Land Surveying III: Traversing, Triangulation and Trilateration
1. Traversing: types, angle checks, latitudes and departures
A traverse is a series of connected lines whose lengths and directions are measured. A closed traverse returns to its start (a loop) or runs between two known points (a link); an open traverse has no check and is avoided for control. Directions come from included angles or deflection angles with a theodolite or total station (or bearings with a compass). Angle check for a closed loop of n sides: the interior angles sum to (2n − 4) × 90° = (n − 2) × 180°, the exterior angles to (2n + 4) × 90°. A hexagon's interior angles total 720°. The angular misclosure is shared equally among the angles if they are equally reliable.
Each leg of length l and whole-circle bearing θ has a latitude L = l cos θ (north +, south −) and a departure D = l sin θ (east +, west −). A 150 m leg at 210° has L = -129.904 m and D = −75.000 m. For a closed loop ΣL = 0 and ΣD = 0 in theory; in practice the residuals give the closing error e = √((ΣL)² + (ΣD)²), with direction tan⁻¹(ΣD/ΣL) taken in the right quadrant, and the relative precision e/Σl, quoted as 1 in (Σl/e). If ΣL = +0.30 m and ΣD = −0.40 m over a perimeter of 1000 m, e = 0.50 m (bearing 306°52′, i.e. N 53°08′ W) and the precision is 1 in 2000.
2. Adjusting a traverse: Bowditch and transit rules, and coordinates
| Rule | Correction to a leg's latitude (departure alike) | Assumes |
|---|---|---|
| Bowditch (compass rule) | −ΣL × (leg length / perimeter) | angular and linear measurements equally precise; errors in direction and length both proportional to √l |
| Transit rule | −ΣL × (|leg latitude| / Σ|latitudes|) | angles measured more precisely than distances (theodolite with tape) |
Worked Bowditch: with ΣL = +0.30 m, ΣD = −0.40 m and perimeter 1000 m, a 250 m leg receives −0.30 × 250/1000 = -0.075 m in latitude and +0.40 × 250/1000 = +0.100 m in departure; every leg's corrections are in the same ratio, so the closing error is removed along its own direction. By the transit rule with Σ|L| = 1200 m, a leg whose latitude is 200 m receives −0.30 × 200/1200 = -0.050 m. Corrected consecutive coordinates are then accumulated from a known station into independent coordinates (Gale's traverse table sets out the whole computation). Omitted measurements — one length and one bearing, two lengths, or two bearings missing — can be recovered from ΣL = ΣD = 0, but then the traverse has no check left.
3. Triangulation
Triangulation covers a region with a network of triangles whose angles are all measured and one or more sides (base lines) measured with great care; the other sides follow from the sine rule, and a check base at the far end tests the chain. Figures: a single chain of triangles (fastest, weakest), chains of braced quadrilaterals (strongest for a narrow strip), and centred polygons (for wide areas). A well-conditioned triangle is one whose shape makes side errors least sensitive to angle errors; no angle should be smaller than about 30° or larger than about 120°, and the ideal isosceles triangle has base angles of 56°14′.
Strength of figure: R = ((D − C)/D) × Σ(δ_A² + δ_Aδ_B + δ_B²), where D is the number of directions observed (excluding those on the known side), C the number of geometric conditions, and δ the change in the sixth decimal of log sine for 1″ of the distance angles; the smaller R, the stronger the figure. The number of conditions is C = (n′ − S′ + 1) + (n − 2S + 3): angle conditions from n′ lines observed both ways with S′ occupied stations, and side conditions from n lines and S stations. For a braced quadrilateral with every line observed both ways, n = n′ = 6 and S = S′ = 4: 3 angle conditions and 1 side condition, C = 4; D = 2 × 6 − 2 = 10, so (D − C)/D = 0.6.
| Correction | Formula | Sign and example |
|---|---|---|
| Temperature | αL(T − T₀) | + when warmer than standard; 30 m, α = 1.2 × 10⁻⁵/°C, 15 °C warmer: +0.0054 m |
| Pull (tension) | (P − P₀)L/(AE) | + when pull exceeds standard; 50 N extra, 30 m, A = 3 mm², E = 2 × 10¹¹ Pa: +0.0025 m |
| Sag | W²L/(24P²), W = weight of the suspended length | always −; W = 6 N, L = 30 m, P = 100 N: −0.0045 m |
| Slope | h²/(2L) (or L(1 − cos θ)) | always −; h = 1.5 m over 30 m: −0.0375 m |
| Reduction to MSL | LH/R (H = height of the base above MSL) | − for a base above MSL; 1000 m at H = 637 m, R = 6370 km: −0.100 m |
Over large triangles the angles are those of a spherical triangle, whose sum exceeds 180° by the spherical excess ε = A/(R² sin 1″) seconds. With R = 6370 km, R² sin 1″ = 196.7 km², so ε is about 1″ for every 196.7 km² of area; a triangle of 590 km² has ε = 3.0″. The misclosure is taken after deducting ε. Stations that cannot be occupied are observed from a nearby satellite station and the angles reduced to centre.
4. Trilateration
Trilateration measures all the sides of the triangles with EDM (or GNSS baselines) instead of the angles. The angles follow from the cosine rule: cos C = (a² + b² − c²)/(2ab). For sides 400, 500 and 600 m, the angle opposite 600 m is cos⁻¹(0.125) = 82.82°. With EDM making long distances cheap and precise, trilateration and combined triangulateration (angles and sides together) have largely replaced pure triangulation; a single triangle measured by its three sides alone has no redundancy, so networks add diagonals (a braced quadrilateral of six measured sides has one redundant measurement).
| Method | Measures | Best when |
|---|---|---|
| Traversing | lengths and angles along a chain | built-up or wooded areas, routes, where long sight lines are impossible |
| Triangulation | angles, with one or two bases | hilly, open country with long intervisible lines |
| Trilateration | sides by EDM or GNSS | precise control with modern distance measurement |
Key takeaways
- Interior angles of a closed n-sided traverse sum to (n − 2) × 180°; latitude = l cos θ, departure = l sin θ.
- Closing error e = √(ΣL² + ΣD²), relative precision 1 in Σl/e.
- Bowditch: correction = −Σ × l/Σl; transit: −Σ × |L|/Σ|L|; corrections oppose the misclosure.
- Triangulation: well-conditioned triangles (ideal 56°14′), strength (D − C)/D with C = (n′ − S′ + 1) + (n − 2S + 3); braced quadrilateral C = 4, (D − C)/D = 0.6; spherical excess ≈ 1″ per 196.7 km².
- Base-line corrections: temperature αLΔT, pull ΔPL/AE, sag −W²L/24P², slope −h²/2L, MSL −LH/R; trilateration uses the cosine rule.
Practice questions (16)
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.
The sum of the interior angles of a closed traverse of six sides, in degrees, should be ____.
Numerical answer — type the value.
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Answer: 720
Σ interior = (n − 2) × 180° = 4 × 180° = 720°, or equivalently (2n − 4) × 90° = 8 × 90°. The exterior angles would sum to (2n + 4) × 90° = 1440°.The five interior angles of a closed traverse add up to 540°01′00″. If the angles are equally reliable, the correction to each angle, in seconds, is ____ (give the signed value).
Numerical answer — type the value.
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Answer: -12
The theoretical sum is (5 − 2) × 180° = 540°, so the misclosure is +1′ = +60″. Shared equally, each angle is corrected by -60/5 = -12″. The sign is negative because the observed sum is too large.A traverse leg is 150.000 m long with a whole-circle bearing of 210°. Its latitude, in metres (to three decimal places, with sign), is ____.
Numerical answer — type the value.
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Answer: -129.904
L = l cos θ = 150 cos 210° = 150 × (−0.866025) = -129.904 m (southward); the departure is 150 sin 210° = −75.000 m (westward). Using the reduced bearing S 30° W gives the same magnitudes, with signs from the quadrant.In a closed traverse of perimeter 1000 m, the sum of latitudes is +0.30 m and the sum of departures is −0.40 m. The closing error, in metres, is ____.
Numerical answer — type the value.
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Answer: 0.5
e = √((ΣL)² + (ΣD)²) = √(0.09 + 0.16) = √0.25 = 0.50 m. The relative precision is 0.50/1000 = 1 in 2000. Adding the magnitudes (0.70 m) is the slip.For the same traverse (ΣL = +0.30 m, ΣD = −0.40 m, perimeter 1000 m), the Bowditch correction to the latitude of a 250 m leg, in metres (with sign, to three decimal places), is ____.
Numerical answer — type the value.
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Answer: -0.075
Bowditch: correction = −ΣL × l/Σl = −0.30 × 250/1000 = -0.075 m. The departure correction is −(−0.40) × 250/1000 = +0.100 m. A positive latitude correction would add to the misclosure instead of removing it.For the same traverse the arithmetic sum of all latitudes, Σ|L|, is 1200 m. By the transit rule, the correction to a leg whose latitude is 200 m, in metres (with sign, to three decimal places), is ____.
Numerical answer — type the value.
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Answer: -0.05
Transit rule: correction = −ΣL × |L|/Σ|L| = −0.30 × 200/1200 = -0.050 m. It distributes by the size of the latitude, not the length of the leg, which is why it suits angles measured more precisely than distances.The Bowditch (compass) rule is most appropriate when
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Answer: A — angles and distances are measured with about equal precision
Bowditch assumes the errors in direction and in length are of equal effect, so it distributes the misclosure in proportion to leg length. When angles are much better than distances, the transit rule is preferred. An open traverse cannot be adjusted, and a blunder must be found, not spread.A 30 m steel tape standardised at 20 °C is used at 35 °C. With α = 1.2 × 10⁻⁵ per °C, the temperature correction per tape length, in metres (to four decimal places), is ____.
Numerical answer — type the value.
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Answer: 0.0054
Ct = αL(T − T₀) = 1.2 × 10⁻⁵ × 30 × 15 = 0.0054 m, positive because the warm tape is longer than its nominal 30 m and so under-reads. Treating it as negative is the usual sign slip.A 30 m tape weighing 6 N between supports is used under a pull of 100 N. The sag correction, in metres (magnitude, to four decimal places), is ____.
Numerical answer — type the value.
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Answer: 0.0045
Cs = W²L/(24P²) = 6² × 30/(24 × 100²) = 1080/240,000 = 0.0045 m, always subtracted, because the chord of a sagging tape is shorter than the tape. Squaring L instead of W is the common algebra slip.A base line measured as 1000.000 m lies at a mean height of 637 m above mean sea level. Taking R = 6370 km, the reduction to mean sea level, in metres (magnitude), is ____.
Numerical answer — type the value.
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Answer: 0.1
C = LH/R = 1000 × 637/6,370,000 = 0.100 m, subtracted, because the same angle subtends a shorter arc at sea level than at the height of the base. Using (R + H) in the denominator changes the answer only in the fifth decimal.Taking R = 6370 km, the spherical excess of a triangle of area 590 km², in seconds of arc (to one decimal place), is ____.
Numerical answer — type the value.
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Answer: 3
ε = A/(R² sin 1″). R² sin 1″ = 6370² × 4.8481 × 10⁻⁶ = 196.72 km², so ε = 590/196.72 = 3.0″. The rule of thumb is about 1″ of excess for every 196 km² of area.In a braced quadrilateral in which every line is observed in both directions, the total number of geometric conditions (angle plus side) is
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Answer: A — 4
With n = n′ = 6 lines and S = S′ = 4 stations: angle conditions n′ − S′ + 1 = 3, side conditions n − 2S + 3 = 1, total 4. Three counts only the angle conditions; the side condition is what checks the scale through the figure.In a trilateration triangle the measured sides are 400 m, 500 m and 600 m. The angle opposite the 600 m side, in degrees (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 82.82
cos C = (400² + 500² − 600²)/(2 × 400 × 500) = (160,000 + 250,000 − 360,000)/400,000 = 0.125, so C = 82.82°. The largest angle is always opposite the longest side; a sign slip in the numerator gives an obtuse angle.Which of the following corrections to a taped base line are always negative?
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Answer: A — Sag correction; B — Slope correction; C — Reduction to mean sea level for a base above sea level
A sagging tape's chord is shorter than the tape, a sloping length is longer than its horizontal projection, and an arc above sea level is longer than its projection on sea level — all three reduce the measured length. The temperature correction is positive above the standardisation temperature and negative below it.Which of the following statements about traverse computations are correct?
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Answer: A — In a closed loop traverse the algebraic sums of latitudes and departures should both be zero; B — The Bowditch rule leaves the bearings of the legs slightly changed after adjustment; C — Two omitted measurements can be computed from the closing conditions, at the cost of any check
A loop returns to its start, so ΣL = ΣD = 0; Bowditch corrections change each leg's latitude and departure, and therefore its bearing and length slightly; and the two closing equations can solve two unknowns but then leave no redundancy. Relative precision is closing error divided by perimeter, quoted as 1 in Σl/e.Which of the following statements about triangulation are correct?
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Answer: A — A smaller value of the strength-of-figure factor R indicates a stronger figure; B — Chains of braced quadrilaterals are stronger than single chains of triangles; C — A check base at the end of the chain tests the accumulated scale error
R measures how much side errors grow through the figure, so smaller is stronger; the extra diagonals of a braced quadrilateral add conditions and redundancy; and a check base compares computed and measured length. Small angles have large log-sine differences per second, so side errors blow up: angles below about 30° are avoided.