Ptolemy now walks us through his supposed method for determining the periodic motions of Saturn. I say “supposed” because, again, Ptolemy’s final result doesn’t quite match what he tells us here. But we’ll follow along anyway.
As usual, he begins with an observation from the Babylonian astronomers:
[I]n the $82^{nd}$ year in the Chaldaen calendar, Xanthikos $5,$ in the evening, the planet Saturn was $2$ digits [i.e., $0;10º$] below [the star on] the southern shoulder of Virgo1. Now, that moment is in the $519^{th}$ year from Nabonassar, Tybi [V] $14$ in the Egyptian calendar [$228$ BCE, March $1$], evening, at which time we found the longitude of the mean sun as $6;10º$ into Pisces.
But the fixed star on the southern shoulder of Virgo had a longitude at the time of our observation of $13 \frac{1}{16}º$ into Virgo; Thus, at the moment of the observation in question, since to the intervening $366$ years corresponds to a motion of the fixed stars of about $3 \frac{2}{3}º,$ its longitude was, obviously, $9 \frac{1}{2}º$ into Virgo. And the planet Saturn had the same longitude, since it was $2$ digits to the south of the fixed star.
By the same argument, since we showed that, in our time, its apogee was at $23º$ into Virgo, at the observation in question, it must have had a longitude of $19 \frac{1}{3}º$ into Virgo.
From this, we conclude that, at the above moment, the apparent distance of the planet from the them apogee, was $290;10º$ of the ecliptic, while the mean sun was $106;50º$ from the same apogee.
Using this information, we’ll produce a new diagram2:
As usual, we have $\overline{AG}$ as our line of apsides on which sit, $Z,$ $D,$ and $E$ which are the centers of mean motion, mean distance, and the center of the ecliptic respectively.
Point $L$ is the mean sun and is connected to $E$. The epicycle’s center is $B$ and the planet is at $\Theta.$
We then produce $\overline{ZB},$ extending it to the opposite side of the epicycle at $H$. We’ll aso draw $\overline{DB}$. We then draw $\overline{E \Theta}$ extending it until it meets a perpendicular dropped from $B$. A parallel line is extended from $D$ until it meets the same perpendicular from $B$ at $X.$ A perpendicular is also dropped from $D$ onto this line which would be parallel to $\overline{XB}.$
Lastly, we draw $\overline{B \Theta}$ which is parallel to $\overline{EM}$ since the planet’s position about the epicycle is always the same as the angle produced for the mean sun from the observer.
Ptolemy told us above that Saturn was observed $290;10º$ after the apogee, which means that $\angle AE \Theta = 69;50º$3. Similarly, we saw that the sun was $106;50º$ after the apogee, which is $\angle AEL$.
We can add these two angles together to determine $\angle \Theta EL = 176;40º.$ This is the same angle as $\angle E \Theta B$. This is easy to see if you imagine extending $\overline{LE}$ and $\overline{B \Theta}$. These two lines are parallel and thus, $\overline{E \Theta}$ would cross both making $\angle \Theta EL$ and $\angle E \Theta B$ alternate angles.
Next, we can look at this angle’s supplement along $\overline{EN}$ which is $\angle B \Theta N = 3;20º.$
We’ll now create a demi-degrees circle about $\triangle B \Theta N$ in which its hypotenuse, $\overline{\Theta B} = 120^p.$ In it, $arc \; BN = 6;40º$ and its corresponding chord, $\overline{BN} = 6;59^p$ although Ptolemy rounds down to $6;58^p.$
This can be converted to the context in which the diameters of the eccentre is $120^p$ since we know the radius of the epicycle in that context:
$$\frac{6;30^p}{120^p} = \frac{\overline{BN}}{6;58^p}$$
$$\overline{BN} = 0;23^p.$$
Now, let’s take a look at $\triangle EDM$. In it, we know that $\angle DEM = 69;50º$ as this is the same angle as $\angle AE \Theta$. Since $\angle EMD$ is a right angle, the remaining angle $\angle EDM = 20;10º.$
We’ll put that on the shelf for a bit and now create a demi-degrees circle about that triangle in which the hypotenuse, $\overline{ED} = 120^p.$ In that context, $arc \; DM = 139;40º$ and the corresponding chord, $\overline{DM} = 112;39^p.$
We can convert that back into our context in which the diameters of the eccentres is $120^p$ since we know the distance between the centers:
$$\frac{3;25^p}{120^p} = \frac{\overline{DM}}{112;39^p}$$
$$\overline{DM} = 3;12^p.$$
Now, because $DMNX$ is a parallelogram, this is the same length as $\overline{XN}.$ So we can add this onto $\overline{BN}$ to determine $\overline{BX} = 3;35^p.$
We’ll now create a demi-degrees context around $\triangle BDX$ in which its hypotenuse, $\overline{DB} = 120^p.$ In our previous context, this was a radius, so hard a measure of $60^p$ which we can use to convert:
$$\frac{120^p}{60^p} = \frac{\overline{BX}}{3;35^p}$$
$$\overline{BX} = 7;10^p.$$
The corresponding arc, $arc \; BX = 6;51º$ although Ptolemy finds $6;52º.$ Thus, the angle this arc subtends on the opposite side of the demi-degrees circle is $\angle BDX = 3;26º.$
Next, $\angle XDM$ is a right triangle, so if we subtract $\angle BDX$ out of it, we find that $\angle BDM = 86;34.$
To this, we can add $\angle EDM$ to find that $\angle BDE = 106;44º.$
This is the supplement to $\angle BDA$ so $\angle BDA = 73;16º.$
We’ll now look at $\triangle DZK$ which contains this angle, creating a demi-degrees circle about it in which the hypotenuse, $\overline{DZ} = 120^p.$ In in, $arc \; ZK = 146;32º$ and the supplement, $arc \; DK = 33;28º.$ The corresponding chords are $\overline{ZK} = 114;55^p$ and $\overline{DK} = 34;33^p.$
We convert this back into our context in which the diameter of the eccentres is $120^p$:
$$\frac{3;25^p}{120^p} = \frac{\overline{ZK}}{114;55^p}$$
$$\overline{ZK} = 3;16^p$$
for which Ptolemy rounds up to $3;17^p.$ Additionally,
$$\frac{3;25^p}{120^p} = \frac{\overline{DK}}{34;33^p}$$
$$\overline{DK} = 0;59^p.$$
We can then subtract this off of $\overline{DB}$ to determine that $\overline{KB} = 59;01^p.$ This is one of the sides of $\triangle ZKB$ in which we also know $\overline{ZK} = 3;16^p$ in this context. Thus, we can use the Pythagorean theorem to determine:
$$\overline{ZB} = 59;06^p.$$
We’ll now create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{ZB} = 120^p$:
$$\frac{120^p}{59;06^p} = \frac{\overline{ZK}}{3;17^p}$$
$$\overline{ZK} = 6;40^p.$$
We can then find the corresponding arc, $arc \; ZK = 6;22º.$ Thus, the angle this arc subtends on the opposite side of the demi-degrees circle, $\angle ZBK = 3;11º.$
We can add this to $\angle ADB,$ which we previously found to be $73;16º,$ to determine $\angle AZB = 76;27º.$
Therefore, at the moment of the above observation, Saturn’s distance from the apogee in mean longitudinal motion was $283;33º,$ i.e., it’s [mean] longitude was $2;53º$ into Virgo.
This was arrived at from the position of apogee being $19;20º$ into Scorpio plus the $283;33º$ minus the full revolution.
And since the mean sun’s position is given as $106;50º,$ if we add the $360º$ of one revolution to the latter, and from the resulting $466;50º$ subtract the $283;33º$ of the longitude [from apogee], we get, for the anomaly at that moment, $183;17º$ from the apogee of the epicycle.
So, since we have shown that, at the moment of the above observation, which is the $519^{th}$ year from Nabonassar, Tybi [V] $14,$ in the evening, [Saturn] was $183;17º$ [in anomaly] from the apogee of the epicycle and, at the moment of the third opposition, which was in the $883^{rd}$ year from Nabonassar, Mesore [XII] $24$, noon, it was $174;44º,$ it is clear that in the interval between the observations, which comprises $364$ Egyptian years and $219 \frac{3}{4}$ days, the planet Saturn had moved $351;27º$ (beyond $351$ complete revolutions in anomaly).
That total motion is $126,711;27º$ over $133,079;45$ days. Dividing that out, I find a mean motion of $0;57,07,43,41,44,18^{º}{day}.$
As with the previous planets, this matches what Ptolemy has in the mean motion tables for the first four sexagesimal places, but diverges for the remaining two. And as with before, I point readers to his Appendix C for further reading.
- γ Vir.
- If you’re interested in seeing a scale drawing of this, I tried that first, but it was quite hard to interpret. Thus, the angles and proportions in this drawing are not at all accurate.
- Ptolemy is immediately jumping into his demi-degrees context despite us not having drawn any demi-degrees circles. Thus, all of his angles are doubled, in case you’re trying to follow along in Toomer.
