In the previous post, we showed that , if a solar or lunar eclipse occurs, it is possible that another may occur six months later. Now, we’ll turn to ask whether or not another lunar eclipse can happen five months after a previous one. To answer this question, we’ll first work out how much the moon would have moved in that time period and then compare that to the eclipse window..
Again turning to the Table of Syzygies we find that the motion of the luminaries in ecliptic longitude over five months is $145;32º$1. Considering the sun first, Ptolemy asks when its true motion is greatest. We can understand why if we look back at the diagram we constructed in the last post for lunar eclipses:
If we consider the eclipse happening at point $B$, we can clearly see that the sun would not make it to point $C$ in five months if its motion is only $145;32º$. However, that motion was for the mean syzygy and thus mean motion, so it’s possible we can extend the motion by maximizing the impact of the anomaly on the motion.
To do so, we need to consider when the sun is moving fastest, which is, of course, near perigee. Thus, we’ll split that $145;32º$ into two equal halves of $72;46º$ on either side of perigee so we’ll consider the anomaly at both of those points, so at $107;14º$ and $252;46º$ from apogee, using our Solar Table of Anomaly.
Doing so, I find that the sun had an anomaly of $2;19$2 at each of these points. But because the first value was found in the first column, it would lag the mean motion, and since the second value was in the second column it would be ahead of the mean motion for a total change in motion due to anomaly of $4;38º$.
If we add that onto the mean motion we get a total true motion of $150;10º$ which is a good sign since it now inches into the next eclipse window. But we can improve those odds even further since this motion only described the mean syzygy. Now we need to determine how much further the sun might move until true syzygy. To maximize the amount of time the sun will have to move from mean to true syzygy, we’ll want the moon to have (over this same interval) moved its least distance meaning it will have a larger gap to catch up to the sun.
In the time to mean syzygy, we’ve already determined the mean moon would have moved $145;32º$ – the same as the sun, less its complete revolutions. So, now let’s consider its anomalistic motion. Again from the Table of Syzygies, we can find its motion about its epicycle in that time period to be $129;05º$. But recall that we’re wanting to minimize its motion, meaning we should consider the moon near apogee instead of perigee.
Therefore, we’ll again divide the motion on the epicycle into two even halves, split around the apogee. Those halves would then be $64;32,30º$ to either side of the apogee or $64;32,30º$ and $295;27,30º$ from apogee. Using our Table of Lunar Anomaly, we find an anomaly of $4;20,06º$ on either side of the apogee, again with one negative and one positive. So when the eclipse around $B$ occurred, the true moon was $4;20,06º$ ahead of the mean moon, and then five lunar months later, it is $4;20,06º$ behind for a total change in lunar position compared to that of the mean motion of $8;40,12º$ which, as we see above, Ptolemy rounded to $8;40º$.
From this, we can get the true motion of the moon in ecliptic longitude over that period by subtracting this from the mean motion:
$$145;32º – 8;40º = 136;52º.$$
Here, we’ve subtracted because we’re arranging things such that the true motion was less than the mean.
We can now compare the true lunar and solar motions and see there’s a difference of $13;18º$ over this interval. In other words, the two aren’t nearly close enough to each other for an eclipse to occur. At least, not at that exact moment. But recall that the Table of Oppositions we used is for mean syzygy. As we’ve seen previously, we’ll need to do some work to determine the true syzygy and the motion of each luminary until then. To do so,
We take $\frac{1}{12}$ of [the distance between the true moon and sun], and get about $1;06º$ of additional motion of the sun before the moon overtakes it.
Here, Ptolemy is asking how far the true moon will have to travel to catch up to the true sun given that it initially has to span the distance of $13;18º$ since the moon’s true motion was less than the sun’s. This is much the same as what we did in this post except there we were only concerned with the distance between the mean and true conjunction. However, the same argument applies: The total extra distance can be determined by dividing the total distance to be spanned by $12$ as we showed in that post. So, the extra distance the moon will have to travel is:
$$\frac{13;18º}{12} = 1;06º.$$
I actually come up with $1;06,30º$ which should round up, but we’ll take Ptolemy’s value so we stay on the same track.
Returning to the question of how far the sun would travel beyond its mean motion, we can now add this extra motion for the catch-up time to the motion caused by anomaly:
$$4;38º + 1;06º = 5;44º.$$
So the sun has travelled $5;44º$ further than it would have based on mean motion alone due to the anomaly and extra time between the mean syzygy and true.
Now let’s turn back to the moon. If we look back at our Table of Mean Syzygies, we can see that in five months, the moon moved on its circle3 by $153;21º$. But, as stated, it must move an additional distance to catch up with the sun. Since the circles of the sun (ecliptic) and moon are only $5º$ inclined with respect to one another, Ptolemy uses the same value as we just calculated for the sun and concludes that the total distance the mean moon must have travelled to the true syzygy is
$$153;21º + 5;44º = 159;05º.$$
The question now becomes whether or not that’s in the eclipse window? If we use the diagram from above, it sure seems like it would be, but recall that that diagram was based on the eclipse window for when the moon was at perigee. In this case, the moon wasn’t at perigee. Rather, we said above that we were splitting the total motion about the epicycle into two even parts of $64;32,30º$ each to either side of apogee. Thus, during this syzygy, the moon would be $\approx 65º$ from apogee which puts a bit shy of its mean distance.
Jumping back a bit, in this post we showed that near perigee, the radius of the earth’s shadow plus the moon’s was $1;03,36º$. We never actually calculated this for the moon at apogee, but we can quickly do so now because I’ve provided the information necessary to do so in the small table towards the top of the post. The sum of these two for apogee is4:
$$0;15;40º + 0;40,40º = 0;56,20º.$$
Doing some really fuzzy math and estimating between these two extremes the sum of the radius of the earth’s shadow plus the radius of the moon at the distance in question would be $\approx 1º$ which happens $\approx 11;30º$ from the nodes on the moon’s circle, thus shrinking the eclipse window. So let’s redraw that diagram:
Here, we can see that the eclipse window has narrowed, being $11;30º$ to either side of the nodes or $23;00º$ in width in both cases, thus making the angle between the eclipse windows (which Ptolemy refers to as the “anecliptic arc”) being $157;00º$.
So let’s compare that to the motions we’ve found above of $159;05º$. Here, we can see that that $159;05º$ is just more than the arc where an eclipse could not occur. This means that it’s possible to have another lunar eclipse five months after the previous, but it can only happen if the first one was $\approx 2º$ before $B$ or $D$. And since it can never pass the opposite node, if it happens, then the eclipse will happen on the same side (top vs bottom) of the moon.
That’s enough for this post. In the next one, we’ll explore whether or not lunar eclipses can occur separated by seven months.
- We find this in the column for the sun in this table, but since this is a table of conjunctions, it is the same for the moon as noted in this footnote.
- Being a bit more careful, I get $2;18,30º$ but Ptolemy rounds, so I shall as well.
- Again, this is the “[Increment in the Argument of] Latitude” column.
- Ptolemy states a value here of $0;56,24º$. Toomer explains how in a footnote stating that he used the formula
$$0;15,40º + (2 \frac{3}{5} \cdot 0;15,40º)$$
wherein the $2 \frac{3}{5}$ came from the ratio of the radius of earth’s shadow to that of the radius of the moon.