Surveying and Levelling: Chains, Traverses, Plane Table, Levels, Contours, Total Station, GPS, Areas and Volumes

The third chapter of Section 4, Soil and Water Conservation Engineering, of the GATE Agricultural Engineering (AG) paper: the sub-heading Surveying and Leveling. Every conservation structure starts as a survey — a contour line to lay a bund on, a longitudinal section to grade a waterway, a cross-section to estimate a dam’s earthwork. The chapter covers the measurement of distance and area, instruments for surveying and levelling, chain surveying, methods of traversing, the measurement of angles and bearings, plane-table surveying, the types of levelling, theodolite traversing, contouring, the total station and an introduction to GPS survey, and the computation of areas and volumes.

1. Measuring distance; chain surveying

Distances are taken with metric chains (20 m and 30 m, with tallies every metre or 5 m), tapes (steel, invar for precision, synthetic for rough work), and electronically by EDM (electromagnetic distance measurement) in a total station. Chain surveying suits small, open, fairly level ground: the area is divided into a network of triangles whose sides are measured, and detail is fixed by offsets (perpendicular or oblique) from the chain lines. The triangles must be well-conditioned — no angle below about 30° or above 120° — so that small errors in length do not shift the plotted points much. Ranging keeps the chain on the straight line between stations, directly or, when the ends are not intervisible, by reciprocal ranging.

A chain that is too long reads too little. If a nominal chain of length L is actually L′, a measured length l corresponds to a true length l × L′/L: a 20 m chain 0.1 m too long gives 500 m measured as 500 × 20.1/20 = 502.5 m true, and areas scale by the square of the ratio. Tape corrections add up the same way — for temperature, pull, sag and slope; a length l measured along a slope with a height difference h is reduced to horizontal by about h²/(2l).

⚠️ A long chain reads short — so the correction adds
Each "20 m" laid down is really 20.1 m, so there are fewer chain-lengths in the line than there should be, and the true distance is more than the reading. Multiplying by 20/20.1 instead gives 497.5 m, the wrong way round.

2. Angles and bearings; traversing with compass and theodolite

A bearing is the horizontal angle of a line from a reference meridian. The whole-circle bearing (WCB) runs from 0° to 360° clockwise from north; the quadrantal (reduced) bearing (QB) is measured from north or south towards east or west, never exceeding 90°. WCB 210° is S30°W, WCB 120° is S60°E, WCB 300° is N60°W. The back bearing differs from the fore bearing by exactly 180°: the back bearing of a line of WCB 75° is 255°. When the fore and back bearings of a compass line do not differ by 180°, a station is under local attraction from iron or electric lines nearby; the included angles computed from bearings at a station are unaffected, which is how the errors are corrected.

A traverse is a series of connected lines whose lengths and directions are measured — with a compass for rough work, with a theodolite (horizontal and vertical angles read on graduated circles, with temporary adjustments of centring, levelling and focusing, and face-left and face-right readings averaged to cancel instrumental errors) for precise work. It is closed when it returns to its start or ends on a known point. Each line’s latitude is L cos θ and departure L sin θ (θ the bearing): a 100 m line at WCB 60° has a latitude of 50 m and a departure of 86.6 m. In a closed traverse ΣL and ΣD should be zero; their residuals give the closing error e = √((ΣL)² + (ΣD)²), and the relative precision is e divided by the perimeter. The error is distributed by the Bowditch (compass) rule, correcting each line in proportion to its length, or by the transit rule. The interior angles of a closed polygon of n sides should sum to (2n − 4) × 90°.

3. Plane-table surveying

In plane tabling the map is drawn in the field on a drawing board set up on a tripod, with an alidade to sight and draw rays, a spirit level, a trough compass for orientation and a plumbing fork to centre the table over a ground point. Because plotting and observation happen together, nothing is forgotten and errors show at once; the price is lower accuracy, dependence on weather and bulky equipment. Orientation — setting the board so that every line on the paper is parallel to its line on the ground — is done by back-sighting along a known line or by the trough compass.

The four methods of plane tabling
MethodWhat is doneUsed for
Radiationrays drawn from one station and distances measured along themsmall areas visible from one point
Intersectionrays to a point from two stations cross at its positioninaccessible points, broken ground
Traversingthe table is moved station to station, back-sighting each timeroads, streams, long narrow strips
Resectionthe table’s own position is found from sights to points already plotted (the two-point and three-point problems)locating a new station quickly

4. Levelling: types, reduction and corrections

Levelling finds heights relative to a datum. A dumpy or automatic level gives a horizontal line of sight, and a graduated staff is read on it. The first reading after setting up is a back sight (BS) on a point of known level (a benchmark); the last before moving is a fore sight (FS); readings between are intermediate sights (IS). A point where both an FS and a BS are taken is a change point. Reduced levels are worked by the height-of-instrument method — HI = RL + BS, then RL = HI − FS (or − IS) — or by the rise-and-fall method, which offers a complete arithmetic check. Either way ΣBS − ΣFS = last RL − first RL. From a benchmark at 100.000 m, a BS of 1.525 m gives HI = 101.525 m, and an FS of 2.340 m puts the next point at 99.185 m.

The types of levelling are simple (one set-up), differential (a series of set-ups between distant points), fly levelling (a quick check run back to a benchmark), profile (longitudinal section) and cross-section levelling for channels and bunds, reciprocal levelling across a river or valley, and trigonometric levelling from vertical angles. Over long sights the level line departs from the horizontal: the curvature correction is −0.0785d² m and the refraction correction +0.0112d² m (one-seventh of curvature), giving a combined correction of −0.0673d² m with d in km — 0.269 m at 2 km. Reciprocal levelling, taking readings from both banks and averaging, cancels curvature, refraction (if the air is the same both times) and any collimation error of the instrument.

🧠 Which way the correction goes
The earth curves away from the horizontal line of sight, so the staff reads too high and curvature is subtracted; refraction bends the ray down towards the earth and gives back a seventh of it. Hence the combined correction is always negative.

5. Contouring, total station, GPS, and the computation of areas and volumes

A contour is a line joining points of equal elevation; the contour interval is the vertical distance between successive contours and is constant on a map. Contours close on themselves, never cross (except at an overhang), are closely spaced on steep slopes and widely spaced on gentle ones, cross ridges and valleys at right angles, and form V-shapes pointing uphill in valleys and downhill on ridges. They are surveyed directly (locating points on each contour in the field — slow but accurate) or indirectly (levelling a grid or radial lines and interpolating). Contour maps give the alignment of contour bunds and graded channels and the capacity of ponds.

A total station combines an electronic theodolite, an EDM and a data recorder: it measures horizontal and vertical angles and slope distance in one sighting to a prism and computes coordinates and elevations directly, which makes detail and contour surveys fast and digital. GPS (GNSS) fixes position by trilateration from satellite ranges; since the receiver clock is an unknown alongside the three coordinates, at least four satellites are needed for a three-dimensional fix. A single receiver is accurate to metres; differential GPS and RTK, correcting against a base station of known position, reach centimetres, enough for land levelling and layout.

Computing areas and volumes
RuleFormula (common interval d)Condition
Trapezoidal (area)d[(O₁ + Oₙ)/2 + O₂ + … + Oₙ₋₁]any number of offsets
Simpson’s one-third (area)(d/3)[(O₁ + Oₙ) + 4(even-numbered) + 2(remaining odd-numbered)]an odd number of offsets
CoordinatesA = ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)|closed traverse with known coordinates
Trapezoidal (end-area) volumed[(A₁ + Aₙ)/2 + A₂ + … + Aₙ₋₁]cross-sections at interval d
Prismoidal volume(d/3)[(A₁ + Aₙ) + 4(even-numbered) + 2(remaining odd-numbered)]an odd number of sections

For offsets of 3, 5, 6, 4 and 2 m at 10 m intervals, Simpson’s rule gives (10/3)[(3 + 2) + 4(5 + 4) + 2(6)] = (10/3)(53) = 176.7 m², against 175 m² by the trapezoidal rule; Simpson is exact for a boundary made of parabolic arcs and needs an odd number of ordinates. Cross-sections of 20, 30 and 44 m² at 10 m spacing give a prismoidal volume of (10/3)(20 + 4 × 30 + 44) = 613.3 m³.

Key takeaways

  • A chain that is too long reads short: true length = measured × (actual chain length/nominal length).
  • Back bearing = fore bearing ± 180°; latitude = L cos θ, departure = L sin θ; closing error = √(ΣL² + ΣD²).
  • Resection — the two- and three-point problems — locates the plane table itself from plotted points.
  • HI = RL + BS, RL = HI − FS; check ΣBS − ΣFS = last RL − first RL; combined curvature and refraction = −0.0673d² m (d in km).
  • Simpson’s rule and the prismoidal formula need an odd number of ordinates or sections; GPS needs four satellites for a 3-D fix.

Practice questions (12)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A line has a whole-circle bearing of 210°. Its quadrantal bearing and its back bearing (as a whole-circle bearing) are:

    1. S30°W and 30°
    2. S30°W and 390°
    3. N30°E and 30°
    4. S60°W and 30°
    Show answer

    Answer: A — S30°W and 30°

    210° lies in the third quadrant, 30° past south towards west, so QB = S30°W; the back bearing is 210° − 180° = 30°. S60°W measures from west instead of south, and 390° forgets to subtract rather than add when the fore bearing exceeds 180°.
  2. A level is set up and a back sight of 1.525 m is taken on a benchmark of reduced level 100.000 m. A fore sight of 2.340 m is then taken on point A. What is the reduced level of A, in m (to three decimal places)?

    Numerical answer — type the value.

    Show answer

    Answer: 99.185

    HI = 100.000 + 1.525 = 101.525 m; RL of A = 101.525 − 2.340 = 99.185 m. A is lower than the benchmark because its staff reading is larger. Adding the fore sight (103.865 m) reverses the sign convention.
  3. What is the combined correction for curvature and refraction for a sight of 2 km, in m (to three decimal places, as a magnitude)?

    Numerical answer — type the value.

    Show answer

    Answer: 0.269

    Combined correction = 0.0673d² = 0.0673 × 2² = 0.2692 m, subtracted from the staff reading. Using the curvature term alone, 0.0785 × 4 = 0.314 m, forgets that refraction gives back one-seventh.
  4. A line measured with a 20 m chain was recorded as 500 m. The chain was later found to be 0.1 m too long. What is the true length of the line, in m?

    Numerical answer — type the value.

    Show answer

    Answer: 502.5

    True length = measured × L′/L = 500 × 20.1/20 = 502.5 m. The chain laid 20.1 m each time it was counted as 20 m, so the true distance is longer. Multiplying by 20/20.1 gives 497.5 m, correcting in the wrong direction.
  5. Offsets of 3, 5, 6, 4 and 2 m are taken at 10 m intervals from a chain line to an irregular boundary. What is the area between the chain line and the boundary by Simpson’s one-third rule, in m² (to one decimal place)?

    Numerical answer — type the value.

    Show answer

    Answer: 176.7

    A = (d/3)[(O₁ + O₅) + 4(O₂ + O₄) + 2O₃] = (10/3)[5 + 4 × 9 + 2 × 6] = (10/3)(53) = 176.7 m². The trapezoidal rule gives 175 m²; swapping the weights 4 and 2 gives (10/3)(5 + 18 + 24) = 156.7 m².
  6. A traverse line 100 m long has a whole-circle bearing of 60°. What is its departure, in m (to one decimal place)?

    Numerical answer — type the value.

    Show answer

    Answer: 86.6

    Departure = L sin θ = 100 × sin 60° = 86.6 m (eastward); latitude = L cos θ = 50 m. Swapping sine and cosine gives 50 m for the departure, the usual slip.
  7. In a closed traverse, the sum of latitudes is +0.3 m and the sum of departures is −0.4 m. What is the closing error, in m?

    Numerical answer — type the value.

    Show answer

    Answer: 0.5

    e = √((ΣL)² + (ΣD)²) = √(0.09 + 0.16) = √0.25 = 0.5 m. Adding the two sums (−0.1) or their magnitudes (0.7) treats a vector error as a scalar one.
  8. Three cross-sections of an embankment, 10 m apart, have areas of 20, 30 and 44 m². What is the volume of earthwork by the prismoidal formula, in m³ (to one decimal place)?

    Numerical answer — type the value.

    Show answer

    Answer: 613.3

    V = (d/3)[A₁ + 4A₂ + A₃] = (10/3)(20 + 120 + 44) = (10/3)(184) = 613.3 m³. The end-area rule gives 10[(20 + 44)/2 + 30] = 620 m³, slightly more, since it over-estimates a prismoid.
  9. In plane-table surveying, the three-point problem is a method of:

    1. resection — locating the table’s own station from three plotted points
    2. intersection — locating an inaccessible point from three stations
    3. radiation from three rays
    4. traversing between three stations
    Show answer

    Answer: A — resection — locating the table’s own station from three plotted points

    The three-point problem finds where the table stands, and orients it, from sights to three points already on the paper — a resection. Intersection fixes a distant point from known stations, the reverse situation.
  10. The minimum number of satellites a GPS receiver needs to fix a three-dimensional position is:

    1. 3
    2. 4
    3. 5
    4. 6
    Show answer

    Answer: B — 4

    There are four unknowns — the three coordinates and the receiver clock offset — so four satellite ranges are needed. Three would suffice only with a perfect clock, which a low-cost receiver does not have.
  11. Reciprocal levelling across a wide river eliminates the errors due to:

    1. curvature of the earth
    2. atmospheric refraction, if conditions are the same for both sets of readings
    3. imperfect adjustment of the line of collimation
    4. a wrongly read staff
    Show answer

    Answer: A — curvature of the earth; B — atmospheric refraction, if conditions are the same for both sets of readings; C — imperfect adjustment of the line of collimation

    Readings from both banks make the curvature, refraction and collimation errors appear with opposite signs, so the mean of the two apparent differences is the true difference. A misread staff is a blunder that no arrangement of the instrument cancels.
  12. What should be the sum of the interior angles of a closed traverse with six sides, in degrees?

    Numerical answer — type the value.

    Show answer

    Answer: 720

    Sum = (2n − 4) × 90° = (12 − 4) × 90° = 720°. The exterior angles would sum to (2n + 4) × 90° = 1440°; using 360° confuses the traverse with the deflection angles, whose algebraic sum is 360°.