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Gale's table for traverse ABCDEA

Short answer

Adjusted lengths/bearings: AB 47.244 m, 41°00'22''; BC 42.509 m, 118°53'02''; CD 67.230 m, 190°01'02''; DE 49.355 m, 279°01'55''; EA 44.031 m, 349°49'10''. Coordinates: A (3051.086 N, 4943.474 E), B (3086.738 N, 4974.473 E), C (3066.205 N, 5011.694 E), D (3000.00 N, 5000.00 E), E (3007.748 N, 4951.257 E).

The following data refers to a traverse ABCDEA. Complete the Gale's table with final adjusted length and bearing of each line.

StationHorizontal AngleDistanceBearing
A128°47'38"47.23 m41°0'39" (AB)
B102°6'18"42.51 m
C108°52'33"67.25 m
D91°0'13"49.36 m
E109°12'08"44.02 m

Coordinate of D is (3000.00 mN, 5000 mE).

Method: check and adjust the angles, find the bearings from the known bearing, compute latitudes and departures, balance them by Bowditch's rule and compute the coordinates from the given station.

Step 1: Angular check and adjustment

For a closed traverse of n=5n=5 sides, the sum of interior angles must be (2n−4)×90∘=540∘(2n-4)\times 90^\circ=540^\circ.

Observed sum = 539°58'50", so the angular error is 0°01'10" (deficiency). The correction is distributed equally: 14.00" (added) per angle.

StationObserved angleCorrectionAdjusted angle
A128°47'38"+14.00"128°47'52"
B102°06'18"+14.00"102°06'32"
C108°52'33"+14.00"108°52'47"
D91°00'13"+14.00"91°00'27"
E109°12'08"+14.00"109°12'22"
Sum539°58'50"540°00'00"

Step 2: Bearings

The traverse is taken as clockwise (interior angles on the right), so

Bearing of next line=Bearing of previous line+180∘−interior angle at the station\text{Bearing of next line}=\text{Bearing of previous line}+180^\circ-\text{interior angle at the station}

Starting from AB = 41°00'39":

LineLength (m)Bearing
AB47.2341°00'39"
BC42.51118°54'07"
CD67.25190°01'20"
DE49.36279°00'53"
EA44.02349°48'31"

Check: carrying the bearing round the traverse brings back AB = 41°00'39", as given.

Step 3: Latitudes, departures and closing error

Lat=Lcos⁡θ,Dep=Lsin⁡θ\text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta
LineLength (m)BearingLatitudeDepartureCorr. latCorr. dep
AB47.23041°00'39"+35.639+30.992+0.0133+0.0066
BC42.510118°54'07"-20.546+37.215+0.0120+0.0059
CD67.250190°01'20"-66.224-11.704+0.0190+0.0094
DE49.360279°00'53"+7.734-48.750+0.0139+0.0069
EA44.020349°48'31"+43.325-7.789+0.0124+0.0061
Sum250.370-0.071-0.035+0.0707+0.0349

Closing error: eN=−0.071e_N=-0.071 m, eE=−0.035e_E=-0.035 m,

e=eN2+eE2=0.0789 m,relative error=eΣL=0.0789250.37=13175e=\sqrt{e_N^2+e_E^2}=0.0789\ \text{m},\qquad \text{relative error}=\frac{e}{\Sigma L}=\frac{0.0789}{250.37}=\frac{1}{3175}

Step 4: Bowditch correction and adjusted lines

Correction to a line =−(total error)×L/ΣL=-\text{(total error)}\times L/\Sigma L. The corrected latitudes and departures give the final length =Lat2+Dep2=\sqrt{\text{Lat}^2+\text{Dep}^2} and bearing =tan⁡−1(Dep/Lat)=\tan^{-1}(\text{Dep}/\text{Lat}):

LineAdj. latAdj. depAdj. length (m)Adj. bearing
AB+35.652+30.99947.24441°00'22"
BC-20.534+37.22142.509118°53'02"
CD-66.205-11.69467.230190°01'02"
DE+7.748-48.74349.355279°01'55"
EA+43.338-7.78344.031349°49'10"

Step 5: Coordinates (D = 3000.000 N, 5000.000 E)

StationNorthing (m)Easting (m)
A3051.0864943.474
B3086.7384974.473
C3066.2055011.694
D3000.0005000.000
E3007.7484951.257

Answer: the final adjusted lengths and bearings are as tabulated above, and the traverse closes with a relative error of 1 in 3175.

✓ Answer checked by Claude Opus 5.5 (high effort)

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