NivaarExam PrepOfficial exam papers ↗

18-Geom-B1 Digital Terrain Modelling · December 2017

Question 5 of 12: Line-of-Sight Visibility Analysis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, December 2017 — 3 hours, closed book (one approved Casio or Sharp calculator permitted). The schedule prints TWELVE questions and states that "10 questions constitute a complete paper": Part A (Q1–Q8) is compulsory, Part B requires ONE of Q9–Q10, and Part C requires ONE of Q11–Q12, for a 100-mark paper. All twelve questions are solved below for completeness (a candidate would answer only Q1–Q8 plus one from each of Parts B and C).

Reference texts: Li, Zhu & Gold, Digital Terrain Modeling — Principles and Methodology (CRC Press, 2005); Maune (ed.), Digital Elevation Model Technologies and Applications: The DEM Users Manual (2nd ed., ASPRS, 2007); Wilson & Gallant, Terrain Analysis — Principles and Applications (Wiley, 2000); Wolf, Dewitt & Wilkinson, Elements of Photogrammetry with Applications in GIS (4th ed., McGraw-Hill, 2014); Isaaks & Srivastava, An Introduction to Applied Geostatistics (Oxford, 1989). Canadian datums throughout (NAD83(CSRS), CGVD2013).

Question 5: Line-of-Sight Visibility Analysis (8 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Viewer elevation $z_v = 10\text{ m}$; target elevation $z_t = 30\text{ m}$; horizontal distance viewer→obstacle $d_v = 200\text{ m}$; obstacle→target $d_o = 350\text{ m}$; obstacle top elevation $z_{ob} = 20\text{ m}$.

Find. Whether the straight sight line from viewer to target clears the obstacle — i.e. is the target visible?

Obstacle Z = 20 mViewer Z_v = 10 mTarget Z_t = 30 mZ_LOS = 17.27 md_v = 200 md_o = 350 m
Terrain profile along the viewer–target line. The dashed red line is the sight line; the obstacle top (20 m) sits above it at the obstacle station, so the target is hidden.

Approach. The sight line is a straight line in the vertical profile from viewer to target; interpolate its elevation at the obstacle’s horizontal position and compare with the obstacle top.

  1. Total horizontal distance viewer→target. $$D = d_v + d_o = 200 + 350 = 550\text{ m}$$
  2. Elevation of the sight line at the obstacle (linear interpolation). The line rises uniformly from $z_v$ to $z_t$ over $D$; at $d_v$ from the viewer, $$z_{\text{LOS}} = z_v + (z_t - z_v)\,\frac{d_v}{D} = 10 + (30-10)\frac{200}{550} = 10 + 7.27$$ $$\boxed{z_{\text{LOS}} = 17.27\text{ m}}$$
  3. Compare with the obstacle top and decide. The obstacle reaches $z_{ob}=20\text{ m}$, which exceeds the sight-line elevation there: $$z_{ob} - z_{\text{LOS}} = 20 - 17.27 = +2.73\text{ m} \;>\; 0$$ Because the obstacle protrudes 2.73 m above the sight line, it blocks the view. $$\boxed{\text{The target is NOT visible to the viewer.}}$$
QuantityValue
Total distance $D$$550$ m
Sight-line elevation at obstacle $z_{\text{LOS}}$$17.27$ m
Obstacle top $z_{ob}$$20$ m
Obstruction ($z_{ob}-z_{\text{LOS}}$)$+2.73$ m (above LOS)
VisibilityTarget NOT visible (blocked)
Check
Earth curvature and atmospheric refraction are neglected (appropriate over 550 m; the combined correction $\approx 0.0675\,D^2$ [km] is only $\sim\!0.02$ m here). Heights are taken as bare eye/target/obstacle-top elevations with no added instrument or target height.