Ptolemy next looks at whether or not it is possible for a solar eclipse to occur five months after a previous one. We’ve already done a fair bit of the heavy lifting for this topic as some of the math we did when considering lunar eclipses separated by five months will still apply. In that post, we determined that the moon would have moved on its inclined circle by $159;05º$ between true conjunctions. This does require we adopt the same assumptions of the sun moving its greatest distance and the moon moving its least.
What we’ll need to focus on for this post is redoing the eclipse limits for the situation in question.
To do so, we’ll again have to assume that the moon is near about its mean distance on the epicycle. This means its radius will be about half way between the two limits I listed in this post, or about $0;16,40º$. Meanwhile, we’ve stated that the solar radius is always assumed to be $0;15,40º$. Therefore, the sum of their radii1 is $0;32,20º$. From that, we can determine the distance from the node that separation would occur by multiplying by $11;30º$ as we discussed in this post. Therefore the maximum distance from the nodes that some sort of solar eclipse could occur would be
$$0;32,20º \cdot 11;30 = 6;12º.$$
Ptolemy then states,
it is clear that, if the moon has no parallax, the event in question [solar eclipses at a $5$-month interval] will be impossible, since the anecliptic arc exceeds the motion over the longest possible $5$-month interval by $8;31º$ counted along the [moon’s] inclined circle, which corresponds to about $0;45º$ on the [great circle] orthogonal to the ecliptic.
He doesn’t show the math here, but it’s quite easy to see how he arrives at this conclusion so we’ll walk through it. Since the distance from the nodes that an eclipse can occur is $6;12º$ that means the window that an eclipse cannot occur is going to be $167;36º$. So our eclipse window diagram would look like this:
Considering the motion of the moon, we’d previously shown its motion would be $159;05º$ over this period. Since this is less than the $167;36º$ by $8;31º$, it would not yet make it to the next eclipse window. To determine how far between the ecliptic and lunar circle it would miss by, we can instead divide by $11;30º$ which suggests we’d miss an eclipse by $0;44,26º$ in ecliptic latitude which Ptolemy evidently rounded up to $0;45º$.
However, at any place where the moon can attain a parallax so great that the parallax at either of the conjunctions at the two ends [of the interval], or the sum of the parallaxes at both conjunctions combined, exceeds $0;45º$, it is possible for the conjunctions at both ends to produce an eclipse at that place.
This statement is rather straightforward: If it’s possible for the total parallax to exceed $0;45º$, then it’s still possible for an eclipse to occur. So how to calculate where that’s possible?
Again referring to our last post on lunar eclipses separated over five months, one of the assumptions we had there was that the sun would be at perigee half way through the five month period. Jumping way back, we’d determined that the solar apogee is at $5;30º$ into Gemini ($65;30º$ ecliptic latitude) which means perigee would be $5;30º$ into Sagittarius which is $245;30º$ in ecliptic longitude.
In addition, we’d split the mean solar motion into two equal segments about that perigee of $72;46º$ plus an additional $2;19º$ of parallax in both directions for a total true motion of $\pm 75;05º$ from the perigee. This means that the first eclipse would happen at $170;25º$ ecliptic longitude ($20;25º$ into Virgo) and $320;35º$ ($20;35º$ into Aquarius).
This tells us where the sun and moon would be on the ecliptic, but we also need to determine where they might actually be in the sky since that’s what actually determines the parallax. To know that, we’ll need to know something about when the eclipses will occur relative to one another. Ptolemy gets us started on this by stating
It is clear, since the period of the mean $5$-month interval is about $147$ days, $15 \frac{3}{4}$ hours, that the period of the longest possible $5$-month interval will be $148$ days and $18$ hours.
This statement seems to come out of nowhere but is actually a consequence of something we stated in the previous post on lunar eclipses separated by five months. There, we discussed how much extra the sun would move, from mean syzygy, until the moon caught up with it and showed it to be $1;06º$. But, because the sun moves about $1º$ per day, this is roughly equal to the amount of time it will take between mean and true syzygy.
Thus, we can determine the length of time between the two syzygies by first finding the length of time between two mean syzygies separated by five months from the Table of Mean Syzygies, which is $147;39,11$ days, and then adding on an extra $1;06$ days, giving a total of $148;45$ days, or $148$ days and $18$ hours as Ptolemy states, after a bit of rounding.
This is the same thing as saying $6$ hours short of $149$ days. Stating that a different way, if we allow the first eclipse to happen at sunset, the second would happen $6$ hours earlier (i.e., near sunrise), $149$ days later, at noon.
So, we have to search for a place and time at which, if the moon is [$\approx 20º$ into] Virgo and also $6$ hours earlier [$\approx 20º$ into] Aquarius, its parallax exceeds the above-mentioned $0;45º$.
In short, we’ll need to consider, for the various latitudes Ptolemy considers the “inhabited world”, if it is possible for the parallax to ever reach $0;45º$. But not only that, we also need to consider if the parallax will be in the right direction. To that end, Ptolemy first considers the case in which the moon was south of the ecliptic. In that case, we could need the parallax to be northwards. As we’ve stated in previous posts, the effect of the parallax is always away from the zenith, This means that to have a northward effect of parallax, the zenith would have to be south of the ecliptic.
At this point, we’re going to run into a bit of an issue following along with Ptolemy because he stops showing any of his work. Quoting at a bit of length:
[W]e find that the moon’s northward parallax never reaches that amount (under the prescribed conditions) in any place in our part of the inhabited world. Hence it is impossible fort he sun to be eclipsed twice in the longest possible $5$-month interval when the moon’s position is to the south of the ecliptic, that is when it is receding from the descending node at the first conjunction and approaching the ascending node at the last.
For this situation, Ptolemy shows neither his work nor does he tell us what amount of parallax is possible for a northwards parallax. He then goes on, discussing the case in which the moon is north of the ecliptic requiring a southwards parallax:
However, [the moon] can achieve a southward parallax of this amount in all regions (beginning almost at the equator, and going northwards), if one takes the combined parallax at both the above signs with a $6$-hour difference. This occurs when $20º$ into Virgo is at the setting-point for the first conjunction, and $20º$ into Aquarius in the meridian at the second conjunction. For, in those situations, we find the following approximate southward parallaxes, for the moon at mean distance (subtracting the solar parallax):
|
|
|
|
|
|
|
|
|
Here, at least, we can see the final results of Ptolemy’s calculations. But there’s still a bit of an issue. The reason is that, to find the total parallax, we need to know the distance of these points on the ecliptic from the zenith. This comes from our Table of Zenith Distances (II.13). However, looking quickly at that table there’s an immediate issue: There is no table for the equator nor “where the longest day is $12 \frac{1}{2}$ hours2.”
Thus, we’re left without an easy way to determine the zenith distance aside from calculating new Zenith tables for these latitudes – an exercise I’m not keen on engaging in3. As such, I think I will attempt to walk through the calculation at the latitude of Meroe, where we do have the necessary tables, as a separate post at some point, but refrain from doing so in this once since it deviates distinctly from what’s being presented in the text.
So for now, we’ll have to take Ptolemy’s word for this.
But before finishing the post out, let’s make sure we fully understand what Ptolemy is stating here because there’s actually another piece of relevant information left out which is how Ptolemy divvied up the six hours short of $149$ days. The example I gave above, such that the first eclipse happens near sunset which the potential second is near noon is just one way we could do it. We could also have the first at noon and the presumptive second at sunrise. We could have the fist three hours after noon and the second three hours before… Ptolemy doesn’t state which distribution he used but Neugebauer indicates that
in order to obtain maximum parallax, we assume that the first conjunction occurs near sunset.
Knowing that this second eclipse is near the horizon means we would know the distance from the zenith to be $\approx 90º$ and then we could look up the total parallax in the Parallax Table, but we’d still need to determine the west angle4 to know how to break it into its components (since we’re after the latitudinal component only).
For the first eclipse, we would need the zenith table for these latitudes5, but wouldn’t need to worry about the angles since objects that are culminating have no longitudinal component.
In addition, Ptolemy would have done the same for the solar parallaxes and subtracted those out as we’ve done before.
In the small table above, we see the results of Ptolemy’s presumptive work, in which the sum of the parallaxes for the two eclipses at the terrestrial equator does not provide sufficient parallax for an eclipse to occur. However, by $8 \frac{1}{2}º$ N latitude, it just barely does and any latitude northwards of there would even more so.
Thus Ptolemy concludes that two eclipses, separated by five months is indeed possible, but only
while the moon’s position is to the north of the ecliptic, that is, when it is receding from the ascending node at the first eclipse and approaching the descending node at the second.
This requirement that it only happen following the ascending node and before the descending node is a direct consequence of the requirement that the moon be to the north of the ecliptic and it would be south if after the descending node and before the ascending.
In the next post, we’ll explore whether or not the possibility exists for two solar eclipses to happen separated by seven months.
- The distance at which an eclipse would just barely not happen.
- This would be at the Avalite Gulf which is $8;25º$ N latitude as we saw in our Rising Time of the Ecliptic Tables.
- Especially since it has been two and a half years since we visited the topic of deriving said tables.
- Which would have come from the Zenith table if we had one for these latitudes.
- Again, which Ptolemy never calculated.