In the last post, we proved that two arcs of the ecliptic that are equidistant from the same equinox rise in the same amount of time. In this post, we’ll prove something similar for what happens with arcs of the ecliptic equidistant from the same solstice.
It’s been awhile since I’ve been able to update the blog with anything from the Almagest. As noted in the last post from the book, this section is not one of the better written ones. Indeed, it’s taken me the better part of a month to really work out how the diagram is put together.
Ultimately the trouble stemmed from the fact that it’s not a single diagram; it’s actually two pasted together1, so instead of throwing it all at you at once like Ptolemy did, let’s work through each piece in turn before pasting it together.
To begin, let’s start with a simple celestial sphere:
Here, I’ve only sketched out the celestial equator and ecliptic. I’ve marked the vernal equinox as this is a point of interest for this diagram. The dot in the center is the Earth.
Let’s consider an observer on that Earth, such that the vernal equinox is below their horizon.
Now I’ve drawn in a great circle representing an observer’s horizon. As I’ve drawn it, consider anything to the left of the plane of the horizon as being above the horizon, and to the right as below. Now we can start labeling a few things, one of which I’ve already done, swapping Θ in for the vernal equinox.
Here, K is the south celestial pole. From the other points I’ve labeled, we can see that we’ve formed a small triangle: $\triangle{EH\Theta}$. Before moving on, let’s define each of these a bit more: E is the intersection of the horizon and celestial equator, H is the intersection of the horizon and ecliptic (which is another way of saying the point of the ecliptic that is rising23), and Θ is an equinox we’re currently considering as a bounding point to form $arc \; H\Theta$.
Now, we’re going to want to keep an eye on $arc \; H\Theta$ because now we’re going to spin the celestial sphere, while keeping the horizon in the same place. To put it another way, we’re going to look at the sky 6 months later when the vernal equinox is on the far side.
Now, we’ve created a new triangle, $\triangle{EHZ}$, where E is again the intersection of the horizon and celestial equator, H is the intersection of the horizon and ecliptic, and now Z is the autumnal equinox, now above the horizon.
You may notice I’ve highlighted $arc \; HZ$ and $arc \; H\Theta$. That’s because these two arcs are the ones fundamental to the setup of the final diagram we’re working up to. In addition, it may feel funny to you to see H in two different places. But while it is odd, it makes a sort of sense, as in both places its being defined the same way: the point of the ecliptic on the horizon4.
I should take a second to note that everything I’ve done thus far is not actually included in Ptolemy’s work. This is all just to help visualize what Ptolemy is actually getting at because in the Almagest, Ptolemy stitches these two diagrams together at point H since it’s a common point when considered in terms of the horizon. So let’s get started on that.
Here, we’ve rotated things a bit. Instead of the horizon being at the steep angle before, I’ve placed it horizontally which means the celestial equator now has a steep angle. I’ve omitted the ecliptic except for the arcs in which we’re interested: $arc \; HZ$ and $arc \; H\Theta$.
We’re going to add a few more things to this, but let’s take a moment to notice a few things. First:
$arc \; E\Theta$ rises with $arc \; H\Theta$, and $arc \; EZ$ with $arc \; ZH$.
Let’s break that down a bit by reminding ourselves what these arcs are. $Arc \; E\Theta$ and $arc \; EZ$ are arcs along the celestial equator, while $arc \; HZ$ and $arc \; H\Theta$ are arcs of the ecliptic. And not just any arcs, but things worked out such that there is a further characteristic of these arcs. To my mind, it falls out of the setup, but Ptolemy actually took care to define it as he set up the problem initially saying these are
two arcs of the ecliptic, equal and equidistant from the winter solstice.
It may not look it in the last picture, but if you look back at the picture where I’d drawn the two arcs in blue, it becomes much more apparent5.
Jumping back to the statement of the arcs rising together, this again comes in my mind, directly from how they’re set up. Consider first those arcs in $\triangle EHZ$. Point Z hits the horizon first and is common to both lines. Then, by definition, since both points H and E must be on the horizon at the same time, the amount of time each took to rise must have been the same. The inverse is true for $arc \; E\Theta$ and $arc \; H\Theta$.
The next point Ptolemy wants to make is that:
the whole arc $\Theta EZ$ is equal to the sum of the rising-times of $arc \; HZ$ and $arc \; H\Theta$ at sphaera recta.
This one, again, seems fairly obvious from what we have just stated above, but Ptolemy takes a slightly roundabout way of demonstrating it. Specifically, he draws in another arc of a great circle:
This arc extends from the south celestial pole to the celestial equator, which is the same thing that the horizon at sphaera recta (on the equator) does. Thus, $arc \; KL$ represents a section of the horizon at sphaera recta.
Taking stock, we can see that this doesn’t alter $arc \; HZ$ or $arc \; H\Theta$; both of those still extend from the point on the (new) horizon to their respective equinoxes. However, this does create new divisions of the arc on the celestial equator. Namely, we now have $arc \; LZ$ and $arc \; L\Theta$.
However, the same sort of argument that we just discussed above still holds here. Now $arc \; HZ$ rises with $arc \; LZ$ and $arc \; H\Theta$ rises with arc $arc \; L\Theta$.
Thus the sum of the $arcs \; (L\Theta + LZ)$ equals the sum of the $arcs \; (E\Theta + EZ)$, and both are comprised in the $arc \; Z\Theta$.
In other words, since the arcs rise together at sphaera recta, and the sums are the same, they must also rise together at all latitudes.
This post was rather muddy, so before closing out, I do want to take one moment to really recap what it was all about. From the way it was presented here, the overall feeling is that we’ve essentially demonstrated that some arc of the ecliptic rises with some arc of the equator. That’s a bit of a duh factor.
But what probably got lost in all the shuffle is the details of which arcs rise together. So to restate the original intent of this proof, here’s what Ptolemy stated:
If two arcs of the ecliptic are equal and are equidistant from the same solstice, the sum of the two arcs of the equator which rise with them is equal to the sum of the rising-times [of the same two arcs of the ecliptic] at sphaera recta.
That’s a bit hard to parse, but we’ll be doing an example that will make use of this in the next post so stay tuned!
- Looking back, this should have been obvious as Ptolemy did the same thing in the last post.
- Extra emphasis that this is the rising point of the ecliptic, not the nearest equinox. I made this mistake a lot which may have contributed to why things weren’t working out in my head for so long.
- It’s not immediately obvious that this is the rising point as opposed to the setting point, but consider that plane of the horizon. North is towards the top, south at the bottom, so east must the be towards us.
- In this most recent diagram, the H which is part of $arc \; H\Theta$ isn’t on the horizon despite looking close. Rather, we’ve maintained the position of H in relation to Θ.
- The winter solstice is not marked but is the lowest point on the ecliptic so you can quite easily see how they are equidistant.