The following FB and BB observed in traversing with a compass in place where local attraction was suspected . find the corrected FB and BB and true bearing of each of the line given that the...

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written 6.2 years ago by

saxena_archit • 760

modified 6.2 years ago by

sanketshingote • 100

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The difference of BB and FB of DE is 180O hence station A and B free from local attraction

FB of DE= 325$^\circ$15’ , BB Of DE= 145$^\circ$15’

FB-BB= 325$^\circ$15’ - 145$^\circ$15’= 180$^\circ$
Observed BB of CD= 207$^\circ$15’

Correct FB of CD= 207$^\circ$15’ - 180$^\circ$=27$^\circ$15’

Observed FB of CD is 25$^\circ$45’

Error is 27$^\circ$15’ - 25$^\circ$45’=1$^\circ$30’
Observed BB of BC is = 278$^\circ$30’

Correct BB of BC= 278$^\circ$30’+ 1$^\circ$30’ = 280$^\circ$

Correct FB of BC = 280$^\circ$- 180$^\circ$= 100$^\circ$

Observed FB of BC=100$^\circ$5’

Error is 100$^\circ$- 100$^\circ$45’ = -0$^\circ$45’
Observed BB of AB = 219$^\circ$15’

Correct BB of AB= 219$^\circ$150 -0$^\circ$45’=218$^\circ$30’

Correct FB of AB = 218$^\circ$30’ - 180$^\circ$ =38$^\circ$30’

Result in tabular form

Line	Observed bearing	correction	Corrected bearing	Remark
AB	38$^\circ$30'	-	38$^\circ$30'	1. Station B and C Are affected by local attraction 2. station D and E are free from local attraction.
BA	219$^\circ$15'	+0$^\circ$45' at B	218$^\circ$30'	-same-
BC	100$^\circ$45'	-0$^\circ$45' at B	100$^\circ$	-same-
CB	278$^\circ$30'	+1$^\circ$30' at C	280$^\circ$	-same-
CD	25$^\circ$45'	+1$^\circ$30' at C	27$^\circ$15'	-same-
DC	207$^\circ$15'	0	207$^\circ$15'	-same-
DE	325$^\circ$15'	0	325$^\circ$15'	-same-
ED	145$^\circ$15'	0	145$^\circ$15'	-same-

To find true bearing

Line	Declination	True F.B	True B.B
AB	10$^\circ$W	28$^\circ$20'	208$^\circ$30'
BC	10$^\circ$W	90$^\circ$	270$^\circ$
CD	10$^\circ$W	17$^\circ$15'	197$^\circ$15'
DE	10$^\circ$W	316$^\circ$15'	135$^\circ$15'

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