--- title: "Corrected bearings of closed traverse PQRST" subject: "Surveying I" chapter: 4 key: "semester_3/surveying_i:closed-traverse-pqrst-correct-bearings" source: https://www.bctengineer.com/q/semester_3/surveying_i/closed-traverse-pqrst-correct-bearings site: BCT Engineering --- # Corrected bearings of closed traverse PQRST Surveying I · Chapter 4: The Compass - Asked in: 2080 Bhadra, 8 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_3/surveying_i/closed-traverse-pqrst-correct-bearings - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4 ## Question Calculate the corrected bearings of legs RS, ST and TP of closed traverse PQRST where the correct bearings of legs PQ and QR are 180°00′ and 90°30′ respectively. The included angles observed at stations R, S, T and P are 30°, 320°30′, 40°30′ and 59°30′ respectively. Take least count of prismatic compass 0°30′. ## Short answer Interior angle sum error +1°00′ distributed equally (−15′) to R, S, T, P. Corrected bearings: RS 300°15′, ST 80°30′, TP 300°45′ (check on PQ closes at 180°00′). ## Answer The bearings of PQ (180°00′) and QR (90°30′) are correct, so the angle at Q is known exactly and the other included angles are corrected to make the total $(2n-4)\times 90^\circ$. #### Step 1: Angle at Q and angle sum At Q, the back bearing of PQ is QP $= 180^\circ + 180^\circ = 0^\circ$ (i.e. 360°00′) and the bearing of QR is 90°30′, so the included angle at Q $= 90^\circ 30'$ (the traverse runs anticlockwise, with the interior on the left). | Station | P | Q | R | S | T | |---|---|---|---|---|---| | Observed angle | 59°30′ | 90°30′ (from correct bearings) | 30°00′ | 320°30′ | 40°30′ | Observed sum $= 59°30′ + 90°30′ + 30°00′ + 320°30′ + 40°30′ = 541°00′$. Required sum $= (2 \times 5 - 4) \times 90^\circ = 540°00′$. Error $= +1°00′$ (excess). The angle at Q is already fixed by correct bearings, so the error is distributed equally over the other four angles: $-60'/4 = -15'$ each. (Least count 30′: the permissible error is $30'\sqrt{4} = 60'$, so the error is just permissible.) | Station | Observed | Correction | Corrected | |---|---|---|---| | R | 30°00′ | −0°15′ | 29°45′ | | S | 320°30′ | −0°15′ | 320°15′ | | T | 40°30′ | −0°15′ | 40°15′ | | P | 59°30′ | −0°15′ | 59°15′ | New sum $= 59°15′ + 90°30′ + 29°45′ + 320°15′ + 40°15′ = 540°00′$ (check). #### Step 2: Corrected bearings For this anticlockwise traverse, the FB of the next line = BB of the previous line + included angle at the station between them. | Line | Working | Corrected FB | |---|---|---| | PQ | given | 180°00′ | | QR | given | 90°30′ | | RS | (90°30′ + 180°) + 29°45′ = 300°15′ | **300°15′** | | ST | (300°15′ − 180°) + 320°15′ = 440°30′ − 360° | **80°30′** | | TP | (80°30′ + 180°) + 40°15′ | **300°45′** | | PQ (check) | (300°45′ − 180°) + 59°15′ | 180°00′ ✓ | **Answer:** RS = 300°15′, ST = 80°30′, TP = 300°45′ (their back bearings are 120°15′, 260°30′ and 120°45′). The check on PQ closes exactly at 180°00′. *Assumption:* the included angles are interior angles of an anticlockwise traverse, the angle at Q (90°30′) being taken from the correct bearings. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the angle at Q left out of the correction? A. It comes from the correct bearings of PQ and QR, so it is already exact B. It is the smallest angle of the traverse C. It was measured with a better compass D. Corrections are only applied to acute angles Answer: A. Q's angle (90°30′) is fixed by the two correct bearings, so the whole error is distributed over the other four angles. (Key point: Angle at Q) 2. The sum of the five angles is 541°00′ but should be (2n-4) x 90° = 540°. What correction is applied to each of R, S, T and P? A. -12′ B. -15′ C. -20′ D. -60′ Answer: B. Error = +1°00′ (excess) = 60′, shared equally among the 4 observed angles: -15′ each. (Key point: Angle sum and correction) 3. With least count 30′, what is the permissible angular error for the four observed angles, and how does the 1°00′ error compare? A. 30′ x sqrt(4) = 60′; the error is just permissible B. 30′ x 4 = 120′; the error is well within it C. 30′ x sqrt(5) = 67′; the error is exceeded D. 30′; the error is exceeded and the work must be repeated Answer: A. Permissible error = L.C. x sqrt(n) = 30′ x sqrt(4) = 60′, equal to the observed error, so it is just permissible. (Key point: Permissible error) 4. For this anticlockwise traverse, what is the corrected bearing of RS? A. 120°15′ B. 270°15′ C. 330°15′ D. 300°15′ Answer: D. FB of RS = BB of QR + angle at R = (90°30′ + 180°) + 29°45′ = 300°15′. 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