Loop Levelling Adjusted by Number of Instrument Setups: 18-Geom-A2 Adjustment of Observations, December 2019, Q7

Question 7: Loop Levelling Adjusted by Number of Instrument Setups (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 closed differential-levelling loop that starts and ends on BM A, running through three turning points. The line returns to BM A reading 200.30 ft against its true 200.00 ft, so the loop misclosure is $+0.30$ ft.

Find. The adjusted elevations of TP1, TP2 and TP3, distributing the misclosure by the number of instrument setups.

Check (assumption). The table lists no per-leg setup count, so — as is standard for this exam prompt — one instrument setup per turning point is assumed (four setups total: BM A→TP1→TP2→TP3→BM A). The correction to each point is then proportional to the cumulative number of setups reached. If a setup tally were supplied it would replace the “1 per leg” weighting directly.

Approach. The misclosure grows with the number of setups (each setup adds an independent reading error), so distribute a correction to each running elevation equal to $-\,e\times(\text{cumulative setups}/\text{total setups})$.

Loop misclosure. $e=H_{A,\text{close}}-H_{A,\text{true}}=200.30-200.00=+0.30$ ft, to be removed over $\sum n=4$ setups.
Per-point correction. Cumulative setups are 1, 2, 3, 4 at TP1, TP2, TP3 and the closing BM A, so $$c_i=-\,0.30\times\frac{n_i}{4}=-0.075,\,-0.150,\,-0.225,\,-0.300\ \text{ft}.$$
Adjusted elevations. Applying each correction to the observed running elevation, $$\boxed{\,H_{TP1}=209.125,\ H_{TP2}=216.290,\ H_{TP3}=211.635\ \text{ft}\,}$$ and the closing BM A returns to $200.30-0.300=200.000$ ft, confirming the loop now closes exactly.

Point	Observed (ft)	Cum. setups	Correction (ft)	Adjusted (ft)
BM A	200.00	0	0.000	200.000
TP1	209.20	1	−0.075	209.125
TP2	216.44	2	−0.150	216.290
TP3	211.86	3	−0.225	211.635
BM A (close)	200.30	4	−0.300	200.000

Concept & Theory

Topic. Ordinary (non-least-squares) adjustment of a single closed level loop by setups.

Key relations. Misclosure $e=H_{\text{close}}-H_{\text{start}}$; correction $c_i=-e\,(n_i/\sum n)$ where $n_i$ is the cumulative number of setups (or, equivalently, cumulative distance) to point $i$.

Why this works. Each instrument setup contributes one independent backsight/foresight error, so the variance of the accumulated height grows linearly with the setup count; distributing the closing error in proportion to that count is the maximum-likelihood correction when all setups are of equal quality. It is the discrete analogue of distributing by distance.

Common pitfall. Applying the full misclosure to every point instead of the cumulative fraction, or getting the sign wrong — a positive closing misclosure (line reads too high on return) requires a negative correction that grows along the run.

Source. Wolf & Ghilani, Elementary Surveying §5.12 (adjusting a level loop by distance/number of setups).

Question 7 of 7: Loop Levelling Adjusted by Number of Instrument Setups

Question 7: Loop Levelling Adjusted by Number of Instrument Setups (20 marks)