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rire

lines for the measurement of which are sketched out as below, are required.

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recede

notes

84

go N. W

24

0 57 40 172 47. 0

Here the double area of the trapezium ABCD is found, as in Problem III, to which the double areas of the several offsets on the lines A B, BC, C D, D A must be added, and the sum will be the double area of the whole.

NOTE. A proof line, as A C should be measured, that the accuracy of the work may be insured :—but often pieces of ground are measured with the chain and cross that are never laid down, in this case great care should be taken in entering the field-notes, and in making the calculation. 5. Required the plan and area of a field from the following notes. To fence.

to OC 80 2024

2310 diag. 1900 to O D

B 990 1810 108 1200

720

1040 D. 66 800

From O A 520

To fence. 60 300

96

1320
130
000

106 1260 to © A.
From
OC

80

1000 To fence.

50

760 50 1230

100 400
70 1100 to O C

124 000
60
800

From OD
20 480
300

Double areas.
30
000

4689300 Trap. A B C D
From OB

392220

AB
To fence.

131920

BC 66 2180

offsets on 325016

go E.

go S.

44

of the

um 7 from

go N.

d.

CD 90 2090 to O B

235280

DA
112 1600
80 1000

2)57-73736
96 600
60 000

28.86868 = 28a. 3r. 19p. Begin At O

The whole area

rement eithe

dicula ee sides

80 W

d, the

NOTE. The field in this case has also four sides, like the one proposed to te measured in the last example. Fields of this kind are often required to be measured in the rural districts, to ascertain the quantity of a growing crop, as grain, hay, turnips, &c., when sold by the acre; also to find the quantity of reaping, mowing, planting, &c.; in these cases the plan is seldom or never required, and the measurement is only taken as far as the growing crop extends, leaving out the hedges, ditches, and all other waste or other ground, not occupied by the crop in question.

6. Give the plan and area of a field from the following notes.

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0 70 88 156

92 170 84

O
From

to O B
2346
2000
1800
1500
1200
800
400
000
O A go w

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PROBLEM VI. TO FIND THE AREA OF A SEGMENT OF A CIRCLE, OR ANY OTHER CURVIL INEAL FIGURE BY EQUIDISTANT OFFSETS OR ORDINATES.

RULE. If a right line AG be divided a b c de f g into any number of equal parts, AB, BC, CD, &c., and at the points of division perpendiculars be erected, Aa, B6, Cc, &c., to the curve a b c d e f g; then to the sum of the first and last offsets, add four

A B C D E F G times the sum of all the even offsets, and twice the sum of all the odd offsets, not including the first and last; multiply the sum by the common distance of the offsets, and one-third of the product will be the area, recollecting that the second, fourth, &c., are the even offsets, and the third, fifth, &c., are the odd offsets.

Note. If any portion of the figure is not included by an even number of offsets, its area must be found separately and added to the area found by the Rule.

EXAMPLES.

1. Required the plan and area of a piece of land measured by equidistant offsets or ordinates, from the following dimensions, (See last figure.)

9 450

=

First and last offsets. Odd offsets. to OG

3rd. = 376 =

Ca 600 G 1st. = 212 = Aa. 5th, = 430 = Ee. f442 500 F

Last = 450 = =Gg. 6 430 400

806 sum. d 406 300 D

662 sum.

2 C 376 200 с Even offsets. 6 306 100 B 2nd. = 306 = Bb

1612 twice sum 4th. = 406 = Dd.

4616 From OA go E. 6th. = 442 = Ff.

662

a 212

000

35 C

1154 sum.

6890 sum total. 4

100 com. distanca [sum. 4616 quadruple

3)6*89000

Area. 2a. Ir. 74 p. = 2•29666 2. Required the plan and area of a piece of land, measured by equidistant offsets, the dimensions being as given below.

175 First and last offsets. to O B

2238 Four times even offsets.
1071

933 Twice odd offsets.
1000 48}d
900
693

3346 sum total.
800
87}

100 common distance.
700 103
600 115

3)3.34600
500 124
400 130

1.11533 Area A adb.
300 132

2864 Trap. a B cdo 200

134 100 131

1.14397 = la. Or. 23p. Area 000 1261 ) From OA go W. Here the area of the trapezoid a Bcd is found

separately and added, agreeable to the preceding

note. 3. Plan and find the area of a piece of ground from the equidistant offsets given below.

96

91

115 126 130 121 110 93 70 From

to OB

600 90
500
400
300

82 200 69 100 43 000 24 DA go E.

Here the piece of ground is curved on two sides, the base line A B pass. ing nearly in the middle space between the two curves ; in this case the sum of the offsets on each side of every distance must be considered as one offset, in finding the arca.

4. Required the areas of two fields, the ends of which ara straight and parallel, and the side curved by the following equidistant offsets.

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Area 3a, 3r. 35p.

Area 2a. 2r. 15p. PROBLEM VII. TO MEASURE A LINE ACROSS A WIDE RIVER. Let the annexed figure be a river, which B is required to be crossed by the chain-line PB. Fix, or cause to be fixed, a pole or mark at B, at or near the margin of the river, in the line to be measured ; erect

D the perpendicular A D, measuring AC А and CD of any equal lengths; at D erect the perpendicular D E; on arriving at E, in the direction of B 0 produced, the distance D E will be equal to A B, the

PI required breadth of the river.

From the arrangement of the lines in the figure, it is evident that the triangles C A B, C D E are equiangular, and since AO was made = 0 D, the triangles are equal in all respects, and consequently A B = D E.

NOTE. We have thus given in sufficient detail the mode of surveying with the cross, which, though not much used by experienced surveyors, is a simple instrument, and its use readily understood by students. This method is, therefore, a proper introduction to the higher branches of surveying; besides, in rural districts, villages, &c., few surveyors use the more expensive instrument, the chain and cross being found quite sufficient to measure the quantities of growing crops, and other such small surveys as may be there required.

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