Notebook·PU-Mu 1050, Box 8, folder 5·June 1934–December 1935

B. H. Hill, I Site levels June 1934–Dec. 1935Bert Hodge Hill

·

Handwritten in English. Every line is transcribed, and the text sits beside the scan.

Scan 25 of notebook PU-Mu 1050, Box 8, folder 5 (B. H. Hill, I Site levels June 1934–Dec. 1935), handwritten in English. The transcription follows.

Scan: Penn Museum Archives, Philadelphia, PU-Mu 1050, Box 8, folder 5, Courtesy of the Penn Museum.

Unnumbered pageScan 25 of 87

Drawing: diagram with points E', E, E', W, T, etc. labels: x E to W of T_IV_ / 90° 22 1/2' / 21.10 / 16.58 / 28.45 / 3.85 / 180° / 195° 5' / 22.97 / (32.02) / 11.52 / E / 76.56 / E' / 4° 26' / (107.80) / W_W_ / W / E' / g / h / 11.70 / 80 in line of asces / of T_o_ T

For p 38-39

$Δ E' g W$ ∠ 15° 5' sides 11.52 + 22.97

log S 15° 5' =9.41535-
- 11.521.06145
8.35390
+ 22.971.36116
S 31° 15' 23" =9.71506
∠ at IV (arg =: 133° 39' 37")
C 11.521.06145
- S 15° 5'9.41535
1.64610
Sin 133° 39.379.85931
= 32.02 =1.50541
line E'-IV
+ Sin 15° 5'9.41535
IV-h = 8.33 =0.92076
Cos 15° 5'9.98477
E'-h = 30.92 =1.49018

In Δ E' E TV \angles 4° 26' (29.15 - 24.49 p 38), 164° 55 sides 76.56 + 32.02

Log 32 02 =1.50541
+ S 164° 05' [written under: 15° 55]9.41535
10.92076
- S 4.° 26'8 88817
2.03259
= 107.79

Margin right: x W. end of Z IV

in Δ E - h - IV E-h = 76.56 + 30.92 = 107.48 h-IV = 8.33 Tan \angleE = 8.33 ÷ 107.48

Log 8.33 =0.92076
107.48 =2.03133
Log Tan8.88943
∠:4° 25' 57"
Sin -8.88814
8.33 +0.92076
2.03262
\therefore E-IV= 107.80

Right page

Drawing: diagram with points C, D, E, II. labels: C / (124.47) / 75.40 / 262.24 / 12.35 / II / (138.27) / 5.15 / E / 117 / 185.74 / D

$Assuming$ \angles II C E + II E C correct as recorded 5° 40' + 5° 15' ∠ II \therefore = 169.5

2.41952
S. 169° 5'9.27734
3.14218
S 5.° 15'8.96143
2.10361 = 126.94
S 17° 15/5° 408.99450
2.13668 = 136.99

Thus if the 5° 40 + the 5° 15 are both correct \angleC D E is out 1^° 23' or more A C D L [illegible] at Damiet 104· 38· 28/67· 38· 30 C = 38.20\ \ D = 80^° 12\ \ h = 61.28

185.74 -136.99 =48.751.68797+
” 185.74 -+ "322.732.50884-
17.459.17913
162.1581.7 1/2a0.80 647
9.98560
44.15.281.68797
125-10.309.99478S 81° 8'
37. 4 301.68275
9.84137S. 44
69.24=1.84038
117 =2.06819
61.289.9437661^° 30'=9.9439061^° 31' = 9.94397
2.124432.124292.12422
38° 209 7925638.20 9.7925638-17 9.79208
82-101.9169982.58 = 1.9168582.47 = 1.91630
80° 129.9936280° 10 99935280 12 9.99362
2.118052.117862.11784
Ch= 131.24= 131.18131.17
h E131.50131.56131.57

Taking 2nd solution of above three If 5° 15' correct at E then ∠ at C must be 5^° 52 1/2 If 5° 40 " C " E " 5^° 6 1/4'

Δ C h II\ \ Ch = 131.18\ \ \angleI = 61° 30\ \ \angleII = 112° 50\ \ C = 5^° 40

131.18 =2.11786
S 112 50/67.109.96456C-II = 2.09720
2.15330S 5° 40 8.99450
S 5° 408.99450II-h = 1.09170 = 12.35
II-h =1.14780 = 14.05C-h" 9.99787
S 61 309 94390209507
C. II =2.09720 = 125.08C-h' = 124.47 \\ h II E = 138.27

Drawing: diagram with points labeled h, II, E, C

65 figures hard to readstruck struck outadded written in above” jug ditto, and what it repeatsword reading not certain

Spotted a wrong reading? Write to us or use alexis@ancientcyprus.com