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LECTURE 24, MARCH 24, 2004

TRIANGULATION j. Figure Adjustment: This involves determination of the most probable values of the

angles in a triangulation figure fulfilling the geometric condition equations. Adjustments are estimated following the least-squares method. For braced quadrilaterals, condition equations include the angle equations and side equations. For a braced quadrilateral illustrated on margin sketch, the angle equations are:

8743

6521

87654321 360

θθθθθθθθ

θθθθθθθθ

+=++=+

°=+++++++

In addition, to ensure closure

4

1

2

7

8

5

6

3

4

1

2

7

8

5

4

1

2

7

4

1

sinsin

sinsin

sinsin

sinsin

sinsin

sinsin

sinsin

sinsin

sinsin

sinsin

θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθ

θθ

DC

BCABADDC

=

===

Restating, sum of logarithm of sine of even subscripted angles is equal to the sum of logarithm of sine of odd subscripted angles. This is the so-called side equation. The constraint equations in terms of the errors are thus:

( ) ( ) 4

8

110

138743

265211

8

1

sinlog1 CCeeee

CeeeeCe

ii

irrrr

rrrri

ri

−=−−=−−+

−=−−+=

∑

∑

=

−

=

θ (6)

where eri is the error in θi, ∑=

−°=8

11 360

iiC θ , 65212 θθθθ −−+=C ,

87433 θθθθ −−+=C and C4 is the sum of logarithms of odd subscripted angles minus the sum of logarithms of even subscripted angles. The corresponding equations in terms of differentials are:

( ) 010

008

1

18743

6521

8

1

=−=−−+

=−−+=

∑

∑

=

−

=

iiri

irrrr

rrrri

ri

feeeee

eeeee

δδδδδ

δδδδδ (7)

where )5.0sin(log)5.0sin(log 1010 ′′−−′′+= iiif θθ . Often both sides of the last equation of Equations (18) are multiplied by 106 before undertaking the least squares

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procedure. Equations (18) may be solved using the method of correlates as illustrated in the following example or using any other appropriate procedure.

Example. Eight observed angles of a geodetic braced quadrilateral after correcting for spherical excess are: θ1 = 71.43433°, θ2 = 53.66517°, θ3 = 31.30293°, θ4 = 23.59779°, θ5 = 89.66956°, θ6 = 35.42974°, θ7 = 14.3008°, and θ8 = 40.60004° (see margin sketch). Assume all observations to be of equal weight.

Solution. Considering the angular errors in seconds of arc, the constraint equations are:

0.481650222.456547638.25982.95950.01214324.81985

3.4625721.548630.70718130.460000000.69000000

27000000.1

87

654

321

8743

6521

87654321

−=−+−+−

+−=−−+

−=−−+−=

+++++++

rr

rrr

rrr

rrrr

rrrr

rrrrrrrr

eeeeeeee

eeeeeeee

eeeeeeee

(A)

The first three are angle equations and the last one is side equation in terms of differences. Note that both sides of the side equation have been multiplied by 106. Note further that it is essential that the constraint equations are formulated in terms of seconds of arc because the difference form of the side equation has been formulated in terms of seconds.

The objective function 28

27

26

25

24

23

22

21 rrrrrrrr eeeeeeee +++++++=φ is

minimized by equating δφ to zero, i.e.,

022 8877665544332211 =+++++++ rrrrrrrrrrrrrrrr eeeeeeeeeeeeeeee δδδδδδδδ (B)

Differentiating the four constraint equations, multiplying by –λ1, –λ2, –λ3, and -λ4 and adding them together one gets:

0)2.456548()8.259800()2.959500()0.012143()4.819850()3.462572(

)1.548630()0.707181(

84317431

64215421

44313431

24211421

=++−+−+−+++−+−+−++−−+−−−+

+−−+−−−

rr

rr

rr

rr

eeeeee

ee

δλλλδλλλδλλλδλλλδλλλδλλλ

δλλλδλλλ

(C)

Adding to Equations (A) and (B) and equating the coefficients of δeri separately to zero the following set of equations are obtained:

8

43184317

42164215

43144313

42124211

2.4565488.2598002.9595000.0121434.8198503.4625721.5486300.707181

λλλλλλλλλλλλλλλλλλλλλλλλ

−−=+−=−−=+−=−+=++=−+=++=

rr

rr

rr

rr

eeeeeeee

(D)

Back-substitution on (C) into the constraint equations (A) yields:

0.48165022121.13647.160532.1059070.65717160.460000007.16053064

0.600000002.1059071427000000.10.65717168

4321

43

42

41

−=+−+=−

−=+−=+

λλλλλλλλλλ

(E)

Hence λ1 = -0.16010897, λ2 = -0.18120967, λ3 = 0.14461473 and λ4 = 0.01654332. Substituting these results into Equations (C), er1 = -0.32961952″, er2 = -0.36693811″, er3 = 0.0417882″, er4 = -0.09523055″, er5 = 0.02130159″, er6 = -0.39027856″, er7 = -0.16807922″ and er8 = -0.34536315″. Thus the corrected angles are: θ1 = 71.43424°, θ2 = 53.66507°, θ3 = 31.30294°, θ4 = 23.59776°, θ5 = 89.66957°, θ6 = 35.42963°, θ7 = 14.30075°, and θ8 = 40.59995°.

The spreadsheet solution for this problem can be obtained by clicking on “Example: Lecture 24” link in Surveying II web page.

Exercise: Following are the measured angles of a quadrilateral ABCD with a central station E:

Triangle Central Angle Left Hand Angle Right Hand Angle AEB 59.052778 61.015000 59.935000 BEC 118.397222 32.065000 29.535000 CED 60.534722 56.466944 62.996944 DEA 122.015278 28.700000 29.283333

Adjust the quadrilateral assuming all observations to be of equal weight.

Hint: The Left Hand Angle for ∆AEB is BAE and the Left Hand Angle for ∆AEB is ABE. There will be one constraint equation per triangle, one constraint equation for the quadrilateral (sum of the central angles at E equal to 360°) and one side equation for the quadrilateral. Thus there will be six correlatives. Use a spread sheet to solve the normal equation for the correlatives.