1. GEOMETRIC DESIGN: VERTICAL ALIGNMENT
CE331 Transportation Engineering

2. Objectives Describe components of vertical alignment
Understand the design criteria for vertical curve
Determine the minimum length of a vertical curve
Calculate the stations and elevations along a vertical curve

3. Vertical Alignment Components Tangent grade Vertical curve
AASHTO guideline recommends maximum grades
Vertical curve
Minimum length (Lmin)
Stations and elevations of points along the curve
Maximum grades are determined by design speed of the roads.
General practice (for all roads):
Gmax = 5% when Vdesign = 70mph; 7-12% when design speed is 30mph. Maximum grades should be used only infrequently.
For short grades <500ft in length and one-way downgrade, the Gmax may be about 1% steeper.
On rural low-volume roads, Gmax may be 2% steeper.
Minimum grades are usually provided to curbed roadways for drainage purpose.

4. Vertical Curve Usually parabolas Crest or sag
Length measured along the horizontal plane

5. Assistant with Target Rod (2ft object height)
Observer with Sighting Rod (3.5 ft)

6. Design Criteria Provide minimum Stopping Sight Distance (SSD)
Required at any point
Provide adequate drainage
Provide comfortable ride
Provide pleasant appearance
Natural topography
Environment/views
Horizontal curves
Balance grades

7. Minimum Length of the Vertical Curve (Lmin) – based on SSD requirement
Crest
Sag
SSD<L
A*SSD2/2158
A*SSD2/( SSD)
SSD>L
2SSD – (2158/A)
2SSD - ( SSD)/A
Parameters are based on the default values:
Crest curve:
The height of the driver’s eye = 3.5 ft
The height of the object = 2.0 ft
Sag curve:
The height of headlight = 2.0 ft
The upward spread of the light beam = 1-degree divergence angle
A = |g2 – g1|*100 (%)
SSD: Stopping Sight Distance (ft)

8. Design Values (K) for SSD
Alternate method …
Design Values (K) for SSD
Speed (km/h)
Crest
Sag 20 1 3 30 2 6 40 4 9 50 7 13 60 11 18 70 17 23 80 26 90 39 38
100 52 45
110 74 55
120 95 63
Rate of curvature
K is length per 1% A
Example
u=80 km/h
Lmin= 5(26)=130 m Crest
Lmin= 5(30)=150 m Sag

9. Example
A vertical curve connects a 3% grade with a 0% grade. The required stopping sight distance is 440ft. What should be the minimum length of the curve?

10. Calculate Stations & Elevations
End points
Sta.PVC = Sta.PVI – L/2
HPVC = HPVI – g1·L/2
Sta.PVT = Sta.PVI + L/2
HPVT = HPVI + g2·L/2
Other points
Define a coordinate system (next slide)
Find H for each x (usually in 100ft increment)

11. Vertical Curves g1 g2 x Y HP= HPVC + g1 x + (g2– g1)x2 / 2L PVI PVC
PVT
L/2 Y
HP= HPVC + g1 x + (g2– g1)x2 / 2L

12. Examples Two types of design problems Unconstrained Constrained
Passing through certain point, or
Maintaining certain clearance from object

13. Example (1/4) = 177.08 Elevation at middle?
Sta
Elev
+3%
-3%
250 m
0.03
250/2 =
HP= HPVC + g1 x + (g2– g1) x2 / 2L
250
175.20

14. Example (2/4) = 176.93 Min clearance 4.5 m?
Sta
Elev
Sta
Elev
+3%
-3%
250 m
0.03
HP= HPVC + g1 x + (g2– g1) x2 / 2L =
250
175.20
Δ= = 3.47 m NO

15. Example (3/4) L for min clearance 4.5 m?
Sta
Elev.
L for min clearance 4.5 m?
Sta
Elev
Sta
Elev
+3%
PVI
Sta /2=
Elev (125) = =
-3%
L m
L/2 35
175.90
0.03
L/2 + 35
HP= HPVC + g1 x + (g2– g1) x2 / 2L
(L/2)

16. Example (4/4) 175.90 = [178.95-0.03(L/2)]+[0.03(L/2+35)]+
136.67L=L2/4+2(L/2)(35)+352
L L+4900=0
L1 = m L2 = m