### CE 2710 Tutorial H and V Alignment Curves April 17

```Introduction to
Transportation Engineering
Instructor Dr. Norman Garrick
Hamed Ahangari
17th April 2014
1
Horizontal Curve
2
Elements

Δ– Intersection angle

Δ
R
Δ
3
Δ = Deflection Angle
PI
Δ
R = Radius of Circular Curve
L
T
L
= Length of Curvature (L = EC – BC)
BC = Beginning of Curve (PC)
EC = End of Curve (PT)
BC
(PC)
PI = Point of Intersection
E
T
EC
(PT)
M
C
R
R
T = Tangent Length (T = PI – BC = EC - PI)
C = Chord Length
M = Middle Ordinate
E = External Distance
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Basic Definitions
• BC/PC: Point of Curvature
• BC = PI – T
Δ
– PI = Point of Intersection
– T = Tangent
BC
(PC)
Δ
• EC/PT: Point of Tangency
• EC = BC + L
EC
– L = Length
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Degree of Curvature
• D used to describe curves
• Arc Method:
– D/ Δ = 100/L
(1)
  (360/D)=100/(2R)
   R = 5730/D
(2)
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Curve Calculations
 Length
L = 100.Δ/D (3)
 Tangent
T = R.tan(Δ /2)
 Chord
C = 2R.sin(Δ /2) (5)
 External Distance
E = = R sec(Δ/2) – R
(6)
 Mid Ordinate
M = R-R.cos(Δ /2))
(7)
(4)
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Example 1
• A horizontal curve is designed with a 2000 ft.
radius. The tangent length is 500 ft. and the EC
station is 30+00. What are the BC and PI
stations?
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Solution
• Since we know R and T we can use
T = R.tan(Δ /2) so Δ = 28.07 degrees
• D = 5730/R. Therefore D = 2.86
• L = 100(Δ)/D = 100(28.07)/2.86 = 980 ft.
• BC = EC – L = 3000 – 980 =2020~20+20
• PI = BC +T = 2020 + 500 = 2520~25+20
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Example 2
A curve has external angle of 20.30’ degrees, a degree
of curvature is 2°30’ and the PI is at 175+00.
Calculate:
Length of Curve
 BC and EC
Chord
External Distance
 Mid Ordinate
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Solution
• Given: D = 2°30’, Δ=22.30’
5730
R
 2291 .83
2.5
• Part ii ) Length of Curve
L  100 
22.5
 900 .00
2.5
• Part iii ) BC and EC
 22.5 
T  2291.38 tan
  455.87'
 2 
BC  (175 00)  (4  55.87)  171 44.13
EC  (171 44.13)  (9  00)  180 44.13
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• Part iv ) Chord
 22.5 
C  2(2291.83) sin
  894.23'
 2 
• Part v ) External Distance

 22.5  
E  2291.83sec(
  1  44.90'
 2  

• Part vi ) Mid Ordinate

 22.5 
M  2291.831  cos
  44.04'
 2 

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Example 3
The central angle for a curve is 30 degrees - the
radius of the circular curve selected for the location
is 2000 ft.
If a spiral with k=3 degrees is selected for use,
determine the
i) length of each spiral leg,
ii) total length of curve
iii) Spiral Central Angle
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Spiral Curves
k = 100 D/ Ls
Δ = Δc + 2 Δs
Δs = Ls D / 200
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Solution
Given: R=2000 , Δ=30, k=3
• D = 5730/2000. Therefore D = 2.86
• L = 100(Δ)/D = 100(30)/2.86 = 1047 ft.
Part i) Ls
k = 100 D/ Ls,  3=100*2.86/Ls
 Ls= 95 ft
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• Part ii) Total Length
Total Length of curve = length with no spiral + Ls
Total Length= 1047+95=1142 ft
Part iii) Spiral Central Angle
Δs = Ls D / 200  Δs = 95*2.86/200= 1.35
 Δs =1.35
 Δc= Δ- 2*Δs=30-2*1.35
 Δc = 27.30 degree
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Vertical Curve
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Horizontal Alignment
Vertical Alignment
Crest Curve
G1
G2
G3
Sag Curve
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Design of Vertical Curves
G1
BVC
Parabolic Curve
EVC
G2
PI
L/2
L/2
L
Change in grade: A = G2 - G1 G is expressed as %
Rate of change of curvature: K = L / |A|
Rate of change of grade: r = (g2 - g1) / L
Equation for determining the elevation at any point on the curve
y = y0 + g1x + 1/2 rx2
where,
y0 = elevation at the BVC
x = horizontal distance from BVC
g = grade expressed as a ratio
r = rate of change of grade (ratio)
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Example 1
• The length of a tangent vertical curve equal to 500 m.
The initial and final grades are 2.5% and -1.5%
respectively. The grades intersect at the station 10+400
and at an elevation of 210.00 m
• Determine:
a)the station and the elevation of the BVC and EVC points
b) the elevation of the point on the curve 100 and 300 meters from
the BVC point
c) the station and the elevation of the highest point on the curve
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PI
BVC
2.5%
EVC
-1.5%
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Solution
• Part a) the station and the elevation of the
BVC and EVC points
– Station-BVC= 10400-250=10150~10+150
– Station-EVC=10400+250=10650~10+650
– Elevation-BVC= 210-0.025*250= 203.75 m
– Elevation-EVC= 210-0.015*250= 206.25 m
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• Part b) the elevation of the point on the curve
100 and 300 meters from the BVC point.
y = y0 + g1.x + 1/2 .r.x^2,
y0= 203.75, g1= 0.025
– r=(g2-g1)/L
– r=(-0.015-(0.025))/500=-0.04/500
r=-0.00008
y = 203.75 + 0.025.x - 0.00004.x^2,
 y(100)=203.75+0.025*100-0.00004*100^2
 y(100) = 205.85
 y(300)= 207.65
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• Part c) the station and the elevation of the
highest point on the curve
– The highest point can be estimated by setting the first
derivative of the parabola as zero.
• Set dy/dx=0,
dy/dx= 0.025-0.0008*x=0
X=0.025/0.00008= 312.5
y(312.5) = 203.75+0.025*31.25-0.00004*(31.25)^2
 y(312.5)= 207.65
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Example 2
• For a vertical curve we know that G1=-4%, G2=-1%,
PI: Station 20+00, Elevation: 200’, L=300’
• Determine:
i)K and r
ii) station of BVC and EVC
iii) elevation of point at a distance, L/4, from BVC
iv) station of turning point
v) elevation of turning point
vi) elevation of mid-point of each curve
vii) grade at the mid-point of each curve
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Solution
-4%
• i) K and r
-1%
– K = L / |A|
– A=G2-G1, A=-4-(-1)=-3
K=300/3=100
– r=(g2-g1)/L
– r=(-0.01-(-0.04))/300=0.03/300
r=0.0001
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• ii) station of BVC and EVC
– L=300’, L/2=150’
– BVC= 2000-150=1850’, ~ 18+50
– EVC= 2000+150=2150’, ~ 21+50
• iii) elevation of point at a distance, L/4, from BVC
– y = y0 + g1.x + 1/2 r.x^2,
– r= -0.0001, g1=0.04
– y0=200+150*0.04=206
y=206-0.04x+0.00005.x^2
L/4~ x=75
 Y(75)=203.28’
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• iv) station of turning point
– at turning point: dy/dx=0
– dy=dx=- 0.04+0.0001x=0
x(turn)=400’
• v) elevation of turning point
– x=400’
– Y (400) = 206-0.04*(150)+0.00005.(150)^2
 Y(400)=198’
This point is after vertical curve
(Turning point is not in the curve)
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• vi) elevation of mid-point
– x=150’
– Y(150)= 206-0.04*(150)+0.00005.(150)^2
y (150)=201.12’
• vii) grade at the mid-point of each curve
– Grade at every points: dy/dx= =- 0.04+0.0001x
– if x= 150  Grade (150)=-0.04+0.0001*150