In the last post, we followed along as Ptolemy determined that the position of “apogee” used for calculating the motion around the epicycle is not the continuation of the line from the center of the ecliptic or center of the eccentre through the center of the epicycle. Rather, motion should be measured from the “mean apogee” which is defined from a third point opposite the center of the ecliptic from the center of the eccentre.
Ptolemy doesn’t give a rigorous proof for this and instead relies on proof by example. So in that last post, we went through one example, but in this post, we’ll do a second one
in order to show that we get the same result at the opposite sides of the eccentre and epicycle.
The observation Ptolemy uses is again one from Hipparchus taken
in Rhodes… [in the] $97^{th}$ year from the death of Alexander, Payni [X] $17$ in the Egyptian calendar [-126 July 7], at $9 \frac{1}{3}$ hours. He says that, while the sun was sighted $10 \frac{9}{10}º$ into Cancer, the apparent position of the moon was $29º$ into Leo. And this was its true position too; for at Rhodes, near the end of Leo, about one hour past the meridian, the moon has no longitudinal parallax.
Again, we’ll also need to determine the position of the mean sun and moon which we do via calculation from epoch. So we’ll need to know the time since the beginning of the epoch which Ptolemy give as $620$ Egyptian years, $286$ days, and $3 \frac{2}{3}$ equinoctial hours.
Going through the calculations to determine the calculated positions, Ptolemy gives them at $12;5º$ into Cancer for the mean sun, $10;40º$ into Cancer for the true sun, and $27;20º$ into Leo for the mean moon, where the true moon is $333;12º$ from apogee on the epicycle1.
From these,we can calculate the elongation between a few values. Ptolemy first does this for the distance between the mean moon and the true sun. Here, we have two values to pick from – the observed position (Cancer $10 \frac{9}{10}º$) and the calculated position (Cancer $10;40º$). Ptolemy uses the latter to determine an elongation between the mean moon and true sun of $46;40º$.
Similarly, we we’ll very shortly need the distance between mean sun and mean moon to determine the position of the center of the eccentre which is
$$147;20º – 102;05º = 45;15º$$
Lastly, we can take the difference between the observed (true) position of the sun and the observed (true) position of the moon:
$$148;60º – 100;54º = 48;06º$$
Using these differences, we can calculate the anomaly which is the difference between the position of the mean moon and true moon, using the true sun as the reference point:
$$48;06º – 46;40º = 1;26º$$
It’s important to note that the true position of the moon was further from the true sun than the mean moon, so this means that we’ll need to put the mean moon counter-clockwise of the true moon by $1;26º$
So let’s start our diagram:
Here again, we’ll start with $\angle{AEB}$ which is the double elongation of the mean moon to mean sun, so $90;30º$.
As usual, we’re going to use the demi-degrees method, so we’ll need a right triangle about which to do so. So we’ll extend $\overline{BE}$ a bit until it it meets a perpendicular line dropped from $D$ at point $K$. As with the last post, this is a tiny angle, so it’s basically indistinguishable at this scale:
In this tiny triangle, we can immediately determine $\angle{AEK} = 89;30º$ as it’s the supplement of $\angle{AEB}$. This means that, in the small circle around $\triangle{DEK}$, $arc \; DK = 179º$.
As a reminder, the hypotenuse of this triangle is $\overline{DE}$ so it divides the circle in half, which allows us to state $arc \; EK = 1º$ as it’s the supplement to $arc \; DK$.
From these two arcs, we can get the corresponding chords:
$$\overline{DK} = 119;59^p$$
$$\overline{EK} = 1;3^p$$
Again, in the context of this small circle, $\overline{DE} = 120^p$, but in the context of the larger diagram, it’s equal to $10;19^p$, so we can use this for our conversion factor.
Converting back to the context of the overall picture $\overline{EK} = 0;5^p$. We can then use the Pythagorean theorem to determine $\overline{DK} \approx 10;19^p$ (as it rounds to be the same as $\overline{DE}$).
Next, we’ll add $\overline{DB}$, which gives us a new right triangle, $\triangle{BDK}$.
In it we just determined $\overline{DK}$ and we know that $\overline{BD} = 49;41^p$ as it’s the previously determined radius of the eccentre. So we can again use the Pythagorean theorem to determine $\overline{BK} = 48;36^p$, from which we can subtract off $\overline{EK}$ to determine $\overline{BE} = 48;31^p$.
Now, we’ll connect the position of the true moon on the epicycle ($H$) to $B$ and then drop $\overline{BL}$ from $B$ onto $\overline{HE}$ such that they’re perpendicular.
We’ve now created a new right triangle, $\triangle{BEL}$2 which contains the angle that describes the equation of anomaly, $\angle{BEL}$, and we just determined it to be $1;26º$. So in a small circle about that triangle, the opposite arc, $arc \; BL = 2;52º$ so its corresponding chord, $\overline{BL} = 2;59^p$. This small triangle also has $\overline{EB}$ as its hypotenuse with length $120^p$. But in the context of the overall picture, $\overline{EB} = 45;31^p$ so we can use for context conversion, allowing us to determine $\overline{BL} = 1;12^p$ in that context.
Now we’ll flip to $\triangle{BHL}$. In it, we know that, in the context of our overall diagram, $\overline{BH} = 5;15^p$ as it’s the radius of the epicycle. But in the context of the small circle about this triangle, it is the hypotenuse with a length of $120^p$.
So again, we can use this to convert contexts and convert $\overline{BL}$ as well, to determine $\overline{BL} = 27;34^p$ in the context of this small circle. It’s arc, $arc \; BL = 26;34º$ which means that the angle that arc subtends, $\angle{BHL}$ is half of this which is to say $\angle{BHL} = 13;17º$.
Let’s pull some of these angles out of the diagram so we can see what we’re working with:
In this diagram, we’ve determined two of the angles in $\triangle{BHE}$. Specifically, $\angle{BHL} = 13;17º$ and $\angle{BEL} = 1;26º$. This means that the third angle, $\angle{HBE}$ is $180º$ minus those two, which is $165;17º$.
From there, we can determine $\angle{ZBH} = 14;43º$ as it’s the supplement of $\angle{HBE}$3 which is the angular distance along the epicycle of the true moon from the apparent apogee from the perspective of the center of the ecliptic.
But what we’re really concerned with is the mean apogee which we’ll now draw in at $M$, connecting it to $N$ on the line through the centers of the ecliptic and center of the eccentre, $\overline{AG}$. In addition, we’ll drop a line from point $E$ onto $\overline{MN}$ such that it’s perpendicular at $X$.
We’ll work more with those new triangles we just created in a moment, but first, we’ll need to figure out some angles in it.
To do so, we know that the true moon at $H$, measured clockwise from $M$ was $333;12º$. This means $arc \; HZM = 26;48º$. We can then subtract out $arc \; HZ$ which we just found to be $14;43º$ to determine $arc \; ZM = 12;05º$ and so too must be the angle that arc subtends, which is to say, $\angle{MBZ} = 12;05º$. And because it’s a vertical angle, so is $\angle{EBX}$.
Now we’ll do another demi-degrees method about the newly created right triangle, $\triangle{EBX}$. So in that circle, $arc \; EX = 24;10º$ and its corresponding chord, $\overline{EX} = 25;7^p$. We also have $\overline{BE} = 120^p$ since it’s the hypotenuse, but in the larger context we previously showed $\overline{BE} = 48;31^p$.
So we can use that to context switch and determine $\overline{EX} = 10;08^p$4.
We’ll set that aside for a moment and pull out a few pieces to see what we’re working with again:
Here, we already know that $\angle{AEB} = 90;30º$ which means $\angle{BEN} = 89;30º$ since it’s the supplement. And since we also know that $\angle{EBN} = 12;05º$, we can determine the remaining angle in $\triangle{EBN}$, $\angle{ENB} = 78;25º$
We’ll do one final demi-degrees circle around $\triangle{ENX}$. And in that small circle $arc \; EX$ is twice the angle it subtends, so $156;50º$, it its corresponding chord, $\overline{EX} = 117;33^p$. Also, the hypotenuse, $\overline{EN} = 120^p$.
We’ve previously shown that, in the overall context, $\overline{EX} = 10;08^p$ so we can use that to convert contexts and state that $\overline{EN} = 10;20^p$. So again, we see that the line, $\overline{MN}$ has the point $N$ very close to equidistant from $E$ as $E$ from $D$. In other words, $\overline{DE} \approx \overline{EN}$.
So that’s two examples for which we’ve demonstrated this now and Ptolemy states
the same ratio results by calculation from a number of other observations.
Thus, these observations confirm the peculiar characteristic of the direction of the epicycle in the hypothesis of the moon: the [uniform] revolution of the center of the epicycle takes place about $E$, the center of the ecliptic, but the diameter of the epicycle, which defines the unchanging point of the epicycle at which the mean epicyclic apogee is located point, not towards $E$, the center of mean motion, but always toward $N$, which is removed in the opposite direction [of $D$ from $E$] by an amount equal to $\overline{DE}$, the distance between the centers [of the epicycle and eccentre].
So at this point, we’ve got a full understanding of the lunar model. From here, the rest of Book V looks to be dedicated to looking at how to use these new complications to calculate lunar position as well as other things that this model will allow us to calculate.
- Toomer and Neugebauer both comment that there are some issues with the calculations here such that the mean moon should be $27;7º$ into Leo and the movement around the epicycle should be $333;12º$ around the epicycle. To arrive at the figures Ptolemy uses $4$ hours instead of the $3 \frac{2}{3}$ equinoctial hours he calculated. This ends up having a notable impact on the final result, giving a far less tidy match.
- As well as $\triangle{BLH}$ which we won’t use just now.
- Ptolemy phrases this calculation as adding of $\angle{BHL} + \angle{BHL}$ which, as we can see, is clearly true, but doesn’t explain why it works. Hence my elaboration here.
- I come up with $10;09^p$.