Compass Surveying

Compass Surveying
• Chain surveying can be used
when the area to be
surveyed is comparatively
small and is fairly flat.
• But when the area is large,
undulated and crowded with
many details, triangulation
(which is the principle of
chain survey) is not
possible. In such an area,
the method of traversing is
adopted.

Traversing
• In Traversing, the framework consist of a number
of connected lines. The length are measured by a
chain or a tape and the directions measured by
angle measuring instruments. In one of the
methods, the angle (direction) measuring
instrument is the compass. Hence, in compass
surveying directions of survey lines are
determined with a compass and the length of the
lines are measured with a tape or a chain. This
process is known as Compass Traversing.

Principle of Compass Surveying
• The Principle of Compass Survey is Traversing;
which involves a series of connected lines the magnetic
bearing of the lines are measured by prismatic compass
and the distance (lengths) of the are measured by chain.
Such survey does not require the formulation of a
network of triangle.
• Compass surveying is recommended when the area is
large, undulating and crowded with many details.
• Compass surveying is not recommended for areas
where local attraction is suspected due to the presence
of magnetic substances like steel structures, iron ore
deposits, electric cables conveying currents, and so on.

Types and Uses of Compass
• Compass: A compass is a small instrument essentially
Consisting of magnetic needle, a graduated circle, and a
line of sight. The compass can not measure angle
between two lines directly but can measure angle of a
line with reference to magnetic meridian at the
instrument station point is called magnetic bearing of a
line. The angle between two lines is then calculated by
getting bearing of these two lines.
• There are two forms of compass available:
• The Prismatic Compass
• The Surveyor’s Compass

Compass Surveying
The Prismatic Compass
The prismatic compass is a
magnetic compass which consists
of the following parts.
Cylindrical Metal Box
Cylindrical metal box is having
diameter of 8 to 12 cm. It
protects the compass and forms
entire casing or body of the
compass. It protects compass
from dust, rain etc.

Pivot
• Pivot is provided at the centre of the
compass and supports freely suspended
magnetic needle over it.
Lifting Pin and Lifting Lever
• A lifting pin is provided just below the
sight vane. When the sight vane is
folded, it presses the lifting pin. The
lifting pin with the help of lifting lever
then lifts the magnetic needle out of pivot
point to prevent damage to pivot head.
Spring Brake or Brake Pin
• To damp the oscillation of the needle
before taking a reading and to bring it to
rest quickly, the light spring brake
attached to the inside of the box is
brought in contact with edge of the ring
by gently pressing inward the brake pin.

• Magnetic Needle: Magnetic needle
is the heart of the instrument. This
needle measures angles of a line
from magnetic meridian a the
needle always remains pointed
towards north and south pole at the
two ends of the needle when freely
suspended on any support.
• Graduated Circle or Ring: This is
an aluminium graduated ring
marked with 0 0 to 360 0 to measure
all possible bearings of lines, and
attached with the magnetic needle.
The ring is graduated to half a
degree.

• Prism
Prism is used to read graduations on ring
and to take exact reading by compass. It
is placed exactly opposite to object vane.
The prism hole is protected by prism cap
to protected by prism cap to protect it
from dust and moisture.
• Object Vane
Object Vane is diametrically opposite to
the prism and eye vane. The object vane
is carrying a horse hair or black thin wire
to sight object in line with eye sight.
• Eye Vane
Eye Vane is a fine slit provided with eye
hole at bottom to bisect the object from
the slit and to take reading simultaneously
from the eye hole. This eye vane is
provided with prism and can be lifted up
and down by the stud to bisect the object
of higher level.

• Glass Cover: It covers the
instrument box from the top such
that needle and graduated ring is
seen from the top.
• Sun Glasses: These are used when
some luminous objects are to be
bisected. These are placed in front
of the eye slit and in bunch of three
or four shades of different colours to
give sharp picture of the object only.
• Reflecting Mirror: It is used to get
image of an object located above or
below the instrument level while
bisection. It is placed on the object
vane.

Working of the Prismatic
Compass
• When the needle of the compass is
suspended freely. It always points
towards the north. Therefore, all
the angles measured with prismatic
compass are with respect to north
(magnetic meridian).
• “The horizontal angle made by a
survey line with reference to
magnetic meridian in clockwise
direction is called the bearing of a
line.’
• While using the compass, it is
usually mounted on a light tripod
which is having vertical spindle in
the ball and socket arrangement to
which the compass is screwed.

Temporary Adjustment of a
Prismatic Compass
• The following procedure should be adopted after the prismatic compass on
the tripod for measuring the bearing of a line:
Centering
Centering is the operation in which compass is kept exactly over the station
from where the bearing is to be determined. The centering is checked by
dropping a small pebble from the underside of the compass. If the pebble
falls on the top of the peg then the centering I correct, if not then the
centering is corrected by adjustment the legs of the tripod.
Levelling
Levelling of the compass is done with the aim to freely swing the graduated
circular ring of the prismatic compass. The ball and socket arrangement on
the tripod will help to achieve a proper lever of the compass. This can be
checked by rolling round pencil on glass cover.
Focusing
The prism is moved up or down in its slide till the graduations on the
aluminium ring are seen clear, sharp and perfect focus. The position of the
prism will depend upon the vision of the observer.

Observing Bearing of a line
• Consider a line AB of which the magnetic bearing is to
be observed.
• Let the ranging rod be fixed at B in line AB and the
compass is centered on A.
• Turn the compass in the direction of line AB.
• When B is bisected by the vertical hair. i.e. when
ranging rod at B comes in line with the slit of eye vane
and the vertical hair. i.e. when ranging rod at B comes
in line with the slit of eye vane and the vertical hair of
the object vane, the reading, under the vertical hair
through prism is taken, which gives the bearing of line
AB. The enlarged portion gives actual pattern of
graduations marked on ring.

The Surveyor’s Compass
• It is similar to a prismatic
compass except that it has
only plain eye slit instead
of eye slit with prism and
eye hole.
• This compass is having
pointed magnetic needle in
place of broad form
needle as in case of
prismatic compass.

The Surveyor’s Compass
Working of Surveyor’s Compass:
• Centering
• Levelling
• Observing the Bearing of a Line
• First two operations are similar to that
of prismatic compass but the method
of taking observation differs from that.
• Observing the bearing of a line. In this
type of compass, the reading is taken
from the top of glass and under the tip
of north end of the magnetic needle
directly. No prism is provided here.

Difference between Prismatic Compass
and Surveyor’s Compass

Bearing
• The bearing of a line is the horizontal angle which it
makes with a reference line (meridian) depending upon
the meridian, there are four types of bearings.
True Bearing
The true bearing of a line is the horizontal angle
between the true meridian and the survey line. The true
bearing is measured from the true north in the clockwise
direction.
Magnetic Bearing
The magnetic bearing of a line is the horizontal angle
which the line makes with the magnetic north.

Bearing
Grid Bearing
The grid bearing of a line is the horizontal angle
which the line makes with the grid meridian.
Arbitrary Bearing
The arbitrary bearing of a line is the horizontal angle
which the line makes with the arbitrary meridian

Designation of Bearings
• The bearing are designated in the following
two systems.
• Whole Circle Bearing System (W.C.B)
• Quadrantal Bearing System ( Q.B.)

Whole Circle Bearing System
(W.C.B)
• The bearing of a line measured with respect to
magnetic meridian in clockwise direction is
called magnetic bearing and its value varies
between 0 0 to 360 0.
• The Quadrants start from North and Progress
in a clockwise direction as the first quadrant is
0 0 to 90 0 in clockwise direction, 2nd 90 0 to
180 0 , 3 rd 180 0 to 270 0 , and up to 360 0 is 4th
one.

Whole Circle Bearing System
(W.C.B)

Quadrant Bearing System (Q.B.)
• In this system, the bearing of survey lines are
measured with respect to north line or south line
which ever is the nearest to the given survey line
and either in clockwise direction or in
anticlockwise direction.
• The bearing of lines which fall in Ist and IV th
Quadrant are measured with respect to north line
is nearer than south line, and bearing of lines fall
in II nd and IIIrd quadrants are measured from
south line as south is the nearer line. The
surveyor’s compass measures the bearing of lines
in the quadrant system.

Reduced Bearing (RB)
• When the whole circle bearing of a line is
converted into quadrantal bearing it is termed
as ‘Reduced Bearing’. Thus, the reduced
bearing is similar to the quadrantal bearing. It’s
value lies between 0 0 to 90 0, but the
quadrants should be mentioned for proper
designation.

The Following Table Should be Remembered
for Conversion of WCB to RB

The Following Table Should be Remembered
for Conversion of RB to WCB

Fore Bearing and Back Bearing
• The bearing of a line measured in the forward
direction of survey line is called the ‘Fore
Bearing’ (FB) of that line.
• The bearing of the line measured in the
direction opposite to the direction of the
progress of survey is called the ‘Back Bearing’
(BB) of the line.

Fore Bearing
Fore Bearing
• The bearing of a line measured in the forward
direction (i.e. along the progress of survey) is
known as fore bearing.
• ForeBearing = Back Bearing ± 180°

Back Bearing
• The bearing of a line measured in the Backward
direction is known as Back Bearing.
• BB= FB 180 0
• + sign is applied when FB is < 180 0
• - sign is applied when FB is > 180 0
• In the quadrantal bearing (i.e. reduced bearing)
system the FB and BB are numerically equal but
the quadrant are just opposite.
• For example if the bearing of AB is N 60 0 E, then
its BB is S 60 0 W.

Examples
• Convert the following WCB into Reduced
Bearing.
• 49 0
• 240 0
• 133 0
• 335 0

Examples
49 0
• Since the line falls in the first quadrant
therefore the nearer pole is the north pole and
is measured from North towards E as 49 0
• There fore RB = N 49 0 E

Examples
240 0
• Since the line falls in the third quadrant
is measured from North towards S as 0
• RB = WCB- 180 0
• RB = 240 0 - 180 0 = 60 0
• RB= S 60 0 W

Examples
240 0
is measured from South towards W as 0
• RB = WCB- 180 0
• RB = 240 0 - 180 0 = 60 0
• RB= S 60 0 W

Examples
133 0
• Since the line falls in the second quadrant
therefore the nearer pole is the south pole and
is measured from South towards E as 0
• RB = 180 0 - Ө
• RB = 180 0 – 133 0 = 47 0
• RB= S 47 0 E

Examples
335 0
is measured from North towards W as 0
• RB = 360 0 – WCB
• RB = 360 0 – 335 0
• RB= N 25 0 W

Examples
Convert the following WCB into RB
• 190 0
• 260 0
• 315 0

Examples
Soln
190 0
• RB= WCB – 180 0
• RB = 190 0- 180 0
• RB = S 10 0 W
260 0
• RB = WCB-180 0
• RB= 260 0 – 180 0
• RB = S 80 0 W

Examples
Soln
315 0
• RB = 360 0 – WCB
• RB = 360 0 – 315 0
• RB = N45 0 W

Examples
• Convert the following reduced bearings into
whole circle bearings:
• N 65° E
• S 43° 15′ E
• S 52° 30′ W
• N 32° 42′ W

Examples
Let ‘θ’ be whole circle bearing.
(i) Since it is in NE quadrant,
θ = α = 65 Ans.
(ii) Since it is in South East quadrant
43 15′ = 180 – θ
or θ = 180 – 43 15′ = 136 45′ Ans.

Examples
(iii) Since it is in SW quadrant
52 30′ = θ – 180
or θ = 180 + 52 30′ = 232 30′
(iv) Since it is in NW quadrant,
32 42′ = 360 – θ
or θ = 360 – 32 42′ = 327 18′

Examples
• The following fore bearings were observed for
lines, AB, BC, CD, DE, EF and FG respectively.
Determine their back bearings:
• 148
• 65
• 285
• 215
• N 36 W
• S 40 E

Examples
Solution:
• The difference between fore bearing and the
back bearing of a line must be 180 . Noting
that in WCB angle is from 0 to 360 ,
• we find Back Bearing = Fore Bearing 180
• + 180 is used if θ is less than 180 and
• – 180 is used when θ is more than 180

Examples
Hence,
• BB of AB = 145 + 180 = 325
• BB of BC = 65 + 180 = 245
• BB of CD = 285 – 180 = 105
• BB of DE = 215 – 180 = 35
• In case of RB, back bearing of a line can be
obtained by interchanging N and S at the same
time E and W. Thus
• BB of EF = S 36 E
• BB of FG = N 40 W.

Example
The Fore Bearing of the following lines are
given Find the Back Bearing.
(a) FB of AB= 310 0 30’
(b) FB of BC= 145 0 15’
(c) FB of CD = 210 0 30’
(d) FB of DE = 60 0 45’

Example
Solution
(a) BB of AB = 310 0 30’ – 180 0 0’ = 1300 30’
(b) BB of BC = 145 0 15’ + 180 0 0’ = 3250 15’
(c) BB of CD = 210 0 30’ – 180 0 0’ = 30 0 30’
(d) BB of DE = 600 45’ + 180 0 0’ = 240 0 45’

Example
FB of the following lines are given, find the
BBs.
(a) FB of AB = S 300 30’ E
(b) FB of BC = N 400 30’W
(c) FB of CD= S 600 15’W
(d) FB of DE = N 45030’ E

Example
Solution
• (a) BB of AB = N 30 0 30’ W
• (b) BB of BC = S 40 0 30’ E
• (c) BB of CD = N 60 0 15’ E
• (d) BB of DE = S 45 0 30’ W

Example
• The fore bearing of the lines AB, BC, CD
and DE are 45 0 30’, 1200 15’, 200 0 30’ and
280 0 45’ respectively, find angles B,C,D

Example
• Interior angle B= BB of AB-FB of BC
= (45 0 30’ + 180 0 0’) -120 0 15’
= (225 0 30’ – 120 0 15’) = 105 0 15’
• Interior angle C= BB of BC – FB of CD
= (120 0 15’ + 180 0 0’) – 200 0 15’
= 300 0 15’ – 200 0 15’ = 100 0 0’
• Exterior angle D = FB of DE- BB of CD
= 280 0 45’ – (200 0 30’ – 180 0 0’)
= 280 0 45’ – 20 0 30’ = 260 0 15’
• Interior angle D= 360 0 0’ – 260 0 15’ = 99 0 45’

Computation Of Angles
• Observing the bearing of the line of a closed
traverse, it is possible to calculate the included
angles, which can be used for plotting the
traverse.
• At the station where two survey lines meet,
two angles are formed, an exterior angle and
an interior angle. The interior angle or
included angle is generally the smaller angle
(< 180 0).

• While calculating the interior or included angles,
it is strongly recommended that a rough sketch of
the traverse must be drawn for the purpose of
calculating the interior angles or bearing from
included angles. A sketch always gives a better
idea for calculations.
• At any survey stations generally FB of one line
and BB of another line are measured. Then
difference of these two bearings will give you
either an interior angle or an exterior angle
depending upon the station position.

• In a closed traverse the following bearings
were observed with a compass. Calculate the
interior angles.

We find,
Back Bearing = Fore Bearing 180
+ 180 is used if θ is less than 180 and
– 180 is used when θ is more than 180

Referring to Figure:
∠A = 150 00′ – 65 00′ = 85 00′
∠B = 245 00′ – 125 30′ = 119 30′
∠C = 305 30′ – 200 00′ = 105 30′
∠D = (360 – 265 15′) + 20 00′ = 114 45′
∠E = (360 – 330 00′) + 85 15′ = 115 15′

• The angles observed with a surveyor compass
in traversing the lines AB, BC, CD, DE and EF
are as given below.
• Compute the included angles and show them
in a neat sketch.

In case of RB, back bearing of a line can be obtained by
interchanging N and S at the same time E and W

• Referring to the figure, we find
• ∠B = 55 30′ + 63 30′ = 119 00′.
• ∠C = 63 30′ + 70 00′ = 133 30′.
• ∠D = 70 00′ + 45 30′ = 115 30′.
• ∠E = 45 30′ + 72 15′ = 117 45′.

• The following Bearing were observed for a
closed traverse.
• Calculate the interior angle of a traverse

∠ A= 180 0 – (FB of DA+ BB of DA)
= 180 0 – (45 0 30’ +75 0 45’)
= 180 0 – 121 0 15’
= 58 0 45’
∠ B= BB of AB + FB of BC
= 45 0 30’ + 60 0 0’
= 105 0 30’

∠ C = 180 0 – (BB of BC + FB of CD)
= 180 0 – ( 60 0 0’ + 10 0 30’)
= 180 0 – 70 0 30’
= 109 0 30’
∠ D= BB of CD + FB of DA
= 10 0 30’ + 750 45’
= 86 0 15’
Check Sum of Interior angle (2N-4) 90 0 =360 0
Now,
∠ A + ∠ B + ∠ C + ∠ D
= 58 0 45’ + 105 0 30’ + 109 0 30’ + 86 0 15’ = 360 0

• The following bearing were taken in a closed
traverse find out the interior angles of a
traverse
Line FB BB
AB 45 0 00’ 225 0 00’
BC 123 0 30’ 303 0 30’
CD 181 0 00’ 1 0
DA 289 0 00’ 109 0 00’

Calculation of Interior Angle:
∠ A= BB of DA- FB of AB
= 109 0 0’ – 45 0 0’
= 64 0 0’
∠ B= BB of AB – FB of BC
= 225 0 0’ – 123 0 30’
= 101 0 30’
∠ C= B B of BC – FB of CD
= 303 0 30’ – 181 0 0’
= 122 0 30’

Exterior angle ∠ D = FB of DA – BB of CD
= 289 0 0’ – 1 0
= 288 0
Interior Angle ∠ D = 360 0 – 288 0
= 72 0
Sum of Angles = ∠ A + ∠ B + ∠ C + ∠ D
= 64 0 + 101 0 30’ = 122 0 30’ + 72 0 0’
= 360 0
Check = ( 2N -4) x 90 0
= ( 2 N – 4) x 90 0
= 4 x 90 0
= 360 0 O.K.

Magnetic Declination
• The horizontal angle between the magnetic
meriadian and true meridian is known as
‘Magnetic declination’
• When the north end of the magnetic needle is
pointed towards the west side of the true meridian
the position is termed as ‘Declination West (ӨW).
• When the north end of the needle is pointed
towards east side of the true meridian the position
is termed as ‘Declination East (Ө E)

Determination of True bearing and
and Magnetic Bearing
• True Bearing = Magnetic Bearing
Declination
• Use + sign when declination is towards East
• Use – sign when declination is towards West
• Magnetic Bearing = True Bearing
Declination
• Use + sign when declination is towards West
• Use – sign when declination is towards East

and Magnetic Bearing
Example
• The magnetic bearing of a line AB is 135 0 30’.
What will be the true bearing, if the
declination is 5 0 15’ W
• The true bearing of a line CD is 2100 45’, what
will be its be its magnetic bearing of the
declination is 8 0 15’ W

Magnetic Bearing
• True Bearing of AB = Magnetic Bearing –
Declination
= 135 0 30’ – 5 0 15’ = 130 0 15’
• Magnetic Bearing of AB = True bearing – Declination
= 210 0 45’ + 80 15’ = 219 0

Example
• True bearing of line AB is 357 and its
magnetic bearing is 1 30′. Determine the
declination. Also find the true bearing of AC
which has magnetic bearing equal to 153
30′.

Example
• Magnetic Declination = 1 30′ + (360 – 357 )
= 4 30′ W.
• Magnetic bearing of AC = 153 30′.
• ∴ True bearing of AC = 153 30′ – 4 30′
=149

Local Attraction
• North end of a freely suspended magnetic needle
will always point towards the magnetic north, if it
is not influenced by any other external forces
except the earth’s magnetic field. It is common
experience that the magnetic needle gets deflected
from its normal position, if placed near magnetic
rocks, iron ore, cables carrying currents or iron
electric poles., therefore , not reliable unless these
are checked against the presence of local
attraction at each station and their elimination.

Local Attraction
Detection of Local Attraction
• The presence of local attraction at any station may
be detected by observing the fore and back
bearing of the line. If the difference between fore
and back bearing is 180 0, both end station are
free from local attraction. If not, the discepancy
may be due to
• An error in observation of either fore and back
bearing or both
• Presence of Local Attraction at either station
• Presence of local Attraction at both the stations

Local Attraction
• It may be noted that local attraction at any station affects all
the magnetic bearings by an equal amount and hence, the
included angles deduced from the affected bearing are
always correct.
• In case the fore and back bearing of neither line of traverse
differ by the permissible error of reading, the mean value of
the bearing of the line least affected may be accepted. The
correction to other stations, may be made according to the
following methods.
• By calculating the Included Angles at the affected
stations
• By checking the required correction, starting from the
unaffected bearing.

Local Attraction
• Methods of elimination of local attraction by
included angles.
The following steps are followed:
• Computing the included angle at each station
from the observed bearing, in case of a closed
traverse.
• Starting from the unaffected line, run down the
correct bearing of successive sides.

• The following bearing were taken for a closed
traverse compute the interior angle and correct them
for observational error assume the bearing of line CD
to be correct adjust the bearing of remaining sides.

∠ A = BB Bearing of AE – FB of AB
= 130 0 15’ – 80 0 10’
= 50 0 5’
∠ B = BB of AB – FB of BC
= 259 0 -120 0 20’
= 138 0 40’
∠ C = BB of BC – FB of CD
= 301 0 – 170 0 50’
= 131 0 0’

∠ D = BB of CD – FB of DE
= 350 0 – 230 0 10’
= 120 0 40’
∠E = BB of DE- FB of EA
= 49 0 30’ – 310 0 20’ + 360 0
= 99 0 10’
∠A + ∠B + ∠ C + ∠D + ∠E
= 50 0 5’ + 138 0 40’ + 131 0 0’ + 120 0 40’ + 99 0 10’
= 549 0 35’
Theoritical Sum = (2N-4) 90 0 = 540 0
Therefore Error = - 25’

Error = -25’
Hence a correction of + 5’ is applied to all the
angles
There fore the corrected angles are
∠ A = 50 0 10’
∠B = 138 0 45’
∠C = 131 0 5’
∠D = 120 0 45’
∠E = 99 0 15’

Starting with the corrected bearing of CD all other
bearings can be calculated as under
FB Bearing of DE= Bearing of DC - ∠ D
= 350 0 50’ – 120 0 45’ = 230 0 5’
BB of Bearing of DE = 230 0 5’ – 180 0
= 50 0 5’

FB Bearing of EA = BB of Bearing of DE - ∠ E
= 50 0 5’ – 99 0 15’ + 360 0
= 310 0 50’
BB Bearing of EA = 310 0 50’ – 180 0
= 130 0 50’
FB Bearing of AB = BB of Bearing of EA- ∠ A
= 130 0 50’ – 50 0 10’
= 80 0 40’

BB of Bearing of AB = 80 0 40’ + 180 0
= 260 0 40’
FB Bearing of BC = BB of Bearing of AB- ∠ B
= 260 0 40’ – 138 045’
= 1210 55’

BB of Bearing of BC = 1210 55’ + 180 0 = 301 0 55’
FB of Bearing of CD = BB of Bearing of BC - ∠ C
= 301 0 55’ – 131 0 5’
= 170 0 50’
BB of Bearing of CD = 170 0 50’ + 180 0
= 350 0 50’ (Check)

Examples
• The following are the observed bearing of the line of
a traverse ABCDEA with the compass in a place
where local attraction was suspected.
• Find the correct bearing of the lines.
Line FB BB
AB 191 0 45’ 13 0 0’
BC 39 0 30’ 222 0 30’
CD 22 0 15’ 200 0 30’
DE 242 0 45’ 62 0 45’
EA 330 0 15’ 147 0 45’

Traverse ABCDEA
(Anticlockwise)
B
N
N
N
N
A
C
D
N
E

Line FB BB Difference
AB 191 0 45’ 13 0 0’ 178 0 45’
BC 39 0 30’ 222 0 30’ 183 0
CD 22 0 15’ 200 0 30’ 178 0 15’
DE 242 0 45’ 62 0 45’ 180 0
EA 330 0 15’ 147 0 45’ 182 0 30’

Calculation of Interior Angle
Interior ∠ A= FB of AB – BB of EA
= 191 0 45’ – 147 0 45’
=44 0 00’
Interior ∠ B= FB of BC – BB of AB
= 39 0 30’ – 13 0 00’
= 26 0 30’
Exterior ∠ C= BB of BC – FB of AB
= 222 0 30’ – 22 0 15’
= 200 0 15’

Interior ∠ C= 360 0 00’ – 200 0 15’
= 159 0 45’
Interior ∠ D= FB of DE – BB of CD
= 242 0 45’ – 200 0 30’
= 42 0 15’
Interior ∠ E= FB of EA – BB of DE
= 330 0 15’ – 62 0 45’
= 267 0 30’
Sum of Interior Angle
= 44 0 00’ + 26 0 30’ + 159 0 45’ + 42 0 15’ + 267 0 30’
= 540 0 00’
Which is equal to ( 2N-4) x 90 0 = 540 0

Calculation of corrected bearing
The Line DE is free from local attraction,
So,
FB of DE = 242 0 45’ (Correct)
FB of EA = 330 0 15’ (Correct)
FB of AB = BB of EA + ∠ A
= (330 0 15’ + 180 0 0’) + 44 0 00’
= 150 0 15’ + 44 0 00’
= 194 0 15’

FB of BC = BB of EA + ∠ B
= (194 0 15’ – 180 0) + 26 0 30’
= 14 0 15’ – 26 0 30’
= 40 0 45’
FB of CD = BB of BC – Exterior ∠ C
= (40 0 45’ + 180 0 0’) - 200 0 15’
= 220 0 45’ – 200 0 15’
= 20 0 30’
FB of DE = BB of BC + ∠ D
= ( 20 0 30’ + 42 0 15’)
= 200 0 30’ + 42 0 15’
= 242 0 45’ (Checked)

Line FB BB
Corrected
AB 194 0 15’ 14 0 15’
BC 40 0 45’ 220 0 45’
CD 20 0 30’ 200 0 30’
DE 242 0 45’ 62 0 45’
EA 330 0 15’ 147 0 45’

Second method- Directly applying Correction
Procedure
(a) On verify the observed bearing it is found that
the FB and BB of line DE differ by exactly
180 0. So, the station 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 action BB of EA should be
330 0 15’ – 180 0 = 150 0 15’
But the observed bearing is 147 0 45’
So the correction of ( 150 0- 147 0 45’)
= + 2 0 30’ at Station A

(d) Correct FB of AB = 191 0 45’ + 2 0 30’ = 194 0 15’
Therefore, the actual correct BB of AB should be
194 0 15’ – 180 0 00’ = 14 0 15’
But observed = 13 0 0’
So a correction of (14 0 15’ – 13 0 0’)
= +1 0 15’ At Station B

(e) Correct FB of BC = 39 0 30’ + 1 0 15’ = 40 0 45’
Therefore Correct BB of BC should be
= 40 0 45’ + 180 0
= 220 0 45’
So a correction of
=( 220 0 45’ – 222 0 30’) = - 1 0 45’At Station C
(f) Correct FB of CD = 22 0 15’ – 1 0 45’ = 20 0 30’
which tallies with the observed BB of CD
So, D is free from local attraction, which also tallies with
remark made at the beginning

Line Observed Correction Corrected Bearing Remark
FB BB FB BB
AB 191 0 45’ 13 0 00’ + 2 0 30’ at A 194 0 15’ 14 0 15’
BC 30 0 30’ 2220 30’ + 1 0 15’ 40 0 45’ 220 0 45’
CD 22 0 15’ 200 0 30’ -1 0 45’ at C 20 0 30’ 200 0 30’ St D and
E are free
from local
attraction
DE 242 0 45’ 62 0 45’ 0 0 at D 242 0 45’ 62 0 45’
EA 330 0 15’ 147 0 45’ 0 0 at E 330 0 15’ 150 0 15’

Sources of Error in Compass Survey
The errors may be classified as
(i) Instrumental Error
(ii) Error of manipulation and sighting
(iii) error due to external influence

Instrumental Errors
• Needle not being perfectly straight
• The pivot being bent, i.e. not being at the centre of the
graduated circle.
• The needle being sluggish, i.e. the needle having lost its
magnetism
• The pivot point being dull
• The needle neither moving horizontally nor moving freely
on the pivot due to the dip of the needle.
• The plane of sight not passing through the centre of the
graduated ring
• The vertical hair being too thick or loose.

Error due to Manipulation and Sighting
• Inaccurate centring of the compass over the station occupied
• Inaccurate leveling of the compass box when the instrument is
set up
• Imperfect bisection of the ranging rods at station or other
objects
• Carelessness is reading the needle or in reading the graduate
circle through the prism in a wrong direction.
• Carelessness in recording the observed reading.

Error due to External Influences
• Magnetic changes in the atmosphere on a cloudy
or stormy day.
• Irregular variation due to magnetic storms,
earthquakes, sun spots, lunar perturbations etc.
• Variation in declination, viz, secular, annual and
diurnal.
• Local attraction due to proximity of steel
structure, electric lines.

Precaution to be taken in Compass
Surveying
The following precaution should be taken conducting
a compass traverse
• The centring should be done perfectly
• To stop the rotation of the graduation ring, the break pin
should be pressed very gently and not suddenly.
• Reading should be taken along the line sight and not
from any side.
• When the compass has to be shifted from one station to
other, the sight vane should be folded over the glass
cover. This is done to lift the ring out of the pivot to
avoid unnecessary wear of the pivot.

Precaution to be taken in compass
Surveying
• The compass box should be tapped gently before taking
the reading. This is done to find out whether the needle
rotates freely.
• The station should not be selected near magnetic
substances.
• The observer should not carry magnetic substances.
• The glass cover should not be dusted with a
handkerchief, because the glass may be charged with
electricity and the needle may be deflected from its true
direction. The glass cover should be cleaned with a
moist finger.

References
• “Surveying and Levelling” Vol- I
Kanetkar and Kulkarni (2011)
• “ Surveying” Vol- I
Dr. B.C. Punamia

Compass Surveying

More Related Content

What's hot

Viewers also liked

Similar to Compass Surveying

More from GAURAV. H .TANDON

Recently uploaded

Compass Surveying