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Link traverse B to M, coordinates X and Y

Short answer

Angular misclosure +75” distributed equally. Bowditch adjustment of linear misclosure 0.043 m (1:2117). Final coordinates: X (10.014 N, 59.980 E), Y (29.932 N, 69.929 E).

A link traverse was run between main traverse stations "B" and "M". Clockwise angles of link traverse taken were: ABX = 135°00'30", BXY = 98°07'55", XYM = 209°45'02" and YMN = 64°39'33" respectively. Lengths of link traverse are: BX = 31.612, XY = 22.260 m and YM = 36.153 m respectively. The coordinates of main traverse stations given are: A (42.360 mN, 18.820 mE), B (20.000 mN, 30.000 mE) and M (50.000 mN, 100.000 mE), N (70.600 mN, 65.6670 mE) respectively. Calculate the final coordinates of stations X and Y. Adjust any misclosure by Bowditch method.

Reference bearings from the coordinates of the control stations:

θAB=tan⁡−130.000−18.82020.000−42.360=153∘26′06′′,θMN=tan⁡−165.667−100.00070.600−50.000=300∘57′50′′\theta_{AB}=\tan^{-1}\frac{30.000-18.820}{20.000-42.360}=153^\circ 26'06'',\qquad \theta_{MN}=\tan^{-1}\frac{65.667-100.000}{70.600-50.000}=300^\circ 57'50''

(The signs of the differences place AB in the south-east and MN in the north-west quadrant.) At B the reference line is BA, with bearing 333°26'06".

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: A (42.360 N, 18.820 E), B (20.000 N, 30.000 E), M (50.000 N, 100.000 E), N (70.600 N, 65.667 E).

Step 1: Bearings and angular misclosure

The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line BA and using the observed angles, the bearing of the last line comes out as 300°59'06", while its known value is 300°57'50" (M-N from the coordinates).

Angular misclosure = computed − known = 300°59'05.8" − 300°57'50.4" = +75.4" ≈ +0°01'15".

The error is spread equally over the 4 angles: correction per angle = −0°00'19". The k-th bearing therefore receives k times this correction.

StationAngleObservedCorrectedBearing of line ahead
BA-B-X135°00'30"135°00'11"108°26'17"
XB-X-Y98°07'55"98°07'36"26°33'53"
YX-Y-M209°45'02"209°44'43"56°18'36"
MY-M-N64°39'33"64°39'14"300°57'50"

Step 2: Latitudes and departures

Lat=Lcos⁡θ,Dep=Lsin⁡θ\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta
LineLength (m)BearingLatitudeDeparture
BX31.612108°26'17"-9.9982+29.9892
XY22.26026°33'53"+19.9100+9.9549
YM36.15356°18'36"+20.0540+30.0812
Sum90.025+29.9658+70.0253

Starting from B (20.000 N, 30.000 E), the unadjusted coordinates of M are (49.966 N, 100.025 E), whereas the known coordinates are (50.000 N, 100.000 E).

Step 3: Linear misclosure

eN=49.966−50.000=−0.034 meE=100.025−100.000=+0.025 me=eN2+eE2=0.043 m,eΣL=0.04390.025=12117\begin{aligned} e_N&=49.966-50.000=-0.034\ \text{m}\\ e_E&=100.025-100.000=+0.025\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.043\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.043}{90.025}=\frac{1}{2117} \end{aligned}

Step 4: Adjustment by Bowditch's rule

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

Clat=−eNLΣL,Cdep=−eELΣLC_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}
LineCorr. latCorr. depAdj. latAdj. dep
BX+0.0120-0.0089-9.9862+29.9804
XY+0.0085-0.0062+19.9185+9.9486
YM+0.0137-0.0101+20.0677+30.0710
Sum+0.0342-0.0253+30.0000+70.0000

Step 5: Final coordinates

StationComputed NComputed EFinal N (m)Final E (m)
B20.00030.00020.00030.000
X10.00259.98910.01459.980
Y29.91269.94429.93269.929
M49.966100.02550.000100.000

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

Answer: X (10.014 N, 59.980 E); Y (29.932 N, 69.929 E) (linear misclosure 0.043 m, 1 in 2117).

✓ Answer checked by Claude Opus 5.5 (high effort)

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