Almagest Book V: The Difference at Syzygies – Lunar Apogee and Perigee

In the last post, we looked at how much the total equation of anomaly would change during syzygy due to the eccentre we added to the lunar model in this book, when the moon was at its greatest base equation of anomaly. As Ptolemy told us, it wasn’t much. However, there was a second effect that can also change the equation of anomaly, which was based on where we measure the movement around the epicycle from. Namely, the mean apogee instead of the true apogee. This has its maximum effect when the moon is near apogee or perigee so in this post, we’ll again quantify how much.

Let’s start off by building our diagram:

In this setup, we’re still taking the sun at the same position: where it’s at its greatest (additive) equation of anomaly. However, for the moon, we’re taking it when it’s at mean perigee on the epicycle, which is to say, as measured from point $Z$. In this case, that’s at point $L$.

As with in our last post, we know that the direction of apogee, $A$ is twice the angle between the mean sun and mean moon. However, this time, there is no component from the lunar equation of anomaly. Rather, the distance between the mean sun and mean moon is entirely caused by the solar equation of anomaly which was $2;23º$. Thus, the double elongation, which gives the angle between the mean moon and apogee of the eccentre, $\mathrm{\angle }AEB$, is $4;46º$.

Due to this smaller angle, I think this diagram is too squished, so I’ll draw another one keeping the relative orientation of the points roughly the same to help us:

Much nicer. But we’ll need some demi-degrees circles soon, so let’s build a few right triangles to help us. First, I’ll add in $\overline{EL}$ and then from $L$, drop a perpendicular onto $\overline{EH}$ at point $N$. Next, I’ll draw in $\overline{DB}$ and drop another perpendicular from $D$ onto $\overline{EH}$ at point $M$. Lastly, I extend $\overline{BE\$}$ until I can drop a perpendicular onto it from $Z$ at $X$.

Now, let’s consider a demi-degrees circle around $\mathrm{△}EDM$. In that circle, we have $\mathrm{\angle }DEM$ which is the same angle as $\mathrm{\angle }AEB$ which was the double elongation or $4;46º$. Thus, the arc opposite it in this small circle, $arcDEM=9;32º$. Again, $\overline{DE}$ is the hypotenuse, thus bisecting the circle which means $arcEM=170;28º$ as it’s the supplement.

Interestingly, this triangle, $\mathrm{△}DEM$ is the same triangle as $\mathrm{△}EXZ$. We know this because $\mathrm{\angle }DEM$ is a vertical angle to $\mathrm{\angle }XEZ$. Both also have a right angle, meaning the third angle must also be the same which gets us to the two triangles at least being similar. But we can also observe that $\overline{DE}=\overline{EZ}$ since they’re both radii of the orbit of the center of the eccentre and given they’re both the hypotenuse, the triangles must also be the same.

This will allow us to give some quick equivalences. So we’ll look up the chords for the arcs we just discussed and then we can easily equate them to this other triangle. Thus, we can say:

$$\overline{DM}=\overline{XZ}=9;{58}^p$$

$$\overline{EM}=\overline{EX}=119;{35}^p$$

As always, the hypotenuses $\overline{DE}$ and $\overline{EZ}$ are both ${120}^p$ in the context of these small circles. But we know in the context of the larger picture, they’re both $10;{19}^p$. Using that ratio, we can context switch to the larger picture to state

$$\overline{DM}=\overline{XZ}=0;{51}^p$$

$$\overline{EM}=\overline{EX}=10;{17}^p$$

Next, we can use the Pythagorean theorem on $\mathrm{△}BDM$ where we know $\overline{BD}=49;{41}^p$ as it’s the radius of the eccentre.

$$0;{51}^2+{\overline{BM}}^2=49;{41}^2$$

$${\overline{BM}}^2=49;{41}^2–0;{51}^2=2468;26–0;43=2467;43$$

$$\overline{BM}=49;{41}^p$$

From that, we can add on $\overline{EM}$ which we just determined to state that $\overline{BE}=59;{58}^p$.

We can extend that even further by adding on $\overline{EX}$ to state that $\overline{BX}=70;{15}^p$.

Next, Ptolemy does something odd and states:

Therefore, by the same argument, hypotenuse $\overline{BZ}$ [of $\mathrm{△}BZX$ will be approximately the same size.

I’m not sure what “same argument” he’s referring to here, but it’s easy enough for us to check since $\mathrm{△}BZX$ is a right triangle, and we now know two sides, so we can again use the Pythagorean theorem:

$${\overline{BZ}}^2=70;{15}^2+0;{51}^2=4935;47$$

$$\overline{BZ}=70;15$$

Sure enough, it works out because it’s such a skinny triangle¹.

Now, consider the small triangle inside this triangle: $\mathrm{△}BLN$. These are similar triangles which allows us to set up some ratios:

$$\frac{\overline{BZ}}{\overline{ZX}}=\frac{\overline{BL}}{\overline{LN}}$$

$$\frac{\overline{BZ}}{\overline{BX}}=\frac{\overline{BL}}{\overline{BN}}$$

as they are the similar sides. Now let’s consider that small triangle. In it, $\overline{BL}=5;{15}^p$ as it’s the radius of the epicycle. Using that in our above ratios, we can determine $\overline{LN}=0;4^p$ and $\overline{BN}\approx 5;{15}^p$.

With this information, we can subtract $\overline{BN}$ from $\overline{BE}$ to determine $\overline{NE}=54;{43}^p$.

We can now consider $\mathrm{△}ELN$ which is also a right triangle. We know the two sides and can determine the the hypotenuse, $\overline{EL}$, via the Pythagorean theorem again, but again, this is such a skinny triangle, Ptolemy just notes it’s “not noticeably different from” the longer side, $\overline{NE}$ and just calls $\overline{EL}=54;{43}^p$.

Now we’ll dive into a demi-degrees circle about this triangle. In it, we can use the hypotenuse, which we just stated is $54;{43}^p$ as the conversion to the demi-degrees context in which it has a measure of ${120}^p$. Using that ratio on $\overline{LN}$ we determine that $\overline{LN}=0;8^p$. Looking up the corresponding arc, we get $arcLN=0;8º$. The angle it subtends, $\mathrm{\angle }LEN$ must then be half or $0;4º$.

Let’s stop and consider what this angle means. Point $L$ is the position of the moon at mean perigee. Point $N$ isn’t actually on the epicycle, but it is still along $\overline{EB}$ which is the line of sight to the true perigee. So this angle represents the angle, viewed from earth, between the true and mean perigee. Thus, it’s the difference in the equation of anomaly between the two which is what we were looking for. Again, this is quite small and introduces

no noticeable error in our previous demonstrations using lunar eclipses when we used the [simple hypothesis], and not that supplemented by using the eccentre.

And that’s it for that chapter. In the next one, we’ll continue working on concepts that we’ll need to eventually predict eclipses. Specifically, the moon is close enough that it has noticeable parallax, so we’ll discuss that!

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1. Much easier to see in the first diagram in this post.