Having determined the eccentricity of Mars’ model as well as the line of apsides,
Our next task is to demonstrate the ratio of the size of the epicycle.
For this task, Ptolemy adds a new observation which was
about three days after the third opposition, that is, in the second year of Antoninus, Epiphi [XI] $15/16$ in the Egyptian calendar [$139$ CE, May $30/31$], $3$ equinoctial hours before midnight. [That was the time,] for the twentieth degree of Libra [i.e., Libra $19º-20º$] was culminating according to the astrolabe, while the mean sun was $5;27º$ into Gemini at that moment. Now, when the star on the ear of wheat [Spica] was sighted in its proper position [on the instrument], Mars was seen to have a longitude of $1 \frac{3}{5}º$ into Sagittarius. At the same time, it was observed to be the same distance ($1 \frac{3}{5}º$) to the rear of the moon’s center.
Since we’re going to be discussing Mars’ position with respect to the moon, Ptolemy also tells us where the moon was located:
Now, at that moment, the moon’s position was as follows1:
Mean Longitude: $4;20º$ into Sagittarius
True Longitude: $29;20º$ into Scorpio
(for its distance in anomaly from the epicycle apogee was $92º$)
Apparent Longitude: $0º$ into Sagittarius
Ptolemy then adds the $1 \frac{3}{5}º$ that Mars was from the moon to the moon’s apparent longitude to state that Mars’ apparent longitude was $1 \frac{3}{5}º$ ($1;36º$) into Sagittarius.
You’ll recall that, in the last post, we stated that perigee was $25;30º$ into Capricorn ($295;30º$ in ecliptic longitude). Since Ptolemy is telling us that Mars was viewed $1;36º$ into Sagittarius ($241;36º$ ecliptic longitude), that’ puts Mars at $53;54º$ prior to perigee.
Next, Ptolemy looks at the interval of time between the third opposition and this new observation.
Looking back at the original opposition observations, the third was on Epiphi $12/13$, $2$ equinoctial hours before midnight. That means this observation was $2$ days and $23$ hours after that one.
We can use this interval to plug into the mean motions table for Mars and determine how much it moves in longitude (i.e., about the eccentre) and anomaly (i.e., about its epicycle).
For longitude, I get $1;33º$ and in anomaly, I get $1;22º$. Ptolemy’s values are slightly lower than this at $1;32º$ and $1;21º$ respectively, but Toomer notes that this is Ptolemy likely applying the equation of time to the interval (see previous footnote) which shortens the interval slightly.
If we add the latter to the positions at the opposition in question… we get, for the moment of this observation
Distance of Mars in longitude from the apogee of the eccentre: $137;11º$
Distance in anomaly from the apogee of the epicycle: $172;46º$
Checking these figures; in the last post we said that Mars’ mean position about the eccentre was $135;39º$, so adding the $1;32º$ Ptolemy found indeed gets $137;11º$.
Similarly, we stated that Mars was $171;25º$ about its epicycle, so adding $1;21º$ we get Ptolemy’s $172;46º$.
With these pieces of information in hand, Ptolemy produces a new diagram. It’s going to get a bit messy, so I’ll take it slowly.
To begin, we’ll use almost reuse the diagram from the last post. However, everything is relabeled here, and we’ve also created $\overline{DB}$. Note that the point on the epicycle’s intersection with this line is not labeled.
Now, let’s create a few right triangles:
Here, we’ve dropped perpendiculars from $E$ and $E$ onto $\overline{ZB}$ at points $L$ and $D$ respectively. Nothing we haven’t seen before.
But now we need to position Mars about the epicycle. We’ll put it at $N$. And once that’s done, we’ll extend a line from $E$ through it, extending it far enough that we can drop a perpendicular on it from $B$ at $X$. Additionally, we’ll extend $\overline{NB}$:
And that’s everything. A bit crowded, but we’ll manage.
To start, we’ll recall that above, we determined that Mars’ mean position about its eccentre from apogee was $137;11º$. This is $\angle AZB$. We can then find its supplement, $\angle BZG = 42;49º$.
We’ll now concentrate on $\triangle DZM$, creating a demi-degrees context about it. In it, $\angle MZD$ is the same one we just mentioned, and its hypotenuse, $\overline{ZD} = 120^p$.
In this context, $arc \; MD = 85;38º$ and its supplement, $arc \; ZM = 94;22º$.
Looking up the corresponding chords, I find $\overline{MD} = 81;34^p$ and $\overline{ZM} = 88;01^p$. These are also the values Ptolemy finds.
We can now convert that back to the context in which the diameter of the eccentre is $120^p$ in which we found $\overline{DZ} = 6^p$.
$$\frac{6^p}{120^p} = \frac{\overline{MD}}{81;34^p}$$
$$\overline{MD} = 4;05^p$$
and
$$\frac{6^p}{120^p} = \frac{\overline{ZM}}{88;01^p}$$
$$\overline{ZM} = 4;24^p.$$
Next, let’s focus on $\triangle DBM$. In this, $\overline{DB} = 60^p$ since it’s a radius, and we just found $\overline{MD}$, which means we can find $\overline{BM}$ using the Pythagorean theorem.
$$\overline{BM} = \sqrt{60^2 – 4;05^2} = 59;52^p.$$
As with before, $\overline{ZM} = \overline{ML}$ and $\overline{EL} = 2 \cdot \overline{DM}$.
We can use the first of these statements to cut $\overline{ML}$ off of $\overline{BM}$:
$$\overline{BL} = \overline{BM} – \overline{ML}$$
$$\overline{BL} = 59;52^p – 4;24^p = 55;28^p.$$
Using the second of these statements, we can say that $\overline{EL} = 8;10^p$.
That gives us two sides of $\triangle BEL$. So we can use the Pythagorean theorem to find the hypotenuse, $\overline{EB}$:
$$\overline{EB} = \sqrt{55;28^2 + 8;10^2} = 56;04^p.$$
We’ll now jump into a demi-degrees context about the same triangle so we can determine the angles. In it, $\overline{EB} = 120^p$ which we can use in our conversion factor for the other pieces:
$$\frac{120^p}{56;04^p} = \frac{\overline{EL}}{8;10^p}$$
$$\overline{EL} = 17;29^p.$$
Ptolemy rounds down to $17;28^p$ which I’ll adopt.
We can then look up the corresponding arc which I find as $arc EL = 16;44º$.
This means that $\angle LBE = 8;22º$.
Ptolemy now zooms out remind us that
the apparent distance of the planet Mars in advance of the perigee $G$, [i.e.,] $\angle GEX$ is given as $53;54º$.
We have also shown that $\angle BZG = 42;49º$.
We can use that to state2:
$$\angle GEB = \angle ZBE + \angle BZG$$
$$\angle GEB = 8;22º + 42;49º = 51;11º.$$
Next, we can subtract to find:
$$\angle BEX = \angle GEX – \angle GEB$$
$$\angle BEX = 53;54º – 51;11º = 2;43º.$$
We’ll now create a demi-degrees context about $\triangle BEX$. In it, the hypotenuse, $\overline{BE} = 120^p$ and $arc \; XB = 5;26º$.
Looking up the corresponding chord, we find that $\overline{XB} = 5;41^p$.
We can now convert back to our context in which the diameter of the eccentre is $120^p$ using $\overline{EB}$ as our conversion factor:
$$\frac{56;04^p}{120^p} = \frac{\overline{XB}}{5;41^p}$$
$$\overline{XB} = 2;39^p.$$
Now let’s focus on point $N$. Measuring counter clockwise from $H$, the apogee of the epicycle, $N$ is $172;46º$ around, which is to say that $\angle HBN = 172;46º$. Thus, the remaining angle for Mars to get to perigee is $\angle KBN = 7;14º$.
However, we also know that $\angle KB \Theta = 8;22º$ as it’s the same as $\angle LBE$ which we found above.
We can then subtract:
$$\angle NB \Theta = \angle KB \Theta – \angle KBN$$
$$\angle NB \Theta = 8;22º – 7;14º = 1;08º.$$
Ptolemy then adds this to $\angle BEX$ to find:
$$\angle XNB = \angle NB \Theta + \angle BEX$$
$$\angle XNB = 1;08º + 2;43º = 3;51º$$
We’ll now focus on the small triangle in the epicycle, $\triangle BNX$, creating a demi-degrees context about it in which the hypotenuse, $\overline{BN} = 120^p$.
In it, $arc \; XB = 7;42º$. We can look up the corresponding chord to find that $\overline{XB} = 8;03^p$.
However, we know this piece in our main context, in which the diameter of the eccentre is $120^p$, to be $2;39^p$, so we can use this in our conversion factor to convert $\overline{BN}$:
$$\frac{2;39^p}{8;03^p} = \frac{\overline{BN}}{120^p}$$
$$\overline{BN} = 39;30^p.$$
Therefore the ratio of the radius of the eccentre to the radius of the epicycle is $60 : 39;30$.
And that’s it for this chapter. But we’re not quite done with Book X yet. In the next chapter, we’ll look at an ancient observation of Mars in which Ptolemy will discuss how he improved the period of anomaly from his predecessors. Lastly, we’ll use everything we’ve learned to be able to determine the position of Mars at epoch.
That will be the end of Book X on Mars, after which, we’ll be repeating much of this for Jupiter!
- Toomer checks these values and finds they are correct if they were calculated for $8;37$ pm. This indicates Ptolemy used the equation of time to subtract $23$ minutes, although $25 \frac{1}{2}$ minutes would have been more accurate.
- If you’re following along with Ptolemy, he does these calculations all in a demi-degrees context in which the angles are all doubled.