Parametric Adjustment of a Six-Section Level Net: 18-Geom-A2 Adjustment of Observations, December 2018, Q4

Question 4: Parametric Adjustment of a Six-Section Level Net (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. Six levelling sections joining four stations, with $H_A=0$ fixed and each observed difference stated in its direction of increasing elevation (all positive). Section lengths give the weights $w_i=1/L_i$.

Find. The parametric (least-squares) adjusted elevations of B, C and D.

Figure 4.1 — Local levelling net: A (fixed) with sides to B, C, D and the interior tie A–D. Labels are section number and observed rise (m); arrows point uphill.

Approach. Choose the three station elevations $[H_B,H_C,H_D]$ as parameters, write each section as $H_{\text{to}}-H_{\text{from}}=\Delta+v$ with A carried to the right-hand side, weight by $1/L$, and solve the weighted normal equations.

Observation equations. For $x=[H_B,H_C,H_D]^{\mathsf T}$ the six rows are: $H_C=6.16$; $H_D=12.57$; $H_D-H_C=6.41$; $H_B=1.09$; $H_D-H_B=11.58$; $H_C-H_B=5.07$ — i.e. $A$-rows $[0,1,0],[0,0,1],[0,-1,1],[1,0,0],[-1,0,1],[-1,1,0]$.
Weights. $W=\operatorname{diag}(1/4,1/2,1/2,1/4,1/2,1/4)$; the 2 km sections are trusted twice as much as the 4 km sections.
Solve the normal equations. $$\hat{x}=N^{-1}A^{\mathsf T}W\ell\;\Rightarrow\;\boxed{\,H_B=1.05,\ H_C=6.16,\ H_D=12.59\ \text{m}\,}$$ (with $H_A=0$ held fixed).
Fit statistics. Residuals are $0,+20,+20,-40,-40,+40$ mm; with $n=6$, $u=3$ (redundancy $r=3$) the reference standard deviation is $s_0=\pm0.026$ m, giving station precisions $\sigma_{H_B}=\sigma_{H_C}=\pm33$ mm, $\sigma_{H_D}=\pm28$ mm — consistent with a modest local loop where the longest (4 km) sections carry the misclosure.

Station	A	B	C	D
Elevation (m)	0.000*	1.05	6.16	12.59
Std. dev. (mm)	—	±33	±33	±28

*datum, held fixed.

Concept & Theory

Topic. Parametric least-squares adjustment of a redundant differential-levelling network.

Key relations. $H_{\text{to}}-H_{\text{from}}=\Delta+v$; $w_i=1/L_i$; $\hat{x}=(A^{\mathsf T}WA)^{-1}A^{\mathsf T}W\ell$; $s_0^2=v^{\mathsf T}Wv/(n-u)$.

Why this works. “Parametric” means the unknowns are the station elevations themselves, so each observation writes down directly as a height difference — the design matrix is just the incidence matrix of the graph with $\pm1$ entries. Holding A at zero fixes the datum; the three independent loops (redundancy 3) let least-squares distribute the 4 cm of accumulated misclosure in inverse proportion to line length.

Common pitfall. Forgetting to weight (an unweighted solve here would over-trust the long 4 km lines) or misreading an uphill arrow and flipping a section’s sign.

Source. Ghilani, Adjustment Computations §12.1–12.4 (parametric adjustment of level nets).

Question 4 of 7: Parametric Adjustment of a Six-Section Level Net

Question 4: Parametric Adjustment of a Six-Section Level Net (20 marks)