Question 3 of 7: Closed-Polygon Traverse — Compass-Rule Balancing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Geom-A2, Adjustment of Observations and Data Analysis. Three-hour, CLOSED-BOOK exam; 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. Least-squares adjustments are carried out in the Canadian survey frame (heights on CGVD2013, horizontal on NAD83(CSRS)); US-foot units are retained in Q4 wherever the printed question uses them.
Reference texts: Ghilani, Adjustment Computations: Spatial Data Analysis (6th ed., Wiley, 2017); Wolf & Ghilani, Elementary Surveying: An Introduction to Geomatics (15th ed., Pearson); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981).
Method note. Every weighted least-squares result below was formed from the normal equations $N\hat{x}=A^{\mathsf T}W\,\ell$; the reference (unit-weight) standard deviation is $s_0=\sqrt{v^{\mathsf T}Wv/(n-u)}$ with $n$ observations and $u$ unknowns.
Given. A three-sided closed loop A→B→C→A with computed departures and latitudes for each side and a perimeter $P=11\,988.689$ m.
Find. Linear misclosure and relative precision; then the compass-rule-adjusted departures, latitudes, lengths and bearings.
Figure 3.1 — Closed polygon traverse A–B–C (plan view, north up), plotted from the given departures and latitudes. The loop should return to A; the small gap is the linear misclosure.
Approach. Sum the departures and latitudes to get the closure components, combine them into a linear misclosure and a relative precision, then distribute the corrections in proportion to side length (compass / Bowditch rule) and recompute lengths and bearings from the balanced components.
Closure components. Summing the columns,
$$\textstyle\sum\text{Dep}=+0.073\ \text{m},\qquad \sum\text{Lat}=+0.166\ \text{m}$$
For a closed loop these should be zero, so the departure and latitude misclosures are $c_D=-0.073$ and $c_L=-0.166$ m.
Linear misclosure and relative precision.
$$e=\sqrt{(\Sigma\text{Dep})^2+(\Sigma\text{Lat})^2}=\sqrt{0.073^2+0.166^2}=\boxed{0.181\ \text{m}}$$
$$\text{Relative precision}=\frac{e}{P}=\frac{0.181}{11\,988.7}=\boxed{1{:}66\,100}$$
Compass-rule corrections. Each side's correction is proportional to its length: $\delta\text{Dep}_i=-(\Sigma\text{Dep})\,L_i/P$ and $\delta\text{Lat}_i=-(\Sigma\text{Lat})\,L_i/P$. Applying these and re-summing gives $\Sigma=0$ to $10^{-3}$ m.
Adjusted lengths and bearings. From the balanced components, $L'=\sqrt{\text{Dep}'^2+\text{Lat}'^2}$ and azimuth $\alpha=\operatorname{atan2}(\text{Dep}',\text{Lat}')$:
$$\boxed{L_{AB}=2120.23\,\text{m},\ \alpha_{AB}=288^\circ12'09''\ (\text{N}71^\circ47'51''\text{W})}$$
$$L_{BC}=4662.43\,\text{m},\ \alpha_{BC}=200^\circ48'46''\ (\text{S}20^\circ48'46''\text{W})$$
$$L_{CA}=5209.04\,\text{m},\ \alpha_{CA}=44^\circ48'17''\ (\text{N}44^\circ48'17''\text{E})$$
Check. The tabulated side lengths (e.g. $AB=2119.287$ m) differ by up to $\sim2.1$ m (BC; AB by $0.94$ m, CA by $0.001$ m) from $\sqrt{\text{Dep}^2+\text{Lat}^2}$ of the same row, so the printed departures/latitudes and lengths are not perfectly self-consistent. The misclosure and relative precision depend only on the coordinate sums and are unaffected; the adjusted lengths above are recomputed from the balanced departures and latitudes, which is the quantity the compass rule actually delivers.
