Land Surveying II: Theodolite, Total Station and COGO, Tacheometry and Trigonometric Levelling
1. The theodolite
A theodolite measures horizontal and vertical angles. Its four axes must satisfy the conditions of permanent adjustment: the plate-level axis perpendicular to the vertical axis; the line of collimation perpendicular to the horizontal (trunnion) axis; the horizontal axis perpendicular to the vertical axis; and the vertical circle reading zero (or 90° zenith) when the line of sight is horizontal (no index error). Observing each angle on both faces — face left and face right, transiting the telescope between — and taking the mean eliminates collimation error, trunnion-axis (horizontal-axis) error and index error, and averages eccentricity of the circles. It does not eliminate an error in the vertical axis itself (plate levels not adjusted), which tilts both faces the same way.
- Least count of a vernier = value of one main-scale division ÷ number of vernier divisions (when n vernier divisions equal n − 1 main divisions). Main divisions of 10′ with a 30-division vernier: 10′/30 = 20″.
- Repetition method: an angle is added on the circle several times (the lower plate clamped between) and the total divided by the number of repetitions, reducing reading and graduation errors; used for single angles of high precision.
- Reiteration (direction) method: from one station, the directions to several targets are read in turn, closing on the first (closing the horizon), on both faces and with the circle zero changed between sets; used in triangulation.
- Vertical angles are measured from the horizontal (elevation +, depression −) or as zenith angles z from the vertical: vertical angle = 90° − z.
2. Total station, EDM and the COGO functions
A total station integrates an electronic theodolite, an electronic distance meter (EDM) and a microprocessor with data storage. The EDM measures by phase comparison of a modulated infrared or laser carrier: the distance is D = (nλ + Δλ)/2, the fractional part from the phase difference and the integer n from several modulation wavelengths. Its accuracy is quoted as ±(a mm + b ppm) — a fixed part and a part proportional to distance: ±(2 mm + 2 ppm) over 1500 m is 2 + 3 = 5 mm. The measured distance needs corrections for the prism constant and for atmospheric temperature and pressure (which change the speed of light). From the slope distance S and zenith angle z it reduces HD = S sin z and VD = S cos z; 250.000 m at z = 85° gives HD = 249.049 m and VD = 21.789 m. The RL of the target is Z_A + hi + VD − ht (instrument and target heights).
| Function | Input → output | Formula or use |
|---|---|---|
| Inverse | two coordinates → distance and bearing | D = √(ΔE² + ΔN²), bearing = atan2(ΔE, ΔN); A(1000, 1000) to B(1300, 1400): 500 m at 36°52′ |
| Forward (radiation) | station, bearing, distance → coordinates | E = E₀ + D sin θ, N = N₀ + D cos θ |
| Area | boundary coordinates → area | A = ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)| (the shoelace formula) |
| Resection (free station) | observations to known points → the instrument's own coordinates | set up anywhere convenient |
| Stakeout (setting out) | design coordinates → bearing and distance to walk the prism onto | building corners, road and curve points |
| Missing line (MLM), remote elevation (REM), offsets | distance and height difference between two remote points; height of an inaccessible point above a prism | cables, bridge soffits, building heights |
Setting out a simple circular curve is the stakeout a geomatics engineer does most. For radius R and deflection (intersection) angle Δ between the tangents: tangent length T = R tan(Δ/2), curve length L = πRΔ/180°, long chord 2R sin(Δ/2), external distance E = R(sec(Δ/2) − 1), mid-ordinate M = R(1 − cos(Δ/2)). Chainage of the first tangent point T₁ = chainage of the intersection point − T; of the second, T₂ = T₁ + L (not the intersection point + T). By Rankine's deflection-angle method, a chord c subtends a tangential angle δ = 1718.9 c/R minutes at T₁. A curve's degree D is the angle subtended by a 30 m arc, so R = 1718.9/D (D in degrees). Worked: R = 300 m, Δ = 40°, intersection point at chainage 1250.00 m: T = 109.19 m, L = 209.44 m, long chord 205.21 m, E = 19.25 m, M = 18.09 m; T₁ at 1140.81 m, T₂ at 1350.25 m; a 20 m chord has δ = 114.59′ = 1°54′35″.
3. Tacheometry
A tacheometer is a theodolite with stadia hairs above and below the central hair. With the line of sight horizontal and the staff vertical, the horizontal distance is D = k·s + C, where s is the staff intercept between the stadia hairs, k = f/i the multiplying constant (usually 100) and C = f + d the additive constant (made zero by an anallactic lens). For an inclined sight at vertical angle θ with the staff held vertical: D = k s cos²θ + C cos θ and V = k s sin 2θ/2 + C sin θ; the RL of the staff station = RL of instrument station + hi + V − h, where h is the central-hair reading. Worked: k = 100, C = 0, hairs 0.800, 1.400, 2.000 (s = 1.200), θ = +8°: D = 120 cos²8° = 117.68 m, V = 60 sin 16° = 16.54 m; with the instrument station at RL 200.00 and hi 1.50, the staff station is at 200.00 + 1.50 + 16.54 − 1.40 = 216.64 m.
- Finding the constants: observe s at two known distances. s = 0.495 at 50 m and 0.995 at 100 m give k = (100 − 50)/(0.995 − 0.495) = 100 and C = 50 − 100 × 0.495 = 0.5 m.
- Tangential tacheometry: two vertical angles to two vanes a known distance apart on the staff, with no stadia hairs.
- Subtense bar: a horizontal bar of known length b (usually 2 m) subtends a horizontal angle θ; D = (b/2) cot(θ/2), independent of slope. A 2 m bar subtending 1° is at 114.59 m.
4. Trigonometric levelling
Trigonometric levelling finds heights from vertical angles and distances. Base accessible (the horizontal distance D to the foot is measurable): h = D tan α above the instrument's line of sight; a tower 50 m away seen at 30° rises 28.87 m above the trunnion axis. Base inaccessible, two instrument stations in line with the object, b apart, with angles α₁ (near) and α₂ (far) at the same instrument height: h = b tan α₁ tan α₂/(tan α₁ − tan α₂), and the distance from the near station is h/tan α₁. With b = 20 m, α₁ = 45° and α₂ = 30°: h = 20 × 1 × 0.5774/(1 − 0.5774) = 27.32 m. If the two stations are at different heights, or not in line with the object, the solution uses the triangle formed in plan.
Key takeaways
- Face left and face right cancel collimation, trunnion-axis and index errors, but not a vertical-axis error; vernier least count = one main division ÷ number of vernier divisions.
- Total station: HD = S sin z, VD = S cos z; EDM accuracy ±(a mm + b ppm); COGO inverse gives √(ΔE² + ΔN²) and atan2(ΔE, ΔN).
- Circular curve: T = R tan(Δ/2), L = πRΔ/180, LC = 2R sin(Δ/2), E = R(sec(Δ/2) − 1), M = R(1 − cos(Δ/2)); T₂ = T₁ + L; δ = 1718.9c/R minutes.
- Stadia: D = ks + C (k = 100, C = 0 with an anallactic lens); inclined: D = ks cos²θ, V = ks sin 2θ/2.
- Trigonometric levelling: h = D tan α for an accessible base; h = b tan α₁ tan α₂/(tan α₁ − tan α₂) for two stations in line.
Practice questions (15)
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 main scale of a theodolite is divided into 10′ divisions, and 30 divisions of the vernier coincide with 29 main-scale divisions. The least count, in seconds, is ____.
Numerical answer — type the value.
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Answer: 20
Least count = one main division ÷ number of vernier divisions = 10′/30 = 600″/30 = 20″. Equivalently it is the difference between one main and one vernier division, 10′ − (29 × 10′/30) = 20″.Averaging face-left and face-right observations of a horizontal angle does NOT eliminate
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Answer: A — error due to the vertical axis not being truly vertical
Collimation, trunnion-axis and index errors reverse sign when the telescope is transited, so they cancel in the mean. A tilted vertical axis tilts the horizontal axis the same way on both faces, so the error is identical on both and survives averaging; only careful levelling of the plate bubbles removes it.A total station measures a slope distance of 250.000 m at a zenith angle of 85°00′00″. The horizontal distance, in metres (to three decimal places), is ____.
Numerical answer — type the value.
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Answer: 249.049
HD = S sin z = 250.000 × sin 85° = 250.000 × 0.996195 = 249.049 m. The vertical component is S cos z = 21.789 m. Using cos z for HD treats the zenith angle as a vertical angle and gives 21.789 m.Point A has coordinates (E, N) = (1000, 1000) m and point B (1300, 1400) m. The whole-circle bearing of AB given by the COGO inverse function, in degrees (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 36.87
ΔE = 300, ΔN = 400, both positive, so the line is in the north-east quadrant. Bearing = arctan(ΔE/ΔN) = arctan(0.75) = 36.87°, and the distance is √(300² + 400²) = 500 m. Taking arctan(ΔN/ΔE) = 53.13° measures from east instead of north.From a station at (E, N) = (500.000, 500.000) m a point is radiated at a whole-circle bearing of 120° and a horizontal distance of 200.000 m. Its easting, in metres (to three decimal places), is ____.
Numerical answer — type the value.
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Answer: 673.205
E = E₀ + D sin θ = 500 + 200 sin 120° = 500 + 200 × 0.866025 = 673.205 m; N = 500 + 200 cos 120° = 500 − 100 = 400.000 m. Swapping sine and cosine gives an easting of 400 m, which would put the point west of the station.A plot has corners at (0, 0), (60, 0), (80, 40) and (20, 50) m, in order. Its area by the coordinate (shoelace) method, in m², is ____.
Numerical answer — type the value.
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Answer: 2800
Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ) = (0·0 − 60·0) + (60·40 − 80·0) + (80·50 − 20·40) + (20·0 − 0·50) = 0 + 2400 + 3200 + 0 = 5600. Area = 5600/2 = 2800 m². Forgetting the ½ gives 5600.An EDM is specified as ±(2 mm + 2 ppm). Taking the two parts as additive, the accuracy of a 1500 m measurement, in millimetres, is ± ____.
Numerical answer — type the value.
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Answer: 5
2 ppm of 1500 m = 2 × 10⁻⁶ × 1,500,000 mm = 3 mm; adding the constant 2 mm gives ±5 mm. The proportional part dominates on long lines and the constant part on short ones.A simple circular curve of radius 300 m joins two straights that deflect by 40°. The tangent length, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 109.19
T = R tan(Δ/2) = 300 tan 20° = 300 × 0.36397 = 109.19 m. Using tan 40° gives 251.73 m, which forgets that the tangent length uses half the deflection angle.For the same curve (R = 300 m, Δ = 40°), the point of intersection is at chainage 1250.00 m. The chainage of the second tangent point, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 1350.25
T₁ = 1250.00 − 109.19 = 1140.81 m. Curve length L = πRΔ/180 = π × 300 × 40/180 = 209.44 m. T₂ = T₁ + L = 1140.81 + 209.44 = 1350.25 m. Adding T to the intersection chainage (1359.19 m) measures along the tangent, which the road does not follow.For a circular curve of radius 300 m set out by Rankine's method, the tangential (deflection) angle for a 20 m chord, in minutes of arc (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 114.59
δ = 1718.9 c/R minutes = 1718.9 × 20/300 = 114.59′ = 1°54′35″. The constant comes from δ = c/(2R) radians × (180 × 60/π) = 1718.9 c/R. Omitting the factor 2 gives the full central angle of the chord, 229.18′.A tacheometer with k = 100 and C = 0 reads stadia hairs of 0.800, 1.400 and 2.000 on a vertical staff at a vertical angle of +8°. The horizontal distance, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 117.68
s = 2.000 − 0.800 = 1.200; D = k s cos²θ = 120 × cos²8° = 120 × 0.98063 = 117.68 m. Using cos θ instead of cos²θ gives 118.83 m; the second cosine comes from the staff being vertical rather than normal to the line of sight.For the same observation (s = 1.200, θ = +8°, central hair 1.400, k = 100, C = 0), the instrument station is at RL 200.00 m and the height of the instrument is 1.50 m. The RL of the staff station, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 216.64
V = k s sin 2θ/2 = 120 × sin 16°/2 = 60 × 0.27564 = 16.54 m. RL = 200.00 + 1.50 + 16.54 − 1.40 = 216.64 m. Forgetting to subtract the central-hair reading gives 218.04 m.Two instrument stations A and B, 20 m apart, lie in line with a chimney, A the nearer. The angles of elevation of the chimney top from A and B are 45° and 30°, observed at the same instrument height. The height of the top above the line of sight, in metres (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 27.32
With D the distance from A: h = D tan 45° = (D + 20) tan 30°, so D = 20 tan 30°/(tan 45° − tan 30°) = 11.547/0.42265 = 27.32 m and h = 27.32 m. Equivalently h = b tan α₁ tan α₂/(tan α₁ − tan α₂) = 20 × 1 × 0.57735/0.42265 = 27.32 m.Which of the following statements about stadia tacheometry are correct?
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Answer: A — An anallactic lens makes the additive constant zero; B — With a horizontal sight, D = k s + C; C — For an inclined sight with the staff vertical, the horizontal distance involves cos²θ
The anallactic lens moves the apex of the stadia cone to the instrument centre, so C = 0; D = ks + C is the horizontal-sight formula; and inclined sights on a vertical staff give D = ks cos²θ + C cos θ. k = f/i, focal length over stadia interval, normally 100 — the stated ratio is inverted.Which of the following are standard COGO or onboard functions of a total station?
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Answer: A — Resection to compute the instrument's coordinates from known points; B — Missing-line measurement between two remote prisms; C — Stakeout of design coordinates
Resection (free station), missing-line measurement and stakeout are all computed from angles and distances with coordinate geometry onboard. Integer ambiguity belongs to GNSS carrier-phase processing, not to a total station's EDM, which resolves its own range by several modulation wavelengths.