PTOLEMY DAY 7 BUILDING A TABLE OF CHORDS AND ARCS: PART 1

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Natalie Elliott
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1 PTOLEMY Y 7 UILING TLE OF HORS N RS: PRT 1 (1) WHY 360 N 120? Ptolemy divides his standard circle s diameter, R, into 120 parts, in keeping with his sexagesimal ways this way the radius of the circle is 60 of those 120 th -parts of the diameter, and each of those is called a R O part of the diameter s 120. nd he divides the arc of the circle (and hence the angles subtending them at the center of the circle) into 360 equal arcs, each one called a degree. s cumbersome as this is for us, there were some advantages to this for Ptolemy, who did not have decimal calculators at his disposal. (a) For one thing, 120 = 5! = , so 120 has lots of integral factors. 360 is triple that, so it also has lots of factors. This makes for more whole numbers of degrees when we cut the circumference of the circle into some whole number of parts; e.g. half a circle, a third of a circle, a quarter of a circle, a fifth of a circle, will all have whole number degree-values, as will a tenth, a twelfth, a fifteenth, and so on. (b) 360 : 120 = 3 : 1, a rough approximation of pi. In other words, one degree of arc is almost equal to one part of the diameter. (c) With this system, the equilateral triangle on the radius has 3 angles of 60 and 3 sides of 60 parts. That s rather pretty. (d) There are 365 days in a year; so now we have the Sun moving about 1 a day on its circle (just a little less, as we shall see). (2) HORS RE NOT S RS. To see the need for this table of chords and arcs, we need to see that chords do not have the same ratios as the arcs they subtend. We easily fall into the mistake of thinking that any two chords in this circle will have to each other the same ratio as the arcs of the circle they cut off. If you are tempted to think this, banish it from your thought! It is not true. nd it is easy to forget that this is not true, even for some veterans of trigonometry. One must burn this annoying fact into memory. If chords were as arcs, then it would be much easier to build up the Table! For instance, since the chord of 180 is 120 parts, it would follow that the chord of 90 must be 60 parts, and so on proportionally. las, it is not so. very quick proof of this: raw a square inscribed in our circle. Obviously the chord of 90 is the side of that square, and the chord of 180, the diameter, is the diagonal. ut the diagonal is not double the side, even though the arc is double the arc. 46

2 Or draw a regular HEXGON in a circle, letting,, be three consecutive sides. Then is the diameter of the circle, and is double. ut the arc (i.e. ) is not double the arc, but triple it! gain, M If M = M, and so arc M = arc M so arc M = ½ arc M is chord M = ½ chord? E No. E = ½ and M > E (hypotenuse) (3) PLN FOR UILING THE TLE: So we must find other inroads into filling out the entries on our table. Plan of attack: [1] FIN HORS OF 36 & 72. = items (4) (5) below. [2] FIN HORS OF 60, 90, 120 and note that the chord of the supplement to an angle whose chord is given, is given. = item (6). [3] FIN HORS OF 144, 108. = item (7). [4] PROVE THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN. = items (8) (9) [5] PROVE THT HORS OF RS WHIH RE HLVES OF RS WITH GIVEN HORS RE GIVEN. = item (10) [6] PROVE THT HORS OF RS WHIH RE SUMS OF RS WITH GIVEN HORS RE GIVEN. = item (12) [7] FIN HORS OF 1 N 1½. = items (13) (15) 47

3 [8] INTERPOLTION OF SIXTIETHS = item (16) [9] RELTION OF TLE OF HORS TO TLE OF SINES = item (17) NOTE: 36, 72, 60, 90 are all found directly. 120, 144, 108 are found as supplements. 12 is found by subtraction, i.e. by [4] above. 6, 3, 1½, ¾ are found by bisection, i.e. by [5] above. ll multiples of 1½ are found by addition, i.e. by [6] above. 1 and ½ are found by approximation. We will cover Steps [1] through [6] today, i.e. items (4) through (12), and we will cover Steps [7] through [9], i.e. items (13) through (17), in ay 8. (4) PRELIMINRY TO FINING THE HORS OF 36 & 72. First, Ptolemy shows that if we have a circle of diameter, perpendicular radius, and bisect radius at E, and join E, and draw a circle with center E, radius E, cutting radius at F, and join F, then F = side of regular inscribed pentagon F = side of regular inscribed decagon F E He uses Euclid s Elements, ook 13 Proposition 10, for this. So now we know that F is the chord of 72, and F is the chord of 36. This does not give us a numerical value for them yet, in terms of the 120 parts of diameter, but it will enable us to do that in the next step. 48

4 (5) FINING THE HORS OF 36 & 72. Now let s find numerical values, as precise as we like, for the chords of 36 and 72 : F E = 120 so E = ¼ = 30 and = ½ = 60 [given] so E = 2 E + 2 = = 4500 = so EF = E = so F = EF E = = i.e. hord 36 = so F 2 = but 2 = 3600 so F 2 = F = = so F = so hord 72 = NOTES: (a) These are decimal values, but are easily translated into sexagesimal values matching those on Ptolemy s Table. (b) Ptolemy speaks of getting or finding chords, or of chords being given. What he means is to find a way of determining a numerical value for their lengths, in units of one-hundred-twentieth parts of the diameter, to any desired degree of precision. We do this by beginning with chords whose exact values are known for geometrical reasons, then by showing how the sought chord is the result of a known operation on the known chords. 49

5 (6) FINING THE HORS OF 60, 90, 120. hord 60 = Radius = 60 so 2 = 2 2 = = 7200 so = so hord 90 = (NOTE: I will stop putting the ellipsis in (...) just for simplicity. I will just truncate the expressions at an arbitrary place.) T If we now let T = 120, so that T must be 60, we know: T 2 = 2 T 2 = = = so T = = so hord 120 = NOTE: WE N FIN HORS OF SUPPLEMENTS. We can now find the chord for any angle which is the supplement of an angle whose chord is known. In the example above, we knew T, the chord of 60, and we knew, the diameter. That, together with the Pythagorean Theorem, was all we needed in order to compute T, the chord of the supplement of 60. There was nothing special about 120 and 60. So now, if we know the chord of any angle, we will also be able to compute the value of the chord of its supplement. (7) FINING THE HORS OF 144 & 108. So using the very same technique, we can find the chords of 144 and 108, since these are the supplements of 36 and 72 respectively, and the chords of those arcs are known. 50

6 (8) LEMM (FOR THE UPOMING PROOF THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN). This may be called the iagonal ross-product Theorem. The Theorem states that if is a cyclic quadrilateral (i.e. a quadrilateral inscribed in a circle), then = + or, put verbally: The rectangle contained by the diagonals is equal to the sums of the rectangles contained by the pairs of opposite sides. 6 E 5 7 gain, since this is just a matter of going through the steps, we will assume it is true and use it, but not bother proving it together. (9) PROOF THT HORS OF RS WHIH RE IFFERENES OF RS WITH GIVEN HORS RE GIVEN (i.e. calculable). Given: Prove: arcs & in degrees; chords & in 120 th parts of diameter chord is also given in 120 th parts of (i.e. can be calculated) Since & are given, thus is given [Euclid 1.47] 51

7 Since & are given, thus is given [1.47] Q.E.. ut = + [by the cross-product Lemma] and all terms in that equation are given except for. Hence is also given. NOTE: We are not just thinking of. (for example) as a rectangle, but as a product of two numbers. There s plenty of philosophically discussible matter there. EXMPLE of the use of this Theorem for our Table: If arc = 72 and arc = 60 then arc = 12. nd since the chords of 72 and 60 are already given, it follows by this Theorem that the chord of 12 is also given now, or calculable to any degree of accuracy we please. (10) PROOF THT HORS OF RS WHIH RE HLVES OF RS WITH GIVEN HORS RE GIVEN. Ptolemy next shows that if we know the chord of a known arc, then we can also calculate the value of the chord of half that arc. That will help fill in a whole lot of entries on the table! If we know arc in degrees (and its midpoint is ), and chord in 120 th parts of diameter, then we will also be able to calculate the value of chord in those units. E F 52

8 Given: Prove: Make: arc in degrees chord in 120 th parts of the diameter arc = arc is given (i.e. the chord of half the given arc ) E = F perpendicular to = E common = E [Euclid 3.27; they stand on equal arcs] so r r E so = E i.e. = E so r EF r F so F = EF i.e. F = ½ E so F = ½ [ E] F = ½ [ ] so F is given [since is given, is supplement of given] Now : = : F [r similar to r F; 6.8] so F = 2 so 2 is given [ & F are given] so is given Q.E.. EXMPLES: Since we had the chord of 12, now we have the chords of 6, 3, 1½, ¾ (11) RS WHOSE HORS RE NOW GIVEN: irect Geometry: Supplements: 36 (side of decagon) 144 = (radius, side of hexagon) 120 = (side of pentagon) 108 = (side of square) 180 (diameter) Subtraction: Supplements: 53

9 24 = = = = = = = = = = = = = = = = isection: Supplements: 42 = = = = = = Now we have all multiples of 6 isection: Supplements: 3 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Now we have all multiples of 3, and by isection: 1½ = 3 2 (12) HORS OF RS WHIH RE SUMS OF RS WITH GIVEN HORS RE GIVEN (this is like addition ) Ptolemy adds this Theorem that will help to fill in new entries on our Table: Given: rcs and are given in degrees hords and are given in diameter-parts Prove: is also given in diameter-parts F 54 E

10 Make: iameters FE and F E is given is given E is given [supplement of ] [supplement of ] [supplement of ] ut E = E + E [quadr. multiplication Lemma, item (11) above] so is given [all other terms given in there] so is given [supplement of ] Q.E.. PPLITION: We can add 1½ to itself thus, getting 1 out of every 3 terms on the table. Now we have 1½ 3 4½ 6 etc. NOTE: The use of given here means can be calculated to whatever degree of accuracy you please, e.g. to as many decimal or sexagesimal places you want. Sometimes we need numerical exercises to get the gist of this. For instance, we have seen that hord 36 = and hord 1½ = so now, using this Theorem, see if you can determine the numerical value for hord 37½. 55

Definitions. (V.1). A magnitude is a part of a magnitude, the less of the greater, when it measures

Definitions. (V.1). A magnitude is a part of a magnitude, the less of the greater, when it measures hapter 8 Euclid s Elements ooks V 8.1 V.1-3 efinitions. (V.1). magnitude is a part of a magnitude, the less of the greater, when it measures the greater. (V.2). The greater is a multiple of the less when