Geomatics Engineering

Syllabus

Principles of surveying; errors and their adjustment; maps (scale, coordinate systems); distance and angle measurement (levelling, theodolite traversing, triangulation); total station; horizontal and vertical curves; photogrammetry and remote sensing (scale, flying height, relief displacement); basics of GIS and GPS.

Key concepts and formulas

Levelling. Height of instrument HI = RL of benchmark + backsight; RL of a point = HI − foresight. Check: Σ BS − Σ FS = last RL − first RL. Corrections over a distance d (in km), in metres:

• curvature: −0.0785d²
• refraction: +0.0112d²
• combined: −0.0673d²

Reciprocal levelling eliminates these errors and collimation error.

Errors. Probable error, weights inversely proportional to the variance; the most probable value of a quantity measured with weights is the weighted mean.

Traversing. Latitude = L cos θ, departure = L sin θ (θ = whole circle bearing). The closing error is √((ΣL)² + (ΣD)²). The Bowditch rule distributes it in proportion to the line lengths; the transit rule in proportion to the latitudes and departures. The sum of interior angles of an n-sided closed traverse is (2n − 4) × 90°.

Tacheometry. Horizontal distance D = Ks + C (stadia, line of sight horizontal), usually with K = 100 and C = 0. With an inclined line of sight at angle θ: D = Ks cos²θ.

Simple circular curves (radius R, deflection angle Δ):

• tangent length T = R tan(Δ/2)
• curve length l = πRΔ/180
• long chord = 2R sin(Δ/2)
• degree of curve (30 m arc): D ≈ 1719/R

Photogrammetry. Scale of a vertical photograph S = f/(H − h), where f is the focal length, H the flying height and h the ground elevation. Relief displacement d = rh/H, where r is the radial distance of the image from the principal point.

Remote sensing and GPS. Spatial, spectral, radiometric and temporal resolution; GPS needs signals from at least four satellites for a 3D position (x, y, z and the receiver clock error).

Worked example

Find the combined curvature and refraction correction for a sight of 2 km. For a curve with R = 400 m and Δ = 60°, find the tangent length and the curve length. Find the photo scale for f = 150 mm, H = 3000 m and h = 600 m.

import math

d = 2.0
print(f"combined correction = {-0.0673 * d * d:.4f} m")
R, D = 400, 60
print(f"T = {R * math.tan(math.radians(D / 2)):.3f} m, curve length = {math.pi * R * D / 180:.3f} m")
f, H, h = 0.150, 3000, 600
s = f / (H - h)
print(f"scale = 1 : {1 / s:.0f}")

combined correction = -0.2692 m
T = 230.940 m, curve length = 418.879 m
scale = 1 : 16000

Practice questions