32.46160 Area of the survey, 128 A. 2 R. 32 P.

EXAMPLE 3.

Required the area of a meadow from the following field-notes.

Left-hand off sets on the 2nd side.

Sta. Dist. Off-sets.

Ch. No. Ch. Ch.
1. N. 41°1 E. 14.35 1. 0.00 0.38
2. S. 421 E. 14.71 2. 2,65 2.35
3. S. 54 W. 16.32 3. 3.80 1.70
4. N. 324 W. 11.50 4. 6.00 2.75

5. 7.50 1.40
6. 9.60 3.20
7. 12.38

2.72
8. 14.71 0.42
Ans. Area 22 A. 3 R. 27 P.

EXAMPLE 4!

The following field-notes are given, to find the area of the survey

Left-hand off-sets.
On the 1st side. On the 2nd side.

Sta. Dist. Off-sets. Sta. Dist. Off-sets.

Ch. No. Ch. Ch. No. Ch. Ch. 1. S. 69°į E. 16.14 1. 0.00 0.44 1. 0.00 0.31 2. S. 28 E. 9.38 2. 3.80 2.00 2.

2. 2.67 2.94 3. S. 321 W. 21.20 3. 7.04 3.79 3. 6.20 2.62 4. N. 48 W. 22.47 4. 9.87 2.34 4. 9.38 0.39 5. N. 26 E. 19.00 5. 13.24 3.00

6. 16.14 0.31

Ans. 56 A. 2 R. 19 P.

a

When all the angles of a field, or small tract of land, can be seen from two stations, either within or without it, the area may be found by means of intersections. The method of doing this will be best explained by an example.

EXAMPLE 1.

Let ABCDEFGA, Fig. 84, represent a field, all the angles of which can be seen from two stations H and I without it. The bearing and distance of the stations, and the bearings of all the angles of the field, from each station, being as follow, it is required to find the area.

The station H bears from the station I, North, dist. 28 Ch.

Bearings.

Bearings.

Draw HI according to the given bearing and distance, and from the points H and I, draw HA, HB, HC, &c.

T

and IA, IB, IC, &c. according to the given bearings; then will the intersections A, B, C, &c. of the corresponding bearings HA and IA, HB and IB, HC and IC, &c. be the angular points of the field.

Calculation.

In each of the triangles IHA, IHB, IHC, &c. we have the side IH, and from the bearings of the sides, we have all the angles, to find the sides IA, IB, IC, &c.

Then in each of the triangles, IAB, IBC, ICD, &c. we have two sides and the included angle; whence the areas may be found by prob. III, sect. 1.

From the sum of the areas of the triangles IAB, IBC, ICD, and IDE, which is equal to the area IABCDEI, subtract the sum of the areas of the triangles IAG, IGF and IFE, which is equal to the area IAGFEI, the remainder will be the area of the field ABCDEFGA.

Note. In working the proportions for finding the sides IA, IB, &c., it will be unnecessary, when the area only is required, to take out the natural numbers corresponding to the logarithms of those sides; because in the proportions for finding the areas it will be sufficient to know the logarithms of the sides, without knowing their real lengths.

To find the log. of IA.

As sin. HAI 70° 00'
: sin. AHI 81 30
:: IH 28

9.97299
999520
1.44716

11.44236