Now that we’ve gotten a few symmetry rules developed, we can return to the main objective of calculating the angle between the ecliptic and meridian at different points along the ecliptic. Specifically, Ptolemy sets out to do this at the first point in every sign. But thanks to the previously derived symmetries, we’ll save ourselves a bit of work.
First Ptolemy does some very short proofs for these angles at the meridian and solstice, and then a slightly more complex one for the signs between them.
Before beginning, Ptolemy has a new spherical geometry trick he’s going to be using quite a bit here. Specifically, we’ll be finding the angle made by the intersection of two great circles by looking at the arc it subtends. I’ve spent a fair amount of time trying to research this and thinking about it, but I haven’t really gotten a good determination of when this can be done. It’s not a rule that I can find anywhere in sources on spherical geometry, so I suspect it arises from the specific way that Ptolemy defines his spheres or the specifics of the problem.
There are a few clues that I haven’t been able to figure out. Particularly, in the problem set up, Ptolemy states that the additional arcs we’re drawing in have a pole at the angle we’re concerned with and “radius the side of the [inscribed] square.” I haven’t been able to make heads or tails of this statement.
Another is that the translation states that the first of the proofs we’ll go through is “derivable from Theodosius Sphaerica II 9.” As we’ll see, about the only step that really happens in that proof is equating the angle to the arc, I imagine this must be what it’s referring to, but there doesn’t seem to be any translations of Sphaerica that I’ve been able to track down.
UPDATE (8/27/2022): Big thanks to reader Eric Piphany who found a document that contains many of the theorems from Sphaerica. In particular, the one to which Ptolemy is referring to here states:
In a sphere, if two circles cut off arcs of each other and a great circle is drawn through their poles, then it bisects the arcs cut off.
In this case, it doesn’t seem that Ptolemy’s statement is “derivable” from this one, but rather, follows directly from it.
My best guess is that it only works when the auxiliary great circle we’ll be drawing in for each of these, has the vertex of the angle as one of its poles as that seems to be done in all of the derivations. Regardless, we’ll just have to take Ptolemy’s word for when this works for now.
Also, a quick note before we get started, each time we do another derivation, Ptolemy changes what symbols are representing, so pay careful attention.
Solstices
So to get started, first Ptolemy wants to look at the angle between the ecliptic and meridian at the solstices. As such, we’ll need a diagram with a meridian (ABGD) and the ecliptic (AEG) such that A is the winter solstice. He then draws in BED such that it’s perpendicular to both making A and G its poles as well as AEG having poles B and D.
This immediately completes the problem because the angle we’re looking for, is $\sphericalangle DAE$, which subtends $arc \; DE$. and since E is on AEG and D is its pole, it is, by definition, 90º as is the angle.
Since we said that point A was the winter solstice, this means that it’s the first point in Capricorn. And by the rule that we derived in the last post this, plus the first point in Cancer (which is the summer solstice) must add up to 180º. Therefore, the angle at the first point in Cancer (at the summer solstice) is also 90º.
Equinoxes
Next, we’ll go for the equinoxes. To define the equinox, we’ll need the celestial equator (AEG) and ecliptic (AZG) in addition the the meridian (ABGD). Now, we’ll position the now autumnal equinox on the meridian and draw in arc BZED such that A is one of its poles. This time, our objective will be to find $\sphericalangle DAZ$ which subtends $arc \; DZ$.
As with before, $ard \; DE = 90º$ so we just need to add on $arc \; EZ$. Similarly $arc \; AZ = 90º$ which means we can look up $arc \; EZ$ in our table of inclinations between the celestial equator and ecliptic for when the arc of the ecliptic is 90º. Thus, $arc \; EZ = 23;51,20º$ and $arc \; DZ = 113;51,20º$ which Ptolemy just rounds off to 113;51º. Thus, $\sphericalangle DAZ = 113;51º$.
Again, A was the autumnal equinox which is the first point in Libra. Thus, by our the proof in our last post, the vernal equinox, the first point in Aries, must be $180º – 113;51º = 66.9º$.
Intermediate Points
So now we’ve figured out four of the points around the ecliptic. Next we’ll need to get the ones between the solstices and equinoxes.
As with before, we’ll have our meridian circle as ABGD. We’ll include the celestial equator (AEG) and ecliptic (BZD) such that Z is the autumnal equinox1. And again we’ll have an extra great circle in there, HEK, that has B at its pole.
This time, the objective is to find $\sphericalangle KB \Theta$.
To start, we’ll do the first point in Virgo, which means that this point will need to be at B, therefore making $arc \; BZ = 30º$ as that’s the arc from the first point in Virgo to the autumnal equinox, Z (i.e., it’s the full sign of Virgo).
There’s also a few other things that fall out of this setup. First off, because B is a pole of HEK, this makes $arc \; BH$ and $arc \; B \Theta$ 90º2. Although it’s not explicitly stated in the problem set up, AEG has its pole on the meridian as well (at some point to the left of K). This is important because it means that point E is on two great circles (AEG and HEK) that both have their poles on another (the meridian) which can only happen if it’s the pole of that second. Thus, $arc \; EH = 90º$ is as well.
As so often happens, we’ve now got a Menelaus configuration. This allows us to state:
$$\frac{Crd \; arc \; 2BA}{Crd \; arc 2AH} = \frac{Crd \; arc \; 2BZ}{Crd \; arc \; 2Z \Theta} \cdot \frac{Crd \; arc \; 2E \Theta}{Crd \; arc 2EH}$$
As usual, let’s take stock of what we know.
Since we stated above $arc \; EH = 90º$, this means that $Crd \; arc \; 2EH = 120$.
Next up, let’s look at $arc \; BA$. This is the arc between the celestial equator and ecliptic when the arc of the ecliptic is 30º. For this, we can refer to our table for such values to determine $arc \; BA = 11;39,59$. From there we can state $Crd \; arc \; 2BA = 24;15,57$ which Ptolemy rounds off to 24;16 in this section. We can subtract $arc \; BA$ from $arc \; BH$ which we above said was 90º to determine that $arc \; AH = 78;20º$ and thus, $Crd \; arc \; 2AH = 117;31$.
Since $arc \; BZ$ is the whole of Virgo, or 30º, we can say $Crd \; arc \; 2BZ = 60$.
And because $arc \; BZ = 30º$ we can subtract that from $arc \; B\Theta$ which we again said above was 90º to determine $arc \; Z\Theta = 60º$ and therefore $Crd \; arc \; 2Z \Theta = 103;55,23$
That gives us everything we need to solve for $arc \; E\Theta$.
$$\frac{24;16}{117;31} = \frac{60}{ 103;55,23} \cdot \frac{Crd \; arc \; 2 E\Theta }{120}$$
Solving for $Crd \; arc \; 2 E\Theta$:
$$Crd \; arc \; 2 E\Theta = 120 \cdot \frac{24;16}{117;31} \cdot \frac{ 103;55,23}{60} $$
Multiplying that out we get $Crd \; arc \; 2 E\Theta = 42;55$3. Therefore $arc \; 2E \Theta = 41;55º$ and $arc \; E \ Theta = 20;57º$ which Ptolemy again just rounds off to 21º.
Now we should recall that what we’re really after is $arc \; K\Theta$ which is $arc \; E\Theta + arc \; KE$. And since $arc \; KE = 90º$ this means $arc \; K\Theta = \sphericalangle KB\Theta = 111º$.
This calculation was for the first point in Virgo, but we have two handy rules to apply.
The first was for points equidistant from the same equinox which produce the same angle. Since we did the first point in Virgo, this will be the same angle for the first point in Scorpio.
The next rule was that the angles made by two points equidistant from the same solstice add up to 180º. This means the first point in Taurus and the first point in Taurus is $180º – 111º = 69º$.
And again returning to the first theorem, the first point in Pisces must also form an angle of 69º.
Next, we’ll repeat this calculation for the first point in Leo so that $arc \; BZ = 60º$. This changes a few of our values.
First off, $arc \; BA$ is now the arc between the celestial equator and ecliptic when the arc of the ecliptic is 60º, so $arc \; BA = 20;30,9º$ and $Crd \; arc \; 2BA = 42;2$.
Next, $arc \; AH$ was $90º – arc \; BA$ so $90º – 20;30,9º = 69;29,51º$ and $Crd \; arc \; 2 AH = 112;24$.
$Arc \; BZ$ is now 60º giving $Crd \; arc \; 2BZ = 103;55,23$.
$Arc \; Z\Theta$ is 30º giving $Crd \; arc \; 2 Z\Theta = 60;0,0º$
and $arc \; EH$ remains 90º.
So plugging everything in and solving again:
$$Crd \; arc \; 2E \Theta = 120 \cdot \frac{42;2}{112;24} \cdot \frac{60}{103;55,23}$$
$$Crd \; arc \; 2E \Theta = 25;55$$
Therefore, $arc \; E \Theta = 12;28º$ which Ptolemy rounds off to 12;30º. We again add this to 90º to get the angle between the ecliptic and meridian in this case to be 102;30º.
Again, applying our rules, this also implies that the first point in Sagittarius is also 102;30º and that the first point in Gemini and Aquarius must be the supplement: 77;30º.
- If one considers the figure as drawn from outside the celestial sphere, then Z would be drawn as the vernal equinox. However, if we consider it being drawn from inside the celestial sphere, it is correct. Either way the math works just fine.
- As well as some others but these are the ones we’ll be using.
- Ptolemy somehow comes up with 42;58.