--- title: "Check Gale's table ABCDEA, D at 2500 N" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:gale-table-abcdea-2081c" source: https://www.bctengineer.com/q/semester_4/surveying_ii/gale-table-abcdea-2081c site: BCT Engineering --- # Check Gale's table ABCDEA, D at 2500 N Surveying II · Chapter 1: Traversing - Asked in: 2081 Chaitra, 10 marks - 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-2081c - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question The following data refers to a traverse ABCDEA. Check the accuracy of work and complete the Gale's table up-to co-ordinate calculation. Coordinate of D is (2500.00 mN, 4500 mE). | Station | Leg | Horizontal Angle | Distance | Bearing | |---|---|---|---|---| | A | - | 128°48' | - | | | B | AB | 102°7' | 47.23 m | 40° | | C | BC | 108°52' | 42.51 m | | | D | CD | 91°0' | 67.25 m | | | E | DE | 109°13' | 49.36 m | | | A | EA | - | 44.02 m | | ## Short answer Interior angles sum 540° exactly, no correction. Bearings: BC 117°53', CD 189°01', DE 278°01', EA 348°48'. Lat/dep errors eN=-0.054 m, eE=-0.034 m, relative error 1/3928, acceptable. After Bowditch adjustment, coordinates: A (2550.086 N, 4442.585 E), B (2586.276 N, 4472.950 E), C (2566.404 N, 4510.530 E), D (2500.000 N, 4500.000 E), E (2506.894 N, 4451.129 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 = 540°00', so there is no angular error and no correction is needed. | Station | Observed angle | Correction | Adjusted angle | |---|---|---|---| | A | 128°48' | 0°00' | 128°48' | | B | 102°07' | 0°00' | 102°07' | | C | 108°52' | 0°00' | 108°52' | | D | 91°00' | 0°00' | 91°00' | | E | 109°13' | 0°00' | 109°13' | | **Sum** | 540°00' | | 540°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 = 40°00': | Line | Length (m) | Bearing | |---|---|---| | AB | 47.23 | 40°00' | | BC | 42.51 | 117°53' | | CD | 67.25 | 189°01' | | DE | 49.36 | 278°01' | | EA | 44.02 | 348°48' | Check: carrying the bearing round the traverse brings back AB = 40°00', 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 | 40°00' | +36.180 | +30.359 | +0.0102 | +0.0064 | | BC | 42.510 | 117°53' | -19.881 | +37.575 | +0.0092 | +0.0057 | | CD | 67.250 | 189°01' | -66.419 | -10.540 | +0.0145 | +0.0091 | | DE | 49.360 | 278°01' | +6.884 | -48.878 | +0.0106 | +0.0067 | | EA | 44.020 | 348°48' | +43.182 | -8.550 | +0.0095 | +0.0060 | | **Sum** | 250.370 | | -0.054 | -0.034 | +0.0540 | +0.0338 | Closing error: $e_N=-0.054$ m, $e_E=-0.034$ m, $$ e=\sqrt{e_N^2+e_E^2}=0.0637\ \text{m},\qquad \text{relative error}=\frac{e}{\Sigma L}=\frac{0.0637}{250.37}=\frac{1}{3928} $$ This is well within the usual limit for theodolite traversing (about 1 in 3000 or better), so the work is acceptable. #### Step 4: Bowditch correction and adjusted lines Correction to a line $=-\text{(total error)}\times L/\Sigma L$ (tabulated above). 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 | +36.190 | +30.365 | 47.242 | 40°00' | | BC | -19.872 | +37.580 | 42.511 | 117°52' | | CD | -66.404 | -10.530 | 67.234 | 189°01' | | DE | +6.894 | -48.871 | 49.355 | 278°02' | | EA | +43.191 | -8.544 | 44.028 | 348°49' | #### Step 5: Coordinates (D = 2500.000 N, 4500.000 E) Starting from D and adding the adjusted latitudes and departures in the order DE, EA, AB, BC, CD: | Station | Northing (m) | Easting (m) | |---|---|---| | A | 2550.086 | 4442.585 | | B | 2586.276 | 4472.950 | | C | 2566.404 | 4510.530 | | D | 2500.000 | 4500.000 | | E | 2506.894 | 4451.129 | Check: C plus line CD returns to D (2500.000 N, 4500.000 E). **Answer:** relative error 1 in 3928 (accurate work); coordinates A (2550.086 N, 4442.585 E), B (2586.276 N, 4472.950 E), C (2566.404 N, 4510.530 E), D (2500.000 N, 4500.000 E), E (2506.894 N, 4451.129 E). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a closed traverse of 5 sides the sum of interior angles should be: A. 360° B. 720° C. 900° D. 540° Answer: D. (2n - 4) x 90° = 6 x 90° = 540°; the observed sum here is also 540°, so no angular correction. (Key point: Angular check) 2. For a clockwise traverse with interior angles, the bearing of the next line is: A. Previous bearing + 180° - interior angle B. Previous bearing + interior angle C. Previous bearing - 180° - interior angle D. Previous bearing - interior angle Answer: A. Interior angles on the right: next bearing = previous + 180° - interior angle; e.g. BC = 40° + 180° - 102°07' = 117°53'. (Key point: Bearing propagation) 3. Closing error e = 0.0637 m and perimeter 250.37 m. The precision is: A. 1 in 393 B. 1 in 3928 C. 1 in 39280 D. 1 in 250 Answer: B. 250.37/0.0637 = about 3928, so the relative error is 1/3928. (Key point: Relative error) 4. Coordinates of the other stations are computed by: A. Adding raw latitudes without correction B. Using bearings only C. Adding adjusted latitudes and departures of successive lines to a known station (D) D. Taking the average of observed angles Answer: C. After Bowditch balancing, independent coordinates are got from the known station D by adding adjusted lat/dep successively. 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