We’re almost to the end of book II. There’s really 2 chapters left, but the next one is almost entirely a table laying out the values we’ve been looking at here recently, so this is the last chapter in which we’ll be working out anything new.
In this chapter, we’ll tackle the angle between the ecliptic and a “circle through the poles of the horizon”. If you imagine standing outside, the zenith is directly overhead which is the pole for your local horizon. Directly opposite that, beneath you, is the nadir. If these two points are connected with a great circle, that’s the great circle we want to find the angle of with respect to the ecliptic. Because we measure upwards, from the horizon, along an arc of these great circles, to measure the altitude of a star, these are often called altitude circles.
But while we’re at it, Ptolemy promises that we’ll also determine “the size of the arc…cut off between the zenith and…the ecliptic.” In other words, because the ecliptic is tilted with respect to the horizon, the arcs between the two will be different.
To get us started, Ptolemy begins with the following diagram.
Here, we have ABG as a segment on the meridian where B is the zenith (a pole of the horizon) and G as a pole of the equator. Point A will lie on the ecliptic and be a solstice. From there, we will draw two points to either side of A that are equidistant, Z and D, as well as a pair of reference points further out from A, H and E respectively. Lastly, $arc \; GZ$ and $arc \; GD$ are drawn from the pole of the equator, as well as $arc \; BZ$ and $arc \; BD$ drawn from the zenith.
From this, Ptolemy sets out to prove two things. First, that $arc \; BZ = arc \; BD$ and second, $\sphericalangle BDE + \sphericalangle BZA = 180º$.
To do so, Ptolemy first states that $\sphericalangle BGD = \sphericalangle BGZ$. The reason for this is that the arcs subtended by these angles ($arc \; AD$ and $arc\; AZ$ respectively) are equal, and therefore the angle must be too.
Next, Ptolemy states that $arc \; GD = arc \; GZ$ because D and Z are equidistant from the same solstice. Since spherical trangles $BGD$ and $BGZ$ both share $arc \; BG$, we can apply Side-Angle-Side symmetry and thus state that their bases, $arc \; BD$ and $arc \; BZ$ are equal as well as are their other angles, specifically, $\sphericalangle BZG$ and $\sphericalangle BDG$.
Jumping back to chapter 10, we demonstrated that two angles, formed by the intersection of great circles going through the pole of the celestial equator and intersecting the ecliptic at points equidistant the same solstice sum to 180º. This means that $\sphericalangle GZA + \sphericalangle GDE = 180º$.
Next, what we’ll do is subtract $\sphericalangle BZG$ from $\sphericalangle GZA$ and add $\sphericalangle BDG$ to $\sphericalangle GDE$. Since we said above that $\sphericalangle BZG = \sphericalangle BDG$ this doesn’t change their sum. Thus:
$$\sphericalangle BZA + \sphericalangle BDE = 180º$$
The next thing we want to demonstrate is that
if we take the same point of the ecliptic at two positions equidistant from the meridian (as measured in time-degrees) on opposite sides of it, the great-circle arcs from the zenith to these two positions are equal, and the sum of the two angles [between altitude circle and ecliptic] east and west [of the meridian] is equal to twice the angle formed by the same point [of the ecliptic] at the meridian, provided that for both positions [i.e. when the point is east and west of the meridian] the points [of the ecliptic] which are [then] culminating are either both north or both south of the zenith.
Translating that a bit, recall that the meridian is defined by the Earth which rotates. It’s a great circle that goes through the points north and south on the horizon, but will therefore also go through the north and south celestial poles sweeping across the celestial sphere. In this case, what we’re wanting to do is look at a single point on the ecliptic. It will be some distance from the meridian at that point in time. Then we’ll let time pass, until the meridian is the same distance from that point, but on the opposite side. What Ptolemy is setting about to show is that if you take the angles formed at those points by the ecliptic and a great circle drawn through the zenith, it is twice the angle formed if that point were on the meridian, between the ecliptic and meridian, so long as both of the points are either north or south of the zenith (not one on either side), which means we’re going to need to do several proofs to fully illustrate where this rule can be applied. Let’s tackle where they’re both south of the zenith first.
So let’s draw that out1. Here, I’ve just drawn the ecliptic with a point, E on it. Then I drew in a meridian to the left of that point which goes through the north celestial pole at D, the zenith at G and connects to the ecliptic at A. Then I’ve just connected G and E with a piece of a great circle. Then just so we have another reference point on the ecliptic, we’ll just add another point Z.
Now I’ll repeat that with the meridian on the other side. However, to keep things from getting confused, I’ll call the point on the ecliptic H but they’re still the same point. This time, the meridian intersects the ecliptic a bit closer to the north celestial pole, so we’ll call that point B instead of A this time as these really are different points. Again, we’ll connect our point of interest to G.
That completes the setup2. Our ultimate goal now is to prove that:
$$\sphericalangle GEZ + \sphericalangle GHB = 2 \cdot \sphericalangle DHB = 2 \cdot \sphericalangle DEZ$$
First, let’s note that spherical triangles GDE and GDH are equal. This is because arcs connecting to the points on the ecliptic ($arc \; DE$ and $arc \; DH$) are actually the same arc. Similarly, $arc \; DG$ is common to both, just rotated around. Lastly, the angle between those two sides must also be equal because they are subtended by $arc \; AE$ and $arc \; HB$ which are the same. So we’ve demonstrated side-angle-side similarity3. Thus, since the spherical triangles are equal, $arc \; GE = arc \; GH$ and $\sphericalangle GED = \sphericalangle GHD$.
We can also state that $\sphericalangle DEZ = \sphericalangle DHB$4 since E and H are actually the same point and Z and B are both on the same line.
Because of the arrangement, we can also state
$$\sphericalangle GHD + \sphericalangle GHB = \sphericalangle DHB$$
Now let’s do a quick substitution to that from what we stated above. We can swap out $\sphericalangle GHD$ for $\sphericalangle GED$.
$$\sphericalangle GED+ \sphericalangle GHB = \sphericalangle DHB$$
Now let’s add these two equations together to get
$$\sphericalangle GHD + \sphericalangle GHB + \sphericalangle GED + \sphericalangle GHB = 2 \cdot \sphericalangle DHB$$
I’ve purposely not collected the $\sphericalangle GHB$ terms here on the left side. To understand why, let’s look at the first three terms. The first two, $\sphericalangle GHD + \sphericalangle GHB = \sphericalangle DHB = \sphericalangle DEZ$. Now we can add on the third term, $\sphericalangle GED$ to form a new angle, $\sphericalangle GEZ$. Swapping that in for the first three terms we get
$$\sphericalangle GEZ+ \sphericalangle GHB = 2 \cdot \sphericalangle DHB$$
which is what Ptolemy set out to prove. And we can easily swap out $2 \cdot \sphericalangle DHB$ for $2 \cdot \sphericalangle DEZ$ which gives us the other inequality.
But that was only for points on the ecliptic south of the zenith. Now we need to place the zenith further down. Again, in my first drawing, we’ll let D be the north celestial pole, G the zenith, E the point of interest, A the intersection of the meridian and the ecliptic, and Z as a reference point along the ecliptic to help define angles. We’ll again connect G to E with an arc of a great circle, but since we’ll need to refer to some more angles, we’ll continue that arc a bit longer to point K.
We’ll let the meridian rotate around a bit and rename E to H and A to B, with $\Theta$ as our new reference point and L our point on the great circle connecting G to H.
This time our goal is to prove that:
$$\sphericalangle KEZ + \sphericalangle LHB = 2 \cdot \sphericalangle DEZ$$
To start off, we can immediately see that $\sphericalangle DEZ = \sphericalangle DHB$.
We can also state that spherical triangles DHG and DEG are the same using the same reasoning as we did in the last section, which means that $\sphericalangle DEG = \sphericalangle DHG$. But consider the supplemental angles of those along $arc \; GK$ and $arc \; GL$ respectively. Those supplements must also be equal. Therefore $\sphericalangle DEK = \sphericalangle DHL$.
Now for some equalities:
$$\sphericalangle DHB + \sphericalangle DHL = \sphericalangle LHB$$
I’ll immediately substitute in $\sphericalangle DEZ$ for $\sphericalangle DHB$ to get:
$$\sphericalangle DEZ + \sphericalangle DHL = \sphericalangle LHB$$
We can also state:
$$\sphericalangle DEK + \sphericalangle DEZ = \sphericalangle LHB$$
Now let’s add these last two equations together to get:
$$ 2 \cdot \sphericalangle DEZ + \sphericalangle DEK + \sphericalangle DHL = 2 \cdot \sphericalangle LHB$$
I’ll move some terms around and unfactor $2 \cdot \sphericalangle LHB$ to get:
$$\sphericalangle LHB + \sphericalangle LHB – \sphericalangle DHL – \sphericalangle DEK = 2 \cdot \sphericalangle DEZ$$
Now let’s look at $\sphericalangle LHB – \sphericalangle DHL$. That’s just $\sphericalangle DHB$ so the equation now becomes:
$$\sphericalangle LHB + \sphericalangle DHB – \sphericalangle DEK = 2 \cdot \sphericalangle DEZ$$
But $\sphericalangle DHB = \sphericalangle DEZ$ so let’s swap that in to get:
$$\sphericalangle LHB + \sphericalangle DEZ – \sphericalangle DEK = 2 \cdot \sphericalangle DEZ$$
And now let’s look at $\sphericalangle DEZ – \sphericalangle DEK$ which is just $\sphericalangle KEZ$ which leaves our equation as:
$$\sphericalangle LHB + \sphericalangle KEZ = 2 \cdot \sphericalangle DEZ$$
Which is what we were trying to prove this time.
So at this point we’ve demonstrated the theorem for points A and B both being north, or both south of the zenith. But now Ptolemy wants to consider what will happen if:
the culminating point on the segment [of the ecliptic] east [of the meridian], namely A, should be south of the zenith G, while the culminating point on the segment west [of the meridian], namely B, should be north of the zenith.
Turns out we can derive a nifty equation for that as well. But first the diagrams which I won’t explain as they’re so similar to the previous diagrams with the exception of where G is located.
This time our goal is to prove:
$$\sphericalangle GEZ + \sphericalangle LHB = 2 \cdot \sphericalangle DEZ + 180º$$
As with before, some of our angles are equal to one another. First, $\sphericalangle DEZ = \sphericalangle DHB$, and $\sphericalangle DEG = \sphericalangle DHG$ because they’re both made of of the intersection of the same lines in each instance.
We can notice that $\sphericalangle DHL + \sphericalangle DHG = 180º$ which we can then substitute in $\sphericalangle DEG$ for $\sphericalangle DHG$ to get $\sphericalangle DEG + \sphericalangle DHL = 180º$.
Now let’s make some equalities:
$$\sphericalangle GEZ = \sphericalangle DEG + \sphericalangle DEZ$$
$$\sphericalangle LHB = \sphericalangle DHL + \sphericalangle DHB$$
Adding those together we get:
$$\sphericalangle GEZ + \sphericalangle LHB = (\sphericalangle DEG + \sphericalangle DEZ) + (\sphericalangle DHL + \sphericalangle DHB)$$
I’ve added some parentheses around the terms to help you keep track of what was what, but we don’t really need them because addition is commutative. So let’s do some rearranging:
$$\sphericalangle GEZ + \sphericalangle LHB = (\sphericalangle DEZ + \sphericalangle DHB) + (\sphericalangle DEG + \sphericalangle DHL)$$
All I’ve done here is moved the terms on the right hand side around. I’ve kept the parentheses in place because if you look at the second set, it’s something we found above. Thus we can substitute in to get
$$\sphericalangle GEZ + \sphericalangle LHB = (\sphericalangle DEZ + \sphericalangle DHB) + 180º$$
And lastly, there’s another substitution from what we’ve stated above allowing us to substitute in for $\sphericalangle DHB$ to get:
$$\sphericalangle GEZ + \sphericalangle LHB = (\sphericalangle DEZ + \sphericalangle DEZ) + 180º$$
Collecting like terms:
$$\sphericalangle GEZ + \sphericalangle LHB = 2 \cdot \sphericalangle DEZ + 180º$$
Which is what we set about to prove this time.
But we’re still not out of this one yet. What happens if A and B are flipped because we’re looking at a point on the other side of the ecliptic?
This time we’ll prove that $\sphericalangle KEZ + \sphericalangle GHB = 2 \cdot \sphericalangle DEZ – 180º$.
Again, here’s the diagrams:
In this diagram, much of what we’ve said previously still holds true:
$$\sphericalangle DEZ = \sphericalangle DHB$$
$$\sphericalangle DEG = \sphericalangle DHG$$
$$\sphericalangle DHL + \sphericalangle DHG = 180º$$
For which we can do a quick substitution to get:
$$\sphericalangle DEG + \sphericalangle DHL = 180º$$
From the setup we get that:
$$\sphericalangle KEZ = \sphericalangle DEZ – \sphericalangle DEK$$
$$\sphericalangle GHB = \sphericalangle DHB – \sphericalangle DHG$$
Adding those together we get:
$$\sphericalangle KEZ + \sphericalangle GHB = \sphericalangle DEZ + \sphericalangle DHB – (\sphericalangle DEK + \sphericalangle DHG)$$
Doing the last bit of substitution and we get:
$$\sphericalangle KEZ + \sphericalangle GHB = 2 \cdot DEZ – 180º$$
which is what we set out to prove.
I’m going to close things out for this post because it’s already getting quite long. However, Ptolemy isn’t done with the chapter yet. As usual, Ptolemy wants to build a giant table with derived values for everything we’ve worked on in the past several sections collected for different latitudes at different signs, but all we’ve done so far in this chapter is laid out some relationships between various things. In my next post, we’ll explore how to actually get hard numbers out of what we’ve done here instead of just relationships.
- This is not the way I would have labeled points. Because we’re again splitting between two drawings, things become somewhat confusing given that variables are being reused. So pay close attention.
- The translation I’m using doesn’t draw the diagrams quite the same. Instead, the images are rotated and overlapped at the meridian. While this certainly works, I feel it’s much more natural to visualize things this way. Despite having to flip back and forth between the two pictures and keep straight which one you’re referring to at a given time, that’s not hard to do from context. The benefit is that it’s much easier to understand why some of the angles we’re about to say are the same, really are. When rotating things and then overlapping as done in the book, it breaks the ecliptic and makes things look funny in my mind.
- Ptolemy only states this as the “same reasoning as before” referring to the last proof above.
- This is quite obvious when looking at the two drawings but is far less obvious in the diagram in the book.