Volume 35, 1902
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### Example No. 1.—The Adjustment of Four Plane Tri-Angles.

For the purposes of comparison with the usual method of adjusting triangulation (in which one-third of the triangular error is applied to each angle of the triangle) an easy example is selected embracing the adjustment of four plane triangles.

The side P4 P in the figure is a side of the existing triangulation, and is to be adopted as correct both in bearing and length. From this side it is desired to extend the triangulation so as to include the points P1 P8 and P2. The angles are all observed, and are shown in column No. 2 of the schedule. No observation was possible between P1 and P8.

#### I. Adjustment as Four Separate Triangles.

The angles in each triangle are adjusted by applying to each angle one-third of the triangular error of the triangle.

The adjustment is shown in the schedule in columns-1 to 4.

In column 1 the names of the angles are entered as follows (see figure):—

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 A1 = P P1 P4 A2 = P P2 P1 B1 = P1 P4 P B2 = P2 P1 P C1 = P4 P P1 C2 = P1 P P2 A3 = P P3 P2 A4 = P P4 P3 B3 = P3 P2 P B4 = P4 P3 P C3 = P2 P P3 C4 = P3 P P4

Column No. 2 contains the observed angles.

In column No. 3 one-third of the triangular error of each triangle is applied to each angle.

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Column No. 4 gives the sums of the angles (seconds only) of columns Nos. 2 and 3.

With these angles (from column 4) the triangles are calculated, and the following results are obtained:—

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First Pair of Triangles—P1 P4 P and P2 P1 P.
Side. Bearing. Distance. Remarks.
P P4 53° 03′ 07″ 27833.3 Adopted as base.
P4 P1 20° 13′ 46.″ 29069.9
P1 P 130° 51′ 38.7″ 16120.5
P1 P2 67° 42′ 42.7″ 34843.4
P2 P 220° 09′ 19.7″ 31089.0
Second Pair of Triangles—P3 P2 P and P4 P3 P.
P P4 53° 03′ 07″ 27833.3 Adopted as base.
P4 P3 30° 49′ 18.7″ 40207.6
P3 P 174° 43′ 37.3″ 17874.2
P3 P2 74° 37′ 29.0″ 22499.4
P2 P 220° 09′ 24.7″ 31093.3

Comparing the bearing and length of the side P2 P as. obtained from the two sets of triangles, we have—

 220° 09′ 19-7″ 31089-0 links; and 220° 09′ 24-7″ 31093-3 "

giving differences of 5" and 4–3 links.

The application of the ordinary adjustment, resulting as it does in these differences, is therefore very unsatisfactory, and the question arises as to whether it is desirable in this and in similar cases to adopt some further adjustment to the observed angles so as to eliminate the discrepancies shown above.

Before discussing the further adjustment it may be as well to remark that the ordinary procedure would be to adopt the mean values of the bearing and distance of P P2. None of the other sides, however, would receive any correction; consequently if the calculation is repeated, using the mean value of P P2 as base, an entirely different set of values will be obtained for all the other sides of the triangles.

As the need for further adjustment is obvious, the method of applying it will now be indicated.

The problem to be solved is: Given the observed angles of the four triangles, corrected as shown in I., by applying one-third of the error of each triangle to each angle, what further corrections must be made to these angles so as to eliminate the discrepancies found above?

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It is evidently desirable that the corrections should be as small as possible so that no undue alterations are made to the angles: this condition is satisfied when the sum of the squares of the corrections is a minimum.

The application of this condition is shown on the schedule, and is briefly as follows:—

In column No. 5 the natural sines of the angles in column No. 4 are given.

If the sines in No. 5 were correct we should have—

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Sin A1 sin A2 sin A3 sin A4 / Sin B1 sin B2 sin B3 sin B4 = 1.

This equation shows that the length of P P2 calculated from P P4 by the first pair of triangles should be the same as the length calculated by the second pair of triangles.

This is not usually the case, so put Sin A1 sin A2 sin A3 sin A4 / Sin B1 sin B2 sin B3 sin B4 = 1 + ε where the sines are taken from column No 5 and ε is n radians.

To convert ε into seconds multiply the value in radians by 206265 (= number of seconds in 1 radian).

The calculation is shown on the schedule, giving, in this particular example, ε = + 28″·01.

(Note.—Attention must be paid to the sign of ε.)

The other necessary condition is that the sum of the angles C1 and C2 should equal the sum of the angles of C3 and C4, or—

C1 + C2 = C3 + C4.

This is not usually the case, so put

C1 + C2 = C3 + C4 + ε0, where the angles are taken from column No. 4 and ε0 is in seconds. This gives ε0 = — 5″ in this example (see schedule).

(Note.—Attention must be paid to the sign of ε0.)

In column No. 6 the natural cotangents of the angles are inserted.

In column No. 7 twice the cotangents are entered.

 Let α1 = cot A1 α2 = cot A2 " β1 = cot B1 β2 = cot B2

and similarly for the other angles.

 Let a1 = 2α1 + β1 a2 = 2α2 + β2 " b1 = − α1 − 2β1 b2 = − α2 − 2β2 " c1 = − α1 + β1 c2 = − α2 + β2

and similarly for a3 b3 c3 : a4 b4 c4.

In column No. 8 the values of a1, b1, c1, & c are given, and a check is obtained by noting that a1 + b1 + c1 = 0.

Square all the values in column No. 8 and add them. This

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is done readily on the Brunsviga calculating-machine without any intermediate record, the result in this example being—

Σ(a2 + b2 + c2) = + 126−790.

Let k = ⅙ Σ(a2 + b2 + c2)

h = c1 + c2 − c3 − c4(from column No. 8).

i = the number of triangles.

Next form the following equations:— hP + 2iQ + ε0 = 0

2kP + hQ + ε = 0

and solve them for P and Q.

With these values of P and Q calculate the corrections to the observed angles thus—

• x1 = a1P − Q

• y1 = b1P − Q

• z1 = c1P + 2Q

• x2 = a2P − Q

• y2 = b2P − Q

• z2 = c2P + 2Q

• x3 = a3P + Q

• y3 = b3P + Q

• z3 = c3P − 2Q

• x4 = a4P + Q

• y4 = b4P + Q

• z4 = c4P − 2Q

where x1, y1, z1, & c., are the corrections in seconds, and the corrected angles are—

 A1 + x1 A2 + x2 A3 + x3 A4 + x4 B1 + y1 B2 + y2 B3 + y3 B4 + y4 C1 + z1 C2 + z2 C3 + z3 C4 + z4

where the angles A1, B1, & c., are taken from column No. 4.

Columns 9, 10, and 11 show the calculation of the corrections.

Column No. 12 gives the final angles (seconds only), and is equal to column 4 + column 11.

Column No. 13 gives the natural sines of the angles in Column No. 12.

This completes the calculation of the least-square corrections.

In practice it is always desirable to check the results obtained, consequently the two following checks are applied:—

(a.) By forming the products of—

Sin A1 sin A2 sin A3 sin A4 (from column 13); and Sin B1 sin B2 sin B3 sin B4 which are equal (see schedule).

(B.) By comparing the values of—

C1 + C2 (from column 12); and C8 + C4 which are equal.

The triangles are now solved, using the angles from column 12, and the results are—

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 Differences. Side. Bearing. Distance. Bearing. Distance. Links. Links. P P4 53° 03′ 07″ 27833.3 P4 P1 20° 13′ 44.6″ 29070.1 + 1.7″ − 0.2 P1 P 130° 51′ 39.0″ 16120.8 − 0.3″ − 0.3 P1 P2 67° 42′ 41.7″ 34845.1 + 1.0″ − 1.7 P2 P 220° 09′ 22.1″ 31090.6 − 2.4″ − 1.6 Second Pair of Triangles. P P4 53° 03′ 07″ 27833.3 P4 P3 30° 49′ 22.2″ 40206.6 − 3.5″ + 1.0 P3 P 174° 43′ 36.9″ 17873.0 + 0.4″ + 1.2 P3 P2 74° 37′ 28.7″ 22497.3 + 0.3″ + 2.1 P2 P 40° 09′ 22.1″ 31090.6 + 2.61″ + 2.7

The columns headed “Differences” between the least-square values and the values obained in I.

A comparison of the values of P2 P as calculated from each pair of triangles shows that the bearing and distance agree exactly.

The process of adjustment here described completely satisfies the geometrical conditions of the figure, and it does so by making the sums of the corrections the least possible.

For the theory of the adjustment reference must be made to any of the treatises on least squares. See in particular “Geodesy,” by Colonel A. R. Clarke, C.B., Oxford, 1880, pp. 217-225. The method here outlined differs from that given in Clarke, inasmuch as the triangular error is applied before the condition equations are derived, thus lightening the subsequent work very considerably, and thereby lessening the risk of numerical slips.

This method also permits of comparison between the ordinary triangular adjustment and the least-square adjustment, as will be seen by comparing columns 3 and 11, where column 11 shows the additional corrections necessary to satisfy the geometrical conditions of the figure.

In this example the calculation has been carried to two decimal places of a second, not because the observations justify so much refinement, but to avoid an unequal distribution of the errors, as, for instance, would occur in distributing an error of 2″ among three angles. If this is done to the nearest second, then two angles would receive a correction of one second each and the third angle would remain unaltered. This would not have been consistent with the theory of the adjustment, which provides that exactly one-third of the triangular error must be applied to each angle.

The whole of the calculations have been done on the Brunsviga calculating-machine with ease, rapidity, and certainty.

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[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

 (1.) (2.) (3.) (4)=(2)+(3). (5)= Sines of (4). (6.) (7.) (8.) (9)=(8)xP. (10.) (11)=(9)+(10). (12)=(4)+(11) (13.) Angles. Observed Angles. Correction ⅓ Error of δ. Corrected Angles. Sines of Corrected Angles. Cot. of (2). Twice Cot of (2). a,b,c Q and 2 Q. Least-square Corrections to Angles. Final Angles. Sines of (12). ° ′ ″ ″ ″ ″ ″ ″ ″ A1 69 22 7 +0.67 7.67 0.9358678 α1=+0.376 +0.752 a1=+2.302 −1.501 −0.560 x1=−2.061 5.61 0.9358643 B1 32 49 20 +0.66 20.66 0.5420369 β1=+1.550 +3.100 b1=−3.476 +2.266 −0.560 y1=+1.706 22.37 0.5420438 C1 77 48 31 +0.67 31.67 0.9774483 c1=+1.174 −0.765 +1.120 z1=+0.355 32.02 0.9774487 179 59 58 A2 27 33 24 −1.00 23.00 0.4626214 α2=+1.916 +3.832 a2=+4.338 −2.828 −0.560 x2=−3.388 19.61 0.4626068 B2 63 8 57 −1.00 56.00 0.8921832 β2=+0.506 +1.012 b2=−2.928 +1.909 −0.560 y2=+1.349 57.35 0.8921862 C2 89 17 42 −1.00 41.00 0.9999243 c2=−1.410 +0.919 +1.120 z2=+2.039 43.04 0.9999244 180 0 3 A1 100 6 13 −4.67 8.33 0.9844961 α3 =−0.178 −0.356 a3 =+1.100 −0.717 +0.560 x3 =−0.157 8.18 0.9844962 B3 34 28 9 −4.66 4.34 0.5659440 β=+1.456 +2.912 b3 =−2.734 +1.783 +0.560 y3 =+2.343 6.67 0.5659534 C3 45 25 52 −4.67 47.33 0.7123913 c3 =+1.634 −1.065 −1.120 z3 =−2.185 45.15 0.7123839 180 0 14 A4 22 13 51 −2.67 48.33 0.3783270 α4 =+2.446 +4.892 a4=+6.263 −4.083 +0.560 x4=−3.523 44.81 0.3783112 B4 36 5 44 −2.67 41.33 0.5891232 β4=+1.371 +2.742 b4=−6.188 +3.383 +0.560 y4=+3.943 45.28 0.5891387 C4 121 40 33 −2.66 30.34 0.8510394 c4=−1.075 +0.701 −1.120 z4=−0.419 29.91 0.8510405 180 0 8 Σ(a2+b2+c2)=+126.790
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Sin A1 sin A2 sin A3 sin A4 (from column 5) = 0.1612581 / 0.1612362 = 1 + ε

Sin B1 sin B2 sin B3 sin B4

∴ 1 + ε = 1.0001358

∴ ε = + 0.0001358 radians.

∴ ε = + 28″.01.

C1 + C2 = C3 + C4 + ε(from column 4).

∴ 31″067 + 41″.00 = 47″.33 + 30″.34 + ε0

∴ ε = − 5″.00.

k = ⅙ Σ (a2 + b2 + c2) = + 21.130 (from column 8).

h = c1 +c2 −c3 −c4 = − 0.795 ″

i = 4.

The equations for P and Q are— hP + 2iQ + ε0 = O.

2KP + hQ + ε = O.

Substitute the values of h, k, i, ε0 and ε and the equations become—

− 0.795P + 8 Q − 5.00 = O.

+ 42.264 P − 0.795 Q + 28.01 = O.

Solving these equations for P and Q, we obtain—

P = − 0−652.

Q = + 0−560.

###### Checks on Final Angles.

Sin A1 sin A2 sin A3 sin A4 = 0-1612457 (from column 13).

Sin B1 sin B2 sin B3 sin B4 = 0-1612457

C1 + C2 = 75″-06 (from column 12). C3 + C4 = 75″-06