Adjustment of a Simple Levelling Circuit: 18-Geom-A1 Surveying, December 2013, Q1

Question 1: Adjustment of a Simple Levelling Circuit (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. A level line is run from benchmark $A$ (known $H_A$) through turning points $P_1$ and $P_2$ to benchmark $B$ (known $H_B$). The observations are the three height differences $h_1$ (A→P₁), $h_2$ (P₁→P₂), $h_3$ (P₂→B) and their section lengths $d_1,d_2,d_3$. No numeric values are attached to the figure, so the procedure is demonstrated symbolically exactly as asked.

Find. The adjusted (final) elevations $H_{P_1}$ and $H_{P_2}$.

Figure 1 — A "link" (connecting) level line tied to two fixed benchmarks $A$ and $B$; $P_1,P_2$ are turning points to be determined.

Approach. This is a link line between two fixed benchmarks, so the adjustment is driven by the misclosure between the two knowns, distributed to each section in proportion to its length (the least-squares model for a single level line).

Compute the misclosure. Running $A\to B$, the observed differences should reproduce the known elevation difference $H_B-H_A$. The misclosure is the observed closing value of $B$ minus its known value: $$w = \left(H_A + h_1 + h_2 + h_3\right) - H_B$$ A well-run circuit keeps $w$ within the allowable tolerance $\pm c\sqrt{\Sigma d}$, with $c$ the specified constant in mm/√km.
Distribute the correction in proportion to distance. Levelling error accumulates with the length of run, so the total correction $-w$ is apportioned to each height difference in proportion to its section length: $$v_i = -\,w\,\frac{d_i}{d_1+d_2+d_3}\qquad(i=1,2,3)$$ The three corrections sum to $-w$, which is exactly what closes the line. The adjusted differences are $\bar h_i = h_i + v_i$.
Propagate the adjusted differences to the unknowns. Add the running adjusted differences to the fixed start: $$H_{P_1} = H_A + \bar h_1 = \boxed{H_A + h_1 - w\,\frac{d_1}{\Sigma d}}$$ $$H_{P_2} = H_{P_1} + \bar h_2 = \boxed{H_A + h_1 + h_2 - w\,\frac{d_1+d_2}{\Sigma d}}$$ As a check, one more step must reproduce the fixed value exactly: $H_{P_2} + \bar h_3 = H_B$.

Quantity	Result
Misclosure	$w = (H_A+h_1+h_2+h_3)-H_B$
Section correction	$v_i = -w\,d_i/\Sigma d$
Elevation of $P_1$	$H_A + h_1 - w\,d_1/\Sigma d$
Elevation of $P_2$	$H_A + h_1 + h_2 - w\,(d_1+d_2)/\Sigma d$

Concept & Theory

Topic. Adjustment of a link (connecting) differential-levelling line between two fixed benchmarks.

Key relations. Misclosure $w=\sum h - (H_B-H_A)$; distance-weighted correction $v_i=-w\,d_i/\Sigma d$; propagation $H_{k}=H_{k-1}+\bar h_k$.

Why this works. In differential levelling the random error of a measured height difference grows with the number of set-ups, hence with distance, so its variance is $\sigma_{h_i}^2 \propto d_i$. Least squares with weights $p_i\propto1/\sigma^2\propto1/d_i$ yields a correction proportional to $d_i$ — precisely the distance-weighting rule. Distributing by distance is therefore the minimum-variance solution, not an arbitrary convention.

Common pitfall. Weighting the correction by $1/d_i$ instead of $d_i$; the weight is inverse-distance but the correction is direct-distance. Also forgetting the arithmetic check that the line must close exactly on $H_B$.

Source. Wolf & Ghilani, Elementary Surveying §5 (Leveling — theory & methods) and §16.6 (adjustment of level nets); Canadian vertical control referenced to CGVD2013.

Question 1 of 7: Adjustment of a Simple Levelling Circuit

Question 1: Adjustment of a Simple Levelling Circuit (20 marks)