--- title: "Gale's table for traverse ABCDEA" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:gale-table-abcdea-2078b" source: https://www.bctengineer.com/q/semester_4/surveying_ii/gale-table-abcdea-2078b site: BCT Engineering --- # Gale's table for traverse ABCDEA Surveying II · Chapter 1: Traversing - Asked in: 2078 Baisakh, 10 marks - Video: [Compass Surveying - Problem No 3 ( Included Angles & Error Correction of a Closed Traverse )](https://www.youtube.com/watch?v=G8asq0xv2zw) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/gale-table-abcdea-2078b - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question The following data refers to a traverse ABCDEA. Complete the Gale's table with final adjusted length and bearing of each line. | Station | Horizontal Angle | Distance | Bearing | |---|---|---|---| | A | 128°47'38" | 47.23 m | 41°0'39" (AB) | | B | 102°6'18" | 42.51 m | | | C | 108°52'33" | 67.25 m | | | D | 91°0'13" | 49.36 m | | | E | 109°12'08" | 44.02 m | | Coordinate of D is (3000.00 mN, 5000 mE). ## 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). ## Answer 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 $n=5$ sides, the sum of interior angles must be $(2n-4)\times 90^\circ=540^\circ$. Observed sum = 539°58'50", so the angular error is 0°01'10" (deficiency). The correction is distributed equally: 14.00" (added) per angle. | Station | Observed angle | Correction | Adjusted angle | |---|---|---|---| | A | 128°47'38" | +14.00" | 128°47'52" | | B | 102°06'18" | +14.00" | 102°06'32" | | C | 108°52'33" | +14.00" | 108°52'47" | | D | 91°00'13" | +14.00" | 91°00'27" | | E | 109°12'08" | +14.00" | 109°12'22" | | **Sum** | 539°58'50" | | 540°00'00" | #### Step 2: Bearings The traverse is taken as clockwise (interior angles on the right), so $$ \text{Bearing of next line}=\text{Bearing of previous line}+180^\circ-\text{interior angle at the station} $$ Starting from AB = 41°00'39": | Line | Length (m) | Bearing | |---|---|---| | AB | 47.23 | 41°00'39" | | BC | 42.51 | 118°54'07" | | CD | 67.25 | 190°01'20" | | DE | 49.36 | 279°00'53" | | EA | 44.02 | 349°48'31" | Check: carrying the bearing round the traverse brings back AB = 41°00'39", as given. #### Step 3: Latitudes, departures and closing error $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ | Line | Length (m) | Bearing | Latitude | Departure | Corr. lat | Corr. dep | |---|---|---|---|---|---|---| | AB | 47.230 | 41°00'39" | +35.639 | +30.992 | +0.0133 | +0.0066 | | BC | 42.510 | 118°54'07" | -20.546 | +37.215 | +0.0120 | +0.0059 | | CD | 67.250 | 190°01'20" | -66.224 | -11.704 | +0.0190 | +0.0094 | | DE | 49.360 | 279°00'53" | +7.734 | -48.750 | +0.0139 | +0.0069 | | EA | 44.020 | 349°48'31" | +43.325 | -7.789 | +0.0124 | +0.0061 | | **Sum** | 250.370 | | -0.071 | -0.035 | +0.0707 | +0.0349 | Closing error: $e_N=-0.071$ m, $e_E=-0.035$ m, $$ e=\sqrt{e_N^2+e_E^2}=0.0789\ \text{m},\qquad \text{relative error}=\frac{e}{\Sigma L}=\frac{0.0789}{250.37}=\frac{1}{3175} $$ #### Step 4: Bowditch correction and adjusted lines Correction to a line $=-\text{(total error)}\times L/\Sigma L$. The corrected latitudes and departures give the final length $=\sqrt{\text{Lat}^2+\text{Dep}^2}$ and bearing $=\tan^{-1}(\text{Dep}/\text{Lat})$: | Line | Adj. lat | Adj. dep | Adj. length (m) | Adj. bearing | |---|---|---|---|---| | AB | +35.652 | +30.999 | 47.244 | 41°00'22" | | BC | -20.534 | +37.221 | 42.509 | 118°53'02" | | CD | -66.205 | -11.694 | 67.230 | 190°01'02" | | DE | +7.748 | -48.743 | 49.355 | 279°01'55" | | EA | +43.338 | -7.783 | 44.031 | 349°49'10" | #### Step 5: Coordinates (D = 3000.000 N, 5000.000 E) | Station | Northing (m) | Easting (m) | |---|---|---| | A | 3051.086 | 4943.474 | | B | 3086.738 | 4974.473 | | C | 3066.205 | 5011.694 | | D | 3000.000 | 5000.000 | | E | 3007.748 | 4951.257 | **Answer:** the final adjusted lengths and bearings are as tabulated above, and the traverse closes with a relative error of 1 in 3175. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 5 sides, observed sum is 539°58'50''. What is done? A. Subtract 14'' from each angle B. Add 14'' to each angle, since the sum should be 540° and the deficiency 1'10'' is spread equally C. Add 1'10'' to the last angle only D. Leave angles unchanged Answer: B. (2n-4) x 90° = 540°; 70'' / 5 = 14'' per angle added. (Key point: Angular adjustment) 2. For interior angles on the right (clockwise traverse), what is the bearing of the next line? A. Bearing of previous line + interior angle B. Bearing of previous line - 180° + interior angle C. Bearing of previous line + 180° - interior angle at the station D. Back bearing of previous line + interior angle Answer: C. BC = 41°00'39'' + 180° - 102°06'32'' = 118°54'07''. (Key point: Bearing rule) 3. How is the closing error distributed by Bowditch's rule? A. Correction to each line proportional to its length, with sign opposite to the error B. Equally among all lines C. Proportional to the line's latitude D. Added wholly to the longest line Answer: A. C = -(total error) x L/SigmaL. (Key point: Bowditch rule) 4. Closing error e = 0.0789 m and perimeter 250.37 m. What is the precision? A. About 1 in 317 B. About 1 in 31,750 C. About 1 in 250 D. About 1 in 3175 Answer: D. 250.37/0.0789 is about 3175. 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