18-Geom-A2 Adjustment of Observations · December 2019
Question 5 of 7: Parametric Adjustment of a Six-Section Level Net
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2019 — 18-Geom-A2, Adjustment of Observations and Data Analysis. Three-hour, CLOSED-BOOK exam; an approved Casio or Sharp calculator is permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Adjustments are carried out in the Canadian survey frame (heights on CGVD2013, horizontal work on NAD83(CSRS)); US-foot units are retained in Q7 wherever the printed question uses them.
Reference texts: Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley, 2017); Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson, 2015); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981).
Method note. Every weighted least-squares result below is formed from the normal equations $N\hat{x}=A^{\mathsf T}W\ell$; the reference (unit-weight) standard deviation is $s_0=\sqrt{v^{\mathsf T}Wv/(n-u)}$ with $n$ observations and $u$ unknowns, and station precisions come from $\Sigma_{xx}=s_0^2 N^{-1}$.
Question 5: Parametric Adjustment of a Six-Section Level Net (20 marks)
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 5.1 — 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: loops A–B–C and A–C–D close exactly as observed, so the whole 0.10 m misclosure of loop A–B–D is shared among the sections of that loop and its neighbours.
