Syzygy is one of those words that has popped up very little in the Almagest so every time it does, I’m always thrown off a bit1. Especially when Ptolemy is going to spend an entire chapter discussing a topic that has scarcely even come up. But here we have Ptolemy spending the entirety of chapter $10$, to demonstrate that these modifications we’ve made to the lunar model have a negligible effect because he fears readers might think it does since
the centre of the epicycle does not always … stand exactly at the apogee at those times, but can be removed from the apogee by an arc [of the eccentre] of considerable size, because location precisely at the apogee occurs at the mean syzygies, whereas the determination of true conjunction and opposition requires taking the anomalies of both luminaries into account.
To begin, let’s consider a generic diagram of our complete lunar model:
Here, we have the eccentre, circle $ABG$, centered on $D$, which rotates around the observer at $E$. Point $B$ is the center of the epicycle, circle $H \Theta KL$ with the moon at $\Theta$. Point $Z$ is the point opposite the observer from the center of the eccentre. Points $K$ and $L$ are the intersection of $\overline{HE}$ and $\overline{BZ}$ with the epicycle, respectively.
Before continuing, let’s pause and understand what “differences” Ptolemy is interested in here: It’s the “size of the [equation of] anomaly” which in this diagram is $\angle{BE \Theta}$. In our first lunar model, this was strictly determined by the position of the moon on its epicycle. But thanks to the changes we’ve made in the updated lunar model, this angle is affected in two different ways:
1) When the center of the epicycle is closer to the observer, for example, at the perigee of the eccentre at point $G$, the angle would appear larger. Conversely, at apogee at point $A$, it would appear smaller.
2) The direction from which we measure the apogee of the epicycle is no longer through $E$, but instead goes through $Z$2.
These two changes each have their maximum effect for different positions of the moon along the epicycle. For the first, it is largest when the the equation of anomaly is at a maximum3. In short, moving the center of the epicycle closer or further with increase or decrease the equation of anomaly respectively, but when there is no equation of anomaly, because the moon is along the same line of sight as the center of the epicycle, moving it closer or further has no effect. However, when the moon is at its maximum equation, that’s more equation to increase or decrease.
For the second, the effect is most pronounced at the exact opposite time: When the moon is closest to apogee and perigee. To see why, let’s look at a few illustrations. First, let’s do an example where the moon is at the mean apogee we derived earlier in this book. As a reminder, this is the point on the epicycle directly opposite point $Z$ in this drawing which wasn’t previously drawn in, but we’ll draw it in now and label it point $X$.
Recall that $H$ is the true apogee. In this case, there’s an effect on the equation of anomaly would be $\angle{HEX}$. Now let’s let the moon progress a little over $90º$, both as it we measured from the true apogee and the mean apogee and look at things again:
In this case, $\angle{HEX}$ is tiny. So small we can’t even distinguish it from the thickness of the lines as I’ve drawn it here.
Regardless, both of these can have an impact on the equation of anomaly, but Ptolemy wants to demonstrate that at syzygies, it’s negligible. But what a syzygy means is rather complicated. As Ptolemy explains:
[T]he true syzygy can differ from the mean by the sum of the equations of the two luminaries, if one is additive and the other is subtractive.
Let’s consider this case first, where the sun’s equation of anomaly is additive, meaning the position of the true sun when viewed along the ecliptic is an advance4 of the mean sun. Then we’ll need to consider the moon’s equation being subtractive, or rearwards.
To make this as clear as possible, Ptolemy takes both at their absolute maximum. For the Sun, this means $2;23º$ as we can see from our table of the equations of anomaly for the sun. Recalling that values in the first column are subtractive and the ones in the second column are additive, that means this will occur when the sun is about $267º$ after solar apogee. Solar apogee was $5 \frac{1}{2}º$ into Gemini or $65;30º$ ecliptic longitude), so this means that at its greatest additive equation of anomaly, the sun would be at $332 \frac{1}{2}º$ ecliptic longitude which is $2 \frac{1}{2}º$ into Pisces. Thus, the true sun would be $2;23º$ in advance of that, or $4;57º$ into Pisces.
For the moon, the maximum equation of anomaly is $5;1º$. Since the definition of a syzygy is that the sun and moon be at the same ecliptic longitude, this means that the mean moon would then be $5;1º$ ahead of the position of the true sun so around ecliptic longitude or
So let’s draw that in, bare minimums as we still haven’t said much about the position of the eccentre5 or where the epicycle is on the eccentre.
Next, we’ll need to determine the position of the eccentre. To do so, we need to make use of the definition of the double elongation from this post, which noted, “the angle between the mean sun and mean moon is always equal and opposite the angle made between the sun an the center of the eccentre.” I’ll draw that in and then explain better.
First, let’s recall a few things about the lunar eccentre: It’s true center is at $D$ which is offset from the center of the ecliptic, $E$. The continuation of $\overline{ED}$ points to the apogee of the eccentre, $A$. We were able to determine the position of $D$ and thus $A$ due to what we said above: “the angle between the mean sun and mean moon is always equal and opposite the angle made between the sun an the center of the eccentre.”
In this case the distance angular between the mean sun and mean moon is the sum of their respective equations of anomaly for a total of $7;24º$. That means that the direction of the center of the eccentre, $D$, and thus the position of apogee of the eccentre, $A$, must be $7;24º$ from the mean sun in the opposite direction
Here, I’ve drawn in the epicycle at the actual size it should be to produce an angle of $5;1º$ so this diagram is quite cramped. I have, however, still allowed the size of the orbit of the center of the eccentre to be exaggerated, although it’s not much help. Readers may find it much more useful to use the original diagram at the beginning of this post which retains the same relative positioning of points, but lays things out such that the angles are much easier to see.
Anyway, because of the above, $\angle{AEB}$ is twice the angle between the mean sun and moon for a value or $14;48º$. Again, this is essentially the definition of the double elongation.
Now, in preparation for a demi-degrees method, we’ll drop a perpendicular from $D$ onto $\overline{EB}$ to create point $M$. I’ll also go ahead and conned $D$ to $B$
In the demi-degrees circle about $\triangle{DEM}$, $\angle{DEM}$ retains that same measure of $14;48º$ and the arc opposite it would have twice the measure which is to say, $arc \; DM = 29;36º$. Recalling that $\overline{EM}$ is the hypotenuse and thus bisects the circle, this means $arc \; EM = 150;24º$ as it’s the supplement.
If we look up the corresponding chords, this means that $\overline{DM} = 30;39^p$ and $\overline{EM} = 116;1^p$. We also know that $\overline{EM} = 120^p$ in this circle as it’s the hypotenuse.
We can context switch to the larger picture because we know that, in that context, $\overline{DE} = 10;19^p$. Switching to that context, $\overline{DM} = 2;38^p$ and $\overline{EM} = 9;59^p$.
And now that we know $\overline{DM}$ in that context, we can use the Pythagorean theorem on $\triangle{BDM}$ to determine $\overline{BM}$, as we already know $\overline{BD}$, the radius of the eccentre which was $49;41^p$6:
$$49;41^2 = 2;38^2 + \overline{BM}^2$$
$$\overline{BM}^2 = 2468;26 – 6;56 = 2641;30$$
$$\overline{BM} = 49;37^p$$
To that, we can add $\overline{EM}$ to state that $\overline{BE} = 59;36^p$.
Now we’ll do another demi-degrees circle around $\triangle{BE \Theta}$. In it, $\overline{BE}$ is the hypotenuse so has a length of $120^p$ in that context. Since we just determined its length in the larger context, we can use that to context switch and state $\overline{B \Theta} = 10;34^p$.
That can then be looked up in our table of chords to determine $arc \; B \Theta = 10;6º$. Thus, the angle it subtends, $\angle{BE \Theta}$ is half that or $5;3º$.
Compare this to the maximum anomaly from our first model of $5;1º$7. There’s only a $0;02º$ difference which is quite small. Certainly below the measurement threshold of instruments at the time. But more importantly, let’s remember the context of this chapter: We’re discussing syzygys. Why? Because eclipses only happen at syzygys. And since the angular diameter of the sun and moon are each about $0;30º$, this means that the difference caused by this effect won’t throw off the upcoming calculations of the eclipse sufficiently enough to prevent one from happening, although it may change the characteristics ever so slightly.
So that covers the first of the two cases in which our second model could skew the results. In the next post, we’ll take a look at how they are impacted when the moon is near apogee or perigee on the epicycle.
- If you’ve forgotten what it means, it’s a conjunction or opposition of two celestial objects, but is most frequently used in relation to the sun and moon.
- As we showed in Chapter $5$.
- As we’ve shown previously, this happens when $\overline{E \Theta}$ is tangent to the epicycle as shown in the diagram above.
- I.e., counter-clockwise.
- Which we define by its apogee.
- As shown in this post.
- Which still occurs if the moon is at its greatest epicyclic equation at apogee.