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Bowditch balancing of link traverse XABCY

Short answer

Compute latitudes and departures from given lengths and bearings. Misclosure from known coordinates is 133.196 m (1:3.5), exceeding permissible 1:500. For adjustment, apply Bowditch correction C = -error × (line length / ΣL). Adjusted coordinates: A (511.884 N, 254.808 E), B (488.781 N, 346.933 E), C (394.648 N, 352.630 E).

Balance the coordinates of a link traverse XABCY using Bowditch's rule. The coordinates of stations X and Y are (562.510 N, 175.250 E) and (443.610 N, 443.610 E) respectively and permissible closing error is 1:500. Other observed data are given below:

LinesXAABBCCY
Length (m)120.00111.50132.4097.60
Bearings135°00'119°30'175°00'77°30'

Method: compute latitudes and departures from the given WCBs, find the closing error against the known coordinates of Y, check it against the permissible limit, then distribute it by Bowditch's rule and compute the adjusted coordinates.

Step 1: Latitudes and departures

Lat=Lcos⁡θ,Dep=Lsin⁡θ\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta

Step 2: Closing error

Required change in coordinates from X to Y:

ΔNtrue=443.610−562.510=−118.900 mΔEtrue=443.610−175.250=268.360 m\begin{aligned} \Delta N_{true}&=443.610-562.510=-118.900\ \text{m}\\ \Delta E_{true}&=443.610-175.250=268.360\ \text{m} \end{aligned}

Computed: ΣLat=−250.530\Sigma \text{Lat}=-250.530 m, ΣDep=288.723\Sigma \text{Dep}=288.723 m.

eN=−250.530−(−118.900)=−131.630 meE=288.723−(268.360)=20.363 me=eN2+eE2=133.196 m\begin{aligned} e_N&=-250.530-(-118.900)=-131.630\ \text{m}\\ e_E&=288.723-(268.360)=20.363\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=133.196\ \text{m} \end{aligned}

Perimeter (total length) ΣL=461.50\Sigma L=461.50 m, so the relative error is e/ΣL=1:3.5e/\Sigma L = 1:3.5.

The permissible closing error is 1:500, i.e. at most 0.923 m, but the error here is 133.196 m (about 1:3.5). The given coordinates of Y and the field data do not agree, so in a real survey this traverse would be rejected and re-measured. The Bowditch adjustment is shown below only to complete the procedure.

Step 3: Bowditch correction

Clat=−eNLΣL,Cdep=−eELΣLC_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L}
LineLength (m)WCBLatitudeDepartureCorr. latCorr. depAdj. latAdj. dep
XA120.00135°00'-84.853+84.853+34.227-5.295-50.626+79.558
AB111.50119°30'-54.905+97.045+31.802-4.920-23.103+92.125
BC132.40175°00'-131.896+11.539+37.763-5.842-94.133+5.697
CY97.6077°30'+21.125+95.286+27.838-4.307+48.962+90.980
Sum461.50-250.530+288.723+131.630-20.363-118.900+268.360

Step 4: Adjusted coordinates

Start from X and add the adjusted latitude and departure of each line:

StationNorthing (m)Easting (m)
X562.510175.250
A511.884254.808
B488.781346.933
C394.648352.630
Y443.610443.610

Check: the computed coordinates of Y agree with the given Y (443.610 N, 443.610 E).

Answer: adjusted coordinates A (511.884 N, 254.808 E), B (488.781 N, 346.933 E), C (394.648 N, 352.630 E); closing error 133.196 m (1:3.5).

✓ Answer checked by Claude Opus 5.5 (high effort)

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