Before moving on, Ptolemy describes how he arrived at the periodic motions of Jupiter, beginning with an ancient observation.
[I]t is declared that in the $45^{th}$ year of the calendar of Dionysius, on Parthenon $10$, that the planet Jupiter occulted the southernmost [of the two] Aselli at dawn. Now, the moment [of the observation] is in the $83^{rd}$ year from the death of Alexander, Epiphi [XI] $17/18$ in the Egyptian calendar [$240$ BCE Sept. $3/4$], dawn.
For that time, we find the longitude of the mean sun as $9;56º$ into Virgo. But, the star called ‘the southern Asellus’1 among those surrounding the nebula in Cancer had a longitude, at the time of our observation [of it], of $11 \frac{1}{3}º$ into Cancer. Hence, obviously, its longitude at the observation in question was $7;33º$ [into Cancer], since to the $378$ years between the observations corresponds to [a precessional motion of] $3;47º.$
Therefore, the longitude of Jupiter at that moment (since it had occulted the star) was also $7;33º$ into Cancer. Similarly, since the apogee was $11º$ into Virgo in our times, it must have had a longitude of $7;13º$ into Virgo at the observation. Hence, it is clear that the distance of the apparent planet from the then apogee of the eccentre was $300;20º,$ while the distance of the mean sun from that same apogee was $2;43º.$
Using this information, Ptolemy sketches a new diagram:
In this diagram, we have the line of apsides as $\overline{AG}$. Points $Z$, $D$, and $E$ remain as the equant, center of mean distance, and observer respectively.
The epicycle is centered on $B$ with Jupiter at $\Theta$. The mean sun is at $L.$ We’ll create $\overline{ZB}$ which gets extended to the opposite side of the epicycle at $H$, as well as lines $\overline{DB}$, and $\overline{E \Theta}$.
We then drop perpendiculars from $Z$ onto $\overline{DB}$ at $K$ and from $D$ onto $\overline{E \Theta}$ at $M$.
We’ll also extend a perpendicular from $B$ onto $\overline{E \Theta}$ at $N$, extending it as far as $X$ where it meets a line created from $D$ that is parallel to $\overline{E \Theta}$, thus forming parallelogram $DMNX.$
To get started, we’ll consider $\angle AE \Theta$. This is the apparent distance of the planet before apogee. We stated above that the planet was observed to be $300;20º$ after the apogee, which means that $\angle AE \Theta$ is a full circle minus this amount of $59;40º.$
Additionally, $\angle AEL$ is the distance between the apogee and the mean sun from the point of view of the observer. Ptolemy has calculated this to be $2;43º$ for us.
We can add these two angles together to determine $\angle LE \Theta = 62;23º.$
Now, recall that $\overline{LE} \parallel \overline{\Theta B}$ because the angle of the planet about the epicycle is always equal to that of the mean sun from the point of view of the observer. This makes $\angle E \Theta B$ an alternate angle, giving it the same value as $\angle LE \Theta$ or $62;23º.$
We can then focus on $\triangle B\Theta N,$ creating a dem-degrees circle about it. In it, the hypotenuse, $\overline{B \Theta} = 120^p$ and $arc \; BN = 124;46º.$ The corresponding chord, then, $\overline{BN} = 106;20^p.$
Since we know the size of this (the radius of the epicycle) in the context in which the diameter of the eccentres is $120^p$, we can use that to convert:
$$\frac{11;30^p}{120^p} = \frac{\overline{BN}}{106;20^p}$$
$$\overline{BN} = 10;11^p.$$
Ptolemy rounds up to $10;12^p.$
We’ll now look at $\triangle DEM$. In it, we know that $\angle DEM = 59;40º$ which means that the remaining non-right angle, $\angle MDE = 30;20º$. We’ll now create a demi-degrees context about this triangle in which the hypotenuse, $\overline{DE} = 120^p$. Additionally $arc \; DM = 119;20º.$ Thus, the corresponding chord, $\overline{DM} = 103;34^p$.
And since we know that $\overline{DE} = 2;45^p$ in the context in which the diameters of the eccentres is $120^p$, we can convert to that context:
$$\frac{2;45^p}{120^p} = \frac{\overline{DM}}{103;34^p}$$
$$\overline{DM} = 2;22^p.$$
Ptolemy finds it to be $2;23^p.$
Since $DMNX$ is a parallelogram, this is the same as $\overline{NX}$ which we can add to $\overline{BN}$ to find that $\overline{BX} = 12;35^p.$
Next, we can focus on $\triangle BDX,$ creating a demi-degrees context about it wherein the hypotenuse $\overline{BD} = 120^p.$ We can use this to convert into the demi-degrees context:
$$\frac{120^p}{60^p} = \frac{\overline{BX}}{12;35^p}$$
$$\overline{BX} = 25;10^p.$$
We can the find the corresponding arc, $arc \; BX = 24;13º$ although Ptolemy finds it to be $24;14º.$ In either case, this means that the angle this arc subtends on the opposite side of the demi-degrees circle $\angle BDX = 12;07º.$
Next, take a look at $\angle XDB$. This is a right angle. Thus, we can subtract $\angle BDX$ out of it which gives us $\angle BDM = 77;53º.$
We can then add $\angle MDE$ to find $\angle BDE = 108;13º$.
We’ll then find the supplementary angle to this along $\overline{AG}$ to find $\angle ADB = 71;47º.$
Now we’ll focus on that skinny triangle, $\triangle ZDK,$ creating a demi-degrees circle about it in which the hypotenuse, $\overline{ZD} = 120^p.$
In that, $arc \; ZK = 143;34º$ and its supplement, $arc \; DK = 36;26º.$ We can then find the corresponding chords, $\overline{ZK} = 113;59^p$ and $\overline{DK} = 37;31^p.$
This can be converted into the context in which the diameter of the eccentres is $120^p$:
$$\frac{2;45^p}{120^p} = \frac{\overline{ZK}}{113;59^p}$$
$$\overline{ZK} = 2;37^p$$
and
$$\frac{2;45^p}{120^p} = \frac{\overline{DK}}{37;31^p}$$
$$\overline{DK} = 0;52^p.$$
We’ll then subtract $\overline{DK}$ off of $\overline{DB}$ to find that $\overline{KB} = 59;08^p.$
We can then focus on $\triangle ZBK$ in which we know two of the sides. Thus, we can find the remaining side:
$$\overline{ZB} = \sqrt{59;08^2 + 2;37^2} = 59;11^p$$
which Ptolemy rounds up to $59;12^p.$
We’ll now create a demi-degrees context about this triangle in which the hypotenuse, $\overline{ZB} = 120^p.$ Converting into that context:
$$\frac{120^p}{59;12^p} = \frac{\overline{ZK}}{2;37^p}$$
$$\overline{ZK} = 5;18^p.$$
We can then find the corresponding arc $arc \; ZK = 5;04º.$ Thus, the angle that this arc subtends on the opposite side of the circle, $\angle ZBD = 2;32º.$
If we add this to $\angle ADB$ we find
$$\angle AZB = 71;47º + 2;32 = 74;19º.$$
This next part is a bit hard to see, so let play with our diagram a bit.
In this diagram, I’ve exaggerated the angles of $\overline{LE}$ and since that’s parallel to $\overline{\Theta B},$ that as well.
I’ve also extended $\overline{HZ}$ until it hits $\overline{LE}$, creating a new point at $T$ and I’ve created $\overline{ES}$ which is parallel to $\overline{ZH}$.
Then, I’ve marked out the angles that are equal:
$$\angle LTZ = \angle LES = \angle \Theta BH$$
and
$$\angle AZH = \angle AES.$$
Now, let’s look at $\angle LES$. This is quite clearly $\angle LEA + \angle AES$.
However, due to the angles we just stated were equal, we can do some substituting to write:
$$\angle \Theta BH = \angle LEA + \angle AZH.$$
We know $\angle LEA = 2;43º$ as this was the angle of the mean sun past apogee. We just found $\angle AZH = 74;19º.$
Thus, we can determine $\angle \Theta BH = 77;02º.$2.
This is the angle of Jupiter about its epicycle from the apogee of the epicycle.
And we had [already] shown that, at the moment of the third opposition, its distance from the apogee of the epicycle was $182;47º.$ Thus, in the interval between the two observations, which comprises $377$ Egyptian years and $128$ days less approximately $1$ hour, its motion was $105;45º$ beyond $345$ complete revolutions.
That is, again, very nearly the same increment in anomaly as one derives from the [tables for] mean motions which we constructed. For, it was from these very same elements that we derived the daily [mean motion in anomaly], by dividing the number of degrees contained in the complete revolutions plus the increment by the number of days contained in the time-interval.
Switching to decimal briefly, that’s a total increase of $124,305.75º$ in $137,732.96$ days. Dividing that out, I get $0;54,09,02,45,08,48^{\frac{º}{day}}.$
So we can see that Ptolemy saying “very nearly” is again accurate as what he actually recorded in the mean motions table for the anomaly of Jupiter was $0;54,09,02,46,26,00^{\frac{º}{day}}.$ The values match until the last three sexagesimal places which is still reasonably good agreement. But again, Ptolemy has clearly done something he hasn’t explained and, as Toomer puts it, “remains obscure.”
Toomer again discusses this in Appendix C of his translation, but this is beyond the scope of what I’m intending with this blog. So I’ll close out here and, in the next post, we’ll determine the epoch positions for Jupiter.
