PROBLEM III. To determine the perpendicular distance of an object below a given horizontal plane. 91. Suppose C to be directly over the given object, and A the point through which the horizontal plane is supposed to pass. Measure a horizontal base line AB, and at the stations A and B conceive the two horizontal lines AC, BC, to be drawn. The oblique lines from A and B to the object will be the hypothenuses of two right-angled triangles, of which AC, BC, are the bases. The perpendiculars of these triangles will be the distances from the horizontal lines AC, BC, to the object. If we turn the triangles about their bases AC, BC, until they become horizontal, the object, in the first case, will fall at C', and in the second at C". Measure the horizontal angles CAB, CBA, and also the angles of depression C'AC, C"BC. Let us suppose that we have AB=672 yards BAC 72° 29' found ABC-39° 20' C'AC 27° 49' C"BC 19° 10' First: In the triangle ABC, the horizontal angle To find the horizontal distance AC. Hence also, CC-CC"=242.06-239.93=2.13 yards; which is the height of the station A above station B. REMARK. In measuring a base line, if great accuracy is required, the theodolite should be placed at one extremity, and the telescope directed to the other, and the alignment of the staves made by means of the intersection of the spider's lines. If the highest degree of accuracy is necessary, the base line should be measured with rods, which admit of being adjusted to a horizontal position by means of a spirit level. APPLICATIONS. 1. Wanting to know the distance between two inaccessible objects, which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and are found to be, of the nearest 57°, of the most remote 25° 30' required the distance between them. Ans. 173.656 feet. 2. In order to find the distance between two trees A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distances of a third point C from each of them were measured, and also the included CB=672 yards CA=588 yards ACB 55° 40'; required the distance AB. Ans. 592.967 yards. 3. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible. hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45'; required the height of the tower. Ans. 83.998 feet. 4. Wanting to know the horizon tal distance between two inaccessi-TE ble objects E and W, the following measurements were made, A Ans. 939.527 yards. D other, equal to 200 yards; from the former of these points A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D a distance DE equal to 200 yards, and the following angles taken, Ans. AB=345.467 yards. 92. It often happens that instruments for the measurement of angles cannot be easily obtained; we must then rely entirely on the tape or chain. We now propose to explain the best methods of determining distances, without the aid of instruments for the measurement of horizontal or vertical angles. PROBLEM I. To trace, on the ground, the direction of a right line, that shall be perpendicular at a given point, to a given right line. FIRST METHOD. 93. Let BC be the given right line, and A the given point. Measure from A, on the line BC, two equal distances AB, AC, A one on each side of the point A. Take a B portion of the chain or tape, greater than AB, and place one extremity at B, and with the other trace the arc of a circle on the ground. Then remove the end which was at B, to C, and trace a second arc intersecting the former at D. The straight line drawn through D and A will be perpendicular SECOND METHOD. 94. Having made AB=AC, take any portion of the tape or chain, considerably greater than the distance B between B and C. Mark the middle point of it, and fasten its two extremities, the one at B and the other at E C. Then, taking the chain by the middle point, stretch it tightly on either side of BC, and place a staff at D or E: then will DAE be the perpendicular required. THIRD METHOD. 95. Let AB be the given line, and C the point at which the perpendicular is to be drawn. From the point C measure a distance CA equal to 8. With C as a centre, and a radius equal to 6, describe an arc on either side of AB: then, with A as a centre, and a radius equal to 10, describe a second arc intersecting the one before described at E: then draw the line EC, and it will be perpendicular to AB at C. REMARK. Any three lines, having the ratio of 6, 8 and 10, form a right-angled triangle, of which the side corresponding to 10 is the hypothenuse FOURTH METHOD. 96. Let AD be the given right line, and D the point at which the perpendicular is to be drawn. Take any distance on the tape or chain, and place one extremity at D, and fasten the other at some point as E, between the two lines which are to form the right angle. Then, having stationed a person at D, remove the extremity of the chain and carry it round until it ranges on the line Place a staff at E. |