Almagest Book II: Angle Between Ecliptic And Horizon – Calculations

We’ll continue on with our goal of finding the angle the ecliptic makes with the horizon. Fortunately, this task is simplified by the symmetries we worked out in the last post meaning we’ll only need to work out the values from Aries to Libra. Unfortunately, this value will change based on latitude as well as the position on the ecliptic, but we’ll still only do this for one location. And for that location, Ptolemy again uses Rhodes.

First we’ll start with angles at the equinoxes:

In this figure, we have AGD as the meridian, AED as the eastern horizon, and $a r c E Z$ on the celestial equator. It’s not included in Ptolemy’s workup, but I’m going to add one more point, P, which we’ll take as the north celestial pole. We’ll consider the vernal equinox first, so to do that, we’ll also add in the ecliptic as it would appear at that time which would contain $a r c E G$ . Our objective will be to find $∢ D E G$ which we can do by finding $a r c D G$ .

This is very straightforward as we know that $a r c P Z$ is 90º. From there, $a r c P D$ is the angle of the north celestial pole above the horizon, which is always the same as the latitude, or 36º in this case. This means that $arc \; DZ is 54º.

Similarly, $a r c G Z$ is just the obliquity of the ecliptic which Ptolemy takes as 23;51º. We can subtract that out from $a r c D Z$ to determine that $a r c D G$ and thus $∢ D E G = 30; 9 º$ .

Next, we can repeat this process for the angle between the horizon and ecliptic for the autumnal equinox by drawing that ecliptic in which will now contain $a r c E B$ .¹.

Now our objective is to find $∢ D E B$ by finding $a r c B D$ . Above, we said that $a r c D Z = 54 º$ so we just need to add on $a r c B Z$ which again, is the obliquity of the ecliptic, so we’ll add 23;51º to determine that $a r c B D$ and thus $∢ D E B$ is 77;51º.

That’s two out of the way. For the remaining points, we’ll need a new diagram to work from.

In this one, I’ve rotated things a bit so the meridian, ABGD isn’t on the perimeter. BED will be the horizon and AEG the ecliptic such that E is the beginning of Taurus.

We’ll also draw in an additional segment of a great circle that intersects the horizon and ecliptic on the far side at points $Θ$ and H respectively, such that E would be the pole of the full great circle, making $a r c E Θ$ and $a r c E H$ both 90º which would require that any point on great circle BED be 90º away from Z as well (i.e., Z is a pole of BED)².

Although it’s hard to see wrapped around the side of the celestial sphere, there’s a Menelaus configuration. This allows us to state:

$\frac{C r d a r c 2 G D}{C r d a r c 2 D Z} = \frac{C r d a r c 2 G E}{C r d a r c 2 E H} \cdot \frac{C r d a r c 2 H Θ}{C r d a r c 2 Z Θ}$

What we’re looking for this time around will be $∢ H E Θ$ which is subtended by $a r c H Θ$ so we’ll need to determine everything else. Let’s start with some tricky ones.

First off, let’s tackle $a r c E G$ which is the arc of the ecliptic from the first point in Taurus (which is on the horizon at E) to the point on the meridian below the horizon (point G), which is known as the lower culmination. While we haven’t talked about lower culmination directly, we have seen how to determine the point at upper culmination. Since the points at upper and lower culmination are opposite each other, we can find the length of the arc of the ecliptic to upper culmination (which would be at A) and add or subtract 180º from it. Since we’ve only seen the example worked through in that section, let’s walk through this one more slowly.

To start, we need to know the rising time for the first point in Taurus at Rhodes which comes from our rising time tables. But remember that what we really need to look at is the last point in Aries which gives a total rising time of 19;12º. From a physical perspective, what this means is that it has taken 19;12º for the first 30º of the ecliptic (from the vernal equinox to the first point in Taurus) to rise. Subtracting 90º from that to get what point is at upper culmination we discover that point would have a rising time of 289;12º. Subtracting 180º from that to get the lower culmination point, it has a rising time of 109;12º. Looking this up on the table for sphaera recta³ we determine this is is Cancer. Specifically, it’s 8;17º past the 10º mark. The difference in rising times between the 10º mark and the 20º is 10;47º, so the point at lower culmination is $\frac{8; 12 º}{10; 47 º} = 0.76 = 76$ into that 10º interval, which means 7;36º past the 10º mark at 17;36º⁴.

From there, we can recall that Cancer is the 4th constellation in the zodiac, so this point is 90º + 17;41º = 107;41º from the vernal equinox. However, what we were concerned with was the point from E which was 30º along the ecliptic already, so $a r c E G = 77; 41 º$ . Thus, $a r c 2 E G = 155; 22 º$ and $C r d a r c 2 E G = 117; 14$ .

Next, let’s work on $a r c G Z$ . Looking back at the figure, we have the horizon and ecliptic drawn in, but what we don’t have is the celestial equator. It would simply make the diagram too busy. But, without drawing it in, let’s concern ourselves for a moment with where it will intersect $a r c G Z$ because that’s actually something we can find because we know point G is 107;41º along the ecliptic and back in Book I, we derived a table relating the arc along the ecliptic to the arc of the meridian between the ecliptic and the celestial equator. That table stops at 90º, but then repeats, going backwards. So 107;41º is 17;41º past 90º which would be the same as 90º – 17;41º or 72;19º.

The table only gives the arcs in hole degree increments so we’ll need to do some extrapolation. Specifically, we’re $\frac{19}{60}$ the way between 72º and 73º which is .316 between 22;37,17º and 22;45;11º, the difference of which is ~0;8º. So 31.6% of that is 0;3º past the start point to get 22;40º. Again, this is the arc of the meridian between the ecliptic (at G) and some undrawn point on the celestial equator. But is that undrawn point between G and Z, or G and D? Because we’re talking about a point in Cancer, that means that the point on the ecliptic is north of the celestial equator meaning the undrawn point must be between G and Z.

That might not seem helpful by itself, but we can also figure out the arc between that undrawn point and Z to complete $a r c G Z$ . To do so, recall that Z is the pole of the horizon. Just as the pole of the celestial equator will be the same angle from the horizon as the latitude, the pole of the horizon must be that distance from the celestial equator. Thus, the arc from the undrawn point to Z is 36º. Adding these two pieces together, we determine that $a r c G Z = 58; 40 º$ .

But wait, that’s not actually something we needed! However, we did need $a r c D G$ which we can now find because $a r c D Z$ (which we also need) is from the horizon to a pole meaning it’s 90º. From that, we can subtractout $a r c G Z$ we find $a r c D G = 31; 20 º$ which means $a r c 2 D G = 62; 40 º$ and $C r d a r c 2 D G = 62; 24$ .

From the setup of the problem, we also stated that $a r c E Θ$ and $a r c E H$ are both 90º which means twice that will be 180º and their $C r d a r c s$ 120, which should be everything we need. So let’s get to plugging in.

$\frac{62; 24}{120} = \frac{117; 14}{120} \cdot \frac{C r d a r c 2 H Θ}{120}$

Rearranging that equation:

$C r d a r c 2 H Θ = 120 \cdot \frac{62; 24}{120} \frac{120}{117; 14}$

And doing the arithmetic:

$C r d a r c 2 H Θ = 63; 52$

Which gives $a r c H Θ = 32; 10 º$ as is $∢ H E Θ$ since it is subtended by this arc.

That finishes out this section, as this was merely an example calculation and Ptolemy doesn’t show the work for the calculation of every arc of the ecliptic at every latitude. But fear not! We’re working up to another massive table.

In the book I’m working from, these two figures are superimposed on one another with the parts of the ecliptic and equator not being directly used cut out, but I have chosen to present it this way as I think it’s a bit clearer.
The Toomer translation I’m using presents this figure differently, with points $Θ$ and H rotated around the backside instead of A and B, but then “peeled off”. It took me awhile to decipher what was going on there which is why I’ve rotated things.
I somehow missed that it was supposed to be on the sphaera recta table when I was working through this and it took me nearly a month to figure out why things weren’t working. That, and Gulf Wars.
Ptolemy comes up with a slightly different value of 7;41º which is what I’ll use going forward.