Now that we’ve introduced a bit about Ptolemy’s math we can take a look at his derivation of the chord tables. Exactly what those are we’ll get into later, once we have played around with some geometry that will hopefully clarify that issue.
To start, Ptolemy considers the following figure.
There’s several things Ptolemy notes about the drawing:
- $ABG$ is a semicircle with center $D$ and diameter $\overline{AG}$
- $\overline{BD}$ is perpendicular to $\overline{AG}$ at $D$
- $\overline{DG}$ is bisected at $E$
- $\overline{BE}$ is equal in length to $\overline{EZ}$
For those that aren’t familiar with mathematical setups like this and are wondering why all of the above is true, you’re overthinking it. It’s true because that’s the way Ptolemy constructed the scenario and by doing things in this specific way, other things, things that he’s interested in, will fall out after some math.
The next step he takes is easier if we temporarily give variables to some of these line segments, so I’ve redrawn the diagram with the extra variables added.
To explain quickly, there’s really 2 variables I’ve employed here: a and x. I’ve called $\overline{DZ}$ x because that’s a really unknown quantity. One that we’re ultimately going to be after. The value of a is really known since it’s half of the radius. Which is why both $\overline{DE}$ and $\overline{EG}$ can have the same value; they’re both half of the radius. And recall that $\overline{BE} = \overline{EZ}$ from our definitions above. So since $\overline{EZ}$ is composed of x+a, so too must $\overline{BE}$ which is how that got there.
Before getting into anything more, I’m going to take a quick deviation and state a mathematical truism:
$$(a + x)^2 = a^2 +2ax + x^2$$
Again, if you’re not mathematically inclined and wondering where this came from all of a sudden, it’s just a mathematically true statement. All I’ve done here is factored the right side to become the left side. Since it’s true that they’re equal, anything that fits into this equation must also be true. But before we start trying to fit our line segments into this, let’s do one more thing:
$$(a+x)^2 = a^2 + (2a+x)x$$
All I’ve done here is factored out the x from the last two terms. But what that does is allow us to start substituting in various arcs for pieces of this equation:
$$(a + x) = \overline{EZ} \implies (a+x)^2 = \overline{EZ}^2$$
$$a = \overline{DE} \implies a^2 = \overline{DE}^2$$
$$(2a + x) = \overline{GZ}$$
$$x = \overline{DZ}$$
Which all beautifully substitutes beautifully into our equation to become:
$$\overline{EZ}^2 = \overline{DE}^2 + \overline{GZ} \cdot \overline{DZ}$$
But recall that above we stated $\overline{BE} = \overline{EZ}$. That’s going to give us another substitution:
$$\overline{BE}^2 = \overline{DE}^2 + \overline{GZ} \cdot \overline{DZ}$$
And check out triangle $BDE$. That’s a right triangle. And even though Ptolemy was some time ago, Pythagoras was even longer which means we can apply the Pythagorean theorem (only thing most of us remember from HS trig. (#amirite?)
$$\overline{BE}^2 = \overline{BD}^2 + \overline{DE}^2$$
And substituting that in we get:
$$\overline{BD}^2 + \overline{DE}^2 = \overline{DE}^2 + \overline{GZ} \cdot \overline{DZ}$$
And now we get one of my favorite things in math: A like term on both sides goes away. See ya, $\overline{DE}$! The equation now simplifies to:
$$\overline{BD}^2 = \overline{GZ} \cdot \overline{DZ}$$
But $\overline{BD}$ is a radius, so it’s equal to all other radii. Ptolemy only cares about $\overline{DG}$, but it’s technically also equal to $\overline{AD}$. Still, following what Ptolemy did here we get:
$$\overline{DG}^2 = \overline{GZ} \cdot \overline{DZ}$$
Factoring that so we can more easily see what’s about to happen:
$$\overline{DG} \cdot \overline{DG} = \overline{GZ} \cdot \overline{DZ}$$
$$\frac{\overline{DG}}{\overline{DZ}} = \frac{\overline{GZ}}{\overline{DG}}$$
That might not look like much, but it demonstrates that the figure shown fits another very special definition from Euclid’s Elements, Book VI, Definition 3 which states:
A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.
That’s a mouthful, but what it’s saying is, if you have a line ($\overline{GZ}$), and it’s divided at a point ($D$), and you divide that line by the longer of the two segments ($\overline{DG}$) and that ratio is the same as the longer segment of the line ($\overline{DG}$) divided by the shorter line ($\overline{DZ}$), then it fits this very special definition of being “cut in extreme and mean ratio.”
Honestly, it’s a phrase that in all my time doing math I’ve never once encountered, but Euclid rather liked it because he went on to prove that all sorts of things fit this definition. Flipping that around, if you have lines that fit that definition, it can tell you about other things those lines could potentially be parts of.
Specifically, Ptolemy is wanting to show that $\overline{DZ}$ is a side of a decagon (a 10-sided polygon) that could be inscribed in a circle with the same radius as the one we’ve been examining, and also that $\overline{BZ}$ is a side of a pentagon under the same conditions. He’s just taking a really long way to get there because the trigonometric functions that would have made this way easier haven’t been developed yet.
Fortunately, for the first of these, Euclid happened to have related decagons and hexagons in circles to lines cut in extreme and mean ratio in Book XIII, Proposition 9, which suggests that of $\overline{DG}$ or $\overline{DZ}$, one must be a side of a decagon, and the other must be the side of the pentagon.
However, $\overline{DG}$ is the radius of the circle and from Elements Book IV, Proposition 15’s corollary, we know that the sides of a hexagon are equal to the radius, this means that $\overline{DZ}$ must be the side of a decagon.
As far as the pentagon, Euclid also has something to say on this as well in Book XIII, Proposition 10:
If an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.
If you’re not understanding what that means here’s a picture that helps (although it’s not related to how Euclid did his proof):
It’s not immediately obvious from the picture, but all three of these shapes share the same radius (the center to one of the corners. So conceptually, you could draw then all with coincident centers and all the corners would lay on the circumference of the same circle, which is the requirement from Euclid. They’ve simply been rearranged here so that you can see that they form a triangle. A right triangle in fact. Which means that, since we just figured out which pieces in our original diagram are equal in length to the sides of hexagon (blue) and decagon (green), we can use the Pythagorean theorem again to get the side of the pentagon (red).
So we concluded that $\overline{DG}$ was the pentagon side. But it’s a radius, so it’s the same as $\overline{BD}$. And $\overline{DZ}$ was the side of the decagon. And if you look for those two line segments ($\overline{BD}$ and $\overline{DZ}$), you’ll see they’re already two of the legs of a right triangle, BDZ. Funny how things work out like that.
Thus far we’ve been looking at things purely in abstract geometry. Time to start putting numbers on things.Recall that what we’re needing here is hard numbers for $\overline{DZ}$ and $\overline{BD}$ so we can apply the Pythagorean thorem to get $\overline{BZ}$.
In my previous post on Ptolemy’s math, I noted that Ptolemy defines his circles with a diameter of 120 points which means a radius of 60 points.
That immediately gives us $\overline{BD}$ since it’s a radius. $\overline{DZ}$ is going to be a little harder to get at. To do so, we’ll need to remember that $\overline{BE}$ is equal in length to $\overline{EZ}$. So let’s first try to get $\overline{BE}$.
This is pretty easy to do since it’s also on a right triangle: BDE. As we just stated, $\overline{BD}$ is a radius, so 60 points, and $\overline{DE}$ is half a radius, so 30 points. Apply Pythagorean theorem:
$$60^2 + 30^2 = \overline{BE}^2$$
$$\overline{BE}^2 = 4500$$
$$\overline{BE} = 67;4,55$$
Recall $\overline{BE} = \overline{BZ}$ and we have what we’re looking for ($\overline{DZ}$) with a bit extra ($\overline{DE}$) tacked on. Fortunately we know the length of $\overline{DE}$ (30 points) and can thus subtract it of getting:
$$\overline{DZ} = 37;4,55$$
Now, with the two legs for triangle BDZ in hand, we can calculate the length of the final piece of the puzzle, $\overline{BZ}$ via the Pythagorean theorem:
$$\overline{BZ}^2 = 60^2 + (37;4,55)^2$$
$$\overline{BZ}^2 =4975;4,15$$
$$\overline{BZ}^2 =70;32,3$$
Many readers are probably a bit lost right now as I haven’t really given much clarity on the purpose of all this math, so I think it’s a good time to try to look at what we’ve learned so far as we’ve done enough to have a few points to discuss.
First off, for a circle with radius of 60 points, we’ve established the length of the chords for a few different shapes: The decagon, the pentagon, and from Euclid, the hexagon as well. Here’s the decagon drawn out below:
Here we know the central angle is 36º because there’s 10 parts evenly breaking up the 360º of the full circle. And the whole point of the derivation above was to determine the chord length of the corresponding side (37;4,55). So now we have a relationship between the angle and the chord.
Similarly we have a relationship between the central angle of a pentagon (72º) and its chord length (70;32,3), as well as the central angle of a hexagon (60º) and its chord length (60;0,0). Let’s put that in a table quickly:
| Central Angle | Chord Length |
| 36º | 37;4,55 |
| 60º | 60;0,0 |
| 72º | 70;32,3 |
Essentially what Ptolemy has started doing is building a list of relationships between angle and chord lengths. So far we’ve only looked at a few special angles, but what Ptolemy is working towards is expanding this table to every central angle, from 0º to 180º in half degree increments.
The reason for this is that mathematicians at the time were brilliant with geometry, but trigonometry hadn’t caught up yet. From what I’ve read, the trig functions (sin, cos, tan) started being developed in the first or second century BCE, but they didn’t start seeing widespread use until about the fifth century CE. So Ptolemy just didn’t use them. Which is sad because this entire exercise would have been trivialized by the use of a trig function.
But without them, Ptolemy can’t really discuss angles very easily since they’re hard to work with in the absence of trig functions. So instead, he works with the chord lengths since, as we’ve seen, he can work all sorts of geometric magic with. Still, Ptolemy wants to know what the angles are since that’s the quantity you actually measure on the sky. Hence why he wants a table full of conversions.
Before closing out this post, let’s look at a few more special angles/shapes which he uses to then derive everything else (I promise these ones are much easier!):
First a square:
Notice the line from the center to the edge of the square is equal to the half length of the chord. This is because the dashed line bisects the angle and thus the chord making this a 45-45-90 triangle and the two legs are equal.
So we can again apply the Pythagorean theorem:
$$x^2 + x^2 = 60^2$$
$$2x^2 = 3600$$
$$x^2 = 1800$$
$$x = 42;25,35$$
But this is only half of the chord length so doubling it to get the full chord we get the chord length corresponding to a 90º angle is 84;51,10.
Next up, the triangle:
This one is slightly less straight forward than the square, but what we can notice immediately is that, the angle being 120º is bisected by the dashed line to 60º. That triangle also has a right angle which means the bottom right corner is 30º (since they must add up to 180º being a triangle and all).
If you’ve forgotten HS math, this is know as a 30-60-90 triangle and these have the special properties of the hypotenuse being 2 times the length of the shorter side. Which is why I was able to mark the dashed line’s length as 30 points.
So again we have a right triangle with two sides known so Pythagorean theorem again:
$$30^2 + x^2 = 60^2$$
$$900 + x^2 = 3600$$
$$x^2 = 2700$$
$$x = 51;57,41$$
Again, this is only half the full chord length so doubling that, we again get the chord length for an angle of 120º is 103;55,23.
Before closing out this post, there’s a few more relatively simple ones we can get based on what we already have. Let’s return to the decagon diagram:
Here, I’ve dropped all the wedges except one at the bottom. Then I’ve extended the diameter all the way across the circle, so its length is 120. Then I’ve drawn in the chord connecting the two points. Since the diameter is a straight line and the wedge from the decagon cuts 36º from it, that means the remaining angle is 144º and there’s a new chord arc across from it I’ve labeled x.
And good news! It’s a right triangle too. We know this because the angles in a triangle add up to 180º. We’ve already used 36º of that, so the remaining angles must be the remaining 144º. And since they’re equal (since the legs are the same length), that means the angles are equal so each is 72º. Applying the same reasoning to the corners of the triangle with the 144º angle, we get the angles of each of those is 18º. 72º + 18º = 90º. Boom. Right angle.
So let’s apply the Pythagorean theorem yet again:
$$34;4,55^2 + x^2 = 120^2$$
$$1375;4,15 + x^2 = 14400$$
$$x^2 = 14400 – 1375;4,15$$
$$x^2 = 13024;55,45$$
$$x = 114;7,37$$
We can apply the same reasoning to the supplemental angle for the pentagon. I’ll spare you the details this time and jump straight to the conclusion and tell you the chord length is 97;4,56.
We could again repeat this for the hexagon, but that has a central angle of 60º, so its supplement is 120º which we’ve already found. So no point in doing it again.
But let’s go ahead and add these two new values to our table:
| Central Angle | Chord Length |
| 36º | 37;4,55 |
| 60º | 60;0,0 |
| 72º | 70;32,3 |
| 90º | 84;51,10 |
| 108º | 97;4,56 |
| 120º | 103;55,23 |
| 144º | 114;7,37 |
From here, there’s no quick way to get the chord lengths (as if what we’ve done thus far is quick). As such, Ptolemy had to do another derivation to derive what’s known as Ptolemy’s theorem which we’ll explore next time.
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