Question 5 of 7: Departures, Latitudes, Misclosure and Relative Precision of a Closed Traverse

Question 5: 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:

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

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.

1. Departures and latitudes. With departure $=L\sin(\text{Az})$ and latitude $=L\cos(\text{Az})$:

   Course	Departure (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.