We have finally reached the final in this run of eclipse timing feasibility checks. In it Ptolemy wants to demonstrate that it is impossible to have two eclipses separated by one month
even if one assumes a combination of conditions which could not in fact all hold true at the same time, but which may be lumped together in a vain attempt to provide a possibility of the event in question happening.
In short, we’re going to assume an overly ambitious “best case” scenario which can’t actually happen because some of these best case conditions contradict one another.
In particular, we’ll be assuming
that the moon is at least distance (to make its parallax large), that the month is the shortest possible (so that the amount by which the month’s motion in latitude exceeds the distance between the sun’s ecliptic limits be as small as possible); and that we use, without analysis [of whether it is a possible situation], those times and zodiacal signs in which the moon’s apparent parallax is greatest.
How do these contradict one another? Well, firstly, the moon at its least distance means that the moon is at perigee on the epicycle1. However, to have the shortest possible month it must be at symmetrical positions away from perigee to maximize the negative anomaly.
We’ll again begin by using our Table of Oppositions, from which we see that the sun and moon will have advanced $29;06º$ in ecliptic longitude and the moon will have moved $25;49º$ about the epicycle.
To produce the least amount of motion, we’ll need to take that $29;06º$ and distribute it evenly2 about the apogee which was at $56;30º$ ecliptic longitude which means our first eclipse3 will be at $41;57º$ ecliptic longitude of $11;57º$ into Taurus. The second potential eclipse would then occur at $71;03º$ ecliptic longitude or $11;03º$ into Gemini.
Turning towards our Table of the Sun’s Anomaly, we can determine the sun’s anomaly at $\pm 14;33º$ with regards to the apogee which is $0;34º$ in each situation which means the sun’s true motion will be $1;08º$ less than its mean over this interval4.
Meanwhile, we’ll also distribute the $25;49º$ of moon’s motion on its epicycle about its perigee or $12;55º$ to either side or at $167;05º$ and $192;55º$ from apogee on the epicycle. If we look this up in our Table of Lunar Anomalies, we find that it will result in an anomaly of $\pm 1;14º$ for a total change in anomaly of $2;28º$, this time, in addition to the mean motion.
Let’s take a moment to sketch things out as they stand now and see what things look like so far for the second eclipse:
Here, I’ve drawn the mean conjunction. The sun is $1;08º$ before this and the moon has already passed it by $2;28º$. Our next step will be to determine the position that true conjunction happened. We can immediately see that the moon is $3;36º$5 ahead of the sun but we can’t simply state the position of true conjunction as the mean position minus $1;08º$ since the sun will have moved in that interval of time. As such, we need to determine by how much.
As we’ve seen before, we can divide the total distance between the two luminaries ($3;36º$) by $12$ to find out how much the sun would have moved in that time. It’s an additional $0;18º$. I’ll draw that out again:
Now we can see the position of the true conjunction in relation to the mean. It’s $1;08º + 0;18º = 1;26º$ before the mean. So we’ll need to subtract that amount from the distance the sun/moon would travel between mean conjunctions in one month. So $29;06º – 1;26º = 27;40º$ is how much the two would have moved in ecliptic longitude over that interval.
Now let’s look at latitude. Again looking to our Table of Oppositions, the the increment in the argument of latitude is $30;40º$. Recall that the argument of latitude is essentially a measure of motion along the ecliptic as well, so we’ll need to subtract this same amount from that argument. Thus,
$$30;40º – 1;26º = 29;14º$$
which is our new argument of latitude. As with the previous discussion in this post regarding the sun’s motion with respect to apogee and the moon’s motion on its epicycle about perigee, we’ll take this motion and split it about the node, so $14;37º$ to either side of it. For example, if the first eclipse happened with the moon slightly before the descending node (i.e., with an argument of $75;23º$) then the second would be slightly after the descending node (at $104;37º$ for an argument). We can then look up these arguments in our Table of Lunar Anomaly. Doing so, I find it to be $\pm 1;16º$ for each, or a total change in latitude of $2;32º$6. Another way of putting this is that the moon would be $1;13º$ above the ecliptic for the first eclipse and $1;13º$ below it for the second7.
However, without parallax, the maximum distance the moon would be from the ecliptic is $0;33º$8 unless there is parallax involved. So, $0;33º$ above and below for a total distance of $1;06º$. Thus, parallax will need to account for the remaining $1;27º$.
Ptolemy states this can be done three ways:
[I]f the sun is to be eclipsed twice at an interval of $1$ month, it would be absolutely necessary either for the moon to have no parallax at one conjunction and more than $1;27º$ at the other, or, secondly, for the parallax at both conjunctions to be in the same direction and for the difference between the parallaxes to be greater than $1;27º$, or, [thirdly], for the parallax at one conjunction to be towards the north and the parallax at the other to be towards the south, while their sum exceeded that amount [$1;27º$].
Ptolemy immediately rules out the first scenario, stating,
[N]owhere on earth does the moon at syzygy, even at its least distance, have a latitudinal parallax of more than $1º$ (when the solar parallax is subtracted). Therefore it will not be possible for a solar eclipse to occur twice at the interval of the shortest month either when the moon has no parallax at one conjunction or when its parallax in the same direction at both conjunctions.
He then discusses the third option in which
the two parallaxes are in opposite directions, and that the sum of both exceed $1;27º$. This could happen for parts of the inhabited zones in different [parts of the earth], since it is possible for the southward parallax of the moon in regions north of the equator, in our part of the inhabited world, and the northward parallax in the regions south of the equator, among the so-called ‘antipodes’, to reach as much as $1º$ (with subtraction of the solar parallax).
In short, what Ptolemy is stating here is that it would be possible for two solar eclipses to occur, separated by one month, if we consider the case for observers at two locations: One in the northern hemisphere, and one in the southern. But this doesn’t really satisfy the criteria we’ve been using throughout this chapter in which we want to determine whether or not two eclipses separated by a certain amount of time would be visible to the same observer.
This only leaves the second option in which the parallax was in opposite directions (since the moon would be on opposite sides of the ecliptic). While the direction of parallax can change between the tropics9, it’s not by enough. Ptolemy states
for those situated directly beneath the equator, the maximum parallax of the moon, both to the north and to the south, does not exceed $0;25º$ and for those at the extreme north or extreme south the parallax in the opposite direction does not exceed the above-mentioned $1º$, so that, even in this case, the sum of the parallaxes is still less than $1;27º$. And since both opposite parallaxes become progressively much smaller in regions between the equator and the other extreme, the impossibility becomes even greater for such regions.
Here, Ptolemy first considers the parallax at the equator. There, he states, the moon can have a maximum parallax in either direction of $0;25º$ which isn’t enough. As we move further away, this can move the moon further from the zenith for one eclipse, but it would then be less for the other. Thus, it can’t solve the issue.
Therefore, it is impossible for the sun to be eclipsed twice in one month for the same observers anywhere on earth, or for different observers in the same part of the inhabited world.
Case closed.
And we’re finally out of Chapter $6$. In the next chapter, we’ll begin working on the Eclipse Tables.
- Note that Ptolemy is not trying to imply this is one of the situations in which the change we made for the second model has pulled the moon closer since that only happens at quadrature when an eclipse is not possible anyway.
- So $14;33º$ to either side.
- I should offer a reminder here that we are talking about mean conjunction at this point. Once we have this worked out, we’ll need to deal with how to determine the position of true conjunction.
- Recall that values in the second column, i.e. before the sun is at apogee, means the anomaly is additive and the position of the true moon is ahead of the mean moon by $0;34º$. The opposite is true for the second column. So for the second potential eclipse, the mean moon would be ahead of the apparent.
- The sum of the two anomalies.
- Ptolemy comes up with 2;33º$.
- Or vice versa if at the ascending node.
- This is the sum of the radius of the moon at perigee on its epicycle ($0;17,40º$) and the radius of the sun ($0;15,40º$).
- Technically, even a bit further beyond the tropics since the tropics are defined by where the ecliptic can cross the zenith. However, since the moon’s path is tilted by $5º$, the moon can cross even further. But since we’re considering eclipses, when the moon is quite near the ecliptic, this region is a decent approximation.