Back in IX.3, Ptolemy noted that the planetary periods
have been corrected by us, on the basis of the comparison of their positions which became possible after we had demonstrated their anomalies, as we shall explain at that point.
As a reminder, what he was telling us then, was that he had made corrections to the periods provided to him by Hipparchus. We’ll now go over those corrections for Mars.
To do so, Ptolemy makes use of
one of the ancient observations, in which it is declared that, in the $13^{th}$ year of the calendar of Dionysius, Aigon $25$, at dawn, Mars seemed to have occulted the northern [star in the] forehead of Scorpius. The moment of this observation is in the $52^{nd}$ year from the death of Alexander, i.e., in the $476^{th}$ year from Nabonassar, Athyr [III] $20/21$ in the Egyptian calendar [$271$ BCE, Jan $17/18$], dawn1.
Having established the date, Ptolemy then calculates the position of the mean sun, finding it to be $23;54º$ into Capricorn.
For the star in question (β Sco), Ptolemy refers to the star catalog and gives its position as $6 \frac{1}{3}º$ into Scorpio in his time. However, due to precession, he adjusts the coordinates for the interval since the observation.
Ptolemy reckons that it has been $409$ years since the observation. So, at a rate of $1º$ per century, that means the position needs to be adjusted by $4;05º$. Thus, the correction position Ptolemy gives is $2 \frac{1}{4}º$ into Scorpio. Since Mars was said to occult this star, Mars would, therefore, have the same ecliptic longitude.
Similarly, the position of apogee also needs to be adjusted by this amount. Instead of being at $25;30º$ into Cancer2, apogee would have instead been located at $21;25º$ into Cancer at the time of the given observation.
Thus, the difference between apogee and Mars’ apparent position is:
$$212;15º – 111;25º = 100;50º$$
after the apogee.
We can also determine the sun’s position relative to the same apogee:
$$293;54º – 111;25º = 182;29º$$
after the apogee, putting it $2;29º$ past the perigee of Mars.
With that information, Ptolemy sets up a new diagram.
You may need to enlarge this image as it’s a bit thinner than normal because I drew the image to scale (yes, the epicycle really is that big!), and there’s a few small triangles that I had to make sure were visible.
To describe the layout, $\overline{AG}$ is the line of apsides with $A$ as apogee and $G$ as perigee. As usual, we have $Z$ as the center of motion (equant), $D$ as the center of mean distance, and $E$ as the observer on Earth.
The epicycle is centered on $B$ with Mars at $\Theta$.
We’ll extend a line from $Z$ through $B$ to $H$, the apogee of the epicycle.
We’ll also extend $\overline{DB}$ as well as $\overline{E \Theta}$.
The sun’s mean position will be at $L$, which we’ll extend a line to from $E$.
Then, we’ll draw a few right triangles. First off, we’ll drop $\overline{BN}$ onto $\overline{E \Theta}$ so that it’s perpendicular at $N$. Note that this point looks like it’s on the epicycle, but it’s not.
We’ll also extend $\overline{E \Theta}$ until it meets a perpendicular dropped from $D$ at $M$.
Lastly, we’ll drop a perpendicular from $Z$ onto $\overline{DB}$ at $K$.
We’ll also extend a line from $D$ onto $\overline{BN}$ at $X$ such that it’s parallel to $\overline{E \Theta}$ also making it perpendicular to $\overline{BN}$.
From the setup above, we stated that the apparent position from apogee of Mars, $\angle AE \Theta = 100;50º.$
Additionally, we stated that the sun was $2;29º$ past Mars’ apogee which is $\angle GEL.$
We can then find $\angle \Theta EL$. This is:
$$\angle \Theta EL = 180º – \angle AE \Theta + \angle GEL$$
$$\angle \Theta EL = 180º – 100;50º + 2;29º = 81;39º.$$
This is the same as $\angle B \Theta E$. This is because $\overline{B \Theta} \parallel \overline{EL}$ making $\angle B \Theta E$ an alternate angle to $\angle \Theta EL$.
We can now focus on $\triangle{B \Theta N}.$ In that context, the hypotenuse, $\overline{B \Theta} = 120^p$.
In it, $arc \; BN = 163;18º$. So we can look up the corresponding chord, $\overline{BN}$. I find it to be $118;44^p$ but Ptolemy rounds down to $118;43^p$.
However, we know the size of the radius in the context in which the diameter of the eccentre is $120^p$. We found it to be $39;30^p$ in the last post. Thus, we can convert $\overline{BN}$ to that context as well:
$$\frac{39;30^p}{120^p} = \frac{\overline{BN}}{118;43^p}$$
$$\overline{BN} = 39;05^p.$$
Ptolemy again rounds down and comes up with $39;03^p.$
Next, take a look at $\overline{\Theta M}$ where it crosses with $\overline{AG}$. We know that $\angle AE \Theta = 100;50º$. Thus, it’s supplement along $\overline{\Theta M}$, $\angle DEM = 79;10º$.
We’ll then enter a demi-degrees context about $\triangle DEM$. In it, the hypotenuse, $\overline{DE} = 120^p$ and $arc \; DM = 158;20º$. Looking up the corresponding chord, I find $\overline{DM} = 117;52^p$.
However, we know the size of $\overline{DE}$ in the context of the eccentre having a diameter of $120^p$ in which it is the eccentricity of $6^p$. So we can convert $\overline{DM}$:
$$\frac{6^p}{120^p} = \frac{\overline{DM}}{117;52^p}$$
$$\overline{DM} = 5;54^p.$$
Because shape $XNMD$ is a rectangular parallelogram, this is also the length of $\overline{XN}$ which we can now subtract off of $\overline{BN}$.
$$\overline{BX} = \overline{BN} – \overline{XN}$$
$$\overline{BX} = 39;03^p – 5;54^p = 33;09^p.$$
Now, we’ll focus on $\triangle BDX$, scaling into it using $\overline{BD}$ as our piece to scale on since it’s a radius in our main context, and the hypotenuse in the demi-degrees context:
$$\frac{120^p}{60^p} = \frac{\overline{BX}}{33;09}$$
$$\overline{BX} = 66;18^p.$$
We can then look up the corresponding arc. Doing so, I find $arc \; BX = 67;04º$ which means that $\angle BDX = 33;32º.$
To that, we can add $\angle XDM$, which was a right angle, to determine that $\angle BDM = 123;32º$.
Next, let’s look at $\angle EDM$. This is part of $\triangle EDM$. In it, we’ve found $\angle DEM = 79;10º$ and $\angle EMD$ is a right angle, so the remaining angle, $\angle EDM = 10;50º$.
We can then subtract this from $\angle BDM$ to find $\angle BDE = 112;42º$.
Therefore, its supplement, along $\overline{AG}$, $\angle BDA = 67;18º$.
This allows us to focus on $\triangle DZK$, creating a demi-degrees context about it in which the hypotenuse, $\overline{DZ} = 120^p$.
We can then state that $arc \; ZK = 134;36º$ and its corresponding chord, $\overline{ZK} = 110;42^p$.
Similarly, $arc \; DK = 45;24º$ and its corresponding chord, $\overline{DK} = 46;18^p$.
We can then convert this back to our overarching context:
$$\frac{6^p}{120^p} = \frac{\overline{ZK}}{110;42^p}$$
$$\overline{ZK} = 5;32^p$$
and
$$\frac{6^p}{120^p} = \frac{\overline{DK}}{46;18^p}$$
$$\overline{DK} = 2;19^p.$$
Then, we can subtract $\overline{DK}$ off of $\overline{BD}$ which was a radius. Thus, $\overline{BK} = 57;41^p$.
We’ll now look at $\triangle BZK$. We now know two sides of it – $\overline{BK}$ and $\overline{KZ}$ so we can use the Pythagorean theorem to state:
$$\overline{BZ} = \sqrt{57;41^2 + 2;19^2}$$
$$\overline{BZ} = 57;57^p.$$
This allows us to enter a demi-degrees context around this triangle in which the hypotenuse $\overline{BZ} = 120^p$:
$$\frac{120^p}{57;57^p} = \frac{\overline{ZK}}{5;32^p}$$
$$\overline{ZK} = 11;27^p$$
which Ptolemy rounds up to $11;28^p.$
We can then look up the corresponding arc. I find this to be $arc \; ZK = 10;58º$. Therefore, $\angle KBZ = 5;29º$.
Ptolemy then tells us (without a proof) that $\angle KBZ + \angle BDA = \angle BZA$.
This is quite easy to prove as both sides of this equation are $180º – \angle BZD$. The left hand side because of the angle in the triangle, and the right because it’s the supplement along $\overline{AG}$.
Plugging in, we find that $\angle BZA = 5;29º + 67;18º = 72;47º$.
Therefore, the mean position in longitude of the planet (i.e., of $B$, the center of the epicycle) at the moment of the observation in question was $72;47º$ from the apogee. Hence, its [mean] longitude was ]$21;25º$ into Cancer $+ 72;47º =$] $4;12º$ into Libra.
In other words, if the center of Mars’ epicycle, $B$ was viewed from the equant, it would appear to be $4;12º$ into Libra.
Next, we can look at $\angle ADL$ measured counter-clockwise. This is $180º$ of the semi circle, plus $\angle GEL$ which we stated previously was $2;29º$ or $182;29º.$
Ptolemy then tells us that this
equals the sum of the mean longitude, $\angle AZB$, and the [mean] anomaly (i.e., the [mean] motion of the planet of the epicycle) $\angle HB \Theta$. This follows from the condition of $\overline{B \Theta}$ having to remain parallel to $\overline{EL}$.
Writing that out:
$$\angle AZB + \angle HB \Theta = 182;29º.$$
We just found \angle AZB, so we can solve for $\angle HB \Theta$:
$$\angle HB \Theta = 182;29º – 72;47º = 109;42º.$$
Therefore, the distance of the planet in anomaly from the apogee of the epicycle at that same moment of the observation above was $109;42º$ which was what we had to determine.
However, this still needs to be compared to the position of Mars from the third opposition:
Now, we had [already] shown that, at the moment of the third opposition, the distance [of Mars] in anomaly from the apogee of the epicycle was $171;25º$. Therefore, in the interval between the observation, which comprises $410$ Egyptian years, and $231 \frac{2}{3}$ days (approximately), the planet moved $61;43º$ beyond $192$ complete revolutions. That is practically the same increment [in anomaly] which we find from the tables for Mars’ mean motion we constructed. For our [mean] daily motion was derived from these very data, by dividing the number of degrees obtained from the complete revolutions plus the increment by the number of days computed from the interval between the two observations.
The word “practically” here is doing some work as Toomer notes. In an appendix, Toomer states (referring to this calculation for all of the planets) that “if one does the computations implied… using Ptolemy’s numbers, in no case does one find agreement with the mean daily motions in anomaly which he actually lists”.
The results are quite close, and Toomer states that “[t]he worst of these discrepancies, that for Jupiter, does not produce an error of as much as one minute of arc in $400$ years”, but it’s still not exact. For Mars, the difference only shows up in the last two sexagesimal places Ptolemy gives, i.e., the $60^{-5}$ and $60^{-6}$ places. So clearly, Ptolemy is doing something he’s not telling us.
Toomer explores this in his Appendix C, but that’s beyond the scope of what I’m willing to entertain in this post, so I’ll end here since this is the end of Chapter $9$.
- Toomer expresses skepticism over this observation. He notes that there are many values given for the date in Aigon in the surviving texts, making the intended date uncertain. Other authors have given the $26^{th}$. However, Toomer notes that modern calculations suggest the occultation described should have “occurred two days earlier than the date Ptolemy gives.”
- In the previous post, we discussed perigee as $25;30º$ into Capricorn. Thus, apogee is diametrically opposite this at $25;30º$ into Cancer.