Weighted Adjustment of a Level Loop: 18-Geom-A1 Surveying, December 2013, Q2

Question 2: Weighted Adjustment of a Level Loop (20 marks)

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Given. A closed level loop $A\to B\to C\to D\to A$ with observed differences and section lengths as tabulated, and fixed $H_A = 236.891$ m.

Section	$\Delta H$ (m)	Distance (km)
A→B	$-15.632$	4
B→C	$+32.458$	6
C→D	$+38.214$	5
D→A	$-55.025$	3

Find. Adjusted elevations $H_B$, $H_C$, $H_D$.

Figure 2 — Closed levelling loop; the sum of observed differences should return to zero at $A$.

Approach. Sum the observed differences to get the loop misclosure, distribute it to each section in proportion to length, then carry the adjusted differences forward from the fixed elevation of $A$.

Loop misclosure. A closed loop returns to its start, so the observed differences should sum to zero: $$w=\sum \Delta H = -15.632+32.458+38.214-55.025 = +0.015\ \text{m}$$ This $+15$ mm must be removed. Against a typical $12\sqrt{\Sigma d}$ mm tolerance with $\Sigma d=18$ km the allowable is $\approx 51$ mm, so the loop is acceptable.
Distance-weighted corrections. Each observed difference receives $v_i=-w\,d_i/\Sigma d$ with $\Sigma d=18$ km: $$v_{AB}=-3.3,\ v_{BC}=-5.0,\ v_{CD}=-4.2,\ v_{DA}=-2.5\ \text{(mm)}$$ The corrections sum to $-15$ mm, closing the loop exactly.
Adjusted elevations. Carry the adjusted differences forward from the fixed $H_A$: $$H_B = 236.891 + (-15.632-0.0033) = \boxed{221.256\ \text{m}}$$ $$H_C = 221.256 + (32.458-0.0050) = \boxed{253.709\ \text{m}}$$ $$H_D = 253.709 + (38.214-0.0042) = \boxed{291.918\ \text{m}}$$ Check: $H_D+(-55.025-0.0025)=236.891\ \text{m}=H_A$, so the loop closes back on the fixed benchmark.

Quantity	Value
Loop misclosure $w$	$+0.015$ m ($+15$ mm)
Adjusted $H_B$	$221.256$ m
Adjusted $H_C$	$253.709$ m
Adjusted $H_D$	$291.918$ m

Concept & Theory

Topic. Distance-weighted adjustment of a single closed levelling loop on one fixed control point.

Key relations. Loop misclosure $w=\sum\Delta H$; correction $v_i=-w\,d_i/\Sigma d$; forward propagation $H_k=H_{k-1}+\bar{\Delta H}_k$.

Why this works. Two proportionalities are easily confused: the weight of an observed difference is inverse-distance ($p_i\propto1/d_i$) because longer runs are less certain, while the correction applied is direct-distance ($v_i\propto d_i$) because more error accumulates over a longer section. Both follow from $\sigma_i^2\propto d_i$. The single fixed datum $A$ absorbs no error; the four measured differences absorb it all.

Common pitfall. Dividing the misclosure equally among sections (ignoring length), or applying $+w$ instead of $-w$ so the "adjusted" loop no longer closes.

Source. Wolf & Ghilani, Elementary Surveying §16.6 (level-net adjustment); Ghilani, Adjustment Computations §12 (weighted observations).

Question 2 of 7: Weighted Adjustment of a Level Loop

Question 2: Weighted Adjustment of a Level Loop (20 marks)