18-Geom-A2 Adjustment of Observations · December 2018
Question 5 of 7: Traverse Departures, Latitudes, Closure and Balancing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 04-Geom-A2, Adjustment of Observations and Data Analysis. Three-hour, CLOSED-BOOK exam; an approved Casio or Sharp calculator is permitted. Format: seven questions are given and any five (20 marks each) constitute a complete paper — all seven are solved below for completeness. Adjustments are carried out in the Canadian survey frame (heights on CGVD2013, horizontal work on NAD83(CSRS)); US-foot units are retained in Q6 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, 2015); Mikhail & Gracie, Analysis and Adjustment of Survey Measurements (Van Nostrand, 1981).
Method note. Every weighted least-squares result below is 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, and station precisions come from $\Sigma_{xx}=s_0^2 N^{-1}$.
Question 5: Traverse Departures, Latitudes, Closure and Balancing (20 marks)
Given. A closed five-sided traverse A–B–C–D–E–A with the length and (already-adjusted) azimuth of each course listed above; perimeter $\textstyle\sum L=9272.012$ m.
Find. Departures and latitudes of every course, the linear error of closure and relative precision, and the balanced (Bowditch/compass-rule) departures and latitudes.
Figure 5.1 — Closed five-sided traverse plotted from the computed departures and latitudes (north up); the loop returns to A within 96 mm.
Approach. Resolve each course into departure $=L\sin(\text{Az})$ and latitude $=L\cos(\text{Az})$, sum the columns to get the misclosures, form the linear error of closure and precision, then distribute the misclosures by the compass rule in proportion to course length.
Departures and latitudes. With $\text{Dep}=L\sin\text{Az}$, $\text{Lat}=L\cos\text{Az}$ the five courses give the values in the table below (e.g. AB: $\text{Dep}=1352.562\sin245^\circ16'24''=-1228.550$, $\text{Lat}=1352.562\cos245^\circ16'24''=-565.763$).
Misclosures. Summing the columns,
$$\textstyle\sum\text{Dep}=+0.049\ \text{m},\qquad \sum\text{Lat}=+0.082\ \text{m}.$$
Error of closure and precision.
$$e=\sqrt{(\textstyle\sum\text{Dep})^2+(\sum\text{Lat})^2}=\sqrt{0.049^2+0.082^2}=0.096\ \text{m},$$
$$\text{precision}=\frac{\sum L}{e}=\frac{9272.012}{0.096}\approx\boxed{1{:}96{,}900}.$$
Balance by the compass rule. Each course receives a correction proportional to its length, $c_{\text{Dep},i}=-\dfrac{L_i}{\sum L}\sum\text{Dep}$ and $c_{\text{Lat},i}=-\dfrac{L_i}{\sum L}\sum\text{Lat}$. The balanced values (which now sum to zero, so the traverse closes exactly) are collected below.
