What are the two stages of adjusting a link traverse?

Balance the angles using the known last bearing, then balance coordinates using the known last station

Balance coordinates first, then the lengths

Adjust only the lengths, then the bearings

Apply the transit rule then the compass rule

How is the correction applied to the bearings of successive lines?

The k-th bearing receives k times the per-angle correction

Only the last bearing is corrected

Corrections are proportional to line length

How does Bowditch's rule distribute the closing error?

Proportional to the length of each line

Proportional to the latitude only

Proportional to the angle at each station

2075 Bhadra · 10 marks

Link traverse Q-m1-m2-X, Gale's table

Short answer

Angular misclosure +16" corrected by -4" per angle. Bowditch adjustment gave linear misclosure 1:3209. Final coordinates: m1 (63192.407 N, 7470.863 E), m2 (63239.826 N, 7414.348 E).

A link traverse Qm₁m₂X was run between two known stations Q and X where Q and X are starting point for the legs QP and XY respectively and following observations were taken.

Known stn	Easting (m)	Northing (m)	Unknown stn	Mean Hz. Angle	Length (m)
P	7432.293	63115.158
Q	4792.189	63148.201	m₁	∠PQm₁ = 93°06'51"	Qm₁ = 49.085
X	7439.604	63321.618	m₂	∠Qm₁m₂ = 155°45'25"	m₁m₂ = 73.787
Y	7358.014	63348.321		∠m₁m₂X = 247°09'37"	m₂X = 85.590
				∠m₂XY = 90°58'47"

Calculate the coordinates in Gale's Table of m₁ and m₂, adjust them by Bowditch's rule and compute easting and northing coordinates.

Data note: the easting of Q is printed as 4792.189 m; it is read as 7492.189 m, since only this value agrees with the line PQ from the coordinates of P (7432.293 E, 63115.158 N). Reference bearings from the coordinates (E, N):

\theta_{PQ}=\tan^{-1}\frac{7492.189-7432.293}{63148.201-63115.158}=61^\circ 06'56'',\qquad \theta_{XY}=\tan^{-1}\frac{7358.014-7439.604}{63348.321-63321.618}=288^\circ 07'21''

A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station.

Known data: P (63115.158 N, 7432.293 E), Q (63148.201 N, 7492.189 E), X (63321.618 N, 7439.604 E), Y (63348.321 N, 7358.014 E).

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of the line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line QP (back bearing 241°06'56") and using the observed angles, the bearing of the last line comes out as 288°07'36", while its known value is 288°07'21" (X-Y from coordinates).

Angular misclosure = computed − known. Using the unrounded bearings from the coordinates (PQ = 61°06'56.4", XY = 288°07'20.6"), it is 288°07'36.4" − 288°07'20.6" = +15.8" ≈ +0°00'16".

The error is spread equally over the 4 angles: correction per angle ≈ -0°00'04". The k-th bearing therefore receives k times this correction.

Station	Angle	Observed	Corrected	Bearing of line ahead
Q	P-Q-m1	93°06'51"	93°06'47"	334°13'43"
m1	Q-m1-m2	155°45'25"	155°45'21"	309°59'04"
m2	m1-m2-X	247°09'37"	247°09'33"	17°08'38"
X	m2-X-Y	90°58'47"	90°58'43"	288°07'21"

Step 2: Latitudes and departures

\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta

Line	Length (m)	Bearing	Latitude	Departure
Qm1	49.085	334°13'43"	+44.2029	-21.3412
m1m2	73.787	309°59'04"	+47.4142	-56.5369
m2X	85.590	17°08'38"	+81.7871	+25.2294
Sum	208.462		+173.4041	-52.6487

Starting from Q (63148.201 N, 7492.189 E), the unadjusted coordinates of X are (63321.605 N, 7439.540 E), whereas the known coordinates are (63321.618 N, 7439.604 E).

Step 3: Linear misclosure

\begin{aligned} e_N&=63321.605-63321.618=-0.013\ \text{m}\\ e_E&=7439.540-7439.604=-0.064\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.065\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.065}{208.462}=\frac{1}{3209} \end{aligned}

Step 4: Adjustment by Bowditch's rule

Bowditch's rule: the correction to each line is proportional to its length:

C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}

Line	Corr. lat	Corr. dep	Adj. lat	Adj. dep
Qm1	+0.0030	+0.0150	+44.2059	-21.3262
m1m2	+0.0046	+0.0225	+47.4187	-56.5144
m2X	+0.0053	+0.0261	+81.7924	+25.2555
Sum	+0.0129	+0.0637	+173.4170	-52.5850

Step 5: Final coordinates

Station	Computed N	Computed E	Final N (m)	Final E (m)
Q	63148.201	7492.189	63148.201	7492.189
m1	63192.404	7470.848	63192.407	7470.863
m2	63239.818	7414.311	63239.826	7414.348
X	63321.605	7439.540	63321.618	7439.604

The computed coordinates of X now equal its known coordinates, so the traverse is balanced.

Answer: m1 (63192.407 N, 7470.863 E); m2 (63239.826 N, 7414.348 E) (linear misclosure 0.065 m, 1 in 3209).

✓ Answer checked by Claude Opus 5.5 (high effort)

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