Almagest Book VI: Lunar Eclipses Separated by Seven Months

So far in this chapter, we’ve demonstrated that it’s quite possible that, if either a solar or lunar eclipse occurs, there will be another one six months later. Then, in the last post we showed that, if a lunar eclipse happens, it’s possible (although unlikely) that there can be another one five months later. In this post, we’ll explore whether or not, for lunar eclipses, you can have two separated by seven months.

The procedure will be the same as in the last post. First we’ll determine the motion of the luminaries, taking into account the discrepancies due to the anomalies, as well as the differences between mean and true syzygy, and then compare those to the eclipse windows which we’ll again recalculate for this situation.

To begin, we’ll again turn to our Table of Mean Syzygies, and find that the motion of the sun in ecliptic longitude over seven months is $203;45º$ which will also be true for the moon. In the same time, the moon will move $180;43º$ about its epicycle.

Let’s again consider the diagram we constructed for lunar eclipses in the first post for this chapter:

Here, we can see that if the sun was at $A$ and it moved $203;45º$, it would be between $C$ and $D$, making an eclipse tentatively possible. However, if we started the sun at B, then $203;45º$ later would put it well post point $D$ and outside any possible eclipse window. Thus, we’ll want to do the opposite of what we did in the last post and minimize the solar motion over this time period.

Thus, we’ll split the $203;45º$ of motion into two equal parts of $101;52,30º$ on either side of apogee or at $101;52,30º$ and $258;07,30º$. Popping those values into our Solar Anomaly Table, we find an anomaly of $2;21,02º$  in each direction for a total difference in anomaly of $4;42,04º$ which I’ll round to a nice even $4;42º$. Again, we’ve minimized the solar motion here, so this subtracts from the mean motion.

Similarly, we can determine the anomalistic motion of the moon. In this case, we’ll want to maximize the lunar motion, so we’ll want to choose the moon to be near perigee and again dividing its motion on the epicycle into two even pieces of $90;21,30º$ which would mean positions at $89;28,30º$ and $270;21,30º$

Looking that up in our Table of Lunar Anomaly and interpolating, I come up with an anomaly of $4;58,44º$ in either direction which leads to a total difference from the mean motion of $9;57,28º$ which Ptolemy rounds up to $9;58º$.  In this case, we’re maximizing the lunar motion, so this is the increment of motion in addition to the mean motion.

Taking these effects of anomaly together, the moon will be ahead of the mean position by $9;58º$ and the sun short of the mean position by $4;42º$ for a total separation between them of $14;40º$ when the mean conjunction happens.

From that, we can determine how much earlier the true conjunction would have happened by dividing that gap by $12$ as we’ve seen previously. Doing so, I find that the true conjunction should have happened $1;13,20º$ earlier. The sum of the sun’s anomaly and this distance, is the total distance before the mean conjunction that the true conjunction would have occurred. In other words $5;55º$ before the mean.

We will next apply that same amount of shortfall to the motion on the moon’s circle which we again get from the Table of Mean Syzygies where we see that the moon would have moved $214;41,39º$. Subtracting out the $5;55º$, we find that the moon would have moved $208;46,39º$ which Ptolemy rounds up to $208;47º$.

This again needs to get compared to the eclipse window and, as with the last post, we know that the moon won’t be at apogee during the potential window and will again be somewhere closer to its mean distance. As such, Ptolemy again uses the eclipse windows we discussed in the last post:

In this case, the distance from $A$ to $D$ is

$$23;00º + 157;00º + 23;00º = 203;00º$$

Thus, we can clearly see that the second true conjunction would not happen within the eclipse window as it would be $\approx 5 \frac{1}{2}º$ past it. Therefore, Ptolemy concludes that two lunar eclipses cannot happen separated by seven months.