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Local attraction with quadrantal bearings

Short answer

All lines show FB−BB≠180°, so all stations are suspected. The included-angle method is used; interior angles sum 540° with corrections. Starting from the most reliable line BC (mean N67°00′E), corrected fore bearings are AB S12°36′W, BC N67°00′E, CD N32°39′E, DE S84°48′W, EA S54°12′W.

The following bearings are observed in a compass traverse survey.

LineABBCCDDEEA
Fore BearingS11°30′WN67°30′EN32°15′ES82°45′WS50°15′W
Back BearingN13°00′ES66°30′WS30°30′WN80°45′EN53°00′E

Apply necessary checks and determine the corrected bearings.

The bearings are first converted to whole circle bearings (FB of AB = S11°30′W = 191°30′, BC = 67°30′, CD = 32°15′, DE = S82°45′W = 262°45′, EA = S50°15′W = 230°15′). FB − BB = 178°30′, 181°00′, 181°45′, 182°00′ and 177°15′, so no line is free and all stations are suspected. The included-angle method is used, and the results are shown in quadrantal form.

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
ABS 11°30′ WN 13°00′ E178°30′Differs by −1°30′
BCN 67°30′ ES 66°30′ W181°00′Differs by +1°00′
CDN 32°15′ ES 30°30′ W181°45′Differs by +1°45′
DES 82°45′ WN 80°45′ E182°00′Differs by +2°00′
EAS 50°15′ WN 53°00′ E177°15′Differs by −2°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 AB138°30′138°24′
BBA and BC54°30′54°24′
CCB and CD145°45′145°39′
DDC and DE52°15′52°09′
EED and EA149°30′149°24′

Sum observed = 540°30′; error = +0°30′; correction per angle = −0°06′ (equal distribution, since each angle is measured with the same care).

No line has FB − BB = 180°. Line BC has the smallest difference (+1°00′), so it is the most reliable; its adopted bearing is the mean of the FB and (BB ± 180°) = N 67°00′ E.

Step 3: Corrected bearings from the corrected angles

Rule used (in WCB): FB of next line = BB of previous line + interior angle (traverse run anticlockwise). Starting from BC = 67°00′ (BB 247°00′): CD = 247°00′ + 145°39′ − 360° = 32°39′, and so on round the traverse.

LineObserved FBCorrectionCorrected FBObserved BBCorrected BB
ABS 11°30′ W+1°06′S 12°36′ WN 13°00′ EN 12°36′ E
BCN 67°30′ E−0°30′N 67°00′ ES 66°30′ WS 67°00′ W
CDN 32°15′ E+0°24′N 32°39′ ES 30°30′ WS 32°39′ W
DES 82°45′ W+2°03′S 84°48′ WN 80°45′ EN 84°48′ E
EAS 50°15′ W+3°57′S 54°12′ WN 53°00′ EN 54°12′ E

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

Corrections are in the whole-circle sense (+ is clockwise).

Answer: Corrected bearings (FB): AB = S 12°36′ W, BC = N 67°00′ E, CD = N 32°39′ E, DE = S 84°48′ W, EA = S 54°12′ W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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