--- title: "Local attraction and true bearings, 1°30′E" subject: "Surveying I" chapter: 4 key: "semester_3/surveying_i:compass-closed-traverse-declination-1-30-east" source: https://www.bctengineer.com/q/semester_3/surveying_i/compass-closed-traverse-declination-1-30-east site: BCT Engineering --- # Local attraction and true bearings, 1°30′E Surveying I · Chapter 4: The Compass - Asked in: 2075 Asoj, 8 marks - Video: [4.6 Compass Surveying: Local Attraction Correction by Bearings (Complete Numerical)](https://www.youtube.com/watch?v=JvOphyhwV44) — Civil Modules - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/compass-closed-traverse-declination-1-30-east - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-4 ## Question The following observations were taken with a compass in case of a closed traverse. Calculate the angles and correct the bearings for local attraction, if any. Calculate the true bearings, if declination is 1°30′ East. | Line | FB | BB | Declination | |---|---|---|---| | AB | 51°30′ | 230°00′ | 1°30′ | | BC | 182°45′ | 2°30′ | | | CD | 4°00′ | 284°45′ | | | DE | 165°15′ | 345°45′ | | | EA | 251°30′ | 71°30′ | | ## Short answer Line EA has FB−BB=180°, making it free from local attraction. The included angle sum is 539°30′ (error –0°30′), corrected by adding 0°06′ to each angle; starting from EA, corrected magnetic FB bearings are AB 51°24′, BC 184°03′, CD 5°27′, DE 165°51′, EA 251°30′. With 1°30′ E declination, true FB bearings are respectively 52°54′, 185°33′, 6°57′, 167°21′, 253°00′. ## Answer **Reading of the data:** the back bearing of CD is printed as 284°45′; with FB = 4°00′ this would give FB − BB = 79°15′, which is not a possible local-attraction error, so it is read as **184°45′** (the interior angles then add up to 540° within 0°30′). All other values are used as given. (Another possible reading is given in the note at the end.) Line EA has FB − BB = 180°00′, so stations E and A are free from local attraction. Correcting the stations one by one does not close (the correction found for E comes out as +0°30′ although E is free), so the included-angle method is used. 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 | |---|---|---|---|---| | AB | 51°30′ | 230°00′ | 181°30′ | Differs by +1°30′ | | BC | 182°45′ | 2°30′ | 180°15′ | Differs by +0°15′ | | CD | 4°00′ | 184°45′ | 179°15′ | Differs by −0°45′ | | DE | 165°15′ | 345°45′ | 179°30′ | Differs by −0°30′ | | EA | 251°30′ | 71°30′ | 180°00′ | Free, difference = 180° | #### 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 | |---|---|---|---| | A | AE and AB | 20°00′ | 20°06′ | | B | BA and BC | 47°15′ | 47°21′ | | C | CB and CD | 358°30′ | 358°36′ | | D | DC and DE | 19°30′ | 19°36′ | | E | ED and EA | 94°15′ | 94°21′ | Sum observed = 539°30′; error = −0°30′; correction per angle = +0°06′ (equal distribution, since each angle is measured with the same care). Line EA has FB − BB = 180°, so its bearing is free from local attraction and is taken as the starting line. #### Step 3: Corrected and true bearings Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise). Declination is 1°30′ East, so true bearing = magnetic bearing + 1°30′. | Line | Observed FB | Correction | Corrected FB | Observed BB | Corrected BB | True FB | True BB | |---|---|---|---|---|---|---|---| | AB | 51°30′ | −0°06′ | 51°24′ | 230°00′ | 231°24′ | 52°54′ | 232°54′ | | BC | 182°45′ | +1°18′ | 184°03′ | 2°30′ | 4°03′ | 185°33′ | 5°33′ | | CD | 4°00′ | +1°27′ | 5°27′ | 184°45′ | 185°27′ | 6°57′ | 186°57′ | | DE | 165°15′ | +0°36′ | 165°51′ | 345°45′ | 345°51′ | 167°21′ | 347°21′ | | EA | 251°30′ | 0 | 251°30′ | 71°30′ | 71°30′ | 253°00′ | 73°00′ | Check: corrected FB − corrected BB = 180° for every line. #### Interior angles from the corrected bearings | Station | Interior angle | |---|---| | A | 20°06′ | | B | 47°21′ | | C | 358°36′ | | D | 19°36′ | | E | 94°21′ | Sum = 540°00′ (as required). **Answer:** Corrected bearings (FB): AB = 51°24′, BC = 184°03′, CD = 5°27′, DE = 165°51′, EA = 251°30′. True bearings (FB): AB = 52°54′, BC = 185°33′, CD = 6°57′, DE = 167°21′, EA = 253°00′. *Note on the data:* the printed pair for CD (FB 4°00′, BB 284°45′) is inconsistent, and with the reading used above the lines CB and CD almost overlap (angle at C = 358°30′). Another reading also fits: FB of CD = **104°00′** with BB = 284°45′. It gives the same observed angle sum (539°30′) and the same corrections. The angles at C and D then become 258°30′ and 119°30′ (corrected 258°36′ and 119°36′), and the corrected bearing of CD becomes 105°27′ (BB 285°27′; true 106°57′). All the other bearings stay the same. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Line EA has FB - BB = 180°00′. What does this tell us? A. Stations E and A are free from local attraction B. Only station A is free C. All stations are free D. The traverse is closed without error Answer: A. A line with exactly 180° difference has no attraction at either end, so E and A are free and EA is the starting line. (Key point: Free station) 2. Why is the included-angle method used instead of correcting stations one by one? A. Station-by-station correction needs a theodolite B. Station-by-station correction does not close (closing error 0°30′) C. There are too few lines D. Declination must be corrected first Answer: B. Correcting station after station around the traverse gives a 0°30′ discrepancy, so the angle method with equal distribution is used. (Key point: Choice of method) 3. The BB of CD is printed 284°45′ but read as 184°45′. Why? A. Back bearings must be below 200° B. FB and BB must be equal C. With 284°45′, FB - BB would be 79°15′, too large to be local attraction D. 184°45′ makes the line parallel to AB Answer: C. The line must differ from 180° only slightly; 184°45′ gives 179°15′, a plausible local-attraction error. (Key point: Misread data) 4. With declination 1°30′ East, how is the true bearing of AB (corrected 51°24′) found? A. Subtract the declination: 49°54′ B. Add 180° to it: 231°24′ C. Keep it unchanged: 51°24′ D. Add the declination: 52°54′ Answer: D. True bearing = magnetic bearing + easterly declination = 51°24′ + 1°30′ = 52°54′. (Key point: Declination) ## 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, bearings AB 305°30′](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-ea-bb-44-00) — Surveying I, chapter 4, asked 2× - [Local attraction and true bearings, 7°W](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-true-bearings-7w) — Surveying I, chapter 4 - [Corrected and true bearings, 5°E declination](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-true-bearings-5-deg) — 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 in closed traverse PQRSTP](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alocal-attraction-pqrstp-188-45) — Surveying I, chapter 4