576. PROBLEM IX.-Having determined the sides and angles of the primary triangles of a system of triangulation lying nearly in the direction of the meridian, to determine the length of that portion of the meridian over which it extends. A Let ABCDEGFDH be a system of primary triangles, whose angles and sides are known as in the preceding problem; find the azimuth & of the station C when observed from A, then angle 6180-(x+y) nearly, y denoting angle ACD; and therefore the spherical excess of the triangle ACM can be found (566); and hence its B true angles can be found, and also its sides AM, MC (573.) But CD is known; hence, in triangle DMR, the side DM = CD — CM is known, and angle D CDE + EDG + GDF, and also angle M = AMC. Therefore, angle R is approximately known; and hence the spherical excess for this triangle can be found, and its true angles and sides can be computed; and hence MR can be found. H D R M F Proceeding in this manner, the lengths of the successive portions AM, MR,...of the meridian can be computed; and hence their sum, or the length of the portion of the meridian comprehended between the extreme stations, can be determined. The latitudes of the extreme stations can be determined by the method in article 476, and then the difference of latitude is known, which is just the number of degrees in the arc of the meridian; and the length of one degree can then be found by a simple proportion. It may be observed here, that though the angles at M are equal in the two triangles ACM, DMR, yet the corresponding angles of their equal-sided plane triangles are not equal, for the spherical excesses of these triangles are not equal, unless their areas are so. EXAMPLE. Let the azimuth a 16° 46′ 27′′-59, ACM = y=143° 13′ 41′′-52, and the side AC = 27458-6 metres; also angle MDR = 134° 11′ 14"-78, and CD = 35164.08 metres, to find the portions AM, MR, of the meridian. The spherical excess for triangle AMC is found to be = 0"96; hence, Mß: = 180° 0' 0":96 (x + y) = 19° 59′ 51′′-85. = Hence, the angles of the equal-sided plane triangle, found by taking 0"-32 from the angles a, B, y, are a = 16° 46′ 27′′-27, B′ = 19° 59′ 51′′-53, y'′ = 143° 13′ 41′′-2. The other two sides found by means of this plane triangle (573) are AM = 48065 63 metres, CM = 23170m.97. Hence, in triangle MDR, DM CD CM 35164.08 23170-9711993-11 metres, M = ß = 19° 59′ 51′′-85, D = 134° 11′ 14′′-78; and computing the spherical excess for this triangle, and then finding the angles of the equal-sided plane triangle, and computing its sides, it is found that MR = 19745.9. Hence, AR AM + MR = 67811.53 metres. The reader is requested to perform the computations in this example that are not given above. By the method explained in this example, the length of the meridian, traversing France from Dunkirk to the parallel of Montjouy near Barcelona, was computed. The arc was 9° 40′ 24′′-24, and its length was = 1075059 metres.* In the preceding figure, the station A is at Dunkirk, B at Watten, C at Cassel, D at Fiefs, E at Bethune, G at Mesnil, F at Sauti, and H at Bennières. 577. PROBLEM X.-Given the length of an arc of, the meridian, and the latitude of its extremities, or the number of degrees in the arc, to find the length of one degree at the middle latitude. 'The number of degrees in the given arc is to one degree, as the length of the given arc to the length of an arc of one degree.' *See Francœur's "Geodesie." EXERCISE. The length of an arc of the meridian between Dunnose and Clifton was found, in the trigonometrical survey of England, to be 1,036,337 feet, and the amplitude of the arc was 2° 50′ 23′′-5; required the length of a degree at the middle latitude, which is 52° 2′ 20′′. = Ans. 364,925-25 feet. In this manner the length of a degree of the meridian at any latitude is found. The length of a degree at the middle latitude between Dunnose and Ardbury Hill was found to be 60,864 fathoms, which is about 54 fathoms more than it ought to be, according to the usual elements of the figure of the earth. This anomalous result is commonly referred to a deflection of the plumb-line of the sector employed for measuring the angles, occasioned by local attraction. 578. The following table presents the results of various surveys for determining the length of a degree in different latitudes : 579. The computations of the angles and distances in a geodetic survey can also be performed by means of the chordal triangles, that is, the triangles formed by the chords of the sides of the spherical triangles. The same data are required in this method as in the former, or there must be previously known all the angles of the spherical triangles by observation, and at least one side must be accurately determined. This method was adopted in the survey of England. Let A, B, C, and a, b, c, be respectively the correct angles and sides of one of the spherical triangles; A1, B1, C1, and a1, b1, C1, the corresponding observed angles and the approximate sides of the same triangle; and let A', B', C', and a, b, c, be the corresponding angles and sides of the triangle formed by its chords. If c is the side whose length is accurately known, then is c' = 2 sin c. 1 Find also the sides a, b, approximately, that is, a and b, by plane trigonometry from the angles A1, B1, C1, and the side c thus: sin A1 sin C1 sin B, 1.sin c; sin C1 and in finding a1, b1, five decimal places in the logarithms are sufficient. 580. Let b 2, C2, be the angular excesses in seconds, that is, the excesses of the correct angles A, B, C, above the chordal angles A', B', C', respectively, then 16a, = R" (1+1) tan ‡ A ̧ — R′′ r b. ( cot A1, where R" 206264"-8, or 206265′′ the number of seconds in an arc equal to the radius, and r is the same quantity as in art. 562, or 20888000. Find in the same manner b2 and c2, and then it is evident that the spherical excess is E=a2+b2+c2, for A' + B' + C′ = 180. And if E' the observed spherical excess = (A1+B1+ C1 - 180) (art. 568), and E1 the sum of the errors of the observed angles, or E1 E-E', then 3 E, is to be applied with its proper sign to A1, B1, C1, in order to give 11. 2 A 1 A, B, and C, supposing that there is an equal chance of error in each; then is 19 1 1 1 A=A1+} E1, B=B1+} E1, and C = C1+3E,; and AA-a2, B'B-b2, and C' —C—C2. = The angles of the chordal triangle are now known, and if correct, then is A'B'C' 180, and A+B+C-180=E. 581. The angles and the side c' of the chordal triangle being now known, its other sides can be computed by plane trigonometry; in which computation, however, logarithms with at least seven decimal places must be used. Having found these chords a', b', the accurate values a, b, c, of the sides of the spherical triangle can now be found thus:— b'3. a'3 a = a' + 2472' and b = b' + 24 72' where r has the same value as in the preceding article. Any of the exercises formerly given under the problem in article 573 can be solved by this method, and will therefore serve as an illustration of it. 582. The distribution of the error E1, as also of E in article 569, must be made according to the relative probabilities of accuracy of the observations, agreeably to the remark in that article. The weight of each observation must be computed, and then the error is to be distributed among the three angles, in the proportion of the reciprocals of the corresponding weights. When an angle is determined by only one observation, its weight must be estimated according to the judgment of the observer. It must be remarked, that by the observed value of an angle is to be understood the mean value obtained, that is, the quotient arising from dividing the sum of the observed values by their number; and the observed error of an observation is its difference from the mean value.* 583. PROBLEM XI.-To find the reciprocal of the weight of the observed value of an angle. 'Find the mean of the observations of the angle, and then * For an example of the computation of the parts of a chordal triangle, consult the "Encyclopædia Britannica," article Trigonometrical Surveying. |