The bearings observed in a traverse survey, at a place where local attraction was suspected are given below. At what stations do you suspect local attraction? Find the corrected bearings of the lines.
| Line | Fore bearing | Back bearing |
| PQ | 124° 30′ | 304° 30′ |
| QR | 68° 15′ | 246° 00′ |
| RS | 310°30′ | 135° 15′ |
| SP | 200° 15′ | 17° 45′ |
SOLUTION :
In this Problem we Can Slove Two methods
1. Direct method
2. Indirect method
1. Direct method
| Line | Fore bearing | Back bearing | Difference in FB and BB |
| PQ | 124° 30′ | 304° 30′ | 180°00′ |
| QR | 68° 15′ | 246° 00′ | 177° 45′ |
| RS | 310°30′ | 135° 15′ | 175° 15′ |
| SP | 200° 15′ | 17° 45′ | 182° 30′ |
By observing the above Table , we can understood that, the FB and B.B of line PQ differ by exactly 180°. So the stations Pand Q are free from local attraction and the observed FB and B.B of line PQ are correct
Also, the FB of QR observed from station Q and B.B of SP observed from station Pare correct
Starting from the correct FB of line
QR = 68° 15′
The actual B.B of line QR should be
= 68° 15 ‘+ 180°= 248° 15′
But, the observed bearing is 246° 0′
Hence, the error at station ‘R’= Observed Bearing – Actual bearing
= 246° 0′ – 248° 15′ = – 2°15′
correction at station R’ = 2° 15′
Correct FB of RS Observed FB of RS + correction at
station ‘R’
= 310° 30′ + 2° 15′
= 312° 45′
Therefore, the actual B.B of RS should be 3
=312°45′ – 180°00′
=132° 45′
But the observed bearing is 135° 15′
Hence, the error at station ‘S’
Observed bearing – Actual bearing
= 135° 15′ – 132° 45′ =+ 2° 30′
Correction at station’ S’ = – 2° 30′
Correct FB of SP= observed FB of SP + correction at station ‘S’
=200° 15′ -2° 30′
= 197° 45′
Check:
B.B of SP = 197° 45′ – 180° 00′
= 17° 45 ‘ check is OK.
The results are tabulated in Table
| Line | F. B or B. B | observed bearing | correction | correction bearing |
| PQ | F. B | 124°30′ | 0° at p | 124° 30′ |
| PQ | B. B | 304°30′ | 0° at Q | 304°30′ |
| QR | F. B | 68°15′ | 0° at Q | 68′ 15′ |
| QR | B.B | 246’00’ | + 2°15′ at R | 248°15′ |
| RS | F.B | 310°30′ | +2° 15′ at R | 312°45′ |
| RS | B. B | 135°15′ | -2° 30′ at S | 132° 45′ |
| SP | F.B | 200°15′ | -2°30 at S | 197° 45′ |
| SP | B.B | 17° 45′ | 0° at p | 17°45′ |
2. Included Angle Method:
Calculation of included angles with reference to Fig
Interior∠P=FB of PQ – B.B of SP
=124° 30′-17° 45′
= 106° 45′
Interior ∠Q = FB of QR – BB of PQ
=68° 15′ – 304° 30′
=-236° 15′
From Fig. it is evident that the above calculated∠Q is a exterior angle. Hence interior angle∠Q =360°-236° 15′ =123° 45′
Interior ∠ R = F.B of RS – B.B of QR
=310° 30′ – 246° 00′
=64° 30′
interior ∠S = F.B of SP – B.B of RS
= 200° 15′ – 135° 15′
= 65° 00′
Check:
Sum of interior angles =∠ P+∠Q +∠R +∠S
=106° 45′ + 123° 45′ + 64° 30′ + 65° 00′
= 360° 00′
Should be equal to (2n – 4) × 90 = 360° 00′
Hence, the calculated interior angles are correct.
Calculation of Corrected Bearings :
By observing the Table , we found that the difference in FB and B.B of line PQ are exactly differ by 180°. Hence, the other bearings observed from stations P and Q are correct.
i .e. FB of QR = 68° 15′ deg and B.BofSP = 17° 45′ are correct
therefore B.B of QR should be: FB of QR + 180° = 68° 15′ + 180°
= 248° 15′
F.B of RS = B.B of OR + ∠R
= 248° 15′ + 64° 30′
= 312° 45′
B.B of RS = FB of RS -180°
=312° 45′ – 180°
=132° 45′
F.B of SP =B.B of RS + ∠S
= 132° 45′ + 65° 0’0
= 197° 45′
Check:
B.B of SP = F.B of SP – 180°
= 197° 45′ – 180°
=17° 45′
Check is OK.
corrected values
| Line | Fore bearing | back bearing |
| PQ | 124° 30′ | 304°30′ |
|---|---|---|
| QR | 68° 15′ | 248° 15′ |
| RS | 312° 45′ | 132°45′ |
| SP | 197° 45′ | 17°45′ |