Next, we’ll turn our attention to determining the size of Saturn’s epicycle. As he did for Jupiter, Ptolemy introduces a new observation.
[W]e took an observation which we made in the second year of Antoninus, Mecchir [VI] $6/7$ in the Egyptian calendar [$138$ CE, December $22/23$]. It was $4$ equinoctial hours before midnight, for according to the astrolabe, the last degree of Aries was culminating, while the longitude of the mean sun was $28;41º$ into Sagittarius. At that moment, the planet Saturn, sighted with respect to the bright star in the Hyades1, was seen to have a longitude $9 \frac{1}{15}º$ into Aquarius, and was about $\frac{1}{2}º$ to the rear of the center of the moon (for that was its distance from the noom’s northern horn).
Ptolemy then calculations the moon’s positions as a reference point:
Mean longitude: $8;55º$ into Aquarius
Anomaly: $174;15º$ from the apogee of the epicycle
hence, its true longitude must have been $9;40º$ into Aquarius
and its apparent longitude at Alexandria, $8;34º$ into Aquarius2.
This is used as a check for Saturn’s position against the astrolabe position.
Thus, from these considerations too, the planet Saturn must have had a longitude of $9 \frac{1}{15}º$ into Aquarius (since it was about $\frac{1}{2}º$ to the rear of the moon’s center.
He then give the mean position with respect to the apogee:
And its distance from the apogee of the eccentre (which was [in] the same [position as at the third opposition], since its shift over so short an interval is negligible), was $76;04º.$
The calculation is not shown here but is the mean ecliptic longitude of Saturn ($9;04º$ into Aquarius or $309;04º$) minus the position of apogee ($23;00º$ into Scorpio or $233º$).
Next, Ptolemy gives the interval of time between the observation of the third opposition and this new observation as
$2$ Egyptian years, $167$ days, $8$ hours.
Using that interval, the mean motions are calculated from the mean motions tables. The increases are
in longitude: $30;03º$
in anomaly: $134;24º.$
Ptolemy describes these as being “calculated roughly” but the values he gives agree to the nearest minute with my calculations.
From there, we can add these to our positions from the third anomaly calculated in the last post, to determine that the positions of Saturn are:
in [mean] longitude: $86;33º$
in anomaly: $309;08º$
This gives us what we need to get started with the diagram to determine the size of the epicycle:
In this diagram we have the line of apsides, $\overline{AG}$ on which sit our three centers: the center of mean motion, $Z$, the center of mean distance, $D$, and the center of the ecliptic. $E$.
The epicycle is centered on $B$ with Saturn at $K$. We connect $\overline{BK}$ and $\overline{EK}$ to it.
We also connect $\overline{DB}$ and $\overline{ZB}$, extending it so that it his the other side at $H$. Additionally, it is extended in the other direction until it meets a perpendicular dropped from $E$ at $L$. Another perpendicular is dropped from $D$ onto that line at $M$. We’ll also draw $\overline{EB}$ extending it as well, until it hits the opposite side of the epicycle at $\Theta.$
Lastly, a perpendicular is dropped from $B$ onto $\overline{KE}$.
To begin, we know that $\angle AZB = 86;33º$ as we established above. Thus, its vertical angle, $\angle DZM$ is as well.
This allows us to create a demi-degrees circle about $\triangle DZM$ in which the hypotenuse, $arc \; DM = 173;06º$. The supplement, $arc \; ZM = 6;54º.$ The corresponding chords are $\overline{DM} = 119;47^p$ and $\overline{ZM} = 7;13^p.$
We can then convert this into the context in which the diameter of the eccentres is $120^p$:
$$\frac{3;25^p}{120^p} = \frac{\overline{DM}}{119;47^p}$$
$$\overline{DM} = 3;25^p$$
and $$\frac{3;25^p}{120^p} = \frac{\overline{ZM}}{7;13^p}$$
$$\overline{ZM} = 0;12^p.$$
We then find $\overline{BM}$ using the Pythagorean theorem:
$$\overline{BM} = \sqrt{60^2 – 3;25^2} = 59;54^p.$$
We also know that $\overline{ZM} = \overline{ZL}$ and $\overline{EL} = 2 \cdot \overline{DM}.$ This allows us to determine that $\overline{BL} = 60;06^p$ and $\overline{EL} = 6;50^p.$
That’s two sides of $\triangle BEL,$ so we can use the Pythagorean theorem to find the remaining side:
$$\overline{BE} = \sqrt{60;06^2 + 6;50^2} = 60;29^p.$$
We’ll now create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{BE} = 120^p.$
$$\frac{120^p}{60;29^p} = \frac{\overline{EL}}{6;50^p}$$
$$\overline{EL} = 13;33^p.$$
We can then look up the corresponding arc, for which I find $arc \; EL = 12;58º.$ This means that the angle this arc subtends on the opposite side of the demi-degrees circle, $\angle EBL = 6;29º.$
Recalling that $\angle AZB = 86;33º,$ we can subtract $\angle EBL$ from this to determine $\angle AEB = 80;04º.$
Ptolemy then tells us to recall that we determined $\angle AEK,$ the apparent distance of the planet from apogee, to be $76;04º$ above. We can subtract to determine:
$$\angle KEB = \angle AEB – \angle AEK = 4;00º.$$
We then focus on $\triangle BEN$ which contains this angle, drawing a demi-degrees circle about it in which the hypotenuse, $\overline{BE} = 120^p.$ In that context, $arc \; BN = 8;00º$ and the corresponding chord, $\overline{BN} = 8;22^p.$
We’ll convert this back into our context in which the diameters of the eccentre are $120^p$:
$$\frac{60;29^p}{120^p} = \frac{\overline{BN}}{8;22^p}$$
$$\overline{BN} = 4;13^p.$$
We’ll now recall that we determined the planet, at $K$ was $309;08º$ about its epicycle from its apogee, at $H$. Thus, $\angle HBK = 50;52º.$
We’ll also recall that we determined $\angle EBL$ to be $6;29º$ above. This is a vertical angle with $\angle HB \Theta$ so we can subtract to determined:
$$\angle \Theta BK = \angle HBK – \angle HB \Theta = 44;23º.$$
Next, let’s look at $\triangle KEB$. In it, we know $\angle KEB$ which we found above to be $4;00º.$ Additionally, we can quickly find $\angle KBE$ as it’s the supplement of $\angle \Theta BK,$ so $135;37º.$ Thus, the remaining angle,
$$\angle BKE = 180º – 135;37º – 4;00º = 40;23º.$$
Now, we’ll focus on $\triangle BKN$ which contains this angle, creating a demi-degrees circle about in in which the hypotenuse, $\overline{BK} = 120^p.$ In that context, $arc \; BN = 80;46º$ and the corresponding chord, $\overline{BN} = 77;45^p.$
However, we also know $\overline{BN} = 4;13^p$ in our context in which the diameters of the eccentre is $120^p,$ so we can use that to convert.
$$\frac{4;13^p}{77;45^p} = \frac{\overline{BK}}{120^p}$$
$$\overline{BK} = 6;30^p.$$
This is the radius of the epicycle.
Ptolemy quickly sums up what we’ve covered:
Thus, we have computed the following:
[R]ound about the beginning of the reign of Antoninus, the longitude of Saturn’s apogee was $23º$ into Scorpio;
where the radius of the eccentre carrying the epicycle is $60^p,$
the distance between the centres of the ecliptic and the eccentre which produces the uniform motion is $6;50^p,$
and the radius of the epicycle $6;30^p.$
- α Tau.
- Toomer has a long note here that it is uncertain how Ptolemy did these calculations as they aren’t a good match with or without the equation of time being applied. Additionally, Toomer states that Ptolemy’s value for the parallax is notably off which leads to a $\frac{1}{4}º$ error in the final result.