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 d b, 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.

If the breadth of the river be every where nearly the same, its breadth taken in different places, by the next Problem, or by Problem 10, Part III. will suffice; 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 B A, set up a pole, suppose at C; measure the distance B C. Return to B, and measure in any direction, making an angle with B C, suppose to D; then set up a pole in a direct line with D A, as at E. Measure the lines D E and E C,

and also

the diagonal C D; these will enable you to construct the trapezium B C E D.

The lines B C and D E, produced, will evidently meet at A.

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

NOTE. This method may be well applied to measuring the breadth of a river, or the distance of any inaccessible object; and any person, acquainted with Trigonometry, may easily find the correct distance, after measuring the lines before mentioned.

B

PROBLEM VII.

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

A

Measure from m, upon the line m n continued, until you are clear of the impediment, as at c; then 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 the building, as at h. Again erect the perpendicular h e, upon which measure till you have made hp equal to m c; and you will then be in a direct line with m A. Erect the perpendicular p x, which (if you have conducted the work with correctness,) will be in a right line with B. Measure the distance p B; then A m, added to c h (= m p), and p B, will give the whole length of the line A B.

PROBLEM VIII.

.

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

of

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 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. 16. Part I.)

THE METHOD OF MEASURING

HILLY GROUND.

A LINE measured upon the acclivity or declivity of a hill, will evidently exceed one measured upon the horizontal base; consequently, if a plan be laid down by the hypothenusal lines, every part will be thrown out of its true situation; so that the boundaries of a mountainous lordship would appear distorted and unnatural; and the estate would scarcely be recognised by its own inhabitants.

Surveyors, therefore, agree in their opinions concerning the necessity of reducing hypothenusal to horizontal lines, for the purpose of planning; but they differ with regard to the modes of finding the area; some contending that it should be computed according to the hypothenusal, and others according to the horizontal lines.

The advocates for the horizontal measure assert, that no more corn, trees, &c. can grow upon the surface of a hill, than upon a space equal in area to its base, admitting both to be of the same quality ; and that hilly ground, in general, is less productive than plains, and its cultivation attended with more expense. The advocates on the other hand state, that the surveyor has nothing to do with the quality of the land; and that it is his duty to return the measurement of the surface, and leave the value to those whom it more nearly concerns.

The horizontal measure, however, is now generally adopted, except for paring, reaping, &c., in which cases the hypothenusal measure is very justly preferred. (See Deut. xxiv. 14, 15; and Prov. xxii. 16.)

Methods used by Practical Surveyors to reduce hypothenusal 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 hypothenuse 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.

NOTE. - For the principles of Trigonometry, the reader is referred to the works of Simpson, Emerson, Vince, Horsley, Keith, Bonnycastle, and the Rev. W. Wright, on that subject; and for the history, construction, and use of Logarithms, to Dr. Hutton's Mathematical Tables.

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.

In the right-angled triangle A B C, are given the hypothenuse А С 2550, and the angle B A C = 16° 15', to determine the base A B. Or A D = = 100, and the angle E A D = 16° 15′ to find

A E.

As Radius

10.00000

Is to the hypoth. A D = 100 links........

2.00000

So is the co-sine of the angle E A D = 16° 15′ ..

9.98229

To AE 96 links

1.98229

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

PROOF. - As 1 2550 :: .96005 (the nat. co-sine of 16° 15′) : 2448.1275 links

=

A B.