ce426 block2 roadway geometric design 1

7/30/2019 CE426 Block2 Roadway Geometric Design 1

1/52

12/28/2

CE 426Highway Design

Dr. Khaled Hamad

Deptartment of Civil Engineering

Block 2:

Roadway Geometric Design

Outline

Geometric Design Concept

Road Design

Horizontal Alignment

Vertical Alignment

2


2/52

12/28/2

GEOMETRIC DESIGN

3

4

Design Concepts

Geometric alignment is a 3D problem broken

down into two 2D problems:

Horizontal Alignment (plan view)

Vertical Alignment (profile view)


3/52

12/28/2

5

3D Perspective

2D Plan

(Horizontal)

2D Profile

(Vertical)

6

Stationing: Distance Referencing Along HorizontalAlignment


Vertical Alignment

Each roadway point is uniquely defined by stationing along the x-axis. E.g., station 345+60or @ the 345th 100 ft-station and 60 ft toward the 346th 100 ft-station)


4/52

12/28/2

7

Road Plan and Profile

8

Geometric DesignConsists of the following:

1. Horizontal Design Horizontal Alignment (HA)

2. Vertical Design Vertical Alignment (VA)

3. Cross-sectional Design Right-of-way, roadway features, Pavement

4. Junction Design Intersections & Interchanges (Layouts and traffic control)

5. Pavement Design

6. Terminal Facilities Parking Lots, Garages, etc.

8


5/52

12/28/2

9

HORIZONTAL ALIGNMENT

Roadway Geometric Design

9

10


Definition:

Series of straight segments of road (tangents)

connected by suitable curves (horizontal curves)

Objective: Establish geometry of directional transition to ensure

safety and comfort

Primary challenges Transition between two directions using tangents and

curves

Relationship between design speeds and curvature

Relationships with superelevation and side friction


6/52

12/28/2


Tangents Curves

11

Tangents & Curves

Tangent

Curve

Tangent toCircular Curve

Tangent to

Spiral Curve to

Circular Curve12


7/52

12/28/2

Motion on Circular Curves

dt

dva

t

Rvan

2

13

coscossin ns amWfW

cos)(cossin2

R

v

g

WWfW s e

tan

cos

sin

gR

vfe s

2

Motion on Circular Curves

gR

vfs

2

tan

W =weight of vehiclefs =coefficient of side frictiong =acceleration of gravityv = vehicle speedR =radius of curve

angle of inclination

e =tan (rate of superelevation)

14

)(

2

sfeg

vR


8/52

12/28/2

Determine Radius of CurveRadius of curve,R, can be calculated as follows:

15

Where:

v= design speed

e= rate of superelevation

0.04 0.12 depending on type of facility

and local environment

fs = coefficient of side friction

0.20 at low speed to 0.08 at high speed

g= acceleration of gravity = 9.81 m/s2 =32.2 ft/s2

)(

2

sfeg

vR

)(15

2

sfe

vR

For v in mph and Rin ft, you can use:

)(127

2

sfe

vR

For v in kph and Rin m, you can use:

Minimum Radius of a Circular Curve Minimum value of radiusRmin can be calculated as follows:

UsingR > Rmin allows the superelevation rate and side friction

to be smaller than maximum.

See selection ofeand fsin next slides

16

)(maxmax

2

min

sfeg

vR


9/52

12/28/2

Side friction factor (fs) It is a function of vehicle speed, pavement/roadway surface,

weather condition, tire condition It decreases as speed increases (less tire/pavement contact)

The maximum side friction factor is the point at which the tiresbegin to skid.

See table for range of values:

17

Design

Speed (mph)

Coefficient of

side friction, fs

Design

Speed (km/h)

20 0.20 30

30 0.18 50

40 0.16 65

50 0.14 80

60 0.12 9570 0.10 110

80 0.08 130

Superelevation e Controlled by following factors:

Climate conditions (frequency and amount of ice and snow)

Terrain (flat, rolling, mountainous)

Adjacent land use and type of area (rural or urban)

Frequency of slow moving vehicles who might be influenced byhigh superelevation rates

Constructability

Highest in common use is 10% Never exceed 12% which could be suitable for roads with no

ice and snow on low volume gravel-surfaced roads

8% is logical maximum to minimize slipping by stoppedvehicles, considering snow and ice Typically use 4 or 6% for urban design areas

Can be omitted on low-speed urban streets

Higher values are permitted on freeways as for arterial roads

For consistency use a single rate within a project or on ahighway

18


10/52

12/28/2

19

AASHTO

Minimum

Radius Table

20

Design Superelevation Rates - AASHTOGreen Book provides charts for 4% to 12% superelevations


11/52

12/28/2

Simple Curve Fundamentals

PC = Point of curvature PT = Point of tangency

PI = Point of intersection

= Central angle

R = Radius of curve

L = Length of curve

T = Tangent length

C = Chord length

E = External distance

M = Middle ordinate

21

Simple Curve Fundamentals

DDR

5729.6180100

2tan

RT

RL180

LD

100

For US units, degree of curvature, D :2sin2 RC

22

1

2cos

1RE

2cos1 RM

Sta. of PC = Sta. of PI T

Sta. of PT = Sta. of PC + L


12/52

12/28/2

23

Horizontal Curve Design Procedure

1. Select tangents, PIs, and general curves making

sure you meet minimum radius criteria2. Determine radius of the curve (R)

3. Measure angle () between tangents

4. Determine length of the curve (L) and Tangent (T)

5. Determine station alignment: stations for PC, PI,PT

6. Calculate spiral (transition) lengths (Ls), if needed

7. Check for widening of curves (Check SSDrequirements)

8. Develop edge profiles (superelevation runoffs)

9. Add information to plans

23

Example 1A horizontal curve is designed with a 1,500 ft radius. The

tangent length is 400 ft and the PT station is 20+00. What

are the PI and PC stations?

Solution:

Since we know R and T:

T = R tan(/2) => = 29.86

L = x R x / 180 = (3.14)(1500)(29.86)/180 = 781 ft

Sta. of PC = Sta. of PT L = 2000 781 = 1219.00 = 12+19.00

Sta. of PI = Sta. of PC + T = 1219.0 + 400 = 1619.00 = 16+19.00

24


13/52

12/28/2

Example 2 A section of a road is being designed as a high-speed divided

highway. The design speed of 60 mph. What is the minimumcurve radius for safe vehicle operation assuming: e is 8% and4%

For 60 mph, use fs = 0.12

Foremax = 8%:R

min= V2/g(fs+e)

= (60)2/15(0.08 + 0.12)

= 1200 ft

Foremax = 4% (urban situation) :R

min= V2/g(fs+e)

= (60)2

/15(0.04 + 0.12)= 1500 ft

25

Example 3Given: Simple horizontal curve with = 50.25, PC = Station (331+38.75)

design speed = 70 mph; and e = 0.07.

Design this horizontal curve (find radius of the curve and Station of PT)

Solution:

For 70 mph, use max f = 0.10

Rmin = v2/g(fs+e)

= (70 x 1.47)2/32.2(0.10 + 0.07) = 1910.33 ft => Use R = 1950 ft

L = 3.14 R /180 = (1950 x 50.25 x 3.14) /180 = 1,710.20 ftSta. of PT = Sta. of PC + L = (331+38.75) + (17+10.20) = 348+48.95

T = R tan(/2) = 1950 tan (50.25 /2) = 914.49 ft

Sta. of PI = Sta. of PC + T = (331+38.75) + (9+14.49) = 340+53.24

26


14/52

12/28/2

Setting Out Horizontal Curve in Field1. Locate PI in the field

2. Using R and , determine T and L (if not shown on plan)3. Measure out T distance from PI to locate BC and EC4. Determine deflection angles for required stations

5. Set up at PC, sight PI and measure out /2 angle

6. If /2 angle line-of-sight hits the PT mark, layout the curveusing the deflection angles.

7. If the /2 angle line-of-sight does not come acceptably closeto the PT mark, check the computations for T and then re-measure the locations of PC and PT if necessary

27

/2 = l/ 2L

Setting Out Horizontal Curve in Field

FindL = 3.14R /180 = 89.71 m

Then calculate = l/L

Deflection angle = /2

Calculate chord for each station

using

28

Station l /2

0+196.738

0+200 3.262 3.262 0 28' 2" 0 14' 1"

0+220 20 23.262 3 19' 55" 1 39' 58"

0+240 20 43.262 6 11' 49" 3 5' 54"

0+260 20 63.262 9 3' 42" 4 31' 51"

0+280 20 83.262 11 55' 35" 5 57' 48"

0+286.448 6.448 89.71 12 51' 0" 6 25' 30"

Given:

= 12 51'

R = 400 m

PC Sta. = 0+196.738

Determine information to stake out

this curve.

2sin2 RC


15/52

12/28/2

More Horizontal Curve Types

Compound Curves: multiplecurves connected directly togethergo from large radii to smaller radii(use with caution)

Reverse Curves: two curves inopposite direction (requireseparation typically forsuperelevation attainment)

Broken-Back Curves: two curvessame direction (should avoid)

Simple Curves with Spirals

29

Compound Circular Curve Ratio of flatter (larger) radius to

sharper (smaller) radius < 2:1

Each curve should pass minlength requirement by AASHTO

30


16/52

12/28/2

Reverse Curves

Seldom used Cause

discomfort andsafety problem

Preferabledesigns: 2 simple curves

w/ sufficienttangents

2 simple curvesseparated byspirals

31

Transition Curves (Spirals) Spiral curves are curves with a continuously changing radius

Design considerations: Safety & Comfort When vehicles enter a curve, the gain or loss of centrifugal forces is not instantaneous

Advantages: Provides natural and easy-to-follow path for drivers (promotes more uniform speeds and

lateral force increases and decreases gradually)

Provides location for attainment of superelevation runoff

Adds flexibility when widening a horizontal curve

Aesthetics, pleasant view

Disadvantages

Involve complex geometry Require more surveying

Are somewhat empirical

If used, superelevation transition should occur entirely within spiral

AASHTO: There is no definitive evidence that transition curves are essentialto safe operation of roadways; therefore, they are not used by many agencies

3232


17/52

12/28/2

33

Transition Curves (Spirals)

No Spiral

Spiral

Length of Spiral Curves US Customary Units Metric Units

Min spiral length, Ls,min is the larger of:

Max spiral length,

Ls,min = minimum length of spiral

Ls,max = maximum length of spiral V = vehicle speed, mph or km/hr

R = radius of the curve

C = rate of increase of radial acceleration (4 ft/sec3 or 1.2 m/sec3)

pmin = minimum lateral offset between tangent an circular curve (0.66 ft or 0.20 m)

pmax = maximum lateral offset between tangent an circular curve (3.3 ft or 1.0 m)

3434

RC

VL

RpL

s

s

3

min,

minmin,

15.3

)(24

RpLs )(24 maxmax,

RC

VL

RpL

s

s

3

min,

minmin,

0214.0

)(24


18/52

12/28/2

35

Desirable Spiral Lengths

Length correspond to travel time of 2 seconds at design speed

Insertion of Transition Curves(Spirals)

Euler (clothoid) curves are used

Radius varies from infinity attangent to radius of circularcurve

Spirals are not used when: Speed < 60 km/hr

Radius larger than that selected foremax of 2%

36


19/52

12/28/2

Transition Curves Fundamentals

37

38

Attainment of Superelevation(Superelevation Transition):

On a tangent, road cross-section is crowned to provideeasy drainage of storm water.

On a curve, cross-section slope (normal crown) shouldchange to that of super-elevation of road, i.e., obtain fullyelevated section.

Change should be gradually achieved over sufficient lengthbecause: Safety and comfort (gentle lateral acceleration and introduction of

centripetal force)

Pleasing appearance (pavement not distorted)

Transition should be achieved along: Spiral curve, if exists

Tangent-to-curve, if spiral does not exist.

38


20/52

12/28/2

39

Superelevation Transition

Superelevation transition

length is summation of: Tangent Runout: Length of

roadway needed toaccomplish a change inoutside lane cross-sectionfrom normal crown to flat(zero slope), or vice versa.

Superelevation Runoff:Length of roadway neededto accomplish a change inoutside lane cross-section

from flat (zero slope) to fullsuperelevation, or viceversa.

39

40


Normal Crown controlpoints:

OE: Outer Edge

CL: Center Line

IE: Inner Edge


21/52

12/28/2

Length of Superelevation Runoff, Lr

41

Length of Tangent Runout, Lt

42


22/52

12/28/2

43

Length of Superelevation Transition

Example For a 4-lane divided highway with cross-section

rotated about centerline, design superelevationrate = 4% and NC = 2%. Design speed is 50mph. What are the minimum length ofsuperelevation runoff (ft) and tangent runout?

.

44

ft144

005.0

75.004.02121

wdr

bewnL

ft7214404.0

02.0 r

d

NCt L

e

eL


23/52

12/28/2


Superelevation runoffis all attained in spiralcurve if used

If no spiral curve isused: Superelevation runoff

must be attained overa length that includesthe tangent and thecurve

Typically, use 66-70%

on tangent and 33-30% on curve

45

46



24/52

12/28/2

Methods of Super-

elevation Attainment

1. Pavement revolvedabout center line (CL)

Centerline is point ofcontrol

2. Pavement revolvedaround inner edge (IE)

Inner edge is point ofcontrol

3. Pavement revolvedaround outer edge (OE)

Outer edge is point ofcontrol

47

48

Curve Widening

A traffic lane on a curve mustbe widened because:

Off-tracking: Rear wheels donot follow exactly the samepath/trajectories of front wheels.

Vehicles front overhangrequires an additional lateralspace.

Difficulty of driving on curvesjustifies wider lateral clearance.

Tendency of drivers to steeraway from the pavements edge.


25/52

12/28/2

49

Curve Widening

Metric

50

ExampleCalculate the widening required for passenger

cars on a curve with radius R =570 ft and

design speed v = 40 mph. The roadway has

two lanes and is 22 ft wide on the tangent

section.

ft3Aft,11Lft,7uft,2.5Cft,22Wn

ft7.11U

115705707U

LRRuU

22

22

ft0.07F

5703)113(2570F

RA)A(2LRF

A

2

A

2

A

ZFC)2(UW Ac

nc WWno widening is needed for

passenger carsft20.11.680.072.5)2(7.11W

ZFC)2(UW

c

Ac

ft1.68570

40Z

R

vZ


26/52

12/28/2

51

Sight Distance on Horizontal Curve

Sight Distance onhorizontal curve iscompromised bypresence of obstructions: Walls

Cut slopes

Trees

SSD is measured aroundmiddle of inside lane

Line of sight is a chord ofcurve

Ms is sight obstructionoffset

Stopping Sight Distance

where:

v = initial velocity when brakesare applied

tP/R = time to perceive/react

f= coefficient of friction

g = acceleration of gravity = 9.81

m/s2 = 32.2 ft/s2

G = grade (decimal)

Distance to stop vehicle = Distance during P/R + Braking Distance

Gfgv

vtSSD RP

2

2

/

This is distance traveled by a vehicle between the time driver observes

an object in vehicles path and the time the vehicle actually comes to

complete stop.

52

Gfv

vtSSD RP

30

47.12

/

Gfv

vtSSD RP

254

278.02

/

Speed in mph and SSD in ft:

Speed in kph and SSD in m:


27/52

12/28/2

53

Sight Obstruction on Horizontal Curves

v

sR

SSD

180

DRSSD ssv

100180

v

vsR

SSDRM

90cos1

v

svv

R

MRRSSD 1cos

90

54

Sight Obstruction on Horizontal Curves


28/52

12/28/2

55

Example

Horizontal curve with 800 ft radius of 35 mph posted speedlimit. Determine horizontal sightline offset that a largebillboard can be placed from the centerline of inside lanewithout reducing required SSD. Assume: e=0%; t=2.5 sec;f=0.35.

Determine SSD:

Determine M:

.

Check value on monograph.

ft9.245

035.030

355.23547.1

3047.1

22

Gf

vvtSSD

ft6.9

80014.3

29.24590cos1800

90cos1

v

vs

R

SSDRM

56

VERTICAL ALIGNMENTRoadway Geometric Design

56


29/52

12/28/2

57

Vertical Alignment

G1 G2

G1 G2

Crest Vertical Curve

Sag Vertical Curve

Definition:

Straight segments of highway (tangents or grades)

connected by vertical curves.

Purpose: Select suitable grades for tangent sections and

appropriate lengths for vertical curves.

58

Vertical Alignment Objectives:

Provide a gradual change from one tangent grade toanother to allows vehicles to run smoothly on road

Determine elevation to ensure proper drainage andacceptable level of safety

Primary Challenges Topography of area has significant impact (following

existing surfaces to minimize earthwork cut & fill) Grades can significantly impact vehicles

Main Criteria: Minimum stopping distance Adequate drainage Comfort Pleasant appearance Minimize cut and fill


30/52

12/28/2

Grades

Grade: measure of inclination or slope (rise over the run)

Grades can significantly impact vehicles, especially trucks(performance increase on downgrades & decrease onupgrades) Cars: negotiate 4-5% grades without significant speed reduction

Trucks: significant speed changes

Maximum grade: depends on terrain type, road functionalclass, and design speed Max grade of 5% for design speed of 110 km/hr (70 mph)

Max grade in range of 7% to 12% for design speed of 50 km/hr (30mph)

AASHTO Recommendation (next slide)

Minimum Grade: Typically 0.5%

Flatter grades sometimes are justified

59

60

AASHTOMaximum

Grades


31/52

12/28/2

Critical Length of Grade

Maximum length which aloaded truck can travel

without unreasonable

speed reduction

Based on speed distance

curve of a typical heavy

truck (with 10 mph speed

reduction as threshold

Critical Length of Grade

General

Design

Speed

Reduction


32/52

12/28/2

63

Types of Vertical Curves

63

Type I

Type II

Type IIIType IV

Vertical Curve PVI = Point of Vertical

Intersection PVC = Point of Vertical

Curvature PVT = Point of Vertical

Tangency L = Length of vertical curve

Y = Offset of curve from initialgrade line

Ym = mid-curve offset Yf= end-curve offset

G1 = Grade of initial tangent, % G2 = Grade of final tangent, %

64


33/52

12/28/2

Properties of Vertical Curve

65

Parabolic function:

yis roadway elevation xdistance from beginning ofcurve.

Characteristics: Constant rate of change of slope.

Implies equal curve tangents:Horizontal distance PVC to PVI = Horizontal distance PVI to PVT= Curve length

Length of vertical curve is the horizontal projectionof the curve and not the length along the curve.

cbxaxy 2

Vertical Curve Fundamentals

10:PVCAt the Gbdxdyx

PVCofElev.0:PVCAt the cyx

L

GGa

L

GGa

dx

yd

22:Anywhere 1212

2

2

cbxaxy 2

PVCofElev.2

Therefore, 1212

xGx

L

GGy

66

Choose Either:

G1, G2 in decimal form, L in ft or m G1, G2 in percent, L in stations

(i.e., divide distance by 100)


34/52

12/28/2

Vertical Curve Fundamentals

2

2x

L

AY

12 GGA Algebraic difference

in grade =

Sta. of PVC = Sta. of PVI - L/2

Offset =

Sta. of PVT = Sta. of PVI + L/2 = Sta. of PVC + L

Elevation of PVC = Elevation of PVI - G1 L/2

Elevation of PVT = Elevation of PVI + G2 L/2

Elevation at any point x = Y +G1 x +Elev. of PVC

67

G1

G2

PVI

PVTPVC

x

Y

8

ALYm

2

ALYf

Mid-curve offset:

Final

offset:

Choose Either: G1, G2 in decimal form, L in ft or m

G1, G2 in percent, L in stations

(i.e., divide distance by 100)

68

Vertical Curve Procedure1. Determine the minimum length of curve to satisfy sight

distance requirements and other criteria.

2. Determine from the layout plans the station andelevation of the PVI (the point where the gradesintersect).

3. Compute the elevations of the PVC (or BVC) and endof vertical curve (EVC or PVT).

4. Compute the offsets, Y, from the tangent to the curve atequal distances or stations.

5. Compute elevations on the curve for each station as:elevation of the tangent offset from tangent, Y. Forcrest curves the offset is (-) and for sag curves theoffset is (+).

6. Compute the location and elevation of highest/lowestpoint on curve.

68


35/52

12/28/2

69

Criteria for Minimum Length of Curve

Minimum length of crest vertical curve shouldbe based on stopping sight distance.

Other (passing or decision) sight distances couldalso be considered if required.

Minimum length of sag vertical curve shouldbe based on:

1. Night sight distance

2. Comfort

3. Appearance4. Drainage

Stopping Sight Distance

where:

v = initial velocity when brakesare applied

tP/R = time to perceive/react

f= coefficient of friction

g = acceleration of gravity = 9.81

m/s2 = 32.2 ft/s2

G = grade (decimal)

Distance to stop vehicle = Distance during P/R + Braking Distance

Gfgv

vtSSD RP

2

2

/

This is distance traveled by a vehicle between the time driver observes

an object in vehicles path and the time the vehicle actually comes to

complete stop.

70

Gfv

vtSSD RP

30

47.12

/

Gfv

vtSSD RP

254

278.02

/

Speed in mph and SSD in ft:

Speed in kph and SSD in m:


36/52

12/28/2

71

AASHTO Stopping Sight Distance

Decision Sight Distance (DSD) For unexpected stimulus or when driver has to make

unusual maneuvers AASHTOs recommends using

decision sight distance (DSD).

DSD depends on

Type of maneuver required

Road location (Urban/Rural)

SSD is often inadequate when drivers must make

complex/instantaneous decisions; DSD is longer thanSSD

Exhibit 3.3 from AASHTOs Green Book is used in

design

72


37/52

12/28/2

Decision Sight Distances for Different Design Speeds and

Avoidance Maneuvers

73

74

Passing Sight DistanceSight distance is determined for a single vehicle passing a single vehicle with the

assumption that cover majority of situations observed in the real-world conditions.

4321 ddddPSD

d1 is the distance traveledduring the P-R time and theinitial acceleration until thepoint where the vehiclebegins to enter the passinglane

d2 is distance traveled whilethe vehicle is passing in thepassing lane

d3 is the distance betweenthe passing vehicle andopposing vehicle

d4 is the distance theopposing vehicle moveswhile the passing vehicle isin the passing lane


38/52

12/28/2

Passing Sight Distance (PSD)

75

Where

t1 = time of initial maneuver, sec

t2 = time the passing vehicle occupies leftlane, sec

a = average acceleration, mph/sec

v = average speed of passing vehicle,mph

m = difference in speed of passed andpassing vehicle, mph

247.11

11

at

mvtd

22 47.1 vtd

d3 = 110 to 300 ft

d4 = 2d2 /3 3/2

m9030

278.0

2278.0

24

3

22

1

11

dd

d

vtd

atmvtd

a = average acceleration, km/h/s

v = average speed of passing vehicle,

km/h

m = difference in speed between

passing and passed vehicles, km/h

U.S. Customary Metric

Passing Sight Distance on Two-lane Highways

76


39/52

12/28/2

77

Minimum Length of Crest Curve

G1 G2

PVI

PVTPVC

h2h1 L

S

2

21

2

min

200

hh

SAL

A

hhSL

2

21

min

2002

For S < L For S > L

Line of Sight

Use , where G1 & G2 in percent, not in decimal form12 GGA

78

Minimum Length of Crest Curve AASHTO Values for h1 and h2:

where:SSD (Stopping Sight Distance)

PSD (Passing Sight Distance)

Parameters (US Units) SSD (ft) PSD (ft)

Height of drivers eye (h1) 3.5 3.5

Height of object (h2) 2.0 3.5

Parameters (Metric) SSD (m) PSD (m)

Height of drivers eye (h1) 1.08 1.08

Height of object (h2) 0.60 1.08


40/52

12/28/2

79

Minimum Length of Crest Curve

The length of crest curve should not be lessthan:

Lmin = 3v(design speed in mph and L in ft)

Lmin = 0.6v(design speed in km/h and L in m)

Generally, minimum lengths of crest curves

are about 100 to 325 ft (30 to 100 m).

80

Minimum Length of Sag Curve

G1 G2

PVI

PVTPVC

h2=0h

L

Headlight Sight Distance (S)

1tan200

2

minSh

SAL

A

ShSL

1tan2002min

For S < L For S > L

headlight beam (diverging by 1 degrees)

h = headlight height

= 2.0 ft (or 0.6 m)


41/52

12/28/2

81

Other Criteria for Minimum Length of Sag Curve

Comfort Criteria: A comfortable ride will be provided if the radial

acceleration is not greater than 1 ft/sec2

or 0.3 m/sec2

General Appearance Criteria: The rule of thumb for the general

appearance criteria is as follows:

Drainage Criteria(only when curve is curbed): The maximum length inwhich a minimum slope of 0.30% be provided within 50 ft (15 m) oflowest/level point.

mph)inft where(in5.46

2

min vAv

L

ft)(in100min AL m)(in30min AL

km/h)inwherem(in395

2

min vAv

L

82

Example 1For a design speed of 50 mph andgrades shown below, determine theminimum length of a vertical curvethat meets the criteria for safepassing.

Solution:

%32 G1%1 G

%4)1(3GGA 12

:

ft939,1)1tan18002(200

18004 2

L

The answer is correct

because the SL condition

is met.

LS)tan(200

2

Sh

SAL

S = 1,800 ft


42/52

12/28/2

Example 2 Given:

A sag vertical curve of a highway with -3% grade joining

+3% grade

Design speed = 55 mph

Perception-reaction time = 2.5 sec

Determine the minimum length of the curve?

Solution:

Determine the SSD required for the design conditions.

Because the grade change constantly on a vertical curve,

the worst-case value for G is used to determine the braking

distance:

ft

GfvvtSSD rp

41.51926.31715.202

)03.035.0(30555.25547.1

)(3047.1

22

/

83

Solution:This is a sag curve.

How we determine if S>L or SL:

The condition is not appropriate since 519.41


43/52

12/28/2

8585

Solution:

Check for other criteria:

Rider comfort:

General Appearance:

Rule of thumb:

So, the minimum length to satisfy all criteria is 731 ft

ftAv

L 3905.46

556

5.46

22

ftAL 6006100100

ftvL 165)55(33min

86

Finding High or Low Point on Vertical Curve

High or low point on a vertical curve is of interest tothe designer to investigate: Drainage conditions

Clearance beneath overhead structures

A

GLy

GKGG

LG

GG

LGx

GxL

GG

dx

dy

xGxL

GGy

LowHigh

LowHigh

2-PVCofElev.

:isPointHigh/LowofElevation

02

2:curveon thepointHigh/LowoflocationfindTo

PVCofElev.2

:usingfoundbecancurveverticalaonpointanyofElevation

21

/

112

1

121/

112

1212


44/52

12/28/2

Example 3

A 400 ft equal-tangent crest vertical curve has a PVC station of 100+00 at

59.00 ft elevation. The initial grade is 2.0% and the final grade is -4.5%.Determine the elevation and stationing of PVI, PVT, and the highest point of

the curve.

PVI

PVT

PVC: STA 100+00

EL 59 ft.

87

Solution:

PVI STA is 102+00 and PVT STA = 102+00 + 200 = 104+00

Elevation of the PVI = 59 + 0.02(200) = 63 ft.

Elevation of the PVT = 63 0.045(200) = 54 ft.

High point elevation requires figuring out the equation for a vertical curve:

y=ax2+bx+c:

At x = 0, y = c = 59 ft

At x = 0, dy/dx = b = G1 = +2% = 0.02

a = (G2 G1)/2L = (-0.045 0.02)/(2*400) = - 0.00008125

Therefore, y = -0.00008125x2 + 0.02x + 59.

Highest point is where dy/dx = 0. That is, dy/dx = -0.0001625x + 0.02 = 0.

Solve for x, you get x = 123 ft (i.e., 1.23 stations).

Find elevation at x = 123 stations:

y = -0.00008125(123)2 + 0.02(123) + 59 = 60.23 ft

88


45/52

12/28/2

Another Solution:

You could obtain the same results using grades in percentage and length

of curve in stations:

The equation for the vertical curve is y=ax2+bx+c:

At x = 0, y = c = 59 ft

At x = 0, dy/dx = b = G1 = +2.0%

a = (G2 G1)/2L = (-4.5 2)/(2(4)) = - 0.8125

Therefore, y = -0.8125x2 + 2x + 59.

Highest point is where dy/dx = 0.

That is, dy/dx = -1.625x + 2 = 0. Solve for x, you get x = 1.23 stations.

Find elevation at x = 1.23 stations:

y = -0.8125(1.23)2

+ 2(1.23) + 59 = 60.23 ft

89

90

Length of Crest & Sag Curves Based on KFactors

Rate of Vertical Curvature, K, is the length of thevertical curve per percent algebraic difference inintersecting grades:

Kis a function of design speed (see tables); thus,

it could be viewed as a design control It is a shortcut for computing minimum length of

curves if known for design speed:

ALK

AKL


46/52

12/28/2

91

Design Controls for Crest Vertical Curves Based on SSD

92

Design Controls for Crest Vertical Curves Based on PSD


47/52

12/28/2

93

Design Controls for Sag Vertical Curves

94

Design Controls for Crest Vertical Curves (Metric)


48/52

12/28/2

95

Design Controls for Crest Vertical Curves (US Units)

96

Design Controls for Sag Vertical Curves (Metric)


49/52

12/28/2

97

Design Controls for Sag Vertical Curves (US Units)

98

Example 4A 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%

and an exiting grade of -2.0%. How long must the vertical curve be?

Solution:

A = |-2 3.2| = 5.2%

For 45 mph, we get K=61.

Therefore, L = KA = (61)(5.2) = 317.2 ft


50/52

12/28/2

99

Example 5A 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% and the exiting grade is

4.0%. A tree has fallen across the road at approximately the PVT. Assumingthe 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?

Solution:

Assume that S>L (it usually is not but for example well do it this way):

S = 146.23 ft < L

Must use S


51/52

12/28/2

101

Sight Distance at Undercrossings

L = Minimum curve length (m or ft)

S =Sight distance (m or ft)

A = |G2-G1|, grade change in %

C= Vertical clearance (m or ft)

h1 = Height of drivers eye (usually 8 ft or 2.4 m)

h2 = Height of object (usually 2 ft or 0.6 m)

2800

21

2

hhC

SAL

A

hhC

SL

2

800

2

21

For S < L For S > L

102

Example 6A +2% grade intersects with a -1% grade at station (535+24.25) at an elevation

of 300 ft. If the design speed is 65 mi/h, determine:

the stations and elevations of the BVC and EVC

the elevation of each 100-ft station

Solution:

A = |2 (-1)| = 3%

For 65 mph, we get K=193

Therefore, L = KA = (193)(3) = 579 ft


52/52

12/28/2

103

Example 6 Continue

END OF BLOCK 2

ce426 block2 roadway geometric design 1

Documents