Next, we’ll determine the size of Jupiter’s epicycle. To do so, we’ll use another observation that Ptolemy has made.
The observation was on
the second year of Antoninus, Mesore [XII], $26/27$ in the Egyptian calendar [$139$ CE, July $10/11$], before sunrise, i.e., about $5$ equinoctial hours after midnight (for the mean longitude of the sun was $16;11º$ into Aries, and the second degree of Aries [i.e., $1-2º$ of Aries] was culminating according to the astrolabe). At that moment, Jupiter, when sighted with respect to the bright star in the Hyades1, was seen to have a longitude of $15 \frac{3}{4}º$ into Gemini.
That gives the apparent position with respect to α Tau.
Next, Ptolemy gives us the calculated position of the moon because, he says, Jupiter
had the same apparent longitude as the center of the moon, which lay to the south of it.
For that moment2 we find, by means of the calculations [previously] explained:
Moon’s mean longitude: $9;00º$ into Gemini
Moon’s [mean] anomaly counted from the epicycle apogee: $272;05º$Hence its true position: $14;50º$ into Gemini
and its apparent position at Alexandria: $15;45º$ into Gemini.Thus, from these considerations too, Jupiter’s longitude was $15 \frac{3}{4}º$ into Gemini.
Ptolemy then gives the time interval from the third opposition to the above observation as $1$ Egyptian year and $276$ days.
This allows him to calculate the change in the longitude and anomaly as:
in longitude: $53;17º$
and in anomaly: $218;31º.$
My calculations from the planetary mean motions table agrees with this.
We then add these increases over this interval to the positions stated in the last post. Thus, we find the positions at the time of this observation were:
in longitude: $263;53º$ from the apogee
and in anomaly: $41;18º$ from the apogee of the epicycle
We now produce a diagram of this configuration:
Quickly explaining this diagram, we have $\overline{AG}$ as the line of apsides. Then points $Z$, $D$, and $E$ are the center of mean motion, mean distance, and the observer, respectively.
All three of these are then connected to $B$, the center of the epicycle with $\overline{ZB}$ being extended to the far side of the epicycle at $H$, and $\overline{EB}$ extended to the far side at $\Theta$.
Jupiter will be at point $K$ and we’ll create $\overline{EK}$ onto which we’ll drop a perpendicular from $B$ at $N$.
Perpendiculars are also dropped from $D$ and $E$ onto $\overline{ZH}$ at $M$ and $L$ respectively.
Then, since the mean position in longitude from the apogee of the eccentre is $263;53º$, $\angle GBZ = 83;53º$.
We can then look at $\triangle DZM$ which contains this angle and create a demi-degrees context about it in which the hypotenuse, $\overline{DZ} = 150^p.$ In that context, $arc \; DM = 167;46º$ and its supplement, $arc \; ZM = 12;14º$.
We can then find the corresponding chords: $\overline{DM} = 119;19^p$ and $\overline{ZM} = 12;47^p.$
This, then, gets converted to the context in which the diameters of the eccentres is $120^p$:
$$\frac{2;45^p}{120^p} = \frac{\overline{DM}}{119;19^p}$$
$$\overline{DM} = 2;44^p$$
and
$$\frac{2;45^p}{120^p} = \frac{\overline{ZM}}{12;47^p}$$
$$\overline{ZM} = 0;18^p.$$
We’ll now focus on $\triangle DMB$ in which we know two sides. Thus, we can find the third side using the Pythagorean theorem:
$$\overline{MB} = \sqrt{60^2 – 2;55^2} = 59;56^p.$$
Then, we can subtract off $\overline{ML}$ which has the same length as $\overline{ZM}$ to determine that $\overline{LB} = 59;38^p.$
As we’ve seen, $\overline{EL} = 2 \cdot \overline{DM} = 5;28^p.$
This means we know two sides of $\triangle ELB$, so we can find the remaining side:
$$\overline{EB} = \sqrt{59;38^2 + 5;28^2} = 59;53^p$$
although Ptolemy gets $59;52^p$ likely due to having computed with greater precision.
We’ll now create a demi-degrees circle about this triangle in which the hypotenuse, $\overline{EB} = 120^p$ and we’ll convert $\overline{EL}$ into that context:
$$\frac{120^p}{59;53^p} = \frac{\overline{EL}}{5;28^p}$$
$$\overline{EL} = 10;57^p.$$
Ptolemy comes up with $10;58^p.$
We then find the corresponding arc, $arc \; EL = 10;30º.$
Thus, the angle opposite it on the demi-degrees circle, $\angle EBL = 5;15º.$
Next, recall that $\angle GZB = 83;53º.$
If we add these two, we get $\angle BEG = 89;08º.$
Furthermore, since the approximate longitude of the perigee, $G$, is $11º$ into Pisces, and the apparent longitude of the planet, as viewed along $\overline{EK}$, was $15;45º$ into Gemini, $\angle KEG = 94;45º.$
We can then subtract $\angle BEG$ from this to determine $\angle KEB = 5;37º.$
We’ll now create a demi-degrees circle about $\triangle BEN$ where the hypotenuse, $\overline{EB} = 120^p$. In it, $arc \; BN = 11;14º.$ As a quick note for later, this also means that the angle this arc subtends, $\angle BEN = 5;37º.$
We can then find the corresponding chord, $\overline{BN} = 11;45^p$ although Ptolemy rounds down to $11;44^p$.
This gets converted back to our context in which the diameter of the eccentre equals $120^p:$
$$\frac{59;52^p}{120^p} = \frac{\overline{BN}}{11;45^p}$$
$$\overline{BN} = 5;52^p.$$
Ptolemy comes out a bit low at $5;50^p.$
Next, we can state that $arc \; HK = 41;18º$ as this is the distance we determined that Jupiter was about its epicycle from the apogee3. Thus, so is $\angle HBK$.
We also determined that $\angle EBZ = 5;15º$ which we can subtract off of $\angle HBK$ to determine $\angle KB \Theta = 36;03º.$
Additionally, we showed that $\angle \Theta EK = 5;37º.$
We can then subtract:
$$\angle BKN = \angle KB \Theta – \angle \Theta EK = 30;26º.$$
We’ll then create a demi-degrees context about $\triangle BNK$ in which the hypotenuse, $\overline{BK} = 120^p$. We can then state that $arc \; BN = 60;52º$ and the corresponding chord, $\overline{BN} = 60;47^p.$
We can use this to convert $\overline{BK}$, the radius of the epicycle, back to our context in which the diameter of the eccentres is $120^p$:
$$\frac{5;50^p}{60;47^p} = \frac{\overline{BK}}{120^p}$$
$$\overline{BK} = 11;31^p.$$
Ptolemy rounds down to $11;30^p$.
You may well note that I’ve called out a number of instances in which Ptolemy did a bit of rounding. Toomer notes that this has caused a moderate amount of accumulated error as a precise value during these calculations would have let to a result of the radius of the epicycle being $11;38^p.$ Toomer speculates that Ptolemy was likely aiming for a nice round number, as seems to be his preference4.
That concludes Chapter $2$!
- α Tau.
- Toomer notes that this calculation looks to have been computed, not for $5$ am as Ptolemy states, but for $4:42$ am, thus including the equation of time with respect to the epoch of era Nobonassar.
- Recalling that apogee is taken from the view of the equant, or center of mean motion.
- See: His interesting roundings that led to the precession of the equinoxes being exactly $1º$ per century.