highway notes

Circular curves - types
Simple circular curves
Compound curves
Reverse curves R2
R2
R
R
R1
R1
R1
R2

PROCEDURE SETTING OUT
Circular curves may be set out in a variety of ways, depending on the accuracy
required, its radius of curvature and obstructions on site.
Methods of setting out are as follows:
Using one theodolite and a tape by the tangent angle method.This method can be
used on all curves, but is necessary for long curves of radius unless they are set out
by coordinates.
Using two theodolites.This method can be used on smaller curves where the
whole length is visible from both tangent points and where two instruments are
available.
Using tapes only by the method of offsets from the tangent. This method is used
for minor curves only.
Using tapes only by the method of offsets from the long chord. This method is used
for short radius curves.

OFFSET FROM TANGENT LINE
Given
Radius, R 600m
Deflection
angle ,
18024'
Offset 20m
Chainage
intersection
point, I
2140m
Draw the table from the formula:
Ofset R R2Y2R2-Y2
formula 22
Y
R
R
X
22
Y
R
2
2
Y
R
R
X
IP
)
Tangent line
Y2 X2
X1
Y1
T1
circular arc T2
P
R
O
C
E
D
U
R
E
Tangent length
= 600 tan (18 24 /2) = 97.20m
Chainage T1= chainage I tangent length
= 2140.00 - 97.20 = 2042.80m
Arc length = 2 x R x
360
= 2 x 600 x 18o= 192.68m
360
Chainage T2= chainage T1+ arc length
= 2042.80 + 192.68 = 2235.48m
Ofset R R2Y2R2-Y2
(1) (2) (3) = (2)2(4) = (1)2(5) = (3) (4) (7) = (2) (6)
0
600 6002=
360000
02 = 0 360000 0 =
360000
=
600.000
600 600 =
0.000
20 202 = 40 360000 40=
359600
=
599.667
600 599.667
=
0.333
40 402 = 1600 360000 1600 =
358400
=
598.665
600 598.665
=
1.335
60 3600 356400 596.992 3.008
80 6400 353600 594.643 5.357
97.20 9447.84 350969.109 592.426 7.574
22
Y
R
2
2
Y
R
R
X
97.20

OFFSET FROM LONG CHORD LINE
Given
Radius, R 600m
Deflection
angle ,
18024'
Offset 20m
Chainage
intersection
point, I
2140m
Draw the table from the formula:
Ofset R R2Y2R2-Y2
formula
2
2
Y
R
2
2
)2/( WR
22
)2/(
22
WRYRX
2
2
)2/(
22
WRYRX
P
R
O
C
E
D
U
R
E
Long chord length
= 2 x 600 sin (18 24 /2) w= 191.857m w/2 = 95.929 m
Tangent length
= 600 tan (18 24 /2) = 97.20m
Chainage T1= chainage I tangent length
= 2140.00 - 97.20 = 2042.80m
Arc length = 2 x R x
360
= 2 x 600 x 18o= 192.68m
360
Chainage T2= chainage T1+ arc length
= 2042.80 + 192.68 = 2235.48m
Ofset R R2Y2R2-Y2w/2 2x
(1) (2) (3) = (2)2(4) = (1)2(5) = (3) (4) (7) (7) (9) =(6) (8)
0
600 6002=
360000
02 = 0 360000 -0
=
360000
=
600.000
95.9292=
9202.277
(360000
9202.277)
=
592.282
600 -
692.282
= 7.785
20 202= 40 360000 40 =
359600
=
599.667
599.667
592.282 =
7.385
40 402= 1600 358400 598.665 6.383
60 3600 356400 596.992 4.710
80 6400 353600 594.643 2.361
95.929 9202.277 350797.723 592.282 0.000
22
Y
R
2
2
)2/( wR
/
=
=
=
9
5
.
9
2
9
m
95.929

DEFLECTION ANGLE METHOD
Given
Radius, R 600m
Deflection angle
,
18024'
Offset 20m
Chainage
intersection
point, I
2140m
R
C
x
ree
60
9
.
1718
)(deg
1
formula
R
Cx
ute
9.1718
)(min
1
Draw the table form for
deflection angle method
Stn. Chainage Chord
length
Deflection
angle,
(0 )
Setting out
angle,
(0 )
Tangent length
= 600 tan (18 24 /2) = 97.20m
Chainage T1= chainage I tangent length
= 2140.00 - 97.20 = 2042.80m
Arc length = R x x 2
360
= 600 x 18o2= 192.684m
360
Chainage T2= chainage T1+ arc length
= 2042.80 + 192.68 = 2235.48m
P
R
O
C
E
D
U
R
E
Stn. Chainage Chord length, CDeflection angle,
(0 ) use formula
Setting out angle,
(0 ) cumulative deflection angle
T1 2042.821 0 00 00
1 2060 17.179 00 00
2 2080 20.000 00 10
3 2100 20.000 00 20
4 2120 20.000 00 30
5 2140 20.000 00 40
6 2160 20.000 00 50
7 2180 20.000 00 60
8 2200 20.000 00 70
9 2220 20.000 00 80
T2 2235.506 15.506 00 90
= 192.684 = 90 / 2 = 180 0
600
60
17.179
9
.
1718
x
x
600
60
000.209.1718
x
x
600
60
506.159.1718
x
x
+=
+=
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f
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u
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T1
T
2