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AN EDUSAT LECTURE ON
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T1
A
B
C
T2
O
Fig. 1 . A CURVE
CURVES
Curves are regularbends provided in
the lines of
communication like
roads, railways and
canals etc. to bring
about gradual
change of direction.
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T1
A
B
C
T2
O
Fig. 2. A CURVE
CURVES
They enable the
vehicle to pass fromone path on to another
when the two paths
meet at an angle. Theyare also used in the
vertical plane at all
changes of grade to
avoid the abrupt
change of grade at the
apex.
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HORIZONTAL CURVESCurves provided in the horizontal plane to have
the gradual change in direction are known ashorizontal curves.
VERTICAL CURVES
Curves provided in the vertical plane to obtain
the gradual change in grade are called as
vertical curves.
Curves may be circular or parabolic. Curves
are generally arcs of parabolas.Curves are laid out on the ground along the
centre line of the work.
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NEED OF PROVIDING CURVES
Curves are needed on Highways, railwaysand canals for bringing about gradual change
of direction of motion. They are provided for
following reasons:-i) To bring about gradual change in
direction of motion.
ii) To bring about gradual change in grade
and for good visibility.
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NEED OF PROVIDING CURVES Contd
iii) To alert the driver so that he may not fall
asleep.
iv) To layout Canal alignment.
v) To control erosion of canal banks by the
thrust of flowing water in a canal.
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CLASSIFICATION OF CIRCULAR CURVES
Circular curves are classified as :
(i) Simple Curves.
(ii) Compound Curves.
(iii) Reverse Curves.
(iv) Deviation Curves.
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T1
A
B
C
T2
O
Fig. 3. A SIMPLE CURVE
i) Simple Curve:
A simple curve
Consists of a
single arc of
circle connecting
two straights. It
has radius of the
same magnitudethroughout.
8
RR
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ii) COMPOUND CURVE
A compound Curve consists of two or
more simple curves having different radii
bending in the same direction and lying on
the same side of the common tangent. Their
centres lie on the same side of the curve.
A
T1
M P N
C
O1
O2
Fig.4 Compound Curve
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R2
R1
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iii) REVERSE OR SERPENTINE CURVE
A reverse or serpentine curve is
made up of two arcs having equal
or different radii bending in
opposite direction with a common
tangent at their junction .
Fig. 5. A Reverse or Serpentine Curve.
MTheir centres lie on
opposite sides of the curve.
Reverse curves are used
when the straights areparallel or intersect at a
very small angle.
N
O2
O1
A
T1
T2
p
B
10
R1
R2
R2
R1
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REVERSE OR SERPENTINE CURVE
Fig.6 A Reverse or Serpentine Curve.
They are commonly used
in railway sidings andsometimes on railway
tracks and roads meant
for low speeds. Theyshould be avoided as far
as possible on main lines
and highways where
speeds are necessarily
high.
A
T1
T2O2
O1
M N
B
P
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iv) DEVIATION CURVE
T1 T2
O2
O1 O3
Building
A deviation curve is
simply a combinationof two reverse curves.
it is used when it
becomes necessary to
deviate from a given
straight path in order
to avoid intervening
obstructions such asbend of river, a
building , etc.
Fig. 7 A Deviation Curve
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B
B
T1 T2
O
R
CA
E
F
I
/2
Fig. 8 SIMPLE CIRCULAR CURVE
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NAMES OF VARIOUS PARTS OF CURVE
(i) The two straight lines AB and BC which
are connected by the curve are called thetangentsor straightsto the curve.
(ii) The point of intersection of the two
straights (B) is called the intersection pointor the vertex.
(iii) When the curve deflectsto the right side of
the progress of survey ,it is termed as righthanded curve and when to the left , it is
termed as lef t handed curve.
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NAMES OF VARIOUS PARTS OF CURVE
(iv) The lines AB and BC are tangents to the
curve. AB is called the f irst tangent or therear tangent . BC is called the second
tangent or the forward tangent.
(v)The points ( T1and T2) at which thecurve touches the tangents are called
the tangent points. The beginning of
the curve ( T1) is called the tangent
curve pointand the end of the curve
(T2) is called the curve tangent point.
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NAMES OF VARIOUS PARTS OF CURVE
(vi) The angle between the lines AB and BC
(ABC) is called the angle of intersection(I).(vii) The angle by which the forward tangent
deflects from the rear tangent (BBC) iscalled the deflection angle () of the curve.(viii) The distance from the point of intersection
to the tangent point is called tangent length( BT1and BT2).
(ix) The line joining the two tangent points (T1
and T2) is known as the long chord.16
( ) Th T1FT2 i ll d th l th f
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(x) The arc T1FT2is called the length of curve.
(xi) The mid point(F) of the arc (T1FT2) is called
the summit or apex of the curve.
(xii) The distance from the point of intersectionto the apex of the curve BF is called the
apex distance.
(xiii) The distance between the apex of the curveand the mid point of the long chord (EF) is
called versed sine of the curve.
(xiv) The angle subtended at the centre of the
curve by the arc T1FT2 is known as
central angle and is equal to the deflection
angle () .17
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ELEMENTS of a Simple Circular Curve
(i) Angle of intersection +Deflection angle = 1800.
or I + = 1800
(ii)
T1OT2 = 180
0- I =
i.e the central angle = deflection angle.
(iii)Tangent length = BT1 =BT2= OT1tan /2
= R tan /2
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ELEMENTS of a Simple Circular Curve
(iv) Length of long chord =2T1E
=2R sin/2(v) Length of curve = Length of arc T1FT2
= R X (in radians)
= R/1800(vi) Apex distance = BF = BOOF
= R sec. /2 - R
= R (1cos /2)=R versine/2
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A curve may be designated either bythe radius or by the angle subtended at the
centre by a chord of particular length.
In India, a curve is designated by the
angle (in degrees)subtended at the centre by a
chord of 30 metres (100 ft.) length. This angle
is called the degree of curve (D).
The degree of the curve indicates thesharpness of the curve.
DESIGNATION OF CURVE
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DESIGNATION OF CURVES
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DESIGNATION OF CURVES.
In English practice , a curve is defined
by the radius of the curve in terms of chains,such as a six chain curve means a curve having
radius equal to six full chains, chain being 30
metres unless otherwise specified.
In America,Canada,India and some
other countries a curve is designated by the
degree of the curve. For example a 40 curve
means a curve having angle of 4 degrees at the
centre subtended by a chord of 30m length
unless otherwise specified.
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RELATION between the Radius of curve and
Degree of Curve.
The relation between the radius
and the degree of the curve may
be determined as follows:-
Let R = the radius of the curve in metres.
D = the degree of the curve.MN = the chord, 30m long.
P = the mid-point of the chord.
In OMP,OM=R,MP= MN =15m
MOP=D/2
Then, sin D/2=MP/OM= 15/R
MN
O
D
D/2
R R
Fig.9 Degree of Curve
P
PTO
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RELATION between the Radius of curve and
Degree of Curve.Then,sin D/2=MP/OM= 15/R
Or R = 15
sin D/2
But when D is small, sin D/2 may be
assumed approximately equal to
D/2 in radians.Therefore:
R = 15 X 360
D= 1718.87
D
Or say , R = 1719
D
MN
O
D
D/2
R R
Fig. 10 Degree of Curve
P
This relation holds good up to 50
curves.For higher degree curves the
exact relation should be used.
(Exact)
(Approximate)
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METHODS OF CURVE RANGING
A curve may be set out
(1) By linear Methods, where chain and tapeare used or
(2) By Angular or instrumental methods,
where a theodolite with or without a chain isused.
Before starting setting out a curve by any
method, the exact positions of the tangents
points between which the curve lies ,must be
determined. Following procedure is adopted:-
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METHODS OF SETTING OUT A CURVE
Procedure :-
i) After fixing the directions of the straights,produce them to meet in point (B)
ii) Set up the Theodolite at the intersection
point (B) and measure the angle ofintersection (I) .Then find the deflection
angle ( ) by subtracting (I) from 1800 i.e
=1800I.
iii) Calculate the tangent length from the
following equation
Tangent length = R tan/2
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Instrumental Methods
(i) Tape and Theodolite Method (Rankine's
method) and
(ii) Two Theodolite Method
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RankinesMethod
In this method, both the linear and angularmeasurements are carried out simultaneouslyto stake points along which curve will be set
out. A tape is used for the linear measurements,
whereas a theodolite is used for the angularmeasurements.
This method is quite accurate and iscommonly used in practice
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RankinesMethod
In this method, curves are staked out by use of deflectionangles turned at the point of curvature from the tangent topoints along the curve.
The curve is set out by driving pegs at regular intervalequal to the length of the normal chord.
Usually, the sub-chords are provided at the beginning andend of the curve to adjust the actual length of the curve.
The method is based on the assumption that there is nodifference between length of the arcs and their
corresponding chords of normal length or less. The underlying principle of this method is that the
deflection angle to any point on the circular curve ismeasured by the one-half the angle subtended at thecentre of the circle by the arc from the P.C. to that point
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Let points a, b, c, d, e are to be identifiedin the field to layout a curve between T1and T2to change direction from thestraight alignment AV to VB.
To decide about the points, chords ab, bc,
cd, de are being considered havingnominal length of 30m. To adjust theactual length of the curve two sub-chordshave been provided one at the beginning,T1a and other, eT2at the end of the curve.
The amount of deflection angles that are
to be set from the tangent line at the P.C.are computed before setting out thepoints. The steps for computations are asfollows:
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let the tangential angles forpoints a, b, c, be d1, d,, d, dnand their deflection angles (fromthe tangent at P.C.) be Da, Db, .., Dn.
Now, for the first tangentialangle d1, from the property of acircle
Arc T1a = R x 2d1radians
Assuming the length of the arc issame as that of its chord, if C1 isthe length of the first chord i.e.,chord T1 a, then
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Thus, the deflection angle for any point on the
curve is the deflection angle upto previous
point plus the tangential angle at the previous
point.
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Procedure : RankineSMethod1. A theodolite is set up at the point of curvature T1, and get it temporary adjusted.
2. The vernier A is set to zero, and get the upper plate clamped. After opening the lowerplate main screw, sight the point of intersection, V. Then the lower plate main screwgets tightened and get the point V bisected exactly using the lower plate tangent screw.Now the line of sight is in the direction of the rear tangent T1V and the vernier A readszero.
3. Open the upper plate main screw, and set the vernier A to the deflection angle D a. Theline of sight is now directed along the chord T1a. Clamp the upper plate.
4. Hold the zero end of the tape of a steel tape at T1. Note a mark equal to the first chordlength C1on the tape and swing an arrow pointed at the mark around a' till it isbisected along the line of sight. The arrow point then indicates the position of the firstpeg a'. Fix the first peg at a'.
5. Unclamp the upper plate, and set the vernier A to the deflection angle Db. The line ofsight is now directed along T1b.
6. With the zero end of the tape at a, and an arrow at a mark on the tape equal to thenormal chord length C, swing the tape around b until the arrow is bisected along theline of sight. Fix the second peg at the point b at the arrow point.
It may be noted that the deflection angles are measured from the tangent point T1butthe chord lengths are measured from the preceding point. thus, deflection anglesobserved are cumulative in nature but chord lengths swung are individual in nature.
7. Repeat steps (5) and (6) till the last point is reached. The last point so located must
coincide with the tangent point T2already fixed from the point of intersection. 43
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METHODS OF SETTING OUT A CURVE
Procedure :-
iv) Measure the tangent length (BT1)
backward along the rear tangent BA from
the intersection point B, thus locating theposition of T1.
vi) Similarly, locate the position of T2 by
measuring the same distance forward
along the forward tangent BCfrom B.
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METHODS OF SETTING OUT A CURVE
Procedure (contd) :-
After locating the positions of the tangentpoints T1 and T2 ,their chainages may be
determined. The chainage of T1is obtained by
subtracting the tangent length from the knownchainage of the intersection point B. And the
chainage of T2 is found by adding the length
of curve to the chainage of T1.
Then the pegs are fixed at equal intervals
on the curve.The interval between pegs is
usually 30m or one chain length. ...............45
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METHODS OF SETTING OUT A CURVE
Procedure (contd) :-
This distance should actually be measured alongthe arc ,but in practice it is measured along
the chord ,as the difference between the chord
and the corresponding arc is small and hencenegligible. In order that this difference is
always small and negligible ,the length of the
chord should not be more than 1/20th of the
radius of the curve. The curve is then obtained
by joining all these pegs. ...............
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METHODS OF SETTING OUT A CURVE
Procedure (contd) :-
The distances along the centre line of thecurve are continuously measured from the
point of beginning of the line up to the end .i.e
the pegs along the centre line of the workshould be at equal interval from the beginning
of the line up to the end. There should be no
break in the regularity of their spacing in
passing from a tangent to a curve or from a
curve to the tangent. For this reason ,the first
peg on the curve is fixed .47
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METHODS OF SETTING OUT A CURVE
Procedure (contd) :-
at such a distance from the first tangent point(T1) that its chainage becomes the whole
number of chains i.e the whole number of peg
interval. The length of the first sub chord isthus less than the peg interval and it is called a
sub-chord. Similarly there will be a sub-chord
at the end of the curve. Thus a curve usually
consists of two sub-chords and a no. of ful l
chords.
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Example : A simple circular curve is to have a radius
of 573 m .the tangents intersect at chainage 1060 m
and the angle of intersection is 1200. Find,
(i) Tangent Distance.
(ii) Chainage at beginning and end of the curve.
(iii) Length of the long chord.
(iv) Degree of the curve.(v) Number of full and sub chords.
Solution: Please see fig.11
Given,The deflection angle,= 18001200 =600
Radius of curve = R = 573 mPTO
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Fig.11
1060 m
O
729.15 1329.15
T1 T2
600
1200
330.85
R=573m
=
L=600m
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(i) We know ,tangent length = R tan /2
= 573 x tan 300
= 573x 0.5774
= 330.85 m (Ans.)
(ii) Length of curve is given by: R1800
= x 573x600
1800
= 600 m (Ans.)
Chainage of first tangent point (T1)
= Chainage of intersection pointtangent length.= 1060330.85= 729.15 m (Ans.)
PTO
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(iii) The length of long chordis given by:
L = 2R sin /2
= 2 x 573 x sin 300
= 573 m ( Ans.)
(iv) Degree of Curve
We know the relation , R= 1719D
or D = 1719
R=30
Therefore , degree of curve is =30(Ans.)PTO
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(v) Number of Full and sub chords:
Assuming peg interval =30mChainage of T1= 729.15 m = 729.15
30
= 24 full chain lengths + 9.15 m
Chainage of Ist peg on the curve should be 25 full chain lengths.
The length of Ist sub chord= (25+00)(24 + 9.15)
= 20.85 mChainage of T2= 1329.15 Chain lengths.
30
= 44 full chain lengths + 9.15 m.
Chainage of last peg on the curve =44 full chains.
Therefore length of last sub chord = (44+9.15)(44+00)= 9.15m
PTO
Chain lengths.
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No. Of full chords = chainage of last pegchainage of Ist peg= 4425 = 19
So, there will be 19 full chords and two sub chords.
Check:Length of full chords = 19x30 =570.00m
Ist sub chord = 20.85m last sub chord = 9.15m
Total length of all chords = 600.00m
PTO
(Same as length of curve)
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LINEAR METHODS of setting out Curves
The following are the methods of setting outsimple circular curves by the use of chain
and tape :-
(i) By offsets from the tangents.(ii) By successive bisection of arcs.
(iii) By offsets from chords produced.
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LINEAR METHODS of setting out Curves
1. By offsets from the tangents. When the
deflection angle and the radius of thecurve both are small, the curves are set out
by offsets from the tangents.
Offsets are set out either(i) radially or
(ii) perpendicular to the tangents
according as the centre of the curve is
accessible or inaccessible
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LINEAR METHODS of setting out Curves
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B
B
T1T2
O
R
CA
Fig. 12 By Radial Offsets
LINEAR METHODS of setting out Curves
Oxx P
P1900
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LINEAR METHODS of setting out Curves
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B
By Radial Offsets
LINEAR METHODS of setting out Curves
Offsets is given by :
Ox= R2+x2R .. (Exact relation.)
When the radius is large ,the offsets may be
calculated by the approximate formulawhich is as under
Ox= x2 (Approximate )
2R
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LINEAR METHODS of setting out Curves
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B
O
(ii) By offsets perpendicular to the Tangents
LINEAR METHODS of setting out Curves
Oxx
P
P1
P2
B
A
B
T2T1
Fig. 13.
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LINEAR METHODS f tti t C
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LINEAR METHODS of setting out Curves
1. (ii) By offsets perpendicular to the Tangents
Ox= R R2x2 (Exact)
Ox= x2 (Approximate )
2R
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LINEAR METHODS of setting out Curves
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LINEAR METHODS of setting out Curves
By offsets from the tangents: Procedure
(i) Locate the tangent points T1and T2.
(ii) Measure equal distances , say 15 or 30 m
along the tangent fro T1.
(iii) Set out the offsets calculated by any of
the above methods at each distance ,thus
obtaining the required points on the
curve.
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LINEAR METHODS of setting out Curves
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LINEAR METHODS of setting out Curves
By offsets from the tangents: Procedure.
(iv) Continue the process until the apex of
the curve is reached.
(v) Set out the other half of the curve from
second tangent.
(vi) This method is suitable for setting out
sharp curves where the ground outside
the curve is favourable for chaining.
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Example. Calculate the offsets at 20m intervals along
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p gthe tangents to locate a curve having a radius of
400m ,the deflection angle being 600.
Solution . Given:
Radius of the curve ,R = 400m
Deflection angle,= 600
Therefore tangent length = R. tan/2= 400 x tan 600
= 230.96 m
Radial offsets. (Exact method)
Ox= R2+ x2 - R (Exact)
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Radial offsets. (Exact method)
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( )
Ox= R2+ x2 - R (Exact)
O20 = 4002+202 - 400 = 400.50 - 400 = 0.50 m
O40 = 4002+402 - 400 = 402.00 - 400 = 2.00 m
O60 = 4002+602 - 400 = 404.47 - 400 = 4.47 m
O80 = 4002+802 - 400 = 407.92 - 400 = 7.92 m
O100 = 4002+1002- 400 = 412.31 - 400 = 12.31 m
And so on.
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B) Perpendicular offsets (Exact method)
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Ox = R R2x2 (Exact)
O20= 400 - 4002- 202 = 400 -399.50 = 0.50 m
O40= 400 - 4002- 402 = 400 -398.00 = 2.00 m
O60= 400 - 4002- 602 = 400 -395.47 = 4.53 m
O80= 400 - 4002- 802 = 400 -391.92 =8.08 m
O100= 400 - 4002-1002 = 400 -387.30 =12.70 m
And so on..65
B By t e approx mate Formu a(Both radial and perpendicular offsets)
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(Both radial and perpendicular offsets)Ox =
2R
Therefore O20 = 202 = 0.50 m2x400
x2
O40 = 402 = 2.00 m
2x400
O60 = 602 = 4.50 m2x400
O80 = 802 = 8.00 m
2x 400O100 = 1002 = 12.50 m
2 x 400
and so on.66
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