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

2. Geometric Design of Highways
The engineering aspects of alignment design is usually referred to as geometric design
Highway alignment is in reality
a three-dimensional problem

3. Components of The Alignment
Horizontal Alignment
Plan View
Vertical Alignment
Profile View
Design & construction is difficult in 3-D so highway design is typically treated as three 2-D problems: Horizontal alignment, vertical alignment, cross-section
Cross-section

4. safety.fhwa.dot.gov
wikipedia

5. Horizontal Curve

7. Horizontal Alignment Objective: Primary challenge :
Safety
Comfort
Primary challenge :
The critical design feature of horizontal alignment is the horizontal curve that transitions the roadway between two straight (tangent) sections. Δ
Transition between two directions

8. Key Elements
Δ– Intersection angle PI Δ
R - Radius L PC PT R R

9. Key Elements Length L = R. Δ  L=. R. Δ /180 Tangent (tan(Δ /2)= T/R)PI Δ T L PC PT
Length L = R. Δ 
L=. R. Δ /180
Tangent (tan(Δ /2)= T/R)
T= R.tan(Δ /2) R R
Δ/2 Δ T R
Δ/2

10. Stationing Reference System
2+00
3+00
4+00
1+00
0+00
Horizontal Alignment

11. R = Radius of Circular Curve L = Length of Curvature
Δ = Deflection Angle
R = Radius of Circular Curve
L = Length of Curvature
BC = Beginning of Curve (PC)
EC = End of Curve (PT)
PI = Point of Intersection
T = Tangent Length
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

12. Basic Definitions L BC/PC: Point of Curvature BC = PI – T TΔ
BC/PC: Point of Curvature
BC = PI – T
PI = Point of Intersection
T = Tangent
EC/PT: Point of Tangency
EC = BC + L
L = Length T L PC
(BC) PT
(EC) R R
Δ/2 Δ 12

13. 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)

14. 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)

15. 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?

16. 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)

17. Example 2
A curve has external angle of 22.30’ 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

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

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

20. Example 3
In a horizontal curve the intersection angle is 60 degree. We also know that the minimum internal sight distance (Mid Ordinate) for this road is 100 ft. If the BC of the curve is in Calculate:
Radius
Length of Curve
PI and EC

21. Solution L=. R. Δ /180 Given: Δ= 60 , M= 100 ft and BC= 10+50
Part i) Radius:
M = R-R.cos(Δ /2)
 R = ft
Part ii ) Length of Curve
L=. R. Δ /180

22. Solution
Given: Δ= 60 , M= 100 ft and BC= (10+50)
Part iii ) PI and EC