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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...
written 6.6 years ago by gravatar for saxena_archit saxena_archit • 760 modified 5.4 years ago by gravatar for sanketshingote sanketshingote • 100
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written 6.6 years ago by gravatar for saxena_archit saxena_archit • 760 modified 6.5 years ago by gravatar for sanketshingote 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

  1. FB of DE= 325∘15’ , BB Of DE= 145∘15’

    FB-BB= 325∘15’ - 145∘15’= 180∘

  2. Observed BB of CD= 207∘15’

    Correct FB of CD= 207∘15’ - 180∘=27∘15’

    Observed FB of CD is 25∘45’

    Error is 27∘15’ - 25∘45’=1∘30’

  3. Observed BB of BC is = 278∘30’

    Correct BB of BC= 278∘30’+ 1∘30’ = 280∘

    Correct FB of BC = 280∘- 180∘= 100∘

    Observed FB of BC=100∘5’

    Error is 100∘- 100∘45’ = -0∘45’

  4. Observed BB of AB = 219∘15’

    Correct BB of AB= 219∘150 -0∘45’=218∘30’

    Correct FB of AB = 218∘30’ - 180∘ =38∘30’

    Result in tabular form

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

To find true bearing

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