Let us take as a general basis for our examples the parallel circle to the equator through Rhodes, where the elevation of the pole is 36º, and the longest day $14 \frac{1}{2}$ equinoctial hours.
Immediately starting the second chapter, we’re given a lot to unpack. First off, Ptolemy chooses to work through this problem by means of an example, selecting Rhodes, a city in Greece as the exemplar. I’m assuming that the “elevation of the pole” is the latitude, as Rhodes’ latitude is 36.2º. But what of these equinoctial hours?
Fortunately, these are just hours as we think of them. However, Ptolemy has made a distinction here because, until the 14th century, the more common method of measuring time was something often referred to as “unequal hours”.
In both cases, there are still 24 hours in a day1. But in the unequal hours system the length of each hour was unequal such that there were always 12 hours of daylight and 12 hours of sunlight2. These would be equal on the equinoxes, but the night hours would be shorter than the day hours in summer, and longer in the summer.
But, as noted above, we’re working with hours in the common sense of $\frac{1}{24}$ of a solar day. Exactly how Ptolemy determined that the longest day in Rhodes is $14 \frac{1}{2}$ isn’t specified, but I have a feeling that it was through direct observation, although this raises the question of how well it was measured as I’m not terribly familiar with instruments for measuring time in antiquity.
So with that out of the way, back to the diagrams! To start, here’s what the book presented.
It’s the same diagram we’ve seen a few times now. But previously, these lines have represented the celestial equator and the ecliptic. But Ptolemy wants to change things up this time, so I think it’s good to slow down and work through things again. Especially since this time, we’re going to introduce something we haven’t really run into before in one of our diagrams: The horizon.
This sounds funny at first because the horizon is on the Earth and not the celestial sphere. But as with the equator, we look at its projection on the celestial sphere. The trouble is that it’s different depending on where you’re observing from. So I’ll walk through my method in visualizing it.
First, let’s just start off looking at Rhodes on the globe:
Here’s a shot from Google Earth which I’ve done a bit of editing to. You’ll notice that I’ve rotated things so Rhodes is way off to the left side. The reason for this is that it will help align with how things are drawn in the above diagram. So let’s drop this into the center of a sketched out celestial sphere.
There we go. I’ve extended the Earth’s equator out to show the celestial equator.
The first thing to note here, is that this is obviously not to scale. The Earth is infinitely small compared to the celestial sphere, but drawing it enlarged helps visualize some important points that I’ve drawn in. The first one, extending from the center to the upper left, is the line pointing to the zenith for an observer in Rhodes. The long line is perpendicular to that, which means it’s in the plane of the horizon. However, it’s just a line in the plane. So let’s draw out the plane and make it a bit easier to see.
All in all, it doesn’t look much different from previous sketches I’ve done of the celestial equator and ecliptic, which is why we can essentially use the same diagram despite the lines representing different things. The angle that the horizon is tilted compared to the celestial equator is different, but all in all, the relative positioning of all the points will remain the same.
From this diagram now, we can see cardinal directions on the plane of the horizon: North is the upper right. South the lower left. East is the intersection of the celestial equator and the horizon facing us while west is the intersection on the opposite side where the two dashed lines meet. This is important because it means that, as I’ve drawn it, the meridian (connecting north to south along the celestial sphere) is the edge of the circle. And sure enough, the zenith is on it. And that’s the real reason I drew the diagram this way: It puts things in the same position as in the original diagram.
But before moving on, let’s let the Earth rotate 1/4 of the way around (6 hours) so we can visualize how things move.
Here, I’ve kept the point of the zenith as well as the line extending to it. I’ve left out the others that connect to the north and south points as this is drawn sufficiently face on it would all look like one single line. But since we’re now looking down on this, this reads like a normal map: North is the top of the dashed line of the horizon. South the bottom. East at right and west to the left.
I hope this digression has helped because this really is a very tricky thing to wrap your head around. Especially because we haven’t even worried about the ecliptic. It’s quite difficult to visualize that and the equator holding still while the plane of the horizon spin under them once a day.
One of the things that really helped me understand this was a celestial sphere model like the one pictured below.
Unfortunately, this one isn’t produced anymore which is a shame because this is the best I’ve ever seen3. The reason is that it includes the horizon ring, the blue ring around the Earth in the center. This can be tilted for whatever latitude is necessary as well as rotated for longitudes. Right now, the one in the picture is nearly horizontal, which means it’s set for an observer near the north pole.
Anyway, that’s enough of a digression. Back to the original diagram we drew and let’s start putting some points on it.
Let’s quickly review: AEG is the celestial equator and the circumference of the circle (ABGD) is the meridian. BED is the horizon for Rhodes and Z is the south celestial pole.
Remember when I said that trying to include the ecliptic really confused things? Well, we’re going to do it briefly.
Here, I’ve added the Sun, located appropriately for the winter solstice which is how Ptolemy set up this problem. Right now, this is drawn such that the Sun is below the horizon. But now let’s get rid of the ecliptic. I just sketched it in so we could see where it is. Instead, let’s look at the path the Sun will take in a single day. During that time, it hardly moves along the ecliptic, so we can get a very close approximation by pretending it traces a small circle as shown below:
You’ll note that I’ve moved the Sun around its path a bit4 such that it’s now on the horizon. In the previous diagram, the Sun was still below the horizon, but now, it’s on the horizon. And if we recall, E is the east side of the horizon, which means it’s rising. We’ll call that point H.
Before moving on, let’s take a moment to really think about what this is saying. We see the small circle that will be the path of the Sun for the day. It rises in the east, peaks towards the south, and sets on the opposite side in the west. Just as it should. It’s also important to note that the arc of the path for that day that’s above the horizon is quite short. This is why days in the winter are short. It’s simply the portion of the Sun’s path that is above the horizon is so short. Were the Sun’s position at the summer solstice, it’d be flipped across the celestial equator and the portion above the horizon would be much longer.
Now, let’s draw in an arc. Specifically, this will be an arc of a great circle, starting from the south celestial pole, through H, until it hits the celestial equator. We’ll call it Θ.
That completes our diagram. So what are we after? As the title suggests, Ptolemy wants to find the distance along the horizon between the celestial equator and the ecliptic. Point E is also the intersection between the celestial equator and the horizon and H is a point on the horizon that’s also on the ecliptic, so we’re looking for $arc \; EH$.
I’ll go ahead and drop the Sun’s path since we only needed it to see where the Sun was rising. Instead, I’ll highlight something important.
That’s a Menelaus configuration which means we’re going to be invoking one of Menelaus’ theorems here because, as Ptolemy put it:
the two great circle arcs EB and ZΘ have been drawn to meet the two great circle arcs AE and AZ, and intersect each other at H.
In other words, we’ve got 4 great circle arcs intersecting. As I noted in this post, whenever you have three (or more), we should start thinking Menelaus’ theorems. Specifically, we’re invoking Menelaus’ Theorem I since the other does not have what we’re looking for in it. In addition, that form would also involve having to know $arc \;BH$, which we don’t. However, Menelaus’ Theorem I allows us to solve for $arc \;BH$ so that’s the one we’re going to go for.
So we can state the following:
$$ \frac{Crd \; arc \; 2 A \Theta}{Crd \; arc \; 2AE} = \frac{Crd \; arc \; 2 \Theta Z}{Crd \; arc \; 2ZH} \cdot \frac{Crd \; arc \; 2HB}{Crd \; arc \; 2BE}$$
Again, we’re after $arc \; EH$. But that’s not in there. So we’ll first find $arc \; BE$ and $arc \; BH$ since the difference of the two is what we’re after. As always, to do that means we’ll need to know all the other variables in the equation.
Let’s tackle the AΘ bit first as it’s the hardest5. In fact, to get there, we’ll first need to find EΘ and subtract that from EA.
But let’s take a moment to return to the path the Sun will be tracing along the sky. I know I just took it out, but Ptolemy’s done something funny, and it takes a bit to really understand.
Here, what I’ve done is moved the Sun a few more hours. What’s important to notice, is that as H moves to H2, Θ moves to Θ2. This may seem rather trivial, but what it really means is that H’s motion can be tracked by Θ’s. And the reason we want to do that is because Θ is the point on the celestial equator and part of that AΘ that we’re so curious about6.
To state this another way, Θ (which is the projection of H onto the celestial equator), will rotate around with H, but it always stays on the celestial equator.
Or, as Ptolemy put it:
Thus the time from rising of H to its upper culmination is given by the equatorial arc ΘA, and the time from its lower culmination to its rising is given by GΘ.
In other words, the amount of time it would take H to move from the horizon to the meridian (which you’ll recall the circumference of the circle of the great sphere is the meridian), is the same as the time it would take Θ to move to A. Since H to the meridian is half of the length of the day so too is $arc \; \Theta A$. Thus, the path of the Sun for the entire day is $arc \; 2 \Theta A$.
Now, let’s recall that H rises at E (which is directly east on the horizon) on the equinox. So $arc \; E \Theta$ is half the difference between the length of the day on the equinox (12 hours) and the length of the winter solstice. Half because there’s another $arc \; E \Theta$ around the other side.
And you may notice Ptolemy did something else funny here. We’ve worked with this in terms of the winter solstice. Yet we were given the length of the longest day which is the summer solstice. However, the length of the winter solstice is just 24 hours minus the length of the longest day. So the absolute value comes out the same either way.
So we take 14.5 hours – 12 hours (since it’s the equinox) to get 2.5 hours which is $arc \; 2 E \Theta$. Dividing that in half, $arc \; E\Theta$ is 1.25 hours if we measure it in hours. Converting that to degrees is $arc \; E\Theta = 18;45º$ 7.
Whew. Now we can note that $arc \; EA = 90º$ (because it’s from the equinox at E to the point of a solstice projected onto another great circle) so subtracting out EΘ we get that $arc \; \Theta A = 71;15º$. So $arc \; 2 \Theta A = 142;30º$8. Getting the corresponding chord from our chord table to get $Crd \; arc \; 2 \Theta A = 113;37,54$.
Thankfully the rest from here are pretty straightforward.
Next, since $arc \; AE = 90$º, $arc \; 2AE = 180$ and its corresponding chord is 120.
Similarly, $arc \; \Theta Z$ is from the celestial equator to the north celestial pole which is 90º so twice that is 180º and again its correspond chord is 120.
Getting $arc \; ZH$ took me a bit of thinking. Ultimately, the argument for it has already been laid out here. However, if you follow that link, it’s talking about $arc \; ZB$ and not $arc \; ZH$. But the same reasoning applies: Consider first $arc \; Z \Theta$. This is from the pole to the celestial equator which is 90º. So $arc \; ZH = arc \; Z \Theta – arc H \Theta$. Since H is a point on the ecliptic, specifically a solstice, we know that it must be the obliquity of the ecliptic. Since Menelaus’ theorem always uses 2 times that arclength, Ptolemy often gives the value in terms of 2 times the obliquity of the ecliptic as $\frac{11}{83}$. So subtracting that out and then doubling $arc \; ZH$ we get $arc \; 2ZH = 132;17,20$ and the corresponding chord length as 109;44,53
Lastly, $arc \; BE$ is again 90º as it’s south to east on the horizon. Thus, $arc \; 2BE = 180$º, and the corresponding chord, $Crd \; arc \; 2BE = 120$.
So with all of this in hand, we can plug everything into our equation and solve for 2HB:
$$ \frac{113;37,54}{120} = \frac{120}{109;44,53} \cdot \frac{Crd \; arc \; 2HB}{120}$$
Let’s go ahead and simplify that a bit right now by cancelling out those 120’s on the right side and combining.
$$ \frac{113;37,54}{120} = \frac{Crd \; arc \; 2HB}{109;44,53}$$
Now we’ll move the denominator to solve for our variable:
$$Crd \; arc \; 2HB = \frac{113;37,54 \cdot 109;44,53}{120} $$
$$Crd \; arc \; 2HB = 103;55,26$$
But again, this is the Crd arc, so popping back to our chord table and getting the actual arc, we find that $arc \; 2HB \approx 120$º so $arc \; HB \approx 60$º.
Subtracting that from $arc \; BE$ which was 90º, we get $arc \; EH = 30$º.
NOTE: This is an updated version of the original post. Much of the original post remained intact. However, the original used the original diagram from the book the entire time. I felt that being able to see things upright, on the celestial sphere made things much easier and as such, I have replaced all the pictures and edited the text to match.
- Careful readers of this blog will have noted that we shouldn’t even take for granted the definition of a day as there’s a difference between a solar day and a sidereal day.
- Obviously this wouldn’t work well at extreme latitudes near the arctic circles.
- I’d love to get my hands on one as it would help immensely for myself as well as trying to teach this to others. If anyone ever finds one for sale at a reasonable price, please let me know!
- I’ve also obviously shrunk the Sun so we can see what’s going on.
- I apologize in advance if this post is hard to follow. I originally wrote it directly following Ptolemy’s work which didn’t state we’d be using Menelaus’ configuration until after finding AΘ. But once I finished the post, I felt it would be much better to preview where we were going with all this math (as it’s one of the longer posts I’ve written), so the structure got significantly changed. I tried to make sure the logic flowed the way I intended, but this is such a long post, I may have missed something.
- The angle they sweep out at Z is equal. However, their actual path lengths are going to be different because the circle H, the Sun is on, is a small circle, and the rule about the length of an arc being equal to its central angle only applies for great circles.
- This is related to a value we’ll be seeing a lot. Specifically, this will be part of several Menelaus configurations down the road, so we might as well note right now that $arc \; 2E\Theta = 37;30$º and thus $Crd \; arc \; 2E\Theta = 38;34,22$.
- I feel like we could also have arrived at this by noting that 2ΘA is the length of the shortest day at this latitude which is 24 – 14.5 = 9.5 hours or 142;30º.