buiding services compass survey

8/19/2019 Buiding Services Compass Survey

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Definitions

Defect or Error

• Stress in member exceeds limiting value permanently

Failure

• Failure means the structure may collapse or totally

fails to perform the designed function

Decay:

• Development of defect due to aging


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Repair:

• Technical aspect of rehabilitation

Rehabilitation:

• Restoring structure to service level

Damage:

• Sequence of defects

Distress:

• Visible or Hidden signs of damage

Contd…


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Causes of Failure

• Inadequate maintenance

• Bad use of building

• Poor workmanship in construction

• Defective construction materials

• Defective construction practices

• Wear and tear

• Weathering


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• Regular maintenance

• Stopping leakage in:

– Roof

– Walls

– Soil pipes

– Plumbing

– Waste water pipes

– Rain water pipes

Remedies to arrest Decay ( Failures )


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Use of:

1. Specified materials

2. Good workmanship during construction

3. Protection against weathering agencies

4. Protection against aggressive environment

5. Avoid overloading

Contd…


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Classification of Failures

Failures can be broadly classified

• Structural failures

• Non structural failures


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Structural Failures

Causes :-

These are due to

1. Incorrect design

2. Faulty construction

3. Over loading

Example : Extensive cracking of the RCC beam.


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Non-structural failures

• These are due to

– Internally induced stresses in building materials

– Penetration of moisture

– Weathering action

Causes :-


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• The non- structural cracks are not a threat to the safety

of building. But they – Look ugly

– Creates impression faulty work

– May give a feeling of instability

Example:

• Vertical cracks in a long compound wall due to – Shrinkage

– Thermal movement .

Contd…


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Failures in Concrete

• Concrete being most widely used, we deal with failures

in concrete. They are

– Cracking

– Spalling

– Disintegration


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Classif ication of Cracks (Based on Width of

Crack )

• Fine: less than 0.5 mm

• Thin : 0.1 mm to 0.3 mm

• Wide: 0.7 to 2.0 mm

• Very wide: 0.7 to 2.0 mm

C td


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Photo1. Vertical Cracks in RCC Railing due to drying shrinkage

of concrete and thermal movement

Contd…


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Photo 2. Vertical crack at the junction of two blocks of residential

flats, each block being about 20m in length

Contd…


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Fig.1.Temperature cracking in Building

Contd…


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Photo.3 Photo.4

Contd…

Contd


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Photo.5 Photo.6

Contd…


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Effects of Over Loading

• Excessive cracking due to :

• Flexure

• Shear

• Diagonal tension

• Compression

• Shear bond failure


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• Column surface spalling due to:

• Excessive compressive stress

• Large deflection

• Cracking in slabs

• Settlement of foundation

Contd…


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Types of Structural Cracks

Due to

• Pure flexure

• Pure tension

• Shear

• Torsion

• Bond

• Concentrated load


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Common forms of failure

• Corrosion of reinforcement

• Alkali and sulphate attack

• Displacement of the system

• Excessive deformation

• Malfunctioning of expansion joints


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Effects of Failures

• Plastic Deformation

• Shrinkage Cracks

• Contraction Cracks

• Cracks Due To Reinforcement Corrosion

• Alkali Aggregate Reaction

• Frost And Sulphate Attack Cracking


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Fig.2 Environment soil structure interaction and resultant

pattern of cracking

Contd…


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Defects in Masonry Structures

• Dampness

• Condensation

• Efflorescence

• Structural cracks


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Strengthening with

• Mild steel plates

• Jacketing techniques

• Fiber reinforced concrete

• Shotcrete

• Ferrocement

• Steel reinforcement and grouting

Strengthening of Members Failed Under

Torsion


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Procedure For Applying FRP SheetsPhoto 7

Contd…


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Procedure For Applying FRP SheetsPhoto 8

Contd…


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Photo.9 Procedure For Applying FRP Sheets

Contd…


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Contd…


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Contd…


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Contd…


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Photo.4


Contd…


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Contd…


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Contd…


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Glass Fibre

Carbon Fibre

Retrofitting

Technique

Photo.17

Photo.18

Contd…


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Strengthening of column using concrete jacketing

Fig.3

Contd…


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Strengthening of column using steel jacketing Photo.19

Contd…


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Strengthening of column using CFRP wrap Photos.20 and 21

Contd…


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Strengthening of beam using steel jacketing Fig.4

Contd…


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• Application of bonding agent over the old surface

• Covering with welded mesh

• Preparation of mortar

• Plastering

• Curing

Procedure For Applying Ferrocement


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• Preparation of form work to cover all sides of beam

• Application of bonding agent

• Preparation of micro concrete

• Filling form work with micro concrete and finishing

• Curing

Procedure For Applying Microconcrete


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• Trees should not be grown in expansive soils and clay

adjacent to buildings

• Trees like ashoka and coconut do not cause distress and

harm less

• Roots of trees like jack fruit spread horizontally: harmful tostructures

Precautions To Prevent Cracks Due To

Vegetation


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Failures of Foundations

Main causes of failure of foundations

1. Unequal settlement of the sub soil.

2. Unequal settlement of the masonry.

3. Withdrawal of moisture from the sub soil

4. Lateral pressure on the superstructure

5. Horizontal movement of the earth

6. Transpiration of trees and shrubs

7. Atmospheric action


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Construction Failures Repairs Maintenance

And Building Services


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Non-structural Failures

Failure of Masonry Structures


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Principal causes of cracks

a) Moisture changes,

b) Thermal variations,

c) Elastic deformation,

d) Creep,

e) Chemical reaction,

f) Foundation movement, and settlement of soil, and

g) Vegetation.


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Defects in Masonry Structures

• Dampness

• Condensation

• Efflorescence

• Structural cracks


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• Surface cracks

• Stains

• Defects in plastering

• Defects in repair work

Contd…


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• Cracks due to tensile stresses

• Flexural tension at lintel level

• Shear cracks

Cracks in Masonry Walls


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Tensile crack in a masonry wallFig.1

Contd…

Contd…


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Flexural Tension Crack in Wall MasonryFig.2

Contd…

Contd…


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Shear crack in Masonry wallFig.3

Contd…


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Shear crack in masonry pillar due to

expansion of RCC beam

Fig.4

Contd…


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Cracking Due to Expansion of BrickworkFig.5

Contd…


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Cracking in Top most storey of a Load Bearing Structure

Fig.6

Contd…


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Cracking due to expansion of brickworkFig.7

Contd…


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RCC Roof Slab Over Masonry WallFig.8

Contd…


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Cracking in Cladding and Cross Walls qf a FramedStructure

Fig.9

Contd…


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Fig.10 Arching up and Cracking of Coping Stones of a

Long Garden Wall

Contd…


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Fig.11 Horizontal Crack at the Base of Brick Masonry

Parapet


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• Shear cracks in masonry pillar at beam support

• Cracks due to expansion of brick work

• Cracks in topmost storey of load bearing structure


Contd…


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• Diagonal cracks in cross walls of multi storeyed load

bearing structure

• Vertical cracks at window openings

• Horizontal cracks in wall at supports

• Cracking of compound wall due to roots

Contd…


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• Cracks in walls due to trees close to buildings

• Horizontal cracks in top most storey due to

shrinkage and deflection of slab

• Diagonal cracks at corners of building due to

foundation settlement

Contd…


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Diagonal cracks in cross walls of multi-storied load bearing

structures

Fig.12

Contd…


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Fig. 13 Vertical Cracks in multi-storied buildings having

window openings in load bearing wall

Contd…


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Fig. 14 Horizontal cracks in a wall at supports due to

excessive deflection of a slab of large span

Contd…


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Cracking of a compound wall due to growing roots under the

foundation

Fig.15

Contd…


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Trees growing close to a building on shrinkable soil may cause

cracks in the walls

Fig.16


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Fig.17


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Trees Growing Close To A Building On Shrinkable Soil May

Cause Cracks In Walls

Fig.18


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Horizontal cracks in top-most storey below slab due to

shrinkage and deflection of slab

Fig.19


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Horizontal Cracks In Brick Panels Of Framed Structure

Fig.20


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Fig.21 Horizontal Crack at the Base of Brick Masonry

Parapet


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Fig.22 Vertical Cracks at the Junction of RCC Column And

Wall


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Photo.1 Horizontal Crack In Masonry Wall Due To Elastic

Deformation, Shrinkage In Canopy Slab


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Photo.2 Horizontal cracks in compound wall of brickwork,

having horizontal reinforcement


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Photo 3 Diagonal crack in the corner of a building due to

foundation settlement


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Fig.23 Horizontal Cracks In Brick Panels Of Framed

Structure


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Fig. 24 Vertical Crack In Brick Panel Wall Due ToExpansion Of Brick Work


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Fig. 25 Vertical crack in brick panel wall of framed structure

due to expansion of brick work


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Fig.26


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Fig.27


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Fig.28


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Fig.29


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Fig.30


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Fig.31


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Fig.32


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Photo 4


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Fig.33


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Fig.34


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PHOTO 5


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Fig.35



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Remedy:

• Strengthening of foundation

• By underpinning

• Grouting

• Strengthening of members

• Ferrocement for repair and rehabilitation

y


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• Do not directly affect the structural safety

• Penetration of moisture through cracks:

• Leads to structural failure

• Spoiling of internal finish

Non - Structural Failures

Causes of Non - Structural Cracks


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• Moisture changes

• Thermal variations

• Creep

• Chemical reactions

• Settlement of soil

• Vegetation

• Shrinkage at the time of initial drying

• Absence of expansion/contraction joints

Precautions / Remedies to Prevent Non-


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• Proper grading of concrete

• Compaction by vibrator

• Maintenance of water cement ratio

• Proper curing

• Cleaning of aggregates

• Avoid bleeding of concrete

• Covering concrete with wet canvas

Precautions / Remedies to Prevent Non

structural Cracks

Contd…


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• Plastering masonry –after drying of curing

Drying for about 3 months

• Providing shrinkage reinforcement

• 0.25% of cross sectional area of concrete

– use small bars at closer intervals

• Plastering of concrete structures immediately afterremoval of shuttering

• Provide strip joints between slab and supporting wall inload bearing structures

Contd…


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• Adequate insulation to roof slabs

• Expansion joints in long walls @ 15 m intervals

• Masonry construction over RCC slabs – 15 days

after removal of form work

Contd…


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• In RCC framed structure:

• completion of frame work before construction of

masonry walls

• Construction of partition walls commences from top

Contd…


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• In masonry partitions having length to height ratio >

2: provide horizontal reinforcement

• If soluble sulphates present in soil, ground water, or

clay bricks: use rich mix and sulphate resistant

cement

• Do not use bricks containing more than 1% soluble

sulphates

Contd…


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• Provide under reamed pile foundation in expansive

soils

• Provide adequate cover to reinforcement

• Horizontal extension of building separated from the

old by expansion joint right from bottom to top

• No excavation below the foundations of old buildings


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Precautions to Prevent Cracks due to


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• Trees should not be grown in expansive soils and

clay adjacent to buildings

• Trees like ashoka and coconut do not cause distress

• Roots of trees like jack fruit spread horizontally:

harmful to structures

Vegetation

Cracks In Masonry


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Causes:

• Poor quality of materials

• Bad workmanship

• Uneven settlement of foundation

• Heavy concentration of loads

• Impact loads

• Vibrations

• Improperly designed lintels and beams

Repairs to Cracks in Masonry


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Cement grouting to minor cracks:

• Clean cracked surface

• Wash with water

• Fill crack with cement grout by cement gun

p y

Contd…


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• Repair to wider cracks using expanding cement slurry

• Clean cracked surface

• Wash with water

• Fill crack with expanding cement grout by cement gun

• Cement expands after setting

Failure in plastering


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p g

1. Blistering of plastered surface

2. Cracks

3. Efflorescence

4. Flaking

5. Peeling

6. Popping

7. Rust-stains

8. Softness

9. Uneven surface

Remedies


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1. Plastering should be carried out in the last

workmanship manner.

2. Bond of brick work should be properly maintained.

3. Efflorescence is removed by rubbing brushes on the

damaged surface. A solution of (one part of Hcl or H2

SO4 + five parts of clean water)

4. Bricks of superior nature be used

5 W t f f lt b d f b i k k d

Contd…


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5. Water free from salts be used for brick work and

plastering work.

6. Surface to be plastered be well-watered

7. Excessive trowelling be avoided.

8. Damp- proof courses be provided at convenient place

in the building.

Leakage of R.C.C Roofs


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Causes:

• Harsh and poorly designed mix

• Poor compaction

• Excess water cement ratio

• Excess fine aggregate

• Use of fine sand

Contd


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• Improper curing

• Thermal expansion

• Failure of expansion joints

• Improper removal of farm work

Contd…

Prevention is better than cure. Therefore water proofing of


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roof is made by employing any one of the following

methods:

• Finishing of roof surface is done at the time of laying

cement concrete. The C.M proportion is 1:4

• Bedding concrete and flooring

• Application of Mastic asphalt and jute cloth

But despite the fact that all the care is taken still leakage

may occur

Prevention of Leakage in R.C.C Roofs


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• Using well designed mix

• Application of:-

– Membrane water proofing

– Cement base paints

– water repelling agents

Contd…


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• Application of surface coat with mixture of ordinary

portland cement and any one of :-

– Hydrated lime

– Hygroscopic salts like calcium chloride, sodium

chloride

– Titanium oxide


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Fig.1 Water Proofing of Terrace

Spalling


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Causes:-

1. Corrosion of reinforcements

2. Sudden changes in temperature

3. Thermal expansion

4. Improper cover

5. Segregation


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Photo.1 Underside of slab deterioration due to water leakage

through slab crack


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Photo 2 Concrete deterioration due to water leakage through

slab crack


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Repair of Spalling


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1. Defective area should be scarified

2. Cleaned of all loose materials

3. Surface kept moist for fourteen hours

4. Strengthening by mat reinforcement

5. Guniting dense concrete

6. Curing

Repair of Spalling

Failures in Flooring


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g

• The common types of failures in flooring• Pitting

• Cracking

• Spalling at joint

• The reasons are

• Wear and tear

• Accident

Pitting


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g

• Pitting consists of small holes on the surface due to

– Aggregate becoming loose

– Dents caused by accidents

• The remedial measure is applying

– Epoxy resins with matching colours and cements

Cracking

• This may occur in concrete and terrazzo flooring


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This may occur in concrete and terrazzo flooring

• Reasons are

– Deflection due to load ( upper floors )

– Due to settlement ( ground floor )

• Remedial measures

• In upper floors a cushion layer of lime concrete is

suggested

• In case of ground floor

– Break the flooring totally at the damaged part – Refill properly and relay the floor

• Can also be repaired by expansive grout or by epoxy

Spalling


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p g

• Generally it happens near joints due to weak mortar.

• Remedial measures

• Depth of spall is less than 25mm use epoxy resins.

• The large chunks are found

– Cut the spalling area in a regular way

– Reconcrete with a binder like araldite , epoxy etc.,


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COMPASS SURVEYING 125

COMPASS SURVEYING

RAMINENI VENU,

M.TECH(STRUCTURES)

Prepared by:

Purpose and Principle of Compass

C A l i i t t (B i )


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Compass :Angle measuring instrument (Bearing)

Principle: The principle of compass surveying is

traversing

Designation of compass

– The size of the compass is expressed in terms of its

diameter

– The compass 12 cms(120mm) means the diameter of box

is 12 cms(120mm)


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Traversing:

It is a series of connected lines.

Types of traversing:-

(a) Open traverse : continuous lines, not returns

(b) Closed traverse : forms closed circuit

Contd…


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Fig.1 Sketch of Open Traverse

Contd…


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Fig.2 Sketch of Closed Traverse

Methods of Traversing


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g

Chain traversing (by chain angle)

Compass traversing (by free needle)

Theodolite traversing (by fast needle)

Plane table surveying (by plane needle)

When is Compass surveying recommended ?


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Compass Survey is recommended when

A large area is to be surveyed

The course of a river or coast line is to be surveyed

The area is crowded with many details and triangulation is not

possible

p y g

When is Compass surveying not recommended ?


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•Where local attraction is suspected due to magnetic material

steel , iron & electric cables


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Types of compass

There are two types of Compass

1. Prismatic Compass(0 to 360 Degrees)

2. Surveyor’s Compass(0 to 90 Degrees)

Prismatic Compass


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Fig.3 Sketch of Prismatic Compass


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Fig.4


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Parts of Prismatic Compass

Compass Box:

– Circular 8 to 10 cm (80mm to 100mm) metallic

– At the center of box pivot is provided.

Magnetic Needle and Graduated Ring: – On pivot the magnetic needle rests

– Aluminium graduated ring is attached to needle

– Graduated ring has 0 degree to 360 degrees clockwise

– i.e. 00 at south ,900 at west ,1800at north ,2700 at east

– Least count is 30minutes


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Contd…


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Sight vane and Object vane Of Compass:

– Sight vane and Object vane are fixed diametrically opposite tothe each other.

– Object vane consists of vertical hair attached to the frame

– Sight vane ( or eye slit )consists of vertical slit cut into upperassembly of the prism unit

– Prism unit is hinged to the box


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Dark Glasses

–Red glasses Sighting luminous objects.

–Blue glasses Reduces strain to observer.

Adjustable Mirror

Contd…


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j

– Provided with sight vane horizontally vertically and Inclined

manner .

Brake Pin

– At the base of sight vane. If pressed, oscillation is stopped

Lifting Pin

– Below the sight vane

– Lifts magnetic needle

Adjustments of Prismatic Compass


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j p

Fixing the compass with tripod stand.

Centering.

Leveling.

Adjustment of prism.

Observation of bearing.

Definitions:-


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Bearing of the line:- The horizontal angle made by the

survey line with some fixed reference direction is bearing of

line.

The fixed reference is known as Meridian

The meridians are

– True meridian

– Magnetic meridian

– Arbitrary meridian

True Meridian: It is the line passing through the pointand joining the North and South poles of the earth


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and joining the North and South poles of the earth.

Q

P

A

Line PAQ is the True Meridian

Fig.3

True Bearing: It is he horizontal angle made by the surveyline with the true meridian


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Fig.4

Magnetic Meridian: The direction shown by a freely


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suspended well balanced Magnetic Needle at any point is

termed as Magnetic Meridian

Fig.5

Magnetic Bearing: Horizontal angle made by the surveyline with the magnetic meridian.


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Fig.6


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Arbitrary Meridian: Any arbitrarily assumed directionchosen to measure the bearings is termed as Arbitrary Meridian

Arbitrary Bearing: The horizontal angle made by the

survey line with the arbitrary meridian


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Bearings are designated by two systems

– Whole Circle Bearing ( WCB )

– Quadrantal Bearing or reduced bearings( QB or RB )


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Whole circle bearing (W.C.B.) :

The bearing of a line measured from the North in clockwise

direction.

The value of WCB varies from 00 to 3600

Sketch Of Whole Circle Bearing (W.C.B.)


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O

D

C

A

B

WCB of OA = Ө1

WCB of OB = Ө2

WCB of OC = Ө3

WCB of OD = Ө4

Fig.1


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Quadrantal Bearing Or Reduced Bearing (Q.B. or R.B.)

The bearing of a Line observed either from the North or South

Measured in Clockwise (or) Anticlockwise direction

Towards East or West

The value of QB ranges between 0

0

to 90

0

Sketch of Reduced (Quadrantal) Bearing


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R.B of OA = N θ10 E

R.B of OB = S θ20 E

R.B of OC = S θ30 E

R.B of OD = N θ40 E

o

D

C

A

B

Rules For Conversion Of Bearing FromW.C.B. To Q.B. (R.B.)


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Table.1

Line lying in

The quadrant

W.C.B lying between Rule for conversion

to Q.B.

Quadrant

I. 00 to 900 R.B. is same as

W.C.B.

N.E.

II. 900 to 1800 RB = 1800 – WCB S.E.

III. 1800 to 2700 RB = WCB – 1800 S.W.

IV. 2700 to 3600 RB = 3600 – WCB N.W.

Reduced Bearing (Q.B.) To W.C.B.:


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Fig.2

Quadrant in

which surveyline lies

W.C.B lying

between R.B.

Rule for conversion to Q.B.

Rule for conversion W.C.B.

Quadrant

W.C.B. lying

between

I. N.E. W.C.B same as R.B. 00 to 900

II. S.E. W.C.B = 180degrees – R.B. 900 to 1800

III. S.W. W.C.B = R.B.+180 degrees 1800 to 2700

IV. N.W. W.C.B = 360 – R.B. 2700 to 3600

Declination: The horizontal angle subtended between the


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Declination: The horizontal angle subtended between thetrue meridian and the magnetic meridian.

Declination West Declination East

Fig.7

Contd…


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If the Magnetic Meridian is towards East of True

Meridian, the declination is East

If the Magnetic Meridian is towards West of True

Meridian, the declination is West

Fig.8

Contd…


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West Declination East Declination

Fig.9

The relation between the true bearing and magnetic bearing of a

line can be written as

Contd…


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line can be written as

True Bearing = Magnetic Bearing ± Declination.

For East declination, +ve Sign is to be used where as –ve sign

is used for West declination.

Magnetic Bearing = True Bearing ± Declination

For East declination, –ve sign is to be used where as -ve sign

is used for West declination.

Local Attraction


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A magnetic needle indicates the north direction when freely

suspended or pivoted.

Some times the needle comes near some magnetic substances

such as iron ore, steel structures, electric cables conveying

current etc.

Local Attraction


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Then it is found to be deflected from its true direction and does

not show the actual magnetic north.

This disturbing influence of magnetic substances is known as

Local Attraction.

Contd…


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If all the lines of a traverse are found to be affected by local

attraction

– The line with minimum error is identified

– The fore bearing and back bearing of this line are

adjusted by distributing the error equally

– Then starting from this adjusted line, the fore and back bearings of other lines are corrected

Problems in W.C.B


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1. Convert the following whole circle bearings into quadrantal

bearings.

a) 780201

b) 130030

1

c) 280030

1

d) 232030

1

IIV

N

B

Contd…


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78020l it is in first quadrant

W.C.B = Q.B

78020l = N 78020l E

IIIII

E

S

W A

78020l

IIV

NContd…


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1300

30l

is in second quadrantQ.B = 1800 – W.C.B = SE

Q.B = 1800 - 130030l = S 490301 E

II

III

E

S

W A

B

130030l

IIV

NContd…


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280030l is in fourth quadrant

Q.B = 3600 – W.C.B = 3600 - 280030’

=N 790301 W

IIIII

E

S

W A

B

280030l

IIV

NContd…


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232030l is in third quadrant

Q.B = W.C.B – 1800 = 232030l – 1800

S 52030l W

IIIII

E

S

W A

B

232030l


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2. Convert the following Whole circle bearing into Reduced

bearing.

a) 420151

b) 2230151

c) 3220301

d) 1420301

IIV

N BContd…


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W.C.B = Q.B

Q.B = N 42015l E

IIIII

E

S

W A

42015l

IIV

NContd…


169/577


2230

15l

is in third quadrantQ.B = W.C.B – 1800 = 223015l - 1800

Q.B = S 430151 W

II

III

E

S

W A

B

223015l

IIV

NBContd…


170/577



Q.B = 3600 – W.C.B = 3600 - 322030l

N 370301 W

IIIII

E

S

W A322030l

IIV

NContd…


171/577


142030l is in second quadrant

Q.B = 1800 – W.C.B = 1800 – 142030l Q.B = S 37030l E

IIIII

E

S

W A

B

142030l

3. Convert the following Whole circle bearing into Reduced


172/577


g g

bearing.

a) 2870

b) 1070451

c) 120

451

d) 1930151

IIV

NContd…


173/577



Q.B = 3600 – W.C.B

Q.B = 3600 - 287000l

Q.B = N 730 W

IIIII

E

S

W AB

287000l

IIV

NContd…


174/577



Q.B = 1800 – W.C.B

Q.B = 1800 – 107045l

Q.B = S 72015l E

IIIII

E

S

W A

B

1070

45l

IIV

NBContd…


175/577



W.C.B = Q.B

Q.B = N 12045l E

IIIII

E

S

W A

12045l

IIV

NContd…


176/577



Q.B = W.C.B – 1800

Q.B = 193015l - 1800

Q.B = S 130151 W

II

III

E

S

W A

B

193015l

4 C h f ll i h l i l b i i d d


177/577


4.Convert the following whole circle bearings into reduced

bearings.

a) 500

b) 2300

c) 1450301

d) 325030

1

IIV

N

B

500

Contd…


178/577


500 it is in first quadrant

W.C.B = Q.B

Q.B = N 500 E

IIIII

E

S

W A

500

IIV

NContd…


179/577


2300 is in third quadrant

Q.B = W.C.B – 1800

Q.B = 2300 – 1800

Q.B = S 500 W

IIIII

E

S

W A

B

2300

IIV

NContd…


180/577



Q.B = 1800

– W.C.B = SEQ.B = 1800 - 145030l

Q.B = S 340301 E

IIIII

E

S

W A

B

145030l

IIV

NBContd…


181/577



Q.B = 3600 – W.C.B

Q.B = 3600 - 325030l

Q.B = N 340301 W

IIIII

E

S

W A325

0

30l


182/577


5. Convert the following whole circle bearing into reduced

bearing.

a) 2890271

b) 1090271

c) 2190541

d) 390451

IIV

N

289027l

Contd…


183/577



Q.B = 3600 – W.C.B

Q.B = 3600 - 289027l

Q.B = N 700331 W

IIIII

E

S

W A

B 289027l

IIV

NContd…


184/577



Q.B = 1800 – W.C.B = SE

Q.B = 1800 - 109027l

Q.B = S 700331 E

IIIII

E

S

W A

B

109027l

IIV

NContd…


185/577



Q.B = W.C.B – 1800

Q.B = 2190541 – 1800 Q.B = S 390541 W

IIIII

E

S

W A

B

2190541

IIV

N

BContd…


186/577



W.C.B = Q.B

Q.B = N 390451 E

IIIII

E

S

W A

390451

6 Convert the following whole circle bearing into quadrantal


187/577


6.Convert the following whole circle bearing into quadrantal

bearing.

a) 870531

b) 2670531

c) 780161

d) 2580161

IIV

N

BContd…


188/577



Q.B = W.C.B

Q.B = N 870531 E

IIIII

E

S

W A

870531

IIV

NContd…


189/577



Q.B = W.C.B – 1800

Q.B = 2670531 – 1800 Q.B = S 870531 W

IIIII

E

S

W AB 2670531

IIV

N

B780161

Contd…


190/577



W.C.B = Q.B

Q.B = N 780161 E

IIIII

E

S

W A

B780161

IIV

NContd…


191/577



Q.B = W.C.B – 1800

Q.B = 2580161 – 1800

Q.B = S 780161 W

IIIII

E

S

W A

B

2580161

7. Convert the following W.C.B to Q.B.S


192/577


g Q

a) W.C.B of AB = 45030

1

b) W.C.B of BC = 125045

1

c) W.C.B of CD = 222015

1

d) W.C.B of DE = 3200301

N

IIV

BContd…


193/577


EW

S

Line AB is in first quadrant

In first quadrant the value of WCB = QB

QB values are always w.r.t to N & S

450301 N 450301E

QB of AB = N 45030

1E

IIIIIA

N

W.C.B of BC = 125045

1

Contd…


194/577


EW

S

Line BC is in second quadrant i.e. SE

QB = 1800 – W.C.B

= 1800

- 1250

451

= 540151

W.C.B of 1250451 = QB S540151E

I

IIIII

IV

B

C

N

W.C.B of CD = 222015

1

I QuadrantIV Quadrant

Contd…


195/577


EW

SLine CD is in third quadrant

If the line is in third quadrant

QB = W.C.B – 1800

= 222015

1 – 1800

= 410151

QB = S410151W

Q

II Quadrant

III Quadrant

IV Quadrant

C

D

N

W.C.B of DE = 320030

1

E

Contd…


196/577


EW

SLine DE is in fourth quadrant

W.C.B = 3200301

QB = 3600 – W.C.B

QB = 3600 - 320

030

1

= 390301

QB = N 390301 W

I Quadrant

II QuadrantIII Quadrant

IV Quadrant

D


197/577


8.Convert the following W.C.B to Quadrantal bearing

a) W.C.B of 450301

b) W.C.B of 1220301

c) W.C.B of 2250001

d) W.C.B of 3050301

N

IIV

BContd…


198/577


EW

S

ø = 450301 is in first quadrant

WCB = QB = N450301E

IIIIIA

N

W.C.B of 122030

1Contd…


199/577


EW

S

QB = 1800 – W.C.B = S E

= 1800 - 122030

1

= S770301E

QB = S770301E

I

IIIII

IV

B

C

N

W.C.B of 225000

1

I Q d t

Contd…


200/577


EW

S

I Quadrant

II Quadrant

III Quadrant

IV Quadrant

C

D

ø3 = 2250001 is in third quadrant

QB = W.C.B – 1800 = SW

= 2250001 – 1800 = S450001W

QB = S450001W

N

W.C.B of 305030

1

I Q d t

E

Contd…


201/577


EW

S

QB = 3600 – W.C.B = NW

QB = 3600 - 305

030

1

= N550301W

W.C.B of 305030

1 = QB of N 550301 W

I Quadrant


IV Quadrant

D

9. Convert the following whole circle bearings into reduced


202/577


bearings.

a) 46010

1

b) 226010

1

c) 1190201

d) 298040

1

IIV

NB

46010l

Contd…


203/577



W.C.B = Q.B

46010l = N 46010l E

IIIII

E

S

W A

IIV

NContd…


204/577



Q.B = W.C.B – 1800 = SW

Q.B = 2260

10l

– 1800

Q.B = S 460101 W

IIIII

E

S

W A

B

226010l

IIV

N

119020l

Contd…


205/577



Q.B = 1800 – W.C.B

Q.B = 1800 – 119020l

Q.B = S 600401 E

IIIII

E

S

W A

B

119020l

IIV

N

B

Contd…


206/577



Q.B = 3600 – W.C.B 1800

Q.B = 3600 – 298040lQ.B = N 61020l W

IIIII

E

S

W A298040l

10. Convert the following whole circle bearing into reduced


207/577


g g

bearing.

a) 1690301

b) 3510101

c) 2800201

d) 990201

IIV

N

169030l

Contd…


208/577


169030l it is in second quadrant

Q.B = 1800 - W.C.B

Q.B = 1800 - 169030l

Q.B = S 10030l E

IIIII

E

S

W A

B

169 30

IIV

NBContd…


209/577



Q.B = 3600 – W.C.B

Q.B = 3600 – 351010l

Q.B = N 80501 W

II

III

E

S

WA

351010l

IIV

N

B

Contd…


210/577



Q.B = 3600 – W.C.B

Q.B = 3600 - 280020l

Q.B = N 790401 W

IIIII

E

S

W A

B

280020l

IIV

NContd…


211/577



Q.B = 1800 – W.C.B

Q.B = 1800 – 99020l Q.B = S 80040l E

IIIII

E

S

W AB99

020l

11. Convert the following whole circle bearing into quadrantal


212/577


bearing.

a) 440301

b) 1240301

c) 181001

d) 2890301

IIV

N B

44030l

Contd…


213/577



W.C.B = Q.B

Q.B = N 44030l E

IIIII

E

S

W A

44 30

IIV

N

124030l

Contd…


214/577



Q.B = 1800 – W.C.B

Q.B = 1800 – 124030l

Q.B = S 55030l E

IIIII

E

S

W A

B

124030l

IIV

NContd…


215/577



Q.B = W.C.B – 1800

Q.B = 18100l - 1800

Q.B = S 1001 W

II

III

E

S

W A

B

18100l

IIV

NContd…


216/577



Q.B = 3600 – W.C.B

Q.B = 3600 – 289030l

Q.B = N700301 W

II

III

E

S

WA

B 289030l

Problems

1.Convert the following Q.Bs into W.C.B.


217/577


1.Convert the following Q.Bs into W.C.B.

a) N450301W

b) N600401W

c) N30201W

d) S56001E

IIV

N

B

45030l

Contd…


218/577



Q.B = 3600 – W.C.B

Q.B = 3600 - 45030l

Q.B = N 314030lW

IIIII

E

S

W A

IIV

N

B

60040l

Contd…


219/577



W.C.B = 3600 – 60040l

Q.B = N 299020lW

IIIII

E

S

W A

IIV

NB

3020l

Contd…


220/577


The line is in fourth quadrant

Q.B = 3600 – W.C.B

Q.B = 3600 - 3020l

Q.B = N 356040lW

IIIII

E

S

W A

IIV

NContd…


221/577


The line is in second quadrant

W.C.B = 1800 – Q.B

Q.B = 1800 – 5600lQ.B = S 12400l E

IIIII

E

S

W A

B

5600l

2.Convert the following Q.B into W.C.B.


222/577


a) S260141E

b) N780181W

c) N69091E

d) S320151W

IIV

NContd…


223/577



W.C.B = 1800 – Q.B

Q.B = 1800

– 260

14l

Q.B = S 153046l E

IIIII

E

S

W A

B

26014l

IIV

N

B

78018l

Contd…


224/577



W.C.B = 3600 – Q.B

Q.B = 3600 - 78018l

Q.B = N 281042lW

IIIII

E

S

W A

IIV

NB

6909l

Contd…


225/577



W.C.B = Q.B

Q.B = N 6909l E

IIIII

E

S

W A

IIV

NContd…


226/577



W.C.B = Q.B + 1800

Q.B = 320

15l

+ 1800

Q.B = S 2120151 W

IIIII

E

S

W A

B

32015l



227/577


a) S460241W

b) S220181E

c) N82091W

d) N320221E

IIV

NContd…


228/577



W.C.B = Q.B + 1800

Q.B = 460

24l

+ 1800

Q.B = S 2260241 W

IIIII

E

S

W A

B

46024l

IIV

NContd…


229/577



W.C.B = 1800 – Q.B

Q.B = 1800 – 22018l

Q.B = S 157042l E

IIIII

E

S

W A

B

22018l

IIV

N

B

8209l

Contd…


230/577



W.C.B = 3600 – Q.B

Q.B = 3600 - 8209l

Q.B = N 277051lW

IIIII

E

S

W A

IIV

N B

32022l

Contd…


231/577



W.C.B = Q.B

Q.B = N 32022l E

IIIII

E

S

W A



232/577


a) S370301E

b) S430151W

c) N73001W

d) N120451E

IIV

NContd…


233/577



W.C.B = 1800 – Q.B

W.C.B = 1800 – 37030lW.C.B = S 142030l E

IIIII

E

S

W A

B

37030l

IIV

NContd…


234/577



W.C.B = Q.B + 1800

W.C.B = 430

15l

+ 1800

W.C.B = S 2230151 W

IIIII

E

S

W A

B

43015l

IIV

N

B

7300l

Contd…


235/577



W.C.B = 3600 – Q.B

W.C.B = 3600 - 7300lW.C.B = N 28700lW

IIIII

E

S

W A

IIV

N B

12045l

Contd…


236/577



W.C.B = Q.B

W.C.B = N 12045l E

IIIII

E

S

W A



237/577


a) N840301W

b) S38001E

c) S410301W

d) N15001W

IIV

N

B

84030l

Contd…


238/577



W.C.B = 3600 – Q.B

W.C.B = 3600 - 84030lW.C.B = N 275030lW

IIIII

E

S

W A

IIV

NContd…


239/577



W.C.B = 1800 – Q.B

W.C.B = 1800 – 3800lW.C.B = S 14200l E

IIIII

E

S

W A

B

3800l

IIV

NContd…


240/577



W.C.B = Q.B + 1800

W.C.B = 410

30l

+ 1800

W.C.B = S 2210301 W

IIIII

E

S

W A

B

41030l

IIV

N

B

1500l

Contd…


241/577



W.C.B = 3600 – Q.B

W.C.B = 3600 - 1500lW.C.B = N 34500lW

IIIII

E

S

W A



242/577


a) S150451W

b) N430451E

c) N210301E

d) S630301W

IIV

NContd…


243/577



W.C.B = Q.B + 1800

W.C.B = 150

45l

+ 1800

W.C.B = S 1950451 W

IIIII

E

S

W A

B

15045l

IIV

N B

43045l

Contd…


244/577



W.C.B = Q.B

W.C.B = N 43045l E

IIIII

E

S

W A

IIV

N B

21030l

Contd…


245/577



W.C.B = Q.B

W.C.B = N 21030l E

IIIII

E

S

W A

IIV

N

A

Contd…


246/577



W.C.B = Q.B + 1800

W.C.B = 63

0

30

l

+ 180

0

W.C.B = S 2430301 W

IIIII

E

S

W A

B

63030l



247/577


a) N460101E

b) S460101W

c) S600

401

E

d) N610201W

IIV

N

EW

B

46010l

Contd…


248/577



W.C.B = Q.B

W.C.B = N 46010l E

IIIII

E

S

W A

IIV

N

EW A

Contd…


249/577



W.C.B = Q.B + 1800

W.C.B = 46

0

10

l

+ 180

0

W.C.B = S 2260101 W

IIIII

E

S

W A

B

46010l

IIV

NContd…


250/577



W.C.B = 1800 – Q.B

W.C.B = 1800 – 60040lW.C.B = S 119020l E

IIIII

E

S

W A

B

60040l

IIV

N

EW

B

61020l

Contd…


251/577



W.C.B = 3600 – Q.B

W.C.B = 3600 - 61020lW.C.B = N 298040lW

IIIII

E

S

W A



252/577


a) N 470301 E

b) S 65001 E

c) S 410301 W

d) N 20001 W

IIV

N

EW

B

47030l

Contd…


253/577



W.C.B = Q.B

W.C.B = N 47030l E

IIIII

E

S

W A

IIV

NContd…


254/577



W.C.B = 1800 – Q.B

W.C.B = 1800 – 6500lW.C.B = S 11500l E

IIIII

E

S

W A

B

6500l

IIV

N

EW A

Contd…


255/577



W.C.B = Q.B + 1800

W.C.B = 41030l + 1800

W.C.B = S 2210301 W

IIIII

E

S

W A

B

41030l

IIV

N

EW A

B

2000l

Contd…


256/577



W.C.B = 3600 – Q.B

W.C.B = 3600 - 2000lW.C.B = N 34000lW

IIIII

E

S

W A

9.Convert the following QB into W.C.B

Contd…


257/577


a) QB of AB = S360301W

b) QB of BC = S430301E

c) QB of CD = N260451E

d) QB of DE = N400151W

N


Contd…


258/577


EW

S

II Quadrant

III Quadrant

A

B

Third quadrant S.W W.C.B = QB + 1800

= 360301 + 1800 = 2160301

W.C.B = 2160301

N

Q.B of BC = S430301E

IIV

Contd…


259/577


EW

IIIII

B

C

Second quadrant Q.B = S.E = W.C.B = 1800- 43030

1

W.C.B = 1360

301

S

N

IIV

D

Q.B of CD = N26045

1E

Contd…


260/577


EW

S

I

IIIII

IV

C

W.C.B = NE = WCB = QB

W.C.B = 260451

N

Q.B of N400151W


D

Contd…


261/577


EW

S


C

Angle is in Fourth quadrant

W.C.B = 3600-Q.B

= 3600 - 400151

W.C.B = 3190451

10.Convert the following QB into W.C.B


262/577


a) N400151E

b) S180141W

c) N240201Wd) S34031E

N

IIV

B

Contd…


263/577


EW

S

I

IIIII

IV

A

AB line is in First Quadrant

W.C.B = QB = 40

0

15

1

N


Contd…


264/577


EW

S

II Quadrant

III Quadrant

B

C

BC is in Third quadrant

W.C.B = QB + 1800

= 18014

1 + 1800

W.C.B = 1980141

N

N240201W


D

Contd…


265/577


EW

S


C

The DC is in Fourth quadrant

W.C.B = 3600-Q.B

= 3600 - 240201

W.C.B = 3350401

N

S34031E

IIV

Contd…


266/577

COMPASS SURVEYING 2669C-105.40 266

EW

S

IIIII

D

E

The DE is in Second quadrant

W.C.B = 1800- Q.B

= 1800- 34031

W.C.B = 1450571


267/577

Contd…

The bearing in the direction from A to B is Fore bearing of AB ,

say α


268/577


say α .

This is simply called bearing of AB

The bearing of line AB in the direction from B to A is Back

bearing of AB , say β

Contd…


269/577


FIG.1

Referring to Fig 1

Let the fore bearing of a line AB = αо

Contd…


270/577


Back bearing of BA= βo

(Or)

β = (1800 + ∟SBA)

= ∟BAN’

= α + 1800

Back bearing = Fore bearing + 1800 … (i)

Contd…


271/577


Fig 2

Now , consider the bearing of BA as Fore bearing = β

Contd…


272/577


Then , α = (1800 ~ ∟S’AB )

= (1800 ~ ∟ABN)

= 1800 ~ (3600 ~ β)

= ( β ~ 1800)

(or )

Back bearing = Fore bearing ~ 1800 … (ii)

Equations (i) and (ii) may be combined into one equation

i.e.

Contd…


273/577


Back bearing = Fore bearing ± 1800

Use + ve sign if the Fore Bearing is less than 1800

Use – ve sign if the Fore Bearing is more than 1800

The rule is applicable for W.C.B System only

Contd…


274/577


If Q.B’s are given , convert them into W.C.B’s

And apply the above rule and reconvert into Q.B’s

(or)

In case of Q.B’s the numerical value of FB’s and BB’s are

same.

But the cardinal directions are reversed as follows

Contd…


275/577


For North, substitute South and vice-versa

For East, substitute West and vice-versa

The following example illustrates all the above cases

Example 1:

The following are the observed Fore bearings of some lines in a

traverse. Find their Back bearings.


276/577


1) AB 420 30’

2) BC 1350

15’

3) CD 2120 30’

4) DE 2850 45’

Solution

1) F.B of AB = 420 30’

B B f AB (420 30’+1800)


277/577


B.B of AB = (420 30’+1800)

= 2220 30’

2) F.B of BC = 1350 15’

B.B of AB = (1350 15’+1800 )

= 3150 15’

Contd…


278/577


Fig.3A Fig.3B

3) F.B of CD = 2120 30’

. . B.B of CD = (2120 30’ ~ 1800)

Contd…


279/577


... B.B of CD (212 30 180 )

= 320 30’

4) F.B of DE = 2850 45’

... B.B of DE = (2850 45’ ~ 1800)

=1050 15’

Contd…


280/577


Fig.4A Fig.4B

Example :2

The F.B of AB is N 300 E .Find its B.B


281/577


Solution:

The F.B of AB is N 300 E

The B.B of AB is S 300

W.

Solution by converting to WCB system

The F.B of AB is N 300 E


282/577


The F.B of AB is in WCB = 300

The B.B of AB is in WCB = 300+1800

= 2100

By reconverting into Q.B The B.B of AB is S 300 W.

Calculation of included anglesCase 1:-

When the bearings of two line measured from same station

(From their point of inter section)


283/577


( p ) Let F. B of AB is θ1 and AC is θ2 as shown below

Angle ∟BAC

= (Bearing of BC ~ Bearing of AB)

= (θ2 - θ1 )= (Bearing of preceding line ~ bearing of succeeding line)

A

B

C

N

θ2θ1

Fig.5

Case:2

When the bearings of two lines measured from different

stations

Contd…


284/577


A

B

θ1

θ2

C

Fig.6

In this case express the bearings of two

– Lines as if measured from single station

– And apply the formula

Contd…


285/577


pp y

Thus find B.B of AB

Then ∟B = F.B of BC ~ B.B of AB

Note:-

1) When two lines meet at a point two angles

– The interior and exterior angles are formed

– The sum of these angles = 3600

Contd…


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g

– ... If the calculated interior angle is >1800 it is exterior

– The interior angle then is = 3600 ~ exterior angle.

2) If the bearing are given in Q.B . System convert them to

W.C.B system and then apply the above rules

– The following numerical examples illustrates the above

cases.

Contd…

Note :-


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“ ~ ”This symbol indicates difference.

Therefore whether we subtract bigger bearing from smaller

bearing

or

vice-versa

we should take only numerical value into consideration

Example 3

Find the angle between the lines OA &OB given their bearings as

below

350 30’ d 1500 45’


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350 30’ and 1500 45’

Solution :

Draw the sketch.

From sketch ∟AOB = Bearing of OB ~ bearing of OA

= 1500 45’ ~ 350 30`

= 1150 15’

Contd…


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Fig.7

Example 4


below

200 15’ and 3230 30’

Solution :


290/577


Draw the sketch.

From sketch ∟AOB = bearing of OB ~ bearing of OA

= 3230 30’ ~ 200 15’

= 3030 15’>1800

= Indicates exterior angle

= 560

45’Therefore interior ∟AOB = 3600 ~ 3030 15

Contd…


291/577


Fig.8

Example 5


below


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1150 15’ and 2500 45’

Solution :

Draw the sketch.

From sketch ∟AOB = Bearing of OB ~ bearing of OA

= 2500 45’ ~ 1150 15’

= 1350 30’

Contd…


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Fig.9

Example 6

The bearings of AB and BC are given below. Find the angle

between the lines

1520 15’ and 700 30’


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Solution :

In this problem the bearings of lines are not from point of

intersection

Therefore express the bearings as if measured from point of

intersection

Then apply the rule.

Contd…


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Fig.10

Bearing of AB = 1520 15`

Bearing of BA = 1520 15` + 1800

= 3320 15`

Contd…


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(Note that Bearing of BA is nothing but B.B of AB)

Bearing of BC = 700 30’

Included ∟B = 3320 15 ~ 700 30’

= 2610 45’ > 1800 Indicates exterior angle.

Therefore Included ∟B = 3600 ~ 2610 45

= 980 15’

Example 7

The bearings of AB and BC are given below. Find the angle

between the lines

650 30’ and1150 15`

Solution :


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Bearing of AB = 650 30’

Bearing of BA = 650 30’+1800

= 2450 30’

(Note that Bearing of BA is nothing but B.B of AB)

Bearing of BC = 1150 15’

Included ∟B = 2450 30’ ~ 1150 15’

= 1300 15`

Contd…


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Fig.11

Example - 8

Find the include angle B given the following bearings

Bearing of AB, N 150 15` E and bearing of AC, N. 870 10` E


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Solution

Step(1)

Convert the given Q. Bs into W. C. Bs

Step(2)

Apply the formula as usual

W.C.B of AB = 15015`

W.C.B of AC = 870

10`

Contd…


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∟BAC = (W.C.B of AC ~ W.C.B of AB)

= 87010` ~ 15015`

= 71055`

Find the include angle B given the following bearings:

Bearing of AB, N 120 24` E and bearing of AC, S. 520 30` E

Contd…

Example - 8


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g , g ,

Solution

Step(1):

Convert the given Q. Bs into W. C. Bs

Step(2):

Apply the formula as usual

W.C.B of AB = 12024`

W.C.B of AC = 1800 ~ 52030`

Contd…


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= 127030`

∟BAC = (W.C.B of AC

~ W.C.B of AB)

= 12024` ~ 127030`

= 11506`

Example - 9

The bearing of line AB is N.620 45` E ;

CA is N.280 15`E Find angle BAC

Solution

C t th b b i i t W C B' d l th l


303/577


Convert the above bearings into W.C.B's and apply the rule

W.C.B if AB = 620 45`

Bearing of CA = N.280 15`W

Bearing of AC = S 280 15`WW.C.B of AC = 280 15` + 1800

= 2080 15`

angle BAC = Bearing of AB ~ Bearing of AC

= 620 45` ~ 2080 15`

= 1450 30`

From the above two examples ,we can conclude that

In a traverse angle between two successive lines

Contd…


304/577


In a traverse angle between two successive lines

= (Back bearing of preceding line ~ bearing of succeeding line)

Procedure to calculate included angles of a Traverse

Draw the required sketch

Carefully enter the readings


305/577


If only fore bearings are given, find corresponding back

bearings.

If the bearings are given in Q. Bs, convert them into W. C. Bs

and find the included angles

Apply the check that the sum of interior angles

Contd…


306/577


= (2n-4) x 900

Where ‘n’ is the no. of sides of the traverse

Example - 1

The following bearings were taken in a closed traverse

ABCD

Line Fore bearing Back bearing


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Calculate the interior angles of the traverse

Line Fore bearing Back bearing

AB 45015` 2250 15`

BC 1230 15` 3030 15`

CD 1810 00` 10 00`

DA 2890 30` 1090 30`

Solution :

Calculation of the interior angles of the closed traverse ABCD


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Fig.1

Contd..

A = (Fore bearing of AB – Back Bearing of DA)

= (450 15` -1090 30`)


309/577


( )

= 640 15`

B = (Fore Bearing of BC – Back Bearing of AB)

= (1230 15 - 2250 15`0 )

= 1020 00`

Contd…

Check : ∟A + ∟B+ ∟C+ ∟D

= 640 15` + 1020 00`+ 1220 15`+ 71030`

= 3600 O.K

A= 640 15`


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A= 640 15

B= 1020 00`

C=1220 15`

D=710

30`

Contd…

The arrow marks indicated below in the table may be followed

for calculation of included angles


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Line Fore bearing Back

bearing

AB 45015` 2250 15`

BC 1230 15` 3030 15`

CD 1810 00` 10 00`

DA 2890 30` 1090 30`

Example - 2 The bearing of the sides of a closed transverse ABCDEA are as

follows

Side F.B B.B

AB 1070 15` 2870 15`


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Compute the interior angles of the traverse and exercise necessary

checks

AB 1070 15 2870 15

BC 220 00` 2020 00`

CD 2810 30` 1010 30`

DE 1810 15` 10 15`

EA 1240 45` 3040 45`

Contd…


313/577


Fig.2

Solution

(i) The included ∟A = (FB of AB ~ BB of EA )

= (1070 15` ~ 3040 45`)


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= (1070 15 ~ 3040 45 )

= 197030` > 1800 (Exterior)

= 3600 ~ 197030` = 162030`

Contd..

(ii) Included angle at B:

= ( FB of BC ~ BB of AB)

= (220 00` ~ 2870 15`)= (2650 15`>1800 ) (Exterior)


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= (2650 15 >1800 ) (Exterior)

= (3600 ~ 265015`)

= 940 45`

(iii) Included angle at C:

= (FB of CD ~ BB of BC)

= (2810

30` ~ 2020

00`)= 790 30`

(iv) Included angle D

= (FB of DE ~ BB of CD)

= (1810 15` ~ 1010 30`)

Contd..


316/577


= 790 45`

(v) Included angle E

= (FB of EA ~ BB of DE)

= (1240 45` ~ 10 15)

= 1230 30`

Check

Sum of the included angles of a pentagon

= (2x

buiding services compass survey

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