First off, what’s a gnomon?
Apparently it’s the part of a sundial that casts shadows. Now you know.
To start this next chapter, Ptolemy dives straight into a new figure, but I want to take a moment to justify it first. To begin, let’s start with a simple diagram. Just a side view of the meridian, the horizon, and the north celestial pole. The zenith is also marked (A).
Here, I’ve drawn the NCP to the right, which means this view is effectively looking west, so south is on the left, and east is behind us. We’ve chosen 36º again because we’ll again be using Rhodes as our example problem as we did previously.
From there we can draw in a line to the celestial equator because we know it’s always 90º from the NCP. Since the sun is on the celestial equator on the equinox, let’s go ahead and give it a label too. We’ll use B.
I’ve split that 90º up to show the angles involved with respect to the zenith as that will become important shortly.
On the solstices, the sun is 23;51,20º away from that point, so we can also draw lines for those in, using L for the winter solstice and H for the summer solstice. And we’ll also call the center of the circle E.
Looking pretty good. But let’s go ahead and extend the important lines all the way across the circle. And for good measure, we’ll label all the opposing points as well.
That’s the majority of the setup for the problem. Except for that part about the gnomon. Here, Ptolemy states:
since the whole earth has, to the senses, the ratio of a point and a center to the sphere of the sun, so that the center E can be considered as the tip of the gnomon, let us imagine GE to be the gnomon.
The first part, “the ratio of a point and a center” is a phrase borrowed from the Greek astronomer Hipparchus in his work On Sizes and Distances. It’s really just a way of saying the sun is sufficiently far away that we don’t need to consider it. For example, if we’re observing the sun, and we take a step to the left or right, a tree in the foreground seems to move in relation to the sun due to parallax. But if we measured the altitude and the azimuth of the sun before and after a small move, it would remain unchanged unless we measured it with extremely precise instruments.
Similarly, we can place our imaginary gnomon wherever we’d like on the landscape, but Ptolemy chooses to do it such that it will have a height of $\overline{GE}$ because this will make things work out well. In particular, the rays extending from the sun at the equinox and solstices will strike the tip of the gnomon at E and continue to trace the paths we have already drawn out.
One small problem is that the shadow of a straight object appears straight. They can’t terminate on the meridian since that’s an imaginary line. So we’ll need to draw in a more realistic shadow.
And there we have the setup for this problem. So let’s start applying some numbers. First, we can see that $\angle DEG$ is 36º because it’s formed by extensions of lines $\overline BE$ and $\overline AE$ and the angle between them was 36º. And since Ptolemy defines his circles such that their perimeters are also 360º, this means $arc \; DG$ is 36º as well.
Similarly, since $\angle{DE \Theta}$ and $\angle{DEM}$ are formed from extensions of the lines we drew for the solstices, which were 23;51,20º from $\overline{BE}$, we know $\angle{DE \Theta}$ and $\angle{DEM}$ are 23;51,20º as are $arc \; D\Theta$ and $arc \; {DM}$. Thus by subtraction we can determine $arc \; G\Theta$ is 12;8,40º. And if we add, $arc \; GM$ is 59;51,20º.
Before moving on, let’s take note of what Ptolemy is after here. As the title suggests, we’re looking for the ratio of the shadows on the equinox and solstices, not in relation to each other, but in relation to the gnomon itself, $\overline{GE}$. So we want $\frac{\overline{GK}}{\overline{GE}}$, $\frac{\overline{GZ}}{\overline{GE}}$, and $\frac{\overline{GN}}{\overline{GE}}$.
If we could use a trigonometry function, this would be easy, but since Ptolemy didn’t apparently know what a cosine was, we’re stuck doing it a much harder way. To understand this method, let’s start with $\frac{\overline{GZ}}{\overline{GE}}$.
First, let’s note that both $\overline{GZ}$ and $\overline{GE}$ are part of the right triangle, $\triangle {EGZ}$. And then let’s draw a circle around that triangle such that all three points lie on the circumference.
To try to keep things from looking too cluttered, I’ve turned down the transparency for everything we’re not using at the moment and we’re just left with a triangle in a circle in which each of the sides happen to be chords.
As usual, we’ll use Ptolemy’s convention of the circle having a radius of 60. Immediately, you may be worried about this because we have another circle (or original one) and what happens if we need to do something with that circle since they obviously can’t have the same radius given they’re obviously different sizes!
Fortunately, things are going to work out since we’re really just looking for ratios meaning any odd units we switch to will end up cancelling out anyway.
Next we’ll draw in the center of the circle. I’ll call it X.
So now we have $\overline{GZ}$, a chord of the circle we’ve drawn in, and it happens to be subtended by two different angles: $\angle{GEZ}$ and $\angle{GXZ}$1. It’s been awhile since we’ve referred to Euclid’s Elements, but Book III, Proposition 20 notes that,
In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.
That’s really poorly worded but essentially what it’s saying is that if you have a triangle inscribed in a circle like above, the angle on the circumference that is subtended by a base, is half the angle at the center subtended by the same base.
Above we noted that $\angle {DEG}$ is 36º. Since that’s the same angle as $\angle {GEZ}$, it too is 36º and from the theorem of Euclid, that implies that $\angle {GXZ}$ must be twice that, or 72º.
But X is a point I introduced to illustrate the method. Ptolemy wants to keep this in terms of $\angle {GEZ}$, so he introduces a new unit: demi-degrees, denoted by two degree symbols (ºº) which is essentially a substitution that really just notes he’s using this method. So by that notation, $\angle {GEZ}$ is 72ºº.
Regardless, we’re curious about the $\angle {GXZ}$ because if we have a chord in a circle where we know the central angle, we can head to the chord tables. So if $\angle {GXZ} = 72º$, then that means $arc \; GZ$ is also 72º based on Ptolemy’s convention of the circumference of his circles having 360 units. Looking up the corresponding chord, we get $Crd \; arc \; GZ = 70;32,3$.
Now, before, we could have said that $\overline{GE} = 60$ because it’s a radius in that big circle. But since we’re currently working in the context of the small circle, we can’t do that, and need to find it in terms of $\angle{EXG}$. Fortunately, as I noted in a footnote above, X falls on $\overline{EZ}$ which means we can state $\angle{EXG} = 180º – 72º = 108º$ as does its corresponding arc. Thus, $Crd \; arc \; EG = 97;4,55$.
Taking the ratio we get $\frac{\overline{GZ}}{\overline{GE}} = 0.727$2. Putting that back into the context of the large circle where we redefine $\overline{GE}$ to be 60 units again, that means that $\overline{GZ} = 43;35,33$ which Ptolemy rounds off to 43;36.
We can follow the same method for the other two angles as well to determine $\overline{GK} = 12;55$ and $\overline{GN} = 103;20$.
Before closing out the chapter, Ptolemy notes that this method could easily be applied in reverse; By knowing the ratio of the length of a shadow to the height of the gnomon, we could calculate the altitude of the sun on the solstices and equinox. While I would think of this as a way to determine one’s latitude, Ptolemy puts it to the use of being a method of finding the angle between the ecliptic and celestial equator. This is a value that we’ve repeatedly taken for granted throughout the posts thus far, but I suppose this is Ptolemy showing where the value can come from in addition to the solar angle dial we saw previously.