In order to determine the relationship between the true distance to the sun and moon, one of the key pieces of information we’ll need is their apparent angular diameters at syzygy. I’ve mentioned several time that both are around half a degree. For this to work, we’ll need to be more accurate than that, but measuring small angles like this is especially tricky. Ptolemy mentions a few ways astronomers prior to him tried to tackle the issue which include
measuring [the flow of] water1 or by the time [the sun and moon] take to rise at the equinox.
However, he rejects these stating that they are not sufficiently accurate. Instead, he states he used a dioptra which is a surveying instrument. While he doesn’t get into the detail of its construction or use, he does give a summary of some of his key findings:
First, he states that the sun’s apparent diameter does not appear to change, but the moon’s apparent diameter does. According to Ptolemy, it has the same angular distance as the sun only when it is at its maximum distance2 which he states disagrees with his predecessors who claimed that the moon’s diameter matched that of the sun only at mean distance. Ptolemy’s position is easily refuted as the existence of annular eclipses by necessitates a smaller angular diameter of the moon than the sun.
The second conclusion Ptolemy gives is that the angular diameters he has determined are “considerably smaller than those traditionally accepted.” Although he doesn’t give any of those previous values, Toomer notes that Hipparchus had a value of “a six hundred and fiftieth of its circle” which is about $0;33,14º$. Ptolemy also doesn’t give away his figure just yet either, but does indicate that it didn’t actually come from use of the dioptra he just mentioned, but “on certain lunar eclipses.”
So in this post, we’ll explore that method, but as a warning, while Toomer praises this section as “elegant and theoretically correct” this section has many problems.
As Neugebauer states,
The chapters on apparent diameters, geocentric distances, and parallax teach us the very valuable lesson that it is futile to try to establish “the” source of an error in the determination of important parameters of the Ptolemaic theory. The same holds for all astronomers before Flamsteed, Cassini, Halley, etc. i.e. before the end of the $17^{th}$ century. In all ancient astronomy, direct measurements and theoretical considerations are so inextricably intertwined that every correction at any one point affects in the most complex fashion, countless other data, not to mention the ever present numerical inaccuracies and arbitrary roundings which repeatedly have the same order of magnitude as the effects under consideration.
In short, we’re dealing with some very small angles so the slightest errors in calculations or measurements as well as rounding3 will suddenly have large impacts. And, as we’ll see, there’s a big question mark surrounding some of the numbers in this chapter.
But moving on, this method turns to a pair of eclipses. I’ll give the details and then discuss where we’re going with them.
Ptolemy describes the first of these eclipses as beginning
in the fifth year of Nabopolassar, which is the $127^{th}$ year from Nabonassar, Athyr [III] $27/28$ in the Egyptian calendar [$-620$ Apr. $21/22$], at the end of the eleventh hour in Babylon, [which is] when the moon began to be eclipsed; the maximum obscuration was $\frac{1}{4}$ of the diameter from the south. Now, since the beginning of the eclipse occurred $5$ seasonal hours after midnight, and mid-eclipse about $6$ [seasonal hours after midnight], which correspond to $5 \frac{5}{6}$ equinoctial hours at Babylon on that date (for the true position of the sun was $27;3º$ into Aries), it is clear that mid-eclipse, which is when the greatest part of the diameter is immersed in shadow, occurred $5 \frac{5}{6}$ equinoctial hours after midnight in Babylon, and exactly $5$ [hours after midnight] in Alexandria.
Ptolemy then gives the time from epoch as $126$ Egyptian years, $86$ days and $16 \frac{3}{4}$ equinoctial hours4. Using that Ptolemy calculates5 a mean lunar position of $25;32º$ into Libra ($205;32º$ ecliptic longitude), a true lunar position of $27;05º$ into Libra ($207;05º$ ecliptic longitude), a distance around the epicycle from apogee of $340;07º$ and a distance [in latitude] from the northern limit of $80;40º$. If the moon is $80;40º$ from the northern limit, this implies it’s $9;20º$ before the descending node.
Meanwhile, the second eclipse occurred
in the seventh year of Kambyses, which is the $225^{th}$ year from Nabonassar, Phamenoth [VII] $17/18$ in the Egyptian calendar [$-522$] July $16/17$], $1$ equinoctial hour before midnight at Babylon, [when] the moon was eclipsed half its diameter from the north. Thus, this eclipse occurred about $1 \frac{5}{6}$ equinoctial hours before midnight at Alexandria6.
He then lists the time from epoch as $224$ Egyptian years, $196$ days and $9 \frac{5}{6}$ equinoctial hours, giving the position of the sun as $18;12º$ into Cancer. His calculated positions for the mean moon is then $20;22º$ into Capricorn and $18;14º$ into Capricorn for the true moon. The distance around the epicycle from apogee as $28;05º$7, and it was $262;12º$ from the northern limit, putting it $7;48º$ from the ascending node.
So why select these two eclipses?
To understand, let’s draw out each of these eclipses from the point of view of an observer on earth:
Here’s the first one. I’ve drawn the moon as the white circle and the earth’s shadow as the shaded one. It is drawn such that the earth’s shadow is covering $\frac{1}{4}$ of the moon as observed at mid eclipse. We determined above that the moon was $9;20º$ before the descending node8. We also know that the moon’s path is inclined to the ecliptic by $5º$.
Now let’s look at the same diagram for the other eclipse:
In this one, the moon has inched closer to a node, now being $7;48º$ away. In doing so, now half of the moon is covered.
In each of these images, I’ve also drawn
the great circle drawn through the moon’s center at right angles to the inclined circle (which is the situation at which the greatest obscuration occurs).
That’s what we need to solve for first.
You may be attempted to apply some simple trigonometry to this since we have a side and an angle9, but keep in mind, this is not a triangle on a plane; it is a spherical triangle so those methods would not work. Instead, the only tool we have available to us is Menelaus’ theorem which we haven’t seen in a good while.
Let’s zoom out and see how that looks from outside the celestial sphere for the first of these two eclipses:
In principle, this problem is very similar to I.14 in which we found the arc between the celestial equator and the ecliptic but instead of having the known arc being the hypotenuse, it’s now one of the angles adjacent to the right angle.
I’ve highlighted a potential Menelaus configuration10 In addition, I’ve added in some measures for other arcs that are effectively knowns: $arc \; AB = arc \; AF = 90;00º$ since they’re both from the pole of the lunar path to the path itself. Also, $arc \; BE = 90º$ since it’s from the northern limit to the node which allows us to subtract out $arc \; FE$ to determine $arc \; BF = 80;40º$. Lastly, we know $arc \; BC = 5;00º$ since that’s the angle between the two.
However, this Menelaus configuration doesn’t work11. No matter how I seem to create a Menelaus configuration, I always seem to need $arc \; DE$ or $arc \; CD$, neither of which are knowns, and I can’t seem to solve for them either.
Somehow, Ptolemy comes up with a value of $0;48;30º$ for the first eclipse, and $0;40,40º$ for the second. Unfortunately, he doesn’t show his work and despite my best efforts, I can’t figure out what he’s done. And it appears I’m in good company. Neugebauer (HAMA, 107) has a commentary on this that looks at a few possibilities. Here’s the table from him:
The first column is Ptolemy’s values. The second is the same values from a $4^{th}$ century commentator, Pappus of Alexandria who, according to Neugebauer, was able to use Menelaus’ theorem. Neugebauer doesn’t show Pappus’ work either and I’ve been unable to locate a source for this to see if Pappus did. But interestingly, I have been able to reproduce Pappus’ exact figures (which agrees with Ptolemy’s for the first eclipse) if I incorrectly use a great circle from the pole of the celestial equator instead of from the pole of the lunar path. This means that the arc in question would not be perpendicular to the lunar path as required. As such, I have a suspicion that Pappus may have performed the incorrect calculation and Neugebauer took the values from Pappus without thorough consideration12.
Moving on, the third column is if you were to take the chords of each of the arcs, essentially converting the spherical triangle to a planar one, and then doing a demi-degrees method. As you can see, this is further off.
The third column considers what would happen if Ptolemy used our lunar model to find the ecliptic latitude. This isn’t the same arc we’re looking for since it uses the ecliptic as the other bound instead of the celestial equator, but Neugebauer considers it nonetheless and shows it too doesn’t correspond to Ptolemy’s values.
Lastly, he considers a modern solution which, while ultimately giving the best agreement, still doesn’t agree with Ptolemy’s values. As such, how Ptolemy arrived at his number is an unsolved mystery. So unfortunately, we’re going to have to accept them and move on.
Getting back to the problem, Ptolemy considers the difference in these two arcs which is $0;07,50º$. And when the moon moves that distance, this was a difference of $\frac{1}{4}$ of its diameter. Thus, its full diameter must be four times this or $0;31,20º$.
Ptolemy then considers the apparent diameter of earth’s shadow which is easy to get if you consider the second eclipse drawing. In it, we can clearly see the earth’s shadow is tangent to the lunar path, so that arc is the radius of the earth’s shadow. He then takes the ratio of this to the just determined lunar radius13 to be $\approx 2 \frac{3}{5}$ as large.
Ptolemy states that he performed this same sort of calculation on several other eclipses to verify it and notes we’ll be making use of these as we turn to the solar distance and start predicting eclipses.
UPDATE: The title of this post has been changed. In part because the title ignored the fact that we also found the diameter of the Earth’s shadow, but I also added language stressing that this chapter was specifically looking at eclipses near apogee and at syzygy where the effect of the second anomaly is minimal. This is because we’ll be doing a very similar calculation in Book VI in which we find the lunar diameter near perigee and I want to ensure that the index is easily searchable.
- I.e., water clocks.
- I.e., at apogee on the epicycle.
- Which we know Ptolemy is particularly fond of.
- Toomer notes that this should be closer to $16 \frac{2}{3}$ equinoctial hours.
- But does not show the work. And at this point we’ve done two examples, so I won’t either and we’ll just accept Ptolemy’s values.
- Toomer notes that this time for the eclipse is incorrect and there is actually an extant cuneiform tablet that lists this eclipse as well but with notably different details. Particularly, the amount of the moon eclipsed was described to be near total.
- Toomer has yet another note here indicating that this was calculated incorrectly using the unadjusted hours (which I don’t tend to list) which was $10 \frac{1}{6}$ equinoctial hours.
- Recalling that the moon moves “rearwards” or right to left when viewed from inside the celestial sphere.
- Or, since modern trig hadn’t been invented in Ptolemy’s time, use a demi-degrees circle about this triangle.
- And here is another possible configuration.
- Nor does the other one I highlighted.
- To be fair, I can’t claim to check every calculation of Ptolemy’s. My goal is to make sure I understand the method, so on many of these routine calculations, I simply assume Ptolemy did it right. It’s only when we haven’t used a technique in awhile that I redo it to make sure I understand. Hence why I caught the discrepancy in this post.
- We actually determined the lunar radius, but converting is trivial.
