We’ve come a long way in this book establishing a working model for solar motion. In fact, we’ve explored two models and derived a table that shows how far the sun would be away from its mean motion based on the mean position. However, at this point, everything has been done in terms of apogee and perigee.
In this chapter, we’ll be defining the “epoch“. What that means is that Ptolemy is going to pick a point in time, and define where the sun was on that date. Then, applying the tools we have developed in this book, we’ll be able to determine where the sun is at any other given date using the epoch as the starting point. For those that like things in a bit more mathy terms, it’s the location of the sun on the eccentre at time = 0, wherein Ptolemy will decide what that date is. To get there, we’ll first establish precise point on the eccentre at a known point in time, and then use the methods from this chapter to go backwards until we get to the chosen epoch date.
To begin, Ptolemy sets up another diagram based on the eccentric model. Here, we’ve extended $\overline{BD}$ until it meets a perpendicular dropped onto it from Θ.
We’ll now make use of the point B that I called out in a previous post but we did not use then. This point traces the projection of the object on the ecliptic, so we’ll define it as the autumnal equinox which is located at the beginning of Libra1.
We’ll also make use of point G, the perigee. Although we didn’t explicitly call it out previously, we know this is $5 \frac{1}{2}º$ into Sagittarius which is 245;30º ecliptic longitude. We know this because we previously found the point of apogee which was $5 \frac{1}{2}º$ into Gemini (or 65;30º ecliptic longitude) and apogee and perigee are diametrically opposite.
This means that from point B, the sun will pass through all of Libra, all of Scorpio, and the first $5 \frac{1}{2}º$ of Sagittarius to reach point G. Since each constellation is defined to be 30º wide, that’s a total of $65 \frac{1}{2}º$. Thus, $arc \; BG = 65 \frac{1}{2}º$ as does the angle it subtends, $\angle{BDG}$. However, this angle is also equal to $\angle{\Theta DK}$ as they are vertical angles.
We’ll use the demi-degrees method on $\triangle{\Theta DK}$ with diameter $\overline{D \Theta}$. Since $\angle{\Theta DK}$ is on the circumference of this small circle, it means that the arc opposite it will be twice that measure in the context of that small circle, which is to say, $arc \; \Theta K = 131º$. From the chord table, we find the chord, $\overline{\Theta K} = 109;12$.
Ptolemy skips a step here, but I’ll walk through it. First, we’ll put this triangle back in the context of the overall diagram. We’ll do this by relating $\overline{D \Theta}$:
$$\frac{\overline{\Theta K}}{\overline{D \Theta}} = \frac{109;12}{120} = \frac{\overline{\Theta K}}{\overline{2;30}}$$
Solving:
$$\overline{\Theta K} = 2;30 \cdot \frac{109;12}{120} = 2;16,30$$
Now we’ll use a second demi-degrees circle around $\triangle{\Theta KZ}$ with $\overline{Z \Theta}$ as the diameter.
$$\frac{\overline{\Theta K}}{\overline{Z \Theta}} = \frac{2;16,30}{60} = \frac{\overline{\Theta K}}{120}$$
Solving:
$$\overline{\Theta K} = 120 \cdot \frac{2;16,30}{60} = 4;33$$
That catches us back up to where Ptolemy skipped to.
From there, we can use the chord table to look up the corresponding arc, $arc \; \Theta K$ which would be 4;20º, thus, the angle this arc subtends on the circumference, $\angle{\Theta ZK}$, is half that, or 2;10º.
Now let’s turn our attention to the eccentre. When looking the the ecliptic, we were concerned with $arc \; BG$, but on the eccentre, the corresponding arc is $arc \; ZH$ which we’d like to solve for. That arc subtends $\angle{Z \Theta H}$ which is less than $\angle{ZDH}$ by $\angle{\Theta ZK}$ which we just found. Since we previously stated that $\angle{BDG} = 65;30º$, we’ll subtract 2;10º from it to get $\angle{Z \Theta H} = 63;20º$. This is the angle before perigee of the mean motion or the angle after perigee and gives the position of the sun on the autumnal equinox.
That takes care of where the sun is along the eccentre on that date.
But the goal of this chapter was to determine the epoch, which is essentially a zero point for the model. Ptolemy refers to an observation he made on Sept 25, 132 CE of the autumnal equinox. He found it to occur about 2 equinoctial hours after noon. So based on what we just calculated above, at this time, the Sun was 116;40º past the apogee on the eccentre (i.e. its mean motion).
While I don’t see a particular reason that Ptolemy couldn’t use this point as the definition for $t = 0$ (i.e. the starting point for the epoch), it has no special significance that would help people remember it. In addition, in future books, we’ll be determining an epoch for the moon and all the planets, and it would be much better to use a common date for all.
So instead, Ptolemy uses a date he discussed a few chapters ago when we set up a table for the mean motion of the sun. He chooses the date as the beginning of I Nabonassar2. Ptolemy walks through various points in time since then but ends up determining that his zero point, the beginning of said era, was some 879 years, 66 days, and 2 hours in the past3.
So where was the mean sun at that time?
To figure that out, we turn to the table of the mean motion of the sun. I believe this is the first time we’ve actually made use of this table4, so I’ll walk through this slowly.
First, let’s tackle the years. The closest we can get on that table is 810 years, which has a mean motion of 163;4,12º. To that, we can add a period of 54 years (346;52,16º) and another period of 15 years (356;51,11º).
Adding all those together we get 866;47,39. However, we can subtract out the 2 complete revolutions of 360º each, so we’re left with 146;17,39º, for the years.
Now we can tackle the 2 months. Recall that the Egyptian months are always 30 days5, so we just need a period of 60 days which is 59;8,17º.
And 6 days is 5;54,49º.
Lastly we’ll add on 2 hours which is 0;4,55º
Combining all that, we get a total of 211;24,40º.
Let’s recap what that means briefly. This figure of 211;24,40º is the amount the sun has moved around the ecliptic since Ptolemy’s $t = 0$ date at the beginning, after dropping off all the full 360º loops its made.
Therefore, if to the 116;40º, which is the [sun’s] distance from the apogee of the eccentre at the above autumnal equinox, we add the 360º of on revolution and subtract from that result the 211;25º of the increment in mean motion over the interval [in question], we find for the epoch in mean motion in thee first year of Nabonassar, Thoth I in the Egyptian calendar, noon, that the sun’s distance in mean motion is 265;15º to the rear of apogee.
Let’s write that out in actual math so we can determine the distance from apogee:
$$116;40º – 211;24,40º + 360º = 265;15º$$
As a reminder, this is the position of the mean sun on the eccentre after the apogee. So as a final step, we need to add this to the position of apogee. As we stated above, perigee was at 245;30º which means apogee is 180º from that or at 65;30º ecliptic longitude. Thus, the mean sun’s ecliptic longitude6 at the beginning of the epoch was
$$65;30º + 265;15º = 330;45º$$
which is 0;45º into Pisces.
In the next chapter, we’ll look at an example of how to use this starting point to be able to calculate forwards from the start of the epoch and calculate the position of the sun.
- A few notes here. The first is the perpetual reminder that this is no longer the case due to precession of the equinoxes. Now the sun is in Virgo on this date.
Second, Ptolemy simply drops the note that, on the autumnal equinox, the sun was just entering Libra. For a moment, this confused me as, if the autumnal equinox was 6 months after the vernal equinox, then the sun would have moved 180º on the eccentre, but it would not have moved 180º in apparent position. Thus, it should not be just entering Libra (the beginning of which was 180º ecliptic longitude).
What I had momentarily forgotten and I want to remind readers is that the autumnal equinox is not exactly 6 months after the vernal; it’s a few days later. We touched on that some in this post in which we noted that the differing times between equinoxes and solstices.
- If you need a refresher on Egyptian calendars, refer to this post.
- The two hours comes from where Ptolemy stated above that he found the equinox to be 2 hours after noon on that date which, to us, seems like it would be 14 hours after the beginning of the day, but we reckon our days starting at midnight where as Ptolemy reckons the start of a day from noon, so his observation of the Autumnal equinox being 2 hours past noon puts it 2 hours into the day.
- Or if it hasn’t been, it’s been 6 months since I looked at it.
- And extra days get tacked on at the end of the year as necessary to make things work.
- In other words, what the ecliptic longitude would be if the center of the eccentre was coincident with the ecliptic and centered on earth.