Claudius Ptolemy: Commentary

Claudius Ptolemy

Commentary on the text

1. The title of the work is technically Syntaxis Mathematica (Mathematical Collection). See the introduction for an explanation of the variance in the title of this work.

2. Ptolemy alludes to Book vi of Aristotle's Metaphysics, where this three-fold categorization of sciences appears.

3. The sun, moon and planets all move along a great circle in the sky, which astronomers call the ecliptic. (Being the path of the sun and moon, it is the place in the heavens where eclipses occur.) The position of the sun in the sky (as for the other "planets") depends not only on the time of day, but on the time of year and also the location of the observer on the earth. Ptolemy's reference to finding the position of the ecliptic, then, refers to the calendric problem of telling time and the geographical problem of locating the observer at a particular place. This corresponded to the astronomer to the problem of predicting where on the horizon the sun would rise on a given day, or where a particular planet would rise on a particular night.

4. Here is the first hint of a description of the geocentric cosmology (that is, an earth-centered model of the universe) that was more commonly referred to as the Ptolemaic system.

5. Ptolemy is using a sexagesimal system of calculation here because this was the tradition handed down by Babylonian astronomers. Angles are measured in units of degrees such that 360 degrees corresponds to a full circle, and distances are measured so that the radius of the circle corresponds to 1 unit, or 60 parts. Parts will be subdivided into sixtieths and further into 3600ths, etc.

6. There is also the practical benefit that sexagesimal arithmetic is often easier than decimal arithmetic!

7. In practice, this means that Ptolemy will record his calculations to three sexagesimal places, giving accuracy of one part in 216,000, certainly "negligible to the senses" of any astronomer using the tools of his day.

8. Ptolemy constructs his table of chords by starting with the chords that are easiest to determine, namely the sides of regular polygons that are inscribed in the circle. In the next few paragraphs then, he will determine the lengths of the sides of the inscribed regular equilateral triangle, square, regular pentagon, regular hexagon, and regular decagon (10 sides). These are determined by basic geometrical means. This will allow him to calculate the chords of angles of degree measure 120 = 360/3, 90 = 360/4, 72 = 360/5, 60 = 360/6, and 36 = 360/10. He begins here with the pentagon and decagon.

9. Here is the statement of a proposition regarding the construction he has just outlined which he will now justify with a proof. In general, his propositions are not labeled as such in the text; he simply makes assertions and then provides proofs.

10. sq. EF = (EF)(EF) = (FD + (1/2)CD)(CF - (1/2)CD) = (FD + ED)(CF - ED) = (CF)(ED) + (CF - FD)(ED) - ED² = (CF)(ED) + (2ED)(ED) - ED² = rect. CF, ED + ED².

11. We have here that (CF)(FD) = CD². This can be rewritten in the form of a proportion: CF : CD : : CD : FD. In other words, the line segment CF is cut by D into two pieces so that the whole CF is to the larger piece CD as this piece CD is to the smaller piece FD.

12. See the link to the cited proposition in Euclid to understand how this piece of the argument is deduced.

13. Here is another self-contained proof of the Euclidean proposition that Ptolemy is quoting here, that in a circle, the square on the side of the inscribed regular pentagon is the sum of the squares of the sides of the inscribed regular hexagon and inscribed regular decagon. The impact in this context is to show that (crd 72°)² = (crd 60°)²+ (crd 36°)².

    Suppose AB is the side of the inscribed pentagon and AC the side of the inscribed decagon in the circle with center O. Bisect angle AOC to strike AB in D, then join CD. It follows that angle AOB = 72°, angle AOC = 36°, and angle AOD = 18°. Also, angle OAB = angle OBA = 54° = angle DOB; and angle CAO = angle ACO = 72° = angle CBO = angle OCB. So angle CAB = angle CBA = 18° = angle ACD. From all this we conclude that triangles ABC and ACD are similar. So we have the proportion AB : AC : : AC : AD. Cross multiplication gives AC² = (AB)(AD).
    Next, we observe that triangles AOB and ODB are similar, so AB : OB : : OB : DB. Here cross multiplication yields OB² = (AB)(DB).
    Putting this together, AC² + OB² = (AB)(AD) + (AB)(DB) = AB(AD + DB) = AB². But this is precisely what the theorem states, so we are done.

14. He applies the Pythagorean Theorem here to right triangle BDE.

15.

16. At this point, he has completed the first part of the process of constructing his table of chords: he has computed values of chords for five special angles, 36°, 60°, 72°, 90°, and 120°. See note 8. Now he begins the second stage, in which he will prove "as few theorems as possible" as needed to fill in the rest of the table.

17. The first of his theorems deals with supplementary angles: the sum of the squares of the chords of an angle and its supplement equals the square of the diameter, 120² = 14400. Therefore,

18. This result is known today as Ptolemy's Theorem: In a cyclic quadrilateral (this term is used to describe any quadrilateral whose vertices lie on the circumference of a circle), the product of the diagonals equals the sum of the products of the pairs of the opposite sides.

19. Ptolemy now uses the theorem he has just proved to demonstrate how, given the values of the chords of two angles, the chord of their difference can be found. Refering to his diagram, AC = crd a and AB = crd b are given and BC = crd(a- b) is to be found. But CD = crd(180° - a),BD = crd(180° - b), and AD = 120^p. By the theorem, (AB)(CD) + (AD)(BC) = (AC)(BD). Substituting the values and solving for crd(a- b) gives a formula for the chord of a difference of angles:

20. Next, he shows how, given the chord of some angle, to find the chord of half the same angle.

21. AC is given since it is the diameter of the circle, and he has just shown that CF = (AC - AB)/2, where AB is the chord of the supplement of the given chord, hence this can also be computed from the given data.

22. The "formula" he gives for finding the chord of the half of the angle is

since CF = half(AC - AB) and AC = 120^p.This gives us a usable formula: by putting crd a = AC and therefore crd a/2 = CD, we get

23. These computations will be verified in the exercises.

24. Now that he has procedures for finding the chords of a supplement of an angle, the difference of angles, and the half of an angle, he will demonstrate how to find the chord of the sum of two angles.

25. If AB = crd a and BC = crd b, then AC = crd(a + b).

26. Using the notation from the previous note, Ptolemy knows how to find CE = crd(180° -b), BD = crd(180° - a) and DE = crd(180° - (180° - a)) = crd a. By the theorem on cyclic quadrilaterals, (BC)(DE) + (CD)(BE) = (BD)(CE), so (CD)(BE) = (BD)(CE) - (BC)(DE). Combining these relationships with the fact that BE = 120^p, we get that

But AC and CD are supplements, so we also have that

Therefore

and this formula can be combined with the formula for supplementary angles to compute the chord of a sum of angles.

27. Ptolemy alludes here to another of the famous construction problems of Greek geometry, the trisection of the angle. The problem is simple to state: given any angle, construct the angle one third as large. Like the quadrature of the circle and the duplication of the cube, this problem was unsolved for centuries, and found in the nineteenth century to be impossible to solve.

28. Ptolemy faces a bit of a dilemma here. He knows of no way to determine the chord of an arc of 1° exactly. So he resorts to an approximation method that will use the values of the chords of nearby arcs of ³/₄° and 1¹/₂°. He ensures, however, that the approximation is accurate to one sixtieth of one sixtieth of a part, which is his unit of measure in the tables he is constructing.

29. By this (componendo) is meant that 1 is added to both sides of the inequality, on the left in the form AE : AE, and on the right in the form angle EDA : angle EDA. Since we have common denominators on both sides, we can add numerators to get the new proportion.

30. This (separando) is the opposite of componendo: 1 is subtracted from both sides of the inequality, on the left in the form angle EDA : angle EDA. Since we have common denominators on both sides, we can subtract numerators to get the new proportion.

31. Of course, Ptolemy is not really saying that the desired value is both larger than and smaller than the same number; this is impossible. What he means is that the upper bound and lower bound of the desired chord round off to the same value, to three sexagesimal place accuracy. Thus the chord has been determined to the same accuracy.

32. By the half-angle formula.

33. Ptolemy is describing what is called interpolation. For chords of angles intermediate between given arcs in the table, he supplies in the "Sixtieths" column incremental values for obtaining accurate results. For example, the table records that crd 25° = 25^p58'22''. Suppose that want crd 25°5' (here 5' means five minutes of arc; 1 minute is, of course, one sixtieth of a degree). Since we are adding 5 sixtieths of a degree, the value of the chord will increase by 5 times the amount 0^p1'1''19''' that appears in the sixtieths column of the table in the row for 25°. Now 5 x 0^p1'1''19''' = 0^p5'6''35''' (remember that we are calculating in sexagesimal!), which rounds to 0^p5'7'', so adding this to 25^p58'22'' yields the computation crd 25°5' = 26^p3'29''.

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