The cross-staff enables us to raise perpendiculars and to test right angles; but it is frequently necessary in surveying to measure angles of various sizes. The essential parts of an instrument which will serve this purpose are— I. A graduated circle or semicircle. 2. A stand upon which this circle can be mounted and placed in a horizontal plane. 3. A movable arm furnished with sights, which always determine a straight line passing through the centre of the graduated circle. If the movable arm in this instrument is a telescope, and the graduated circle is complete, the instrument is termed a theodolite. A perfect theodolite, however, is so constructed that it will not only measure angles in a horizontal plane, or "azimuths," * but also angles in a vertical plane, or altitudes and depressions. (Fig. 148.) Description of the Theodolite.-The telescope, A B, consists of two tubes, one sliding within the other. The outer tube has at its further end, A, the object-glass; and the inner tube has at its nearer end, B, the eye-piece. With most theodolites the object is seen inverted. By moving the inner tube inwards and outwards by the screw b, the object-glass and eye-piece are brought to the same focus, and the image is made distinct. The spirit-level, c, enables the observer to set the telescope horizontal. When the instrument is adjusted, the axis, C, of the telescope is perfectly horizontal; so that the telescope moves in a vertical plane. The supports, DD, are usually high enough to admit of the telescope being turned completely over in a vertical plane; but sometimes they are lower, and the telescope is then turned by lifting it out of its bearings or Y's. The vertical circle, E, is fixed on the axis. It is divided into four quadrants, each of which is divided into 90°, the A FIG. 148. E The azimuth of an object, with regard to a station in a horizontal plane, is the number of degrees, minutes, and seconds in the angle between the line drawn due north from the station and the line drawn to the object. two o's being at the ends of the diameter parallel to the line of sight (or line of collimation) of the telescope and the two 90's at the ends of. the diameter perpendicular to this line. The indices (with verniers), which show through how many degrees the telescope is turned, are the extremities of a horizontal bar, h. The vernier-plate, G, is circular; and when the instrument is adjusted it moves in a horizontal plane, carrying with it the bearings of the telescope, DD, a magnetic compass, and a pair of spirit-levels, s s. The horizontal circle, H, is graduated on the edge or limb from o° to 360°. The angle through which the vernier-plate is turned is read on the horizontal circle by means of a small microscope, g, attached to the vernier-plate above an index or vernier*. A vernier is a small movable scale running parallel with the fixed scale of any instrument and having the effect of subdividing the divisions of that instrument into more minute parts. Suppose that it is required to subdivide the divisions on a scale into 10 parts. According to the ordinary construction of the vernier, its total length will be 10 - 1, or 9 parts of the scale, and it will be divided into 10 equal parts; so that each part of the vernier will be less than each part of the scale by I-10th. 32 Let us call the divisions on the scale degrees, and suppose that the index is between the 12th and 13th degree. The zero of the vernier is brought in a line with the index, and the line of the vernier which coincides or is nearest to a line of the scale is noted. Suppose it to be the fourth (see Fig. 150), then the index marks 4-10ths less than 13 degrees-that is to say, 12.6. If the vernier is to divide each degree into 6oths, its length must be 59 degrees, and it must be divided into 60 parts To Measure the difference of Level between Two Points.-The instruments used are a level or theodolite (Figs. 29 and 148), and the levelling-staff. The latter is a rectangular wooden rod, having a face about two inches and a half broad, on which is painted a scale of feet divided into tenths and hundredths. The horizontal tine determined by the level is termed the line of sight. Two observations are taken with the level in each position; one towards A, for example (Fig. 149), termed the back-sight, the other towards F, termed the fore-sight. = = Let S, S', S", &c., be the positions of the level. Let b, c, d, e, f be the heights observed in passing from A to F; that is to say, the foresights, and a', b', c', d', e' the back-sights. It is evident that the difference of level between A and the height of the instrument at S": the height a'b' —b + c —c; and the difference of level between the point F and the same position of the instrument — d—d' + e - e' +ƒ. Hence the rule: the difference of level between two points is found by taking the sum of the fore-sights and the sum of the back-sights, and subtracting one from the other. If the sum of the back-sights is the greater, the point of departure is lower than the terminal point; but if less, the first point is the higher. The following are the dimensions taken in observing the difference of level between A and F. (Fig. 149.) Arithmetical Exercises. I. Find the area of a four-sided field, the diagonals of which are at right angles and measure respectively 77 and 50 yards. 2. What is the area of a pentagon whose side measures 185 inches and the radius of the inscribed circle 127}? 3. Find the area of a heptagon whose side measures 2 feet 11 inches and the perpendicular from the centre 3 feet. 4. Find the area of a circle whose diameter is 14 yards and circumference 44 yards. 5. Draw the plan and find the area of a field from the following dimensions : Find the areas of fields the dimensions of which are given below in links. (a.) Left 372 346 223 318 246 0 Base Line A 266 345 465 560 718 790 987 1015 1325 B 162 134 58 136 (c.) The longest side, A B, of a field is a straight line. Passing from A to B, the offsets on the left are, at A, o; at 248 from A, 34, at 342, 73; at 412, 139; at 464, 113; at 5c2, 142; at 602, 0. Find the area. SECTION IV.-The Relation between the Area of a Triangle and the Lengths of its Sides. When the lengths of the sides can be expressed in numbers the theorems in Section II. are equivalents of the following well-known algebraical forms, if we suppose a and b to be numbers representing the units of length in the two parts of a line : Theorem XLVIII. (a + b)2 = a2 + b2 + 2 ab; Theorem XLIX. (a + b)2 + b2 = a2 + 2 b (a + b); Theorem L. (2 a − b) b + (a — b)2 — a2. = Thus the product of two algebraical quantities of the first degree has a rectangle for its geometrical equivalent. On the Area of a Triangle.-The area of a triangle may be found when the three sides only are given, but the proof of the rule is algebraical rather than geometrical. Let c, b, and a stand for the lengths of the sides opposite respectively to the angles A, B, C, and let h be the height of the triangle B (Fig. 151). D FIG. 151 c2)2 4 a2 b2 = (a + b + c) (a + b — c) (a − b + c) (b + c — a). a2 + b2 and CD= hence h262 (a2 + b2 h2 = 4 a2 Hence the rule is :—Find half the sum of the sides and from it take each side separately. Multiply together the three remainders and the half sum, and take the square root of the last product. M 2 |