Surveying and Geomatics Flashcards
7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Surveying and Geomatics flashcards as text
A surveyor measures a slope distance of 200.0 ft along a 5% grade. What is the horizontal distance?
Answer: 199.75 ft
HD = SD × cos(θ) where θ = arctan(0.05) = 2.862°; HD = 200 × cos(2.862°) ≈ 199.75 ft.
Convert the bearing S 45° W to an azimuth.
Answer: 225°
S 45° W means 180° (due south) plus 45° toward west, giving an azimuth of 225°.
A backsight of 8.42 ft is taken on a benchmark with elevation 100.00 ft. What is the height of instrument (HI)?
Answer: 108.42 ft
HI = Benchmark Elevation + Backsight = 100.00 + 8.42 = 108.42 ft.
A closed traverse has a linear misclosure of 0.42 ft and a total perimeter of 1,260 ft. What is the precision ratio?
Answer: 1:3,000
Precision = 1:(perimeter/misclosure) = 1:(1,260/0.42) = 1:3,000.
A horizontal curve has R = 600 ft and a central angle Δ = 30°. What is the arc length?
Answer: 314.2 ft
L = (Δ/360°) × 2πR = (30/360) × 2π × 600 = 314.16 ft ≈ 314.2 ft.
A vertical curve connects a +3% grade to a −2% grade over a length of 500 ft. What is the rate of grade change per station (100 ft)?
Answer: 1.0%/sta
Rate = |g₂ − g₁| / (L/100) = |−2 − 3| / (500/100) = 5/5 = 1.0%/station.
In stadia surveying on a level sight, the stadia interval is 1.50 ft, the stadia constant K = 100, and instrument constant C = 1.0 ft. What is the horizontal distance?
Answer: 151 ft
H = K × s + C = 100 × 1.50 + 1.0 = 151.0 ft.