--- title: "Link traverse B to M, coordinates X and Y" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:link-traverse-bowditch-x-y-between" source: https://www.bctengineer.com/q/semester_4/surveying_ii/link-traverse-bowditch-x-y-between site: BCT Engineering --- # Link traverse B to M, coordinates X and Y Surveying II · Chapter 1: Traversing - Asked in: 2073 Bhadra, 12 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/link-traverse-bowditch-x-y-between - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question A link traverse was run between main traverse stations "B" and "M". Clockwise angles of link traverse taken were: ABX = 135°00'30", BXY = 98°07'55", XYM = 209°45'02" and YMN = 64°39'33" respectively. Lengths of link traverse are: BX = 31.612, XY = 22.260 m and YM = 36.153 m respectively. The coordinates of main traverse stations given are: A (42.360 mN, 18.820 mE), B (20.000 mN, 30.000 mE) and M (50.000 mN, 100.000 mE), N (70.600 mN, 65.6670 mE) respectively. Calculate the final coordinates of stations X and Y. Adjust any misclosure by Bowditch method. ## Short answer Angular misclosure +75” distributed equally. Bowditch adjustment of linear misclosure 0.043 m (1:2117). Final coordinates: X (10.014 N, 59.980 E), Y (29.932 N, 69.929 E). ## Answer Reference bearings from the coordinates of the control stations: $$ \theta_{AB}=\tan^{-1}\frac{30.000-18.820}{20.000-42.360}=153^\circ 26'06'',\qquad \theta_{MN}=\tan^{-1}\frac{65.667-100.000}{70.600-50.000}=300^\circ 57'50'' $$ (The signs of the differences place AB in the south-east and MN in the north-west quadrant.) At B the reference line is BA, with bearing 333°26'06". A link traverse is checked at both ends, so the adjustment has two stages: (1) balance the angles using the known bearing of the last line, (2) balance the coordinates using the known position of the last station. Known data: A (42.360 N, 18.820 E), B (20.000 N, 30.000 E), M (50.000 N, 100.000 E), N (70.600 N, 65.667 E). #### Step 1: Bearings and angular misclosure The angles are measured clockwise. Bearing of a line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line BA and using the observed angles, the bearing of the last line comes out as 300°59'06", while its known value is 300°57'50" (M-N from the coordinates). Angular misclosure = computed − known = 300°59'05.8" − 300°57'50.4" = +75.4" ≈ +0°01'15". The error is spread equally over the 4 angles: correction per angle = −0°00'19". The k-th bearing therefore receives k times this correction. | Station | Angle | Observed | Corrected | Bearing of line ahead | |---|---|---|---|---| | B | A-B-X | 135°00'30" | 135°00'11" | 108°26'17" | | X | B-X-Y | 98°07'55" | 98°07'36" | 26°33'53" | | Y | X-Y-M | 209°45'02" | 209°44'43" | 56°18'36" | | M | Y-M-N | 64°39'33" | 64°39'14" | 300°57'50" | #### Step 2: Latitudes and departures $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ | Line | Length (m) | Bearing | Latitude | Departure | |---|---|---|---|---| | BX | 31.612 | 108°26'17" | -9.9982 | +29.9892 | | XY | 22.260 | 26°33'53" | +19.9100 | +9.9549 | | YM | 36.153 | 56°18'36" | +20.0540 | +30.0812 | | **Sum** | 90.025 | | +29.9658 | +70.0253 | Starting from B (20.000 N, 30.000 E), the unadjusted coordinates of M are (49.966 N, 100.025 E), whereas the known coordinates are (50.000 N, 100.000 E). #### Step 3: Linear misclosure $$ \begin{aligned} e_N&=49.966-50.000=-0.034\ \text{m}\\ e_E&=100.025-100.000=+0.025\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.043\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.043}{90.025}=\frac{1}{2117} \end{aligned} $$ #### Step 4: Adjustment by Bowditch's rule Bowditch's rule: the correction to each line is proportional to its length: $$ C_{lat}=-e_N\frac{L}{\Sigma L},\qquad C_{dep}=-e_E\frac{L}{\Sigma L} $$ | Line | Corr. lat | Corr. dep | Adj. lat | Adj. dep | |---|---|---|---|---| | BX | +0.0120 | -0.0089 | -9.9862 | +29.9804 | | XY | +0.0085 | -0.0062 | +19.9185 | +9.9486 | | YM | +0.0137 | -0.0101 | +20.0677 | +30.0710 | | **Sum** | +0.0342 | -0.0253 | +30.0000 | +70.0000 | #### Step 5: Final coordinates | Station | Computed N | Computed E | Final N (m) | Final E (m) | |---|---|---|---|---| | B | 20.000 | 30.000 | 20.000 | 30.000 | | X | 10.002 | 59.989 | 10.014 | 59.980 | | Y | 29.912 | 69.944 | 29.932 | 69.929 | | M | 49.966 | 100.025 | 50.000 | 100.000 | The computed coordinates of M now equal its known coordinates, so the traverse is balanced. **Answer:** X (10.014 N, 59.980 E); Y (29.932 N, 69.929 E) (linear misclosure 0.043 m, 1 in 2117). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At B, which line is used as the starting reference for carrying bearings? A. BA, with bearing equal to the bearing of AB reversed by 180° (333°26'06") B. AB with bearing 153°26'06" used directly as the line behind C. BX D. MN Answer: A. The first clockwise angle is A-B-X, measured from BA, so the back line BA (AB ± 180°) is the reference. (Key point: Reference line at the start) 2. The computed bearing of MN is 300°59'06" and the known is 300°57'50". With 4 angles, what correction applies per angle? A. +19" to each angle B. -19" to each angle C. -76" to the last angle only D. -5' to each angle Answer: B. Misclosure = computed - known = +1'16" (about 76"), so -76"/4 = -19" per angle, and the k-th bearing gets k times that. (Key point: Angular misclosure) 3. In Bowditch's rule, the correction to the latitude of a line is proportional to what? A. The size of the latitude of the line B. The length of the line, C = -e_N x L/ΣL C. The number of stations D. The departure of the line Answer: B. Bowditch distributes errors in proportion to line length, and the sign is opposite to the misclosure. (Key point: Bowditch rule) 4. Computed M is (49.966 N, 100.025 E) and the known M is (50.000 N, 100.000 E). What are e_N and e_E, and what is the result of the final adjustment? A. e_N = +0.034, e_E = -0.025; M stays at the computed value B. e_N = -0.034, e_E = +0.025; corrections sum to +0.034 and -0.025 so computed M equals known M C. e_N = -0.034, e_E = +0.025; corrections sum to -0.034 and +0.025 D. e_N = 0.043 and e_E = 0; only the latitudes are corrected Answer: B. Misclosure = computed - known, and corrections are of opposite sign so they sum to +0.034 in latitude and -0.025 in departure, closing on M. 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