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

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

Text·

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

Scan 33 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.

Page 66Scan 33 of 87

	△	_EF_IV__	(drawing)
	∠ IV · E · E'	4^° 26
	F · E · E'	1· 25
	IV· E\cdotF	5^° 51

	∠ IVFE	12^°· 35	(p\ 34\ 40· 50 - 28· 15)
		[Take 46.80 as compromise estimate]

	-	Log	46.76	=	- 1.66987	5.26 - 41
	+		5^° · 26^′ · 41^′′	+	8.97720	12.38 - 90-
					7.30733	18.04-50-
	-	Log	107.80		2.03262	161· 55· 10
	F =	+ 13^° 34^′· 42^′′ =			9.33995	Sin = + 1.66987
		12· 38· 9	an error	of 1^° at	F?	+ 8.97720
		12· 38· 9·				$7.69267$
				Sin	161· 55· 10	9.49160
						$7.18427$
				152.85	= EF =	152.85

	△	_DF_IV__	∠ D =	2^° 18^′	[42.43 - 40· 25	p 36]
		_D_IV__ =	$265.72 =$	Log	$2.42442$
		IV· E	$46.76$	” Log	$1.66987$
	$1.66987$				0.75455
	8.60349	}	2^° 18	ls	8.60349
	$3.06638$	13^° 10^′ 56^′′	$=$	9.35804
	9.42641	164\ 31\ 4
	$2.49279$	= 311.02	= DF

	△	_GE_IV__	7942 the	measured	∠ at G =	$18.40$
		GF_IV_	makes F_IV_	too long	. solver △
	[4 \angles four sides $326.90$, $306.96$ & $46.76$
	[using formula for 1/2 Sin A = sqrt{(s-b)(s-c)bc}
	Result:	G =	7^° 39^′	20^′′	IV =	111^° 21^′ to^′′	F =	60^° 59^′ 30^′′
							13	27

	∠ FGE \therefore =	19· 14· 42	(EG_IV_	p 65)
		7· 39· 20	(IVGE	above)
		26^°· 54^′· 2^′′

Right page

	△ FGE	EG =	$252.60$	FG	$326.90$
		∠ G =	26^° 54^′\ 2^′′	[fort	p \ 66]
	$326.90$	-	$252.60$	= 74.30	(Log \ 1.87099)
		+		$= 579.50$	("\ 2.76305)

	$180 - 26.54.10 =$	$153-5-50$	1/2 =	76· 32· 59
	$74.30$	$1.87099$		$1.87099$
	Tan	+ 0.62131	+ S =	9.98492
		$2.49230$		$1.85891$
	$579.50 -$	2.76305	- Sin	9.67439
	Tan	$9.72925$		$2.18452$
		28^°· 11· 46	EF =	152.96
		- 76· 32· 59
	∠ FEG	104· 44· 45	15
	∠ EFG	48· 21· 13	$13$ or $12$

	△ DEF	DE	$185.74$	(Log	$2.26887$)	(p $44$ oe)
		DF	$311.07$	” (Log	$2.49286$	(opp. page)
		EF	$152.97$	” (Log	$2.18461$	(above)
	∠ D	FD_IV_ \ (42.43 - 40.25\ p 36) = 2^° 18^′ \ (p 66) \\ _IV_DE 13.27 + 5.16 \ (p\ 64-65) = 18.43
			21^° 1^′
	[4 measured 20· 55; but 20· 1 results as a
	"bm" for solution as below for separate △
	Give $185.74$ $152.96$ & 21^° 1^′ solve for
	\angles & side DF

	\therefore S\ 21· 1	$9.55466$
	: $152.96$ -	2.18452
	EF	$7.37014$
	$185.74 +$	2.26887
	Sin	$9.63901$
	∠ F =	25^° 49^′\ 7^′′

	$2.18452$	$21-1-0$
	- 9.55466	25\ 49\ 7
	$2.62986$	$46-50-7$
	+ 9.86296	Sin	133· 9· 53
	$2.49282$
	311.04	= DF

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

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