Question 4 of 7: Closed Link (Connecting) Traverse
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.
Question 4: Closed Link (Connecting) Traverse (20 marks)
Given. A traverse that begins at the fixed station A and closes on the fixed station E, oriented by two azimuth marks; five angles to the right and four measured lengths (US survey feet).
Find. Preliminary and adjusted azimuths; departure/latitude misclosures and relative precision; compass-rule-balanced coordinates of B, C, D; and final lengths and azimuths of AB, BC, CD, DE.
Figure 4.1 — Link traverse from fixed A to fixed E (triangles), oriented by backsight AzMk1 and foresight AzMk2. Arrows show the run of the traverse; BS/FS mark the orienting rays.
Check (orientation). The printed adjusted azimuth $310^\circ17'20''$ sits on the line between AzMk1 and A. Taken as az A→AzMk1, propagating the angles from it places the traverse to the south-west, which contradicts the fixed coordinates (E lies north-east of A). Using the reverse, az A→AzMk1 $=130^\circ17'20''$, reproduces both the fixed closing azimuth E→AzMk2 and the known coordinates of E, so the backsight azimuth is taken as $130^\circ17'20''$ (the tabulated value is the mark-to-station direction AzMk1→A, read top-to-bottom like the closing entry E→AzMk2). The two readings give the same angular misclosure, but only $130^\circ17'20''$ lands the traverse on E — the other misses it by more than 1000 ft.
Approach. Carry azimuths through the traverse by adding each angle-right to the back azimuth of the previous line; check against the fixed closing azimuth, distribute the angular misclosure equally, then reduce to departures/latitudes, balance by compass rule against the fixed endpoint coordinates, and recover coordinates, lengths and azimuths.
Preliminary azimuths. Starting from az A→AzMk1 $=130^\circ17'20''$ and applying $\alpha_{\text{next}}=\alpha_{\text{back}}+\text{angle right}$ (mod $360^\circ$):
$$\alpha_{AB}=42^\circ57'20'',\;\alpha_{BC}=130^\circ24'48'',\;\alpha_{CD}=37^\circ27'19'',\;\alpha_{DE}=327^\circ02'58''$$
Angular misclosure. Carrying one more angle to the closing ray gives az E→AzMk2 $=57^\circ32'27''$ against the fixed $57^\circ32'42''$, a misclosure of $-15''$ over five angles, i.e. $+3''$ applied to each angle.
Adjusted azimuths. With $+3''$ per angle,
$$\boxed{\alpha_{AB}=42^\circ57'23'',\ \alpha_{BC}=130^\circ24'54'',\ \alpha_{CD}=37^\circ27'28'',\ \alpha_{DE}=327^\circ03'10''}$$
which close exactly on the fixed azimuth.
Departures, latitudes and linear misclosure. Using $\text{Dep}=L\sin\alpha,\ \text{Lat}=L\cos\alpha$ and comparing the sums with the fixed coordinate differences $\Delta X=2058.94,\ \Delta Y=2233.58$ ft:
$$c_D=+0.242\ \text{ft},\quad c_L=+0.277\ \text{ft},\quad e=\sqrt{c_D^2+c_L^2}=0.368\ \text{ft}$$
$$\text{Relative precision}=\frac{0.368}{5357.77}=\boxed{1{:}14\,600}$$
Compass-rule balancing and coordinates. Distributing $-c_D,-c_L$ in proportion to length and accumulating from A gives
$$\boxed{B=(643\,822.573,\ 90\,066.063)}$$
$$\boxed{C=(644\,873.207,\ 89\,171.305)},\quad \boxed{D=(645\,621.160,\ 90\,147.554)\ \text{ft}}$$
and the accumulated position closes on the fixed $E=(644\,905.59,\ 91\,251.50)$.
Final lengths and azimuths. Recomputed from the balanced coordinates (inverse of each line):
$$L_{AB}=1432.14,\;L_{BC}=1380.01,\;L_{CD}=1229.84,\;L_{DE}=1315.58\ \text{ft}$$
$$\alpha_{AB}=42^\circ57'23'',\;\alpha_{BC}=130^\circ25'08'',\;\alpha_{CD}=37^\circ27'27'',\;\alpha_{DE}=327^\circ02'56''$$
