Differential-Levelling Notes, Misclosure and Compass-Rule Adjustment: 18-Geom-A1 Surveying, May 2019, Q4

Question 4: Differential-Levelling Notes, Misclosure and Compass-Rule Adjustment (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 levelling loop that starts on B.M.1 (height $225.412$ m), passes through benchmarks B.M.2 and B.M.3 and seven turning points, and closes back on B.M.1, with ten instrument setups.

Find. The completed notes ($\Delta H$, adjusted $\Delta H$ and heights), the arithmetic checks, the loop misclosure, and the compass-rule adjusted heights of B.M.2 and B.M.3.

Closed differential-levelling loop BM.1 → … → BM.1 through benchmarks BM.2, BM.3 and seven turning points.

Approach. Reduce the notes by rise and fall: each setup gives $\Delta H = \text{B.S.} - \text{F.S.}$ to the next point. Apply the page check, compare the loop’s net change with zero to get the misclosure, then distribute it by the compass rule. No section lengths are given, so assume equal sight lengths, which makes the correction proportional to the number of setups.

Height differences. Each setup’s backsight minus the next foresight gives, for example, B.M.1→TP1: $3.150 - 3.346 = -0.196$ m, and TP4→TP5: $0.089 - 3.736 = -3.647$ m. Carried down the loop, these give the completed field notes:

Station	B.S.	F.S.	ΔH (m)	Unadjusted height (m)	Adjusted ΔH (m)	Adjusted height (m)
B.M.1	3.150	—	—	225.412	—	225.4120
TP1	2.831	3.346	−0.196	225.216	−0.2032	225.2088
TP2	4.104	2.725	+0.106	225.322	+0.0988	225.3076
TP3	2.654	3.008	+1.096	226.418	+1.0888	226.3964
B.M.2	0.368	3.208	−0.554	225.864	−0.5612	225.8352
TP4	0.089	1.534	−1.166	224.698	−1.1732	224.6620
TP5	2.863	3.736	−3.647	221.051	−3.6542	221.0078
B.M.3	3.356	0.100	+2.763	223.814	+2.7558	223.7636
TP6	2.781	1.662	+1.694	225.508	+1.6868	225.4504
TP7	3.365	0.111	+2.670	228.178	+2.6628	228.1132
B.M.1	—	6.059	−2.694	225.484	−2.7012	225.4120
Σ / check	25.561	25.489	ΣΔH = +0.072	e = +0.072	Σ = 0.0000	closes

Customary arithmetic (page) check. $$ \Sigma\text{B.S.} - \Sigma\text{F.S.} = 25.561 - 25.489 = +0.072\text{ m} = \Sigma\Delta H = H_{\text{end}} - H_{\text{start}} = 225.484 - 225.412. $$ All three agree, so the reduction arithmetic is correct.
Misclosure. The loop starts and ends on B.M.1, so its true net change is zero. Therefore $$ \boxed{e = \Sigma\Delta H - 0 = +0.072\text{ m}} $$ The unadjusted closing height of B.M.1 is 225.484 m against the known 225.412 m. For scale, if the ten setups cover about 1 km, the Canadian allowances would be 24 mm (Third Order) and 120 mm (Fourth Order), so a 72 mm closure is Fourth-Order work. That is acceptable for construction or topographic control but not for benchmarks of a higher order.
Compass-rule distribution. With equal setups, each $\Delta H$ receives $$ c = -\frac{e}{n} = -\frac{0.072}{10} = -0.0072\text{ m per setup}, $$ which gives the adjusted $\Delta H$ column. The adjusted $\Delta H$ values sum to zero, and the running heights return exactly to 225.412 m at B.M.1. The adjusted benchmark heights are $$ \boxed{\text{B.M.2} = 225.835\text{ m}}, \qquad \boxed{\text{B.M.3} = 223.764\text{ m}}. $$

Quantity	Value
$\Sigma$ Backsights / $\Sigma$ Foresights	$25.561$ / $25.489$ m
Loop misclosure $e$	$+0.072$ m
Correction per setup (compass rule, 10 setups)	$-0.0072$ m
Adjusted height, B.M.2	$225.835$ m
Adjusted height, B.M.3	$223.764$ m
Closing height, B.M.1 (after adjustment)	$225.412$ m

Concept & Theory

Topic. Differential-levelling note reduction, the page check, and misclosure adjustment of a closed loop.

Key relations. $\Delta H = \text{B.S.} - \text{F.S.}$; page check $\Sigma\text{B.S.} - \Sigma\text{F.S.} = \Sigma\Delta H = H_{\text{end}} - H_{\text{start}}$; compass-rule correction $c_i = -e\,L_i/\Sigma L$, which becomes $-e/n$ per setup when the lengths are equal.

Why this works. Every backsight and foresight pair measures one rise or fall, so their sum is the net change in height. Around a closed loop that change must be zero, and anything left over is accumulated random error. The compass rule assumes the error grows with the distance (or number of setups) levelled, so it spreads the correction in that proportion.

Common pitfall. Pairing a backsight with its own station’s foresight instead of the next station’s, and applying the correction with the same sign as the misclosure. The page check catches arithmetic slips only. Field blunders, such as a mis-read staff, show up only as a large loop misclosure.

Source. Wolf & Ghilani, Elementary Surveying, 15th ed., Ch. 5 (differential-levelling notes, arithmetic checks and adjustment of simple level circuits).

Question 4 of 9: Differential-Levelling Notes, Misclosure and Compass-Rule Adjustment

Question 4: Differential-Levelling Notes, Misclosure and Compass-Rule Adjustment (20 marks)