Surveying and levelling

Surveying determines the relative positions of points on, above or below the earth's surface by measuring distances, heights and angles, and then plots them as maps and plans. Every road, irrigation canal, hydropower scheme and building in Nepal starts with a survey. The Survey Department of Nepal maintains national control points and maps using the Modified UTM (MUTM) projection on the Everest 1830 spheroid; GNSS work uses WGS-84. This note covers fundamentals, measurements, levelling, topographic survey, simple circular curves and GPS/GIS.

Key concepts

Fundamentals

• Plane surveying ignores earth's curvature (areas < ~250 km²); geodetic surveying accounts for it.
• Two principles: work from whole to part (establish control first, then fill details, to limit error accumulation) and locate a new point by at least two independent measurements from known points.
• Classification by instrument: chain, compass, plane table, theodolite, total station, GNSS; by purpose: topographic, cadastral, engineering, hydrographic, route.

Linear measurement

• Direct: chain (20 m/30 m), steel/invar tape; indirect: tacheometry, EDM / total station.
• Corrections to tape: temperature $C_t = \alpha L(T - T_0)$, pull $C_p = (P-P_0)L/AE$, sag $C_s = -\dfrac{w^2L^3}{24P^2}$ (always negative), slope $C_h \approx -h^2/2L$, standardisation.

Vertical measurement and levelling

• Reference: mean sea level (MSL), via benchmarks (GTS, permanent, temporary).
• Instruments: dumpy, tilting, auto level, digital level; staff.
• Terms: backsight (BS) – first reading on a known point; foresight (FS) – last reading before moving; intermediate sight (IS); change point (CP); height of instrument (HI) = RL + BS.
• Reduction methods: Height of instrument (faster, fewer checks on IS) and Rise and fall (complete arithmetic check including intermediate sights).
• Arithmetic check: $\sum BS - \sum FS$ = Last RL − First RL.
• Combined curvature and refraction correction: $C = 0.0673 d^2$ m ($d$ in km), subtractive.
• Reciprocal levelling eliminates collimation, curvature and refraction errors across rivers/valleys.
• Contours: lines of equal elevation; closely spaced = steep slope; contours never cross (except overhanging cliff); V pointing uphill = valley, V pointing downhill = ridge.

Angles and directions

• Bearings: whole circle bearing (0–360° from north) or reduced/quadrantal bearing (N θ E etc.).
• Meridians: true, magnetic, grid, arbitrary. Magnetic declination = angle between true and magnetic north. Local attraction is detected when fore and back bearings differ by ≠ 180°.
• Theodolite/total station: horizontal and vertical angles; traversing (open/closed); closing error adjusted by Bowditch's rule (proportional to side length) or transit rule.
• Closed traverse checks: sum of interior angles = $(2n-4)\times90^\circ$; ΣLatitudes = 0, ΣDepartures = 0.

Topographic survey

• Establish horizontal and vertical control, then pick up details and spot heights with total station or GNSS, plot contours.
• Methods of contouring: direct (locating points of equal RL) and indirect (grid, cross-section, tacheometric).
• Applications: route alignment, reservoir capacity, earthwork, site planning, catchment delineation.

Simple circular curves

• Used to connect two straight tangents. Elements for radius $R$ and deflection angle $\Delta$:

Element	Formula
Tangent length	$T = R\tan(\Delta/2)$
Length of curve	$L = \pi R\Delta/180$
Long chord	$C = 2R\sin(\Delta/2)$
External distance	$E = R(\sec(\Delta/2) - 1)$
Mid-ordinate	$M = R(1 - \cos(\Delta/2))$
Degree of curve (30 m arc)	$D = 1718.9/R$ degrees

• Chainage of PC = chainage of PI − $T$; PT = PC + $L$.
• Setting out: Rankine's deflection angle method ($\delta = 1718.9\,c/R$ minutes for chord $c$), offsets from long chord or tangents, two-theodolite method.

GPS/GIS

• GPS/GNSS: satellite positioning (GPS, GLONASS, Galileo, BeiDou) using signals from at least 4 satellites to solve 3-D position and receiver clock error. Modes: absolute (single receiver, metre-level), differential/DGPS, RTK (cm-level), static (mm–cm for control).
• GIS: system to capture, store, analyse and display spatially referenced data. Data models: vector (points, lines, polygons) and raster (grid cells). Uses: land-use planning, cadastral mapping, disaster (landslide, flood) risk mapping, road network, utility management.

High-yield points

• Principles: work from whole to part; fix each point by two independent measurements.
• Sag correction is always negative; curvature–refraction correction $0.0673d^2$ is subtractive.
• Closed traverse: sum of interior angles $=(2n-4)\times90^\circ$; Bowditch's rule distributes error in proportion to side length.
• Contours never cross (except overhangs); a V pointing uphill marks a valley.
• GPS needs at least 4 satellites for a 3-D fix; RTK gives cm-level accuracy.

Formulas

• HI = RL + BS; RL = HI − FS; check: $\sum BS - \sum FS$ = last RL − first RL.
• Curve: $T = R\tan(\Delta/2)$, $L = \pi R\Delta/180$, $D = 1718.9/R$ (30 m arc).
• Sag correction $C_s = -w^2L^3/(24P^2)$.

Worked example

Two tangents meet at chainage 1250.00 m with deflection angle 40°. Radius 300 m. Find T, L, and chainages of PC and PT.

• $T = 300\tan20^\circ = 300\times0.3640 = 109.19$ m
• $L = \pi\times300\times40/180 = 209.44$ m
• PC = 1250.00 − 109.19 = 1140.81 m
• PT = 1140.81 + 209.44 = 1350.25 m

Exam tips

• Check in levelling: ΣBS − ΣFS = last RL − first RL; rise-and-fall checks intermediate sights too.
• Sag correction is always negative; curvature-refraction correction 0.0673d² is subtractive.
• Fore and back bearing differ by 180°; unequal difference indicates local attraction.
• GPS needs at least 4 satellites for 3-D fix; RTK gives cm accuracy.
• Chainage of PT = PC + curve length (not PI + T).

Exam-style questions

From Calmon's question bank, which has 25 questions on this topic.

1. The magnetic bearing of a line is 45° and the magnetic declination at the place is 3° West. The true bearing of the line is

• A. 42°
• B. 48°
• C. 45°
• D. 225°

Answer: A. True bearing = magnetic bearing − west declination $= 45^\circ - 3^\circ = 42^\circ$.

2. On a topographic map with a contour interval of 2 m, two adjacent contours are 50 m apart horizontally. The average ground gradient between them is

• A. 1 in 100
• B. 1 in 25
• C. 1 in 2
• D. 1 in 50

Answer: B. Gradient $= \frac{2}{50} = \frac{1}{25}$.

Practise ACiE01 questions →