--- title: "Link traverse Gale's table M2 to M8" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:link-traverse-gale-m2-m8" source: https://www.bctengineer.com/q/semester_4/surveying_ii/link-traverse-gale-m2-m8 site: BCT Engineering --- # Link traverse Gale's table M2 to M8 Surveying II · Chapter 1: Traversing - Asked in: 2077 Chaitra, 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/link-traverse-gale-m2-m8 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question The following observations were recorded during link traversing. The independent coordinates of M2 is (1000 mN, 1000 mE) and M8 is (1095 mN, 1042 mE). Compute minor traverse in Gale's Table, indicating balancing of coordinates by Bowditch method. | Station | Horizontal Angle (degrees) | Leg | Length (m) | Bearing (Degrees) | |---|---|---|---|---| | M2 | 110 | M1-M2 | - | 105 | | m1 | 200 | M2-m1 | 35.50 | | | m2 | 90 | m1-m2 | 55.00 | | | M8 | 250 | m2-M8 | 42.35 | | | | | M8-M9 | - | 36 | ## Short answer Angular misclosure of -1° was distributed equally (+0°15' per angle); Bowditch method adjusted coordinates. Final coordinates: m1 (1028.951 mN, 1020.494 mE), m2 (1060.041 mN, 1065.829 mE); linear misclosure 0.150 m (1:883). ## Answer 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: M2 = (1000 N, 1000 E), M8 = (1095 N, 1042 E); bearing M1-M2 = 105°, bearing M8-M9 = 36°. #### 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 M2-M1 (back bearing 285°00', since M1-M2 = 105°) and using the observed angles, the bearing of the last line comes out as 35°00', while its known value is 36°00' (M8-M9). Angular misclosure = computed − known = 35°00' − 36°00' = -1°00'. The error is spread equally over the 4 angles: correction per angle = +0°15'. The k-th bearing therefore receives k times this correction. | Station | Angle | Observed | Corrected | Bearing of line ahead | |---|---|---|---|---| | M2 | M1-M2-m1 | 110°00' | 110°15' | 35°15' | | m1 | M2-m1-m2 | 200°00' | 200°15' | 55°30' | | m2 | m1-m2-M8 | 90°00' | 90°15' | 325°45' | | M8 | m2-M8-M9 | 250°00' | 250°15' | 36°00' | #### Step 2: Latitudes and departures $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ | Line | Length (m) | Bearing | Latitude | Departure | |---|---|---|---|---| | M2m1 | 35.500 | 35°15' | +28.9908 | +20.4887 | | m1m2 | 55.000 | 55°30' | +31.1523 | +45.3269 | | m2M8 | 42.350 | 325°45' | +35.0061 | -23.8348 | | **Sum** | 132.850 | | +95.1492 | +41.9808 | Starting from M2 (1000.000 N, 1000.000 E), the unadjusted coordinates of M8 are (1095.149 N, 1041.981 E), whereas the known coordinates are (1095.000 N, 1042.000 E). #### Step 3: Linear misclosure $$ \begin{aligned} e_N&=1095.149-1095.000=+0.149\ \text{m}\\ e_E&=1041.981-1042.000=-0.019\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.150\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.150}{132.850}=\frac{1}{883} \end{aligned} $$ #### 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 | |---|---|---|---|---| | M2m1 | -0.0399 | +0.0051 | +28.9509 | +20.4938 | | m1m2 | -0.0618 | +0.0079 | +31.0906 | +45.3349 | | m2M8 | -0.0476 | +0.0061 | +34.9585 | -23.8287 | | **Sum** | -0.1492 | +0.0192 | +95.0000 | +42.0000 | #### Step 5: Final coordinates | Station | Computed N | Computed E | Final N (m) | Final E (m) | |---|---|---|---|---| | M2 | 1000.000 | 1000.000 | 1000.000 | 1000.000 | | m1 | 1028.991 | 1020.489 | 1028.951 | 1020.494 | | m2 | 1060.143 | 1065.816 | 1060.041 | 1065.829 | | M8 | 1095.149 | 1041.981 | 1095.000 | 1042.000 | The computed coordinates of M8 now equal its known coordinates, so the traverse is balanced. **Answer:** m1 (1028.951 N, 1020.494 E); m2 (1060.041 N, 1065.829 E) (linear misclosure 0.150 m, 1 in 883). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What two stages are used to balance a link traverse? A. Balance angles using the known last bearing, then balance coordinates using the known last station B. Balance coordinates first, then angles C. Only angles are balanced D. Only the first and last lines are balanced Answer: A. Both ends are known so each gives a check. (Key point: Two-stage adjustment) 2. Computed last bearing is 35°00' and known is 36°00', with 4 angles. What corrections apply? A. +1° to the last angle only B. -15' per angle C. +15' per angle; the k-th bearing receives k x 15', so the last gets +60' D. +4' per angle Answer: C. Misclosure = 35° - 36° = -1°; correction is +60'/4 = +15' per angle, cumulative along bearings. (Key point: Angular correction) 3. Unadjusted M8 is (1095.149, 1041.981) and known M8 is (1095.000, 1042.000). What is e_N, e_E? A. e_N = -0.149 m, e_E = +0.019 m B. e_N = +0.149 m, e_E = -0.019 m C. e_N = 0.149 m, e_E = 0.019 m D. e_N = 0.019 m, e_E = 0.149 m Answer: B. Error = computed - known. (Key point: Linear misclosure) 4. With e_N = +0.149, what is the latitude correction to M2m1 (35.5 m of 132.85 m)? A. About +0.040 m B. About -0.149 m C. About +0.149 m D. About -0.040 m Answer: D. C = -e_N x L/SigmaL = -0.149 x 35.5/132.85 = -0.0398 m. 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