written 5.7 years ago by
saxena_archit
• 760
|
modified 4.5 years ago
by
sanketshingote
• 100
|
At what station do you suspect local attraction? Find correct bearing of line?
|
written 5.7 years ago by
saxena_archit
• 760
|
modified 5.6 years ago
by
sanketshingote
• 100
|
| 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- |
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