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18-Geom-A1 Surveying · December 2016

Question 1 of 7: True/False Statements with Corrections

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

Notes on this paper

National Exams — December 2016 — 04-Geom-A1 Surveying. Closed-book; any non-communicating calculator permitted. Format: Question 1 is compulsory and any four of Questions 2–7 (20 marks each) constitute a complete paper — all seven are solved below for completeness. Where a datum is implied, elevations are referenced to the Canadian vertical frame (CGVD2013) and azimuths to NAD83(CSRS); US-foot stationing is retained wherever the printed question uses it.

Reference texts: Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley); Hofmann-Wellenhof et al., GNSS — Global Navigation Satellite Systems (Springer, 2008).

Question 1: True/False Statements with Corrections (20 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. Ten statements spanning traverse-angle tolerance, GNSS height systems, geodetic-surface relationships, degree of curve, bearing/azimuth reversal, trigonometric levelling, horizontal- and vertical-curve stationing, traverse classification, and random-error propagation.

Find. A True/False verdict for each statement, with the correcting statement supplied for every false one.

Approach. Judge each statement against the governing definition or formula; where it is false, state the smallest correction that makes it true. The three quantitative statements (1, 4 and 10) are settled by direct computation.

  1. Statement 1 — T. The Leica TPS1205 is a $5''$ instrument (its DIN 18723 / ISO 17123-3 angular standard deviation is $5''$). The allowable angular misclosure of a closed traverse is $c\sqrt{n}$, where $c$ is the instrument accuracy and $n$ the number of measured angles. For the nine angles of a 9-sided traverse, $5''\sqrt{9} = 5''(3) = \boxed{15''}$, so the stated tolerance is correct.
  2. Statement 2 — F. GPS observes geometric range to the satellites, so it delivers heights referred to the reference ellipsoid. Correction: GPS heights are ellipsoidal heights $h$ (measured with respect to the ellipsoid), not orthometric heights referred to the geoid; a geoid model $N$ is required to convert them.
  3. Statement 3 — F. The correct relation among the three surfaces is $h = H + N$. Correction: the ellipsoidal height $h$ equals the orthometric height $H$ plus the geoidal height (undulation) $N$ — not "geoidal height $=$ ellipsoidal $+$ orthometric."
  4. Statement 4 — F. By the chord definition, $D_c = 2\arcsin\!\left(\dfrac{50}{R}\right) = 2\arcsin\!\left(\dfrac{50}{900}\right) = 6^\circ22'10''$, whereas by the arc definition $D_a = \dfrac{5729.578}{R} = \dfrac{5729.578}{900} = 6^\circ21'58''$. The printed value is the arc-definition result. Correction: $6^\circ21'58''$ is the degree of curve by the arc definition; the chord-definition value is $\boxed{6^\circ22'10''}$.
  5. Statement 5 — T. Azimuth $235^\circ$ lies in the third quadrant, so its bearing is $S(235^\circ-180^\circ)W = S55^\circ W$ ✓; the back azimuth is $235^\circ-180^\circ = 55^\circ$, whose bearing is $N55^\circ E$ ✓. The statement is internally consistent.
  6. Statement 6 — F. The instrument height $h_i$ is needed only when the total station is set up over one of the two points, where $\Delta\text{Elev} = h_i + S\sin\alpha - h_r$ ($S$ slope distance, $\alpha$ vertical angle, $h_r$ reflector height). In the usual "leapfrog" trigonometric levelling the instrument is set up at an arbitrary, unmarked point roughly midway between the two successive points and sights a reflector of the same height on each. The line-of-sight elevation is then common to both sights and cancels: $\Delta\text{Elev}_{1\to2} = S_2\sin\alpha_2 - S_1\sin\alpha_1$, so $h_i$ is never measured. Correction: the surveyor does not need to know the instrument height — setting up between the two points (as in differential levelling) eliminates it; $h_i$ is required only for a set-up over a station.
  7. Statement 7 — F. The first half is right (PC $=$ PI $-\,T$) but the second is wrong: the PT is reached from the PC along the arc, not out to the PI and back. Correction: the station of the PT equals the station of the PC plus the curve length $L$ (PT $=$ PC $+\,L$), which differs from PI $+\,T$ because $L \neq 2T$.
  8. Statement 8 — F. An equal-tangent vertical curve is symmetric about the PVI, so each end lies half the length away. Correction: PVC $=$ PVI $-\,L/2$ and PVT $=$ PVI $+\,L/2$ (half the curve length, not the full length).
  9. Statement 9 — F. A traverse that begins on one known point and ends on a different known point is still closed (a closed connecting or link traverse), because it can be checked against control. Correction: such a traverse is a closed (connecting/link) traverse; an open traverse is one that ends at a point of unknown position with no closure check.
  10. Statement 10 — T. Random (accidental) errors propagate as the root-sum-square, so for $n$ equal measurements the total is $E = e\sqrt{n} = 0.006\sqrt{36} = 0.006(6) = \boxed{\pm0.036\ \text{m}}$. The statement is correct.
StatementVerdictCorrection (if false)
1 — 9-sided traverse, TPS1205, $\pm15''$T— ($5''\sqrt{9}=15''$)
2 — GPS heights w.r.t. geoidFw.r.t. the ellipsoid (ellipsoidal height $h$)
3 — geoidal $=$ ellipsoidal $+$ orthometricF$h = H + N$ (ellipsoidal $=$ orthometric $+$ geoid)
4 — chord-def $D = 6^\circ21'58''$Fthat is the arc def; chord def $=6^\circ22'10''$
5 — back azimuth/bearing of BAT—
6 — must know instrument heightFnot needed: set up midway between the points and $h_i$ cancels
7 — PT $=$ PI $+ T$FPT $=$ PC $+ L$
8 — PVC/PVT $=$ PVI $\mp L$FPVI $\mp L/2$ (half length)
9 — point-to-point is "open"Fit is a closed (connecting) traverse
10 — total error $\pm0.036$ mT—
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