will be known to determine the area which lies without the station line AB. The points b, d and g, of the fence which runs from A to B, are also determined.

Erect, in a similar manner, offsets to the other courses, and determine the areas which lie without the station lines. These several areas being added to the area within the' station lines, will give the entire area of the ground.

If the offsets fall within the station lines, the corresponding area must be subtracted from the area which is bounded by the station lines.

II. To determine the bearing and distance from one point to another, when the points are so situated that one cannot be seen from the other.

34. Let A and C be the two points, and AB a meridian passing through one of them. From either of them, as A, measure a course A2, of a convenient length in the direction towards C, and take the bearing with the com· pass. At 2, take the bearing of a second course, and measure the distance to 3. At 3, take a third bearing and measure to 4. 4, take the bearing to C, and measure the distance from 4 to C.

At

B

Then, the difference between the sum of the northings and the sum of the southings will be represented by AB, and the difference between the sum of the eastings and the sum of the westings by BC. The base AB, and the perpendicular BC of the right-angled triangle ABC, are then known. The angle at the base, BAC, is the bearing from A to ; or the equal alternate angle at C is the bearing from C to A, and the hypothenuse AC is the distance.

35. Having measured the bearings and courses on the field, form a table, and find the base and perpendicular

Hence, the bearing and distance are both found.

OF SUPPLYING OMISSIONS IN THE FIELD NOTES.

36. The last problem affords an easy method of finding

when the bearings and lengths of all the others are known. It may be necessary to use this method when there are obstacles which prevent the measuring of a course, or when the bearing cannot be taken. Indeed, two omissions may in general be supplied by calculation. It is far better, however, if possible, to take all the notes on the field. For, when any of them are supplied by calculation, there are no tests by which the accuracy of the work can be ascertained, and all the errors of the notes affect also the parts which are supplied.

1. In a survey we have the following notes:

What is the bearing and distance from station 3 to 4?

Ans. {Bearing, S 38° 52' EL.

2. In a survey we have the following notes:

7.03 ch.

What is the bearing and distance from 3 to 4?

Ans.

Bearing, N 34° 47' E.

III. To determine the angle included between any two courses, when their bearings are known.

REMARK. The above principles are determined, under the supposition that the two courses are both run from the

two courses run in the ordinary way, as we go around the field, the bearing of one of them must be reversed before the calculation for the angle is made.

1. The bearings of two courses, from the same point, are N 37° E, and S 85° W: what is the angle included between them?

Ans. 132°.

2. The bearings of two adjacent courses, in going round a piece of land, are N 39° W, and S 48° W: what is the angle included between them?

Ans. 87°.

3. The bearings of two adjacent courses, in going round a piece of land, are S 85° W, and N 69° W: what is the angle included between them?

Ans. 154°.

4. The bearings of two adjacent courses, in going round a piece of land, are N 55° 30′ E, and S 69° 20′ E: what is the angle included between them?

OF DIVIDING LAND.

Ans. 124° 50'.

that it is difficult

It is by practice branch of survey

38. Fields are so variously shaped to give rules that will apply to all cases. alone that facility is obtained in that ing relating to the division of estates. We shall add only a few examples that may serve as general guides in the application of the principles of Plane Geometry to such cases as may arise.

I. To run a line from the vertex of a triangular field which shall divide it into two parts, having to each other the ratio of M to T.

39. Let ABC be any triangular field.

Divide the side BC into two

parts, such that (Geom., Bk. IV.,

Prob. 1.)

BD : DC :: m : and draw the line AD:

n;

B