Question 5 of 9: Error of Misclosure of a Closed-Loop Traverse

Question 5: Error of Misclosure of a Closed-Loop Traverse (10 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 four-course closed loop A–B–C–D–A with the azimuth and horizontal length of each course. Starting coordinates are $X_A = 1\,984\,400.612$ m and $Y_A = 518\,430.033$ m, with $X$ as easting and $Y$ as northing.

Find. The linear error of misclosure, its direction, and the relative precision of the traverse.

Approach. The azimuths are given directly, so no angle balancing is needed. Resolve each course into a departure ($L\sin\theta$) and a latitude ($L\cos\theta$). For a closed loop both sums must be zero, so the sums themselves are the misclosure components.

1. Departures and latitudes. For example, course BC gives $\text{Dep} = 164.988\sin 94^\circ03' = +164.576$ m and $\text{Lat} = 164.988\cos 94^\circ03' = -11.653$ m. All four courses:

   Course	Azimuth	Length (m)	Departure (m)	Latitude (m)	Dep. corr’n (m)	Lat. corr’n (m)
   AB	0°42'	372.242	+4.548	+372.214	+0.381	+0.235
   BC	94°03'	164.988	+164.576	−11.653	+0.169	+0.104
   CD	183°04'	242.458	−12.971	−242.111	+0.248	+0.153
   DA	232°51'	197.165	−157.152	−119.069	+0.202	+0.125
   Σ		976.853	−0.999	−0.618	+0.999	+0.618

2. Misclosure components. $$ \Sigma\text{Dep} = -0.999\text{ m}, \qquad \Sigma\text{Lat} = -0.618\text{ m}. $$
3. Linear error of misclosure. $$ e = \sqrt{(\Sigma\text{Dep})^2 + (\Sigma\text{Lat})^2} = \sqrt{(-0.999)^2 + (-0.618)^2} \;\Rightarrow\; \boxed{e = 1.175\text{ m}} $$ The misclosure vector (from A to the computed closing point A′) has azimuth $\tan^{-1}(\Sigma\text{Dep}/\Sigma\text{Lat}) = 238^\circ16'00''$, pointing south-west.
4. Relative precision. $$ \text{precision} = \frac{e}{\Sigma L} = \frac{1.175}{976.853} \;\Rightarrow\; \boxed{1 : 830} $$
5. Coordinates (showing the misclosure). Carrying the unadjusted departures and latitudes from A shows directly that the loop fails to return to A by $(-0.999,\ -0.618)$ m:

   Station	X (m), unadjusted	Y (m), unadjusted
   A (start)	1984400.612	518430.033
   B	1984405.160	518802.247
   C	1984569.736	518790.595
   D	1984556.765	518548.484
   A′ (computed close)	1984399.613	518429.415

Assessment. A precision of about 1:830 is well below what a closed traverse normally achieves (1:5000 or better for most engineering work, and 1:10 000 or better for control). The azimuths are recorded only to the nearest minute, and one minute over a 372 m course is already 0.11 m, but that alone cannot explain an error of more than a metre. Before adjusting, a surveyor would look for a blunder. A useful check is to find the course whose azimuth lies near the misclosure direction ($238^\circ16'00''$) or its reverse, which here is DA at $232^\circ51'$, since a distance blunder shows up along that line. If the result is accepted, the compass-rule corrections in the table ($-\Sigma\text{Dep}\,L_i/\Sigma L$ and $-\Sigma\text{Lat}\,L_i/\Sigma L$) remove the misclosure in proportion to course length.