In my last post, I mentioned that entered a paper based on the rising sign calculations presented in this post into an A&S competition. This was a very interesting piece to do because it showed how well woven the roots are, as doing so made use of almost every section we’ve gone through previously. As such, it felt like a good capstone for book II. But it doesn’t end there.
Rather, Ptolemy decides to go on for several more chapters as this book is focused on the great circles on the celestial sphere. While we’ve covered the ecliptic and celestial equator pretty extensively, we have done less with the horizon and meridian which is where Ptolemy seeks to go for the last few chapters in this book. Specifically, we’ll be covering:
- The angles between the ecliptic and meridian
- The ecliptic and horizon
- The ecliptic and an arc from horizon to the zenith (an altitude circle)
All followed by another summary chapter at various latitudes. As the title of this post may have indicated, we’ll be covering the first of these in this post1.
But first, Ptolemy does something that we’ve previously avoided: Defining angles on the surface of the celestial sphere. Part of the reason Ptolemy takes the time to do so is because that whenever you have two lines intersecting on a plane, it creates four angles (two pairs), and Ptolemy wants to be clear that the one we’ll concern ourselves with is the one that
lies to the rear of the intersection of the circles and to the north of the ecliptic.
We’ll return to this at the end of this Book when we’ll develop the aforementioned table and it becomes more important. However, for now, just think of this as the angle that’s “up and left” when viewed from inside the celestial sphere2.
To get at the angle between the ecliptic and meridian, Ptolemy leads us on a bit of a wild ride. Before getting into the meat of things, we’re going to quickly develop two new rules. Then we’ll get into some other diagrams, all of which use the same letters to represent different things. I’ll try to break things up so it’s clear when we switch things over, but take special care as we proceed. And ultimately, the goal here will be to determine the angle between the ecliptic and meridian at the borders between each zodiacal constellation.
The first new rule that we want to develop is to
show that the points on the ecliptic equidistant from the same equinox produce angles … equal to each other.
Here’s the diagram we’ll use.
Here, we have celestial equator,(ABG) and the ecliptic (DBE) with the north celestial pole at Z. From there, he draws two arcs to these great circles such that $arc \; BH = arc \; B\Theta$.
What we want to show is that $\sphericalangle KHB = \sphericalangle Z \Theta E$ in this diagram.
I’m going to break with the proof here for a moment because I want to take a good look at which angles these are. Both are angles with a vertex on the ecliptic. If we consider the north celestial pole to be towards the top, they’re to the right. And that’s what we’re going to do for a bit: Look at how two of these upper right angles are related for different points along ecliptic. In this case, we’re starting by looking at two angles equidistant from the same equinox.
In addition to knowing that $arc \; BH = arc \; B\Theta$ (which was given), we also know that $\sphericalangle HBK = \sphericalangle LB\Theta$ because these angles are vertical angles which are congruent. Lastly, we know $arc \; HK = arc \; L\Theta$ because these are arcs of great circles through the pole (i.e. declinations) equidistant from the same equinox and thus must be due to symmetry3.
So now we have two sides equal and the non-included angle which means we have Side-Side-Angle similarity. This means that the related angles within these two triangles, $\sphericalangle KHB$ and $\sphericalangle B\Theta L$ are equal and since $\sphericalangle Z \Theta E$ is a vertical angle to $\sphericalangle B\Theta L$, it too must be equal4.
The next goal will be to,
prove that the sum of the angles between ecliptic and meridian at points on the ecliptic equidistant from the same solstice is equal to two right angles.
Here’s the diagram we’ll use for that.
Here, we have ABG as the ecliptic and B as the solstice. Again, Z is the celestial pole and we’ll draw two arcs from it down the the ecliptic such that $arc \; BD = arc \; BE$ where B is the solstice.
In this case, we want to prove that $\sphericalangle ZDB + \sphericalangle ZEG = 180º$5. Whereas the previous rule was about angles equidistant from the same equinox, now we’re developing one for angles equidistant from the same solstice. Happily, this is a very straightforward proof.
Since ABG is a great circle (which is akin to a straight line for spherical geometry) we can state:
$$\sphericalangle ZEB + \sphericalangle ZEG = 180º$$
Next, $\sphericalangle ZDB = \sphericalangle ZEB$6. Thus, we can substitute to state $\sphericalangle ZDB + \sphericalangle ZEG = 180º$ which is what we set out to prove: That the “up and right” angles, equidistant from a solstice sum to 180º.
So now we’ve developed two symmetry rules which will make our upcoming work much quicker as we’ll only need to calculate the angle between the ecliptic and meridian at the beginning of each sign in one quadrant as these rules will complete the rest for us.
- Ptolemy even does a bit of motivation for these topics stating that these calculations will “play a very important part in the requirements for lunar parallax.”
- Assuming the northern hemisphere since that’s where Ptolemy did his work from.
- This is something we addressed in this post where they were called $arc \; H\Theta$ and $arc \; KX$.
- Ptolemy actually approaches this through Side-Side-Side, which works just as well, because $arc \; BK = arc \; BL$ as we proved in this post where they were called $arc \; E\Theta$ and $arc \; EZ$.
- NOT that they are both right angles.
- We know this again because we can draw an arc from the pole to the solstice, B, which makes two triangles which are identical, just mirrored across that line, again due to SSS similarity.