--- title: "Bowditch link traverse XABCY (X 1877.51 N)" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:bowditch-link-traverse-xabcy-1877" source: https://www.bctengineer.com/q/semester_4/surveying_ii/bowditch-link-traverse-xabcy-1877 site: BCT Engineering --- # Bowditch link traverse XABCY (X 1877.51 N) Surveying II · Chapter 1: Traversing - Asked in: 2070 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/bowditch-link-traverse-xabcy-1877 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question Balance the following coordinates and compute the total coordinates for a link traverse XABCY using Bowditch's rule. The given coordinates of X and Y are (1877.51 mN, 1290.20 mE) and (1626.50 mN, 1578.87 mE) respectively. If permissible closing error is 1:500, justify your assessment. Other observed data are as follows: | Lines | XA | AB | BC | CY | |---|---|---|---|---| | Length (m) | 120.00 | 111.50 | 132.40 | 97.60 | | Bearing | 135°00' | 119°30' | 175°00' | 77°30' | ## Short answer Compute latitudes and departures from bearings, find closing error against known Y, then apply Bowditch corrections proportional to line length. Adjusted coordinates are A (1792.532 N, 1375.039 E), B (1737.511 N, 1472.071 E), C (1605.477 N, 1483.595 E); closing error 0.483 m (1:955) is within 1:500. ## Answer 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 $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ (The values for each line are listed in the table in Step 3.) #### Step 2: Closing error Required change in coordinates from X to Y: $$ \begin{aligned} \Delta N_{true}&=1626.500-1877.510=-251.010\ \text{m}\\ \Delta E_{true}&=1578.870-1290.200=288.670\ \text{m} \end{aligned} $$ Computed: $\Sigma \text{Lat}=-250.530$ m, $\Sigma \text{Dep}=288.723$ m. $$ \begin{aligned} e_N&=-250.530-(-251.010)=0.480\ \text{m}\\ e_E&=288.723-(288.670)=0.053\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.483\ \text{m} \end{aligned} $$ Perimeter (total length) $\Sigma L=461.50$ m, so the relative error is $e/\Sigma L = 1:955$. Assessment: the permissible error is $461.50/500=0.923$ m, and the actual error of 0.483 m (1:955) is smaller, i.e. better than the permissible 1:500. The traverse is therefore acceptable and the error can be distributed. #### Step 3: Bowditch correction $$ C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L} $$ | Line | Length (m) | WCB | Latitude | Departure | Corr. lat | Corr. dep | Adj. lat | Adj. dep | |---|---|---|---|---|---|---|---|---| | XA | 120.00 | 135°00' | -84.853 | +84.853 | -0.125 | -0.014 | -84.978 | +84.839 | | AB | 111.50 | 119°30' | -54.905 | +97.045 | -0.116 | -0.013 | -55.021 | +97.032 | | BC | 132.40 | 175°00' | -131.896 | +11.539 | -0.138 | -0.015 | -132.034 | +11.524 | | CY | 97.60 | 77°30' | +21.125 | +95.286 | -0.102 | -0.011 | +21.023 | +95.275 | | **Sum** | 461.50 | | -250.530 | +288.723 | -0.480 | -0.053 | -251.010 | +288.670 | #### Step 4: Adjusted coordinates Start from X and add the adjusted latitude and departure of each line: | Station | Northing (m) | Easting (m) | |---|---|---| | X | 1877.510 | 1290.200 | | A | 1792.532 | 1375.039 | | B | 1737.511 | 1472.071 | | C | 1605.477 | 1483.595 | | Y | 1626.500 | 1578.870 | Check: the computed coordinates of Y agree with the given Y (1626.500 N, 1578.870 E). **Answer:** adjusted coordinates A (1792.532 N, 1375.039 E), B (1737.511 N, 1472.071 E), C (1605.477 N, 1483.595 E); closing error 0.483 m (1:955), within the permissible 1:500. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the closing error of the link traverse XABCY found? A. Compare ΣLat and ΣDep with the true differences in coordinates between X and Y B. Add all latitudes and expect zero C. Compare the sum of bearings with 360° D. Compare ΣL with the straight-line XY distance only Answer: A. Since X and Y are known, ΣLat should equal N_Y - N_X (-251.010) and ΣDep should equal E_Y - E_X (288.670). (Key point: Required coordinate change) 2. With e_N = 0.480 m, e_E = 0.053 m and ΣL = 461.50 m, what is the relative error? A. About 1:955 B. About 1:500 C. About 1:96 D. About 1:9550 Answer: A. e = sqrt(0.480^2 + 0.053^2) = 0.483 m, and 461.5/0.483 is about 955. (Key point: Closing error value) 3. Is the traverse acceptable if the permissible error is 1:500? A. No, 1:955 is worse than 1:500 B. Yes, 1:955 is better (smaller error) than 1:500, so it is distributed C. Cannot be decided D. Only after doubling the lengths Answer: B. A larger denominator means a smaller relative error, so 1:955 is within the 1:500 limit. (Key point: Acceptance test) 4. What are the signs and basis of Bowditch corrections here? A. Positive, proportional to latitude B. Negative on latitudes and departures (opposite to the positive errors), proportional to line length C. Equal for all lines D. Positive on departures only Answer: B. e_N and e_E are positive, so corrections are -e x L/ΣL, giving a total of -0.480 and -0.053. (Key point: Bowditch corrections) 5. How is the adjusted traverse verified? A. Adjusted coordinates computed from X reach the given coordinates of Y B. The bearings sum to 360° C. All lines have equal length D. Latitudes sum to zero Answer: A. After correction ΣLat and ΣDep equal the true differences, so the computed Y equals the given Y. 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