--- title: "Link traverse A to X, Y by transit method" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:link-traverse-transit-a-x-y" source: https://www.bctengineer.com/q/semester_4/surveying_ii/link-traverse-transit-a-x-y site: BCT Engineering --- # Link traverse A to X, Y by transit method Surveying II · Chapter 1: Traversing - Asked in: 2073 Magh, 10 marks - Video: [Traverse Calculations // Surveying](https://www.youtube.com/watch?v=mMDcODSe_ic) — engiNERDing - 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-transit-a-x-y - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question A link traverse was run between stations A and X. The co-ordinates of the controlling stations at the ends of the traverse are as follows: | Stn. | E(m) | N(m) | Clockwise Angle | Length (m) | |---|---|---|---|---| | A | 1769.15 | 2094.72 | 115°37'00" | A-1 = 208.26 | | B | 1057.28 | 2492.39 | | | | X | 2334.85 | 1747.32 | 173°31'00" | 4-X = 224.79 | | Y | 2995.85 | 1616.18 | | | | 1 | | | 168°19'10" | | | 2 | | | 281°12'40" | 1-2 = 193.47 | | 3 | | | 242°53'40" | 2-3 = 326.71 | | 4 | | | 80°26'20" | 3-4 = 309.15 | Calculate coordinates of stations 1, 2, 3 and 4. Adjust any mis-closure by transit method. ## Short answer Angular misclosure ‒2’08” distributed equally over 6 angles. Linear misclosure 0.247 m (1:5121) adjusted by transit rule. Coordinates: 1 (2214.74 N, 1939.40 E), 2 (2355.93 N, 2071.72 E), 3 (2090.46 N, 2262.17 E), 4 (1815.65 N, 2120.64 E). ## Answer Coordinates are listed as E, N in the question; in the working below they are written as (N, E). The reference bearings come from the coordinates of the control stations: $$ \begin{aligned} \theta_{AB}&=\tan^{-1}\frac{E_B-E_A}{N_B-N_A}=\tan^{-1}\frac{-711.87}{+397.67}\ \Rightarrow\ 299^\circ 11'20''\\ \theta_{XY}&=\tan^{-1}\frac{E_Y-E_X}{N_Y-N_X}=\tan^{-1}\frac{+661.00}{-131.14}\ \Rightarrow\ 101^\circ 13'18'' \end{aligned} $$ 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 (2094.72 N, 1769.15 E), B (2492.39 N, 1057.28 E), X (1747.32 N, 2334.85 E), Y (1616.18 N, 2995.85 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 AB (299°11'20" from the coordinates) and using the observed angles, the bearing of the last line comes out as 101°11'10", while its known value is 101°13'18" (X-Y from the coordinates). Angular misclosure = computed − known = 101°11'10" − 101°13'18" = -0°02'08". The error is spread equally over the 6 angles: correction per angle = +0°00'21". The k-th bearing therefore receives k times this correction. | Station | Angle | Observed | Corrected | Bearing of line ahead | |---|---|---|---|---| | A | B-A-1 | 115°37'00" | 115°37'21" | 54°48'41" | | 1 | A-1-2 | 168°19'10" | 168°19'31" | 43°08'13" | | 2 | 1-2-3 | 281°12'40" | 281°13'01" | 144°21'14" | | 3 | 2-3-4 | 242°53'40" | 242°54'01" | 207°15'15" | | 4 | 3-4-X | 80°26'20" | 80°26'41" | 107°41'56" | | X | 4-X-Y | 173°31'00" | 173°31'21" | 101°13'18" | #### Step 2: Latitudes and departures $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ | Line | Length (m) | Bearing | Latitude | Departure | |---|---|---|---|---| | A1 | 208.260 | 54°48'41" | +120.0137 | +170.2027 | | 12 | 193.470 | 43°08'13" | +141.1795 | +132.2838 | | 23 | 326.710 | 144°21'14" | -265.4949 | +190.3993 | | 34 | 309.150 | 207°15'15" | -274.8293 | -141.5718 | | 4X | 224.790 | 107°41'56" | -68.3398 | +214.1500 | | **Sum** | 1262.380 | | -347.4708 | +565.4639 | Starting from A (2094.720 N, 1769.150 E), the unadjusted coordinates of X are (1747.249 N, 2334.614 E), whereas the known coordinates are (1747.320 N, 2334.850 E). #### Step 3: Linear misclosure $$ \begin{aligned} e_N&=1747.249-1747.320=-0.071\ \text{m}\\ e_E&=2334.614-2334.850=-0.236\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.247\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.247}{1262.380}=\frac{1}{5121} \end{aligned} $$ #### Step 4: Adjustment by the transit rule The transit rule: the correction to latitude (or departure) of each line is proportional to the size of that latitude (or departure): $$ C_{lat}=-e_N\frac{|\text{lat}|}{\Sigma|\text{lat}|},\qquad C_{dep}=-e_E\frac{|\text{dep}|}{\Sigma|\text{dep}|} $$ | Line | Corr. lat | Corr. dep | Adj. lat | Adj. dep | |---|---|---|---|---| | A1 | +0.0098 | +0.0474 | +120.0234 | +170.2500 | | 12 | +0.0115 | +0.0368 | +141.1910 | +132.3206 | | 23 | +0.0216 | +0.0530 | -265.4733 | +190.4523 | | 34 | +0.0224 | +0.0394 | -274.8069 | -141.5325 | | 4X | +0.0056 | +0.0596 | -68.3342 | +214.2096 | | **Sum** | +0.0708 | +0.2361 | -347.4000 | +565.7000 | #### Step 5: Final coordinates | Station | Computed N | Computed E | Final N (m) | Final E (m) | |---|---|---|---|---| | A | 2094.720 | 1769.150 | 2094.720 | 1769.150 | | 1 | 2214.734 | 1939.353 | 2214.743 | 1939.400 | | 2 | 2355.913 | 2071.636 | 2355.934 | 2071.721 | | 3 | 2090.418 | 2262.036 | 2090.461 | 2262.173 | | 4 | 1815.589 | 2120.464 | 1815.654 | 2120.640 | | X | 1747.249 | 2334.614 | 1747.320 | 2334.850 | The computed coordinates of X now equal its known coordinates, so the traverse is balanced. **Answer:** 1 (2214.743 N, 1939.400 E); 2 (2355.934 N, 2071.721 E); 3 (2090.461 N, 2262.173 E); 4 (1815.654 N, 2120.640 E) (linear misclosure 0.247 m, 1 in 5121). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the bearings of the reference lines AB and XY obtained in this link traverse? A. From the coordinates of the control stations using tan^-1(dE/dN), with the quadrant fixed by the signs B. By measuring them with a compass in the field C. By averaging the observed clockwise angles D. By assuming AB is due north Answer: A. The control stations have known coordinates, so bearings follow from tan^-1 of the difference in easting over the difference in northing, with signs deciding the quadrant. (Key point: Reference bearings from coordinates) 2. The computed bearing of the last line is 101°11'10" against a known 101°13'18", with 6 angles. What is done? A. Add 21" to each angle (the k-th bearing gets k times 21"), so the last bearing equals the known value B. Add 128" to the last angle only C. Subtract 21" from each angle D. Ignore it because it is under 1 degree Answer: A. Computed is less than known by 2'08", so the correction is positive, +128"/6 = +21" per angle, accumulated bearing by bearing. (Key point: Angular misclosure and its distribution) 3. Under the transit rule, how is the correction to the latitude of a line decided? A. Proportional to the length of the line B. Proportional to the size of that line's latitude relative to the sum of all latitudes (absolute values) C. Equal for every line D. Proportional to the angle at the station Answer: B. Transit rule: C_lat = -e_N x |lat|/Σ|lat|, and likewise departures use |dep|/Σ|dep|. (Key point: Transit rule) 4. With e_N = -0.071 m and e_E = -0.236 m over ΣL = 1262.38 m, what is the linear misclosure and precision? A. 0.165 m, 1 in 7650 B. 0.247 m, 1 in 5121 C. 0.307 m, 1 in 4112 D. 0.247 m, 1 in 512 Answer: B. e = sqrt(0.071^2 + 0.236^2) = 0.247 m, and 0.247/1262.38 = 1/5121. 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