1. Geometric Design
CEE 320 Anne Goodchild

2. Outline Concepts Vertical Alignment Horizontal Alignment
Fundamentals
Crest Vertical Curves
Sag Vertical Curves
Examples
Horizontal Alignment
Superelevation
Other Non-Testable Stuff

3. 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

4. Stationing Horizontal Alignment Vertical Alignment
Therefore, roads will almost always be a bit longer than their stationing because of the vertical alignment
Draw in stationing on each of these curves and explain it

5. From Perteet Engineering
Typical set of road plans – one page only
From Perteet Engineering

6. Vertical Alignment

7. Vertical Alignment Objective: Primary challenge
Determine elevation to ensure
Proper drainage
Acceptable level of safety
Primary challenge
Transition between two grades
Vertical curves
Sag Vertical Curve G1 G2 G1 G2
Crest Vertical Curve

8. Vertical Curve Fundamentals
Parabolic function
Constant rate of change of slope
Implies equal curve tangents
y is the roadway elevation x stations (or feet) from the beginning of the curve

9. Vertical Curve Fundamentals
PVI G1 δ
PVC G2
PVT
L/2 L x
Choose Either:
G1, G2 in decimal form, L in feet
G1, G2 in percent, L in stations

10. Relationships Choose Either: G1, G2 in decimal form, L in feet
G1, G2 in percent, L in stations
Relationships

11. Example
A 400 ft. equal tangent crest vertical curve has a PVC station of 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.
PVI
PVT
G1=2.0%
G2= - 4.5%
PVC: STA
EL 59 ft.

12. PVI G1=2.0% PVT G2= -4.5% PVC: STA 100+00 EL 59 ft.
400 ft. vertical curve, therefore:
PVI is at STA and PVT is at STA
Elevation of the PVI is 59’ (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)) =
y = x2 + 2x + 59
High point is where dy/dx = 0
dy/dx = x + 2 = 0
x = 1.23 stations
Find elevation at x = 1.23 stations
y = (1.23)2 + 2(1.23) + 59
y = ft

13. Other Properties G1, G2 in percent L in feet G1 x PVT PVC Y Ym G2 PVIYf
Last slide we found x = 1.23 stations

14. Other Properties K-Value (defines vertical curvature)
The number of horizontal feet needed for a 1% change in slope
G is in percent, x is in feet
G is in decimal, x is in stations

15. Crest Vertical Curves For SSD < L For SSD > L SSD h2 h1 L
PVI
Line of Sight
PVC G1
PVT G2 h2 h1 L
For SSD < L
For SSD > L

16. Crest Vertical Curves Assumptions for design Simplified Equations
h1 = driver’s eye height = 3.5 ft.
h2 = tail light height = 2.0 ft.
Simplified Equations
Minimum lengths are about 100 to 300 ft.
Another way to get min length is 3 x (design speed in mph)
For SSD < L
For SSD > L

17. Crest Vertical Curves
Assuming L > SSD…

18. Design Controls for Crest Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

19. Design Controls for Crest Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

20. Light Beam Distance (SSD)
Sag Vertical Curves
Light Beam Distance (SSD) G1
headlight beam (diverging from LOS by β degrees) G2
PVC
PVT h1
PVI
h2=0 L
For SSD < L
For SSD > L

21. Sag Vertical Curves Assumptions for design Simplified Equations
h1 = headlight height = 2.0 ft.
β = 1 degree
Simplified Equations
What can you do if you need a shorter sag vertical curve than calculated? Provide fixed-source street lighting
Minimum lengths are about 100 to 300 ft.
Another way to get min length is 3 x design speed in mph
For SSD < L
For SSD > L

22. Sag Vertical Curves
Assuming L > SSD…

23. Design Controls for Sag Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

24. Design Controls for Sag Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

25. 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?
Assume S<L (this is the case more often than not)
Using the S<L equation, it’s a quadratic with roots of ft and ft.
The driver will see the tree when it is feet in front of her.
Available SSD is ft.
Required SSD = (1.47 x 30)2/2(32.2)( ) + 2.5(1.47 x 30) = 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.

26. Sag Vertical Curve
Assume S<L, try both, but this is most often the case
Equation specific to sag curve which accommodates headlight beam
L and S in horizontal plane and comparable (150 and 146 ft)
Required SSD = ft assumes 0 grade
Text problem versus design problem.

27. Light Beam Distance (S)
Sag Vertical Curves
Light Beam Distance (S) G1
diverging from horizontal plane of vehicle by β degrees G2
PVC
PVT h1
PVI
h2=0 L
Daytime sight distance unrestricted

28. 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?
Assume S<L (this is the case more often than not)
A = 6.4
224.9 ft. This is > l (which is 150 ft) so the assumption is wrong.
Okay, since that didn’t work: S>L, therefore SSD = ft. which is greater than L
The driver will see the tree when it is feet in front of her.
Available SSD = ft.
Required SSD = (1.47 x 30)2/2(32.2)( ) + 2.5(1.47 x 30) = 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.

29. Crest Vertical Curve
Assume S<L, try both, but this is most often the case
Equation specific to crest curve which accommodates sight over hill
L and S in horizontal plane and comparable (150 and 243 ft)
Required SSD = ft assumes 0 grade
Text problem versus design problem.

30. Crest Vertical Curves
SSD
PVI
Line of Sight
PVC G1
PVT G2 h2 h1 L

31. 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) = ft.

32. Trinity Road between Sonoma and Napa valleys
Horizontal Alignment

33. Horizontal Alignment Objective: Primary challenge Fundamentals
Geometry of directional transition to ensure:
Safety
Comfort
Primary challenge
Transition between two directions
Horizontal curves
Fundamentals
Circular curves
Superelevation Δ

34. Horizontal Curve Fundamentals
D = degree of curvature (angle subtended by a 100’ arc) PI T Δ E M L PC
Δ/2 PT
D = degree of curvature (angle subtended by a 100’ arc) R R
Δ/2
Δ/2

35. Horizontal Curve FundamentalsPI T Δ E M L PC
Δ/2 PT R R
Δ/2
Δ/2

36. Example 4
A horizontal curve is designed with a 1500 ft. radius. The tangent length is 400 ft. and the PT station is What are the PI and PT stations?
Since we know R and T we can use T = Rtan(delta/2) to get delta = degrees
D = /R. Therefore D = 3.82
L = 100(delta)/D = 100(29.86)/3.82 = 781 ft.
PC = PT – PI = 2000 – 781 =
PI = PC +T = =
Note: cannot find PI by subtracting T from PT!

37. Superelevation Rv ≈ Fc α
Fcn
Fcp α e W
1 ft Wn Ff Wp Ff α

38. e = number of vertical feet of rise per 100 ft of horizontal distance = 100tan
Superelevation
Divide both sides by Wcos(α)
Assume fse is small and can be neglected – it is the normal component of centripetal acceleration
This is the minimum radius that provides
for safe vehicle operation
Rv because it is to the vehicle’s path

39. Selection of e and fs Practical limits on superelevation (e)
Climate
Constructability
Adjacent land use
Side friction factor (fs) variations
Vehicle speed
Pavement texture
Tire condition
The maximum side friction factor is the point at which the tires begin to skid
Design values of fs are chosen somewhat below this maximum value so there is a margin of safety
Design values of fs are chosen somewhat below this maximum value so there is a margin of safety

40. Minimum Radius Tables

41. WSDOT Design Side Friction Factors
For Open Highways and Ramps
from the 2005 WSDOT Design Manual, M 22-01

42. WSDOT Design Side Friction Factors
For Low-Speed Urban Managed Access Highways
from the 2005 WSDOT Design Manual, M 22-01

43. Design Superelevation Rates - AASHTO
There is a different curve for each superelevation rate, this one is for 8%
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

44. Design Superelevation Rates - WSDOT
emax = 8%
from the 2005 WSDOT Design Manual, M 22-01

45. Example 5
A section of SR 522 is being designed as a high-speed divided highway. The design speed is 70 mph. Using WSDOT standards, what is the minimum curve radius (as measured to the traveled vehicle path) for safe vehicle operation?
For the minimum curve radius we want the maximum superelevation rate.
WSDOT max e = 0.10
For 70 mph, WSDOT f = 0.10
Rv = V2/g(fs+e) = (70 x 1.47)2/32.2( ) = ft.
This is the radius to the center of the inside lane. Depending upon the number of lanes and perhaps a center divider, the actual centerline radius will be different.

46. Example 5
A section of SR 522 is being designed as a high-speed divided highway. The design speed is 70 mph. Using WSDOT standards, what is the minimum curve radius (as measured to the traveled vehicle path) for safe vehicle operation?
For the minimum curve radius we want the maximum superelevation rate.
WSDOT max e = 0.10
For 70 mph, WSDOT f = 0.10
Rv = V2/g(fs+e) = (70 x 1.47)2/32.2( ) = ft.
This is the radius to the center of the inside lane. Depending upon the number of lanes and perhaps a center divider, the actual centerline radius will be different.
For the minimum curve radius we want the maximum superelevation.
WSDOT max e = 0.10
For 70 mph, WSDOT f = 0.10

47. Stopping Sight Distance
SSD (not L)
Looking around a curve
Measured along horizontal curve from the center of the traveled lane
Need to clear back to Ms (the middle of a line that has same arc length as SSD) Ms
Obstruction
Basically it’s figuring out L and M from the normal equations Rv Δs
Assumes curve exceeds required SSD

48. Stopping Sight Distance
SSD (not L) Ms
Obstruction
Basically it’s figuring out L and M from the normal equations Rv Δs

49. Example 6
A horizontal curve with a radius to the vehicle’s path of 2000 ft and a 60 mph design speed. Determine the distance that must be cleared from the inside edge of the inside lane to provide sufficient stopping sight distance.

50. Superelevation Transition
FYI – NOT TESTABLE
Superelevation Transition
from the 2001 Caltrans Highway Design Manual

51. Spiral Curves FYI – NOT TESTABLE No Spiral Spiral
Ease driver into the curve
Think of how the steering wheel works, it’s a change from zero angle to the angle of the turn in a finite amount of time
This can result in lane wander
Often make lanes bigger in turns to accommodate for this
Spiral
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004

52. FYI – NOT TESTABLE
No Spiral
I-90 past North Bend looking east

53. Spiral Curves WSDOT no longer uses spiral curves
FYI – NOT TESTABLE
Spiral Curves
WSDOT no longer uses spiral curves
Involve complex geometry
Require more surveying
Are somewhat empirical
If used, superelevation transition should occur entirely within spiral

54. Operating vs. Design Speed
FYI – NOT TESTABLE
Operating vs. Design Speed
85th Percentile Speed vs. Inferred Design Speed for 138 Rural Two-Lane Highway Horizontal Curves
According to NCHRP Report
85th Percentile Speed vs. Inferred Design Speed for Rural Two-Lane Highway Limited Sight Distance Crest Vertical Curves