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Circular curves
Horizontal curves
Definition
Horizontal, circular or simple
curves are curves of constant
radius required to connect
two straights set out on the
ground.

Usage
roads, railways, kerb lines,
pipe lines and may be set out
in several ways, depending on
their length and radius.
Circular Curve Geometry
FORMULA
θ
T = R tan
2
θ
C = 2 R sin
2
 ∆ 
E = R sec − 1
 2 

θ

M = R1 − cos 
2


θ
L = 2πR
360
δ

1 (deg ree )

1718.9 x C
=
60 R

δ1(min ute ) =

1718.9 x C
R
Chainage T1 = Chainage I + tangent
length
Chainage T = Chainage T – arc length
2

1
Setting out method
Offset From The Tangent Line
Offset From The Long Chord
Line
Deflection Angle Method
Sub Chords Line Method
OFFSET FROM TANGENT LINE
Given

)θ

Radius, R

600m

Deflection
angle , θ
Offset

18 0 24'

Chainage
intersection
point,
formul I

a

IP
Tangent line

Y2

X2
X1
Y1
T1

20m

circular arc

2140
m

X = R − R −Y
2

2

Ofset

T2

Draw the table from the formula:
R R2 Y2 R2-Y2

R −Y
2

2

X = R− R − Y
2

2
PROCEDURE

Tangent length

= R tan θ/2
= 600 tan (18°24′/2)

= 97.20m

Chainage T 1

= chainage I – tangent length
= 2140.00 - 97.20
= 2042.80m

Arc length

= 2π x R x θ
360
= 2π x 600 x 18 o 24’
360

Chainage T 2

=
=

= 192.68m

chainage T 1 + arc length
2042.80 + 192.68
= 2235.48m

Ofset

R

R2

Y2

(1)

(2)

(3) = (2)2

(4) = (1)2

R2-Y2

R −Y
2

2

X = R− R − Y
2

2

(5) = (3) – (4)

(6) = √(5)

(7) = (2) – (6)

3600

360000 – 0 =
360000
360000 – 40=
359600
360000 – 1600 =
358400
356400

√ 360000 =
600.000
√ 359600 =
599.667
√ 3584000 =
598.665
596.992

600 – 600 =
0.000
600 – 599.667 =
0.333
600 – 598.665 =
1.335
3.008

80

6400

353600

594.643

5.357

97.20

9447.84

350969.109

592.426

7.574

0

02 = 0

20

202 = 40

40
60

600

6002 =
360000

402 = 1600
OFFSET FROM LONG CHORD LINE
Given
Radius, R

600m

Deflection
angle , θ
Offset

18 0 24'

Chainage
intersection
point,
formul I

2140m

20m

a
X = R − Y − R − (W / 2)
2

Ofset

2

R

2

R2

Y2 R2-Y2

Draw the table from the formula:
2

R −Y
2

2

R − (W / 2)
2

2

X = R − Y − R − (W / 2)
2

2

2

2
PROCEDURE

Long chord length

= 2R sin θ/2

= 2 x 600 sin (18 °24′/2)

Tangent length

w

= 191.857m

= R tan θ/2
= 600 tan (18°24′/2)

= 97.20m

Chainage T 1

= chainage I – tangent length
= 2140.00 - 97.20

Arc length

= 2π x R x θ
360
= 2π x 600 x 18 o 24’
360

Chainage T 2

=
=

R

R2

Y2

(1)

(2)

(3) = (2) 2

(4) = (1) 2

0

02 = 0

20

202 = 40

40

6002 =
360000

= 2042.80m

= 192.68m

chainage T 1 + arc length
2042.80 + 192.68

Ofse
t

R2-Y2
(5) = (3) – (4)

w/2 = 95.929 m

R −Y
2

= 2235.48m
2

(6) = √(5)

w/2

2

(7)

360000 -0 = √ 360000 =
360000
600.000
360000 – 40 = √ 3596000 =
359600
599.667
95.929 2 =
9202.277

R − (w / 2)
2

2

(8) = √(3) – (7)

x
(9) =(6) – (8)

600 - 692.282
= 7.785
599.667 –
592.282 =
√(360000 –
7.385
6.383
9202.277)=
592.282
4.710

402 = 1600

358400

598.665

3600

356400

596.992

80

6400

353600

594.643

2.361

95.929

9202.277

350797.723

592.282

0.000

60

600
DEFLECTION ANGLE
METHOD
Radius, R
600m
Deflection
Given
angle , θ
Offset

18 0 24'

Chainage
intersection
point, I

2140m

20m

Draw the table form for
deflection angle method

formul
a

δ
δ

1 (deg ree )

1 (min ute )

1718.9 x C
=
60 R
1718.9 x C
=
R

Stn
.

Chainag
e

Chord
length

Deflection
angle,δ
(0 ‘ “)

Setting out
angle, δ
(0 ‘ “)
Tangent length

= R tan θ/2
= 600 tan (18°24′/2)

PROCEDUR
1718.9 x 17.179
δ =
E60xx600.000
1718.9 20
= 97.20m

Chainage T 1

= chainage I – tangent length
= 2140.00 - 97.20
= 2042.80m

Arc length

= R x θ x 2π
360
= 600 x 18 o 24’ x 2π
360

Chainage T 2

Stn.

=
=

Chainage

δ =

chainage T 1 + arc length
2042.80 + 192.68
=

Chord length, C

60 x600
1718.9 x 15.506
δ =
2235.48m
60 x600

= 192.684m

Setting out angle, δ
(0 ‘ “) – cumulative deflection

Deflection angle,δ
(0 ‘ “) – use formula

+
+

angle

=
=

T1

2042.821

0

00 0’ 0”

00 0’ 0”

δ1

2060

17.179

00 49’ 12”

δ2

2080

20.000

00 57’ 18”

10 46’ 30”

δ3

2100

20.000

00 57’ 18”

20 43’ 48”

δ4

2120

20.000

00 57’ 18”

30 41’ 6”

δ5

2140

20.000

00 57’ 18”

40 38’ 24”

δ6

2160

20.000

00 57’ 18”

50 35’ 42”

δ7

2180

20.000

00 57’ 18”

60 33’ 0”

δ8

2200

20.000

00 57’ 18”

70 30’ 18”

δ9

2220

20.000

00 57’ 18”

80 27’ 36”

T2

2235.506

15.506

00 44’ 25”

90 12’ 1”

Σ = 192.684

Σ = 90 12’ 1”

00 49’ 12”

θ / 2 = 180 24’ / 2 = 90 12’ 1”
Sub chords line method
Radius, R

600m

Deflection
angle , θ
Offset

18 0 24'

Chainage
intersection
formul
point, I

2140m

• Given

a
a
Ofset =
2R
b(b + a)
Ofset =
2R
b
Ofset =
R
2

1

2

2

n

Ofset =
n −1

c (c + b)
2R

20m

Draw the table form for
sub chord line method
Stn
.

Chainag
e

Chord
length

Offset
Tangent length
Chainage T 1
Arc length

Chainage T 2

= R tan θ/2
= 600 tan (18°24′/2)

PROCEDUR
17.179
Ofset =
2x
E600
= 97.20m

2

= chainage I – tangent length
1
= 2140.00 - 97.20
= 2042.80m

20(20 + 17.179)
Ofset
= 192.684m =
2 x600
20
= chainage T
+ arc length
Ofset
= 2042.80 + 192.68
= 2235.48m =
600
15.506(15.506 + 20)
Chord
Offset
Ofset =
length
2 x600
0
= R x θ x 2π
360
= 600 x 18 o 24’ x 2π
360

2

2

1

n

Stn
.
T1
δ1
δ2
δ3
δ4
δ5
δ6
δ7
δ8
δ9
T2

Chainag
e
2042.821
2060
2080
2100
2120
2140
2160
2180
2200
2220
2235.506

n −1

a = 17.179
b = 20.000
20.000
20.000
20.000
20.000
20.000
20.000
20.000
c =15.506
Σ = 192.684

0.246
0.620
0.667
0.667
0.667
0.667
0.667
0.667
0.667
0.459
-

First Offset
Second Offset

Other Offset

Last Offset
PROCEDURE ways, depending on the accuracy
SETTING
Circular curves may be set out in a variety of
required, its radius of curvature and obstructions on site.
OUT
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.
FIELD WORK

OPTICAL SQUARE
Optical squares are simple sighting instruments
used to set out right angles. They can be
provided either with mirrors or with one or two
prisms. Because of practical difficulties in using
squares with mirrors, they have been replaced
by squares with prisms: "prismatic squares".
There are two major types of prismatic squares:
single prismatic squares and double prismatic
squares; both will be dealt with in the sections
which follow.

OFFSET LINE FROM BASELINE USED
OPTICAL SQUARE
STEP 1

STEP 2

OFFSET LINE FROM BASELINE
STEP 1

STEP 2

STATION AT BASELINE
End of sub -topic
• Exercise
• Practical 6 & 7

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Topic 4 curve lesson 2

  • 2. Horizontal curves Definition Horizontal, circular or simple curves are curves of constant radius required to connect two straights set out on the ground. Usage roads, railways, kerb lines, pipe lines and may be set out in several ways, depending on their length and radius.
  • 4. FORMULA θ T = R tan 2 θ C = 2 R sin 2  ∆  E = R sec − 1  2  θ  M = R1 − cos  2  θ L = 2πR 360 δ 1 (deg ree ) 1718.9 x C = 60 R δ1(min ute ) = 1718.9 x C R
  • 5. Chainage T1 = Chainage I + tangent length Chainage T = Chainage T – arc length 2 1
  • 6. Setting out method Offset From The Tangent Line Offset From The Long Chord Line Deflection Angle Method Sub Chords Line Method
  • 7. OFFSET FROM TANGENT LINE Given )θ Radius, R 600m Deflection angle , θ Offset 18 0 24' Chainage intersection point, formul I a IP Tangent line Y2 X2 X1 Y1 T1 20m circular arc 2140 m X = R − R −Y 2 2 Ofset T2 Draw the table from the formula: R R2 Y2 R2-Y2 R −Y 2 2 X = R− R − Y 2 2
  • 8. PROCEDURE Tangent length = R tan θ/2 = 600 tan (18°24′/2) = 97.20m Chainage T 1 = chainage I – tangent length = 2140.00 - 97.20 = 2042.80m Arc length = 2π x R x θ 360 = 2π x 600 x 18 o 24’ 360 Chainage T 2 = = = 192.68m chainage T 1 + arc length 2042.80 + 192.68 = 2235.48m Ofset R R2 Y2 (1) (2) (3) = (2)2 (4) = (1)2 R2-Y2 R −Y 2 2 X = R− R − Y 2 2 (5) = (3) – (4) (6) = √(5) (7) = (2) – (6) 3600 360000 – 0 = 360000 360000 – 40= 359600 360000 – 1600 = 358400 356400 √ 360000 = 600.000 √ 359600 = 599.667 √ 3584000 = 598.665 596.992 600 – 600 = 0.000 600 – 599.667 = 0.333 600 – 598.665 = 1.335 3.008 80 6400 353600 594.643 5.357 97.20 9447.84 350969.109 592.426 7.574 0 02 = 0 20 202 = 40 40 60 600 6002 = 360000 402 = 1600
  • 9. OFFSET FROM LONG CHORD LINE Given Radius, R 600m Deflection angle , θ Offset 18 0 24' Chainage intersection point, formul I 2140m 20m a X = R − Y − R − (W / 2) 2 Ofset 2 R 2 R2 Y2 R2-Y2 Draw the table from the formula: 2 R −Y 2 2 R − (W / 2) 2 2 X = R − Y − R − (W / 2) 2 2 2 2
  • 10. PROCEDURE Long chord length = 2R sin θ/2 = 2 x 600 sin (18 °24′/2) Tangent length w = 191.857m = R tan θ/2 = 600 tan (18°24′/2) = 97.20m Chainage T 1 = chainage I – tangent length = 2140.00 - 97.20 Arc length = 2π x R x θ 360 = 2π x 600 x 18 o 24’ 360 Chainage T 2 = = R R2 Y2 (1) (2) (3) = (2) 2 (4) = (1) 2 0 02 = 0 20 202 = 40 40 6002 = 360000 = 2042.80m = 192.68m chainage T 1 + arc length 2042.80 + 192.68 Ofse t R2-Y2 (5) = (3) – (4) w/2 = 95.929 m R −Y 2 = 2235.48m 2 (6) = √(5) w/2 2 (7) 360000 -0 = √ 360000 = 360000 600.000 360000 – 40 = √ 3596000 = 359600 599.667 95.929 2 = 9202.277 R − (w / 2) 2 2 (8) = √(3) – (7) x (9) =(6) – (8) 600 - 692.282 = 7.785 599.667 – 592.282 = √(360000 – 7.385 6.383 9202.277)= 592.282 4.710 402 = 1600 358400 598.665 3600 356400 596.992 80 6400 353600 594.643 2.361 95.929 9202.277 350797.723 592.282 0.000 60 600
  • 11. DEFLECTION ANGLE METHOD Radius, R 600m Deflection Given angle , θ Offset 18 0 24' Chainage intersection point, I 2140m 20m Draw the table form for deflection angle method formul a δ δ 1 (deg ree ) 1 (min ute ) 1718.9 x C = 60 R 1718.9 x C = R Stn . Chainag e Chord length Deflection angle,δ (0 ‘ “) Setting out angle, δ (0 ‘ “)
  • 12. Tangent length = R tan θ/2 = 600 tan (18°24′/2) PROCEDUR 1718.9 x 17.179 δ = E60xx600.000 1718.9 20 = 97.20m Chainage T 1 = chainage I – tangent length = 2140.00 - 97.20 = 2042.80m Arc length = R x θ x 2π 360 = 600 x 18 o 24’ x 2π 360 Chainage T 2 Stn. = = Chainage δ = chainage T 1 + arc length 2042.80 + 192.68 = Chord length, C 60 x600 1718.9 x 15.506 δ = 2235.48m 60 x600 = 192.684m Setting out angle, δ (0 ‘ “) – cumulative deflection Deflection angle,δ (0 ‘ “) – use formula + + angle = = T1 2042.821 0 00 0’ 0” 00 0’ 0” δ1 2060 17.179 00 49’ 12” δ2 2080 20.000 00 57’ 18” 10 46’ 30” δ3 2100 20.000 00 57’ 18” 20 43’ 48” δ4 2120 20.000 00 57’ 18” 30 41’ 6” δ5 2140 20.000 00 57’ 18” 40 38’ 24” δ6 2160 20.000 00 57’ 18” 50 35’ 42” δ7 2180 20.000 00 57’ 18” 60 33’ 0” δ8 2200 20.000 00 57’ 18” 70 30’ 18” δ9 2220 20.000 00 57’ 18” 80 27’ 36” T2 2235.506 15.506 00 44’ 25” 90 12’ 1” Σ = 192.684 Σ = 90 12’ 1” 00 49’ 12” θ / 2 = 180 24’ / 2 = 90 12’ 1”
  • 13. Sub chords line method Radius, R 600m Deflection angle , θ Offset 18 0 24' Chainage intersection formul point, I 2140m • Given a a Ofset = 2R b(b + a) Ofset = 2R b Ofset = R 2 1 2 2 n Ofset = n −1 c (c + b) 2R 20m Draw the table form for sub chord line method Stn . Chainag e Chord length Offset
  • 14. Tangent length Chainage T 1 Arc length Chainage T 2 = R tan θ/2 = 600 tan (18°24′/2) PROCEDUR 17.179 Ofset = 2x E600 = 97.20m 2 = chainage I – tangent length 1 = 2140.00 - 97.20 = 2042.80m 20(20 + 17.179) Ofset = 192.684m = 2 x600 20 = chainage T + arc length Ofset = 2042.80 + 192.68 = 2235.48m = 600 15.506(15.506 + 20) Chord Offset Ofset = length 2 x600 0 = R x θ x 2π 360 = 600 x 18 o 24’ x 2π 360 2 2 1 n Stn . T1 δ1 δ2 δ3 δ4 δ5 δ6 δ7 δ8 δ9 T2 Chainag e 2042.821 2060 2080 2100 2120 2140 2160 2180 2200 2220 2235.506 n −1 a = 17.179 b = 20.000 20.000 20.000 20.000 20.000 20.000 20.000 20.000 c =15.506 Σ = 192.684 0.246 0.620 0.667 0.667 0.667 0.667 0.667 0.667 0.667 0.459 - First Offset Second Offset Other Offset Last Offset
  • 15. PROCEDURE ways, depending on the accuracy SETTING Circular curves may be set out in a variety of required, its radius of curvature and obstructions on site. OUT 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.
  • 16. FIELD WORK OPTICAL SQUARE Optical squares are simple sighting instruments used to set out right angles. They can be provided either with mirrors or with one or two prisms. Because of practical difficulties in using squares with mirrors, they have been replaced by squares with prisms: "prismatic squares". There are two major types of prismatic squares: single prismatic squares and double prismatic squares; both will be dealt with in the sections which follow. OFFSET LINE FROM BASELINE USED OPTICAL SQUARE
  • 17. STEP 1 STEP 2 OFFSET LINE FROM BASELINE
  • 18. STEP 1 STEP 2 STATION AT BASELINE
  • 19. End of sub -topic • Exercise • Practical 6 & 7