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Use Deja Vu Serif Condensed, 10 point/6 point
Direct translation by J. Holly DeBlois
*****TITLE*****
1 Title:
1-Greek
1-unicode
1-direct trans.
1-Berggren trans.
This is Archimedes' “Dimensions of A Circle,” ed. Heiberg, pp 258
ΚΥΚΛΟΥ ΜΕΤΡΗΣΙΣ
Of Circles Measure
MEASUREMENT OF A CIRCLE.
*****PROP1*****
1.1 Proposition 1:
0-Greek
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0-translation
0-translation Berggren
1.1.1.1
Sentence 1:
1-Greek
1-unicode
1-direct translation
1-translation Berggren
ά·
a.
Proposition 1.
Πα̃ς 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 right-angled triangle in which one of the sid
Stop here.
*****PROP2*****
1.2 Proposition 2:
0-Greek
0-unicode
0-translation
0-translation Berggren
1.2.1.1
Sentence 1:
1-Greek
1-unicode
1-direct translation
1-translation Berggren
̃β΄·
03b2 0384(tonos) 0387(ano teleia)
b.
Proposition 2.
̃Ο ϰύϰλος …
039f(Greek capital letter omicron)
03f0 03cd 03f0 03bb 03bf 03c2 …
The circle pros to apo whose diameter squared logon is, on ia pros id.
The area of a circle is to the square on its diameter as 11 is to 14.
Stop here.
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*******PROP3*********
1.3 Proposition 3:
0-Greek
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p93
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1.3.1.1
Section 1:
Sentence 1:
1-Greek
1-unicode
1-direct translation
1-translation Berggren
p93
*****
Note: I'm anglicizing Gamma as G not C.
p262
̃γ΄·
03b3(Greek small letter Gamma) 0384(tonos) 0387(ano teleia)
c.
Proposition 3.
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
The ratio of the circumference of any circle to its diameter is less than 3 1/7 bu
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 one-half 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 (C4-correction 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
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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 one-half of one side of the 12-sided 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.(C3-clearer
37 We know hyp/ZG and r/ZG from (C5-addition: line 14) the ratios in the example
38 That means hyp +r/ZG = hyp/opp + r/opp = (hyp + r)/opp = (307+265)/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 one-half 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 (1530+307): 571 or 1836:571 or 3 12
45 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.
1.3.2.1
Section 2:
Sentence 1:
p264
1-Greek
ἔστω ϰύϰλος, ϰαὶ διαμετρος ἡ ΑΓ, ϰαὶ ϰέντρον το Ε, ϰαὶ ἡ ΓΛΖ ἑ
1-unicode
1-direct translation
So we have this circle, and also diameter AG, and also center E, and also line G
1-translation Berggren
p93
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
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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
1000-9999 use iota sub or superscript with letters for 1-9
10000=M, and higher numbers put the number in front of M (or on top of it)
27 symbols: numbers 1-10, tenths 20-100, hundredths 200-900, greek letters plus 3-4 phoen
11-19: ia, ibeta, etc.; 100900: 347 tau mu zeta
1.3.2.2
Sentence 2:
2-Greek
2-unicode
ἡ ΕΖ ἂρα πρὸς ΖΓ λόγον ἒχει, ὃν τζ προς ρνγ΄.
2-direct 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
2-translation Berggren
p94
1.3.2.3
Berggren reverses it:
The OA:AC is >265:153...(1), and OC:CA = 306:153...(2). [OA:AC=
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:
3-Greek
ἡ δὲ ΕΓ πρὸς τὴν ΓΖ λόγον ἒχει, ὃν σξέ προς ρνγ΄.
3-unicode
3-direct translation
sigma xi epsilon to 153
But the [other side] EG to [vertical] GZ surely is, as which 265::15
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the horiz.
the vert. going up
The ratio of the radius EG to tangent segment GZ is square root 3
1.3.2.4
1.3.2.5
3-translation Berggren
And for this line: The OA:AC is >265:153...(1) [OA:AC=square roo
Sentence 4:
4-Greek
4-unicode
4-direct trans.
4-Berggren trans.
DRAW NEXT LINE (Bisect 30 degrees, get 15 degrees)
τετμήσδω οὖν ἡ υπο ZEΓ δίχα τη EH.
Sentence 5:
5-Greek
5-unicode
5-direct trans.
5-Berggren trans.
Euclid: VI: 3:
1.3.2.6
Sentence 6:
6-Greek
6-unicode
6-direct trans.
6-Berggren trans.
Bisect therefore the angle ZEG and [draw] EH to divide it [evenly]
First, draw OD bisecting the angle AOC and meeting AC in D.
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
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
CO+OA:CA=OA:AD.
ZE + EG:ZG=EG:GH
Not often used.
1/4=2/8 so 1/2=4/8
We know: ZE/ZG=2/1 and EG/ZG=sq.root 3/1.
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By the steps above we reach: (ZE +EG)/ZG = EZ/HG.
Compute (ZE + EG)/ZG = ZE/ZG + EG/ZG = 2/1+root3/1 = 306/15
1.3.2.7
Sentence 7:
7-Greek
ωστε ή ΓΕ προς ΓΗ μείζονα λογον εχει, ήπερ φοά προς ρνγ.
7-unicode
7-direct trans.
571
EG:GH greater now, bigger as:
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
7-Berggren trans.
p94
1.3.2.8
Therefore [by (1) and (2)] OA:AD .571:153....(3).
EG:GH
Sentence 8:
19,450 measures what?
Could the M be 340,000?
13,404 measures what?
Could M be 20,000?
9,450 measures what?
it's in Bergren!
3,404 measures what?
3,409?
Line 68-(7) has:So hyp2 squared/base2 squared = 349,450/23,409
8-Greek
8-unicode
8-direct 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 9+450 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 non-doubled base2 in hyp2/base2. So doublin
Which gives: 295/153 NO! The answer 591/153 is next sentence.
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8-translation Berggren
p94
1.3.2.9
Sentence 9:
9-Greek
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!
Hence OD sq :AD sq [(OA squared + AD squared: AD squared >(57
Is EH sq:HG sq
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!!
9-unicode
9-direct translation
9-translation Berggren
p94
lengthen thus, the from before ?
The ratio of slant line EH to vertical HG is thus lengthened to 591/
so that OD: DA> 591 1/8: 153....(4)
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 one-half of one side of the 24-sided 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/
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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
New part added:
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 + 153 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 153.
69 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 32-37 above, which concluded (line 37):
We know hyp/ZG and r/ZG from (C5-addition: 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 153 + 571/
(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 one-half 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 (3060+612): 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?
1.3.2.10 Sentence 10:
10-Greek
10-unicode
10-direct translation
10-translation Berggren
p94
Sentence 9 yes, but sentence
DRAW NEW LINE.
πάλιν δίχα ἡ υπο ΗΕΓ τή ΕΘ.
Again, once more, on the other hand, bisect as far as angle HEG b
Secondly, let OE bisect the angle AOD, meeting AD in E.
ET
HEG
[Then DO:OA=DE:EA, so that DO+OA:DA = OA:AE]
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1.3.2.11 Sentence 11:
11-Greek
11-unicode
11-direct translation
διά τά αύτα άρα ή ΕΓ προς ΓΘ μείζονα λογον εχει,
Because the same ratio EG: GT all the moreso, lengthens so is 116
11-translation Berggren Therefore OA:AE[>(591 1/8 + 571): 153, by (3) and (4)]
p94
>1162 1/8: 153 ...(5)
EG:GT
Both are radius to new GT base.
1.3.2.12 Sentence 12:
12-Greek
12-unicode
12-direct translation
ή ΘΕ άρα προς ΘΓ μείζονα λογον εχει, ή όν
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 + 153 sq}: 153 sq
>(1350534 33/64 + 23409): 23409
>1373943 33/64: 23409]
12-translation Berggren Thus OE:EA > 1172 1/8: 153...(6)
p95
ET:TG
hyp to new base
Per Berg we now have the two ratios needed: the r/GT AND the wi
1.3.2.13 Sentence 13:
13-Greek
13-unicode
13-direct translation
13-translation Berggren
p95
DRAW NEW LINE.
έτι δίχα ἡ υπο ΘΕΓ τη ΕΚ.
Moreover, bisect the angle TEG with line EK.
Thirdly, let OF bisect the angle AOE and meet (vertical segment) A
1.3.2.14 Sentence 14:
14-Greek
14-unicode
ή ΕΓ άρα προς ΓΚ μείζονα λογον εχει, ή όν βτλδ΄ δ΄΄ προς ρνγ.
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14-direct translation
Because the same ratio EG: GK (radius to new base) all the moreso
14-translation Berggren We thus obtain the result [corresponding to (3) and (5) above] that
p95
OA:AF [>(1162 1/8 + 1172 1/8): 153]
EG:GK >
>2334 1/4:153 .... (7)
[Therefore OF sq:FA sq >{(2334 1/4)sq + 153 sq}:153 sq
>5472132 1/16: 23409.]
1.3.2.15 Sentence 15:
15-Greek
15-unicode
could get old theta char from
ή ΕΚ άρα προς ΓΚ μείζονα, ή όν βτλθ΄ δ΄΄ προς ρνγ΄.
15-direct translation
Because the same ratio EK: GK all the moreso, is 2339 to 153.
slant line EK to vertical seg. GK
15-translation Berggren Thus, OF:FA >2339 1/4: 153 ...(8)
p95
EK : KG
1.3.2.16 Sentence 16:
16-Greek
16-unicode
16-direct translation
16-translation Berggren
DRAW NEW LINE.
έτι δίχα ἡ υπο ΚΕΓ τη ΛΕ.
Moreover, bisect the angle KEG with line LE.
Fourthly, let OG bisect the angle AOF, meeting AF in G.
p95
1.3.2.17 Sentence 17:
17-Greek
17-unicode
17-direct translation
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
ή ΕΓ άρα προς ΛΓ μείζονα [μήκει] λογον εχει,
Because the same ratio EG: LG all the moreso, is greater being 4,6
EG horizontal to LG vertical segment (smallest above ho
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17-translation Berggren We have then OA:AG [>(2334 1/4+2330 1/4): 153, by means of (7)
p95
EG:GL >4673 1/2: 153.
1.3.2.18 Sentence 18:
18-Greek
18-unicode
18-direct translation
ἐπεὶ οὖν ἡ ὑπὸ ΖΕΓ τρίτον οὖσα ὀρϑᾔς τέτμηται τετράϰις
epei?
? consequently, the first angle ZEG, one-third of a right angle, bise
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
18-translation Berggren Now the angle AOC, which is one-third of a right angle,
p95
has been bisected four times, and it follows that
angle AOG = 1/48 (of a right angle).
Angle GEL
1.3.2.19 Sentence 19:
ϰείσϑω οὖν αὐτἤ ίση προς τὦ Ε ἡ ὑπὸ ΓΕΜ.
19-Greek
19-unicode
19-direct translation
Make the self same angle to E as angle GEM (dips below horizonta
19-translation Berggren Make the angle AOH on the other side of OA equal to the angle AO
p95
and let GA produced meet OH in H.
1.3.2.20 Sentence 20:
ἡ ἄρα ὑπὸ ΛΕΜ ὀρϑᾔς ἐστι ϰδ΄ʹ
20-Greek
20-unicode
20-direct translation
And angle LEM is 24 to 3 of a right angle.
20-translation Berggren Then angle GOH= 1/24 (a right angle).
p95
must be one of Phoenician chars ϚϜϞϠ : 6,6,90,900
stigma, digamma, koppa, sampi
1.3.2.21
Sentence 21:
21-Greek
21-unicode
21-direct translation
21-translation Berggren
p95
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.
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1.3.2.22 Sentence 22:
22-Greek
ἐπεὶ οὖν ἡ ΕΓ προς τὴν ΓΛ ἐδείχϑη μείζονα λογον
τή̓ς μἐν ΕΓ διπλἥ ἡ ΑΓ, τή̓ς δὲ ΓΛ διπλασίων ἡ ΛΜ, ϰαὶ ἡ ΑΓ
περίμετρον μείζονα λογον εχει, ήπερ δΧογ΄∟ ΄΄ προς Μ ͵δχπηʹ.
22-unicode
22-direct 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.
22-translation Berggren And, since OA:AG > 4673 1/3 : 153,
p95
while AB:2OA, GH = 2AG.
it follows that AB:(perimeter of polygon of 96 sides)[>4673 1/2:15
>4673 1/2:14688.
1.3.2.23 Sentence 23:
23-Greek
23-unicode
23-direct translation
και ἐστίν τριπλασία, καἰ ὐπερέχουσιν ͵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=.1428 4/7.
23-translation Berggren But 14688/4673 ½ = 3 + 667 1/2/4673 ½ = [< 3 + 667 1/2/4672 ½
p96
<3 1/7
1.3.2.24 Sentence 24:
24-Greek
24-unicode
24-direct translation
ὤστε τὸ πολυγώνον τὸ περὶ τὸν κύκλον της͂ διαμέτρου ἐστί τριπλ
Thus the polygon to around the circle of its diameter is threefold b
24-translation Berggren no sentence- some subtleties are being argued?
p95
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1.3.2.25 Sentence 25:
25-Greek
25-unicode
25-direct trans.
25-Berggren trans.
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.
*****
1.3.3.1
Section 3:
Sentence 1:
1-Greek
1-unicode
1-direct trans.
1-Berggren trans.
1.3.3.2
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 one-third
of a right angle. Join BC.
Sentence 2:
B
right angle
adj=AB
GB=opp
30 deg angle
G
A
hyp=AG
2-Greek
ἡ AB ἄρα πρὸς BΓ ἐλάσσονά λόγον ἒχει, ἡ ὃν ͵ατναʹ προς ψπ
[ἡ δὲ AΓ ἄρα πρὸς ΓB, ὃν ͵αφξʹ προς ψπʹ].
2-unicode
2-direct trans.
2-Berggren trans.
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 1560
Then AC:CB [=root 3:1] <1351:780.
*****
Section 4:
p268
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1.3.4.1
Sentence 1:
1-Greek
1-unicode
1-direct trans.
1-Berggren trans.
1.3.4.2
Sentence 2:
2-Greek
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
ἐπεὶ ίση ἐστιν ἡ υπο BAΗ τή υπο ΗΓB, ἀλλἁ
ϰαὶ ή υπο ΗΓB τή ἱπο
B
H
x
x
x
G
2-unicode
2-direct trans.
2-Berggren trans.
1.3.4.3
1.3.4.4
Sentence 3:
3-Greek
3-unicode
3-direct trans.
3-Berggren trans.
Sentence 4:
4-Greek
4-unicode
4-direct trans.
4-Berggren trans.
1.3.4.5
Sentence 5:
5-Greek
5-unicode
5-direct trans.
5-Berggren trans.
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 τὤ ΓΗZ τρἱγώνω.
Similar are the AHG to GHZ triangles.
It follows that the triangles ADB, [AC d], BD d are similar.
(AHG, [ABZ], GHZ)
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1.3.4.6
Sentence 6:
6-Greek
6-unicode
6-direct trans.
6-Berggren trans.
ἐστιν ἄρα, 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.
1.3.4.7
Sentence 7:
7-Greek
7-unicode
7-direct trans.
7-Berggren trans.
1.3.4.8
1.3.4.9
Sentence 8:
8-Greek
8-unicode
8-direct trans.
8-Berggren trans.
Sentence 9:
9-Greek
9-unicode
9-direct trans.
9-Berggren trans.
ἀλλ̓ 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!
ϰαὶ συναμφοτερος ἄρα ή ΒΑΓ προς ΒΓ, ή
But together … angle BAG (30 deg) in ratio to BG (base), and AH (
Haven't got to above yet still!
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!
1.3.4.10 Sentence 10:
10-Greek
ἐλάσσονά λόγον ἒχει, ἤπερ ͵βϠια΄ προς ͵ψπʹ,
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ἡ δὲ ΑΓ προς τὴν ΓΗ ἐλάσσονά,
ἢ ὃν ͵γιγ΄ ∟΄΄ δ΄΄ προς ψπʹ.
10-unicode
10-direct trans.
Being on the lesser side,
10-Berggren trans.
Therefore AD:DB < 2911: 780 … (1)..(p97) (and from above: = BD
1.3.4.11 Sentence 11:
11-Greek
11-unicode
11-direct trans.
11-Berggren trans.
δίχα ἡ υπο ΓAΗ τή AΘ.
1.3.4.12 Sentence 12:
12-Greek
12-unicode
12-direct trans.
12-Berggren trans.
1.3.4.13 Sentence 13:
13-Greek
13-unicode
13-direct trans.
13-Berggren trans.
1.3.4.14 Sentence 14:
14-Greek
14-unicode
14-direct trans.
14-Berggren trans.
1.3.4.15 Sentence 15:
15-Greek
15-unicode
15-direct trans.
15-Berggren trans.
1.3.4.16 Sentence 16:
16-Greek
16-unicode
16-direct trans.
16-Berggren trans.
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1.3.4.17 Sentence 17:
17-Greek
17-unicode
17-direct trans.
17-Berggren trans.
1.3.4.18 Sentence 18:
18-Greek
18-unicode
18-direct trans.
18-Berggren trans.
1.3.4.19 Sentence 19:
19-Greek
19-unicode
19-direct trans.
19-Berggren trans.
1.3.4.20 Sentence 20:
20-Greek
20-unicode
20-direct trans.
20-Berggren trans.
1.3.4.21 Sentence 21:
21-Greek
21-unicode
21-direct trans.
21-Berggren trans.
1.3.4.22 Sentence 22:
22-Greek
22-unicode
22-direct trans.
22-Berggren trans.
*****
1.3.5.1
Section 5:
Sentence 1:
1-Greek
1-unicode
1-direct trans.
p270
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1-Berggren trans.
1.4 Words Used:
Greek:
ἀλλ
ἀλλἁ
αρα
αύτα
αὐτἤ
δε
διά
διπλἥ
διχα
δυνάμεί
ἔβδομον
ἐλάσσονά
ἐλάττονά
έναλλάξ
ἐπεὶ
εχει
ἔχοντος, ἔχο
ἐστι
έστω
έστω
έτι
εὐϑεἴα
ήπερ
ἱπο
ίση
και
ϰείσϑω
λογον
ϰοινή
μἂλλον
μείζονα
μείζονα λογον
μέρει
μήκει
ουν
ον
Thus the ratio of the circumference to the diameter is < 3/1/7 but
English:
change
other, else, rest, next
then therefore so then (seems like it is 'ratio')
same
self
but and (text has δὲ)
for, because
double
form of divide
strengthen
seventh
less than (elasson)
less than (elasson)
ίση ἐστιν
alternately
has
to be worth
albeit (the text has a different accent)
v.i. to be
still, moreover
straight, direct
hyper, more
same
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
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πάλιν
περίμετρος
πλευρὰ
πολὺ
προς
ρνγ
σξε
συναμα
συναμφοτερος
τά
τετμήσδω
τετράϰις
την
τζ
τοῦτο
τρἱγώνω
τριπλασία
τριτη
ὐπερέχουσιν
upo
ώς
ωστε
again, once more, on the other hand
perimeter
side
much, very or long
from before
not found(ρνγ΄in text)
not found
same time, together
the,them
not found
four times
not found
not found
this
triangle
threefold
exceeding
under, secretly, slightly
as far as, until (acc.), adv. ?
consequently adv.
5.2.6 The
Measurement of the
Circle
~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~
The Measurement of
the Circle is made
up of the following
folios.
+--------------+---------------+
| Undertext
| Prayer Book
|
+===============+==
=============+
| Arch68r
| 0000-171v
|
+--------------+---------------+
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| Arch68v
| 0000-171r
|
+--------------+---------------+
| Arch69r
| 177r-172v
|
+--------------+---------------+
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oint/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??
right triangle, where distance from the center is the same as one side from the right angle, and the perimeter is same as base.
ngled 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.
uared logon is, on ia pros id.
diameter as 11 is to 14.
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ιαμέτρον τριπλασίων εστί, και έτι υπερέχει
ιαμέτρον, μείζονι δε ή δέκα έβδομηκοστομόνοις.
bf(small omicron) 0384(tonos) 03c2(small final sigma)
03f0 03cd 03f0 03bb 03bf 03c2 …
its diameter is triple, but still measures more by on the low side a seventh of its diameter, by on the high side ten seventifirsts.
3 1/7=3 10/70, 10/71 sts is less.
e to its diameter is less than 3 1/7 but greater than 3 10/71.
n the circumference C, radius EG of length r and diameter D.
t the line from Z to center of circle E creates an angle ZEG of 30 deg.
he 30 degree angle at E.
e ZEG, so the (unequal) sides are ZE and GE and the base is GZ.
E as hyp for hypotenuse and length ZG as opp for side opposite central angle ZEG. The third side length is r.
ygon of six sides that circumscribes the circle of radius r.
:1 (or cosecant of angle ZEG).
1 (or cotangent of angle ZEG).
irst estimate of C/D as 6*(1:square root 3) and is larger than true C/D.
ted as 1.5, so 1/1.5 = 2/3, so 6*2/3 = 12/3 = 4, we get C/D = 4.
polygon 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.
d 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.
des, bisect angle ZEG.
H on tangent line.
he circle and between Z and G, EH > r and EH < hyp from triangle ZEG.
ght angle HGE and angle HEG = 15 degrees.
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le HEG, so the (unequal) sides are HE and GE and the base is GH.
E as hyp2 and base length GH as opp2 (side opposite top angle HEG).
riangle HGE to base GZ of right triangle ZGE. (C1-corrected)
nts GH and HZ.
al 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.
E has ZE > r, base segments ratio ZH to HG will have ZH > HG.
o estimate the polygon side using the new (shorter) right triangle HGE.
12-sided polygon determined by angle HEG being 15 degrees.
triangle HGE, but we need to know the length of HG. (C2-shortened)
s of the known ratios hyp/opp (or hyp/base) and r/opp (or r/base).
le ZGE, unequal sides ratio hyp/r = unequal base segments ratio ZH/HG.
e both pieces, keeping the denominator the same.
ich is hyp + r/r = base ZG/base segment HG.
r 1/4=3/12, it is also true that 1/3=4/12, so use that!
ich is hyp + r/ZG = r/HG.(C3-clearer to write base instead of ZG?)
ion: line 14) the ratios in the example: hyp/ZG=hyp/opp=307/153 and r/ZG=r/opp=265/153.
= (hyp + r)/opp = (307+265)/153 = 571/153.
larger than the starting ratio r/ZG=265/153.
which 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.
ngth of a side of a polygon with 12 sides and r is half the diameter D.
io of the circumference to the diameter is 12*HG/r.
(1530+307): 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.
AG, and also center E, and also line GLZ tangent, and also angle ZEG a third of a right angle.
ts centre, AC the tangent at A; and let the angle AOC be one-third of a right angle.
sliterated 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
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C
…
…
D
…
…
A
A
O
B
stigma, digamma,koppa,sampi
,8=η,9=θ,10=ι
00=Ϡ sampi
(or on top of it)
00, greek letters plus 3-4 phoenician symbols
τζ προς ρνγ΄.
Remember: 10 is iota, 100 is rho; 6, 90 and 900 are special.
Numbers as letter combinations:
Ratio occurrence 0th:
τζ προς ρνγ΄.
tau=300,zeta=7;rho=100,nu=50;gamma=3
s which 307 to 153.
nt line from center E to point Z on vertical tangent to length of vertical tangent from Z to G on diameter. It is 30
sions of a Triangle” written before!
f a right angle or 30 degrees. Why not exactly 2:1??
OC:CA = 306:153...(2). [OA:AC=square root 3:1; OC:CA=2:1].
slant to the vertical going down
153...(2). [OC:CA=2:1].
306 is incorrect; forgot digamma.
Ratio occurrence 1st:
ὃν σξέ προς ρνγ΄.
σξέ προς ρνγ΄.
sigma 200,chi 60, epsilon 5
sigma xi epsilon to 153
265::153
] GZ surely is, as which 265::153 [measures less].
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nt segment GZ is square root 3 to 1 or 265:153.
65:153...(1) [OA:AC=square root 3:1].
ees, get 15 degrees)
d [draw] EH to divide it [evenly].
AOC and meeting AC in D.
προς ΗΓ [καί έναλλάξ καί συνδεντι].
G [and also alternately and also connected]
Picture:
C or Z
D or H
A or G
O or E
B or A
it's above
it's added to explain
It's covered next sentence
ρος ΖΓ, ή ΕΓ προς ΓΗ.
o EG in ratio to ZG is the same as EG in ratio to GH.
EG in ratio to ZG and EG in ratio to GH.
al as is radius to lower section of vertical.
Not often used.
sq.root 3/1.
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+EG)/ZG = EZ/HG.
EG/ZG = 2/1+root3/1 = 306/153 + 265/153 = 571/153.
Ratio occurrence 2nd:
φοά προς ρνγ. 571::153
Phi 500,omicron 70,alpha 1, rho,nu,gamma 571::153
Also see notebook sketch 1.3.2.7.2
ν εχει, ήπερ φοά προς ρνγ.
note: no initial number sign, why?
to
153
more so has, more now as 571 to 153.
us EG to tangent segment GH is bigger than before, 1.3.2.3.
oot 3 to 1, and is now: 571:153.
571:153....(3).
Ratio occurrence 3rd: MYSTERY!
Μ ͵ϑυν΄ προς Μ ͵γυϑ΄ .
9,450 measures what?
theta 9, upsilon 400, nu 50;
it's in Bergren!
450 is in Bergren
3,404 measures what?
Then gamma 3, upsilon 400, delta 4
404? B has: 409
What does it mean to put smaller # before larger?
means thousands, right?
ase2 squared = 349,450/23,409 so the denominator is square of 153.
role of M? M is below horizontal in graph!
εχει, όν Μ ͵ϑυν΄ προς Μ ͵γυϑ΄ .
ϑ
is Greek special theta U+03d1
HG (or ratio EH:HG) strengthens all the moreso
0 in ratio to considering M 3+404?? role of M unclear! #s off?
ontal) from before M ?.
se2 HG ratio increases to ?459/434 or 10,000+459/10,000+434
he hypotenuse hyp2 in triangleHGE to the doubled base?
base2 in hyp2/base2. So doubling base decreases ratio by half.
swer 591/153 is next sentence. This is something prior.
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from this!?!
g squares against the sides of the triangle.
knew the Pythagorean relationship.
side of the circumscribed polygon or its a number with no number sign.
's significance either!
mbers for Bergren's squares!
Bergren's squares:
+ AD squared: AD squared >(571sq + 153 sq): 153 sq]
>(571 sq + 153 sq): 153 sq
>349450:23409
ike upside down 'h'
ϜϞϠ : 6,6,90,900
Ratio occurrence 4th:
φ ? ά ή΄ προς ρνγ΄
num: φ ? ά: phi 500 ? a 1
phi 500,koppa 90, alpha 1=591
? It must be!!
denom: ρνγ΄
Rho 100, nu 50, gamma 3=153
Ratio: 591/153
l HG is thus lengthened to 591/153.
ion using point theta (called T):
f sides, bisect angle HEG.
on tangent line.
he circle and between H and G, ET > r and ET < hyp from triangle HEG.
ht angle TGE and angle TEG = 7.5 degrees.
le TEG, so the (unequal) sides are TE and GE and the base is GT.
E as hyp3 and base length GT as opp3 (side opposite top angle TEG).
riangle TGE to base GH of right triangle HGE.
nts GT and TH.
al parts, but segments GT and TH are not equal.
top of a triangle, the segments of the base have the same ratio as the lengths of the sides.
E has HE > r, base segments ratio HT to TG will have HT > TG.
o estimate the polygon side using the new (shorter) right triangle TGE.
4-sided polygon determined by angle TEG being 7.5 degrees.
triangle TGE, but we need to know the length of TG.
s of the known ratios hyp/opp (or hyp/base) and r/opp (or r/base).
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le HGE, unequal sides ratio hyp/r = unequal base segments ratio HT/TG.
e both pieces, keeping the denominator the same.
ch is hyp + r/r = base HG/base segment TG.
r 1/4=3/12, it is also true that 1/3=4/12, so use that!
ch is hyp + r/HG = r/TG. Call baseHG base2 and hyp of HGE=hyp2,opp=opp2.
m the prior data, but so far we only know r/opp2=r/base2=571/153.
07/153 and r/base ZG=r/opp=265/153; so how to calculate?
and the known ratio r/base2=571/153, we compute hyp2/base2 as follows:
d is a ratio of two sides of a triangle.
res of the other two sides gives ratio: (r squared + base2 squared)/base2 squared
571 squared + 153 squared)/153 squared=(326,041 + 23,409)/23,409
50/23,409 so the denominator is square of 153.
tio hyp2/base2= 591 1/8 over 153.
PI source book stops with this.
p/opp + r/opp = (hyp + r)/opp = (307+265)/153 = 571/153.
nd hyp2/base2=591 1/8 over 153 and we seek r/TG:
uded (line 37):
ion: line 14) the ratios in the example: hyp/ZG=hyp/opp=307/153 and r/ZG=r/opp=265/153.
+ r/base2 = 591 1/8 over 153 + 571/153 = 1,162 1/8 over 153=7.6
s larger than the starting ratio r/HG=571/153 (initial one was 265/153).
which gives us that the ratio of the tangent segment TG to the radius r is smaller than before.
ratio of the circumference to the diameter as follows.
ngth of a side of a polygon with 24 sides and r is half the diameter D.
io of the circumference to the diameter is 24*TG/r.
(3060+612): 1162 or 3672:1162 or 3 186/1162 or 3 1/6 approximately.
e(3 1/5), less than first estimate C/D=4 and greater than the proposition 3 result 3 1/7 for the circumscribed case.
Sentence 9 yes, but sentence 7 no.
d, bisect as far as angle HEG by line ET (theta)
OD, meeting AD in E.
OA:DA = OA:AE]
Ratio occurrence 5:
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αρξβ΄ ή προς ρνγ.
αρξβ΄ =1,100,60(xi),2=1162
Ρνγ=100,50,3=153
ζονα λογον εχει, ή όν αρξβ΄ ή προς ρνγ.
the moreso, lengthens so is 1162 to 153.(line 70 above)
: 153, by (3) and (4)]
the front alpha
means thousand!
Ratio occurrence 6:
αροβ΄ ή προς ρνγ΄.
1,100,70,2=?
100,50,3=153
εχει, ή όν αροβ΄ ή προς ρνγ΄ .
e moreso has, a being 1172 to 153.
62 1/8 sq + 153 sq}: 153 sq
34 33/64 + 23409): 23409
33/64: 23409]
This is hypotenuse to new base in Bergren. Wrong?
Should be upper base seg to lower base seg per Greek?!!
os needed: the r/GT AND the will be useful next hyp/GT.
E and meet (vertical segment) AE in F.
??where?
HOORAY: from below?
͵δχπηʹ has lower/upper numeral signs
Ratio occurrence 7:
βτλδ΄ δ΄΄ προς ρνγ.
βτλδ΄ δ΄΄ or?2,300,30,4=2334
ρνγ or 100,50,3=153
χει, ή όν βτλδ΄ δ΄΄ προς ρνγ.
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dius to new base) all the moreso, is 2334 to 153.
onding to (3) and (5) above] that
radius to base
4)sq + 153 sq}:153 sq
hyp to base
Ratio occurrence 8:
βτλθ΄ δ΄΄ προς ρνγ΄.
what is the delta?
βτλθ΄ δ΄΄ or 2, 300,30,9=2339
ρνγ΄ or 100,50,3=153
could get old theta char from above!
τλθ΄ δ΄΄ προς ρνγ΄.
βτλθ΄ δ΄΄ προς ρνγ΄.
the moreso, is 2339 to 153.
hyp to base
hyp to base
OF, meeting AF in G.
only approximate!
iz is wiggly.
ϜϞϠ : 6,6,90,900
Could rewrite using:
deja vu san condensed angle: ∠
or measured angle: ∡
Ratio occurrence 9:
δΧογ΄∟΄΄ προς ρνγ΄.
δΧογ΄∟΄΄ or 4,(chi)600,70,3
ρνγ΄ or 100,50,3=153
No meaning given to wiggly L??
λογον εχει, ήπερ δΧογ΄∟ ΄΄ προς ρνγ΄.
the moreso, is greater being 4,673 to 153. (wiggly L not translated).
ical segment (smallest above horiz.) THIS IS radius to base.
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2330 1/4): 153, by means of (7) and (8).]
Number other:
mi μη΄ʹ 1/48?
m is 40, eta is 8
do the two '' give us fractions?
ρϑᾔς τέτμηται τετράϰις δίχα, ἡ ὑπὸ ΛΕΓ ὀρϑᾔς ἐστι μη΄ʹ.
mi? That much smaller?
, one-third of a right angle, bisected four times, is angle LEG ?.
which is 48 from 3. (another ratio!!)
6 1/12 1/24 1/48 of 90 deg or going from 30 deg to 1 7/8 degrees.
hird of a right angle,
t follows that
ὑπὸ ΓΕΜ.
ngle GEM (dips below horizontal).
side of OA equal to the angle AOG,
ϜϞϠ : 6,6,90,900
Number also:
Κδ
kappa is 20, delta is 4
Could the orthis be fraction??
The Greeks did do fractions, but how?
Number also:
ҷς ΄ʹ
koppa 90,stigma 6
κύκλον ἐστι πολυγώνου πλευρὰ πλευρὰς ἔχοντος ҷς ΄ʹ
the circle is polygon side worth 96 sides.
loygon of 96 sides
Ratio occurrence 10:
δΧογ΄∟΄΄ προς ρνγ΄.
δΧογ΄∟΄΄ or 4,(chi)600,70,3
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ρνγ΄ or 100,50,3=153
(like sentence 17)
μείζονα λογον ἔχουσα, ήπερ δΧογ΄∟΄΄ προς ρνγ΄, ἀλλἁ
διπλασίων ἡ ΛΜ, ϰαὶ ἡ ΑΓ ἄρα προς τὴν τὸὕ ҷς ΄ πολυγώνου
ρ δΧογ΄∟ ΄΄ προς Μ ͵δχπηʹ.
COULD THAT BE lower/upper numerals?? YES!
sentence 17:δΧογ΄∟΄΄ προς ρνγ΄: 4673 to 153
Ratio occurrence 11 (above):
δΧογ΄∟ ΄΄ προς Μ ͵δχπηʹ.
δΧογ΄∟ ΄΄ or 4,600,70,3=4673
Μ ͵δχπηʹor 10000,4000,600,80,8
o to the GL all the more so is worth 4673 to 153,
eter AG, then GL (the half side) doubled gives LM (the side),
the 96 sided perimeter of polygon is greater than
ygon of 96 sides)[>4673 1/2:153 x 96]
Number also:͵χξζʹ ∟΄΄
Chi 600, xi 60, zeta 7=667
Number also:δΧογ΄∟
4000,600,70,3=4673
σιν ͵χξζʹ ∟΄΄, ἅπερ τὤν ͵δΧογ΄∟ ΄΄ ἐλάττονά ἐστίν ἡ τὸ ἔβδομον.
should be two lamdas?
y ratio of 667 to 4,673 less than the threefold plus seventh.
4673) whereas 1/7=.1428 4/7.
673 ½ = [< 3 + 667 1/2/4672 ½]
κλον της͂ διαμέτρου ἐστί τριπλασίον και ἐλάττονι ἡ τᾠ ἔβδομᾠ μέρει μείζον.
cle of its diameter is threefold but less than to a seventh greater than.
eing argued?
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ὺ μἂλλον ἐλάσσων ἐστὶν ἤ τριπλασίον και ἔβδομᾠ μέρει μείζον.
circle is much more less than threefold but measures more than ? Seventh.
circle(being less than
ortiori less than 3 1/7 times
ἡ ὑπὸ ΒΑΓ τρίτον ὀρϑᾔς.
BA is the right angle.)
meter AG, also of the angle BAG a third of a right angle.
circle, and let AC
ngle CAB equal to one-third
Ratio occurrences 12th/13th:
͵ατναʹ προς ψπʹ
alpha=1,tau=300,nu=50;alpha=1
Psi=700, pi=80
1,351 to 780
͵αφξʹ
alpha=1,phi=500, xi=60
156 to 780
ONE DIGIT DROPPED
add last zero
ον ἒχει, ἡ ὃν ͵ατναʹ προς ψπʹ
pp side) is on the lesser side than, being 1,351 to 780
ide) being 156 to 780].
2*780=1560
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G by line HA (eta alpha).
and meet BC in d and the circle in D. Join BD.
ΗΓB, ἀλλἁ ϰαὶ τή υπο ΗAΓ,
ΗΓB τή ἱπο ΗAΓ ἐστιν ίση .
B or HAG (15 degrees)
Could ἱπο be a typo??
Angle GHA is 90 deg.
It's in triangles GHA, GHZ.
Use sum of angles is 180 deg.
This is well-known, says Offner.
υπο AΓΗ ίση .
he big angle AGH.
75 deg.
dBD, and angles at D, C are both right angles.
yes.
AC d], BD d are similar.
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This is:
Segs of base same ratio as sides.
ή ΓΗ προς ΗZ, ϰαὶ ή AΓ προς ΓZ. adj/opp, adj/opp, ?
G, GH to HZ and AG to GZ are the same ratios.
This is the above
this too is above
This is below.
diam plus 30line: nextbase+upseg1stbase
AG+AB:GZ+ZB (Greek not seen yet)
υναμφοτερος ή ΓΑΒ προς ΒΓ.
ratio to GZ (lower half of base),
ngle – 30 deg) in ratio to BG (base).
α ή ΒΑΓ προς ΒΓ, ή ΑΗ προς ΗΓ .
) in ratio to BG (base), and AH (new line) in ratio to HG (seg of base).
to the HG (its base).
Ratio occurrence 14:
͵βϠια΄ προς ͵ψπʹ,
beta=two thousand
sampi=nine hundred
iota alpha=ten+one=11
Psi=700, pi=80
..2911: 780
͵βϠια΄ προς ͵ψπʹ,
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1)..(p97) (and from above: = BD:Dd [=AC:Cd])
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e to the diameter is < 3/1/7 but > 3 10/71.
it is 'ratio')
ίση ἐστιν
pa and accent)
word, so surely?)
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wing pages in Latin.
meter is same as base.
e circumference, of the circle.
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h side ten seventifirsts.
0/71 sts is less.
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and no lines for the circle's edge, diameter or tangent:
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to G on diameter. It is 307 to 153, which is roughly two to one.
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