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2. Let the foregoing figure represent a river, a plan of which is required.

Begin at a, and measure to c, taking offsets to the river's edge. as you proceed. From c measure to d, and there take the tie or chord-line db, which will enable you to lay down the first and second lines. Continue the second line to n, and from m measure to r, at which place take the tie-line rn; and thus proceed until you come to the end of your survey at x.

The breadth of a river, if it be everywhere nearly the same, may be taken in different places by the next problem or by Problem X. Part III.; but if it be very irregular, dimensions must be taken on both sides, as above.

When the area is required, it must be found from the plan, by dividing the river into several parts, and taking the necessary dimensions by the scale.

NOTE.-Any bog, marsh, mere, or wood whatever may be its number of sides, may be measured by this problem.

PROBLEM VI.

Taking Distances by the Chain and Scale.

EXAMPLE.

Required the distance of an object at A, from B.

First, make a station at B; then, in a direct line with BA, set up a pole, suppose at c; measure the distance BC. Return to B, and measure in any direction, making an angle with BC, suppose to D; then set up a pole in a direct line with DA, as at E. Measure the lines DE and EC, and also the diagonal CD; these will enable you to construct the trapezium BCED.

The lines BC and DE, produced, will evidently meet at A.

Measure the line BA with the same scale by which you have constructed the trapezium, and it will be the distance required.

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B

A

PROBLEM VII.

To erect a Perpendicular by the Chain, or to measure Lines upon which there are Impediments.

EXAMPLE.

Suppose CDEF to represent the base of a building, through which it is necessary a line should pass from A, to an object at B, the direction AmB on the side a being given.

Measure from A to m; and from m, measure back to a, 40 links. Let one end of the chain be kept fast at a, and the eightieth link at m; take hold of the fiftieth link, and stretch the chain so that the two parts an and mn may be equally tight then will mn be perpendicular to

am.

For mn will be 30, am 40, and an 50 links; or the sides of the right-angled triangle amn will be in proportion to each other as 3, 4, and 5. (See Prob. XVIII. Part I.)

Measure from m, upon the line mn continued, until you are clear of the impediment, as at c; than continue the line 40 links farther, to b. Find by the above process the perpendicular cd; and proceed in that direction till you are beyond

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the building, as at h. Again erect the perpendicular he, upon which measure till you have made hp equal to me; and you will then be in a direct line with ma. Erect the perpendicular px, which (if you have conducted the work with correctness) will be in a right line with B. Measure the distance pB; then Am, added to ch (=mp), and pB, will give the whole length of the line AB.

PROBLEM VIII.

Having the Plan of a Field, and its true Area, to find the Scale
by which it has been constructed.

RULE.-By any scale whatever measure such lines as will give you the area of the figure: then say, as this area is to the square of

the scale by which it was found, so is the true area to the square of the scale required.

Example.

Suppose the true area of a field, the plan of which is given, to be 9a. 1r. 32p.; and that by a scale of 2 chains to an inch, I find the area to be 4a. Or. 32p.; required the scale by which the plan was

constructed.

First, 9a. 1r. 32p. = 945000 square links; and 4a. Or. 32p. =420000 square links; then, as 420000: 4:: 945000: 9. Hence it appears the plan was constructed by a scale of 3 chains to an inch.

NOTE. The principle of this process is, that the areas of similar figures are to each other as the square of their homologous sides. (Theo. XVI. Part I.)

HILLY GROUND.

The survey of a hill is taken conjointly upon the horizontal and vertical planes. Both planes are generally required to determine the area as represented upon the plan. The details of the vertical survey are shown by sections.

There are three methods practised in finding the area of a hill, viz., the horizontal, the hypotenusal, and the superficial.

The first way is that which determines the horizontal area. It accords with trigonometry on the horizontal plane, and is that which is generally practised in Land-surveying.

The second way gives the hypotenusal area, and also the angle which the hypotenusal plane makes with the horizon. Surveys of this kind are frequently ordered for plantation and other grounds, and for whole estates when they lie on the sloping sides of large hills.

The third way returns the superficial area. It is generally adopted for paring and burning, trenching, ploughing, reaping, and other agricultural works.

Methods for reducing Hypotenusal to Horizontal Lines.

METHOD I.

When the hill is of a regular slope, take its altitude with a theodolite, or with a quadrant; then, by a trigonometrical canon, in which the hypotenuse may be counted 100 links, determine the number of links in the base. These deducted from 100 will show the number of links by which each chain must be shortened, for the purpose of planning.

Example.

Suppose the altitude of a hill to be 16° 15', and the length of a line measured upon its surface to be 2550 links; required the length of the line that must be used in planning.

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In the right-angled triangle ABC are given the hypotenuse AC = 2550, and the angle BAC = 16° 15', to determine the base AB.

AD = 100, and the angle EAD = 16° 15′, to find AE.

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Or

Hence it appears that 4 links must be subtracted from each chain; consequently (25 × 4 + 2 = ) 102 links must be taken from AC; hence AB 2448 links, the line required.

=

PROOF.-AS, 1: 2550 :: 96005 (the nat. co-sine of 16° 15') : 2448-1275 links = AB.

NOTE.-Surveyors, when serving an apprenticeship, acquire a knowledge of how to drive the chain in hilly ground; and those who master this branch of their profession approximate very closely to the length of horizontal lines found by trigonometry.

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