1. Introduction to Transportation Engineering
Instructor Dr. Norman Garrick
Hamed Ahangari
17th April 2014

2. Horizontal Curve

3. Elements
Δ– Intersection angle
R - Radius Δ R Δ

4. R = Radius of Circular Curve L = Length of Curvature (L = EC – BC)
Δ = Deflection Angle
R = Radius of Circular Curve
L = Length of Curvature (L = EC – BC)
BC = Beginning of Curve (PC)
EC = End of Curve (PT)
PI = Point of Intersection
T = Tangent Length (T = PI – BC = EC - PI)
C = Chord Length
M = Middle Ordinate
E = External Distance PI Δ T E T L EC
(PT) BC
(PC) M C R R
Δ/2 Δ

5. Basic Definitions BC/PC: Point of Curvature BC = PI – T
PI = Point of Intersection
T = Tangent
EC/PT: Point of Tangency
EC = BC + L
L = Length Δ BC
(PC) EC Δ

6. Degree of Curvature D used to describe curves D defines Radius
Arc Method:
D/ Δ = 100/L (1)
  (360/D)=100/(2R)
   R = 5730/D (2)

7. Curve Calculations Length L = 100.Δ/D (3) Tangent T = R.tan(Δ /2) (4)
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)

8. Example 1
A horizontal curve is designed with a 2000 ft. radius. The tangent length is 500 ft. and the EC station is What are the BC and PI stations?

9. Solution Since we know R and T we can use
T = R.tan(Δ /2) so Δ = 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 = = 2520~25+20

10. Example 2
A curve has external angle of 20.30’ degrees, a degree of curvature is 2°30’ and the PI is at Calculate:
Radius
Length of Curve
BC and EC
Chord
External Distance
Mid Ordinate

11. Solution Given: D = 2°30’, Δ=22.30’ Part i) Radius:
Part ii ) Length of Curve
Part iii ) BC and EC

12. Part iv ) Chord
Part v ) External Distance
Part vi ) Mid Ordinate

13. 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

14. Spiral Curves k = 100 D/ Ls Δ = Δc + 2 Δs Δs = Ls D / 200

15. 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

16. Part iii) Spiral Central Angle
Part ii) Total Length
Total Length of curve = length with no spiral + Ls
Total Length= =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 = degree

17. Vertical Curve

18. Horizontal Alignment
Vertical Alignment
Crest Curve G2 G3 G1
Sag Curve

19. Design of Vertical Curves

20. Parabolic Curve
BVC G1 G2
EVC 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 g = grade expressed as a ratio
x = horizontal distance from BVC r = rate of change of grade (ratio)

21. 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 and at an elevation of 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

22. PI
EVC
BVC
2.5%
-1.5%

23. Solution
Part a) the station and the elevation of the BVC and EVC points
Station-BVC= =10150~10+150
Station-EVC= =10650~10+650
Elevation-BVC= *250= m
Elevation-EVC= *250= m

24. 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= , g1= 0.025
r=(g2-g1)/L
r=( (0.025))/500=-0.04/ r=
y = x x^2,
 y(100)= * *100^2
 y(100) =
 y(300)=

25. 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= *x=0
X=0.025/ = 312.5
y(312.5) = * *(31.25)^2
 y(312.5)=

26. 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

27. Solution i) K and r K = L / |A| A=G2-G1, A=-4-(-1)=-3 K=300/3=100
-4%
i) K and r
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/ r=0.0001
-1%

28. ii) station of BVC and EVC
L=300’, L/2=150’
BVC= =1850’, ~ 18+50
EVC= =2150’, ~ 21+50
iii) elevation of point at a distance, L/4, from BVC
y = y0 + g1.x + 1/2 r.x^2,
r= , g1=0.04
y0= *0.04=206
y= x x^2
L/4~ x=75
 Y(75)=203.28’

29. v) elevation of turning point
iv) station of turning point
at turning point: dy/dx=0
dy=dx= x=0
x(turn)=400’
v) elevation of turning point
x=400’
Y (400) = *(150) (150)^2
Y(400)=198’
This point is after vertical curve
(Turning point is not in the curve)

30. vi) elevation of mid-point
x=150’
Y(150)= *(150) (150)^2
y (150)=201.12’
vii) grade at the mid-point of each curve
Grade at every points: dy/dx= = x
if x= 150  Grade (150)= *150
 grade(150)=