Now that we’ve laid out how the two hypotheses work and explored how they sometimes function similarly, it’s time to use one of them for the sun. But which one? Or do we need both?
For the sun, Ptolemy states that the sun has
a single anomaly, of such a kind that the time taken from least speed to mean shall always be greater than the time from mean speed to greatest.
In other words, the sun fits both models, but only requires one. Ptolemy chooses the eccentric model due to its simplicity.
But now it’s time to take the hypothesis from a simple toy, in which we’ve just shown the basic properties, and to start attaching hard numbers to it to make it a predictive model. To that end, Ptolemy states,
Our first task is to find the ratio of the eccentricity of the sun’s circle, that is, the ratio of which the distance between the center of the eccentre and the center of the ecliptic (located at the observer) bears to the radius of the eccentre.
Fortunately for Ptolemy, this was something worked out well before his time by the Greek astronomer Hipparchus nearly 400 years earlier, having found that the radius of the eccentre is approximately 24 times greater than the separation of these centers. But for good measure, Ptolemy walks us through the proof anyway.
To start he begins with the observation that the “interval from spring equinox to summer solstice is 94.5 days, and that the time from summer solstice to autumnal equinox is 92.5 days.” Ptolemy affirms these observations of Hipparchus having observed this interval himself in the years 139 and 140 CE.
Now to set up the diagram. First we’ll set up a circle, ABGD which is the ecliptic, centered on the observer at E. We’ll go ahead and draw in some diameters, $\overline{AG}$ and $\overline{BD}$ at right angles to one another and let point A be the spring equinox, B the summer solstice, G the autumnal equinox, and D the winter solstice.
Next, we’ll need to place the circle of the eccentric (the eccentre). As a reminder, this is the circle that the sun actually travels on in this model, and its center will be offset from the observer. But which quadrant to put it in? Because of the lengths of the intervals, the only place it works, is in the upper left quadrant,
located between lines EA and EB. For semi-circle ABG comprises more than half of the length of the year, and hence cuts off more than a semi-circle of the eccentre; and quadrant AB too comprises a longer time and cuts off a greater arc of the eccentre than quadrant BG.
Let’s draw in the eccentre and that might make a bit more sense.
Here, I’ve added the eccentre, circle NPOS on center Z1. We’ve also put in diameters $\overline{PS}$ and $\overline{NO}$, which are perpendicular to one another and parallel to the similar diameters in the ecliptic. Lastly, I’ve labeled all points where these two circles intersect.
To understand why this location is chosen, consider the sun starting at Θ, which is in the direction of the vernal equinox. The amount of time it takes to travel from that apparent position to K (when its apparent position is at the summer solstice), is greater than the time from K to L (apparent position at autumnal equinox). What’s not mentioned thus far in the chapter but is still important to this placement, is that the autumnal equinox to winter solstice (L to M) is a little over 88 days, while winter solstice to vernal equinox (M to Θ) is just over 90 days. So the appropriate placement is the one that gets the ratios of these arcs in the correct proportion to the year.
Before diving into the math, we’ll add a few more lines to help us with things.
Specifically, we’ve added in $\overline{HE}$, $\overline{QK}$, and $\overline{Y \Theta}$ as well as a few more points of intersection. What we’re really after here is $\overline{EZ}$ as this is the distance between the two centers. So let’s get started.
Above we stated that the time it takes to get from point Θ to K was 94.5 days. This means $\frac{94 \frac{1}{2}}{365 \frac{1}{4}} \cdot 360º = 93;9º = arc \; \Theta K$2. Similarly, the time from K to L is 92.5 days which gives us 91;11º for $arc \; KL$. Adding those two together to get $arc \; \Theta L = 184;20º$.
However, we can see that from point N to O is 180º, so we can subtract that out to get $arc \; \Theta N + arc \; OL = 184;20º – 180º = 4;20º$. However, each of these arcs is the same length which means each one is half that value or $2;10º$.
Now let’s look at $arc \; YN \Theta$. It’s twice $arc \; \Theta N$ so we’re right back to $4;20º$. Taking the Crd arc of that3 we get $Crd \; arc \; \Theta Y = \overline{\Theta Y} = 4;32$. And we can divide that in half to get $\overline{T \Theta} = 2;16$4.
This also gives us a few other lines which are all bounded by the same parallel lines, $\overline{AG}$ and $\overline{NO}$, specifically $\overline{RZ}$ and $\overline{EX}$ which are both equal to $2;16$.
Next up, recall that $arc \; \Theta PK = 93;9º$. We’ll subtract from that $arc \; NP = 90º$ and $arc \; \Theta N$ which we just showed to be $2;10º$ leaving us with $arc \; PK = 0;59º$. This is half of $arc \; KPQ$. So $arc \; KPQ = 1;58$. Taking its Crd arc, we get $Crd \; arc \; KPQ = \overline{KQ} = 2;4$.
Again, this can be divided in half to get $\overline{KF} = \overline{XZ} = \overline{ER} = 1;2$.
Now we’ve gotten both $\overline{XZ}$ and $overline{EX}$ which means we can use the Pythagorean theorem to state $\overline{EZ} = \sqrt{(1;2)^2 + (2;16)^2} = 2;29:30$. Again, the radius of the eccentre is 60 units, so $\overline{EZ}$ is approximately $\frac{1}{24}$ of the radius of the eccentre, in agreement with Hipparchus.
Next, Ptolemy wants to figure out where apogee (H) is in relation to the equinoxes and solstices. Unfortunately, Ptolemy’s method uses the unusual “demi-degrees” that, so far, we’ve only encountered in this post. I did my best to explain there, but since we’ve used them so infrequently I’ll explain again here.
The first thing that Ptolemy does is add yet another circle:
Here, I’ve drawn in the circle going through points Z, X, E, and R. If, as usual, we define the diameter of the circle as 120 parts, we can redefine the length of $\overline{XZ}$ in terms of this new circle.
Previously the length of $\overline{EZ}$ was 2;29:30 and the length of $\overline{XZ} = 1;2$. Thus, the ratio of $\frac{\overline{XZ}}{\overline{EZ}} = \frac{1;2}{2;29:30}$.
However, we’re going to scale that up so $\overline{EZ} = 120$. Thus to determine the length of $\overline{XZ}$ we set up the equation:
$$\frac{\overline{XZ}}{120} = \frac{1;2}{2;29:30}$$
Solving:
$$\overline{XZ} = \frac{1;2}{2;29:30} \cdot 120 = 49;46$$
This is a chord, so looking up the corresponding arc, we get $arc \; XZ \approx 49º$.
Here’s where I’m going to add in one more line that Ptolemy doesn’t to help make sense of the next step5:
Now that we’ve determined $arc \; XZ$ we can relate this to the central angle of this new circle to state that $\angle{XIZ} \approx 49º$.
Next, take a look at $\triangle ZEX$. This is a triangle inscribed in a circle with it’s hypotenuse as the diameter. In the previous post where we saw demi-degrees, we learned that the angle on the circumference is half of the angle at the center for the same base. In this case, the base is $\overline{XZ}$ and the central angle is $\angle{XIZ}$. This means that $\angle{ZEX}$ is half of the central angle or 24;30º.
Ptolemy skips these intermediate steps and uses the demi-degrees notation stating $\angle{ZEX} = 49ºº = 24;30º$. Again, the demi-degrees (ºº) symbol just means he’s invoking this method.
So why were we after this angle? It’s because it’s vertex, E, is the center of the ecliptic and it tells us $arc \; BH$ is also $24;30º$, which is the angular distance between apogee (H) and the summer solstice. Similarly, that also means that $arc \; AH = 65;30º$ which is the angular distance between spring equinox and apogee6. Putting that in the context of the zodiac, since the spring equinox is at the beginning of Aries, and each constellation is 30º in width, this means apogee occurs at 5;30º into the third constellation, Gemini.
So far, so good. But we still have only dealt with the left side of the circle. We haven’t yet addressed the right side which is the autumnal equinox (G) to winter solstice (D) and from the winter solstice to the vernal equinox.
To address that, we’ll start by again noticing that $arc\; OS = arc \; SN = 90º$.
Above, we showed that $arc \; \Theta N = 2;10º$, but this is equal to $arc \; OL$ Similarly, $arc \; PK = arc \; MS = 0;59º$.
Doing a bit of subtraction, we can get $arc \; LM = 86;51º$ and $arc \; M \Theta = 88;49º$.
For $arc \;LM$, we’ll determine how long that period is by dividing that arc by the full 360º in the circle, and then taking that ratio as the proportion of the year.
$$\frac{86;51º}{360º} \cdot 365 \frac{1}{4} = 88;7 days$$
This is the number of days it should take from autumnal equinox to winter solstice and Ptolemy rounds it off to $88 \frac{1}{8}$ days.
Repeating the calculation for $arc \; M \Theta$ we get 90;7 days which Ptolemy again rounds to get $90 \frac{1}{8}$ days. Again, this is in accordance with observed durations.
So here we’ve established the basic parameters of the eccentric model for the behavior of the sun. In the next post, we’ll apply this model to investigate the differences between mean and anomalistic motion.
As a side note, I’ve put together some data on my progression in the Almagest and as a bit of a record, and motivation for myself, I’ll add it to the end of each post on the subject. Here’s where things stand as of the end of this post.
- As is done in the translation I’m using, I’ve drawn it smaller than the ecliptic whereas previously we’ve seen it drawn the sameish size. Ultimately it doesn’t matter as the angles are all preserved.
- It’s been awhile since I’ve used sexagesimal, so if you need a bit of a refresher, refer to this post.
- We haven’t seen a Crd arc in awhile either! Refer to Book 1 if you need a refresher. Specifically the Ptolemy’s Table of Chords.
- Things are a bit funny here because Ptolemy always defines circles with a radius of 60 units (and a diameter of 120). But we have two circles which are entirely different sizes. Thus, it’s important to distinguish that this is being done in the context of the smaller circle, the eccentre since that’s what we’re looking at the arcs on.
- Holy hell, what letter’s next… Uh…. we haven’t used Iota (I).
- File this away in the back of your head as we’ll see this pop-up again in a few chapters.