Vertical AlignmentConcepts
Alignment is a 3D problem broken down into two 2D problems
Horizontal Alignment (plan view)
Vertical Alignment (profile view)
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
CEE 320 Spring 2008
CEE 320 Spring 2008
Implies equal curve tangents
(or feet) from the beginning of the curve
CEE 320 Spring 2008
G1, G2 in percent, L in stations
CEE 320 Spring 2008
G1, G2 in percent, L in stations
CEE 320 Spring 2008
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%
G2= - 4.5%
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
y = -0.8125x2 + 2x + 59
dy/dx = -1.625x + 2 = 0
y = -0.8125(1.23)2 + 2(1.23) + 59
CEE 320 Spring 2008
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
CEE 320 Spring 2008
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)
CEE 320 Spring 2008
Design Controls for Crest Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
CEE 320 Spring 2008
Design Controls for Crest Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
CEE 320 Spring 2008
CEE 320 Spring 2008
β = 1 degree
For SSD < L
For SSD > L
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
CEE 320 Spring 2008
Design Controls for Sag Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
CEE 320 Spring 2008
Design Controls for Sag Vertical Curves
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
CEE 320 Spring 2008
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 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.
CEE 320 Spring 2008
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 = 196.53 ft assumes 0 grade
Text problem versus design problem.
CEE 320 Spring 2008
Daytime sight distance unrestricted
CEE 320 Spring 2008
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 = 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.
CEE 320 Spring 2008
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 = 196.53 ft assumes 0 grade
Text problem versus design problem.
CEE 320 Spring 2008
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.
CEE 320 Spring 2008
CEE 320 Spring 2008
Safety
Comfort
D = degree of curvature (angle subtended by a 100’ arc)
D = degree of curvature (angle subtended by a 100’ arc)
CEE 320 Spring 2008
Example 4
A horizontal curve is designed with a 1500 ft. radius. The tangent length is 400 ft. and the PT station is 20+00. What are the PI and PT stations?
Since we know R and T we can use T = Rtan(delta/2) to get delta = 29.86 degrees
D = 5729.6/R. Therefore D = 3.82
L = 100(delta)/D = 100(29.86)/3.82 = 781 ft.
PC = PT – PI = 2000 – 781 = 12+18.2
PI = PC +T = 12+18.2 + 400 = 16+18.2.
Note: cannot find PI by subtracting T from PT!
CEE 320 Spring 2008
for safe vehicle operation
Rv because it is to the vehicle’s path
e = number of vertical feet of rise per 100 ft of horizontal distance = 100tan
Divide both sides by Wcos(α)
Assume fse is small and can be neglected – it is the normal component of centripetal acceleration
CEE 320 Spring 2008
Climate
Constructability
Vehicle speed
Pavement texture
Tire condition
Design values of fs are chosen somewhat below this maximum value so there is a margin of safety
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
CEE 320 Spring 2008
from the 2005 WSDOT Design Manual, M 22-01
For Open Highways and Ramps
CEE 320 Spring 2008
from the 2005 WSDOT Design Manual, M 22-01
For Low-Speed Urban Managed Access Highways
CEE 320 Spring 2008
Design Superelevation Rates - AASHTO
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
There is a different curve for each superelevation rate, this one is for 8%
CEE 320 Spring 2008
Design Superelevation Rates - WSDOT
emax = 8%
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
Rv = V2/g(fs+e) = (70 x 1.47)2/32.2(0.10 + 0.10) = 1644.16 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.
CEE 320 Spring 2008
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.
WSDOT max e = 0.10
For 70 mph, WSDOT f = 0.10
For the minimum curve radius we want the maximum superelevation rate.
WSDOT max e = 0.10
Rv = V2/g(fs+e) = (70 x 1.47)2/32.2(0.10 + 0.10) = 1644.16 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.
CEE 320 Spring 2008
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)
Assumes curve exceeds required SSD
Basically it’s figuring out L and M from the normal equations
CEE 320 Spring 2008
SSD (not L)
Basically it’s figuring out L and M from the normal equations
CEE 320 Spring 2008
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.
CEE 320 Spring 2008
FYI – NOT TESTABLE
Spiral
from AASHTO’s A Policy on Geometric Design of Highways and Streets 2004
FYI – NOT TESTABLE
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
CEE 320 Spring 2008
CEE 320 Spring 2008
Involve complex geometry
Require more surveying
Are somewhat empirical
FYI – NOT TESTABLE
85th Percentile Speed
vs. Inferred Design Speed for 138 Rural Two-Lane Highway Horizontal Curves
85th Percentile Speed
vs. Inferred Design Speed for Rural Two-Lane Highway Limited Sight Distance Crest Vertical Curves
FYI – NOT TESTABLE