# Geometric Design CEE 320 Anne Goodchild.

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Geometric Design CEE 320 Anne Goodchild

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

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

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

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

Vertical Alignment

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

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

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

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

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.

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

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

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

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

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

Crest Vertical Curves Assuming L > SSD…

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

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

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

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

Sag Vertical Curves Assuming L > SSD…

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

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

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.

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.

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

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.

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.

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

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.

Trinity Road between Sonoma and Napa valleys
Horizontal Alignment

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

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

Horizontal Curve Fundamentals
PI T Δ E M L PC Δ/2 PT R R Δ/2 Δ/2

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!

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

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

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

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

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

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

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

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.

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

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

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

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.

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

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

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

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

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