In the last post, we explored various lunar cycles from astronomers predating Ptolemy in which the moon reset its ecliptic longitude and anomalistic motion to define a full lunar period. These ancient astronomers did this by studying pairs of lunar eclipses1but Ptolemy notes that this method
is not simple or easy to carry out, but demands a great deal of extraordinary care
The reason for this difficulty is that, without careful consideration there can essentially be false positives of eclipses separated equally in time, but do not in fact, result in the moon returning to the same ecliptic longitude or same speed.
One of the reasons is that the conditions necessary to produce a lunar eclipse are also dependent on the sun, which has anomalistic motion. As such, it could be entirely possible that the moon could not have yet returned to the same ecliptic longitude as a previous eclipse, but the sun’s anomaly could cause an eclipse anyway. Thus, a pair of eclipses may be equally separated in time, but
this is no use to us unless the sun too exhibits no effect due to anomaly, or exhibits the same [anomaly] over both intervals: for if this is not the case, but instead, as I have said, the equation of anomaly has some effect, the sun will not have travelled equal distances over [the two] equal time intervals, nor, obviously, will the moon.
To illustrate this, Ptolemy starts with an example.
Ptolemy blows through it pretty quickly and without diagrams, but I’m going to try to slow it down. Although we haven’t discussed it specifically, we’re now up to two celestial spheres since we’ve added the moon. The lunar sphere is the one closest to Earth with the sun’s sphere outside that. So let’s start by drawing that, but remembering from Book III that Ptolemy models the sun’s sphere with an eccentric, so the Earth will be offset from the center of the solar sphere, but the lunar sphere will still be centered on the Earth:
Here, I’m following some notation from Neugebauer’s History of Ancient Mathematical Astronomy. A is the apogee and P the perigee of the solar sphere making this larger circle the solar sphere. M is the center of the solar sphere and O is the observer or Earth. The unlabelled circle is the lunar sphere.
Now, let’s draw in a pair of eclipses:
First, let’s consider an eclipse to happen when the sun is at E. Because the moon must be opposite the Earth, this means the moon would be at L. Then we let time pass until another eclipse occurs, this time with the sun at E’ and the moon at L’. Now, because the periods of the sun and moon aren’t the same, this wouldn’t happen in a simple 6 months, but some integer number of years plus six months for this example. Subtracting out the integer number of revolutions, we would have that the moon would then have travelled from L to L’ going counter-clockwise, upwards in this diagram.
We can be even more specific and put some numbers on this using our table of the sun’s anomaly as the positions here are chosen where E is 90º from A and E’ is 270º around. In both of these cases, the anomaly is 2;23º, but when it’s at E, it’s negative, and at E’, it’s positive. This means there’s a difference of 4;46º2. Thus, the angular distance the moon would have travelled along the ecliptic3 is 180º plus 4;46º for a total of 184;46º.
Next, consider a second pair of eclipses, separated by an equal amount of time as the one we just described, but this time, the first one when the sun is at E’ and the moon at L’, and the second one with the sun at E and the moon at L. The same logic applies, but this time the moon will have travelled from L’ to L, again going counter-clockwise, but over the shorter path which is 180º minus 4;46º for a total distance of 175;14º travelled.
So we see that just because the time between pairs of eclipses is equal it doesn’t mean the eclipse periods were in fact similar because the position of the sun can throw things off.
Thus, we must consider only cases in which the anomaly of the sun isn’t a factor. But how to do that? Ptolemy lists 4 instances in which the sun’s anomaly would not impact the position of the moon between cycles:
[1] [The sun] must complete an integer number of revolutions
[2] [The sun must] traverse the semi-circle beginning at the apogee over one interval and the semi-circle beginning at the perigee over the other
[3] [The sun must] begin from the same point [of the ecliptic] in each interval
[4] [The sun be at] the same distance from apogee (or perigee) at the first eclipse of one interval as it is at the second eclipse of the other interval, [but] on the other side
Ptolemy offers no explanation for these, but I’ll try my best. Basically, in each case, we’ll have two pairs of eclipses. As we demonstrated above, if the equation of anomaly is different between them, it impacts the position of the moon. So each of these represents a situation in which the differences in the equation of anomaly are the same for each pair of eclipses.
To assist, I’ll refer to a series of images from Neugebauer’s History of Ancient Mathematical Astronomy. Here’s his diagram for the first instance:
Before proceeding, let’s figure out what we’re looking at here. These images are essentially the same as the ones I drew above, but here, Neugebauer hasn’t drawn in the position of the moon and he’s using π instead of P for the perigee, but otherwise things are pretty much as I defined them above.
However, because we’re really wanting to compare pairs of eclipses, that’s why there’s two columns. Hence, we now have E1 and E2.
As demonstrated above, what we’re really needing to demonstrate is that for all four cases, both the time between pairs of eclipses is equal at the same time the change in position from the point of view of the observer is equal since the lunar sphere is centered on the Earth at O.
So how do those requirements translate into the drawing? The time component is pretty easy, it’s the change in ecliptic longitude for the mean motion (i.e. when viewed from M) which is represented by $\Delta \bar{\lambda}$. Since this is being treated as a given for the pairs of eclipses, you’ll notice Neugebauer didn’t need to add a subscript to this. They’re simply equal.
As far as the change in position being the same in each pair, that’s shown when $\Delta_1 \lambda = \Delta_2 \lambda$4. However, since it’s only the equation of the anomaly that would throw this off, demonstrating that the difference in the equation of anomaly for each pair is the same will be sufficient to prove the point.
So returning to this first instance, we don’t care what the starting position is for each pair of eclipses. It only matters that the second eclipse in each pair has the sun at the same ecliptic longitude. This means that the equation of anomaly between E and E’ will be the same. Thus, there difference will be zero.
Let’s take an example. Let’s let $E_1 = 330º$5 which is about what it looks like in the image. Using the table of the sun’s anomaly, we determine that the equation of anomaly for that position to be 1;9º.
However, since the sun has made an integer number of rotations, this means $E’_1 = 330º$ as well, and thus its equation of anomaly is also 1;9º. Subtracting the first from the second, we see the difference is zero. We can choose any other position along the ecliptic for $E_2 = E’_2$ but since the equations of anomaly for the pair will be the same, it will again equal zero. Thus, we have demonstrated that the effect of the equation of anomaly in both pairs of eclipses would be identically zero.
Here’s situation 2:
In this situation, the sun starts at apogee and ends at perigee or vice versa. This is a very simple case, because at apogee and perigee, the equations of anomaly are zero. Thus the equation of anomaly is the same in all instances: zero. As such, the differences between them will again, always be zero.
Here’s the third case:
The requirement here was that $E_1 = E_2$6. Since the amount of time that passes between the first and second eclipse is both pairs will be the same, this also implies that $E’_1 = E’_2$. This means that the difference in the equation of anomalies for each pair will be the same as well, although non-zero in this case.
Lastly, the fourth case:
This time was that the requirement was that the angular distance before (or after) the apogee for $E_1$ be the same distance after (or before) the apogee for $E’_2$. I think this is best illustrated with an example. So let’s let $E_1 = 24º$ before the ecliptic. This means it’s at 336º. Again using the table of the sun’s anomaly, that means its anomaly will be 0;56º7. Now, we’ll let the sun move to 102º after the apogee for $E’_1$ which is a total motion of 126º. From the table of the sun’s anomaly, we get that 102º after the ecliptic give an equation of anomaly of -2;21º8
Here, let’s calculate the difference taking the anomaly at $E’_1 – E_1 = -2;21º – 0;56º = -3;17º$.
Now, let’s work on the second pair of eclipses. Here, the position of $E’_2$ is 24º after the ecliptic, so its equation of anomaly is -0;56º. The position of $E_2$ was 126º before that which is 102º before apogee or 258º after. We again use the table of anomaly to determine that its equation of anomaly would then be +2;21º.
Again, we’ll take the difference of the anomaly at $E’_2 – E_2 = -0;56º – 2;21º = -3;17º$.
In both cases, the difference in equation of anomaly is the same. Therefore, again it will not have an impact.
As such, we have demonstrated that Ptolemy’s 4 cases for when the sun’s anomaly is negligible to be true.
For only under one of these conditions will there be no effect due to the anomaly, or the same effect over both intervals, so that the arc traversed beyond complete revolutions over one interval is equal to that traversed over the other, or even equal to the mean motion of the sun [over the intervals] as well.
It must have been quite challenging for ancient astronomers to dig through all the records of eclipses to find pairs fitting just these criteria!
In the next post, we’ll explore several more special cases like these, but this time, related to the moon’s varying speed.
- For reasons we established in chapter 1 of this book.
- Which Ptolemy rounds off to $4 \frac{3}{4}º$.
- Excluding the integer number of rotations.
- We’re using some modern notation here with Δλ, so if you’ve forgotten what that is because you’ve totally immersed yourself in Greek math, it’s the change in λ. So the ecliptic longitude at the second position in each pair minus the ecliptic longitude at the first position in each pair. Again, stating that mathematically, $\Delta \lambda = \lambda(E’) – \lambda(E)$.
- When measured counter-clockwise from apogee from the position of M as we did in Book III.
- Technically, the ecliptic longitude is measured from the vernal equinox which is why Neugebauer drew it in as well.
- Note that this is positive since we’re drawing from the second column.
- Negative because we’re drawing from the first column.