In the last post, we explored how to make use of the table of mean syzygies to calculate the true syzygies. However, that chapter focused mostly on finding the time when the moon and sun would have either the same or exactly opposite ecliptic latitude. But what got left by the wayside was the lunar ecliptic latitude. We did a bit of work on calculating the argument of it but, aside from my mention of it in the afterword of the post, we never really completed that calculation. And eclipses of either type cannot truly occur unless the lunar ecliptic latitude is reasonably close to zero.
So we could calculate the ecliptic latitude of the moon for every conjunction and opposition but instead, Ptolemy decides we should first do a bit of a sanity check before getting any more involved. To do so, Ptolemy wants to examine how far from a node is it even possible for the ecliptic longitude of the syzygy to occur and still have an eclipse. If it’s outside of these limits, then no further calculation is necessary. To do this, Ptolemy is going to need to know some additional values. In this post, we’ll explore the angular diameter when the moon is at the perigee of its epicycle at syzygy1 as well as determining the width of Earth’s shadow at that distance.
Calculating the lunar diameter is something we have some experience with, having calculated it back in V $14$ where we found the angular diameter at the moon’s greatest distance (i.e., at apogee) to be $0;31,20º$. This was the greatest distance value because the eclipses we used to determine this were near the apogee of the epicycle. But what about at perigee?
Since we have a fully worked out lunar model in which we know the radius of the deferent, epicycle, and offset for the second model, my immediate thought would be to simply calculate this. Yet Ptolemy doesn’t. Likely because, as we’ve noted, doing so would return lunar diameters so obviously at odds with observations, that it could scarcely be taken seriously. Indeed, Ptolemy even quips,
it is safer to demonstrate this kind of parameter from the actual phenomena.
As such, he chooses another pair of eclipses where the moon should be near perigee and does a similar set of calculations. For the first eclipse
In the seventh year of Philometor, which is the $574^{th}$ year from Nobonassar, on Phamenoth [VII] $27/28$ in the Egyptian calendar [$173$ BCE May $0/1$], from the beginning of the eigth hour till the end of the tenth in Alexandria, there was an eclipse of the moon which reached a maximum obscuration of $7$ digits from the north. So mid-eclipse occurred $2 \frac{1}{2}$ seasonal hours after midnight, which corresponds to $2 \frac {1}{3}$ equinoctial hours, since the true position of the sun was $6 \frac{1}{4}º$ into Taurus. And the time from epoch to mid-eclipse is $573$ Egyptian years, $206$ days and $14$ equinoctial hours reckoned in meal solar days. At this moment, the position of the center of the moon was as follows:
- Mean longitude: $7;49º$ into Scorpio
- True longitude: $6;16º$ into Scorpio
- Distance [in anomaly] from the apogee of the epicycle: $163;40º$
- Distance from the northern limit on the inclined circle: $98;20º$
I’ll pause there so we can sketch this out.
As with before, the earth’s shadow I’ve drawn in as the gray circle, and the moon as the clear one. The moon’s position was given as $98;20º$ from the northern limit and since the descending node is $90º$ from the northern limit, this means that the moon’s center should have been $8;20º$ past the node as I’ve drawn. In addition, Ptolemy stated that the eclipse was “$7$ digits”. As a reminder, a digit is considered $\frac{1}{12}$ so the eclipse covered just past half of the lunar disc, from the north.
For the second eclipse, Ptolemy describes one
in the thirty-seventh year of the Third Kallippic Cycle, which is the $607^{th}$ year of Nabonassar, Tybi [V] 2/3 in the Egyptian calendar [$140$ BCE Jan $27/28$], at the beginning of the fifth hour [of night] in Rhodes, the moon began to be eclipsed; the maximum obscuration was $3$ digits from the south.
Here, then, the beginning of the eclipse was $2$ seasonal hours before midnight, which corresponds to $2 \frac{1}{3}$ equinoctial hours in Rhodes and in Alexandria, since the true position of the sun was $5;08º$ into Aquarius. And mid-eclipse, at which the greatest obscuration occurred, was about $1 \frac{5}{6}$ equinoctial hours before midnight. The time from epoch to mid-eclipse is $606$ Egyptian years, $121$ days and $10 \frac{1}{6}$ equinoctial hours. At this moment, the position of the center of the moon was as follows:
- Mean longitude: $5;16º$ into Leo
- True longitude: $5;08º$ into Leo
- Distance [in anomaly] from the apogee of the epicycle: $178;46º$
- Distance from the northern limit on the inclined circle: $280;36º$
Again sketching this out.
Here, I’ve sketched it so that the shadow is obscuring about $\frac{1}{4}$ of the lunar diameter, on the south side. We also know that the center of the moon is $10;36º$ past the ascending node which is, by definition, $270º$ from the northern limit.
Next, we need to determine how far the center of the moon is in each situation, from the respective node in ecliptic longitude since we currently have the distance along the lunar path and not the ecliptic. And as with the previous time Ptolemy did this, he doesn’t explain how he does this but comes up with a value of $0;43,03º$ for the first eclipse and $0;54,50º$ for the second. The difference in these two ecliptic longitudes is $0;11,47º$ which results in a change of obscuration of $\frac{1}{3}$ of the lunar diameter. Thus, the lunar diameter is $0;35,21º$ when at perigee during syzygy or a radius of $0;17;40º$.
Lastly, using this information we can also determine the width of earth’s shadow at that distance. Since the distance at perigee is closer than at apogee and this is closer to the base of earth’s shadow cone, we should expect this to be larger than the previous situation.
This time, Ptolemy looks at the second eclipse in which the moon was obscured by $3$ digits or $\frac{1}{4}$ of its diameter. In that eclipse, he calculates the angular distance between the moon and the center of earth’s shadow to be $0;56,50º$. Again, the amount of the moon obscured was $\frac{1}{4}$ of its diameter which, based on the number we just arrived at for the diameter is $0;08,50º$. Similarly, the angular distance from the edge of the shadow to the moon’s center is also $\frac{1}{4}$. Thus, we can determine the radius of earth’s shadow by subtracting that distance from the total distance between the centers, which gives us that the earth’s shadow at the distance of perigee has a radius of $0;46,00º$ or roughly $2 \frac{3}{5}$ the diameter of the moon itself.
This gives us a nice stopping point for this post as it directly mirrors the previous post in which we did these sorts of calculations. In the next post for this chapter, we’ll look at how the sun fits in to determine the limits on eclipses I described at the beginning of this post.
But before closing out this post, I want to again consider some criticisms. The first is obviously the suspect calculation we’ve already addressed in which Ptolemy determines the ecliptic longitude of the moon from the node.
A second one, that I did not discuss in the previous post, is the estimation of how much of the moon was obscured. Ptolemy offers no assurance how how this was determined, mentioning no instruments. Thus, it was likely a simple visual estimation. And claiming to be able to estimate to $\frac{1}{12}$ of the lunar diameter is rather unreasonable without instrumentation. To illustrate, let’s round the lunar diameter to $\frac{1}{2}º$ in which case $\frac{1}{12}$ of that is $0;02,30º$ or two and a half minutes of arc. This is only slightly higher than the human eye’s limiting resolution of about $0;01º$. Theoretically, we could see it, but to claim to measure with accuracy is suspect. Consider the famous double star in the handle of the big dipper: Mizar and Alcor. These two stars are $0;11,08º$ apart and many people have trouble distinguishing them, let alone trying to estimate that distance.
So what impact would even a $\frac{1}{12}$ difference make in this estimation? Well, if we considered the first eclipse here to be $8$ digits instead of $7$ and the second to be $2$ instead of $3$, then we’d come up with a lunar diameter of only $0;23,34º$. Conversely, if we changed the first to $6$ and the second to $4$ we’d get a lunar diameter of $1;10,42º$. Minor misestimates result in huge swings of the final value. My suspicion is that Ptolemy arrives at a value reasonably close to the modern one due to repeating this procedure for many different sets of eclipses and selecting an example that represents the average best.
Lastly, unlike the previous calculation for the lunar diameter, this one hid another potential source of error. Namely, one eclipse was from Alexandria and the other from Rhodes. While we didn’t show the step, Ptolemy had to calculate the position of the moon at the same time, so there was a step of converting the time in Rhodes to Alexandrian time and the difference in longitude was not well known.
So again, we see that this discussion of the moon and eclipses is ultimately not quantitatively satisfying. Rather, I echo Neugebauer who states,
What is really admirable in ancient astronomy is its theoretical structure, erected in spite of the enormous difficulties that beset the attempts to obtain reliable empirical data. Without the cinematic theories of the Almagest, it would have been impossible to introduce, on the basis of better observational techniques, those improvements which found their explanation in Newton’s celestial mechanics.
