--- title: "Local attraction in closed traverse PQRSTP" subject: "Surveying I" chapter: 4 key: "semester_3/surveying_i:local-attraction-pqrstp-188-45" source: https://www.bctengineer.com/q/semester_3/surveying_i/local-attraction-pqrstp-188-45 site: BCT Engineering --- # Local attraction in closed traverse PQRSTP Surveying I · Chapter 4: The Compass - Asked in: 2075 Chaitra, 8 marks - Video: [local attraction problem in surveying || compass survey|| survey engineering|| #local_attraction](https://www.youtube.com/watch?v=i_3QTIOjHek) — Aim 2 Civil Class - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/local-attraction-pqrstp-188-45 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4 ## Question The following table gives the FB and BB of the sides of a closed compass traverse PQRSTP. | Line | PQ | QR | RS | ST | TP | |---|---|---|---|---|---| | FB | 188°45′ | 119°15′ | 346°30′ | 337°00′ | 293°30′ | | BB | 7°45′ | 298°15′ | 168°30′ | 158°30′ | 113°00′ | Check the bearings for local attraction. Correct the bearing by the method of included angles. ## Short answer No line shows FB – BB = 180°, so all stations have local attraction. Using included angles, the error of –1°00′ is distributed equally (+12′ per angle), and starting from the most reliable line TP (adopted bearing 293°15′), corrected FB bearings are: PQ 189°12′, QR 120°54′, RS 349°21′, ST 338°03′, TP 293°15′. ## Answer Here no line has FB − BB = 180°, so **every station (P, Q, R, S, T) is suspected of local attraction**. The included-angle method is used because the angle at a station is not affected by local attraction there. Local attraction at a station is detected by comparing the fore bearing (FB) and back bearing (BB) of each line: if there is no local attraction at either end, $\text{FB} - \text{BB} = 180^\circ$ exactly. #### Step 1: Check each line | Line | FB | BB | FB − BB | Remark | |---|---|---|---|---| | PQ | 188°45′ | 7°45′ | 181°00′ | Differs by +1°00′ | | QR | 119°15′ | 298°15′ | 181°00′ | Differs by +1°00′ | | RS | 346°30′ | 168°30′ | 178°00′ | Differs by −2°00′ | | ST | 337°00′ | 158°30′ | 178°30′ | Differs by −1°30′ | | TP | 293°30′ | 113°00′ | 180°30′ | Differs by +0°30′ | #### Step 2: Included (interior) angles The included angle at a station uses the FB of the line leaving it and the BB of the line arriving at it, both observed at the same station, so local attraction cancels in the angle. The sum of the interior angles should be $(2n-4)\times 90^\circ = 540^\circ$. | Station | Angle between | Observed | Corrected | |---|---|---|---| | P | PT and PQ | 75°45′ | 75°57′ | | Q | QP and QR | 111°30′ | 111°42′ | | R | RQ and RS | 48°15′ | 48°27′ | | S | SR and ST | 168°30′ | 168°42′ | | T | TS and TP | 135°00′ | 135°12′ | Sum observed = 539°00′; error = −1°00′; correction per angle = +0°12′ (equal distribution, since each angle is measured with the same care). No line has FB − BB = 180°. Line TP has the smallest difference (+0°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 293°15′. #### Step 3: Corrected bearings from the corrected angles Rule used: FB of next line = BB of previous line + interior angle (traverse run anticlockwise). Starting from TP = 293°15′ (BB 113°15′): PQ = 113°15′ + 75°57′ = 189°12′, and so on round the traverse. | Line | Observed FB | Correction | Corrected FB | Observed BB | Corrected BB | |---|---|---|---|---|---| | PQ | 188°45′ | +0°27′ | 189°12′ | 7°45′ | 9°12′ | | QR | 119°15′ | +1°39′ | 120°54′ | 298°15′ | 300°54′ | | RS | 346°30′ | +2°51′ | 349°21′ | 168°30′ | 169°21′ | | ST | 337°00′ | +1°03′ | 338°03′ | 158°30′ | 158°03′ | | TP | 293°30′ | −0°15′ | 293°15′ | 113°00′ | 113°15′ | Check: corrected FB − corrected BB = 180° for every line, and working round the traverse returns to TP = 293°15′. **Answer:** Corrected bearings (FB): PQ = 189°12′, QR = 120°54′, RS = 349°21′, ST = 338°03′, TP = 293°15′. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is local attraction detected from FB and BB? A. If FB equals BB, the line is affected B. If FB + BB equals 180°, the line is affected C. If FB - BB is not 180°, one or both ends of the line are affected D. If FB - BB is 90°, the line is affected Answer: C. With no local attraction at either end, FB - BB = 180° exactly; any difference signals attraction (or error). (Key point: Detecting local attraction) 2. Why is the included-angle method used when no line has FB - BB = 180°? A. Included angles need no compass reading at all B. An angle measured at a station is not affected by local attraction at that station C. Included angles are always larger than bearings D. It avoids computing back bearings Answer: B. Both bearings forming the angle are read at the same station, so the same local error cancels in the difference. (Key point: Why included angles) 3. The observed interior angles sum to 539°00′ for a 5-sided figure. What correction is added to each angle? A. +0°12′ B. +0°10′ C. -0°12′ D. +1°00′ Answer: A. Required sum is 540°, error -1°00′, so +60′/5 = +12′ per angle. (Key point: Angle correction) 4. Line TP (difference +0°30′) is the most reliable. How is its adopted bearing obtained? A. The FB alone, 293°30′ B. The BB alone, 113°00′ C. The sum of FB and BB D. Mean of the FB and (BB ± 180°), giving 293°15′ Answer: D. Mean of 293°30′ and 113°00′ + 180° = 293°00′ is 293°15′. (Key point: Most reliable line) ## Related in Surveying I - [Local attraction: AB 69°30′ compass traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-69-30-reordered) — Surveying I, chapter 4 - [Local attraction, BC 79°30′ and DE 82°15′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ab-191-30-bc-79-30) — Surveying I, chapter 4 - [Local attraction and true bearings, 1°30′E](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Acompass-closed-traverse-declination-1-30-east) — Surveying I, chapter 4 - [Local attraction in closed traverse ABCDEFA](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-abcdefa-216-30) — Surveying I, chapter 4 - [Local attraction, bearings AB 305°00′ (EA 45°)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-305-00) — Surveying I, chapter 4, asked 2× - [Local attraction, S 40°30′ W traverse](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-quadrantal-s40-30w) — Surveying I, chapter 4 ## Related in other subjects - [Design main transmission pipes PQ and QR](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c70o) — Water Supply Engineering, chapter 7