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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
  1. In the deflection angle method for horizontal curve layout, what is the total deflection angle from PC to PT if the central angle Δ = 40°?

    Answer: 20°

    The total deflection angle from the tangent to the final chord at the PT equals Δ/2 = 40°/2 = 20°.

  2. Using the coordinate method, what is the area of a triangle with vertices at (0, 0), (6, 0), and (3, 4) in feet?

    Answer: 12 sq ft

    Area = 0.5 × |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| = 0.5 × |0 + 24 + 0| = 12 sq ft.

  3. A six-sided closed traverse has a total angular misclosure of 1°30". Distributing the error equally, what is the correction applied to each interior angle?

    Answer: 15"

    1°30" = 90"; correction per angle = 90" / 6 = 15" per interior angle.

  4. What does a low Geometric Dilution of Precision (GDOP) value indicate in GPS surveying?

    Answer: Favorable satellite geometry yielding higher positional accuracy

    A low GDOP indicates that satellites are well spread across the sky, which minimizes the amplification of ranging errors into position errors.

  5. A benchmark has an elevation of 450.25 ft. A backsight of 4.78 ft and a foresight of 6.12 ft are recorded at the next point. What is the elevation of the foresight point?

    Answer: 448.91 ft

    HI = 450.25 + 4.78 = 455.03 ft; Elevation = HI − FS = 455.03 − 6.12 = 448.91 ft.

  6. Which of the following is classified as a systematic error in tape surveying?

    Answer: Steel tape length error due to temperature variation from the calibration temperature

    Temperature-induced tape length change is systematic because it follows a predictable formula (αLΔT) and can be fully corrected.

  7. A horizontal curve has R = 500 ft and central angle Δ = 60°. What is the tangent length T?

    Answer: 288.7 ft

    T = R × tan(Δ/2) = 500 × tan(30°) = 500 × 0.5774 = 288.7 ft.