We’re almost finished with chapter 6. All that’s left is to determine the position of the mean moon during one of the eclipses which will tell us the equation of anomaly at that point. To do so, we’ll add a few more points to the image we ended the last post with:
Here, $A$ is the first eclipse in this set of three. $B$ is the second. We’ve omitted the third. We also have the apogee and perigee at $L$ and $M$ respectively and $K$ is the center of the epicycle. From that point, we drop a line that’s perpendicular to $\overline{AD}$ at $N$ and extending to $X$ on the circumference. Point $E$ is also on the circumference on $\overline{AD}$.
We’ve already figured out many of these pieces1. We’ll begin in the context of the epicycle having a radius of $60^p$. There:
$$\overline{DK} = 689;8^p$$
$$\overline{DE} = 643;36,39^p$$
$$\overline{AE} = 88;40,17^p$$
$$arc \; AE = 95;16,50º$$
We can then state that
$$\overline{NE} = \frac{1}{2} \cdot \overline{AE} = \frac{1}{2} \cdot 88;40,17^p = 44;20,8^p$$
That can then be added to $\overline{DE}$:
$$\overline{DN} = \overline{DE} + \overline{NE} = 643;36,39^p + 44;20,8^p = 687;56,47^p$$
Now we’ll draw $circle \; DKN$. In it, $\overline{DK}$ is the hypotenuse so has a measure of $120^p$ which we can use to convert $\overline{KN}$ as well:
$$\frac{\overline{DN}}{687;56,47^p} = \frac{120^p}{689;8^p}$$
$$\overline{DN} = 687;56,47^p \cdot \frac{120^p}{689;8^p} = 119;47,36^p$$
From this chord, we can look up the corresponding arc, $arc \; DN$ which comes out to be $173;17º$. If we then look at the angle opposite that arc, $\angle{DKN}$, it’s half the measure or $86;38,30º$.
Now looking at that angle in the context of the epicycle, it is subtended by $arc \; MEX$ which will have the same measure. Thus, the supplement, $arc \; LAX = 93;21,30º$.
We can also state that2
$$arc \; AX = \frac{1}{2} \cdot arc \; AE = \frac{1}{2} \cdot 95;16,50º = 47;38,30º$$
This can then be subtracted from $arc \; LAX$:
$$arc \; AL = arc \; LAX – arc \; AX = 93;21,30º – 47;38,30º = 45;43º$$
This gives us the distance that the first eclipse, $A$, was away from apogee. However, we can also quickly determine the distance the second eclipse, $B$, was since we know that $arc \; AB = 110;21º$. Subtracting out $arc \; AL$ we determine $arc \; LB = 64;38º$.
Now that we know where the moon actually was on the epicycle with respect to apogee, we can turn towards finding the equation of anomaly. To do so, we’ll solve for $\angle{KDN}$ since we know the other two angles in that right triangle:
$$\angle{KDN} = 180º – 90º – 86;38º = 3;22º$$
That’s the equation of anomaly for the middle of the first eclipse. But again, Ptolemy is more interested in the second eclipse, so we’ll use this, plus our previous determination that $\angle ADB = 7;42º$ to find the difference:
$$\angle{KDB} = 7;42º = 3;22º = 4;20º$$
Since the epicycle in the lunar model rotates clockwise, this would have a subtractive effect on relation to the true position. When we first introduced these Alexandrian eclipses, Ptolemy stated that the stated that the sun was $25 \frac{1}{6}º$ into Libra. The moon, being opposite the sun then would have been $25 \frac{1}{6}º$ into Aries. However, that’s the true position of the moon. To determine the position of the mean moon, we’ll need to undo the subtractive effect of the anomaly which tells us that, at the time of mid-eclipse for this event, the moon should have been at $29;30º$ in Aries.
- In the last post, I diverged somewhat from Ptolemy’s proof, skipping the unnecessary demi-degrees. This resulted in some compounded differences in rounding which led to slightly different results. So in this post, I’ll adopt Ptolemy’s values in the hopes that it makes things easier for the reader to follow along if they’re using the Toomer translation.
- The actual value is $47;38,25º$, but Ptolemy rounds this up.