CIVL361 Transportation
Engineering
Horizontal and Spiral Curves
Dr. Mehmet M. Kunt
04/11/2013
Horizontal Curve Types
•Simple
•Compound
•Reverse Curve
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AZIMUTH AND BEARING
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Azimuth and Bearing
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EQUATIONS
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Equations for Horizontal Curves

T  R tan 
2
Tangent
Example:
A horizontal curve has a radius of 320
m, and deflection angle of Δ = 24 o.
Please compute L, T, M, E, C.
Solution:
T = 320 * tan( 24 /2.)= 68.02 m
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Equations for Horizontal Curves

M  R  R cos  Middle distance
2
Example:
A horizontal curve has a radius of 320
m, and deflection angle of Δ = 24 o.
Please compute L, T, M, E, C.
Solution:
M = 320 - 320 *cos( 24 /2.)= 6.99 m
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Equations for Horizontal Curves
E
R
R

cos 
2
External Distance
Example:
A horizontal curve has a radius of 320
m, and deflection angle of Δ = 24 o.
Please compute L, T, M, E, C.
Solution:
E = 320 /cos( 24 /2.) - 320 = 7.15 m
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Equations for Horizontal Curves

C  2R sin 
2
Chord
Example:
A horizontal curve has a radius of 320
m, and deflection angle of Δ = 24 o.
Please compute L, T, M, E, C.
Solution:
chord = 2* 320 *sin( 24 /2.)= 133.06 m
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Equations for Horizontal Curves
L  Rc  rad
Length
Example:
A horizontal curve has a radius of 320
m, and deflection angle of Δ = 24 o.
Please compute L, T, M, E, C.
Solution:
L = 320 *( 24 /(180./pi))= 134.04 m
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Example
For a horizontal curve with radius of 350 meters and deflection angle of
24 degrees please obtain the T, M, E, C and L values.
SOLUTION
>>> hcall(350,24,0)
Horizontal Curve Parameter Computations
INPUT
======
Radius
= 350 m
Deflection Angle = 24 degrees
COMPUTATIONS
============
Tangent, T
= 74.4 m Function: T = hct(Radius,delta)
Middle distance, M = 7.6 m Function: M = hcm(Radius,delta)
External distance, E = 7.8 m Function: E = hce(Radius,delta)
Chord, C
= 145.5 m Function: C = hcchord(Radius,delta)
Length of the curve,L = 146.6 m Function: L = hcl(Radius,delta)
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Plot of a Horizontal Curve
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L
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T
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M
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E
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Example
For a horizontal curve with radius of 350 meters and
deflection angle of 24 degrees please obtain the T, M, E, C
and L values.
SOLUTION
>>> hcall(350,24,1)
Radius = 350
Deflection angle = 24
Tangent = 74.4
Middle distance = 7.6
External distance = 7.8
Chord = 145.5
Length of the curve = 146.6
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Example
For a horizontal curve with radius of 350 meters and
deflection angle of 24 degrees please obtain the T, M, E, C
and L values.
SOLUTION
>>> hcall(350,24,2)
Input: Radius, deflection angle
350 24
T,M,E,C,L
74.4 7.6 7.8 145.5 146.6
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Deflection Angle
 x 
dx  

 2 R  RAD
cx  2R sin d x
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Example
• Prepare a table giving chords and deflection angles
for staking out a 400-m radius circular curve with a
total deflection angle of 30o . The TC point is at
station 44+60. Give the deflection angles and chords
for 0.25L, 0.5L, 0.75L and L.
 x 
dx  

 2 R  RAD
L  Rc  rad
cx  2R sin d x
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Example
Prepare a table giving chords and deflection angles for staking
out a 400-m radius circular curve with a total deflection angle
of 30o . The TC point is at station 44+60. Give the deflection
angles and chords for 0.25L, 0.5L, 0.75L and L.
 x 
dx  

 2 R  RAD
cx  2R sin d x
L  Rc  rad
SOLUTION
X values are equal to 0.25 L, 0.5L, 0.75L and L
L = 400*30/57.295 = 209.44 m
X = 0.25L =52.36 m
dx = (52.36/(2*400)) = 0.0654
cx = 2*400*sin(0.0654) = 52.32 m
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Example
Prepare a table giving chords and deflection angles for staking
out a 400-m radius circular curve with a total deflection angle
of 30o . The TC point is at station 44+60. Give the deflection
angles and chords for 0.25L, 0.5L, 0.75L and L.
 x 
dx  

 2 R  RAD
cx  2R sin d x
L  Rc  rad
SOLUTION
And all of the calculations can be computed as
follows:
Length of the
Station X
44+60
0.0
45+12
52.36
45+64
104.72
46+16
157.08
46+68
209.44
curve =
dx
0.0000
0.0654
0.1309
0.1963
0.2618
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209.44 m 2.09 sta
cx
0.000
52.323 (computed in previous
104.421
156.073
207.056
page)
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Stopping Sight Distance on
Horizontal Curves
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SUPERELEVATION
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Superelevation
V2
Rc 
127 f  e 
Rc = Radius of the curve
V = Speed of the vehicle, km/h
f = coefficient of side friction
e = superelevation rate
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Table 4.5 Values of side friction
recommended by AASHTO
Design Speed,
km/h
30
40
50
60
70
80
90
100
110
120
Maximum side
friction factor
0.17
0.17
0.16
0.15
0.14
0.14
0.13
0.12
0.11
0.09
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Table 4.6 Recommended minimum radius of
curvature
Design
Speed, km/h
30
40
50
60
70
80
90
100
110
120
Minimum curve
radius, m
35
60
100
150
215
280
375
490
635
870
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SPIRAL CURVES
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Spiral of Theodorus
http://en.wikipedia.org/wiki/Spiral_of_Theodorus
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Spiral curve parameters
Ls
s 
2 Rc
A  Ls Rc
L5s
X s  Ls 
40A 4
L3s
L7s
Ys 

2
6
6A
336 A
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;
Spiral curve parameters
s 
Ls
2 Rc
A  Ls Rc
X s  Ls 
L5s
L3s
L7s
Y


40A 4 s 6 A 2 336 A 6
p  Ys  Rc 1 cos  s 

T '  Rc  p  tan 
2
k  X s  Rc sin  s
Lc  Rc rad  Ls
c
X 2 Y 2
Y 
d  tan  
X
1
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Problem
• A roadway goes from tangent alignment to a
250-m circular curve by means of a
80-m-long spiral transition curve. The
deflection angle between the tangents is 45°.
Use formulas to compute Xs Ys p, and k.
Assume that the station of the P.L,
measured along the back tangent, is 250 + 00,
and compute the stations of the TS,
SC, CS, and ST.
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Solution
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