1. Introduction to Transportation Engineering
Instructor Dr. Norman Garrick
Tutorial Instructor Hamed Ahangari
April 2016

2. Horizontal Alignment Spiral Curve

3. Circle Curve Lc Ls Ls
Spiral Curve Δc Δs Δs

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

5. Circle Curve Spiral Curve 1- Ltotal = LC + 2 Ls = Lcircle + Ls
2- Δ = Δc + 2 Δs
3- Δs = Ls D / 200
4- Δc = Lc D / 100
5- k = 100 D/ Ls

6. Example 4
The central angle for a curve is 40 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

7. Solution Given: R=2000 , Δ=40, k=3 D = 5730/2000. Therefore D = 2.86
L = 100(Δ)/D = 100(40)/2.86 = 1398 ft.
Part i) Ls
k = 100 D/ Ls,  3=100*2.86/Ls
 Ls= 95 ft

8. Part iii) Spiral Central Angle
Part ii) Total Length
Total Length of curve = length with no spiral + Ls
Total Length= =1493 ft
Part iii) Spiral Central Angle
Δs = Ls D / 200  Δs = 95*2.86/200= 1.35
 Δs =1.35
 Δc= Δ- 2*Δs=40-2*1.35
 Δc = degree

9. Example 5
In a spiral curve the Δc is 33 degree. If the degree of curvature is 6 and the rate of increase in degree of curvature is 4.
determine the
length of each spiral leg,
the intersection angel,
Radius
total length of curve

10. Solution Given: Δc =33, D=6, k=4 i) Ls ? k = 100 D/ Ls,  4=100*6/Ls
 Ls= 150 ft
ii) Δ ?
Δ = Δc + 2.Δs
Δs = Ls D / 200  Δs = 150*6 / 200
 Δs = 4..5
Δ = *2= 42

11. iii) Radius
R=5730/D  R =5730/6= 955 ft
iv) total length of curve
Ltotal = LC + 2 Ls = Lcircle + Ls
Method a) Ltotal =Lcircle + Ls
L = 100(Δ)/D = 100*(42)/6 = 700 ft.
Ltotal = =850 ft
Method b) Ltotal = LC + 2 Ls
Δc = Lc D / 100  Lc = Δc *100 / D =33*100/6=550
Ltotal = *150 =850 ft

12. Vertical Curve

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

14. Design of Vertical Curves

15. 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)

16. 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

17. PI
EVC
BVC
2.5%
-1.5%

18. 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

19. 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)=

20. 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) = * *(312.5)^2
 y(312.5)=

21. 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 the curve
vii) grade at the mid-point of the curve

22. Solution i) K and r K = L / |A| A=G2-G1, A=-4-(1)=-5 K=300/5=60
-4%
i) K and r
K = L / |A|
A=G2-G1, A=-4-(1)=-5
K=300/5=60
r=(g2-g1)/L
r=(-0.01-(-0.04))/300=0.03/ r=0.0001
-1%

23. 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 + g1x + 1/2 rx2,
r= , g1=0.04
y0= *0.04=206
y= x x^2
L/4~ x=75
 Y(75)=203.28’

24. 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)

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