Question 1 of 7: Differential-Levelling Field Notes and Page Check
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geom-A1 Surveying. Closed-book; any non-communicating calculator permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Where a datum is implied, elevations are referenced to the Canadian vertical frame (CGVD2013) and azimuths to NAD83(CSRS); US-foot stationing is retained wherever the printed question uses it.
Reference texts: Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley); Hofmann-Wellenhof et al., GNSS — Global Navigation Satellite Systems (Springer, 2008).
Question 1: Differential-Levelling Field Notes and Page Check (20 marks)
Given. A differential-levelling circuit from BM1 (Elev $=88.00$ ft) through three turning points to BM2:
Station
BS (ft)
FS (ft)
BM1
2.45
—
TP1
5.43
6.53
TP2
3.18
4.91
TP3
4.22
7.42
BM2
—
6.11
Find. The completed field-note table (HI and Elevation at every station), the elevation of BM2, and the arithmetic page check.
Figure 1 — Differential-levelling run BM1 → TP1 → TP2 → TP3 → BM2; each set-up takes a backsight (BS) to the known point and a foresight (FS) to the next.
Approach. March the height of instrument (HI $=$ Elev $+$ BS) and elevation (Elev $=$ HI $-$ FS) alternately down the line, then verify with the page check $\Sigma\text{BS} - \Sigma\text{FS} = \text{Elev}_{\text{last}} - \text{Elev}_{\text{first}}$.
Complete the field notes. Starting from BM1 (Elev $=88.00$, HI $=88.00+2.45=90.45$) and applying HI $=$ Elev $+$ BS and Elev $=$ HI $-$ FS in turn:
Station
BS
HI
FS
Elevation (ft)
BM1
2.45
90.45
—
88.00
TP1
5.43
89.35
6.53
83.92
TP2
3.18
87.62
4.91
84.44
TP3
4.22
84.42
7.42
80.20
BM2
—
—
6.11
78.31
$\Sigma$
15.28
24.97
Elevation of BM2. The last elevation carried down the line is
$$\text{Elev}_{\text{BM2}} = \text{HI}_{\text{TP3}} - \text{FS}_{\text{BM2}} = 84.42 - 6.11 = \boxed{78.31\ \text{ft}}$$
Page check. The sums of the backsights and foresights must reproduce the net elevation change:
$$\Sigma\text{BS} - \Sigma\text{FS} = 15.28 - 24.97 = -9.69\ \text{ft}$$
$$\text{Elev}_{\text{BM2}} - \text{Elev}_{\text{BM1}} = 78.31 - 88.00 = -9.69\ \text{ft}\ \checkmark$$
The two agree, so the arithmetic of the field notes is verified.