Before going on the several post detour in developing several theorems and making extra sure we understood a few, we stated the goal was to determine the length of an arc at various points between the celestial equator and ecliptic1.
Fortunately, after all the complexity of the last post’s spherical geometry, this will remain relatively simple as the above picture already has pretty much everything we need to approach the problem. However, Ptolemy’s diagram is at a slightly different angle so I’ll reproduce it for the sake of consistency.
Here, we’ve rotated things so the ecliptic ($arc \; BED$) and equator ($arc \; AEG$) are the two lines running roughly vertical. The $arc \; H \Theta$ is what we’re after and we’ve extended it up to $Z$ which is at a pole. $E$ here is the point of the spring equinox.
Translating that back to my drawing, it looks like this:
Here, we’ve just rotated the chord we’re looking for around to the same side as the vernal equinox. Which is fine, because what Ptolemy is after is all of the chords between two ecliptic and celestial equator. So might as well choose that one instead of the one I’d originally sketched in.
So to get us started, Ptolemy begins with this chord being 30º around the ecliptic. Since Ptolemy breaks the circumference of his spheres into 360 parts, that means $arc \; EH$ is 30 parts.
Notice we have three arcs of great circles now. As I’ve discussed previously, this should be a prime indication that we need to apply one of Menelaus’ theorems. Ptolemy invokes theorem 13.6 to state:
$$\frac{Crd \; arc \; 2ZA}{Crd \; arc \; 2AB} = \frac{Crd \; arc \; 2 \Theta Z}{Crd \; arc \; 2 \Theta H} \cdot \frac{Crd \; arc \; 2HE}{Crd \; arc \; 2EB}$$
We’ll come back to that in a moment once we have a few more thing to plug into it.
Before diving in, recall that we’re after $arc \; H \Theta$. Since we only have one equation, that means we need to determine the other five pieces of the equation ($Crd \; arc \; 2ZA$, $Crd \; arc \; 2AB$, $Crd \; arc \; 2 \Theta Z$, $Crd \; arc \; 2HE$, and $Crd \; arc \; 2EB$) to solve for it. So let’s get started.
First, we’ll notice that $arc \; ZA$ is from the pole to the celestial equator. This is by definition 90º which means that $arc \; 2ZA = 180$º. We can look up the corresponding chord length for that to get that $Crd \; arc \; 2ZA = 120$ parts.
Similarly, point $E$, being an equinox, is 90º along the celestial equator from the solstices meaning $arc \; EB = 90$º and thus $arc \; 2EB = 180$º which means $Crd \; arc \; 2EB = 120$ parts.
That same reasoning applies for $arc \; Z \Theta$: Z is a pole which by definition is 90º away from the celestial equator on which $\Theta$ lies, so $arc \; Z \Theta$ is again 90º making $arc \; 2Z \Theta = 180$º and thus $Crd \; arc \; 2Z \Theta = 120$ parts. This is coming along quite nicely.
Next up, let’s notice that $arc AB$ is the arc subtended by $\angle AEB$ and $\angle AEB$ is the angle between the ecliptic and celestial equator. That’s a figure we discussed earlier in this book. Since the value is really $2AB$, Ptolemy uses the approximation $\frac{11}{83}$ which is 47.71º or 47;42,36 in sexagesimal (implying AB = 23.85º which, as noted previously, was a bit high from today’s value of 23.43º). So again, we can look up the chord which corresponds to this angle to get $Crd \; arc \; 2AB = 48;31,55$ parts.
Next up, the distance around from the equinox we’re using for this particular calculation: $arc \; HE$ which Ptolemy decided to do as 30º first, which means $arc \; 2HE = 60$º and its corresponding chord would then be $60$ parts.
Finally, with all that in hand we can plug into our equation and solve for $arc \; H \Theta$:
$$\frac{120}{48;31,55} = \frac{120}{Crd \; arc \; 2 \Theta H} \cdot \frac{60}{120}$$
Let’s do a bit of algebra to solve for what we’re after:
$$Crd \; arc \; 2 \Theta H = 48;31,55 \cdot \frac{120}{120} \cdot \frac{60}{120}$$
I structured this in a particular way because it makes it extra obvious that the $\frac{120}{120}$ cancels to be 1, and $\frac{60}{120} = \frac{1}{2}$ making the whole thing reduce to:
$$Crd \; arc \; 2 \Theta H = 48;31,55 \cdot \frac{1}{2}$$
Finishing up the math here:
$$Crd \; arc \; 2 \Theta H = 24;15,57 parts$$
But that’s a chord length and we’re looking for the length of the arc that subtends it. So back to the chord table to look up what arc (or angle) corresponds. This is the first time we’ve gone backwards like this, so I’m going to go through it slowly as an example. First, we need to place some bounds on it. We see that 24;15,57 falls between $23$º and $23 \frac{1}{2}$º.
So we’ll start with 23º.
The third column is the sixtieths column, so each time we add that figure to the second column, we increase the first column by 0;1. We could sit here and add it on one at a time until we get as close as possible to the figure we’re looking for, but we should know better math than that. So let’s subtract the chord length for 23º from our chord length to get the difference:
$$24;15,57 – 23;55,27 = 0;20,30$$
Now divide that result by the increment to determine how many times we would have to add it on2:
$$0;20,30 \div 0;1,1,33 = 19;59$$
So that’s the number of sixtieths (or arcseconds) we need to add to 23º. Thus, the actual value is $arc \; 2 H \Theta = 23;19,59$º.
But that’s still twice the arc. So dividing by 2 we get $arc \; H \Theta = 11;39,59$º.
So that’s one value of the arc length. As with the chord tables, Ptolemy repeats this math for every angle, this time in 1º increments from 0º to 90º.
I’m not going to repeat the math, but let’s explore it a bit because we don’t need to really repeat everything we’ve walked through in this post. The reason is that our big equation needed five variables to solve for what we were after. However, only one of those variables ($Crd \; arc \; 2HE$) ever changes and even then, that was a very easy value to look up in the chord tables.
So for each angle, we would double it, look up the corresponding chord length, drop it into the equation, plug and chug, then reverse lookup and divide to get the answer. It’s still quite a bit of work, but at least we don’t have to do five variables every time.
So why did we do all this? Ptolemy states that the goal is to:
find how long, in equinoctial time-degrees, it takes a given section of the ecliptic to cross the meridian at any point on earth and the horizon at sphaera recta (for only in that situation does the horizon pass through the poles of the equator).
First off, let’s start with what “sphaera recta” is. To me, this is easiest to understand by comparing it with other forms:
Each of these diagrams is meant to represent our view of half of the celestial sphere. I think it’s easiest if we think of it as facing east. Then the zenith (the point straight up from you) is at the top of the diagram). The light portion of the diagram is the sky, and the dark portion is the ground.
In sphaera recta3, the celestial equator (the middle of the 3 lines) rises perpendicular to the horizon, passing through the zenith. The poles lie on the horizon. This is how the celestial sphere appears when you are on the actual equator.
In sphaera obliqua, things are slightly different. The celestial equator rises at an angle and the pole is somewhere in the sky between the horizon and zenith4.
In sphaera parallela the celestial equator is parallel to your horizon and the pole is at the zenith. This happens when you are at the Earth’s poles.
So why does this matter? Consider the red and green dots. In sphaera recta, as the celestial sphere seems to turn, they will rise at the same time. However, in sphaera obliqua, two objects rising at the same time are not at the same angle from the equinox5.
So Ptolemy’s goal with all of this is to begin to be able to determine actual times objects will rise and set, starting with the simple case of sphaera recta.
- If you need a quick refresher on what these are, check this post.
- I’ll be completely honest. I don’t do the math in sexagesimal. I convert to base ten decimal and then convert back.
- Literally, “the upright sphere”
- In actuality, it corresponds to your latitude.
- In other words, they have different right-ascension. I touched on RA in this post, but we’ll discuss that more shortly.