6. Required the plan and area of a piece of ground from the

7. Required the plan and area of a field from the following equi

Answer.

727 the first and last ordinates 4384 four times the sum, &c.

1540 twice the sum, &c.

6651 sum total

100 the common distance

3)665100

221700

36288 trapezoid at the end 2-57988 area in square links

4

2.31952

40

12-78080 Area 2a. 2r. 12 p.

NOTE.-Whenever the rule given in this Problem can be applied, it will be found more easy, expeditious, and accurate, in finding the areas of offsets and of narrow pieces of land, than the rules for triangles and trapezoids.

PROBLEM X.

To find the Breadth of a River.

EXAMPLE.

Let the following figure represent a river, the breadth of which is required.

Fix upon any object B, close by the edge of the river, on the side opposite to which you stand. By the help of your cross, make AD perpendicular to AB; also make ACCD, and erect the perpendicular DE; and when you have arrived at the point E, in a direct line with CB, the distance DE will be = AB, the breadth of the river; for by Theo. 1, Part I., the angle ACB = DCE, and as AC = CD, and the angles A and D are right angles, it is evident that the triangles ABC and CDE are not only similar but equal.

NOTE 1.-The distance between A and the edge of the river must be deduced from DE, when it is not convenient to fix a close by the river's edge.

2. This Problem may also be well applied in measuring the distance of an inaccessible object; for let ac equal 8, CD equal 2, and DE equal 10 chains; then, by similar triangles, as CD: DE: AC: AB; that is, as 2: 108: 40 chains = =AB. (See Theo. 11, Part I.)

PROBLEM XI.

Lines upon which there are Impediments not obstructing Sight.

Suppose mn to represent a deep pit or water, and A and B two

objects, the direct distance of which is required.

At the verge of the impediment, having fixed the mark m, in a right line with A and B, measure from A to m; and at m, by the help of your cross, erect the perpendicular ma, which measure to the outside of the interposed obstruction, as at c. Then on

the other side, as at n, in a line with A and B, erect the perpendicular ne; and make nb equal to me.

Measure bc, which will be equal to mn; and from n measure the distance nB; then bc added to Am and NB will give the whole distance

AB.

B

m

PROBLEM XII.

Lines upon which there are Impediments obstructing Sight.

EXAMPLE.

Suppose CDEF to represent the base of a building which obstructs the sight, and through which it is necessary that a straight line

beyond the building, as at b. Erect the perpendicular bd; and make bn equal to me, at which point you will be in a direct line with mA. Erect the perpendicular nB, which measure; then bc, added to Am and nB, will give the whole distance AB.

NOTE-In the tenth and previous editions of the work the range am was evidently understood in this problem as given, and also in Problem VII. Part IV. p. 139, although not so expressed. This involves an omission; for the range or direction of a line must always be known before it can be measured. The error is now corrected in both examples. And as impediments obstructing sight are of frequent occurrence in surveying, two propositions have been added to find the line when the range am is not given-the one with the chain, p. 89, and the other with the theodolite and chain, Part IX.

PART IV.

THE METHOD OF SURVEYING WITH THE CHAIN ONLY; AND OF MEASURING MERES, WOODS, DISTANCES, LINES UPON WHICH THERE ARE IMPEDIMENTS, AND HILLY GROUND.

MISCELLANEOUS INSTRUCTIONS.

THE method of surveying with the chain only is adopted by most practical surveyors, and is certainly preferable to that by the chain and cross; because it is not only as accurate, but generally more expeditious.

Whatever be the form of the field or ground to be surveyed, measure as many lines as will enable you to plot it with accuracy. The plan being drawn, you may then divide the figure into trapeziums, triangles, &c., and measure the diagonals, perpendiculars, &c., with your plotting-scale.

It is better, however, to divide small pieces and single fields into trapeziums and triangles, by measuring the diagonals and bases during the survey; so that to find the area, you will have only the perpendiculars to measure with the scale.

You should also measure, in some convenient direction, a proofline to each trapezium and triangle.

NOTE 1.-The offsets must be treated according to the directions in Part III. Prob. VI. Or you may reduce the crooked sides to straight ones, by including as much of what does not belong to the field under your survey as you exclude of what does, in the following manner:-Apply to the crooked line in question the straight edge of a clear piece of lantern horn, so that the small parts cut off by it from the crooked figure may be equal to those which are taken in (of this equality you will presently be able to judge very correctly by a little practice); then, with a pencil, draw a line by the edge of the horn. The sides being thus successively straightened, the content may be easily found.

2. A slender bow of cane or whalebone, strung with a silk thread, may be substituted for the horn. The thread must be applied to the crooked fence, and two marks made by which to draw a straight line.

3. The sides may also be straightened by a parallel ruler; but the operation is generally tedious, and must be performed with the greatest care, or it will not be more correct than the foregoing method.

4. When the three sides of a triangle are given, the area may be found as fol