Having established that two solar eclipses separated by the five months from the same location are just barely possible, Ptolemy then works on whether it will be possible for the same to occur over a period of seven months concluding that it is possible, provided it happen in the “shortest $7$-month interval”1.
When we looked at lunar eclipses separated by seven months, one of the things we determined is that the moon’s motion along its circle would be $208;47º$.
As an initial sniff test, we can compare this to the eclipse limits for the solar eclipses we found in the last post.
In doing so, we see that, unless parallax plays a role, eclipses would not be able to happen seven months apart because the distance from $A$ to $D$ (as well as $C$ to $B$) are both less than the lunar motion over that interval. Thus, this motion should overshoot the eclipse window by $16;23º$ along the moon’s inclined circle.
Again, as we’ve seen before, we can approximate the arc between the moon’s circle and the ecliptic by dividing the overshoot arc by $11;30º$. Doing so give an arc of $1;25º$ which is how much the moon and sun would miss each other by2 that far from the node and how much the parallax will need to make up for in order for an eclipse to occur.
To determine what terrestrial latitudes at which that may be possible, we’ll again need two pieces of information for each of the eclipses: where they are along the ecliptic and the variance in timing. These will allow us to determine the zenith distances and necessary angles between the ecliptic and altitude circle3 for each.
Ptolemy first determines the ecliptic positions, not showing the work. However, we can quickly determine this as we did in the last post where we previously investigated the solar motion. If we take the mean motion and subtract the difference due to anomaly, we find that the sun’s true motion was
$$203;45º – 4;42º = 199;03º$$
over this time period. We’ll again divide that into two equal segments of $99;32º$ each, and split them on either side of solar apogee which occurs at $65;30º$ ecliptic longitude. Thus, the eclipses would happen at $325;58º$ and $165;02º$ ecliptic longitude or $25;58º$ through Aquarius and $15;02º$ through Virgo, respectively. Ptolemy declines to give this level of precision and instead refers to this as being from
the end of Aquarius to the middle of Virgo.
Ptolemy then addresses the length of time between the eclipses, noting that we previously determined that, when mean conjunction happens, the true moon has already overshot the true sun $14;40º$. Again, we can determine how much earlier true conjunction happened by dividing this amount by $12$ which tells us it happened $1;16,31$ days or $\approx 1$ day and $5$ hours earlier.
If we again turn to our Table of Oppositions, we can see that the mean conjunction over seven months happens in $206;42,51$ days. Thus, the true conjunction happened $205;26,20$ days after the first. Ptolemy evidently rounds this off to $205;30$ days, or $205$ days and $12$ hours.
So we have to search for a place and time at which the moon’s parallax can exceed $1;25º$, either at one of [the eclipses] or at both [eclipses] combined, when the two situations are separated by $12$ hours, i.e. one sign is setting and the other rising (for otherwise it will be impossible for both eclipses to be visible at the same time.
Here, we’re actually a bit more constrained by the scenario than we were previously. In the last post, the eclipses were only separated by $6$ hours different than a full day which you can distribute nearly anywhere in the sky. Here, we have the constraint that they’re separated by $12$ hours which means they must be on opposite horizons since anything else would require one being below the horizon.
As with the last post, we need to consider the location of the moon, north or south of the ecliptic. If the moon is south of the ecliptic, Ptolemy immediately dismissed this possibility, stating that
it is impossible for the moon to achieve a northward parallax of the required amount for any region in our part of the inhabited world, since, even for those living directly below the equator, the [northward] parallax in latitude at the [moon’s] mean distance never exceeds $0;23º$. Hence, it is impossible for the sun to be eclipsed twice in the shortest $7$-month interval when the moon’s position is to the south of the ecliptic, i.e., when it is approaching the ascending node at the first conjunction and receding from the descending node at the last conjunction.
Basically, Ptolemy is stating that if the first eclipse occurs near $A$ and a second conjunction near $D$, both of which put the sun south of the ecliptic, then it simply won’t be possible for a second eclipse to occur since the portion of the ecliptic where this would occur is not far enough to the north of the zenith to create $1;25º$ of parallax. Again, this is hard to check since we don’t have the necessary tables for the zenith distances as the terrestrial equator.
Ptolemy then moves on to the situation in which the moon is north of the ecliptic.
[W]e find that a southward parallax of [the necessary amount] is achieved [for regions north of a latitude which is] approximately the parallel through Rhodes, when the end of Aquarius is rising beneath the same parallel, at both of the above situations the parallax of the moon at the mean distance (with the solar parallax subtracted) is about $0;46º$ southwards. Thus, already in these regions the sum of the parallaxes at both conjunctions is greater than $1;25º$. And, since for regions yet farther north than this parallel, the southward parallax is greater, it is obvious that for the inhabitants of those regions, an eclipse of the sun can be observed twice in the shortest $7$-month interval.
Here, Ptolemy is stating that it is possible for two eclipses to be visible, seven months apart, but only around the latitude of Rhodes ($36;00º$ N latitude) or north thereof. Here, finally, is something we can check as we have all the necessary tables. So let’s do that as a bit of practice.
As a reminder, for the full breakdown of how to calculate lunar parallax, we should refer to this post, but we’ll quickly see there’s several things that are going to let us do some simplification to the process. So, let’s get started.
First off, we’ll need to decide which point is on which horizon. In short, which is rising and which is setting. Ptolemy tells us that he considered the case in which
the end of Aquarius is rising and the middle of Virgo is setting.
With that in hand, the first step (following along with our previous example) is to enter the time until (or since) the point in question has crossed the meridian into the Table of Zenith Distances and Ecliptic Angles. However, the point of doing that was to find the arc (between the zenith and the point in question). However, in this case, we already know what that’s supposed to be: If the points in question are on the horizon then, by definition, they’re $90º$. However, what we do need is the East angle for the one rising (Aquarius) and the West angle for the one setting (Virgo). Interpolating between the tables4, I find the East angle is $122;59º$for the eclipse in Aquarius and the West angle is $58;52º$ for the eclipse in Virgo.
As such, we can jump straight to the next step in which we’d enter this into the parallax table and record the values from column $2$ for the sun and columns $3$ and $4$5.
Doing so, we find the following:
| Arc | 2 | 3 | 4 |
| $90º$ | $0;02,51$ | $0;53,34$ | $0;10,17$ |
Next is the part where we would find the adjustments based on the anomaly and elongation. Fortunately, because we’re talking about solar eclipses, the elongation is, by definition, $0º$ and thus, there is no need to concern ourselves with column $9$ as it will be $0º$ as well.
However, let’s examine what we know about the anomaly. In the same way as we attempted to minimize the solar motion, we’re also wanting to minimize the lunar motion. Which means we need to split its motion about the apogee, which we have not yet done. As we discussed in the post on lunar eclipses separated by seven months, the moon moves $180;43º$ about its epicycle in that period. We’ll split that into two intervals of $90;22º$ on either side of the epicycle.
Since we’re wanting to minimize the motion, the first eclipse needs to be before the apogee and the second after6. This means that at the first eclipse in Aquarius, the moon is $269;38º$ about the epicycle from its apogee or $89;38º$ past perigee7. This agrees well with Ptolemy’s statement that the moon is at its mean distance (i.e., half way between apogee and perigee). In fact, it’s close enough I don’t think it’s necessary to work through everything to precisely determine the adjustment, and instead, we’ll just average columns $3$ and $4$ to get a total parallax of $0;58,44º$. We’ll also want to subtract out the sun’s parallax8 for a net parallax of $0;55,53º$.
Let’s take a quick look at what that looks like in Stellarium:
Now we need to break that into its components so I’ll draw a more schematic diagram so we can see what’s going on with them.
As always, the angle in question is the upper left angle of the ones at $L_T$, which is the supplement of the angle in our parallax triangle. Thus, the angle in the triangle at $L_T = 57;01º$. Breaking the net parallax into components using a bit of trig, I find that the latitudinal parallax, $p_\beta = 0;46,53º$.
As a reminder, we’re trying to get to a total parallax of $1;25º$ or more between the two eclipses. So this is a good start as this value is just more than half of what we need. So let’s see if the eclipse in Virgo can make up the other half.
Again, we turn to our Table of Zenith Distances and Ecliptic Angles to find the West angle to be $58;52º$9.
Since the moon and sun will be the same distance from the zenith, we can reuse the net parallax from above $0;55,53º$. Now let’s take a look at what our parallax triangle looks like, first in Stellarium:
Now, let’s convert that into a more schematic diagram so we can see what’s going on with the parallax since it would be under the horizon:
In this case, our west angle is a vertical angle with the one in our parallax triangle, so no need to take a supplement. Using it, we can determine that the latitudinal parallax, $p_\beta = 0;47,50º$.
Ptolemy states that the parallaxes are each “about $0;46º$ southwards” so we’re definitely in the right ballpark. The sum of these two parallaxes is then $1;34,43º$ which is about $10$ arcmin more than we needed, thus confirming Ptolemy’s statement that there is sufficient parallax at this location to allow for an eclipse to happen on this interval.
Next up, we’ll determine whether it’s possible for two solar eclipses to happen one month apart!
- This was the same assumption we had for the seven month lunar eclipse post in which we assumed the moon was near perigee.
- In ecliptic longitude.
- Assuming, of course, that we have the necessary tables.
- Since each is part way through their respective constellation and the tables are for the first point in each.
- Why just these two columns? Recall that columns $3 – 6$ correspond to the four positions we found in this post. The positions starting with $1$ were when the mean moon was at the distance from the first model which happens at syzygy, whereas the positions starting with $2$ are when the moon is pulled closer due to the second model and thus are for quadrature. Since we’re talking about eclipses, specifically solar eclipses which happen at conjunction, we need to be concerned with the first case and thus, columns $3$ and $4$.
- Recalling that the epicycle for the moon rotates clockwise which would mean this would produce a negative net anomaly over this period.
- Remember in my new year’s recap when I said I was stuck on something? This was that something. I wasn’t taking into account the position of the moon about the epicycle and was therefore adding columns $3$ and $4$ like we have done so many times before without thinking about why!
- Another thing I was forgetting to do when I was stuck.
- Again interpolated between the tables for Virgo and Libra since the point in question was part way though Virgo.