2. A = 49° 12′ 13′′·02, A' 49° 12′ 12′′-21. A' 65° 14' 572. When the observed angles of any of the spherical triangles have been observed with equal accuracy, the angles of the equal-sided plane triangle can be obtained exactly in the same manner as from the true angles, by deducting onethird of the observed spherical excess from the observed angles. The exercises given above will exemplify this rule, if the angles A, B, C, be considered to be observed angles. Since the angles of the equal-sided plane triangle can be thus obtained from the observed angles, if one side of the spherical triangle is accurately known, the other two sides, being equal to those of the plane triangle (570), can be easily computed by means of it. The computation of the spherical excess is therefore of little practical utility, except as a means of testing the accuracy of the observed angles; and when they are thus verified, the surveyor is confident of the accuracy of his operations. 573. PROBLEM VIII.-Given three parts of a spherical triangle, one of them being the correct value of at least one side, but the given angles being only observed angles, to find accurately the parts of the triangle. By means of the parts given, compute the spherical excess; then find the true values of the angles, and then the angles of the equal-sided plane triangle; and in the latter triangle, by means of its angles and the given side or sides, find its other side or sides, and they will also be the required side or sides of the spherical triangle." When the observed angles are considered to be equally accurate, the angles of the plane triangle can be obtained from them (572), and the required sides calculated. EXAMPLE.-Given the three observed angles of a spherical triangle, namely, A, 35° 23′ 13"-87, B, 83° 26' = 23" 6, and C, 61° 10′ 24′′-18, and the side a = 84383.12 = feet, to find the true angles and the other two sides of the triangle. There is only one side given; hence, compute another side, supposing the triangle plane (572), or use the formula sin Asin B (565) T=c2. sin (A+B)' which here becomes L. sin A' L. a 180 0 1.65 1 sin B1sin C1 = 9-7627511 L. 2 T L. 2 L. T L. E L. 2T E=0"-84. = 1 B1 =83 26 23.6 To find E. Also (571) A' = 35° 23′ 13′′-32 = 10-02948 0.30103 = And (569) A = 35° 23′ 14′′·16 C61 10 24.47 = 180 0 0 To calculate the sides b, c, of the plane triangle A', B', C'. L. sin A' = = L. sin c' L. C And c 9.72845 9.32677 ⚫40168 180 0 2.52 9-7627511 9.9425445 4.9262556 14.8688001 = 5.1060490 127658-29. The sides b, c, are also sides of the spherical triangle, so that its angles and sides are now all accurately known. 574. Since the observed angles are supposed to be equally accurate, the angles of the plane triangle A', B', C', could have been obtained from them by deducting from each of them E' = 3 × 1′′·65 =0′′·55. EXERCISES. In the following exercises, one side of the triangle is given, and all the angles by observation, except in the third and fourth, in which only two angles are given. In the third, two angles of the plane triangle are found by deducting one-third of the spherical excess from the two given angles, and its third angle is then easily found; and in the same manner, the fourth exercise is calculated. The elements required are-the spherical excess, the true angles of the spherical triangle, the angles of the plane triangle, that is, the angles for calculation, and the other two sides of the spherical triangle, which are just those of the plane triangle. The spherical excess for the third exercise has been already calculated in the exercises in article 566.* In the first line of each exercise, the accurate value of one side is given, and in the second column are the observed angles; these are the data in each exercise: the required parts are-the spherical excess given in the third column, the angles of the equalsided plane triangle given in the fourth column, and the other two sides given in the last column, and placed opposite to the angles to which they are opposite in the triangles. 1. Severndroog Castle, from Leith Hill Station, 144760.96 feet. * These exercises are selected from Captain Kater's "Account of Trigonometrical Operations for determining the difference of Longitude between Paris and Greenwich." 2. Wrotham Station, from Severndroog Castle, 75014.27 feet. Observed Sphe- rical Angles for Distances Names of Stations. Westminster Abbey, Crowborough, 84°59′ 56" 81 Stede Hill, ... 3. Hanger Hill, from Leith Hill Station, 127658.21 feet. ... ... ... 179 59 59 33 097 180 0 0 17 40 36.85 ... 1":2 4. Westminster Abbey, from Leith Hill, 130366.27 feet. Severndroog Castle, 62°33′ 57′′-67 ... ... 16°35' 2"00 149 26 13 58 44 42 63488-87 ... ... 13:58 133640 58 5. Leith Hill, from Wrotham, 158844:37 feet. ... 77°17'27"-37 84 59 56 41 130366.27 17 42 36 22 39809.02 180 1'35 180 0 0 ... 87° 5′15′′01 38 56 55 95 180 0 2.09 3" 03 180 0 0 ... 0 62°33′57′′-22 99 45 26.38 144759.97 17 40 36 40 44601.10 6. Wrotham, from Crowborough, 99982:55 feet. 44°44' 52" 83 180 0 3.31 2" 24 180 87° 5' 14" 32 38 56 55 25 99982.55 53 57 50 43 128615 26 M The com The triangles for the first, fifth, and sixth exercises, are given in the adjoining figure; the letters at the angular points are the initials of the names of the stations, except S', which is Stede Hill. The bearing of Leith Hill, from Severndroog Castle, that is, angle LSN, being 43° 5' 51" SW., MSR will be the direction of the meridian. putation of the system of triangles is continued in the same manner to any extent. Sometimes the distance of some two of the stations is calculated L by means of several triangles of which it is a common side, and the results serve as a means of verifying each other. Thus, for instance, the distance between Chingford station and Severndroog Castle was calculated by means of five triangles of which it was a side, and the extremes of the computed distances differ by only 1.04 feet; and the greatest difference between any of them and the mean, which is 63489-32 feet, or about 12 miles, is only 0.49 of a foot. It may be observed here, as a proof of the extraordinary precision with which the original base of a geodetic survey is measured, that the base of the survey in Ireland, which is on a level plane near Londonderry, is more than seven miles long, and that the greatest possible error on its length is considered to be within 2 inches. N 575. When the object of the survey is merely to determine the length of a degree of the meridian, the system of triangulation is extended as nearly as possible in the direction of the meridian. The method of computing this length is explained in the next problem. The triangles composing a system of triangulation are called primary triangles; and other triangles, formed within the system for determining the positions of objects or other purposes, are called secondary triangles. D R W |
0 Reviews Reviews aren't verified, but Google checks for and removes fake content when it's identified Write review |