--- title: "Link traverse Q-m1-m2-X, Gale's table" subject: "Surveying II" chapter: 1 key: "semester_4/surveying_ii:link-traverse-qm1m2x-gale-bowditch" source: https://www.bctengineer.com/q/semester_4/surveying_ii/link-traverse-qm1m2x-gale-bowditch site: BCT Engineering --- # Link traverse Q-m1-m2-X, Gale's table Surveying II · Chapter 1: Traversing - Asked in: 2075 Bhadra, 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-qm1m2x-gale-bowditch - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-1 ## Question A link traverse Qm₁m₂X was run between two known stations Q and X where Q and X are starting point for the legs QP and XY respectively and following observations were taken. | Known stn | Easting (m) | Northing (m) | Unknown stn | Mean Hz. Angle | Length (m) | |---|---|---|---|---|---| | P | 7432.293 | 63115.158 | | | | | Q | 4792.189 | 63148.201 | m₁ | ∠PQm₁ = 93°06'51" | Qm₁ = 49.085 | | X | 7439.604 | 63321.618 | m₂ | ∠Qm₁m₂ = 155°45'25" | m₁m₂ = 73.787 | | Y | 7358.014 | 63348.321 | | ∠m₁m₂X = 247°09'37" | m₂X = 85.590 | | | | | | ∠m₂XY = 90°58'47" | | Calculate the coordinates in Gale's Table of m₁ and m₂, adjust them by Bowditch's rule and compute easting and northing coordinates. ## Short answer Angular misclosure +16" corrected by -4" per angle. Bowditch adjustment gave linear misclosure 1:3209. Final coordinates: m1 (63192.407 N, 7470.863 E), m2 (63239.826 N, 7414.348 E). ## Answer **Data note:** the easting of Q is printed as 4792.189 m; it is read as 7492.189 m, since only this value agrees with the line PQ from the coordinates of P (7432.293 E, 63115.158 N). Reference bearings from the coordinates (E, N): $$ \theta_{PQ}=\tan^{-1}\frac{7492.189-7432.293}{63148.201-63115.158}=61^\circ 06'56'',\qquad \theta_{XY}=\tan^{-1}\frac{7358.014-7439.604}{63348.321-63321.618}=288^\circ 07'21'' $$ 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: P (63115.158 N, 7432.293 E), Q (63148.201 N, 7492.189 E), X (63321.618 N, 7439.604 E), Y (63348.321 N, 7358.014 E). #### Step 1: Bearings and angular misclosure The angles are measured clockwise. Bearing of the line ahead = bearing of the line behind reversed (back bearing) + the clockwise angle at the station. Starting from the line QP (back bearing 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 from coordinates). Angular misclosure = computed − known. Using the unrounded bearings from the coordinates (PQ = 61°06'56.4", XY = 288°07'20.6"), it is 288°07'36.4" − 288°07'20.6" = +15.8" ≈ +0°00'16". 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°59'04" | | 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" | +44.2029 | -21.3412 | | m1m2 | 73.787 | 309°59'04" | +47.4142 | -56.5369 | | m2X | 85.590 | 17°08'38" | +81.7871 | +25.2294 | | **Sum** | 208.462 | | +173.4041 | -52.6487 | 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 $$ \begin{aligned} e_N&=63321.605-63321.618=-0.013\ \text{m}\\ e_E&=7439.540-7439.604=-0.064\ \text{m}\\ e&=\sqrt{e_N^2+e_E^2}=0.065\ \text{m},\qquad \frac{e}{\Sigma L}=\frac{0.065}{208.462}=\frac{1}{3209} \end{aligned} $$ #### 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 | +0.0030 | +0.0150 | +44.2059 | -21.3262 | | m1m2 | +0.0046 | +0.0225 | +47.4187 | -56.5144 | | m2X | +0.0053 | +0.0261 | +81.7924 | +25.2555 | | **Sum** | +0.0129 | +0.0637 | +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 3209). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the two stages of adjusting a link traverse? A. Balance the angles using the known last bearing, then balance coordinates using the known last station B. Balance coordinates first, then the lengths C. Adjust only the lengths, then the bearings D. Apply the transit rule then the compass rule Answer: A. Angles are first closed on the known bearing of the last line, then the coordinates are closed on the known position of the last station. (Key point: Two-stage adjustment of link traverse) 2. How is the correction applied to the bearings of successive lines? A. The same -4" to each bearing B. The k-th bearing receives k times the per-angle correction C. Only the last bearing is corrected D. Corrections are proportional to line length Answer: B. Each corrected angle carries into every later bearing, so the k-th bearing gets k times the correction. (Key point: Cumulative bearing correction) 3. How does Bowditch's rule distribute the closing error? A. Equally to every line B. Proportional to the length of each line C. Proportional to the latitude only D. Proportional to the angle at each station Answer: B. Correction to each line = -(error) x L / sum L. (Key point: Bowditch rule) 4. With e_N = -0.013 m, e_E = -0.064 m and sum L = 208.462 m, what is the relative precision? A. about 1 in 320 B. about 1 in 3200 C. about 1 in 32000 D. about 1 in 1600 Answer: B. e = sqrt(0.013^2 + 0.064^2) = 0.065 m; 208.462/0.065 is about 3209, so 1 in 3209. 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