--- title: "Link traverse Q to X, Gale's table m1, m2" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:link-traverse-qx-gale-m1-m2" source: https://www.bctengineer.com/q/semester_4/surveying_ii/link-traverse-qx-gale-m1-m2 site: BCT Engineering --- # Link traverse Q to X, Gale's table m1, m2 Surveying II · Chapter 1: Traversing - Asked in: 2079 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/link-traverse-qx-gale-m1-m2 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question A link traverse was run between two known station Q and X where Q and X are the ending and starting points for the known traverse legs, PQ and XY respectively. The following observations were taken: | Known stn. | N (m) | E (m) | Link stn | Hz. angles to right | Line | Length (m) | Bearing | |---|---|---|---|---|---|---|---| | P | - | - | | ∠PQm₁ = 93°6'51" | | | PQ = 61°06'56" | | Q | 63148.201 | 7492.189 | m₁ | ∠Qm₁m₂ = 155°45'25" | Qm₁ | 49.085 | | | X | 63321.618 | 7439.604 | m₂ | ∠m₁m₂X = 247°9'37" | m₁m₂ | 73.787 | | | Y | - | - | | ∠m₂XY = 90°58'47" | m₂X | 85.590 | XY = 288°07'21" | Calculate the independent coordinates of m₁ and m₂ in Gale's table by adjusting the coordinates using Bowditch's method. ## Short answer After distributing the +15" angular misclosure equally, linear misclosure (0.065 m, 1:3211) is adjusted by Bowditch's rule. The adjusted coordinates are m1 (63192.407 N, 7470.863 E) and m2 (63239.826 N, 7414.348 E). ## 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: Q (63148.201 N, 7492.189 E), X (63321.618 N, 7439.604 E); bearing PQ = 61°06'56", bearing XY = 288°07'21". #### 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 QP (back bearing of PQ = 241°06'56") and using the observed angles, the bearing of the last line comes out as 288°07'36", while its known value is 288°07'21" (X-Y). Angular misclosure = computed − known = 288°07'36" − 288°07'21" = +0°00'15". The error is spread equally over the 4 angles: correction per angle = -0°00'04". The k-th bearing therefore receives k times this correction. | Station | Angle | Observed | Corrected | Bearing of line ahead | |---|---|---|---|---| | Q | P-Q-m1 | 93°06'51" | 93°06'47" | 334°13'43" | | m1 | Q-m1-m2 | 155°45'25" | 155°45'21" | 309°59'05" | | m2 | m1-m2-X | 247°09'37" | 247°09'33" | 17°08'38" | | X | m2-X-Y | 90°58'47" | 90°58'43" | 288°07'21" | #### Step 2: Latitudes and departures $$ \text{Lat}=L\cos\theta,\qquad \text{Dep}=L\sin\theta $$ | Line | Length (m) | Bearing | Latitude | Departure | |---|---|---|---|---| | Qm1 | 49.085 | 334°13'43" | +44.2028 | -21.3412 | | m1m2 | 73.787 | 309°59'05" | +47.4142 | -56.5369 | | m2X | 85.590 | 17°08'38" | +81.7871 | +25.2295 | | **Sum** | 208.462 | | +173.4040 | -52.6486 | Starting from Q (63148.201 N, 7492.189 E), the unadjusted coordinates of X are (63321.605 N, 7439.540 E), whereas the known coordinates are (63321.618 N, 7439.604 E). #### Step 3: Linear misclosure $$ \begin{aligned} e_N&=63321.605-63321.618=-0.013\ \text{m}\\ e_E&=7439.540-7439.604=-0.064\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.065\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.065}{208.462}=\frac{1}{3211} \end{aligned} $$ #### 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 | |---|---|---|---|---| | Qm1 | +0.0031 | +0.0150 | +44.2059 | -21.3262 | | m1m2 | +0.0046 | +0.0225 | +47.4187 | -56.5144 | | m2X | +0.0053 | +0.0261 | +81.7924 | +25.2556 | | **Sum** | +0.0130 | +0.0636 | +173.4170 | -52.5850 | #### Step 5: Final coordinates | Station | Computed N | Computed E | Final N (m) | Final E (m) | |---|---|---|---|---| | Q | 63148.201 | 7492.189 | 63148.201 | 7492.189 | | m1 | 63192.404 | 7470.848 | 63192.407 | 7470.863 | | m2 | 63239.818 | 7414.311 | 63239.826 | 7414.348 | | X | 63321.605 | 7439.540 | 63321.618 | 7439.604 | The computed coordinates of X now equal its known coordinates, so the traverse is balanced. **Answer:** m1 (63192.407 N, 7470.863 E); m2 (63239.826 N, 7414.348 E) (linear misclosure 0.065 m, 1 in 3211). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A link traverse is adjusted in what order? A. Coordinates first, then angles B. Only coordinates C. Angles using the known final bearing, then coordinates using the known final station D. Only angles Answer: C. Both ends are known, so angles are first balanced against the known last bearing, then coordinates against the known X. (Key point: Two-stage adjustment) 2. Computed last bearing 288°07'36 sec, known 288°07'21 sec. Misclosure is +15 sec spread over 4 angles. Correction per angle? A. +15 sec B. -15 sec C. -4 sec (about) D. +4 sec Answer: C. Computed exceeds known by 15 sec, so subtract about 15/4 = 4 sec from each angle; the k-th bearing gets k times this. (Key point: Angular correction) 3. Unadjusted X is (63321.605, 7439.540) and known X is (63321.618, 7439.604). Misclosure e and precision? A. 0.065 m, 1 in 3211 B. 0.013 m, 1 in 16000 C. 0.650 m, 1 in 321 D. 0.064 m, 1 in 100 Answer: A. e = sqrt(0.013^2 + 0.064^2) = 0.065 m; 0.065/208.462 = 1/3211. (Key point: Linear misclosure) 4. Bowditch corrections to latitude are +0.0130 m in total. Why positive? A. Corrections are always positive B. Because the lines are long C. Because the latitude sum is positive D. They are opposite in sign to e_N = -0.013 m, so the adjusted X matches the known value Answer: D. Correction = -e x L/sum L; with e_N negative, corrections are positive and sum to +0.013. 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