the sum of the southings will be denoted 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 C; or the equal alternate angle at C is the bearing from to A, and the hypothenuse AC, is the distance. Having measured the bearings and courses on the field, form a table, and find the base and perpendicular of the rightangled triangle, in numbers; after which, find the bearing and distance. To find the angle BAC, or the bearing from A to C. Hence, the bearing and distance are both found. NOTE 1.-Had any of the courses run south, AB would have been equal to the sum of the northings, minus the sum of the southings. NOTE 2. The last problem affords an easy method of finding the bearing and length of one of the courses of a survey, 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. EXAMPLES. 1. In a survey we have the following notes: What is the bearing distance from station 3 to 4? 2. In a survey we have the following notes: What is the bearing and distance from 3 to 4? II. To determine the angle included between any two courses, AC is N 26° W AH is N 65° W CAH = 39° PRINCIPLES. When the meridional letters are alike, and those of departure also alike, the difference of the bearings is the angle between AB is N 46° E САВ = 72° AC is N 26° W AD is S 66° W CAD 180° — 92° 88° AC is N 26° W CAF 180° — 40° = 140° = When the meridional letters are alike, and those of departure unlike, the sum of the bearings is the angle between the courses. When the meridional letters are unlike, and those of departure alike, the angle between the courses is equal to 180°, minus the sum of the bearings. When the meridional letters are unlike, and those of departure also unlike, the angle between the courses is equal to the difference of the bearings taken from 180°. NOTE. The above principles are deduced, under the supposition that the two courses are both run from the same angular point. Hence, if it be required to apply these rules to 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. EXAMPLES. 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? Ans. 124° 50'. LAYING OUT AND DIVIDING LAND. 103. The surveyor is often required to lay off a given quantity of land, in such a way that its bounding lines shall form a particular figure, viz., a square, a rectangle, a triangle, &c. He is also often called upon to divide given pieces of land into parts containing given areas, or, into areas bearing certain relations to each other. The manner of making such divisions must always depend on a skilful and judicious application of the principles of geometry and trigonometry to the particular case. For example, if it were required to lay out an acre of ground, in a square form, it would be necessary to find, by calculation, the side of such a square, and then trace, on the ground, a figure bounded by four equal sides, at right angles to each other. PROBLEM I. 104. To lay out a given quantity of land in a square form. RULE. Reduce the given area to square chains, or square rods: then extract the square root, and the result will be the side of the required square. This square being described on the ground, will be the figure required. 1. To trace a square which shall contain 15 A. 0 R. 12 P. 15 A 60 R = 2400 P; hence, First, 12 P 15 A 0 R 12P = 2412 P; the square root of which is 49.11, nearly. Therefore, if a square be traced on the ground, of which the side is 49.11 rods, it will be the required figure. |