756 
first. The double of the 
Triangle Agb = 185 x HO = 20350 
Trapezoid ghcb = 250 x 160 — 40000 
Do. bide = 120 X 180 = 21600 
Do. iked = 32.5 X 190 = 61750 
Do. kfe — 300 x 349 = 104700 
Triangle 11 if = 630 X 289 = 182070 
"j430470 
2.15235 
.60940 
Content, 2 acr. 0 rds. 24 perches. 24.3 7 60~ 
Secondly. The double of the triangle 
ABC = AB X AC = 1810 X 23b = >430780 
2.1539 
, .6156 
24.624 
Content, 2 acr. 0 rds. 24 perches, as before. 
From whence it appears, that the content of 
the new triangle is the same as the aggregate 
contents of all the original triangles and trape- 
zoids, to within the decimal of a perch. 
In working with a chain and its olF-set staff, a 
measurer does well in making a rough sketch in 
his field-book, large enough to admit his writing 
down the lengths of all the necessary lines, 
whether for planning, or for casting off the 
content without a plan. 
Where there is a general base line, with seve- 
ral perpendiculars raised thereon, it may be 
best to continue the reckoning throughout that 
line - and, by subtraction, find the intermediate 
distances between one perpendicular and an- 
other. 
Example. Suppose from the sketch and dimen- 
sions of fig. 19, a true plan and the content of 
the field were required. 
Beginning at A, draw a line towards the tree 
at the upper end, and thereon prick off the dis- 
tances, as in the sketch. 
A; the proper points erect the perpendicu- 
lars, according to their respective lengths ; and 
the true figure will be as fig. 20. The whole 
content may be foi nd. by seeking the separate 
content of each triangle and trapezoid, from the 
dimensions given in fig. 19, thus : 
Of the trapez. b = 350 -f- 260~ X 200 = 122000 
c = 260 + 400 X 60 = 39600 
d — !00 4 - 350 X 490 = 367500 
e — 850 + 200 X 250 = 137500 
. / — 300 + 200~x 409 — 220000 
g = 250 + 400 X 500 = 325000 
of the triangle h = 2C0 x 400 = 80000 
2)1326600 
6.63300 
4 
~2.5S2 
40 
The content 6 ac. 2 rds. 21 perches. ~2h28 
Hitherto we hare supposed all the measuring- 
lines to be taken within the fields ; but a mea- 
surer may sometimes meet with fields so circum- 
stanced. by woody ground, meres of water, &c. 
as not to admit’ of the necessary internal lines 
being taken. Such pieces of land may, how- 
ever, be measured, by taking surrounding lines, 
making one or more right angles with each 
other, and raising perpendiculars from those 
lines to the angular points of the fields ; by 
which a true plan maybe constructed, and from 
thence the content found, either by equalizing 
the sides by the parallel ruler, or by deducting 
the contents of the small parts without, from the 
general content of the trapezium or surround- 
ing figure : see fig. 21. 
£.xiw?ple, A plan of the piece of woody ground 
SURVEYING. 
\’ 2 ’ 4, 5, 6, 7, 8, 9, 10, 11, 12, and IS, being 
drawn by a 6-chain scale, the content thereof is 
required. 
I be /_ A being (by the cross) made a right 
one, and the sides BA and AD being measured, 
the diagonal BD is readily found by construc- 
tion , or else, by extracting the root of AB X 
This diagonal being a base to the triangle 
BCD, and the other sides BC and CD measured, 
that triangle also is readily constructed, and 
the trapezium completed. 
The off-sets being made on the lines of the 
trapezium, the figure of the piece of wood may 
be correctly drawn. 
Its content may then be obtained, either by 
equalizing (with the parallel ruler, or other- 
wise) the lines of the wood, and thus reducing 
it to a trapezium ; or by deducting the content 
of all the small trapezoids without, from the 
general content of the outer trapezium, fig 22. 
First, by reducing the figure to the trape- 
zium, EFGH. 
Ihe lines being straightened, as before direct- 
ed, the diagonal of this new trapezium, found 
by the scale, will be 1070, and the sum of the 
perpendiculars, 1065. 
eg X eV+1R io 32 x 1060 
Then, ~L — — — =516 
2 2 
X 1060 = 546910= 5 acres, I rood, 34 perches, 
for the answer. 
Secondly, by finding the content of the sur- 
rounding trapezium, (fig. 21,) and from thence 
deducting the aggregate of the outer trapezoids. 
To find the triangle ABD, we have, ~~ 
910 x 930 
= = 455 X 930 = 423150. 
The diagonal BD = yAB 2 -f- AD 2 = 
+ 910 2 + 930 2 = 1301. 
Then, the triangle BCD is found thus : 
V s X s — d X s — b x s — c, where a, l, c , 
stand for the sides, and s for the half sum of 
those sides. 
a — 1301 
b = 970 
c = 830 
2)3101 
1550 = s, the half sum of the sides, 
■f — a = 249 
s — b = 580 
j- — c = 720 
Then, +1550 X 249 X 580 x 720 = 
61 172720000 = 401463. 
The triangle ABD = 4231 50 
The triangle BCD = 401463 
824613 = surrounding 
trapezium ABCD. 
Sum of 
perpend. Lengths. 
The trapezoid, No. 1 
= 129 
X 
250 = 30000 
2 
= 110 
X 
520 = 57200 
3 
= 100 
X 
160 = 16000 
4 
= 210 
X 
250 = 52500 
5 
= 160 
X 
210 = 33600 
6 
= 130 
X 
320 = 41600 
7 
= 80 
X 
50 = 4000 
8 
= 150 
X 
380 = 57000 
9 
= 160 
X 
280 = 44800 
10 
= 250 
X 
160 = 40000 
11 
= 410 
X 
110 = 45100 
12 
= 150 
X 
460 = 69000 
= 380 
X 
170 = 69600 
2)556400 
— (the aggregate of all the small trapezoids; 
which taken from 824613 (the content of the 
surrounding trapezium), leaves 546410, = 5 
acres, 1 rood, 34 perches., the content, as before. 
Thus far we have applied ourselves to single 
fields only; but we will now proceed to the 
measuring of two or more, lying contiguously 
to each other 
Example. From the dimensions given in the 
sketch, fig. 23, the contents of the fields A and 
B are required. 
acres, 3 roods, 11 perches, for the measure 
thereof. 
Field B. = 185000 = 1 acre, 3 
roods, 16 perches, for the content of that field. 
As in measuring single fields, various methods 
are pursued for obtaining the contents ; such as 
general lines with normals, or triangles com- 
bined with normals, <kc. ; so also may the con- 
tent of each respective field, contained in an 
estate, be found by like means. 
The estate, fig. 24, may be measured by a ge- 
neral line, with normals erected thereon, in 
manner following ; viz. 
Beginning at the southern end, a measurer 
takes his principal line from A towards the tree 
on the northern limits of his work ; and at 
every necessary point in that line, he sets off 
such perpendiculars as will lead him to the 
corners of each field, as in the figure. The di- 
mensions taken in each field being as here given, 
the content of each may be found in manner 
following : 
DIMENSIONS TAKEN. 
Double 
Bases. Normals. Operations. Areas. 
I-A a X ±L~ 74 0 X 25 = 14S00 
Aa X Ab -f- a2 = 360 X 80— 28800 
Ab X Aa + bo = 790 X 1500= 1185000 
ac X a2 4 c/ = 490 x_ 120= 58800 
bo X be = 760 x 60 = 45600 
2)1333000 
6.665 
2.66 
26. 
IF cf X 
+ 
4 
II 
CO 
Cn 
O 
X 
860 = 
731000 
dc X 
r/4 = 488 
X 
130 = 
62400 
c6 X 
cb 4 6,2 = 350 
X 
90 = 
31500 
6/ X 
6,2 = 500 
X 
40 = 
20000 
fi X 
f q = 470 
X 
40 = 
18,800 
2)863700 
4.3185 
1.274 
10.96 
III. Im X _4/ = 820 X 40 = 32800’ 
gh X gm 4 hk — 590 X 1290 = 761100. 
ik X ri = 770 X 30 = 23100 
gb X hi 4 gl = 590 X 270 = 159300 
)976300 
4.8815 
3.5260 
21.04* 
278200 
