The following bearing where taken in running and open traverse with a compass in a place where local attraction was suspected?

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written 6.3 years ago by

saxena_archit • 760

modified 6.1 years ago by

sanketshingote • 100

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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