Problem 8 The following are the bearings observed in traversing.

with a
compass, an area where local attraction was suspected. Calculate
the interior angles of the traverse and correct them if necessary
Line FD BB
AB 150°0 330°0
BC 230 30 48°0
CD 306°15 12745
DE 298°00 120 00
EA 49°30' 229°30

Solution

E B

Fig. P-3.10

(a) Interior ZA =
BB of EA -

FB of AB
=
229°30' -

150°0' =79°30
 Compass Traversing 103
(b) Interior 2B = BB of AB FB of BCC
=
330°0' -

230°30' =
99°30'
Exterior 2C =
FB of CD -

BB of BC
306°15'- 48°0 = 258°15"
(c) Interior 2C = 360°0 - 258°15' = 101°45'

Exterior ZD =
FB of DE -

BB of CD
=

298°00 -

127°45' =170°15
(d) Interior 2D = 360°0' - 170°15' = 189°45'
(e) Interior 2E = BB of DE - FB of EA

=
120°0' -

49°30' =
70°30'
Check Sum of interior angles =ZA + ZB + 2C + ZD + ZE
= 541°0

But, the sum of angles should be (2N 4)-90° 540°0'

Here, Error 541° 540=+ 1°
Correction per angle =
60-12
5
The error should be equally distributed among all the angles.

Angle Calculated value Correction Corrected value

LA 79°30 -12' 79°18'

LB 99 30 -12 99°18
4C 101°45 -12 101°33
ZD
18945 -12' 189°33
LE 7030 -12 70°18
Total 541°0 540°00'

3.14 PROBLEMS ON LOCAL ATTRACTION

Problem 1 The following are the observed bearings of the lines of a traverse
ABCDEA with a compass in a place where local atraction
was suspected.

Line FB BB
AB 19/°45 130
BC 39030' 222°30
CD 22°15 200°30
DE 242°45 62045
EA 330°15 14745
Find the correct bearings of the lines.
104 Surveying and Levelling

Solution First Method-By Calculating Interior Angles

Fig. P-3.11

(a) Calculation of Interior Angle

Interior A = FB of AB - BB of EA = 19145' - 14745' = 44°00"
Interior B = FB of BC - BB of AB 39°30 - 13°00 = 26°30
Exterior C = BB of BCc- FB of CD =222°30 - 22°15' 200°15"
Interior LC = 360°00 - 200°15' = 159°45'

Interior D =FB of DE BB of CD
= 242°45 200°30' = 42°15"
Interior E = FB of EA - BB of DE

330°15' - 62°45' = 267°30'

Sum of interior angles = 44°00' + 26°30'+ 159°45' + 42°15' + 267°30

= 540°00'

which is equal to (2N 4) x 90° = 540°00'

So, the calculated angles are

correct.
(b) Calculation of Corrected Bearing
e line DE is free from local attraction. So,

FB of DE =
242°45' (correct)
and FB of EA = 330°15' (correct)

FB of AB BB of EA + 2A
=
(330°15' -

180°0') + 44°00
=
150°15' + 44°00 =194°15'
FB of BC BB of AB + ZB
=
(194°15' -

180°0') + 26°300'
=
14°15' + 26°30 40°45' =

FB of CD = BB of BC - exterior 2 C

(40°45+ 180°00') - 200°15

220°45 - 200°15' =20°30'
 Compass Traversing 105
FB of DE =BB of CD + ZD
=
(20°30' + 180°0') + 42°15"
=
200°30' + 42°15'
= 242°45
(checked)
The result is tabulated as follows.

Correctecd
Line
FB BB

AB 194°15' 14°15
BC 40°45 220°45
CD 20°30 200°30'
DE 242°45" 62 45
EA 330°15' 150°15'

Second Method-Directly Applying Correction

Procedure
(a) On verifying the observed bearing, it is found that the FB and BB of line
DE differ by exactly 180°. So, the stations D and E are free from local
attraction and the observed FB and BB of DE are correct.
(b) The observed FB of EA is also correct.
(c) The actual BB of EA should be
330°15' 180°0' = 150°15"

But the observed bearing is 147 45'.

So, a correction of (150°15 - 147°45') = + 2°30' should be applied at A.

(d) Correct FB of AB =
191°45' + 2°30' 194°15' =

Therefore, the actual correct BB of AB should be

194°15'- 180°00' = 14°15'
But Observed bearing = 13°0

So, a correction of (14°15 - 13°0) = +1°15' should be applied at B.

39°30' + 1°15' =
40°45'
(e) Correct FB of BC
=

Correct BB of BC should be = 40°45'+ 180°0 = 220°45"

Observed bearing of BC = 222030'
But
So, a correction of (220°45-222°30)=-1°45 should be applied atC
(1) Correct FB of CD 22°15' 1°45' 20°30
=
= -

Therefore, the BB of CD should be

20°30' + 180°0' = 200°30'

which tallies with the observed BB of CD.

 Surveying and Levelling
106
which also tallies with the remark
So, D is free from local attraction,
made at the beginning.
The result is tabulated as follows:

Observecd Correction Correct

Remarks
Line
FB BB FB BB

191°45' 13°00' +2°30' at A 194°15" 14°15

AB

BC 39030' 222°30 +1°15 at B 40°45 220°45

CD 22°15' 200°30'-1°45' at C 20°30 200 30'
242 45 62°45 Station D is free from
DE 242045 62 45 0° at D
local attraction

330°15' 147°45' 0° at E 330°15' 147°45 Station E is also free

EA
from local attraction