This document provides information on geometric design concepts for highways, with a focus on vertical alignment and vertical curves. It includes definitions of terms like gradient, ruling gradient, limiting gradient, minimum gradient, and critical length of grade. It describes factors that influence grades like vehicle speed, acceleration and comfort. It also covers vertical curve fundamentals, including equations for crest and sag vertical curves based on stopping sight distance and headlight sight distance. Examples are provided for calculating sight distances and lengths for different grade change scenarios.

1. CEE320
Winter2006
Geometric Design
CEE 320
Steve Muench

2. CEE320
Winter2006
Outline
1. Concepts
2. Vertical Alignment
a. Fundamentals
b. Grades
c. Crest Vertical Curves
d. Sag Vertical Curves
e. Examples

3. CEE320
Winter2006
Concepts
• Alignment is a 3D problem broken
down into two 2D problems
– Horizontal Alignment (plan view)
– Vertical Alignment (profile view)
• Stationing
– Along horizontal alignment
– 12+00 = 1,200 ft.
Piilani Highway on Maui

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Winter2006
Stationing
Horizontal Alignment
Vertical Alignment

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Winter2006
Vertical Alignment

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Winter2006
VERTICAL ALIGNMENT
• Follow the general topography or profile of land.
• Profile not always
• V.A = Elevation of C.L. profile
• Consists of Grades and vertical curves
• Grades = The rate of rise or fall along the length of road to achieve uniform
speed
• Expressed in Ratio 1: x or 1: n or s% or G%
• Ascending: Rise in direction of movement
• Descending: Fall in direction of movement
Influence of Grades:
– On vehicle speed
– Acceleration and deceleration rate
– S.S.D
– Passenger comfort

7. CEE320
Winter2006
Categories of Gradients or
control grades
Categories of Gradients:
• Ruling or Design Gradient:
• Maximum gradient, within that the vertical profile is
designed
• Difficult to fix because depends on
– Type of terrain
– The length of the grade ( Change in speed affected by the
length)
– The design speed (classification of roads)
– Pulling power of the Vehicles
– Presence of horizontal curve ( provide flatter gradient)
3% for plain, 5% for Rolling, 7 % for Hilly

8. CEE320
Winter2006
Categories of Gradients:
• Limiting Gradient:
• Steeper than the ruling gradient
• Provide due to topographic constraints
• Extra care required
– Place a level stretch or easier grade between longer limiting
grades
– 5% for Plain and Rolling, 7 % for Hilly,
• Exceptional Gradient:
• Provided in extreme difficult situations
• Steeper than the limiting
• For only shorter stretches ( not > 60 m in one Km)
• 7% for Plain and Rolling, 8% for hilly

9. CEE320
Winter2006
Minimum Gradient
• Provided to drain out the water along the
side drains and depends
– Surface of the drains ( earthen, R.C.C ….)
– Rainfall Run-off
– Type of soil
– Topography and site condition
• 0.2% for used generally
• 1% for earthen or open drains

10. CEE320
Winter2006
Critical Length of Grade
• Maximum length of ascending gradient on
which a loaded heavy vehicle can operate
without undue reduction in speed ( not >
15 km/h or not > 10mph )
Depends on
– gradient and
– horse power of truck

11. CEE320
Winter2006
Vehicles on Grades
• Passenger cars
– The upgrades up to 3% have a slight impact on passenger
cars.
– On steeper grades this effect becomes more pronounced.
– No special consideration is needed.
• Trucks
– The effect of grades on truck speeds can be quite strong (Exh.
3-53, 3-60).
– The truck with the mass/power ratio 120 kg/kW 0r 200 lb/hp) is
selected to represent heavy vehicles (conservative
assumption).
• Recreational vehicles
– Consideration of recreational vehicles on grades can be
justified for recreational roads with the low percent of trucks
(for example, to consider an additional lane).

12. CEE320
Winter2006
Speed Profiles for Trucks Decelerating
on Grades
Crawling
speed

13. CEE320
Winter2006
Speed Profiles for Trucks
Accelerating on Grades

14. CEE320
Winter2006
Truck Speed Reduction vs. Safety

15. CEE320
Winter2006
Critical Lengths of Grades

16. CEE320
Winter2006
• What is critical length for grade of a
highway with design speed of 85 kmph
with an upgrade of 4 %.
• If we follow 15kmph reduction as given
then critical length for grade will be
around 350 meters.

17. CEE320
Winter2006
Highway plan and profile

18. CEE320
Winter2006

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Winter2006
Vertical Curves
Objective:
• Provided at intersection of two gradients
• Safe transition between two grades
• To smoothen the vertical profile
• To ease of the changes from one grade to
another
G1 G2
G1
G2
Crest Vertical Curve
Sag Vertical Curve

20. CEE320
Winter2006
Vertical Curves
• Summit / crest curves: whose convexity
upward
• Valley / sag curves: whose convexity
downward
• Mostly parabola curve is used because
– It offers very comfortable riding
– Easy to compute the ordinates
G1 G2
G1
G2
Crest Vertical Curve
Sag Vertical Curve

21. CEE320
Winter2006
Types of Vertical Curves

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Winter2006
Vertical Curve Fundamentals
• Mostly Parabolic curves are used for
Vertical Curves
• Parabolic function
– Constant rate of change of slope
– involves equal curve tangents
• y is the roadway elevation x stations
(or feet) from the beginning of the curve
cbxaxy ++= 2

23. CEE320
Winter2006
Vertical Curve Fundamentals
G1
G2
PVI
PVT
PVC
L
L/2
δ
cbxaxy ++= 2
x
Choose Either:
• G1, G2 in decimal form, L in feet
• G1, G2 in percent, L in stations

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Winter2006
General Consideration in
design of vertical curves

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Winter2006
Other Properties
• K-Value (defines vertical curvature)
– The number of horizontal feet or meter needed
for a 1% change in slope
A
L
K =
1./ GKxptlowhigh =⇒

26. CEE320
Winter2006
Crest Vertical Curves
• Maximum deviation angle, when
ascending gradient meets with
descending gradient.
• C.F. on crest curves for fast moving
vehicles acts upward against the gravity
force
• But produce insignificant discomfort,
because the pressure on tires or springs
is relaxed.

27. CEE320
Winter2006
Crest Vertical Curves
• Comfort of driver not considered in design
of summit curve.
• But SSD is obstructed on summit as well
as on valley curves, hence used in design
from safety point of view.

28. CEE320
Winter2006
Crest Vertical Curves
G1
G2
PVI
PVTPVC
h2
h1
L
SSD
( )
( )2
21
2
22100 hh
SSDA
L
+
= ( )
( )
A
hh
SSDL
2
21200
2
+
−=
For SSD < L For SSD > L
Line of Sight

29. CEE320
Winter2006
Crest Vertical Curve
( )
( )
m.inheightobject
m,insurfaceroadwayaboveeyeofheight
%,inchangegrade
m,indstancesight
m,incurveoflenghtminimum
:where
200
2For
22100
For
2
1
2
21
2
21
2
=
=
=
=
=
+
−=≥
+
⋅
=<
h
h
A
S
L
A
hh
SLLS
hh
SA
LLS

30. CEE320
Winter2006
Stopping Sight Distance
Consideration
height.objectm,0.60
surface,roadwayaboveeyeofheightm,08.1
:settingafter
658
2For
)
658
(or
658
For
2
1
22
=
=
−=≥
=
⋅
=<
h
h
A
SLLS
S
K
SA
LLS

31. CEE320
Winter2006
Crest Vertical Curves
• Assumptions for design
– h1 = driver’s eye height = 3.5 ft.
– h2 = tail light height = 2.0 ft.
• Simplified Equations
( )
2158
2
SSDA
L = ( )
A
SSDL
2158
2 −=
For SSD < L For SSD > L

32. CEE320
Winter2006
Appearance and Drainage
Considerations
Another way to get min length is 3 x (design speed in
mph)
Minimum lengths are about 100 to 300 ft.

33. CEE320
Winter2006
Stopping Sight Distance
Consideration

34. CEE320
Winter2006
Passing Sight Distance
Consideration
height.objectm,33.1
surface,roadwayaboveeyeofheightm,08.1
:settingafter
864
2For
)
864
(or
864
For
2
1
22
=
=
−=≥
=
⋅
=<
h
h
A
SLLS
S
K
SA
LLS

35. CEE320
Winter2006
Passing Sight Distance
Consideration
Design controls for passing sight
distance typically are not used due
to high costs of fitting the
alignm
ent to the terrain

36. CEE320
Winter2006
Sag Vertical Curves
• Valley / sag curves: whose convexity
downward
• Three factors considered
– Safety, comfort and appearance.
• Safety
– During day time SSD already available
– But during night SSD is limited to head light
– Hence design based on availability of SSD
under head light.

37. CEE320
Winter2006
Sag Vertical Curves
G1
G2
PVI
PVTPVC
h2=0h1
L
Light Beam Distance (SSD)
( )
( )βtan200 1
2
Sh
SSDA
L
+
= ( ) ( )( )
A
SSDh
SSDL
βtan200
2 1 +
−=
For SSD < L For SSD > L
headlight beam (diverging from LOS by β degrees)

38. CEE320
Winter2006
For S < L
L
A S
h S tn
=
⋅
+ ⋅
2
200( )β
For S ≥ L
L S
h S tn
A
= −
+ ⋅
2
200( )β
Assuming h=0.6 m, headlight height; and a β = 1°,
upward divergence of the light beam,
For S < L
L
A S
S
=
⋅
+
2
120 35.For S ≥ L
L S
S
A
= −
+
2
120 35.
Sag Vertical Curves
Headlight Sight Distance

39. CEE320
Winter2006
For S < L
L
A S
h S tn
=
⋅
+ ⋅
2
200( )β
For S ≥ L
L S
h S tn
A
= −
+ ⋅
2
200( )β
Assuming h=0.6 m, headlight height; and a β = 1°,
upward divergence of the light beam,
For S < L
L
A S
S
=
⋅
+
2
120 35.For S ≥ L
L S
S
A
= −
+
2
120 35.
Sag Vertical Curves
Headlight Sight Distance

40. CEE320
Winter2006
The centrifugal acceleration should not exceed 0.3
m/s2. The minimum length of the sag curve is:
where:
A = change in grade, V = speed in km/h.
This requirement is only about 50% of the headlight
sight distance.
Sag Vertical Curves
Comfort Consideration
395
2
AV
L =

41. CEE320
Winter2006
The minimum length is
L=0.6 V
(as for crest curves)
In addition, one may consider applying K at least
equal to 30 to small and moderate changes in grade.
Longer curves on high-type highways seem to be
more appropriate.
Sag Vertical Curves
Appearance Consideration

42. CEE320
Winter2006
Sag Vertical Curves

43. CEE320
Winter2006
Sag Vertical Curves
• Assumptions for design
– h1 = headlight height = 2.0 ft.
– β = 1 degree
• Simplified Equations
( )
( )SSD
SSDA
L
5.3400
2
+
= ( ) ( )





 +
−=
A
SSD
SSDL
5.3400
2
For SSD < L For SSD > L

44. CEE320
Winter2006
Design Controls for Sag Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2001

45. CEE320
Winter2006
Example 1
A car is traveling at 30 mph in the country at night on a wet road
through a 150 ft. long sag vertical curve. The entering grade is -2.4
percent and the exiting grade is 4.0 percent. A tree has fallen across
the road at approximately the PVT. Assuming the driver cannot see
the tree until it is lit by her headlights, is it reasonable to expect the
driver to be able to stop before hitting the tree?

46. CEE320
Winter2006
SOLUTION to Problem #1
• Assume that S>L (it usually is not but for example we’ll do it this
way), therefore S = 146.23 ft. which is less than L
• Must use S<L equation, it’s a quadratic with roots of 146.17 ft and
-64.14 ft.
• The driver will see the tree when it is 146.17 feet in front of her.
• Available SSD is 146.17 ft.
• Required SSD = (1.47 x 30)2
/2(32.2)(0.35 + 0) + 2.5(1.47 x 30) =
196.53 ft.
• Therefore, she’s not going to stop in time.
• OR
• L/A = K = 150/6.4 = 23.43, which is less than the required K of 37
for a 30 mph design speed
• Stopping sight distance on level ground at 30 mph is
approximately 200 ft.

47. CEE320
Winter2006
Example 2
Similar to Example 1 but for a crest curve.
A car is traveling at 30 mph in the country at night on a wet road
through a 150 ft. long crest vertical curve. The entering grade is 3.0
percent and the exiting grade is -3.4 percent. A tree has fallen across
the road at approximately the PVT. Is it reasonable to expect the
driver to be able to stop before hitting the tree?

48. CEE320
Winter2006
Solution Problem #2
• Assume that S>L (it usually is), therefore SSD = 243.59 ft.
which is greater than L
• The driver will see the tree when it is 243.59 feet in front of
her.
• Available SSD = 243.59 ft.
• Required SSD = (1.47 x 30)2
/2(32.2)(0.35 + 0) + 2.5(1.47 x 30)
= 196.53 ft.
• Therefore, she will be able to stop in time.
• OR
• L/A = K = 150/6.4 = 23.43, which is greater than the required
K of 19 for a 30 mph design speed on a crest vertical curve
• Stopping sight distance on level ground at 30 mph is
approximately 200 ft.

49. CEE320
Winter2006
Example 3
A roadway is being designed using a 45 mph design speed. One
section of the roadway must go up and over a small hill with an
entering grade of 3.2 percent and an exiting grade of -2.0 percent.
How long must the vertical curve be?
For 45 mph we get K=61, therefore L = KA = (61)(5.2) = 317.2 ft

50. CEE320
Winter2006
Solution Problem #3
• For 45 mph we get K=61, therefore L = KA
= (61)(5.2) = 317.2 ft.

51. CEE320
Winter2006
Example
A 400 ft. equal tangent crest vertical curve has a PVC station of
100+00 at 59 ft. elevation. The initial grade is 2.0 percent and the
final grade is -4.5 percent. Determine the elevation and stationing of
PVI, PVT, and the high point of the curve.
G1
=2.0%
G
2 = - 4.5%
PVI
PVT
PVC: STA 100+00
EL 59 ft.

52. CEE320
Winter2006
G1
=2.0%
G
2 = -4.5%
PVI
PVT
PVC: STA 100+00
EL 59 ft.
400 ft. vertical curve, therefore:
PVI is at STA 102+00 and PVT is at STA 104+00
Elevation of the PVI is 59’ + 0.02(200) = 63 ft.
Elevation of the PVT is 63’ – 0.045(200) = 54 ft.
High point elevation requires figuring out the equation for a vertical curve
At x = 0, y = c => c=59 ft.
At x = 0, dY/dx = b = G1 = +2.0%
a = (G2 – G1)/2L = (-4.5 – 2)/(2(4)) = - 0.8125

53. CEE320
Winter2006
• y = -0.8125x2 + 2x + 59
• High point is where dy/dx = 0
• dy/dx = -1.625x + 2 = 0
• x = 1.23 stations
• Find elevation at x = 1.23 stations
• y = -0.8125(1.23)2 + 2(1.23) + 59
• y = 60.23 ft

54. CEE320
Winter2006
Other Properties
G1
G2
PVI
PVT
PVC
x
Ym
Yf
Y
2
200
x
L
A
Y =
800
AL
Ym =
200
AL
Yf =
21 GGA −=
•G1, G2 in percent
•L in feet
Last slide we found x = 1.23 stations