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The angle sought is NAE = angle PZS = arc NE. Now, in the spherical triangle PZS, PZ is the co-latitude of the point A, which must be known.

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Soiving this tri

sin polar dist.

or sin Z =

cos lat.

From this, the angle Z becomes known, and, accordingly, it may be formed on the west side of the line AE, and thus the direction of the meridian AN determined.

On AN, thus found, let a substantial stake be set a hundred yards or more from A, and we have a permanent meridian with which we may compare the magnetic meridian at any time, and thus determine the declination of the needle.

The declination of the needle is the angle which the magnetic meridian makes with the astronomical meridian.

Every surveyor should have a standard meridian, as a means of knowing as near as possible the declination of the needle in his locality.

SCH.-For the purpose, simply, of finding the declination of the needle, it is sufficient to lay out on the ground the line of direction of the star at one of its elongations, and then, knowing the bearing of this line as shown by the needle, and the corresponding azimuth of the star, the declination of the needle is readily computed.

Thus, let a = azimuth, ± b

=

bearing, and ± d = declination, accordingly as they are east or west. Then ±d = ± a − (± b).

RULE.-Subtract the bearing from the azimuth.

In applying the Rule, due regard is to be had to the algebraic signs.

73. Field Work.-This is the same as in Chain Surveying, with the additional matter of observing and recording the bearings of lines.

Lines traversed on bearings are called Courses.

The following are examples of forms of Field Notes:

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S. 40° W. 8

3

9.88

2

11.74

N. 30° E.

S. 40° W.

9.88

O (2)

N. 30° E.

11.74

O (1)

The first form admits of writing offsets at the right and the left of the column as heretofore done. In practice, the bearings in this form are written obliquely across the column, so as to be easily seen without turning the paper.

74. The following is a specimen of the form of Field Notes used in the United States Survey:

THIRD STANDARD PARALLEL NORTH

through

Range No. 21 East

of the

PRINCIPAL BASE AND MERIDIAN

in the

TERRITORY OF MONTANA

as surveyed by

JAMES M. PAGE,

U. S. Deputy Surveyor,

On the night of August 22, 1880, I took an observation on Polaris in accordance with instructions contained in the Manual of Surveys, and drove pickets on the line thus established.

Survey commenced August 23, 1880, with a Burt's improved solar compass.

Before commencing this survey, I test my compass on the line established last night, and find it correct.

I begin at the standard cor. to townships 13 north, ranges 20 and 21 east, which is a post 4 inches square, marked

S. C. T. 13 N. on N.;

R. 21 E., S. 31 on E.; and

R. 20 E., S. 36 on W. faces, with 6 notches on N., E., and W. faces, and pits N., E., and W. of post, 6 ft. distant, and mound of earth around post.

Thence I run

Chains

18.00 40.00

57.00

80.00

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A point about 200 ft. above township cor., top of ridge.
Set a sandstone 18 x 8 x 5 in., 12 in. in the ground, for stàndard
4 sec. cor. marked S. C. 4 on N. face, dug pits 18 x 18 x 12
in. E. and W. of stone, 5 ft. dist., and raised a mound
of earth 11⁄2 ft. high, 31⁄2 ft. base, alongside; thence over

high, rolling prairie.

Enter pine timber.

Set a sandstone 21 x 10 x 7 in., 18 in. in the ground, for a standard cor. to Secs. 31 and 32, marked S. C., with 5 notches on E and 1 notch on W. edges; from which

A pine 12 in. diam. bears N, 77 E., 41 lk. dist., marked

T. 13 N., R. 21 E., S. 32 B. T.

A pine 18 in. diam., bears N. 50° W., 20 lk. dist., marked T. 13 N., R. 21 E., S. 31 B. T.

A pine 7 in. diam., bears S. 30° W., 119 lk. dist., marked T. 13 N., R. 21 E., S. 5 B, T.

Land high, mountainous, hilly and rolling.

oil sandy, gravel, and rocky; 4th rate.

Timber pine, 23 ch.; mostly dead and fallen.

As will be noticed, this form consists of two columns, reading from the top. In the first, are entered the distances from the point of beginning; and in the second, the direction and description of the line run, the monuments planted, with the witnesses of the same, the features of the ground, timber, soil, etc.

It is a commodious and convenient form for use, and ought to be adopted by surveyors throughout the country.

Finding Area. The manner of finding the area of fields surveyed with the Compass is developed in the solution of the following problems.

75. Prob. 1. Given two sides and the included angle of a triangle, to find the area.

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ABC =

FIG. 59.

CX CD

= bc sin A

2

Solution.-Let ABC be

a triangle in which are known the sides b and c and the angle A. Let CD be a perpendicular upon AB, from the opposite anBgle.

Geometrically, the area

; but CD=b sin A, whence area ABC

Example.-To find the area of a triangular field, there were measured the bearings and lengths of two sides, from the same corner, as follows: (1) N. 20° E., 12 ch.;

(2) N. 72° E., 30.10 ch.

Solution.-Let b = 12 and c = 30.10. The included angle A 72° 20°

= 52°.

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Area

142.8

log

2.154723

2 ar. co. log

or Area = 14.28 A. = 14 A. 1 R. 4.8 P.

Let the student present an example from the "field." 76. Prob. 2. Given two angles and the included side of

a triangle, to find the area.

Solution.-Let A and B be the given angles, and c the given side, (Fig. 59).

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Example.-To find the area of a triangular field one of whose corners was marked by a stake in a pond, there were measured the opposite side 12.50 ch. and at one end were taken the bearings of sides N. 30° E. and N. 78° E., and at the other end, the bearing of the third side N. 40° W.

Solution.-Let A be the angle at which the first two bearings were taken, and B the angle at which the third 'was taken.

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Then, A = 48°, and B = 180° − (78° + 40°) = 62°.
Applying logarithms c = 12.50

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log

1 096910

log

1.096910

log sin 9.871073

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77. Prob. 3. Given two angles and an opposite side of a triangle, to find the area.

Solution.-Let A and B be the given angles, and a the given side.

Now, sin C = sin (A + B), and c: a :: sin C: sin A, a sin (A + B)

We also have CD a sin B.

whence, c =

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Whence, area ABC =

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Example.-Given a = 10 ch., A = 25°, and B = 40°,

to find the area of the triangle.

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