, as he did for Mars,
from the fact that the apparent intervals [in longitude] of the planet derived from the ratios we have thus found turn out to be the same as those observed.
To do so, we’ll lay out a diagram for the first opposition:
This diagram is largely the same as the one we used in this post, except that we only include the circle of mean distance.
To begin, we determined that $\angle LZA = 77;15º$1. Thus, $\angle DZH$ is as well since it’s a vertical angle.
We can then create a demi-degrees context about $\triangle DZH$ in which the hypotenuse, $\overline{DZ} = 120^p$. In it, $arc \; DH = 154;30$ and its supplement $arc \; ZH = 25;30º$. Thus, the corresponding chords I find to be $\overline{DH} = 117;02^p$ and $\overline{ZH} = 26;29^p$.
We’ll then convert this to the context in which $\overline{ZD} = 2;45^p$ and the radius of the eccentre is $120^p$.
$$\frac{2;45^p}{120^p} = \frac{\overline{DH}}{117;02^p}$$
$$\overline{DH} = 2;41^p$$
and
$$\frac{2;45^p}{120^p} = \frac{\overline{ZH}}{26;29^p}$$
$$\overline{ZH} = 0;36^p.$$
We now know two sides of $\triangle ADH$. Thus, we can determine the remaining side:
$$\overline{AH} = \sqrt{\overline{AD}^2 – \overline{DH}^2}$$
$$\overline{AH} = \sqrt{60^2 – 2;41^2} = 59;56^p.$$
We can then add on $\overline{H \Theta}$ (which is the same length as $\overline{ZH}$ to determine $\overline{A \Theta} = 60;32^p.$
Similarly, $\overline{E \Theta} = 2 \cdot \overline{DH} = 5;22^p$.
This gives us two sides of $\triangle AE \Theta$. So we can again use the Pythagorean theorem to find the remaining side:
$$\overline{AE} = \sqrt{60;32^2 + 5;22^2} = 60;46^p.$$
We’ll then create a demi-degrees circle about this triangle and convert into it using $\overline{AE}$ as our conversion piece since it’s the hypotenuse in this triangle, so will have a measure of $120^p$ in the demi-degrees context:
$$\frac{120^p}{60;46^p} = \frac{\overline{E \Theta}}{5;22^p}$$
$$\overline{E \Theta} = 10;36^p.$$
We can then find the corresponding arc, which I find to be $arc \; E \Theta = 10;08^p$. Thus, the angle opposite it on the other side of the circle is $\angle EA \Theta = 5;04º.$
We can then subtract this from $\angle LZA$ to determine $\angle LEA = 72;11º.$
That, then, was the distance in the ecliptic [longitude] of the planet from its apogee at the first opposition.
However, this alone doesn’t give us anything to check since our initial observations didn’t include the line of apsides. Thus, we’ll need to repeat this for the other oppositions and take the differences to check them.
For the second opposition, we’ll produce a similar diagram:
We’ll again start by recalling that our calculations indicated that $\angle BZM = 2;50º.$
We can then draw a demi-degrees circle about $\triangle ZHD$ in which the hypotenuse, $\overline{ZD} = 120^p$. Then, $arc \; HD = 5;40º$ and its supplement, $arc \; HZ = 174;20º.$
We then find the corresponding chords:
$$\overline{HD} = 5;56^p$$
although Ptolemy comes up with $5;55^p$, and
$$\overline{HZ} = 119;51^p.$$
We now convert these to our general context in which the diameters of the eccentres is $120^p$:
$$\frac{2;45^p}{120^p} = \frac{\overline{HD}}{5;56^p}$$
$$\overline{HD} = 0;08^p$$
and
$$\frac{2;45^p}{120^p} = \frac{\overline{HZ}}{119;51^p}$$
$$\overline{HZ} = 2;45^p$$
We know that $\overline{BD} = 120^p$ since it’s the radius, which gives us two sides of $\triangle BHD$. Thus, we can find the remaining side:
$$\overline{BH} = \sqrt{60^2 = 0;08^p} \approx 60^p.$$
We can then subtract $\overline{H \Theta}$ off of this to find that $\overline{B \Theta} = 57;15^p.$
Next, we’ll look at $\triangle EB \Theta$, of which we again know two sides. Thus, we can find the remaining side:
$$\overline{EB} = \sqrt{57;15^p + 0;16^p} = 57;15^p$$
We’ll now convert into a demi-degrees context about this triangle:
$$\frac{120^p}{57;15^p} = \frac{\overline{E \Theta}}{0;16^p}$$
$$\overline{E \Theta} = 0;34^p$$
although Ptolemy rounds down to $0;33^p$.
Then, we find the corresponding arc, $arc \; E \Theta = 0;32º$ meaning the angle it subtends on the opposite side of the circle, $\angle = EB \Theta = 0;16º$.
We add this to $\angle BZM$ to find that $\angle BEM = 3;06º$.
Therefore, the distance of the planet in advance of the perigee at the second opposition was $3;06º$. And we showed that, at the first opposition, it was $72;11º$ to the rear of apogee. Thus, the computed apparent interval from first to second oppositions is the supplement [of $3;06º + 72;11º$], $104;43º$, in agreement with the interval derived from the first observations.
The value calculated matches what we started with indicating that things check out.
We’ll now double check it for the second interval which requires us to repeat this for the third opposition:
To begin, we determined that $\angle MZG = 30;36º.$
We first create our demi-degrees circle about $\triangle DZH$ in which $\overline{DZ} = 120^p$ and $arc \; DH = 61;12º$. Its supplement, $arc \; HZ = 118;48º.$
We then find the corresponding chords, $\overline{DH} = 61;05^p$ which Ptolemy rounds to $61;06^p$. Additionally, $\overline{HZ} = 103;17^p.$
We now convert back to the context in which the diameter of the eccentres is $120^p$:
$$\frac{2;45^p}{120^p} = \frac{\overline{DH}}{61;05^p}$$
$$\overline{DH} = 1;24^p$$
and
$$\frac{2;45^p}{120^p} = \frac{\overline{HZ}}{103;17^p}$$
$$\overline{HZ} = 2;22^p.$$
We then look at $\triangle DHG$ in which we know two sides and thus find the remaining side using the Pythagorean theorem:
$$\overline{GH} = \sqrt{60^2 – 1;24^2} = 59;59^p.$$
Next, we subtract off $\overline{H \Theta}$ to determine $\overline{G \Theta} = 57;37^p.$
Additionally, $\overline{E \Theta} = 2;48^p$ which means we know two sides of $\triangle EG \Theta.$ Thus, we can find the remaining side:
$$\overline{EG} = \sqrt{2;48^2 + 59;59^2} = 57;41^p.$$
We now draw a demi-degrees circle about this triangle and convert into that context:
$$\frac{120^p}{57;41^p} = \frac{\overline{E \Theta}}{2;48^p}$$
$$\overline{E \Theta} = 5;49^p$$
which Ptolemy rounds to $5;50^p.$
We then look up the corresponding arc for which I find $arc \; E \Theta = 5;34º$. Thus, the angle on the opposite side of the circle, $\angle EG \Theta = 2;47º.$
This gets added to $\angle MZG$ to find that $\angle MEG = 33;23º.$
That, then, was the distance of the planet to the rear of perigee at the third opposition. And we showed that, at the second opposition, its distance in advance of the same perigee was $3;06º$. Therefore, the apparent interval [in ecliptic longitude] from the second to the third oppositions is computed as the sum, $36;29º$, once again, in agreement with the observed interval.
Thus, Ptolemy shows we’re able to recover our original observations from the values we derived.
But before closing out this chapter, Ptolemy calculates the position of Jupiter about its epicycle. We’ll begin with another diagram:
In this, we have determined that $\angle MZG = 30;36º$. Additionally we showed above that $\angle EGZ = 2;47º$. Thus, $arc \; \Theta K$ is as well. Thus, measuring from $H$ on the epicycle, from $H$, going counterclockwise, $arc \; HK = 182;47º.$
Therefore, at the moment of the third opposition, namely, in the first year of Antoninus, Athyr [III] $20/21$ in the Egyptian calendar, $5$ hours after midnight, the planet Jupiter had the following mean positions:
in longitude: $210;36º$ from the apogee of the eccentre (i.e., its mean longitude was $11;36º$ in Aries)
in anomaly: $182;47º$ from the apogee of the epicycle, $H.$
That concludes this chapter. In the next few chapters on Jupiter, we’ll determine the size of Jupiter’s epicycle, examine the corrections to the mean motion, and determine the epoch positions.
