Previously, we looked at how Ptolemy made corrections to the anomalistic motion for his lunar model. In this post, we’ll be doing something similar for the mean motion of lunar latitude.
Ptolemy explained that the value we originally noted was in error,
because we too adopted Hipparchus’ assumptions that [the diameter of] the moon goes approximately 650 times into its own orbit, and $2 \frac{1}{2}$ times into [the diameter of] the Earth’s shadow, when it is at mean distance in the syzygies.
In short, Hipparchus’ figures were a good starting point but now we can do better by
using more elegant methods which do not require any of the previous assumptions for the solution of the problem.
To use this new method, Ptolemy again turned to eclipses, looking for a pair, separated by as much time as possible, that fits four criteria:
- The amount of obscuration was equal
- The eclipses took place near the same node
- The eclipse was from the same side
- The moon was roughly the same distance from the Earth
To understand, let’s look at the two eclipses:
The first eclipse we used is the one observed in Babylon in the thirty-first year of Darius I, Tybi [V] 3/4 in the Egyptian calendar, [-490 Apr. 25/26] at the middle of the sixth hour [of night]. It is reported that at this eclipse the moon was obscured 2 digits1 from the south.
The second eclipse we used is the one observed in Alexandria in the ninth year of Hadrian, Pachon [IX] 17/18 in the Egyptian calendar [125 Apr. 5/6], $3\frac{3}{5}$ equinoctial hours before midnight. At this eclipse too, the moon was obscured $\frac{1}{6}$ of its diameter from the south.
The position of the moon in latitude was near the descending node at each eclipse.
Since the amount of obscuration was equal, this fulfills the first, second, and third criteria.
Thinking about these three criteria together is also rather important. Consider that if both eclipses have the same amount of obscuration and that obscuration is on the same side (top or bottom), this will mean that either the eclipses were in roughly the same place with relation to the same node2 or the same amount before one node as after the other. Thus, the third criteria, that they’re near the same node, implies it’s the first case and are thus, essentially at the same position and therefore latitude.
Now that we have those dates, we can go through the process of determining the lunar position from the epoch and tables of mean motion. But first, we’ll need to know the interval which Ptolemy gives as $256$ years, $122$ days, and $10\frac{1}{4}$ hours from the beginning of the epoch for the first, and $871$ years, $256$ days, and $8\frac{1}{12}$ hours for the second.
Although Ptolemy doesn’t show the math here, he takes these intervals, looks them up in the mean motion tables and determines the advance in mean motion and anomalistic motion.
Let’s walk through one of these as an example looking at the anomaly:
| º | ‘ | ” | ”’ | ”” | ””’ | ””” | |
| 252 years | 37 | 7 | 24 | 29 | 10 | 27 | 0 |
| 4 years | 354 | 52 | 29 | 54 | 44 | 55 | 40 |
| 4 months (120 days) | 127 | 47 | 52 | 35 | 43 | 58 | 0 |
| 2 days | 26 | 7 | 47 | 52 | 35 | 43 | 58 |
| 10 hours | 5 | 26 | 37 | 28 | 27 | 26 | 40 |
| 0.25 hours | 1 | 21 | 39 | 22 | 6 | 51 | 40 |
| TOTAL | 192 | 43 | 51 | 42 | 49 | 22 | 58 |
This gives the distance the moon advanced around the epicycle from the beginning of the epoch. To get the true position, we’ll need to recall that the moon was $268;49º$ after the apogee at epoch, so if we add on the advance and subtract out the full revolution, we determine that the moon is $101;33º$ past apogee during the first eclipse.
Going through the same exercise for the second eclipse, we would show that the sun advanced about $343;1º$ during this interval. Again adding that to the position at epoch and subtracting out the full revolution, it’s around $251;50º$ past apogee.
Let’s sketch that out, putting the two eclipses along the same line of sight:
Here, the first eclipse is on the left, and the second on the right. As you can see, the moon is quite close to the same distance3, being “a little closer to the perigee than the mean distance” despite being on different sides of the mean. This satisfies the fourth criteria.
So from this, Ptolemy asks what is that difference as compared to the mean position? In other words, what is the equation of anomaly for each?
To answer that, I’m quite tempted to just head towards a table for the equation of anomaly… but we don’t have one yet. This is because the incline of the lunar orbit will impact this, so we’ll need to finish getting this worked out before completing such a table. So instead, we’ll have to work through things geometrically as we did for the Babylonian and Alexandrian eclipse triples. Again, Ptolemy skips this math, but we can walk through one as an example. Fortunately, since the eclipse was on closer to perigee, there’s a simpler method to solve than what we did for the previous ones. Here’s a diagram to get us started:
Here, we have the apogee at $L$, the center of the epicycle at $K$, the eclipse at $A$, perigee at $M$, and the observer at the center of the deferent at $D$. I’ve dropped a perpendicular from $A$ onto $\overline{LD}$.
We can start by recalling that the first eclipse happened when $A$ was $101;33º$ from apogee, which means $\angle{LKA} = 101;13º$ as well. Thus, its supplement, $\angle{XKA} = 78;27º$, which we can use to determine $\angle{KAX} = 11;33º$.
Now we’ll jump into a demi-degrees circle around $circle \; KAX$. Now that $\angle{KAX}$ is on the circumference, we double its measure for demi-degrees so $\angle{KAX} = 23;06ºº$ which then has the same measure as the arc opposite it, $arc \; KX$. We can then look up the length of $\overline{KX}$ to be $24;2^p$.
And while we’re in the context of this circle, we might as well determine $\overline{XA}$ which we can do without the demi-degrees method since it’s a right triangle and we can use the Pythagorean theorem. Doing so determines $\overline{XA} = 117;34^p$.
Now we’ll swap contexts to the context of the deferent, where $\overline{KD} = 60^p$. We’ll need to convert the two sides of $\triangle{KAX}$ we just found. Which we can do because we know the length of $\overline{KA}$ in both contexts since it’s the radius of the epicycle which we’ve determined twice now. I’ll adopt the value of $5;14^p$ which I noted is what Ptolemy got from the Alexandrian eclipses.
$$\frac{\overline{KX}}{24;2^p} = \frac{5;14^p}{120^p}$$
$$\overline{KX} = 24;2^p \cdot \frac{5;14^p}{120^p} = 1;3^p$$
And doing the same for $\overline{XA}$:
$$\frac{\overline{XA}}{117;34^p} = \frac{5;14^p}{120^p}$$
$$\overline{XA} = 117;34^p \cdot \frac{5;14^p}{120^p} = 5;8^p$$
Next, we’ll be wanting to context switch into $circle \; DAX$ because $\overline{XA}$ will be the chord subtending the angle we’re after, $\angle{KDA}$. But before we can do that, we’ll need to determine the length of some line that’s common to both our current context and that circle. Since the only line in that circle we know is the diameter (which will be $\overline{DA}$, that’s the one we’ll shoot for.
Fortunately, that triangle will be a right triangle, so if we can determine $\overline{DX}$ we can use the Pythagorean theorem again. And we’re in luck because $\overline{DX} = \overline{KD} – \overline{KX}$. Writing that out:
$$\overline{DX} = 60^p -1;3^p = 58;57^p$$
Popping that into the Pythagorean theorem we determine $\overline{DA} = 59;10^p$.
Now we’re ready to context swap into $circle \; DAX$ in which the only chord we’ll care about is $\overline{XA}$:
$$\frac{\overline{XA}}{5;8^p} = \frac{120}{59;10^p} = 10;25^p$$
Using the chord table to convert this back to the corresponding arc we determine $arc \; XA = 9;56º$. Thus, $\angle{XDA} = 9;56ºº$ and the same angle in regular degrees is half the measure, so the equation of anomaly for the first eclipse is $4;58º$. Ptolemy must have done some slightly different rounding because he comes up with an even $5º$.
In this scenario, the moon was rearwards of the mean moon.
We can repeat this same method for the second eclipse in which case Ptolemy came up with an equation of anomaly of $4;53º$ in advance of the mean moon. If we add these two figures together, we get a total of $9;53º$.
So let’s consider what this actually represents. In this time period, we have demonstrated that the moon returned to the same latitude because it was the same distance from the same node. Thus, it completed an integer number of returns in latitude. However, it did not complete a full return of the mean motion. It fell short by $9;53º$.
Now, the amount of time between the first of these eclipses and the second Ptolemy gives as $615$ years, $133$ days and $21 \frac{5}{6}$ days. If we then use Hipparchus value the mean motion in latitude of $13;13,45,39,40,17,19^{\frac{º}{day}}$ over that period4, we would predict that the moon’s mean position should have changed by $10;2º$. In other words, Hipparchus’ prediction was off by $0;9º$ in that period. Impressively accurate, but now that Ptolemy knows of this error, he calculates a new correction factor to Hipparchus’ value by dividing that error among the interval to get the error in degrees per day of $0;0,0,0,8,39,18^{\frac{º}{day}}$ and then subtracts that from the mean motion given by Hipparchus to get a corrected value in the mean motion in latitude of $13;13,45,39,48,56,37^{\frac{º}{day}}$, which is what was used in the lunar table of mean motions.
I’ll go ahead and wrap up there. But before I close out completely, I’ll return to a statement Ptolemy made at the beginning of this chapter that I’ve saved to preserve the flow a bit better. Now that we’ve discussed these corrections Ptolemy made to the venerable Hipparchus’ work, it feels appropriate to include the quote on Ptolemy’s opinions on being open to correction. He welcomed it:
For those who approach this science in a true spirit of enquiry and love of truth ought to use any new methods they discover, which give more accurate results, to correct not merely the ancient theories, but their own too, if they need it. They should not think it disgraceful, when the goal they profess to pursue is so great and divine, even if their theories are corrected and made more accurate by others beside themselves.
- As explained in this post, a digit is $\frac{1}{12}$ of the lunar diameter.
- Either equidistant before or after.
- Along the dotted line which represents the line of sight.
- Which Ptolemy gives as $224609$ days.