horizontal alignment of roads

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Horizontal Alignment

CE 453 Lecture 16

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Horizontal Alignment

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Horizontal Alignment1. Tangents2. Curves3. TransitionsCurves require superelevation (next lecture) Reason for super: banking of curve, retard

sliding, allow more uniform speed, also allow use of smaller radius curves (less land)

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Curve Types1. Simple curves with spirals (why spirals)2. Broken Back – two curves same direction

(avoid) 3. Compound curves: multiple curves connected

directly together (use with caution) go from large radii to smaller radii and have R(large) < 1.5 R(small)

4. Reverse curves – two curves, opposite direction (require separation typically for superelevation attainment)

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Highway plan and profile

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Important Components of Simple Circular Curve = deflection angle ( ), PC = point of curve D = curve degree (100 m/ or 100 ft /), PT = point of tangent, L = curve length (m), PI = point of intersection R = radius of circular curve (m, ft), T = tangent length (m, ft), E = external distance (m, ft).

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Important Components of Simple Circular Curve See: ftp://165.206.254.150/dotmain/design/dmanual/English/e02a-01.pdf 

1.     PC, PI, PT, E, M, and 2.     L = 2()R()/3603     T = R tan (/2)

Source: Iowa DOT Design Manual

Direction of st

ationing

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Horizontal Curve Calculations

R R

/2

ET T

/2

PC PT

PI

= deflection angle ( ), PC = point of curveD = curve degree (100 m/ or 100 ft /), PT = point of tangent, L = curve length (m), PI = point of intersectionR = radius of circular curve (m, ft),T = tangent length (m, ft),E = external distance (m, ft).

1

2cos

1

2tan

3.57

6.5729

RE

RT

RL

RD

8CE

E 3

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te

r 2

00

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H o r i z o n t a l C u r v e F u n d a m e n t a l s

R

T

P C P T

P I

M

E

R

Δ

Δ / 2Δ / 2

Δ / 2

RRD

000,18

180100

2tan

RT

DRL

100

180

L

9CE

E 3

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H o r i z o n ta l C u r v e F u n d a m e n ta l s

1

2cos1RE

2

cos1RM

R

T

P C P T

P I

M

E

R

Δ

Δ / 2Δ / 2

Δ / 2L

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Horizontal Curve Example Deflection angle of a 4º curve is 55º25’, PI at

station 245+97.04. Find length of curve,T, and station of PT.

D = 4º = 55º25’ = 55.417º D = _5729.58_ R = _5729.58_ = 1,432.4 ft R 4

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Horizontal Curve Example D = 4º = 55.417º R = 1,432.4 ft L = 2R = 2(1,432.4 ft)(55.417º) =

1385.42ft 360 360

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Horizontal Curve Example D = 4º = 55.417º R = 1,432.4 ft L = 1385.42 ft T = R tan = 1,432.4 ft tan (55.417) = 752.29 ft

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Stationing ExampleStationing goes around horizontal curve.

For previous example, what is station of PT?

First calculate the station of the PC:

PI = 245+97.04

PC = PI – T

PC = 245+97.04 – 752.29 = 238+44.75

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Stationing Example (cont)PC = 238+44.75

L = 1385.42 ft

Station at PT = PC + L

PT = 238+44.75 + 1385.42 = 252+30.17

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Example 4A 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 PC stations?

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Solution

• Since we know R and T we can use • T = R tan(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 – L = 2000 – 781 = 12+18.2• PI = PC +T = 12+18.2 + 400 = 16+18.2.

• Note: cannot find PI by subtracting T from PT!

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Design of Horizontal AlignmentDetermination of Minimum Radius Length of curve Computation of offsets from the tangents

to the curve to facilitate the setting out of t6hye curve

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Horizontal Curve design Design based on appropriate relationship

between design speed and curvature and their relationship with side friction and superelevation

Turning the front wheels, side friction and superelevation generate an acceleration toward center of curvature (centripetal acceleration)

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Motion on Circular Curves

dtdvat

Rvan

2

20

coscossin ns amWfW

coscos)(cossin2

WRv

gWWfW s

e tan

cossin

gRvfe s

2

Motion on Circular Curves

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Vehicle Stability on Curves

where:

gRvfe s

2

(ft/s), speeddesign v(-),t coefficienfriction sidesf

).ft/s (32on acceleratigravity 2g

(-),tion superelevae

(ft), radiusRAssumed

Design speed (mph)

Maximum design

fs max 20 0.1770 0.10

Must not be too short

(0.12) 0.10 - 0.06max e

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Radius CalculationRmin = ___V2______

15(e + f) Where:Rmin is the minimum radius in feet

V = velocity (mph)e = superelevation f = friction (15 = gravity and unit conversion)

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Radius Calculation• Rmin uses max e and max f (defined by AASHTO, DOT,

and graphed in Green Book) and design speed • f is a function of speed, roadway surface, weather

condition, tire condition, and based on comfort – drivers brake, make sudden lane changes, and change position within a lane when acceleration around a curve becomes “uncomfortable”

• AASHTO: 0.5 @ 20 mph with new tires and wet pavement to 0.35 @ 60 mph

• f decreases as speed increases (less tire/pavement contact)

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Max e Controlled by 4 factors:

Climate conditions (amount of ice and snow) Terrain (flat, rolling, mountainous) Type of area (rural or urban) Frequency of slow moving vehicles who

might be influenced by high superelevation rates

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Max e Highest in common use = 10%, 12% with no

ice and snow on low volume gravel-surfaced roads

8% is logical maximum to minimize slipping by stopped vehicles, considering snow and ice

Iowa uses a maximum of 6% on new projects For consistency use a single rate within a

project or on a highway

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Source: A Policy on Geometric Design of Highways and Streets (The Green Book). Washington, DC. American Association of State Highway and Transportation Officials, 2001 4th Ed.

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Radius Calculation (Example)Design radius example: assume a maximum e

of 8% and design speed of 60 mph, what is the minimum radius?

fmax = 0.12 (from Green Book)

Rmin = _____602________________

15(0.08 + 0.12)

Rmin = 1200 feet

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Radius Calculation (Example)For emax = 4%? (urban situation)

Rmin = _____602________________

15(0.04 + 0.12)Rmin = 1,500 feet

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Radius Calculation (Example)For emax = 2%? (rotated crown)

Rmin = _____602________________

15(0.02 + 0.12)Rmin = 1,714 feet

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Radius Calculation (Example)For emax = -2%? (normal crown, adverse

direction)

Rmin = _____602________________

15(-0.02 + 0.12)Rmin = 2,400 feet

Sight Distance for Horizontal Curves

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Sight Distance for Horizontal Curves Location of object along chord length that blocks

line of sight around the curve m = R(1 – cos [28.65 S]) RWhere:m = line of sightS = stopping sight distanceR = radius

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Sight Obstruction on HorizontalCurves Assume S is the SSD equal to chord AT Angle subtended at centre of circle by arc AT

is 2θ in degree then Total circumference perimeter (πR) S / πR = 2θ / 180 S = 2R θπ / 180 θ = S 180/ 2R π = 28.65 (S) / R R-M/R = cos θ M = R [1 – cos 28.65 (S) / R ]

θR

MA TB

T

O

O

θ

B

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Line of sight is the chord AT

Horizontal distance traveled is arc AT, which is SD.

SD is measured along the centre line of inside lane around the curve.

See the relationship between radius of curve, the degree of curve, SSD

and the middle ordinate

S

R

M

O

TA

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Sight Distance ExampleA horizontal curve with R = 800 ft is part of a 2-lane

highway with a posted speed limit of 35 mph. What is the minimum distance that a large billboard can be placed from the centerline of the inside lane of the curve without reducing required SSD? Assume p/r =2.5 and a = 11.2 ft/sec2

SSD = 1.47vt + _________v2____ 30(__a___ G)

32.2

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Sight Distance ExampleSSD = 1.47(35 mph)(2.5 sec) + _____(35 mph)2____ = 246 feet 30(__11.2___ 0)

32.2

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Sight Distance Examplem = R(1 – cos [28.65 S]) Rm = 800 (1 – cos [28.65 {246}]) = 9.43 feet 800

(in radians not degrees)

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Horizontal Curve Example Deflection angle of a 4º curve is 55º25’, PI at

station 245+97.04. Find length of curve,T, and station of PT.

D = 4º = 55º25’ = 55.417º D = _5729.58_ R = _5729.58_ = 1,432.4 ft R 4

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Horizontal Curve Example D = 4º = 55.417º R = 1,432.4 ft L = 2R = 2(1,432.4 ft)(55.417º) =

1385.42ft 360 360

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Horizontal Curve Example D = 4º = 55.417º R = 1,432.4 ft L = 1385.42 ft T = R tan = 1,432.4 ft tan (55.417) = 752.29 ft

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Stationing ExampleStationing goes around horizontal curve.

For previous example, what is station of PT?

First calculate the station of the PC:

PI = 245+97.04

PC = PI – T

PC = 245+97.04 – 752.29 = 238+44.75

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Stationing Example (cont)PC = 238+44.75

L = 1385.42 ft

Station at PT = PC + L

PT = 238+44.75 + 1385.42 = 252+30.17

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Suggested Steps in Horizontal Design1. Select tangents, PIs, and general curves making

sure you meet minimum radius criteria2. Select specific curve radii/spiral and calculate

important points (see lab) using formula or table (those needed for design, plans, and lab requirements)

3. Station alignment (as curves are encountered)4. Determine super and runoff for curves and put in

table (see next lecture for def.)5. Add information to plans

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HOMEWORK Your team is responsible for the design of

a small roadway project in Iowa. Your individual task is to design a horizontal curve to the right using an even value radius slightly larger than the minimum radius curve. Use a design speed of 55 mph and a superelevation rate of 4%. Assume the PI has a station of 352+44.97; the Δ (delta) of the curve is 35° 24’ 55”.

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HOMEWORK Referring to the Iowa DOT Design

Manual you find guidance at ftp://165.206.203.34/design/dmanual/02a-01.pdf on the design of horizontal curves.

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HOMEWORKYour assignment: Reading carefully the design guidance

with special attention to the items to be included on the plans, calculate all of the values to be shown on a plan set for this curve. Be sure to calculate the stations of the PC and PT in addition to the values listed.

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HOMEWORKYour assignment: At this time do not concern yourself with

superelevation runoff; just use the design speed and superelevation rate to determine the minimum radius curve allowable. It may help you to create a list of the items to be included before doing your calculations.

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HOMEWORKYour assignment: Now use a design speed of 60 mph and

the same 4% superelevation rate to calculate the curve. Your PI station and Δ will remain unchanged.