of the vernier coincides with a line of the limb, the vernier will have been moved 30', or half a degree; and the point will at the same time coincide with a line of the limb, and show 10° 30'. The general rule for reading the angle for any position of the vernier may now be stated. When the line of the vernier coincides with a line of the limb, the arc is easily read from the limb; but when it falls between two lines, note the degrees and half degrees up to the line on the right; then pass along the vernier till a line is found coinciding with a line of the limb: the number of this line from the o point, indicates the minutes which are to be added to the degrees and half degrees, for the entire angle. To measure a horizontal angle with the theodolite. Place the axis of the instrument directly over the point at which the angle is to be measured. This is effected by means of a plumb, suspended from the plate which forms the upper end of the tripod. Having made the limb truly level, place the o of the vernier at 0 or 3600 of the limb, and fasten the clamp-screws of the vernier plate. Then, facing in the direction between the lines which subtend the angle to be measured, turn the limb with the outer spindle, until the telescope points to the object on the left, very nearly. Clamp the limb with the clamp-screw K, and by means of the tangent screws L and 7, bring the intersection of the spider's lines to coincide exactly with the object. Having loosened the clamp-screw Q of the lower telescope MN, direct it with the thumb-screw P to the same object at which the upper telescope is directed ; then tighten the clampscrew Q. This being done, loosen the clamp-screw S of the vernier plate, and direct the telescope to the other object : the arc passed over by the point of the vernier, is the measure of the angle sought. The lower telescope having been made fast to the limb, will indicate any change of its position, should any have taken place; and, as the accuracy of the measurements depends on the fixedness of the limb, the lower telescope ought to be often examined, and if its position has been altered, the limb It is not necessary to place the o point of the vernier at the o point of the limb, previously to commencing the measurement of the angle, but convenient merely; for, whatever be the position of this point on the limb, it is evident that the arc which it passes over is the true measure of the horizontal angle. If, therefore, its place be carefully noted for the first direction, and also for the second, the difference of these two readings will be the true angle, unless the vernier shall have passed the point of the limb, in which case the greater reading must be subtracted from 360°, and the remainder added to the less. To measure a vertical angle. In Fig. 3, AB represents a view of the vertical limb opposite the thumb-screw Z, and ED is the vernier. The o point of this vernier is at the middle division line, and fifteen spaces lie on each side of it. The relation which exists between the spaces on the limb and those of the vernier, is the same as that between the divisions of the horizontal limb and its vernier, and the degrees and half degrees are read in the same manner: the angles of elevation being read from the o of the limb towards the right, and those of depression in the contrary direction. For the minutes, we pass along the verpier in the direction in which the degrees are counted, and if we reach the extreme line, which is the fifteenth, without finding a coincidence, we must then pass to the other extremity of the vernier, and look along towards the o point till two lines are found to coincide : the number of the line on the vernier will show the minutes. The lines of the vernier are numbered both ways from the o point, and marked 5, 10, 15, to one extremity, and correspondingly from the other extremity 15, 20 and 25, to the o point again. The upper range shows the minutes for angles of elevation, and the lower range for those of depression. The vernier in Fig. 3 stands at 2° 15' of depression. Had the 15th line at the left, passed the short line with which it now coincides, we should pass to the line 15, on the lower range to the right, and then count towards the o to the left. The first thing to be done, is to ascertain the point of the vertical limb at which the point of the vernier stands, when the line of collimation of the upper telescope, together with its attached level, is truly horizontal. This is called the true O of the limb. If the instrument be accurately constructed, and the parts have not been disarranged, this point is the o point of the limb. This, however, is easily ascertained by turning the limb till the o's correspond, and then examining if the upper level be truly horizontal. If not, direct the telescope to a distant and elevated object, and read the degrees on the vertical limb. Turn the vernier plate 1809, reverse the telescope, direct it a second time to the same point, and read the arc on the vertical limb. The half difference of these two readings, counted from the Q point of the limb, in the direction of the greater arc read, gives the true a point of the vertical limb; that is, the point at which the Q of the vernier stands when the line of collimation is horizontal. Suppose for example, that we had directed the telescope to a point and found the o of the vernier to stand at 10° of elevation. If we now reverse the telescope, it ought to incline at an equal angle of depression. If then we turn the whole 180°, and then raise the depressed end of the telescope with the thumb-screw Z, until it is directed to the same point as before, the o ought to stand at 10°. If it shows a less arc, the true o is between the of the limb and the first arc read; if a greater, it is on the other side, and the difference divided by two will indicate the exact o point. The half difference thus found is called the correction. When the true 0 falls between the marked o and the eyeglass, the correction is to be subtracted from the are read, for angles of elevation, and added, for angles of depression; and the reverse when it falls on the other side. The eyeglass is supposed to be over the thumb-screw Z, as in the plate. These preparatory steps being taken, let the axis of the telescope be directed to any point either above or below the plane of the limb, and read the arc indicated by the o of the vernier. To the arc so read apply the proper correction, if any, and the result will be the true angle of elevation or de. pression. 87. Having explained the preliminary principles, it only remains to apply them to the measurement of Heights and PROBLEM I. To determine the horizontal distance to a point which is inacces sible by reason of an intervening river. 88. Let C be the point. Measure along the bank of the river a horizontal base line AB, and select the stations A and B, in such a manner that each can be seen from the other, and the point C from both of them. Then measure the horizontal angles CAB and CBA. B В Let us suppose that we have found AB=600 yards, CAB= 570 35' and CBA=64° 51'. The angle C=180° - (A+B)=57° 34'. To find the distance BC. As sin c 570 34' . ar. comp. 0.073649 : sin A 570 35' 9.926431 AB 600 2.5778151 : BC 600.11 yards. 2.778231 To determine the altitude of an inaccessible object above a given horizontal plane. FIRST METHOD. D 89. Suppose D to be the inaccessible object, and BC the horizontal plane from which the altitude is to Be be estimated : then, if we suppose DC to be a vertical line, it will re E Measure any horizontal base line, as BA; and at the extremities B and A, measure the horizontal angles CBA and CAB. Measure also, the angle of elevation DBC. Then in the triangle CBA there will be known, two angles and the side AB; the side BC can therefore be determined. Having found BC, we shall have, in the right-angled triangle DBC, the base BC and the angle at the base, to find the perpendicular DC, which measures the altitude of the point D above the horizontal plane BC. Let us suppose that we have found BA=780 yards, the horizontal angle CBA=410 24', the horizontal angle CAB=46° 28', and the angle of elevation DBC=10° 43'. In the triangle BAC, to find the horizontal distance BC. The angle BCA=180°-(41° 24'+96° 28') =42° 08'=C. 42° 08' 0.173369 : sin A 96° 28' 9.997228 2.892095 BC. 1155,29 3,062692 . . . . AB. • 780 . : In the right-angled triangle DBC, to find DC. As R ar. comp. 0.000000 : tan DBC. 10° 43' 9.277043 BC 1155.29 3.062692 : DC . 218.64 2.339735 . . Remark I. It might, at first, appear that the solution which we have given, requires that the points B and A should be in the same horizontal plane, but it is entirely independent of |