Now that we’ve done a bit of preliminary theory, we’re ready to jump into an example problem. As with before, Ptolemy selects Rhodes at a latitude of 36º. Since we’ve been seeing several of the same diagrams here, I’m going to forego the buildup and give a mostly completed diagram:
Here we again have the celestial sphere. While things look quite complicated this is actually no different than a setup we had in the last section. The only difference is that I’ve labeled a few more intersections. So let’s go through things.
First off, AEG is the celestial equator. The ecliptic is ZHΘ. Our horizon here is BED, drawn for Rhodes. Lastly, I’ve drawn this such that the meridian for this horizon is the perimeter, ABGD. Point H is the vernal equinox and is above the horizon. Point K is the north celestial pole.
We will take $arc \; HL$ to be the sign of Aries1, which is to say, the arc of the ecliptic that passes through the sign of Aries. Functionally, this means that $arc \; HL = 30$ as Ptolemy divides the ecliptic into 12 equal parts for the zodiac as noted a few posts ago.
The ultimate goal here is:
given $arc \; HL$, find the arc of the equator which rises with it, which is to say $arc \; EH$.
To complete our diagram, we’ll need to sketch in one more line extending from the north celestial pole to the celestial equator, passing through L, the point the ecliptic crosses the horizon2.
Now let’s highlight a few segments. Specifically, we have some great circle arcs, $arc \; ED$ and $arc \; KM$ which are being bounded by two other great circle arcs, $arc \;EG$ and $arc \; GK$ as shown below.
It doesn’t much look like it because it’s so mushed up, but that’s actually a Menelaus configuration. Let’s pop it out so we can see what we’re working with more clearly:
Here’s the Menelaus configuration pulled off the sphere, and flipped left to right. This gets it in the proper configuration to state:
$$\frac{Crd \; arc \; 2DK} {Crd \; arc \; 2DG} = \frac{Crd \; arc \; 2KL}{Crd \; arc \; 2LM} \cdot \frac{Crd \; arc \; 2EM}{Crd \; arc \; 2EG}$$
As usual, we need to take stock of what we know so we can solve for something new.
To start, $arc \; DK$ is the angle of the north celestial pole above the horizon. This is the same as the latitude of Rhodes which means $arc \; 2DK = 72º$ and thus $Crd \; arc \; 2DK = 70;32,3$3.
Next, $arc \; GK = 90º$ because it’s the arc from a pole to the celestial equator. We just stated $arc \; DK = 36º$ so we can subtract that out to get $arc \; DG = 54º$. This means $arc \; 2DG = 108º$ and $Crd \; arc \; 2DG = 97;4,55$4.
Let’s next look at $arc \;LM$. This is the arc between the ecliptic and the equator. Back in book 1, we had an entire post looking at how to determine the length of such arcs5 which was a prelude to an entire table for these values6. While the points are labeled differently, the example done there is actually the same because the arc along the ecliptic is 30º there as well. So reusing that result, we can state that $arc \; LM = 11;39,59º$, $arc \; 2LM = 23;19,59º$, and thus $Crd \; arc \; 2LM = 24;15,57$.
This gives us $arc \; KL$ because $arc \; KM$ is from a pole to the equator, meaning it’s 90º. Thus, subtracting out $arc \; LM$ we can determine $arc \; KL = 78;20,1º$, $arc \; 2KL = 156;40,1º$, and $Crd \; arc \; 2KL = 117;31,15$.
Lastly, we can state that $arc \; EG = 90$. It certainly doesn’t look it as drawn due to the perspective, but let’s again remind ourselves what these points are: E is the intersection of the horizon and the celestial equator which is either due east or west (east in this case). Meanwhile, G is on the meridian, Since the meridian is always 90º in altitude from a point due east or west, we can state $arc \; EG = 90º$, $arc \; 2EG = 180º$, and $Crd \; arc \; EG = 120$.
Plugging everything in:
$$\frac{70;32,3} {97;4,55} = \frac{117;31,15}{24;15,57} \cdot \frac{Crd \; arc \; 2EM}{120}$$
Solving the equation for our unknown and going through the math:
$${Crd \; arc \; 2EM} = {120} \cdot \frac{70;32,3} {97;4,55} \cdot \frac{24;15,57}{117;31,15}$$
$${Crd \; arc \; 2EM} = 18;0,7$$
Looking that up in our chord table we get that $arc \; 2EM = 17;16º$ and $arc \; EM = 8;38º$.
We’re not quite there yet. Recall, we’re after $arc \; EH$. We could get at it if we subtracted what we just found ($arc \; EM$) from $arc \; HM$. So do we know $arc \; HM$? First off, let’s look at exactly what that arc is: It’s the section of the celestial equator directly adjacent to the equinox that rises with 30º of the ecliptic.
Let’s digress for a moment and consider arc KM. This arc is from the north celestial pole to the celestial equator making them perpendicular. This only happens at sphaera recta, so we can consider arc KM to be a segment of the horizon at sphaera recta. At that location, point H would first rise on the horizon, and then points L and M would be on the horizon simultaneously. This means, at sphaera recta, the rising time for arc HL = arc HM.
This is something we found at the end of book 1 as well. The rising time of the first arc of the ecliptic is 30º is 27;50º7.
Thus, $arc \; EH = 27;50º – 8;38º = 19;12º$.
Now to put that to a bit of use. A couple Almagest posts ago, 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 found the rising time for Aries. This butts right up against the spring equinox, which means that the sign on the opposite side of that equinox, Pisces rises in the same amount of degrees.
Meanwhile, in the last Almagest post, we proved that
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 was a mouthful, but now we can try to make more sense of it since we have an example to work with. To help visualize things, let’s again refer to this chart:
Let’s start with Aries. Aries is equally distant from the solstices as Virgo. Thus, the amount of time it would take the sections of the equator rising with Aries + Virgo to rise at Rhodes is the same amount of time it would take for the arcs of the ecliptic of Aries + Virgo to rise at sphaera recta.
Above, we found the length of the equator rising with Aries, which means if we know the length of the ecliptic for Aries + Virgo, we could subtract off the equator from that sum to get the rising time for Virgo. And as a bonus, we’d also get the rising time for Libra, since Libra is opposite an equinox from Virgo.
Fortunately, we do know the rising time for both of those sections of the ecliptic. They are listed in the diagram above but you can refer back to this post if you need a refresher on where those came from. Both are 27;50º which means the sum of their rising times is 55;40º. Subtracting off the 19;12º for the rising time of the equator rising with Aries at Rhodes, we determine that the rising time of the equator rising with Virgo is 36;28º (as is Libra).
So to quickly recap what we’ve done in the past several posts:
1) We proved that arcs of the ecliptic equidistant from the same equinox rise in the same amount of time
2) We proved that the sum of arcs of the equator equidistant from solstices equal the sum of the complimentary arcs of the ecliptic at sphaera recta.
3) We did an example at Rhodes using this for Aries that also netted us Pisces, Virgo, and Libra.
But that still leaves out a lot of signs. And we’ve only yet done this for a specific latitude. So what we’ll be working towards is repeating this for all of the signs, at many different latitudes and we’ll organize it all in a nice table, which I’d hinted at at the very end of this post in book 1.
- While this works perfectly well for when Ptolemy was writing, things aren’t quite the same today to the precession of the equinoxes. Now, the vernal equinox is no longer at the border between Aries and Pisces. It has drifted almost all the way through Pisces and is almost in Aquarius. Astrological signs are based to this day based on the position of the vernal equinox from ~2,000 years ago and not an accurate representation of the actual position of the sun on the date of your birth. Yet another reason that astrology makes no sense.
- Here’s where my limited drawing skills are getting me into a bit of trouble as everything’s squished up on the right side of the diagram. Ideally, I’d like to rotate this so point E is facing us, but that’s trickier to draw for me and I can visualize things perfectly well on this diagram. If it’s too cluttered for you, here is the image as presented in the text. Do note that they’ve rotated things so the horizon is horizontal. I dislike this image as it makes the ecliptic look as if it is not part of a great circle, hence why I have created a diagram from scratch.
- The translator’s footnote here states that the Greek and Arabic translations of the Almagest have this value slightly incorrect as 70;32,4 which is not what is listed in the chord table. Thus, the translator suggests that there may have been an earlier version of the chord table.
- The translation here too has a discrepancy, listing the value as 97;4;56, but does not give any explanation.
- Thanks to Padraig for helping me remember that we’d done this. I was completely stumped as to where Ptolemy had pulled this number from.
- Which we will be making more use of in the next few posts.
- In that section Ptolemy called it $arc \; E\Theta$.
