The following table gives the fore and back bearings of sides of a closed compass traverse. Test and correct the values of the bearings for local attraction
| Line | fore bearing | Back bearing |
| AB | N 55°00′ E | S 54°00′ W |
| BC | S 67° 30′ E | N 66°00′ W |
| CD | S 25°00′ W | N25°00’E |
| DE | S77°00′ W | N75°30’E |
| EA | N 64°30′ W | S63°30′ E |
SOLUTION :
The above Problem can be sloved by two methods discussed below
1. Applying the correction by direct method without changing the bearing into WCB System
2. By converting all the given bearings into WEB system and applying Correction.by direct method
FIRST METHOD :
In the given Problem the F. B and B. B of a line CD are numerically Equal but their quadrant are just opposite. So, the F.B and B.B Of a line CD are correct. and also the other bearing observed From station ‘c’ and ‘D’ are correct
Hence FB and ‘DE’ and B.B of ‘BC’ are correct
Therefore FB of line DE = S77°00’W
The actual B. B of line DE should be N77° 00′ E
But the observed B. B Of line DE is N75°30′ E
Hence Error at station
‘E’ = observed bearing – Actual bearing
= 75°30′ -77°00′
= 1°30′
∴ correction at station
‘E’= +1°30’ For clockwise bearing
= -1° 30′ For anti clockwise bearing
correct FB of line
EA= N 64°30′ W – 1° 30′
= N 63°00’W
The actual B. B of line EA should be S 63°00’E
But the observed B.B of line EA is S 63° 30′ E
Hence, Error at station
‘A’= observed bearing – Actual bearing
= 63°30′ -63°00′
= +0°30′
∴ correction at station
‘A’= +0°30’ clockwise
= -0°30′ anti clock wise
correct FB of line
AB= N55°00′ E + 0°30′
= N55°30′ E
The actual B.B of line AB should be S 55°30’W
But the observed B.B of line AB is S 54°00’w
Hence, error at Station
‘B’= observed bearing – actual bearing
= 54°00′ -55°30′
= -1°30′
∴ correction at station
‘B’= +1°30′ c lockwise and -1° 30’ ante clock wise
correct F. B of line
BC= S 67°30’E – 1° 30′
= S 66° 00′ E
check:
. B.B of line BC= N66°00′ W.
check is ok
| Line | F.B and B.B | observed bearing | correction | corrected bearing |
| AB | F.B | N55°0’E | +0°30 at A | N 55° 30′ E |
| AB | B.B | S 54°00′ W | +1°30′ at B | S 55°30′ W |
| BC | F.B | S 67°30′ E | -1° 30′ at B | S66°00′ E |
| BC | B. B | N 66°00′ W | – | N66°00′ W |
| CD | F. B | S 25°00’W | _ | S 25°00′ W |
| CD | B.B | N25°00′ E | _ | N25°00’E |
| DE | F.B | S77°00′ W | . . _ | S77°00′ E |
| DE | B.B | N75°30’E | +1° 30′ at E | N77°00’W |
| EA | F.B | N64°30′ W | -1°30’at E | N63°00’W |
| EA | B.B | S 63° 30′ E | -0°30’atA | S63°00’E |
SECOND METHOD :
In this method First all the reduced bearing are converted into whole circle bearing show the conversion of RB to WCB
| Line | F.B | Rule for conversion | F.B in WCB | B.B | Rule For conversion | B.B in WCB | Diff. in F. B and B.B |
| AB | N55°00’E | 55°00′ | S54°00’W | (180°+ θ) | 234°00′ | 179°00′ | |
| BC | S67°30’E | (180°- θ) | 112°30′ | N66°00’W | (360°- θ) | 294°00′ | 181°30′ |
| CD | S25°00’W | (180°+ θ) | 205°00′ | N25°00’E | – | 25°00′ | 180°00′ |
| DE | S77°00’W | (180°+θ) | 257°00′ | N75°30’E | – | 75°30′ | 181°00′ |
| EA | N64°30W | (360°- θ) | 295°30′ | S 63°30’E | (180°- θ) | 116°30′ | 179°00′ |
From the above Table , we can understand that the difference in F.B and B.B of line CD was exactly differ by 180°. Hence, the other bearing observed from stations ‘C’ and ‘D’ are also correct that is the F.B Of DE and B.B Of BC are also correct
Therefore starting from
FB of DE= 257°80′(Correct)
Actual BB of line DE Should be
= 257°00′-180°00′
= 77°00′
But the observed bearing is 75°30′
Error at Station ‘E’ = observed bearing- actual bearing
E = 75°30′ – 77° 00′
‘E’= -1°30’
correction at station ‘E’= +1°30’
The corrected FB of line
EA = observed bearing + correction at station E
EA= 295°30’+1°30′
EA = 297°00′
Actual BB of line EA should be
=297°00′-180°
=117°00′
But the observed BB of line EA Is 116°30′
Error at station ‘A= observed bearing – Actual bearing
= 116 °30′ -117°00′
= -0°30′
correction at station ‘A’ = +0°30′
The corrected FB of line
AB= observed bearing + correction at station A
= 55°00’+0°30′
= 55°30′
Actual BB of line AB should be
= 55°30′ +180°
= 235°30′
But the observed BB of line AB is 234°00′
Errorat station’ B =observed bearing – Actual bearing
B= 234°00′ – 235°30′
= -1°30′
correction al station. ‘ B’=+1°30’
The correction FB of line
BC=observed bearing + correction at station B
BC= 112°30’+ 1°30′
= 114°00′
check :
Actual B. B of line BC should be
= 114°00’+180°
= 294°00′
check is ok
The results are tabulated in table
| Line | F.B or B.B | Observed Bearing | correction | corrected WCB | conversion | Reduce bearing |
| AB | F.B | 55° 00′ | +0°30’atA | 55°30′ | N θ E | N55°30’E |
| AB | B.B | 234° 00′ | +1°30′ atB | 235°30′ | S(θ-180°)W | S55°30’W |
| BC | F.B | 112°30′ | +1°30′ atB | 114°00′ | S(180°-θ) E | S66°00′ E |
| BC | B.B | 294°00′ | _ | 294°00′ | N(360°-θ)w | N66°00’W |
| CD | F.B | 205°00′ | – | 205°00′ | S(θ-180)W | S25°00W |
| CD | B.B | 25°00′ | _ | 25°00′ | N θ E | N25°00’E |