1Berggren trans. MEASUREMENT OF A CIRCLE. 03a0 03b1 0xxx 03c2 03f0 03cd 03f0 03bb 03bf 03c2. 03b2 0384(tonos) 0387(ano teleia)


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1 Use Deja Vu Serif Condensed, 10 point/6 point Direct translation by J. Holly DeBlois *****TITLE***** 1 Title: This is Archimedes' Dimensions of A Circle, ed. Heiberg, pp Greek ΚΥΚΛΟΥ ΜΕΤΡΗΣΙΣ 1unicode 1direct trans. Of Circles Measure 1Berggren trans. MEASUREMENT OF A CIRCLE. *****PROP1***** 1.1 Proposition 1: 0Greek ά 0unicode 0translation a. 0translation Berggren Proposition Sentence 1: 1Greek 1unicode 1direct translation 1translation Berggren Πα ς ϰύϰλος 03a0 03b1 0xxx 03c2 03f0 03cd 03f0 03bb 03bf 03c2 All circles [area] measures the same as a right triangle, where distance from th The area of any circle is equal to a rightangled triangle in which one of the sid Stop here. *****PROP2***** 1.2 Proposition 2: 0Greek β 0unicode 0translation b. 0translation Berggren Proposition 2. 03b2 0384(tonos) 0387(ano teleia) Sentence 1: 1Greek 1unicode Ο ϰύϰλος 039f(Greek capital letter omicron) 03f0 03cd 03f0 03bb 03bf 03c2 1direct translation The circle pros to apo whose diameter squared logon is, on ia pros id. 1translation Berggren The area of a circle is to the square on its diameter as 11 is to 14. Stop here. Page 1
2 *******PROP3********* Note: I'm anglicizing Gamma as G not C. 1.3 Proposition 3: p262 0Greek γ 0unicode 0direct translation c. 0translation Berggren Proposition 3. p93 ***** 03b3(Greek small letter Gamma) 0384(tonos) 0387(ano teleia) Section 1: Sentence 1: 1Greek 1unicode 1direct translation p262 Παντος κύκλον ή περίμετρος της διαμέτρον τριπλασίων εστί, και έ ελάσσονι μεν ή έβδόμω μέρει τής διαμέτρον, μείζονι δε ή δέκα έβδ 03a0(capital pi) 03b1(small alpha) 03bd(small nu) 03c4(small tau) 03bf(small omicron) 0384(tonos) 03c2(small final sigma) In any case, the perimeter of the circle to its diameter is triple, but still measur 1translation Berggren The ratio of the circumference of any circle to its diameter is less than 3 1/7 bu p93 ***** Gist of argument: 1 Consider a circle with center E, point G on the circumference C, radius EG of le 2 Place a vertical line tangent to the circle at G. 3 Let Z be the point on the tangent such that the line from Z to center of circle E 4 Triangle ZEG has a right angle at G and the 30 degree angle at E. 5 Consider the top of the triangle to be angle ZEG, so the (unequal) sides are ZE 6 In right triangle ZGE, designate length ZE as hyp for hypotenuse and length ZG 7 Length opp is onehalf of the side of a polygon of six sides that circumscribes th 8 By Euclid VI.3, the ratio of hyp to opp is 2:1 (or cosecant of angle ZEG). 9 Also, the ratio of r to opp is square root 3: 1 (or cotangent of angle ZEG). 10 Note bene: the ratio of opp to r gives the first estimate of C/D as 6*(1:square ro 11 To calculate, with square root of 3 estimated as 1.5, so 1/1.5 = 2/3, so 6*2/3 = 1 12 Now, increase the number of sides of the polygon by bisecting the angle ZEG re 13 As an example, use a circle of radius r (C4correction was r=153,now r=265/15 14 In the example, hyp/opp=2/1=307/153 and r/opp=square root 3/1=265/153, so 15 Note that these two formulas use all three sides of right triangle ZEG and know 16 To make the first increase in number of sides, bisect angle ZEG. 17 Construct line EH from center E to point H on tangent line. 18 Because H is on the tangent line outside the circle and between Z and G, EH > 19 Let HGE be the new right triangle with right angle HGE and angle HEG = 15 d Page 2
3 20 Consider the top of this triangle to be angle HEG, so the (unequal) sides are HE 21 In right triangle HGE, designate length HE as hyp2 and base length GH as opp2 22 As before, the third side has length r. 23 Now compare (shorter) base GH in right triangle HGE to base GZ of right trian 24 Base GZ is bifurcated by H, giving segments GH and HZ. 25 The angle ZEG was bisected into two equal parts, but segments GH and HZ are 26 By Euclid VI:3, for a bisected angle at the top of a triangle, the segments of the 27 Because sides ratio ZE to r in triangle ZGE has ZE > r, base segments ratio ZH 28 Now use the construction for angle HEG to estimate the polygon side using the 29 Its base GH is onehalf of one side of the 12sided polygon determined by angle 30 We know the radius r which is one side of triangle HGE, but we need to know th 31 We seek to express the length HG in terms of the known ratios hyp/opp (or hyp 32 By Euclid above for the larger right triangle ZGE, unequal sides ratio hyp/r = u 33 Now expand the ratio numerator to include both pieces, keeping the denominat 34 This gives ratio two sides/r = base/hg which is hyp + r/r = base ZG/base segm 35 Next swap two 'middle' components, ie for 1/4=3/12, it is also true that 1/3=4/1 36 This gives ratio two sides/base = r/hg which is hyp + r/zg = r/hg.(c3clearer 37 We know hyp/zg and r/zg from (C5addition: line 14) the ratios in the example 38 That means hyp +r/zg = hyp/opp + r/opp = (hyp + r)/opp = ( )/153 = 5 39 Therefore r/hg=571/153, which is larger than the starting ratio r/ 40 Now work to get C/D by flipping the ratio which gives us that the ratio of the ta 41 This gives us a next approximation for the ratio of the circumference to the diam 42 The tangent segment HG is onehalf the length of a side of a polygon with 12 si 43 Thus, the second approximation of the ratio of the circumference to the diamete 44 HG/r is 153:571, so what is 12*HG/r? It is ( ): 571 or 1836:571 or This value is less than the first estimate C/D=4 and greater than the proposition 46 Then we repeat for more bisections of the angle at E. Section 2: Sentence 1: 1Greek 1unicode 1direct translation 1translation Berggren p93 p264 ἔστω ϰύϰλος, ϰαὶ διαμετρος ἡ ΑΓ, ϰαὶ ϰέντρον το Ε, ϰαὶ ἡ ΓΛΖ ἑ So we have this circle, and also diameter AG, and also center E, and also line G I. Let AB be the diameter of any circle, O its centre, AC the tangent at A; and le Diagram 1 Descriptions: Labels The Greek version of the picture with transliterated letters and no lines for the Page 3
4 Z H L G E Greek numbers used by Archimedes: ϚϜϞϠ source:http://www.mlahanas.de/greeks/counting.htm 1=α,2=β,3=γ,4=δ,5=ε,6={digamma,Ϝ, or stigma,ϛ },7=ζ,8=η,9=θ,10=ι 11=ια,...,20=ϰ, 21=ϰα,..., 30=λ,40=μ,50=ν,60=ξ,70=ο,80=π,90=Ϟ 100=ρ,200=σ,300=τ,400=υ,500=φ,600=χ,700=ψ,800=ω,900=Ϡ sampi use iota sub or superscript with letters for =M, and higher numbers put the number in front of M (or on top of it) 27 symbols: numbers 110, tenths , hundredths , greek letters plus 34 phoen 1119: ia, ibeta, etc.; : 347 tau mu zeta Sentence 2: 2Greek 2unicode ἡ ΕΖ ἂρα πρὸς ΖΓ λόγον ἒχει, ὃν τζ προς ρνγ. 2direct translation The EZ therefore to ZG surely is, as which 307 to 153. This is the ratio of the length of slant line from center E to point Z Note: there may have been Dimensions of a Triangle written bef It is 2:1 because angle ZEG is 1/3 of a right angle or 30 degrees. W Berggren reverses it: 2translation Berggren The OA:AC is >265:153...(1), and OC:CA = 306:153...(2). [OA:AC= p94 the horiz. to the vert. the slant to the vertical going down So for this line: The OC:CA = 306:153...(2). [OC:CA=2:1] Sentence 3: 3Greek ἡ δὲ ΕΓ πρὸς τὴν ΓΖ λόγον ἒχει, ὃν σξέ προς ρνγ. 3unicode sigma xi epsilon to direct translation But the [other side] EG to [vertical] GZ surely is, as which 265::15 Page 4
5 the horiz. the vert. going up The ratio of the radius EG to tangent segment GZ is square root 3 3translation Berggren And for this line: The OA:AC is >265:153...(1) [OA:AC=square roo Sentence 4: DRAW NEXT LINE (Bisect 30 degrees, get 15 degrees) 4Greek 4unicode 4direct trans. τετμήσδω οὖν ἡ υπο ZEΓ δίχα τη EH. Bisect therefore the angle ZEG and [draw] EH to divide it [evenly] 4Berggren trans. First, draw OD bisecting the angle AOC and meeting AC in D Sentence 5: 5Greek 5unicode 5direct trans. 5Berggren trans. Euclid: VI: 3: έστιν άρα, ώς ή ΖΕ προς ΕΓ, ή ΖΗ προς ΗΓ [καί έναλλάξ καί συνδ Therefore it is, as far ZE to EG, as ZH to HG [and also alternately and also conn Now CO:OA=CD:DA, [Eucl.VI.3] If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle; and, if segments of the base have the same ratio as the remaining sides of the triangle, then the straight line joining the vertex to the point of section bisects the angle of the triangle. so that [CO+OA:OA=CA:DA, or] CO+OA:CA=OA:AD. Translating letters in picture: ZE:EG = ZH:HG, it's above [ZE + EG:EG = ZG:HG or] it's added to explain ZE + EG:ZG = EG:GH It's covered next sentence Sentence 6: 6Greek 6unicode 6direct trans. ώς άρα συναμφότερος ή ΖΕ, ΕΓ προς ΖΓ, ή ΕΓ προς ΓΗ. As far as therefore ZE connected to EG in ratio to ZG is the same a Therefore same ratios are ZE plus EG in ratio to ZG and EG in rati ie slant side plus radius to vertical as is radius to lower section o 6Berggren trans. CO+OA:CA=OA:AD. ZE + EG:ZG=EG:GH 1/4=2/8 so 1/2=4/8 Not often used. We know: ZE/ZG=2/1 and EG/ZG=sq.root 3/1. Page 5
6 By the steps above we reach: (ZE +EG)/ZG = EZ/HG. Compute (ZE + EG)/ZG = ZE/ZG + EG/ZG = 2/1+root3/1 = 306/ Sentence 7: 7Greek ωστε ή ΓΕ προς ΓΗ μείζονα λογον εχει, ήπερ φοά προς ρνγ. 7unicode EG:GH greater now, bigger as: direct trans. Consequently, GE is to GH all the more so has, more now as 571 to Consequently, the ratio of the radius EG to tangent segment GH is Was 265:153, which is square root 3 to 1, and is now: 571:153 7Berggren trans. Therefore [by (1) and (2)] OA:AD.571:153...(3). p94 EG:GH Sentence 8: 19,450 measures what? 9,450 measures what? Could the M be 340,000? it's in Bergren! 13,404 measures what? 3,404 measures what? Could M be 20,000? 3,409? Line 68(7) has:so hyp2 squared/base2 squared = 349,450/23,409 8Greek 8unicode 8direct translation ή ΕΗ άρα προς ΗΓ δυνάμεί λογον εχει, όν Μ ϑ The EH slant line thus from before HG (or ratio EH:HG) strengthe April16: to be considering M in ratio to considering M 3+4 Earlier: with being M (below horizontal) from before M?. In triangle HGE, the hyp2 EH to base2 HG ratio increases to?459/ Longshot: Archimedes is ratioing the hypotenuse hyp2 in triangleh I got 591/153 for the nondoubled base2 in hyp2/base2. So doublin Which gives: 295/153 NO! The answer 591/153 is next sentence. Page 6
7 Earlier: How do you get a squaring from this!?! The Greeks did squaring by drawing squares against the sides of t They then compared the areas and knew the Pythagorean relation So either M is about the complete side of the circumscribed polygo Did the 10,000 M need the number sign? April16: we don't know the 3 and 9's significance either! Not very likely that it is the big numbers for Bergren's squares! 8translation Berggren Hence OD sq :AD sq [(OA squared + AD squared: AD squared >(57 p94 Is EH sq:hg sq Sentence 9: 9Greek μήκει άρα, φ? ά ή προς ρνγ. second digit in first number looks like upside down 'h' must be one of Phoenician chars ϚϜϞϠ : 6,6,90,900 stigma, digamma, koppa, sampi CAN WE PROVE THIS IS A KOPPA? It must be!! The ratio of slant line EH to vertical HG is thus lengthened to 591/ 9unicode 9direct translation lengthen thus, the from before? 9translation Berggren so that OD: DA> 591 1/8: (4) p94 see this worked out below at line 68 onward! This is for Sentence 10: next bisection using point theta (called T): 47 To make the second increase in number of sides, bisect angle HEG. 48 Construct line ET from center E to point T on tangent line. 49 Because T is on the tangent line outside the circle and between H and G, ET > 50 Let TGE be the new right triangle with right angle TGE and angle TEG = 7.5 de 51 Consider the top of this triangle to be angle TEG, so the (unequal) sides are TE 52 In right triangle TGE, designate length TE as hyp3 and base length GT as opp3 53 As before, the third side has length r. 54 Now compare (shorter) base GT in right triangle TGE to base GH of right triang 55 Base GH is bifurcated by T, giving segments GT and TH. 56 The angle HEG was bisected into two equal parts, but segments GT and TH are 57 By Euclid VI:3, for a bisected angle at the top of a triangle, the segments of the 58 Because sides ratio HE to r in triangle HGE has HE > r, base segments ratio HT 59 Now use the construction for angle TEG to estimate the polygon side using the 60 Its base GT is onehalf of one side of the 24sided polygon determined by angle 61 We know the radius r which is one side of triangle TGE, but we need to know th 62 We seek to express the length TG in terms of the known ratios hyp/opp (or hyp/ Page 7
8 New part added: 63 By Euclid above for the larger right triangle HGE, unequal sides ratio hyp/r = u 64 Now expand the ratio numerator to include both pieces, keeping the denominat 65 This gives ratio two sides/r = base/tg which is hyp + r/r = base HG/base segme 66 Next swap two 'middle' components, ie for 1/4=3/12, it is also true that 1/3=4/1 67 This gives ratio two sides/base = r/tg which is hyp + r/hg = r/tg. Call basehg 68 We want to know hyp2/base2 and r/tg from the prior data, but so far we only k Earlier, we got: hyp/base ZG=hyp/opp=307/153 and r/base ZG=r/opp=265/153 Using Euclid I.47 (Pythagorean theorem) and the known ratio r/base2=571/153 Consider that hyp2 squared/base2 squared is a ratio of two sides of a triangle. Rewrite hyp2 squared by the sum of squares of the other two sides gives ratio: We know r and base2 so this gives ratio: (571 squared squared)/153 squa So hyp2 squared/base2 squared = 349,450/23,409 so the denominator is squar Taking square root of numerator gives ratio hyp2/base2= 591 1/8 over Before (line 38) we had: (hyp +r)/zg = hyp/opp + r/opp = (hyp + r)/opp = (307 So now we know both r/base2=571/153 and hyp2/base2=591 1/8 over 153 and This is like lines above, which concluded (line 37): We know hyp/zg and r/zg from (C5addition: line 14) the ratios in the example So we use the new data for the two ratios: Here we use: (hyp2+r)/base2=hyp2/base2 + r/base2 = 591 1/8 over / (Also see pictures for April 16 in notebook) 70 Therefore r/tg=1162/153, which is larger than the starting ratio r 71 Now work to get C/D by flipping the ratio which gives us that the ratio of the ta 72 This gives us a next approximation for the ratio of the circumference to the diam 73 The tangent segment TG is onehalf the length of a side of a polygon with 24 sid 74 Thus, the second approximation of the ratio of the circumference to the diamete 75 TG/r is 153:1162, so what is 24*TG/r? It is ( ): 1162 or 3672:1162 or 3 76 This value is less than the second estimate(3 1/5), less than first estimate C/D= 77 Then we repeat for more bisections of the angle at E. So do the Greek numbers match? Sentence 9 yes, but sentence Sentence 10: DRAW NEW LINE. 10Greek πάλιν δίχα ἡ υπο ΗΕΓ τή ΕΘ. 10unicode 10direct translation Again, once more, on the other hand, bisect as far as angle HEG b 10translation Berggren Secondly, let OE bisect the angle AOD, meeting AD in E. p94 ET HEG [Then DO:OA=DE:EA, so that DO+OA:DA = OA:AE] Page 8
9 Sentence 11: Sheet1 11Greek διά τά αύτα άρα ή ΕΓ προς ΓΘ μείζονα λογον εχει, 11unicode 11direct translation Because the same ratio EG: GT all the moreso, lengthens so is translation Berggren Therefore OA:AE[>(591 1/ ): 153, by (3) and (4)] p94 >1162 1/8: (5) EG:GT Both are radius to new GT base Sentence 12: 12Greek ή ΘΕ άρα προς ΘΓ μείζονα λογον εχει, ή όν 12unicode 12direct translation The TE ratio from before TG all the moreso has, a being 1172 to 1 [It follows that OE sq: EA sq >{(1162 1/8 sq sq}: 153 sq >( / ): > /64: 23409] 12translation Berggren Thus OE:EA > /8: (6) p95 ET:TG hyp to new base Per Berg we now have the two ratios needed: the r/gt AND the wi Sentence 13: DRAW NEW LINE. 13Greek έτι δίχα ἡ υπο ΘΕΓ τη ΕΚ. 13unicode 13direct translation Moreover, bisect the angle TEG with line EK. 13translation Berggren Thirdly, let OF bisect the angle AOE and meet (vertical segment) A p Sentence 14: 14Greek 14unicode ή ΕΓ άρα προς ΓΚ μείζονα λογον εχει, ή όν βτλδ δ προς ρνγ. Page 9
10 14direct translation Because the same ratio EG: GK (radius to new base) all the moreso 14translation Berggren We thus obtain the result [corresponding to (3) and (5) above] that p95 OA:AF [>(1162 1/ /8): 153] EG:GK > >2334 1/4: (7) [Therefore OF sq:fa sq >{(2334 1/4)sq sq}:153 sq > /16: ] Sentence 15: 15Greek 15unicode could get old theta char from ή ΕΚ άρα προς ΓΚ μείζονα, ή όν βτλθ δ προς ρνγ. 15direct translation Because the same ratio EK: GK all the moreso, is 2339 to 153. slant line EK to vertical seg. GK 15translation Berggren Thus, OF:FA >2339 1/4: (8) p95 EK : KG Sentence 16: DRAW NEW LINE. 16Greek έτι δίχα ἡ υπο ΚΕΓ τη ΛΕ. 16unicode 16direct translation Moreover, bisect the angle KEG with line LE. 16translation Berggren Fourthly, let OG bisect the angle AOF, meeting AF in G. p Sentence 17: Note: there is one character that is only approximate! the character looks like L but the horiz is wiggly. must be one of Phoenician chars ϚϜϞϠ : 6,6,90,900 stigma, digamma, koppa, sampi 17Greek ή ΕΓ άρα προς ΛΓ μείζονα [μήκει] λογον εχει, 17unicode 17direct translation Because the same ratio EG: LG all the moreso, is greater being 4,6 EG horizontal to LG vertical segment (smallest above ho Page 10
11 17translation Berggren We have then OA:AG [>(2334 1/ /4): 153, by means of (7) p95 EG:GL >4673 1/2: Sentence 18: 18Greek 18unicode 18direct translation ἐπεὶ οὖν ἡ ὑπὸ ΖΕΓ τρίτον οὖσα ὀρϑᾔς τέτμηται τετράϰις epei?? consequently, the first angle ZEG, onethird of a right angle, bise 18translation Berggren p Sentence 19: 19Greek 19unicode 19direct translation 19translation Berggren To do it out: 1/3 of 90 deg with 4 bisections: 1/6 1/12 1/24 1/48 of 90 deg or go Now the angle AOC, which is onethird of a right angle, has been bisected four times, and it follows that angle AOG = 1/48 (of a right angle). Angle GEL ϰείσϑω οὖν αὐτἤ ίση προς τὦ Ε ἡ ὑπὸ ΓΕΜ. Make the self same angle to E as angle GEM (dips below horizonta Make the angle AOH on the other side of OA equal to the angle AO p95 and let GA produced meet OH in H Sentence 20: 20Greek 20unicode 20direct translation 20translation Berggren p95 ἡ ἄρα ὑπὸ ΛΕΜ ὀρϑᾔς ἐστι ϰδ ʹ And angle LEM is 24 to 3 of a right angle. Then angle GOH= 1/24 (a right angle) Sentence 21: 21Greek 21unicode 21direct translation 21translation Berggren p95 must be one of Phoenician chars ϚϜϞϠ : 6,6,90,900 stigma, digamma, koppa, sampi ϰαὶ ἡ ΛΜ ἄρα εὐϑεἴα τὸὕ περί τὸν κύκλον ἐστι πολυγώνου πλευρὰ Thus LM straight side to go around the circle is polygon side worth Thus GH is one side of a regular poloygon of 96 sides circumscribed to the given circle. Page 11
12 Sentence 22: 22Greek Sheet1 ἐπεὶ οὖν ἡ ΕΓ προς τὴν ΓΛ ἐδείχϑη μείζονα λογον τή ς με ν ΕΓ διπλἥ ἡ ΑΓ, τή ς δὲ ΓΛ διπλασίων ἡ ΛΜ, ϰαὶ ἡ ΑΓ περίμετρον μείζονα λογον εχει, ήπερ δχογ προς Μ δχπη ʹ. 22unicode 22direct translation Consequently, the radius EG in ratio to the GL all the more so is w then double radius EG giving diameter AG, then GL (the half side) and the ratio of the diameter AG to the 96 sided perimeter of polyg 4,673 to 14,688. it follows that AB:(perimeter of polygon of 96 sides)[>4673 1/2:15 22translation Berggren And, since OA:AG > /3 : 153, p95 while AB:2OA, GH = 2AG. >4673 1/2: Sentence 23: 23Greek 23unicode 23direct translation και ε στίν τριπλασία, και υ περέχουσιν χξζʹ, But this is threefold, exceeding it by ratio of 667 to 4,673 less than 667/4673=.1427 (remainder 1729/4673) whereas 1/7= /7. 23translation Berggren But 14688/4673 ½ = /2/4673 ½ = [< /2/4672 ½ p96 <3 1/ Sentence 24: 24Greek 24unicode 24direct translation 24translation Berggren p95 ὤστε τὸ πολυγώνον τὸ περὶ τὸν κύκλον της διαμέτρου ε στί τριπλ Thus the polygon to around the circle of its diameter is threefold b no sentence some subtleties are being argued? Page 12
13 Sentence 25: 25Greek 25unicode 25direct trans. 25Berggren trans. ***** Section 3: Sentence 1: 1Greek 1unicode 1direct trans. 1Berggren trans. Sheet1 ἡ τοὔ κύκλου ἄρα περίμετρος πολὺ μἂλλον ἐλάσσων ἐστὶν ἤ τριπ Therefore the circumference of the circle is much more less than t Therefore the circumference of the circle(being less than the perimeter of the polygon) is a fortiori less than 3 1/7 times the diameter AB. p266 ἔστω ϰύϰλος, ϰαὶ διαμετρος ἡ ΑΓ, ἡ ὑπὸ ΒΑΓ τρίτον ὀρϑᾔς. (Archimedes assumes that angle GBA is the right angle.) So we have this circle, and also diameter AG, also of the angle BAG II. Next let AB be the diameter of a circle, and let AC meeting the circle in C, make the angle CAB equal to onethird of a right angle. Join BC Sentence 2: B right angle adj=ab GB=opp 30 deg angle G A hyp=ag 2Greek 2unicode 2direct trans. ἡ AB ἄρα πρὸς BΓ ἐλάσσονά λόγον ἒχει, ἡ ὃν ατναʹ προς ψπ [ἡ δὲ AΓ ἄρα πρὸς ΓB, ὃν αφξʹ προς ψπʹ]. The ratio of AB to BG (adj side to opp side) is on the lesser side tha [the ratio of AG to GB (hyp to opp side) being 156 to 780]. SHOULD BE Berggren trans. Then AC:CB [=root 3:1] <1351:780. ***** Section 4: p268 Page 13
14 Sentence 1: 1Greek 1unicode 1direct trans. 1Berggren trans Sentence 2: 2Greek 2unicode 2direct trans. 2Berggren trans Sentence 3: 3Greek 3unicode 3direct trans. 3Berggren trans Sentence 4: 4Greek 4unicode 4direct trans. 4Berggren trans Sentence 5: 5Greek 5unicode 5direct trans. 5Berggren trans. Sheet1 δίχα ἡ υπο BAΓ τή AΗ. Divide or bisect as far as angle BAG by line HA (eta alpha). First, let AD bisect the angle BAC and meet BC in d and the circle ἐπε ὶ ίση ἐστιν ἡ υπο BAΗ τή υπο ΗΓB, ἀλλἁ ϰαὶ ή υπο ΗΓB τή ἱπο B H x G x x A The same are angles BAH and HGB or HAG (15 degrees) and angles HGB and HAG are the same. See below. ϰαὶ ϰοινή ἡ υπο AΗΓ ὀρϑᾔ. The common angle AHG is a right angle. See below. ϰαὶ τρἱτη ἄρα ἡ υπο ΗZΓ τρἱτη τή υπο AΓΗ ίση. The big angle HZG is the same as the big angle AGH. Then angle BAD=angle dac=angle dbd, and angles at D, C are bo GAH = ZAB = ZGH,.. H,B ἰσογώνιον ἄρα τὸ AΗΓ τ ὤ ΓΗZ τρἱγώνω. Similar are the AHG to GHZ triangles. It follows that the triangles ADB, [AC d], BD d are similar. (AHG, [ABZ], GHZ) Page 14
15 Sentence 6: 6Greek 6unicode 6direct trans. 6Berggren trans. ἐ στιν ἄ ρα, ώς ή AΗ προς ΗΓ, ή ΓΗ προς ΗZ, It follows that the ratio of AH to HG, GH to HZ and AG to GZ are th Therefore AD:DB = BD:Dd [=AC:Cd]...=AB:Bd [Eucl. VI. 3]...=AB + AC: Bd + Cd...=AB + AC: BC or BA + AC : BC = AD:DB Sentence 7: 7Greek 7unicode 7direct trans. 7Berggren trans. ἀλλ ὡς ἡ ΑΓ πρὸς ΓZ, ϰαὶ συναμφοτερος Change then the AG (diameter) in ratio to GZ (lower half of base), but together the GAB (central angle 30 deg) in ratio to BG (ba Haven't got to above yet! Sentence 8: 8Greek ϰαὶ συναμφοτερος ἄρα ή ΒΑΓ προς ΒΓ, ή 8unicode 8direct trans. But together angle BAG (30 deg) in ratio to BG (base), and AH ( 8Berggren trans. Haven't got to above yet still! Sentence 9: 9Greek 9unicode 9direct trans. 9Berggren trans. διὰ τοῦτο οὐν ή ΑΗ προς τὴν ΗΓ. for this therefore the AH (new line) to the HG (its base). Haven't got to above yet still! Sentence 10: 10Greek ἐ λάσσονά λόγον ἒχει, ἤπερ βϡια προς ψπ ʹ, Page 15
16 10unicode 10direct trans. 10Berggren trans Sentence 11: 11Greek 11unicode 11direct trans. 11Berggren trans. Sheet1 ἡ δὲ ΑΓ προς τὴν ΓΗ ἐλάσσονά, ἢ ὃν γιγ δ προς ψπʹ. Being on the lesser side, Therefore AD:DB < 2911: 780 (1)..(p97) (and from above: = BD δίχα ἡ υπο ΓAΗ τή AΘ Sentence 12: 12Greek 12unicode 12direct trans. 12Berggren trans Sentence 13: 13Greek 13unicode 13direct trans. 13Berggren trans Sentence 14: 14Greek 14unicode 14direct trans. 14Berggren trans Sentence 15: 15Greek 15unicode 15direct trans. 15Berggren trans Sentence 16: 16Greek 16unicode 16direct trans. 16Berggren trans. Page 16
17 Sentence 17: 17Greek 17unicode 17direct trans. 17Berggren trans. Sheet Sentence 18: 18Greek 18unicode 18direct trans. 18Berggren trans Sentence 19: 19Greek 19unicode 19direct trans. 19Berggren trans Sentence 20: 20Greek 20unicode 20direct trans. 20Berggren trans Sentence 21: 21Greek 21unicode 21direct trans. 21Berggren trans Sentence 22: 22Greek 22unicode 22direct trans. 22Berggren trans. ***** Section 5: Sentence 1: 1Greek 1unicode 1direct trans. p270 Page 17
18 1Berggren trans. Sheet1 Thus the ratio of the circumference to the diameter is < 3/1/7 but 1.4 Words Used: Greek: English: ἀλλ change ἀλλἁ other, else, rest, next αρα then therefore so then (seems like it is 'ratio') αύτα same αὐτἤ self δε διά but and (text has δὲ) for, because διπλἥ double διχα form of divide δυνάμεί strengthen ἔβδομον ἐλάσσονά seventh less than (elasson) ἐλάττονά less than (elasson) ίση ἐστιν έναλλάξ alternately ἐπε ὶ εχει ἔχοντος, ἔχο has to be worth ἐστι έστω albeit (the text has a different accent) έστω v.i. to be έτι still, moreover εὐϑεἴα straight, direct ήπερ hyper, more ἱπο ίση same και ϰείσϑω λογον ϰοινή μἂλλον μείζονα μείζονα λογον μέρει μήκει ουν ον and also (the text has different kappa and accent) not found not found (another said man of his word, so surely?) common more, rather, to some extent greater all the more so portion lengthen therefore, consequently as which or being, creature Page 18
19 πάλιν again, once more, on the other hand περίμετρος perimeter πλευρὰ side πολὺ much, very or long προς from before ρνγ not found(ρνγ in text) σξε not found συναμα same time, together συναμφοτερος τά the,them τετμήσδω not found τετράϰις four times την not found τζ not found τοῦτο this τρἱγώνω triangle τριπλασία threefold τριτη υ περέχουσιν exceeding upo under, secretly, slightly ώς as far as, until (acc.), adv.? ωστε consequently adv The Measurement of the Circle ~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~ The Measurement of the Circle is made up of the following folios Undertext Prayer Book +===============+== =============+ Arch68r v Page 19
20 Arch68v r Arch69r 177r172v Sheet1 Page 20
21 int/6 point Character map Group by unicode subrange Greek A Circle, ed. Heiberg, pp 258,260,262,264,266,268,270 (Greek) and alternating following pages in Latin. Remaining questions: *In Prop 3, Section 4, s2: 156 should be 1560 How do Greeks do zero for *10?? ight triangle, where distance from the center is the same as one side from the right angle, and the perimeter is same as base. gled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference, of the circle. ared logon is, on ia pros id. iameter as 11 is to 14. Page 21
22 αμέτρον τριπλασίων εστί, και έτι υπερέχει αμέτρον, μείζονι δε ή δέκα έβδομηκοστομόνοις. f(small omicron) 0384(tonos) 03c2(small final sigma) 03f0 03cd 03f0 03bb 03bf 03c2 ts diameter is triple, but still measures more by on the low side a seventh of its diameter, by on the high side ten seventifirsts. to its diameter is less than 3 1/7 but greater than 3 10/ /7=3 10/70, 10/71 sts is less. the circumference C, radius EG of length r and diameter D. the line from Z to center of circle E creates an angle ZEG of 30 deg. e 30 degree angle at E. ZEG, so the (unequal) sides are ZE and GE and the base is GZ. as hyp for hypotenuse and length ZG as opp for side opposite central angle ZEG. The third side length is r. gon of six sides that circumscribes the circle of radius r. 1 (or cosecant of angle ZEG). 1 (or cotangent of angle ZEG). rst estimate of C/D as 6*(1:square root 3) and is larger than true C/D. ed as 1.5, so 1/1.5 = 2/3, so 6*2/3 = 12/3 = 4, we get C/D = 4. olygon by bisecting the angle ZEG repeatedly, increasing opp/r, bettering C/D. correction was r=153,now r=265/153=1.7) and right triangle ZEG with hyp and opp as the other sides. r/opp=square root 3/1=265/153, so r/opp is less than hyp/opp as expected for the circumscribed polygon case. sides of right triangle ZEG and knowing r means all lengths of sides are known. es, bisect angle ZEG. on tangent line. e circle and between Z and G, EH > r and EH < hyp from triangle ZEG. ht angle HGE and angle HEG = 15 degrees. Page 22
23 e HEG, so the (unequal) sides are HE and GE and the base is GH. as hyp2 and base length GH as opp2 (side opposite top angle HEG). iangle HGE to base GZ of right triangle ZGE. (C1corrected) ts GH and HZ. l parts, but segments GH and HZ are not equal. top of a triangle, the segments of the base have the same ratio as the lengths of the sides. has ZE > r, base segments ratio ZH to HG will have ZH > HG. estimate the polygon side using the new (shorter) right triangle HGE. 2sided polygon determined by angle HEG being 15 degrees. triangle HGE, but we need to know the length of HG. (C2shortened) of the known ratios hyp/opp (or hyp/base) and r/opp (or r/base). e ZGE, unequal sides ratio hyp/r = unequal base segments ratio ZH/HG. both pieces, keeping the denominator the same. ch is hyp + r/r = base ZG/base segment HG. 1/4=3/12, it is also true that 1/3=4/12, so use that! ch is hyp + r/zg = r/hg.(c3clearer to write base instead of ZG?) on: line 14) the ratios in the example: hyp/zg=hyp/opp=307/153 and r/zg=r/opp=265/153. = (hyp + r)/opp = ( )/153 = 571/153. larger than the starting ratio r/zg=265/153. hich gives us that the ratio of the tangent segment HG to the radius r is smaller than before. ratio of the circumference to the diameter as follows. gth of a side of a polygon with 12 sides and r is half the diameter D. o of the circumference to the diameter is 12*HG/r ): 571 or 1836:571 or 3 123/571 or 3 1/5 approximately. D=4 and greater than the proposition 3 result 3 1/7 for the circumscribed case. ϰαὶ ϰέντρον το Ε, ϰαὶ ἡ ΓΛΖ ἑφαπτομένη, ϰαὶ ἡ ὑπὸ ΖΕΓ τρίτον ὀρϑᾔς. G, and also center E, and also line GLZ tangent, and also angle ZEG a third of a right angle. s centre, AC the tangent at A; and let the angle AOC be onethird of a right angle. literated letters and no lines for the circle's The Berggren edge, diameter version or of tangent: the picture with transliterated letters and no lines for the circle's edg Page 23
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