- 2073 Shrawan · 8 marks
Local attraction: AB 69°30′ compass traverse
Short answer
No line has FB−BB=180°, so all stations are suspected. The sum of included angles is 537°30′; error –2°30′ is distributed equally, adding +30′ per angle. Taking BC line bearing (192°15′) as most reliable, the corrected FB bearings are AB 67°45′, BC 192°15′, CD 229°00′, DE 258°15′, EA 29°15′.
The following bearing was observed in a compass traverse.
| Line | FB | BB |
|---|---|---|
| AB | 69°30′ | 246°30′ |
| BC | 191°30′ | 13°00′ |
| CD | 230°15′ | 53°00′ |
| DE | 262°45′ | 80°45′ |
| EA | 32°15′ | 210°30′ |
At which of these stations would local attraction be suspected? Find the corrected bearing of the lines.
The differences FB − BB are 183°00′, 178°30′, 177°15′, 182°00′ and 181°45′, so none of the lines is free from local attraction and all stations A, B, C, D, E are suspected. 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, exactly.
Step 1: Check each line
| Line | FB | BB | FB − BB | Remark |
|---|---|---|---|---|
| AB | 69°30′ | 246°30′ | 183°00′ | Differs by +3°00′ |
| BC | 191°30′ | 13°00′ | 178°30′ | Differs by −1°30′ |
| CD | 230°15′ | 53°00′ | 177°15′ | Differs by −2°45′ |
| DE | 262°45′ | 80°45′ | 182°00′ | Differs by +2°00′ |
| EA | 32°15′ | 210°30′ | 181°45′ | Differs by +1°45′ |
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 .
| Station | Angle between | Observed | Corrected |
|---|---|---|---|
| A | AE and AB | 141°00′ | 141°30′ |
| B | BA and BC | 55°00′ | 55°30′ |
| C | CB and CD | 142°45′ | 143°15′ |
| D | DC and DE | 150°15′ | 150°45′ |
| E | ED and EA | 48°30′ | 49°00′ |
Sum observed = 537°30′; error = −2°30′; correction per angle = +0°30′ (equal distribution, since each angle is measured with the same care).
No line has FB − BB = 180°. Line BC has the smallest difference (−1°30′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = 192°15′.
Step 3: Corrected bearings from the corrected angles
Rule used: FB of next line = BB of previous line − interior angle (traverse run clockwise). Starting from BC = 192°15′ (BB 12°15′): CD = 12°15′ − 143°15′ + 360° = 229°00′, and so on round the traverse.
| Line | Observed FB | Correction | Corrected FB | Observed BB | Corrected BB |
|---|---|---|---|---|---|
| AB | 69°30′ | −1°45′ | 67°45′ | 246°30′ | 247°45′ |
| BC | 191°30′ | +0°45′ | 192°15′ | 13°00′ | 12°15′ |
| CD | 230°15′ | −1°15′ | 229°00′ | 53°00′ | 49°00′ |
| DE | 262°45′ | −4°30′ | 258°15′ | 80°45′ | 78°15′ |
| EA | 32°15′ | −3°00′ | 29°15′ | 210°30′ | 209°15′ |
Check: corrected FB − corrected BB = 180° for every line.
Interior angles from the corrected bearings
| Station | Interior angle |
|---|---|
| A | 141°30′ |
| B | 55°30′ |
| C | 143°15′ |
| D | 150°45′ |
| E | 49°00′ |
Sum = 540°00′ (as required).
Answer: Corrected bearings (FB): AB = 67°45′, BC = 192°15′, CD = 229°00′, DE = 258°15′, EA = 29°15′.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Local attraction, BC 79°30′ and DE 82°15′ · Surveying I, chapter 4
- Local attraction, bearings AB 305°00′ (EA 45°) · Surveying I, chapter 4 · asked 2×
- Local attraction ABCDEA by included angles · Surveying I, chapter 4 · asked 2×
- Local attraction, bearings AB 305°30′ · Surveying I, chapter 4 · asked 2×
- Local attraction and true bearings, 1°30′E · Surveying I, chapter 4
- Local attraction, S 40°30′ W traverse · Surveying I, chapter 4
- Traverse missing length CD, bearing EA · Surveying II, chapter 1
- Closed traverse missing EA bearing, BC length · Surveying II, chapter 1
- Bearings of BC and CD in traverse ABCDE · Surveying II, chapter 1 · asked 2×
- Missing bearings of BC and DE in traverse · Surveying II, chapter 1