Almagest Book X: Position of Venus About the Epicycle on Oct 10/11 271 BCE

Continuing with our process of finding the position of Venus about the epicycle at two widely discrepant points in time, we’ll now look at one from Timocharis.

[$2$] From the ancient observations, we selected one which is recorded by Timocharis as follows. In the thirteenth year of Philadelphos, Mesore [XII] $17/18$ in the Egyptian calendar [October $11/12$ $271$ BCE], at the twelfth hour, Venus was seen to have exactly overtaken the star opposite Vindemiatrix. That is, the star which, in our descriptions, is the one following the star on the tip of the southern wing of Virgo, and which had a longitude of $8\frac{1}{4}º$ in Virgo in the first year of Antoninus. Now, the year of the observation is the ${476}^{th}$ from Nabonassar, while the first year of Antoninus is $884$ [years] from Nabonassar; to the ${408}^{th}$ years of the interval corresponds to a motion of the fixed stars and the apogees of about $4\frac{1}{12}º$. Hence, it is clear that the longitude of Venus was $4\frac{1}{6}º$ into Virgo, and the longitude of the perigee of its eccentre $20\frac{11}{12}º$ into Scorpio. And, here too, Venus was past its greatest elongation as morning star; for $4$ days after the above observation, on Mesore $21/22$, as one can deduce from what Timocharis says, its longitude was $8\frac{5}{6}$ in Virgo, according to our coordinates, and the mean position of the sun was $17;03º$ into Libra at the first observation and $20;59º$ in Libra at the next: this its elongation at the first observation comes to $42;53º$ and, at the next, $42;09º$.

To determine the position about the epicycle, we’ll produce a diagram almost the same as the previous one, but flipped across the line of apsides and without one side of the epicycle crossing that line.

As with before, $\overline{AE}$ is the line of apsides¹ again with $A$ as the point $25º$ into Taurus (apogee) and $E$, the point $25º$ into Scorpio (perigee). Point $B$ is the point about which the eccentre rotates and $G$ its actual center. $D$ is the observer on earth.

The epicycle is centered on $Z$ with Venus at $K$, again, sometime after greatest elongation.

From $B$ and $D$ we extend $\overline{BH}$ and $\overline{D\mathrm{\Theta }}$ respectively, through $Z$. We’ll also create $\overline{GZ}$ and $\overline{DK}$.

Again, we’ll drop perpendiculars from $G$ and $D$ onto $\overline{BH}$ at $L$ and $M$ respectively as well as from $Z$ onto $\overline{DK}$, perpendicular at $N$.

Let’s get started.

First, we’ll find $\mathrm{\angle }EBZ$ which is the angle of the mean sun before the perigee. However, we must recall that this observation, as Ptolemy told us in his summary of it, was $408$ years ago and thus, the line of apsides had progressed $4\frac{1}{12}º$. Thus, $E$, at that time, was $20;55$ into Scorpio instead of $25º$. Therefore the difference between that and the position of the mean sun at that time (which Ptolemy gave us as $17;03º$ into Libra) is $33;52º$.

We’ll then turn our attention to $\mathrm{△}BGL$ creating a demi-degrees context about it. In that context, the hypotenuse, $\overline{BG}={120}^p$. The arc opposite $\mathrm{\angle }GBL$, $arcGL=67;44º$ and thus, the corresponding chord $\overline{GL}=66;{52}^p$.

Using either the Pythagorean theorem or some more angle/chord relationships, we can also determine that $\overline{BL}=99;{38}^p$.

But again, in the larger context for Venus’ model, we’ve determined that $\overline{BG}=1;{15}^p$ which we can use to switch back to that context.

Doing so, I find that $\overline{GL}=0;{42}^p$ and $\overline{BL}=1;{02}^p$.

Next, let’s take a look at $\mathrm{△}ZGL$. We just determined $\overline{GL}$ in this, and $\overline{GZ}={60}^p$. Thus, we can use the Pythagorean theorem to determine $\overline{ZL}=59;59,{45}^p$ which Ptolemy just rounds to $\approx 60$ again.

Using the same reasoning as from the last post, we can state that $\overline{BL}=\overline{LM}$ and that $\overline{DM}=2\cdot \overline{GL}$.

This allows us to subtract:

$$\overline{ZM}=\overline{ZL}–\overline{LM}={60}^p–1;{02}^p=58;{58}^p$$

and

$$\overline{DM}=1;{24}^p.$$

This gives us two sides of $\mathrm{△}ZDM$ allowing us to use the Pythagorean theorem to determine $\overline{ZD}=58;{59}^p$.

We’ll now create a demi-degrees circle about this triangle.

In it, the hypotenuse, $\overline{ZD}={120}^p$ which gives us our avenue for converting the rest of the triangle to this context. Specifically, $\overline{DM}=2;{51}^p$ which we can then use to determine the corresponding $arcDM=2;43º$ for which Ptolemy, evidently, rounded up again, getting $2;44º$.

Thus, the opposing angle, $\mathrm{\angle }MZD=1;22º$².

Next, we’ll add³:

$$\mathrm{\angle }EDZ=\mathrm{\angle }BZD+\mathrm{\angle }EBZ=1;22º+33;52º=35;14º.$$

Turning back to the observation, Ptolemy reported that it was $4;10º$ into Virgo, as observed from earth at the time of observation. This is $76;45º$ before the perigee⁴ which is represented by $\mathrm{\angle }EDK$.

We can then subtract $\mathrm{\angle }EDZ$ from this to determine that $\mathrm{\angle }ZDK=41;31º$.

We’ll now set up a new demi-degrees context about $\mathrm{△}DZN$. In it, the hypotenuse, $\overline{DZ}={120}^p$ and $arcZN=83;02º$ for which the corresponding chord, $\bar{Z}N=79;{32}^p$. Ptolemy again rounds up to $79;{33}^p$.

We can now convert back to the previous context.

Doing so, I find that $\overline{ZN}=39;{06}^p$. And again, Ptolemy has evidently rounded up to $39;{07}^p$⁵

Next, we’ll focus on $\mathrm{△}ZKN$, creating a demi-degrees circle about it. In this context, the hypotenuse, $\overline{ZK}={120}^p$. However, this is the radius of the epicycle which we know to be $43;{10}^p$, allowing us to convert to this context.

Doing so, I find that $\overline{ZN}=108;{45}^p$. Looking up the corresponding chord, I find that $arcZN=129;59º$ which Ptolemy rounds to $\approx 130º$ which makes the opposite angle, $\mathrm{\angle }NKZ=65º$.

This, then, gets added to $\mathrm{\angle }ZDK$ (which we found to be $41;31º$ to determine $\mathrm{\angle }\mathrm{\Theta }ZK=106;31º$.

To this we can add $\mathrm{\angle }\mathrm{\Theta }ZK$ (which is the vertical angle of $\mathrm{\angle }MZD$ to determine $\mathrm{\angle }HZK=107;53º$. This is the angle of Venus before apogee. Taking it the other way around, we find that Venus was $252;07º$ after its last pass through the apogee, which is what we set about to determine.

With this result, and the one from Ptolemy’s time, we can then determine how far Venus advanced about the epicycle during the interval between the observations.

The interval between the two observations, Ptolemy tells us, was $409$ Egyptian years and roughly $167$ days.

If we look back at the chapter on periodic motions for the planets, we see that there are $5$ full returns in anomaly every $8$ years. Thus, there would have been $255$ complete returns in that period.

That does leave some additional time on the table as that would only take us to $408$ years, but Ptolemy states:

the remaining year plus the additional days do not complete the period of one revolution.

This is correct as a full return in anomaly occurs every 1.6 years which we’re not going to be able to get to with what was left on the table.

To this, we’ll add the amount in which the anomaly increased based on our two observations.

In the historic observation from Timocharis we explored in this post, Venus was $252;07º$ past apogee. In the observation from Ptolemy’s time, it was $230;32º$ past apogee, which means that Venus had advanced $338;25º$ in addition to the full revolutions in that time.

That’s a total advance $25,35,38;25º$ in a period of $41,30,52;00$ days.

If we divide that out, I find it comes out to $0;36,59,25,49,08,{51}^{\frac{º}{d}}$.

This agrees with Ptolemy’s value in the mean motions table up to the third sexagesimal place.

Toomer notes this as well and considers one possible reason for why the values don’t match. Specifically, Ptolemy claims that the interval was an integer number of days. However, in the observation from his time, the observation was around $4:45$am. For this observation, it was given as the “twelfth hour” which is, presumably, of the night, which would put this observation at $6:00$am. This is a difference of over $1\frac{1}{4}$ hours or, if the equation of time is taken into consideration, $1\frac{1}{2}$ hours.

However, while taking this improves the value (with respect to the value Ptolemy reports in the mean motions table), it still doesn’t quite match. Thus, Ptolemy’s exact methods are unknown here.

In the next post, we’ll use this information to determine Venus’ epoch position which will complete the discussion of Venus for now⁶.

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1. Truncated of course.
2. Ptolemy again continues his calculations in the demi-degrees context in which all angles are doubled.
3. The proof for this can be found in the last post.
4. Again, recalling that the perigee was $20;55$ into Scorpio instead of $25º$ at the time of observation.
5. Toomer, again, notes these rounding errors and states that the accumulated rounding error amounts to $0;04º$ for the final result.
6. We still have to discuss the latitudinal motion of the planets, which Ptolemy will return to later.