In the last post, we explored how to calculate parallax if the distance to an object is known and its distance from the zenith. This was done for the sun and the moon at four different distances. However, because the moon varies so widely in distance in Ptolemy’s model, we need a way to estimate between those positions and we’ll begin by looking at the effect the epicycle has on distance for various points throughout its cycle. To help us, we’ll start with a new diagram:
Here, the circle at top is the lunar epicycle on center $E$. Down at the we have point $Z$ which we’ll take as the center of the earth as well as the center of the ecliptic. This means that $A$ will be the true apogee.
We’ll position the moon at $B$ and draw a line from $Z$ to $B$ also including point $G$ where this line intersects the epicycle on the near side and connect both of those points to $E$. In addition, from each of them, we’ll drop perpendiculars onto $\overline{AD}$ forming $\overline{BH}$ and $\overline{G \Theta}$.
For our first example, we’ll consider when the moon is $60º$ from apogee indicating that $arc \; AB = 60º$. as does the central angle it subtends, $\angle{AEB}$
Now, let’s consider a demi-degrees circle about $\triangle{BEH}$. In it, $arc \; BH = 120º$ as it’s opposite the angle we just discussed. Thus, $arc \; EH$, it’s supplement, is $60º$.
Converting those to their respective chords, $\overline{BH} = 103;55^p$ and $\overline{EH} = 60^p$. We also know that, in this context, the hypotenuse, $\overline{EB} = 120^p$.
We can now context swap this back to the overall diagram since we know that, in that reference frame, $\overline{EB} = 5;15^p$. And notice here that we’re using the abstract unit of parts instead of earth radii. That indicates that the distance from the center of the earth, $Z$, to the mean moon, $E$ is $60^p$. So converting to that context, $\overline{BH} = 4;33^p$ and $\overline{EH} = 2;38^p$.
We can now add $\overline{EH}$ to $\overline{EZ}$ to determine $\overline{HZ} = 62;38^p$. That gives us one side of the right triangle, $\triangle{HBZ}$ of which we also just determined another side: $\overline{HB}$. As such, we can use the Pythagorean theorem to determine the hypotenuse, $\overline{BZ}$ to be $62;48^p$.
We’ll pause on that track of thought for a moment to consider the first limit – what we called $1a$ in the last post. That would be the case where the moon was at $A$ and thus, the distance from the earth would be $\overline{ZA}$ which, in the context we’re considering would have a length $60;00^p + 5;15^p = 65;15^p$.
And we just determined the distance to the moon for our scenario to be $62;48^p$. Thus, the moon is $2;27^p$ closer than at the first limit, out of a possible $10;30^p$ (the diameter of the epicycle). Expressing that as a ratio, this is $\frac{2;27}{10;30}$1.
But $10;30^p$ is sort of an odd number to use as a denominator. So Ptolemy instead expresses this as sixtieths, which comes out to $\frac{14;00}{60;00}$. Ptolemy phrases this as $14$ sixtieths which is the same thing, although it looks slightly different.
I want to pause here to consider why Ptolemy did this as it certainly escaped me when I first went through it2. Essentially what Ptolemy has done here is consider the first and second limits as the extremes. At apogee, the moon is at the first limit. At perigee, it’s as close as it can possibly be due to the epicycle. If we were expressing this using modern math terms, we’d say at apogee it’s $0$% of the distance closer and at perigee it’s $100$% of the distance closer. Ptolemy is really saying the same thing. It’s just Ptomeny doesn’t work in percents, because that’s base $10$ math. Rather, he works in base $60$, so he instead expresses that proportion using that base. So at apogee it’s $\frac{0}{60}$ of the distance and at perigee it’s $\frac{60}{60}$. Same reasoning as percentage; just a different way of expressing it.
Moving on, Ptolemy then states that this number will be entered into the table we’re working on constructing, but I’ll save discussion of that until we get start looking at the table to prevent us from breaking the flow of the calculations.
Next, Ptolemy considers the case in which the moon is at $G$, $60º$ before perigee on the epicycle, which would be $120º$ after apogee. In other words, when $arc \; GD = 60º$. He skips the initial math as it’s the same as we did above, drawing a demi-degrees circle about $\triangle {EG \Theta}$, determining the parts. In doing so, we would get that $\overline{G \Theta} = 4;33^p$ and $\overline{E \Theta} = 2;38^p$. Note that these are the exact same as the previous values we determined for $\overline{BH}$ and $\overline{G \Theta}$ to the level of precision we’re discussing here. In our diagram, we can see that $\overline{G \Theta}$ is actually the shorter of the two, but our drawing is not to scale. In reality, $\triangle{HBZ}$ is sufficiently long that the difference between $\overline{BH}$ and $\overline{G \Theta}$ is negligible to the scale of minutes.
Subtracting $\overline{E \Theta}$ from $\overline{ZE}$ we determine that $\overline{Z \Theta} = 57;22^p$ and again use the Pythagorean theorem to determine $\overline{ZG} = 57;33^p$.
Taking that as the difference from the first limit of $65;15^p$ again, we determine that this puts the moon $7;42^p$ closer than the first limit. If we instead express that as sixtieths as we did before, it’s $44;00$.
Now we’ve taken two intermediate examples between the limits when the mean moon is at apogee on the eccentre. But our other two lunar cases were when the mean moon is at perigee on the eccentre. We could go through the math again, but Ptolemy takes a bit of a shortcut to skip the first few steps involving the demi-degrees circle.
Instead, he refers back to a bit of a side note we’d looked at in this post. There, he said that at perigee, the ratio of the distance between the earth and mean moon to the radius of the epicycle was about $\frac{60}{8}$. In the context of our current drawing, that’s the ratio $\frac{\overline{ZE}}{\overline{EB}}$ when at perigee on the eccentre. Contrast that with the discussion we just went through for apogee in which that ratio was $\frac{60}{5;15}$. This essentially gives us a scale factor which we can apply. The $60$’s will cancel out and we’re just left with a ratio of $\frac{8}{5;15}$. Thus, in the context of perigee,
$$\overline{BH} = \overline{G \Theta} = 4;33^p \cdot \frac{8}{5;15} = 6;56^p$$
$$\overline{EH} = \overline{E \Theta} = 2;38^p \cdot \frac{8}{5;15} = 4;00^p$$
Again, referring to the previous post Ptolemy used for this, this was in the context in which the distance between the mean moon and earth was $60^p$, so this is still in a context in which $\overline{ZE} = 60^p$. Thus, we can add and subtract to determine $\overline{ZH} = 64;00^p$ and $\overline{Z \Theta} = 56^p$. Then using the Pythagorean theorem again, we can determine $\overline{ZB} = 64;23^p$ and $\overline{ZG} = 56;26^p$.
Ptolemy stresses that this is for the context in which the third limit (which we previously called $2a$), has $\overline{ZA} = 68^p$3, and the diameter of the epicycle is $18^p$. So again, Ptolemy determines the difference between these calculated distances for $30º$ after apogee and before perigee, and the total distance of the third limit. This gives us $3;37^p$ for the third limit ($2a$) and $11;34^p$ for the fourth ($2b$). Again, expressing this in sixtieths instead of eighteenths, we get a value of $13;33$ and $43;24$ respectively. This too will get entered into the table we’re constructing and I’ll discuss where once we look at the table.
I’ll wrap up this post here, but before doing so, let’s review what exactly we calculated here: It was the proportion closer to the earth the moon was, expressed as a sixtieth of the diameter of the epicycle. It was not the actual change in parallax. We’ll come to that in time. But before doing so, we’ll also want to consider how the eccentre affects the lunar distance as this will impact parallax as well. And that will be the topic of the next post.
- You’ll notice that we’ve dropped the units here. This is because we’re now considering it as a proportion of the diameter of the epicycle and the diameter of the epicycle also has units of parts. Thus, the units cancel and we’re left with a dimensionless result.
- I originally thought of it along the lines of the context switching we regularly do for the demi-degrees method.
- This is the $60^p$ we used for $\overline{ZE}$ plus the $8^p$ for the radius of the epicycle in that context.