Weighted Adjustment of a Level Net with Four Fixed Benchmarks: 18-Geom-A2 Adjustment of Observations, December 2018, Q2

Question 2: Weighted Adjustment of a Level Net with Four Fixed Benchmarks (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five level runs tie two unknown points A and B to four fixed benchmarks. Reading the arrows in the figure, courses 1 and 2 run into A (BM1→A, BM2→A), course 3 runs A→B, and courses 4 and 5 run into B (BM3→B, BM4→B). Line lengths are 2, 2, 0.5, 1 and 1 km respectively, so the levelling weight is $w_i=1/L_i$.

Find. Adjusted elevations of A and B and their standard deviations.

Figure 2.1 — Level net: fixed benchmarks BM1–BM4 (triangles) and unknown points A, B (circles). Labels give course number and length; arrows show the levelling direction (all runs proceed toward an unknown).

Check (source ambiguity). The list of differences gives course 1 as $+10.997$ m, but a later sentence quotes “the observed elevation difference was 10.970 m.” The enumerated value $+10.997$ is used here because it agrees with course 2 at A to within 12 mm (using 10.970 opens a 39 mm gap); adopting 10.970 instead would move $H_A$ by only $\approx7$ mm and $H_B$ by $\approx3$ mm, and does not change the method.

Approach. Write one observation equation per course in the two unknowns $[H_A,H_B]$, moving each fixed benchmark to the right-hand side, weight by $1/L$, and solve the weighted normal equations; the adjusted-benchmark precisions come from the cofactor matrix.

Reduce each run to a station equation. Courses into a fixed benchmark become direct height estimates: $$H_A=785.232+10.997=796.229,\quad H_A=805.410-9.169=796.241,$$ $$H_B=794.881+4.858=799.739,\quad H_B=801.930-2.202=799.728,$$ and the connecting run gives $H_B-H_A=3.532$. Rows of $A$ (for $x=[H_A,H_B]$) are $[1,0],[1,0],[-1,1],[0,1],[0,1]$.
Weights. $W=\operatorname{diag}(1/2,\,1/2,\,1/0.5,\,1/1,\,1/1)=\operatorname{diag}(0.5,0.5,2,1,1)$ — the short 0.5 km connector A→B is the most trusted run.
Solve. $$\hat{x}=N^{-1}A^{\mathsf T}W\ell\;\Rightarrow\;\boxed{\,H_A=796.218\ \text{m},\quad H_B=799.742\ \text{m}\,}$$
Precision of the adjusted benchmarks. With $n=5$, $u=2$ (redundancy $r=3$), the residuals are $-10.8,-22.8,-8.4,+2.9,+13.9$ mm, giving $$s_0=\sqrt{\tfrac{v^{\mathsf T}Wv}{n-u}}=\pm0.0148\ \text{m}\ (\text{i.e. }\pm14.8\ \text{mm}/\!\sqrt{\text{km}}),$$ and from $\Sigma_{xx}=s_0^2N^{-1}$, $\boxed{\sigma_{H_A}=\pm10.5\ \text{mm},\ \ \sigma_{H_B}=\pm9.1\ \text{mm}}$.

Adjusted benchmark	Elevation (m)	Std. dev. (mm)
A	796.218	±10.5
B	799.742	±9.1

Concept & Theory

Topic. Weighted least-squares levelling with multiple fixed data (an over-constrained network).

Key relations. $w_i=1/L_i$ (variance of a level run $\propto$ length); normal equations $A^{\mathsf T}WA\,\hat{x}=A^{\mathsf T}W\ell$; benchmark covariance $\Sigma_{xx}=s_0^2(A^{\mathsf T}WA)^{-1}$, so $\sigma_{\text{stn}}=s_0\sqrt{Q_{ii}}$.

Why this works. Four fixed benchmarks over-determine two unknowns, so the four independent height estimates for A and B disagree; least-squares returns the weighted mean that minimises $\sum w_iv_i^2$, and the same normal-equation inverse that produces the heights also propagates their variances — no separate error analysis is needed.

Common pitfall. Mis-reading an arrow and writing a run in the wrong sense; a single flipped sign here throws $H_B$ off by metres. Always confirm the direction of levelling from the figure before assigning the sign of $\Delta$.

Source. Ghilani, Adjustment Computations §12.5 (level nets with several fixed benchmarks) and §12.7 (station-coordinate standard deviations).

Question 2 of 7: Weighted Adjustment of a Level Net with Four Fixed Benchmarks

Question 2: Weighted Adjustment of a Level Net with Four Fixed Benchmarks (20 marks)