Mathematical spheres and solid globes

Transcription

1 ancient reek : Mathematical spheres and solid globes Nathan Sidoli (joint work with Ken Saito) I Paris, February 2010

2 Instrumental practices of reek geometry xamples of constructions in Practice s I I s

3 in reek mathematics side from some abacus boards and papyri containing calculations, we have no material evidence for practices in reek mathematics. Nevertheless, there is textual evidence for various instruments. Some examples: Plato s and uratothenes devices for analog calculation of two mean proportionals. Nicomedes device for producing a conchoid for use in neusis constructions. iocles description of a bone ruler for drafting a parabola. Ptolemy s analemma board and Heron s local hemisphere, for analog calculations. I

4 The geometry of instrumental practice ertain domains of geometric theory may have been, in some sense, delimited by certain instrumental practices. The classic example is uclid s elementary geometry, which is based on the postulation of operations equivalent to the use of an unmarked ruler and a collapsing compass. Pappus s ollection VIII gives constructions using an unmarked ruler and a fixed compass. The analemma techniques rely on the use of a set square and a non-collapsing compass. I will argue in this talk that ancient spherics was a geometry constructed with an unmarked ruler and a non-collapsing compass. I

5 in reek astronomy We have no surviving observational instruments, but we have many descriptions. Mostly from Ptolemy: the meridian quadrant, the equatorial ring, the armillary sphere, etc. nd others as rchimedes, eminus and Proclus: the dioptra, the angular size dioptra, the armillary sphere, the gnomon, etc. We have a fair number of objects related to the general cultural interest in astronomy and astrology. Sundials (fixed, portable, geared), the ntikythera mechanism, the parapegmata and other astronomical inscriptions, etc. strologer s boards and markers, inscribed horoscopes, etc. I

6 vidence for reek globes We have both material and textual evidence for globes in reek antiquity. Some examples: Ptolemy (lmagest) describes the construction of an elaborate star globe, Leontius describes a demonstration globe. eminus refers to inscribed globes and Ptolemy (Plansiphere) talks about the lines that are found on struck globes. We know of three ornamental globes the Farnese tlas and two simple celestial spheres inscribed with the principle circles and decorated with images of the constellations (all 1 st or 2 nd century). I

7 The propositions of elementary treatises, like the lements or the, are divided into two types: Theorems iven some set of initial objects, a theorem asserts some property that is true of these objects. ( If... then... ) Problems iven some set of initial objects, a problem shows how to do something (say, to find, draw, set out) and then demonstrates that what has been done is satisfactory. ( To do such-and-such... ) I Some theorems can be intelligibly expressed as problems and the converse. (We will see that theorem Spher. I 8 might very well have been a problem.)

8 The components of a reek proposition Proclus (5 th ) put forward the following six parts of a reek proposition. (In fact, they are only ever found so complete and clearly divided in uclid s lements.) Introductory components 1 1 nunciation (prìtasic): general statement of what is to be shown (done). 2 xposition (êkjesic): statement setting out the given objects with letter names. 3 Specification (diorismìc): restatement in terms of the specific objects of (1) what is to be shown (theorem), or of (2) what is to be done, including any conditions of solvability (problem). I 1 Introductory division not distinguished by Proclus.

9 The components of a reek proposition Proclus (5 th ) put forward the following six parts of a reek proposition. (In fact, they are only ever found so complete and clearly divided in uclid s lements.) ore components 1 4 onstruction (kataskeuă): Statements about the production of new objects that will be required in the proof. (Relies on posts. 1-3 and problems.) 5 Proof (Ćpìdeixic): logical argument that the proposition holds (has been done). (Relies on the other assumptions and theorems.) 6 onclusion (sumpèrasma): restatement, in general terms, of what has been shown (done). 2 I 1 ore division not distinguished by Proclus. 2 Very rare, except in the lements.

10 Using constructions to prove theorems In a geometric theorem, it is sometimes the case that the properties of the objects stated in the enunciation are sufficient to demonstrate the proposition, but more often than not we have to introduce new objects and use their properties in the argument. These new objects are introduced using constructions. (In fact, the geometer asserts that these objects must have been produced, using the same grammatical constructions, and sometimes even the same verbs as are used when the initially given objects are set out. 3 ) I 3 In Spher. I 8, we will see a case where the exposition is undistinguishable from the construction, except on the basis of the proof structure.

11 Using constructions to solve problems problem is solved by producing a specific geometric object. problem shows how to produce the object using constructions and then uses deductive argumentation to show why this object is correct. This deductive argumentation sometimes requires new constructions in the same way as a theorem. That is, the geometric object that solves the problem together with the initial objects stipulated in the problem are sometimes not sufficient to show that this object is a solution. In such cases, we must construct new objects for the proof. I

12 xample theorem, I 1 1 If a plane cuts a spherical surface, the curved line in the spherical surface is a circle. 2 Let a plane cut a sphere and make curved line. 3 I say, is a circle. I Upper: manuscript figure Lower: reconstruction

13 xample theorem, I 1 ore [ase 1] 5 1 If the plane goes through the center of the sphere, then every line from the center of the sphere to is equal (definition of a sphere), so that curved line is a circle. I Upper: manuscript figure Lower: reconstruction

14 xample theorem, I 1 ore [ase 2] 4 2 If the plane does not go through the center of the sphere, let the center of the sphere be [?], a draw to plane [l.11.11], join,, and (P.1). I 5 2 Use l.1.47 to show that =. This could be done for any other two points on curved line. Hence, is a circle with center. Upper: manuscript figure Lower: reconstruction a How we locate is passed over with a simple use of êstw. ompare with the next proposition.

15 xample problem, I 2 H Z K 1 To find the center of a given sphere. 2 Let there be a given sphere. a 3 I say, it is necessary to find its center. I H a Notice there are no letter names. K Z Upper: manuscript figure Lower: reconstruction

16 xample problem, I 2 ore [solving the problem] H Z H K 4 Let some plane cut it, making circle [?;Sph.1.1]. a Find as the center of [l.3.1], draw Z circle to meet the sphere at Z, [l.11.12;p.2]. isect Z at H [l.1.10]. 3 const I say, H is the center of the sphere. I K Z Upper: manuscript figure a We are not told how this is done. Perhaps we can imagine some sort of postulate, To pass a plane through a solid figure. t any rate, such a construction was regularly performed. Lower: reconstruction

17 xample problem, I 2 H ore [verifying the solution] Z H K 5 ssume, if possible, that there is some other center, say. 4 proof Let a perpendicular be drawn from to circle at K [l.11.11]. Then, K is a center of [Sph.1.1], and so is, which is impossible. I K Z Upper: manuscript figure Lower: reconstruction

18 omponents of a eometry Problem 1 p nunciation: eneral statement of what is to be done. 2 p xposition: Statement of what is given, usually using specific, letter names. 3 p First Specification: Specific statement of what is to be done, often with qualifications. 4 p Solution: onstruction of the geometric object which satisfies the requirements of the problem. 5 p Second Specification: Specific statement of what is to be shown. 1 6 p onstruction: onstruction of any new objects necessary to the proof. 1 7 p Proof: rgument that the solution meets the requirements of the proposition; if necessary, using the new objects of the construction. I 1 May be absent.

19 The role of constructions in I The book opens with a number of propositions that use lines internal to the sphere to demonstrate the properties of great circles and sets of parallel circles and their relations. For most of the book, constructions serve the role of supplying objects which are necessary to proofs. We then have two problems that show us how to draw these internal lines outside of the sphere, into a space where we can use them. (From this point on, we can solve problems by actually drawing some object.) I The book ends with two problems that show how to draw a great circle through two points and how to find the pole of a circle working entirely outside of the sphere.

20 Using internal constructions, I 8 Z H Z H Upper: manuscript figure 1 If a circle is in a sphere and a perpendicular is drawn from the center of the sphere and extended in both directions, it will fall on the poles of the circle. 2 Let be a sphere, let its center,, be taken [Sph.1.2], let be drawn [l.11.11], and extended to meet the sphere at Z and H [l.p.2]. 3 I say, Z and H are the poles of. I Lower: reconstruction

21 Using internal constructions, I 8 Z ore 4 Let,, Z, Z, H and H be drawn [l.p.2]. H H Z 5 =, the s at are right and is common, so Z = Z. We could make the same argument for any other point on. Likewise with respect to H. So Z and H are the poles of circle. 6 If a circle is in a sphere, from the center of the sphere, and so on. I Upper: manuscript figure Lower: reconstruction

22 ringing out a diameter, I 18 1 p To set out (âkjèsjai) the diameter of a given circle in a sphere. 2 p Let the given circle be. I 3 p So, it is necessary to set out its diameter.

23 ringing out a diameter, I 18 ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. I

24 ringing out a diameter, I 18 F ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. Let Z be put together from (sunestĺtw) three lines (such that equals that from to, etc.) [?]. I Upper: a plane diagram, outside the sphere Lower: a solid sphere

25 ringing out a diameter, I 18 H F ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. Let Z be put together from (sunestĺtw) three lines (such that equals that from to, etc.) [?]. t point, draw H [l.1.11], I Upper: a plane diagram, outside the sphere Lower: a solid sphere

26 ringing out a diameter, I 18 H F ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. Let Z be put together from (sunestĺtw) three lines (such that equals that from to, etc.) [?]. t point, draw H [l.1.11], t point Z, draw ZH Z [l.1.11], I Upper: a plane diagram, outside the sphere Lower: a solid sphere

27 ringing out a diameter, I 18 H F ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. Let Z be put together from (sunestĺtw) three lines (such that equals that from to, etc.) [?]. t point, draw H [l.1.11], t point Z, draw ZH Z [l.1.11], and join H [l.p.1]. I Upper: a plane diagram, outside the sphere Lower: a solid sphere

28 ringing out a diameter, I 18 H F ore [solving the problem] 4 p Let three random points,, and, be taken on the circumference. Let Z be put together from (sunestĺtw) three lines (such that equals that from to, etc.) [?]. t point, draw H [l.1.11], t point Z, draw ZH Z [l.1.11], and join H [l.p.1]. I Upper: a plane diagram, outside the sphere 5 p <I say, line H is equal to the diameter of circle.> Lower: a solid sphere

29 ringing out a diameter, I 18 ore [verifying the solution] F 6 p raw the diameter of circle [?; l.3.1&p.1], H I Upper: a plane diagram, outside the sphere Lower: a solid sphere

30 ringing out a diameter, I 18 ore [verifying the solution] F 6 p raw the diameter of circle [?; l.3.1&p.1], and join,, and [l.p.1]. H I

31 ringing out a diameter, I 18 ore [verifying the solution] F 6 p raw the diameter of circle [?; l.3.1&p.1], and join,, and [l.p.1]. H 7 p Now, =, = and = Z, = Z [l.1.4]. nd = while Z = HZ [l.3.27], so = HZ. ut = ZH = R and =, so = H. Therefore, H is equal to the diameter of the circle. I

32 ringing out a diameter, I 19 Summary of the solution Let there be a sphere. I

33 ringing out a diameter, I 19 Summary of the solution Let there be a sphere. We take two random points, and. I

34 ringing out a diameter, I 19 Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. I

35 ringing out a diameter, I 19 F H Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. Then it is possible set out the diameter of this circle as F [Sph.1.18], I Upper: a plane diagram, outside the sphere Lower: a solid sphere

36 ringing out a diameter, I 19 F H Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. Then it is possible set out the diameter of this circle as F [Sph.1.18], so that we can put together FH [?;le.1.22*]. I Upper: a plane diagram, outside the sphere Lower: a solid sphere

37 ringing out a diameter, I 19 F H Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. Then it is possible set out the diameter of this circle as F [Sph.1.18], so that we can put together FH [?;le.1.22*]. We draw H H and F F [le.1.11], meeting at point. I Upper: a plane diagram, outside the sphere Lower: a solid sphere

38 ringing out a diameter, I 19 F H Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. Then it is possible set out the diameter of this circle as F [Sph.1.18], so that we can put together FH [?;le.1.22*]. We draw H H and F F [le.1.11], meeting at point. We join [le.p.1]. I

39 ringing out a diameter, I 19 F H Summary of the solution Let there be a sphere. We take two random points, and. With as pole and as distance, we draw circle [?]. Then it is possible set out the diameter of this circle as F [Sph.1.18], so that we can put together FH [?;le.1.22*]. We draw H H and F F [le.1.11], meeting at point. We join [le.p.1]. I K We draw the internal lines and show that line is equal to the diameter of sphere.

40 n interlude I 19 is never explicitly used in the text. F H In a number of propositions, however, it is required to let a circle be drawn with a given point as pole and a pole-distance equal to the side of a square described in a great circle. This construction can be effected using I 19. I We set out the diameter of the sphere, as. We construct a square on as diameter, and use one of its sides, say H, as the pole-distance of the great circle.

41 rawing a great circle through two points, I 20 Summary of the solution, 1 Let and be two given points. I

42 rawing a great circle through two points, I 20 Summary of the solution, 1 Let and be two given points. [ase 1:] If and are the endpoints of a diameter of the sphere, there are an infinite number of great circles through them [?]. I

43 rawing a great circle through two points, I 20 Summary of the solution, 2 Let and be two given points. [ase 2:] With as pole, we draw great circle [Sph.1.19 (*inter.)]. I

44 rawing a great circle through two points, I 20 F H Summary of the solution, 2 Let and be two given points. [ase 2:] With as pole, we draw great circle [Sph.1.19 (*inter.)]. With as pole, we draw great circle FH [Sph.1.19 (*inter.)], intersecting great circle at points and. I

45 rawing a great circle through two points, I 20 F H Summary of the solution, 2 Let and be two given points. [ase 2:] With as pole, we draw great circle [Sph.1.19 (*inter.)]. With as pole, we draw great circle FH [Sph.1.19 (*inter.)], intersecting great circle at points and. With, or, as pole, we draw a circle passing through either or [le.p.3*]. I

46 rawing a great circle through two points, I 20 F H Summary of the solution, 2 Let and be two given points. [ase 2:] With as pole, we draw great circle [Sph.1.19 (*inter.)]. With as pole, we draw great circle FH [Sph.1.19 (*inter.)], intersecting great circle at points and. With, or, as pole, we draw a circle passing through either or [le.p.3*]. I We argue that = are equal to the side of a square inscribed in a great circle.

47 Finding the pole of a given circle, I 21 Summary of the solution, 1 Let be a given circle. I

48 Finding the pole of a given circle, I 21 Summary of the solution, 1 Let be a given circle. We take a random point,. I

49 Finding the pole of a given circle, I 21 Summary of the solution, 1 Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. I

50 Finding the pole of a given circle, I 21 Summary of the solution, 1 Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. a F I a If we set out the diameter of, we can transfer the arcs = and bisect the arc using le.3.30.

51 Finding the pole of a given circle, I 21 F Summary of the solution, 1 Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 1:] If circle is not a great circle, we draw great circle F [Sph.1.20]. I

52 Finding the pole of a given circle, I 21 Summary of the solution, 1 H F Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 1:] If circle is not a great circle, we draw great circle F [Sph.1.20]. We bisect arc HF at point H [?]. a I We argue that all the lines drawn from H to circle are equal. a We set out the diameter of the sphere [Sph.1.19] and draw a great circle. Then we set out the diameter of circle [Sph.1.18], fit it into the great circle [le.4.1] and bisect the resulting cord [le.3.30].

53 Finding the pole of a given circle, I 21 Summary of the solution, 2 Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 2:] If circle is a great circle, we bisect arc F at [?]. a I F a We set out the diameter of the sphere [Sph.1.19], draw a great circle, and bisect a semicircle [le.3.30].

54 Finding the pole of a given circle, I 21 Summary of the solution, 2 F Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 2:] If circle is a great circle, we bisect arc F at [?]. With as pole, we draw a great circle through and F [Sph.1.19 (*inter.)]. I

55 Finding the pole of a given circle, I 21 Summary of the solution, 2 H F Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 2:] If circle is a great circle, we bisect arc F at [?]. With as pole, we draw a great circle through and F [Sph.1.19 (*inter.)]. We bisect arc F at H [?]. a I a gain, we reproduce the arc outside the sphere, bisect it and then transfer the distance back to the sphere.

56 Finding the pole of a given circle, I 21 Summary of the solution, 2 H F Let be a given circle. We take a random point,. We cut off arcs = [le.p.3*]. We bisect arc at F [?]. [ase 2:] If circle is a great circle, we bisect arc F at [?]. With as pole, we draw a great circle through and F [Sph.1.19 (*inter.)]. We bisect arc F at H [?]. I We argue that all the lines drawn from H to circle are equal.

57 rawing a tangent great circle, II 14 Summary of the solution Let there be a given lesser circle with point given on it. I

58 rawing a tangent great circle, II 14 Summary of the solution Let there be a given lesser circle with point given on it. We find the pole of circle, as point [Sph.1.21]. I

59 rawing a tangent great circle, II 14 Summary of the solution Let there be a given lesser circle with point given on it. We find the pole of circle, as point [Sph.1.21]. We draw a great circle through and [Sph.1.21], I

60 rawing a tangent great circle, II 14 Summary of the solution Let there be a given lesser circle with point given on it. We find the pole of circle, as point [Sph.1.21]. We draw a great circle through and [Sph.1.21], and cut off arc equal to a quadrant [Sph.1.19 (*inter.)]. I

61 rawing a tangent great circle, II 14 Summary of the solution F Let there be a given lesser circle with point given on it. We find the pole of circle, as point [Sph.1.21]. We draw a great circle through and [Sph.1.21], and cut off arc equal to a quadrant [Sph.1.19 (*inter.)]. With pole, we draw great circle F [le.p.3*]. I

62 rawing a tangent great circle, II 14 Summary of the solution F Let there be a given lesser circle with point given on it. We find the pole of circle, as point [Sph.1.21]. We draw a great circle through and [Sph.1.21], and cut off arc equal to a quadrant [Sph.1.19 (*inter.)]. With pole, we draw great circle F [le.p.3*]. I The proof follows immediately from the properties of tangency [Sph.2.3-5].

63 rawing a tangent great circle, II 15 Summary of the solution, 1 Let be a given lesser circle and a given point, not on it. a I a Point must be between and the circle equal and parallel to it.

64 rawing a tangent great circle, II 15 Summary of the solution, 1 Let be a given lesser circle and a given point, not on it. We find the pole of circle, as point [Sph.1.21]. I

65 rawing a tangent great circle, II 15 F Summary of the solution, 1 Let be a given lesser circle and a given point, not on it. We find the pole of circle, as point [Sph.1.21]. With as pole, we draw lesser circle F [le.p.3*]. I

66 rawing a tangent great circle, II 15 F Summary of the solution, 1 Let be a given lesser circle and a given point, not on it. We find the pole of circle, as point [Sph.1.21]. With as pole, we draw lesser circle F [le.p.3*]. We draw great circle, meeting circle at point [Sph.1.20], I

67 rawing a tangent great circle, II 15 Summary of the solution, 1 F Let be a given lesser circle and a given point, not on it. We find the pole of circle, as point [Sph.1.21]. With as pole, we draw lesser circle F [le.p.3*]. We draw great circle, meeting circle at point [Sph.1.20], and cut off quadrant [Sph.1.19 (*inter.)]. I

68 rawing a tangent great circle, II 15 Summary of the solution, 1 F H Let be a given lesser circle and a given point, not on it. We find the pole of circle, as point [Sph.1.21]. With as pole, we draw lesser circle F [le.p.3*]. We draw great circle, meeting circle at point [Sph.1.20], and cut off quadrant [Sph.1.19 (*inter.)]. With as pole, we draw great circle H [le.p.3*]. I

69 rawing a tangent great circle, II 15 M F N H Summary of the solution, 2 We draw two great circles, H and, meeting circle at points N and M [Sph.1.20]. I

70 rawing a tangent great circle, II 15 Summary of the solution, 2 M F N H K L We draw two great circles, H and, meeting circle at points N and M [Sph.1.20]. We cut off arcs HL and K equal to arc [le.p.3*]. I

71 rawing a tangent great circle, II 15 Summary of the solution, 2 O M F N H K X L We draw two great circles, H and, meeting circle at points N and M [Sph.1.20]. We cut off arcs HL and K equal to arc [le.p.3*]. With L as pole and distance LN, we draw circle XN, and with K as pole and KM as distance, we draw circle OM [le.p.3*]. I

72 rawing a tangent great circle, II 15 Summary of the solution, 2 O M F N H K X L We draw two great circles, H and, meeting circle at points N and M [Sph.1.20]. We cut off arcs HL and K equal to arc [le.p.3*]. With L as pole and distance LN, we draw circle XN, and with K as pole and KM as distance, we draw circle OM [le.p.3*]. I We draw in the internal lines and show that the are all equal to the side of a great square.

73 l-harawī on One sees he considers inadequate in his treatise On Spheres and thinks that the way he followed is other than satisfactory, since there is difficulty in it, and the setting out of many lines, and he did not adhere, in it, to the properties of the figures that occur on the sphere, namely the conditions of the angles that arise from the intersection of the circles. Upon my life, has easily shown everything that proved in that book and he intends that the proof be straightforward ( ) 4 without using straight lines. I 4 That is by the correct way, the by the straight route. Maybe direct, as opposed to indirect.

74 To construct equal angles, Spher. I 1 1 p We want to make an angle at a given point on the circumference of a great circle that is equal to a given angle. 2 p Let be the given point on great-circle arc and let the given angle be angle. I 3 p So, it is necessary to make an angle at equal to angle.

75 To construct equal angles, Spher. I 1 ore [solving the problem] 4 p With as pole, we draw arc with whatever distance ( ). I

76 To construct equal angles, Spher. I 1 ore [solving the problem] 4 p With as pole, we draw arc with whatever distance ( ). With as pole, we draw arc Z with the same distance, I Z

77 To construct equal angles, Spher. I 1 ore [solving the problem] 4 p With as pole, we draw arc with whatever distance ( ). With as pole, we draw arc Z with the same distance, and we cut off arc Z = arc. I Z

78 To construct equal angles, Spher. I 1 ore [solving the problem] Z 4 p With as pole, we draw arc with whatever distance ( ). With as pole, we draw arc Z with the same distance, and we cut off arc Z = arc. We join great-circle arc Z. I

79 To construct equal angles, Spher. I 1 ore [solving the problem] Z 4 p With as pole, we draw arc with whatever distance ( ). With as pole, we draw arc Z with the same distance, and we cut off arc Z = arc. We join great-circle arc Z. 5 p I say, angle Z is equal to angle. I

80 To construct equal angles, Spher. I 1 ore [verifying the solution] Z 6 p Two great circles, &, pass through the pole of circle, so their planes are to circle [TSph.1.15]. The two intersections of great circles & with circle, pass through the center of circle, while intersection of circles & is to circle at its center, hence the intersections of circles & with circle are each the intersection of circles &... I

81 To construct equal angles, Spher. I 1 ore [verifying the solution] Z 6 p... The same argument can be made for the internal objects of triangle Z. ut circle Z = circle and arc Z = arc, so the inclination of plane Z on plane is equal to the inclination of plane on plane. Therefore, Z = [def. of equal inclination]. I

82 ll of the constructions used to solve problems in ancient spherics can be carried out through physical manipulations of an unmarked ruler and a non-colapsing compass. In the texts on spherics, an instrument like a compass must be used to transfer distances. The use of practical constructions appears to have guided not only the solution to problems but also to have influenced the overall, theoretical approach. moves from considerations of internal lines to considerations of the objects on the surface of the sphere. attempts to restrict his attention to the objects on the surface. (Of course, he still uses internal lines, but he does not name them; moreover, he uses s theorems that make use of internal lines.) I