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Mathematics of the Almagest

Jon Chesey

Thank you everyone for coming to this class. The purpose of this class is to give an introduction to the mathematics that Claudius Ptolemy uses in his 2nd century astronomical text, the Almagest.

The thing I frequently tell people is that the math isn’t hard. It’s just different. At worst, a bit messy as you can see in this image. However, the messiness can be overcome by taking the time to break down diagrams and understand what they’re representing.

Ultimately, the intent of the mathematics will even be familiar - largely solving triangles. But because the mathematical techniques we learn in high school trigonometry weren’t always developed, Ptolemy has to use different methods to achieve the same goal. While Ptolemy’s toolkit is limited, he makes exceptional use of it, but ultimately uses the same tools over and over which is why we can cover many of the tools and tricks he uses in this single class.

Specifically, we’ll be really covering only three things in this class:

The sexagesimal system in which Ptolemy and astronomers today still frequently express numbers
A method known as the demi-degrees method for solving right triangles
And Menelaus’ theorem for solving triangles and circles on spheres

However, we’ll ground some of these concepts in math you’re probably more familiar with, and not even particularly challenging math. As such, I encourage everyone taking this class to approach it from that fresh perspective. Don’t worry about whether or not math has made sense to you previously. In this class, we’re going to start over entirely. The only tools we can carry with us are basic functions (addition, subtraction, multiplication, division) and basic concepts (ratios, angles, arcs, exponents…). Beyond that, we will be starting mostly from scratch, beginning with the very depiction of numbers themselves.

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What even is a number?

Part 1:

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Numbering Systems

745.13

7 x 100

4 x 10

5 x 1

1 ÷ 10

3 ÷ 100

7 x 10²

4 x 10¹

5 x 10⁰

1 x 10^-1

3 ÷ 10^-2

Base 10 - Decimal

2,15;07,35

2 x 60¹

15 x 60⁰

7 x 60^-1

35 x 60^-2

Base 60 - Sexagesimal

Greek

Let’s begin by taking a look at a number. Say 745.13. In our system of numbering it is written as 7-4-5 point 1-3.

�In this, the seven is in the hundreds place, the four in the tens, the five in the ones, the one in the tenths, and the three in the hundredths. This is something we’re taught in elementary school, but rarely do we examine what this means, but let me break this out mathematically and it might become more apparent:

�7*100 + 4*10 + 5*1 + 1/10 + 3/100

�Here, I’ve just put each numeral as a coefficient of the place it’s in. But let me take one more step and express those places as exponential notation:

�7*10² + 4*10¹ + 5*10⁰ + 1*10^-1 + 3*10^-2

�Expressed this way, we can see a pattern - Each place to the left is an additional power of ten - Ten is the base so we call our numbering system “base ten”. But numbers don’t have to be this way. Indeed, computers are built on binary, where the base is two. In truth, we can choose any base we’d like and math still works correctly, although it will feel very funny for us to work in initially.

�Because much of the Almagest deals with circles and circles have historically been defined to have 360º in them, choosing a base that works well with this will make sense. As such, the Babylonian astronomers used a base 60 system as 60 shares many common factors with 360. A base 60 system is known as sexagesimal. In translations of the Almagest, we see numbers expressed this way:

�2,15;07,35

�Here, we can apply the definition of each place and see the same pattern: Each place to the left increases the power of the base by 1, but we’re using a base of 60 instead of 10. In addition, translations use commas to denote each place and a semi-colon to note where it changes to fractions.

As a quick note, this isn’t exactly how the Greek astronomers wrote things as the numbers we’ve put into each place are Arabic numerals which the Greeks obviously didn’t use. Rather, they used alphanumeric, using letters for numbers. Alpha = 1, Beta = 2, etc… for the one’s place, and the 10’s place started with Iota = 10, Kappa = 20, etc…. In addition, the Greeks didn’t use commas and semi-colons and instead just used well defined columns to define the places and a bar instead of the semi-colon. However, translations of the Almagest use the format I’ve shown for ease of readability so I won’t explore this further.

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= 135.1264…

Converting: Sexagesimal 🡪 Decimal

2,15;07,35

2 x 60¹

15 x 60⁰

7 x 60^-1

35 x 60^-2

120

15

.1166…

.0097…

+

2 x 60

15 x 1

7 ÷ 60

35 ÷ 3600

Apply the definition:

Instead, let’s work through this quickly to translate this number into something a little more familiar to us because, truth be told, I hate working in sexagesimal. Again, we can begin by splitting it up into the places and bases using the semi-colon to indicate where the exponents change to the negative.

�2*60¹ + 15*60⁰ + 7*60^-1 + 35*60^-2

�If we simply apply the definition and carry out the math here, we find out this number is equal to

�120 + 15 + 7/60 + 35/3600

�135.1264….

�Because I think better in base 10 and that’s what all calculators use, when working with numbers I will almost always convert them when doing anything complex. However, it’s not strictly necessary to do so since, as noted before, basic mathematical operations still work just fine. But let’s do a quick example to prove that to ourselves.

�

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135.1264…

Math Still Works with Different Bases

2,15;07,35

Example:

Sexagesimal

Decimal

1,50;50,30

+

3,65;57,65

4,05;58,05

1 x 60 + 50 x 1 + 50 ÷ 60 + 30 ÷ 3600 =

110.8417…

+

4 x 60 + 5 x 1 + 58 ÷ 60 + 5 ÷ 3600 =

245.9681…

To save a bit of time, I’ll use the number we just did:

�2,15;7,35

�Then we’ll add another number in sexagesimal:

1,50;50,30

�We can follow the same procedure to convert this to decimal as well. Now let’s add both ways and confirm they end up giving equivalent answers.

To do so, we add up each place as a column. For the sexagesimal, we get 3,65;57,65. However, since this is base 60, we’re not allowed to go over 59 in any place the same way we can’t go past 9 in any place in base 10. Thus, we should do some carrying where we increment the column to the left by one and reduce any column with 60 or more by 60. This gives us: 4,05;58,05. If we do the math for the decimal, we get 245.9681…

Now let’s convert the sexagesimal solution to decimal to ensure it’s the same and they do indeed return the same answers.

We could do an example with subtraction but would need to make sure we borrowed properly. Similarly, multiplication and division get more complicated due to how much carrying needs to be done, but ultimately math still works.

�

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Converting: Decimal 🡪 Sexagesimal

Example:

Not over 3600, so the first place will be the 60’s. Therefore, divide by 60.

245.9681 ÷ 60 ≈ 4.0995

The whole number tells you how many 60’s there are in the number. Thus, it will go in the 60’s place.

Then take the remainder and multiply by 60.

Work

Reasoning

Solution

.0995 x 60 ≈ 5.9681

4,XX;XX,XX

4,05;XX,XX

Repeat to desired precision.

.9681 x 60 ≈ 58.0860

4,05;58,XX

.0860 x 60 ≈ 5.1600

4,05;58,05

So now that we’ve seen how to convert from sexagesimal, let’s discuss how to convert back.

We first have to decide where to start. We can quickly see that this is not over 3600 which is 60², so we won’t need to start in that place and can only worry about how the 60¹place, so we’ll divide by 60 too see how many 60’s will go in that place. When we do so we get 4.099513888... The 4 here tells us the coefficient that should be in the 60’s place.

The remainder will go into the next place so we then need to subtract off its coefficient and multiply what’s left by 60 to determine the next place which would be 5.9708333.... Again, the 5 goes in the first place and we subtract it off and multiply by 60 again to get 58.25. Put the 58 in the 60ths palace and repeat to get 15 which goes in the 60^-2 place. Putting that all back together:

�4,05;58,05

�Which is exactly what we started with.

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Triangles are a thing

Part 2:

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Why triangles?

Consider an object on an orbit:

It moves from A to B.

Connect A to B and it forms a

Triangle.

Describing Triangles:

Side:

Notation: AB
Measured in: “Parts” (^p)

Angles:

Notation: <BCA
Measured in Degrees (^o)

To see, think of an object on an orbit. At some point in time, it’s at point A. After some time, it’s at point B. From the point of view of an observer at the center of the circle at C, it’s made an angular change. Then, if you connect the position of the object at two points in time, you naturally get a triangle. Thus, triangles become a central theme of Ptolemaic astronomy. So, let’s remind ourselves some notation quickly.

First off, triangles are created by line segments between three points. In the Almagest, they get described by the two points that are the ends of each segment with a line over them.

Since Ptolemy often doesn’t begin by knowing the scale of his solar system, he can’t actually assign units to anything. Thus, he simply uses arbitrary units he calls “parts” denoted by a superscript p.

The angles within a triangle are denoted by the points of the triangle with the point where the angle is, known as the vertex, as the middle one with an angle sign before it. So <BCA is this central angle here. Angles are measured in degrees which we’ll explore more in a bit. For now, let’s remind ourselves of some concepts relating to triangles.

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Pythagorean Theorem

A² + B² = C²

A & B are sides
C is the hypotenuse
Only works for right Triangles

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Similar Triangles

Triangles that have same shape (i.e. same angles)
Can be rotated or flipped without losing similarity
Useful because sides are increased or decreased by same ratio

3

7

11

6

22

14

Tests for Similarity

At least two equal angles (AAA)
Two sides and the enclosed angle (SAS)
The ratio of the hypotenuse and a leg is equal in both (HL)

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Coming Around to Circles

Part 3:

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Definitions

Circles have 360^o

Each degree divided into 60 minutes (‘)
Each minute divided into 60 seconds (“)

Radius of 60^p unless otherwise defined
A chord is a line between two points on the perimeter of a circle
An arc is the section of the perimeter of a circle between two points

Always has the same measure as the central angle it subtends

First off, a circle is defined to have 360º. What’s not often taught in high school math is that each one degree is divided into 60 minutes and each minute into 60 seconds which is where the sexagesimal we just covered starts to come in and why our methods of timekeeping to this day have minute and seconds in it - It’s has to do with the fact that time is measured based on astronomical principles.

We can also talk about the radius of a circle. Quite often, Ptolemy doesn’t begin actually knowing what the radius of a circle is. For example, he doesn’t presume to know the distance to the moon on its circle. Thus, Ptolemy just choses 60 to be the radius but doesn’t define that radius as a defined distance. Rather, it’s just 60 parts and he’ll work out what a part is later.

Next, a chord is defined as a line segment between two points on the perimeter of a circle and the arc is the section of the perimeter between two points. Ptolemy describes the perimeter of his circles in degrees (again having 360º) instead of units of length, so any arc will have the same number of degrees as the central angle it subtends.

.

Next, let’s look at the intersection between triangles and circles because this is where things start getting interesting.

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Chords and Angles

Central angle of 180^o

Chord of 120^p

Central angle of 90^o

Chord of 84;51,50^p

Let’s say we have a circle, and we draw in the diameter. Here, the central angle is 180º. Since the chord is the diameter which is two radii, the corresponding chord is 120^p. Next, consider having a triangle that has a right triangle at the center. In this case, both legs are 60^p and we could use the Pythagorean theorem to determine that the third side is 84;51,50^p.

Without belaboring the point, we can quickly see that there’s a relationship between the central angle and the length of the chord. As such, Ptolemy spends an entire chapter in the first book of the Almagest deriving the length of the chord created in such a circle, for each ½ of a degree arc along the circle which he lays out in a large table of chords. Obviously, he only needs to go up to 180º since after that, it repeats in the opposite direction.

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Inscribed or Central Angle Theorem

The measure of an angle formed by two points on the perimeter of a circle with a vertex on the perimeter is half the angle formed by the same two points with a vertex at the center.

Doesn’t matter where the vertex is on perimeter

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Consequence of Central Angle Theorem

If the angle on the perimeter subtends the diameter of a circle, it is a right angle

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Putting it all together

Solving Triangles with Circles

Part 4:

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The Demi Degrees Method

Givens:

<BDA = 90;00^o
<DBA = 9;32^o
BA = 9;58^p

Find other sides

To begin, let’s start with a right triangle. In it, we know one of the other angles to be 9;32º and that its hypotenuse is 9;58^p. Our objective is to find the length of the other sides. If we could use modern math, we would apply some trigonometry. We’d remember “soh cah toa” and see that we can use the fact that the sin of the angle we know is the opposite, unknown side over the known hypotenuse and then simply solve for the unknown side. But Ptolemy couldn’t do that because this sort of math wouldn’t be developed until a few hundred years after Ptolemy’s death.

Instead, Ptolemy begins by drawing a circle around the triangle. As we just learned, this means that the hypotenuse is now the diameter of the circle. In addition, Ptolemy’s circles have a diameter of 120^p. I’m coloring this in red here because it means we’ve changed the context of the triangle we’re working in. In other words, we’re temporarily creating a similar triangle. The angles will all be the same, but the sides will be different for now.

With that in mind, we can draw in the central angle, recalling from our central angle theorem that this angle will be twice the angle on the perimeter, thus having a measure of 19;04º. And now that we have the central angle, we can use the table of chords Ptolemy created to look up the length of the chord. However, we quickly run into trouble since that table is only in half degree increments and this is somewhere between two entries. Thus, we’ll have to do a process known as interpolation which just means estimating between the two entries.

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Central Angle	Chord Length
18;30^o	19;17,21^p
19;00^o	19;48,21^p
19;30^o	20;19,19^p
20;00^o	20;50,01^p

19;04^o

19;30^o

19;00^o

20;19,19^p

19;48,21^p

-

0;30^o

18

78

79

0;30,58^p

Interpolation Interlude

Differences

0;30,58^p ÷ 0;30^o = 0;01,01,56^p/’

0;01,01,56^p/’ x 4’ = 0;04,07,44^p ≈ 0;04,08^p

0;04,08^p + 19;48,21^p = 19;52,29^p

19

To begin, here’s a segment of Ptolemy’s chord table which has the range we need. Let’s start by looking at the differences between both the central angle and chord length for the entries on either side of what we need. For the difference for the angle, it’s easy: half a degree or 30 minutes.

It’s a bit more complicated for the chord length. When we try to subtract, we see we can’t take 21 from 19, so we’ll need to borrow. We’ll also see we won’t be able to subtract 48 from 18, so we’ll need to borrow again. Now we can do the math and see that, in this range, the half degree increase in the angle results in a 0;30,58^p increase in the length of the chord.

But let’s break that down to see what each minute increase results in by dividing that difference in the chord length. When we do, we see that each minute increase results in a 0;01,01,56^p increase in the length of the chord. And the amount we’re over the 19º mark is 4 minutes so we need to multiply that by 4. Doing so we get that we need to add another 0;04,08^p to the length of the chord from 19º. Thus, the length of the chord should be 19;52,29^p in the context where the diameter is 120^p. So let’s pop back and add that to our picture.

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The Demi Degrees Method

Givens:

<BDA = 90;00^o
<DBA = 9;32^o
BA = 9;58^p

Find other sides

BD² + 19;52,29² = 120²

BD = √120² - 19;52,29²

BD = 118;20,34^p

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The Demi Degrees Method

Givens:

<BDA = 90;00^o
<DBA = 9;32^o
BA = 9;58^p

Find other sides

9;58^p

120^p

= 0.0831…

118;20,34^p x 0.0831… = 9;50,03^p

19;52,29^p x 0.0831… = 1;39,06^p

If we look at side BA, we know its measure in both contexts, so we can use that to take the ratio and determine by how much it’s scaled up or down. Doing so, we see that the true context we’re after is 8.31% of what we just calculated. So, if we just apply that to all the other sides, we can get them correct.

Doing so we get that side BD is 9;50,03^p which we can add to our diagram. Similarly, side DA is 1;39,06^p which we can again add to the diagram.

Thus, we have completely solved this triangle using the same method Ptolemy does. That’s all I’m going to cover for now. I’d intended to cover a bit more as it relates to doing geometry on the surface of a sphere, but 1) I didn’t get finished in time and 2) I want to see how long this much material takes to cover before adding more. So thanks for attending.

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On the Sphere

Part 5:

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The Celestial Sphere

Apparent sphere around the Earth which rotates once per day
Planets move around this sphere on a plane which defines a circle

Each object has their own plane
Circles intersect at nodes

Each circle has a pole

The last topic I want to dip into is that of spherical geometry. This becomes important because the cosmos appears to be a sphere of which we appear to be the center. I’m not going to get into details in this class, but the sun and planets all move with respect to the stars on the celestial sphere and trace out paths along its surface that create circles. Hence why learning about circles and triangles was so important.

If we’re only worrying about the path of one object, then it creates a great circle about the sphere. Since a circle is a plane, we can use the math we’ve just discussed to solve it.

However, if we’re trying to find parts of these circles in relation to other circles on the celestial sphere, new math will be needed. And for this, Ptolemy has one equation he uses which is a spherical version of a theorem known as Menelaus’ theorem.

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Menelaus’ Theorem

Describes relationship between arcs of 4 great circles when they form this shape
Two versions:

This theorem requires that you have four great circles. This may sound scary and complicated, but it’s not nearly as bad as it seems.

The reason is that there are two major great circles that we become extremely familiar with early in the Almagest. They are the ecliptic (the path of the sun against the background stars throughout the year), and the celestial equator (Earth’s own equator extended into the celestial sphere, 90º away from the Celestial poles around which all stars seem to rotate).

Furthermore, we can always draw another great circle known as an altitude circle which goes through the celestial poles and is generally used to measure the angle above or below the celestial equator, which is known as the altitude - hence the name. This is especially convenient because this then moves us towards getting a spherical right triangle since the altitude circle will always be perpendicular to the celestial equator.

Thus, with those three pieces, most applications of this theorem boil down to only one really different great circle if even that. Many of the early applications involve multiple altitude circles as Ptolemy is trying to solve sections and angles regarding the ecliptic and celestial equator.