Now that we’ve demonstrated a parallax of $1;07º$ for the moon, we can use that do determine a distance to the moon. As a forewarning, some of the math may seem suspect here, so I’ll do my best to explain it.
To being, let’s start off with a new drawing:
In this image, for the first time, we’re considering the earth as something larger than a point. As such, $circle \; AB$ is the earth. Next, $circle \; GD$ is the meridian at the distance of the moon1, viewed side on, with $G$ as the direction to the zenith and $D$ as the moon. The outer circle, $circle \; EZH \Theta$ represents the celestial sphere. And that’s where things start getting funny because the celestial sphere is imagined as infinitely far away. Or, at least so far away that the size of the earth is negligible2. But obviously, if I drew it that way, we wouldn’t be able to observe anything else. So we’re going to have to do some flexible thinking when we consider this circle. On this circle, $E$ is the zenith, $H$ is the projection of the lunar position onto the celestial sphere if there were no parallax, and $\Theta$ is the projection of the lunar position onto the celestial sphere from Alexandria. Next, $\overline{AZ}$ is simply a line parallel to $\overline{KH}$ but at Alexandria instead of the true center of all these circles, which would be $K$. We’ve also dropped a perpendicular from $A$ only $\overline{KH}$.
In the previous post, we calculated the true distance of the moon, $D$ from the zenith $G$ to be $49;48º$. Thus, $arc \; GD = 49;48º$ as does the central angle it subtends, $\angle{GKD}$.
Now let’s consider the parallax. We said that the observed angle from the zenith at Alexandria, $\angle{GAD} = 50;55º$. In addition, the parallax angle was $1;07º$ which is $\angle{ADK}$ as well as its vertical angle, $\angle{HD \Theta}$ and the arc that subtends that angle on the celestial sphere, $arc \; H \Theta$.
And here’s where the celestial sphere is going to start getting funny. Although it doesn’t look like it here, $arc \; Z \Theta \approx arc \; H \Theta$. If you imagine shrinking the size of the Earth until it was imperceptibly small, point $A$ converges to point $K$, and $Z$ converges to $H$. However, the arc on the celestial sphere, $arc \; H \Theta$ would remain unchanged. This means that the apparent difference of those arcs in this drawing is really only due to the drawing itself. This allows us to state that $\angle{ZA \Theta} = 1;07º$.
We’ll set that aside and jump into a demi-degrees circle around $\triangle{ADL}$. We previously stated that $\angle{ADL} = 1;07º$ as well so the arc opposite in this circle will have twice its measure giving us, $arc \; AL = 2;14º$. Taking its chord, we get $\overline{AL} = 2;21^p$ in the context of this circle in which the hypotenuse, $\overline{AD} = 120^p$
Next, we’ll need to do a bit more fuzzy thinking. Again, consider zooming out until the earth becomes a single point. We already noted that $A$ converges with $K$, but so to do $L$ and $B$ converge. So in that respect, we can also state that $\overline{LD} \approx \overline{AD} \approx 120^p$.
I should note that this is some really fuzzy thinking here because considering these points to converge only works in the context of the celestial sphere. However, the triangle we’re considering right now doesn’t have any points on the celestial sphere. Rather, it only extends to the circle of the moon. Thus, Ptolemy’s logic here isn’t particularly sound3, as there will be a non-negligible difference between the length of these two lines. But where Ptolemy goes, so must we follow.
Without really resolving much in that demi-degrees circle we were just looking at, Ptolemy now jumps into another one, this time about $circle \; ALK$. In it, $\angle{AKL} = 49;48º$ since it’s the same angle as $\angle{GKD}$. Thus, the arc opposite it in this circle, $arc \; AL$ will have a length of twice that, or $99;36º$. Its supplement, $arc \; LK$ then will be $80;24º$. Finding the corresponding chords, we then have $\overline{AL} = 91;39^p$ and $\overline{LK} = 77;27^p$ in the context of this circle in which the hypotenuse $\overline{AK} = 120^p$.
At this point, we’d normally context switch back to the big picture but, as of yet, we don’t have the measure of any of these in any other context. So we’ll make one up and define a new context in which the radius of the earth, $\overline{AK}$ is our base unit. In other words, $\overline{AK} = 1^p$. We can then use that to context switch this triangle to state that $\overline{AL} = 0;46^p$ and $\overline{KL} = 0;39^p$.
Now we’ll return to the previous demi-degrees circle. In it, we stated that $\overline{AL} = 2;21^p$ and $\overline{LD} = 120^p$. Since we now have $\overline{AL}$ in the context in which the radius of the earth is $1^p$, we can use that to context switch this triangle and determine $\overline{AL} = 0;46^p$ and $\overline{LD} = 39;06^p$.
We can then add $\overline{KL} + \overline{LD} = 0;39^p + 39;06^p = 39;45^p$ which gives the distance to the moon as $39;45$ earth radii.
However, that’s not a constant. The moon moves closer and further on the eccenter as well as on its epicycle. As such, that was only the distance at a specific point in time! We’ll need to do some more work to work out those various components, which I’ll save for the next post.
Gotta love the Christmas/New Years bump!
- Although we don’t normally think of it this way, the meridian is actually a plane which has on its surface, north, south, and the zenith. We typically only discuss it in terms of its intersection with the celestial sphere, but now we’re also discussing the context of where it intersects the sphere of the moon.
- As Ptolemy puts it, “to which the earth bears the ratio of a point.”
- As I noted in this post, Neugebauer calls out the calculation of lunar parallax and distance as particularly poor. This seems like an excellent example of why.