Correction of bearings affected by local attraction./ How to correct bearings affected by local attractions?

 For the closed traverse ABCD, the following bearings were taken by compass. Find out which station is affected by local attraction & correct the bearings of the lines.

Sl. No.	Line	FB	BB
1.	AB	145°15'	325°45'
2.	BC	282°30'	102°0'
3.	CD	202°45'	22°45'
4.	DA	345°	165°0'

 

Let us draw the closed traverse ABCD from the above-given data.

As you can observe in the above drawing, the traverse is not closed properly at station A, due to the errors in the bearings affected by local attraction.

Calculation:

In the WCB system, the difference between the FB & BB should be 180°

So, let us find out the affected bearings by finding the differences between FB & BB for each line.

FB & BB difference:

1. For line AB:

 FB - BB = [ 145°15' - 325°45']

               = - 180° 30'

Error = [ 180°30' - 180°] = +30'

The error exceeds 30' above the 180° 

So, there is a correction of  -30' 

2. For line BC:

FB - BB = [ 282°30' - 102°0']

               = 180° 30'

Error = [ 180°30' - 180°] = +30'

The error exceeds 30' above the 180° 

So, there is a correction of  -30' 

3. For line CD:

FB - BB = [ 202°45' - 22°45']

               = 180° 

Error = [ 180° - 180°] = 0°

There is no error for line CD

Hence there is no correction.

 4. For line  DA:

FB - BB = [ 345°0' - 165°0']

               = 180° 

Error = [ 180° - 180°] = 0°

There is no error for line DA

Hence there is no correction.

Line	Local attraction	Error	Correction
AB	Yes	+30°	-30°
BC	Yes	+30°	-30°
CD	No	0°	0°
DA	No	0°	0°

 

From the above calculation, you will find that there is no local attraction at stations D & C as there is a 0° error for line CD & DA.

So, The observed bearing of line CD = 202°45' is correct

       The observed bearing of line CB = 102°0' is correct

        The observed bearing of line DC = 22°45' is correct

        The observed bearing of line DA = 345°0' is correct

         

For station A

The observed bearing of line AD - observed bearing of line DA should be equal to 180°

i.e. [165°- 345°] = -180°

As there is no local attraction in station D & the difference in the bearings is 180°, which means there is no local attraction in station A.

Therefore FB of line AB = 145°15' is correct.

Now, 

For line AB

 BB - FB should be equal to 180°

BB = [180°+ FB]

      = 180° + 145°15'

       = 325°15'

    The corrected bearing of line BA = 325°15'

But the observed bearing of line BA = 325°45'

Correction for station B 

= [corrected bearing - observed bearing]

= [325°15' - 325°45'] = -30'

Observed FB of line BC = 282°30'

To correct the FB of line BC, we have to deduct -30' from the bearing as there is a correction of -30' for station B.

So corrected FB of line BC

    = [282°30'-30'] = 282°

Check:

As there is no local attraction in station C, the difference between the corrected FB & observed BB of line BC should be 180°

i.e. [282° - 102°] = 180° ✔

Now the traverse is closed correctly.

Line	Observed bearing	Correction	Corrected bearing
AB	145°15'	0°	145°15'
BA	325°45'	-30'	325°15'
BC	282°30'	-30'	282°0'
CB	102°0'	0°	102°0'
CD	202°45'	0°	202°45'
DC	22°45'	0°	22°45'
DA	345°0'	0°	345°0'
AD	165°0'	0°	165°0'

 

To understand A to Z of surveying, click here.

Thank you for going through these calculation steps❤. Have a good day 😄.

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