Field Notes. 1st operation.

1815. Nov. 10. went on the top of the sharp or South peak of Otter, & from thence made these observations.

		°	′
	the meridian altitude of the sun by sextant	71–	8
	− error of instrument		1–	30
		71–	6–	30
		35	33	151

magnetic rhumb of North or flat peak N. 35.° 50′ E.

(the varian2 of needle had been before observed to be 2° East of N. i.e. 2° to the right of the true rhumb in all points)3

∠ of elevation of N. peak from apex of S. peak 52.′

from the apex of the sharp peak

	the highest point of the main ridge of Willis’s mountains makes an ∠ with the course of4 the summit of N.5 peak of 43°–37′
	magnetic rhumb of same point in Willis’s mount N. 80. E making the error of the needle 33′
	∠ of depression of the same point in Willis’s ridge from the apex of the S. peak is 46′

6 

 

1815. Nov. 11. 2d opern

this Fig. is entirely in the plane a.b.h.

at c. observe it’s position

	the vertical ∠ acg. + 7°–15′
	magn. rhumb c.g. N. 46. W
	height of instr. above water. ch 8.f
	vertical ∠s icd + 21′7
		ch  l
	measure c.d. to bank of Otter	25–52
	to publick road	17–
at d. observe		42–52	= 2806.32 f

the vertical ∠ adf 8°–12′8

 

both vertical & horizontal angles refer to the horizon.9

the rhumb is magnetical. varian 2.° E. of North.

measures in chains of 4. po.10 & 100. links, except where otherwise mentioned11

 

Notes for corrections

	Fig. 2.	in the vertical ∠ i.c.d. the elevn of 21.′ pointed 18 I. above the true parallel to the inclined plane. c.d.
		note that in all the operns the plane of the theodolite was 3 f 8 I. above the ground.
	Fig. 3.	in the vertical ∠ t.n.m. the elevn of 27′ pointed 11.f. above the true parallel of the inclined plane n.m.
	Fig. 2.	where the measured line c.d. crossed the main Otter river the plane of the theodolite was 3 f–8 I + 4 f–4 I = 8. f above the surface of the water.
	Fig. 3.	at the close of the measured line n.m. of 399.33 po. the main Otter crossed it at 8. po. from m. and twice more in the 12. po. preceding the last crossing.12

 

1815. Nov. 11. 3d opern

at n. observe it’s position

the	vertical ∠	l.n.p. 8°–20′
	horizl ∠	t.n.p. 90°–1313
	magn. rhumb.	n.p. N. 56. W
	vertl14 ∠	tnm 27′
	rhumb	nmS 32° 25′ W

	height of instrum. above the water NS. 3 f–8 I + 4 f–4 I = 8 f
	the altitude of summit of N. peak L 8°–27′

	it horizl ∠	with n.p. 27°–11′15
		with n.m 117°–24′16
	it’s magn. rhumb. N 29°–52½′ W

measure n.m. 399.33 po. = 6589.f

at m. observe

	the	vertical ∠ lmo 8°–2′17
		horizl ∠ omu 70°–41
		rhumb of mo N. 37–45 W
	alt.	of summit of N18 peak L 7°–13′19
		it’s horizl ∠ with mo 21°–53′
		mu 48°–48′20
		it’s magnetic rhumb. N. 15°–45′ W21

 

Nov. 12. a repetition of the observns from the station m.

 

the	∠ between mn. & the N. peak. 48°–48′  u.m.L.
	rhumb of the N. peak N. 15°–45′ W.  m.L.
	verticl ∠ of altitude of N. peak 7°–13′
	horizl ∠ between that & S. peak 21°–53′ o.m.L.
	rhumb of S. peak. N. 37°–45′ W.  m.o.
	verticl ∠ of altitude of S. peak 8°–2′

Fig.22 2.

in the vertical23 △	cdi. to reduce cd. to ci
	R : cd :: d : ci
	R. : 2806.32 :: S. 89°–39′ : ci		=	2806.261
	Log. 2806.32 =	3.4481372
	L.S. 89°–39′	9.9999910
−	Rad.	3.4481291.	=	2806.261
	reqd d.i.
	R  : cd   ::  c :  di
	Rad : 2806.32 :: S. 21′ : di		=	17.1427
	Log 2806.32 =	3.4481372
	L.S. 21′  =	7.7859427
−	Rad.	1.2340799	=	17.1427
but this is to be corrected by subtracting				1.5	see page 3.
gives the true length of di. =				15.6427

now to reduce the ∠ icd. to the corrected length of d.i.

in the △ cdi. given	ci = 2806.261
	di = 15.6427

 reqd ∠ icd

ci	:	di	::	Rad	:	T. dci
2806.261	:	15.6427	::	Rad	:	T. dci	= 19′–9″

	Log. 15.6427 =	1.1943118
+	Rad	10.
		11.1943118
−	Log 2806.261	3.4481279
		7.7461839	= 19′–9″

then in the △ cdi as reduced & corrected

	the ∠ icd	=	19′–9″
	ci	=	2806.261
	di	=	15.642724

---

 

in the vertical25 △ dik, to find i.k. in order to reduce ci. to ck.

 given di. and all the angles

S. k	:	di	::	S d	:	ik
S. 8°–12′	:	15.6427	::	S. 81°–48′	:	ik	= 108.5521

Log. 15.6427	1.1943118
L.S. 81°–48′	9.9955370
	11.1898488
− L.S. 8°–12′	9.1542076
	2.0356412	= 108.552

then ck = ci − ik = 2806.261 − 108.552 = 2697.709

in the vertical26 △	ack to find ak.
	given ck	=	2697.709
	∠ c	=	7°–15′
	∠ k	=	180° − 8°–12′ = 171°–48′
	∠ a	=	180° − c27 − k =  57′

 reqd ak.

S. a	:	ck	::	S. c.	:	ak
S. 57′	:	2697.709	::	S. 7°–15′	:	ak	= 20533.8

Log. 2697.709	3.4309952
L.S. 7°–15′	9.1010558
	12.5320510
− L.S. 57′	8.2195811
	4.3124699	= 20533.8

in the right ∠d vertical28 △ akg. given ak	=	20533.8
∠ k	=	8° –12′
∠ a	=	81–48
reqd ag. and gk.

Rad	:	ak	::	S. k	:	ag.
Rad	:	20533.8	::	S. 8°–12′	:	ag.	=	2928.711
Log. 20533.8			4.3124699				+	4.333
L.S. 8°–12′			9.1542076					2933.044	= ab
− Rad.			3.466677529				=	2928.711

---

Rad	:	ak	::	S. a	:	gk.
Rad	:	20533.8	::	S. 81°–48′	:	gk.	=	20323.9
Log. 20533.8			4.3124699
L.S. 81°–48′			9.9955370
− Rad.			4.308006930				=	20323.9
							+	2697.709	= ck
								23021.609	= cg31

 

Fig. 3.

in the r. ∠d vertical32 △ mnt to reduce nm. to nt

 and to find mt.

	given nm	=	6589. feet.
	∠ n	=	27′
	∠ m	=	89°–33′
	reqd nt. and mt.

Rad.	:	nm	::	S. m	:	nt
Rad	:	6589	::	S. 89°–33′	:	nt.	=	6588.8
Log. 6589			3.8188195
L.S. 89°–33′			9.9999866
− Rad.			3.818806133				=	6588.8

---

for mt. say Rad.	:	nm	::	S. n.	:	mt
Rad	:	6589	::	S. 27′	:	mt	= 51.7493

	Log. 6589.	3.8188195
	L.S.  27′	7.8950854
−	Rad.	1.713904934	=	51.7493
to be corrected by subtracting				11.	see page 3.
true length of mt =				40.7493

---

next reduce the ∠ n. to the corrected length of mt.

in the r. ∠d △ mnt. given nt	=	6588.8
mt	=	40.7493
reqd ∠ n.

nt	:	mt	::	Rad.	:	T. n
6588.8	:	40.7493	::	Rad.	:	T. n	= 21′–11″

	Log. 40.7493 + Rad. =	11.6101202
−	Log. 6588.8	3.8188061
		7.7913141.	= T. 21′–11″

---

in the horizontal △ npt. given nt.	=	6588.8
		°   ′
∠ tnp	=	90–13
∠ ptn	=	70–41
∠ npt	=	19– 6
reqd pt. and pn.

for pt. say S. p.	:	nt	::	S. n	:	pt
S. 19°–6′	:	6588.8	::	S. 90°–13′	:	pt.	= 20135.61

	Log. 6588.8	3.8188061
	L.S. 90°–13′ = 89°–47′	9.9999969
		13.8188030
−	L.S. 19°–6′	9.5148371
		4.3039659	= 20135.61 = pt35

 

for pn. say S. p	:	nt	::	S. t.	:	pn.
S. 19°–6′	:	6588.8	::	S. 70°–41′	:	pn.	= 19002.2

	Log. 6588.8	3.8188061
	L.S. 70°–41′	9.9748361
		13.7936422
−	L.S. 19°–6′	9.5148371
		4.2788051	= 19002.2 = pn

---

in the r. ∠d verticl △ lmo given mo.	=	20135.61
∠ m.	=	8°–2′
∠ l	=	81–58

reqd lo.

S. l.

:

mo

::

S. m.

:

lo.

S. 81°–58′

:

20135.61

::

S. 8°–2′

:

lo.

=

2841.83

oq.

=

48.7493

Log. 20135.61

4.3039659

lq

=

2890.5793

L.S. 8°–2′

9.1453493

op

=

mt

=

40.7493

13.4493152

pq

=

tr

=

8.

−

L.S. 81°–58′

9.9957172

oq.

=

48.7493

3.4535980

= 2841.8336

---

let LOPQ. be points in the axis of the North peak corresponding horizontally with l.o.p.q. in yt of the S. peak; and S. the summit of the North peak.

we have still to find	the horizontl distance of the two Axes
	the height of S. by the observns at	l.
	at	m.
	at	n.

reduce the inclined37 trapezium lmnL. fig. 4. to it’s corresponding horisontal38 one p.t.n.P. in the horizontal plane of n

 

	In the horisontal39 △ Ptn.
	given nt	=	6588.8
			° ′	° ′
	∠ Pnt	=	117. 24	or 62–36
	∠ Ptn	=	48–48
	∠ tPn	=	13–48
	reqd Pn.
	Pt.

to find Pt. S. P.	:	nt	::	S. Pnt	:	Pt
S. 13°–48′	:	6588.8	::	S. 62°–36′	:	Pt.	=	24523.3

	Log. 6588.8	3.8188061
	L.S. 62°–36′	9.9483227
		13.7671288
−	L.S. 13°–48′	9.3775493
		4.3895795	= 24523.3

---

to find Pn. S. P	:	nt	::	S. Ptn	:	Pn.
S. 13°–48′	:	6588.8	::	S. 48°–48′	:	Pn	=	20783.21

	Log. 6588.8	3.8188061
	L.S. 48°–48′	9.8764574
		13.6952635
−	L.S. 13°–48′	9.3775493
		4.3177142	= 20783.21

---

In the horizl40 △ Ptp. given Pt =		24523.3
pt =		20135.61
∠ Ptp =		21°–53′
reqd pP.41

 

	Pt + pt	:	Pt − pt	::	T. tpP. + pPt⁄2	:	T.	tpP − pPt⁄2
					°  ′ ″			° ′   ″
	44658.91	:	4387.69	::	T. 79–3–30	:	T.	26–55–26
	Log. 4387.69				3.6422278			79– 3–30
	L.T. 79°–3′–30″				10.7134430			105–58–56	= tpP.
					14.3556708			52– 8– 4	= tPp.
−	Log. 44658.91				4.6499081
					9.7057627		= 26°–55′–26″ = T d/2

---

for pP. S. tPp		:	pt	::	S. Ptp	:	pP
	S. 52°–8′–4″	:	20135.61	::	S. 21°–53′	:	pP. = 9506.451
	Log. 20135.61		4.3039648
	L.S. 21°–53′		9.5713802
			13.8753450
−	L.S. 52°–8′–4″		9.8973264
			3.9780186		= 9506.451

---

In the horizl42 △ Pnp. given Pn	=	20783.21
pn	=	19002.2
∠ Pnp	=	27°–11′	reqd pP.

	Pn + pn	:	Pn − pn	::	T. npP + nPp⁄2	:	T.	n.pP − nPp⁄2
					° ′   ″			° ′   ″
	39785.41	:	1781.	::	T. 76–24–30	:	T.	10–28–2343
	Log. 1781.				3.2506639			76–24–30
	L.T. 76°–24′–30″				10.6165949			86–52–53	= npP.
					13.8672588			65–56– 7	= nPp.
−	Log. 39785.41				4.5997239
					9.2675349		= T. d/2 = 10°–28′–23″

---

for pP. S. nPp		:	pn	::	S. Pnp		:	pP.
	S. 65°–56′–7″	:	19002.2	::	S. 27°–11′		:	pP. 9507.27
	Log. 19002.2		4.2788039
	L.S. 27°–11′		9.6597633
			13.9385672
−	L.S. 65°–56′–7″		9.9605114
			3.9780558		=	9507.27
						9506.451
						19013.721
by mean of 2. operns			pP. =			9506.86 f.	= 1.8 mile
						9506.45
						19013.31
	mean length of p.P.					9506.6544

 

By way of proof of these operations, we may at this stage collate the magnetic rhumb of m.n. or t.n.45 observed Nov. 11. and transferred by calculation to p.P. fig. 5. with the rhumb of l.S. or p.P. observed from l. Nov. 10.

from p. in the horizontal trapezium ptnP. Fig. 5. draw pv. parallel with tn. this parallelism makes the ∠ tpv the supplement to ptn.

				°   ′
we found the	∠	ptn	=	70–41
then is		tpv	=	109–19
but		tpP	=	105–58–56
and consequently		P.pv	=	3–20– 4
by observn Nov. 11. tn =		pv	is	N. 32–25– 0	E
add the	∠	Ppv		3–20– 4
gives by calculn		pP		N. 35–45– 4
by observn of Nov. 10		pP.		N. 35–50– 0
difference				0– 4–56

altho’ the needle is not to be relied on for exact precision, the magnetic rhumbs of the lines were taken as general checks on the other observations, and particularly to indicate whether the corresponding angles observed were to the right or left. the rhumbs now collated are so far a confirmation of the correctness of the intermediate work.

---

for the height S. above l. that is, the height of the North peak above the South, by the observns at l.

in the right ∠d vertical △ lLS.	Fig. 1.
given ∠ SlL.	= 52′
lL	= pP. = 9506.86
reqd LS

	Rad	:	lL	::	T. SlL	:	LS
	Rad	:	9506.86	::	T. 52′	:	LS	= 143.482

Log. 9506.86		3.9770371
L.T. 52′		8.1797626
−	Rad.	2.1567997	46 = 143.482

---

for lS Si. of lSL	:	lL	::	Rad	:	lS
S. 89–8′	:	9506.65	::	Rad	:	lS	= 9507.73

Log. 9506.65 + Rad.	13.9780275
L.S. 89°–8′	9.9999503
	3.9780772	= 9507.73 = lS.47

for the height of S. the summit of the N. peak by the observations

  from m.

in the horizontal △ Omu. Fig. 4.

	given mu	=	6588.8
			° ′
	∠ umO	=	48–48
	muO	=	117–24 or 62°–36′
	mOu	=	13–48
	reqd mO
	uO.

	S. O	:	mu	::	S. u	:	mO.
	S. 13°–48′	:	6588.8	::	S. 62°–36′	:	mO.	= 24523.3

	Log. 6588.8	3.8188061
	L.S. 62°–36′	9.9483227
		13.7671288
−	L.S. 13°–48′	9.3775493
		4.3895795	= 24523.3

---

for uO, S. O	:	mu	::	S. m	:	uO
S. 13°–48′	:	6588.8	::	S. 48°–48′	:	uO	= 20783.21

	Log. 6588.8	3.8188061
	L.S. 48°–48′	9.8764574
		13.6952635
−	L.S. 13°–48′	9.3775493
		4.3177142	= 20783.2148

 

In the right ∠d vertical △ SmO
given mO.	=	24523.3
		° ′
∠ m	=	7–13
∠ S	=	82–47
reqd SO

	S. S	:	mO	::	S. m	:	SO
	S. 82°–47′	:	24523.3	::	S. 7°–13′	:	SO.	=	3105.261
	Log. 24523.3			4.3895795			+ oq	=	48.749
	L.S. 7°–13′			9.0990651			SQ	=	3154.01
				13.4886446
−	L.S. 82°–47′			9.9965459
				3.4920987			= 3105.261

---

In the right ∠d vertical △ SnP.
	given nP	=	uO = 20783.21
			° ′
	∠ n	=	8–27
	∠ S	=	81–33
	reqd SP

	S. S	:	nP	::	S n.	:	SP
	S. 81°–33′	:	20783.21	::	S. 8°–27′	:	SP.	=	3087.54
	Log. 20783.21			4.3177142			+ pq	=	8.
	L.S. 8°–27′			9.1671586			SQ	=	3095.54
				13.4848728
−	L.S. 81–33			9.9952597
				3.4896131			= 3087.5449

 

Proceeding to the result of the observations of altitude, they do not come out so fortunately correct as the horizontal angles did. for, taking a mean of the two altitudes of the S. peak, and of the two of the North peak, that of 143 f resulting from the observation at l. will not fill up the whole space of their difference. to bring them together we must distribute the whole error equally among the several vertical angles which enter into the operations.

3. of these are in Fig. 2. viz. acg. dci. & adf.

2. are in Fig. 3. mnt & lmo.

1. in Fig. 1. SlL.

2. in the altitudes of the N. peak, the one taken

8.   from m. the other from n.

the following equation gives the portion of error in altitude to be ascribed to each of these vertical angles.

2933 + 3x + 2890 + 2x⁄2 + 143 + x	=	3154 − 2x + 3095 − x⁄2.
3054.5 + 3.5x	=	3124.5 − 1.5x
5x	=	70
x	=	14

there must be a correction then of 14.f in altitude for every vertical angle entering into each result, as follows.

the height of S. peak
	from Fig. 2. is 2933. add 14 × 3. ye error for 3 ∠s gives	2975
	from station m. 2890 + 14 × 2. for 2. ∠s	2918
		5893
	the mean may be considd the height of S. peak	2946.5
the height of N. peak
	from station m. 3154. subtract 14 × 2. error for 2. ∠s	3126.
	from station n. 3095. − 14 error of 1. ∠	3081
		6207
	the mean height of the N. peak	3103.5
	to the height of S. peak 2946.5 add 143 + 14 = 157 for 1. ∠	3103.5
	from the height of N. peak 3103.5 subtract 157	2946.5

this error of 14.f. in altitude on every vertical observn amounts to about 2¼′ which with an instrument whose Nonius indicates to 3.′ only, is perhaps as near as we may generally count on coming; at least as near as eyes of 72. may come. add to ys too the diameter of the crosshair which, magnified as it is produces sensible uncertainty.50

 

To recapitulate

the mean height of the sharp or S. peak		f
	above the surface of Otter river is	2946.5
	of the North peak	3103.5
	their difference of height	157.

the distance of the 2 summits nearly 1.8. mile exactly 9507.73

the magnetic bearing of the summit of the North from that of the S. peak is N. 35–50. E from which 2.° must be subtracted for the present variation of the needle.

the base lines measured, the one of 2806.33 feet or .55 of a mile, the other of 6589 f. or 1¼ mile, were on the plains of Otter river51 belonging to Christopher Clarke esq. & to Donald’s heirs, near the mill of the latter; the former line in exact direction to the axis of the S. peak, the latter nearly parallel with the bearing of the one peak from the other.

the distance of the base lines measured, from the points in the bases of the mountains vertically under their summits was, the nearest 19002.2 f. the farthest 24523.3 f. or about 4. miles generally.

supposing the radius of the earth 3965. miles and the height of the52N. Peak 3103.5 feet = .5876893 mile then it may be seen over a level country to the distance of 68.2083 miles, which will include the whole or a part of the counties of Amherst, Nelson, Albem.Fluvanna, Buckingh. CumbldFranklin, BedfdCampbllPr. EdwdCharlottePatrick, Henry, Pittsylva, Halifax

  and over the tops of the intervening mountains in Rockbridge & Botetourt, the following is the proof.

in a right ∠d △ given the hypoth. b.c. = 3965.588 miles a leg. ac = 3965. required the other leg. ab.

	bc	:	Rad	::	ac	:	S. b.
	3965.588	:	Rad.	::	3965	:	S. b	= 89°–0′–52″

	Rad. + Log. 3965	13.5982432
−	Log. 3965.588	3.5983075		°   ′   ″
		9.9999357	=	89– 0–52	= ∠ b
			and	59– 8	= ∠ c53

	Rad	:	bc	::	S c	:	ab
	Rad	:	3965.588	::	S. 59′–8″	:	ab	68.2083

Log. 3965.588		3.5983075
L.S. 59′–8″		8.2355299
−	Rad.	1.8338374	54 = 68.2083

vertical angles		corresponding bases		corresponding perpendiculars
	° ′
SlL.	0–52	lL	9506.86	SL.	143.482
icd.	21	ci	2806.261	di	17.1427
akg	8–12	gk.	20323.9	ag.	2928.711
acg.	7–15	cg	23021.609	ag.	2928.711
tnm.	27	nt.	6588.8	mt.	51.7493
lmo	8– 2	mo.	20135.61	lo.	2841.83
S.mO	7–13	tP.	24523.3	SO.	3105.261
SnP.	8–27	nP.	20783.21	SP.	3087.54

to estimate loosely the average of error in the vertical observations the average of the perpendiculars is 1888. of which 14 f. is the 1⁄135 part the average of the vertl angles is 5°–6′–7½″ = 18360.″ 135th part is 136″ = 2′–16″55

 

Map of the ground. scale 1000. f = ¼ inch

The56 form of the Peaks as seen from mr Clark’s