On chain line.

Offsets.

Ag 150
Ah=323

gC: = 141 to left
hE = 180 to right

=

Ai 597 iD=167 to left
Ak624 kF=172 to right

AB=769

A

Twice the area of the

=

triangle AgC Ag gC=150 x 141 trapezoid

The double of the areas of the component triangles and trapezoids are found, in order that there may be only one division by 2, namely, that of their sum.

E

gi

Ai-Ag=447, iB AB-Ai 172, and hk= Ak -Ah301, Bk - AB-Ak = 145.

= 21150

CgiD=gi (Cg+ Di) = 447 × (141 + 167) = 137676 triangle DiB=Bi iD=172 x 167 triangle AhE=Ah hE=323 × 180

= 28724

= 58140

trapezoid

And area = 188291 = 1 ac. 3 ro. 21.26 pls.

=

hEFk hk (hE+kF) = 301 (180+172) = 105952 triangle B/F Bk kF 145 x 172 = 24940

Twice area 376582

20. Instead of writing the measurements as above, they are usually registered in a tabular form, called a field-book, as below. The beginning of the field-book is at the lower end of the table, as this arrangement suggests more readily the direction of the measurements. The middle column of the field-book contains the lengths measured on the chain lines, and the columns to the right and left of it contain respectively the right and left offsets.

The station from which the measurements are begun is called the first station; that next arrived at, the second; and so on.

The field-book of the measurements of a field similar to that of the last example, is given below in the following exercise; in which A is 0, and B is 02.

1

EXERCISE.

Find the area of a field, the dimensions of which are given in the following field-book :

Left Offsets.

334

To fence 282

Ans. 7 ac. 2 ro. 5-04 pls.

Example 2.-Find the area of the subjoined field from the following measurements:—

460

291

Chain Lines.

AO 291 links

Chain Line.

...

...

Right Offsets.

344

360 to road

210 25

Offsets.

Bn=155

DO=160

ps

30

gr 25

gh

= 50

ik 55

1

•

= 9000

Twice the areas of quadrilateral ABCDAC (Bn+DO) = 450 (155+160)=141750 triangle Agh Ag gh=180 × 50 trapezoid gikh = gi (ik+gh) = 230 (55+50) = 24150 triangle Bik Bi ik=50 x 55 triangle Dps Dp ps=65 x 30

·

2750

1950

=

7975

trapezoid pqrs
triangle Aqr

pq (ps+qr) = 145 (30+25) =
Ag gr=115 x 25

= 2875

=

Left Offsets.

=

=

And area = 95225 sq. lks. = 0 ac. 3 ro. 32-36 pls.

EXERCISE.

Find the area of a field similar to the preceding, from the measurements given in the subjoined field-book.

Ans. 0 ac. 3 ro. 39.72 pls.

Chain Line.

Twice area = 190450

360 to 02

344

248

From 01

192

to 04

To 03 276

21. The initial letters R and L are used for right and left, to denote the direction in which a line is to be measured. Sometimes the marks and are used to denote respectively a turning to the right and left. The expression in the above field-book " From og on L of 02," means that a

chain line is to be measured from the third station, and that it is situated to the left of the second station, in reference to the direction in which the first chain line, AC, is measured; so "From 01 to L of 02," means that the next chain line extends from o, to a point on the left of 02, namely, to o1

1

When the field to be surveyed is not very extensive, or the measurements not complex, they may be marked on a rough sketch of the field instead of in a field-book, as in the figure to Ex. 2.

22. On the left of the numbers denoting the left offsets, and to the right of those denoting the right offsets, lines are sometimes made, to represent in a general way the form of the boundary to which the offsets are drawn.

Example 3.- Find the area of the adjoining field ABCIHDFGE from the measurements in the following field-book, A, B, C, D, E, being respectively the 1st, 2d, 3d, 4th, and 5th stations.

Ans. 4 ac. 3 ro. 23.6 pls.

Find the areas of the principal triangles ABC, ACE, and CDE, in the above exercise by article 260 in Mensuration of Surfaces; then find the areas of the triangles and trapezoids composing the offset spaces EGFD and DHIC, the former of which is to be added to the areas of the principal triangles, and the latter to be deducted, in order to give the area of the given field ABCIHDFGE. Or, take the difference between the offset spaces, and add it to or subtract it from that of the principal triangles, according as the external space EGFD, or the internal one CIHD, is the greater.