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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At what station do you suspect local attraction? Find correct bearing of line?

1 Answer
Line Observed Bearing
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$12'
  1. Difference between F.B and B.B of line DE is exactly equal to 180$^\circ$ hence D and E are free from local attraction and bearing observed from D and E is correct.

  2. F.B and B.B of DE is 320$^\circ$12’ and 140$^\circ$12’

  3. Observed B.B of CD= 212$^\circ$02’

    Correct F.B of C.D= 212$^\circ$02’- 180$^\circ$= 32$^\circ$2’

  4. Observed F.B of CD=30$^\circ$40’

    The error is =32$^\circ$2’- 30$^\circ$40’=1$^\circ$22’

  5. 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’

  6. Observed F.B of BC is 96$^\circ$20’

    The error is =96$^\circ$20’- 95$^\circ$40’=0$^\circ$40’

  7. 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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