Art. VIII.—The Adjustment of Triangulation by Least Squares.—Part II

Transactions and Proceedings of the Royal Society of New Zealand, Volume 37, 1904, Page 195

Art. VIII.—The Adjustment of Triangulation by Least Squares.—Part II By G. E Adams, B.Sc (Honours), N.Z. Univ.; A.I.A. (Lond.); late Engineering Entrance Scholar and Engineering Exhibitioner, Canterbury College; late Senior Scholar in Physical Science, N.Z. Univ. [Read before the Wellington Philosophlcal Society, 10th October, 1903.] Example No. 2.—The Adjustment of a Polygon. Following the method adopted in example No. 1,* Trans. N.Z. Inst., vol. xxxv., p. 201. the usual adjustment as practised in New Zealand will be first described,

and then the least-square adjustment will be considered and the practical application of it explained. In the figure the sides P P1 and P P4 are derived from the existing triangulation, and are to be adopted as correct both in bearing and distance. The triangulation is to be extended to include the stations P2 and P3, and to this end all the angles of the triangles are equally well observed. These observed angles are shown in column 2 of the schedule. I. The Ordinary Adjustment. The first correction to the observed angles consists in applying one-third of the triangular error in each triangle to each angle, and is shown in (3). The corrected centre angles are entered in (4) and added to the given angle P1 P P4, and the sum of these angles should be 360°. This is not usually the case, so the difference between the sum and 360° is distributed equally among the centre angles: thus each centre angle receives a further correction of ε0/i where ε0 = the difference between the sum of the angles in (4) and 360°, and i = the number of triangles. To keep the sum of the angles of each triangle equal to 180° it is necessary to apply half this correction to each of the base angles. These corrections are shown in (5), and the corrected angles appear in (6). This completes the adjustment of the centre angles, and the subsequent corrections affect the base angles only. The length of the side P P4 is calculated from P P1 using the angles in (6)—the sines of these angles appear in (11)—and the length so obtained is compared with the true length. As the two values of P P4 do not usually agree, a further correction to the base angles becomes necessary, and is found thus: If l = true length of P P4 and l1 = length of P P4 calculated from P P1, using the angles in (6), then ε = l-l1/l radians. In (9) the cotangents of the base angles are given and the sum ∑(cot A + cot B) obtained. Then the correction to each base angle is = ε/∑(cot A + cot B) radians. The calculation of the correction is shown on the schedule, and gives 4″ 06 in this example. In (7) the correction is applied, and the final angles appear in (8). At this stage the work is checked by calculating P P4 from P P1, using the final angles in (8), and, as shown on the schedule, the calculated value of P P4 agrees with the true value, thus proving the correctness of the work. In (10) the value of 1″ for each of the base angles is given. The products of (7) and (10) give in (12) the corrections to be

applied to (11) to obtain the sines of the final angles shown in (13). The corrections to the sines in (11) are shown in (12), and are equal to (7) × (10). The triangles are now solved, using the final angles in (8), and the results are given in (15). The bearings in (16) are obtained by applying the final angles in (8) to the given bearing of P P1, checking on the bearing of P P4. The rectangular co-ordinates of P2 and P3 are calculated from the bearings and distances in (16) and (15). This completes the adjustment of the polygon. It will be seen that all the geometrical conditions of the figure are completely satisfied, but, as will be shown subsequently, the corrections applied to the observed angles are considerably larger than those required by least squares. It will also be noticed that the centre angles are corrected independently of the side adjustment, which is thus not allowed to influence the adjustment of the centre angles in any way. II. The Least-square Adjustment. The application of this adjustment is given on the schedule. Columns (1), (2), (3), and (4) are the same as in the ordinary adjustment. The angles of the first computation in (5) are not corrected for the error of the centre angles, but are equal to (2) + (3). (9) contains the sines of (5), and with these sines the three triangles are solved, and the length of P P4 obtained by calculation from P P1. Comparison of this value with the true value gives ε, while e0 is obtained from (4). Column (10) gives a1, b1, c1 &c., and ∑ (a2 + b2 + c2); from this column h = c1 + c2 + c3 and 2 k ⅓ ∑ (a2 + b2 + c2) are obtained The equations for P and Q are now formed and solved. With these values of P and Q the corrections to the observed angles are calculated, thus:— x1=a1 P—Q x2=a2 P—Q x3=a3 P—Q y1=b1 P—Q y2=b2 P—Q y3=b3 P—Q z1=c1 P + 2Q z2=c2 P + 2Q z3=c3 P + 2 Q and the values are entered in column (6) and applied to the angles in (5), giving the final angles as shown in (7). A check is obtained at this stage by noting that P1 P P4 + C1 + C2 + C3 (from (7)) = 360°G (11) contains the sines of the angles in (7), and a check computation of P P4 from P P1 proves the correctness of the work. The triangles are solved using the sines in (11), and the results given in (13). (14) contains the bearings obtained by

applying the final angles in (7) to the bearing of P P1. The rectangular co-ordinates are obtained from the bearings and distances in (14) and (13). III. Comparison of Results. The corrections to the angles, after adjusting each triangle to 180°, are obtained from (5) and (7) of, I., and are given below; the corresponding least-square corrections are also given. Angles. Ordinarv Collections = (5) + (7)of I. Least square Collections = (G) of II. " " A1 −4.84 −3.80 B1 + 3.29 −0.17 C1 + 1.55 + 3.97 A2 −4.84 −3.14 B2 + 3.28 + 2.17 C2 + 1.56 + 0.97 A3 −4.84 −4.10 B3 + 3.29 + 4.38 C3 + 1.55 −0.28 — — Total corrections ir respective of sign 29.04 22.98 In the angles C1 and B3 the least- square correction is larger than the ordinary correction; but the total connections to all the angles of the figure are, by least squares, much less than those obtained in the ordinary way. Thus, the total ordinary correction is 29.04 or 9.68 per triangle, while the total least square is 22.98 " 7.66 " thus giving a difference of 6 06 " 2.02 " in favour of the least-square adjustment A comparison of the bearings and lengths shows the following differences:— Side Bearings, Ordiuaiy (from I) Leat Square (from II) Differences Lengths, Ordinaiy (from I) Least Squaie (from II) Differences. ° ′ ″ ″ ″ Links Links Links PP2 113 28 19.78 17 36 + 2.42 96148 79 90148.54 + 0.25 PP368 21 30.22 28.39 + 1.83 73459.60 78459.02 + 0.58 P1P2 59 27 34.95 31.49 + 8.46 49764.48 49765.26 −0.78 P2P3 343 3 50.06 46.53 + 3.52 68352.69 68352.29 + 0.40 P3P4 282 46 40.17 39.44 + 0.73 61867.90 61867.06 + 0.84 These differences in bearings and lengths result from the methods of adjustment adopted, and are entirely independent

of the observed angles, as the same observed angles have been used in each case. Each adjustment gives a consistent geometrical figure, and preference is given to the least-square adjustment because it gives this consistent figure with the least alteration to the observed angles, as shown in this example, and as would also appear by comparing any other process of adjustment therewith. The notation used in the least-square adjustment is similar to that used in example No. 1, and is here repeated for convenience of reference. Let l1 = length of P P4 calculated from P P1, using the angles from (6) Schedule 1, (5) Schedule 2; " l = true length of P P4: then ε = l - l1/l radians ε0 = sum of angles at P (from (4)) - 360° α1 = cot A1 β1 = cot B1 a1 = 2α1 + β1 b1 = - α1 - 2β1 c1 = - α1 + β1 2 k = ⅓ ∑ (a2 + b2 + c2) h = c1 + c2 + c3 i = the number of triangles. The equations for P and Q are— h P + 2i Q + ε0 = 0 2 k P + h Q + ε = 0 The corrections to the angles are— x1 = a1 P - Q y1 = b1 P - Q z1 = c1 P + 2 Q, &c., and are given in (6); and the corrected angles are— A1 + x1 B1 + y1 C1 + z1, &c., where the corrections are applied to the values of the angles in (5). For the theory of the adjustment reference must be made to any of the treatises on least squares. The method here used is described by Colonel Clarke in his “Geodesy,” but differs in the application of the triangular error, which is applied before the condition - equations are derived. This shortens the numerical work considerably, and thereby lessens ihe risk of numerical slips. As in the case of example No. 1, all the calculations have been performed on the Brunsviga calculating-machine.

Schedule No. 1.—Ordinary Computation of Polygon with Brunsviga Calculating-Machine. PP1 side, 144° 30′ 46″, 78084.3 links; PP4 side, 11° 1′ 36″, 41542.2 links. Adjustment of Angles. Adjustment of Sides. (1.) (2.) (3) (4.) (5.) (6.) (7.) (8.) (9) (10.) (11.) (12.) (13.) (14.) (15) (16.) Ls. Observed. Corr Centre Ls Corr. Angles of First Computation Corr. Final Ls. Cot Val. 1.″ Nat. Sines of (6) Corr. Final Sines of (8). Sides. Links. Bearings ° ′ ″ ″ ° ′ ″ ″ ′ ″ ″ ′ ″ P1PP4 226 30 50 00 ° ′ ″ A1 54 0 29 + 0.67 −0 78 28.89 −4 06 24.83 + 0.726 + 28.49 0.8090993 −116 0.8090877 PP1 78084 30 144 30 46.00 B1 94 57 5 + 0 66 −077 4 89 + 4.06 8.95 −4.18 0.9962686 −17 0.9962669 PP2 96148.79 113 28 19.78 C1 31 2 24 + 0.67 31 2 24.67 + 1.55 26.22 26 22 0.5156456 P1P2 49764 48 59 27 54 95 — 179 59 58 — A2 85 17 47 −2 −0 78 44 22 −4 06 40 16 + 0 082 + 3 98 0 9966311 −16 0 9966295 B2 49 35 29 −2 0 78 26 22 + 4.06 30 28 + 0 851 + 31 43 0 7614321 + 128 0 7614449 PP3 73459 60 68 21 30 22 C2 45 6 50 −2 45 6 48.00 + 1.56 49.56 49.56 0 7085093 P2P3 68352 69 343 3 50.06 — 180 0 6 — A3 88 15 0 + 0 66 −077 14 59.89 −4.06 14 55.83 + 0031 + 1.48 0 9995336 −6 0 999.330 B3 34 25 6 + 0 67 −0.78 5.89 + 4 06 9 95 + 1 459 + 39.99 + 0 5652305 + 163 0.5652468 PP4 41542.20 11 1 36.00 C3 57 19 52 + 0 67 57 19 52 67 + 1 55 54.22 54.22 0.8418098 P3P4 61867.90 282 46 40.17 — — — 179 59 58 359 59 55 34 + 3.062 — — 360 0 0.00 — ε0 =0 0 4.66 — Check from (13). Sin A1 sin A3 (from column (11)) = 0 8059974,2 Sin A1 sin A2 sin A3 = 0 8059841,0 Sin B1 sin B2 sin B3 (from column (11)) = 0.4287787,3 Sin B1 sin B2 sin B3 = 0.4287975,2 78084 3 × 0 4287787,3/0 8059974,2 = 41539695 415422/2505 780843 × 0.4287975,2/08059841,0 = 41542.20,2 2 505 × 206265/41542.2 × 3 062 = 4″062

Schedule No. 2.—Least-Square Computation of Polygon with Brunsviga Calculating-Machine. PP1 side, 144° 30′ 46″, 78084.3 links; PP4 side, 11° 1′ 36″, 41542.2 links. (1.) (2.) (3.) (4.) (5) (6) (7) (8.) (9.) (10.) (11.) (12.) (13.) (14.) Ls. Observed. Corr. Centre Ls Angles of First Computation Corr. Final Ls. Nat. Sines of (5). a, b, and c. Final Sines Sides. Links. Bearings. ° ′ ″ ″ ° ′ ″ ″ ″ ′ ″ ° ′ ″ P1PP4 226 30 50 00 A1 54 0 29 + 067 29 67 −3.80 25.87 0.8091015 a1= + 1.365 0 8090907 PP1 78084.3 144 30 46 B1 94 57 5 + 0 66 5.66 −0 17 5.49 0.9962680 b1=−0.552 0.9962680 PP2 96148.54 113 28 17.36 C1 31 2 24 + 0 67 24 67 24 67 + 3 97 28 64 c1=−0.813 0.5156556 P1P2 49765.26 59 27 51.49 — 179 59 58 — A2 85 17 47 −2 45 00 −3.14 41.86 0 9966314 a2= + 1.015 0 9966302 B2 49 35 29 −2 27 00 + 217 29.17 0 7614346 b2=1 784 0 7614414 PP3 73459.02 68 21 23 39 C2 45 6 50 −2 48 00 48 00 + 097 48 97 c2= + 0769 0 7085074 P2P3 68352 29 343 3 46.53 — 180 0 6 — A3 88 15 0 + 066 0.66 −4 10 14 56.56 0 9995337 a2= + 1 521 09995331 B3 34 25 6 + 067 6.67 + 4 38 11 05 0.5652337 b3= + 2 949 0.5652512 PP4 41542.20 11 1 36.00 C3 57 19 52 52 67 52.67 −0.28 52 39 c3= + 1.428 0.8418050 P3P4 61867.06 282 46 39.44 — — 179 59 58 55 34 — 60 00 Σ(a2 + b2 + c2) = + 20.682 — ε0= − 4 66 — Checks. Sin A1 sin A2 Sin A3 (from (9)) = 0.8059999,5 Sin B1 sin B2 Sin B3 (from (9)) = 0.4287822,4 78084.3 × 0.4287822,4/0.8059999,5 = 41539 903 41542 2/+ 2.297 ε = 2.297/41542 = + 0 00005529 radians = + 11″.40 h = c1+c2+c3 = 1.384 (a) 226 30 50.00 from (4) 2k = ⅓ ∑ (a2+b2+c2) = + 6 894 28 64}from (7). i = +3 48.97 ε0 = - 466 5239 ε = +11.40 0.00 1.384P + 6Q - 4.66 = 0} (β) Sin A1 sin A2 Sin A3 (from (11)) = 0 8059877,4 6 894 P + 1.384Q + 11.40 = 0 Sin B1 sin B2 Sin B3 (from (11)) = 0.4287993,9 P = - 1 897 Q = + 1.214 780843 × 04287993,9/0.8059877,4 = 41542.198 [For diagram see Schedule 1.