Art. XIV.—The Adjustment of Triangulation by Least Squares.
[Read before the Wellington Philosophical Society, 18th November, 1902.].
It is proposed in the following paper to select examples of the ordinary methods of adjusting triangulation as practised in New Zealand and to apply to them the leastsquare adjustment, so as to compare the relative results obtained, and to show by actual examples that this method of adjustment alters the observed angles less than any other method. It will also be shown that the leastsquare adjustment can be simply and readily applied to most cases that occur in practice.
It is hoped that this treatment of the subject will be of use to the practical computer, and that it will enable him to see the advantages of the leastsquare adjustment by comparing its results with those usually obtained.
To make the treatment as simple as possible it will be assumed that all the angles are equally well observed.
Example No. 1.—The Adjustment of Four Plane TriAngles.
For the purposes of comparison with the usual method of adjusting triangulation (in which onethird 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 P_{4} 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 P_{1} P_{8} and P_{2}. The angles are all observed, and are shown in column No. 2 of the schedule. No observation was possible between P_{1} and P_{8}.
I. Adjustment as Four Separate Triangles.
The angles in each triangle are adjusted by applying to each angle onethird of the triangular error of the triangle.
The adjustment is shown in the schedule in columns1 to 4.
In column 1 the names of the angles are entered as follows (see figure):—
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A_{1} = P P_{1} P_{4}  A_{2} = P P_{2} P_{1} 
B_{1} = P_{1} P_{4} P  B_{2} = P_{2} P_{1} P 
C_{1} = P_{4} P P_{1}  C_{2} = P_{1} P P_{2} 
A_{3} = P P_{3} P_{2}  A_{4} = P P_{4} P_{3} 
B_{3} = P_{3} P_{2} P  B_{4} = P_{4} P_{3} P 
C_{3} = P_{2} P P_{3}  C_{4} = P_{3} P P_{4} 
Column No. 2 contains the observed angles.
In column No. 3 onethird of the triangular error of each triangle is applied to each angle.
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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Side.  Bearing.  Distance.  Remarks. 

P P_{4}  53° 03′ 07″  27833.3  Adopted as base. 
P_{4} P_{1}  20° 13′ 46.″  29069.9  
P_{1} P  130° 51′ 38.7″  16120.5  
P_{1} P_{2}  67° 42′ 42.7″  34843.4  
P_{2} P  220° 09′ 19.7″  31089.0  
Second Pair of Triangles—P_{3} P_{2} P and P_{4} P_{3} P.  
P P_{4}  53° 03′ 07″  27833.3  Adopted as base. 
P_{4} P_{3}  30° 49′ 18.7″  40207.6  
P_{3} P  174° 43′ 37.3″  17874.2  
P_{3} P_{2}  74° 37′ 29.0″  22499.4  
P_{2} P  220° 09′ 24.7″  31093.3 
Comparing the bearing and length of the side P_{2} P as. obtained from the two sets of triangles, we have—
220° 09′ 197″  310890 links; and 
220° 09′ 247″  310933 " 
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 P_{2}. None of the other sides, however, would receive any correction; consequently if the calculation is repeated, using the mean value of P P_{2} 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.
II. The Leastsquare Adjustment.
The problem to be solved is: Given the observed angles of the four triangles, corrected as shown in I., by applying onethird 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?
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 A_{1} sin A_{2} sin A_{3} sin A_{4} / Sin B_{1} sin B_{2} sin B_{3} sin B_{4} = 1.
This equation shows that the length of P P_{2} calculated from P P_{4} 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 A_{1} sin A_{2} sin A_{3} sin A_{4} / Sin B_{1} sin B_{2} sin B_{3} sin B_{4} = 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 C_{1} and C_{2} should equal the sum of the angles of C_{3} and C_{4}, or—
C_{1} + C_{2} = C_{3} + C_{4}.
This is not usually the case, so put
C_{1} + C_{2} = C_{3} + C_{4} + ε_{0}, where the angles are taken from column No. 4 and ε_{0} is in seconds. This gives ε_{0} = — 5″ in this example (see schedule).
(Note.—Attention must be paid to the sign of ε_{0}.)
In column No. 6 the natural cotangents of the angles are inserted.
In column No. 7 twice the cotangents are entered.
Let α_{1} = cot A_{1}  α_{2} = cot A_{2} 
" β_{1} = cot B_{1}  β_{2} = cot B_{2} 
and similarly for the other angles.
Let a_{1} = 2α_{1} + β_{1}  a_{2} = 2α_{2} + β_{2} 
" b_{1} = − α_{1} − 2β_{1}  b_{2} = − α_{2} − 2β_{2} 
" c_{1} = − α_{1} + β_{1}  c_{2} = − α_{2} + β_{2} 
and similarly for a_{3} b_{3} c_{3} : a_{4} b_{4} c_{4}.
In column No. 8 the values of a_{1}, b_{1}, c_{1}, & c are given, and a check is obtained by noting that a_{1} + b_{1} + c_{1} = 0.
Square all the values in column No. 8 and add them. This
is done readily on the Brunsviga calculatingmachine without any intermediate record, the result in this example being—
Σ(a^{2} + b^{2} + c^{2}) = + 126−790.
Let k = ⅙ Σ(a^{2} + b^{2} + c^{2})
“h = c_{1} + c_{2} − c_{3} − c_{4}(from column No. 8).
“i = the number of triangles.
Next form the following equations:— hP + 2iQ + ε_{0} = 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—

x_{1} = a_{1}P − Q

y_{1} = b_{1}P − Q

z_{1} = c_{1}P + 2Q

x_{2} = a_{2}P − Q

y_{2} = b_{2}P − Q

z_{2} = c_{2}P + 2Q

x_{3} = a_{3}P + Q

y_{3} = b_{3}P + Q

z_{3} = c_{3}P − 2Q

x_{4} = a_{4}P + Q

y_{4} = b_{4}P + Q

z_{4} = c_{4}P − 2Q
where x_{1}, y_{1}, z_{1}, & c., are the corrections in seconds, and the corrected angles are—
A_{1} + x_{1}  A_{2} + x_{2}  A_{3} + x_{3}  A_{4} + x_{4} 
B_{1} + y_{1}  B_{2} + y_{2}  B_{3} + y_{3}  B_{4} + y_{4} 
C_{1} + z_{1}  C_{2} + z_{2}  C_{3} + z_{3}  C_{4} + z_{4} 
where the angles A_{1}, B_{1}, & 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 leastsquare 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 A_{1} sin A_{2} sin A_{3} sin A_{4} (from column 13); and Sin B_{1} sin B_{2} sin B_{3} sin B_{4} which are equal (see schedule).
(B.) By comparing the values of—
C_{1} + C_{2} (from column 12); and C_{8} + C_{4} which are equal.
The triangles are now solved, using the angles from column 12, and the results are—
Differences.  
Side.  Bearing.  Distance.  Bearing.  Distance. 
Links.  Links.  
P P_{4}  53° 03′ 07″  27833.3  
P_{4} P_{1}  20° 13′ 44.6″  29070.1  + 1.7″  − 0.2 
P_{1} P  130° 51′ 39.0″  16120.8  − 0.3″  − 0.3 
P_{1} P_{2}  67° 42′ 41.7″  34845.1  + 1.0″  − 1.7 
P_{2} P  220° 09′ 22.1″  31090.6  − 2.4″  − 1.6 
Second Pair of Triangles.  
P P_{4}  53° 03′ 07″  27833.3  
P_{4} P_{3}  30° 49′ 22.2″  40206.6  − 3.5″  + 1.0 
P_{3} P  174° 43′ 36.9″  17873.0  + 0.4″  + 1.2 
P_{3} P_{2}  74° 37′ 28.7″  22497.3  + 0.3″  + 2.1 
P_{2} P  40° 09′ 22.1″  31090.6  + 2.61″  + 2.7 
The columns headed “Differences” between the leastsquare values and the values obained in I.
A comparison of the values of P_{2} 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. 217225. 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 leastsquare 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 onethird of the triangular error must be applied to each angle.
The whole of the calculations have been done on the Brunsviga calculatingmachine with ease, rapidity, and certainty.
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(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.  Leastsquare Corrections to Angles.  Final Angles.  Sines of (12).  
° ′ ″  ″  ″  ″  ″  ″  ″  
A_{1}  69 22 7  +0.67  7.67  0.9358678  α_{1}=+0.376  +0.752  a_{1}=+2.302  −1.501  −0.560  x_{1}=−2.061  5.61  0.9358643 
B_{1}  32 49 20  +0.66  20.66  0.5420369  β_{1}=+1.550  +3.100  b_{1}=−3.476  +2.266  −0.560  y_{1}=+1.706  22.37  0.5420438 
C_{1}  77 48 31  +0.67  31.67  0.9774483  c_{1}=+1.174  −0.765  +1.120  z_{1}=+0.355  32.02  0.9774487  
179 59 58  
A_{2}  27 33 24  −1.00  23.00  0.4626214  α_{2}=+1.916  +3.832  a_{2}=+4.338  −2.828  −0.560  x_{2}=−3.388  19.61  0.4626068 
B_{2}  63 8 57  −1.00  56.00  0.8921832  β_{2}=+0.506  +1.012  b_{2}=−2.928  +1.909  −0.560  y_{2}=+1.349  57.35  0.8921862 
C_{2}  89 17 42  −1.00  41.00  0.9999243  c_{2}=−1.410  +0.919  +1.120  z_{2}=+2.039  43.04  0.9999244  
180 0 3  
A_{1}  100 6 13  −4.67  8.33  0.9844961  α_{3} =−0.178  −0.356  a_{3} =+1.100  −0.717  +0.560  x_{3} =−0.157  8.18  0.9844962 
B_{3}  34 28 9  −4.66  4.34  0.5659440  β=+1.456  +2.912  b_{3} =−2.734  +1.783  +0.560  y_{3} =+2.343  6.67  0.5659534 
C_{3}  45 25 52  −4.67  47.33  0.7123913  c_{3} =+1.634  −1.065  −1.120  z_{3} =−2.185  45.15  0.7123839  
180 0 14  
A_{4}  22 13 51  −2.67  48.33  0.3783270  α_{4} =+2.446  +4.892  a_{4}=+6.263  −4.083  +0.560  x_{4}=−3.523  44.81  0.3783112 
B_{4}  36 5 44  −2.67  41.33  0.5891232  β_{4}=+1.371  +2.742  b_{4}=−6.188  +3.383  +0.560  y_{4}=+3.943  45.28  0.5891387 
C_{4}  121 40 33  −2.66  30.34  0.8510394  c_{4}=−1.075  +0.701  −1.120  z_{4}=−0.419  29.91  0.8510405  
180 0 8  Σ(a^{2}+b^{2}+c^{2})=+126.790 
Sin A_{1} sin A_{2} sin A_{3} sin A_{4} (from column 5) = 0.1612581 / 0.1612362 = 1 + ε
Sin B_{1} sin B_{2} sin B_{3} sin B_{4}
∴ 1 + ε = 1.0001358
∴ ε = + 0.0001358 radians.
∴ ε = + 28″.01.
C_{1} + C_{2} = C_{3} + C_{4} + ε(from column 4).
∴ 31″067 + 41″.00 = 47″.33 + 30″.34 + ε_{0}
∴ ε = − 5″.00.
k = ⅙ Σ (a^{2} + b^{2} + c^{2}) = + 21.130 (from column 8).
h = c_{1} +c_{2} −c_{3} −c_{4} = − 0.795 ″
i = 4.
The equations for P and Q are— hP + 2iQ + ε_{0} = O.
2KP + hQ + ε = O.
Substitute the values of h, k, i, ε_{0} 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 A_{1} sin A_{2} sin A_{3} sin A_{4} = 01612457 (from column 13).
Sin B_{1} sin B_{2} sin B_{3} sin B_{4} = 01612457
C_{1} + C_{2} = 75″06 (from column 12). C_{3} + C_{4} = 75″06