So far in this chapter, we’ve reviewed how to calculate the lunar parallax for certain limits of the lunar position and looked at what’s necessary to estimate the effects for lunar positions away from those limits due to the epicycle. Now, we need to discuss the impact of the eccentre and how that we can estimate the effect on parallax due to it bringing the moon closer and further.
So let’s set up a generic diagram of our eccentric model, ignoring the epicycle and only concerning ourselves with the mean moon:
Here, we have the center of the eccentre at $E$, and earth, at the center of the ecliptic at $Z$. This puts apogee of the eccentre at $A$ and perigee at $G$. For our first example, we’ll place the moon at $B$, $60º$ past apogee1 as viewed from earth. Thus, $\angle{AZB} = 60º$ as does $\angle{GZD}$ as it’s a vertical angle. We’ve also dropped a perpendicular from $E$ onto $\overline{BD}$ at point $H$, and connected $B$ and $D$ to $E$.
We’ll first consider the small triangle, $\triangle{EHZ}$ and imagine a demi-degrees circle. In it, $arc \; EH$ is opposite the angle we just discussed, so its measure is double, or $120º$. Then, its supplement, $arc \; HZ = 60º$.
We can find the corresponding chords to state that $\overline{EH} = 103;55^p$ and $\overline{HZ} = 60^p$. Since it’s the hypotenuse, we also know that $\overline{EZ} = 120^p$ in this context, we can use to switch back to the larger context in which $\overline{EZ} = 10;19^p$. In that context, $\overline{EH} = 8;56^p$ and $\overline{ZH} = 5;10^p$.
Now let’s focus on $\triangle{BHE}$. In it, we know $\overline{BE} = 49;41^p$ as it’s the radius of the lunar eccentre and we just determined $\overline{EH}$, so we can use the Pythagorean theorem to determine $\overline{BH} = 48;53^p$.
To that, we can add $\overline{HZ}$ which gives the distance to the mean moon from the earth, $\overline{ZB} = 54;03^p$.
Ptolemy then reminds us of the context we’re working in where $\overline{ZA} = 60^p$ and $\overline{ZG} = 39;22^p$ for a difference of $20;38^p$. He brings this difference up because he then does the same thing he did in the last post. There, when considering the amount closer and further the epicycle carried the true moon, that distance was expressed as a sixtieth of the diameter of the epicycle. Here, he will express the impact as a sixtieth of the difference between apogee and perigee of the eccentre.
So first, in the situation we’re examining, when the moon is at $B$, or $60º$ from apogee2, the amount closer the moon is brought it $60;00^p – 54;03^p = 5;57^p$. To convert that to the sixtieth of the difference we equate the ratios:
$$\frac{5;57}{20;38} = \frac{x}{60}$$
$$x = 17;18$$
Now, let’s look at $\triangle{HED}$. This triangle is actually the same triangle as $\triangle{HEB}$, just mirrored, which we know because they are both right triangles with two sides equal. Specifically, they both share the side $\overline{HE}$ and both have a radius of the eccentre as their hypotenuse. Since both are right triangles, we could use the Pythagorean theorem to show the third side is equal thus proving SSS congruence. As such, $\overline{BH} = \overline{HD} = 48;53^p$, from which we can subtract $\overline{HZ}$ to determine $\overline{ZD}$ the distance from the earth to the mean moon when the moon is $60º$ past perigee is $43;43^p$.
If we take the difference between that and the distance to the mean moon at apogee, we get a difference of $16;17^p$ which, expressed as sixtieths of the aforementioned difference is $47;21$ sixtieths.
Again, Ptolemy explains how this will be entered into our table, and again I’ll save that until we have the table in front of us so we can see it more easily. Fortunately, that table is coming in the next post!
