Question 2 of 7: True/False Statements with Corrections
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geom-A1 Surveying. Closed-book; any non-communicating calculator permitted. Format: seven questions are given and any five (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 2: True/False Statements with Corrections (20 marks)
Given. Ten statements spanning GNSS height systems, geodetic surface relationships, horizontal- and vertical-curve stationing, bearing/azimuth reversal, 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 two quantitative statements (4 and 10) are settled by direct computation.
Statement 1 — F. Three satellites determine only the three position unknowns and leave the receiver-clock bias unsolved. Correction: a minimum of four satellites is required — three for the $X,Y,Z$ position and a fourth to solve the receiver-clock error.
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 $h$ (relative to the ellipsoid), not orthometric heights referred to the geoid; a geoid model $N$ is needed to convert them.
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."
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 $\boxed{6^\circ22'10''}$.
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 consistent throughout.
Statement 6 — F. For a parabola the tangent (vertical) offset from the tangent grows with the square of the distance. Correction: the tangent offsets vary as the square of the distance from the point of tangency ($y \propto x^2$), not linearly with distance.
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 $L \neq 2T$.
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).
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.
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.
Statement
Verdict
Correction (if false)
1 — three satellites fix position
F
needs four (fourth solves the clock bias)
2 — GPS heights w.r.t. geoid
F
w.r.t. the ellipsoid (ellipsoidal height $h$)
3 — geoidal $=$ ellipsoidal $+$ orthometric
F
$h = H + N$ (ellipsoidal $=$ orthometric $+$ geoid)