Ex19-2 Following are the observed magnetic bearings of the traverse legs:

At what stations local attraction is suspected? Determine the correct bearings of the traverse legs and also calculate the included angles.

Solution :

Figure Ex19-2

The FB of the lines are given in reduced bearing. Their equivalent WCB are

The FB and BB of any line differs exactly by 180°, if the stations are free from local attraction. In the given observation, the FB and BB of the line AB differs by 180° and thus stations A and B are free from local attraction. The bearing of the lines observed at stations A and B may be considered to be correct.

Given, FB of BC = 78° 15'

Therefore Correct, BB of BC = 78° 15' + 180° = 258° 15'

But, observed BB of BC = 256° 00'

Therefore Error at C = 258° 15' - 256° 00' = - 2°15'

Correction at C = 2° 15'

Observed FB of CD = 300° 30'

Correction at C = + 2° 15'

Therefore corrected FB of CD = 302° 45' - 180° = 122° 45'

and corrected BB of CD = 125° 15'

Error at D = + 2° 30'

or Correction at D = - 2° 30'

observed FB of DA = 210° 15'

Therefore corrected FB of DA = 210° 15' - 2° 30' = 207° 45'

and corrected BB of DA = 207° 45' - 180° = 27° 45' (Checked)

Line	Corrected FB	Corrected RB
AB	120° 30'	300° 30'
BC	78° 15'	258° 15'
CD	302° 45'	122° 45'
DA	207° 45'	27° 45'

Included angle

As the traverse is running anti-clockwise the included angle will be the interior angles.

Angle at A = F.B. of AB - B.B. of DA = 120° 30' -27° 45' = 92° 45'

B = F.B of BC - B.B of AB = 78° 15' - 300° 30' = - 222° 15' + 360° = 137° 45'

C = F.B of CD - B.B of BC = 300° 30' - 256° 00' = 44° 30'

D = F.B of DA - B.B of CD = 210° 15' - 125° 15' = 85° 00'

Calculation of Bearing (2nd method)

Bearing of the line AB = 120° 30' (correct)

B = 137° 45'

-----------------------

Bearing of the line BC = 258° 15' - 180° = 78° 15' (since traverse is anti-clockwise)

C = + 44° 30'

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122° 45'

Bearing of the line CD = 122° 45' + 180° = 302° 45'

D = + 85° 00'

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387° 45'

Bearing of CD = 387° 45' - 180° = 207° 45'

A = + 92° 45'

------------------------

300° 30'

Bearing of AB = 300 30' - 180 = 120 30' (checked)

Line	FB	BB	Angle
AB	120° 30'	300° 30'	A = 92° 45'
BC	78° 15'	258° 15'	B =137° 45'
CD	302° 45'	122° 45'	C = 44° 30'
DA	207° 45'	27° 45'	D = 85°

[Note : In any traverse, running anti-clockwise, included angle at any stations = F.B. of the forward line - B.B. of the backward line].

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