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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.

LineFBBB
AB69°30′246°30′
BC191°30′13°00′
CD230°15′53°00′
DE262°45′80°45′
EA32°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, FB−BB=180∘\text{FB} - \text{BB} = 180^\circ exactly.

Step 1: Check each line

LineFBBBFB − BBRemark
AB69°30′246°30′183°00′Differs by +3°00′
BC191°30′13°00′178°30′Differs by −1°30′
CD230°15′53°00′177°15′Differs by −2°45′
DE262°45′80°45′182°00′Differs by +2°00′
EA32°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 (2n−4)×90∘=540∘(2n-4)\times 90^\circ = 540^\circ.

StationAngle betweenObservedCorrected
AAE and AB141°00′141°30′
BBA and BC55°00′55°30′
CCB and CD142°45′143°15′
DDC and DE150°15′150°45′
EED and EA48°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.

LineObserved FBCorrectionCorrected FBObserved BBCorrected BB
AB69°30′−1°45′67°45′246°30′247°45′
BC191°30′+0°45′192°15′13°00′12°15′
CD230°15′−1°15′229°00′53°00′49°00′
DE262°45′−4°30′258°15′80°45′78°15′
EA32°15′−3°00′29°15′210°30′209°15′

Check: corrected FB − corrected BB = 180° for every line.

Interior angles from the corrected bearings

StationInterior angle
A141°30′
B55°30′
C143°15′
D150°45′
E49°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)

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