Now read the arc by the same index as before, noting the minutes and seconds by the other verniers. Take the mean of the minutes and seconds of each position for the true reading. Then the true reading in the first position taken from that in the second will give the angle required. It is convenient to have a table prepared, with the requisite number of columns, in which to set down the readings of the different verniers. Thus, suppose there were three verniers, 120 degrees apart: rule a table, with six columns, as below: 75° 8' 15" 8' 0" 8' 30" 75° 8' 15" The first column is the occupied station; the second, the observed station; the next three the readings of the verniers, and the sixth the mean. In the case above, the angle BAC would be 75° 8' 71'. The instrument is supposed to read to 30", the 15′′ being taken when two lines on the vernier appear equally near coincidence. 196. Repetition of Angles. The following method of observation is sometimes employed. Suppose the angle ABC is to be measured, A being the left-hand object: direct to A, and turn to B as above directed. Clamp the vernierplate and loosen below, and bring the telescope again to A. Clamp below, loosen the vernier, and bring the telescope again to B. The index has now traversed an arc measuring twice ABC. The operation may be repeated as often as desired, noting the number of whole revolutions the telescope has made. Then divide the whole number of degrees by the number of repetitions. The result will be the degrees of the angle required. If there is a watch-telescope, it should be set carefully before each observation. When this is done, and proper care is taken to avoid deranging the instrument, the result may be depended on as more accurate than any single reading. Any error in the final reading, being divided by the number of observations, will affect the result by but a small part of its value. 197. Verification of the Angles. When it is possible to do so, all the angles of a triangle should be measured. If their sum does not make 180°, there must be an error somewhere. Should the error be considerable, the work ought to be reviewed. But if it does not exceed two or three minutes, providing the instrument only reads to -minutes, it may be distributed equally among the three angles, should there be no reason to suppose one is more accurate than another. But if more observations have been taken for some angles than for others, their determination should be most depended on, and a proportionally less part of the correction assigned to them. Suppose, for example, the angle A is the mean of five observations, B of three, while at C but one was taken, the error being 1' 45": we would proceed thus:-As ++1::: 1′ 45′′: 14", the correction for A. In the same manner the correction for B would be found to be 23", and for C, 1' 08". 198. Reduction to the Centre. Where the object that has been observed is a spire or other portion of a building, it is impossible to set the instrument underneath the signal. In such cases, the observed angle must be reduced to what it would have been had the station been at the proper point. Thus, let C (Fig. 69) be the correct station, and D the occupied station, which should be taken as near as possible to C. Take the angle ADB. Then if A, C, D, and B are all in the D circumference of a circle, this will be equal to ACB. The station should C E Fig. 69. B be assumed as near this as possible. Calculate BC and AC from the distance AB and the angles observed at A and B. Also measure DC, either directly or by trigonometrical methods to be explained hereafter, and take ADC. Then, (Art. 139,) As CA: CD:: sin. ADC: sin. CAD. And as CB: CD :: sin. BDC: sin. CBD. Hence, ACB becomes known. Example. Let CA 9647 ft.; CB 8945 ft.; ADB = = 97° 37'. = A. C. 6.015608 2.176091 199. Angles of Elevation. In measuring angles of elevation, the instrument must first be levelled; the telescope being then directed to the object, the reading of the vernier corrected for the index-error will be the angle of elevation. SECTION VI. MISCELLANEOUS PROBLEMS TO ILLUSTRATE THE RULES OF PLANE TRIGONOMETRY. Problem 1. Being desirous of determining the height of a fir-tree standing in my garden, I measured 100 feet from its base, the ground being level. I then took the angle of elevation of the top, and found it to be 47° 50' 30". Required the height, the theodolite being 5 feet from the ground. Solution. Make AB (Fig. 70) equal to 100 feet; draw AD and BC perpendicular to AB, making the former five feet from the same scale. Draw DE parallel to AB, and make EDC 47° 50', the given angle. Then will CB be the height of the tree. Calculation. = D Fig. 70. 110.45 feet; As rad. : tan. EDC :: DE: EC Problem 2. One corner C (Fig. 71) of a tract of land being inaccessible, to determine the distances from the adjacent corners A and B, I measured AB = 9.57 chains. At A, the angle BAC was 52° 19' 15", and at B, the angle ABC was 63° 19' 45". Required the distances AC and BC. B Calculation. B As sin. ACB (64°21′): sin. A (52° 19′ 15′′) :: AB (957) : BC = 840.2 links. As sin. ACB (64° 21′): sin. B (63° 19′ 45"): AB AC = 948.7 links. Fig. 72. D C B Problem 3. In measuring the sides of a tract of land, one side AB (Fig. 72) was found to pass through a swamp, so that it could not be chained. I therefore selected two stations, C and D, on fast land, and took the distances and angles as follows,viz.: AC 37.56 chains; CD = 50.25 chains; BAC = 65° 27' 30"; ACD 123° 46' 20"; CDB 107° 29′ 15′′: the corner B being inaccessible, the distance BD could not be measured. Required AB. The angle CDA could not be taken, owing to obstructions. = = Solution. Join AD. Then, from the triangle ACD, we have, (Art. 140,) As CD + CA (87.81): CD — CA (12.69): : tan. CAD + CDA 2 CAD - CDA (28° 6′ 50′′): tan. 6' =4° 24' 54"; 2 whence CAD = 28° 6′ 50′′ +4° 24′ 54′′ = 32° 31′ 44′′, and CDA 28° 6' 50" 4° 24′ 54′′ = 23° 41′ 56′′; then, sin. CDA sin. ACD :: AC AD = 77.68. : Now, in ADB we have AD=77.68, the angle DAB = CAB - CAD = 32° 55′ 46′′, and the angle ADB = BDC — ADC 83° 47′ 19′′, to find AB; thus, = As sin. B: sin. ADB :: AD: AB = 86.455 chains. Problem 4. To determine the position of a point D on an island, I ascertained the distances of three objects on the main land as follows:-AB = 248.75 chains, BC = 213.25 chains, and AC = 325.96 chains. At D the angle ADB was found to be 29° 15', and BDC 20° 29' 30". Required the distance of D from each of the objects. 1. In ABC we have the three sides to find the angle BAC = 40° 51′ 30′′. 2. In CAE we have the angles and side AC to find the side AE = 208.705. 3. In BAE we have BA, AE, and the included angle BAE, to find ABE = 50° 55' 48", AEB = 67° 43′ 12′′. |