written 6.2 years ago by | modified 5.0 years ago by |
At what station do you suspect local attraction? Find correct bearing of line?
written 6.2 years ago by | modified 5.0 years ago by |
At what station do you suspect local attraction? Find correct bearing of line?
written 6.2 years ago by | modified 6.1 years ago by |
Line | Observed Bearing | |
---|---|---|
F.B | B.B | |
AB | 44$^\circ$40' | 225$^\circ$20' |
BC | 96$^\circ$20' | 274$^\circ$18' |
CD | 30$^\circ$40' | 212$^\circ$02' |
DE | 320$^\circ$12' | 140$^\circ$12' |
Difference between F.B and B.B of line DE is exactly equal to 180$^\circ$ hence D and E are free from local attraction and bearing observed from D and E is correct.
F.B and B.B of DE is 320$^\circ$12’ and 140$^\circ$12’
Observed B.B of CD= 212$^\circ$02’
Correct F.B of C.D= 212$^\circ$02’- 180$^\circ$= 32$^\circ$2’
Observed F.B of CD=30$^\circ$40’
The error is =32$^\circ$2’- 30$^\circ$40’=1$^\circ$22’
Observed B.B of BC is 274$^\circ$18’
Corrected B.B of BC =274$^\circ$18’+ 1$^\circ$22’ = 275$^\circ$40’
F.B of BC =275$^\circ$40’ - 180$^\circ$= 95$^\circ$40’
Observed F.B of BC is 96$^\circ$20’
The error is =96$^\circ$20’- 95$^\circ$40’=0$^\circ$40’
Observed BB of AB is 225$^\circ$20’
Correct BB of AB= 225$^\circ$20’ - 0$^\circ$40’ = 224$^\circ$40’
Required FB of AB = 224$^\circ$40’ -180$^\circ$= 44$^\circ$40’
Line | Observed Bearing | Correction | Corrected Bearing | remark |
---|---|---|---|---|
AB | 44$^\circ$40' | 0 | 44$^\circ$40' | Station B and C are affected by local attraction and station D and E are free from local attraction. |
BA | 225$^\circ$20' | -0$^\circ$40' at B | 224$^\circ$40' | -same- |
BC | 96$^\circ$20' | -0$^\circ$40' at B | 95$^\circ$40' | -same- |
CB | 274$^\circ$18' | +1$^\circ$22' at C | 275$^\circ$40' | -same- |
CD | 30$^\circ$40' | +1$^\circ$22' at C | 32$^\circ$2' | -same- |
DC | 212$^\circ$2' | 0 | 212$^\circ$2' | -same- |
DE | 320$^\circ$12' | 0 | 320$^\circ$12' | -same- |
ED | 140$^\circ$12' | 0 | 140$^\circ$12' | -same- |