In the last post, we started in on the angle between the ecliptic and an altitude circle, but only in an abstract manner, relating various things, but haven’t actually looked at how this angle would be found. Which is rather important because Ptolemy is about to put together a huge table of distances of the zenith from the ecliptic for all sorts of signs and latitudes. But to do so, we’ll need to do a bit more development of these ideas. So here’s a new diagram to get us going.
Here, we have the horizon, BED. The meridian is ABGD, and the ecliptic ZEH. We’ll put in the zenith (A) and nadir (G) and connect them with an altitude circle, AEG1. Although it’s not important at this precise moment, I’ve drawn it such that AEG has E at the point where the ecliptic is just rising.
Now, Ptolemy lists off several things we can find in this diagram although not all of them are directly related to the topic of this chapter, but we’ll be using them shortly. First Ptolemy notes that we’re able to find $\sphericalangle DZE$ as this is the angle between the meridian and ecliptic which was the subject of II.10.
Next, we do get to something that is applicable to this chapter which, in this diagram would be $\sphericalangle AEZ$ or finding $arc \; AZ$ which define the angle between the ecliptic and the altitude circle, AEG. Fortunately, this is quite simple, but to do so, we’ll need to make use of the celestial equator which Ptolemy hasn’t drawn in yet. So I’ll go ahead and add it myself adding a point, $\Theta$, at the intersection of the ecliptic and meridian. Now $arc \; A \Theta$ is simply the angle between the celestial equator and the zenith, which is your latitude and $arc \; Z \Theta$ is the angle between the ecliptic and the celestial equator at that point, which what the topic of I.12. Hence we can add or subtract them as necessary to determine $arc \; AZ$.
Now, for when AEG is drawn such that E is the point where the ecliptic is rising, this means $arc \; AE = 90º$ since it’s from the zenith to a horizon and similarly, $\sphericalangle AED = 90º$. And since $\sphericalangle DEH$ is the angle between the horizon and ecliptic, which is something we determined in II.11, we could add that on to $\sphericalangle AED$ to determine $\sphericalangle AEH$.
So why have we done all this?
[I]f we compute, for each latitude, just the angles and arcs before [i.e. to the east of] the meridian and just for the signs from the beginning of Cancer to the beginning of Capricorn, we will simultaneously have found the angles and arcs for the same signs [Cancer to Capricorn] after the meridian too, and also the angles and arcs both before and after the meridian for the remaining signs.
As usual, we’ll use Rhodes for the example, starting with the beginning of Cancer is one equinoctial hour to the east of the meridian. Let’s start by setting up our figure.
In this diagram, ABGD is the meridian, BED the horizon, And ZHΘ the ecliptic such that H is the first point in Cancer. This means that Z is the point at upper culmination, and Θ is the point that is currently rising. We’ll also draw AHEG in as an altitude circle which means that A is the zenith which is the pole of the horizon. Our goal will be to solve for $arc \; AH$. Lastly, point T will be the intersection of the celestial equator and meridian2.
To find this, we’ll use a Menelaus configuration:
$$\frac{Crd \; arc \; 2ZB}{Crd \; arc \; 2BA} = \frac{Crd \; arc \; 2Z \Theta}{Crd \; arc \; 2 \Theta H} \cdot \frac{Crd \; arc \; 2HE}{Crd \; arc \; 2EA}$$
You’ll immediately notice that $arc \; AH$ isn’t in this setup, but because $arc \; AE = 90º$ we can find $arc \; EH$ and subtract it out to find $arc \; AH$. Let’s get started.
First, let’s fine $arc \; Z \Theta$. To do so, we’ll need to figure out exactly where Z and $\Theta$ are along the ecliptic. Unfortunately, we don’t have the information we really need to use either method described in this post, but we can still figure it out. To do so, let’s take a look at what this actually looks like in the sky.
Here, we can see that the beginning of Gemini has already past the meridian as it is to the west. But the first point in Cancer is to the east by an hour as defined in the setup meaning the point currently at upper culmination is somewhere in Gemini3. An hour is $\frac{1}{24}$ of 360º or 15º of the celestial equator, so our goal is to find what proportion of the ecliptic rose in that same amount of time.
For this, we turn to our rising times table. We’ll need to use the table at sphaera recta to start. There, the first point in Cancer (or the last point in Gemini) rises in the same amount of time as 90º of the celestial equator. But that point is still 15º away from where we’re concerned with, so we need the point $90º – 15º = 75º$ accumulated rising time (in the Acc column) and we’ll need to figure out how many degrees of the ecliptic (the 10º column) would rise in that time. So we’re kind of using the table backwards.
So looking at the sphaera recta table, 75º falls between the 10º and 20º divisions in Gemini. Our goal is to figure out exactly how far. We can determine how far past the 10º mark by subtracting the corresponding rising time (68;18º) from it to get that we’re 6;42 rising time degrees past the 10º. But the width of that 10º interval is 10;47 rising time degrees, so we need to take the proportion. So,
$$\frac{6;42º}{10;47º} = .621$$
which is to say that we’re 62.1% of the way through the 10º interval or 6.21º past the 10º mark which is 16.21º, or converting that to sexigesimal, 6;12º which is what Ptolemy came up with. That’s a 76;12º total segment of the ecliptic. So that’s Z. Now to go after Θ.
For this one, we just need to remember that the point on the horizon is 90º further (in rising time degrees) from what’s currently culminating. Since 75º was culminating, the rising point is 165º which we can look up in the tables again, but this time, using the table for Rhodes. This obviously puts the point somewhere in Virgo.
Doing the same steps as before, 165º is 9;15 rising time degrees past the 10º mark in Virgo. The division is 12;9 rising time degrees, so we’re proportionally 76.1% the way through the 10º interval for a total into Virgo of 17.61º which corresponds to 17;37º in sexagesimal which is a 167;37º segment of the ecliptic.
Subtracting these, we can determine that
$$arc \; Z \Theta = 167;37º – 76;12º = 91;25º$$
And the corresponding Crd arc of twice the angle:
$$Crd \; arc \; 2Z \Theta = 119;58$$
From this, we can quickly determine $arc \; H \Theta$ since H is the first point in cancer (90 rising time degrees). Therefore, we can subtract that from Θ to get
$$arc \; H \Theta = 167;37º – 90º = 77;37º$$
and
$$Crd \; arc \; 2H \;Theta = 117;12$$
Next, let’s turn our attention to $arc \; ZB$ which we’ll need to do some intermediate steps to solve. $arc \; AB = 90º$ because it’s from the horizon to the zenith, so if we can find $arc \; ZA$, we can subtract that out. To find that, we’ll need to look at point T. Again, this point is the intersection of the celestial equator with the meridian.
From there, we can state that $arc \; AT = 36º$ which is the latitude of Rhodes4.
But that’s too much. We only needed $arc \; ZA$ which means we’d need to subtract out $arc \; TZ$ which is the arc between the ecliptic and celestial equator. Fortunately, we have a whole table of those values to look it up on. We’ll recall that point Z is 76;12º along the ecliptic from the vernal equinox, so doing a bit of extrapolation, we can determine that $arc \; TZ = 23;7º$.
Subtracting
$$ arc \; AZ = arc \; AT – arc \; TZ = 36º – 23;7º = 12;53º$$
Which we can then subtract from $arc \; AB$ (which was 90º) to state
$$arc \; ZB = 90º – 12;53º = 77;7º$$
So
$$Crd \; arc \; 2ZB = 116;59$$
And that’s all the hard ones. The two remaining arcs we need for the Menelaus configuration, $arc \; BA$ and $arc \; EA$ are both 90º since they’re from the horizon to the zenith, thus their Crd arcs are both 120.
Plugging into our Menelaus equation:
$$\frac{116;59}{120} = \frac{119;58}{117;12} \cdot \frac{Crd \; arc \; 2HE}{120}$$
Solving:
$$Crd \; arc \; 2HE = 120 \cdot \frac{116;59}{120} \cdot \frac{117;12}{119;58} = 114;17$$
Looking up the corresponding arc from our chord tables:
$$arc \; 2EH = 144;26º$$
which implies
$$arc \; EH = 72;13º$$
As noted at the outset, this needs to be subtracted from $arc \; EA$ (90º) to determine what we’re really after, $arc \; AH$. Thus,
$$arc \; AH = arc \; AE – arc \; EH = 90º – 72;13º = 17;47º$$
Good effort, but we’re still not there as we’re not looking for an arc; we’re looking for the angle between the altitude circle and the ecliptic which, in this case, is $\sphericalangle AH \Theta$.
To get there, we’ll almost use the same diagram as before, but we’ll change point K to N so we can use K as part of arc KLM which is a piece of a great circle, such that that great circle would have a pole at H
Ptolemy has a cute little trick here, wherein he states:
since circle AHE is drawn through the poles of EΘM and KLM, both EM and KM are quadrants5.
The rule is that, if the pole of two great circles lies on another great circle, then the arcs of the original great circles that lie between that third great circle and the intersection of the two original great circles are each 90º.6
No matter how you arrive at it, this new great circle arc we’ve drawn in creates a new Menelaus configuration, so as usual, we can write it out:
$$\frac{Crd \; arc \; 2 HE}{Crd \; arc \; 2EK} = \frac{Crd \; arc \; 2 HΘ}{Crd \; arc \; 2ΘL} \cdot \frac{Crd \; arc \; 2 LM}{Crd \; arc \; 2KM}$$
As with before, this won’t immediately get us to what we’re looking for. As a reminder, we’re after $\sphericalangle AHΘ$. Previously, we’ve gotten this by finding a great circle arc subtended by this angle, but in this diagram there isn’t one. However, $\sphericalangle AHK = 180º$, so our goal will be to subtract out $\sphericalangle LHK$. This is also something we won’t be getting directly at, but because we said above that $arc \; KM = 90º$, we can find $arc \; LM$ (which is in the Menelaus equation) and subtract it out. A lot of steps, but a pretty clear plan. So let’s get started:
We already stated above that $arc \; 2EH = 144;26º$, so $Crd \; arc \; 2EH = 114;16$.
We can get at $arc \; EK$ by subtracting $arc \; HE$ from $arc \; HK$. We stated above that $arc \; EH = 72;13º$, and $arc \; HK = 90º$ since it’s from a great circle to its pole which means $arc \; EK = 17;47º$, $arc \; 2EK = 35;34º$, $Crd \; arc \; 2EK = 36;38$.
Similarly, we can find $arc \; L\Theta$ by subtracting $arc \; H\Theta$ out of $arc \; HL$. From above, we have that $arc \; H \Theta = 77;37º$, so $arc \; L \Theta = 12;23º$, $arc \; 2L \Theta = 24;46º$, and $Crd \; arc \; 2L \Theta = 25;44$.
We already found that $Crd \; arc \; 2H \Theta = 117;12$ and $arc \; KM = 90º$ so $Crd \; arc \; 2KM = 120$, which gives us everything we need to plug in. Here we go:
$$\frac{114;16}{36;38} = \frac{117;12}{25;44} \cdot \frac{Crd \; arc \; 2 LM}{120}$$
Solving and doing the math:
$$Crd \; arc \; 2 LM= 120 \cdot \frac{114;16}{36;38} \cdot \frac{25;44}{117;12} = 82;11$$
Converting that back to the arc we get $arc \; 2LM = 86;28º$. Thus $arc \; LM = 43;14º$ and $arc \; LM = 90º – 43;14º = 46;46º$ as does the angle it subtends, $\sphericalangle LHK$. As outlined in our plan above, we can now subtract that from $\sphericalangle AHK$ which is 180º. Doing that:
$$\sphericalangle AH \Theta = \sphericalangle AHK – \sphericalangle LHK = 180º – 46;46º = 133;14º$$
And with that, we’re done with all the math for Book II. Ptolemy does the rest behind the scenes, repeating the calculations we just did for all zodiacal constellations differing times before the the meridian, at 7 different latitudes. It’s a massive table, but it’s also the last chapter in Book II. We’ll see in the next post before moving on to Book III which is all about the motion of the Sun.
- Technically, AZB is an altitude circle too.
- This point was not included in the original diagram, but I include it for clarity although I’ve omitted the celestial equator itself to prevent the diagram from being unnecessarily cluttered.
- Bear in mind that the RA shown in the info at upper right is not correct for the problem given there’s been 2000 years of precession of the equinox.
- Quick version of the proof: From the horizon to the north celestial pole (NCP) is the latitude (36º), so the supplement to the zenith must be 54º since the horizon to the zenith is a right angle. From the NCP to the ecliptic is also a right angle, and we just showed the angle between them was 54º, so it’s supplement, the angle between the zenith and the ecliptic must be 36º.
- Where a quadrant = 90º.
- I’m not sure of the origin or a proof for this. I suspect it could be found in Euclid’s Elements, but I’m not about to go hunting for it. Regardless, it’s not difficult to see how it’s true without a rigorous proof. For the case of $arc EM$, consider $arc HL$ which we know to be 90º because H is defined as the pole of the great circle HLM. If you consider Θ an axis you’re rotating around, the point H rotates to E, L moves to M. Similarly, we know that $arc HK$ is 90º, and if that arc is mirrored across Θ, it will be M.