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18-Geom-A1 Surveying · December 2016

Question 4 of 7: Departures, Latitudes, Misclosure and Relative Precision of a Closed Traverse

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 4: Departures, Latitudes, Misclosure and Relative Precision of a Closed Traverse (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. A closed traverse $A$–$B$–$C$–$D$–$E$–$A$ with course lengths and azimuths:

CourseLength (m)Azimuth
AB1352.562$245^\circ16'24''$
BC1999.670$147^\circ06'37''$
CD1329.127$95^\circ33'20''$
DE2427.328$23^\circ45'21''$
EA2163.325$274^\circ01'46''$

Find. (1) departure and latitude of each course, (2) linear misclosure $e$, (3) relative precision $e/\text{perimeter}$, and the balanced (adjusted) departures and latitudes.

ABCDEAN
Figure 3 — Closed traverse plotted from the computed departures and latitudes (north arrow shown); it returns almost exactly to $A$.

Approach. For each course, departure $=L\sin(\text{Az})$ and latitude $=L\cos(\text{Az})$; the algebraic column sums are the closure components, whose resultant is the linear misclosure, and its ratio to the perimeter is the relative precision. The compass (Bowditch) rule then distributes the misclosure in proportion to course length.

  1. Departures and latitudes. With departure $=L\sin(\text{Az})$ and latitude $=L\cos(\text{Az})$:
    CourseDeparture (m)Latitude (m)
    AB$-1228.550$$-565.763$
    BC$+1085.868$$-1679.157$
    CD$+1322.884$$-128.674$
    DE$+977.825$$+2221.662$
    EA$-2157.977$$+152.015$
    $\Sigma$$+0.049$$+0.082$
    The column sums are the closure components $C_D = +0.049$ m and $C_L = +0.082$ m.
  2. Linear misclosure. The resultant of the two closure components: $$e = \sqrt{C_D^2 + C_L^2} = \sqrt{0.049^2 + 0.082^2} = \boxed{0.096\ \text{m}}$$
  3. Relative precision. With perimeter $\Sigma L = 9272.012$ m, $$\frac{e}{\Sigma L} = \frac{0.096}{9272.012} \approx \frac{1}{96{,}900} \approx \boxed{1{:}96{,}000}$$ This comfortably exceeds the $1{:}5{,}000$–$1{:}10{,}000$ typical of ordinary boundary traverses, indicating high-quality field work.
  4. Balancing (compass / Bowditch rule). Each correction is $-C\,(L_i/\Sigma L)$ applied to the departure and latitude, so the corrections sum to $-C_D$ and $-C_L$ and drive both column totals to zero. The per-course corrections (metres) are:
    Course$\delta$Dep$\delta$LatAdj. DepAdj. Lat
    AB$-0.007$$-0.012$$-1228.557$$-565.775$
    BC$-0.011$$-0.018$$+1085.857$$-1679.175$
    CD$-0.007$$-0.012$$+1322.877$$-128.686$
    DE$-0.013$$-0.021$$+977.812$$+2221.641$
    EA$-0.011$$-0.019$$-2157.988$$+151.996$
    $\Sigma$$-0.049$$-0.082$$0.000$$0.000$
    The adjusted departures and latitudes now close exactly, so the traverse is balanced.
QuantityValue
Closure in departure $C_D$$+0.049$ m
Closure in latitude $C_L$$+0.082$ m
Linear misclosure $e$$0.096$ m
Perimeter $\Sigma L$$9272.012$ m
Relative precision$\approx 1{:}96{,}000$
Balanced departures / latitudesclose to $0.000$ / $0.000$ m (compass rule)