As a general rule, I try to stay away from using too much modern math as I work through the Almagest. The goal of this project is to try to understand how astronomers worked in a historical context – not simply examining their work through a modern lens.
However, Ptolemy’s discussion around Mercury has been greatly frustrating me. There’s several reasons for this. A large one is certainly that the language Ptolemy used is clunky which is challenging for a model that is so complex.
Therefore, I want to dig deeper into what’s happening with the double perigee and make sure I fully understand it. In particular, I previously showed a diagram from Pedersen which looked at the path the center of the epicycle would trace out for various eccentricities. This graphically showed the distance from earth over time, but I wanted something more quantitative, so in this post, we’ll derive an equation to determine the distance between the center of Mercury’s epicycle and earth as a function of the angle from apogee.
This method comes straight from Pedersen but I’ll be doing it in the context of Toomer’s translation1.
As usual, let’s start with a diagram:
By and large, this image is what we’ve seen before. $\overline{AG}$ is the line of apsides. Points $Z$, $D$, and $E$ are the same as they were when we set up Mercury’s model. In other words, $Z$ is the point around which the eccentre that carries the epicycle rotates clockwise, $H$ is its geometric center, $D$ is the point about which Mercury itself rotates counter-clockwise, and $E$ is the observer on earth.
As we discussed when we set up Mercury’s model, $\overline{ED} = \overline{DZ} = \overline{ZH}$. Instead, of typing these out every time, we’ll just use a shorthand and call that length $e$ as it is a constant in this model.
Similarly, we’ll also call $\overline{KH}$ (the radius of the eccentre the epicycle rides on) $R$ as this is also a constant.
Lastly, $\angle AZH$ is always equal and opposite $\angle ADK$. We’ll call these angles $\Theta$.
We’ve connected points $H$ to $K$ and $K$ to $E$. We’ve also extended $\overline{KD}$ to point $B$ where we’ve dropped a perpendicular on it from $H$.
So, let’s get started on the derivation.
First, we’ll concentrate on $\triangle ZDH$. In this triangle, $\angle DZH = 180º – \Theta$ as it’s the supplemental angle to $\Theta$.
This is also an isosceles triangle since two of its sides $\overline{DZ}$ and $\overline{ZH}$ are equal. This means that, within this triangle, $\angle ZDH = \angle ZHD$.
We can solve for these angles since the angles in a triangle add up to $180º$ and we know what the third angle is.
Thus:
$$\angle ZDH + \angle ZHD = 180º – (180º – \Theta)$$
$$\angle ZDH + \angle ZHD = \Theta$$
Since these angles are equal,
$$\angle ZDH = \angle ZHD = \frac{\Theta}{2}$$
Next, we’ll start looking at the angles about $D$. We already stated that $\angle ZDK = \Theta$ as this is part of the hypothesis/model. We also just solved for $\angle ZDH$. So this means the only piece left on $\overline{KB}$ is $\angle{HDB}$ and the sum of these three angles must be $180º$ since they add up to the straight line.
Thus, we can write:
$$\angle DZK + \angle ZDH + \angle HDB = 180º$$
And substituting in:
$$\Theta + \frac{\Theta}{2} + \angle HDB = 180º$$
Solving for $\angle HDB$ and simplifying:
$$\angle HDB = 180 – \frac{3 \Theta}{2}$$
Next, we’ll want to start working on $\triangle DHB$, but we’ll start by returning to $\triangle ZDH$ and find the shared side, $\overline{ZH}$.
Again, this triangle is isosceles, so we’ll divide it in half about $Z$. This allows us to state:
$$cos(\frac{\theta}{2}) = \frac{\frac{1}{2} \overline{DH}}{e}$$
Solving:
$$\overline{DH} = 2e \cdot cos(\frac{\theta}{2})$$
Now, we know both an angle and the hypotenuse in $\triangle DHB$, so we can go ahead and solve the other two sides. I’ll start with $\overline{BH}$:
$$sin(180 – \frac{3 \Theta}{2}) = \frac{\overline{BH}}{\overline{DH}}$$
$$\overline{BH} = \overline{DH} \cdot sin(180 – \frac{3 \Theta}{2})$$
And substituting in for $\overline{DH}$:
$$\overline{BH} = 2e \cdot cos(\frac{\theta}{2}) \cdot sin(180 – \frac{3 \Theta}{2})$$
Next, we’ll solve for $\overline{DB}$ in the same triangle:
$$cos(180 – \frac{3 \Theta}{2}) = \frac{\overline{DB}}{\overline{DH}}$$
$$\overline{DB} = \overline{DH} \cdot cos(180 – \frac{3 \Theta}{2})$$
$$\overline{DB} = 2e \cdot cos(\frac{\theta}{2}) \cdot cos(180 – \frac{3 \Theta}{2})$$
Next, we’ll turn our attention to $\triangle KBH$.
First, $\overline{KB} = \overline{KD} + \overline{DB}$. We don’t know $\overline{KD}$ currently, so let’s work on that.
This triangle is a right triangle, so applying the Pythagoren theorem we have:
$$\overline{KH}^2 = R^2 – \overline{BH}$$
Substituting in some:
$$(\overline{KD} + \overline{DB})^2 = R^2 – [2e \cdot cos(\frac{\theta}{2}) \cdot sin(180 – \frac{3 \Theta}{2})]^2$$
$$\overline{KD} + \overline{DB} = \sqrt{R^2 – [2e \cdot cos(\frac{\theta}{2}) \cdot sin(180 – \frac{3 \Theta}{2})]^2}$$
And solving for $\overline{KD}$:
$$\overline{KD} = \sqrt{R^2 – [2e \cdot cos(\frac{\theta}{2}) \cdot sin(180 – \frac{3 \Theta}{2})]^2} – \overline{DB}$$
And substituting in for $\overline{DB}$:
$$\overline{KD} = \sqrt{R^2 – [2e \cdot cos(\frac{\theta}{2}) \cdot sin(180 – \frac{3 \Theta}{2})]^2} – 2e \cdot cos(\frac{\theta}{2}) \cdot cos(180 – \frac{3 \Theta}{2})$$
We can do a bit of simplification using the identities $cos(180º – \Theta) = – cos (\Theta)$ and $sin(180º – \Theta) = sin(\Theta)$.
$$\overline{KD} = \sqrt{R^2 – [2e \cdot cos(\frac{\theta}{2}) \cdot sin(\frac{3 \Theta}{2})]^2} + 2e \cdot cos(\frac{\theta}{2}) \cdot cos(\frac{3 \Theta}{2})$$
Finally, we can turn to $\triangle KDE$. In it, we know that $\angle KDE = 180º – \Theta$ as it’s the supplement to $\angle ADK$. We also just solved for one side, $\overline{KD}$.
It’s not a right triangle, so we’ll need to use the law of cosines to solve for $\overline{KE}$ which is what we’re really after here as it’s the distance from earth at any given value of $\Theta$.
$$\overline{KE} = \sqrt{\overline{KD}^2 + e^2 – 2 \cdot \overline{KD} \cdot e \cdot cos(180 – \Theta)}$$
I could substitute in for $\overline{KD}$ but this is already messy enough2 . So I’ll just simplify the final $cos$ term again.
$$\overline{KE} = \sqrt{\overline{KD}^2 + e^2 + 2 \cdot \overline{KD} \cdot e \cdot cos(\Theta)}$$
I’ve turned this into a Google sheet to see how it behaves3. There’s an obvious question of what values to use for $e$ and $R$, so I’ll jump ahead and tell you now that Ptolemy will get a value of $e = 3^p$ when $R = 60^p$.
We can plot that, and in doing so, get the following:
While there is clearly a double minimum here (indicating a double perigee), we can’t tell directly from the graph where it is. However, since the points I calculated were $0.1º$ apart, we can look at the data set to get a pretty good understanding.
Doing so, I find the minimum values at $120.5º$ and $239.5º$ (from apogee). This means that, if apogee is $10º$ into Libra, which places the nearest points in Gemini and Aquarius and is consistent with what Ptolemy reported in the last post.