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Lecture 18

The document discusses computing bearings from interior angles in a closed traverse. It provides an example with four survey stations where the interior angles and one initial bearing are given. It then shows the step-by-step workings to calculate the bearings of the remaining sides by using the formula of bearing of one side equals bearing of previous side minus the interior angle.

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26 views10 pages

Lecture 18

The document discusses computing bearings from interior angles in a closed traverse. It provides an example with four survey stations where the interior angles and one initial bearing are given. It then shows the step-by-step workings to calculate the bearings of the remaining sides by using the formula of bearing of one side equals bearing of previous side minus the interior angle.

Uploaded by

YOUR FATHER
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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use under special circumstances of Online Education due to COVID-19 Lockdown
situation and may include copyrighted material - the use of which may not have
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Use of any such copyrighted material as provided in globally accepted law of many
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class being conducted by the presenter.
Taxonomy
CLO Description PLO
level
1 Discuss various survey equipment and techniques
to be used for linear and angular measurements 1
and for computing the areas of plots C2 (Engineering
Knowledge)
Computation of Bearings from Angles
Example
A,B,C and D are four survey stations of a closed
traverse. Following are the interior angles

∠A = 140o 10’
∠ B = 90o 08'
∠ C = 60o 22'
∠ D = 69o 20'

If the observed bearing of AB is 60o 00’, then


compute the bearings of remaining sides?
the observed bearing of AB is 60o 00’

600 22’

Check for closed traverse = Sum of int. angles = (2n-4)x900


 1400 10’ + 900 08’ + 600 22’ + 690 20’ = (2*4 – 4)x900

Solution A

Bearing of BC

= Bearing of BA - ∠ B C

= B.B of AB - ∠ B

= (F.B of AB ± 180o) - ∠ B
= (60o 0' + 180o ) - 90o 08'
Bearing of BC =1490 52’
B

Solution
600 22’

D
Bearing of CD

= Bearing of CB - ∠ C
= B.B of BC - ∠ C

= (F.B of BC ± 180o) - ∠ C
= (1490 52’ + 180o ) - 60o 22'

Bearing of CD =2690 30’


A

Solution

C
Bearing of DA

= Bearing of DC - ∠ D
= B.B of CD - ∠ D

= (F.B of CD ± 180o) - ∠ D
= (2690 30’ - 180o ) - 69o 20'

Bearing of DA =200 10’


B

Solution

Bearing of AB (Check)

D
= Bearing of AD - ∠ A
= B.B of DA - ∠ A

= (F.B of DA ± 180o) - ∠ A
= (200 10’ + 180o ) - 140o 10'

Bearing of AB =600 0’ Checked


Adjustment of error in traverse angles
and bearings

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