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Local attraction in closed traverse PQRSTP

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′.

The following table gives the FB and BB of the sides of a closed compass traverse PQRSTP.

LinePQQRRSSTTP
FB188°45′119°15′346°30′337°00′293°30′
BB7°45′298°15′168°30′158°30′113°00′

Check the bearings for local attraction. Correct the bearing by the method of included angles.

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, FB−BB=180∘\text{FB} - \text{BB} = 180^\circ exactly.

Step 1: Check each line

LineFBBBFB − BBRemark
PQ188°45′7°45′181°00′Differs by +1°00′
QR119°15′298°15′181°00′Differs by +1°00′
RS346°30′168°30′178°00′Differs by −2°00′
ST337°00′158°30′178°30′Differs by −1°30′
TP293°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)×90∘=540∘(2n-4)\times 90^\circ = 540^\circ.

StationAngle betweenObservedCorrected
PPT and PQ75°45′75°57′
QQP and QR111°30′111°42′
RRQ and RS48°15′48°27′
SSR and ST168°30′168°42′
TTS and TP135°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.

LineObserved FBCorrectionCorrected FBObserved BBCorrected BB
PQ188°45′+0°27′189°12′7°45′9°12′
QR119°15′+1°39′120°54′298°15′300°54′
RS346°30′+2°51′349°21′168°30′169°21′
ST337°00′+1°03′338°03′158°30′158°03′
TP293°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′.

✓ Answer checked by Claude Opus 5.5 (high effort)

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